text:
Which bond - C-C or C-H - is stronger?
I've read that nonpolar bonds are stronger than polar bonds. If it is true, why $\ce{C-H}$ bond entalpy ($\pu{413 kJ/mol}$) is more than $\ce{C-C}$ ($\pu{348 kJ/mol}$)? I'm very confused at the moment.

The "strength" of a bond can't be evaluated just based on its enthalpy.
According to this relation
$$ \Delta H = \sum \Delta H_\text{bonds broken} - \sum \Delta H_\text{bonds created} $$
the strength of your bond will depend on the reactive nearby the bond. If the molecule is likely to create bonds with higher enthalpy after the bond's breakdown, then the enthalpy variation will be negative. The process will then be spontaneous and the initial bond will breaks "easily".
There is also some steric consideration, the hydrogen atom is smaller, the bond is then more easy to be "attacked".
Finally, for a polar bond, the electronic cloud is delocalised closer to one atom rather than the other (according to their electronegativity) so you have "less perturbation to add" to fully remove the covalent electron from the less electronegative atom.

Thanks! Could you tell me why in asymmetric alkenes the carbon with more hydrogen is more electronegative that the other?

What do you mean by "will depend on the reactive nearby the bond"? What is "reactive"? Did you mean to write something else?

Moreover, there are a few anomalies: "If "the hydrogen atom is smaller, the bond is then more easy to be "attacked"", then shouldn't its energy be lower instead, because other atoms are able to attack it and break the bond "easily" as you say? Similarly, "you have "less perturbation to add" to fully remove the covalent electron from the less electronegative atom." This again hints at the C-H bond (the more polar one) being weaker. Both are contrary since $\ce{C-H}$ has a higher bond enthalpy. Could you please explain? Thank you!

I disagree with the currently accepted answer, and believe that there are other reasons at play.
First of all, the difference in the electronegativities of carbon (2.55) and hydrogen (2.2) is roughly 0.35. The $\ce{C-H}$ bond is indeed slightly more polar than the $\ce{C-C}$ bond, but this slightly increased polaity can in no solely way account for the approximately 20% increase in bond energy. In fact, Wikipedia writes: "Because of this small difference in electronegativities, the C−H bond is generally regarded as being non-polar."
I believe the real reason here is atomic size. The hydrogen atom is much smaller than the carbon atom. Smaller bonds lead to higher bond energy, therefore $\ce{C-H}$ bond has higher bond enthalpy than the $\ce{C-C}$ bond.


text:
List in order of increasing polarity?
I am doing some exam preparation and I came across this Question which I am finding challenging:
List from least polar to most polar, the following molecules:
Naphthalene, Phenol, 3-aminophenol and chlorobenzene.
I believe I know that Naphthalene is least polar as the benzene ring is very stable however, the confusion arises with chlorobenzene as the substituent Chlorine is very electronegative hence, it is strongly inductively electron-withdrawing thus is less polar than a simple benzene ring. So which is less polar, chlorobenzene or naphthalene?
Also which is most polar? 3-aminophenol or phenol? Phenol has a lone pair on oxygen in the oH group which can donate electrons to the ring, although with 3-aminophenol, both the NH2 and OH donate electrons, so wouldn't 3-aminophenol be more electron rich and hence more polar?
Please give the order of the molecules from least to most polar, explaining the order itself.
Thank you!

I would expect that the amino phenol is more polar than the phenol. This due to the fact that $\ce{NH2}$ donates electrons to the Benzene ring, making the difference in charge between the Benzene Carbon and the Oxygen stronger.
The Chlorine atom is less electro negative than oxygen (if I'm not mistaken) so that would mean: Naphatlene < Chlorobenzene < Phenol < Amino Phenol.

I guess that 3- amino phenol is more polar than phenol because both OH and NH2 groups have a (+)-M-effect (donate electrons to the ring) so amino group will support electron donation to the ring.

See that's what I thought, I felt that the 3-aminophenol has 2 substituents which donate to the benzene ring and make it more polar as opposed to phenol which only has one substituent donating its electrons to the ring. What is you opinion on the polarity of chlorobenzene and naphthalene? Which one is least polar and why?

If you draw the C-H dipole moments of Naphthalene, and add them together, what do you see?

I think chlorobenzene is more polar than naphtalene.
notice that Cl group has n-electrons around it which increase polarity.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What is a water blank?
What is a water blank?
I am doing some research and I am suppose to prepare a solution. I should mix 10g of soil with 95ml of water blank in order to create a soil suspension.
My problem is I am not English and I have no idea what a water blank is. I have a feeling the type of water is pretty important here though. (can't find what it is on the internet).
Help please. A simple definition/description will do.

Generally a blank means no sample and just solution. So it would not contain the 10g of soil in this case. I think for soil analysis it's used as a control so you can have something to compare to.
Here's some links I found by doing an internet search.
http://asl.irri.org/lims/index.jsp?page=about&feature=soilanalysis
http://www.soiltest.vt.edu/PDF/lab-procedures.pdf

Thanks. I'm basically just trying to isolate some microorganisms from the soil in order to then incubate them, so they grow on a solid medium (BHI agar). And I don't want to kill them off by accident or do something wrong.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Differences between alkyl halides and alkyl tosylates
What are the functional differences between alkyl halides alkyl tosylates? Both halides and tosylates are good leaving groups, yes. But is there chemistry that one will do and the other won't? Etc.? I'm not worried about their relative leaving group abilities. My prof treats the two as equivalent but I can't help but wonder if this is an oversimplification. Wikipedia doesn't point out any differences. The only difference I can think of is that an alkyl tosylate is an ester ... and this opens up the world of ester chemistry - i.e. hydrolysis with hot concentrated acid. Where as I can't see alkyl halides reacting in hot concentrated acid.

Functional differences in general? There are plenty of reactions specific to the different functional groups. For instance the preparation of a grignard is specific to a alkyl halide.

Both can be leaving groups. In general, TsO- are better leaving groups, thats why they are often specifically used as so. Halides can be used in many more reactions, like preparation of organometallic compunds.
TsO- more common function is to transform an alcohol group in a good leaving group. Formation of an halide, usually is not the way to go in this case.
But their basic reactivity is very similar.
If you dive into more complex reactivity, you can complicate things to a very very large extent...


text:
What's the strongest known organic acid?
What's the strongest known organic acid? Is it maybe trifluoromethanesulfonic acid or tautomer of pentacyanocyclopentadiene?
EDIT: Since the question was reactivated I thought about formalising it. I can think of a few categories of compounds: neutral (actually created or only studied in silico) and cationic (also two options). Second Dave's answer and ANM's one do the trick for two groups, but two other options remain (still prefer answers without boro/silico/etc. stuff).

This has been discussed recently.  Enter "strongest acid" in the search box up above and you'll find your answer.

I'm asking specifically for organic acid not inorganic like carboborane or hexafluoroantimonic.

I wouldn't consider triflic acid to be an organic molecule, it behaves more like the derivative of sulfuric acid than methane.

Well, sulfonic acids are organic compounds. That there's only one carbon and there are 3 fluors doesn't change this.

According to "The Strongest Isolable Acid" Angewandte Chemie International Edition 43 (40): 5352–5355, carborane acids are the strongest, with $\ce {H(CHB11Cl11)}$ being the strongest.
Depending upon ones preferred definition of "organic" this might not be considered organic, but the derivative $\ce{H(CHB11Me5Br6)}$ is strong enough to fully protonate benzene.
$\ce{H(CHB11Me11)}$ is stronger than $\ce{H(CHB11Me5Br6)}$ and only slightly weaker than $\ce {H(CHB11Cl11)}$, so if $\ce {H(CHB11Cl11)}$ is disregarded as no being organic, then $\ce{H(CHB11Me11)}$ is strongest.  If $\ce{H(CHB11Me11)}$ still isn't organic enough, then protonated benzene or a protonated alkane.
(A later article and the article in Nicolau's comment report $\ce {H(CHB11F11)}$ and $\ce {H(CEtB11F11)}$ which may be even stronger than the above)

The same group has since isolated the fluorinated derivative of carborane superacid, $\ce{H(CHB11F11)}$, which is even stronger still.

@NicolauSakerNeto Cool!

Also, protonated methane (methanium, $\ce{CH5+}$) is a significantly stronger acid than protonated benzene (comparing proton affinities; the less positive the value, the stronger the acid), though it quickly tends to decompose into the methylium cation ($\ce{CH3+}$) and hydrogen gas. The cation $\ce{CH6^{2+}}$ is also known, and is arguably an even stronger acid. Of course, none of these can be isolated in a neutral compound, as they will protonate essentially any anionic conjugate base.

I have just one more comment. To be clear, the picture shown is only for the conjugate base, $\ce{CHB11Cl11^{-}}$. The acidic proton is not shown (i.e. it's not the hydrogen atom attached to the carbon). Presumably it stays somewhat spread over the entire ion, staying only weakly coordinated to the chlorine atoms.

Like I commented earlier, I wanted tu exclude carboborane acids... Also non-cationic species would be welcome unless you know something that would be stronger than triprotonated methane and still stable :)

@Mithoron I added a new answer excluding carboborane acids, based upon new research.

I think the recent article by Zenaidee et al.1 answers this question well.
They managed to generate highly protonated cytochrome c molecules under MS-ESI conditions, which are acidic enough to protonate noble gases such as Ar according to the equation:
$$\ce{[cyt $c$, 23H]^23+ (g) + Ar (g) -> [cyt $c$, 22H]^22+ (g) + ArH+ (g)}$$
Of course such protein molecules can not be isolated under normal conditions (not yet at least). More details can be found in the article. The abstract is as follows:

The basicity of highly protonated cytochrome c (cyt c) and myoglobin (myo) ions were investigated using tandem mass spectrometry, ion–molecule reactions (IMRs), and theoretical calculations as a function of charge state. Surprisingly, highly charged protein ions (HCPI) can readily protonate non-polar molecules and inert gases,including Ar, O2, and N2 in thermal IMRs. The most HCPIs that can be observed are over 130 kJ mol−1 less basic than the least basic neutral organic molecules known (tetrafluoromethane and methane). Based on theoretical calculations, it is predicted that protonated cyt c and myo ions should spontaneously lose a proton to vacuum for charge states in which every third residue is protonated. In this study, HCPIs are formed where every fourth residue on average is protonated. These results indicate that protein ions in higher charge states can be formed using a low-pressure ion source to reduce proton-transfer reactions between protein ions and gases from the atmosphere.

Zenaidee, M. A.; Leeming, M. G.; Zhang, F.; Funston, T. T.; Donald, W. A. Highly charged protein ions: the strongest organic acids to date. Angew. Chem. Int. Ed. 2017, 56 (29), 8522–8526. DOI: 10.1002/anie.201702781.

This may be a good answer to the question, but posting bare references with a screen capture of the abstract is frowned upon on the site. It's hard to read, and completely unsearchable.  Could you replace it with some sort of summary of your own?  You could absolutely quote the key bits.

Oh, agreed, I don't have a plagiarism concern here. The image of the paper contents is not searchable. Were you to type up a description, including quotations, then the quoted material and your commentary would be searchable by the site's search feature and other engines like Google, Bing, etc. As it is, the only searchable text in your answer is the citation itself, in which the only real keywords are the authors' names...not particularly topical.

Even a sentence or two like what you put in your comment would strongly improve the post -- you could copy-paste part or all of the abstract into a blockquote, too. Wouldn't take long.

Btw, please don't take the above critique the wrong way: it's a great find, and a great answer to the question. Just needs some buffing. :-)

No, not at all. Just not sure i quite understand that is all. I have edited it a bit into what i think addresses the comment but do let me know if anything else is required.

That works! The idea is that the post itself will show up in searches. Before, if someone searched 'protonated cytochrome c' on Google or wherever, it would never be in any results list. Now, it can be.

Ohhhh, ok now i see what you mean by searchable. I thought you refered to the reference and the concept in the answer rather than the post itself (hence why i was puzzled) . Makes sense now, will take it into account for any other answers, thank you.

Yeah, I need to find a better word than 'searchable'.

Shout out to the PI that made the Zenaidee paper possible W. Alexander Donald and the acid used in the process fluoroantimonic  acid  that has a Hammett acidity function of -31.3

There is a recent article Design of the strongest organic Brønsted acids in gas phase Chemical Physics Letters (2017) Volume 681, Pages 50-55.

In this work, a new class of superacids containing poly enol groups was designed and their acidities were examined theoretically in gas phase. The conjugate bases of the superacids were stabilized due to formation of an aromatic ring and a hydrogen bonding network. To enhance the acidity, the electron withdrawing groups, F and CN, were substituted into the structures so that some cyano-derivative acids with $ΔH_\mathrm{acid} = \pu{223–243}$ and $ΔG_\mathrm{acid} = \pu{215–236 kcal/mol}$ were obtained. These thermodynamic data show that these compounds are the strongest organic acids in gas phase that have already been reported. Although $\ce{CB11(CF3)11H1H}$ and $\ce{B12(CF3)12H2}$ are stronger than the proposed acids they are not organic compounds.

In Table 2 the article lists theoretical $\mathrm{p}K_\mathrm{a}$ values in DMSO, the lowest of which is $−28.4$ for this class of compounds.

A relatively strong acid that does not employ borane, cyano, or halogen functions, containing only carbon, hydrogen and oxygen, is croconic acid.  The structure of this acid from Wikipedia is given below:

(From WP user rohnjones) Wikimedia link
Croconic acid has two stages of dissociation, given in Wikipedia as $\pu{pK_{a1}}=0.80, \pu{pK_{a2}}=2.24$.  The second stage dissociation especially stands out, being about two  $\pu{pK_a}$ units or more stronger than most polycarboxylic acids (e.g., oxalic, malonic, tartaric, citric) and also stronger (in both stages) than other "ring" acids such as deltic or squaric.
The relatively strong diprotic acid character of this compound comes from the di-anion having aromatic character (mentioned in the Wikipedia article) plus the ability of a five-membered conjugate ring to take up negative charge, which is also compatible with aromatic coupling.  The latter leaves less excess negative charge on the oxygens than is typical in acids containing carbon, hydrogen and oxygen only.

I wouldn't be surprised if rhodizonic acid is stronger in non-aqeous solutions  https://chemistry.stackexchange.com/questions/158685/acidic-strength-comparison-of-squaric-acid-and-rhodizonic-acid

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is a sacrificial agent?
I came across this reading a journal article. Here I have the context:

Here we report for the first time the synthesis of $\ce{TiO2}$ NPs
  on a layered $\ce{MoS2}$/graphene (MG) hybrid for use in
  photocatalytic H2 production. It is shown that the activity of
  the $\ce{TiO2}$ NPs is significantly enhanced by the presence of this
  layered MG cocatalyst. In this case, ethanol was used as a
  sacrificial agent, as it is a sustainable and renewable source and
  showed very good performance with this photocatalyst;
  however, other sacrificial agents (e.g., glycerol) can be used
  to make this strategy feasible.

Google results came up with similar instances.
My guess is that it would be a chemical consumed in a reaction, but one that doesn't contribute to the structure of the product. Thoughts?

"Sacrificial agent" usually refers to something variable, but with little value, that can be used to produce something specific or carry out a specific function. In this case, an alcohol (ethanol or glycerol) is sacrificed to act as a source of hydrogen gas, while the titanium species is a catalyst that is activated by light to allow generation of hydrogen.


text:
Why is water transparent?
Some substances like copper sulfate for example have vivid colors. But why is water transparent? Does it not emit any visual light from the electromagnetic spectrum?

A portion of the explanation can be found at this site.

@NicolauSakerNeto So it is because of the reflection of the light and the interactions between the photons with light. But however, I thought water was transparent. Take tap water for example. Why transparent? I understand why it is blue (because of reflections and adsorptions) but how & why is transparent?

did you mean "colorless"?

Water is not, in fact, colourless, but is slightly blue. As you can see below, in the visible range, there is slightly more absorption toward the long wavelengths, so more blue light makes it through than redder light. However, the absorption is very low in the visible range so the blue colour is faint and not apparent unless looking through a significant thickness of water. The reason things like $\ce{Cu^2+}$ solutions appear so vividly blue is because they also absorb red light, but very strongly, so a good deal more blue light than red passes through the solution. (see here)
I'm not sure exactly what you mean by transparent here, but if you mean why isn't water opaque, because water doesn't absorb much visible light (as described above), and doesn't contain any non-uniformities (like particles or the fat globules in milk) that can cause scattering, photons can travel through water in a straight line and most make it through without being absorbed, making water colourless (when looking through a small quantity) and transparent, much as air is.

Source

As for the transparency, water is not really transparent, see: https://en.wikipedia.org/wiki/Deep_sea

That was kind of my point. Transparency generally refers to the ability to pass light without scattering. If you have enough of something that it can prevent light from passing due to absorption, it's said to be opaque. If you have less and some light gets through without scattering, then it's partially transparent. Transparency is a property of a certain system, not an intrinsic properly of a material. Even  metals and such can be said to be partially transparent in thin films.

@MichaelD.M.Dryden Thank You for clearing this up.


text:
How can I interpret the colour of flames in terms of spectral series?
I saw different colors of fire burning different metals. But how does this work? The Balmer series requires electrons to drop back down to n = 2. But, not all alkali metals have empty space in their second principle energy level. Does one electron, from the second principle energy level, just jump to somewhere of energy and come back down.
My description is kinda confusing, but all I'm  saying is, where does that space in the second energy level come from?

The Balmer series relates to the electronic transitions in the hydrogen spectrum. In other elements it only requires that the energy difference between different excited states corresponds to a wavelength in the visible spectrum. These will clearly differ depending on the effective nuclear charge of the elemnt concerned.

Just to complete the previous answer, the Balmer serie isn't the only desexcitation serie producing visible light spectrum for atom nucleus.
But if you are speaking about the Hydrogen, for instance you also have the Paschen serie for electron down to the $n = 3$ level.
So it's the same for other elements, there is no special need for an electron to jump down on the precise level $n = 2$ to produce a photon within the visible spectra during its desexcitation.

Pache series does not produce visible light

Well, I don't know about Pache serie. But it's true that Paschen Serie does not,  althought it was just to illustrate the fact an electron has no vital need to jump down to the $ n = 2 $ energy level to produce a photon during the desexcitation.


text:
More acidic - protonated carbonyl or protonated hydroxyl?
Consider two acetic acid derivatives, one with a protonated carbonyl oxygen and one with a protonated hydroxyl oxygen.
An argument for the one with the protonated hydroxyl oxygen being more acidic is as follows:
This structure has a lone pair on its protonated hydroxyl oxygen and the oxygen is adjacent to an $sp^2$ carbon. Therefore we can draw an extreme resonance structure in which the hydroxyl oxygen has a +2 formal charge. No such "extreme" structure exists for the one with the protonated carbonyl oxygen.
Now, we can definitely draw this resonance structure, but its contribution is likely very small and bordering on insignificant. I can't see how delocalization of electrons away from a highly electronegative, positively charged oxygen can help stabilize it.
Therefore: Which is more acidic? The argument that the protonated hydroxyl one should be more acidic seems to be flawed at best.

Protonated ketones are not a good reference point, because they are not able to delocalize positive charge on oxygen as shown on the picture above for protonated carbonyl group of carboxylic acid.

You have outlined the correct reasoning but drawn the wrong conclusion. The protonated carbonyl structure (top left) is stabilized by resonance (bottom left), giving two equivalent resonance structures. Although the protonated hydroxyl (top right) has a resonance structure that can be drawn (bottom right), it should be considered a negligible contributor since there is a double positive charge on a very electronegative atom that would be sp2 hybridized (also toss in the charge separation). Since the protonated hydroxyl is not stabilized by resonance, it is more acidic, i.e., it is more willing to give away the proton (and keep the electrons).

According to this table, protonated ketones have a pKa of ~ -7.3, and the pKa of the hydronium ion is -1.76.  I think the proper argument is that the electronegativity of oxygen in acetone is greater than in water.

I don't think this is a valid comparison because the hydroxyl (of a carboxylic acid) will be strongly electron donating compared to the weaker methyl group (of the ketone).


text:
Do we need to standardize 10M of NaOH?
I made $10~\mathrm{M}$ of $\ce{NaOH}$ solution by diluting $40~\mathrm{g}$ of its pellets in distilled water. I'm not sure if it's really $10~\mathrm{M}$ or at least close enough, but I don't think standardize can be done in such concentration. I've been careful enough in making the solution such as measuring the mass in low humidity room and closing it with plastics/glass.

This is a highly concentrated base, which is very corrosive. I would assume, that its concentration is quite unstable.
It will, for example, quite rapidly react with carbon dioxide from the air:
$$\ce{NaOH + CO2 -> NaHCO3}\tag{1}$$
Also it will corrode glass containers, dissolving minerals like magnesium and calcium from it:
$$\ce{Mg^{2+} + 2NaOH -> Mg(OH)2 v + 2Na+ }$$
A PE bottle should be used to avoid that.
Even if you worked in a very clean and low humidity environment, the surface of your pallets will be corroded since sodium hydroxide is hygroscopic and will further react according to $(1)$, even when stored carefully.
Therefore the concentration you obtain by weighing the pallets can only be quite approximate. If you need to know the exact concentration of your base, you have to calibrate it versus a primary standard, like dried succinic acid, before and after the intended use.
In your case standardisation will indeed be not easy to perform, since your concentration is very high. Most likely there are uncertainties to the procedure that will likely leave you with similar error bars than not doing the calibration.

I am in complete agreement with the answer above. I am just curious as to why you would need to know the exact concentration of such a very concentrated solution. However, If you do need to standardise it with any degree of accuracy, you will need to dilute it quantitively to match the concentration of the standard acid you are using in order that the volumes used match each other as closely as possible.

actually this is for my undergraduate research and since some of my friends asked me whether I have standardized it or not, I became curious about it (and be prepared if my examiners ask). I'm using it to make titania nanotube via alkaline hydrothermal method. thank you for of your answers, both of you :)


text:
How to detect azide group on a primer?
I bought azide-labeled primer for click chemistry, but it didn't work. So I want to prove the existence of azide group. How could I achieve this? I found a method to add PPh3 to Kaiser test reagents, but the concentration of the primer is too low (the protocol needs the solution be higher than 0.1M) and I cannot see the reaction.

Does click reaction work with labeled nucleotide?

No. That's why I would suspect there's problem with azide group :(

One possibility would be to check the IR spectrum.  If there's any azide in there the azide band should show up as a strong band at around 2150 cm-1.  The trouble is of course that this isn't quantitative, you can only prove the absence of azide, not how much is present.
The Staudinger reaction might be a viable option.  PPh3 won't go up in aqueous solution, but sulfonated PPh3 will.  One should be able to distinguish starting material and product by 31P NMR.  With a known quantity of TPPTS you can even determine % azide in the starting material.


text:
Why is the dipole moment of ethyl chloride larger than that of vinyl chloride?
The dipole moment of ethyl chloride is 2.10 D (J. Mol. Struct. 1990, 216, 9–26), and that of vinyl chloride is 1.42 D (J. Chem. Phys. 1960, 32 (1), 205–209).
Why is this the case? I would have thought that hyperconjugation would lead to vinyl chloride possessing a larger dipole moment instead.

I see two explanations for why the dipole in vinyl chloride is smaller than ethyl chloride.

Consider a resonance structure where the chlorine donates electrons to the alkene. This puts a formal positive charge on chlorine:

This would be a very minor contributor given the electronegativities and the mismatch in orbital size. However, this is not possible in ethyl chloride at all.
The more likely explanation is that an sp2-hybridized carbon is more electronegative than an sp3-hybridized carbon. Since the electronegativity of chlorine will be closer to the electronegativity of the carbons in vinyl chloride than the carbons of ethyl chloride, the C–Cl bond will be less polarized in vinyl chloride, giving a smaller dipole.

I like your intuition. But I'm skeptical of saying carbons with different hybridizations have different electronegativities.

There are two factors:

The bonding orbital of carbon in $\ce{C2H3Cl}$ is an $\mathrm{sp^2}$ hybrid orbital while $\ce{C2H5Cl}$’s is an $\mathrm{sp^3}$ one. The former has a higher s-content and a lower p-content resulting in it being closer to the atom resulting in a shorter bond. A shorter bond means less charge separation.
$\mathrm{sp^2}$ carbons are also more electronegative than $\mathrm{sp^3}$ carbons; see the discussion on how to interpret the high acidity of phenol.

Don't forget that dipole moment depends not only on electronegativity differences but also bond lengths. All other factors held equal, the longer the bond, the greater the dipole moment.
Why would dipole moment depend on bond length? Consider an extreme example in which, say, sodium ion were in exactly the same place as fluoride ion. Obviously the two differ tremendously in electronegativities. However, there cannot be a dipole moment; there are no poles of opposite charge since the two are directly on top of each other.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is a good (small) model system for ligand metal charge transfer practises?
I am looking for an easy to calculate model system in which I could see a ligand metal charge transfer. Basically I am looking for a model complex, that might have the metal/ligand in different oxidation states: $\ce{ M^{III}(Lig)}$ versus $\ce{M^{II}(Lig^{\bullet +})}$ or $\ce{M^{IV}(Lig^{\bullet -})}$. I am calling it a charge transfer since I am not sure how else to call it.
I am also quite aware, that the description of these systems is necessarily a multireference case, so ordinary DFT methods will fail. (At least broken symmetry is necessary.)
In principle I was thinking about an iron heme complex that might exhibit that feature.

However, I have the feeling, that this kind of system is already pretty demanding to calculate the electronic structure. Since I would like to use it as an exercise and for a tutorial achieving alternative occupations I would prefer a much simpler case. It should eventually treated with broken symmetry DFT and CAS/MR methods.
Does anyone know of such a system or would suggest one. It does not have to be a real one. Purely hypothetical is just as fine.

AFAIK, permanganate ion is colored because of intense line, corresponding to oxygene-manganese charge transfer. This is noted in wikipedia article on charge-transer complex, but, unfortunately, without quote, and I'm unable to come with one right now
Edit:
Aha. Here there is a MO diagram on $\ce{MnO4-}$ ion and extensive discussion of (mis)understanding of its spectra, including statement I made above.

Well that certainly looks very promising and thank you very much for the reference. I will definitely have a deeper look into it next week - I am at home now and have no access to the paper, but the abstract sounds like a system worth considering.

you can try $\ce{Cu^{2+}--Cys}$ vs $\ce{Cu^+--Cys}$, much smaller and it is a model of $\ce{Cu-S}$ clusters in eg $\ce{Cu}$-binding proteins
I'm interested in these systems too, maybe we can share information?
Goedele

Thank you very much for the suggestion. I will have a closer look at that. It might be suitable for my purpose, since it is organometallic chemistry. You state that it is used as a model for copper binding proteins, do you have any reference for that statement? I would like to read up more about that.


text:
Why formulate drugs as HCl salts when HCl is in stomach acid?
I know there are issues of formulation and industrial processing that make it advantageous to produce many amine containing drugs as their salts rather than as freebases. And if giving the drug intravenously, the increased water solubility is clearly an advantage. But if giving orally, is there not sufficient HCl in the stomach to have this occur in vivo? Is it just that some of these drugs are too non polar to even go into solution in the stomach acid? Or do we just need to avoid neutralizing too much of the stomach acid with a drug not intended for that purpose?

Another reason is that many drugs which are conjugated to hydrochloride salts is they are amines, and amines by themselves often stink. So people might be turned off from taking drugs that smell terrible (fishy).
Now, note that even though a label might say something such as "fexofenadine HCl" that doesn't mean there is HCl in the drug. Fexofenadine is an amine, and amines are somewhat basic - and more than basic enough for a strong acid such as HCl to completely protonate.
So if this is what you had in mind for the structure of fexofenadine, it isn't the most correct structure:

This would be a much better structure. The nitrogen in the ring has been protonated and it now bears a positive formal charge. Chloride ion is the counter ion.

Typically, a neutral drug species containing an amine nitrogen will have a higher melting hydrochloride salt that may be better than the "free base" in that it is easier to crystallize (and hence purify) for use in a drug formulation. But many types of salts are used besides the hydrochloride - it all depends on which salt gives good crystals and also works well from a physiological standpoint. For oral administration in the typical quantities used for drug doses, the counterion used probably will not make much if any difference, as it will be outnumbered by chloride and other species present in digestive juices.


text:
Can you concentrate alcohol by evaporating the aqueous solution at room temp and collecting on a cold surface?
I have a bowl containing cheap wine and an immersible fish tank heater. The heater warms the wine to 30 deg C. The bowl is located in a large closed bucket, where it sits on a support. The bucket is sited out of doors in winter when the air temperature is Ca 1 deg C. Evaporation occurs and the vapour condenses on the inner surface of the bucket. Droplets of condensate run down and form a pool of liquid on the floor of the bucket.
Question: what can be predicted about the composition of the condensate?
Question #2 What effect is achieved by raising or lowering the temperature of the wine on the composition of the condensate. In plain language what temperature of the wine will give me the highest concentration of alcohol in the condensate? How can the strength of the alcohol be measured?
Question #3 What is the effect of different air temperatures outside of the bucket (on the composition of the condensate)?
Question #4 Why is this method not commonly used to concentrate alcohol?

Yes you can do that, this technique is called 'distillation' (it requires a minimum difference of 10 degrees celsius to seperate two liquids with different boiling points, for efficiency), I don't know what temeperature the fish tank heater runs at, but you should know that ethanol i,e. what you are trying to separate evaporates at 37 degrees celsius same as normal body temperature, and the temperature of the cooling plate should be < 27 degrees for efficiency
Edit -
yes, exactly the boiling point of ethanol is 78.3 degrees but ethanol is highly volatile and evaporates very fast at 37 degrees (without taking much water vapour with it) as far as my observation is concerned(with room temperature around, 1 atm pressure). the values I mentioned are the optimal conditions with experimental background. your goal as I can percieve is to avoid as much water as possible, if you take the condenser temperature to below 10 degrees the water vapour present in air will also start to condense at a good rate

I have set the fish tank heater to 30 deg C, but I could set it as low as 20 deg C. How would the lower temperature affect the composition of the condensate? The "cooling plate" (the walls of the bucket) are at, or close to, the external air temp which is currently 1 deg C. So there is a  29 degree difference between the wine and the condensing surface. If the weather gets colder (the differential increases) how does this affect the condensate?

<< ethanol . . . evaporates at 37 degrees celsius>> My understanding is that ethanol boils at 78.3 C. but I am concerned with evaporation, which I think can occur at any temperature above the solid state (frozen ethanol) not solely at the 37 degrees you mention. The point of my question is to see if different conditions can improve the process in the aim of achieving a higher ABV.

This doesn't really answer the question at all.  Can you elaborate as to the other aspects of the questions the OP asked?

37 degrees celsius is a temperature for efficiency under room temperature i.e, 25 degrees. I have myself experienced the values experimentally in lab

the temperature difference mentioned by the questioner should also work efficiently, the only problem here is that if the setup is open to air it might condense water from air as well.

@jonsca, As far I can perceive the questioner is inquiring about distillation

yes exactly the boiling point of ethanol is 78.3 degrees but ethanol is highly volatile and evaporates very fast at 37 degrees (without taking much water vapour with it) as far as my observation is concerned(with room temperature around, 1 atm pressure). the values I mentioned are the optimal conditions with experimental background. your goal as I can percieve is to avoid as much water as possible, if you take the condenser temperature to  below 10 degrees the water vapour present in air will also start to condense at a good rate.

yes testing it out would be really good, you can use 20 degrees plus temperature but less than 27 degrees.

<<37 degrees celsius is a temperature for efficiency under room temperature i.e, 25 degrees>> Please define what you mean by efficiency. I don't understand what the meaning of"under room temperature" might be in this context. Please explain.

Please note as originally stated, this operation takes place in a closed container (the atmosphere is excluded).

<< if you take the condenser temperature to below 10 degrees the water vapour present in air will also start to condense at a good rate>> Can one predict by theory whether condensation of the vapor (the ethanol component) will occur at temperatures above  10 deg, or is experimentation the only way to know?

then you can dip the condenser temperature to 15 degrees any low again the water factor

not by theory not by calculating, its by both

<<then you can dip the condenser temperature to 15 degrees any low again the water factor>> For clarity, please state what you think are the best temperatures for my purpose. 1) Ethanol / water bath. 2) Condenser surface.

ETHANOL/WATER BATH 35 to 45 degrees and                              CONDENSER 15 to 20 degrees

if it did work please leave an update either in comments or in your answer, this helps future users

I have ordered an alcohol meter to measure the results. What I can say currently is that under the conditions of the first experiment (alcohol / water solution 30 C. , condenser 1 C.) concentration occurs but the output cannot be set alight - ie less than proof strength. I will report back when I have new data.

The alcohol hydrometer which I obtained gives only coarse indications for dilute concentrations. It indicated about 18% alcohol which was evaporated and condensed out of a 11% base, which was heated to 30C. Room temperature (the condenser surface) 17C. Production rate about 300 ml in 24 hours with my set-up. I don't have appropriate measuring equipment to see what the effects of varying the temps would be.

The answer to my principal question is that you can concentrate an aqueous ethanol solution by "room temperature distillation". I don't have the right equipment to be able to determine what variation in output quantity and strength might be obtained by varying the temperature of the source solution and the condenser surface. But I can state that the alcohol concentration in the condensate under the conditions I used finally was ca 18% coming from the source solution which was 11%. What I like about the process is that little energy is used for heating(30 watts) and no water is wasted for cooling as in conventional distillation.
Further experiment - from a wash of sugar and beer yeast using the method above I obtained a distillate which measured 36% on my $3 Russian alcohol hygrometer.

When you say "little energy is used" what you mean is that the rate of energy use is low. Given that the low temperature distillation will take a lot longer, I don't think you can comfortably say the total use is lower.


text:
Proper typography of "NOx"
I am a journal typesetter, and recently I came to a problem that I'm not sure I know to solve. The question is how to properly typeset the "NOx" abbreviation for nitrogen oxides. Which is typographically correct:

$\mathrm{NO}_x$ (because $x$ is a numeric variable); or
$\mathrm{NO}_{\mathrm{x}}$ (because all names of chemical compounds are strictly upright)?

I quote the Green Book by IUPAC, 2nd printing (2008), section 1.6, enumeration item 2:

The overall rule is that symbols representing physical quantities or variables are italic, but symbols representing units, mathematical constants, or labels, are roman. [...]

As such, the correct way to write it is $\ce{NO}_x$, because $x$ is a variable. Please note that it is only the sum formula for the compound and not its name. Since the element names still are in roman (upright) font, everything is fine.
The same goes for a formula like $\ce{C_{$n$}H_{$2n+2$}}$, which is typeset exactly like that in the Gold Book (hat tip to @Martin).
Here at StackExchange, you can type it as $\ce{NO_$x$}$ ($\ce{NO_$x$}$) or $\ce{NO_x}$ ($\ce{NO_x}$).

Note that I haven't been able to find a direct reference or instruction by IUPAC, so I might well be wrong here. I'm just combining some rules in a logical way.

Well, the analogy to $\mathrm{C}n\mathrm{H}{2n+2}$ is convincing me quite.

@tohecz The IUPAC goldbook typesets it exactly this way.

The American Chemical Society recommends $\ce{NO}_x$: page 191 of Appendix 10-2, Abbreviations, Acronyms, and Symbols of their style guide (PDF).
Unfortunately, it's just in a list of abbreviations and I've not managed to find any definitive explanation of why it should be typeset that way, elsewhere in the style guide. However, tschoppi's explanation is what I would have said, too.

The IUPAC Nomenclature of Inorganic Chemistry (‘Red Book’) Recommendations 2005 reflects on this subject.

IR-2.9 ITALIC LETTERS
(…)
(g) Other uses of italicized capital letters are as locants in substitutive nomenclature (see, for
  example, Section IR-6.2.4.1), and the letter H for indicated hydrogen (see, for example,
  Section IR-6.2.3.4). Italic lower case letters are used to represent numbers, especially in
  formulae where the numbers are undefined.
Examples:

$\ce{(HBO2)_n}$
$\ce{Fe^{n+}}$

$\ce{NO_x}$ is analogous to the examples mentioned above.
Some loosely related examples with ‘x’ variable:

IR-2.2.3 Parentheses
(…)
(f) To indicate the composition of a non-stoichiometric compound.
Examples:

$\ce{Fe_{3x}Li_{4-x}Ti_{2(1-x)}O_6} ~ (x=0.35)$
$\ce{LaNi5H_x} ~ (0<x<66.7)$

or

IR-11.3.2 Phases with variable composition
(…)
Examples:

$\ce{Cu_xNi_{1-x}} ~ (0 \le x \le 1)$ is equivalent to $\ce{(Cu,Ni)}$.
$\ce{KBr_xCl_{1-x}} ~ (0 \le x \le 1)$ is equivalent to $\ce{K(Br,Cl)}$.


text:
Auto ignition of tert-butyl mercaptan mixed with mineral oil
I am designing an experiment that involves immersing pieces of pipe lining in a mixture of tert-butyl mercaptan (2 %) and white mineral oil (98 %) at 230 °C (the standard temperature for the test). If I was using pure tert-butyl mercaptan I would be concerned with it exploding, due to its auto ignition temperature being 250 °C. However, if the white mineral oil contains only 2 % tert-butyl mercaptan would the risk of autoignition be present at 230 °C?
I figure Mineral oil has vapour pressure of less than 1mm Hg. Tert butyl mercaptan has vapour pressure of 303 mm Hg. For the tert butyl mercaptan (2%) in mineral oil (98%) mixture I calculated that the partial pressure for tert butyl mercaptan would be 6.06 mm Hg and for mineral it would be 0.098 mm Hg. Therefore, the composition of gas above the surface of the mixture would be tert butyl mercaptan 98% and mineral oil 2%. If this is the case then the auto ignition temperature of pure tert butyl mercaptan of 250 °C would apply to the resultant gas and it would not be safe operate at 230 °C. Is this right?

What is "pipe lining"? Are you intending to run this experiment in air? I hope not, but then how should anything auto-ignite?

authors: ธนวัฒน์ ราชแข็ง


text:
How is bismuth(III) able to form 3 bonds with hydroxide molecules in bismuth(III) hydroxide?
Bismuth(III), if I am not mistaken, has only two valence electrons in the 6s orbital. However it is able to form 3 bonds with hydroxide. This goes against my intuition. How is it able to form 3 bonds with 3 hydroxide molecules with only 2 valence electrons?
my best guess is that i am mistaken in thinking that the bismuth(III) is forming 3 covalent bonds when in actually it is forming ionic bonds. As a result, bismuth does not need to offer/share 3 electrons to each hydroxide, but simply take 3. Is this correct?

There isn't a unique sturcture for bismuth hydroxide, but ions such as
$$\ce{[Bi6O4(OH)4]^{6+}}$$
are involved, where the 6 Bi atoms are at the corners of an octahedron, none of the Bi atoms are directly bonded to each other, and O atoms bridge pairs of Bi atoms.
See Solvation of the Bismuth(III) Ion by Water... Inorg. Chem. 2000, 39, 4012-4021 as a starting point.
But supposing the structure was $\ce{Bi(OH)3}$, Bi3+ has two electrons, each OH- has 8 electrons, so Bi could share 2 electrons from each OH- to form an octet.

Bismuth hydroxide has a trigonal planar geometry, meaning it has no lone electron pairs. It would only be borrowing 2 electrons from each hydroxide, giving it only 6 valence electrons, not an octet.

@ByronS If Bi3+ has two valence electrons, and borrows 2 from each of 3 hydroxides, isn't that 8 altogether?

Yessir, but in doing so, it would take on a "pyramidal" shape in terms of geometry (3 covalent bonds, 1 lone electron pair). However, my resource is telling me that bismuth(III) hydroxide takes on a "trigonal planar" geometry, meaning it forms 3 covalent bonds and has no leftover lone electron pairs. Is my resource incorrect?

@ByronS I don't have any resources that even say the such a molecule exists, are you sure it isn't boron instead of bismuth in your resource?

No, it is bismuth. My resource is most likely wrong, considering it is simply a question on an old exam made by my professor. Question asks for the molecular geometry of "Bismuth(III) Hydroxide". According to your logic, which also makes sense to me, it should be forming a pyramidal shape, but my professor claims that it takes on a trigonal planar shape. I'll have to speak with him about it soon. Thank you for the help.

@ByronS I think he's wrong, a very bad question to put on an exam in any case


text:
Where does the molar mass difference between N₂ and CO come from?
The molar masses of $\ce{N2}$ is 28.0134 g/mol, for $\ce{CO}$ it's 28.0101 g/mol (WolframAlpha computation).
Of course, they do have the same mass in atomic units (28). My question is, where does this small discrepancy come from? Different binding energies in the respective nuclei? I know that oxygen is a "double-magic" nucleus from an atomic physics POV, that is, it's binding energy per nucleon is higher than what would be expected. But since normally, you equate mass with energy, shouldn't $\ce{CO}$ be heavier then?

The differences in molecular mass stem from two sources:

Nuclear binding energy

The definition of the atomic mass unit and Avogadro's number (and thus the g/mol), is the mass of one atom of the carbon-12 isotope $\ce{^{12}_6 C}$. 1 amu = $\frac{1}{12}$ the mass of one atom of $\ce{^{12}_6 C}$ and the g/mol unit is based on the definition of Avogadro's Number, which is the number of $\ce{^{12}_6 C}$ atoms in 12 grams of $\ce{^{12}_6 C}$.
Because of the differences in binding energies in the various nuclei, the monoisotopic mass of individual isotopes are not integers. For example, the mass of $\ce{^{16}_8 O}$ is 15.99491463 amu (and not 16). The monoisotopic masses of the most common isotopes of oxygen, nitrogen, and carbon is shown below:
           mass number   % absundance    isotope mass
Carbon     12            98.930          12 (defined)
Nitrogen   14            99.632          14.003074
Oxygen     16            99.757          15.99491463

Isotopes

Generally, molecular masses (unless otherwise specified) are average molecular masses composed of average molecular masses of the elements based on their isotopes. For example, even though the most common isotope of carbon has a mass defined as 12 amu, there is another naturally occurring isotope $\ce{^{13}_6 C}$ with 1.07% natural abundance, leading to carbon having a non-integer average atomic mass of 12.01 (to 4 significant figures). There are two isotopes of nitrogen, and three isotopes of oxygen.
            Average atomic mass
Carbon      12.011
Nitrogen    14.007
Oxygen      15.999

Thus, $\ce{CO}$ has a molecular mass of $12.011 + 15.999 = 28.010$ and $\ce{N2}$ has a molecular mass of $14.007+14.007=28.014$. Even their monoisotopic masses are different:
                    average molecular mass       monoisotopic mass
dinitrogen          28.014                       28.006148
carbon monoxide     28.010                       27.99491463

Just a little nitpick: You are using the abbreviation for atomic mass unit, which was originally defined through $$\mathrm{amu}=\frac{1}{16}m(\ce{{}^{16}_8O}).$$ It has been replaced with the unified atomic mass unit, which is based on the definition you used, hence $$\mathrm{u}=\frac{1}{12}m(\ce{{}^{12}_6C}).$$ While this usually does not make a big difference, the latter should be used exclusively. Se also wikipedia for a short historical summary.

So you are arguing that over all the $\mathrm{CO}$ and $\mathrm{N}_2$ molecules there are in a mol, the mass difference comes from the isotope mixing? What would happen if I looked at two specifically picked $\mathrm{CO}$ and $\mathrm{N}_2$ molecules where I know that I have nitrogen-14 and carbon-12 with oxygen-16? Which one would be heavier then?

@PMPJohn - I calculated the monoisotopic mass in my post already. Due to the nuclear binding energy, $\ce{N2}$ would still be heavier (and in fact the mass difference for the monoistopic version [0.011] is larger than the average atomic mass version [0.004]).

There are three factors to consider:

The natural abundance isotope composition of each element.

Carbon isn't all carbon 12, there is 1.1% Carbon 13.
Oxygen isn't all oxygen 16, etc.

The neutron-proton ratio of each isotope. The mass of a neutron is greater than the mass of a proton plus electron. Isotopes having a higher neutron-proton ratio relative to carbon 12 (which is 1:1) will be relatively heavier per nucleon.
The binding energy of each isotope.

And binding energy is the winner!  I would also add that N-14 does not have a mass of 14.000000 amu (well, because of the binding energy of the nucleus being different per nucleon than for C-12). So N-14 has an official mass of 14003074.00443 micro-amu per the 2012 atomic mass evaluation (AME2012). O-16 is 15994914.61957 micro-amu (so more tightly bound than C-12).  C-12, of course, has a mass defined as 12000000.0 micro-amu.  You can look these up through the National Nuclear Data Center based at http://www.nndc.bnl.gov/ (Brookhaven national Laboratory).


text:
How can a waterproofing material be "water-based"?
I've been trying to determine how a particular waterproofing material, which claims to be "water based", can still be hydrophobic. First of all, the product in question is EPRO ECOLINE-R, which also claims to be an asphalt emulsion. My first guess is that the curing process causes a change in some way to become hydrophobic once applied. But I just can't wrap my head around how a product, which is designed to resist water absorption, can also be water based? I can find no information on the exact chemical makeup, other than "petroleum hydrocarbons".

One might need to have more information about the product you've posted to say what occurs in the specific example. But, the general concept could be similar to Latex (water-based) paint. Where the water proofing chemical is dispersed by an emulsion, and once the water evaporates, the proofing material is restructured by its, now unhindered, intermolecular forces. These forces might be strong enough, once the proper structure exists, to resist water coming back and trying to interrupt the interactions.
Below is the Wikipedia explanation for my example of latex paint.

The term "latex" in the context of paint in the USA simply means an aqueous dispersion; latex rubber from the rubber tree is not an ingredient. These dispersions are prepared by emulsion polymerization. Such paints cure by a process called coalescence where first the water, and then the trace, or coalescing, solvent, evaporate and draw together and soften the binder particles and fuse them together into irreversibly bound networked structures, so that the paint cannot redissolve in the solvent/water that originally carried it.

After reading up on the article you linked and a few others on latex paint, I think you nailed it on the head. Water is the vehicle with which the petroleum product is applied, which is why it can still be applied to a damp surface. The curing process is essentially the water evaporating out of the porous structure. This makes me question, however, how water cannot be reintroduced through the pores after curing. Perhaps as the petroleum product naturally outgasses, the water trapped within gets pushed out to the surface and effectively closes the porous pathway to the interior?

Such a question goes beyond my understanding. Its possible there's a strong lattice structure and chemical bonding, such as oxidation, happens at the edge, preventing a reintroduction of water. We would have to know the formula and structure of the material to really speculate about that.


text:
How to calculate the molar mass of a protein from its osmotic pressure?
I am unsure how to go about solving this question. I know how to get the osmotic pressure, but where do I go from there?
I can't use the gas law since the protein is not a gas.

$\pu{2.38 g}$ of a protein is dissolved in $\pu{100 mL}$ of water at $\pu{25^\circ C}$. The solution rises to a height of $\pu{8.5 cm}$ in an osmometer (a device for measuring osmotic pressure). What is the molar mass, in grams per mole, of the protein?
Data: assuming the water ls pure, then the osmotic pressure $\pi = \rho gh$, where $h$ is the height of the water, $\rho$ is the density of water ($\pu{1 g/cm^3}$),and $g$ is the acceleration due to gravity ($\pu{9.81 m/s^2})$.

UPDATE: I've spoken with a masters student (my TA) about this and it turns out that the way to solve this is to use an osmotic pressure equation which is identical to the ideal gas law, but only because it is an approximation. If this question is unlocked I would be happy to share my answer.

I'd like to answer this question myself now. I didn't want to give my original workings when I asked this question because I did not want to lead people into the same method that I used. The reason this confused me so much was because the osmotic pressure equation used is identical to the ideal gas law--which I had a long discussion with a masters student and have obtained a basic grasp of the concept (though the relation to the ideal gas law is still odd); anyway, if someone unlocks this I would be happy to provide an answer myself.

Use osmotic pressure $Π=iMRT$
From this get your concentration and calculate molarity of solution.
By the way if you are looking more on this topic then you must consider the chapter on solutions and their colligative properties in your chemistry textbooks

It seems to me that you should use Morse equation $\mathrm{\Pi=iMRT}$ - from your data you should get  $\mathrm{M}$ - molarity and then molar mass of the protein. You can skip $\mathrm{i}$ - it's not a problem if it's a dimer or oligomer.


text:
Reactions of tin (IV) iodide with potassium iodide and water
In the reaction of tin (IV) iodide in acetone and water, the solution turns a urine yellow.
Is this simply due to the hydrolysis of tin (IV) iodide:
$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$
why is the acetone needed though?
and for the reaction with $\ce{KI}$:
$$\ce{SnI4 + KI -> ?}$$
I understand it forms iodine, but what happens to $\ce{Sn}$ and $\ce{K}$ - do they react with the acetone for instance?

If iodide is oxidized to iodine (the cause of the yellow color), then something needs to be reduced in turn. Potassium is already in its most stable oxidation state (+I), so tin will be reduced, from $\ce{Sn}$(IV) to $\ce{Sn}$(II).
$$\ce{SnI4 + 2H2O -> SnO2 + 4HI}$$
$$\ce{SnO2 + 2HI -> I2 + SnO + H2O}$$
With $\ce{KI}$, a yellow-brown solution of $\ce{KI3}$ will be formed, which contains the triiodide anion $\ce{I3-}$.
$$\ce{I2 + KI -> KI3}$$
When acetone is added, the yellow color of $\ce{I2}$ will slowly disappear as iodoacetone and $\ce{HI}$ are formed in an acid-catalyzed reaction (source):
$$\ce{(CH3)2CO + I2 -> CH3C(O)CH2I + HI}$$


text:
The toxicity of carbon monoxide in terms of its interaction with haemoglobin
On the basis of its interaction with haemoglobin, explain the toxicity of carbon monoxide.
I understand that ligand substitution is involved and carboxyhaemoglobin is formed meaning that the haem group in the haemoglobin molecule cannot form a ligand with the oxygen molecule, but i would like to know more in depth in terms of its chemistry, why this is the case.

Carbon Monoxide and dioxygen both reversibly bind to the iron atom in hemoglobin.
The two molecules compete for the same binding site.
So if both $\ce{O2}$ and $\ce{CO}$ are present, some of each will bind, but for a given amount of $\ce{O2}$, less $\ce{O2}$ will bind if $\ce{CO}$ is present. The affinity of the Fe for $\ce{CO}$ is much greater than for $\ce{O2}$, so a lower concentration of $\ce{CO}$ results in a given saturation level of hemoglobin compare to $\ce{O2}$.
The same would be true for cyanide which is isoelectronic with carbon monoxide.
A more complete explanation needs to go into the cooperation between the four hemoglobin subunits.  $\ce{CO}$ bound to two of the four subunits causes $\ce{O2}$ to being too tightly bound in the other subunits.  That the oxygen can not be released from the hemoglobin, is another aspect of toxicity.


text:
How to derive the pressure dependency for the Gibbs free energy?
$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$
The above is an exact differential. Firstly what does this mean physically? (I hear about the fact that a change in G is independent of the "path" but I don't understand this argument because I don't know what is meant by this path. Also how would something be dependent on the path?)
It can also be shown that $\mathrm{d}G$ can be written:
$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T $$
So by comparing coefficients:
$$\left(\frac {\partial G}{\partial p}\right)_T=V$$
It is useful to integrate this to see how $G$ varies with $p$:
$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$
$V$ can be expressed in terms of $p$ using the equation $V=\frac{n\mathcal{R}T}p$ Where $n\mathcal{R}T$ are constants. The above equation becomes:
$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$
Integrating the LHS of the above equation is the crux of my question (I don't understand how that can be integrated). I am not sure whether the following step is correct mathematically, please explain why this is correct/incorrect. I cancel out the $\mathrm{d}p$'s on the LHS:
$$\int_{p_1}^{p_2}\mathrm{d}G=n\mathcal{R}T\int_{p_1}^{p_2}\frac{\mathrm{d}p}p$$
Then the result is:
$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$
Or...
$$\Delta G=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$
This result is correct but I don't understand the left hand side. I've been told that the reason it's possible to write "$G(p_2)-G(p_1)$" or "$\Delta G$" is because $G$ is a state function/exact differential but, again, I don't understand this. If $G$ was not a state function/exact differential how would the result change?

This question has a lot of parts! I hope I manage to answer most but probably edits will be required.
$$\mathrm{d}G=\left(\frac {\partial G}{\partial p}\right)_T+\left(\frac {\partial G}{\partial T}\right)_p$$
What does this mean physically?
This is simply an expression that shows the contributions to the change in the Gibb's free energy, which varies with both pressure and temperature (i.e., $G$ is a function of $p$ and $T$). However, it is possible to separate the exact differential of $\mathrm{d}G$ into two terms using the mathematical expression of partial derivatives, first term is the variation of $G$ with pressure at constant temperature, and the second term is the variation of $G$ with temperature at constant pressure.
What does it mean for $\Delta G$ to be independent of the path? (same question: what does it mean for $\Delta G$ to be a state function?)
You can calculate a change in $G$, or equivalently, $\Delta G$, for any reaction. So for example for a reaction like a combustion process for glucose
$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g),}$$
you can calculate $\Delta G$ for this reaction at any particular temperature/pressure that you would like. And in fact, if you don't do the combustion directly but somehow do it via completely dissociated gaseous atoms
$$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 C~(g) + 18 O~(g) + 12 H~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$
You will still end up with the same $\Delta G$ for the overall reaction. Obviously if you calculate the $\Delta G$ for each individual step/arrow in the gaseous atoms case it will be very different, but they will combine to give the same overall $\Delta G$ as the direct combustion. In other words, $\Delta G$ won't be able to tell you whether you have done the combustion directly or done it laboriously via gaseous atoms, it just tells you what is the difference between the two states, first state being glucose + oxygen, second state being carbon dioxide + water. $\Delta G$ for glucose that you have eaten and combined with oxygen you breathe in and breathe out carbon dioxide + water would still the same $\Delta G$ as for combustion, even though now it is all done through cells, Kreb's cycle and a lot of biochemistry.
How would something be dependent on the path?
As you can see from my example above, the amount of effort you need to make $$\ce{C6H12O6~(s) + 3 O2~(g) -> 6 CO2~(g) + 6 H2O~(g)}$$ happen can differ a lot, and for each of the cases, what you "get back" from doing that reaction is different, such as heat released when burning glucose vs keeping your cellular biochemistry going with respiring glucose. In physics/physical chemistry, this is formally "work done". Work is dependent on path.
I am a little unsure whether you went from
$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T \tag1$$
to
$$\left(\frac {\partial G}{\partial p}\right)_T=V \tag2$$
in the right way, so I am going to go through it again.
At constant $T$, $\mathrm{d}T=0$. So $(1)$ becomes
$$ \mathrm{d}G=V\mathrm{d}p \tag3$$
which after rearranging is equivalent to $(2)$.
The integration then should go from $(3)$ to give
$$ \int_{p_1}^{p_2}\mathrm{d}G=\int_{p_1}^{p_2}V\mathrm{d}p \tag4$$
The reason why what you wrote look a little weird

$$\int_{p_1}^{p_2}\left(\frac {\partial G}{\partial p}\right)_T\mathrm{d}p=\int_{p_1}^{p_2}V\mathrm{d}p$$

is that you usually don't just cancel partial differentials like you did. Whereas $(3)$ is an exact differential, which you integrate over in $(4)$, and that is OK. Unfortunately I am not sure whether what you wrote is wrong, it is just something I would not write myself.
The rest of the derivation is fine.
Final question. I believe you are fine with mathematically going from $\int_{p_1}^{p_2}\mathrm{d}G$ to $G(p_2)-G(p_1)$, however, if you think about this result physically, $\Delta G$ does not depend on any other $p_3, p_4, p_5...p_n$ that the system might pass through on the way from $p_1$ to $p_2$. If $G$ is not a state function, then you would expect to have to consider at the very least all these different points of $p$ that you might have gone through, and different ways of doing the reaction passing through different $p$ would change $G$ even if the initial and final $p$ are the same. In that case, an integral like what we have written here would be a very poor way of finding $\Delta G$ because we are missing the contribution from all the intervening $p_n$, and we would have to use some different mathematical tools. In fact, I believe we would have failed to derive the master equation $(1)$ much earlier on, so the rest of the derivation would simply not follow.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Concerning your problem with "cancelling partial differentials" have a look here. Of course, you are right that it has nothing to do with cancelling like in normal products and that this is mathematically much more involved but nevertheless it's a good mnemonic because it usually can simply be treated just like that.

Thanks! I didn't realise you can use mhchem, should probably have tested it. Thanks for fixing my answer. Also thanks for clarifying that you can cancel partial differentials (and finding a German link, though I should probably clarify I have only recently moved to Germany!)


text:
Have Ionic Liquids been used to refine aluminum on an industrial scale?
Ionic liquids have been shown capable of reducing aluminum trichloride to solid aluminum at 3kw-hrs per kg of Al(s) produced instead of the current ~15kw-hr/kg. However, this would seem to neglect the fact that bauxite is aluminum trioxide not trichloride. This is no small detail and leads me to wonder if it is an insurmountable hurdle to the widescale adoption. Has this technology been implemented (or slated for implementation) on an industrial scale? If not, what holds it back?

I wonder what the cost of the ionic liquid and conversion to chloride is relative to the cryolite/alumina method.

I think you are misunderstanding the article linked in the question.
In the process described in the article, the starting material is A360 aluminum metal alloy.
The goal is to make very pure aluminum metal from impure aluminum metal.
The process is an electrolysis process.
$$\ce{Al(anode) + 7AlCl_4^{−} -> 4Al2Cl_7^{−} + 3e-}$$
$$\ce{4Al2Cl_7^{−} + 3e- -> Al(cathode) + 7AlCl_4^{−}}$$
The existing industry techniques referred to by the article as requiring ~15kw-hr/kg are also for refining impure aluminum metal to yield high purity aluminum.  This number is not referring to purification from bauxite.  The "three layer process" is one of the existing refining techniques, and it does involve ionic liquids, but the electrolyte is liquid only at high temperature.

Nice answer. Do you know if refining of Bauxite has been done with ionic liquids?

It isn't being done on an industrial scale, but it is being investigated. http://www.researchgate.net/publication/264695026_Preliminary_investigation_of_ionic_liquids_utilization_in_primary_aluminum_production


text:
Half reaction method justification
I have to say I have never really understood the half reaction method. What bothers me is that it seems to be too artificial. Too many little rules to remember!
Why don't we use the mass conservation law to justify the process and use sistematically a set of linear equations to do the balances?
Thanks in advance!

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

The concept of the half reaction has nothing to do with balancing redox reactions; how one arrives at a balanced reaction is not really important. The interesting thing about redox reactions is that electrons do not necessarily have to be transferred directly between the oxidized and reduced species. In many types of electrochemical cells, the anode and cathode are isolated from one another by a salt bridge and the species in each half-cell never approach one another. In techniques such as potentiometry, often one cares only about one half of the reaction, e.g. in a combination pH electrode, the response from the half reaction that contains hydronium ions produces the analytical signal and the the other half is a reference electrode reaction whose job is to simply produce a constant potential. In things like voltammetry, one is often only concerned about the half reaction occurring at the working electrode and the half reaction that balances it at the counter electrode isn't analytically significant (and may in fact be more than one different reaction).
Because the two half-reactions need not necessarily occur at the same place and one half reaction can generally be substituted without affecting the other, it makes sense to think of them separately. This lets us do things like think about different lithium batteries. There are many types and they all use the $\ce{Li -> Li+ + e-}$ oxidation half-reaction, but it's coupled with different reduction half-reactions based on cobalt, iron, etc.


text:
Electrophilic aromatic substitution with fluorine
Why don't we usually use diatomic fluorine in electrophilic aromatic substitution?
Is it because fluorine is a potent oxidizer and will react with the carbons?
What might happen if we reacted benzene with diatomic fluorine in the presence of some Lewis acid such as boron trifluoride? Would we get the expected product: monosubstituted benzene?

This should cover many aspects of your question.

Fluorine reacts so rapidly with benzene that aromatic fluorination requires special conditions
and special types of apparatus. Even then, it is difficult to limit the reaction to monofluorination


text:
What is the mechanism of a 1,2-hydride shift?
What is the mechanism of a 1,2-hydride shift? I read that there is an aromatic intermediate, in that the intermediate obeys Hückel's rule.
However, Wikipedia doesn't offer any further explanation. Can someone with more knowledge elaborate on the transition state of a 1,2-hydride shift?


text:
Finding the thermodynamics of protein unfolding from temperature and absorbance using fluorescence spectroscopy?
I recently did an experiment on protein unfolding using a Perkin Elmer fluorescence spectrometer. The protein was $\alpha$-chymotrypsin. I was given this paper for my procedure. I am now trying to make sense of the data.
Increasing temperature was used as the denaturant. Temperature and absorbance were plotted, resulting in an essentially sigmoidal curve describing unfolding. The paper gives the following equation for determining an equilibrium constant for the experiment.
$$K_u=\frac {X_N-X}{X-X_D}=\frac {[P_D]_{eq}}{[P_N]_{eq}}$$ Where $K_u$ is the equilibrium constant, $X_N$ is the absorbance for the native state, $X_D$ is the absorbance for the denatured state, and $X$ is any given absorbance. $P_D$ is the denatured protein and $P_N$ is the native protein.
The paper says that the free energy of unfolding can be described by the familiar equation: $$\Delta G = -RT\ln(K_u)$$
I'm fairly sure that I've calculated $K_u$ at each temperature recorded, but I'm not quite sure whether I use each $K_u$ as a point for another plot, or average them, or something else. If I use a Van't Hoff plot to graph each point of $\ln(K_u)$ vs. $\frac {1}{T}$, will the slope and intercept be $\frac{-\Delta H}{R}$ and $\frac{\Delta S}{R}$ respectively? I think so due to the rearrangement described here, and perhaps I've answered my own question as I attempted to ask it. Nevertheless, I would appreciate having my understanding of this calculation confirmed.
Update:
I've plotted $\ln(K_u)$ vs. $\frac {1}{T}$ and the slope of $\ln(K_u)$  vs. $\frac{1}{T}$ is not linear as had been expected. Linear regression analysis gives a trendline with an $R^2$ value of 0.87. Should I separate the data so as to graph $\ln(K_u)$ vs. $\frac {1}{T}$ for points leading up to the sudden unfolding and also for points once unfolding becomes dramatic? This would give me two values for the enthalpy of unfolding--one before it unfolds much and the other for when unfolding is happening quickly. If I do so, will the difference between those enthalpies be the overall enthalpy of unfolding? I'm not sure what the significance of those two values would be.

As I've looked more into the use of the Van't Hoff plot, it seems that only a maximum and minimum temperature are used. Should I pick these as the initial and final temperatures or the ones that are at the bounds of the steep part of the sigmoidal graph of absorbance vs. T?


text:
Melting and boiling points of benzene and fluorobenzene

This species is a derivative of benzene, with a single fluorine atom
  attached. Its melting point is -44 °C, which is lower than that of
  benzene, indicative of the remarkable effect of fluorination on the
  intermolecular interactions as seen throughout organofluorine
  chemistry. In contrast, the boiling points of PhF and benzene differ
  by only 4 °C.

Why is this true? Wikipedia doesn't elaborate on why.
From what I understand the melting point is the point at which the solid and liquid forms are in equilibrium.
Boiling point is the point at which the liquid and gas phases are in equilibrium.
I know that intermolecular forces play an integral role in melting and boiling point. However, why does it seem that the strength of the intermolecular forces in the solid phase is stronger than that in the liquid phase for fluorobenzene?

The answer has everything to do with fluorine.  Dunitz, in his early studies on fluorine interactions found that intermolecular $\ce{C-H..F}$ interactions can be solid-state structure (e.g., crystal packing) directing.  In the case of fluorobenzene, the molecule adopts a solid-state structure that sacrifices long-range order so that these short-range $\ce{C-H..F}$ interactions can be preserved (also see here, and here where it is stated that, "Its [fluorobenzene] melting point is $\pu{-44 ^\circ C}$, which is lower than that of benzene, indicative of the remarkable effect of fluorination on the intermolecular interactions as seen throughout organofluorine chemistry"). The loss of long-range order in the solid-state, as compared to benzene, produces a lower melting point in fluorobenzene $(\pu{-44 ^\circ C})$ than in benzene ($\pu{5.5 ^\circ C}$). Note too that the "size" of fluorine in organic molecules is very similar to that of hydrogen (that is why it has found such widespread use in pharmaceuticals) - it is often referred to as a hydrogen "mimic" (see here). So steric (or symmetry) arguments cannot be used to explain the packing disruption with fluorobenzene.
Once the molecule is in the liquid state the intermolecular $\ce{C-H..F}$ short-range interactions are also diminished due to increased molecular motion; as a result, fluorbenzene and benzene boil relatively close to one another ($\pu{84 ^\circ C}$ and $\pu{80 ^\circ C}$, respectively).

Then why does chlorobenzene have almost exactly the same melting point, -45C? and bromobenzene and iodobenzene also quite low?

@DavePhD  Fluorine plays a determining role in answering the OP's question, take a look at the links.  As to chlorobenzene, perhaps the increased MW (raise mp) roughly balances the lost C-H-F interaction (lowers mp).  Perhaps it is the same reason that the bp of benzene = mp of naphthalene and the bp of naphthalene ~ mp of anthracene - coincidence. The OP's question deals specifically with fluorobenzene and in that case it is known that the C-H-F interaction determines the solid state structure.

@ron What interactions are lost in napthalene which are balanced by the increased molecular weight of napthalene vs benzene?

It's not about fluorine as many of the specific examples show. Toluene also has a much lower melting point than benzene and it is all to do with crystal packing not the specific substituent added to the ring. Most non-substituted benzenes have much lower melting points regardless of the substituent because they all disrupt the crystal packing possible in a neat flat ring.

@matt_black  Yes, it is about crystal packing, but steric arguments cannot be used to explain the packing disruption with fluorobenzene. The size of fluorine in organic molecules is very similar to that of hydrogen (that is why it has found such widespread use in pharmaceuticals) - it is often referred to as a hydrogen "mimic" (see here).  It is the C-H-F interaction that disrupts the packing.

@Dissenter  Sorry, I don't understand what your asking in the above comment.  I only used the benzene-naphthalene-anthracene case as an example of coincidence, otherwise it is not related to the question at hand.

@ron Steric arguments aren't going to explain disruption of packing, but charge arguments may. Benzene has + on the Hs and - charge in the ring and the + and - favorably interact, perfluorbenzene has - on the Fs and + in the ring, and the + and - favorably interact.  Substituting one F for an H would decrease such favorable charge interaction.  The hydrogen bonding effect may compensate for some of the lost interaction.  Also, there are the residual entropy considerations that make the more symmetric structure higher melting point http://pubs.rsc.org/en/content/articlehtml/2004/ob/b407105k

@DavePhD  wrote, "Steric arguments aren't going to explain disruption of packing, but charge arguments may."  Yes, I agree, we appear to disagree on what the actual charge argument is.

The answer really doesn't have much to do with fluorine.  Beacall's 1928 The melting-points of benzene derivatives explains on the first page that it has long been known "that the more symmetrical the constitution of a benzene derivative, the greater the resistance it offers to the passage from the solid to the liquid state".
In the more modern Effect of molecular symmetry on melting temperature and solubility:

As a general rule, symmetrical molecules in crystalline form have higher melting temperatures and exhibit lower solubilities compared with molecules of similar structure but with lower symmetry. Symmetry in a molecule imparts a positive amount of residual entropy in the solid phase (i.e., more possible arrangements leading to the same structure). This means that the entropy of a crystal of symmetric molecules is greater than the entropy of crystal of a similar, but non-symmetric molecule.

Benzene and hexafluorbenzene have almost the same melting point, while monofluoro, monochloro, monobromo benzene all have much lower melting points.
Both benzene and hexafluoro-benezne have a very stable "herringbone" crystal structure, where the edge of one ring approaches the center of another ring in a perpendicular manner.

So realizing that the cyrstal structure of benzene is unusually stablized, it is understandable that the melting point of benzene is higher than monofluorobenzene, but the boiling point is not.

what's stabilizing about the herringbone structure

What is the crystal structure of fluorobenzene? Is it more or less stable than that of benzene,  and why? Without answers to those questions how can you draw a conclusion?

@Dissenter Some describe it as the postive electrostatic potential of the ring edge interacting with the negative electrostatic potential of the ring face, while others describe it as a hydrogen bond interaction.  Even the gas phase dimer has the edge-face T-shape geometry as lowest energy by 2 kcal/mol.

@ron, you are right, it would be a much better answer if I had references for the crystal structures of the various benzene derivatives in the answer


text:
What is the dipole moment of the most stable conformer of 1,1,2,2-tetrafluoroethane?
I want to know about the dipole moment of this molecule. I think the molecule has no dipole moment, but a book says me that it has a dipole moment.

Why do you think it has 0 dipole moment?

The dipole moment is 2.45 Debye. The molecule is dipolar as you have asymmetrical electron distribution around both the carbon atoms.
Source: Complete Structure of Gauche 1,1,2,2-Tetrafluoroethane Determined by Microwave
Spectroscopy, J. Phys. Chem. A, 2000, 104, 9489-9493.

You found data about another isomer that was asked.

I've corrected my answer - apologies.


text:
Why can't all electrophiles act as Lewis acids?
We know that all Lewis acids are electrophiles but I have a confusion whether the reverse is also true. Electrophiles are molecules that are in search of electrons while Lewis acids are short of electrons. So can't all electrophiles act as Lewis acids?

The definition of Lewis acids states:

A molecular entity (and the corresponding chemical species ) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:

(source: iupac.org)

In principle, since all electrophiles can be electron acceptors, they can usually act as Lewis bases, too. However, sometimes sterically demanding moieties stabilise the electrophile, so that it is impossible to form a Lewis adduct. These compound are then one part of a frustrated Lewis pair.

Then, Lewis acids = Electrophiles + FLPs?

@imrobert I do not understand your comment.

Do electrophiles and FLPs represent all Lewis acids or are there other types?

@imrobert FLPs are the combination of a Lewis acid and a Lewis base that do not form an adduct. They are not a type of Lewis acid.

Lewis acids are electron acceptors, the slight asymmetry in the definition arises because accepting means you have to move electrons into a Lewis acid, i.e. H+ and AlCl3. On the other hand the carbonyl carbon is electrophilic, but when it accepts two electrons, it rejects two others. Hence, it is not a Lewis acid.
Electrophilic - desires electrons
Lews Acidis   - desires and is able to take on additional electrons


text:
Are there any sources for tabulated Rf-values, especially 1-phenylazo-2-naphthol?
I  have synthesized 1-phenylazo-2-naphthol in my laboratory tutorial and have to analyze it. I've done a thin layer chromatography (TLC) of it:

solvent: ethanol
eluent: toluene
stationary phase: silica gel plate

My only problem is, I need an theoretical value for comparison but haven't found any source.

Did you react aniline (or its hydrochloride) with sodium nitrite in the presence of 2-naphthol?
I'm afraid there will not be any table with $R_\mathrm{f}$ values.
These values are not only dependent on the compounds and the mobile phases, but also on the stationary phase: e.g. silica, aluminium oxide, etc.
Your best option is to use reference samples and measure the $R_\mathrm{f}$ value yourself for the conditions you want to apply or transfer from TLC to column chromatography.
My suggestion is to collect the crude reaction product (Büchner funnel, etc.) dry it, note the yield and measure the melting point.
For purification, recrystallize from ethanol. You should receive red needles with a melting point of $\pu{134 ^\circ C}$.
NOTE: Wear proper protection when handling the azo dye, which is a cancer suspect.

References:

IOC-Praktikum (PDF in German)
GESTIS-database on hazardous substances

I agree 100%. Even if such tables existed, depending on how accurately you measure the solvents ratio, the $R_\mathrm{f}$ will change slightly. So the best bet is to leave a little room on the TLC plate and spot reference samples for comparison.

$R_\mathrm{f}$ is sensitive to particular stationary phase, up to having different $R_\mathrm{f}$ on different plates.  Traditionally, it is done comparing $R_\mathrm{f}$ of freshly acquired sample and obtained previously in a different manner reference.
However, I agree with Klaus, use melting point of characterisation if possible.


text:
Acetic acid or ethanoic acid - trivial names vs systematic names and correct formula representation
My teacher likes to use household materials to do experiments. Today, we did an experiment that showed the reaction of vinegar and eggshell.
When we were doing the analysis of the experiment, I realized that my teacher used the term acetic acid ($\ce{CH3COOH}$) instead of ethanoic acid. So, my questions are:

What is the accepted usage for the terms acetic acid and ethanoic acid?
Which formula is the accepted representation of acetic acid, $\ce{CH3COOH}$ or $\ce{C2H4O2}$?

Carboxylic acids that can be isolated from a natural source were commonly given trivial names. These trivial names are usually related to the biological origin of the material. Various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations. For example, the trivial name “caproic acid” (for hexanoic acid) was abandoned in the 1979 recommendations, and the name “valeric acid” (for pentanoic acid) was no longer mentioned in the 1993 recommendations.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), various trivial names are still retained, but only for general nomenclature (i.e. they are not preferred IUPAC names). Many of these names may not be used when the structure is substituted at any way, for example “propionic acid” and “butyric acid”.
According to Subsection P-65.1.1.1, only the following five retained names are preferred IUPAC names:

formic acid
acetic acid
benzoic acid
oxalic acid
oxamic acid

Therefore, the retained name “acetic acid” is still the preferred IUPAC name; however, the systematically formed name “ethanoic acid” may be used in general nomenclature.

The choice of formula depends on the context.
Generally, the molecular formula is not suitable to unambiguously identify organic compounds. For example, the molecular formula $\ce{C2H4O2}$ for acetic acid could as well represent methyl formate, hydroxyacetaldehyde, 1,2-dioxetane, or 1,3-dioxetane. However, you could use the molecular formula as shorthand if the concerned compound is clear from the context.
Simple chemical structures can often be represented by a line formula*, such as $\ce{CH3COOH}$. They rely on the fact that many elements have consistent and well-understood bonding patterns. (Recommendations for contracted labels that represent multiple atoms can be found in the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008).)
However, complex chemical structures depend on structural representations exhibiting the atom–atom connectivity, including the order and stereochemistry of the bonds. Therefore, it may be preferable to visualize the structure using a chemical structure diagram.
* According to IUPAC recommendations concerning nomenclature of organic as well as inorganic chemistry, a line formula is a formula that is written on a single line, as is text. Unfortunately, however, there are also other definitions of line formula.

While I agree that the systematic name ethanoic acid makes a lot more sense, the name acetic acid has been used for so long that it's the accepted IUPAC name and is used almost universally. Both refer to the same compound.
Molecular formulas like $\ce{C2H4O2}$ are often seen when doing elemental analysis and the like, but give no structural information. $\ce{CH3COOH}$ makes it clear that it's an acid and shows the carbonyl group.


text:
Etching around corners
In etching of mono silicon, why does the etching process around the corners differ from the etching process around the flat edges? What brings about this difference and is there some way that MEMS engineers could quantify the process around the corners?

Are you asking about anisotropic wet etching or about reactive ion etching? Also, by "around the corners", do you mean perpendicular to the face of the wafer?

anisotropic wet etching and yes perpendicular to the face of the wafer, for example in a <100> wafer, the selectivity of the etching rates across different faces makes it etch along a <111> face, but around a square mask, something different is supposed to happen according to my professor.

Related to your previous question: http://chemistry.stackexchange.com/questions/19618/how-is-the-crystal-direction-important-for-etching-silicon-surfaces?rq=1

What happens is that convex (outside) corners get severely undercut in wet etching. On a <100> cut wafer, the edges leading up to the corner reveal <111> planes which are resistant to etching, but right at the corner, where the two planes intersect, etching reveals a <110> plane, which is etched much more quickly. The difference in etch rates has to do with the number of "dangling" bonds—faces with fewer Si-Si bonds at the surface etch faster. The reason why concave corners don't undercut, despite having the same <111> plane intersection is that there are no dangling bonds in an inside corner so it doesn't etch any faster than the <111> planes.
It should also be noted that adding a small amount of surfactant to the etchant can dramatically decrease undercutting by adsorption to the corners.
This (fortunately open-access) article explains the phenomenon quite nicely.


text:
Why there is charge on the elements, in the dative bonding?
According to the definition of dative bonding electrons are shared which are given by one of the atoms, so if the electrons are shared so how come there is charge on the elements in the dative bonding?

To begin with, the term "dative bond" is not used anymore. It has been replaced by "coordinate bond" :
Let's have an example of this kind of bond.

As you can see, this kind of bond occurs when an atom is giving two electrons from itself to form the bond. The typical coordinate bond is the $\ce {N}$ already bonded with $\ce {3H}$ forming a bond with a proton. Since the proton has no electron, the $\ce {N}$ atom has to fully give one of his electron from his remaining doublet to form the bond. But to respect the charge equality, you had a positive charge at the begining, no electrons were added or removed from the system, so the positive charge should remains on the final molecular edifice.
The question is, where is located this positive charge ? One could start thinking this charge is located on the $\ce {N}$ atom since it is giving an electron. But actually, the $\ce {N}$ atom has a greater electronegativity than the $\ce {H}$ atom, that's say its hability to attract electrons is greater. Therefore the positive charge remaining should be equaly distributed over all the $\ce {4H}$ since they are equivalent.
But, altogether, the net charge of the edifice will be $+1$.

It's formal charge - it often doesn't mean there are full charges like in zwitterions but dative bonds are usually strongly polarised.

Dative bond/coordinate bond is a type of covalent bond where an atom donate electron and another one does not share any but receive the electron.
Note that i bolded covalent bond. Meaning that this bond happened between 2 non-metals, not between metal and non-metal (that's Ionic bond). In ionic bond there's formal charge because there's electrostatic force occuring between the two atom, but in covalent bond there is no electrostatic attraction at all.
In order to keep track between the transfer of electron we use either formal charge or oxidation number, because covalent bond doesn't have any electrostatic attraction we use oxidation number (also known as imaginary formal charge).
So what you meant might be Oxidation Number and not Formal Charge
Hope that helps.

This is incorrect. There can be dative bonds between any type of atom, they usually have high covalent character. A partial charge results from the asymmetry of the nuclear charges and the electrons. There always is electrostatic interactions (which are certainly attractive) in bonds. Oxidation number, formal charge, actual charge are not really related concepts.


text:
Why doesn't Boron form Diborane with water?
Why is it that Boron does not form Diborane with water rather than $\ce{B2O3}$?

I assume the hypothetical reaction you are asking about is:
$\ce{3H2O + 2B -> B2H6 + 3/2O2}$
under standard conditions.
The standard enthalpy of formation of water is −285.8 kJ/mol, whereas that of diborane is 35.4 kJ/mol; this reaction would therefore be very endothermic and hence unfavourable. You're breaking up strong O-H bonds and B-B bonds to form B-H bonds (including two 3c2e bonds, which are very weak) and weak O-O bonds. In contrast the formation of boron trioxide by reaction of boron and water is very exothermic, and has been considered in potential applications of boron as a rocket fuel.

Whether 3c-2e bonds are "very weak" depends on what you're comparing them to.  A 3c-2e bond is weaker than two separate bonds joining the atoms, but stronger than one ordinary pairwise bond would be.


text:
Can HBr add to a non-sp2 hybridized carbon?

Does this reaction make sense? Mechanistically speaking, it does not for me. Is this a special case, or just a typo in the book?

I don't believe it's a typo. Can you describe the mechanism as you see it?

I think I got it. There is a carbocation rearrangement. Basically the H can produce two secondary carbocations. One of those secondary carbocations can do a 1,2-hydride shift, which produces a tertiary carbocation. This carbocation, being tertiary, will react much faster, resulting in being in the major product.

Excellent. You should write that up as an answer and answer your own question!

There is a carbocation rearrangement. Basically, the H can produce two secondary carbocations. One of those secondary carbocations can do a 1,2-hydride shift, which produces a tertiary carbocation. This carbocation, being tertiary, will react much faster, resulting in being in the major product. So, yes, HBr can add to a non-sp2 carbon.

Most likely what happens is that the secondary carbocation rearranges to the tertiary carbocation much faster than bromide can attack the secondary carbocation. The tertiary carbocation is more stable than the secondary carbocation, so all else being equal, the secondary carbocation will react faster than the tertiary carbocation. And that's exactly what we see in this case, except that the reaction isn't addition of a nucleophile, it's the 1,2-hydride shift.

Jerepierre, I think you meant "tertiary carbocation will react faster"

No, I meant that the secondary carbocation will react faster... It's less stable, so it should react faster. The reason the tertiary bromide is observed as the product here is because the secondary carbocation rearranges to the tertiary carbocation faster than the nucleophilic attack.