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What is viscosity in terms of chemistry?
In physics, viscosity is the force applied in a unit area times time or basically Pascal times second. We could say that it is a measure of its resistance to gradual deformation by shear stress or tensile stress.
But what does it mean in the chemistry world?
I believe it is related to the distance between molecules in a solution. When molecules are farther apart then viscosity is lower than when molecules are closer.
Could you tell me if my mature concept about viscosity is on the right track or far too off? Could you also define viscosity in terms of chemistry?

There are intermolecular forces but also molecules are not point masses so they don't just slide past each other smoothly. This would give rise to the resistance to shear.

Viscosity is defined in chemistry and physics pretty much equally. For exact definitions of various viscosities open IUPAC Gold Book and search for "viscosity" (left toolbar) – you will see it's the same physical terms.

I agree with @andselisk a̶s̶ ̶a̶l̶w̶a̶y̶s̶. Perhaps you meant your title to read: "Viscosity: The molecular perspective" or something like that ;)

More or less similar…I will try to put brief concepts of viscosity in terms of pharmaceutics which is somewhat another branch of chemistry, where it is very much applicable in drug design.

Viscosity is an expression of the resistance to flow of a system under an applied stress. The more viscous a liquid is, the greater is the applied force required to make it flow at a particular rate.

Viscosity is defined in terms of the force required to move one plane
  surface past another under specified conditions when the space between
  is filled by the liquid in question. More simply, it can be considered
  as a relative property, with water as the reference material.

Viscosity implies that the liquid flows even under the smallest stress and does not return to its original shape or form once the stress is removed.
Viscous deformation, i.e. viscous flow, occurs in accordance with Newton’s law,
$\ce{σ=ηỳ}$

where the applied stress σ results in flow with a velocity gradient ỳ
  or rate of shear. The proportionality constant η is termed viscosity,
  while its reciprocal is called fluidity.

Viscosity has also been
 described as the internal friction in the fluid as it corresponds to the resistance of the fluid to the relative motion of adjacent layers
 of liquid.
The viscosity of simple liquids (i.e, pure liquids consisting of small molecules and solutions where solute and solvent are small molecules) depends only on composition, temperature, and pressure. It increases moderately with increasing pressure and markedly with decreasing temperature.
There is quite a lot to discuss on viscosity in depth but these are just shallow concepts to understanding the term.

References

Martin’s Physical Pharmacy and Pharmaceutical Sciences : Rheology
Remington: Essentials of Pharmaceutics: Rheology
Ansel’s Pharmaceutical Dosage Forms and Drug Delivery Systems : Special Solutions and Suspensions
Aulton ’s Pharmaceutics The Design and Manufacture of Medicines 4th edViscosity Rheology, and flow of fluids


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Typesetting of the retardation factor (Rf)
What should be the typesetting for the symbol of the retardation factor as used in thin-layer chromatography?
These are the possibilities:

Rf - only f in italics
Rf - both R and f in italics
Rf - only R in italics

On Wikipedia, they have used all sorts of typesetting. There is also RF with a capital F.

According to Nomenclature for Chromatography (IUPAC Recommendations 1993) [1, pp. 843, 845] (also listed in IUPAC Gold Book), retardation factor is denoted as follows:

Column chromatography: $R$ (capital $R$ in italics).

3.7.13 Retardation Factor $(R)$
The fraction of the sample component in the mobile phase at equilibrium; it is related to the retention factor and other fundamental chromatography terms:
$$R = 1/(k + 1)$$

Planar chromatography: $R_\mathrm{F}$ (capital $R$ in italics and upright capital $\mathrm{F}$ in subscript).

3.8.04 Retardation Factor ($R_\mathrm{F}$)
Ratio of the distance travelled by the center of the spot to the distance simultaneously travelled by the mobile phase. Using the symbols of Fig. 2:
$$R_\mathrm{F} = b/a$$
By definition the $R_\mathrm{F}$ values are always less than unity. They are usually given to two decimal places.
In order to simplify this presentation the $hR_\mathrm{F}$ Values may be used: they correspond to the $R_\mathrm{F}$ values multiplied by 100.
Ideally, the $R_\mathrm{F}$ values are identical to the $R$ values (see 3.7.13).

Reference

Ettre, L. S. Nomenclature for Chromatography (IUPAC Recommendations 1993). Pure Appl. Chem. 1993, 65 (4). DOI: 10.1351/pac199365040819.

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Is there any mathematical formula for electron affinity and ionization potential?
I already know about the mathematical formula for electronegativity, but it requires the ionization potential and electron affinity. How can I can easily calculate the ionization potential and electron affinity with a mathematical formula?

You can't.$\mathstrut$

Technically, you could solve the multi-electron Schrodinger equation and compute the energy difference, but I believe that is an intractable problem without at least severe approximations.

@Zhe one such severe approximation is to solve the Schrodinger equation using the Hartree Fock approximation, then assume frozen orbitals for the ionization process. https://en.wikipedia.org/wiki/Koopmans'_theorem

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Is molten potassium nitrate possible?
Is it possible to melt potassium nitrate without it decomposing into potassium nitrite and oxygen? If so when it solidifies does it keep its oxidizing properties?

https://en.wikipedia.org/wiki/Potassium_nitrate gives a melting point of $\pu{334°C}%edit$ without decomposition, but then decomposition sets in at $\pu{400°C}$ so you have a narrow liquid range.
If you avoid decomposition the oxidizing property still works.


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Finding average Mr of a gaseous mixture

A 2 g mixture of cyclohexanol, acetone and pentanal is completely burnt in oxygen. When the gaseous product is passed through $\ce{P2O5}$, the mass of $\ce{P2O5}$ increased by 1.998 g. Find the average molecular mass of the mixture.

$\ce{P2O5}$ absorbs water, so the amount of water absorbed is 0.111 mol. I attempted constructing an equation for the combustion reaction but realized that the proportions and thus the mole ratios of each gas are unknown.

Let's see your balanced reaction equations and your calculated molecular masses.

Are you asking about the average relative molecular mass $M_\mathrm r$ (as in the title), the average molecular mass (as in the text), or the average molar mass $M$ (not mentioned but reasonable)? These are different quantitites with different dimensions.

I suspect one might utilize the fact that all three compounds – cyclohexanol $\ce{C6H12O}$, acetone $\ce{C3H6O}$ and pentanal $\ce{C5H10O}$ – share the same empirical formula $\ce{(CH2)_xO}$, which is also applicable to the mixture of reactants, and so instead of writing a set reaction equations
\begin{align}
\ce{C6H12O + 8.5 O2 & -> 6 H2O + 6 CO2} \tag{R1}\\
\ce{C3H6O + 4 O2 & -> 3 H2O + 3 CO2} \tag{R2}\\
\ce{C5H10O + 7 O2 & -> 5 H2O + 5 CO2} \tag{R3}
\end{align}
one can avoid finding volume fraction $\phi_i$ of each individual component and use a single reaction scheme such as
$$\ce{(CH2)_xO + $\frac{3x-1}{2}$ O2 -> x H2O + x CO2} \tag{R4}$$
and find relation between average molecular weight $\bar{M}$ and $x$ solely from (R4). Lets denote the mass and the total amount of gas mixture components as $m$ and $n$, respectively. Then according to (R4) total amount of gas mixture is $n = \frac{n(\ce{H2O})}{x}$, and the average molecular weight is established as
$$\bar{M} = \frac{m}{n} = \frac{x \cdot m}{n(\ce{H2O})} = \frac{x \cdot m \cdot M(\ce{H2O})}{m(\ce{H2O})} = \frac{x \cdot \pu{2 g} \cdot \pu{18.015 g mol-1}}{\pu{1.998 g}} = (18.033 x) \,\pu{g mol-1} \tag{1}$$
At the same time using molecular weights of the elements:
$$\bar{M} = x \cdot (M(\ce{C}) + 2M(\ce{H})) + M(\ce{O}) = (14.027 x + 15.999)\,\pu{g mol-1} \tag{2}$$
Equating (1) and (2):
$$18.033 x = 14.027x + 15.999 \tag{3}$$
and solving the resulting equation (3) for $x$, one finds that $x = 3.994$. Solving either (1) or (2) with obtained $x$ yields in average molecular weight $\bar{M} = \pu{72.023 g mol-1}$.


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Does graphite burn?
Is it possible to make graphite oxidize in the presence of oxygen?
I know it's possible to make other forms of carbon oxidize and even burn with the production of residual heat in the presence of oxygen.
As far as I know graphite is very resistant to heat, and it isn't possible to make it burn even at high temperatures.
So, my question is: what are the conditions required to make it burn producing carbon dioxide?

You can burn diamonds if you really want to (even higher heats and needing better than atmospheric oxygen)

The autoignition temperature of graphite is $\pu{730^oC}$. This means that pure graphite will ignite at that temperature under standard atmospheric conditions without an external source of ignition. The fraction of carbon that gets oxidized all the way to carbon dioxide at that temperature may not be unity depending on the oxygen concentration, whether the graphite is a fine powder vs. a solid rod, etc. I worked in an analytical laboratory in which we occasionally determined the carbon content of graphitic substances by quantitatively converting the graphite to carbon dioxide at $\pu{950^oC}$ under a pure oxygen atmosphere.

How does the graphite crucible work? Is it not pure graphite? I have seen some videos of metling precious metals like silver and gold which has higher melting point than 730°C

@jnovacho Bulk graphite won't burn well even at its autoignition temperature (thin sheets might). And graphite crucibles are often used in environments where the hottest parts are not in contact with much oxygen. If the rate of burning is very slow, it may not matter much in a bulk object.

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text:
Would a button cell battery contain mercury? If so, why?
I was wondering if an alkaline button-cell battery that says $0~\%\ \ce{Hg}$ is actually completely mercury free or if it is still considered to have traces of mercury. Why do some batteries say no intentionally added mercury? Where would the unintentional mercury come from? Just wondering because I accidentally washed my son's little toy watch in the washing machine and it contained one of these types of batteries. Is it okay to use my washer or could it be contaminated?
Also, I use disinfecting wipes around the house quite a bit that contain dimethyl. Could using these wipes around where the battery was create dimethylmercury?

First take some slow breaths and calm down. EVERTHING contains at least trace amounts of all the elements. Your washer is fine, and using wipes won't create dimethyl mercury.

Button cells can be dangerous for more reasons than that they contain poisonous components. They are a big danger to children if swallowed, though mainly because they set up corrosive currents through tissue. Even mercury containing ones, though, pose no hazard if they are undamaged as they are usually well sealed against leaks.

Anybody care tell me what exactly is dimethyl?

@Karl - Dimethyl benzyl ammonium chloride is evidently a common wipe component, but the dimethyl could be from something else. The OP is obviously clueless about chemicals.

Mecury batteries were indeed once popular. In the charged state, they contained mercury(II) oxide $\ce{HgO}$ and zinc metal as the active components. During use, the overall reaction would be:
$$\ce{\underset{\text{mercury(II) oxide}}{HgO} + \underset{\text{zinc}}{Zn} -> \underset{\text{mercury}}{Hg} + \underset{\text{zinc oxide}}{ZnO}}\tag{1}$$
As you can see from the equation, these batteries would produce mercury while decharging so in the fully decharged state practically all mercury(II) oxide would be converted to elemental mercury. Mercury(II) oxide is toxic on indigestion while elemental mercury is toxic especially when inhaling vapour. The latter can also accumulate in the environment meaning disposing these batteries was a significant issue. This is why the US and the EU banned the advertisement and sale of batteries containing a certain percentage of mercury in the 1990’s as the Wikipedia article explains.
The battery in your son’s toy watch obviously was not a mercury battery since it reads $0~\%\ \ce{Hg}$. Nowadays, various different types of batteries are used in button cells, replacing the formerly popular mercury cell type; usually lithium or alkaline cells. These do still come with their own specific dangers — most notably, batteries should always be disposed of properly and never with household waste for environmental reasons — but do not contain compounds as toxic as the mercury compounds in mercury cells.
Therefore, your washing mashine is not contaminated with mercury and still safe to use.

On to your side questions. I don’t exactly know what the active compound in your ‘dimethyl wipes’ is, but I can assure you that it will not react with mercury to give dimethyl mercury. Chemistry is more complex than just taking word roots and putting them together. A common German chemists’ joke actually plays on the word root thing, causing laughter with those who know how absurd it is thanks to the similarities of the German words for strawberry (Erdbeere), blackberry (Brombeere), the ground or earth (Erde) and bromine (Brom). It suggests throwing a blackberry on the ground so that it reacts with earth to form a strawberry and liberate bromine.
Finally, your unsureness about the ‘intentionally added’. If your analytical instrument is good enough, you can practically detect every element in every sample. This is especially true for samples of similar types; i.e. metals will always contain other metals as trace impurities. However, these contaminants are present in absolutely minute traces. We are talking one part per billion or even less. Among these impurities there is indeed likely mercury. But these concentrations are so low that they do not matter for any practical purposes. I assume that the manufacturer of the battery simply didn’t want any super smart jackarses coming who throw the full blow at one of those to determine $\pu{0.1ppb}$ mercury and proceed to sue the company in the US.

Fünf Mark in die Schlechte-Wortspiel-Kasse für den Schenkelklopfer mit der Brombeere.


text:
Does salt dissolve in fat when heated?
As the question states, does salt dissolve in fat when heated.  If no how can we get it to dissolve in fat and distribute through the fat like it does with water?

No. Adding a bit of 15-crown-5 ether might help, but I have a vague feeling this is not what you are looking for.

‘As the question states’ — which question? I can only see a title.

Heating will only change the viscosity of the fat, but it has little to do with solubility in this case. Solubility is not a function of viscosity, but different interactions including solvation, which wouldn't take place in fat as much as it would in water. I'll tell you more: acetone is less dense, less viscous, and even polar, but NaCl is still only slightly soluble in it.
To do what you ask, about distribution, I would probably dissolve salt in something it is well soluble, and add some surfactants. Fat and water (?) would make a homogenous mixture, and with right choosing of ratios of fat/water/salt, you'll get almost what you want.


text:
Why does cyclooctene have three stereoisomers?
How does cyclooctene have three stereoisomers?
I thought that there are only two stereoisomers, cis- and trans-, but according to my textbook there are three stereoisomers. I want to know how?

Could be helpful https://chemistry.stackexchange.com/questions/35470/why-is-trans-cyclooctene-chiral


text:
How can multiplets in para-disubstituted benzene rings be described?
How should I label the multiplicity of the aromatic signals in an 1H-NMR spectrum with the following substitution pattern?

I have seen lots of people labeling both signals in this kind of system as doublets, probably because the resolution of the spectrum was not high enough, or because we are often told, these look like doublets. However, what looks like a doublet doesn't necessarily have to be one and thus, should not be called doublet, because it doesn't reflect the actual couplings in this system:

The ortho-coupling (3J) for both signals should be about 6.0 and 9.0 Hz.
The meta-coupling (4J) for both signals should be between 0.9 and 3.0 Hz.
The para-coupling (5J) for both signals should be between 0.0 and 1.0 Hz.

(These numbers are from the Hesse-Meier-Zeeh, a German textbook about spectroscopy in organic chemistry).
I am inclined to call these signals multiplets rather than doublets, because I can only guess the coupling constants, especially for the right signal at about 7.36 ppm.
Or is it good practice to call them pseudo-doublets? And what would be the symbol (like s for singlet, d for doublet, etc.)?

First, if R1 and R2  don't produce any kind of steric hindrance in the molecule and they rotate freely around the R-C(Ar) bond, I don't see any reason why H and H' should be different (I mean, should there be any splitting at all for them on them?). Second, if you want to be very precise, you can call them ddd, but that arises the problem with the constant values. Maybe you can try to register this spectrum on an NMR of a higher frequency? But synthetic papers don't really often care about higher coupling constants than 3J.

@wolphram This is true, but as you can see in the spectrum, this arene gives more than just a doublet. Even the signal on the right has visible shoulders. I could call them ddd, but then I would have to show three coupling constants, which I cannot measure.

@wolphram $\ce{H}$ and $\ce{H'}$ may be chemically equivalent but are not magnetically equivalent, the coupling constants of each to $\ce{H_X}$ (one single, specific proton, not the ‘closer neighbour’) are different for $\ce{H_A}$ and $\ce{H_A'}$.

@Jan so you are saying that if that's, say, p-chlorotoluene, they won't be magnetically equivalent?

@wolphram Correct.

@Jan that's interesting. I'll try to think about that more. I mean, as for me, their magnetic surrounding is quite similar, it's not like acyl or N-methyl or something. Maybe you know a specific book to look at? I would appreciate that.

@wolphram Unfortunately no, I am not a good person to ask for book recommendations. But it should be mentioned in all advanced NMR textbooks.

@wolphram Maybe this can help: link. Quote (about AA'XX' systems): Only if the J(AB) coupling is identical to the J(AB') coupling by accident does the system become A2B2 or A2X2, and a first order pattern is seen.

Many people report these as doublets. However, you are correct that these are not doublets - there are second order effects in AA’XX’ spin systems (note that a first-order analysis will fail, the pattern is not a ddd - see https://organicchemistrydata.org/hansreich/resources/nmr/?page=05-hmr-15-aabb%2F for more info).
Using the phrase “apparent doublet” (abbreviate as app d) is probably a good compromise between being technically correct and retaining useful information. You can still report the “coupling constant” between the two largest peaks, but with the caveat that it is not a first-order doublet.
If you want to be very precise, then multiplet is the way to go. However a typical organic chemist would also be interested in the splitting between the two largest peaks in your multiplet (which if I am not mistaken, corresponds to the sum of $^3J_\ce{AX}$ and $^5J_\ce{AX’}$). Since the para coupling is tiny, the splitting is essentially representative of the ortho coupling. In my opinion it is helpful to leave that information in.

At our university, we usually have the sentence "First-order principles were applied in the analysis of the spectra" in our theses. So I guess, I will be fine with "app d". Thanks! Adding coupling constants to a multiplet would seem rather wrong.


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In ESI-MS, why do we assume that the positive charge on the analyte must come from protonation from the solvent?
In ESI-MS, we ionise the molecules using a high voltage. Why wouldn't this high voltage cause the ejection of an electron from the analyte itself, generating a M+ ion. Why do we have to assume the the positive charge comes from protonation of solvent molecules?

The ions have to get into the mass spectrometer, and you need to have a spray, i.e. you need the Raleigh instability.   Take the case of quaternary ammonium ions like tetramethylammonium.  If you ran "electrospray" with no voltage on the ion source, you would see a vastly, vastly attenuated signal for tetramethylammonium, even though it is permanently charged and doesn't need protonation by the solvent for detection.  Sure, electrochemical reactions at the capillary walls

You might be asking two questions here.
(1) Can the high electric field strength induce electron detachment in the vapor? No. Normal electrospray voltages are not high enough to do so. Air, for instance, has a breakdown voltage of around 30 kV/cm, and liquids and aerosol particles are ionized only at even higher electric field strengths. Electrospray voltages are around one-tenth of that.
(2) Can the positive potential at the capillary walls induce oxidation of $M$ to $M^+$? In principle, yes, but only if there are no other species present with higher oxidation potentials than $M$ -- such as the metal walls of the capillary.


text:
How to calculate steric number of methyl free radical?
According to the definition, steric number  = number of atoms it is attached + lone pair.
In that way, I get 4 for methyl free radical which means sp3 hybridization . But that's wrong !
Why is this happening ?

The point is that in methyl radical there is not a lone pair and a direct application of the formula for steric number does not apply, not smoothly. The unpaired electron can reside in a non hybrid p orbital without much repulsion with the bonding paired electrons. As such the C atom is sp2 and the geometry is quasi planar trigonal. However the barrier to a pyramidal molecular geometry is low. If H atoms are substituted by more electronegative atom or radicals (e.g. in CF3 radical) geometry is then pyramidal. Qualitatively, you can see this as if an unpaired electron count about 1/2

Then how do I arrive at sp2 hybridization ? What else technique to use ?

Vsepr gives (with exceptions) reliable predictions but for electron pairs! Is it in its name. The trick of assigning about 0.5 to an unpaired electron can be useful while comparing species which formally differs only in the number of electrons like NO2+ NO2, or to rationalise the case CH3-, CH3 radical, CH4. The case of CH3 radical is indeed a special one, and its geometry cannot be really predicted this way.

Meantime I saw as I expected that basically the same question has been answered https://chemistry.stackexchange.com/questions/19370/why-is-the-methyl-radical-planar         it should be clear in every book that vsepr is a first approximation...


text:
Beer's Law: Are the graphs of concentration/time and absorbance/time supposed to look very similar?
For my lab, I'm trying to figure out if the reaction between NaOH and crystal violet is a zeroth, first, or second reaction. My slope is 5.714*10^4 and y-intercept 0.008016. In logger pro, after finishing collecting my absorbance/time of NaOH and crystal violet, I tried getting the concentration of crystal violet by using that equation. When graphed, it looks exactly the same as my original graph, just on different numbers. I'm confused by this because when I've tried graphing ln(concentration) and 1/concentration, the graph returned no straight lines. I'm wondering if I'm doing something wrong.

Well, since Beer's Law says
$A = \epsilon⋅l⋅c$
And $A$ hence is directly proportional to $c$, as Beer's Law actually represents a line equation.

This is why you do so to get reaction order


text:
Laplacian of the electron density
Is it always the sign of the Laplacian of the electron density a good criterion to decide on the covalent nature of a bond?

Yes, and no, and maybe.
Oftentimes the sign of the Laplacian at the bond critical point in terms of the quantum theory of atoms in molecules (QTAIM) is taken as a measurement for the covalency of a bond. That it is not that simple is quite logical from your question alone.
Just as in bonding in general, i.e. there is a spectrum from pure covalent (ideally, but not achievable) and pure ionic (ditto),* there cannot be a clear yes or no answer here, too. I think the case of $\ce{BCl3}$ versus $\ce{AlCl3}$ and respective dimers, which I have written about, demonstrates this feature and how QTAIM can help understand bonding. As with many things it certainly is hard to draw a line. In that sense it is probably a good criterion to judge the nature of a bond. Judging only based on the sign of the Laplacian, while being a rigorous and unambiguous way to look at things, might lead to the wrong conclusions though.
What does it matter anyway? In many cases such a strict separation is not helpful. It should be clear, that there are covalent and ionic contributions in every bond, and reactivity often does not care about that. You'll never obtain a complete picture, if you are looking at a single property.
* And then there is charge-shift bonding (What is charge shift bonding?), but I personally don't see any value in that.

Forgive me my incompetence but not even $\ce{H2}$ can be considered a purely covalent bond?

@Jan I would say it is probably as close as you can get. But there will always be an electrostatic contribution with it.


text:
Why is triptycene aromatic?
I came across this wonderful compound, triptycene.

Why is triptycene an aromatic compound, even though it is non-planar? I thought planarity was a requirement for aromaticity.

Related:https://chemistry.stackexchange.com/questions/37641/is-buckminsterfullerene-aromatic?rq=1

@Tyberius I don't really see how this is analogous to fullerenes.

Each benzene ring is aromatic so the whole molecule is described as aromatic. I doubt the aromaticity extends over the bridging carbons though.

@bon it gets at misconceptions that the OP is having, like that aromaticity is a binary variable and that planarity is an absolute requirement for aromaticity.

@bon I believed planarity is required as (1)  that's what I was taught (2) that's what I found on Wikipedia (3) third party source (i believed barrelenes to be unrelated here) with these three source confirming each other I had a valid question in mind, hence my post.

@GaurangTandon I would look at Ivan Neretin's answer in the linked post. As to your question, I think bon gets at the right idea that the aromaticity doesn't extend over the whole molecule. For example, something like tricyclobutabenzne is also considered aromatic even though the aromaticity doesn't apply to the cyclobuta groups.

For future reference: Aromaticity is not at all an easy concept and there was/is/will be a lot of discussion especially in ambiguous cases. Hückel's rules (plural!) only strictly apply to monocyclic-hydrocarbons, better not overuse it.

In addition to the reading of all comments and answers (incl. those of the related question) I suggest you to take it easy. Again definition of aromaticity is still subject of research. Simple rules developed for well defined systems can present exceptions, etc. In this case the meaning is simple. Each peripheral ring display a chemistry similar to that of benzene. That for sure. One could look at the behavior of the Hs at the bridge to get an idea. I think (but can't prove) that any chemistry typical of aromatics will take place at the peripheral rings. The molecule is stabilize, so in a sens

It is aromatic, independent of the behaviour of the bridge. C6H5-C6H11 is an aromatic molecule, at the end.

Think also of triphenylamine. Is a different example of complexity and variety of chemistry. You will classified as aromatic at first glance (correctly). But there are no inner ring, "just a common substituent" for each ring. However, triphenylamine is also not pyramidal as the N lone pair delocalises indeed! It has a almost planar geometry but whit a tilt of the rings (propeller shape). However you will say it is aromatic independent of this....

The confusion arises because of a simplistic definition of aromaticity.
What matters for aromaticity is the planarity of the relevant part of the molecule. In triptycene there are three planar benzene groups joined together. Each is individually aromatic, but the whole molecule is not. Similarly, triphenyl-methane has three individually planar benzene groups, each of which is aromatic, but the molecule cannot be planar because of steric interactions between the groups. There are many other such examples of complex molecules containing an aromatic component attached to other things.

First of all, there's one thing you must know, aromaticity is a property of a ring system and not always of the molecule.
Small molecules like benzene and pyridine have only one ring, so it's "ok" to call them as aromatic, since they have only one ring system. But this approach can't be taken if you want to analyze, say, biphenyl.

Here, if you (incorrectly) try to analyze the whole molecule for aromaticity, you'd get 12 $\pi$ electrons, which is clearly a non-Huckel number, and hence you might infer it as non-aromatic. But however that approach is incorrect. However if I ask you Is biphenyl aromatic? what would you say? The answer is that it's meaningless to ask if biphenyl is aromatic or not. You may however ask Does biphenyl have aromatic ring systems in it?, which is a perfectly valid question.


text:
Could muscarine be an antidote to atropine poisoning?
According to Wikipedia,

"Muscarine poisoning is characterized by miosis, blurred vision, increased salivation, excessive sweating, lacrimation, bronchial secretions, bronchoconstriction, bradycardia, abdominal cramping, increased gastric acid secretion, diarrhea, and polyuria. If muscarine reaches the brain it can cause tremor, convulsions, and hypothermia. Cardiac ventricles contain muscarinic receptors that mediate a decrease in the force of contractions leading to a lower blood pressure. If muscarine is administered intravenously, muscarine can trigger acute circulatory failure with cardiac arrest... The specific antidote is atropine."

However, the wikipedia article on atropine also states,

"In overdoses, atropine is poisonous."

Could muscarine be used to treat atropine poisonings, just like atropine is used to treat muscarine poisonings?

In short: no.
It is not that simple. According to Wikipedia, atropine works as a muscarinic antagonist, which blocks the action of acetylcholine at the muscarinic acetylcholine receptor. So, the mode of action as an antidote is to prevent muscarine from binding to the receptor, thus inhibiting its effects on the body.
The poisonous effects of atropine cannot be reversed by muscarine, though. That may be due to the fact, that it is able to displace muscarine and not the other way around.


text:
How can you detect iron hydroxide or iron sulfide in quarried rocks?
We have a quarry that produces some gray rocks that sometimes turn green, then orange and the color bleeds, making the rocks unsuitable for our purposes. But it can take up to 6 months out in the weather to see if the gray rock will end up having this problem. It's been suggested that the compounds causing the problem in the rock are iron hydroxide and iron sulfide. Is there a way to detect these chemicals sooner, or the chemicals that precede them, so that we don't have to wait 6 months to weed out the bad rocks? Would also appreciate knowing any resources that might help figure it out.

I don't understand close- and downvotes. IMO this is a good question. What I would suggest is to rent or buy climate chamber in order to perform fast simulation on your samples, or a portable XRF analyzer for mineralogists to detect elements on the surface.

Thanks @andselisk - we will look into those options. Do you think there might be some kind of an acid test or something we could spray on the rocks to accelerate the reaction?

@SawShop I think yes, some environmental chambers should be able to simulate acidic/basic conditions as well, but I cannot say for sure which model and how much it would cost. I only used a primitive one once long time ago.

Rocks that turn green and then red seems to me like you might have some copper sulfides, like chalcopyrite (CuFeS2). At first, the copper reacts with the atmosphere to form something like malachite (similar to the green patina observed on things like the Statue of Liberty). Next, the iron oxidises to form iron oxy-hydroxides (as you correctly identified).
In order to detect these things rapidly, you have several options:

Use expensive methods like XRD or XRF. I would recommend against, because these only test the very thing you are analysing, and if you would like to test a whole rock face, this would be extremely expensive. It would however be beneficial to give one of the stained samples to a nearby research institute to test what the material is, either using SEM-EDS or XRD. I might be wrong and it would not actually be copper.
Just look at the rock very close, preferably with a hand-lens. Chalcopyrite (and other copper/iron bearing sulfides) and shiny, usually golden or slightly green-purple when tarnished. You can look at a gallery of chalcopyrite to have a rough idea of how it looks like. Note that the gallery has nice specimens, which yo clearly don't have, but try to see how the lustre and colour look like.
Spray bleach (sodium hypochlorite). This would speed up the reaction, better than acid. Note that this could affect other things in the rock and cause staining or dissolution of things that wouldn't have without the bleach.
Try to smell the rock, especially around the places where you see incipient greening. Can you smell a sulfurous smell? Our noses are surprisingly extremely sensitive to sulfur, and can detect it even when some instruments can't.


text:
How to measure ratio of Na to K in table salt?
Some people state, that the salt which we buy at the supermarket contains too much potassium chloride. For me it seems to be a hoax, but I want to have a proof.
Is there any way to measure the percentage of sodium chloride and potassium chloride in salt at home without special equipment?
If it is not possible, what kind of equipment is needed?

It doesn't make much difference, but it's potassium iodide, which is added. Iodide anion concentration would be much easier to find out.

RE: "Some people state..." -- Love it. Reference to vague and unknown authority.

Atomic absorption spectroscopy, if you can get the use of an instrument.

It would be very very easy to do with XRF. You'd just need to press the salt into a pellet in a die.

@Mithoron It could be iodate, not iodide.

@MaxW who states it, does not really matter. The question can be understood and probably answered :)

@GáborLipták - I think you miss the point that I was trying to make. By using "Some people" as the authority it is impossible to backtrack who actually said what. So such a claim could be on the Skeptics forum asking if any recognizable authority has actually stated that "regular salt" has too much potassium.

Two problems: 1) why worry about the amount of potassium? I isn't harmful. 2) Why not just take what the label says: most authorities insist on ingredient labelling and this should help identify any salt with more than traces of potassium (some high potassium salt is sold as a "low sodium" alternative to normal salt).

Flame spectroscopy could be used to quantitate Na:K ratio.

@ELawrence Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. I have converted you answer to a comment and deleted your other answer, which was also a comment. Please only user the answer field for actual answers. On our site we are looking for more detailed and referenced answers, rather stan one-line statements.

@MaxW The question is not asking about an authoritative source, though. The question is asking how to perform an experiment. It would be just as valid of an experiment if you dreamed it, or if you took drugs and a fence told you.

To answer your question: no, you cannot do it at home. Some analytical techniques are mentioned in the comments.
However I decided to post because LIKELY people should not worry about the Potassium content of table salt.
Table salt (normally "just" NaCl) containing as much of 28.5 % of Potassium (might be more perhaps, but this is what I have at hand) can be sold free of medical prescription not only in drugstores/pharmacies but also in supermarkets, where it can be found in the special diets section or alongside standard salt.
It is a dietary salt indicated to limit the Sodium intake (for instance in case of high blood pressure).
Its price is much higher that that of common salt. It can be used to replace salt in all kitchen uses. It is less salty and more bitter than the normal salt, and one has to "get the mouth" to it.
I would therefore infer that is highly unlikely that something sold as table salt NaCl could contain so much potassium to worry about. Especially when some of those worrying of it, might be prone to buy Potassium integrators during summer :)
Disclaimer: it is clear that the potassium content of whatever table salt sold in food stores and supermarkets should not be matter of concerns however, as for any diets and integrators (Fe, K, Mg...) there exists health conditions which are incompatible. Before starting a diet or taking dietetic salts one should seek the advice of his/her physician.


text:
How can I free base cystamine dihydrochloride?
I am using cystamine (free based) to conjugate to a phosphate group on an oligonucleotide. I tried several keyword searches on Google for neutralizing the dihydrochloride at the ends of the cystamine. But, I am unable to find a protocol. I had once found a procedure for the process, but I did not reference it and hence I am unable to find it again.
The procedure I followed was:

Dissolve the cystamine.2HCl in water.

Transfer to a separation funnel and add chloroform.

To this add 1N NaOH.

After shaking vigorously collect the chloroform layer which contains the cystamine.

Repeat extraction couple of times using more chloroform.

Pool fractions - dry with anhyd. MgSO4 - transfer to an RBF.

Evaporate on a RotaVap

Dry on schlenk line and store in an inert environment at 4 deg. C.

However, the extraction was not very efficient. Can anyone refer a protocol for the same?

Two things come to mind: cysteamine is water soluble so you may need to salt out your aq phase and also 1M NaOH may be strong enough to deprotonate the thiol. I would suggest adjusting the pH to close to 7 and adding salt before trying the organic extraction.

Thanks for the tip. I will attempt your method to see if there is any change in the result. But, I have done an NMR on the extracted "cystamine" and it shows a clean spectra with an intact disulfide bond.

Perhaps go with NaHCO3 sat

In that case focus on keeping your water volume down, use more concentrated base and salt the aq phase. Your product is most likely still in the aq phase

First off, I would always use methylene chloride ahead of chloroform, especially under basic conditions. You can 'pep' up your extraction by adding a small amount of a third solvent that dissolves in both water and your extracting solvent, and also dissolves your compound to some extent. Methanol, ethanol or THF normally spring to mind.

Yes I agree use CH2Cl2.

@Beerhunter Could you explain how the third solvent improves the extraction efficiency of the cystamine free base if it dissolves in both the aqueous phase and the extracting - chloroform phase?

@Beerhunter
Waylander and everyone else thanks for the answers. I think if I am able to follow the question in my previous comment, my query in the entirety would be clarified.
Thank you

@Somdeb you said yourself the extraction wasn't efficient. Either switch extraction solvent, augment your extraction solvent or hamper your water solubility. I have no feel for the solubility without data. If polarity is an overriding factor, increase it for your organic phase. I don't mean add so much solvent you get one phase. Start with 5-10%.

If you found an answer useful in helping you, please mark it as accepted. If none of the answers helped you directly, you may post an answer yourself with the information. This is not a forum, so please don't edit the title with meta information. You can take the [tour] to find out more about this site.

Maybe the MgSO4 is absorbing your product. You may want to try another drying agent Try Na2SO4, zeolites, or azeotropic distillation (e.g. with benzene). Maybe NaOH will work, but it may react with cysteamine.

This question answered my query. So I am marking this as an answer. Hence marking my question as SOLVED. I am sorry but I don't require further answers to this question. Can any mod mark this question as CLOSED to answers? @Martin - マーチン♦

Freebase cystamine can be recovered from the dihydrochloride salt in $95%$ or greater yield using the following method:

Dissolve the $\ce{cya.2HCl}$ in pure $\pu{10N}~\ce{NaOH}$. I typically did this with $\pu{10g}$ cyanuric acid at a time, but it will probably work at larger scales.

As the $\ce{cya.2HCl}$ reaches nearly full solubility (you may see a few small flakes still hanging on) the addition of a small amount more of the $\pu{10N}~\ce{NaOH}$ will turn the solution cloudy. This is the freebase cyanuric acid crashing out of solution. This typically happens at around $10-11~\mathrm{pH}$, and I believe it is the result of the high $\mathrm{pH}$ and high concentration of $\ce
{NaCl}$ produced by the acid-base neutralization that salts the product out.

This cloudy solution can then be extracted with DCM. One extraction will get most of the cyanuric acid, but I typically do a couple more just to be sure. Doing this extraction in a separatory funnel tends to form a pretty thick emulsion - just wait it out.

Dry the DCM fractions over anhydrous salt, and reduce to a viscous, slightly opaque oil under vacuum.

As a note, be careful while doing the liquid-liquid extraction. You really don't want to spill a mixture of strong base and DCM all over yourself and your fume hood.


text:
Major contributor for resonance
I was doing some practice problems, and apparently 6 is the main contributor, due to having all full octets, but I'm not seeing how the oxygen with the + formal charge has a full octet in this case. I thought neither of the molecules had full octets for everything, so 6 would be the major contributor because it has more bonds.
What is the reason for 6 being the major contributor?
Thanks!

Don't forget about lone pairs that are not drawn.
In (5), O has 2 lone pairs and a full octet, but C doesn't. Compare that to (6), where O contributes one of its lone pairs to (form an extra)* bond with C. Now C has a full octet and O still has a full octet and a formal charge of +1. The formal charge +1 means there are 5 electrons around O, which includes a lone pair that is not shown.


text:
How to find the vitamin C content of orange juice?
I'm an IB chemistry SL student doing an IA on the vitamin C concentration of orange juice. We haven't learned about the titration process, so here is what I have done so far:
$\ce{5KI + KIO3 + 3H2SO4 -> 3I2 + 3K2SO4 + 3H2O}$
This is the equation for producing the iodine solution. Now using a titration process, I added the iodine solution to the orange juice. I measured the volume of Iodine solution used.
Here is the formula for the reaction between the vitamin C and the Iodine solution.
$\ce{C6H8O6 + I3- + H2O -> C6H6O6 + 3I- + 2H+}$
I'm not sure where the Triiodide comes from. Can someone explain this to me? I'm not a very good chemistry student.

"IB", "SL", "IA" mean nothing to me. Are they relevant to the question?

@TAR86 International Baccalaureate (IB) Standard Level (SL) chemistry student doing an Internal Assessment (IA), see https://owltutors.co.uk/chemistry-ib-ia-ideas-2017-2018/ ... and not relevant to the question ...

Iodine and iodide will form triiodide in aqeous solution. This is a common way to increase the solubility of $\ce{I2}$ in water. For a titration, one will use excess $\ce{KI}$ and control the formation of iodine by adding $\ce{KIO3}$ using a burette.

In response to the comment:
$\ce{KI}$ contains iodide anions. I strongly suspect that an excess of $\ce{KI}$ was used (more than fivefold compared to $\ce{KIO3}$), leaving enough iodide anions to form the triiodide. Also note that effectively a $\ce{I-}$ can be subtracted from either side of the second equation of the question.

Where does the iodide come from? What I'm seeing here is that the Iodine in the first equation is somehow changed into triiodide in the second equation.
I included the course name, so people would understand the topics I have done so far, and explain things I would know about. Again, I'm not a particularly good chemistry student.

Thanks, that helped me. I did use more than five times the amount of KI.  I'm just not sure how the I- can be subtracted from either side of the second equation. I am pretty behind in my course.

Also, how can you measure the mass of triiodide if you have an excess of KI? The reaction is reversible, so I'm not sure how much of the triiodide would be left.

One way to determine the amount of vitamin C in food is to use a redox titration. The redox reaction is better than an acid-base titration since there are additional acids in a juice, but few of them interfere with the oxidation of ascorbic acid by iodine.
Iodine is relatively insoluble, but this can be improved by complexing the iodine with iodide to form triiodide:
$$\ce{I2 + I- <=> I3-}$$
Triiodide oxidizes vitamin C to form dehydroascorbic acid:
$$\ce{C6H8O6 + I3- + H2O → C6H6O6 + 3I- + 2H+}$$
As long as vitamin C is present in the solution, the triiodide is converted to the iodide ion very quickly. However, when all the vitamin C is oxidized, iodine and triiodide will be present, which reacts with starch to form a blue-black complex. The blue-black color is the endpoint of the titration.
This titration procedure is appropriate for testing the amount of vitamin C in vitamin C tablets, juices, and fresh, frozen, or packaged fruits and vegetables. The titration can be performed using just iodine solution and not iodate, but the iodate solution is more stable and gives a more accurate result.`
See: https://www.jacobtuition.co.uk/post/inspirational-ideas-for-your-chemistry-ib-ia

(+1) We used to use a variant of this, using coulometric titration to produce iodine in situ, as described in the last paragraph of my answer here. The solution had excess KI. The endpoint was via the standard starch-iodine complex.

Balance the second equation

Note that you copied an unbalanced equation from the OP.


text:
Ions not attached but included in chemical formula for tartrazine
In some diagrams for tartrazine ($\ce{C16H9N4Na3O9S2}$), I see that $\ce{SO3Na}$ is actually represented as $\ce{SO3^-}$ with a floating $\ce{Na^+}$ next to it. How does this work?

EDIT: Wait, are they just ionicly bonded together?

That would be a yes to your ionic bond theory ;)

Don't worry, there were worse ;)


text:
Finding the wavelength of maximum absorption of Tartrazine using Fieser-Kuhn
How would you use Fieser-Kuhn rules to calculate the wavelength of maximum absorption for tartrazine?

Fieser-Kuhn rules states that:
$\lambda_{max}=114+5M+n(48-1.7n)-16.5\ R_{endo} - 10\ R_{exo}$

What would $M$ be in this example? There aren't any alkyl substituents, but what I'm not sure about ring residues. Would this refer to the $\ce{SO3Na}$ and $\ce{COONa}$?

Would $n$, the number of conjugated double bonds be equal to 12?

Does Pyrazole count as a ring for $R_{endo}$? So $R_{endo}=3$?

Or would Woodward-Fieser rules be more suitable to use in this scenario?

Have a look on mhchem for input of chemical formulae in the body of questions, answers, comments.  Said page includes a test drive to familiarize yourself with this tool, too.

Terminal -SO3Na groups have each two double bonds

Fieser-Kuhn rules needs to be used to calculate the wavelength in this case, since there are more than 4 conjugated double bonds. Woodward-Fieser rules are only applicable for conjugated dienes and polyenes with upto 4-double bonds or less. [1]
M = 6 (SOONa, SOONa, COONa, 2 N atoms connected by triple bond,N connected to the Benzene sulfinate group)

n=12 as illustrated in the question

 = 3

=114+(5*6)+12*(48−1.78*12)−16.5*3−10*0
=425.7nm

This calculation matches the experimental data as seen in [2][3][4] which list the  value of tartrazine as 425nm, 427nm, 428nm respectively.

References:
[1] Mehta, A. (2015) Ultraviolet-visible (UV-vis) spectroscopy – Fieser-Kuhn rules to calculate wavelength of maximum absorption (lambda-max) of polyenes (with sample problems): Analytical Chemistry, PharmaXChange.info. Available at: https://pharmaxchange.info/2013/05/ultraviolet-visible-uv-vis-spectroscopy-%E2%80%93-fieser-kuhn-rules-to-calculate-wavelength-of-maximum-absorption-lambda-max-of-polyenes-with-sample-problems/ (Accessed: March 4, 2023).
[2]Gobara, Mohamed & Elsayed, Mohamed. (2016). Preparation and Characterization of PANI/TiO2 Composite for Photocatalytic Degradation of Tartrazine Dye. The International Conference on Chemical and Environmental Engineering. 8. 370-389. 10.21608/iccee.2016.35131.
[3]  Ayman A. Ali et al. (2020). Spectrophotometric determination of Tartrazine in soft drink powder and
pharmaceutical dosage form. Journal of Basic and Environmental Sciences, 7, https://jbesci.org/published/7.2.7.pdf
[4] Gobara, Mohamed & Baraka, Ahmad. (2014). Tartrazine Solution as Dosimeter for Gamma Radiation Measurement. International Letters of Chemistry, Physics and Astronomy. 33. 106-117. 10.18052/www.scipress.com/ILCPA.33.106.

References 2 and 4 are identical

@Maurice I have now corrected the references. Both the papers 2 and 4 were from the same set of authors so I had pasted same link by mistake.


text:
Can nitrate and sulfate ions be oxidised during electrolysis of their molten compounds?
$$\begin{align}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{S2O8^2- + 2e- \;&<=> 2SO4^2-}\quad\quad &&E^\circ = +2.01\ \mathrm{V}
\end{align}$$
From the above, I know that an aqueous solution of a sulfate salt undergoing electrolysis will not oxidise to peroxydisulfate ions over water being oxidised to oxygen at the anode unless it is at extremely high concentrations.
However, what if the solution is molten? Will the product at the anode be peroxydisulfate?
Also, can nitrate ions be oxidised? Perhaps to peroxynitrate or peroxynitrite?

Peroxysulfates are industrially created in aqueous electrolysis. Nothing extreme is needed, just concentrated solution and high density current. You didn't think about overpotential here, it seems.

Not only they can, but they must get oxidized. There is no other option. This is the only way the electric current can pass between the electrode and the solution.
Whether or not the products will be some peroxyanions is another question. At high temperatures typical of molten salts, I don't think they will survive. Hence I would expect $\ce{O2}$ + disulfate in the first case and $\ce{O2 + NO2}$ in the second.

The proposed anodic products are interesting for another reason.  They would seem to suggest that nitrogen is reduced from +5 to +4 at the anode!
Might we have to abandon our usual oxidation state convention, which is geared to a water-based world, for nitrate solvent systems?

Nitrogen would get reduced because it has nowhere else to go. There are no binitrates (unlike bisulfates), and $\ce{N2O5}$ is not particularly stable. This has little to do with electrolysis.

Note the autoionization of dinitrogen tetroxide given by https://en.wikipedia.org/wiki/Molecular_autoionization.  Oxygen seems to transfer with one negative charge.  Might the effective oxidation state of oxygen in nitrogen oxide/nitrate solvent systems be $-1$ instead of $-2$?  It would give a more parsimonious description of the anodic reaction of nitrate ion in nitrate solvents.

Whatever. Oxidation states are human invention, not a fact of nature; they are all arbitrary anyway.


text:
How to determine which p-type GTO to choose in HF method?
I am trying to understand the working principle of basis sets in quantum chemistry. As far as I understand in the Hartree-Fock method, the basis set is used to calculate the matrix representation of Fock operator in each iteration in the SCF procedure. Usually each basis function in a chosen basis set is represented as a linear combination of several primitive Gaussian type orbital (GTO). I want to use cc-PVDZ basis for C atom which I can take from the EMSL Basis Set Exchange Library. For example in the Gaussian format:

!  cc-pVDZ  EMSL  Basis Set Exchange Library   10/19/17 2:40 AM
! Elements                             References
! --------                             ----------
! H     : T.H. Dunning, Jr. J. Chem. Phys. 90, 1007 (1989).
! He    : D.E. Woon and T.H. Dunning, Jr. J. Chem. Phys. 100, 2975 (1994).
! Li - Ne: T.H. Dunning, Jr. J. Chem. Phys. 90, 1007 (1989).
! Na - Mg: D.E. Woon and T.H. Dunning, Jr.  (to be published)
! Al - Ar: D.E. Woon and T.H. Dunning, Jr.  J. Chem. Phys. 98, 1358 (1993).
! Sc - Zn: N.B. Balabanov and K.A. Peterson, J. Chem. Phys. 123, 064107 (2005),
! N.B. Balabanov and K.A. Peterson, J. Chem. Phys. 125, 074110 (2006)
! Ca     : J. Koput and K.A. Peterson, J. Phys. Chem. A, 106, 9595 (2002).
!

****
C     0
S   8   1.00
   6665.0000000              0.0006920
   1000.0000000              0.0053290
    228.0000000              0.0270770
     64.7100000              0.1017180
     21.0600000              0.2747400
      7.4950000              0.4485640
      2.7970000              0.2850740
      0.5215000              0.0152040
S   8   1.00
   6665.0000000             -0.0001460
   1000.0000000             -0.0011540
    228.0000000             -0.0057250
     64.7100000             -0.0233120
     21.0600000             -0.0639550
      7.4950000             -0.1499810
      2.7970000             -0.1272620
      0.5215000              0.5445290
S   1   1.00
      0.1596000              1.0000000
P   3   1.00
      9.4390000              0.0381090
      2.0020000              0.2094800
      0.5456000              0.5085570
P   1   1.00
      0.1517000              1.0000000
D   1   1.00
      0.5500000              1.0000000
****

In the lines such as S   8   1.00 I know that the letter S is the orbital type, 8 is the number of GTO in this orbital's expansion, but what is 1.00 for?
The GTO for a p-type orbital has the form $x^l y^m z^n \exp(-ar^2)$ where the powers satisfy $l+m+n=1$ and thus there are three possible sets of $\{l,m,n\}$. But for instance in that link for lines with p-type orbitals, which possibilities shall I pick? The same question for d-type orbitals.

The link is broken. I have used my crystal ball to guess that you looked at a Gaussian94-style listing. In this case, the 1.0 refers to a sort of global scaling factor. In general, see reference of the program the format of which you use for a description of the basis set listing format. In terms of which p-orbitals to pick: all of them. You never want just one - imagine your molecule rotates.

Yes you are right it's Gaussian94 style for cc-PVDZ basis set for C atom.
Using example in that link, there are 8 listed orbitals, 3 of them p, one of them d, the rest is s. Does this mean there are in total 4+(3x3)+(1x6)=19 orbitals to be used to represent the Fock operator?

Your calculation is correct for "cartesian" gaussians (noted as 6D in Gaussian03 etc.). Many programs do linear combinations of the cartesian gaussians to yield "spherical" gaussians, that is, 5 d functions and one s function (which I imagine has nodal planes). The s function is then effectively discarded to save computational time.

@nougako I think you confused Dunning's cc-pCVDZ basis set with the cc-pVDZ basis set. Martin has posted the cc-pVDZ basis set as you said, but from your comment I assume you were looking at the core polarizied double zeta basis set (which has an additional sp shell). Could you state which basis set your intented to use?

I intended to use cc-pVDZ for C atom. It's been a while now and I don't remember why I said there are 8 orbitals there while it seems like there are only 6 in my previous comment.

Have a look in the Gaussian manual if something is unclear.
If you want to use an extern basis set in Gaussian you have to respect the gaussian input conventions for the basis set, which you can find here (gaussian.com/gen/). Since you are using the EMSL Basis Set Exchange the server gives you a file that satisfies these conventions. Like:
Type  NGauss  Sc
$α_1\quad d_{1μ}$
$α_2\quad d_{2μ}$
…
$α_N\quad d_{Nμ}$
Type stands for the shell, then you have the number of primitives with NGauss and scaling factor Sc. Afterwards you put your exponents and coeffients of the primitive gaussian functions which then will be contracted to the respective contracted gaussian functions. You might notice that the scaling factor is set to unity for almost all basis sets—I checked some Dunning, Pople and Ahlrichs basis sets and always found it set to unity, which seems resonable. In the TURBOMOLE format the scaling factor is omitted totally, so it seems to be a Gaussian specific hack.
Now to your second question. You are specifing the shell, not a single function, a shell contains always all spherical harmonics belonging to the given quantum number or all cartesian gaussian functions. The spherical harmonics are the solution of the Schrödinger equation for an electron on a surface. You have generally $2l+1$ spherical harmonics where $l$ is the azimudal quantum number specified as 0/s, 1/p, 2/d, 3/f, 4/g, 5/h and so on. There exists also the possibilty to use cartesian gaussians with $(l+1)(l+2)/2$ degenerated functions.
Why am I telling you that? You always choose a whole shell with a azimudal quantum number, how many function belong to this shell depends on the quantum chemistry code you are using. ASAIK, Gaussian supports both.
In case of a p-shell it does not matter, both expressions evaluate to three p-functions (as expected). So you get three p-functions with one p-shell.
The basis set you have chosen would be evaluted as:
(9s4p1d) → [3s2p1d]
Having nine primitive s type gaussians (note that Dunning basis sets use a general contraction scheme), four primitive p type gaussians and one primitive d type gaussian which form a basis set out of three s shells, two p shells and one d shell with 14 contracted spherical harmonic gaussian functions or 15 contracted cartesian gaussian functions for carbon.


text:
What would happen if the volume changes in first order reactions?
I am searching for any piece of hints but couldn't find out anything
What would happen if the volume changes in first order reaction?
For example, suppose there is a first order reaction of ideal gases as below
\begin{align}\ce{A (g) &-> B (g) + C (g)}& v &=k[\ce{A}]\end{align}
Then we all know that
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$ and $t_{1/2}$ is constant if the volume is the same.
What if the reaction is happening in a cylinder which is not fixed?
I suggest that since this is a first order reaction, the proportion of molecules which proceeds in reaction is the same every time, and the extra volume change is independent of the happening reaction, so the concentration change would get more complex, but the change of the total number of molecules would be the same as the concentration change when volume is fixed.
I am not a professional in this field so I might be wrong. If wrong, in what point?

Go back to the definitions of the various free energies and see which one applies and how that would affect your $k$.

May there be something that I don't know? In my knowledge, the constant is affected by temperature and activation energy, which are both provided constant in this situation. sorry if i didn't mention that


text:
Osmium tetroxide reaction with cell membrane
Background: $\ce{OsO4}$ is a widely used staining agent used in transmission electron microscopy (TEM) to provide contrast to the image. In the staining of the plasma membrane, osmium(VIII) oxide binds phospholipid head regions, thus creating contrast with the neighbouring protoplasm (cytoplasm).The osmium tetroxide molecule is tetrahedral and therefore non-polar. This nonpolarity helps $\ce{OsO4}$ penetrate charged cell membranes.

Question:
What would be the product formed by reaction of $\ce{OsO4}$ with polar head of phospholipid of cell membrane?

My attempt , since $\ce{OsO4}$ is non-polar and phospholipid is polar, therefore it must be some sort of covalent interaction, I couldn't find any literature regarding further at wikipedia.

OsO4 is pretty polar.

Its tetrahedral in shape, so all dipoles cancel each other. Could you explain further?

True, but that's irrelevant.

So , is it polar interaction? If yes, then how does the osmium tetroxide diffuse to other side of membrane? As far as I know cell membrane doesn't allow polar molecules to pass through freely.

@IvanNeretin OsO4 is very reactive but it isn't polar. It's a symmetrical tetrahedron and can't be polar. It is very reactive with certain organic functional groups, which is why it stains tissues and polymers.

Symmetry enforces zero dipole moment, but this does not negate polarity.

sigh Not that again... If I've never again heard of zero dipole molecule as "non-polar" I'd be happier. Substances are polar and this has not much to do with any dipoles. Also that is irrelevant to the question.

I really don't understand the reason for OsO4 being polar, all 4 O atoms are at equidistant from central metal atom, so how can it be polar?

I has zero dipole, but that should  not be called "non-polarity". Polar can be solvents and have much different parameters that describe it.

According to the paper The chemical nature of osmium tetroxide fixation and staining of membranes by x-ray photoelectron spectroscopy[1] by White et al. (1976),

The results support a scheme for the reaction of osmium tetroxide with tissues in which the initial reaction site is the double bonds of unsaturated lipids to form $\ce{Os(VI)}$ derivatives. Subsequent hydrolysis and further reduction yield complexes of $\ce{Os(IV)}$ and $\ce{Os(III)}$. A mixture of these three states is present in membrane specimens during microscopic observation. $\ce{Os(VI)}$ and $\ce{Os(IV)}$ could be present as osmate esters and osmium dioxide, respectively; $\ce{Os(III)}$ could be present as an oxo- or amino complex(es).

However, since $\ce{OsO4}$ reacts with biological material, the above list is probably incomplete.
Riemersma[2] suggests that negatively charged colloidally dispersed osmium dioxide hydrate is deposited at the polar ends of lipids that contain a positively charged (phosphoryl)choline moiety (lecithin, phosphatidylcholines, etc). In this case we would have a pure Coulomb interaction.

References

White DL, Andrews SB, Faller JW, Barrnett RJ, Biochim Biophys Acta, 436(3):577-92 (1976).
Jelle C. Riemersma, J. Histochem. Cytochem, 11, 436-442 (1963)


text:
Synthesis of Copper(I) Iodide
Copper(I) iodide is usually made by the reduction of copper(II) salts. An example of using potassium iodide and copper(I) sulphate is shown below.
$$\ce{2CuSO4 + 4KI + 2Na2S2O3 -> 2CuI + 2K2SO4 + Na2S4O6 +2NaI}$$
I did another method of synthesising $\ce{CuI}$ by preparing it directly from its elements, with $\ce{NaI}$ as the solvent for solid $\ce{I2}$:
$$\ce{2Cu + I2 -> 2CuI}$$
May I know why the 2nd method (direct synthesis) is better than the 1st in terms of green chemistry?

According to the United States Environmental Protection Agency, there are 12 principles of green chemistry. The second method of synthesis is more green than the first method of synthesis as it adheres to the first, second and fourth principles while the first method does not. We can easily observe why this is so.
Firstly, the second synthesis method prevents wastage of any reactants, all the reactants go into the making of the product, copper (I) iodide while for the first method, there are other products generated along with the desired product. In other words, there is a wastage of reactants if the first synthesis method is used.
Secondly, the second synthesis method maximises atom economy (i.e. waste few or no atoms). This is similar to the first principle. Atoms are wasted in the first synthesis method in making the other products.
Thirdly, the second synthesis method produces what is desired and does not produce any other chemicals that may be hazardous. According to this and this, sodium tetrathionate and sodium thiosulfate are irritants to the human body.
Hope I have answered your question.


text:
How to calculate the concentration of the elements after dissolution of iron(III) chloride?
I was presented with the following problem:

In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

What is the molarity of the solution?
What is the concentration of iron(III) cation in M?
And what is the concentration of chloride anion in M?

I know how to do the first part of the problem, the answer is $\pu{0.406 M}\ \ce{FeCl3}$. But when I go to balance the equation, the computer system tells me the equation is wrong.
My equation:
$$
2\ce{FeCl3 -> 2 Fe + 3Cl2}
$$
The computer system:
$$
\ce{FeCl3 -> Fe^3+ + 3Cl-}
$$
I don't understand the computer systems' answer. Simply because when by itself, chlorine is a diatomic molecule therefore, should be $\ce{Cl2}$ not just $\ce{Cl}$. Am I wrong to think this? Am I correct or the computer system? If the computer system is correct, why?

The key is the difference in chlorine vs chloride.  Any ideas?

Fe has 3+ charge and Chlorine has  a 1- charge. Is that what you are referring to?

The answer needs charges which you seem to have omitted. It's not $\ce{Cl}$, but it's also not $\ce{Cl2}$ for this question.

I've put in an edit with charges in the correct place.  It may make the answer to the question a bit more obvious.

Okay I think I understand. So when you dissolve the FeCl3 it completely dissociates and yields one Fe +3 ion and 3 Cl -1 ions. Elemental Chlorine gas is diatomic, however, Chloride ion is a single chlorine atom with a negative one charge.  When you have an ionic compound, as in the case of FeCl3, the compound is made up of elements in their ionic states, NOT in their elemental states.

@user213305 In mhchem you can (and should) typeset the reaction equation as a whole: \ce{2FeCl3 -> 2Fe + 3Cl2} not \ce{2FeCl3} \rightarrow 3\ce{Fe} + 3\ce{Cl2}.

When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.

In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

\begin{align}
  m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\
  M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\
  V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\
\end{align}

What is the molarity of the solution?

Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.
\begin{align}
  c &= \frac{n}{V}\\
  n &= \frac{m}{M}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\
  c(\ce{FeCl3 (aq)}) &=
    \pu{0.406 mol//L}
\end{align}

What is the concentration of iron(III) cation in M?
And what is the concentration of chloride anion in M?

You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.
For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is
$$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$
From this you can derive the actual relationships between the ion concentrations.
You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore
\begin{align}
  c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)})
  &&=\pu{0.406 mol//L},\\
  c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)})
  &&=\pu{1.22 mol//L}.
\end{align}

The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:
$$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$

The other equation could also be the electrolysis of the molten salt.

you finish question 1 and you need to finish question 2 and 3
but why are you making a equation for that. it seem like that was what stopped you working in the first place.
but to make it balanced... you said dissolved in volumetric flask. if it dissolved its ionic.
and that means there should be a "cation +" and a "anion -"
hence the $$\ce{fe^3+ + 3cl- }$$
and the $\ce{fe^3+}$ is third oxidation state floating around in solution it could had been a $\ce{fe^2+}$ if you started with $\ce{fe(oh)2}$
so the pc is correct....
1 mole of $\ce{fecl3}$ will yield's 1,5 mole of cl2 therefore it has more $cl$ then $fe$ ions.
concentration of $fe$ ion in $M = 0.406$
concentration of $cl$ ion in $M = 1.218$
$final\ equation\ =$ $$molarity\ of\ \ce{fecl3} * number\ of\ fe\ ion's\ in\ the\ molecule$$
$$\ce{fe^3+}=\  0.406M * 1 = 0.406M$$
$$molarity\ of\ \ce{fecl3} * number\ of\ cl\ ion's\ in\ the\ molecule$$
$$\ce{cl^-}=\    0.406M * 3 = 1.218 M$$

Please use capital letters and please don’t misuse MathJax.


text:
When using LPCVD, is it possible to pattern a silicon wafer with poly-silicon using a mask?
I know it's possible to etch (i.e. pattern) poly-silicon, but I'm curious if one can just use a mask instead. Are there masks that inhibit the growth of poly-silicon?

In my fab experience, the answer is no, you are not going to inhibit deposition of silicon with some other material. You might well change the final grain size, but it will deposit. Furthermore, LPCVD generally has good sidewall coverage, so you can't use a lift-off mask either.
The OP asked for references, so lets dig in to those. First, it is somewhat difficult to find specific references to deposition rates on various substrates that can be broadly compared - reactor specifics, gas flow, temperatures, etc., all make it hard to compare experiment-to-experiment. But, we can pin down a few things.
First, going back to Andy Grove's (yes, Intel's Andy Grove) book "Physics and Technology of Semiconductor Devices" (John Wiley and Sons, New York NY, 1967) one finds no mention of polysilicon deposition. At that early stage in technology it was not used - metal gates were the way to go.
Polysilicon became a thing with poly-Si gates and the LOCOS (local oxidation of silicon) process. Here one might end up depositing poly over a wafer that included single crystal silicon (to make source-drain contacts), oxide (gate and field isolation), and silicon nitride (for the LOCOS process). Looking at cross sections from LOCOS devices (in various more modern text books - one I have is Stephen A. Campbell, "The Science and Engineering of Microelectronic Fabrication" (Oxford University Press, New York NY, 1996)), you see no variation in poly thickness across those different materials.
There is much more literature available around questions like polysilicon grain size vs initial surface roughness, annealing temperature, etc., such as Voutsas and Hatalis, "Surface Treatment Effect on the Grain Size and Surface Roughness of as-Deposited LPCVD Polysilicon Films", J. Electrochem. Soc. 140(1) 282-288 (1993).
Finally, a number of people have done work on the kinetics of LPCVD poly-Si growth, with one early reference being M.L. Hitchman et al., "Polysilicon growth kinetics in a low pressure chemical vapour deposition reactor", Thin Solid Films 59 231-247 (1979).
There are two takeaways from papers like this. First, under LPCVD reactor conditions, the mean free path of a silane molecule (or any other reactant) is quite long. This is, of course, highly desired to have uniform deposition across many wafers in, e.g., a tube furnace. The flip side is that the probability of a silane molecule reacting upon hitting a surface is quite low - they bounce around a lot. And, once a surface has any poly-Si on it, it all looks the same.
Second, the only way to not have poly-Si deposition is to never allow nucleation in the first place. Yet, point one above implies that lots of silane is hitting the surface, and if any one molecule manages to stick you now have a place to grow more. Further, all that seems to be needed to get some silane reactivity is the presence of hydrogen on the surface, and there will be lots of hydrogen around as silane decomposes elsewhere (plus hydrogen on surfaces is pretty much a given - it is by far the easiest surface termination on just about anything, particularly after a chemical process in an aqueous solution, like an acid clean).


text:
State functions and their arguments
$U$ and $H$ are defined as
$$U=\sum E_{pot} +\sum E_{kin} ; dU=dq+dw $$
$$H=U+pV$$
However, it was stated in a lecture note that for a fixed molar number, $U$ and $H$ can be defined using any two properties, example: $U(Temperature,Volume)$ and $H(Temperature, Pressure)$.
However, I cant figure out why only two quantities are enough, even if we consider the above example.
Can somebody show how by using just these two quantities is enough to figure out $U$ and $H$? (Just proving for the above example would be fine)

See derivation within: https://en.wikipedia.org/wiki/Internal_energy#Changes_due_to_temperature_and_volume

Also: https://en.wikipedia.org/wiki/Enthalpy#Other_expressions

It you know temperature and pressure or temperature and density, then you know everything about the state.  So, the internal energy and enthalpy per mole are determined, relative to any other state.

@Zhe To which derivation are you referring? I might've well missed it but didn't find it in the Wiki at a quick glance. Anyhow, this is a good question. Look forward to a proofy answer.

@LinearChristmas I was thinking of the ones that refer to $dU$ in terms of $dT$ and $dV$. And likewise, $dH$ expressed as $dT$ and $dP$.

@Zhe Doesn't the derivation of these results already assume that the state functions are two-dimensional when enplethies are constant? The OP is asking (I think) about the mathematical background or logical soundness of why $U = U(P, T, n_i) $ (or some equivalent) instead of, for instance, $U = U(P,T,S,V,n_i)$. A way to push the question forward would be to say that it works because $S = S(T,P)$, and hence we may leave out entropy (or temperature, or pressure). But it's hardly satisfying, and sadly circular.

Yeah, I guess that's fair @LinearChristmas. Someone who's a real physical chemist should answer instead...

This follows from the first law of thermodynamics; i.e., $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ which identifies $U$ as a function of $S$ and $V$. If in addition we have other forms of work, such as chemical work, then we add a chemical work term $\mu\mathrm{d}N$ to the first law, and introduce an additional dependence of $U$ on $N$.
In addition, we can also replace a given independent variable by its conjugate variable, since, by thermodynamic stability, the independent variable is a monotonic function of its conjugate and vice versa. That is, the following sets of independent variables can all be used to characterize $U$: $$(S, V, N), (T, V, N), (S, P, N), (S, V, \mu), (T, P, N), (T, V, \mu), (S, P, N).$$ Note in particular that $(T, P, \mu)$ cannot be used, because these three variables, all being intensive, are not independent and are related by the Gibbs-Duhem equation.


text:
Zinc Sulfate Purification
In the process for electrolysis for zinc, three things must be done:

Leaching the ZnO
Purifying the ZnSO4 by means of cementation
Performing the electrolysis

However, step 2, as it is presented by Wikipedia, is quite vague. In particular, it says:

The purification process utilizes the cementation process to further purify the zinc. It uses zinc dust and steam to remove copper, cadmium, cobalt, and nickel, which would interfere with the electrolysis process.

However, when looking it the definition of cementation, a crude summary of what happens is that one metal gets oxidized in order to reduce the other ion. I am assuming that this means that the zinc sulfate is getting reduced. However, what is reducing it? What equations are involved? I apologize if this question does not make sense, as chemistry is not particularly my strong suit, but I believe that this has gone unanswered online so far.

I took the liberty of adding the relevant hyperlink.

As with nearly all STEM-related Wikipedia pages, this one reads quite clear and concise... that is, if you already know at least 75% of the stuff involved. Otherwise it may lead you into the deepest muddle, so you did the right thing by asking this question.
Follow the logic. What is electrolysis used for? It reduces a metal (in particular, zinc) from its salts. You don't reduce what's already been reduced. This means zinc was not reduced before the step 3. In particular, it was not reduced during the cementation. Still, cementation implies that one metal is reduced by another. Which of the two is zinc if we've just concluded that zinc does not get reduced on this step?
Since the naturally found compounds are rarely pure, zinc sulfate after the step 1 contains a few percent of other metals (copper, nickel, etc.), also in the form of sulfates. Should we electrolyze it now, these metals will get reduced along with our zinc and remain in it as impurities, which we don't want. Worse yet, they are mostly less active than zinc, so they will reduce before it and thus concentrate in the produced metal. How do we get rid of them?
Now comes the idea: let's reduce them. We vaguely remember this can be done with a more active metal. But it must not be too active, otherwise it will reduce zinc as well. Also, the oxidized form of this new metal must not spoil our $\ce{ZnSO4}$. Just what metal can that be?
Guess what?

 It is zinc.
 Yes, we add metallic zinc (just a few percent, I remind) and it reduces the impurities as follows:
 $$\ce{CuSO4 + Zn -> Cu + ZnSO4}$$
 The only soluble byproduct is $\ce{ZnSO4}$, which does not upset us one bit.

So it goes.

Wow, this is the perfect answer. Thanks so much!


text:
How can an SN1 reaction give non-racemic products?
I'm in organic 1 right now and we assume that SN1 reaction will always have racemic products for the sake of simplicity. But then our professor have a question for us that these solvolysis reactions will not end up with racemic products. Could you help me find the reason why it is not racemization?

Please use Stack Exchange image hosting service for uploading images. Press the little image pictogram above the text box in the editor and upload. The image will be uploaded to a dedicated webspace which is guaranteed to remain existant. (Note that you should not upload copyrighted work.)

The assumption SN1 -> racemic is a useful concept, but the reaction does not have to proceed by SN1 only. Further, consider that the lifetime of certain carbocations may be long enough for geometry changes to more favorable conformations that may have preferred attack paths.

Look up conformations of substituted t-butyl cyclohexanes to get a clue. For the 2nd, find a neighbouring part of the molecule that could "participate"


text:
Do indigo and leuco-indigo have fully conjugated pi systems?
I've just come across a figure indicating the positions of conjugated bonds in indigo and leucoindigo, and I'm quite confused about why both molecules aren't fully conjugated.

My reasoning is as follows:

Both carbons above and below the bond are sp2 hybridized and have π electrons.
The carbon to the left is sp2 hybridized and has π electrons, and the nitrogen to the right has a lone pair.
Same as 2.
Same as 1.

I also want to ask about why leuco-indigo isn't planar. According to Wikipedia:

In indigo white, the conjugation is interrupted because the molecule is nonplanar.

If a carbon atom is sp3 hybridized, it doesn't  have any pi electrons. Other than that, you are mostly right, and the book tends to oversimplify things. Besides, they have their leucoindigo formula downright wrong.

Let's just start with stating the very obvious: Figure 2.23 has more errors than there should be in an entire chapter.
Indigo
With the exception of bonds to hydrogen, all other bonds are nicely conjugated in indigo. The π molecular orbital show this very clearly.[1] The molecule has a fantastic C2h symmetry, which also means it is in one plane. Intramolecular hydrogen bonds certainly further stabilise this conformation. They can be visualised with the help of the Quantum Theory of Atoms in Molecules (QTAIM).[2]
 (click for full image)

Interestingly eleven out of twenty π orbitals are occupied, which leaves the HOMO being anti-bonding with respect to the central carbon-carbon bond, which is 138 pm long.
Going further I will first add electrons, then add protons according to the following scheme:

Indigo2-
I am happy to report, that the molecule stays flat. The central carbon-carbon bond elongates a little to 141 pm. It basically still has comparable orbitals, although the location of the nodal planes already accommodate the divide of the central bond. I am guessing that the intramolecular hydrogen bonds are accountable for this since the molecule is highly charged.

Now the basic question is how do the protons behave, where do they go, will they break the symmetry.
Leucoindigo
A bit to my surprise, the flat version of this compound is also most stable.

Preliminary conclusions
There are several points, where this project could have derailed. Level of theory is only one of them. The evidence presented here is not conclusive enough. In due time I might be able to do some more, but for now, at least the textbook picture is debunked.

Calculated at the DF-BP86/def2-SVP level of theory. Occupied orbitals are blue and orange, virtual orbitals (unoccupied) are red and yellow.
Analysis with MultiWFN 3.3.8, see http://sobereva.com/multiwfn
Nuclei are brown dots, bond critical points are blue dots, ring critical bonds are orange dots. Blue lines are zero-flux surfaces that seoarate the molecule into atomic basins.

Concerning this part of your question:

I also want to ask about why leucoindigo isn't planar. According to
  Wikipedia, "In indigo white, the conjugation is interrupted because
  the molecule is nonplanar

If I combine the images on the Wikipedia page for leuco dye we get:

which shows that the leuco- 'form' has a single (rotatable) central bond, not a double bond. Also the -OH and -NH look like they might clash - but would need more detailed 3D geometry for that, rather than just a 2D image.
Alternatively, there could be intra-molecular hydrogen bonds in indigo, that would not be present in the reduced leucoindigo. These hbonds would tend to stabilise the flat structure : for example see this study ("TD-DFT study on electron transfer mobility and intramolecular hydrogen bond of substituted indigo derivatives" C. Ma et al, Chemical Physics Letters, 2015).

It is not that simple. Clash is equally possible in both structures; so is hydrogen bond.

@IvanNeretin : Yes, I see you are right. I wonder what the optimal geometry of these bonds are. Perhaps the lone pairs on the carbonyl oxygen are in the plane, while those of the hydroxyl oxygen are not?

Lone pairs would readily adjust if that was the problem. No, it's not that simple.

Ok, well I guess I have no idea then. For me, it's interesting that the number of atoms in any intramolecular hbond would be nearly the same as the 'ring' of a protein backbone gamma turn.

Indigo: nice and planar, see crystal structure.

Well, indeed. The suggestion here is that leucoindigo is non-planar. In any case, the crystal structure also has pi- stacking interactions?

The hydrogen bonding argument is supported by the fact that leucoindigo is soluble in water while the purplish blue stuff is not (https://en.wikipedia.org/wiki/Leuco_dye).  Indigo forming internal hydrogen bonds instead of external ones can do that.

yes @OscarLanzi the dye molecules aggregate with each other and remains insoluble. But glucose , sodium dithionite reduce indigo to this leuco indigo and render it soluble.


text:
Why is a comma used to separate hydrogen gas and platinum when using cell notation?
I have come across many online articles and books where I see the following:

Gaseous electrode
Gases such as $\ce{H2}$ and $\ce{Cl2}$ can act as anode and as cathode by loosing electrons ans by accepting electrons, respectively. For example,
$\ce{Pt,H2 | H+}$ (acts as oxidation potential)
  or
  \begin{align}
\ce{0.5 H2 (g)(\pu{1 atm}) &-> H+ (aq) + e-} \\
E_\mathrm{cell} &= E_\mathrm{cell}^\circ -\frac{0.059}{1}\log{[\ce{H+}]} \\
 &= 0 + 0.059\mathrm{pH}
\end{align}

Aren't the two species in different phases? If so, aren't they supposed to be separated by a '|' and not by a comma?

What actually reacts is hydrogen adsorbed on platinum, which I wouldn't really call a different phase. You might argue that if we interpret it like that, then the gaseous hydrogen is missing, but everyone knows it is there anyway, so why bother.

(Without bothering to look at the textbook definition) I think that the "|" denotes the two different sides of the equation (in an abbreviated manner, since Pt would technically belong on both sides, being effectively a structure-providing catalyst).

Since $\ce{Pt}$ is not a reactant and only a catalyst, one only wants to write $\ce{H_2}$ on the reactant side. When it is specified, since it is not involved in the reaction, the | is not warranted since Pt doesn't show up in the reaction as the comment by TAR86 pointed out.
Generally, the electrode material is not entered into the shorhand notation since the material of the electrode doesn't change the thermodynamic potential. In this case, the choice of $\ce{Pt}$ is critical and that is why it is specified. The potential of the electrode would be affected by the details of the measurement device with any other electrode material that is a worse catalyst.

My question isn't why platinum is included. Rather it is about the semantics regarding why there is a comma after platinum.

I edited the answer. It is since the Pt is not showing up on either side of the reaction.


text:
remove inhibitor from monomer
I am working with a commercial monomer (monomethacryloxypropyl terminated polydimethylsiloxane) with an inhibitor of BHT @1%.
I need to remove it for making an ATRP reaction. What is the good way to do this?

What monomer? And what are your ideas?

MCR-M11 from gelest

I first tried to do it with basic alumina column, but I still have some traces..

Picture was too big and non-googlable. It may be difficult to don't have traces, dunno what you could do.

The BHT is only present to prevent small amounts of potential radicals polymerising the monomer. One option is to simply consume it i. e.  Put in one extra per cent of your initiator.
Otherwise,  similarly to basic alumina, use a basic resin. Once you consume the BHT,  your monomer is prone to polymerisation, so you may need to run it into your reactor under your reaction conditions e.g. flushed with nitrogen, pre-cooled, etc

For such a large macromonomer, it would be wise to dilute it before removing BHT. A good option would be dissolving your PDMS-methacrylate in DCM and flush it through a basic alumina column. You should be able to see a slight color change on the top of the column which is the captured phenolate.
The solution provided by Beerhunter does not work for ATRP, although it is typical for conventional radical polymerization. The reason here is that you do not have control when the hydrogen transfer from BHT to your radical chain ends happens. If you use an extra amount of initiator (e.g. alkyl bromide) in your reaction, you are most likely obtaining polymers with a broad molecular weight distribution due to termination by BHT at various stages of the reaction.


text:
How can functional groups be bounded?
When I read about the different kinds of functional groups, they usually have formulas like $\ce{-NH2} $, $\ce{-COOH}$, $\ce {-OH}$, etc.
What does the bar represent? Any bond? Or can it only be a covalent one?

Yes, the bar represents any bond, which in this case can only be a covalent one.

@IvanNeretin Why? What about compunds like $\ce {NaOH}$?

@Qwedfsf because OH is not a "functional" group of "Na".

We don't apply the concept of functional groups in inorganics, that's why. Otherwise things could have been different.

@Qwedfsf You will not see anyone write $\ce{Na-OH}$.

@Hexacoordinate-C Only because $\ce {NaOH}$ is inorganic? Or is there more to it? Is it possible to encounter, say, $\ce {-OH}$ in an organic compound and it not be a hydroxy group?

I don't understand that last question. What do you want it to be? I mean is the hydroxy group that's in a carboxylic acid a hydroxy group? Does a hydroperoxy group have a hydroxy group?

I can't think of any -OH group that wouldn't be a hydroxyl group. A chemist would never consider a -COOH group to be a combination of a -C=O (a ketone) and a -OH group. // I'd also stipulate that "functional groups" are abstractions for organic chemistry. Hence the bar in organic chemistry is used for covalent bonds.

Functional groups are a unifying idea in organic chemistry. There are tens of millions of different organic chemicals but many of their key properties are determined by the functional groups they contain. For example, amines (compounds with the $\ce{-NH2}$ group) are basic and often smelly; carboxylic acids (containing the $\ce{-COOH}$ group) are acidic.
The bar in the symbol represents the bond to the rest of the molecule. Sometimes chemists write this slightly differently by including a specific symbol for the rest of the molecule like this: R-OH where R is the rest of the alcohol. So a chemist might talk about a series of similar alcohols by defining R as $\ce{CH3}$ (methanol) $\ce{CH3CH2}$  (ethanol) or $\ce{(CH3)2CH}$ (isopropanol).
This simplifying terminology allows chemists to talk about large groups of related compounds without having to be too verbose and writing out all the details of every structure.
This is an organic chemistry concept so the functional group attaches covalently to a carbon. Some groups can also occur in inorganic compounds but the terminology would be different. For example, Sodium Hydroxide contains an isolated OH but this isn't an alcohol and wouldn't be written or described using terminology related to functional groups.

So what about N-hydroxyphthalimide? Does that contain the hydroxy functional group or not?

@jan It isn't an alcohol and the functional group would probably be described as something like an "N-hydroxy" or a hydroxyimide. The alcohol functional group is defined by an OH connected to a carbon.


text:
Why does Bi show inert pair effect and not Sb?
The 6s orbital electron pair is strongly attracted by the nucleus due to the poor shielding effect of 4f and 5d orbitals. However there are other filled orbitals that shield the nucleus. In case of 5s orbital, it is nearer to the nucleus and has the other filled orbitals preceding it, in common with 6s orbital, except the 5p, 4f and 5d orbitals. For 5s orbital there are fewer orbitals shielding the nucleus and it is also nearer to the nucleus, in comparision to 6s orbital. So, why Bi shows inert pair effect and not Sb?

@paracetamol I mean it should be more pronounced in Sb because of less no. obstructing orbitals as compared to Bi. So why Bi?

The inert pair effect is a type of relativistic effect, which becomes more pronounced as the atomic number gets higher. I have a related previous answer which touches on Bi in the comments.

Antimony does show the inert pair effect. While antimony(V) oxide is known and bench-stable, heating it to over $\pu{700°C}$ will convert it to a mixture of antimony(III) and antimony(V) oxide showing that release of oxygen by reduction is thermodynamically possible.


text:
Buckminsterfullerene double bonds
I am not now nor have I ever been a chemist.
I'm practicing by drawing molecules in the Chemspider structure search tool. Today I drew buckminsterfullerene, in order also to test my 3D spatial awareness. I chose to draw the double bonds on the hexagonal rings only - the resulting symmetry making this the intuitive choice, I thought.
Chemspider recognised my sketch as C60 right enough, but in its own 3D rendering has the double bonds placed differently, with some in the pentagonal rings, so of course I run to Google to see what is up with that, and learn the following from Wikipedia :

C60 tends to avoid having double bonds in the pentagonal rings, which
  makes electron delocalization poor, and results in C60 not being
  "superaromatic".

So according to this my intuition was correct, but I guess Chemspider just has a different version stored for some reason?
So:
Q0: Any guess as to WHY Chemspider has different configuration?
Q1: Is the extract above actually true? (If you can imagine for a moment a world where Wikipedia might not be 100% accurate and authoritative)
Q2: If so, why? As I said, the more maximally symmetric arrangement seems intuitive. I guess the charge of the mutually repelling (repulsive?) electrons would be more evenly distributed this way, and so this is how it would self-organise, like a soap bubble.
Q2: In the context of the extract, does electron delocalization being 'poor' mean that there is a lot of it (this is my instinct), or not so much?
Q3: Either way, why is the degree of delocalization contingent upon where the double bonds are? My first guess would be that the positively charged nuclei in the hexagons keep the electrons in a less vice-like grip than in the more tightly convex pentagons.
Q4: Are the bonds static? Once you have a C60 molecule with its bonds a certain way, do they stay that way, or do they jump about arbitrarily?
Q5: Does it matter? Does the geometry of the bonds affect the chemistry, or indeed physics, of C60 that much?
If your answer is 'go read a book/paper on C60' then that's fine, if you could recommend a good one for a beginner. There's a lot out there.
Cheers MD.

Well, you have talent for finding dumber lines in Wikipedia, or just bad luck? ;)

That's imo a common misuse of the term "delocalised bond", which really is a 0/1 statement. pi electrons however delocalise even when you only have a chain of conjugated double bonds, where each double bond is perfectly fixed and cannot  change unless you do actual, practical chemistry to it.

All carbon atoms in fullerene are part of one "double bond", as you draw it. In quantum chemical reality, the structure of the fullerene is made up by fixed "single" bonds between those 60 sp2 hybridised carbon atoms, with the residual thirty pairs on pi electrons being somewhat delocalised.
In a sheet of graphene, the pi electrons are delocalised, and every bond lying between two six-rings gets the same share in the electron cloud.
In fullerene, they are as delocalised, but not equally distributed. The bonds between two six-rings are "more equal" (more aromatic) than those between five- and six-rings, and grab themselves a larger part of the electron density.
Problem of the five-rings is that the bond angles are not 120°, as they should be for a proper sp2 hybridised carbon atom. They become more sp2.4 or something, which let's the pi electrons move out of there a bit.


text:
Process Chromatography vs Solid Phase Extraction techniques
Process Chromatography is typically performed by flowing eluent from the top to the bottom of the column. The standard method of solid phase extraction at industrial scale e.g. to extract natural products from botanicals,  is to flow from the bottom up. What is the advantage or rationale for this way of operating?

Are you looking at any physically built processes in particular? Any specifics you can share about the motivation for this question?

Unlike chromatography,  extraction removes material from the "column".  I believe this could reduce efficiency for bottom up flow, unless a fluidised bed effect comes into play. Top down means the botanicals would be more inclined to be under the effect of gravity i. e.  Giving a better "bed" as it shrinks vs bottom up. This is my hypothesis, with no direct evidence, so I wanted to seek wider knowledge


text:
Why do chloroform and benzene exhibit negative deviation from Raoult's law?
Chloroform is a polar molecule, and benzene is nonpolar. Shouldn't the chloroform-chloroform and benzene-benzene intermolecular forces be stronger than chloroform-benzene interactions (like dissolves like), which would result in a positive deviation from Raoult's law?

related https://chemistry.stackexchange.com/questions/6955/deviation-from-raoults-law-because-of-adhesive-and-cohesive-forces

This is because when these liquids are mixed, H-bonding type interactions are formed between hydrogen atom of chloroform (partial positive charge due to 3 Cl atoms) and 'pi' electron cloud of benzene ring.
Thus, chloroform-benzene interactions are stronger than chloroform-chloroform and benzene-benzene interactions.

Then, why does acetone and benzene show positive deviation? Doesn't acetone (being a polar molecule) form hydrogen bonds with the "pi" electron cloud of the benzene ring?

@AdityaKumarPanda such an interaction would require availability of a hydrogen bonded to an electronegative atom, which is absent in acetone.

@ShivanshJaiswal Yeah, I've got it. Thanks!