text:
In helium-2, would each electron experience a single positive charge from the nucleus?
If we have an atom of hydrogen-1, we know that there exists one proton with one electron and in helium-2 two protons with two electrons. My question is that do protons get shared equally with electrons? For example, is it true that each electron in helium experiences the same force from the nucleus as each electron in hydrogen, due to the proton to electron ratio being 1:1 in both atoms?

No, it is not true. If it were, the electron energy levels of helium and hydrogen would be the same. However, it is true that in multi-electron atoms (neutral or otherwise) you have to take into account the 'shielding' of the nuclear charge by the other electrons. However, that shielding is not easy to describe, nor is it perfect in the sense that one electron shields one proton. It just doesn't work that way.

Even in a simplistic (non-quantum) model of the atom, the scale of the distance between the nucleus and the "orbit" of the electron would suggest the nucleus looks just like a point to the electrons so there should be no difference based on the "shape" of the distribution of charge in the nucleus. A quantum view is more complex (some orbitals have a positive probability of electrons being "in" the nucleus), but the effects are usually still small.

Sorry for mentioning Helium-2 as I understand that the isotope does not usually exist, but I mentioned it for the simplicity of the situation, although the neutrons would not really make a difference.

To answer this question, I am going to assume that a $\ce{^2He}$ nucleus actually exists long enough for two electrons to surround it — this is probably not a valid assumption since Webelements.com does not list $\ce{^2He}$ while it does list the isotopes $\ce{^6He}$ and $\ce{^8He}$ (although they admittedly have long half-lifes measured in tenths of seconds). But for our case, let us assume so.
The charge radius of a proton — our closest experimental measurement of its diameter, if ‘diameter’ is even a valid term at subatomic scales — is given on Wikipedia as $\pu{8.75e-16 m}$ or $\pu{0.875fm}$. At this point, take a step back to remember that the Bohr radius, approximately the most probable distance between the electron and the proton in a hydrogen atom, is $\pu{5.29e-11 m}$ or $\pu{52.9pm}$. This is a difference in scale by a factor of $10^5$.
Therefore, a reasonable assumption for the $\ce{^2He}$ nucleus — and indeed for practically any nucleus — is that it can be approximated by a dimensionless point of $+2$ charge and not any sphere, ovaloid, double-sphere or other three-dimensional object. Therefore, if we want to use quantum mechanics to calculate the wave function of $\ce{^2He+}$, we assume a spheric potential caused by a positive charge in the origin of a coordinate system; the value of the positive charge is $2e$ with $e$ being the elementary charge.
A two-electron system cannot, to the best of my knowledge, accurately be calculated. However, we may approximate it as just taking the wave function solutions of $\ce{^2He+}$ and adding another spin-opposite electron. (Then, we slightly adjust the energy values to account for electron-electron repulsion.) This way of approaching the problem shows that the helium atom should really better be considered a central positively charged dot with the outer space occupied by the negatively charged electrons, each of which see a $+2$ charge in the centre and an additional negative charge ‘buzzing around them’ (Note: physically incorrect macroscopic picture).
Replacing $\ce{^2He}$ (as is present in your question) with either of the stable isotopes $\ce{^3He}$ or $\ce{^4He}$ would not change the picture; the size of a neutron (see my note in the second paragraph on subatomic scales) is roughtly that of a proton. If anything, adding one or two neutrons to the two protons will make the nucleus smaller since neutrons increase the intranuclear attractive forces.

In many introductory chemistry books, the core and shell model of an atom and the relevant size differences are exemplified by macroscopic comparisons. My chemistry book offered: if you enlargen an atom so that it would fit the Munich Olympic Tower inside it, the nucleus would inflate to the size of a cherry stone, match head or something of that size. This again shows how good the assumption of poiut-sized charges in the nucleus is.

OP surely meant the elements ordinal number when writing H-1, He-2, not the mass.

@Karl I am not willing to make that assumption but I will note that it does not change the general picture if we add one or two neutrons into the equation.

@Jan So when each electron 'sees' a +2 point positive charge with an additional negative charge orbiting it, does the effect of the positive charge for the electron in question in any way get diminished?

@Coder511 no, that’s what I tried to outline in my answer.

Electrostatic interaction is like gravitation: A spherical charge (or body) gives the same attraction as an equal point charge (or mass) at the centre of the sphere. If you are in the sphere, gravity is the same as for a mass with the radius of your current distance to the centre. All the mass or charge closer to the surface than yourself has zero net interaction with you.
The electrons in helium are in the spherical s-orbital. In every instant, one electron feels the full charge of two, while the other one, which is a bit further out, only feels a charge of one, because the other electron shields the core.
This is a bit oversimplistic (hello quantum mechanics), but each electron feels a core charge roughly around 1.5  in average.

You are arguing that electrons in the same orbital shield each other. I'm not sure that is right in either the classical or the quantum view of the situation. Perhaps it is an approximation for electrons in different orbitals with different spatial distributions.

@matt_black It's a rather good approximation for electrons in different (filled towards partially filled) orbitals. For two electrons in the same orbital it is of course a dead alley towards a realistic model, but the general idea is certainly valid. Perhaps it should say "very simplistic" in the conclusion.

First of all, helium-2 is an extremely short-lived species - if you wanted to speak about its existence at all. So I assume that you had a more stable isotope in mind, e.g. helium-3 or helium-4.

... is it true that each electron in helium experiences the same force ...

Both electrons in a helium atom are indistinguishable from each other. If you wanted to apply the concept of a force exerted on an electron at microscopic scale, then both electrons would experience the same force.

... would each electron experience a single positive charge from the nucleus?

No. The charge experienced is the charge of the two protons and the electrons that shield each other.
The second ionization energy of helium is four times the ionization energy of hydrogen, as predicted by quantum mechanics. The first ionization energy of helium is greater than that of hydrogen and smaller than its second ionization energy. This is in accordance with the statement above and suggests an "effective charge" roughly about $1.3$ to $1.4$.

$\pu{1312.0 kJ/mol}$ Hydrogen 1. ionization energy
$\pu{2372.3 kJ/mol}$ Helium 1. ionization energy
$\pu{5250.5 kJ/mol}$ Helium 2. ionization energy

No, the protons are not "shared" and the attractive force from the nucleus on each electron is about 6 times higher for helium than for hydrogen. This is due to the fact that there are now 2 protons pulling on each electron and they are considerably closer, so each one pulls about 3 times as hard. There is a repelling force between both electrons, but there is no simple analytical expression for it.
In a classical description of this system, the force for each particle-pair is given by Coulomb's law. A single electron around a Hydrogen nucleus feels an attractive force that is proportional to $\frac{1}{r^2}$ and each electron around the Helium-nucleus will feel this force for each proton. The smaller radius of the $\ce{He}$ atom (roughly $\frac{1}{\sqrt{3}}$ times the Bohr radius*)  will lead to a ~3 times higher force per proton, so a ~6 times higher force between the entire nucleus and each electron.
If you wanted to relate this to ionization energy, you can effectively assume that the nuclear attraction force on the departing electron is shielded by the remaining electron. As an approximation, this energy is given by the shielded nuclear charge divided by the radius in Rydberg atomic units : $\frac{Z-1}{r} \approx\sqrt{3}$, or $\pu{24 eV}$. The remaining electron of what is now $\ce{He+}$ will have moved closer to the nucleus, at a distance of $\frac{1}{\sqrt{4}}$, i.e. $\frac{1}{2}$. Its ionization potential is $\frac{Z-0}{r} = 4$, or $\pu{54 eV}$. For completeness sake, in these units the radius of $\ce{H}$ is $1$ and its ionization potential is $\frac{Z-0}{r} = 1$, or $\pu{13.6 eV}$.

* the $\frac{1}{\sqrt{3}}$ expression for the radius of $\ce{He}$ is just a shortcut for ease of calculation. It is within 2% of theoretical/experimental values.


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Difference between Sildenafil and Sildenafil citrate
The data I found seems to show no difference between Sildenafil and Sildenafil citrate. But if they are the same things, why Pfizer use Sildenafil citrate instead of Sildenafil as their product Viagra?

Amines are generally used as their salts.

Different topic, but the answer to this question might be relevant.

Also https://chemistry.stackexchange.com/questions/9499/what-is-the-difference-between-doxycycline-hcl-and-doxycycline-hyclate

Also be aware that several companies and medical sites will frequently refer to their active ingredient without the counterion e.g. for Nimbex,  you'll see cisatracurium rather than cisatracurium besylate. Hence why colloquially,  Viagra 's API is referred to as sildenafil.

In many pharmaceuticals (and many recreational drugs) the choice of form is determined by the convenience of the manufacturing and delivery
Many drugs (not just sildenafil) are manufactured as salts or simple derivatives of the active molecule. For example, the anti-inflammatory steroid fluticasone is usually delivered as the propionate or furoate esters and the recreational drug cocaine is usually delivered as the hydrochloride salt. In all this cases the active ingredient is exactly the same in the body.
But there are two factors that matter as well as the active ingredient.
One is how easy it is to obtain a convenient form that is pure and easy to make. Amines, for example, are much less prone to produce nice crystals than their salts (sildenafil and cocaine are amines). Since the industry needs to have a good route to a verifiably pure compound and since crystallisation is a good way to obtain pure compounds, they often experiment with different salts to find one that gives a satisfactory end result. This is convenient for the manufacturer and makes no difference to the active ingredient which is determined by the pH of the biological environment.
The second reason is the ease of delivery of the drug. The physical form and chemistry of the substance or salt do matter for the bioavailability. If the drug is taken as a pill, it has to be well absorbed in the digestive system (which means it should, usually, dissolve quickly in the stomach). Salts are more likely to be rapidly soluble than simple amines (though this depends on the form as much as the fact they are a salt though the form is more controllable if you have nice crystals to start with in manufacturing).
The illegal drug cocaine is an interesting case. It is easy to maufacture it as a salt and it is commonly available as the hydrochloride. The hydrochloride dissolves in nasal linings giving a high. But some prefer a faster, more intense (and also, unfortunately, more addictive) high obtained by using the vaporised freebased amine. This is easy to make by reacting the hydrochloride with household baking powder. But it can't be sniffed in that form (as a powder), it has to be heated and vaporised for effective absorption. The vapour is absorbed more rapidly directly into mucous membranes, making a faster hit. But this isn't as convenient as sniffing a finely ground powder and would not be a convenient way to distribute legal pharmaceuticals (imagine the police attention if you had to heat a spoon full of fluticasone to relieve your blocked nose).
So, in short, convenience of manufacture and delivery frequently lead to the ideal form of a drug being a salt and not the pure active ingredient.

Thanks! So from my understanding, Amines are easier to purify with their salt form, and the salt form are more soluble than Amines, is it right?

Another beautiful feature about crystals is that a simple visual inspection will often be enough to determine that the reaction proceeded as intended.

The stability of the free base may be lower than the salt form

Sildenafil Citrate is one and the same thing (at least in pharmaceutical marketing) as Sildenafil 1-[4-ethoxy-3-(6,7-dihydro-1-methyl-7-oxo-3-propyl-1H-pyrazolo[4,3-d]pyrimidin-5-yl)-phenylsulfonyl]-4-methylpiperazine citrate.

Tl;dr
In pharmaceutical products there are several considerations that influence salt selection.
Solubilitity factors
Many synthetic substances to be used in solid dosage form are too limited in solubility to be therapeutically effective.[1] Salts are most frequently employed for altering aqueous solubility; however, selected salt forms also influence a range of other properties such as melting point, chemical stability, hygroscopicity, solution pH, dissolution rate, crystal form and mechanical properties. Different salt forms may have different stability characteristics.
Bioavailability considerations
Generally speaking, different salt forms of the same drug rarely differ pharmacologically, unless the dose is sufficiently high such as to render the counterion toxic or pharmacologically active. Rather, the onset and intensity of the drug plasma concentration is that which may be considerably different between different salt forms, as well as between a salt and its free acid or free base. If the improvement in bioavailability of the drug is significant, certain pharmacological responses may be elicited as a result.
Stability considerations
Salt stability can be related to factors such as melting point, solubility, hygroscopicity, and microenvironmental pH.[2] Pharmaceutical salts frequently exist as hydrates, which create difficulties during formulation development, particularly in wet granulation. It often becomes difficult to monitor and control the anhydrate/hydrate form during the wet mixing, fluid bed drying and aqueous film coating processes. This factor has led to the deselection of a particular salt form in further development process many times.[3]
A number of pivotal issues with respect to salt selection for basic drugs take into consideration the molecular and bulk properties of the material and the impact of the material as a salt form to the pharmacokinetics of the molecule. [4] If a salt has the tendency to form many polymorphs, then, unless production of the desired form can be easily controlled, it should be rejected in favour of one that exhibits less polymorphic behaviour.
Manufacturability considerations

The decision of whether to develop a salt form or a free form of a
  drug must also include the cost and feasibility of making the desired
  form. The feasibility and ease of crystallization of a salt form could
  be an important factor in the salt selection decision.

Since the crystallization process serves the important purpose of purifying a drug substance, a salt form with a higher degree of crystallinity will be preferred from the chemical processing perspective.
A less-crystalline salt form may not readily provide the desired chemical purity and could require extensive effort and cost to develop a robust crystallization process. In some cases, however, the most bioavailable form is developed, and appropriate manufacturing and storage conditions are implemented to minimize degradation and conversion to
another form. [2]
The decision to investigate salt forms should be driven by the intended use and properties of the drug, such as the pKa, which will influence the selection of counterions used to make salts and the tendency of the salt form to disproportionate as well as other important factors such as economic viability etc [2]

References

Ahuja, S., & Scypinski, S. (Eds.). (2001). Handbook of Modern Pharmaceutical Analysis (Vol. 3). San Diego: Academic Press.
Adeyeye, Moji Christianah; Brittain Harry, G. (2008). Preformulation in Solid Dosage Form Development (Vol. 178). New York, NY 10017: Informa Healthcare USA, Inc.
Bharate, S. S., & Vishwakarma, R. A. (2013). Impact of preformulation on drug development. Expert Opinion on Drug Delivery, 10(9), 1239-1257.
Mark, G. (2009). Pharmaceutical Preformulation and Formulation (Vol. 199). New York, NY 10017: Informa Healthcare USA, Inc.

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Fleshing out the stochiometric/molar quantities
Any help appreciated, it's been 37 yrs. since I was in univ. chemistry class.
Trying to flesh out the stochiometric/molar quantities (grams) of magnesium oxide to be combined with monoammoniumphosphate (MAP, a common fertilizer) to create ammonium magnesium phosphate x 6 H2O ?
MAP (NH4H2PO4) is an acidic dihydrogen compound, MO is an alkaline oxide, so it appears at 1:1 mix ratio I already have a problem of excess protons - -  not enough hydroxyls.
I'm guessing I can simply heat it the combination in distilled H2O on a magnetic stirrer overnight, then try cooling it to separate by crystals , but I have read that AMP is sparingly soluble in water, so maybe just push pH slightly above 7, cool in fridge, decant and harvest the "sludge" or crystals ?
Combining the two should ultimately yield H2O as the byproduct, plus I have excess protons.
This substance is a common problem in sewage treatment plants where it precipitates on pipes, as well as kidney stones in humans and dogs.

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Is the recipe for my shaving cream complete nonsense?
Preamble
I have only basic education in Chemistry (excelled at Organic chemistry nomenclature, but then subsequently quit when I could not grok the simplest replacement reactions). Also I will note I could not find a tag for retail product labelling.
I bought a lot of shaving cream product that I really like. I have been reading the ingredients and I don't understand why it contains so many acids and bases. Perhaps this is an oversimplification, but I was under the impression any acid and base that were mixed would in time, resolve themselves to a salt and an extra water molecule. Let me share with you the ingredients list verbatum so you can see too:

Aqua (water)
Stearic Acid
Myristic Acid
Potassium Hydroxide
Coconut Acid
Glycerin
Triethanolamine
Theobroma Cacao (Cocoa) Seed Butter
Parfum (Fragrance)
Sodium Hydroxide
Melaleuca Alternifolia (Tea Tree) Leaf Oil
Limonene
Linalool

So it seems to me the major ingredients are acids, bases, water and fats blended together with an emulsifier (Glycerin?).
Question
Wouldn't the acids and bases react with each other shortly after mixing, leaving my product with a different mixture of compounds than in the ingredients list?
Speculation
Here is my attempt at answering, so you can an idea that I did put minimum thought into this.

Are some of these things with "acid" in the name, not really acids as I learned them (pH under 7, with a donor H+ ready to go)? And therefore they do not react?
Considering the vendor spells out cocoa butter with a long latin name, maybe they have arranged the ingredients list to impress the customer? For instance writing out Stearic Acid and KOH instead of, say, "lard" or "wax" or whatever unimpressive name is actually the product of the acid-base reaction?
Perhaps the vendor is legally required to print the reagents rather then the product(s), in some circumstances?
Perhaps there is some process in the mixing of the ingredients, or some catalyst that inhibits the reaction? And to what end? For the reaction to happen on my face when water is mixed into the product?

Fatty acid plus alkali hydroxide is the same as the respective soap in water.

sigh That is ingredients list, just like on any other product - lists things that were used to make product. That's it and it's not supposed to be anything else.

So it's all soap!

My question should have been: What is shown on an ingredients list? and is only tangentially related to Chemistry.

As @Mithoron commented, these are ingredients, not a chemical breakdown of the result.
And, as @Karl commented, traditionally you mix together some rendered fat or oil -- that's a mixture of fats and oils with the fatty acids and glycerin that make them up -- with potassium or sodium hydroxide to make the potassium or sodium salt (conjugate base) of the fatty acids, which is soap.
Here, you've got:

Solvents:

Water
Limonene

Fats, glycerin, and fatty acids:

Stearic Acid
Myristic Acid
Coconut Acid
Glycerin
Theobroma Cacao (Cocoa) Seed Butter
Melaleuca Alternifolia (Tea Tree) Leaf Oil

Base:

Potassium Hydroxide
Sodium Hydroxide
Triethanolamine

Extra smells

Parfum (Fragrance)
Also Limonene
Linalool
Also tea tree oil

The only reason it has so many is that they've mixed together either a few different soaps or a few different fats before saponification (soap-becoming), and each soap will have been (mostly) neutralised.
Edit: note that the cocoa butter and tea tree oil are almost certainly added after saponification, for fragrance, to alter the material properties of the product, and to be left on the skin.

Thank you very much! My wife makes soap at home. I can't believe I didn't see this.

Oh, and Tea Tree oil is almost certainly in the "smell" category (an essential oil) rather than a saponification reagent.

and what makes the foam?

It is not a foamy product. It is more of a cream.

@DouglasHeld You're probably right on the tea tree oil: it has a few different applicable and desirable properties and probably isn't saponified. I'll correct it.


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What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation?

The stability order of the following is:

A. III > II > I > IV
B. II > III > I > IV
C. I > III > IV > II
D. I > III > II > IV

Answer:

 D

I came across a question where we had to determine the stability order of this carbocation compared with three others. This cation (I) was given to be the most stable (note that there's complete conjugation in the ring, shown here with a dotted line). Can someone enlighten me on how this is so?
Although this is a homework question, my point is not to just figure out the answer, I wanted to know why the first carbocation has a higher stability as compared to the rest.
This question is different from the other one because the compounds and functional groups are different. I did take a look at that one before I asked mine, as a result of which my question header is more or less the same. This is, however, different.

It's basically a bunch of homoallyl cations. The big question for ranking III over I is whether or not the phenyl rings are going to be fully coplanar, or if it's too hindered for that to happen. Probably the latter.

Possible duplicate of What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

I have big troubles finding any resemblance of this question in your proposal, @Mithoron.

@Martin-マーチン I could edit this so the point of the question would be more evident.

related https://chemistry.stackexchange.com/a/49307/9961

@Mithoron No you can't. The question you linked to is about a non-classical carbocation, while this is about substituted cyclopropenylium cations. For my taste these are two very different questions, the only link is that they both are centred about carbocations. Even if the substructure would be identical, there is a significant deviation in the substituents, which would warrant it not being an exact duplicate. In this case the linked answer isn't even helpful.

@Martin-マーチン That's the problem, not getting my point... Compare with this  (which might have more obvious connection) : https://chemistry.stackexchange.com/questions/48264/stability-comparison-between-the-tropylium-and-tricyclopropylcarbinyl-carbocatio Also OP has comparison of different substituents on stability of cyclopropenyl cations and asking about the one with cyclopropyls. Is it now more clear?

@Mithoron Oh yes, let's answer this with the number of resonance structures. Apart from this, the questions might be related, but they are definitely not similar enough to be considered duplicates. Even if the question was 'why is the cyclopropanyl-moiety a good pi-donor', and you can find the answer to that, I would very much hesitate to close at duplicate.

Questions like these always appear simple, when in fact they are not. First of all it has to be clear that stability is no absolute concept. In this case one could ask which of the cations is most persistent, in other words is least reactive towards nucleophiles. With the exception of the tert-butyl cation, one could also ask the question, which of the remaining substituted cyclopropenium cations has the largest aromatic stabilisation. One would evaluate this traditionally with the hydrogenation energy. This obviously proves difficult, because it is a cation, it has to be neutral first.
Of course you can go ahead and apply approximations like +/-I or +/-M effect, and hyper- or  ordinary conjugation, or negative conjugation, and the there are steric effects to be considered, and dispersion effects, and don't neglect the solvent, and what about the anion. There are many things to be considered, and just naming them makes this problem a lot more complicated. It is our way of looking at these questions that make them appear simple.
While I wanted to use a more rigorous approach and I am generally interested in such weird molecules, I still have not come up with something proper, that suffices my standards (in a reasonable time frame).
First I thought that could be solved with calculating the ions and comparing them via the following isodesmic reaction scheme:

This is a conformational nightmare, a reaction that produces diastereomers, highly strained rings, etc. I therefore dropped this approach.
The other way to evaluate this is to simply add a hydride ion. While this reduces the complexity a little, it is still a conformational nightmare. It is in principle a little bit too much time to invest in such an exercise, and in the end the answer might not be helpful at all.

Coming back to your original question:

What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation?

In extremely simple terms, it's exceptional stable because of the cyclopropenium core structure, because it is aromatic. Larger substituent moieties usually further stabilise such substructures, because they hinder an attack from nucleophiles. With a large enough counter ion (weakly coordinating ions, not necessarily as large as the ones mentioned on Wikipedia), it will be very persistent.
When you look at all the choices you have been given, the first conclusion should be, that they are all comparatively stable. We are not talking about primary, or secondary carbocations, or other clear cases. There really is only very little use in further analysing it.
For the sake of the argument:

Methyl groups can provide some more stabilisation (than hydrogen) because of hyperconjugation.
Phenyl groups are excellent at delocalising a positive charge when it overlaps with its π system. In that case the cyclopropenium ring is too small and the rings are too large, so they have to twist out of plane and therefore conjugation is not ideal. Here is the optimised structure of a DF-BP86/def2-SVP calculation:

Cyclopropanyl groups are rather small. They will usually attach in a way that the hydrogen is in plane with a π system. The group itself behaves like a π donor. (That's a factoid you might want to remember. See my answer on Why does cyclopropane react with bromine? and there are surely others to find. You can search in ron's answers first.)
To demonstrate this a little further, I have calculated the molecule at DF-BP86/def2-SVP level of theory and the π orbitals are below (click to enlarge).

In summary, the cyclopropanyl moiety has a good size and nice π-donating properties.

I still have to reiterate, that I find these questions not very helpful, especially if the limitations of such model views are not thoroughly discussed.

While the computational level of BP86/def2-SVP is probably sufficient for the semiquantitative reasoning given here, I would suggest to include a modern dispersion correction such that you may obtain better structures.

@TAR86 Meh. I am quite ambivalent towards dispersion corrections. For this exercise, completely unnecessary.

I'm not. I ran the calcs, depending on functional, the difference is 0.8 to 1.3° on average, which is not nothing compared to about 10° deviation from flatness. Of course, the conclusions drawn in this post stand regardless.

@tar86 I agree with what you have said previously, I was a fool not to include dispersion corrections at the time; something I wouldn't run into today. I think the reasoning itself holds, so I think in this case, for simplicity, one can ignore them, but you are absolutely right that I shouldn't have. (A proper benchmarking would even be better, but I don't think I'm going to be able to do that.)

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text:
How real are bond dipoles?
In general chemistry, it is common to teach students to determine a molecular dipole by having them first determine "bond dipoles" which are just based on electronegativity. Then, by adding up these vectors, one determines the direction of the overall dipole.
In quantum mechanics, however, the molecular dipole is found by calculating the expectaion value $\langle\psi|\hat{\mu}|\psi\rangle$. This can be decomposed into $x$, $y$, and $z$ components if one so desires. Regardless, the actually quite useful idea of a bond dipole is nowhere to be found.
I assume that there is not a rigorous way of defining a bond dipole so that a sum of these bond dipoles equals the expectation value above. This is because there is not a well-defined way to partition electron density which is exact. I suppose one could do this using the theory of atoms in molecules.
Nonetheless, other ideas like local mode vibrations are approximate in the sense that a local mode is not an eigenfunction of the polyatomic vibrational Hamiltonian, yet local modes are still very useful and even more accurate in some contexts.
The same is sort of true of bond dipoles in that if one imagines a very long molecule which is polar at one end, a nearby molecule which moves along this large molecule will experience a changing electric field because the dipole of the molecule is only a true point-dipole when the molecule is very far away. So, the field experienced by nearby molecules is more akin to the sum of bond dipoles in the region nearby.
So, is there a rigorous way to determine bond dipoles from first principles? Is it as simple as projecting the total dipole onto a bond axis? What is the form of this projector? Also, in cases where the partitioning of atoms is well-defined, such as the theory of atoms in molecules, are bond dipoles well-defined and do they relate to the total dipole as we expect?
Answers to any of these questions would be welcome.

To be more clear, I am talking about the bond dipoles described on this wikipedia page. They do not provide any formal way of actually calculating the bond dipoles. They mention one can get these dipoles after calculating the total dipole, but I am not sure of the uniqueness of this under unitary transformations.

I am assuming that you are referring to the transition dipole moment. In that case, the expression given in the wiki page (https://en.wikipedia.org/wiki/Transition_dipole_moment) is accurate. Here is a journal that uses it to calculate the permanent dipole: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.68.022501

I believe that the dipoles are real. There is ab initio software that predicts IR spectra. I am not sure if they use another form to calculate the transition dipole moment.

I think you misunderstand. I'm referring to the idea that one that determine the total dipole moment of a molecule by adding up small contributions due the polarity of all the individual bonds.

If you look that deep, then the bonds themselves are not really real.

@IvanNeretin I don't think this is looking that deep really. I think it makes sense that there should be a way of calculating the total dipole approximately by summing up individual small dipoles which are found based on properties of the atoms themselves. I have never heard of this, but I don't see why it shouldn't be possible even if it doesn't give a good answer. Don't you think there's some sense to the idea?

I am not sure if this is what you are looking for, but there's something that does what I think you want. It is called Ruedenberg partition. Here are some journals that use it/illustrate it http://pubs.acs.org/doi/abs/10.1021/ja00756a012?journalCode=jacsat

Once you've uttered the words "first principles" and "electron density", you are looking that deep. There are no more atoms, just ugly lumps of electron density. To even separate one atom from another, you have to come up with some non-obvious and highly sophisticated definition. Ditto for bonds.

If by "bond dipole moment" you mean "electric dipole moment", then the electric dipole moment is the first term in a multipole expansion of the molecular charge distribution. In that respect, an electric dipole moment is an approximation.
For a static charge distribution, we are interested in the resulting electric fields or potentials. For a general charge distribution, this does not have a closed form expression, but can be increasingly approximated by higher order terms in a multipole expansion. The dipole approximation is good enough for many cases. Depending on the experimental setup, the only surviving term might be the dipole term. So in that sense it maps to an experimental value and can be considered "real".

I appreciate the answer. I am aware of this. I'm thinking of something different. Namely when one looks at a molecule we tend to determine whether it has a dipole by looking for the absence of certain symmetries and whether the bonds themselves will be polarized towards a certain atom. The electric dipole moment is a property of the whole charge distribution while the qualitative approach looks at polarity in specific regions (namely bonds). I am wondering if this qualitative view (bond polarity as small dipoles) can be quantified and related to the electric dipole which you describe.

The total dipole moment is the trace of the charge distribution (density matrix) dotted into the dipole moment integrals along each Cartesian axis. Since the trace is a sum, the total dipole moment can be thought of as a sum of smaller dipole moments (sum of MO dipole moments, for example). The trace is invariant to change of basis, so the partitioning is not unique, so you can choose any basis of "small dipoles" you want.

Ya that's what I was thinking would be the case and mention in the edit I just added. Ok so I guess one could localize the orbitals however one pleases and choose these small dipoles to be the bond dipoles. I suppose this could be useful even if it is arbitrary. Ok well that's what I was asking about. Those small dipoles.


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Does Reduction Potential affect rate of electrolysis
I am currently doing a project on electrolysis, and I was wondering if the reduction potential (ie. $E^0$) will affect the rate of electrolysis. In other words, across different cells with differing $E^0$ values, is there a trend in the rate of electrolysis of these cells? Why or why not? I have seen other similar questions but all were talking about the ease of discharge/selective discharge of ions, which is not what I am looking for.
Thanks!

Standard Reduction Potentials represent the free-energy released or required to drive a specific half-reaction to completion. Depending on the species which react/are produced, this value will vary. Since free-energy doesn't hold a direct correlation with kinetics, then a relationship between Standard Reduction Potentials and reaction speed shouldn't be expected. In saying that, however, in the case of an electrolytic cell, increasing the external voltage will typically increase this rate of reaction.

@XeLa so supposing I have a constant external voltage, but the voltage required to drive forward the reaction is lower than this external voltage, will the "surplus voltage" increase the rate of reaction?

After factoring in any cell resistance/heat dissipation, then (in a sense) yes, it should. However, applying a voltage* greater than that necessary for your specific reaction can lead to additional unwanted half-reactions occurring at the electrodes. Just keep that in mind! *Let me just emphasise that its ultimately, however, the current which drives the reaction - voltage is just the result.

@XeLa ah I see, thanks for the clarification!


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Thermochemistry, calculating T given Q
I have the reaction $\ce{2COCl2(g) -> CCl4(l) + CO2(g)}$
And I want to calculate $T$ (temperature) when $Q = 10^{-4}$ at equilibrium.
I tried simply putting $$T = -\frac{\Delta G^{0}}{\Delta S^{0}-R \ln Q}$$
Anyway, since $\Delta G^{0}$ for the reaction ($\pu{42000J/mol}$) is the value at standard condition (25 degrees celsius) I use that $\Delta G = \Delta H^{0} - T\times\Delta S^{0} + RT\ln Q$ where $\Delta G = 0$ since we're at equilibrium.
From this I obtain that $$T = \frac{\Delta H^{0}}{\Delta S^{0}-R\ln Q}$$
But which $\Delta H^{0}$ and $\Delta S^{0}$ values do I use here? I can calculate them from the standard conditions at 25 degrees celsius using SI, but that wouldnt be correct, would it? I'm rather confused by this task, which seems simple at first sight. Hope someone can clarify.
(The answer should be $\pu{476K}$)

Are you familiar with the Van't Hoff equation?

No, not yet. Is this required here?

Your $\LaTeX$ skills are fine, as far as $\Delta{}G^0$ goes. There was another thing that needed correction, but it is specific to this site.

Look it up online and see.

This question was taken from an earlier test I'm practising on, van't Hoff shouldn't be required. Is there a way around it without using van't Hoff?

We know that, at temperature T, $$\Delta G^0(T)=\Delta H^0(T)-T\Delta S^0(T)$$
From Hess's Law, we also know that
$$\Delta H^0(T)=\Delta H^0(298)+???$$
and
$$\Delta S^0(T)=\Delta S^0(298)+???$$
That should be a good enough hint.


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Oxygen required to burn carbon

What volume of oxygen at standard temperature and pressure (STP) is required to burn $2.4 \,\rm{g}$ of carbon to completion?

No one has answered this but it seems within my high school chemistry.
Look up the definition of mole and you should be able to easily figure out how many moles of carbon you have.  The quantity $\pu{2.4 g}$ seems to be have been chosen to make this step easy.  It also suggests that we are assuming that the atomic mass of carbon is 12 (so we are ignoring the small amount of other isotopes).
I expect that you know that the formula of Carbon Dioxide is $\ce{CO2}$.  I presume that we can ignore Carbon Monoxide as we are told that the carbon burns completely.
Now you can calculate how many moles of $\ce{O2}$ are required to burn a mole of $\ce{C}$.  This is very easy.
Finally, look up molar volume to calculate the volume of this quantity of oxygen.

Hi Badjohn, and welcome to our site! You actually did a decent job of answering without spoonfeeding the answer, but the reason we don't answer these questions is they're just closed as homework. Ideally someone would comment under the question, but no one has up to now. We also load the mhchem package with MathJax, so you'd be able to correctly format chemical formulas.

Check out the \pu{} and the \ce{} command.

Thanks.  I recognised it as probable homework so I resisted just giving the answer.  So, the culture here is just to ignore such questions entirely?  I am primarily active in maths and usually just browse here.  So, I am familiar with MathJax but not a chemistry extension.  I'll have a look.


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Electron configuration of indium cation
Just had a lecture there where the lecturer was doing revision on electronic configuration.
He stated that the electronic configuration of $$\ce{In^3+=[Kr] 5s^2 3p^1}$$
He didn't include the $\ce{4d^10}$ orbital. He said it doesn't matter and if we write the full E.C out it's incorrect. It basically goes against everything we studied in 1st year general chemistry. Also I believe he didn't take into account the fact that $\ce{In^3+}$ has $46 \, e^-$ not $49 \, e^-$.
Should the electronic configuration not be $$\ce{[Kr] 5s^0 4d^10 3p^0}= \ce{[Kr] 4d^10}$$
Can someone shed some light on whether he is right.

Yup, it's $[\ce{Kr}]4d^{10}$.  Post-transition ($p$-block) elements rarely if ever ionize or even covalently bond their $d$ subshell electrons.  Although we use the bracketed core designation only for noble gas cores, by the time we get to Group 12 or 13 the $d$ electrons are also effectively part of the core.

for example $\ce{Se^2-}$ would be $\ce{[Kr]}$?

You could do that, although usually when writing electron configurations we go to the next lower noble gas, thus $[\ce{Ar}]3d^{10}4s^24p^6$.


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What are these "certain chemical compounds" denser than osmium?
The Wikipedia entry for density states

Osmium and iridium are the densest known elements at standard
  conditions for temperature and pressure but certain chemical compounds
  may be denser.

I have not been able to turn up what these chemical compounds are.  I can imagine a salt or some molecule in which dense atoms (e.g. of mercury) are packed in a conformation such that these atoms are held closer together than in the pure metal, so giving the molecule a higher density than the pure metal or of higher molecular weight metals like osmium.
I know there are elements with higher molecular weight than osmium and I have read about superdense exotic states of matter.  I am curious about this chemical (presumably comprised of atoms with lower MW that osmium) which is more dense than osmium - presumably at earthly temperatures and pressures.  Or an osmium-containing molecule more dense than metallic osmium would qualify.
If someone can answer this question he or she might also consider editing the Wikipedia article to clarify.

I read once that some alloys of osmium and iridium may be denser than either pure element but when I tried to find it again later, I failed.  At least, there are non-exotic possibilities other than compounds.

The core of the Sun is denser than osmium, but that is "cheating" because we use high pressure.

I edited the wikipedia entry. It now reads "Osmium and iridium are the densest known elements at standard conditions for temperature and pressure but certain chemical compounds[citation needed] may be denser. "

As the set of chemical elements is just a subset of the set of all chemical substances, therefore maximal density within the elements is less or equal to maximal density within all substances.    It was said "may", not "are".

https://chemistry.stackexchange.com/questions/24787/can-alloys-or-ceramics-have-higher-density-than-elemental-solids

This text was added as part of a refactoring/cleanup of the page in this edit in March 2013, where it replaced the text "Osmium and iridium are the densest known elements at standard conditions for temperature and pressure but not the densest materials." This older text was tagged with the note "Which are denser, at standard conditions?" It in turn was added in this edit in May 2011 by an anonymous editor (the IP address used is apparently registered to the Wyoming Community College Commission). The original wording, "Osmium is [...] not the densest material", is ambiguous, but could be read as referring to exotic forms of matter such as neutronium, in spite of the earlier "standard temperature and pressure" caveat.
In short, this likely derives from a well-meaning but misguided editor attempting to note the existence of much higher-density, non-chemical substances in an inappropriate place.

Possibly a dumb question, but what is a non-chemical substance?

Neutronium for example. IDK if "non-chemical substance" was really the best term for it.

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


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Deviation angle for 1st order diffraction

(i) A diffraction grating has 560 lines per mm. Calculate the grating spacing. If the total length of the grating is 2 cm, how many lines are there?
(ii) Light of wavelength 500 nm is incident on the grating with this spacing. Calculate the deviation angle for 1st order diffraction.

I am having difficulty with the (ii) part. For my spacing I got 0.0018 mm. So, then I used
\begin{align}
n \lambda &= d \sin{\Theta} \\
\frac{n \lambda}{d} &= \sin{\Theta} \\
0.27 &= \sin{\Theta} \\
\Theta &= \sin^{-1}{(0.27)} = 15.7^\circ
\end{align}

The grating equation you've used is correct, and in order to find the first order diffraction angle ($n = 1$) you need to use arcsine trigonometric function and carefully place data with respective units:
\begin{align}
n \lambda &= d \sin{\Theta} \tag{1}\\
d &= \frac{\pu{1 mm}}{N}~[\pu{mm}] \tag{2}
\end{align}
where $N$ is number of lines per unit length. Combining (1) and (2) and converting the wavelength from nm to mm:
$$\Theta = \arcsin{\left(\frac{n \lambda}{d}\right)} = \arcsin{\left(\frac{n \lambda N}{\pu{1 mm}}\right)} = \arcsin{\left(\frac{1 \cdot \pu{5e-4 mm} \cdot 560}{\pu{1 mm}}\right)} = 16.26^\circ \tag{3}$$


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Do we use volumes of both solutions if they have different densities to find cocentration of ions
So essentially lets say I am given a mixture of  hexane, water and nickel chloride(masses also given) in a beaker, and am asked to find the concentration of the chloride ions in the aqueous layer of the mixture.
Now, I know what we should first do is find the mols of NiCl2(using the given mass)
mass of NiCl2/molar mass of NiCl2
However, knowing that hexane is less dense than water, hexane will not "mix" and instead float on top of the nickel chloride solution (Sorry for my poor/incorrect terminology!).
So my question here is, when we find the concentration using c=m/v, do we include the volume of the hexane as well? Or only the volume of the water in which the nickel has dissolved in. Thanks in advance!

Your terminology is fine (indeed, hexane will float on top), but your way of thinking is not. Why on Earth would you include the volume of hexane? You could just as well add the volume of air in the room. After all, air is much like hexane: it also kinda "floats" on top of solution and doesn't mix with it.

I guess I assumed we had to include hexane since both were liquids(and got that mixed up with aqueous since it asks for concentration in the aqueous layer, my bad!). But your explaination has cleared my confusion. Thank you!


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Amorphous conductor transparent to XRD
I want to conduct in-situ XRD of a cathode in my battery using this coin cell with a kapton viewing window. I assume this cell configuration is the best I can find for my purpose, and that I can't easily make a better cell. However, the viewing window isn't super helpful if I still need to use a stainless steel current collector, as I have previously.
Thus it seems I have two options:

Make a donut-shaped anode and separator, and XRD through the hole in the donut to hit the cathode. However, I worry that the cathode may not fully lithiate in the center, because it is not as close to in contact with my anode.
Find a current collector which is transparent to XRD. Does anyone have a suggestion for such a current collector?

The link to the cell casing suggests using a graphite current collector if performing XRD.

Yes, but that suggests it for the donut like cell setup, if I understand correctly.

Ah, possibly. Based on the structure image on that page, I'd assumed you were simply trying to replace the Al/Cu ring, but I see that might not be the case. What is the structure of your cell, and which bit are you trying to study with XRD?

I am trying to replace the al/cu ring with a full disc which can distribute current to the entire cathode. This is because I want to make sure that the section I am observing is truly fully lithiated.

In that case I'd think you would be ok with a graphite full disc. They suggest you coat the electrode onto the graphite, which doesn't sound right for a donut structure anyway. HOPG does have an XRD pattern, but it's a simple one that you should be able to overlook if it's present in your data.

Given that my cathode is has carbon-based matrix, I am concerned that this may confound some of my findings for the actual cathode. Also, as someone with significantly more XRD knowledge than me, what is required to be able to "overlook" other xrd peaks? I can find XRD patterns in literature for all manner of things, but that doesn't mean peer reviews would be ok with me slapping anything with characterized XRD peaks ontop of my sample, right?

You might have success with amorphous carbon sheets[1]. It's not cheap, but it's effective. It's the material used in the windows of the AMPIX cell[2] developed at Argonne. It does react at low potentials, though, so might not work for anode studies.
Have you considered aluminum? It's crystalline, but cubic so there aren't very many peaks to worry about.
This paper[3] might be helpful as you work on your experiments. You're right to think that not having a current collector for part of the electrode would affect the degree of lithiation; it's actually worse than that. Those kapton window cells are convenient, but always put an asterisk next to your results.
I'm doing a fair amount of in-situ characterization for my PhD. If you can share more about you goals, I'd be happy to chime in.
[1] http://www.htw-germany.com/products.php5?lang=en&nav0=3&nav1=6
[2] http://scripts.iucr.org/cgi-bin/paper?S0021889812042720
[3] https://pubs.acs.org/doi/abs/10.1021/acs.jpclett.5b00891

This is actually really interesting. The AMPHIX cell looks great, and even like something I could build. However, I require that my electrodes be able to have some form of current collector, rather than begin free standing. Do you have any experience when it comes to XRD that's note free standing?

A little, though not in the way you're trying to do it. I've used pyrolized graphite sheets but above 4.5V it will start intercolating PF5^- ions.

I'm not sure I fully follow but it seems like theres not a great way to do any insitu xrd without the electrode being free standing?

What energy X-rays are you hoping to use?


text:
Setting up a water aspirator pump for vacuum distillation
I have a question about the functioning of a water aspirator in a vacuum distillation setup.
I have read about how the water aspirator is hooked up to the water source and the vacuum is in turn connected to the vacuum adapter, and that there should be a splash guard or tubing on one end of the aspirator as well. I'm trying to figure out what to make of the highly suggested use of a water trap. Supposedly the water trap needs to be connected to the vacuum as well as the system.
What outlet is there left to connect to the aspirator, already being connected to the water source and apparatus, as well as the splash guard underneath? Also, if a water trap should be connected to the system, where should it be connected, is a flask with additional adapters required?

Since you want the distillate, use the following setup.

This way if the water pressure drops then the water is sucked into the trap flask, not the distillate.


text:
Series of reactions

  The compound given above is subjected to the following reaction (in that order):

$\ce{t-BuOK}$;
$\ce{CHCl3}$ and $\ce{CH3ONa}$;
$\ce{H3O+}$ and heat.

Find out the final product.

Source
Attempt:
The question involves exocyclic double bond which is less stable than endocyclic double bond, and so I get that the compound should rearrange to five-membered ring containing a double bond.
In the first reaction tertiary butoxide is a bully base, so it should bring about elimination. Hence $\ce{MeOH}$ will be removed and a double bond will be formed:

However, I could not get how the reaction proceeds. Any ideas?

Disregard previous comment. Carbene formation requires loss of bromide group...

Are you sure you drew that bromide in the right place in the product?

@Zhe I thought that since endocyclic double bond is more stable than exocuclic double bond the ring should expand

You still haven't answer my question. Are you sure that the bromide is in the 3-position and not the 2-position?

@Zhe I  am not sure and I am not able to think of any mechanism of  conversion of exocyclic double bond to endocyclic double bond.

I had some ideas, but I don't have any way of putting the bromide there. That's a bit too weird.

@Zhe please share your ideas.My attempt might be wrong

There were two starting points. One was to the a vinyl carbene by removing HBr. The other was to make a cyclobutene, that opens via a pericyclic reaction.

User 471651: I ask the same question as Zhe. Are you guessing that the cyclopentadiene is the product or was it given as the product of the reaction? Step 1 is most likely a vinyl carbene reaction with ring expansion. Step 2 appears to be dichloro carbene formation. Step 3 looks like acid-catalyzed elimination of methanol. With a response, I will post a mechanism and/or product.

@user5519 I am not aware of the answer.it is just my attempt.please share your ideas.

What is the source of this problem? Is there any information as to the structure of the product? Your answer is a known cmpd, but not made this way. The first step in the reaction is a known ring expansion. Without any info on the source or product, this question is highly suspect.


text:
Predict polarity (TLC, column) using ChemSketch (-Draw)
I'm currently working on a project where I have to make and isolate often quite similar molecules. But, as I change the functional groups the overall electronic situation and polarity of the product and of the side-products often changes. I can just estimate what spot on my TLC might be my product and check later with the NMR but I'm still interested in a software-supported way. So I checked ChemSketch (I don't use ChemDraw) and found some parameters that might be of help:

The polarizability, the density, the molar mass, the index of refraction and (although often not displayed) the dielectric constant.

Now I know this is all just a very rough and simple calculation but sometimes I just need a very basic feeling if the molecule can be considered polar or not.
So I thought perhaps using the equations of Clausius-Mossotti and Lorentz-Lorenz I might be able to determine the dielectric constant (because the software is often unable to predict it).
To my understanding is the dielectric constant a very fundamental parameter here. Polar solvents have a much larger one than non-polar solvents and the dielectric constant is responsible for how well a salt is dissolved in a solvent. For TLC and column chromatography on a polar stationary phase, a more polar substance will move much slower.
So can I just break this down in comparison of two molecules, that the one with the higher dielectric constant will also have a smaller $\mathrm{R_f}$-value?

A low tech alternative: Take some of the sample from a crude reaction and run a prep-TLC. Scrape off the bands, dissolve the compound off the silica with MeOH/DCM, filter and run NMRs. This way you should be able to ascertain which spot to aim for when running a column.


text:
How can I compare liquid concentration to solid concentration?
Total novice here!
I'm trying to determine which is more sensitive between two drug tests, 1 that uses ng/ml and the other that uses pg/mg.  Is there a way to standardize concentrations to compare liquids to solids?
For example determining concentrations of THC: cut-off in liquid is $\pu{50 ng/ml}$ and in solid is $\pu{1pg/mg}$.  I want to know which is a more sensitive test.  Is it just as simple as knowing a picogram is smaller than a nanogram?

What are NG, MG, PG, ML and THC?

In order: nanograms, miligrams, picograms, mililiters and Tetrahydrocannabinol (the high inducing component in marijuana)

You need to mind the capitalization. Nanograms = ng, miligrams = mg, picograms = pg, mililiters = mL.

@andselisk Correction: millilitres = ml. Only in some education systems and journals is the SI convention not followed and nonstandard mL used for millilitres.

@Jan To my knowledge it's exactly the opposite. ACS Style Guide suggests using "L" exclusively, BIPM allows both, but "L" is again preferred over "l" "in order to avoid the risk of confusion between the letter l (el) and the numeral 1 (one)". In IUPAC color books it's also primarily "L", and vendors also use "L" (at least what I saw in Germany and Russia). The only institution I know really like "l" is ISO, but who cares about them:)

@andselisk ‘The litre, and the symbol lower-case l, were adopted by the CIPM in 1879 (PV, 1879, 41). The alternative symbol, capital L, was adopted by the 16th CGPM (1979, Resolution 6; CR, 101 and Metrologia, 1980, 16, 56-57) in order to avoid the risk of confusion between the letter l (el) and the numeral 1 (one).’ Source, directly from the BIPM. Lowercase l is the standard symbol according to BIPM, uppercase is an alternative. This alternative is only really popular in the US, Canada and Australia afaik.

@Jan I have no idea what's going on – this is the same citation I provided, but expanded, and it doesn't reinforce your point whatsoever. BIMP allows both, but the latter amendment in my understanding have higher priority and overall brings up a good point: from the typographical point of view "L" is overall better. Unless there is a solid evidence that "l" must be preferred, I'd say it's just unnecessary nitpicking.

@andselisk There is absolutely nothing from a typographical point of view that makes the uppercase intrinsically better, and it was explicitly added as an alternative not as the standard. Since the BIPM sets the SI standard, their word is the standard and the alternative is not the standard. Also, you never gave a source.

@Jan Your statements about typography contradicts your cited text from BIMP (which I provided a link to with a quoted text – the source is easily googleable, so I thought it shouldn't be a problem for you to follow up) – "l" is very ambiguously font-dependent (good luck distinguishing "l", "I", "|" written with Arial), whereas "L" is clear and font-agnostic. There is a major typographic improvement. BIMP equally allows both "L" and "l".  ChemSE uses ACS style for drawing structures, but you for some reason decide to dodge their recommendations when it comes to units. I don't follow you.

@andselisk Well, the symbol for litres hardly ever appears isolated. I have yet to come across an actual use of the symbol where ambiguity may arise. I for my part cannot make sense out of a milli-one and no unit exists abbreviated I to the best of my knowledge. The style question is easy: I conform to the highest standard where it makes sense and to an arbitrary one where none exist. I repeat that alternative is not equal.

@Jan I don't know the amounts you are dealing with, but I again vastly disagree. Lab solvents are shipped in liters; I synthesize precursors in liter amounts, and good portion of people I deal with use "L" just as often as "mL". To the BIMP part – what makes you think "l" has the priority? They are both listed as equal notations (in text and in Table 6), but as "L" has been suggested later in a form of amendment, it has somewhat higher priority as an improvement of the existing standard (compare with US Constitution and Amendments). And again, what's wrong with ACS standards?

Let us continue this discussion in chat.

Thanks all, like I said total novice.  I was going off of the promotional materials sent to me from the drug testing company and everything was capitalized.  I appreciate the correction!

Assuming the density is ~1, i.e. that of water or dilute solutions, 1 ml weighs about 1 g or 1,000 mg. 1 nanogram, ng, is 10-9 g and 1 picogram, pg, is 10-12 g. Therefore:

1 ng = 1,000 pg
1 ml ≈ 1,000 mg

E.G. if the test can detect 1 ng/ml, that would be approximately 1,000 pg/ 1,000 mg or 1 pg/ml.
Note that this assumes samples are prepared the same way and are tested in the same concentration. If the sensitivity quoted in pg/mg is for a dry sample, the comparison is meaningless.


text:
Pulling Calcium Phosphate from Animal Bones
I am looking to pull Calcium Phosphate out of animal bones (chicken, fish, etc) without destroying the collagen. If I soak animal bones in brown rice vinegar will the vinegar pull out Calcium Phosphate leaving behind the collagen 'bendy bone'? Or is it just pulling out Calcium?

Chances are it will. Why won't you try it? Chemistry is an experimental science, after all.


text:
What is the conversion rate in a chemical reaction?
Consider the following chemical reaction
$$\ce{X <=>[w_1][w_2] Y}$$
where $w_1$ is called the conversion rate of $X$ to molecule $Y$. What is the exact meaning of this conversion rate? Is it the same as probability?
The reference paper is:
Ref

Your formula fails to display correctly - please check it thoroughly.

The coefficients relate to ODEs of the concentrations (or maybe activities). They can also relate to the transition functions between states. Stochiometry makes things complicated - I think they turn the ODE variables into powers of the concentrations, or at least forces them to be multiplied. If you look at the examples at the end the coefficient near the base of the arrow is multiplied with the corresponding concentration.

I think they might be called transition probability rates or propensities too? Don't quote me on that though.

The paper you linked to pertains to birth-death processes. An example of a birth-death process in the physical chemistry context that you're interested in  would be a set of molecules where each molecule is in one of two states, $X$ or $Y$, which are isomers of each other. The model assumes that due to the molecules' environment, each isomer can turn into the other isomer at times that are modeled as being random.
In the paper you provided a link to, $w_1$ is the probability per unit time that within a short period of time $\Delta t \ll 1/w_1$, a given molecule that's in state $X$ will transition to state $Y$. I.e., if a given molecule is in state $X$ at time $t$, then there is a probability of $w_1 \Delta t$ that the molecule will have randomly transitioned to state $Y$ by time $t+\Delta t$.

Thanks for your reply. If we assume that $t=5$, then if I want to calculate this probability for $t=6$, what should I do? In other word, what should I do to calculate  the probability that a molecule will go from X to Y for unit time?

Your omega is what is normally called k (for reaction constant).  https://en.m.wikipedia.org/wiki/Reaction_rate_constant


text:
What is the Difference between Menthol Crystal and Menthol Powder?
I want to buy menthol powder from a chemical supplier but he is selling menthol crystal at half price of menthol powder (he says manufacturing process is the difference) so I am really confused what should I do. Can I buy menthol crystal and just powder them to get the powder form I want it or are crystal and powder menthol totally different things altogether?

Just powder it.

You should be able to tell by the CAS number (or similar identifiers) if the chemical is identical. Note that there may be solubility differences between different crystal modifications. As Ivan suggested, powdering it yourself should be possible.

@IvanNeretin so it means powder form and crystal form is the same thing ?

I suspect so. Powder is made of small crystals, after all.

Be careful with the chirality. Are you sure that both are l-menthol? One isn't a racemic mixture? That's my bet here.

Unless there is something bizarre going on (which should be mentioned more or less clear on the website/in the catalogue because no chemist would expect it), menthol will be menthol as there is only one compound defined as menthol. This compound is chiral, so I would expect a stereodescriptor. In its absence, I would not be sure whether I am dealing with a racemic sample (more likely in the case of a chemical supplier aimed at practicing chemists) or a natural, chiral sample (more likely in the case of a supplier selling to the general public).
Chiral menthol crystallises easily and nicely, and because crystals are easily separated from contaminants, they are often the preferred form of isolation of compounds for sale; especially since the crystals themselves very rarely contain impurities. While a powder can be the better choice for several purposes (e.g., weighing small sample amounts into a flask), it may not be as simple to produce. In the case of menthol, I expect the powder to derive from powdering the crystals. This is an extra step and thus costs more money.
Depending on how important the quality of the powder is to you, it may make more sense to buy the powder directly or to buy crystals and grind them to powder yourself. However, as I said above, I am very sure that they contain the same chemical in a comparable purity.

I believe the major difference is the grade and the purity. powder form is more expensive because it is additionally purified and process where the crystal form requires less processing step, thus cheaper


text:
Accurate way to tell a material is PP with easily available means?
Sometimes even though certain products that I bought come with a "PP 5" marking on it, I suspect it may not be the case, so I wanna know, is there a relatively safe and accurate way to tell if certain material is really Polypropylene, with easily available appliances or no extra equipment at all?
BTW you can damage and sample so said products.

Probably IR. I remember an article recently in the Nachrichten aus der Chemie …

The question is what analytical tools you have at your disposal. In general, nearly all polymers, except for polyethylenes and polypropylenes, drown in water. So it really a matter of distinguishing between PEs and PPs. PP would burn similarly to PE, but burning PP accompanied by streaks (one can pull the thread from the melt). The smell is more acute, similar to the smell of burnt rubber or sealing wax.

@andselisk Thats quite brilliant really, I'm going to hold out a bit longer and wait for more answers. On a slightly related note, PP is the least toxic plastic, right?

@VictorCharlie23 I'd say plastic toxicity greatly depends on what it's being treated with, and also manufacturing process plays important role (leftovers of metal complex catalysts, monomers and plasticizers are usually way more toxic than the plastic itself). But in general yes, PP is not the worst polymer in terms of toxicity.

Haha, thank you, can you recommend a book which I can read on these materials? Even specialized and professional ones will do :)


text:
Hydrochloric acids inability to chlorinate cinnamyl alcohol on 2 sites
According to a reference (J.  Am.  Chem.  SOC.  1985,  107,  2033-2046), cinnamyl chloride can be prepared in quantitative yield by the reaction of cinnamyl alcohol with hydrochloric acid. Looking at the structure, I can see that the hydroxy group will be chlorinated, but I do not understand why the delicate double bond will be left untouched?

Because it’s not really delicate in cinnamate.

Is there any literature describing why the bond on cinnamates is not delicate?

Not sure. But the key point is that the double bond participates in resonance both with the phenyl ring and the carboxy/$\ce{COCl}$ group making it part of one big pi system and rather unprone to traditional reactions of alkenes.

Thanks for the insight - add a reply to get the points for this one.


text:
Why is CH2Br2 more polar than CH3I?
According to my textbook (A Level Chemistry a for OCR Student Book, Pages 84-5) $\ce{CH2Br2}$ is more polar than $\ce{CH3I}$. The molecules are both polar, they both contain different bonded atoms, and the bonded atoms have different electronegativities but why is $\ce{CH2Br2}$ MORE polar than $\ce{CH3I}$?

Why not? Br is clearly more electronegative than I.

As stated in the comments, $\ce{Br}$ is more electronegative than $\ce{I}$ (and that is a factor). Also, you have two polar bonds ($\ce{C}-\ce{Br}$) instead of just one ($\ce{C}-\ce{I}$) and, as a first approximation, dipoles due to bonds tend to add like vectors.


text:
Rate constant units for gravimetric measurements
For gravimetric measurements, why is the reaction rate constant $k$ represented change in mass per time?

Evaporation Rate and Activation Energy
Evaporation changed the slag chemical composition and as a result affected the evaporation rate in the process of measurements. To mitigate this effect, the evaporation rates were analyzed in the first 10 minutes of the TGA measurements. Evaporation rate $k$, $\pu{s^{-1}}$, was calculated as:
$$
\begin{align}
k &= - \frac{\Delta m_i / m_0}{t_i} \\
\Delta m_i &= m_i - m_0
\end{align}
$$
where $\Delta m_i / m_0$ is the flux mass change, and $m_i$, $\pu{g}$, is the flux mass at time $t_i$, $\pu{s}$.

Welcome to Chemistry! Take the tour to get familiar with this site. In the future, please don't paste images; reformat mathematical expressions and equations using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

I guess that this is merely convenient. You can go through the motions of converting it to moles, but that would not yield a more meaningful result. Given the application (thermo-gravimetric analysis), a rate in terms of mass is probably the most useful.


text:
Which (carbonaceous) fuels produce more or less soot in a diffusion flame, and how to predict that?
From what I've read, the production of soot in a flame is still an open research topic. But some hydrocarbon fuels produce much more soot than others. Are there (simple) rules that tell which fuels produce more or less soot? Especially for aliphatics?
Some things that influence soot production that I am aware of:

The amount of oxygen available and if this is pre-mixed with the fuel. I want to limit this question to diffusion flames (non-pre-mixed) with plenty of air available.
According to Wikipedia, naphthalenes produce more soot than benzenes which produce more soot than aliphatics.
Longer carbon chains generally produce more soot than shorter carbon chains.

But now I read that ethanol and diethylene glycol produce less soot than methane, even though both contains two-carbon chains and methane only a single carbon atom. Does it apply in general that hydrocarbons with oxygen atoms produce less soot, or are these an exception? And are there more general rules to predict which fuels produce more or less soot, short of setting up a full chemical simulation?

Generalities, though not rules: more hydrogen, less soot. E.G. ethane has 3 times as much H as sooty acetylene. Because of the H "end-cap", the proportion of H :C decreases for longer alkanes. More oxygen, less soot, also.

Thank you for asking this question. Can you provide some references? I can't track this down: "According to Wikipedia, naphthalenes produce more soot than benzenes which produce more soot than aliphatics." Also "ethanol and diethylene glycol produce less soot than methane", "the production of soot in a flame is still an open research topic". I am interested in this subject, and even though I can't answer the question, I think it would help me if you added more references.

@metamorphic I added a link to the wikipedia article, which itself has some references to chemistry publications. The methane/ethanol statement is based on rules for bioethanol and natural gas fireplace burners. Gas burners require a chimney (unless they pre-mix the gas and air but then you only get blue flames which doesn't look as nice), bioethanol does not because it burns much cleaner. I'll try to find my source on diethylene glycol, can't find it now.

Possible duplicate of https://chemistry.stackexchange.com/questions/78623/is-there-a-flame-which-doesnt-produce-smoke


text:
Platt's Notation/Nomenclature for transitions
How do you read Platt's Notation / Nomenclature? Where can I learn this?
Some examples where I've seen this mentioned:

In Platt's notation the transitions are designated as $\ce{A -> B}$ and $\ce{A -> C}$ transitions
The analysis reveals that the $L_a$ and $B_a$ states of the benzene fragments mix with CT configurations
The second $\pi\pi^*$ state has the largest oscillator strength and it is designated as $L_a$, using Platt's nomenclature

$\ce{A -> B}$ seems to refer to the $\ce{\pi -> \pi^*}$, while $\ce{A -> C}$ seems to refer to $n\ce{-> \pi^*}$, but I'm not sure how to read the notation.

Platt's notation was introduced in the following manuscript from 1949 in reference to classification of hydrocarbons (J. Chem. Phys. 117, 484 (1949), DOI: 10.1063/1.1747293; Classification of Spectra of Cata-Condensed Hydrocarbons). The rules and origin of their notation are described therein.


text:
Difference between chiral and stereogenic centre?
I've seen in some books where they state all chiral centres are stereogenic centres, but not the other way around.
Is "chiral centre" related to only to optical isomerism and "stereogenic centre" related to stereoisomer in a broader sense?

Would you 1) confirm that the definitions used in your books are identical to the ones given on this wiki page 2) point out what remains unclear after reading that page?

So basically chiral center is a type of stereocentre.

So a compound can have different numbers of chiral centres and stereo centres.

Now it’s clear to me

related https://chemistry.stackexchange.com/questions/59124/what-is-the-perfect-definition-for-chirality


text:
How can a conductivity meter be used to determine the equivalence point?
I know how a conductivity meter works. Power is supplied to electrodes and these electrodes change charge. The ions in the solution will be attracted to the oppositely charged electrode. The charge they carry at the electrodes is measured.
The equivalence point may be located graphically by plotting the change
in conductance as a function of the volume of titrant added.
I'm not sure this is an accurate enough answer. Is there something I'm leaving out or that could improve the answer?

I find your post confusing, what exactly you wanna know? As one sentences go this describes it rather OK, but you could probably write a book about conductometry in titration.

I want to know "How can a conductivity meter be used to determine the equivalence point?"

I'm asking about the body of your text which is kinda perplexing.

Let me break it down. In order to sufficiently post a quesiton on this website you need to demonstrate a clear effort at trying to solve the problem/question yourself. You can do this by showing what you know(that is related to the question) which is what I have done here, very clearly in paragraph one and two. Secondly I explain in paragraph three what I think is wrong with my understanding, I think it's not accurate enough for a complete answer to the question. then I finally say in my last sentence, can someone fill in the blanks. So to speak.

Well, it does seem you treat homework policy seriously, even too seriously ;) Please don't start with "I know" there's nigh inevitably something wrong in following statement. Here you don't measure charge, but voltage, and from it you get conductivity.

It would be very helpful, if you could add something like a reaction you are investigating with this methodology.

I’m assuming you’re talking about acid/base titrations.
In these, you typically start with an acid (maybe strong) and slowly add a strong base until you reach the equivalence point (classical, using indicator) or are sufficiently past it (for plotting, using a conductivity meter). Of course, switching acid and base is possible but for the ease of discussion I will consider only acids to be titrated with bases.
Say for example your mixture included hydrochloric acid and acetic acid. Hydrochloric acid (strong) can be considered fully dissociated, acetic acid (weak) mainly undissociated in the initial solution. Thus, your solution contains significant concentrations of:
$$\ce{H3O+\qquad Cl- \qquad CH3COOH}\tag{1}$$
Now you start adding a base such as sodium hydroxide. The most significant macroscopic observation is that the concentration of hydronium ions $\ce{H3O+}$ decreases while additional sodium ions are added. At half-way towards having titrated away all hydrochloric acid, you would have the following species in solution:
$$\ce{\underset{less}{H3O+}\qquad Cl- \qquad Na+ \qquad CH3COOH}\tag{2}$$
At the equivalence point of hydrochloric acid, we can consider the significant portion of hydronium to be consumed; remember that only a small percentage of acetic acid is dissociated, meaning most of it is not. Thus, we have the species:
$$\ce{Cl- \qquad Na+ \qquad CH3COOH}\tag{3}$$
Henceforth, any additional hydroxide added will, in a macroscopic and sighlty simplified view, only serve to deprotonate the acetic acid to acetate. Thus, half-way towards the equivalence point of acetic acid, we can observe the following species in significant concentrations:
$$\ce{Cl- \qquad Na+ \qquad \underset{less}{CH3COOH} \qquad CH3COO-}\tag{4}$$
Finally, we will reach the second equivalence point at which all acetic acid has been consumed to acetate, meaning the solution features:
$$\ce{Cl- \qquad Na+ \qquad CH3COO-}\tag{5}$$
Hitherto, we have not observed any hydroxide because it was immediately quenched. Now, however, all species that would be able to quench hydroxide have been consumed. Thus, any additional sodium hydroxide will remain sodium hydroxide, giving us the following species:
$$\ce{Cl- \qquad Na+ \qquad OH- \qquad CH3COO-}\tag{6}$$
Now that the types of species in each solution has been established, which influence the conductivity and how? Water by itself does not conduct electricity but water with dissolved ions does. Therefore, we would immediately expect the solution’s conductivity to rise from point (3) to point (5) since a non-ion is converted to an ion (and sodium ions are constantly being added). This is indeed observed: a slow but steady rise in conductivity from (3) to (5).
The key, however, is to realise what happens from (1) to (3) (which also influences our understanding of (6)). Here, it may seem that hydronium ions are constantly being replaced by sodium ions, meaning that the total ion count should remain constant. Indeed, it approximately does. But hydronium (and hydroxide) is a much stronger conductor than any other ion.
Typical ions would conduct electricity by being drawn towards the anode or cathode depending on their charge. This is also, in general, what happens to hydronium (or hydroxide) ions. However, because they are defined merely by an excess (or missing) proton added to (or substracted from) a water molecule, they have the ability to move much more rapidly following the mechanism below:
$$\begin{multline}\ce{cathode^-\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-(OH)\bond{...}H-OH2+} \\\ce{-> cathode-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}(OH)-H\bond{...}OH2}\end{multline}\tag{7}$$
Because this proton-exchange mechanism (the Grotthuss mechanism) allows both hydronium and hydroxide travel speeds much more rapidly than diffusion control, these two ions will induce a much higher conductivity if present. Therefore, our overall graph will feature:

a high-conductivity starting poing
a strong, linear decrease in conductivity while hydrochloric acid is consumed
a global minimum
a slow, linear rise (as discussed before)
a turning point
a strong, linear increase in conductivity while hydroxide is added.

To find points 3 and 5, the best method is to determine sufficiently many data points of the linear sequences 2, 4 and 6 and extend the straight lines. The crossing points represent the equivalence points 3 and 5.
What I have outlined here with two acids also, naturally, works for a single acid; depending on the nature of the acid you will either not see 1/2 or not see 4.

I also fail to remember the name and I usually use "proton relay race" instead:)


text:
So, is baking soda actually a strong base?
Based on my recent thoughts, when baking soda ($\ce{NaHCO3}$) is dissolved in water, the following hydrolysis reaction occurs:
$$\ce{NaHCO3 + H2O <=> NaOH + H2CO3}$$
However,
$$\ce{H2CO3 -> H2O + CO2 ^}$$
so the actual one is:
$$\ce{NaHCO3 -> NaOH + CO2 ^}$$
$\ce{CO2}$ escapes, so the reaction is irreversible, all the $\ce{NaHCO3}$ is converted to $\ce{NaOH}$ and the product is a strong base.
The following reaction is possible:
$$\ce{HCO3- <=> H+ + CO3^2-}$$
but it's reversible unless it results in $\ce{H2CO3}$ formation, which is the same as the first one, so it doesn't really matter.
However, I was unable to find any evidence to that but a lot of discussions on whether $\ce{NaHCO3}$ behaves like a weak or a strong base, which put me in doubt. Am I right?

https://chemistry.stackexchange.com/questions/72775/can-sodium-bicarbonate-be-considered-an-acid

There is nothing to discuss (unless you find it necessary to compare two known numbers by discussion). $\ce{NaHCO3}$ is a very weak base.
Really, do you have baking soda around? Throw some in the water and see whether $\ce{CO2}$ actually escapes. It doesn't. It will if you heat the solution to near-boiling temperatures; then you'll have $\ce{Na2CO3}$ which is a base of moderate strength. You can't convert it all to $\ce{NaOH}$ this way.
So it goes.

:) The reference from Slaughterhouse-Five

@JavaScriptCoder True, but it is broader than that.


text:
Why are there only two irreducible representations in the C3 group?
There are three elements in a $C_3$ group and there are three classes in this group cause it is an Abelian group.
\begin{array}{c|ccc|cc}
\hline
C_3 & E & C_3 & C_3^2 & & \varepsilon = \exp(2\pi\mathrm{i}/3) \\ \hline
\mathrm{A} & 1 & 1 & 1 & z, R_z & x^2+y^2, z^2 \\
\mathrm{E} & \left\{ \begin{aligned}1 \\ 1\end{aligned} \right. & \begin{aligned}\varepsilon \\ \varepsilon^* \end{aligned} & \left. \begin{aligned}\varepsilon^* \\ \varepsilon \end{aligned}\right\} & (x, y), (R_x, R_y) & (x^2-y^2,xy), (xz, yz) \\ \hline
\end{array}
According to group representation theory, the number of classes of a group should be the same as the number of irreducible representations of the group, which means there should be 3 irreducible representations for $C_3$ group.
In a character table, this means there should be 3 rows.
Further more, according to $$\sum_{i}l_{j}^2= h$$
in which $l_{j}$ stands for the dimension of the $j-th$ irreducible representation, and $h$ stands for the order of the group.
All of the three irreducible representation of $C_3$ group should be of 1 dimension.
But in the link above, for $C_3$ group, there are only two irreducible representations, and one of the irreducible representation is of 2 dimension, so it seems it violates the basic results of group representation theory.
So, what I have misunderstood?

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=203&option=4 The E representation is two very similar representations.

You got everything right. It is just that two of these representations would use complex numbers ($\varepsilon=e^{2\pi i\over3}$ and $\varepsilon^2$), and we don't like that, so we stick them together into one representation of dimension 2, which makes the imaginary part vanish. Afterwards we treat this representation as "kinda irreducible". In fact it can be reduced, but we just don't want to go that way.
The same applies to other cyclic groups.

Thank you! So actually the representation labeled $E$ in the character table is actually a reducible representation? I observe that the inner product $1+\epsilon \epsilon^* +\epsilon^*\epsilon =3 $.

Well, let's put it this way: it is reducible if you allow complex numbers, and irreducible otherwise.


text:
Why shape of diamond is like diamond?
I know that in diamond carbon atoms occur in 3d geometry but i want to ask that why it is in shape of diamond not like some other 3d geometrical shape like cube or cuboid?

I'm not sure I understood the question, but I suspect there might be a confusion between brilliant
and diamond. Diamonds come in variety of crystal morphologies, but what is casually called "diamond shape" is a brilliant, a diamond after being cut.

I think the OP is asking why a crystal of diamond is an octahedron rather than say a cube.

I think as andselisk

Diamond has the shape of diamond for precisely the same reason why any other crystal has the shape it has. Some crystal faces have lower surface energy than others, so the crystal grows in these directions, as if it wanted to develop these faces and avoid developing others.
This, BTW, is not determined by crystal family alone. Think of pyrite: it also has cubic unit cell, but it mostly develops the face (001) (see Miller indices) and hence makes those nice cubic crystals. On the contrary, diamond prefers the face (111), which together with its symmetry equivalents produces an octahedron. And that's why natural diamonds are typically found in this shape (not counting scratches and fractures).
Because of its low energy, the (111) face is also the hardest face of a diamond, to the point that jewelers deliberately avoid it when cutting brilliants, for it can't be polished quite as good as any other arbitrary plane.
So it goes.


text:
What is the function of PhLi in this reaction?
Can someone please explain the mechanism of the reaction when we add 2 moles of $  \ce{Ph3P}$ and $\ce{PhLi}$ to 1,2-dichloroethane.
I know this is a type of Wittig reaction but I am not able to figure out how the reaction will work out.

I was wrong, this is not a Wittig reaction. Just posting the answer to let anybody who comes in hunt for this in the near future.
On googling for about an hour, I discovered that this is indeed a preparation of phosphonium ylide which is the witting reagent.
On adding $ \ce{Ph_3P} $ to 1,2-dichloroethane , $\ce{PhLi}$ acts as an catalyst and eliminates $\ce{HCl}$ from both the carbons resulting in a phosphonium di ylide.

Pretty sure it's not a catalyst.

Indeed it's just strong base, strong enough to deprotonate it twice. Catalysts aren't permanently destroyed in reactions which they're supposed to catalyse.


text:
Difference between coordination entity and coordination sphere
What’s the difference between a coordination entity and a coordination sphere? Are they the same? Here in my book, the same example is given for both the coordination entity and coordination sphere.

(a) Coordination entity
A coordination entity constitutes a central metal atom or ion bonded
to a fixed number of ions or molecules. For example, $\ce{[CoCl3(NH3)3]}$
is a coordination entity in which the cobalt ion is surrounded by
three ammonia molecules and three chloride ions. Other examples
are $\ce{[Ni(CO)4]}$, $\ce{[PtCl2(NH3)2]}$, $\color{red}{\ce{[Fe(CN)6]^{4–}}}$, $\ce{[Co(NH3)6]^{3+}}$.

(e) Coordination sphere
The central atom/ion and the ligands attached to it are enclosed in
square bracket and is collectively termed as the coordination
sphere. The ionisable groups are written outside the bracket and
are called counter ions. For example, in the complex $\ce{K4[Fe(CN)6]}$, the coordination sphere is $\color{red}{\ce{{[Fe(CN)_6]^{4–}}}}$ and the counter ion is $\ce{K+}$.

Please help.
Source: http://ncert.nic.in/ncerts/l/lech109.pdf
Section 9.2 Definitions of Some
Important Terms Pertaining to Coordination Compounds
Page no. 240-241

There are also coordination moiety, coordination shell and coordination environment terms.

Coordination entity is the part of complex compound having central atom surrounded with the number of secondary linkage.
But,if this entity is written along with charge on it ,then it is called coordination sphere
For example  $\ce{[Fe(CN)6]^{4-}}$ is a coordination sphere while $\ce{Fe(CN)6}$ is the coordination entity


text:
Should data tables be made in pen or pencil?
For a chemistry lab, I have to make a data table. Should data tables be written in pen or pencil? What is done in the real world of Chemists and why is it done in such a way?

For a "professional" lab book the pages should be written in ink and witnessed.

We usually use computers nowadays. For your personal notes anything should be fine. When you have to write a report, you should consult your teacher or adviser about the official guidelines.

Just don't use erasable pens (like Pilot FriXion). The ink degrades very fast upon heating and also over time. I knew a person who lost some lab notes because hot air from the laptop was blowing towards the pile of the notes, and some of them were made with erasable pen.

Also, if you like to do note taking at the bench where the solvents and solutions are stored, I'd recommend using pencil. Inks are not very spill-resistant.

@andselisk you can make it reappear by cooling it down, e.g. in the freezer. Liquid nitrogen works too ;-)

@DSVA I'll pass this wisdom – looking forward to a Dewar with a pack of tied lab notebooks casually "chilling" at the desk  :D

@andselisk What a chill way to deal with invisible ink!

@andselisk funny picture you painted there. Liquid nitrogen is actually too cold (has to be a kinetic thing) but during warming up it reaches the right temperature  for the ink to reappear. It can also be repeated for at least some cycles in my experience.

Perhaps is more a curiosity than an answer. Normally and for patenting purposes a lab journal should be written in permanent ink and witnessed, as said in other answers. Not sure about the validity of digital media , although they are crucial too in case of disputes or investigation. However, in my lab (master thesis), journals were written in pencil as the information is not lost in case of spilling. Water and solvents were always around the lab book, at least searching for eluents etc if not to run the reactions. We students had a lot of space in the labs but not a real office desk.

Frankly, a lab journal should be ‘clean’ and thus never on the bench, in the hood, or anywhere else where there is even the risk of spillage. Furthermore, you should not spill as each solvent spillage can be detrimental to your health and clean working is suggested.

@Jan. Thanks.chemist since the 80s here. Not to mention who preceded me. Ps we were lucky to have new labs and lot of spaces for reacting. Never saw so tidy labs anymore anywhere. Have you ever been in an organic chemistry lab? Have you ever copied your tlc for real or compared the tlc already in a journal to your actual one while running a column? But you are right. What shall I say. Ps2 from a time in which latex gloves and safety goggles were often left alone. And recrystallisation was done without mounting a condenser atop the erlenmayer :)

You obviously didn’t check my profile because I have spent my entire working days in organic labs since 2012 when I did my master’s. But no hard feelings ;) I actually always try to keep my bench and hood clean (meaning also that chaos from the neighbours stops right at a line; if needed, I push stuff back). I copy my TLCs in a transparent foil on the copying machine (very likely not an option in the 80’s). In my previous lab, I walked back to my desk if I wanted to compare, now I have a practically desk-like space (which I keep free of chemicals) right behind my hood.

But usually, I would run the column, record the TLCs and then take the copies to compare with the ones already in my journal. Safe and clean working is possible, you just have to want it and be prepared to live with the consequences ;)

No polemic here too. It was more or less a witnessed story. I keep my organic notes in pencil still nowadays. But I have two journals has my work is synthesis than implementation. In case I feel I hand copy the recipe to the witnessed " physics" applied one. Didn't see your profile but I knew you are of the safety goggles generation. We did improved ;)

In the real world, chemists have a lab journal to document what they did; Good Scientific Practice guidelines state that results should be written there immediately, readably, non-removably, directly and a few other adverbs. ‘Great’ scientific practice or patent-safe documentation would require a witness.
In universities, I have heard of a trend of more and more lab journals being replaced by computer files. In the labs I have worked in, this was not the case and the paper lab journal remained the way to go. Apparantly, if you want to document something in a patent-safe manner, a paper book is mandatory.
Since the entries into a lab journal must not be removable, the only option you have is ink; usually, ball-point pens (biros) are used.

However, the wording of your question (‘chemistry lab’ and ‘data table’) makes me think you could still be a student. In that case, the only advice you should take is that of your teacher/advisor/…

My labs have used electronic lab notebooks for about the last 8 years. TLCs are recorded as .jpgs

@Waylander Especially interesting considering your profile says you work for pharma, which is a patent-sensitive industry! All I can do is repeat what I was told at my previous position (a Helmholtz centre), namely that only hand-written, dated documents signed by oneself and a witness would be accepted in patent disputes. (But the guy who told me was not in the patent department of the centre, so ~ xD)

It took a lot of arguing with the patent attorneys but it was implemented. AFAIK most big pharma now uses ELNs

I would recommend to draw the table itself in pencil and write the data inside it in pen.
Of course, it's best to ask your teacher about the guidelines, just to be sure.


text:
Computing i0 in the Butler-Volmer equation: How to interpret Wikipedia and Wolfram Alpha
So far, when using the Butler-Volmer equation, I had j0 given and could then just use I = A * j0 * [exp(...) - exp(...)], with A the electrode surface area.
Now I have a material from which I don't know j0 but just the rate constant. According to Wikipedia on exchange current density I can compute j0 as

Wolfram-Alpha on Butler-Volmer equation however says:

In this equation, the expression F * k0 would correspond to the above j0. k0 is explicitly defined as the rate constant in m/s on that same page.
Is Wolfram-Alpha wrong by missing out the concentration? The F * k0 does not even result in the units of a current density, and the whole expression will not evaluate as a current.

As I said before, Wolfram Alpha does not know apples from oranges when it comes to chemistry.

@IvanNeretin: So Wolfram Alpha is wrong and Wikipedia is right in that case, just to be sure?

k0 has units of frequency (1/time)
All concentrations are surface concentrations (mol/area)
Sometimes people use k0 with units length/time and volume concentrations (mol/volume)
On Wikipedia: Coxy stands for Ox concentration * at equilibrium *. Cred stands for Red concentration * at equilibrium *. It's for 1 electron. Multiply by n for n electrons.
On Wolfram: [Ox] stands for Ox concentration at electrode, [Red] stands for Red concentration at the electrode. These concentrations are not necessarily at equilibrium.


text:
ammonia, methylammonia, dimethylammonia, trimethylammonia reactions
Perhaps a "basic" question (pun intended), but I have a few questions about the reactivity of ammonia and its methylated cousins.  I come from an engineering background, essentially I use a process in which thermally cracked ammonia releases its nitrogen atom to a surface under temperature.  I suppose in principle any of the other 3 molecules, with 1,2 or 3 methyl groups could do the same.  However, could I definitely determine whether they would do so at higher or lower temperature?  Essentially, is it as simple as comparing the the C-N bond vs N-H bond energy, or more involved? Would steric factors of the N ability to "see" the surface be involved? Finally, would attempting to do this in solution versus in gas state change anything significantly?  Thanks for any pointers and/or literature refs!

Some other factors to consider: What happens to the H of $\ce{NH3}$? It may diffuse through the surface, $\ce{CH3}$ won't. The hydrocarbon species involved may poison the catalyst (if there is one) or just deposit on the surface. Finally, I am not sure that thermal cracking of these compounds is possible in solution - I would assume temperature are too high for that.

Great considerations.  This process is under high vac conditions, ~ 1e-6 to 1e-5 torr, and the surface/subsurface already has hydrogen impurities, so we don't think too much more gets in.  The H likely forms H2. Similar with CH3, although surface residue is not an issue, we could remove later.  This isnt a catalytic reaction per se, what I would like to determine is if any of the methylated versions would more effectively "drive" N into a N deficient structure.  Valid point about temperature in solution, so lets stick to low pressure gaseous.

Sounds sensible, in the end, you may not get an answer without an experiment. I suggest you look into the conditions of the Haber-Bosch process of ammonia formation for similarities to your conditions. I still suspect the surface may act as a catalyst.

Right... In my case this is a treatment for the surface, not a gas phase process, so I didn't think of the surface as a catalyst...but I guess that's a small detail, and maybe just wrong.


text:
Whats the full name of HMOBA?
I'm doing some statisitcal research and I have come across an acronym for a chemical. I have tried googling it but it comes up with very little and I can't find the name.
I only have HMOBA. Can anyone tell me what this stands for?

there is this image from a chinese journal : http://www.jcscp.org/article/2016/1005-4537/1005-4537-36-5-389/img_6.png

You should specify where you came across the acronym. Gilleain has a reasonable answer, but it could also be the HMOBA from this patent which is  N-(2-hydroxy-3-methyl-1-oxobutyl)-β-alanine or from this other patent which mentions 3-hydroxy-4-methoxybenzylamine.

It might be possible to figure it out from the context of where you are seeing this. As it stands, there is not enough information to answer unambiguously.