text:
Analysis of methylphenidate (Ritalin) HCl extended release tablets
For a project I have to determine the drug concentration release profile of self-made methylphenidate HCl (MHCl) extended release tablets (hydrophilic matrix). We are thus doing a dissolution test as described conform the US/Eur. Pharmacopeia. However a problem we are facing is that we are not sure how to analyse the samples that we take from the dissolution test. This is because we are not sure what detection method to use. The problem is that we cannot directly get the right concentration for our primary detection method which is UV spectrophotometry. This is because MHCl has a very low extinction coefficient and thus a relatively higher concentration is needed to get the measurable absorption range of 0.2-0.8, but since the tablet dose is only 18 mg this is difficult to accomplish. We thought of doing two things: a solid phase extraction followed by UV or picric acid ionpairing followed by colorimetry. We asked our professor if the SPE followed by UV was possible, but he said that in this case it wasnt practical cause of the such low amount of drug in the tablets. So that leaves us with the picric acid analysis method. Problem is that we havent ever performed a colorimetric analysis before, though it looks pretty similar to spectrophotometric. From what we understand we need to measure the absorbance at a specific wavelength that the picric acid-MHCl complex is absorbing at. But how can we calculate the MHCl concentration from this? How much picric acid do we need to use to know for sure that all the MHCl has formed pairs with the picric acid? It's kind of confusing still and appreciate it if anyone could point us to some literature or give us insight on how to analyse these samples. Thanks in advance.

@xavier_fakerat I'm not quite following. What do you mean by taking the weight 10 or 20 tablets? What we have done so far is taken 3 tablets and performed a dissolution test with the paddle apparatus, taking around 8 samples per tablet over time period of 3 hours.

Maybe I missed the question I will get back to you, I am thinking of another way..

@xavier_fakerat Isnt the method you're describing to test content uniformity instead of drug release profile over time? We want to individually analyse our 8 samples we took from each extended release tablet and eventually graph the concentration time profile. But the problem is that since the dose is really low, even if hypothetically everything got out of the matrix, only 18 mg would be dissolved in 500 ml water (our paddle setup). Since we only took samples of 5 ml we havent much to work with. Now that I think of it I think we couldve better done this in a smaller volume...

Its somewhat similar though here is summary: USP II Apparatus was used to perform dissolution test. The speed for the paddle was set at 50rpm. 900ml of distilled water was placed in each of the vessels and maintained at a constant temp (37°C).  Three tablets from each of the batches were added to each vessel (one tablet per vessel) at the same time and a timer started. The immediate release tablet was analysed every 5 minutes and the sustained release analysed every 10 minutes using UV-VIS Spectrophotometer at  271nm. The results were used to calculate conc and  (%) released of the 3 batches.

But I do get your point the dose is really low, you are very right, we did with 250mg so this range was more favourable, also we went on to do content uniformity as you noted

@xavier_fakerat We indeed performed the exact same test you described, only we used 500 ml distilled water and took samples at different times. For the UV calculation,  if using the minimum boundary absorption of 0,2, we would need a minimum concentration of: 0,2/9 *1000 = 22 mg/100 ml to analyse with UV (9 = specific absorbance).  Let's assume that if we take a sample at a moment where the whole tablet dose is dissolved in 500 ml water, the concentration would be 18/500 =  0,036 mg/ml = 3,6 mg/100 ml. Which would mean we'd have to concentrate that sample by at least 6x.

@xavier_fakerat And that's assuming the whole tablet dose dissolved. You can imagine that in the early stages where less methylphenidate is dissolved it would need greatly more concentrating to be able to be detected. Thus probably having to resort to the picric acid method or something else if anybody has an idea.

And, of course provided proper calibration, linearity check of concentration and read-out, a recurrent HPLC-MS  is not an option?

@Buttonwood No, only UV as detecting method.

If you are insistent on a picrate salt, I'd suggest first taking your methylphenidate and forming a known concentration of the picrate by adding a known amount of picric acid to a solution in which you know the base would be in excess. This ensures that all of the picric acids forms the picrate salt, and you would therefore know the picrate complex concentration (assuming there are no other compounds in solution that might react with the picrate). You could then measure the absorbances at a wavelength at which picric acid has only a negligible absorbance, provided one exists. Note that if no such wavelength exists, matters become rather more complex, and you will have to look for changes in absorbance on either side of an isosbestic point.
An alternative reaction to the use of picric acid is to instead use 1-fluoro-2,4-dinitrobenzene (FDNB), which reacts with primary and secondary amines at room temperature to form a highly-coloured 1-amino-2,4-dinitrobenzene (ADNB). The spectral overlap between FDNB and ADNB is typically pretty small, so using an excess of FDNB usually doesn't greatly change the absorbance at the maximal wavelength of ADNB. Again, to at least know the absorbance of the methylphenidate ADNB, you would have to first perform the reaction with an excess of methylphenidate, and assume a complete reaction to the ADNB.

Would it be possible to do a liquid liquid extraction to isolate the complex in an organic phase, since the unbound  charged picrate should favor water more? And if this is possible which organic phase?

@user21398 I think it would probably be decently difficult to separate by a normal extraction to a level acceptable for spectroscopy. The methylphenidate picrate salt is probably decently soluble in water (the HCl salt is has a solubility of about 18.6 mg/mL in water), but so is picric acid (about 12.7 mg/mL), resulting in a difficult separation. It may be worth seeing if there is a wavelength at which the complex absorbs a significant amount and at which picric acid does not.


text:
Am I solving this pH problem correctly?
A liter of a solution contains $\pu{0.120 mol}$ of $\ce{C6H5COOH}$ $(K_a = 6.5
×10^{-5})$ and $\pu{0.105 mol}$ of $\ce{C6H5COONa}$. I'm asked to calculate the pH of the solution after adding $\pu{100 mL}$ of a $\pu{1.20 M}$ $\ce{NaOH}$ solution, the pH of the initial solution is $4.13$.
Here's what I've done. Since $\ce{NaOH}$ is a strong base it will disassociate completely, so we will end up with $0.120$ moles of $\ce{OH^{-}}$. Calulating the $\ce{pOH}$ we get $0.92$ and with that we can calculate the $\ce{pH}$ which will be $13.07$. Is that the correct way of doing it ?

Nope.
While $\ce{NaOH}$ is a strong base, the mixture of $\ce{C6H5COOH}$ and $\ce{C6H5COONa}$ is a  buffer solution and tends to resist large changes in pH.
After adding the $\ce{NaOH}$, the extra $\ce{OH-}$ ions would react with the undissociated $\ce{C6H5COOH}$, effectively being removed from the solution. The solution would now be $1.1 \text{L}$ and contain a bunch of ions, like $\ce{Na+}$, $\ce{C6H5COO-}$, $\ce{H+}$, and $\ce{OH-}$. You can calculate the value of $\ce{[Na+]}$ now.
Use the Henderson's buffer equation:
$$\text{pH}=\text{p}K_a + \log{\left(\frac{\ce{[C6H5COO^-] }}{\ce{[C6H5COOH]}}\right)}$$
I think you can continue from here.

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text:
Is “amount of substance” the same thing as “number of moles”?

Background:
Amount of substance is a fundamental physical quantity which has mole (mol) as it's SI unit. Therefore all expressions for amount of substance should have the unit mole on simplification.
If $A$ is the amount of substance (normally in moles), $m$ is the mass of the substance in a particular unit (normally in grams), $M$ is the mass per unit amount of substance (normally in grams/mole), then
\begin{align}
A &= \frac{m}{M}\tag1\label1
\end{align}
If the substance is a molecule then $M$ is known as molar mass of the substance.
But in my textbook and in many websites on the internet, I have encountered the phrase number of moles. I think it refers to the amount of substance or maybe the numerical part in the amount of substance. The formulas given there were strange.
If $n$ is the number of moles, $m$ is the mass of the substance in grams and $x$ is the atomic weight (for atoms of elements) or molecular weight (for molecules of elements and compounds), then,
\begin{align}
n &= \frac{m}{x}\tag2\label2
\end{align}
This equation is not dimensionally correct if I am right.
My questions:
First of all I would like to ask whether amount of substance and number of moles refer to the same thing, or is it that amount of substance has a unit along with a numerical value whereas number of moles does not have a unit and represents the numerical value in the magnitude of the amount of substance.
Secondly, between equations $\eqref1$ and $\eqref2$, which one is completely correct (both in meaning and dimension)?
Thirdly, since in some place I have encountered gram atomic/molecular mass in place of molar mass, I would like to know what are the differences between both and do they have the same units or different units?

See the comment thread here: https://chemistry.stackexchange.com/questions/58582/using-masses-and-and-moles-of-starting-materials-to-calculate-the-amount-of-subs/58588#comment98014_58588

This is also quite thorough and an interesting read: May I treat units (e.g. joules, grams, etc.) in equations as variables?

It is like to say "amount of meters" while speaking about length. Quite a common misuse

"Amount of substance" is one of the seven base quantities of the SI [1]. Its dimension is noted with the symbol N, and its base unit is the mole (mol). Similarly "length" is another base quantity with dimension noted L and with base unit the meter (m).
"Number of moles of [substance X]" rigorously means the numerical value (thus dimensionless) associated with the corresponding "amount of [substance X]" expressed in moles, but from my experience it is often misused for "amount of [substance X]" (thus of the dimension of an "amount of substance"). - Think about how you would respond to the question: "How many meters are there between your house and the nearest bakery?" You could answer "35" or "35 meters", both would be grammatically correct; but strictly speaking the question expects just a numerical answer.
Note that according to the SI, a number of entities (molecules, atoms,..) is dimensionless and the unit of a dimensionless quantity is simply the number one, 1 [1]. My way of reconciling these two views (an "amount of substance" of dimension N and of unit the mole vs. a number of entities, dimensionless and of unit 1) is that the "amount of substance" refers to a combination of an amount (a number) and a given type of entity of a given substance, whereas a number of entity simply refers to the number itself. Why not consider an amount of substance as a dimensionless quantity? I think it is a reminiscence of history, at the time when chemists had to measure amount of substance without knowing how this substance was structured (but I can be wrong).
Regarding the equations, I agree with this other answer by @desc (equation 1 is correct, equation 2 isn't). Importantly: "atomic weight" is dimensionless (it is a ratio between a mass and a reference mass) and should not to be mistaken with "atomic mass". [2]
Finally, regarding the units "gram-atom" or "gram-molecule", these are outdated units which, according to SI were connected with "atomic weights" and "molecular weights" [2]:

Following the discovery of the fundamental laws of chemistry, units
  called, for example, "gram-atom" and "gram-molecule", were used to
  specify amounts of chemical elements or compounds. These units had a
  direct connection with "atomic weights" and "molecular weights", which
  are in fact relative masses. "Atomic weights" were originally referred
  to the atomic weight of oxygen, by general agreement taken as 16. But
  whereas physicists separated the isotopes in a mass spectrometer and
  attributed the value 16 to one of the isotopes of oxygen, chemists
  attributed the same value to the (slightly variable) mixture of
  isotopes 16, 17 and 18, which was for them the naturally occurring
  element oxygen. Finally an agreement between the International Union
  of Pure and Applied Physics (IUPAP) and the International Union of
  Pure and Applied Chemistry (IUPAC) brought this duality to an end in
  1959/60. Physicists and chemists have ever since agreed to assign the
  value 12, exactly, to the so-called atomic weight of the isotope of
  carbon with mass number 12 (carbon 12, $^{12}$C), correctly called the
  relative atomic mass $A_r$($^{12}$C). The unified scale thus obtained gives
  the relative atomic and molecular masses, also known as the atomic and
  molecular weights, respectively.
  [2]

https://www.bipm.org/en/publications/si-brochure/chapter1.html
https://www.bipm.org/en/publications/si-brochure/mole.html

Your answer is trying to reconcile sloppy terminology with actual meaning. While I find the question you pose, "How many meters are there between your house and the nearest bakery?", okay for a casual conversation, I would regard it as wrong (or very sloppy) in the scientific context, where you should ask "What is the distance between ...?". You are quite correct though that the phrase 'number of moles' has incorrectly been used synonymous with 'amount of substance'.

@The Quark Is the same with degrees of freedom? I mean if degrees of freedom refer to a number it doesn't make sense to say 2 or 3 degrees of freedom. Shouldn't we say just degrees of freedom is 2 or 3? Also the "How many..." part in the question of distance can be change to "What is the number of..."? Thanks in advance.

@adosar Saying "the number of degrees of freedom is 3" or "there are 3 degrees of freedom" is the same. "Dimensionless" doesn't mean we cannot talk about a number "of something" or that we should not take care about what we are counting. (...)

(...) Basically: if it is something you can count (1, 2, 3,...) it is dimensionless (but it is still a number "of something" - unless you are in the pure mathematical realm; if it is an amount that has to be quantified compared to another amount of the same kind (dimension) and which you take as a unit, then your amount has the dimension of that amount that you take as a unit. (...)

(...) For the second part of your question: yes, you could replace "How many ...?" by "What is the number of...?". What I wanted to stress in my annswer is that the question "How many meters ...?" is not a fomulation one normally uses. Similarly we should not talk for instance about "number of moles" [of molecular oxygen] but rather about "amount" [of molecular oxygen].

@TheQuark " "the number of degrees of freedom is 3" or "there are 3 degrees of freedom" is the same." But degrees of freedom is defined as the number of indepedent parameters that specify the configuration of a system, so the number is already incorporated in the definition. So isn't pleonasm to say 3 degrees of freedom? Or is my definition wrong?

Your question is mainly semantic. The amount of a substance and number of moles are both unitless quantities (and the same thing). A mole is simply defined as "Avogadro number" of something. That is, 1 mole of (say) carbon atoms is about $6.02\times10^{23}$ carbon atoms. Both have a dimension of 1 (the number one); a mole is conveniently represented as a unit of measurement but doesn't actually hold any physical meaning. Using moles (and derivative quantities, such as molar mass which is the mass of "Avogadro number" items of something) is only a way to avoid extremely large or small numbers.
Another important point is that the variables in an equation only have the meaning you attach to them, so though conventions as to the meaning of each letter in your equations do exist I will answer to your definitions.
Equation (1) is correct - dividing the mass of a sample with its molar mass will give the number of moles.
Equation (2) is incorrect - diving the mass of a sample with the weight of a single item (atom or molecule) will give the number of items, not the number of moles.
Two notes:

Common convention is that $n$, and not $A$, is the number of moles. So the most conventionally correct equation is $$n = {m \over M}$$
where $n=[\mathrm{mol}]$, $m=[\mathrm g]$ and $M=\left[\mathrm{g \over mol}\right]$ (brackets are used to denote the unit of measurement of quantities).
The reason the second equation usually works is a double error - the atomic "weight" is usually given in number of nuclear particles (protons and neutrons) and not in actual weight units. This number is very close to the molar weight in $\mathrm{g \over mole}$ (on purpose, the definition of a mole is meant to be this way), but these are different quantities. Many chemistry books tend to confuse them. The actual atomic or molecular weight (in grams) of a single particle is very small and it is not likely you will actually use it much.

Amount of substance is not a dimensionless quantity.

This answer is not in accordance with the International System of Units (SI). Therefore, for all intents and purposes of this site, this answer is wrong.

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text:
Why group 13 elements don't show +2 as their major oxidation state?

These are the things I understand:
1) Group-13 elements show +1 and +3 as their major oxidation
  states(OS).
2) The general configuration of G-13 elements is $ns^2$ $np^1$.
3) In the excited state, their electronic configuration is $ns^1$
  $np_x^1$ $np_y^1$ .

On losing two electrons from their excited state,
their configuration will become $ns^1$ $np_x^0$ $np_y^0$
I don't understand why will this configuration be unstable?

These electrons excite only when they form covalent bonds. They won't be exciting in free states.

@PrittBalagopal They will lose 2 electrons only in their excited states.

The one p electron is energetically much higher than the two s electrons (and any d electrons, if present). So it is rather easily removed to give a monocation or the $\mathrm{+I}$ oxidation state.
Thereafter, to remove a second electron you need to rip out one of the paired s electrons. This is not going to be easy and requires a lot of energy. But once you have arrived there, at a hypothetical $\mathrm{+II}$ cation, you again have one single lone electron in a relatively (compared to the core orbitals) orbital that should, again, be easily removed. Obviously that can happen leading you quickly to the $\mathrm{+III}$ state.
Indeed, it would be rather easy to see that two $\mathrm{+II}$ atoms might come together and transfer a single electron from one to the other—a disproportionation, giving $\mathrm{+I}$ and $\mathrm{+III}$ from two $\mathrm{+II}$. This would be predicted to be an exothermic process.
This only applies to isolated atoms; where bonds are formed other oxidation states are accessible. For example, boron exhibits the $\mathrm{+II}$ oxidation state in bis(pinacolato)diboron as can be deduced easily.

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text:
Significance & applications of polarizability for nucleophiles
One of the reasons given for the greater nucleophilicity of iodine ion compared to fluorine ion is the greater polarizability of the large electron cloud of iodine. The other reason given is that it is less crowded by polar protic solvent molecules.
However, when the solvent is changed to a polar aprotic solvent the nucleophilicity order is reversed.
If polarizability had any significant role to play, the order would've not been completely reversed. It seems to me that the only reason iodine ion is more nucleophilic than the fluoride ion is because it is less crowded by the solvent.
Now, this leads me to believe that polarizability is not a very important factor and should only be used when comparing nucleophiles of the same basicity, but of different size (and hence polarizability) in a polar aprotic spent.
But why then, for uncharged nucleophiles does polarizability dominate over basicity in determining the order of nucleophilicity (regardless of the solvent used)?

Related: Why do polarizable nucleophiles favour SN2 over E2?

The examples the site you mentioned gave of uncharged nucleophiles were: the hydrides of Group V and VI. They say that the bigger atoms can have a more effective overlap of atomic orbitals and thus, the bigger the uncharged nucleophile, the stronger it is.
I would like to dispute this reasoning and answer your question regarding this trend.
How we should see a nucleophile should always primarily be in terms of electron donation because a nucleophile donates electron density to electrophiles to form bonds. This is a very important consideration.
For bigger atoms, the valence electron pair, which is being donated to the electrophile, is further away from the nucleus and is more shielded from the nuclear charge. Thus, it is easier for donation to take place. This should be the reason why uncharged nucleophiles with bigger atoms are more nucleophilic; that is, they are more capable of donating the electron pair of theirs. My explanation of the trend has nothing to do with polarisability.
The reasoning stated is incorrect because although bond formation is easier for bigger atoms, it does not mean that the overlap of atomic orbitals is more effective. In fact, it is likely less effective because the atomic orbitals used are more diffused (Recall that as the principle quantum number n increases, the orbitals become more diffuse in terms of electron density). Thus, their reasoning is nonsensical.
Hope I have clarified everything you were not sure of.

It seems to me that both trends end up pointing the same best uncharged nucleophiles, that means, polarizability increases down the periodic table (bigger electron clouds have a better overlap in the Sn2 transition state) while basicity decreases (the electron density is more disperse around bigger atoms). So polarizability does not dominate over basicity anyway, because it would be unfair to relate a property which is considerably increasing with another which is decreasing down the PT, it is just that polarizability has a more pronounced effect then basicity in this situation. If you compare analogous alcohols and thiols you can see this trend.


text:
Is there a haloalkane for astatine?
I'm studying organic chemistry, and I noticed that there are fluoro-, chloro-, bromo-, and iodoalkanes. However, there are none for Astatine. Could someone shed some light on this, and what it would be called?

Astatine is so unstable that for most chemical purposes you may just as well consider it non-existent.

The fact that you're unlikely to come across an organoastatine compound doesn't make the question unimportant or uninteresting; astatine derivatives have definitely been made, and there are certainly naming conventions.

There also should more be carbon-hydrogen astatine compounds like for example methyl astatide (CH3At) , dimethyl astatide (CH2At2) and so on. But it is highly radioactive like all astatine compounds,so it might just break down into its elements

Part of the reason why there has been very little research done on organoastatine chemistry is the extreme radioactivity of astatine; the half-life of $\ce{^210At}$, the longest-lived isotope, is about 8 hours. However, the element can be incorporated into organic compounds (e.g., via a destannylation reaction).1
In organic chemistry, the prefix is "astato-", as in astatobenzene or 3-astatotamoxifen, both of which have been synthesized.1,2 This is analogous to "fluoro-", "chloro-", and so on. As a general term, astatine compounds can be referred to as astatides (e.g. aryl astatides, analogous to aryl fluorides, chlorides, etc.).1
The aforementioned synthesis of 3-astatotamoxifen is shown below.

As for inorganic nomenclature, astatine can be denoted with the prefix "astatido-"; this follows from the IUPAC Red Book for nomenclature of inorganic compounds.

Milius, R. A., W. H. McLaughlin, R. M. Lambrecht, A. P. Wolf, J. J. Carroll, S. J. Adelstein, and W. D. Bloomer. Appl. Radiat. Isot. 1986, 37 (8), 799-802.
Samson, G., and A. H. W. Aten. Radiochimica Acta 1970, 13 (4), 220-221.

There's a page for it on Wikipedia. Have a go at reading it.
https://en.wikipedia.org/wiki/Hydrogen_astatide
There's something interesting here. With Astatine being so electropositive, dissociation of $\ce{HAt}$ could form $\ce{H-}$ and $\ce{At+}$. If you tried to add this to propene, you would most probably get 1-astatopropane rather than 2-astatopropane.
Don't push your luck too far. Before you can call your fellow chemistry buddy and say "Hey look here! This is 1-astatopropane!" It's gone. That's because Astatine is highly radioactive:
$$\ce{^{210}_{85}At -> ^{206}_{83}Bi + ^{4}_{2}He}$$
With a half life of 8.1 hours, your beloved 1-astatopropane is going to decay away, leaving only a trail of alpha particles, and a few gamma photons. sob
Actually, for that matter even $\ce{HAt}$ isn't going to be there for long.

The logical follow-up question: what happens to the 1-astatopropane molecule once its astatine has become bismuth?  Could you conceivably use this to wedge bismuth atoms into molecules where they wouldn't normally go?

@zwol A nuclear decay releases so much energy compared to a chemical bond that chemistry can be ignored when considering nuclear decay. In practice you'll probably end up with ionized bismuth, hydrogen and helium after decay.

@ntoskrnl Not really https://chemistry.stackexchange.com/questions/65809/what-happens-to-a-radioactive-carbon-dioxide-molecule-when-its-carbon-14-atom-de


text:
When (R)-2-bromopentane reacts with sodium methoxide, what is the major product?
I know that with a SN2 reaction mechanism this becomes (S)-2-methoxypentane, but with elimination this is trans-2-pentene. I know that elimination happens quickly when you apply heat. In this case we have a secondary alkyl halide so its possible to use SN2 or E2 due to methoxide being a strong nucleophile.

To be honest, I think it is very hard to decide as not enough information is provided. I think if the reaction temperature was given, it would help us find the answer much more easily. But yes, your deduction that it is either (S)-2-methoxypentane  or trans-2-pentene is sound.


text:
Chemical detection of cell division
So I have been working on a project for a while, and there is one mechanism that I need a little bit of advice about.
The project is chemical-biology based, so don't be deterred :)
Theoretically, by this point in the experiment:
I put cells (let's say lymphatic or endometrial) into stasis, through means of lyopreservation, and now, I have brought them out of stasis, through exposure to a sucrose (trehalose)-based medium.
Now I am attempting to chemically detect mitotic growth within the cells, how can I do this?
This is where the chemistry comes in: My first thought was to use a chromatic indicator that would fluoresce or irradiate in response to mitotic acceleration, but I have not yet found a suitable chemical. In research, I found that a similar experiment had been done in bacteria, and the chemical dimyristoylphosphatidylcholine (1,2-Dimyristoyl-sn-glycero-3-phosphocholine) had been used. Would this work in mammalian cells? Another idea was to use a chemical that detects the unnatural growth from the spindle fiber and centriole point of view, reacting when the centrioles entered the phase to replicate. I have been unable to find any chemicals that make use of the replication of centrioles, or the spindle fibers, so I need ideas.
To summarize: Does anybody know of any biochromatic indicators that react through detection of rapid, mammalian, mitotic growth? Not necessarily cancerous and uninhibited, but rapid meaning 'a speed of mitosis that is faster than the norm.' Preferrably, the indicator would be able to stay in stasis for extended periods of time.
If such an indicator is not available (or you do not believe so) then does there exist a biochromatic indicator that reacts in the presence of any sort of cell division?
Is there a way to chemically detect cell division?
I am welcome to all ideas.

(figure source / chemspider.com)

I have a feeling this would do better in Biology.SE with the biochemistry tag.

Possibly, and I might post it there, but since I was specifically asking about the existence of a chemical, I thought I would try Chem first.


text:
Why flame color of salts is determined by metal
If a pure metal is heated up in a flame, we see it giving the flame a characteristic color, e.g. $\ce{Na}$ burns yellow.
If a salt, e.g. $\ce{NaCl}$ or $\ce{Na2SO4}$, is heated up, resulting color to a greater degree still is that of the pure metal. Although the color does change depending on the anion, this effect is usually not much pronounced.
Why is that the case? Shouldn't ionization happen to anions as well, causing electrons (including those pulled from the cation) to "fall back" and give off photons? In other words shouldn't we see a mix of two characteristic colors?

Related: Anions produce flame color?.

I edited your post to make use of the \ce{...} notation to format the chemical formula correctly (upright with correct sub scripts).

Not all flame colours are determined by the metal, some anions also colour flames (see the linked question.)
Also, not all metals impart a notable flame colour (platinum, for example, is often used as a substrate material to do flame tests because it doesn't cause interference: small amounts of the substance of interest are deposited on a platinum wire which is then inserted into the flame).
What actually causes colour are the electron transitions where excited electrons fall back to lower energy levels emitting light in the process. Flames cause enough excitation that this process is common. Some metals have strong electronic transitions that match wavelengths of visible light. The bright yellow characteristic of sodium come mostly from electrons falling from the 3p to the 3s levels in sodium atoms and have a wavelength around 589nm. This is the same bright yellow that is visible in low pressure sodium street lighting (and is caused by exactly the same electronic process).
Other metals have different electronic structures and their emission spectra therefore have different colours. But not all metals have bright lines in the visible (the strongest lines in the spectrum of mercury, for example, are in the UV so won't give the same notable visible colour in a flame test). Many metals do, though, which is why flame tests are often useful. And even when you can't see the colour, instruments can, so spectroscopy is a useful analytical tool even when your eye can't see much.

Also, I found this answer clarifying https://chemistry.stackexchange.com/questions/31476/why-do-s-block-elements-impart-color-to-a-flame?rq=1


text:
Is there an important difference between soap and detergent?
I've read recently somewhere that most "soaps" are actually detergents and the real soap would not be a good idea in everyday applications like washing hands, shower gels and shampoos, and dish washing liquids. But I couldn't find a sufficiently detailed explanation about the differences of the two, just that the detergents are often marketed as soap.
Are they really different? What negative effects does soap have that detergent doesn't? Is there something soap does that detergent doesn't do?

Are you interested in the chemical differences ?

Yes. I wanna know what the negative effects of regular use of soap would be on hands instead of detergent.

Soaps are generally made by treating fats with sodium or potassium hydroxide, so they may be more caustic than detergents? There is some discussion of the differences in the Wikipedia entry for detergents.

Soap is one of many detergents or surfactants. Soap is a salt of a long-chain fatty acid. A detergent is any chemical that is attracted to both polar and non-polar sub substances.  See https://en.wikipedia.org/wiki/Detergent


text:
A basic solution of HCl
So I had a test question the other day, and it essentially went as follows:

A student measures the concentration of a $\ce{HCl}$ solution to be $10^{-9}$M by using a pH meter. Is the meter wrong?

This means that the $\ce{HCl}$ solution has a pH of 9, which seems unreasonable, yet that is not what I said. I said that the meter is not wrong. My reasoning was that you could have an $\ce{HCl}$ solution so incredibly dilute that the pH would in fact be 9. Since molarity is defined as moles per liter, if you hold the moles constant and increase the volume of solution, you will eventually reach a very tiny molarity.
In essence what I was wondering was: Is my reasoning correct?
Is this solution considered basic?
Can this be generalized such that any originally acidic solution is considered basic at a low enough concentration, or a high enough dilution?

also https://chemistry.stackexchange.com/questions/66843/ph-of-aqueous-solution-of-hcl-of-low-concentration

Unfortunately, your reasoning is wrong, because you forgot to take into account the acidity of water. While in most cases we can ignore the acidity of water as the hydrogen ion concentration contribution by $\ce{HCl}$ dominates, at an extremely low concentration of $\ce{HCl},$ water becomes the main $\ce{H+}$ contributor.
It is well known that water has a $\pu{10^{-7} M}$ concentration of hydrogen ions at $\ce{25^\circ C}.$ Therefore, water's effect on the acidity of the solution is significant when looking at solutions of $\pu{10^{-6} M}$ or lower concentrations of strong acids. For example, if there is $\pu{10^{-9} M } \ce{HCl,}$ then the total $\ce{H+}$ concentration would still nearly be $\pu{10^{-7} M}.$ Therefore, no matter how much you dilute the acid, it can never turn into a basic solution.

Your answer is correct. At the same time, I think the OP's reasoning is also correct. You can dilute HCl to whatever concentration you want and therefore the meter is not wrong. What is wrong wtih that reasoning?

The concentration could be anything but it is impossible to measure an HCl concentration of 10^-9 M using a pH meter. So the meter is wrong.

@Paul - You mean, assuming that the HCl is being diluted in water, the meter must be wrong.  Perhaps some other solvent could be used to dilute the acid?  At a minimum, using D2O should allow you to approach a pH of ~7.5.

@aroth It won't, because D2O won't produce hydrogen ions when it dissociates.

Assuming this is an aqueous solution, you cannot have an HCl sample with pH 9. Consider: what is the pH of pure water? Of course, it is pH 7. As you add more and more solvent to your HCl solution, as you say, it becomes more dilute, but the autoionization of water eventually dominates and the pH will asymptote to 7 (from below since you started as an acid). The pH of a sample will always be between the pH's of the pure constituents. A solution with a pH greater than 7 is basic.
Now, let's say you had a sample of DCl (that is, deuterated HCl). If you could monitored the pD (analogous to pH) it could eventually get to 9, but the total acidity of the solution p(D+H) would go to 7, and it would always be acidic or neutral.


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Carbene mesomeric structure of vinyl carbocation?
Why isn't a mesomeric structure with carbon having only two single bonds and a lone pair used for depiction of vinyl carbocation?

The alternative answer to what @levineds said is that you're asking about a carbene. Carbenes are actually reasonably common in chemistry though we generally have them coordinated to metals for stability. There are a few reactions that proceed through putative naked carbenes.

If I understand your drawing correctly, you are asking about the $\ce{C2}$ molecule. It is highly unstable but it does exist (and is a favorite molecule of quantum chemists because the bonding situation is quite controversial).
https://en.wikipedia.org/wiki/Diatomic_carbon
EDIT: To elaborate slightly on what I mean by controversial bonding. Shaik et al have argued that this a quadruple bond 1 on the basis of Valence-Bond calculations. Basic MO theory indicates that this is a bond somewhat intermediate to a double and a triple bond (but close to a double bond). The stretching frequency and bond dissociation energy are intermediate to a double and triple bond as well, and some have argued strongly that this implies that the bond is just a double bond as expected 2. Other sophisticated calculations indicate that the bond is strongly correlated and so the question of bond order is sorta complicated and probably best described as fractional (i.e. not just double or triple or quadruple).
1: Shaik, S., Danovich, D., Wu, W., Su, P., Rzepa, H., et al. Nat. Chem. 4, 195–200 (2012)
2: Hermann, M., Frenking, G. Chem. Eur. J. 22 (2016), 4100

I think it was meant to be the vinyl cation, not carbon dimer.

In the vinyl cation, examine the geometry of the P-orbitals involved in the pi-bond and the empty P-orbital of the carbocation. You will see that the pi-system and the empty P-orbital are orthogonal/perpendicular. This means that the two pi-electrons cannot simply be dumped from the pi-system into the empty P-orbital to plug up the carbocation (orthogonal orbital systems do not normally exhibit resonance).
Now don't get me wrong, the resonance you showed can still occur. However, the empty P-orbital will remain empty, and that carbon's P-orbital previously used for the pi-bond will now be filled with a lone pair. Thus, the carbene will be a singlet carbene, which is indeed net neutral but more accurately thought of as both positive (empty P) and negative (lone pair in another P). This high-energy species might be even more unstable than the vinyl cation, in which case it may not be a significant resonance contributor.
For more, you might investigate the difference between singlet carbenes and triplet carbenes.

In addition to @Zhe 's answer, let me add a bit more information about carbenes.
Carbenes are divalent carbon species, that are highly reactive, and rarely occur as stable compounds.
Recall that carbon has an electronic configuration of $\ce{[He]}2s^2 2p^2$. In most carbon compounds, an electron from the $2s$ orbital excites and lands in the $2p$ orbital. This makes the configuration as $\ce{[He]}2s^1 2p^3$ In this excitation, carbon produces 4 unpaired electrons that form covalent bonds with other atoms.
Carbenes don't excited in the same way. As a result, the have only 2 unpaired electrons, and in addition a vacant $p$ orbital and a lone $s$ pair.
The presence of both empty orbitals and a lone pair makes them liable to both nucleophilic and electrophilic attack. This makes them unstable.

This description of the atomic-like excitations is misleading. There is no excitation necessary (although I know that this is how some people teach hybridization), the adiabatic surface from symmetrically stretching the C-H bonds in methane connects directly to the ground atomic state. Moreover, there are both singlet and triplet carbenes in which the electrons not in a bond are either coupled as a pair or not. So this whole excitation argument is sort of irrelevant for discussing carbenes. They are just molecules in which carbon forms two bonds instead of four.

Ignoring the ideas about excitation of electrons, and ignoring triplet carbenes, the hybridisation description is still slightly incorrect - in (singlet) carbenes the lone pair is not pure 2s in character, there is still hybridisation, as is evidenced by the bond angles.


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What is the importance of iron in complexes and in the process of electron transport?
What is the importance of the iron in complexes and in electron transport? This question is related to synthesis of ATP. I have read about iron-sulfur proteins, but I don't understand how this works.

I think this is a little bit broad - it's possible to write a lot based on this prompt. Are you just asking about how Fe-S clusters work in electron transport? The synthesis of ATP (I assume this refers to cellular respiration in general) involves many heme proteins as well.

and they are also important partners in electron transfer in PSI of plant photosynthesis.

You might have more luck with this on Biology.SE, tagging with biochemistry - there are quite a few actual biochemists around there who know about this kind of chemistry! (Not that I think its necessarily off-topic here, just that you're less likely to get a well informed answer)

The Fe-S proteins are part of the electron-transport chain (together with other complex molecules such as other proteins and coenzymes etc.) In my opinion, I think it would be unfair just to describe the Fe-S complex without understanding bits of the processes and other important molecules involved in this system since it involves a chain of interlinked events.
Components of the Electron-Transport Chain

Many of the proteins embedded in the inner mitochondrial membrane are
  organized into the four respiratory complexes of the
  electron-transport chain. Each complex consists of several protein
  components that are associated with a variety of redox-active
  prosthetic groups with successively increasing reduction potentials.

The mitochondrial respiratory chain consists of a series of sequentially acting electron carriers, most of which are integral proteins with prosthetic groups capable of accepting and donating either one or two electrons. Three types of electron transfers occur in oxidative phosphorylation:

direct transfer of electrons, as in the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$;
transfer as a hydrogen atom ($\ce{H^+ + e_}$); and
transfer as a hydride ion ($\ce{:H^-}$), which bears two electrons.

In addition to NAD and flavoproteins, three other types of electron-carrying molecules function in the respiratory chain: a hydrophobic quinone (ubiquinone) and two different types of iron-containing proteins (cytochromes and iron-sulfur proteins). Ubiquinone (also called coenzyme Q, or simply Q).
The electron-transport chain involves several different molecular species, including:

Flavoproteins, which contain tightly bound FMN or FAD as prosthetic groups
and which may participate in one- or two-electron transfer events.
Coenzyme Q, also called ubiquinone (and abbreviated CoQ or UQ), which can function in either one- or two-electron transfer reactions.
Several cytochromes (proteins containing heme prosthetic groups), which function by carrying or transferring electrons), including cytochromes b, c, c 1, a, and a3. Cytochromes are one-electron transfer agents in which the heme iron is converted from $\ce{Fe^3+}$ to $\ce{Fe^2+}$ and back.
A number of iron–sulfur proteins, which participate in one-electron transfers involving the $\ce{Fe^3+}$ and $\ce{Fe^2+}$ states.
Protein-bound copper, a one-electron transfer site that converts between $\ce{Cu^+}$ and $\ce{Cu^2+}$.

The Electron-Transport Chain Can Be Isolated in Four Complexes
1. Complex I (NADH Dehydrogenase or NADH–Coenzyme Q Oxidoreductase)
Complex I (also called NADH dehydrogenase) passes electrons from NADH to CoQ. Situated in the inner mitochondrial membrane,e contains one molecule of flavin mononucleotide (FMN; a redox-active prosthetic group that differs from FAD only by the absence of the AMP group) and eight or nine iron–sulfur clusters that participate in the electron transport process.
a.    Iron–Sulfur Clusters

These are redox active. They commonly occur as prosthetic groups in iron–sulfur proteins.
There are four common types of iron–sulfur clusters: Those designated [2Fe–2S] and [4Fe–4S]. Clusters consist of equal numbers of iron and sulfide ions and are both coordinated to four protein Cys sulfhydryl groups. The [3Fe–4S] cluster is essentially a [4Fe–4S] cluster that lacks one Fe. One means of identifying these clusters
utilizes the fact that their sulfide ions are acid labile.
The oxidized and reduced states of all iron–sulfur clusters differ by one formal charge regardless of their number of Fe ions. This is because the Fe ions in each cluster form a conjugated system and thus can have oxidation states between the +2 and +3 values possible for individual Fe ions.

The overall equation for the reaction is:
$$\ce{NADH +  H+ +  CoQ -> NAD+  +  CoQH2}$$
b. The Coenzymes of Complex I
FMN and ubiquinone (CoQ), the coenzymes of Complex I, can each adopt three oxidation states. Although NADH can only participate in a two-electron transfer, both FMN and CoQ are capable of accepting and donating either one or two electrons because their semiquinone forms are stable. In contrast, the cytochromes of Complex III, to which reduced CoQ passes its electrons, are only capable of one-electron reductions.
FMN and CoQ, therefore, provide an electron conduit between the two-electron donor NADH and the one-electron acceptors, the cytochromes.
2. Complex II [Succinate Dehydrogenase or Succinate–Coenzyme Q Oxidoreductase (SQR)]
Complex II, which is also the citric acid cycle enzyme succinate dehydrogenase, passes electrons from succinate to CoQ to yield fumarate and CoQH2. It does so with the participation of a covalently bound FAD, a [2Fe–2S] cluster, a [4Fe–4S] cluster, a [3Fe–4S] cluster, and one heme b560.
Complex II is important because it injects these relatively high-potential electrons into the electron transport chain.  It catalyzes the transfer of electrons to coenzyme Q.
However, its source of electrons (in other words, the substance being oxidized) differs from the oxidizable substrate (NADH) acted on by NADH-CoQ oxidoreductase. In this case, the substrate is succinate from the citric acid cycle, which is oxidized to fumarate by a flavin enzyme;
$$\ce{Succinate + E-FAD -> Fumarate +  E-FADH2}$$
The notation E-FAD indicates that the flavin portion is covalently bonded to the enzyme. The flavin group is reoxidized in the next stage of the reaction as another iron–sulfur protein is reduced:
$$\ce{E-FADH2  +  FeOSoxidized -> E-FAD _ Fe-Sreduced}$$
This reduced iron–sulfur protein then donates its electrons to oxidized coenzyme Q, and coenzyme Q is reduced;
$$\ce{Fe-Sreduced  +  CoQ  + 2 H+ ->  Fe-Soxidized  +  CoQH2}$$
The overall reaction is:
$$\ce{Succinate  + CoQ -> Fumarate  + CoQH2}$$
a. Cytochromes

Cytochromes are redox-active proteins that contain heme groups that
  reversibly alternate between their Fe(II) and Fe(III) oxidation states
  during electron transport.

3. Complex III (Coenzyme Q–Cytochrome c Oxidoreductase or Cytochrome bc1 Complex)
Complex III passes electrons from reduced CoQ to cytochrome c. It contains four redox cofactors: two b-type hemes, a c-type heme, and one [2Fe–2S] cluster. This complex -CoQH2-cytochrome c oxidoreductase (also called cytochrome reductase), catalyzes the oxidation of reduced coenzymeQ (CoQH2). The electrons produced by this oxidation reaction are passed along to cytochrome c in a multistep process. The overall reaction is;
$$\ce{CoQH2  +  2 Cyt c[Fe(III)]  -> CoQ  + 2 Cyt c[Fe(II)]  +  2 H+}$$
Cytochrome c
Cytochrome c is a peripheral membrane protein that is loosely bound to the outer surface of the inner mitochondrial membrane. It alternately binds to cytochrome c1 (of Complex III) and to cytochrome c oxidase (Complex IV) and thereby functions to shuttle electrons between them.
Complex IV (Cytochrome c Oxidase)
Cytochrome c oxidase (COX or CcO), the terminal enzyme of the electron transport chain, catalyzes the one-electron oxidations of four consecutive reduced cytochrome c molecules and the concomitant four-electron reduction of one  $\ce{O2}$ molecule to yield $\ce{H2O}$:
$$\ce{4cytochrome c^2+  +   4H^ + O2   -> 4 cytochrome c3+ + 2H2O}$$
Overview:

Summarised version:

The mitochondrial electron-transport chain. The pathways of electron transfer (black) and proton pumping (red) are indicated. Electrons are transferred between Complexes I and III by membrane-soluble CoQ (Q) and between Complexes III and IV by the peripheral membrane protein cytochrome c (Cyt c). Complex II (not shown) transfers electrons from succinate to CoQ
These Fe-S clusters are present in the complexes described above and play a big part in the electron transport chain, as they are redox active hence allow the transport of electrons along the chain through a series of redox reaction yielding $\ce{H2O}$ in the final reduction of $\ce{O2}$ molecule (producing the ATP molecules along the way). Although this is not exhaustive of the whole process it gives the "main" idea of molecules (including the so-called iron-sulfur proteins you mentioned in the question) and processes involved in the ETC. It was just impossible for me to explain only iron-sulfur proteins without getting into detailed processes
Hope this helps you
References
Biochemistry (Campbell and Farrel) 5th ed (Chapter 20)
Biochemistry (Voet and Voet) (Chapter 22)
Biochemistry (Grisham) 4th ed(Chapter 20)
Lehninger  Principles of Biochemistry (Chapter 19)


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Where does the energy of solar radiation hitting the atmosphere go?
I was reading about solar radiation, and there was a part that says that the atmosphere absorbs most of the radiation emitted by the sun. I wonder if when the atmosphere absorbs the energy, where does it go since energy doesn't disappear or that's what I've been taught in high school. I'm guessing it transforms to heat. Correct me if I'm wrong, and a explanation would be very appreciated.

Atmosphere is pretty transparent, as you might have noticed. Most of the radiation reaches us down below and gets absorbed by the plants, soil and whatever else. What little of it is absorbed by the atmosphere does indeed transform to heat.

There is some detail about which molecules in the atmosphere absorb ir radiation and why in the answer to this question https://chemistry.stackexchange.com/questions/73565/why-are-some-molecules-unable-to-absorb-infrared-radiation/73589#73589

The atmosphere also radiates heat.

It is broadly true that some solar radiation ends up heating the atmosphere. But it isn't true that the atmosphere absorbs most of the heat emitted by the sun.
Broadly there are several ways that radiation is absorbed:

Some of it (but not much of it) is absorbed directly by certain molecules in the atmosphere (carbon dioxide and water vapour are biggies), directly heating them.
Some of it is reflected by clouds and ice and other parts of the earth and reemitted to space (quite  a lot of it)
Some radiation is absorbed by plants and used to drive photosynthesis (important but not a lot of absorbed radiation)
A lot of it is absorbed by the surface of the earth (land and sea) which is the part that contributes most of the warming effect of sunlight (of course this warmer sea and earth then warms the atmosphere above it either directly be contact or by being absorbed before it is transmitted back into space, though a lot of it is emitted to space).

So, in the end, a lot of the radiation ends up as heat, but mostly via heating of earth and sea not by directly heating the atmosphere. On the other hand, the atmosphere does end up much warmer as a result. And the absorption by the atmosphere does make the difference between the earth being very cold or very habitable as a planet. The details of this are very complicated which is why people build very big computer models when they want to understand global warming which depends of the complicated details of how the atmosphere absorbs fractions of the heat.

Not sure why you think this is a chemistry problem: see https://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif, for example. Note that this energy diagram ignores both biological energy flow (e.g. photosynthesis, decay, burning) as well as geothermal energy flow (e.g. volcanoes) as well as processes which take hundreds, thousands, and millions of years (cooling of core of Earth). An isolated system is one which, by definition, energy is constant (although the form that that energy takes may and often does change). In common experience, the Law of Conservation of Energy and Mass holds. In more exotic situations we have to include strain or density as also part of the conserved quantity, but unless you're dealing with relativistic velocities, or gravities far heavier than that found on Earth, or with densities far greater than those found on and within several hundred kilometers of the surface of the Earth, you can ignore that and just remember that Mass is energy, and energy is conserved. The other exception is at cosmological scales where space is expanding but our best models suggest that each cubic meter of vacuum (of space) has a constant energy density. This means that as space expands, the total energy of our Universe is increasing. This is theoretical, and is disputed among experts in the field.


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Application of LeChatelier's principle for reactions with solid products

I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.
Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not  really a hard limitation at all.

I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.

Let's get back to the fundamentals of Le Chatelier's principle :

When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract (partially) the effect of the applied change and a new equilibrium is established.

Let's assume you have a reaction $\ce{A(g) <=> 2B(g)}$. When lowering the pressure, the system will adjust itself to counteract this, i. e. increase the pressure. This can be achieved by displacing the equilibrium towards the right, as the direct reaction increases the amount of gas molecules.
In your case, however, there is no increase in the amount of gas compounds. Regarding the solid, as it does not participate to the overall pressure in the system (only gases do), it should not be taken into consideration here. Therefore, a change in pressure wouldn't alter the equilibrium state of your reaction.
That's for the math-less part.
You can also understand this concept using the equilibrium constant. For my example, it would be $K_c = \frac{[\ce{B}]^2}{[\ce{A}]}$. Lowering the pressure corresponds to an increase in volume, thus a decrease in both concentrations. This leads to a ratio that is different from the equilibrium constant, which will force the system to a new equilibrium.
$$K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{\left(\frac{n_\ce{B}}{V}\right)^2}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V}$$
When increasing $V$ to $V'$,
$$\frac{[\ce{B}]^2}{[\ce{A}]} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V'} < K_c$$
As this ratio is lower than $K_c$, the reaction will evolve towards the formation of products.
For your equation,
$$K_c = \frac{[\ce{B}]}{[\ce{A}]} = \frac{\frac{n_\ce{B}}{V}}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}}{n_\ce{A}}$$
there is no dependance on the volume, thus none either on the pressure.
C is not included as it is a pure solid ($a(\ce{C}) = 1$).

Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not  really a hard limitation at all.

It is true that, if you find a way to remove products from the system, you can reach full conversion of your reagents. But that's actually the catch : it can be very hard to remove those products. Take as an example the Haber-Bosch process :
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
As it is a slow reaction, a high temperature is required to get the product in a reasonable time. How would you remove $\ce{NH3}$ from the system, leaving the reagents inside? We then use other ways of increasing the yield (pressure and temperature), but they do not allow a full conversion of the reagents.

Thanks Thomas. I agree that it is not always practically possible to remove products easily (ie like using a simple membrane reactor and so on), but my question was to show that at least theoretically no reaction has an equilibrium limitation such as, the limitation is more about inability to devise practical means to quickly remove products. And may be this where a challenge can be converted into an opportunity, to innovate.

Thomas and anybody else, I actually have this doubt about removing reaction products from reaction space part. In say a gas-solid catalytic reaction, or gas-liquid, isn't the product gas immediately removed from reaction space just by virtue of constant flow through the reactor? If I had  a reaction say A(g)-->B(g) happening on a solid catalyst surface in a fixed-bed reactor, as soon as B is formed it is not going to stay on catalyst surface but come out of the reactor due to gas flow anyways. So are we rather talking about SELECTIVE removal of product alone from reaction space?

@daraj To increase overall yield, you would have to circulate the reaction mixture while continuously removing products. The Haber Bosch process does that.

"Because of increase in moles, you can say lower pressure will favor the reaction going right" - no, in your equation it is the same on both sides and favors neither. But if it had been, it would have applied regardless of solid products. The volumetric influence of 1 unit of gas is significantly higher than 1 unit of solid, and as such the unit of solid will never impact the partial pressure of any of the gases. The best you can argue for it is that it can, in an enclosed reaction chamber, cause a negligible increase in system pressure.
For some combinations of reactants you can see the limitations when you draw the phase diagram - if you apply Gibbs Phase rule and have a mind to check for incompatible coexistence you will find that some reactions are thermodynamically limited - the combination of reactants necessary to produce something will instead produce something else. So if you "remove all products" as you say, you end up producing something you weren't supposed to produce.
edit: An example. (Please, thermodynamic wizards clean up my notation, it is too long since I were in school to remember what was the G$^0$'s and G's and there is probably multiple formal errors in here...)
So, we would like to make Silicon from what the earth gives us, Quartz -  $\ce{SiO2}$, and coal - $\ce{C}$
We put quartz and coal together and heat it up, we know something will happen because
$\ce{SiO2(s) + 2C(s) <=> Si(l) + 2CO(g)}$
has a positive $\Delta$G so heating it up will force something to happen. But, to our surprise, no matter how vigorously we (try to) remove the carbon monoxide and the silicon, we only find silicon carbide $\ce{SiC}$. Turns out that
$\ce{Si(s,l) + C(s) = SiC(s)}$
There are competing reactions happening, disturbing our equilibrium, consuming our product. But, chemists as we are, we blow oxygen into the mix, thinking - now we will get our silicon, because
$\ce{2SiC + O2(g) <=> 2Si(l) + CO}$
and that has a negative $\Delta$G no less, so we will regain our losses! Again, we are stymied, getting only Silicon monoxide gas $\ce{SiO}$.
$\ce{2Si(l) + O2(g) = 2SiO(g)}$
But, we are closer to our target. What if we can react these two different products together? Like...
$\ce{SiO(g) + SiC(s) <=> 2Si(l) + CO(g)}$
Well, now we got our precious Silicon, but the trouble we had to go through. And, we now reach our understanding - it is impossible to produce silicon without also producing the intermediaries. And we have to be careful how we treat the intermediaries because SiO and CO are quite sensitive and there are circumstances (like availability of C(s) or Si(l)) that can drive opposing reactions with the intermediaries.
$\ce{SiO(g) + C(s) <=> SiC(s) + CO(g)}$
It turns out that for this (these) specific reaction(s) you just simply cannot have 100% yield. The conditions that are optimal requires you remove some of the SiO and all of the CO, and you only get a 90-something % yield. This is one example of a reaction that is thermodynamically favorable but impossible to "drive all the way". I am just a sledgehammer chemist, there are certainly organic chemists that can tell you that a 90% yield is worthy of a celebration, many thermodynamically inhibited reactions yield less.

"if you apply Gibbs Phase rule and have a mind to check for incompatible coexistence you will find that some reactions are thermodynamically limited - the combination of reactants necessary to produce something will instead produce something else." you mean the reaction will be spontaneous but in a different direction? i.e it will yield different products? can you elaborate? if this is the case it should be that way from the beginning itself,not just at equilibrium

I'll try to edit in an elaboration

Stian , please also address the question I added above in response to Thomas


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Volume of anion and comparison
Recently, I came across a book in which I was solving the exercises and found out that the solution to a question says that the fluoride ion, $\ce{F^-}$, is smaller in size than the hydride ion, $\ce{H-}$.
As far I know, the hydrogen has only one shell in contrast to fluorine. Addition of a electron to hydrogen adds a electron to the 1s orbital and thus to the first shell. Similarly in fluorine, the electron is added to the 2p orbital and therefore to the same shell.
Anions are a bit bigger than their parent atom, and there is some serious electron repulsion, but how can such a repulsion make the hydride ion bigger than the fluoride ion, given that the fluoride ion also faces some repulsion within itself. Are there some other factors - or is the fact which I reported is very wrong?

The quick answer to which ion, $\ce{H-}$ or $\ce{F-}$, is bigger is "it depends". The general reasoning behind the claimed fact the hydride anion is larger than the fluoride anion typically comes down to charge density. In $\ce{H-}$, there are two electrons for the one proton, giving it a charge ratio of 2, while in $\ce{F-}$, there are ten electrons and nine protons for a charge ratio of roughly 1.1. The significantly higher charge ratio of hydride leads to a much more diffuse orbital, and a potentially larger ionic radius.
However, there are a few issues that arise, including:

How do we measure the ionic radius? Ions and atoms don't have a hard boundary; the probability of finding an electron a certain distance from the nucleus drops, but there's no firm boundary.
Ionic radii depend on the nature of the counterion

One potential way to measure the ionic radius is to look at lattice parameters of solids, determined by X-ray crystallography. This does not give much information about what the ions look like in solution, but this is not of huge importance in this case as formal hydride is unlikely to exist in solution. More importantly, X-ray crystallography gives the distance between the two ions, but it does not give the ionic radii. However, we can directly compare the lattice parameters of two solids sharing the same cation with the idea that a larger lattice parameter would be indicative of a larger anion.
For the alkali metal binary salts, the fluoride ion does typically lead to a smaller lattice constant than hydride (e.g. 4.027 and 4.083 Å for lithium salts, respectively). However, for calcium hydride and fluoride, the latter gives a larger orthorhombic cell, implying that fluoride is a larger anion.1 Indeed, for most of the analogous binary hydrides and fluorides in this reference, the fluorides produce larger lattice constants. Using different approaches to measuring ionic radii, including the soft sphere and hard sphere approaches, Lang and Smith succinctly declare that "the ions maintin the following order of size: $$\ce{Li+}<\ce{H-}<\ce{Na+}<\ce{F-}<\ce{K+}<\ce{Cl-}<\ce{Rb+}<\ce{Br-}<\ce{Cs+}<\ce{I-}$$
All the anions are larger than isoelectronic cations, but the Pauling order of ion size is not maintained."2
Therefore, while it may be true that the hydride anion appears larger than the fluoride anion in a few cases, this is not true in general.

1 J. Solid State Chem., 1970, 2 (2), 144
2 Dalton Trans., 2010, 39, 7786


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Measuring the colloidal nanoparticles stability by dynamic light scattering
In this article the ligand exchange reaction is done for thiol coating of gold nanoparticles.In the first step dicholoromethane (DCM) is used as a extraction solvent to remove CTAC(Cetyl trimethylammonium chloride) from the gold nanoparticle's surface.CTAC acts as a stabilizer for gold nanoparticles.However,only incomplete CTAC removal occurs to ensure that there is still enough CTAC present to maintain colloidal stability.The optimum of solvent extraction is determined by dynamic light scattering.
My question is "How can be measured the the optimum solvent extraction by dynamic light scattering?"

This question is very broad and probably covered by a couple of textbooks of the field. In its current state it asks too much to be a fit for our format and to attract a high quality answer. If you can narrow the scope, for example to a specific problem you face, please [edit] the question. Otherwise I assume this questions might get closed and stays without an answer.

Dynamic light scattering is a sizing technique; that is, it provides an indirect measurement of the size of nanoparticles (details may vary depending on experimental set-up).
However, it's important to note that it doesn't provide a direct measurement of the solid particle, but of its effective size when moving in solution. In other words, it measures the hydrodynamic radius of the nanoparticles - which, by the way, is not a single value but an ensemble average, which you usually measure in statistical fashion.
Now, at the nanoparticle-solution interface, you have several layers of material that is bounded to the surface and will move around with the nanoparticle, contributing to the hydrodynamic radius. In this particular paper, you have a CTAB/C monolayer (at least) which will be more or less tightly bounded to the gold particle, and also counter-ions to provide charge balance and several shells of solvation water (progressively less ordered and more weakly bound as you move away from the particle). It's the size of that whole entity that you measure (indirectly) with DLS.
As you can imagine, reducing the concentration of CTAB/C in the solution will change nanoparticle coverage - and will change the size, shape and strength of the layers surrounding the nanoparticle. As with colloidal systems, when synthesising nanoparticles there's a real concern that they can aggregate and precipitate or otherwise change their properties - which is the reason that you produce them in the presence of an stabiliser such as CTAB/C in the first place. If you look at figure 4, the histograms show hydrodynamic size for DLS measurements in different conditions and you can see they show that this aggregation isn't happening: if it did, you'd see far larger objects moving around in the solution. And as the body of the article explains, the small changes in hydrodynamic radius are probably due to the removal of CTAB/C from the nanoparticle surface, which affects solvation shells and changes the DLS-measured radius.


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Common hydrogen bonded liquids
For some experiments I need liquids with and without hydrogen bond which are pure or highly concentrated. I have only found $\ce{HNO3, CH3COOH}$ concentrated. But i need more liquids. And it would be better if they are common.
Also it's even harder to find non H-bonded liquids. There's oil but common cooking oil is a mixture, I need oils as pure as possible.

What's wrong with water?  Or (diethyl) ether?
Stand back and think about what makes things hydrogen-bonded, or not.


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Does the addition of acid change alkalinity?
I'm attempting to speed up the dissolution of calcium carbonate by the addition of hydrochloric acid. However, I'm using the calcium carbonate as a source of carbonate for a water sample. But would the addition of acid counteractively lower the alkalinity in the process, therefore lowering the carbonate concentration?

What do you mean exactly by "alkalinity" ? $\ce{HCl}$ dissolves carbonates by neutralizing them. If you need a known amount of carbonate in your solution, you cannot add any acid to it. You may want to use other methods to dissolve it, such as possibly complexation of calcium using EDTA.

Sorry, I mean the carbonate alkalinity.

Why aren't you using soda ash? It's alkaline and very soluble. Calcium carbonate is very poorly soluble in water. Check the wiki page link.

@Phanz I'm needing a source of calcium which contains carbonates. I already have a source of sodium through the addition of NaHCO3. I'm also considering using calcium hydroxide as my source of calcium and carbonates.

Calcium carbonate has a solubility of $\pu{13 mg/L}$ at room temperature. Adding acid to this solution will dissolve the calcium carbonate, by destroying the carbonate via the equilibria
$$\ce{CO3^2- <=>[+ H+][- H+] HCO3- <=>[+ H+][- H+] H2CO3 -> H2O + CO2,}$$
with the carbon dioxide having a tendency to escape the solution.


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Why does increasing electronegativity difference result in decreasing density?
I've been collecting data and there's a very strong inverse relationship between electronegativity difference and density ($\pu{R^2}$ value of roughly 0.95 from my data). Is there any chemistry that could explain these results?
This doesn't make much sense to me because I would have thought increasing electronegativity difference would result in more polar bonds and thus more polar molecules, which would result in stronger inter-molecular forces and thus smaller volume (and hence more mass being packed into a smaller volume).

The orbital size of the atoms with a negative dipole will increase relative to electronegativity , right? Since the electrons localize close to that point, I think the electron cloud would need to expand.

What substances are you talking about?


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Why don't alloys self-galvanically corrode?
Total chemistry newbie here (so this may be a fairly obvious question). With that said, I've had a niggling question that I can't quite resolve myself;
Given that galvanic corrosion is driven by the different electronegativities of different elements, why don't alloys made of significantly different elements self-galvanically corrode very quickly?
Essentially, if I were to place two electrically connected lumps in a conducting bath, one of copper and one of tin (with an anodic difference of 0.3V), I would expect them to corrode relatively quickly. Yet if I make these lumps small enough (i.e. make an alloy of them) the resulting bronze mass is fairly resilient. At what size do the individual lumps of copper and tin change from being rapidly corroding to something stable?

They certainly can. Most cases where the alloying compound is more noble this is a hazard. Nickel in Iron is a popular example, you can get corrosion acceleration on the iron by reduction happening on the Ni-surfaces.
Where the alloying element is less noble you usually have superficial corrosion to begin with but in the end the resulting surface is protected by the main element - which is common for bronze(s). If you just have lumps, then you do not have a bronze and slowly but certainly, nature will reclaim the tin.
One has to care about the homogeneity of the alloy.

Ah, so would it be reasonable to say that if I were to magically construct a "bronze" by embedding small spheres of tin in a copper bulk, that I would get similar corrosion behavior even for macroscopic (say, micrometer-sized) tin spheres? Essentially, the tin is selectively corroded eventually leaving a pitted contiguous copper surface? And should the tin content increase, would it be reasonable to expect both an actual bronze and this pseudo-bronze to become more and more sponge-like as the tin corrodes?

Yes, insofar as the "magic" part - excepting that then certainly it is "reasonable". In real life though, when creating the bronze you will during the hot parts of that process always have some solvation. If you quench it so you get $\beta$-tin in it I'm not sure what would happen but it would have poor mechanical properties.

Well, that's what I have my magic atom-at-a-time 3D-printer for =)
It's always nice when "reasonable" mental models predict reality well. Many thanks!


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What is the nature of the mysterious Kw, the ionic product of water?
In pure water, [H$^+$] = 10$^{-7}$ M and [OH$^-$] = 10$^{-7}$ M. So, we conclude pH of water is 7 and so is the pOH. We define the ionic product of water as Kw = [H$^+$][OH$^-$] = 10$^{-14}$ M$^{2}$. Till this, everything is fine.
Now, without any premise, it is concluded that the product [H$^+$][OH$^-$] of all concentrations in solutions of all sorts is the same, i.e. 10$^{-14}$.
This leads to the conclusion that for all acidic, basic and neutral solutions, pH + pOH = 14.
So, why is it concluded that [H$^+$][OH$^-$] = 10$^{-14}$ M$^2$? What is the nature of Kw, the ionic product of water, that mysteriously is applicable to all solutions?

This is just an equilibrium constant, like any other.

We derive it for water. Why is it true for all solutions?

You don't have to derive it using water and anybody who says that Kw can only be determined by measuring the pH of water is wrong. You seem to have been taught something that makes no logical sense.

We start with water as if the experiments with water led to this. Then we extrapolate to all solutions. That's what I don't understand

@ThemysteryOflife all solution have to follow this because anything with $\ce{H+}$ and $\ce{OH-}$ in it has to follow the equilibrium condition for water.

@Tyberius the question is Why?

@ThemysteryOflife: All (aqueous) solutions contain water. At equilibrium, water coexists with $\ce{H+}$ and $\ce{OH-}$ in a ratio determined by the equilibrium constant. This is true regardless of the presence of other solutes.

Very roughly speaking, given that pure water has an unusually high concentration of 55.5 mol of water molecules per litre, to a first approximation you can consider the vast majority of common aqueous solutions (up to about 0.1 mol of solute per litre) to be "pure water plus some rounding error".  Thus the ionic product of water is essentially unchanged. For more concentrated solutions, though, the solute does interfere with the ionization of water. See the wiki page.

Does this answer your question? Why does the ionic product of water remain constant after addition of non-neutral solute?

I think this is the answer you are looking for: https://chemistry.stackexchange.com/questions/7905/ionic-product-of-water

The solution product for the reaction $\ce{H2O -> H+ + OH-}$ is a constant, $10^{-14}$, which we call $K_w = [\ce{H+}][\ce{OH-}]$. To make any aqueous solution, we put a solute into water. Because the dissociation constant (which is basically an equilibrium constant) of a substance only changes with temperature, $K_w$ remains the same. Thus, for any aqueous solution at a given temperature, $K_w$ is constant.
You need to treat $K_w$ as an equilibrium constant. For example, if I have a reaction $\ce{A -> 2B + C}$, $K_{eq} = \frac{[\ce{B}]^2[\ce{C}]}{[\ce{A}]}$; let's say I measured at equilibrium $[\ce{B}] = 1$ M, $[\ce{C}] = 2$ M, and $[\ce{A}] = 0.1$ M . This means $K_{eq} = 20$. Now, if I increased the concentration of $[\ce{A}]$ to $5$ M, $K_{eq}$ wouldn't change: the concentrations would have to adapt to my change. This is the same for water: once I determine $K_w$, it is the same for all concentrations of $\ce{H+}$ and $\ce{OH-}$.

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.

@QingQuan Xia This is not what I meant

Kw remains same, I understand. It is applicable for all solutions, I don't understand


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Why bond energy of acetone is more though it is more polar than formaldehyde?
I was going through the general properties of carbonyl compounds and then found this (in the pic) data, where it says that dipole moment of acetone (if we assume $\ce{R = -CH3}$) is more than acetaldehyde which in turn is greater than that of formaldehyde.
But I think, as $\ce{-CH3}$ is a $+I$ effecting group, so it will satiate the residing positive pole on carbon slightly and hence will make the $\ce{C=O}$ bond less polar and hence the sequence should be reversed!

Please anyone make me correct where am I mistaking with the concept?
Picture source
EDIT :- I got my answer of the above confusion. Thanks to Physicsapproval.
But with that I came to face another confusion. In the next part of the text (see the paragraph just below the pic in the Picture source) I found that "the larger the dipole moment the greater the polar character of the carbonyl group". So that means, $\ce{C=O}$ bond in acetone is more polar and least in the case of formaldehyde! Now in the next part it is mentioned that the bond energies of $\ce{C=O}$ bond in the mentioned compounds are: $$ \ce{H2C=O} \, ; 170~\mathrm{kcal/mol}$$ $$\ce{RHC=O} ~ ; 175~\mathrm{kcal/mol}$$ $$\ce{R2C=O}~; 180~\mathrm{kcal/mol}$$
Now I know that the more polar the bond is, the less is its bond energy. So, in this case why the trend is reversed, i.e., $\ce{C=O}$ bond of acetone being more polar has the highest bond energy compared to other two?

I think that is because -R group behave as electron donating like group, so the two vectors would further add up to make the dipole moment even greater.

Yes, thank you @Physicsapproval. I just forgot to consider that dipole moment is a vector quantity!

the more polar the bond is, the less is its bond energy

I'm not sure where you are getting this from.
Take Acid halides as an example series. As the bond get more polar ( HI (0.4 D) < HBr (0.7 D) < HCl (0.9 D) < HF (1.9 D) ) the bond energies follow accordingly ( HI (295 kJ/mol) < HBr (362 kJ/mol) < HCl (428 kJ/mol) < HF (565 kJ/mol) )
As the charge separation gets larger (and therfore more polar) there is a stronger will to not have that large charge separation, as nature wants to naturally come to an equilibrium - and thus the bonds are more energetic.
Does this clear up your confusion?

Thanks for your response @ChildishJack. As far as I got your point that you want to say, the more the charge separation, the more the bond tends to be ionic and hence the more is its bond energy. Yes I agree and your data too substantiate that logic. But then how would you describe the fact that ClOH is an acid and NaOH is a base, where In case of ClOH, Cl–O bond polarity < O–H bond polarity, so O–H bond dissociates easily and hence it's an acid and in case of NaOH, Na–O bond polarity > O–H bond polarity, so the Na–O bond dissociates easily and hence it's a base. Aren't both data contradict?

Contradict what? Your statement about the correlation between bond polarity and bond energy isn't always true. The opposite statement isn't always true either. There is simply no such rule, one way or another. There is nothing to contradict.

Also, electrolytic dissociation is only distantly related to bond energy.


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Why is the distance between the atoms shorter in the dicarbon anion compared to the cation?
In general, the anions have bigger ionic radii than their corresponding neutral atoms, and the cations have smaller ionic radii then their corresponding neutral atoms. How come that does not apply to $\ce{C2-}$ and $\ce{C2+}$?

What other anion/cation are you talking about? And are they mono-atomic or diatomic?

The distance between two C atoms has nothing to do with the ionic radii in the first place. It is a covalent bond.

@IvanNeretin how do I then determine the distance between the C-atoms? or at least which on is bigger?

You consider the molecular orbitals and deduce the bond order. The higher it is, the shorter is the bond.

I assume you are concerned with the diatomic molecules/ions; could you provide some data for your claim?

Looking up in a current compilation of bond lengths in the International Tables for Crystallography (IT), apparently there is indeed not so much deviation in the distance between two adjacent carbon atoms in a $\ce{C(sp)#C(sp)}$ pattern:

(IT, 2006 edition)
Some hints to render this excerpt easier to access.  The IT indeed discern different bond types (13 for all combinations of $\ce{C-C}$,  and 625 for all organic / organometallic compounds), and consequently, the source offers separate listings for, for example, $\ce{C(sp^3)-C(sp^2)}$ or $\ce{C(sp^2)-C_{ar}}$, too (first column). The second column is about the structure around the pattern in question highlighted in bold. For the $IT$, $\ce{C_{ar}}$ explicitly represents an aryl C in an six-membered ring; $\ce{C^{\#}}$ any $\ce{C(sp^3)}$ - linearly substituted, or not.
The third, fourth and fifth column are the unweighted sample mean $d$ in [Å], sample median $m$, and sample standard deviation $\sigma$, followed by the lower and upper quartile, $q_l$ and $q_u$, as well as the number of observations $n$.  Overall, the compilation is based on a subset of 10 324 X-ray and neutron diffraction data found in the Cambridge Crystal Database by September 1985, judged to be "good enough" (the source offers additional insight about the criteria applied).


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Long-term storage of Ascorbic Acid mixed with Sodium Bicarbonate
Every day I mix two parts of ascorbic acid powder with 1 part of sodium bicarbonate powder and place the resulting powder into an airtight container. I have noticed that if the resulting powder is mixed with water within 30-60 minutes, it fizzles indicating the release of carbon dioxide as a byproduct of the acid-base reaction. However, if I leave it untouched (within a dry and airtight container) for 6 to 8 hours, it hardens. If I crush it (so it turns into powder again) and mix it with water, it no longer fizzles.
My questions are:

Why does it harden and no longer fizzles if mixed with water after a few hours? Does the acid-base reaction take place even without adding any liquids to the mixed powder?
Most importantly, does the reaction that causes the hardening and absence of fizzling cause any kind of degradation (e.g., oxidation) to the ascorbic acid? In other words, can I mix ascorbic acid with sodium bicarbonate (both in powder form), store the resulting powder in an airtight container and consume over a few months? Would that diminish the efficacy of the Vitamin C in any way?

Thank you in advance.
Update:
I have bought sodium bicarbonate from a different manufacturer and now after mixing it with ascorbic acid, nothing happens until water is added. It can sit for 24+ hours, without hardening and it will still fizz when mixed with water. I suspect the sodium bicarbonate from the previous manufacturer was slightly moist  and this one is completely dry.

"Every day I mix two parts of ascorbic acid powder with 1 part of sodium bicarbonate powder and place the resulting powder into an airtight container. " Just curious...why do you do this?

I take ascorbic acid as a vitamin C supplement. Its acidity tends to upset my stomach, so I add sodium bicarbonate to neutralize it.

The reason that your solid mixture is clumping and hardening inside your airtight container is that the reaction between the two compounds is producing water as follows:
$$\ce{C6H8O6 + NaHCO3 <=> NaC6H7O6 + CO2 + H2O}$$
where $\ce{C6H8O6}$ is ascorbic acid.
This reaction proceeds rapidly in the presence of excess water, which allows the reacting species to dissolve and thus have ready access to each other. This results in the observed fizzing from the release of carbon dioxide.
In the solid phase, the initial reaction is limited to the surfaces of the particles coming into contact with each other. As the reaction proceeds, the once purely solid material becomes wet via the acid-base reaction producing water, and the remaining materials can somewhat dissolve, facilitating migration of the reacting species so that on a timescale of hours all of your material has reacted.
Regarding the stability of storing the solid mixture vs. storing the solids separately, you are probably better off storing them separately until use. Both ascorbic acid and sodium ascorbate are susceptible to being oxidized by long-term exposure to air. Although water itself shouldn't act as an oxidant here, it could facilitate reaction with oxygen in the air in a similar manner in which it facilitates the acid-base reaction, as discussed above.

The sodium bicarbonate (a base) is reacting with the ascorbic acid to form sodium ascorbate and carbon dioxide, as well as water. The carbon dioxide is responsible for the fizzing you note when you dissolve the mixture in water. When you mix the two solids, the reaction is relatively slow, since the reaction must take place at the interface of the two solids; however, it appears that allowing it to sit overnight is ample time. The hardening is probably due to the release of water, which causes to solid to aggregate; the solid will no longer bubble in water as it has already reacted and released its carbon dioxide.
As for the fate of the ascorbic acid, the sodium salt is stable, and will be converted back to the free acid in the presence of another acid. Just don't expect it to fizz.

Thank you for your detailed answer. I did not know the reaction also released water. I read that mixing ascorbic acid with water causes it to oxidize in a few hours. Wouldn't the water releases by the reaction oxidize the ascorbic acid (or sodium ascorbate)? Sorry for my ignorance, I'm a complete dummy when it comes to chemistry.

Water isn't (typically) an oxidant; however, the sodium salt of ascorbic acid is likely to be more easily easily oxidized than the free acid. Both ascorbic acid and sodium ascorbate are often used as an antioxidant additive in foodstuffs, where oxygen typically acts as the oxidant.


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How is a metal's relative resistance to oxidation relate to it's position in the reactivity series?
How does a metal's relative resistance to oxidation relate to its position in the reactivity series?

Hello and welcome to Chemistry.SE. Please feel free to visit the [help], and in particular read through the homework policy. You can then [edit] your question accordingly so that you have a chance of getting a good answer. As it stands, the question is likely to be closed.

I have edited my post and removed excess information

The oxidation series is a means by which one can order elements as to how easily they are oxidized or reduced. In the attached picture, we see the blue arrow- "Ease of oxidation increases." What this means is, as you go up the series, from gold up to lithium, elements are more easily oxidized, or they lose their electrons more easily. This is equivalent to their being a better reducing agent. Now, lithium is first on our table. What this means is that, out of all the elements on this specific series that I chose to include, lithium loses electrons the best. You may be wondering why hydrogen is included in this list. Well, hydrogen appears in many single replacement reactions, such as $\ce{Zn}+2\ce{HCl} \ce{->}\ce{ZnCl2}+\ce{H2}$. In this reaction, we see that zinc has replaced hydrogen's position. Now, looking at the activity series, we see that zinc is indeed more easily oxidized than hydrogen. This is the basis for single-replacement reactions. To determine a single-replacement reaction's viability, we see if it is possible for the lone reactant to replace the bonded reactant by each determining their relative reactivities. These are also in the category of redox reactions. To answer your question specifically, as we go up (or down, depending on the chart) the activity series, the ability for an element to be oxidized is increased (or decreased, again, depending on the chart). Now you may wonder why these elements are arranged in such a way: there is no rhyme or reason. That's just how it is. I highly recommend against memorizing the position of every single element on the series, or even more than 5. What you should do is know that the group 1 and 2 elements are very easy to oxidize, and the typical "jewelry metals" are the hardest to oxidize. Indeed, we see applications of this in the real world. In cars, there are apparati called catalytic converters. These typically have platinum components, due to the low reactivity of platinum. Gold is used in many microwaves, within the emitter, again, due to its unreactivity (among other things).
In short, it depends on which table you use.
The activity series I was referring to:


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How can aqueous KOH give alcohol by SN2 of alkyl halide?
I ask this because SN2 requires a strong nucleophile, and I have read $\ce{OH-}$ isn't a strong nucleophile in a polar protic solvent like water. So, how can $\ce{OH-}$ replace, say $\ce{I-}$ from $\ce{CH3I}$ (methyl iodide)?

Even though its weaker in protic solvents, I won't say its  not a strong nucleophile.

With methyl iodide there is no opportunity for E2 elimination. Unhindered primary halides (Cl,Br,I) favor SN2 over E2.

When you are considering a particular SN2 reaction, you have to consider the nucleophilicity as well as the nature of the leaving group of both the groups.
Here as you mentioned, $\ce{OH-}$ and $\ce{I-}$  (and other halide ions, except $\ce{F-}$, which is a very bad nucleophile) have comparable nucleophilicity in protic solvents. But, as for their leaving nature, $\ce{I-}$ (as well as other halides) is far more better leaving group than $\ce{OH-}$, because we know weak bases are better leaving groups. That is why the reaction is feasible.


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Conductive hydrophobic material
I was looking for a conductive hydrophobic material if there is one? If not why? I was researching and I found some material about laser etching on metals that causes superhydrophobicity but that I don't think will work if I wanted a curved surface since its roughness would be smoothed out I think. Please correct me if I'm wrong.

How about electrically conductive polymers?

What's the point? Typically people would just take some conductive material and cover it with some hydrophobic material.

@IvanNeretin "Science isn't about why. It's about why not".

@airhuff thanks that works great! Sorry forgot to reply.

@IvanNeretin I needed the outside of the material to be conductive as well


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how to remove products from reaction space?(to apply LeChatelier's principle)
I thought this question merited  a discussion of its own. In industrial applications, we talk of shifting equilibrium of reaction to the right(or desired products) by removing products quickly from reaction space? I want to know what this means? If you consider a simple fixed-bed/packed-bed reactor with gaseous reactant and product (say A-->B), does it mean I need to remove B quickly from the packed-bed reactor volume to push equilibrium to the right? My point is, isn't that already happening? By virtue of flow through the reactor isn't B already constantly being "removed" from the reactor(along with unreacted A), while the reactor keeps getting hit with fresh A? so what more effort is needed to push equilibrium to the right? are there any timescales involved? or does B have to be removed quicker than A? if so, how quicker? I need to understand this better,please help. This question can apply to other types of reactors as well like fluidized-bed or moving-bed and so on..

Yes, this is precisely what's already happening. You can't remove B quicker than you supply A, nor do you need to.

"You can't remove B quicker than you supply A, nor do you need to" why? If I somehow manage to remove B quicker than my rate of supply of A, wont it be beneficial? I don't understand your point. Membrane reactors sometimes work on this principle by selectively removing products alone from reaction space. (also quicker, means how quicker?)

The scheme you ask about is effectively the same as $\ce{A <=>[k_1][k_{-1}] B ->[k_2] C}$ where $k_2$ measures the effect of removing B making C the removed product. The actual reaction is much more complicated as presumably mixing/diffusion has to be taken into account. Leaving aside these complications  as $k_2$ increases then the amount of B in the reaction reduces and the amount of C increases if measured at a fixed time. As long as $k_2$ is not zero eventually all B is removed, you can see this in the figure which shows the amount of C for different rate constants $k_2$  assuming the initial amount of A = $1$ and $k_1= 1, \, k_{-1}=3$. When $k_2=0$ (top figure) no product is formed and A and B come to equilibrium. The lower figure show the effects on forming C by removing B.

porphyrin, you are answering a different question, not what I asked. The reaction is just A<-> B, that's all. Where did C come from? What Iam asking is a fundamental question applicable to any reversible reaction, for that matter. I am trying to understand what does "removing product quickly from reaction space" means in the context of applying Lechatelier's principle. I believe it has something to do with ensuring that the reactor volume/bed is always concentrated in A and not B as much so as to push equilibrium to right, but I need a more detailed explanation

I think it is what you asked. The C is the removed product (call it B(removed) if you want) so that we can consider the loss of B to C as this removal step. Thus C then is the  accumulated product separated from the reaction mixture. I'm assuming first order removal rate constant, for simplicity, if not first order the detail will differ but the trend will be the same. All the detail is in the rate equations for the reaction  proposed.  You can see as removal rate k2 increases both  [A] and [B] fall.


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Relaxed scan of Cp-M torsion angles of piano stool compounds
As part of a conformation analysis of a library of piano stool complexes to find global minima and TS, a relaxed scan where the legs of the piano stool complex rotated about the Cp-M bond need to be performed (Here a $\ce{[(Me-Cp)RuClH(CO)]}$ is given here to illustrate the task in question) :

However it was found that Gaussian 09 or 16 does not allow dummy atoms in redundant coordinates and cannot build dihedrals where one of the atoms is a dummy atom as far as Generalised Internal Coordinates (GIC) goes, because atoms are not considered an internal coordinate. Also unlike this question, since the rotation has to be done about the Cp-M bond,the workaround in that question will not work.
In addition, none of the existing dihedrals can be added together to give a dihedral where the middle bond will be aligned to the Cp-M axis. This means, the dummy atom for the centre of the Cp is necessary.

What are the possible workarounds that can achieve an outcome equivalent to the relaxed scan of the Cp-M dihedral?

Are there other chemistry packages that allow relaxed scan using dummy atoms or (better) user defined rotation axes?

I tried, I tried very hard. Even with the awesome possibilities of GIC, this does not work. You quickly surpass the number of imposable constraints when trying to fix an axis (and plane) and then use the arccos to scan the angle. The definition worked, but the scanning failed in predicting the next geometry.

I would use the dummy, define the 3 X 10 11 (16 and 18 too) dihedrals in the connectivity matrix, run the optimization and later write a shell script for running subsequent calculations in separate files replacing the value of the previously defined dihedrals. So I guess my solution is only slightly above doing it by hand but so far I haven't found any software that does this and G09 has no workaround.
I hope this helps