text:
Biochemical conversion of ethanol to acetic acid
I have questions about the involvement of the electron transport chain (ETC) in the biochemical conversion of ethanol to acetic acid. In which of the reaction steps is the ETC employed?

I just made a question from your water comment..    as it merits a separate question  http://chemistry.stackexchange.com/questions/69388/in-the-production-of-vinegar-where-does-water-come-from     Hopefully somebody here will know/post an  answer.  You can comment on it if you want, though i'm not sure if 16 points will be enough rep to, but if not and you want to comment then you could do so here as it doesn't require much rep to comment on your question or an answer to your question.

The principal conversion of ethanol to acetaldehyde takes place in the cytosol of liver cells, no ETC involved directly. This is mediated by the enzyme alcohol dehydrogenase (ADH). However, the NADH produced as a consequence would indeed deliver its electrons to the ETS, an electron transport chain.
The acetaldehyde is then metabolized to acetate in the mitochondria. Again, the involvement of the ETS is in accepting electrons from the NADH that is also produced in this step.
Here's the details: https://pubs.niaaa.nih.gov/publications/arh294/245-255.htm

Also further, acetate can react with Co-enzyme A to form acetyl coenzyme A. Which is used in a reaction with oxaloacetate to form citrate. Coenzyme A will form a thioester bond and can later be used in a reaction in conjunction with NAD+ and alpha-ketoglutarate to form succinyl-CoA and NADH.

@Bob. True. I limited my response to the OP's purview.

Of course! I just thought I'd add a little extra since I personally would have also been curious about where the carbon goes.

Many thanks guys for the input. I really appreciate it.

Also There's a water molecule in the second reaction where acetaldehyde goes to form acetate. Where does this water come from as it is not produced in the first step.

Also I notice that link mentions the liver, i.e. human cells doing it. And human cells of course have mitochondria which have an electron transport chain. But does this answer cover the case of for example, the conversion of alcohol that is done in the production of vinegar,  where the conversion is done by the bacteria Acetobacter aceti?

@Lobell this may help http://chemistry.stackexchange.com/questions/69388/in-the-production-of-vinegar-where-does-water-come-from

@Lobell. The water is abundant in the aqueous solutions these reactions occur in.


text:
Liquid state oxygen
In liquid state oxygen molecules are held together by what? Van der Waals forces or covalent bonding?

This should be by 'Van der Waals' forces and no different to other liquids formed by molecules without a dipole. As the molecules have no dipole the interaction should be due to the induced-dipole induced-dipole energy (or dispersion energy) between two similar molecules and which  is proportional to $\alpha^2I/r^6$, where $\alpha$ is the polarisability, I the first ionisation potential and r the separation.
The dispersion interaction is quantum mechanical in origin but can be explained intuitively as follows; although the dipole is zero on each molecule its instantaneous dipole is not. The overall dipole is zero  because the time average of the instantaneous dipole is zero. This dipole is caused by the 'instantaneous' position of electrons about each atom's nucleus. However, if there is a nearby molecule the instantaneous dipole moment in one molecule can generate an instantaneous dipole moment in the other. The resulting interaction gives rise to an attractive force between the molecules whose time average is not zero.

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Charge on Nitrogen in NCl3
What is the charge on $\ce{N}$ in $\ce{NCl3}$? I want to know if it is positive or negative. I feel it should be positive but I am not sure.

Judging by the hydrolysis pathway, it is negative.

@IvanNeretin Can you elaborate?

NCl3 is said to hydrolyze into NH3 and HClO.

But if we look at the structure, Cl being more electronegative, should "pull" electrons towards themselves therefore creating a positive charge on Nitrogen.

Electronegativity is hardly a precise measure of anything.

@Ivan Neretin But electronegativity plays a dominant role in understanding beginner chemistry?

Maybe. So what?

First of all charge of atom is not well defined (maybe except AiM). Formal charge of both is zero, but that does not help us to understand chemistry. Above mentioned reaction shows that nitrogen is more negative, and if you do simple QM computation both Mulliken and NBO gives ~-.44 for N and ~+.147 for Cl. If you want to have a bit feeling about it, think that way: both have similar electronegativity but Cl is bigger so it's easier to polarize.

If we want to determine which side of a bond carries a partial positive and which side a partial negative charge, we can attempt to react the substance in question with a nucleophile or electrophile. Nucleophiles will always react with the partially positive component and vice-versa.
We have two nucleophiles that we could use and the reaction of which with $\ce{NCl3}$ are known: $\ce{Cl-}$ and $\ce{H2O}$. The reaction equations are given below:
$$\begin{align}\ce{NCl3 + Cl- + H+ &-> HNCl2 + Cl2}\tag{1}\\
\ce{NCl3 + H2O &-> HNCl2 + HOCl}\tag{2}\end{align}$$
Both show that chlorine carries a positive partial charge.

Charge on an element is given by the oxidation number hence the total charge on the compound is equal to 0 then the charge on nitrogen + chlorine must be equal to 0.
N+3(valency of $\ce{Cl}$) =0
N+3(-1)=0
N-3=0
N=+3
Hence the charge on nitrogen in $\ce{NCl3}$ is +3

1st of all atomic charge is not an oxidation state. Even if it would be, then your whole argumentation is based on assumption that Cl is -1. Unfortunately both these elements have nearly equal electronegativity so this assumption is out of blue. Sorry world is not black and white. There are a lot of shades of gray.

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What is the seventh plane of symmetry in benzene?
We get the first six planes of symmetry by joining the opposite edges and vertices. But what is the seventh plane of symmetry?

Benzene is flat... (hint)

Have a look at this picture which shows some of the symmetry axes and planes. You should be able to see other similar ones. Taken from www.molecule-viewer.com.(You can also use this site to practice working out point groups.)

Complementary to the answer by @porphyrin, you may train yourself on the manifold ressources about symmetry at Otterbein university.  All you need is a device capable to handle javascript (which is not java) http://symmetry.otterbein.edu/


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Sample preparation for UV analysis of drug content in capsules/tablets
I want to analyse the drug content of a sample of gelatin capsules filled with active drug (declared dose is known) and excipients such as lactose monohydrate, microcrystalline cellulose etc. with UV spectroscopy. However I have a question about the sample preparation of my capsules. So first what I want to do is dissolve the capsule content in an organic solvent (such as ethanol, chloroform etc.) because the solubility of the drug is bad in water, and then dilute the samples to a concentration which is suitable for UV analysis. The problem is that when I dissolve the content of the capsule, I'll likely get a suspension where the excipients won't dissolve in the organic solvent. This would interfere with my UV analysis. So I was wondering if it would be a good idea to centrifuge the samples, or just wait for it to precipitate on its own? And if centrifuge, would it be better to first centrifuge the stock solution and then dilute or first dilute and then centrifuge the acquired diluted solution? Thanks in advance.

This is a non trivial problem because some quantity of solution will be lost with the precipitate. A possible better way to do this is to apply your procedure to a sample where the exact composition is known. If the concentration is low enough, Beer's law will allow you to determine the amount relative to the reference.

I'd agree with Zhe that "this is a non trivial problem." Method development is tricky if you don't have a standard analysis procedure.
When you dissolve the capsule, if you do have suspended particles then yes they will interfere with the UV analysis. So do centrifuge.
But think of the removal of the liquid as an extraction. So add more solvent to the centrifuge tube, shake up the particles and centrifuge again. Assuming that you won't over dilute, I'd do the extraction three times, then dilute to the final volume.
The rub here in my mind is that UV is not very selective. I'd doubt that you're going to get a nice flat baseline with one nice peak. So you'll need a "pure" drug sample to compare to your extract samples to see if there is any other gunk causing interference.

Thank you. And yeah I do have a reference solution of the pure drug form which I am going to measure aswell, forgot to mention.


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Cl-A-Cl bond angle discrepancy
In 2,2-dichloropropane the Cl–C–Cl angle (108.3°) is less than the C–C–C angle (113.0°).
In Me$_2$TiCl$_2$, in contrast, the Cl–Ti–Cl angle (116.7°) is larger than the C–Ti–C angle (106.2°).
How is this discrepancy explained?

This Related to the Electro negativity of atom and bond length character.


text:
Can somebody help me understand this sentence?
"The molar enthalpy of vaporisation of a liquid is the enthalpy change when one mole of the liquid changes into one mole of its gas at the boiling temperature."
 The part which i am not sure of is the one highlighted above. Does it mean that the temperature of the flame that is supplying the energy needed to vaporise a substance is at the boiling temperature of the liquid ? Or is the liquid itself at its boiling point ? Because i cant imagine how the temperature of a liquid could be maintained at its boiling temperature so i guess its the temperature of the flame right ?

The temperature of the flame has absolutely nothing to do with this except that the flame must be hotter than the boiling point of the liquid. // The enthalpy of vaporization is the amount of heat necessary to convert one mole of liquid to one mole of gas given that the liquid is already at the boiling point. In other words the heat to rasie the liquid from room temperature to the boiling point doesn't count in the heat necessary for the liquid to vapor transition.

but if it is already at its boiling point then the liquid is already boiling and doesnt need any energy to turn into vapor so the definition is "the amount of  energy needed to keep a mole of a liquid at its boiling temperature until it is converted to a mole of gaseous molecules"   ?

That is essentially correct. // You have to realize that all of the thermodynamics terminology hangs together. There are all sorts of problems hiding in simple explanations.

It doesn't have to be the boiling point.  The enthalpy of vaporization applies to converting a saturated liquid to a saturated vapor at any temperature and corresponding vapor pressure.

I'll give it a try :-)
There is always a transition of liquid to gas till the system gets to equilibrium by rault's law which is true at every temperature.But at a specific temperature all the liquid will go to the gas phase, that is the boiling point. So, the molar enthlapy of vaporisation of a liquid refers to the enthalpy change of one mole of liquid that changes into gas at the boiling temperature of that specific solution.

I don't think the OP was referring to a chemical mixture.

You need to understand that the thermodynamic quantity being measured (the heat of vaporisation) is a separate idea from the heat capacity of the liquid.
If you took a liquid at well below its boiling point and applied heat until it was turned to vapour, some of the energy would be used to heat the liquid. The energy required to vapourise the liquid would be overestimated if you just measured the total energy input.
The point of the sentence is to separate the two thermodynamic concepts. It take energy to warm a liquid and it takes energy (a different amount) to vaporise the liquid. These quantities are different and should not be confused.
The thermodynamic concepts are ideal concepts and not strictly an explanation of what happens when you boil a liquid (unless you do it very carefully and slowly). In the real world a boiling liquid will not be evenly heated and some of it won't be at the boiling point. the thermodynamic concept is about the ideal experiment where you do everything slowly and ensure perfect mixing. In this case the liquid will reach the boiling point and any extra energy added will cause vaporisation rather than a mixture of warming and vaporisation.
The temperature that matters here is the temperature of the liquid.


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Production of fluorine gas
I have to suggest a method in order to produce $\ce{F2}$. It also says that I should consult the redox potential values. I know that there is a way to produce fluoride which is:
$$\ce{K2[MnF6] + 2 SbF5 -> 2 K[SbF6] + MnF2 + F2}$$
I also know that it can be produced by the electrolysis of non-aqueous halides ($\ce{KF}$).
Why do I have to check the values of redox potential? I know that $\ce{F2}$ can't be produced by the electrolysis of aqueous solutions because it oxidizes water.
Is this method correct?
\begin{align}
\ce{2 K+ + 2 e- &-> 2 K} \\
\ce{2 F- &-> F2 + 2 e-} \\
\hline
\ce{2 K+ + 2 F- &-> F2 + 2 K}
\end{align}

RE:  "Is this method correct?" // You jumped to the ending without telling the story. How do you propose to get $\ce{K+}$ and $\ce{F-}$? (I have real nice method in mind...)

Very related: http://chemistry.stackexchange.com/questions/44296/fluorine-preparationhow-can-this-be-explained-on-basis-of-lewis-acid-base-react

It is a Lewis acid-base reaction.

My question is not about the first reaction. Also, using KF helps increasing conductivity. For example, when we want to achieve industrial production of Cl2 we can use the electrolysis of melt NaCl. I assume that I have to consult the values of redox potential in order to decide who gets the electrons. But I can't figure out what I should use in order to produce F2.

I don't know if the production of fluorine gas has a similar method like the production of Cl2 which is: Cathode(-) : Na+  + e- → Na  and Anode(+) : 2Cl- → Cl2 + 2e-    and after electrolysis we get the Cl2. But I don't know that I will use for anode and cathode.


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Books on Chemical Reaction Theory
I got interested in Chemical Reaction Theory by studying the lecture notes of Feinberg and related papers for my bachelor thesis and want to dive deeper into that subject.
Are there any books you would recommend for doing so? I would really appreciate it. I'm looking for something which emphasizes the mathematical aspects but also gives good actual chemistry examples or references.
So far I found a book by Peter Érdi & Janos Tóth from 1989 called "Mathematical Models of Chemical Reactions: Theory and Applications of Deterministic and Stochastic Models", which does just that. However I fear it is a little outdated as it seems there has been quite some progress in that field recently. So maybe there isn't even a book out yet as the theory seems relatively fragmented.
Do you know more?

The book you mention is tough, besides books just on chemical kinetics (of which there are many) for a more fundamental approach try (a) Steinfeld, Fransisco & Hase, 'Chemical Kinetics & Dynamics' (b) Levine & Bernstein 'Molecular Reaction Dynamics and Chemical Reactivity', (c)Eyring, Lin & Lin, 'Basic Chemical Kinetics', (d) for a statistical approach Gillespie 'Markov Processes' which describes algorithms for one of the best ways to simulate complex chemical reactions.

Thank you. Yes, I'm a mathematician and interested in a more fundamental approach. The book by Steinfeld, Fransisco and Hase looks really good. I will work with it and the book by Èrdi and Tóth.

The keyword you are looking for is "reaction dynamics", that is the name of the branch of physical chemistry that does molecular level modelling. The Levine book mentioned in the earlier answer is a good one.  Another one I'd suggest is Miller's Dynamics of Molecular Collisions. But in fact, what I'd say is to look at research articles in this topic. I really do suggest you run through some selected papers from William Miller - he  is probably the most influental researcher in the modern field of reaction dynamics.

Thanks. I will look into him. Most of the stuff that I know so far is from papers anyways :)

Personally I did not like the Levine book. The writing style was not for me, and the heavy use of really, really, really long footnotes was extremely inconvenient... If you need to put half a page of information in a single footnote, clearly it's important, so just include it in the main text!

If you are up for some fun and want to learn about unimolecular reactions (e.g., rearrangements decompositions after energy absorption) then you should try Theory of unimolecular and recombination reactions by RG Gilbert, SC Smith. Oxford: Blackwell Scientific Publications (1990).


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How does the polarity of the eluent and sample affect the Rf value in thin layer chromatography?
A three-component mixture (X, Y, Z) was separated using normal phase TLC. Three eluents of varying polarity were used. Select the TLC plate that was developed using the least polar eluent (click on the corresponding TLC number 1, 2, or 3 on the bottom of the plate). On TLC plate 3, select the least polar compound (click on X, Y, or Z spot on plate 3).

Does normal phase TLC mean silica gel TLC? If this is the case then plate 3 has the least polar eluent because the compounds are at the top. For the same reason compound Z would be the least polar.
The term eluent means solvent doesn't it.

Hello and welcome to Chemistry.SE. If you have any questions about the site, I recommend taking the short [tour].

Normal phase as opposed to reversed phase where polar substances have higher $R_{f}$ values. Eluent is a fancy name for solvent in chromatography.

@Zhe there was confusion in the answer given below.
We're told each unknown eluent has different polarity, but it's normal phase TLC. Based on that we should be able to infer from the compound positions on the plate. In normal phase TLC the lowest Rf will always be the most polar correct? Compound polarity is now known.
Then for eluent polarity, it's the opposite of compound polarity isn't it? All three compounds in plate 2 is polar, so the eluent should be less polar/nonpolar.

In normal phase TLC, the lowest $R_{f}$ value corresponds to the substance with the highest affinity for the silica substrate. Generally, these are the most polar compounds, but less polar alcohols might also have high affinity due to hydrogen bonding. Furthermore, in general, higher eluent polarity just increases the $R_{f}$ for all eluates. The scaling isn't perfect because eluent-eluate interactions can be complex.

@Zhe Well with the reasoning used, if you check the discussion below, I'd have to say plate 2 has the least polar eluent given all the polar compounds packed together. And on plate 3, compound z is the least polar.

Never in my years in the lab have I seen a more polar eluent that causes a lower $R_\mathrm{f}$ value than a less polar eluent. It also does not make sense theoretically.
TLC 2 is obviously the one with the least polar eluent and compound $z$ is the least polar compound.
I fully disagree with half of MaxW’s proposed answer.

You must learn the terminology to follow along in chemistry. If in doubt Google the term...

Eluent - As Zhe noted "Eluent is a fancy name for solvent in chromatography." It can be a gas or a liquid. For TLC or HPLC a liquid is used. For GC a gas is used.
Normal phase TLC - For normal phase TLC the stationary phase is polar and the mobile phase is typically non-polar or only moderately polar. Polar molecules are attracted to the stationary phase while non-polar compounds are carried along by the mobile phase.
Reverse phase TLC - For reverse phase TLC the stationary phase is non-polar and the mobile phase is typically moderately polar to strongly polar. Non-polar molecules are attracted to the stationary phase while polar compounds are carried along by the mobile phase.

The normal designations is because the first HPLC columns used alumina and silica packing, which are both polar, in columns. With setup The least polar compounds were eluted first and the most polar compounds were eluted last.
Switching to non-polar packing was then noted as reverse phase  since it reversed the order of elution. The most polar compounds were eluted first and the least polar compounds were eluted last.
ANSWER
This is normal phase TLC so the stationary phase is polar.
Assuming the compounds are "relatively" polar
Plate 2 has the LEAST polar eluent because the compounds have lower $\text{R}_{\text{f}}$ values than they do on the other two plates.
Plate 3 has the MOST polar eluent because the compounds have higher $\text{R}_{\text{f}}$ values than they do on the other two plates.
On plate 3 the least polar compound would be Z.
Assuming the compounds are "relatively" non-polar
Plate 2 has the MOST polar eluent because the compounds have lower $\text{R}_{\text{f}}$ values than they do on the other two plates.
Plate 3 has the LEAST polar eluent because the compounds have higher $\text{R}_{\text{f}}$ values than they do on the other two plates.
On plate 3 the least polar compound would be Z.
So the problem would seem to be ambiguous. We don't know that any of the solvents is "absolutely" non-polar like hexane, or that one might be "absolutely" polar like water.

I'm not an expert in TLC, but I think the wording is wrong in the definitions.  Shouldn't normal-phase TLC involve polar stationary phases and non-polar solvents?  (An example: naked silica gel and solvents like hexanes / ethyl acetate.)  And shouldn't reverse-phase TLC involve non-polar stationary phases and polar solvents?  (Example: octadecylsilane-derivatized silica and alcoholic or aqueous solvents.)

@CurtF. - Thank you for the correction. You are absolutely right. To make this more embarrassing I chastised the OP about knowing terminology.  I think I've fixed it now.

@MaxW So if compounds are less polar, like on plate 3 because they have high Rf values, the eluent is more polar? Since lower Rf, more polar compounds. So compound Z on plate 3 should be least polar, since it has the highest Rf.

Hmmm... The polarity of the compounds. I had just assumed that they were polar.

I think this is right now. I obviously don't remember as much about TLC as I thought I did. But that is why I stated fooling around here was to brush up on my chemistry...

@MaxW I don't think anything is missing/ambiguous since we're told each unknown eluent has different polarity, but it's normal phase TLC. Based on that we should be able to infer from the compound positions on the plate.                                                                                In normal phase TLC the lowest Rf will always be the most polar correct? That gets compound polarity down.

@MaxW Then for eluent polarity, it's the opposite of compound polarity isn't it?  All three compound in plate 2 is polar, so the eluent should be less polar/nonpolar

I'm in more trouble than I first realized answering this. I opened another question about this problem. http://chemistry.stackexchange.com/questions/69335/what-inference-can-be-made-from-the-fact-that-there-is-no-reversal-in-how-three

@MaxW In my notes there's an example similar to plate 2(Lab notes are vague, just says solvent wasn't polar enough). The compounds are at the bottom of plate 2, which means the solvent wasn't polar enough. So Plate 2 should have the least polar eluent.

These sort of what-if questions can be over analyzed. But if the compounds are relatively non-polar, I think plate 3 could be with a very non-polar eluent like hexane, and plate 2 could be with a more polar eluent. The difference being that the compounds are particularly soluble in the more polar eluent.

@MaxW Answer: Plate 2 used the least polar eluent (hence more polar compounds)

Jason - I changed my answers to 2. I had the right text about the lowest Rf values and just flupped the plate numbering. I still wonder what would happen if the compounds were relatively non-polar.

I don’t understand how you reach the conclusion that TLC 2 could ever have anything other than the least polar eluent.

The picture wasn't visible in the portion of the window into which I was typing. I just didn't proof read adequately...I was thinking the Rf values in the image were ordered low medium high


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Manganates (II)
I want to ask about character of manganese(II) compounds. I read that $\ce{MnO}$ is basic, just like $\ce{Mn(OH)2}$. I started doing research and I read ions like $\ce{MnO2^{2-}}$ and $\ce{HMnO2^-}$ exist too, but how? Are they related to reactions in solid state? Manganese(II) hydroxide isn't amphoteric like aluminium hydroxide so it's really confusing for me. I also found reaction that "confirms" amphoteric character of manganese(II) oxide
$$\ce{Mn(OH)2 + 2NaOH -> Na2[Mn(OH)4]}$$

Yeah, it is weakly amphoteric. http://onlinelibrary.wiley.com/doi/10.1002/zaac.19512640217/abstract;jsessionid=54189D2A000D76548DB237FCE8105623.f03t02

Manganese(II) hydroxide is not only a strong base but is also amphoteric as it dissolves in strong alkali solution to form hydroxomanganates:

From the solutions of $\ce{Mn(OH)2}$ in 50% Sodium hydroxide solution,
  the hydroxometalates $\ce{Na2[Mn(OH)4)}]$ is isolated.

But before forming hydroxomangantes, intermediate ions like $\ce{MnO2^2-}$ and $\ce{HMnO2-}$ are formed. This paper(*) discuss this fact:

The high solubility of manganese hydroxide in acid solution and its
  low solubility in alkali solutions give a qualitative indication of
  the fact that
  manganese hydroxide is a strong base. This seems to be further verified by the observation that manganese salts do not hydrolyze
  appreciably. The complete verification of this prediction is given in
  the subsequent calculation of the acidic and basic constants of
  manganese hydroxide.
The following equations probably indicate the reaction that manganese
  hydroxide undergoes in water, in acid, and in alkali.
(...)
$$\ce{Mn(OH)2(s) + OH- -> HMnO2- + H2O }$$
    $$\ce{Mn(OH)2(s) + 2OH- ->MnO2^2- + 2H2O}$$
(...)
The Equilibrium in Alkaline Solutions.
If manganese hydroxide is a strong base then we may expect its
  hydrogen ion dissociation to be very slight, particularly the second
  dissociation. Unfortunately, the solubility is so slight in the range
  where it is possible to calculate these acid dissociation values that
  no data were obtained.
(...)
$$\ce{Mn(OH)2 -> H+ + HMnO2- K = 1 \times 10^{-19}}$$

*http://pubs.acs.org/doi/pdf/10.1021/ja01852a001

No hydroxide can be amphoteric and also a strong base.  If something dissociates hydroxide ions readily, why would it turn around and take in more?


text:
Is HCl (l) + H2O (l) equal to HCl (aq)?
The Wikipedia article about aqueous solutions claims that:

An aqueous solution is a solution in which the solvent is water

So basically, is writing $\ce{HCl(aq)}$ equal to: $\ce{HCl(l) + H2O(l)}$?

By definition An aqueous solution is a solution in which the solvent is water. So the Wikipedia article isn't "making a claim," rather it is a truth by definition. // Conversely an non-aqueous solution is any solvent but water.

$\ce{HCl{(l)}}$ is not the correct notation in this case. Arrhenius acids and bases (chemicals that interchange $\ce{H^+}$ with water to form $\ce{H_3O^+}$ or $\ce{OH^-}$) break apart in water, forming ions. In your case, there is an equilibrium $\ce{HCl{(aq)} + H_2O{(l)}<=> H_3O^+{(aq)} + Cl^-{(aq)}}$ which very strongly lies to the right; virtually all of the $\ce{HCl{(aq)}}$ is broken apart in this way (which is why $\ce{HCl}$ is known as a strong acid).
Now, not everything dissolved in water has to dissociate into ions. For example, glucose $\ce{(C_6H_12O_6{(s)})}$ dissolves in water to form $\ce{C_6H_12O_6{(aq)}}$. Weak acids are acids that do not entirely dissociate into ions (for example, $\ce{CH_3COOH{(aq)} + H_2O{(l)} <=> CH_3COO^-{(aq)} + H_3O^+{(aq)}}$ still has some undissociated $\ce{CH_3COOH{(aq)}}$ in solution.)
A final point is that some dissolved species can, in fact, be solvents on their own. In this case, the definition of whether the species is aqueous or liquid is not well-defined and usually depends on the context. For example, a 1:4 mixture of $\ce{MeOH}$ and $\ce{H_2O}$ is typically written as $\ce{MeOH{(l)} + {H_2O{(l)}}}$, but a solution where a small amount of $\ce{MeOH}$ (maybe used as a reactant) would be denoted $\ce{MeOH{(aq)}}$. This has to do with the definition of the thermodynamic activities of the species and if you have questions about this, there's probably someone else on here who is a better expert than I to explain.

The example you give is not quite right. $\ce{HCl}$ is a gas at standard temperature and pressure. It is also a strong acid in water, so a more correct way to express the dissolution of $\ce{HCl}$ would be:
$$\ce{HCl{(g)} + H2O{(l)} <=> H3O+{(aq)} + Cl^-{(aq)}}$$
A simpler example would be glucose, which readily dissolves in water without dissociating into different species:
$$\ce{C6H12O6{(s)} + H2O{(l)} <=> C6H12O6{(aq)} + H2O{(l)}}$$
When a compound dissociates like a salt for example, we write it like this:
$$\ce{NaCl{(s)} + H2O{(l)} <=> Na+{(aq)} + Cl^-{(aq)} + H2O{(l)}}$$
Because water is simply acting as the solvent, you could leave it out of the last two examples, and the fact that water is the solvent would be designated by the $\ce{(aq)}$ designation of the dissolved species.

No. $\ce{HCl}$ is a gas at STP (standard pressure and atmosphere). As such, you can store it in pressurized cylinders, but commonly it is stored as a concentrated aqueous solution $\ce{HCl(aq)}$. In the gas phase, $\ce{HCl}$ is composed of covalently bonded molecules, H bonded to Cl.
In water, the compound ionizes virtually completely to $\ce{Cl-}$ and $\ce{H+}$ (with the understanding that the latter is shorthand for the more complicated reality - the proton is essentially never found alone (except in outer space, or at extreme temperatures (eg. inside the Sun)).
$\ce{HCl(aq)}$ is equivalent to $\ce{Cl- + H+ + H2O}$ , where it is understood that both ions are hydrated (surrounded by oriented water molecules).

I'd quibble that you should make the point that HCl(l) is the liquefied HCl gas. It has a boiling point of −85.05 °C at atmospheric pressure.

No HCl would dissolve in the water and for an aqueous solution to exist the solvent must be water. HCl(aq) + H2O(l) is correct.  Never a liquid with a liquid without it dissociation of the HCl.

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text:
Why does Mn (II) behave as "hard"?
Why, within HSAB, does $\ce{Mn^{2+}}$ behave as a hard acid?
If we go across the third block, then all metals to the left of manganese form soft divalent cations (e.g. $\ce{Ti^{2+}}$, $\ce{V^{2+}}$, $\ce{Cr^{2+}}$) and all to the right are borderline (e.g. $\ce{Fe^{2+}}$, $\ce{Co^{2+}}$, $\ce{Ni^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Zn^{2+}}$).
An example of the effect of its "hardness" on reactivity is that $\ce{Mn^{2+}}$ forms more stable complexes with oxalate ($\beta=10^{3.8}$) than with ethylenediamine ($\beta=10^{2.2}$).

It has to do with the fact that $\ce{Mn^{2+}}$ is a high-spin $d^5$ ion (in most cases) which is afforded special stability, as all frontier $d$ orbitals are half-filled. This reduces affinity for more covalent (soft) interactions that would involve disrupting the $d^5$ state by addition of electrons. So, the character is more ionic.
Editing after some additional searching: This also explains why ions such as $\ce{Fe^{3+}}$ are also hard. However, it seems to contradict in cases like organocuprates, where it would seem that the formation of $\ce{Cu(I)}$ would result in a similarly stabilized $d^{10}$ ion (same with silver). In these cases, is it simply because polarizability of transition metals in that region outweigh the $d^{10}$ stabilization?

Cu(I) has only one positive charge.  That softens things up.

@xahoc Thank you for your answer! But I still don't understand: the orbital sets derived from d-orbitals remain populated by the electrons of the metal only, and the configuration remains (t2g)^3 (eg)^2 in all complexes. Is it something to do with the spin exchange energy decreasing as t2g and eg sets get farther apart in energy? Is it possible to show the spin exchange energy depends on two orbitals being close in energy mathematically?

@xahoc Moreover, even if electrons were added to t2g, the situation would be analogous to adding an electron to Ti2+ d orbitals, i.e. neither increasing nor decreasing the overall pairing energy.

You can consider ligands with soft character to bind covalently with metals in ways that informally affect the electrons on the d-orbitals of the metal. Consider $\ce{d^6 Mo(CO)6}$ versus $\ce{d^6 Mo(H_2O)6}$. The pi backbonding capability of $\ce{CO}$ reduces electron density in the $\ce{d{xy},d{xz},d_{yz}}$ orbitals, lowering their energies. You can imagine the introduction of a pi donor such as $\ce{I^-}$ would increase electron density in those same orbitals. While additional bonding to soft/weak Lewis acid like CO may be OK, covalent bonding to soft Lewis bases is disfavored.

Also, in the case of $\ce{d_1,d_2 (Ti^{2+})}$ ions, there is a vacant $\ce{T_{2g}}$ orbital for electrons to enter, which wouldn't involve a pairing penalty while conversion of high spin $\ce{d_5}$ to $\ce{d_6}$ would have such a penalty. (High spin electron transfer would also require a spin-forbidden $\ce{T_{2g}}$ to $\ce{E_g}$ transition, but this is a kinetic, and doesn't affect the question).

@Xahoc Thank you! Ok, I see now how the effect of pi donors acceptors on the d orbitals would be disfavoured, but what about pure sigma donors? How are going to affect the d orbitals?


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Calculation of the pH of a mixture of a strong acid and weak acid
The question is to find out the $\mathrm{pH}$ of a mixture of weak acid and strong acid. My book just states the formula as $$\mathrm{pH}=-\log
\frac{C_2+\sqrt{C_2^2+4K_\mathrm{a}C_1}}{2}$$
where $C_1$ is the concentration (in mole/litre) of the weak acid (ionisation constant $K_\mathrm{a}$), and $C_2$ is that of the strong acid.
I understand that the strong acid will dissociate completely and the weak acid will dissociate with an ionisation constant. So the $\ce{H+}$ ion concentration from strong acid will be simply $C_2$ and that from weak acid will be $\sqrt{K_\mathrm{a}C_1}$. But I could not derive the above formula for the $\mathrm{pH}$ using these facts. I need help here.

can anyone point out which book that provides the stated equation? thank you in advance

Chemical thermodynamics is easy in that it always has enough equations to derive all the variables involved, and guarantees to produce a unique solution. The derivation itself may be not that easy, but that's another story.
In short, we have some weak acid $\ce{HA}$ and the products of its dissociation. Mind you, $\ce{H+}$ comes also from the strong acid, which dissociates completely. Now, the equations:
$$\ce{[HA] + [A-] = }C_1 \tag1$$
$$\ce{[H+]=[A-]}+C_2 \tag2$$
$${\ce{[H+][A-]}\over\ce{[HA]}} = K_\mathrm{a} \tag3$$
Is the rest clear?

 Use the equation $(2)$ to exclude $\ce{[A-]}$, then use the equation $(1)$ to exclude $\ce{[HA]}$, and end up with a quadratic equation on $\ce{H+}$.

Thanks for your answer.can you please explain how you wrote down first two equations.

Well, the first one is just the conservation law: the summary concentration of protonated and deprotonated forms of HA is always the same, for it can't be created or destroyed. The second is charge conservation: for every H+, there is an A- or the other anion.

Nope. I used A for the anion of the weak acid. The strong acid I didn't identify at all, because why would I?

@IvanNeretin I derive this formula according to your method :$$\mathrm{pH}=-\log
\frac{-(K_a-C_2)+\sqrt{(K_a-C_2)^2+4K_\mathrm{a}(C_1+C_2)}}{2}$$

I try to approximate the above formula as:$$\mathrm{pH}=-\log
\frac{C_2+\sqrt{C_2^2+4K_\mathrm{a}(C_1+C_2)}}{2}$$ Is it right?

First formula is OK. Second formula is OK if Ka is negligibly low compared to C2, but in that case it can be simplified further.

@IvanNeretin thank you.How can be simplified further?

Ignore the dissociation of the weak acid altogether.

-The above workup should be pretty easy to understand.
-The only tough task is to understand the approximation in 3rd last step.

Ivan's answer was sufficient.

He just gave hints.The above answer is complete workup.

The answer was accepted which means the OP was able to do the working himself. It's just about solving a quadratic.

Hey Ritwik, thanks for your answer! Unfortunately though, most of what you wrote is already covered by Ivan's answer. Only a quadratic equation needed to be solved, as Abcd also said. The only new thing I see is a simple statement: "$K_1<C_2$ (magnitude wise)". Perhaps, that was better suited as a comment on Ivan's answer instead? || Also, we support mathjax on our site. You needn't paste photos of math expressions, head over here to learn more. Thanks!

It should be noted that this formula is applicable only if the acid constant of the weak acid is indeed much smaller than the concentration of the H3O+ ions formed in the protolyzation of the strong acid. This is because the formula uses the approximation K1 << C2 in the derivation.
For example, if:
$$ K_1 = 5 \cdot 10^{-2}, $$
and the
$$ C_2 = 0.01 M, $$
then the suggested formula is not valid, which is obviously because the used approximation (K1 << C2) does not hold.


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Stability and resonance structures
In school they ask us to count the number of resonance structures in different organic compounds to determine it's stability. The one with more resonance structures is more stable. Is this notion correct?

Hello and welcome to Chemistry.SE. If you have any questions about the site I suggest starting off with the short [tour]. For information on homework-type of questions in particular, you can read through this discussion. Regarding your question, do you understand why the presence of resonance structures might lead to increased stability of a molecule?

It's generally not that simple. See for example Why does having equivalent resonance structures give more stability?

Here is a good overview about resonance: What is resonance, and are resonance structures real?

Isn't stability in molecules is relative; relative to atoms, bonds, connectivity angles in it?

Stability is a funny word. It gets tossed around all the time, but a molecule's stability really depends on the context you are considering it in. For example, benzene is totally unreactive towards electrophilic addition of $\ce{Br2}$ like most alkenes, but $\Delta_\mathrm{f}H^\circ(\text{benzene}) = 49.0\ \mathrm{kJ/mol}$, while $\Delta_\mathrm{f}H^\circ(\text{cyclohexane}) = -37.8\ \mathrm{kJ/mol}$.
So when we talk about resonance and stability, what are we really talking about? What this really refers to the stabilization energy associated with conjugated systems. For every p orbital that is added to $\pi$-system, it mixes in phase to create a new molecular orbital that is lower in energy than the two starting orbitals, or mixes out of phase to create an orbital that is higher in energy than the two starting orbitals, and in some instances doesn't mix at all. This is illustrated here:

You can't directly compare the relative energies of the molecular orbitals of two unlike molecules simply based on their number of resonance structures, though. Different atoms and substituents will cause deviations and make comparison meaningless, so it is really only useful to compare homologous compounds, like butane, but-1-ene, and buta-1,3-diene.

You cannot use a simple concept such as ‘counting resonance structures’ to determine stability. This has multiple reasons.

Stability has much more aspects to it than just counting resonance structures. For example, I can draw six canonic resonance structures of the benzyl radical. I can only draw two canonic resonance structures for toluene — the same molecular structure only with an additional hydrogen on the side chain. Yet undoubtedly, toluene is more stable than any radical.

You cannot really count resonance structures as that would imply some finite number. What one generally does is merely count the most probable resonance structures. You can draw many, many more less favourable ones; e.g. charge separation, homolytic bond cleavage and all that follows from them.
Thankfully, though, the canonic resonance structures which are typically taken to be the most probable can be counted somewhat.

Resonance as a concept is based on localised bonds. Molecular orbital theory tells us that better descriptions of molecular structures typically do not have localised bonds — although there are often mathematically equivalent descriptions that would.


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Ionisation energy of Nickel
Look for the following data for Nickel
IE1+IE2  2490 kj/mol
IE3+IE4.  8800kj/mol
Total.       11290kj/mol
Hence Nickel can show +2 oxidation state but not +4.
But according to me Ni in +4 oxidation state can acquire 3d6 and 4s0 which has t2g3 configuration.
Why do it has high ionisation energy?

Metals with a $d^8$ oxidation state are stabilized by a square-planar molecular orbital configuration. In this configuration, the $\ce{d_{yz},d_{xz},d_{z^2}}$ orbitals are stabilized, while the $\ce{d_{xy}}$ orbital is slightly destabilized and the $\ce{d_{x^2-y^2}}$ is severely destabilized (however, in $d^8$ species this orbital is unoccupied). The amount of stability provided by the square planar configuration is far greater than the amount of stability provided by the low-spin $d^6$ configuration.
This is a good online powerpoint that shows the stabilization.


text:
What molecules can be eliminated in condensation polymerization?
Is there any specific rule to what kind of molecules can be eliminated in condensation polymerization? From Wikipedia I found that "small molecules" like $\ce{H2O}$ can be eliminated. Is there a more accurate criteria to what these "small moecules" are?

AFAIK there isn't any rule. What they are depends on reaction.

In my school exams, I often see questions asking us to determine which reactions are possible condensation polymerization, usually depending on the eliminated waste like $\ce{H2O}$ or $\ce{HCl}$. So is there at least a way I can be sure if it must be impossible, or if it must be possible?

Well, if there's no side-product in reaction then it's not polycondensation. If there is then it may be polycondensation, you should know mechanisms to be sure. BTW in polycondensation side-product (of main reaction) is produced stoichiometricly.

As an abstract answer, to me it would be whatever is commercially viable to manufacture and sell. Commercial polymerisation processes eliminate ethylene glycol (in the manufacture of PET) which is probably recycled,  methanol in transesterification reactions, ethene or propene in metathesis type,  phenol (recycled) in the use of diphenylcarbonate with bisphenol A, etc. In home applications,  acetoxysilicone sealant gives off acetic acid. So, anecdotally, it seems "simple" molecules of MW around 100 or less might meet your criteria.

MW refers to relative molecular mass?

Yes, I was being a bit lazy

yes , only small molecules such as H2O or CO are eliminated , as such it is no rule . if large and bulky molecules are extracted as waste, it will no longer be called an elimination product but a side-product. i hope you understand now !

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Polar/ non-polar nature of HgI2
Is $\ce{HgI_2}$ polar or non-polar?
The electronegativity difference between Hg and I is about 0.66, so the compound should be almost covalent. However, I cannot find the structure of $\ce{HgI_2}$ as a single molecule, so, I cannot find if it is polar or not, as a whole.

You seem to be using a false dichotomy. A great deal of covalent molecules are polar. Many are even more polar than $\ce{HgI2}$ (which is polar, too).

@IvanNeretin I did not say covalent molecules are non-polar. If a molecule is ionic then it is surely polar, but if it is covalent , we must have its stucture to tell, whether it is polar or not. For example, CCl4 is non-polar as a whole.

Oh, you mean the overall dipole moment, Well, it's $0$, because the molecule is linear, much like CO2.

The electronegativity difference using the Pauling scale is not accurate in heavy transition metals. This is because the Pauling electronegativity involves the measure of bond energies, while the bonding of several metals (including "noble" metals like gold, silver, platinum, mercury, etc.) with large halides are diffuse and do not have sigma character . This is because the frontier orbitals of these elements are in the $d$ block, so molecular orbital interactions, if any, are diffuse especially in larger 4d and 5d metals. In reality, compounds like $\ce{HgI_2}$, as well as ones like $\ce{PbI_2}$ and $\ce{BiBr_3}$ (which I work with) are ionic salts.

The interesting thing about many of these solids is that that despite having ionic character, the polarizability of these compounds are so high (due to their large atomic sizes) that in solid form, they can have semiconductive or conductive properties. If you look at new solar cell materials, you will find that there is a lot of research on ionic materials known as perovskites, which have the formulas similar to  $\ce{CH_3NH_3PbI_3}$ yet conduct electrons very efficiently.

That is quite thoroughly wrong. 1) Pauling isn't particularly wrong for heavy metals, 2) What do you even consider as conventionally bond? 3) Did you hear about HSAB? Its foundation is that compounds like you mention are covalent.

I'll rework the "conventionally bond" wording. From what I understand reading The Electronic Structure and Chemistry of Solids by P.A.Cox, the bonding between elements of high atomic number is diffuse, akin to metallic bonding. This reduces the accuracy in calculating sigma bond dissociation energies from which the Pauling electronegativities are derived. As a result, the Mulliken electronegativity is used for heavy elements.

HSAB uses the elements' polarizability to make determinations on the type of bonding. But, you cannot use only
HSAB. In the specific case of $\ce{HgI_2}$, iodine is a strong pi donor yet there are not orbitals on mercury to accept electron density, so
the bonding becomes ionic. But, this does not mean that the ionicity is not weak or diffuse. Similarly, for $\ce{PbI_2}$ and $\ce{BiBr_3}$,
the stabilizing energy by removing electrons from the 6p orbitals of lead results in a more ionic contribution than would be expected from HSAB alone.


text:
What is the proof for saying all the compounds-of-an-element (except few), to have the same oxidation number?
Following statements are extracted from Puri, Sharma, and Kalia's Principles Of Inorganic Chemistry book:

In all hydrogen compounds (except hydrides of active metals), the oxidation number of hydrogen per atom is +1.
In all oxygen compounds (except hydrogen peroxide and other peroxides and superoxides), the oxidation number of oxygen per atom is -2.

Have Chemists tested the oxidation state of every compound (of hydrogen, Oxygen or any element considered to have fixed oxidation sate) in the universe, to say all the compounds (with few mentioned exceptions) of the considered element to have the fixed oxidation state?
Is there any proof or law for supporting above statements? Or is the author sloppy in not saying "In all the known hydrogen/oxygen compounds"?

It is about oxidation number, which is not a physically defined quantity.

This statement is a generalisation. As every generalisation it makes things simpler to remember but even here they give exceptions. Of course, chemists couldn't test all oxygen or hydrogen compounds. The assumption is based on electronegativity of atoms in the "usual" compounds. In OF2 oxygen is not in -2 oxydation state. Also in BH3 hydrogen is not +1. I would not even go as far as "In all the known hydrogen/oxygen compounds". "In most of known" would be enough.


text:
How are hydrogen atoms formed in hydrogen discharge tube?
I (partially) understand how a breakdown voltage is applied, then the gas is ionized to cations and electrons. Then the electrons will accelerate to the cathode and knock off more electrons from other H2, creating a current ( which is the flow of electrons ). The electrons might excite the gas and the gas emit light as it goes back to normal state. If that the case the shouldn’t in hydrogen discharge tube, the spectral lines observed are those of hydrogen molecules (the gas)? And how does it atomize to give the hydrogen spectral lines?

Because of all those energetic electrons flying around and hitting things. You get excited H2, atomic H, higher order molecules like H3, all can be excited or ionized or whatnot. But, you have lots of H, so lots of H emission lines.

additionally to @Jon Custer's comment, $\ce{H2}$ lowest excited state emission is at approx $110 $ nm which is in the far UV so not detectable by our eyes. However, H atoms emit in the visible; search for Balmer Series to see list of values.


text:
The "hazardousness" of artificially made sweteeners
In school I learned of the "possible hazardity of artificial sweeteners." But that was some time ago, now. I think that science should have caught up by now.
My question is as follows.

Which chemical properties in artifical sweeteners could be potentially cancerogenous?

(This is to say, if there exists cancerogenous properties in artificially made sweeteners from the start).
Please enlighten me.
Examples of sweeteners:

Fig 1 Acesulfame Potassium

Fig 2 Aspartame

Isn't that kind of common knowledge -> just look up wikipedia, for example https://en.wikipedia.org/wiki/Aspartame#Cancer for aspartame and https://en.wikipedia.org/wiki/Acesulfame_potassium#Safety I don't see a good reason to just repeat that here.

From the first look at it, at least acesulfame seems quite reactive. The structure kind of reminds me of 1,3-propanesultone, a known teratogene.

You assume hazards exist. They don't or we would have noticed by now. Lots of people have been using artificial sweeteners for years and no good studies have shown measurable problems. Traditional sugar (sucrose, glucose or fructose) is associated with many diseases (and it makes you fat which itself is a cancer risk).

From the NHS website in the UK:
The US National Cancer Institute conducted a study of nearly half a million people, comparing those who consumed drinks containing aspartame with those who did not. Results of the study found aspartame did not increase the risk of leukaemia, lymphoma or brain cancer."
Also:
In 2013 the European Food Safety Authority (EFSA) conducted a comprehensive review of the evidence (there is a pdf on the website) and concluded that aspartame was safe for human consumption, including pregnant women and children.
Cancer Research UK also have stated that artificial sweeteners are safe.
Rigorous testing is carried out before a sweetener can be approved for use in the EU.
So, no evidence of cancer in humans. Rats, maybe. Humans no.


text:
Why doesn't LiHCO3 exist in solid state?
I have read that $\ce{LiHCO3}$ does not exist in solid state. Is it because of the size of anion-cation concept or is it something else? So why is it only for lithium and not for sodium too?

This site states that it is a solid at room temperature, http://www.guidechem.com/dictionary/5006-97-3.html but information on this compound seems scarce.

Why does beryllium peroxide not exist?  Why is magnesium peroxide only metastable but by the time you get to barium the peroxide is fully stable?  Polarization by small, densely packed cations tends to pull polyatomic anions like peroxide or bicarbonate apart, leaving smaller nations in the compound like oxide or carbonate.

CRC handbook of chemistry and physics (also latest 97th Ed.) includes $\ce{LiHCO3}$ in the table "Physical Constants of Inorganic Compounds" (4-70) as white powder, slightly soluble in water 0_O.

CRC Handbook 62nd ed (1981-2) says LiHCO3 is soluble 5.5 grams/100 mL H20 at 13C. It appears that LiHCO3 becomes more elusive as time passes.

Claiming that a particular chemical compound "doesn't exist" is difficult to prove. The only convincing proof would be to show that neither LiHCO3 nor any of its hydrates can be isolated from (aqueous) solution. Essentially, you would have to show that using any of an almost infinite number of methods, that it couldn't be prepared at any temperature or pressure - not very likely to have ever been done, imho. So, I'll ask you for your source for such a claim, or better yet, I'll ask you to check and verify that your source - which you accept as being accurate - has made stringent efforts that demonstrate the claim as well as having published that work in peer reviewed literature. As an aside, there are plenty of compounds which require "waters of hydration" in their crystal structure. If it really doesn't exist, then the reason is simple: kinetically the reaction 2LiHCO3 → Li2CO3 + H2O + CO2 is spontaneous and energetically favored. It just seems to me that if I placed Li2CO3 in a pressure cell with water and CO2 at a billion gigapascals of pressure, I could make the bicarbonate salt, but perhaps some other structure is more likely...and anyway, generally what they mean by "doesn't exist" is that it "isn't known to exist at standard temperature and pressure".

The source: Concise Inorganic Chemistry Fifth Edition (Wiley - India Edition) By J.D. Lee, page no: 295 - "Lithium is exceptional in that it does not form a solid bicarbonate, though $\ce{LiHCO3}$ can exist in solution."

Due to the small size of lithium, it will polarize the large anion oxygen, the $\ce{C-O}$ bond will weaken and $\ce{Li-O}$ bond will strengthen, they will split, and they will not be in the solid state.

It's is because $\ce{Li}$ is in diagonal relationship with $\ce{Mg}$ in periodic table. As magnesium hydrogen carbonate is not obtained in solid form, $\ce{LiHCO3}$ also cannot be since both have similar properties.

This is because, lithium ion has a very high charge density therefore it distorts the structure of hydrogen carbonate ion and readily converts it to carbonate ion, which is smaller in size as compared to hydrogen carbonate ion.

due to small size of lithium its hydration enthalpy is very high and moreover hydration enthalpy is higher than lattice enthalpy of lithium bicarbonate
hence lithium bicarbonate  exists in solution  state only


text:
Why is this reaction deprotonated with CH3OH?

Why is this reaction deprotonated with CH3OH and not water?

Shouldn't MeOH be more basic than water? Just from the first look at it I would expect higher substituted alcohols to be more basic...

E.g. isopropanol, tert-butanol would be even more basic.

It would generally be helpful to provide the source you are referring to. By the way, I wouldn't post copyrighted material like this ...

It doesn't matter whether methanol or water deprotonates it considering that they have almost the same basicity. If there's more MeOH lying around then it's more likely to be deprotonated by MeOH, if there's more H2O lying around then it's more likely to be deprotonated by H2O.

Pka of water is 15.7
Pka of methanol is 16
In the end, methanol should be protonated somehow. But the truth is, in solution both water and methanol could theoretically donated the proton in the first step. And both water and methanol could accept the proton in the last step.
When I took organic chemistry, as long as we showed the complete electron exchange, water could be excluded or included entirely.


text:
What inference can be made from the fact that there is no reversal in how three components eluate?
I gave an answer this question about TLC. I now realize that I am blundering my way through this question.
The problem statement doesn't specify if the compounds to be eluted are polar or non-polar. So this seems to leave a door open.
Question 1 - Could compounds be either relatively polar or relatively non-polar?
Question 2 - If an absolutely non-polar solvent like hexane were used, and some other more polar solvent, wouldn't a reversal of the order of the elution of the compounds happen?
Question 3 - Have I in fact, like a blind squirrel, finally found the nut?
EDIT 1
The problem specifies that "normal phase TLC" is being used so the stationary phase is polar.
The problem only states that the elutants have different polarity and not that any is actually non-polar like hexane.
EDIT 2
With Question 3 I was trying to be funny and ask if my two answers to the original question are right.

Question 1
Yes, of course! Something that is "relatively" non-polar would travel further with a non-polar mobile phase because it would have less attraction to the stationary phase. While a polar substance would not travel as far with the mobile phase because it would have a greater attraction to the stationary phase.
Question 2
Now, that depends. Are you also changing the stationary phase? Exactly how polar is the new polar-mobile phase? It would really depend on which phase the solutes have a greater attraction to.
Question 3
You are on your way. Quantitative HPLC is the bane of my existence. There's a million factors to consider, such as column temperature, flow rate of the solvent, particle size, polarity and charge. Usually, it takes a bit of experimentaton to reach a perfect combination that allows complete separation of whatever it is you're interested in. Usually, your best bet in identifying a compound at a particular point is to run a little bit of pure sample on its own. In the end, above all else, you need to find a combination that doesn't result in chemicals overlapping with each other.
In qualitative chromatography, as long as you can resolve all possible signals, you've done well enough.

In answering the problem am I correct in believing that because the compounds could be relatively polar, or relatively non-polar that the most polar phase could flip-flop between plates 2 and 3?

Assuming all compounds are polar, the most polar mobile phase would be on plate #3. Because, the mobile phase would be able to "push" compounds out of their binding sites by binding to the silica. The least polar compound is compound Z,  because the mobile phase can more easily break it's bond to the silica. That means the least polar solvent is on plate #2 - because the mobile phase can not replace the binding sites as easily.

They could indeed flip-flop, but I think you'd need to change both mobile and stationary phases.

To be completely honest with you though, chromatography confuses the hell out of me. There's too much theory involved lol. Hence why it's the bane of my existence. But, in applied sciences, I'm pretty sure most of us probably just fiddle around with chemicals til we find something that works.

Understand what you mean about chromatography. Factor A makes it go up, Factor B makes it go down, factor C makes it go left. Factor D makes it go right. So which way does it go?!?

I guess my line of thought was less about chromatography and more about solubility. If the compounds were relatively non-polar and hence rather insoluble in a relatively polar solvent, then it would seem plate 3 could be the least polar solvent. So such behavior would be unusual, but not totally weird.

To what you said about Factor A making it go up, and Factor B making it go down.. I guess it depends which Factor happens to have a bigger impact in a given situation. I suppose when dealing with non-polar compounds, solubility in the mobile phase would indeed be a more important factor in determining how far they travel up the plate since they're unlikely to actually bind to the silica.


text:
is there a filter to seperate glycerol and biodiesel?
i am currently working on a project of manufacturing biodiesel from cooking oils by the process "TRANSESTERIFICATION". at the end of this reaction, after the reaction mixture settles I have two layers 1. biodiesel on top 2.glycerol at the bottom. manually the separation can be done by a separating funnel but is there a way by which I could automate the separation process of this mixture?

What is the scale You are doing this? If You do it on a serious scale, you can use a separator. If You play around, buy a 5l separating funnel. https://de.wikipedia.org/wiki/Separator_(Verfahrenstechnik) The english page is very poor.

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From this: http://dx.doi.org/10.1016/S0961-9534(01)00074-5
And this: https://en.m.wikipedia.org/wiki/Glycerol
Apparently most bodiesels have a BP of 340-375 degrees Celsius, while glycerol has a boiling point of 290 degrees Celsius.
I do not know whether or not heating the mixture to 300 degrees would cause a bad reaction or not. But you could try a small scale distillation and find out.


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A unit cell for graphene
In a study I have run into some questions asking for "How many atoms are contained in the unit cell" for graphene. I have some trouble understanding the unit cell of graphene which is always depicted as a mono-layer of atoms. What is a unit cell in a mono-layer?
The closest I could find is something like this picture:

The upper right image shows something soon-to-be a hexagonal structure. But when is such structure actually appearing?

well considering graphene is a single layer of carbon atoms it's not really appropriate to try and define a unit cell.

I would go for one atom in special position (origin) and a planar hexagonal group p6mm, but as my International Tables for Crystallography are at university I can be mistaken.

The upper-right picture is graphite: https://en.wikipedia.org/wiki/Graphite

As graphene isn't crystalline, what sense would it have to apply rules of crystalline structure to it?

Related: https://mattermodeling.stackexchange.com/q/12023/5

The unit cell for graphene is a two-dimensional rhombus according to the figure shown on page 31 of this paper.$^1$ (also here.) The result is that two atoms are contained per unit cell. The upper right structure actually appearing in graphite, stacked layers of graphene.

$^1$Zhou, J; Huang, R. Internal lattice relaxation of single-layer graphene under in-plane deformation. Journal of the Mechanics and Physics of Solids, 2008, 56(4), 1609-1623. https://doi.org/10.1016/j.jmps.2007.07.013

Bit late,  but a unit cell of graphene is just the 'surface' area of 2 carbon atoms.

This area is 0,052 nm2, which gives us a density of 0,77 mg/m2.
Same principle for all monolayer materials.


text:
Normal Phase TLC using ethyl acetate as eluent

Compounds 1-3 were analyzed by normal phase TLC (shown) using ethyl acetate as the eluent. Assign spots on the TLC plate to compounds 1-3.
1: 2-acetoxybenzoic acid 2: 2-hydroxybenzoic acid 3: benzoic acid

I asked a question about TLC and I received some mixed answers, so I'd like to see if my understanding of TLC is fine.
We're using normal phase TLC so lower Rf compounds will be more polar. $0.47$ is the most polar compound, $0.70$ is the least polar compound.
Ethyl acetate is polar. Does the eluent polarity just determine where the compounds will run on the TLC?

The stationary phase is polar so it will be something like R-O-H or R-(OH)2 where the H's will hydrogen bond. So in terms of affinity for the stationary phase I think it depends on the number of oxygen molecules here. So 1 (blue) >2 (green) >3 (red).

@MaxW Why do you think compound 1 is the most polar?

In part it seemed that 4 oxygen atoms offered more chances to hydrogen bond than 3.

@MaxW " I think it depends on the number of oxygen molecules here." is definitely not correct. An ester is usually less polar than an alcohol or phenol but has more oxygen atoms. The big question here is whether 1 or 3 is less polar, and here it get's quite tricky, since compound 1 can do an intramolecular hydrogen bond.

@DSVA - Shouldn't both 1 and 2 form an intramolecular hydrogen bond? I was flipping a coin based on oxygen count.

Jason - I think that this is one of those situations that once given the answer you can make a compelling argument. So is it H from stationary phase H-bonding to oxygen of molecule, or H of molecule H-bonding to oxygen of stationary phase?

TLC is only about polarity.
Since all three compounds share the benzoic acid functionality, we only have to consider the rest.
Then we have three compounds: $\ce{R-OCOCH3}$, $\ce{R-OH}$ and $\ce{R-H}$.
In general the polarity is:
alkane < ester < alcohol

That is what we have here also: 3 < 1 < 2.
The one with the lowest polarity will have the highest R$_f$.
To show that it works, I have run them on a silica TLC plate with AcOEt:

AcSA: acetylsalicylic acid; SA: salicylic acid; BA: benzoic acid.

It's more accurate to say that TLC is comparing the eluate's differential interactions with the stationary phase and the mobile phase, where the primary interaction is polarity.


text:
Separation of Liquids based on Density
I was doing some practice problems when I came across this one

The answer key says the answer is B but I don't know why. Isn't the difference in the molar masses of the two substances in D greater than that in B? Or am I supposed to look for how many times greater the molar mass of one is than the other?
Thanks.

The ability of liquid mixtures to separate is not based on density, but on polarity of each liquid. $\ce{CCl_4}$ and $\ce{C_6H_{14}}$ are both non-polar; they will mix well in solution and not separate easily. $\ce{CH_3OH}$ is polar due to presence of a polar alcohol group. Mixtures of non-polar and polar compounds will separate easily.
Density comes into play when you need to determine which of the compounds settle on top of the other. So, for (B), you would use the fact that the density of $\ce{CH_3OH}$ (0.79 g/mL) is less than $\ce{CCl_4}$ (1.59 g/mL) to say that the methanol layer would float on top of the $\ce{CCl_4}$ layer.

Ah yes thank you. I don't know why I didn't think of that. Thanks a lot! :)


text:
How does the size of a molecule affect acidity?
Today I learned that $\ce{H2S}$ is more acidic than $\ce{H2O}$ because $\ce{H2S}$ has a larger molecule size. But $\ce{Fe^3+}$ is more acidic than $\ce{Fe^2+}$ and $\ce{Fe^3+}$ is smaller than $\ce{Fe^2+}$.
Likewise, formic acid is more acidic than acetic acid but formic acid is smaller than acetic acid.
How does this go together?

Are you trying to compare Brønsted-Lowry acidity and Lewis acidity?

It is key to learn about the applicability of certain concepts — or why does a concept work in a certain case — along with the concepts themselves. Most notably, atom size (it should not be molecule size, but for $\ce{H2O/H2S}$ it does not matter due to hydrogen’s tinyness) is only of limited help in a few select cases to determine acidity.
When comparing the acidity of two compounds you should always actually look at the following equation and then compare the anionic products of two different acids:
$$\ce{HA + H2O <=> A- + H3O+}\tag{1}$$
Whichever generates the more stable anion is the stronger acid. Thence, a number of concepts have been developed to determine the more stable anion from a given list of alternatives — one of them is the atomic size. It is useful only when directly comparing two otherwise group-identical species — like oxygen and sulfur which are in the same group but different periods or chlorine and iodine. Basically, under otherwise identical conditions (same valency and same charge) the larger anion is able to stabilise the negative charge better. Thus, $\ce{H2S}$ is a stronger acid than $\ce{H2O}$ and $\ce{HI}$ is stronger than $\ce{HCl}$ but we cannot say anything about the difference between $\ce{H2S}$ and $\ce{HCl}$ or even $\ce{H2S}$ and $\ce{HF}$ from that.
All the other pairs you listed do not fit into that difference concept at all. The acidity of $\ce{Fe^n+}$ is not explained simply by looking at the metal ion. And the difference between acetic and formic acid is on a much more subtle scale since both are oxyanions. To reiterate: do not only learn concepts but learn what concepts are for.

Since OP discussed formic and acetic acids, it should also be pointed out that once past about propanoic acid that additional $\ce{-CH2 -}$ groups don't change pKa much. So for organic compounds molecular size means little. For organic compounds it is more about what type of functional group. Carboxylic acid, phenol, alcohol etc. (This is obviously much further down the rabbit hole than the points you raised.)


text:
How many hydrogen bonds are formed by water and by HF?
How many hydrogen bonds are possible for $\ce{H2O}$, given that oxygen has two lone pairs? Is it 4 or 2?
Related: why is it that HF forms only 1 hydrogen bond, given that HF has three lone pairs?

Try drawing a bunch of water molecules. How many ways do they have to connect to others?

If you search online for water hydrogen bonding you will find many images that will answer your question. Same for HF.

Water is capable of participating in 4 hydrogen bonds at once, granted it only does this when it forms a perfect crystal structure. That means that every hydrogen will coordinate to an oxygen lone pair. In the liquid phase, the average number of hydrogen bonds that a water molecule participates in varies greatly with temperature, the average being 3.69 at 0ºC, 3.59 at 25ºC, and decreasing to 3.24 at 100ºC as a result of increased molecular motion and decreased density.$^{[1]}$
Hydrogen fluoride is also capable of participating in 4 bonds at once, but in the solid state forms only 2 hydrogen bonds on average. Though it has three fluorine lone pairs, each of which is capable of accepting a hydrogen bond, there are not enough hydrogen for every fluorine to do so. Similarly, ammonia can also form 4 hydrogen bonds but only forms 2 on average because each molecule has only one lone pair.

$^{[1]}$ Hydrogen Bond, Hydrogen bonds in water


text:
What does teflon give on decomposition?
I have read that teflon coatings undergo decomposition at temperatures above 300 degree Celsius. What do they decompose to? And what other conditions apply for the decomposition? Is this decomposition harmful for humans as cooking goes beyond that temperature?

Have you checked out wikipedia?

Like any other polymer decomposition process, the products of PTFE decomposition depends on the chemical species present while PTFE is undergoing the process and temperatures.
General process is this : Decomposition  is  initiated  by  random-chain
scission, followed by depolymerization. Termination is by
dis-proportionation. And all of this happens rapidly above 600 K (~326 C). There are no other conditions applied for the decomposition to take place hence can happen in dry or aqueous environments, which would give different by products.  Cooking below 200 C, it would be completely safe since the mass loss below 300 C is undetectable. Only above the glass transition temperature (~326 C) is the mass loss significant.
Decomposition products will depend on the environment and usually
Oxygen usually does not enter the cycle directly but through water, to give species like,

Carbonyl fluoride
Carbonyl difluoride

As expected other species like fluorinated alkanes and alkenes are obtained like,

the monomer, tetrafluoroethylene
Hexafluoroethane
Octafluorocyclobutane
Octafluoroisobutylene
Perfluoroisobutylene
Tetrafluoroethylene
and more.

However, the chemicals (species) listed above are for controlled lab experiments. What happens in real life cooking scenario is not (or cannot) be anticipated. But, it can be readily said that the above species can react with other chemicals in food to give fluorinated compounds that will be harmful for humans. As a precaution, one should not cook above 200 C on PTFE coated utensils to be completely safe.
Check this extremely detailed page for abstracts and toxicity remarks from research papers on PTFE decomposition on Fluoridealert.org where the conditions and level of toxicity is reported in an organized manner.
Polymer decompositions- check PTFE section for general comments.

Tetrafluoroethylene is the monomer

Although the question is about harm in humans, it's worth noting that pet birds are very sensitive to PTFE toxicosis.  (The proverbial "canary in the coalmine.")   The dangers of overheated teflon to birds has been public since the 1980s.  See sources for  Teflon Toxicity (PTFE Toxicosis) in Birds.

Carbonyl fluoride is dangerous stuff. It's pretty similar to phosgene, a chemical weapon used in World War 2.

Teflon is a mono polymer.
The monomer is Tetrafluoroethylene — $\ce{C2F4}$

Pyrolysis of PTFE is detectable at 200 °C (392 °F), and it evolves several fluorocarbon gases and a sublimate. An animal study conducted in 1955 concluded that it is unlikely that these products would be generated in amounts significant to health at temperatures below 250 °C (482 °F)

Source: ( Zapp JA, Limperos G, Brinker KC (26 April 1955). "Toxicity of pyrolysis products of 'Teflon' tetrafluoroethylene resin". Proceedings of the American Industrial Hygiene Association Annual Meeting.)

I had, until recently, machined pure Teflon (PTFE) in industry. I know the MSDS sheets I was provided with mentioned, at least under heat of cutting, one thing it breaks down into (among others) was hydrofluoric acid. Stuff you do NOT want to breathe! I know that as a highly poisonous acid that attacks glass.


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Are there types of glue that are designed to be dissolved easily?
I've heard that some types of glue are easily dissolved.  For example, wood glue dissolves in vinegar and super glue dissolves in acetone.  This doesn't mean that wood glue or super glue were necessarily intended to be dissolved.
Are there types of glue/adhesive that are pretty strong and have a cheap solvent that is intended to remove the glue/adhesive later?

White "School glue" immediately comes to mind. More amiable to coming out in clothes wash than "regular" white glue. Kindergarten kids are messy...

"Wood glues" can have several different compositions, chances are if the ones you are working with respond to vinegar they are either protein based or PVA.  In the protein case the solvation is likely accomplished by the vinegar protonating the polymer chains, and therefore changing their attraction to each other. In the case of PVA based glues, vinegar is likely contributing to acid catalyzed hydrolysis.  The softening of cyanoacrylate (super glue) with acetone is a more traditional dissolving mechanic, as no chemical changes are being made.
That being said, all three mechanisms mentioned above can be pretty well represented in the glue world with different polymers.  Based on what you consider cheap solvents, you have a few options. Cyanoacrylate (superglues), some UV-cured adhesives (which are frequently very well crosslinked and strong), PVC, and many others are all softenable and removable by acetone
Additionally, anything that can be taken out by acetone will be solvated in MEK (methyl ethyl ketone).  MEK is less water miscible and more aggressive though, which may be relevant to your needs.
Silicone adhesives are great for flexible applications, but frequently need stronger organic solvents (like toluene) to swell it and then it can be mechanically removed.  There are also commercially available silicone removers that appear to use acidic or basic bond cleavage to break up the adhesive.
Rubber cement is another flexible and relatively strong bonding adhesive mainly composed of solvated polymers, drying and tackifying by solvent evaporation.  Reintroducing the solvents (many commercially available paint thinners should work) will soften the adhesive.
As an aside note, depending on the materials you are bonding together, heat and cold can also be used to weaken joints and destabilize the adhesive bond separately or in tandem with chemical treatments.

I'm not an expert in the field of glue science, but the glue on the back of stickers on plates and such can be easily be dissolved by rubbing a little vinegar on it.  A common misconception is that water is better for this, but that only makes it worse.


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Physical meaning of (negative) pressure in molecular dynamics simulations
Lately I have been simulating a system in an NPT-ensemble where $N$ is constant and $P$ and $T$ are coupled to a Berendsen barostat/thermostat. The system is cubic and periodic in all directions.
I am trying to fathom how the pressure affects my system, why it fluctuates so much and how negative pressure should be interpreted in systems such as the one described above.
I have some knowledge of classical MD, and it is my understanding that (in general three dimensional system), the total pressure of the system is computed with the virial equation
\begin{equation}
P=Nk_{b}TV^{-1}+\frac{1}{3}\bigg<\sum_{i=1}^{N}r_{i}\cdot F_{i}\bigg>V^{-1}\tag{1}
\end{equation}
where $N$ is the number of particles, $T$ is temperature and $k_{b}$ is the Boltzmann constant.
By looking at equation 1, it is not fully clear to me what negative pressure would imply, so I looked how the equation is derived.
The derivation starts from the Clausius virial function
\begin{equation}
W(r_{1},...,r_{n})=\sum_{i}^{N}r_{i}\cdot F_{i}^{TOT} \tag{2}
\end{equation}
where $F_{i}^{TOT}$ is the total force acting on particle $i$.
This can be averaged over the whole MD trajectory, and integrated to obtain
\begin{equation}
\big<W\big> = -\lim_{t\rightarrow 0}\frac{1}{t}\int^{t}_{0}d\tau\sum_{i=1}^{N}m_{i}|r'(\tau)|^{2} \tag{3}
\end{equation}
from which, by using equipartition of energy, one can obtain
\begin{equation}
\big<W\big>=-3Nk_{b}T \tag{4}
\end{equation}
where 3 is the dimensionality, $N$ is the number of particles, $T$ is temperature and $k_{b}$ is the Boltzmann constant.
The force exerted by the container walls can be taken in to account by 'splitting' the $F_{i}^{TOT}$ term in to internal and external part, and then
evaluate $\big<W^{EXT.}\big>$ using:
\begin{equation}
\big<W^{EXT.}\big> = 3L³(-P) = -3PV\tag{5}
\end{equation}
Where $L$ is the length of the cubic box (so that $L_{x}~=~L_{y}~=~L_{z}~=~L$)
Now, since $\big<W\big> = -3Nk_{b}T$, one can rewrite it as:
\begin{equation}
\bigg<\sum_{i=1}^{N}r_{i}\cdot F_{i}\bigg> - 3PV = -3Nk_{b}T\tag{6}
\end{equation}
and get the original virial equation (1).
My questions are:

Is this correct? Mathematics (unfortunately) is not yet my strong suit, and I know that my reasoning (even with great help from different MD-related tutorials and sites) is prone to errors.
Could you explain the negative pressures in MD simulation simply by arguing that the force exerted by the container walls is negative?
If so, can one just argue that because the force is negative, the volume of the container must be increasing?

Thank you for your time.

Looks fine. 2. I'd phrase that more generally as the simulation volume would tend to expand. 3. But can only do so if the ensemble permits it (pressure can be negative in NVT).

The fluctuations are a consequence of making an observation of a macroscopic quantity over small numbers of degrees of freedom, or time. See http://www.gromacs.org/Documentation/Terminology/Pressure


text:
Oxidation state of Halogens
So the question is which is the most common oxidation state of halogens and I have to justify my answer. Is my following thought correct? Do I have to add something?
The most common oxidation state of halogens is $-1$. Halogens are the most electronegative elements of the periodic table. Their outer electron configuration is $n\mathrm{s^2}n\mathrm{p^5}$. If chlorine, for example, gains one more electron, the outer p orbitals are completely filled (resulting in a full octet). Since halogens are electronegative they can easily remove an electron from the nearby atom. So then the halogen becomes isoelectronic with a noble gas and it is more stable in this state.

Whatever you have thought is correct.

@Prakhar No, that is wrong. Please do not confuse people by posting wrong comments.

Well ,we are talking about the most common oxidation state of Halogens, and -1 is the most common oxidation state. I know that they show oxidation states from -1 to +7 (except for fluorine), but that is not what we are being asked here. We are being asked about the oxidation state that Halogens exhibit more frequently or the oxidation state that is observed mostly.

@Prakhar Again, this is wrong. Iodide is not iodine’s most common oxidation state.

I suspect the "book" answer is -1, but that does not make it really right for all halogens especially iodine and heavier.  "Dumbing down" science for introductory courses is something we have to put up with :-( .

One curious case is only a single halogen is known in a compound with zero oxidation state. That would be fluorine in "gold heptafluoride", $\ce{AuF5•F2}$. Everything else with the halogens is known only as $-1$ or positive, and predominantly odd.

There is not one single most common oxidation state across all halogens. While fluorine is the most electronegative element of which compounds are known — meaning that it can only exist in the $\mathrm{-I}$ and $\pm0$ oxidation states — the electronegativity continually reduces down the group to iodine. Iodide in the oxidation state $\mathrm{-I}$ is easily oxidised to $\ce{I2}$ or $\ce{IO3-}$ with the oxidation states $\pm0$ and $\mathrm{+V}$, respectively.
What is true, however, is that all halogens except fluorine show oxidation states from $\mathrm{-I}$ to $\mathrm{+VII}$ as can be predicted from their electronic structure $n\mathrm{s^2}\,n\mathrm{p^5}$.

Hmm, plus/minus zero?

@Mithoron Yes …?

sigh That meant that it has no sense whatsoever in chemistry to put such sign in front of zero (sort of euphemism ;)

@Mithoron I actually learnt to always label the oxidation state as $\pm0$. Because all other oxidation states are signed, that one should be, too. And also, all other oxidation states are written in Roman numerals but there is no Roman numeral zero — having something that stands out makes it clearer we are dealing with an oxidation state. All of my answers that talk about an oxidation state $\pm0$ should have that sign; if not it is a mistake on my behalf and I shall correct it asap ;)

In the wikipedia article, it's written $0$ everywhere. @Jan


text:
In the production of vinegar, where does water come from?
In the production of vinegar, where does the water come from?
I have heard that vinegar production involves these two reactions..
$$\ce{C2H6O {(ethanol)} + NAD+ -> C2H4O {(acetaldehyde)} + NADH + H+}$$
and
$$\ce{C2H4O {(acetaldehyde)} + NAD+ + H2O ->   C2H4O2 {(acetic acid)} + NADH + H+}$$
But I don't see where water comes from as an input into the second reaction. Like if it's added manually or if it's produced by something else.
Note- there was an additional part of the question which i've since removed.. some of that is addressed https://biology.stackexchange.com/questions/56995/do-acetic-acid-bacteria-use-the-electron-transport-chain-when-converting-ethanol

In biochemistry, you typically have plenty of water around.

Vinegar is made in a fermentation process of grape juice (short version). Now take a wild guess where the water comes from...

ok thanks,  i've edited my question accordingly

I think this is relevant  http://www.bbc.co.uk/schools/gcsebitesize/science/triple_edexcel/organic_chemistry/organic_chemistry/revision/6/   ethanol  + oxygen → ethanoic acid + water C2H5OH (l) + O2 (g) → CH3COOH (aq) + H2O (l) <--- (CH3COOH being ethanoic acid  / acetic acid / C2H4O2 which is key to the production of vinegar)   another  related- http://chemistry.stackexchange.com/questions/33135/which-one-is-the-best-notation-for-sodium-acetate

@Jan i've removed the additional aspect.

@Jan please explain to me why as soon as I remove that part of the question that you complained about, the question is then closed for completely different reason to the one you complained about? And how was this question not "Questions relating to observed chemical phenomena"?

I have reopened. See here.

See the timeline. It was closed way before your edit. I voted to reopen as soon as I saw your edit but nobody followed me. Thankfully, @ortho stepped in to perform the task. Unfortunately, unless you’re a diamond moderator, it takes five votes to reopen (but the same goes for closing). Also, any custom message always defaults to the ‘off topic’ close reasen for technical reasons.

@Jan i'm distinguishing between it saying "on hold" and it saying "closed". The timeline might not use the phrase "on hold", and might just call both "closed" regardless.  My edit was when the question was marked "on hold" and not yet marked "closed". See the chat link posted by orthocresol where I mentioned that re "on hold" and "closed" and he replied re that.

@barlop [on hold] and [closed] are two different names for practically the same state; after five days of being [on hold] the system automatically (no user input required) switches it to being [closed]. While a question is [on hold], an edit will automatically push it to a review queue for reopening. Read more in the help centre

@Jan  yeah, otho mentioned much of that in the chat link. It's still not wrong to distinguish the two.

Life on Earth evolved in aquaeous solutions. Cells are made up mostly of water; most cells are approximately $70~\%$ water. Water is thus ubiquitous in biochemical reactions. Only few enzymes go great lengths to explicitly exclude water from active sites to prevent undesired side reactions.
Concerning the production of vinegar, that usually starts from some kind of liquid mixture — e.g. wine — which has a high water content. The initially resulting alcoholic solution is also very aquaeous. As stated above, there is ample water.

Thanks. . I understand there are these two reactions too http://www.bbc.co.uk/schools/gcsebitesize/science/triple_edexcel/organic_chemistry/organic_chemistry/revision/6/   ethanol + oxygen → ethanoic acid + water  and ethanoic acid + ethanol <--> ethyl ethanoate + water   Is the water produced by those reactions a lot less than the water in the wine?

@barlop Yes. That water is at best equimolar to ethanol/acetate. Alcoholic fermentation rarely produces more than $15~%\ \mathrm{v/v}$ and alcohol is less dense and has a higher molecular weight than water so less moles of alcohol give more volume than water.

and by alcoholic fermentation, you mean the reaction(s)/process(es) that takes alcohol as input and produces acetic acid? (as opposed to the one that takes sugar as input and produce alcohol)

@barlop I meant the process that takes sugar and produces alcohol – the process typically used to create the substrate of vinegar fermentation (the one that produces acetic acid). By the way, alcohol concentrations at which the reactions you mention would become significant with regard to water production are those concentrations that no known organism is capable of surviving in.

thanks, also.. I have read https://en.wikipedia.org/wiki/Acetic_acid_bacteria "Some genera, such as Acetobacter, can oxidize ethanol to carbon dioxide and water using Krebs cycle enzymes."   I normally associate krebs cycle with respiration and the electron transport chain. Would I be correct in thinking that this oxidative fermentation reaction  ethanol+O2->acetic acid + water,  is not respiration, it does not use the electron transport chain?


text:
How to determine DDVP quality before appplication
The question is quite straightforward I need to determine the quality of  2,2-dichlorovinyl dimethyl phosphate (commonly abbreviated as DDVP) it is solved in xylene (50%). The purpose is to know if it can kill the insects. The determination must be fast and an a person with few knowledge of chemistry must be capable of carry it out.

Are the people who apply the pesticide getting the 2,2-dichlorovinyl dimethyl phosphate dissolved in xylene from different sources? The general notion being that the  2,2-dichlorovinyl dimethyl phosphate concentration varies depending on the source?

There are three isomers of xylene but density is between 0.86 and 0.88. 2,2-dichlorovinyl dimethyl phosphate has a density of 1.415. So a simple hygrometer test could tell what the % 2,2-dichlorovinyl dimethyl phosphate was. // It seems that this is a nasty pesticide which is not only toxic to insects...


text:
Why is AlCl₃ is known as aluminium chloride while PCl₃ is known as phosphorus trichloride?
Why is $\ce{AlCl3}$ is known as aluminium chloride while $\ce{PCl3}$ is known as phosphorus trichloride?
I think it is due to variable valency if phosphorus. Is it true?

Because there is only one aluminium chloride, but more than one phosphorus chloride.

In case it is not obvious, Loong's answer lists the rules: direct link.

Names of binary molecular compounds (which contain only two different elements, normally nonmetals, and do not contain positive and negative ions) can contain the prefixes mono, di, tri, tetra, penta, etc.
Normally, names of ionic compounds do not use these prefixes to indicate the number of positive or negative ions. $\ce{CaCl2}$ is calcium chloride, not "calcium dichloride".
However, aluminum chloride, $\ce{AlCl3}$, is sometimes called aluminum trichloride which is not incorrect in this case because it actually is a molecular compound (it has very polar aluminum-chlorine covalent bonds) even though it looks like it should be ionic since it contains metal and nonmetal elements typical of ionic compounds.
 $\ce{MnO2}$ is often called manganese dioxide rather than manganese(IV) oxide for the same reason. Usually, it is best to play it safe with compounds like these and use the ionic names:
$\ce{FeCl3}$ is ferric chloride or iron(III) chloride rather than "iron trichloride." $\ce{PCl3}$, a molecular compound (expected since only nonmetallic elements are present), is commonly called phosphorus trichloride, but is sometimes called "phosphorus(III) chloride."


text:
Nucleophilic Substitution Reactions help
Which halogenoalkane reacts fastest with sodium hydroxide?
A. 1-iodobutane
B. 1-chlorobutane
C. 2-chloro-2-methylpropane
D. 2-iodo-2-methylpropane
Iodine has a larger radius, thus less electronegative making it a better leaving group. And since OH is a protic substance it favors Sn1, which is a tertiary, so the answer must be d... but it says that the answer is b.. how????

As others have noted below, this seems line an $S_{N}2$ reaction and iodine is an excellent leaving group, so the answer seems to be "A".  Let us know if you hear back about why "B" is correct or if the answer key was in error.

Yes iodine's good leaving group.  the leaving ability is best described by the Gibbs energy change that occurs.
So since chlorine is a poor  leaving group so that make the reaction slower as less  energy is available.
So I think  that you are right so far.

Iodide is a much better leaving group than chloride.

ohh! sorry about that I just thought the reverse case ;

Also, this looks like a better candidate for $\mathrm{S}_{N}2$ chemistry...


text:
Increase in Viscocity in Gases with Temperature
Why does viscocity increase with temperature in gases? When they gain kinetic energy, they should be more free right?

For more details refer: Temperature dependence of viscosity - Wikipedia

From the basis of the kinetic theory of gases, viscosity in gases increases with temperature because the increase in kinetic energy will increase the frequency of collisions between the molecules. This increased number of collisions decreases the ability of the molecules to diffuse in the desired direction (i.e. the gas doesn't flow down the pipe to a reactor).

A short, qualitative answer:
In gas, the addition of heat increases the average velocities of molecules, which means more intermolecular collisions, which means more intermolecular interactions that limits macroscopic mobility of fluid in certain direction.
In liquid it's the opposite - heating the system increases the average velocities of particles which means less intermolecular interactions with other molecules.


text:
Density of moist air
Why does moist air have lesser density than dry air? When the amount of vapor inside increases, the mass increases and hence won't the density as well increase?

The composition of dry air is about $78\%$ $\ce{N2}$, $21\%$ $\ce{O2}$ and $1\%$ $\ce{Ar}$.  The molecular weights of these compounds are:
$\ce{N2} = \pu{28 g/mol}$
$\ce{O2} = \pu{32 g/mol}$
$\ce{Ar} = \pu{40 g/mol}$
So, the average molecular weight of dry air is given as:
$\pu{(0.78 * 28 g/mol) + ( 0.21 * 32 g/mol) + ( 0.01 * 40 g/mol) g/mol = 29 g/mol}$
The molecular weight of water is only $\pu{18 g/mol}$.
So, a molecule of air being replaced by a molecule of water results in an air molecule with an average molecular weight of $\pu{29 g/mol}$ being replaced by a water molecule having a molecular weight of only $\pu{18 g/mol}$.
This scenario assumes a constant temperature and pressure when humidifying the air.

Important starting principle is that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or atoms in the case of monatomic gases)...

@DJohnM , of course very true. Worth pointing out, thanks.

If you considered the air within a given volume in two rooms, one with moist air and one with dry air, moist air will have some water vapor in place of other molecules that make up air (e.g. oxygen and nitrogen). A Water molecule has a lower mass than either oxygen or nitrogen, thus the volume of moist air will have a lower mass than the volume of dry air and therefore a lower density.


text:
What goes on the NFPA 704 for multiple chemicals?
If you have a chemical that is classified (Health, Fire, Reactivity) 2, 1, 2. And another chemical that is 1, 3, 2. If these two chemicals are stored together, what is put on the storage container/building? What is the rule for this; the largest in each category is displayed, 2, 3, 2. And the specific hazards are listed?

"Section 4.2.3.3 of NFPA 704 provides three different methods to rate multiple chemicals. In addition, professional judgment must still be used to increase or decrease the rating to more accurately assess the degree of hazard, perhaps due to quantities, or synergistic effects of the chemicals, etc. The ratings for a
chemical that contains multiple ingredients should be obtained from data for the mixture as presented on the SDS."
~From this NFPA FAQ

According to Section 4.2.3.3 of NFPA 704:

Where more than one chemical is present in a building or specific
  area, professional judgment shall be exercised to indicate ratings
  using the following methods:
(1) Composite Method. Where many chemicals are present, a single sign
  shall summarize the maximum ratings contributed by the material(s) in
  each category and the special hazard category for the building and/or
  the area.
(2) Individual Method.  Where only a few chemicals are present or
  where only a few chemicals are of concern to emergency responders
  (taking into account factors including physical form, hazard rating,
  and quantity), individual signs shall be displayed. The chemical name
  shall be displayed below each sign.
(3) Composite–Individual Combined Method. A single sign shall be used
  to summarize the ratings via the Composite Method for buildings or
  other areas containing numerous chemicals. Signs based on the
  Individual Method shall be used for rooms or smaller areas within the
  building containing small numbers of chemicals.