text:
How does the HCl-KCl Buffer work?
I have just been studying the $\ce{HCl}$-$\ce{KCl}$ 'buffer', but there are still quite a few things I am uncertain about. I would appreciate any help in clearing up some questions I have.
What I understand (possibly incorrectly) from my research so far:
Water is the weak acid acting as a buffer in this system. The $\ce{HCl}$ and $\ce{KCl}$ in the system are there to increase the solution's ionic strength, which somehow improves water's buffer capacity.
In high ionic strength solutions, the 'standard' equilibrium equation has to be changed to include the activity coefficients. For water's equilibrium, as the ionic strength increases, the activity constants decrease and so the K value decreases.
If anything above is incorrect, please point out my misunderstanding. There are still a few questions I can't find answers to:

Why does the increased ionic strength of the solution improve water's buffer capacity?
Isn't a higher equilibrium constant required to make water a better buffer? Then isn't it bad that the ionic strength decreases K?
Could any other salt have been used to increase the ionic strength, or is there some specific reason $\ce{HCl}$-$\ce{KCl}$ is used?

I'm finding this 'buffer' system to be extremely confusing, so any help answering these questions would be greatly appreciated.

Introduction
Let's define buffer capacity quantitatively as
$$\beta=\cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}=-\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})}$$
that is, the relationship between concentration (in equivalents) of strong base ($c_\mathrm{b}$) or acid ($c_\mathrm{a}$) added to a solution and its change in $\mathrm{pH}$. From now on I'll assume we're adding a monoprotic base or acid, so "equivalents" and "moles" (and their concentrations) can be used interchangeably.
Pure water
Pure water has a very low buffering capacity - its $\mathrm{pH}$ is very sensitive to the addition of acids or bases. For instance, imagine we add a concentration $c_\mathrm{b}$ of base to pure water:
$\ce{H2O + B- <=> OH- + BH}$
The self-ionisation equilibrium of water will be displaced:
$\ce{[OH-]} \approx c_\mathrm{b}$
and therefore
$ c_b \approx \ce{[OH-]} = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}}$
so, taking the derivative,
$\beta_{\ce{OH-}} = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})} = \cfrac{\mathrm{d}}{\mathrm{d}(\mathrm{pH})} 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \ln{10}$
it's easy to show that, similarly, for the addition of acids,
$\beta_{\ce{H+}} = -\cfrac{\mathrm{d}c_\mathrm{a}}{\mathrm{d}(\mathrm{pH})} = 10^{-\mathrm{pH}} \ln{10}$
and, combining the buffer effect of both semi-systems, we get the total buffer capacity of water:
$\beta_{\ce{H2O}} = \left( 10^{-\mathrm{pH}} + 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} \right) \ln{10}$

Weak acid/base pair
1:1 solutions of weak acids/bases are a typical buffer system around $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ for the buffer. To understand why, let's imagine we add a weak acid/base pair, $\ce{HA/KA}$, with total concentration $C_\mathrm{A}$, to which we later add a certain concentration of base, $c_\mathrm{b}$. This weak acid/base pair will be described by the equilibrium
$K_A = \cfrac{\ce{[H3O+] [A-]}}{\ce{[HA]}}$
which, taking into account that $C_\mathrm{A} = \ce{[HA]} + \ce{[A-]}$, implies
$\ce{[A-]} = C_\mathrm{A} \cfrac{K_\mathrm{A}}{\ce{[H3O+]} + K_\mathrm{A}}$
From charge balance:
$\ce{[H3O+]} + \ce{[K+]} = \ce{[OH-]} + \ce{[A-]}$
Taking into account that $\ce{[K+]}$ is equal to the formal concentration of $\ce{KA}$ we added, so $c_\mathrm{b} = \ce{[K+]}$, and the previous expression for $\ce{[A-]}$:
$c_b = \cfrac{K_\mathrm{w}}{\ce{[H3O+]}} - \ce{[H3O+]} + C_\mathrm{A} \cfrac{K_A}{\ce{[H3O+]} + K_\mathrm{A}} = 10^{\mathrm{pH}-\mathrm{p}K_\mathrm{w}} - 10^{-\mathrm{pH}} + C_\mathrm{A} \cfrac{10^{-pK_A}}{10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}}}$
So the buffer capacity of the solution will be:
$\beta = \cfrac{\mathrm{d}c_\mathrm{b}}{\mathrm{d}(\mathrm{pH})}= \left(10^{\mathrm{pH}-\mathrm{p}K_w}  + 10^{-\mathrm{pH}} + C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \right) \ln{10}$
Note that the first two terms are the buffer capacity of water, so the contribution of the acid/base pair is
$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{A}} \right)^2} \ln{10}$
Additive buffer capacity contributions can be calculated this way.
For instance, for a typical acetate buffer ($\ce{HAc} \ 0.2 \mathrm{M} $, $ \ce{NaAc} \ 0.2 \mathrm{M} $; $\mathrm{p}K_\mathrm{a} = 4.76$, $C_\mathrm{A} = 0.4\mathrm{M}$), the buffering capacity looks like this:

Ionic strength
Since buffers are moderately concentrated electrolyte solutions, the assumption that activity coefficients in them are approx. 1 shouldn't be automatic. How does this affect the calculations of buffer capacity we've performed until now?
Let's define an effective equilibrium pseudo-constant, $K'$, operating on concentrations, that is mathematically equivalent to the actual equilibrium constant, $K$, operating on activities:
$K_\mathrm{A} = \cfrac{a_{\ce{H3O+}} a_{\ce{A-}}}{a_{\ce{HA}}} \iff K'_\mathrm{A} = K_\mathrm{A} \cfrac{\gamma_{\ce{H3O+}} \gamma_{\ce{A-}}}{\gamma_{\ce{HA}}} = \cfrac{\ce{[H3O+]} \ce{[A-]}}{\ce{[HA]}}$
So reflecting the impact of ionic strength on buffer capacity becomes a question of evaluating $K'_\mathrm{A}$ and using it to replace $K_\mathrm{A}$ in our previous expressions. For instance, we can use the Debye-Hückel relationship:
$\mathrm{p}K'_\mathrm{A} = \mathrm{p}K_\mathrm{A} + A \left( 2 z_{\ce{HA}} - 1 \right) \left( \cfrac{\sqrt{I}}{1 + \sqrt{I}} - 0.1 I\right)$
where $A$ is a constant ($\approx 0.51$ at room temperature), $z_{\ce{HA}}$ is the charge of the conjugate acid and $I=\frac{1}{2}\sum c_i z^2_i$ is the ionic strength of the solution. Note that $\mathrm{p}K'_\mathrm{A}$ has to be solved recursively, however, as it depends on $I$, which depends on the concentration of the different ions, which depend on $\mathrm{p}K'_\mathrm{A}$.
A similar correction can be applied to $K_\mathrm{w}$ to obtain $K'_\mathrm{w}$.
Applying this ionic strength correction to our acetic/acetate $\pu{0.4M}$ buffer above, we can see the impact of ionic strength:

The $\ce{HCl/KCl}$ buffer
So what about the $\ce{HCl/KCl}$ buffer? Unlike the example of acetic/acetate, $\ce{HCl}$ is a strong electrolyte - so it dissociates completely at all values of $\mathrm{pH}$. So we can treat it in two ways: as simply a water buffer with a different starting $\mathrm{pH}$, or as an acid/base conjugate pair buffer with a $\mathrm{p}K_\mathrm{A} < 0$. Both are equivalent, as the acid/base contribution is negligible compared to the water contribution due to the $\mathrm{p}K_\mathrm{A}$ value.
For instance, for a typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer ($\mathrm{p}K_\mathrm{A} = -6.3$, $\mathrm{pH} \approx 0.7$, $I \approx 0.8$), the contributions to buffer capacity are as follows:
$\beta_{\ce{H2O}} = \left( 10^{\mathrm{pH}-\mathrm{p}K'_\mathrm{w}} + 10^{-\mathrm{pH}} \right) \ln{10} \approx  10^{-\mathrm{pH}} \ln{10} = 0.4594$
$\beta_{\ce{HCl/Cl-}} = C_A \cfrac{10^{-\mathrm{pH}-\mathrm{p}K'_\mathrm{A}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K'_\mathrm{A}} \right)^2} \ln{10} \approx C_\mathrm{A} 10^{\mathrm{p}K'_\mathrm{A}-\mathrm{pH}} \ln{10} = 5.865 · 10^{-8}$
So, for all effects and purposes we have a buffer that behaves just like water. It is, however, a buffer that behaves just like water in a region of the $\mathrm{pH}$ scale which has a relatively large buffer capacity by virtue of being close to $\mathrm{pH}=0$. For instance, this is what buffer capacity looks like around our typical $\ce{HCl}\ 0.2\mathrm{M}$, $\ce{KCl}\ 0.2\mathrm{M}$ buffer (the red dashed line marks the initial $\mathrm{pH}$):

Note that buffer capacity is highly asymmetrical around this point - adding acid increases the buffer capacity of this system, requiring increasingly more acid to decrease $\mathrm{pH}$, while adding base decreases the buffer capacity, making the system's $\mathrm{pH}$ more sensitive to further base additions. This can be easily visualised if we realise that the area under the $\beta$ curve represents the concentration that needs to be added to move in the $\mathrm{pH}$ scale.
Note also that, unlike in weak acid/base pairs, we cannot simply increase concentration to increase buffer capacity while retaining the same $\mathrm{pH}$ - as the concentration-dependent term, corresponding to the $\ce{HCl/KCl}$ base, is negligible. Ionic strength is irrelevant for $\beta$, as well, as $K'_\mathrm{w}$ and $K'_\mathrm{A}$ is not involved - $\beta_{\ce{H3O+}}$ is the only relevant contribution to $\beta$ in this very acidic region, and, by definition, it is not affected by activity concerns (unlike $\beta_{\ce{OH-}}$ or $\beta_{\ce{HCl/Cl-}}$).
Does that mean that $\ce{HCl/KCl}$ buffers with different concentrations all have the same buffer capacity? Not at all: but their different buffer capacity is mediated by $\mathrm{pH}$. In other words, different concentrations produce different $\mathrm{pH}$ which, in turn, produces a different $\beta$ - which is unresponsive to changes in $I$ that do not change $\mathrm{pH}$.
Two final plots to illustrate this. Imagine we prepare solutions of $\ce{HCl/KCl}$ (1:1) at different total concentrations $C_{\ce{Cl-}}=c_{\ce{HCl}}+c_{\ce{KCl}}$. If we compare the two contributions to $\beta$, it's clear that $\ce{HCl/KCl}$ increases buffer capacity almost exclusively through $\mathrm{pH}$; $\beta_{\ce{H3O+}}$ is proportional to $\ce{[H3O+]}$ which increases linearly with $c_{\ce{HCl}}$ as $\ce{HCl}$ is a strong electrolyte:

Conversely, what if we control for $\mathrm{pH}$ and change only $I$? Let's imagine a series of solutions containing a fixed concentration of $\ce{HCl}$ and varying amounts of a non-reacting, strong electrolyte without common ions - such as $\ce{NaBr}$ - so we can change the ionic strength of the solution without affecting neither the $\mathrm{pH}$ nor the $\ce{HCl/Cl-}$ equilibrium. The impact of $I$ on $\beta$ would look like this (note that the minimum $I$ corresponds to a solution with $\ce{HCl}$ and no $\ce{NaBr}$):

This is because although the $\beta_{\ce{HCl/Cl-}}$ and $\beta_{\ce{OH-}}$ terms do depend on $I$, these terms are many, many orders of magnitude lower than the $\beta_{\ce{H3O+}}$ term, which doesn't.
So, after this discussion, we can now directly respond to your questions.

TL;DR
1) Why does the increased ionic strength of the solution improve water's buffer capacity?
It doesn't. The buffer capacity of solutions of strong acids are dominated by the $\beta_{\ce{H3O+}}$ term, which (by definition of $\mathrm{pH}$) is not affected by ionic strength.
There are, however, other reasons to desire an ionic strength in that range - in biological systems, many molecules, particularly proteins, have a privileged ionic strength stability gap - above or below which they tend to denaturalise or precipitate, and outside of which their biological activity can be inhibited or poisoned. Although different systems will require different ionic strengths, buffers are typically prepared with ionic strengths in the range $0.1-1\mathrm{M}$ for that reason.
2) Isn't a higher equilibrium constant required to make water a better buffer? Then isn't it bad that the ionic strength decreases $K$?
The $\beta_{\ce{H3O+}}$ term, by construction, does not depend on $K'_\mathrm{w}$, so it is unaffected by ionic strength. The $\beta_{\ce{OH-}}$ term does, and as you mention, it is decreased at moderate ionic strength - although it is increased at higher $I$. In general, the impact of ionic strength on buffers is as follows: increasing ionic strength begins by lowering $\beta$, then, as $I$ keeps increasing, $\beta$ raises again.
3) Could any other salt have been used to increase the ionic strength, or is there some specific reason $\ce{HCl-KCl}$ is used?
Yes, any other salt could have been used to increase the ionic strength. Adding $\ce{KCl}$ affects the acid-base pair contribution to $\beta$, but as we've seen, that contribution is completely negligible - as will be the case with strong acids, which operate mainly through the $\ce{H3O+}$ contribution.

Thank you for this answer! This is the first time I've seen an actually expression formulated for buffer capacity -- I fully intent to steal it for use elsewhere!


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What are the products A and B in the following reaction?

What are the products A and B in the following reaction?

According to me, the $\ce{NH2^-}$ anion being a strong base would form an alkoxide ion. But since there are 3 $\ce{OH}$ groups it will form the most stable alkoxide ion and since a primary alkoxide ion is most stable the alkoxide ion will be formed at the $\ce{-CH2OH}$ site. This ion will undergo a nucleophilic substitution reaction forming an ether. But this is not the correct answer. If anyone could point out where I am wrong, please?
According to the textbook the correct structure of compound B is

I wonder why do you think that primary alkoxides are most stable.

@mykhal Since secondary and teritary have more alkyl groups attached to the alpha carbon.Alkyl grps are electron releasing in nature and thus will increase the electron density on the negative oxygen atom and thus making it unstable

Such differences are minor. In practice I doubt there would be much selectivity over which alkoxide is formed

@Waylander But in any case ether will be formed at one of the sites.As far as I know Kmno4 would not affect ethers.So Product B will have am ether group .But none of the options of this question satisfy this condition

I think there is an error in your textbook. MeI is going to react with any alkoxide to give a methyl ether, and you are correct that KMnO4 does not react with ethers, so there has to an ether in the product.  I am already suspicious of the quality of a textbook that suggests that there will be any selectivity in the formation of the Methyl ether.

@Waylander For some curiosity, I looked up for the starting material in the OP, and the suggested product B from the OP.  Yet both appear as unknown (so far) to Reaxys, and CA ...

@Waylander Thanks for clearing this thing up.Mistakes like these are very annoying because I will always think that I am missing something.

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How to turn acetaldehyde into acetic acid using one of the five chemicals on the list?
From the Russian State Chemistry Test:

It's easy to see $X$ is $\ce{H2O}$: water turns acetylene into acetaldehyde.
But what can turn acetaldehyde into acetic acid?
I checked: $\ce{NaOH}$ would turn it into sodium acetate, $\ce{CuO}$ would not react (you'd need $\ce{Cu(OH)2}$), ethanol would turn it into either an acetal or hemiacetal.
Would acetaldehyde react with formic acid to produce acetic acid? I tried to google for it, and found no mention.
Would water in the presence of a heat source somehow react with acetaldehyde? I googled and discovered that it might form hydrates, not acetic acid.
So I'm at a loss. If you consider the question too easy, you could give just a hint.

If you Google CuO oxidation of aldehydes you do get hits

I don't think it's a stretch to think that two copper(II) reagents would both do the trick, considering one is essentially just the hydrated form of the other...

https://chemistry.stackexchange.com/questions/50642/can-oxidation-of-ethanol-by-copperii-oxide-yield-copperi-oxide-as-one-of-the

Oxidation of aromatic aldehydes to acids by air oxidation catalysed by CuO is reported in Molecules 2008 13 948
None of the other reagents can oxidise aldehyde to acid.

Acetaldehyde isn't aromatic.

I know!! But given that CuO oxidises aromatic aldehydes it is not such a stretch to think it probably will oxide acetaldehyde. Plus none of the other options look even remotely right

This paper says that acetaldehyde gives high yield of acetic anhydride when copper salts are used as catalyst.

Perhaps, but remember that we are supposed to solve the conceptual difficulties, not eliminate options like in a competitive examination. However I agree that $\ce{CuO}$ has a better chance of oxidizing.

A friend who teaches chemistry has just said to me that CuO will first turn into Cu(OH)2 and only then the reaction will proceed. I don't know whether that's true.

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text:
How to convert micromolar concentrations into molar quantities, when the volume is also known
\begin{array}{c|cccccc}
\text{Sample} & \text{Ammonia} & \text{Phosphate} & \text{Silicate} &\text{Nitrate + nitrite} & \text{Nitrite} & \text{Mn} \\
 & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu g L-1}) \\
\hline
\text{Water} & 0.02 & 0.51 & 9.92 & 9.93 & 0.00 & 0.2 \\
\text{Core 1} & 189.74 & 6.13 & 24.56 & 0.73 & 0.73 & 320.2
\end{array}
The samples are from sediment cores. Different cores have different volume; I need to work in quantities rather than concentrations. I need to calculate the mole (actually micromole) quantities of the $\ce{Mn}$, $\ce{NH3}$, $\ce{NO3-}$ and $\ce{NO2-}$ from their concentrations. The volume for core 1 that is provided here is $\pu{6.86 L}$.
Perhaps I'm being silly, but I am not sure how to do this and how to present this data that is to plotted as follows: molar quantities of the nutrients and metals ($y$-axis) against the oxygen molar concentration ($\pu{\mu M}$) of the water ($x$-axis).
I am also supposed to explain the results in terms of redox chemistry and try to determine approximately at what oxygen concentration in the overlying water these redox reactions in the underlying sediment are initiated.

Molar concentration, or molarity, is moles per unit volume. If you know concentration and sample volume, you can calculate the moles of a particular component in the sample. Since your concentrations are already in micro-molar, when you multiply by the sample volume (in liters), you will have the micro-moles of each component.
For the Mn result, you will get the mass of Mn (in micrograms) when you multiply the test result by the sample volume, so you can convert that to moles using the molar mass of Mn.

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Why titrating strong base with weak acid doesn't end at pH 7?
If I understand this titration correctly, if we titrated 50mL 1.00M acetic acid with 1.00M NaOH, the equivalence point would happen at a pH greater than 7, and less than 50mL of NaOH would need to be used.
What I'm curious about is why the titration does not take 50mL of both and why it does not end at pH of 7.  I can accept the numbers above, but I would like to know why.
The reason I struggle with this is because since the sodium hydroxide is a strong base and will ionize almost entirely, its hydroxide ions will react with whatever hydrogen ions have been ionized from the acetic acid.  This neutralization will force more acetic acid to be ionized, and the process will repeat.  Then, the same volume and molarity of each will be used, and the titration will end with complete neutralization at pH 7.  Why is this not the case?  Can weak acids and bases not be forced to ionize in such a way?  Is there some point at which a weak base or acid will not ionize any more, no matter what?

Related:https://chemistry.stackexchange.com/questions/47705/buffer-formation-in-weak-acid-strong-base-titration

@Tyberius Thanks for that.  It definitely is related, although it is not the same question I'm asking.  I had not thought of the buffer effect in there, but it does make sense, yes.


text:
Why doesn't this β-hydroxycarboxylic acid undergo dehydration upon heating?

This is a β-hydroxyacid, but why doesn't it get dehydrated to give an α,β-unsaturated acid on heating?
The mechanism which I think should occur is:

What you've drawn isn't a dehydration, it's a retro-aldol reaction. You're losing formaldehyde instead of water. And your product isn't an alpha,beta-unsaturated acid.

@orthocresol I'm not sure of the mechanism that causes the conversion from beta hydroxy acid to alpha beta unsaturated acid. Could you give me a link? The mechanism in the image is just a guess...

Isn't the carbocation more likely to be captured by the carboxy group giving a 4-membered lactone?

@Waylander Which carbocation?

Depends on which O you think gets protonated first. If the carboxy group protonates the the alcohol OH can attack similar to acid-catalysed ester formation. The resulting B-lactone is a strained species,yet B-lactones exist.

@Waylander Why can't we just protonate the OH, and then decarboxylate leading to simultaneous loss of H2O, creating a new exocyclic double bond?

@SpaceSkiier  The decarboxylation will not proceed because there is no alpha proton to the carboxyl group

@Waylander To make things clearer, the final product I am talking about is methylenecyclohexane, which can be formed (at least theoretically) by simultaneous decarboxylation and dehydration of the last but one structure in Martin's answer. Kindly tell me why this cannot occur.

In order to obtain an α-β-unsaturated carbon acid from a β-hydroxy acid, you need a secondary or tertiary carbon in α position, because otherwise you would have to break a carbon-carbon bond, which requires a lot more energy then simply heating.
The mechanism is basically self catalysed, which is why this may happen easily with a little bit of heat. The acid may protonate intramolecular (or intermolecular) the hydroxy group. You have water as a nice leaving group, creating a carbocation, and abstraction of a proton in α-position leads to the olefin.
If you do not have that α-hydrogen, you are stuck with the carbocation. You might be able to obtain a lactone,like Waylander suggests, but I doubt this would easily happen with a little bit of heating, as spiro-compounds with four-membered rings tend to be quite unstable.

In your attempt you choose the less acidic hydrogen to transfer and then break a carbon-carbon bond resulting in a very unstable molecule which would rearrange to cyclohexanecarboxylic acid. I'd suspect you would need insane temperatures to overcome the activation barrier.

In the last but one structure you have drawn, we could've even removed CO2 and H2O in a single step, forming a new exocyclic double bond, leading to an alkene. Can you please explain why this would be incorrect?

@SpaceSkiier No I cannot. Sorry. You might want to [ask] a follow up question and someone else answers that.


text:
Relate phosphamic acid with phosphoric acid
It is a simple question, but I am not able to figure it out how the formula of phosphamic acid, which is developed from phosphoric acid. In other words, I want to know how to relate it with phosphoric acid? Also in naming some acids we use meta like in, trimetaphosphamic acid, my question is meta is used for what?
Please help me understand. Thanks.

Not able to understand from this link

I edited question now.

https://chemistry.stackexchange.com/questions/40506/what-does-the-prefix-meta-mean-in-inorganic-chemistry

From Merriam-Webster:

An acid $\ce{PONH2(OH)2}$ derived from phosphoric acid by replacement
  of one hydroxyl group by an amido group is called  phosphamic acid.
  It is also called phosphoramidic acid

For the 'meta' part, see this - What does the prefix meta mean in inorganic chemistry?

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


text:
Resonance energy
Which resonating structures are more stable . How to comment about stability among resonating structures . What are its criterea ?  Explain with simple examples with rule?

You should provide more information, for example by adding  the structures you are trying to compare and your attempt at reasoning the answer out.

This sounds like a homework question.

I have already mention tag of homework question

There is no hard and fast rule for comparing resonance structures, but it is always useful to follow the following steps:

the structure with more covalent bonds is more stable
the structure in which all atoms have an octet is more stable
the structure with less unlike charge separation is more stable
the structure with charges on carbon atoms are unstable
structures with positive charge on electronegative atoms are unstable.

Ty Daenys for u answer

Please help me to how to post pic on this site


text:
Why is the product of the reaction between iron and steam is iron (II, III) oxide and not iron (II) oxide or iron (III) oxide?
The reaction between Iron and steam in my textbook is given as this:
$$\ce{3Fe + 4H_2O <=> Fe_3O4 + H_2} $$
My question is why these following reactions are not possible or do not happen.
$$\ce{Fe + H_2O <=> FeO + H2}$$
$$\ce{2Fe + 3H_2O <=> Fe_2O_3 + 3H2}$$
I mean, how do the atoms know to form iron (II,III) oxide instead of Iron (II) oxide or iron (III) oxide in this reaction?

Recall that $\ce{Fe3O4}$ is basically just an equimolar mixture of $\ce{FeO}$ and $\ce{Fe2O3}$. So, in that way, both of the reactions that you mentioned do infact happen, and the result is what was mentioned in your textbook.


text:
Oxidation of zinc in NaCl solution
I'm working with a zinc sheet in water + Nacl solution, I have adjusted the pH to 10 to develop a ZnO layer and prevent all the faradaic reactions (passivation state). I've removed the oxygen with a pump (I've added nitrogen too). In this system, what are the reactions that could appear? How can I prevent them?


text:
Isn't 11-trans-retinal more stable than 11-cis-retinal?
11-cis-retinal can be found as a part of rhodopsin and it is the non-excited form of retinal. By absorbing light, retinal goes to a higher energy state and converts to 11-trans-retinal. At least I understand it so.
If all above is correct, I just don't get one thing. Shouldn't the cis form be less stable than trans (due to steric hindrance)?

In general yes. However, retinal doesn't exist on its own, it is bound to a protein, so that might well lead to a different preferred stereochemistry.

Also why exact enthalpies would matter photochemical reaction? Photon gives energy for specific reaction.

The chromophore is fixed in the protein as a retinylidine molecule and as such is also distorted  by amino acid groups  in such a way that it is on the pathway to isomerisation, i.e. is in a high energy state. The initial isomerisation process is very rapid, sub picosecond.  The photo-induced isomerisation starts a series of reactions in which the chromophore goes through several stages (intermediates) as it returns to its resting state. The effect of this is to allow an ion channel to open in the protein which hence acts as a transducer of photons to chemical signal.

Cis-retinal is less stable than the trans isomer.  The carotenoids ingested by the body are all trans.  Enzymes in the body convert them to trans-retinoic acid, then turn them into a fatty ester which is then converted by the enzyme retinoid isomerohydrolase to cis-retinal.  The higher energy state of the cis isomer makes it prone to photoisomerization back to the lower energy trans isomer.

The answer below is in very simplified terms and does not take the enzyme into account.
The photoexcitation will effectively move an electron from a π to a π* orbital. Thereby, a π bond is effectively broken and the molecule can rotate around what is now effectively only a σ bond. Thus, this rotation will lead to a thermodynamic equilibrium between cis and trans until the electron relaxes back to the ground state.
We do indeed expect the trans form to be more stable; therefore it is more likely for a cis-retinal molecule to be converted to trans-retinal than vice-versa. It is entirely possible that nature found this out as well — through random evolution — and selected the process that would yield a slightly higher probability of a photon inducing isomerisation. This process would lead to a slightly higher light sensitivity or the ability to better see in the dark.


text:
How do P-H bonds in phosphorus acids impart reducing properties?
My textbook states:

The oxoacids of Phosphorous which contain $\ce{P-H}$ bond have strong reducing properties. Thus, hypophosphorous acid is a good reducing agent as it contains two $\ce{P-H}$ bonds and reduces, for example, $\ce{AgNO3}$ to metallic silver.
These $\ce{P-H}$ bonds do not ionize to give $\ce{H+}$ and do not play any role in basicity.

On the basis of what's given in the text; how do these $\ce{P-H}$ bonds account for the reducing properties of phosphorus acids?

The $\ce{P-H}$ bond imparts an oxidation state of -1 per bond to the acid. Upon oxidation, these are converted into $\ce{P-OH}$ bonds, which imparts a +1 oxidation state per bond. You can clearly see this in the reaction:
$$\ce{H3PO2 + 4AgNO3 + 2H2O -> 4Ag + H3PO4 + 4HNO3}$$
There are two $\ce{P-H}$ bonds in $\ce{H3PO2}$, both of which are converted into $\ce{P-OH}$ bonds in $\ce{H3PO4}$.
Moreover the $\ce{P-H}$ bond has an enthalpy of 322kJ, while the $\ce{P-O}$ bond has an enthalpy of 335kJ/mol. This slight increase in bond energies drives the reaction in the forward direction. Resonances may also be a factor that contributes as well.

Answer seems convincing. So, all it has to do with the $\ce{P-H}$ bonds is with comparision with the product, right?

Yes, that right @Reeshab Ranjan

First of all, the electronegativity of hydrogen (2.20) is greater than phosphorous (2.19), so in the P-H bond, the P atom actually gets a +1 oxidation state, whereas hydrogen gets a -1 oxidation state.
It is a well known fact that negative oxidation state hydrogens are highly reducing in nature; hence, the H atoms attached to the phosphorous atom contribute to the high reducing nature of phosphorous acids having P-H bonds.


text:
What are the chemical process responsible for the warping of wood?
Background:
According to this Wikipedia article:

Wood warping costs the wood industry in the U.S. millions of dollars per year. Straight wood boards that leave a cutting facility sometimes arrive at the store yard warped. This little-understood process is finally being looked at in a serious way. Although wood warping has been studied for years, the warping control model for manufacturing composite wood hasn't been updated for about 40 years.

Factors contributing to the warping of wood include plant species, temperature and temperature changes, humidity and humidity changes and wet-dry cycles.
Discussion:
I'm not interested in which species are more prone to warping, how to prevent warping by treatments, etc. Nor am I interested in the biological processes of living or freshly cut wood. I'm interested in the processes happening over time on a molecular level that lead to macro-scale warping of wood. Water, in particular, seems to play a major role in the warping of wood, implying that possibly the lignin and cellulose matrix of wood is disturbed by the formation of hydrogen bonds. But if this is the case, or whatever the case, why would the microscale effects not cancel out rather than causing macro scale warping in a particular direction? This may be more of a physical than a chemical question, however.
Question:
What are the chemical process responsible for the warping of processed (as opposed to living or freshly cut) wood?
Edit 1:
A link to an interesting and relevant video was given in a comment, in which several types of wood were placed in a chamber which was evacuated of air, then pressurized with ammonia gas. The video also gave a link to a patent for this process. The following is an excerpt from the patent:

The rigidity of wood is the result of crosslinking between adjacent cellulose chains by water molecules which hydrogen bond between sites on adjacent cellulose chains. Anhydrous ammonia is extremely reactive with water, and it is believed that anhydrous ammonia scavengers water from wood, breaking the crosslinking between cellulose chains, and permitting the cellulose chains to slide, which renders the wood very flexible so that it can readily be bent or twisted. When the ammonia volatilizes from the treated wood, water vapor reestablishes crosslinking and rigidity.

The reason for this edit is to suggest that the disruption of hydrogen bonding between adjacent cellulose chains, when the wood is dried then rehydrated, for example, may also play a role in "natural" warping. I'm going to look into this more and I hope it gives somebody else a clue to a good answer.

It's probably both -- the swelling of wood when wet is more of a physical process, but the properties that lead to permanent deformations on subsequent drying likely have some manner of chemical explanation.

Related: https://www.youtube.com/watch?v=9Z0SsAyHKzc

It's rather a biological question, behavior of dying wood depends on it's structure as a whole not directly on single molecules.

@Mithoron , thx for the input. I clarified my question to read that I'm interested in warping of processed (dead) wood. I do think there is clearly some non-biological process happening at the molecular level that causes warping, on repeated wetting and drying of a board for example.

@hBy2Py , I agree that the swelling of wood upon wetting may not require a chemical change. It does seem that the permanent deformation would require some change at the molecular level. There are examples where this is not the case, however, like bending a steel rod. Thus it may be that physical processes dominate the warping of a wood board, though my feeling is that's not the case...that's why I've asked ;)

Great video @CurtF. (I recommend everyone take a look). Here's what the linked patent said: "The rigidity of wood is the result of crosslinking between adjacent cellulose chains by water molecules which hydrogen bond between sites on adjacent cellulose chains. Anhydrous ammonia is extremely reactive with water, and it is believed that anhydrous ammonia scavengers water from wood, breaking the crosslinking between cellulose chains, and permitting the cellulose chains to slide, which renders the wood very flexible so that it can readily be bent or twisted." Then it hardens when water returns.

Since this, as u say, has already been studied at length I don't think I can give a better answer. As u say its the water that cross links the cellulose chains. My speculation is that once wood dries considerably the cellulose fibers are released to deform into a form that they would have grown into to compensate for the stresses they were exposed to in nature. I believe there is a lot of unevenness in the cellulose chains in the wood. Actually, I believe rived wood doesn't warp because it is split along the grain of the wood. Sawn boards warp because they are cut across the grain in many plac

Above 320°C cellulose goes from being crystalline to amorphous.  It looks like the mentioned process simply alters the form of some of the cellulose.  More amorphous cellulose = more flexibility... more crystalline cellulose = more rigidity

Warping (be it a bow, crook, cup, or twist); as well as loosening of knots, checking, splitting, and collapse; of wood is caused by uneven shrinkage.
To explain uneven shrinkage chemically, consider wood as 50% cellulose.  Cellulose, which is a polymer consisting of linked glucose monomers, is arranged into fibrils.  Fibrils are part of larger structures which make up the cell walls of wood fiber.  Binding these cell walls together is lignin, which is a phenylpropanol polymer.  Cellulose and lignin (along with hemicellulose) are the major structural compounds in wood.
Cellulose in wood is a tangle of rigid crystalline cellulose and flexible amorphous cellulose.  While the structure of crystalline cellulose makes it water insoluble, amorphous cellulose has sites like its base monomer glucose which are capable of absorbing water. When water is absorbed both the shape and volume of the amorphous cellulose change.  While amorphous cellulose changes its shape, the crystalline cellulose (as well as lignin) retains its shape and volume.
Cellulose may also change shape and volume due to other factors such as temperature fluctuation and biological degradation.  Because (1) wood is not uniform; (2) moisture, temperature, and biological changes are rarely uniform throughout the bulk of a piece of wood; and (3) most wood is exposed to some degree of these changes over time; uneven shrinkage occurs.  Because of uneven shrinkage in wood, it warps.

These phenomena affects degradation of wood fibers in historical documents as well; however, I figured this is outside the scope of the question.


text:
How to effectively remove chloramine from tap water
I am looking for a way to remove or substantially reduce chloramine from tap water. From what I learned chloramine is created when ammonia is added to chlorinated water.
It is now a method of choice to disinfect water by many municipalities and from what I read it is much harder to remove than free chlorine itself. I read that activated carbon filter needs more time to remove chloramine than chlorine and that catalytic activated carbon filter can do it faster. Both of these apparently remove only chlorine and still leave ammonia? Is that correct?
How to remove ammonia then? I read that ion exchange can remove it. But then the water will be stripped of all dissolved solids and become acidic? To make it back pH neutral or alkaline I guess there should be another element that adds back minerals.
I basically want to achieve whole house water filtering that will give water free of chloramines and harmful organic compounds and suitable for general use (bath, shower, etc) and ideally also for drinking.

Small amounts of don't matter, so it's likely safe to bathe and drink. What are you worried about?

https://chemistry.stackexchange.com/questions/26968/test-for-chlorine-and-chloramine-in-city-water-supply

You can use Campden tablets (potassium metabisulfate) to remove chloroamine from water.
The recommended amount is one tablet for 20 gallons of water.


text:
Use of ethanol versus DMSO as a solvent

My professor has had the reaction shown in DMSO and ethanol as solvents but made clear these are used for different circumstances. I did not understand what the difference in circumstances are. Does it have something to do with being polar and aprotic and what alkene is being reacted with the cyanide?

The alkene being reacted with hydrocyanic acid here goes by benzylideneacetone, or 4-phenylbut-3-en-2-one in the language of the IUPACians. But you shouldn't view this molecule as an alkene, you should view it as an "enone", or $\alpha,\beta$-unsaturated carbonyl.

Enones are notable for having "regioselective" reactions where nucleophiles add to either the beta-carbon (via 1,4-addition, vinylogy chemistry) or the carbonyl carbon (via 1,2-addition, standard carbonyl chemistry) regions of the molecule.

In your question, you see we have a 1,4-addition of $\ce{H+}$ and $\ce{CN-}$. (After cyanide attacks the 4-position, the proton adds to O, the "1-position", and the enol intermediate then tautomerizes to a ketone.)
The choice of solvent can have some effect on influencing whether 1,2- or 1,4- addition dominates. Aprotic solvents can encourage 1,4-addition (conjugate addition) to be favored over 1,2-addition (carbonyl addition) (Source).
Your ethanol is a polar, protic solvent because the hydroxyl is capable of hydrogen bonding or donating the proton (its Bronsted-Lowry acidity is about that of water, pKa 15.7 vs 16). DMSO is a polar, aprotic solvent because it does not contain hydrogens that can H-bond nor are (appreciably) acidic. Use of DMSO can encourage 1,4-addition (source).
Simply changing the protic-aprotic nature of the solvent may not always be enough to effectively toggle the dominant regiochemistry between 1,2- and 1,4- addition motifs. It may depend on the other reaction conditions (such as temperature), the structural qualities of the particular enone substrate at hand, and the nature of the nucleophile. In the Hard-Soft Acid-Base Model, "hard" Lewis bases tend to add directly to the carbonyl (1,2) whereas "soft" Lewis bases tend to perform conjugate addition (1,4). For nucleophiles (Lewis bases) not on either extreme end of the hard-soft spectrum, changing the solvent type can sometimes be enough to change the regioselectivity.
I recommend you ask your professor whether there is a rule of thumb he wants you to use to determine the solvent's effect (such as "polar aprotic always gives 1,4"), or whether you should memorize specific examples (because there are many exceptions).


text:
Long lasting dry ice
I have a chemistry question but I'm not sure which subtopic it'd go under, so I'm asking it here. I'd really appreciate any help I can get, if you have an answer, or if you can tell me where I can find the answer. I should also mention I don't know anything about chemistry, so I'm sorry if my question sounds stupid :-)
I was wondering if it's possible for dry ice (the industrial ones or those that are sold at stores) to stay cold/frozen in a rather warm/hot room temperature for around 2-4 hours. Let's say blocks of 3x3x4 inches. I just want to know if that's feasible or not.

Are these blocks of dry ice allowed to be contained in something? Or do they have to be able to survive open to the room?

Yes, I'd like them to survive in the open room. I don't know the difference between dry ice and the packages you buy from grocery stores by the way. All I'm looking for is ice (in smaller size 3x3x4 in) that can stay cold for 3-4 hours. I just want to know if that's possible.  Thank you so much for your help.

Put them in a container. It could be open at the top, but should be enclosed in the bottom.
The dry ice will insulate itself with a layer of cold air/CO2. As long as they are not under direct heating illumination or the convection of air in the room is high, this should be sufficient for a suitable mass of dry ice.
We routinely keep dry ice in "open" unpowered freezers at my lab (with no refrigeration) for the duration of the day. If your room is very humid you can get buildup of ice / condensed water, so some insulation around the container might be prudent.

As your dry ice is open to the air I hope that your lab is well ventilated!

Be very careful about putting dry ice in closed containers. If the seal isn't tight or the lid is left slightly ajar, it might be fine (though you still have to worry about ventilation), but if it does from a good seal (which can even happen accidentally from water/condensation freezing the lid shut), in the worst case scenario you are looking at a dry ice bomb. These can be dangerous even in something as innocent as a plastic bottle; I've seen people with mauled fingers from the shrapnel.

Make the container from expanded polystyrene and this will work for an extended period, possibly days if you have enough dry ice to start with.

@porphyrin Yeah, I do not work in unventilated laboratories =) There are numerous alarms on numerous gases. As for OP's question I assume he understands the risks of carbon dioxide and death by respiratory acidosis, because he should have gotten (will get) that information where he got (will get) the dry ice.


text:
Good collection of images showing periodicity of the Table?
Mendeleev's recognition of the periodicity of atomic trends was a milestone of human understanding, the detection of a pattern in what had appeared to be chaotic data. I think the significance and sheer beauty of these patterns is lost on many people (layfolk and students alike). I am wondering whether anyone knows of a coffee-table book or some other collection of images that evoke the predicted nature of atomic properties such as atomic radius, ionization energy, melting point, etc. Sure, the information is available via interactive software, but I'm looking for something more aesthetic... if it exists yet.
Particularly, my search is provoked by this striking arrangement of oxidation state information. Were it to be cleaned up a bit, it would make a beautiful poster.

I have an entire chapter regarding periodicity in my textbook. Have a look at the images in it:
Atomic Radius:

Ionization Enthalpies:

A summary of periodic trends:

These are the pictures in my textbook. As a bonus, I'll even add a few examples from our day-to-day life:
Alkali Metals:
Lithium:

Sodium:

Potassium:

Rubidium: (its enclosed in glass because it spontaneously combusts in air)

Cesium:

You can notice that properties of elements periodically repeat. All the alkali metals have silvery appearance, with cesium having a little bit of yellow-golden in it. They are soft, and have exceptionally high reactivity.
Halogens:
Fluorine:

Chlorine:

Bromine:

Iodine:

You can see that all the halogens are quite colored, and most have low melting and boiling points. Thats why fluorine and chlorine are gases, while bromine is a liquid with thick vapors. Iodine, however is a solid due to higher attractions due to increased molecular mass. All the halogens are quite reactive, with fluorine and chlorine violently so.
Transition Metals (a few examples):
Iron:

Cobalt:

Nickel:

Chromium:

You can see how the transition metals are quite similar to each other. They are lustrous in appearance, are quite hard, and also display quite a few catalytic properties. For example, iron is used in the manufacture of ammonia, nickel is used to hydrogenate vegetables oils, palladium is used in catalytic converters, et cetera.

Added Halogens :)

Added Transition Elements!

Congratulations on your 1k rep. Keep contributing to the site like this.

@NilayGhosh Thank you. Definitely StackExchange and me won't be leaving each other for a while.


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Why do α-hydroxy ketones give Tollens' test?
I know that tautomerism causes the $\alpha$-hydroxy ketones to give the all the tests of an aldehyde. But how does it exactly tautomerise and what's the intermediate?
Does benzoin gives Tollens' test? If not, why? I'm really confused with this. I read somewhere that benzoin doesn't give Tollens' test as it does not have an alpha carbon to form enediol intermediate. Please help.

https://chemistry.stackexchange.com/questions/64363/why-do-alpha-hydroxy-ketones-respond-positively-to-tollens-fehlings-benedict?s=15|0.2154

Benzoin does give positive tollens  test.

The Tollens' test is done in base, which mediates tautomerization through the enolate:

The misconception that you have is that all alpha-hydroxy ketones can give a positive Tollens' test. Aldehydes give a positive test, and so only alpha-hydroxy ketones that can tautomerize to aldehydes will give a positive Tollen's test. My example above is 1-hydroxy-2-propanone, which tautomerizes to 2-hydroxypropanal. Most alpha-hydroxy ketones with names that match the pattern 1-hydroxy-2-alkanone will give a positive Tollens' test.
Benzoin will not give a positive test, since it cannot tautomerize to an aldehyde. What does benzoin form upon tautomerization?

Can you put the answer to your final question in spoiler blockquotes (>)? Or maybe just state it here? Is the answer  to "What does benzoin form upon tautomerization?" simply benzoin itself.

@GaurangTandon - Yes. Benzoin tautomerises to benzoin.

This question directly contradicts your final statement that benzoin does not give a positive test. I hope you may want to have a look here.

Please consider revising this answer, as Benzoin does give Tollens' test.

@BenNorris benzoin in basic medium gets oxidised to benzil through SET(Single Electron Transfer) and in this process converts $\text(Ag)^+$ ions into $\text(Ag)$ solid... So technically it gives tollens' test!

-1 because of wrongly saying that benzoin does not give tollen test https://youtu.be/xlaKvKfNidA

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


text:
Color transition
We have to determine which method is true in order to find which color is absorbed and which one is transmitted?

I have read in an example that $\ce{Cu^{2+}}$ absorbs red light but transmits blue.This is true if we see pyramid but not with respect to circle....How could we make jugdement in exams while solving mcq related to it?

The problem with your color wheel is that it presents each color as an equal wedge. They are not. Note that red is 180 nm in width and orange is 40 nm in width. Here is a better one that I made:

As you can see, there is a range of red wavelengths that can be absorbed to get transmittance in blue.

...then what do you think about pyramidal cycle....is it correct?

It is better than the circle, but it is still flawed in that it represents each color as equal width.


text:
While filling electrons, we follow Aufbau principle, but not while removing them. Why is this so?
I recently came across a question Why is the vanadium(3+) ion paramagnetic?, where the asker is wondering how $\ce{V^{3+}}$ is paramagnetic (he used Aufbau in reverse to remove the electrons), while the correct answerer to that question remarked that removing electrons must follow the order from outer shell to inner shell. For example, in the case of $\ce{V^{3+}}$ the electronic configuration of $\ce{V}$ is $\ce{[Ar] 3d^3 4s^2}$. The asker used reverse-Aufbau, which is $\ce{[Ar] 3d^0 4s^2}$; while the actuality was $\ce{[Ar] 3d^2 4s^0}$.
My question: Why does filling electrons follow a certain rule, while removing them follows a different rule?

Well in our textbook it is written that "while removing electrons always remove electrons from the last shell". AFAIK, The exceptions to this rule are few and dont occour till 4d series.

Aufbau works for the wrong reasons. The 4s orbital is not below 3d in energy. http://pubs.acs.org/doi/abs/10.1021/ed055p2 Many people have written on this already. https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium

The whole idea of "adding" or "subtracting" an electron from the electron configurations of different atoms, or even the same atom in different ionization states, is fundamentally wrong. The overall energy for every possible electronic configuration must be determined separately for every combination of the number of electrons and number of protons in an atom, and whichever has the lowest energy is the ground state, irrelevant what the result is for the element or ion next to it.

Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from $\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$ to $\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not only an electron but also a proton to your atom.
When you remove electrons to get to a cation, you only remove electrons. Thus it is a different situation, with different interactions between nucleus and electrons, which affects the energy of the orbitals.
Also keep in mind, that the orbital picture is just a simplification. Strictly speaking (and from a purely theoretical point of view) orbitals do not exist, they are merely a mathematical crutch to solve the Schrödinger Equation. Although in practice this picture works surprisingly well, hence people use it a lot.
A proper quantum chemical calculation will give you the correct ground state, but you might have problems identifying it as something like $\ce{[Ar] 4s^2 3d^0}$. It will be a mixture (linear combination) of many electron configurations, although $\ce{[Ar] 4s^2 3d^0}$ might be the most important one for $\ce{V^{3+}}$.


text:
What is the proton NMR spectrum of p-methoxyphenol?

I think there will be two pairs of doublets, since the hydrogens adjacent to the $\ce{OCH3}$ and those adjacent to the $\ce{OH}$ group will be in different environments. The answer, however, is that there'll be a singlet for 4 Hydrogens.
To give some context, I am extremely unused to the proton NMR of benzene derivatives, and any (introductory) links telling me a bit about it would be appreciated. I'm aware of the theory behind NMR spectrosocopy, the 'n+1' rule, and the ways to tell if two Hydrogens are magnetically equivalent.

Two peaks won't get separated, because of similar environment, I guess; you might get sth more detailed then singlet with hi-res though.

But one has OCH3 and the other has an OH..

And that is very similar.

I see.. but if it were something like a OCH2CH2Ch3? What about a Fluorine directly attached?  Basically, at what 'point' do I know if it is similar enough for the 2 doublets to become a singlet?

@Mithoron  According to a very rough estimation (http://chat.stackexchange.com/rooms/57692/p-methoxyphenol-nmr); around 600 MHz (and above) might offer a separation.

@Saad In absence of experimental data and empirical rules, quantum chemical calculations at different levels of sophistication may provide a prediction.  Not only in NMR spectroscopy only the experiment changes a "state of guessing" into the one of "now I know".

In theory, this structure is expected to provide four different signals:

a broad singlet, due to the potential exchange of the phenolic OH
a singlet accounting for three protons of the methyl group
and signals for the remaining four aryl protons ... which may be tricky

The chemical shift of the aryl protons is influenced by the two, para standing towards each other, subsitutents; here: OH and methoxy.  With their electronic properties, they influence the electron density of the aromatic ring, and eventually alter the location of the signals of the aryl H's. The centre of theses signals may be predicted by empirical increment rules which many textbooks about NMR include, like Silverstein, for example. Depending on the experimental conditions, accidentally they may simplify to a signal which only looks like a singlet:

(source, recorded in $\ce{CDCl3}$ at a Lamor frequency of 89.56 MHz)
Addition: The distance between the aryl H's and the phenol or the methoxy group (at least three bonds away) and their very similar electronic influence on the aromatic ring may render the aryl H's very similar.  So it takes a spectrometer with higher measurement resolution (measurement points per ppm) and higher magnetic field to discern them.  Rayment et al. published more recently in J. Org. Chem., 2012, 77, 7052–70602012 (doi 10.1021/jo301363h) data recorded again in $\ce{CDCl3}$, on an instrument with stronger magnetic field.  To quote:

1H NMR (400 MHz, CDCl3) δH 3.78 (3H, s), 4.77 (1H, br s),
  6.76−6.82 (4H, m)

Even here, some discern, yet likely the two doublets overlap a bit, as a prediction by ChemDoodle suggests:

(the other doublet corresponds to the second pair of aryl H's).  Note the spread of the scale, too.  While I imagine even more advanced NMR spectrometer may eventually resolve the doublets (600 MHz is not so uncommon, and a few of about 1 GHz are around, too), I doubt someone will invest the more precious measurement time on these to delve more into detail and publish the then very high resolution 1H NMR, instead of collecting 1H NMR data of proteins and other natural products.
Addendum and correction:  Following the comment by @long, I reread the relevant pages in  Silverstein again.  Stating the for aryl H's would yield two doublets is wrong.
There are three types of coupling each of the aryl protons in p-methoxy phenol participates which are illustrated below:

(Source for the values of coupling constants: Silverstein, loc. cit., 6th edition, 1998, p. 212)
In blue is the proton which signal is of concern.  In the top row, it is the left in ortho position towards the methoxy group.  This proton couples with the close ortho standing proton (only three bonds away, hence $^3J$), typically with a coupling constant of $\pu{9 Hz}$, hence yielding according the $(n + 1)$ rule a doublet.  Simultaneously the proton marked in blue couples with the meta standing one with a typical coupling constant of $\pu{3 Hz}$; hence the previous doublet would become a doublet of doublet (dd).  Even farther distant, the para standing proton, hence coupling with an even lesser coupling constant which not every NMR spectrometer is capable to resolve -- but in principle yielding ddd, a doublet of doublet of doublet -- solely for this proton, up to $2 \times 2 \times 2 = \pu{8 signals}$ may become observable.
Following the theory further, the proton marked in blue of the bottom row is not identical to the one marked in the top row, and consequently, this proton may yield up to 8 observable signals, its own ddd.  This already adds up to 16 lines, which however is not (yet) the end of nitpicking theory as -- according to Silverstein's textbook -- the aryl Hs on the left hand are not equivalent to the ones on the right and a total of 32 lines possibly to observe.
With the graphical example provided by the Silverstein book for p-chloro nitrobenzene

which will serve for a follow-up question by mine, to record the aryl H's in a signal in shape like a singlet was by chance.

But why do two doublets appear as a singlet? The resolution isn't high enough?

Thank you. It makes some sense now, but could you read my comment on the question? Also, this may be unrelated, but why does the strength of the magnetic field affect the 'separation' of the doublets?

@Saad  better here : http://chat.stackexchange.com/rooms/57692/p-methoxyphenol-nmr

Suffice to say that the chemical shifts are coincidental within the natural linewidth. Sometimes that just happens. Under different conditions (different Bo, solvent, temperature, concentration etc), there may be more of a discernible difference. Note that this is an AA'BB' spin system, not an AB spin system. Also the splitting diagrams taken from Libretexts do not represent an AB spin system - they are for an AX system. Only when the chemical shift difference is comparable to the coupling constant (Δ δ>~3J) does the system begin to show second order effects.

Also, the expected splitting for a p-disubstituted benzene is not a pair of doublets as you suggest - AA'BB' are classically stereotypical and even at lower fields typically display 6 lines. You should not ignore these extra lines as they are important diagnostic features of the spin system that can help characterise the molecule.

I'm extremely confused now... So there won't be 2 doublets? Why will there be 6 peaks? Please keep in mind that I'm only a highschooler, so you'll have to simplify things

@Saad On first sight it may look like doublets, but their appearance is misleading.  The answer was rectified.

I am 100% sure that I am not expected to know this stuff. In the actual question, we were given a 4H singlet at d=6.8, and were also given the peaks for the -OCH3 protons and OH proton. I worked out that it was going to be a benzene ring with a OCH3 and OH attached, but the answer said that 'the pattern fits 1,4 substitution', and placed the OCH3 and OH as shown. How would we look at a 4H singlet and then judge whether the 2 groups are (1,4), (1,3) or (1,2)?

Surprisingly in-depth answer. +1


text:
Oxidation of alkyl side chain of a benzene ring
We do know that the side alkyl chain of benzene ring is oxidized by hot KMnO4 to benzoic acids. But how exactly it happens. I couldn't get a proper mechanism for this reaction.
Please help me out

https://chemistry.stackexchange.com/questions/33735/oxidation-of-alkyl-side-chain-using-kmno4

Oh...I hadn't noticed that this question has already been put up....thank you for that....


text:
The best method(s) to identify/verify the presence of a small organic molecule
What is the fastest or the most definite method (preferentially a single unexpensive method) for verifying the presence of a certain small orgnic molecule which is expected to be of >90% purity in a "powdry" sample (known solvent is also availble)?
Thank you in advance

Proton NMR, if the molecule is 90% or better should be fairly clear assuming it has protons in the structure


text:
Production of ozone from oxygen and Le Chatelier's principle
My NCERT textbooks says as follows:

Since the formation of ozone from oxygen is an endothermic process, it is necessary to use a silent electrical discharge in its preparation to prevent its decomposition.

Reaction is:
$3\ce{O2}\ce{->}2\ce{O3}$ $\triangle H$=+142 kJ mol $^{-1}$
And somewhere on the net, I have learnt that silent sparkless discharge is needed as sparks may increase the temperature of the vessel/setup environment.
My question is, isn't the text contradicting the very principle of le Chatelier for endothermic processes, i.e the temperature should be as high as reasonably viable and allowable (for kinetics reason)?

Explain the downvote please.

I didn't down vote, just want to know the context of your question. Could it be that in the context of what you read "somewhere on the net", ozone formation is not the only process occurring so the higher temps were adversely affecting some other factor?

@J.Ari I just gave what I have learnt on the web, from some Q and A site which just cannot explain more for my question.

Seems you're not getting that you explicitly don't want to get to equilibrium here. That's why discharge is used.


text:
Reaction of potassium cyanide with 2-(chloromethyl)furan
What would the mechanism for the reaction between 2-(chloromethyl)furan and potassium cyanide be, as detailed below?

I know that for 6-membered rings like pyridine, a $\mathrm{S_NAr}$ reaction can occur. However, I'm not sure as to how the mechanism differs between the two in a way that the cyanide group is added at the 5-position instead of the methyl group.

google "vinylogous SN2 reaction"

I had a look and I'm not sure how to relate it to the mechanism in question

Probably SN1 not SN2, as outlined in Blaise's answer.

Cyanide ion attacks the 5 position of the furan. The 4,5 double bond migrates to 3,4. The 2,3 double bond migrates to the Me group kicking out Chloride and giving the exomethylene (double bond outside tbe ring) intermediate. Then the bonds reorganise to re-aromatise giving the product shown.
I hope this makes sense. I have no access to a drawing package.

Although the answer was already appropriately provided, for anyone interested in further reading, the reaction is covered in detail by Divald and coworkers:1

Essentially, the proposed mechanism starts from the resonance structures of 2-chloromethyl furan, where the chlorine bond is treated as ionic and the electron-donating property of oxygen stabilizes the shared positive charge similar to a benzyl cation. The 5' position is therefore electrophilic (and is presumably further activated by the inductive effects of a neighboring oxygen), resulting in a 60/40 yield of 5/2-methyl substitution.
It is however mentioned early in the paper that 5' substitution is only seen in protic solvents. This could be explained in the proposed mechanism, where after addition to the 5' position, elimination of the geminal hydrogen and tautomerization proceed through exchange with a protic solvent.

Reference

Divald, S.; Chun, M. C.; Joullié, M. M. Chemistry of 2-(chloromethyl)furans. Reaction of 2-(chloromethyl)furans with aqueous potassium cyanide and other nucleophiles. J. Org. Chem. 1976, 41 (17), 2835–2846. DOI: 10.1021/jo00879a010.


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Oxidation of alkenes and alkynes by Chromic Oxide?
As far as I know, $\ce{CrO3}$ doesn't effect the $\mathrm{\pi}$ bonds in an inert medium like pyridine. Would $\ce{CrO3}$ in acidic medium reduce alkynes to have the same products as acidified, warm $\ce{KMnO4}$ would have? Or does it produce diketones similar to reductive ozonolysis?


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Proof of Dynamic Nature of Equilibrium
Here is how my textbook proves that dissolution of solid in liquid is dynamic in nature. It considers the example of a saturated solution of sugar. It says that,

Though the solution is in equilibrium, the process of dissolution of sugar and crystallization of sugar takes place continuously and also at the same rate. This can be confirmed with the help of radioactive sugar. If we drop some radioactive sugar into saturated solution of non-radioactive sugar, then after some time radioactivity is observed both in the solution and in the solid sugar. Initially there were no radioactive sugar molecules in the solution but due to dynamic nature of equilibrium, there is exchange between the radioactive and non-radioactive sugar molecules between the two phases.

It also proves the dynamic nature of chemical equilibrium in a similar way.
From this proof, my argument is that dynamic nature of equilibrium is proved only by adding another substance. This disturbs the equilibrium. It is possible that the saturated solution was in static equilibrium but addition of this substance has led to the dynamic nature.
I do not find it logical to prove that it is dynamic by the process of disturbing it.
Is this proof invalid or am I going wrong in my arguments?

For my clarification, why do you say that another substance is being added? To me, the description of the experiment says sugar (with radioactive carbon) is added to a saturated solution; sugar is being added to a sugar solution.

In that statement, I was referring to how it is proved generally  (in chemical equilibrium as well). When I said another substance is added, I meant that - always something (same or different) is introduced from outside. And still isn't sugar with radioactive carbon different from that of normal carbon?

The importance is in how it is different. In this case, the different isotope of carbon will not be so different as to perceptibly affect the dissolution process.

One would have the same effect by having two identical solutions (saturated, some solid sugar at the bottom), except that one is of radioactive sugar, with a removable wall in between. Some time after the removal, one would find radioactive sugar at the bottom of the container that initially only contained non-radioactive sugar.

@TAR86 I am not sure whether on removal of the wall, the combined system (of two equilibria) is immediately in equilibrium. If this is false, then according to your example isn't this a disturbance? If the systems were assumed to be in static equilibrium, then due to this disturbance, it can be argued that the reactions would activate and this would lead radioactive sugar to be part of other container.

@J.Ari It maybe same. By adding radioactive sugar, aren't we increasing the mass of the precipitate and disturbing the original mass that was in equilibrium with the dissolved sugar?

@DoubtExpert The concentration is the same on either side of the wall before and after the removal - so I do not think there is a disturbance on a chemical level. Obviously, since the radioactive sugar is missing on one side (diffusion takes time), the solution is not homogenous, but that is the whole point.

Digestion and growing of the precipitate crystals is another hint, if not a proof.

You are technically correct in saying that radioactive sugar is another species in that it is possible to distinguish it from 'normal' sugar, although chemically identical. But think about it this away. Some of the added radioactive sugar solution (which could be micromolar or less) finds its way into the solid sugar at the bottom of the flask, but some remains in solution so that (dynamic) equilibrium is reached with respect to the radioactive sugar. You could do it the other way round, add solid radioactive sugar and it will appear in the solution.  What is so special about the radioactive sugar that it can do this and 'normal' sugar cannot? Well, there is nothing special so it is entirely reasonable to suppose that the normal sugar does this also. By inference, other molecules also do this as we know of no experimental reason why sugar is peculiar among all the other types of molecules known.
If you are still not convinced an experiment could be done with a coloured salt such as copper sulfate, so only one species. Make a saturated solution in which there will be solid in equilibrium with water, which will be blue due to the copper ions. Next, filter off the water and retain the solid and (gently) add more water, initially this will be clear but becomes increasingly blue as the sulfate dissolves. You will see equilibrium being established again as the sulfate dissolves, and provided there is some solid left equilibrium solid to solution will be achieved.
There is an experimental technique called temperature jump that is used specifically to study the time it takes to 'relax' back to equilibria and so obtain rate constants for the process. In this method, a short duration laser pulse is used to rapidly heat a solution (typically by $\ll 1$ C) and change its equilibrium a little. The return to equilibrium is followed by optical/spectroscopic methods so nothing has to be taken from the solution. Many reactions have been studies this way e.g. $\ce{H3O+ + OH^- <=> 2H2O}$.

This experiment doesn't disturb the equilibrium and is a good way to demonstrate its dynamic nature
The key point in this experiment is that it doesn't disturb the equilibrium. Radioactive sugar is–for all practical reasons that matter to the experiment–chemically identical to the sugar in the solution. So, from a chemical point of view, the equilibrium is not disturbed at all by the addition of radioactive solid sugar.
If the equilibrium were static (meaning that no molecules were exchanged between the solid and the liquid once the solution is fully saturated) then there would be no exchange between the solid and the liquid. If the equilibrium is dynamic, then there will be. If we tried to test this with identical sugar molecules then we would be stuck as there is no way to tell whether the molecules exchanged between the solid state and the solution (because all sugar molecules are identical). Using radioactive sugar allows us to track the molecules but doesn't disturb the chemistry in any way (being radioactive doesn't change the key chemical properties of the molecules relevant to the dynamic equilibrium but does allow us to label the molecules that start in the solid state.
This trick means we can now tell whether there is any exchange between the solid and the liquid even when the solution is saturated. The fact that exchange does happen and radioactive sugar appears in solution means we know the equilibrium is dynamic: this cannot possibly happen if the equilibrium between a solid and a saturated solution is static.

If the system you describe is in static equilibrium, the solution is saturated and by definition would not accept any more sugar into solution. Since the sugar you're adding is radioactive, this means that you'd observe radioactivity in the solid sugar only.
You don't provide any mechanism by which adding more sugar would disturb this equilibrium. It is true that radioactive sugar will have slightly different properties than normal sugar, but in this case the difference would be imperceptible.

At equilibrium, we know that the concentration of the substances involved remain a constant. By adding radioactive sugar, we increase the concentration of sugar. Isn't this disturbing the equilibrium?

OK, now I see your confusion. Saturated, by definition, means that the sugar has reached its maximum possible concentration. So you aren't increasing the concentration of the sugar in solution, you're simply increasing the mass of the precipitate.

By adding radioactive sugar, aren't we increasing the mass of the precipitate and disturbing the original mass that was in equilibrium with the dissolved sugar?

Adding solid sugar does not disrupt the concentration of solid sugar. The concentration of solid sugar (mass or moles per volume) is a constant - the density of the sugar. Having 200 grams or a 1,000,000 grams does not change the concentration of the solid.


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How to test a sample of cigarette tobacco for its nicotine content?
I was wondering how one would go about testing samples of different cigarette tobaccos for their nicotine content. Is there a kit of some kind available online to do this?

A cursory search didn't reveal anything along the lines of a "home kit," but check out this commercial product and accompanying brochure for details pertaining to an industrially-market analysis tool suite.

You could do a bio-assay. Nicotine is a poison to insects.

Unfortunately, nicotine analysis is prohibitively complicated and expensive for a home kit. The presence of other alkaloids similar to nicotine pretty much necessitates that somewhat expensive equipment (i.e. a gas chromatograph) is used to first separate out the nicotine prior to analysis (typically by mass spectroscopy).
The only practical means for a layperson to test tobacco products for nicotine content is to out-source the work to a GC-MS equipped laboratory that routinely performs such analysis.


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Is there any difference between balancing redox reactions in acidic and basic medium?
For balancing redox reactions in an acidic medium I follow this procedure:
1. Writing the half reaction
2. Balance the atoms other than $\ce{O}$ and $\ce{H}$
3. Balance the oxygen by adding water
4. Balance the hydrogen by adding protons
5. Make the number of electrons equal and add the half reaction.
Is there any difference if I change the medium if so what are the differences ?

You're good. That procedure works in any and every case.

The five points to pay attention are correct; and I would like to add a sixth one: depending on the reaction condition, namely if the reaction occurs in acidic, neutral, or basic solution, the outcome and the reaction product(s) of the half reaction may differ.
Take potassium permanganate, for example.  Under aqueous acidic conditions, this is a very powerful oxidiser, and formally $\ce{Mn^{7+}}$ is reduced all the way down to $\ce{Mn^{2+}}$.  This half reaction, starting from purple coloured solution of $\ce{KMnO4}$ to practical colourless $\ce{MnSO4}$ (assuming $\ce{H2SO4}$ as the acid present) may be written as
$$ \ce{MnO4^-  + 5e^- + 8H^+ ->  Mn^{2+} + 4H2O} $$
Under neutral conditions, the product however is manganese dioxide (formally $\ce{Mn^{4+}}$), which is poorly soluble in water and precipitates
$$ \ce{MnO4^-  + 3e^-  + 2H2O  ->  MnO2 + 4OH^-} $$
Eventually under (very) basic conditions, the reaction will stop already at a green colour in form of oxidation state $\ce{Mn^{6+}}$:
$$ \ce{MnO4^-  + e^- ->   MnO4^{2-}} $$
Paying attention to this has practical, preparative relevance, too.  It offers one way to attune the oxidation power of potassium permanganate.  A moderated reactivity often correlates with selectivity of the reactions.
This pH-dependence of oxidation-reduction reactions correlates with electrode potentials which often are already tabulated, like here or in the CRC Handbook of Chemical and Physics (keyword Nernst equation).
As a side note: These equations sometimes do not tell the complete story.  $\ce{MnO2}$, for example, may act / may be used itself as oxidiser. The simple sum formula however hides well that the oxidising power of this material depends how this material was prepared, too.


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Acetaldehyde tautomer name

Thought it may have been ethylene oxide, but it's not.

Ethylene oxide has its oxygen atom attached to both carbon atoms making a saturated three-membered ring.  See https://en.wikipedia.org/wiki/Ethylene_oxide.

Learn a bit of IUPAC nomenclature first. Then you dont have to worry about naming.

The IUPAC name is ethenol. It is also commonly called vinyl alcohol.

I don't think it's such a good idea to answer such questions. It will probably get closed, it is already downvoted a lot. I don't think that it will be of much help for future visitors, as I doubt it will be found ever again. In the long run it's just a question that dilutes the content. But that is only my personal point of view. (Your positive scoring answer prevents the system from cleaning up this question.)

@Martin-マーチン , I guess that is a pretty useless question to anybody else, not even searchable. I'll try to think harder about questions that shouldn't be committed to eternity and just ignore them or leave a comment for the OP. Thanks for the heads up...


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Why is the rate law typically written in terms of reactants whilst equilibrium constants are written in terms of products*?
That is, equilibrium constants are written in terms of products/reactants or for Ksp, in terms of products only.
Why doesn't the rate law also focus on products?
I usually see it written in terms of reactants.

Because products do not influence the rate of a non-equilibrium reaction, while products are the reactants of the reactants in the backward reaction in equilibrium

If the reaction is $A \rightarrow B$ then  you can write $dA/dt=-k_1A$ or $dB/dt=+k_1A$, to solve you also have to know how much A and B are present initially, if in equilibrium $A = B$;  $K_e=[B]_e/[A]_e$


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Mass spectrum analysis
When finding ions which correspond to certain fragments on a mass spectrum; there are multiple possible ions that can correspond to the same fragment;
How should I know which one is the correct one?

see if you can get your sample onto a Time of Flight - Reflectron spectrometer  as this will have superior mass resolution.

Generally speaking: knowing the history of your sample (e.g. earlier steps of synthesis allowing / excluding elements that could numerically fit into the list of suggested fragments) and additional spectroscopic characterisation (UV-Vis, IR, Raman, NMR, etc.) may be helpful.
The problem you describe may be attenuated by using a mass spectrometer of higher resolving power.  Instead of submitting the sample to the detector behind a smaller LC-MS coupled analysis, providing a purified sample to a larger, tuneable sector mass spectrometer, for example.
An important alternative is coupling of several mass spectrometers along one line, where after initial ionisation primary fragments are separated from each other, and their corresponding subsequently generated daughter fragments are eventually characterised, like in a tandem MS/MS:

(source)
An overview about the later provides this wikipedia entry.


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Why are both products observed due to Molecular Orbital theory

I can reason out the first product as the highest occupied orbital of one excited molecule is approached by the lowest unoccupied orbital of the ground state molecule. I cannot rationalize the second product; am I missing an aspect of MO bonding that is important?

Simply turn one of them around.

Oh, wow, that clears it up.

This is probably worth buffing up into a full answer, @Mithoron


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Is this photochemical [2+2] cycloaddition enantioselective?

How is it deduced that one enantiomer is preferentially formed in the reaction? When I think of the MOs of the reactants, both enantiomers are viable as products.

If there is no source of chirality present in the reaction mixture, then there is no reason for the reaction to be enantioselective. In this case there are two possible sources of chirality: firstly the light, which could be circularly polarised (not a common thing to do), and secondly the catalyst, which can be chiral.
Although the exact catalyst is not specified, the above scheme seems to be one of the earlier reports of transition metal photoredox catalysis, from the lab of Tehshik Yoon.1 The catalyst used is $\ce{[Ru(bpy)3]Cl2}$, the cation of which is chiral (it has point group $D_3$). These cations exist in two enantiomeric forms, labelled Λ and Δ. [Thanks to Zhe for pointing this out.] Upon absorption of visible light, the excited $\ce{[Ru(bpy)3]^2+}$ complex can transfer an electron to the aryl enone, the LUMO of which is lowered by coordination to $\ce{Li+}$. This generates a radical anion which adds into methyl acrylate.

However, there is no report of the product being formed enantioselectively – only diastereoselectively. There is no indication that the authors used a single enantiomer of the catalyst. [Exactly why it is diastereoselective is a slight mystery, to me at least.] Note that this cycloaddition reaction is not concerted, so the MO-based arguments for pericyclic reactions such as the concerted Diels–Alder are not valid here.
In any case, transition metal photoredox chemistry is generally very poor at asymmetric induction, because the role of the catalyst is only to generate transient radical intermediates; the catalyst does not play any significant role in the actual bond formation. It is for this reason that photoredox catalysis is often combined with other modes of catalysis, e.g. organocatalysis, for highly enantioselective reactions to take place.
Apart from this, there are no other reports of this particular cycloaddition in the literature. All in all, I think it is safe to say that the 92% ee is not real.
Reference

Du, J.; Yoon, T. P. Crossed Intermolecular [2+2] Cycloadditions of Acyclic Enones via Visible Light Photocatalysis. J. Am. Chem. Soc. 2009, 131 (41), 14604–14605. DOI:10.1021/ja903732v.


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Reaction between bromobenzene and methyllithium to yeld toluene: possible?
I was solving some reactivity problems, and I was asked to prepare toluene from bromobenzene.
While the suggested solution of the problem involves the preparation of a Grignard reagent, followed by a reaction with methyl bromide, I was wondering:
Could the same compound be prepared with the straightforward reaction between bromobenzene and methyl lithium?
I tried to search some similar reactions, but I found nothing for both the possible ways.

Probably not, bromobenzene might not be electrophilic enough.

MeLi certainly does lithium-halogen exchange and in the presence of MeI takes bromobenzenes to toluenes   Synth. Comm. 2001 2323
If you're asking whether the MeBr produced by the exchange can act as the methylating agent? I have never seen this reported but maybe someone else can pull up a reference.


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Brownian motion of colloids at high density
Colloidal particles in dilute suspensions tend to perform a universal type of Brownian motion at finite temperature, according to the Stokes Einstein relation $D\approx\frac{k_BT}{6\pi \eta r}$ in the case of spherical colloids at low density. When the concentration is high enough and the colloids start to aggregate, however, the diffusion constant of colloids within the aggregate will change. This is true not only because of confinement of colloids by their neighbors, but also because of the fact that proximity of colloids to a stationary aggregate should actually change the diffusion constant from its value when colloids are isolated. This can be seen from the fact that the liquid phase is essentially incompressible, and also that the effects of viscosity would be amplified. My questions are, is this effect well understood, and if so, what would be some illuminating resources to learn about it in more detail? Full answers would also be much appreciated.

Most of the colloid stuff is old but of interest again because of nano-particles. One place to start might be Collins & Kimball, Journal of Colloid Science, Volume 4, Issue 4, August 1949, Pages 425-437 which gives what is now the standard theory of diffusion controlled reactions so deals with the topics you mention.

Thanks for the reference. I read the paper, but couldn't find where it addresses my specific question. Note that I'm asking how the diffusion constant of colloids changes when they are very close to a large aggregate. There are no reactions involved, just Brownian motion and hydrodynamics.

Have a look at Derjaguin approximation for the forces between two spheres, there is some discussion on this in Israelachvilli 'Intermolecular and Surfaces Forces'

Again, the question isn't about the forces between two spheres (at least, not the standard depletion, electrostatic/ionic, and Van der Waals forces), but rather about how the proximity of two particles influences their individual types of Brownian motion. I assume that when particles are close enough to an aggregate, the approximations made in the derivation of Stokes-Einstein break down. I'm asking about an indirect and somewhat unusual interaction between particles that influences the rate and amplitude of their meandering.

so you are looking at brownian motion with sphere of differing sizes, sounds as if you are getting towards percolation where small particles move along channels formed by larger slow moving or static particles.

Yes, that is related. My question is about how mere proximity of particles influences their relative mobility in a viscous medium.

Here's one paper that seems to address the problem: https://arxiv.org/pdf/1410.8651.pdf


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Chemically generating a high vacuum?
Suppose I fill a vacuum chamber with 99% oxygen, then pull a vacuum on it down to 10 pascal absolute. If I then vaporize aluminum (for vapor deposition), will the aluminum bind with the remaining oxygen and create say a 0.1 pascal vacuum?
I'm not a chemist or a physicist, but it seems logical that the heavier (and high melting point) aluminum oxide molecules would collect on the walls and floor of the container resulting in a high vacuum mainly containing vaporized aluminum.
I am looking to experiment with aluminum vapor deposition and want to avoid buying a diffusion or turbo molecular pump. Seems like this should work, but I have not found any documentation on it. I'm not looking for optics quality here, so maybe high vacuum is not an issue in the first place.
Let me know if you guys think I should post this on the Physics SE instead. Thanks!

It is not all that easy to vaporize aluminum in the first place. Nor is it necessary, come to think of it. Then again, do you really need a high vacuum, or just an oxygen-free environment? These are different things, and they are achieved by different means. Your title suggests the former, but your method suggests the latter.

In principle yes you could do it, but first bake the system to get rid of water /organics on all the surfaces. Then work out how many oxygen molecules there are left and how many Al ions you are likely to generate; the numbers may be so different that it may take years to do. I recall form some distant memory that a 'getter' may do a similar job.

This is essentially how a titanium sublimation pump works. They tend to be operated at orders of magnitude lower pressure, though, to help achieve ultra high vacuum. It may take a while (and quite a lot of power) to deal with 10 Pa of oxygen (plus outgassing water, leaks etc).

@IvanNeretin, I am not sure if nitrogen and other impurities cause issues when preforming aluminum vapor deposition, but all the literature I have read insists on a high vacuum. I'm not sure what the vaporization temperature of aluminum is in a vacuum, but adding heat to aluminum is the easy part ;-)

@porphyrin, I think I can increase the aluminum vapor to a relatively high concentration which should speed up the reaction. +1 on the "getter" vocab word; I will research more on this.

@AndyW, I will look into that titanium sublimation pump, thanks.


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In a multi equilibrium system, how does change in one equilibrium affect the other?
Suppose in a reaction vessel of fixed volume we have the following equilibria established (existing simultaneously in the vessel),
\begin{align}
\ce{A (g) &<=> B (g) + C (g)}\tag{i}\label{i}\\
\ce{D (g) &<=> E (g) + C (g)}\tag{ii}\label{ii}
\end{align}
Frankly speaking I do know that when we increase, say, $\ce{A (g)}$ concentration in $\eqref{i}$ then indeed the concentration of $\ce{C}$ or $\ce{B}$ is going to increase as per Le Chatelier's Principle. However if we increase the concentration of $\ce{E (g)}$ in the vessel what impact will it have on the concentration of $\ce{C}$ and $\ce{A}$?
From my perspective, when $\ce{E}$ is increased in concentration, it will now combine with $\ce{C}$ to give $\ce{D}$ so that reaction goes backwards, and so more of $\ce{C}$ needs to be formed as a result $\ce{A}$ concentration in equilibrium $\eqref{i}$ goes down. However I am not convinced with my own explanation as I feel I am missing out on something. Please feel free to help me out.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Ahh, I was sumultaneously editing at the same time @Martin !

Increasing $\ce{[E]}$ lowers $\ce{[C]}$ by a simple application of Le Chatelier's Principle. Regarding the effect of increasing $\ce{[E]}$ on $\ce{[A]}$, your understanding is correct. To make your observation more obvious, we may reverse reaction (ii) and add it to reaction (i): $$\ce{A(g) + E(g)<=>B(g) + D(g)}$$
So we can, once again, simply apply Le Chatelier's Principle and see that increasing $\ce{[E]}$ lowers $\ce{[A]}$.

But does the decrement of A inturn imply an increment in C as per equilibrium (i) ?

@DIPANJANChowdhury, no. Increasing $\ce{[A]}$ increases $\ce{[C]}$, as you've already stated in your post.


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Why don't elements having the same valency as Carbon bond so readily?
I haven't been in touch with Chemistry as a subject of study for several years now. Still- if I remember correctly, Si, Ge, Sn, Pb all have same number of electrons as Carbon in their outer orbits.
So why is Carbon so ubiquitous? I believe the number of electrons an element can give or accept decides how readily it bonds with other elements. For instance, inert gases have no room for extra electrons and hence, are inert.
So why don't other elements in the column of Carbon bond as easily? In other words, why aren't we made up of Pb primarily instead of C?
I remember asking this question to my high school chemistry teacher, but as far as I can remember, he wasn't able to give a satisfactory answer. Forgive me for asking this , if this question is too naive. But I genuinely wondered a lot about it back in my school days and still don't know the answer.

They do bond as easily, and maybe even easier. For example, Pb slowly oxidizes in the air without heating, while carbon just sits there indefinitely. This has nothing to do with their abundance in nature, though.

It's not that they don't bond readily, it's that they can't form as stable of networks as carbon. Carbon has many different allotropes, while silicon has only a few. As you can see, carbon forms uniform hexagonal sheets with massive $\pi$ systems as in graphene and many of its other allotropes, while silicene is incapable of remaining flat and is bumpy. This is a result of the bond distances and p-orbital overlap.
$$
\begin{array}{|c|c|}\hline
\text{Bond} &\ \Delta H ^\circ_f \text{ (kcal mol}^{-1}) &\ \text{Bond Length (pm)} \\
\hline
\ce{C-C} &\ 82.70 &\ 154 \\
\hline
\ce{Si-Si} &\ 53.06 &\ 233 \\
\hline
\ce{Ge-Ge} &\ 44.93 &\ 241 ^{[1]} \\
\hline
\end{array}
$$
As you can see, as the atoms get larger, the bond length increases. $\pi$ bonding is completely reliant on the overlap of adjacent p-orbitals, so the farther away these orbitals are, the weaker the bonds.

Bond dissociation enthalpy also increases with atom size. This leads to an overall decrease in the stability of larger silicon or germanium compounds. Tin and lead are metals, which have much different chemistry altogether on account of their electronegativity and occupied d-orbitals, in addition to their size.

${[1]}$ Bond Enthalpy Table

I have had the same doubt myself, thank you for clearing it up! +1


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Why does this compound have two benzene rings?
If we look at carbons 1 and 2, they have 5 electrons. 3 are directly involved in bonding, and 2 are involved in delocalisation to create the 2 benzene rings. This would mean that they had 5 valence electrons, which isn't possible, and thus there should be only one benzene ring.

These are not benzene rings, and even calling them aromatic because of Huckel's rule is a bit of a stretch. Yet still they are rings, and they contain delocalized bonds, so why not draw them with a circle.

If you really need a Lewis structure, use the one without rings, with 5 double bonds, and have on mind that it os one of may so called resonance structures that are in reality "averaged".

Chemists are already accustomed to this type of formula, even though it is not absolutely correct. More exact would be drawing all bonds with double solid/dashed bonds to express that the bond order is between 1 and 2, approximately 1.5, but is almost never done

Rarely can be seen formula with the two circles merged (and maybe dashed also), like

But the electrons are in fact delocalized across all bonds of the polycycle structure in "electron cloud" that looks somewhat like the infinity sign ∞ in this case (naphthalene) above and below the planar molecule. What matters is that the count of π (those from the "second bond" of the double bond) electrons is 4n+2 for n is some natural number (here it is n=2). It is a special feature called aromaticity, which has something to do with advanced topic of quantum chemistry; and it greatly stabilizes the molecule. Any of these formulas is unable to describe reality.

Do the two rings in this behave like benzene rings?

@Saad yes, they have similar properties, but not exactly.. thanks to symmetry, there are there three types of carbon centers (some even do not bear hydrogen atom), while in the benzene there is only one type.

So the friedel-crafts reactions would work with this?

@Saad, yes, naphthalene is quite aromatic. Did you at least try to use google? HOWGH

No, I only just learnt that it's called napthalene.

Instead of the historical "4n+2 pi electrons", I recommend thinking of Huckel's Rule as "an odd number of pi-electron pairs." Much more intuitive and accessible for new students.

Well, to nitpick, this one may carry the wrong impression that every carbon on the ring has equal electron density. I don't really think there's a way to be 100% accurate at describing the structure of naphthalene, without drawing out two resonance forms. So your concluding statement is certainly important.

I find this representation to be kind of sloppy for the reason you mentioned. The $\pi$ system is delocalized throughout the molecule, but that doesn't mean it's two aromatic systems.
I find this representation much better:

This one carries the wrong impression that the rings are different, which they are not.

@Ivan Neretin Then how would one represent it? Two 'circles' leave no room to differentiate it from two benzene rings. Also, if these rings aren't benzene rings, then it implies that the one with the OH group attached cannot be called a phenol group?
Do these rings behave like benzene rings though?

Two circles are fine. There is no such thing as benzene ring. These rings are delocalized; arguably, they are aromatic, albeit to a lesser extent than benzene. In some reactions naphthalene behaves like benzene, and in some it behaves differently.

@IvanNeretin Arguably we have a problem because our representations all have drawbacks because there is no way to capture the complexity with a few straight lines or circles. Your criticism is nevertheless totally valid.

Any representation of anything is a simplification and misses some of the complexity. As long as we remember that, we'll be all right.

Actually, if you treat it as one of two resonance structures, it's good depiction, showcased in Clar's rule.

Also to @IvanNeretin's concern about lack of symmetry--this doesn't stop anyone from drawing a Kekule structure for benzene. Also, naphthalene under goes Diels-Alder, so this representation is better for that: http://pubs.acs.org/doi/abs/10.1021/ja9107275 I would rather just draw out all 6 double bonds, which, looking back at 5 volumes of lab notes, is what I actually did.


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What type of catalyst would this be?
We were discussing types of catalysts, homogeneous and heterogeneous.
Our question is: If you have two reactants, one reagent is aqueous and the other is a gas, and the catalyst is aqueous, what type of catalyst is it?
My guessing is that it is a homogeneous catalyst because the gas will dissolve in the solution at some point so that the reaction can occur. What is your guess?

There is no point in guessing. You don't know whether the reaction actually occurs in the solution or on the surface. It can turn out either way.


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Solubility of Aluminium Halides by Fajans rule
According to Fajan's rule polarizability of anion is directly related to the size and hence  the covalent character would be more.
Now going by this definition aluminium chloride having greater covalent character would be less soluble than aluminium fluoride, but in reality it's just the opposite. How do I account for this?

Also why isnt fajans rule applicable here and where else does it not hold true ?

You should consider all parameters before trying to conclude something. Have you considered the fact that fluoride is a small anion, and that increases lattice energy?

Fajans rule for solubility is applicable only in polar solvent . In nonpolar solvent it is reverse the order in polar So check the solvent


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Strength of OH bond
This question might seem a little general, but how can we know the strength of the OH bond of third period element hydroxides? Does it increase across the period or does it decrease?But more importantly, on what should we rely in order to understand how the strength of the bond changes?