text:
How do I calculate the pH of this buffered solution?

Calculate the $\mathrm{pH}$ of a buffered solution containing $\pu{0.5 M}$ ammonia and $\pu{0.5 M}$ ammonium chloride when $\pu{0.15 M}$ $\ce{HCl}$ is added into it. The $\mathrm{p}K_\mathrm{b}$ of ammonia is $4.75$.

This is what I think shoud be going on in the buffer solution:
\begin{align}
\ce{NH3(aq) +H2O(l) &<=> NH4+(aq) +OH-(aq)} \\
\ce{NH4Cl(aq) &-> NH4+(aq) +Cl-(aq)}
\end{align}
This is where I get stuck thinking that the question told me that I have a buffered solution containing $\pu{0.5 M}$ $\ce{NH3}$ and $\pu{0.5 M}$ $\ce{NH4Cl}$ solution, but is that even possible? How can you make a $\pu{0.5 M}$ ammonia solution? Would not that ammonia exist as ammonium $\ce{NH4+}$ and whether the equations I wrote are correct. I am just so confused!
Sorry for the delay but I'm having exams. Okay, I'll try to do this now by not using the Henderson equation.
Firstly, ammonia will refer to ammonium hydroxide, i.e. $\ce{NH3 -> NH4OH}$:
\begin{array}{cccc}
&\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\
&\mathrm{I}: &\pu{0.5 M} & &\pu{0 M} & & \pu{0 M} \\
&\mathrm{E}: &(0.5 - x)\,\pu{M} & &x\,\pu{M} & &x\,\pu{M}
\end{array}
Now for the salt:
\begin{array}{cccc}
&\ce{&NH4Cl &<=> &NH4+ &+ &Cl-} \\
&\mathrm{I}: &\pu{0.5 M} & &x\,\pu{M} & &\pu{0 M} \\
&\mathrm{E}: &\pu{0 M} & &(0.5 + x)\,\pu{M} & &\pu{0.5 M}
\end{array}
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH4OH}]} = \frac{(x + 0.5)x}{(0.5 - x)}$$
and since $K_\mathrm{b} = 1.778 \times 10^{-5}$, I can find the value of $x$ to be $1.7778 \times 10^{-5}$.
$$[\ce{H+}][\ce{OH-}] = 10^{-14} \to [\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$$
and $x = [\ce{OH-}]$. So
$$[\ce{H+}] = 5.624 \times 10^{-10}$$
Now I add the $\ce{HCl}$:
\begin{array}{cccc}
&\ce{&HCl &<=> &H+ &+ &Cl-} \\
&\mathrm{E}: &\pu{0 M} & &\pu{0.15 M} & &\pu{0.15 M}
\end{array}
I will ignore any common ion effect since it will be negligible (I think).
The pH will be the total $\ce{H+}$ concentration, so $[\ce{H+}] = 5.624 \times 10^{-10} + 0.15$. This is a pH of about 0.823 which is totally wrong. What am I doing wrong? Also, the volumes of the buffer solution or of the acid solution have not been given.
Okay, so I now know what I was doing wrong (thanks to someone pointing it out). I was adding in $\ce{HCl}$ without thinking that it would actually react with the ammonia in equilibrium in solution (which is just me being either dumb or ignorant).
So I have the situation that:
\begin{array}{cccc}
&\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\
&\mathrm{E_\text{(no HCl)}}: &(0.5 - 1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} \\
&\mathrm{E_\text{(HCl)}}: &()\,\pu{M} & &()\,\pu{M} & &()\,\pu{M}
\end{array}

You are right, consider ammonia as ammonium hydroxide. Use equilibrium calculations and find out hydrogen ion or hydroxide ion concentrations. That should let you find the pH.

Ammonia does exist as NH4+ , which is essential in a buffer solution as it depends on common ion effect. Use equilibrium concepts to figure out the concentration of the compounds in the solution and apply the Henderson-Hasselbalch equation. Hope this helps.

Okay, but would I assume the initial concentration of ammonia to be of NH3 or NH4+ from the first equation? Because the question says NH3 so assuming the initial concentration of NH3 makes sense to me but at the same time it introduces more variables. So I can't solve it if I assume the initial concentration to be of NH3.

Don't ever consider it as $\ce{NH3}$. Assume all the ammonia to exist as $\ce{NH4OH}$, although ammonia essentially only contributes to $\ce{[OH^-]}$, since much of the$\ce{[NH4^+]}$ is obtained from the $\ce{NH4Cl}$.

In order to help you, what is the volume of the buffer and what volume of 0.15 M HCl was added?

It will be easier for you if you could find mole number of Salt and Acid then use Handerdson Haselberg Equation. N.B >My answer doesnot represent full answer I have just given a hint.

You can think of $\ce{NH4OH}$ as a hydrate: $\ce{NH3_{(aq)}}$; no need for complicated reactions interconverting the two.
Concerning the problem you described, at these concentrations all subtleties are negligible, and you basically only need to calculate the nominal concentrations of $\ce{NH4+}$ and $\ce{NH3}$, and plug them into the Henderson-Hasselbach equation.
You start with $\ce{0.5 M NH3}$ and $\ce{0.5 M NH4Cl}$, and add $\ce{HCl}$ to a total nominal concentration of $\ce{0.15 M}$, without altering the total volume (the question is written ambiguously, but that's what it means, most likely). As $\ce{HCl}$ is a very strong acid, it will convert all $\ce{NH3}$ to $\ce{NH4Cl}$ (actually $\ce{NH4+}$, because we are in solution). Therefore your final situation is: $\ce{(0.5-0.15)=0.35 M NH3}$ and $\ce{(0.5+0.15)=0.65 M NH4+}$. This should give you a pH of about 8.98 (lower than the one you would have without the HCl, consistently with the addition of an acid).
So your main error above was that you dissociated $\ce{HCl}$ (correctly), but then just added the resulting $\ce{H+}$ to the total, forgetting that there was a ton of $\ce{NH3}$ around that would react with it.
Another example is when you add a solution of some particular transition metal salt to a solution of a weak acid (could be iron sulfate + hydrogen sulfide for instance). Despite the fact that the salt solution isn't acidic at all, the pH drops considerably compared to the solution of the weak acid alone, simply because the metal and the anion form a very insoluble salt that precipitates, subtracting weak acid anions from the solution and forcing the previously non-dissociated acid to dissociate.
In general, my advice is stick to what you know, don't go looking for complications: if they did not teach you to do very exact/sophisticated pH calculations, they probably don't expect you to pull them out of a magic hat at the exam. And as my stoichiometry teacher used to say, always write down all species you have in solution.

No, actually I do just fine with acid buffers involving $\ce{HCl}$, it's just these base buffers involving $\ce{NH3}$ that are confusing me. I know that it's easy using the Henderson equation, but I'm just trying really hard to understand what is actually happening in the solution. And the volumes not given is also confusing me. How can I calculate the change in pH of a solution when the volumes of the solutions have not been given. You mentioned something about not altering the total volume?

The basic concept is that acids will react with bases. $\ce{HCl}$ is a strong acid, $\ce{NH3}$ is a base, the two will react to give $\ce{NH4Cl}$ quantitatively, within the approximate conditions we're applying. I don't really see the difference with adding $\ce{HCl}$ to an acetate/acetic acid buffer: there too you remove acetate anion (a base) by reacting it with $\ce{HCl}$ and turning it to its conjugate acid $\ce{CH3COOH}$. It's just a matter of recognising this general mechanism, I suppose.

Oh, and about the volume: yes, I am making an assumption by deciding that 0.15 M HCl means that we add concentrated HCl so that the final nominal concentration of HCl is 0.15 M, but without changing the volume significantly. In this specific case, the total volume would not matter, as it's a buffer solution under the approximate conditions we discussed. However I agree with you, it is confusing when questions/problem are ambiguously formulated. I saw plenty of exam papers that, as far as I am concerned, would constitute sufficient evidence to fire the teacher on the spot.

Well, first of all, as you have written before, the main reaction is the following one,
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
In addition, you know the ammonia constant: $K_\mathrm{b}=10^{-4.75}$
(Note that if the question does not have a mention about equilibrium, it is because you don't need use the equilibrium in order to make the calculations)
So, let's start.

Initially you have:
$$[\ce{NH3}] = [\ce{NH4+}] = \pu{0.5 M}$$
Then, just taking into account the previous reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
you can get the result easily:
\begin{align}
K_\mathrm{b} &= \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \\
[\ce{OH-}] &= K_\mathrm{b} \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
[\ce{OH-}] &= K_\mathrm{b} = \pu{10^{-4.75} M}
\end{align}
and finally the pH would results as
$$\mathrm{pH} = 14 - \mathrm{pOH} \overset{\mathrm{pOH} = 4.75}{\to} \mathrm{pH} = 9.25$$
You add $\pu{0.15 M}$ $\ce{HCl}$ (acid), where it reacts with, obviously, the ammonia (base) which is found at the equilibrium, so
\begin{array}{ccccc}
& \ce{&NH3 &+ &H2O &<=> &NH4+ &+ &OH-}\\
&i) &0.5 & &0.15 & &0.5 & &/ \\
&f) &0.35 & &/ & &0.65 & &/
\end{array}
Now, you may observe that both concentrations have changed, so we could obtain, using the same strategy as before we did:
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \to [\ce{OH-}] = K_\mathrm{b} \frac{[\ce{NH3}]}{[\ce{NH4+}]} = 10^{-4.75}\frac{0.35}{0.65} \to [\ce{OH-}] = \pu{10^{-5.01} M}$$
To conclude, the $\mathrm{pH}$ is given by
$$\mathrm{pH} = 14 - \mathrm{pOH} \overset{\mathrm{pOH}=5.01}{\to} \mathrm{pH} = 8.98$$
Observe that the buffer turns into more acid as much more acid is introduced in it.

I mainly want to rectify one point that you are confused with and that all the other answers fail to point out explicitly. You say in the beginning that ammonia should react with water according to $(1)$.
$$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{1}$$
Obviously, this is an acid-base reaction. We can deduce where the equilibrium will be just by checking out the $\mathrm pK_\mathrm a$ values of the two Brønsted acidic species, water and ammonium:
\begin{array}{lc}\hline
\text{compound} & \mathrm pK_\mathrm a\\ \hline
\ce{H2O} & 14\\
\ce{NH4+} & 9.25\\ \hline\end{array}
As we can see, ammonium is more acidic than water by almost five logarithmic units. Therefore, it is safe to assume that in a solution of ammonia in water only a very neglegible amount of ammonia is protonated. Thus, when mixing a $\pu{0.5M}$ nominal solution of ammonia (which is very close to an actual $\pu{0.5M}$ solution) to a $\pu{0.5M}$ solution of ammonium chloride, we can indeed just use the Henderson-Hasselbalch equation to shortcut the way to the $\mathrm{pH}$ value:
\begin{align}\mathrm{pH} &= \mathrm pK_\mathrm a + \lg\frac{[\ce{NH3}]}{[\ce{NH4+}]}\\
&= 9.25 + \lg 1\\
&= 9.25\end{align}
When we then in the second step add $\pu{0.15M}$ of a strong acid, we can simply identify the strongest base in solution and consider $\pu{0.15M}$ of that base additionally protonated. The strongest base around to any appreciable extent is ammonia, so we now have a new concentration of $\pu{0.65M}$ of ammonium and $\pu{0.35M}$ of ammonia. Thus, our new $\mathrm{pH}$ value becomes:
\begin{align}\mathrm{pH} &= \mathrm pK_\mathrm a + \lg\frac{[\ce{NH3}]}{[\ce{NH4+}]}\\[0.5em]
&= 9.25 + \lg \frac{\pu{0.35M}}{\pu{0.65M}}\\[0.5em]
&= 9.25 + \lg 0.54\\
&= 9.25 - 0.269\\
&= 8.98\end{align}


text:
Organic Synthesis

Anybody has any idea how to do this? I tried converting the ketone to a alcohol using Grignard, but I have no idea how to proceed afterwards. I also tried starting with the Wittig reaction by converting to an alkene, but likewise, I dont know how to proceed. I have also tried working backwards, but to no avail as well. Anybody mind helping?

I'd say we have to functionalize that methyl group first, using the C-H acidity provided by the ketone, and only then destroy the latter.

@IvanNeretin wow! totally didn't think of that! but do you have any suggestions on how to destroy it afterwards?

Is this a homework problem or an actual synthesis question? If it's homework, I like @Waylander's answer. If it's for real, I think that won't work, and it's going to be tough with that highly substituted quaternary center.

React the enolate of pinacolone (t-Bu Methyl ketone) with methyl vinyl ketone. It will first add 1,4 then the resulting enolate will do an aldol on the pinacolone ketone to give 3-tBu-cyclohexenone. React this with Phenyl cuprate and, despite the steric hindrance, it will add Michael-wise to give you the desired product.

My first thought was exactly the same, but I find it highly unlikely that you're going to add the phenyl group to a carbon bearing a t-butyl group...

I share that concern but I have no Scifinder access to check. If you can't do it that way then it becomes a far more complex problem.

I'm in the same boat regarding a lit search. I really want your answer to be viable because I can't think of an easy way to do this otherwise, either.

@Zhe, Waylander, I had a look at this on Reaxys today, the exact compound with that heavily hindered quaternary stereo centre isn't found, so I suspect the question is hypothetical

Thank you. This increases my suspicion that the enone is too hindered to undergo the cuprate addition.

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sort: votes


text:
How was it concluded that the H3O+ rather than H+ is the "acid" ion?
I suspect that initially, scientists believed that the acid ion was $\ce{H^+}$ since $\ce{H2}$ is released through electrolysis, right? But what experiment was done to change the standpoint to assume that it is instead the $\ce{H3O^+}$ ion?
Or perhaps there's both $\ce{H^+}$ and $\ce{H3O^+}$ ions?

Nice question! Related : https://en.wikipedia.org/wiki/Hydronium#Acids_and_acidity

Related - https://chemistry.stackexchange.com/questions/14117/does-h3o-donate-a-h-to-water/14119#14119

Also related https://chemistry.stackexchange.com/questions/20076/how-should-the-hydrated-proton-be-represented-in-chemical-equations/20081#20081

According to Wikipedia, the actual structure of hydronium was discovered using infrared spectroscopy.
It should be noted that $\ce{H+}$ notation is still useful for reaction balancing, since it's much easier to count atoms in $\ce{H+}$ instead of $\ce{H3O+}$, even if the former ion may not really exist.

Actually, the initial theories before Lewis suggested that $\ce{H+}$ is the cause of acidity. However, it soon turned up that an ion as small as the nucleus of hydrogen (you may simply call it a proton) can't be created in low energy reactions due to its high polarising power. So, $\ce{H+}$ is though the cause of acidic nature in aqueous solutions, $\ce{H+}$ never exists as $\ce{H+}$ but as $\ce{H3O+}$, $\ce{H9O4+}$, etc. It is because the proton is heavily hydrated. Though we are not very sure of what is the actual hydrated form, we usually refer it to as $\ce{H3O+}$ or $\ce{H+ (aq)}$.

It's not that we don't know what the hydrated form is, but there simply is no definite form, but a highly dynamic equillibrium between several states.

As stated here, the notion that (pure) water disassociates into some kind of ions is actually very misleading. Instead of thinking of liquid water as existing in separate molecules, it is far better to consider it as a vast network of hydrogen and oxygen atoms, where bonds are not in a binary state of 'formed' or 'broken' only, but have continuously varying strength depending more or less on the distance. It is then easy to understand how a charge imbalance can be very rapidly transferred to another location by just slight shifts in the positions of the atoms in the network. Unfortunately, this process is still called "proton transfer" despite the fact that there is no actual movement of a single proton across a long distance.

Of course, for the purpose of analyzing reactions it is sufficient to use the "H3O+" ion since it has pretty much the same properties as the local configuration of the hydrogen and oxygen atoms that are involved in the reaction, namely the slightly positive H and the slightly negative O that can accept up to 1,4 'bonds' (electron pairs) each.

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text:
Why only a specific pressure for the Haber's process?
This question, in my opinion is different from Why is the Haber process carried out at such high temperatures?
Referring to the above link, I understand that why high temperature of 700 K is chosen, i.e, to overcome the activation energy barrier as well as kinetic considerations. However, why is the pressure 200 atm only. Why not 300?
According to Le Chatelier's principle, there larger the pressure, more ammonia will be formed.

I think this is an interesting question. I would like a better verified answer, but for now I can direct you to a quote from the Wikipedia page for the Haber Process: "Economically, pressure is an expensive commodity. Pipes and reaction vessels need to be strengthened, valves more rigorous, and there are safety considerations of working at 20 MPa. In addition, running pumps and compressors takes considerable energy. Thus the compromise used gives a single pass yield of around 15%."(The claim doesn't have a citation)


text:
What are the molecular requirements for condensation polymerization?
I know that in order to have addition polymerization, the monomers must have a C=C double bond. However, I don't know the molecular requirements for condensation polymerization. Must the monomers contain a functional group at each and?
Any help is appreciated. Thanks.

You are misapplying the categories. Addition polymerisations often involve CC double bonds, but that is by no means a requirement. For condensations, it is hard to imagine a mechanism without two functional groups, but that is not the point!

Monomers must contain a functional group at each side (at least bifunctional) and these functions  have to react with each other (like amine and carboxylic acid to yield amide group or alcohol and carboxylic acid to yield ester group,...) .

We can start with two types of monomers that hold identical functional groups at each side.

One example of that is the reaction of 1,6-diaminohexane with hexandioic acid to yield Nylon-6,6.

We can also start from one type of monomers with two different functional groups at each side.

One example of that is the reaction of 11-aminoundecanoic acid to yield Nylon-11.
The condition that the reaction should give off a small molecule is no longer valid. The counter example is the formation of polyurethane by reaction of diisocyanates and diols without giving off  small molecules. We prefer name this type of polymerization "a step growth polymerization" in opposite to "chain growth polymerization" (known before as polyaddition).

Condensation polymerization can be shortened by having two monomers which are bifunctional that can react giving off small molecules like $\ce{H2O}$ or methanol.
This can be exemplified by amino acids which have a carboxyl group ($\ce{-COOH}$) and an amine group ($\ce{-NH2}$). When making the polymer the carboxylic group reacts with the amine group of the neighboring moleucle giving a dimer which is bonded via an amide bond ($\ce{-CONH-}$). This dimer has an amine group on one side and a carboxylic group on the other making it susceptible for further reaction thus making a trimer and a tetramer... and so on.

Carbon-carbon double bonds are not necessary for condensation polymerisation. The reaction usually involves the elimination of water. When two different functional groups are present on a molecule a polymer results, for example with $\omega$-hydroxyhexanoic acid repeated condensation yields a polyester,
$$\ce{HO(CH2)COOH \rightarrow HO-(CH2COO)_n-H}$$
With two different bifunctional molecules are used such as an amine and an acid, such as adipic acid and hexamethylenediamine,  nylon can be made;
$$\ce{ HOOC(CH2)_4COOH + H2N(CH2)_6NH2 \rightarrow HOOC(CH2)_4CONH(CH2)_6NH2 + H2O}$$
and this is followed by further steps where the initial (acid -amine) product reacts with more adipic acid.
Polyesters, polyamides, polyurethanes and silicones are made this way. If more that two functional groups are present insoluble, infusible polymers such as Bakelite are formed.
The kinetics do not follow that of chain reactions, there is no initiation or termination and the general mechanism is $\ce{M_j + M_k \rightarrow M_{j+k}}$. The kinetics can be analysed by assuming that only the end groups are important, thus the rate constant is assumed to be independent of molecular size then the rate is found to be bimolecular, (at least  for acid catalysed reactions)  i.e. $\ce{d[A]/dt=-$k$[A][B]}$ for reacting species A and B.


text:
Gas liquid transition

This is isotherm of a real gas is represented as shown in diagram at three temperature . In my book it is written that in ACIH gas liquid transition will involve only one phase throughout .
But I could not understand how this can be possible as at A it would be gas while at H it would be liquid .

What's unclear about that? Do you know what a critical point is (and where is it on your picture, BTW)?

@IvanNeretin point F

Good. Then again, what's unclear? You have a field with a fence, and the fence is not infinite; it has an end somewhere far away. You want to get to the other side of the fence. To climb over it, or to walk around it, that's the question.

@IvanNeretin I mean at H ,I it will be liquid while at A ,C it will be gas . So in the book how they have written there is no phase transition.

They have written so because this is true. If you go A-D-E-G, you climb over the fence. If you go A-C-I-G, then you walk around the fence.

@IvanNeretin but in that there is phase change that is initially it was gas and latter it is liquid

So what? Again, it is good that you know where is critical point, but do you know what is it?

@IvanNeretin under that curve both liquid and gas exist

What curve? I thought we were talking about the critical point.

@IvanNeretin by curve I mean EFD . critical point is the point above which there is no phase change

Well, yeah, you may define it like that. Now what's the question? You walk around the critical point, and you get from gas to liquid without a phase transition.


text:
Can we synthesize alcohols by the photolysis of hydrogen peroxide?
As I was going through my revision of organic chemistry, I came across free radical substitution of alkanes with halogens. In a similar process where we use hydrogen peroxide, instead of chlorine, so that photolysis would yield a hydroxide radical $\ce{OH.}$ instead of chlorine radical $\ce{Cl.}$. The rest of the steps would be the regular mechanism of This way we can get an alcohol instead of alkyl chloride. Here is what I mean:
\begin{align}
  \ce{H-O-O-H &->[h\nu]H-O. + .O-H}\\
  \ce{CH3-CH3 + .O-H &-> CH3-CH2. + H2O}\\
  \ce{CH3-CH. + .O-H &-> CH3-CH2-OH}\\
\end{align}
Does my hypothesis work in reality?
I am aware of the fact that dialkyl peroxides cannot be used since they are highly explosive. However, hydrogen peroxide isn't that explosive, so I think that its safe to use it. Since $\ce{H2O2}$ is a liquid, while alkanes and $\ce{Cl2}$ are gases, the reaction with $\ce{H2O2}$ could be done at low pressure. We could photolyse the $\ce{HO-OH}$ bond using light of a sufficient frequency, which probably should initiate a free radical reaction.
I did a few calculations and got the maximum wavelength of light needed to photolyse the $\ce{O-O}$ bond is $\pu{0.0084m}$, which is pretty low. So I think it's possible to photolyse $\ce{H2O2}$ and carry on the reaction.

Two problems. 1. You should consider what happens to both chlorine atoms in the alkyl chloride synthesis. 2. Dialkylperoxides are notoriously explosive.

@Zhe I have added a picture, see if that tells you what happens to the other OH radical; it becomes H2O similarly to the other Cl becoming HCl.

@Zhe if the dialkylperoxides are too explosive, then we can perform a photolysis at a lower temperature and pressure. That should keep it from going too violent as well as keeps it in the gaseous phase.

Got too hasty with my reply... Alcohols have an $\alpha$ hydrogen that is very easy to abstract. You will probably make ethanal instead or continue to acetic acid.

I think your hypothesis should work as in the presence of hv H2O2 splits into radical and there begins chain initiation.

$\ce{CH3-CH2.}$ is highly unstable, I'm not sure if its formation would be favorable.

@BerryHolmes it occurs as an intermediate in free-radical halogenation, so why not?

I do not believe so.
The mechanism of decomposition of $\ce{H2O2}$ is:
$\ce{H2O2(g) -> 2HO.(g)}$
$\ce{H2O2(g) + .OH(g) -> H2O(g) + HOO.(g)}$
$\ce{HOO.(g) + .OH(g) -> H2O(g) + O2(g)}$
$\ce{.OH(g), HOO.(g)}$ are both radical compounds.
If you take a look at the first step in the mechanism, it is the rate limiting step and it is the same as your first step in the mechanism, the photolysis step.  Therefore if $\ce{H2O2}$ undergoes partial photolysis all that will happen is that the sample will undergo decomposition quicker.
Additionally, your mechanism of oxidation has been proposed before.  You can take a look at this article, in the section Mechanism of Oxidation by Peroxides: Radical or not Radical, That is the Question, in the end of subsection 4.1 and 4.2 they refer to the mechanism you proposed.  They end up proving that it proceeds through a concerted mechanism that requires the presence of a suitable metal catalyst.  It involves oxidative insertion or some type of $\ce{C-H}$ insertion by the metal followed by oxidation of the product of that reaction.
But say the generation of hydroxyl radicals does work out, the corresponding alcohol may not be the only product that forms.  Hydroperoxides have a tendency of forming as well.  This is because hydrogen peroxide tends to decompose into oxygen and water, the oxygen can interact with the alkyl radical to form hydroperoxides.  Along with the fact that the alcohol formed will tend to undergo oxidation in the presence of hydrogen peroxide and hydroxyl radicals.
Also if you are still going to try the reaction, the bond dissociation energy of the $\ce{O-O}$ single bond is around 146 kJ per mole. So the energy per molecule is $\ce{2.433\times10^{-19} J}$.  So the maximum wavelength of light to ionize this is $\ce{8.21\times10^{-7} m}$.  Your calculation is a little off, it is in the near infrared range (you may have forgot to convert from kilojoules to joules or something like that).  This is why hydrogen peroxide is stored in dark bottles and told to not be put in the presence of light.


text:
Relative energies of singlet and triplet states in Jablonski diagrams
In a Jablonski diagram, we have singlet and triplet excited states where spin is conserved or changed, respectively. My book says any singlet excited state is found to have higher energy than the corresponding triplet state from the same electron configuration. Why is that?

What is the physical basis for Hund's first rule?

It is true experimentally that the triplet state arising from the same configuration as the singlet is of lower energy, the answer to this question gives some explanations https://chemistry.stackexchange.com/questions/58625/what-is-the-physical-basis-for-hunds-first-rule/58654#58654


text:
Can ethanol be oxidized by hydrogen peroxide?
Is it possible to oxidize ethanol to acetic acid with hydrogen peroxide and if yes then
under what circumstances? I tried it in room temperature but either concentration was too small (of hydrogen peroxide(3%)) or I couldn't quite precisely read the the results of the universal indicator. I also tried heating it up, but it didn't change anything. The color stayed the same (of indicator). Can someone please explain me if the mistake was in my experiment (if the reaction can happen in room temperature) or the reaction needs some specific catalyst or other conditions.
The reaction would be:
\begin{align}
\ce{\underset{(ethanol)}{C2H6O} + H2O2 &->
  \underset{(aldehyde)}{C2H4O} + 2H2O}\\
\ce{\underset{(aldehyde)}{C2H4O} + H2O2 &->
  \underset{(acetic acid)}{C2H4O2} + H2O}
\end{align}
Or the reaction without the middle part (since aldehyde will try to oxidize faster then ethanol) would be:
$$\ce{C2H6O + 2H2O2 -> C2H4O2 + 3H2O}$$

What are you trying to oxidize it to? Ethanal? Acetic acid? Carbon dioxide?

Oxidizing ethanol to acetic acid with H2O2 is pretty much like throwing a lump of bubble-gum from the roof of a building in the hope that it would stick to the wall at the height of 3rd floor, no more, no less. With great chances, it would either stick at some other height, or fall all the way down to the ground.

Indeed, that doesn't seem good idea. At very least you'd need much more concentrated and probably acidified solution.

First things first: Don't mess with higher concentrations of hydrogen peroxide unless you are a trained and well-equipped chemist. We're talking "steel-reinforced gauntlets" here. This chemical is as volatile as nitroglycerine.
At 3% you have mostly de-ionized water and for good reason. Hydrogen peroxide is extremely, violently unstable, and is just as likely to oxidize anything else than what you actually intend to react it with.
Anything above a concentration of 30% is very dangerous. Above 60% is suicidal in ill-equipped scenarios. 100% is a hypothetical, and is in essence a potential component of rocket fuel.
So, yes it will work. You may lose a hand, but it will work.

Listerine Total Care Stain Remover AntiCavity Mouthwash contains 21% ethanol (from the label) and hydrogen peroxide (lower down on the ingredients list; I would hazard a guess of about 0.75 - 1.0% from the amount of foaming in use).
The product is stable for months - perhaps years. My bottle is way past its expiration, yet still foamable. Although the concentrations are not extreme, the long term stability suggests that H2O2 doesn't react significantly with ethanol, at least not at these concentrations.
Higher concentrations are more reactive: On 16 July 1934, in Kummersdorf, Germany, a tank containing a mixture of hydrogen peroxide and ethanol exploded during a test, killing three people.
Fenton's Reagent, which is hydrogen peroxide with a catalytic amount of ferrous sulfate, is used to oxidize waste water containing organic compounds, but as indicated in a comment above, getting it to stop at acetic acid may be tricky.

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text:
Does the reaction of sulfur and potassium nitrate involve production of sulfur trioxide?
I've heard that oxidising sulfur with potassium nitrate yields $\ce{SO3}$ gas, but can't find a definitive answer anywhere.
Is sulfur trioxide produced? More generally, what's the full reaction?

There is nothing definitive about it. Oxidizing S with KNO3 is a pretty violent reaction, and it may involve a lot of things.

The gun powder was thoroughly studied by Liebig and coworkers in the beginning of the $20$th century. It was shown that, in absence of air the reaction  goes as Nilay Ghosh states, producing $\ce{K2S}$ as main solid stuff. But in the presence of air, $\ce{K2S}$ is quickly oxidized. With few oxygen, it is transformed  into sulfur +  $\ce{K2O}$ (quickly transformed into $\ce{K2CO3}$). And if more oxygen is available, $\ce{K2S2O3}$ is produced. And with still more oxygen, the final product is $\ce{K2SO4}$

If we talk about the reaction of only sulfur and potassium nitrate, then the reaction yields potassium oxide, sulfur dioxide and nitrogen.
$$\ce{4KNO3 + 5S -> 2K2O + 5SO2 + 2N2}$$
However, when carbon is added in the form of charcoal, then the resulting mixture is called gunpowder. The overall chemical reaction of gunpowder is as follows:
$$\ce{2 KNO3 + S + 3 C -> K2S + N2 + 3 CO2}$$
$$\ce{10 KNO3 + 3 S + 8 C -> 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2}$$
There are plenty of questions in chem.SE related to gunpowder:

What is the oxidation mechanism of gunpowder?
What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder?
Why can sulfur act like a catalyst?
How to write the balanced equation of the reaction of potassium nitrate, carbon, and sulfur?
What are the reaction products of sulfur and potassium nitrate?

i believe OP is trying to make sulfuric acid. this is too irrelevant

@FrancisL. Where did you find sulfuric acid? I only see SO3 being mentioned.

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text:
Solid state and packing
In NaCl there is a simple cubic unit cell and the coordination number is 6. So i am confused how are both the ions arranged in such a lattice, with such a cordination no as in simple cubic unit cell there are 8 atoms on the edge of a cube with a contribution of 1/8 by each of them to each unit cell. Also how does it manage to have itself satisfied its valency in such a way to have a chemical formula of NaCl.

There is no such thing as valency. As for the unit cell, it is by no means simple. Imagine a cubic unit cell with the size twice as much as you imagined before, with sodium ions at the corners and at the centers of all faces, and with chlorine atoms at all mid-edges and at the center of the entire cell (or vice versa, that doesn't matter). That would be it.

But in a unit cell an atom always lies on the corners of the cube.so where does this mid-edges come into play!

You have it all backwards. A unit cell may contain one, or ten, or 100, or 1000 atoms. Whether or not an atom lies at the corners is but a matter of convention. Sometimes it does, sometimes it doesn't. In my case it does, so what's the problem?

Oh yes i get it but how does each atom have a cordination number of 6.Also with regard to valency in NaCl each ,Na+ ion has an electrovalent bond with Cl- ion. So each Na + ion must have only 1 effective bond with all the chlorine ions in its vicinity.I also want to know if a unit cell must effectively  show only one molecule or a atom as it says unit cell or there is a different story there.

Also tell me if a molecule has a simple cubic crystal it can have all type of cubic  unit cells like fcc, bcc,etc or not

Build a 3D model out of matchsticks and bubble-gum (or any other means), and you'll get the idea of it. As for the rest, there is no such thing as effective bond. Also, the unit cell is by no means constrained to contain only one molecule. Also, there is no such thing as NaCl molecule, since NaCl doesn't consist of molecules. Also, fcc or bcc is by definition not simple.

https://commons.wikimedia.org/wiki/File:Sodium-chloride-3D-ionic.png

An unit cell is a periodic structure. There are two types of unit cells: conventional unit cells and primitive unit cells. The conventional unit cell is the one that is used most often. Nevertheless, I suggest that you take a look at what is the difference between the two. There are plenty of resources that explain the difference.
Now with that out of the way. Let's look at the unit cell of NaCl. According to Chem Libre, the unit cell is:

Recall that ionic solids are held together by electrostatic forces. So what matters is that the charges in a unit cell cancel. In other words, we want the net charge to be 0.  Let's count the number of sodium atoms and chloride atoms to see whether the charges cancel in this unit cell.
At the top plane, there are 4 chloride atoms that contribute $\frac{1}{8}$ of an atom. There is also 1 chloride atom that contributes $\frac{1}{2}$. On the other hand, there are 4 Na atoms that contribute $\frac{1}{4}$ of an Na atom. Now, each Na atom has a 1+ charge whereas each Cl atom has a 1- charge. Therefore, at the top plane we have that:
$$(+1)(4\,\cdot\,\frac{1\,\,\text{Na atoms}}{4})+(-1)(4\,\cdot\,\frac{1\,\,\text{Cl atoms}}{8})+(-1)(1\,\cdot\,\frac{1\,\,\text{Cl atoms}}{2})=0$$
So we can see that the overall charge for the top layer is 0.
The same is true for the plane in the middle. Work through it. It is a good exercise.
Now, lets look at the coordination number. What do we mean by coordination number? The coordination number is simply the number of nearest neighbors. That is all that there is to it. It doesn't matter whether a single sodium ion is in contact with 6 chloride ions. This doesn't imply that the sodium ion will end up with -5 negative charge. This is because even though the sodium ion is in contact with 6 chloride ions, each of these chloride ions is in contact with other sodium ions. As a result, the sodium ion does not really accumulate negative charge.
What determines the coordination number is a bit more complicated. The coordination number is dependent on the size of each atom, the charges, and even the spin. One simple rule that helps you determine the coordination number of binary solids is the radius ration rule. I suggest that you take a look at it: http://minerva.mlib.cnr.it/mod/book/view.php?id=269&chapterid=111


text:
Sodium Silicide (NaSi) Synthesis, Reversability
I'm trying to understand a) the reaction that would produce $\ce{NaSi}$ from $\ce{Na}$ and $\ce{Si}$, as well as how to reverse the reaction of $\ce{NaSi}$ with $\ce{H2O}$. The product of the latter would be $\ce{Na2Si2O}$ (aq), and I want to understand the reaction which would transform it back to $\ce{NaSi}$ (including heat inputs).
Where could I find this information? Is there an obvious repository of chemistry information where I could look this up?
Thanks you!

https://en.wikipedia.org/wiki/Sodium_silicide - Wikipedia does seem as obvious repository...

The reaction of $\ce{NaSi}$ with water is both highly enthalpically and entropically favored—the reaction produces 5 molecules of $\ce{H2}$ gas for every 2 $\ce{NaSi}$ molecules that react, and the enthalpy of reaction is –175 $\mathrm{kJ\ mol^{-1}}$. The reverse reaction would require ludicrous amounts of energy and is likely not even kinetically feasible.
It would be especially pointless to attempt the reverse reaction when $\ce{NaSi}$ is generated easily enough from $\ce{Na}$ and silica gel. Liquid sodium-potassium alloy absorbed onto silica gel results in a fine black powder that spontaneously ignites in humid air. Heating $\ce{Na}$ and silica gel to 165ºC with continual agitation produces a shelf stable, yet still highly reactive $\ce{NaSi}$. When slowly heated to 400ºC, air stable $\ce{NaSi}$ may be produced, though with less reducing power than those produced via the other two methods. It will, however, still react with water to produce $\ce{H2}$ and $\ce{ Na2Si2O5}$.$^{[1]}$

$^{[1]}$ Alkali Metals Plus Silica Gel:  Powerful Reducing Agents and Convenient Hydrogen Sources


text:
Where is my antimony going?
I work in a trace metals laboratory operating an ICP-OES. I've been having some trouble with the recovery of antimony in a weak $\ce{HCl}$ solution post-digestion.
The antimony is part of a mix of elements in ~$\ce{5\% HNO3}$, 1 mL of which is added to a beaker with ~$\pu{50 mL}$ of $\ce{0.5N HCl}$, then heated at a low temperature for ~10-15 minutes. There is a filtering process afterward using simple glass fibre filters. I've been getting between 60-75% recovery of the antimony, and at least 80-90% recovery of the other elements.
I've been able to find some information regarding issues with antimony and $\ce{HCl}$, and possibly with natural light affecting the stability of the solution, but nothing definitive or on-topic.
Any ideas?

I'm pretty sure I remember the folks from our ICP lab discussing problems with Sb volatilizing, though I thought that was during higher temperature digestion of solids. Anyway, SbCl3 is fairly volatile (BP ~220C) and ferric chloride will complex it and reduce it's volatility. If I knew the details of how to do this right though, I'd have written a formal answer ;)

And, the antimony trichloride is accessible because you have the nitric acid there to oxidize the antimony to the appropriate oxidation state.

@Oscar - Would the little amount of nitric involved be a player? It's 1mL of 5%. Though I guess it doesn't take much with how strong nitric can be.

@airhuff - I'd have to check the actual temperature of the hotplate. It's on one of the lowest settings, and the 0.5N HCl solution involved never reaches boiling. The powdery substrate this test is for does cake up somewhat at the bottom and cause some sputtering.
I don't have the option of adding ferric chloride, must follow a pretty straightforward process.

@Max also probably not a lot of antimony.

@Oscar - No, it's only 1 ppm levels. The same standard is used in calibration at the same level and I get great recovery there. However that solution is a different acid matrix of 10% HCl / 2% HNO3.


text:
What is the molarity of a solution made by dissolving 25.63 grams of naphthalene (MM = 128.17g) in 250.0 grams of benzene?
I don't really understand how to find the amount of liters with benzene in order to find the molarity.

Do you know how to find the number of moles of naphthalene in 25.63g ?

Question wasn't framed correctly. You need to know the density, either must be given in question or remembered beforehand


text:
Which steps are most critical in driving glycolysis foward?
I'm assuming that the non-reversible steps (1,3,10 on the chart below) in glycolysis are the most critical in driving it forward, but I have a feeling it might be something else.  If so, why are they the most critical?
I've been using this chart as a reference.

The second half of my question is how mass-action ratio (Q) affects these reactions.

Please try to focus on one question at a time, would be happy to assist you with the first part only

Those steps you mentioned are not merely critical steps, in glycolysis but are however the control points of the process and I would prefer to call them rate controlling steps. There is nevertheless one critical step which is believed to be the first committed step in glycolysis (not really sure if there are more).

There are three irreversible steps in glycolysis, and the differences
between glycolysis and gluconeogenesis are found in these three
reactions. The first of the glycolytic reactions is the production of
pyruvate (and ATP) from phosphoenolpyruvate. The second is the
production of fructose-1,6-bisphosphate from fructose-6-phosphate, and
the third is the production of glucose-6-phosphate from glucose.
Because the first of these reactions is exergonic, the reverse
reaction is endergonic. Reversing the second and third reactions would
require the production of ATP from ADP, which is also an endergonic
reaction.

Having said that, there are factors whih affect the process of glycolysis:
Enzymes
Control of hexokinase
The enzyme that catalyses the first reaction is hexokinase. The substrate of hexokinase is not necessarily glucose; rather, it can be any one of a number of hexoses, such as glucose, fructose, and mannose. Glucose-6-phosphate inhibits the activity of hexokinase; this is a control point in the pathway.  Hexokinase is inhibited by high levels of its product, glucose-6-phosphate. When glycolysis is inhibited through phosphofructokinase, glucose-6-phosphate builds up, shutting down hexokinase. (However, the liver contains a second enzyme that phosphorylates glucose-glucokinase).
Control of phosphofructokinase
The phosphorylation of fructose-6-phosphate is highly exergonic and irreversible, and phosphofructokinase, the enzyme that catalyses it, is the key regulatory enzyme in glycolysis. Phosphofructokinase is a tetramer that is subject to allosteric feedback regulation. fructose-2,6-bisphosphate an important allosteric activator of phosphofructokinase (PFK), the key enzyme of glycolysis; (it is also an inhibitor of fructose bisphosphate phosphatase (FBPase), which plays a role in gluconeogenesis.)
Control of Pyruvate Kinase
The final step of glycolysis is also a major control point in glucose metabolism. Pyruvate kinase (PK) is allosterically affected by several compounds. ATP and alanine both inhibit it. (It would be wasteful to breakdown more glucose is there is already vast amounts of ATP). Alanine is essentially pyruvate with an amino group. (converted to pyruvate via a transaminase enzyme). Therefore, a high level of alanine indicates that a high level of pyruvate is already present, so the enzyme that would make more pyruvate can be shut down. Fructose-1,6-bisphosphate allosterically activates pyruvate kinase so that the incoming products of the first reactions of glycolysis can be processed.
Substrate
The reaction in which fructose-6-phosphate is phosphorylated to give fructose- 1,6-bisphosphate is the one in which the sugar is committed to glycolysis. Glucose-6-phosphate and fructose-6-phosphate can play roles in other pathways, but fructose-1,6-bisphosphate does not. After fructose-1,6-bisphosphate is formed from the original sugar, no other pathways are available, and the molecule must undergo the rest of the reactions of glycolysis.
It is frequently observed that control is exercised near the start and end of a pathway, as well as at points involving key intermediates such as fructose-1,6-bisphosphate. This is the summarised version of control factors of glycolysis:

Remarks
To sum up, the reason I explained your mentioned reactions not to be critical is because (with reference to reaction 1 and 3):

Even if hexokinase is inhibited, glycolysis can still take place via catalysis by glucokinase in liver.

There are other feeder pathways for glycolysis even if the glucose sugar is not present, the saccharides (maltose, lactose, trehalose, and sucrose, mannose and galactose) can still be used to produce fructose 6-phosphate.

e.g $$\ce{Fructose + ATP  (Mg2+) → fructose 6-phosphate + ADP}$$

The last reaction would be bypassed by transaminase (is not important since the reaction would have been completed):

So these are major controlling steps, meaning glycolysis can still take place  (if committed step is reached) though at reduced rates until optimum conditions are restored. The second part of your question can’t not answer it because it is a different concept. Hope this helps
References

Harpers illustrated Biochemistry
Lehninger Principles of Biochemistry
Biochemistry (Campbell and Farrell)

The committed enzyme in glycolysis is considered to be PFK-1, and is under complex control.  While the hexokinase/glucokinase steps are of course critical for getting glucose into the cell, the product, glucose 6-phosphate has multiple fates (on to glycogen or the pentose pathway), so is not unique to the glycolytic pathway.
In muscle, the key activator of PFK-1 is AMP, signalling a drop in the energy charge of the cell and therefore a need to switch on ATP generating pathways.
In the liver, the key activator is Fructose 2,6 bisphosphate, which rises in response to insulin.


text:
What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD?
What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD ?

I already know the mechanism the question is the water molecules that is attacked by the isocyanide is it already present in the solution or is the one formed within the reaction in situ?, secondly If we are given the reaction is carried out in Br2 and KOD then do i have to assume that it is actually  KOD in D2O because my teacher says there would be no crossover product that if the reaction is carried out in aforesaid conditions the product would be RND2but i think if the water molecule in the reaction mechanism are  the ones generated in situ then there is a fair chance of other products like RNHD or even RNH2.
So the main question are :-
1) Are the water molecules the ones generated in situ ?
2) Is KOD equivalent to KOD in D2O ?
3) Would RNHD and RNH2 also form in the given conditions ?

@orthocresol if the reaction is carried out in some non polar solvent as DMS  then would the replacement be by the H2O or HOD formed in situ

This isn't the most well-thought-out scenario, because if you dissolve $\ce{KOD}$ in $\ce{H2O}$ all the deuteriums will be lost immediately anyway.
$$\ce{KOD + H2O <=>> KOH + HOD}$$
Assuming that the amount of $\ce{KOD}$ is much less than the amount of $\ce{H2O}$ (after all $\ce{H2O}$ is the solvent) then the position of this equilibrium will lie very, very far to the right. So, at the end of the day, if you are using $\ce{H2O}$ as the solvent, the use of $\ce{KOD}$ won't have any effect. If you want to see any difference in deuteration, you have to at least use $\ce{KOD/D2O}$. However, there is a second problem with this:
If you dissolve an amine in $\ce{D2O}$, then all the amine hydrogens will be very rapidly replaced with deuteriums. It does not even matter mechanistically what the pathway is. Regardless of whether the initial product is $\ce{RNH2}$, $\ce{RNHD}$, or $\ce{RND2}$, as long as there is $\ce{D2O}$ in the mixture, all three will become $\ce{RND2}$. The same can be said of dissolving them in $\ce{H2O}$.

To answer your questions:
(1) No. In general, protonation can and will occur by any solvent molecule that happens to be close by. If you are running the reaction in $\ce{D2O}$ solvent, then there is much, much more $\ce{D2O}$ than there will ever be $\ce{H2O}$ produced. So, protonation by $\ce{H2O}$ is statistically close to impossible.
(2) None of us can answer that, as we cannot presume what your teacher was thinking of when they set the question.
(3) Depends on what your solvent is, see comments above on dissolving amines in $\ce{D2O}$ or $\ce{H2O}$.


text:
Enthalpy definitions. What are their main differences?
The definition of enthalpy is properly described in here.
The principal objective of this question is mainly academic. In order to any students with internet connection can find a definition and an example of the enthalpy concept applications.
What are the main differences of:

Standard enthalpy
Enthalpy of atomization
Enthalpy of formation
Enthalpy of hydration
Enthalpy of reaction
Enthalpy of combustion
Enthalpy of solution
Enthalpy of lattice dissociation
Enthalpy of lattice formation
Enthalpy of mixture
Enthalpy of excess mixture
Enthalpy of (any phase transition)
Law of Hess
Law of Kirchhoff

Any other definition not listed here

Explanation of notation:
$H$ is the enthalpy of the system.
$\Delta$ means change of, so $\Delta H$ means change of the enthalpy

This symbol means standard condition, standard condition is defined as a pressure of $100\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\text{M}$. However, since LaTeX doesn't have this symbol, I'll substitute it for $^\ominus$.
Enthalpy changes:
Enthalpy of dilution – The enthalpy change when a solution containing one mole of a solute is diluted from one concentration to another.
Enthalpy of ($n\text{th}$) electron affinity – The enthalpy change when $n$ electrons are added to one mole of gaseous atoms.
$$\ce{Li(g) + e-(g) -> Li-(g) +60\ \text{kJ}}$$
$$\ce{F(g) + e-(g) ->  F-(g) +328\ \text{kJ}}$$
Enthalpy of ($n\text{th}$) ionization – The enthalpy change when $n$ electrons are removed from one mole of gaseous atoms. It is always positive.
$$\ce{Li(g) +520\ \text{kJ} -> Li+(g) +e-(g)}$$
$$\ce{He(g) +2372\ \text{kJ} -> He+(g) +e-(g)}$$
Enthalpy of lattice dissociation – The enthalpy change when one mole of an ionic lattice dissociates into isolated gaseous ions.
An example for sodium chloride which have an enthalpy of lattice dissociation, also known as lattice energy, of $787\ \mathrm{kJ/mol}$
$$\ce{NaCl(s) +787\ \text{kJ} -> Na+(g) + Cl-(g)}\qquad\Delta H=787\ \text{kJ/mol}$$
Enthalpy of lattice formation – The enthalpy change when one mole of solid crystal is formed from its scattered gaseous ions.
$$\ce{Na+(g) + Cl-(g) -> NaCl(s)} +787\ \text{kJ}\qquad\Delta H=-787\ \text{kJ/mol}$$
Which means Enthalpy of lattice dissociation$=-$Enthalpy of lattice formation
Enthalpy of hydration($\Delta_\text{hyd}H^\ominus$) – Enthalpy change when when one mole of ions undergo hydration.
Enthalpy of mixing – The enthalpy change from a substance when mixed.
Enthalpy of neutralisation – The enthalpy change when an acid is completely neutralised by a base.
For a strong acid, like $\ce{HCl}$ and strong base, like $\ce{NaOH}$, they disassociate almost completely $\ce{Cl-}$ and $\ce{Na+}$ are spectator ions so what is actually happening is $$\ce{H+(aq) + OH-(aq) -> H2O(l) + 58\ \text{kJ/mol}}$$
However, using a weak acid/base will have a lower enthalpy of neutralisation as normally most of the acid/base does not disassociate.
For example, mixing ethanoic acid and potassium hydroxide only has a enthalpy of neutralisation of $-11.7\ \text{kJ/mol}$
Enthalpy of precipitation – The enthalpy change when one mole of a sparingly soluble substance precipitates by mixing dilute solutions of suitable electrolytes.
Enthalpy of solution – Enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.
Enthalpy of solution can be positive or negative as when a ionic substance dissolves, the dissolution can be broken into three steps

Breaking of solute-solute attraction (endothermic)
Breaking solvent-solvent attraction (endothermic), eg. hydrogen bonds, LDF
Forming solvent-solute attraction (exothermic)

An example of a positive enthalpy of solution is potassium chlorate which has an enthalpy of solution of $41.38\ \text{kJ/mol}$
Enthalpy of (Solid$\rightarrow$Liquid: $\Delta_\text{fus}H^\ominus$, Liquid$\rightarrow$Solid:$\Delta_\text{freezing}H^\ominus$, Liquid$\rightarrow$Gas: $\Delta_\text{vap}H^\ominus$, Gas$\rightarrow$Liquid: $\Delta_\text{cond}H^\ominus$, Solid$\rightarrow$Gas: $\Delta_\text{sub}H^\ominus$, Gas$\rightarrow$Solid: $\Delta_\text{deposition}H^\ominus$) – The enthalpy change from providing energy, to a specific quantity of the substance to change its state.
$$\Delta_\text{fus}H^\ominus=-\Delta_\text{freezing}H^\ominus,\Delta_\text{vap}H^\ominus=-\Delta_\text{cond}H^\ominus,\Delta_\text{sub}H^\ominus=-\Delta_\text{deposition}H^\ominus$$
Standard enthalpy of atomization($\Delta_\text{at}H_T^\ominus$) – Change when a compound's bonds are broken and the component atoms are reduced to individual atoms at at $T^\circ K$.
$$\ce{S_8 -> 8S}\qquad\Delta_{at}H^\ominus=278.7\ \text{kJ/mol}$$
Standard enthalpy of combustion($\Delta_\text{c}H_T^\ominus$) – The enthalpy change which occurs when one mole of the compound is burned completely in oxygen at $T^\circ K$ and $10\ \mathrm{kPa}$.
$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)}+572\ \text{kJ}\qquad\Delta_cH^\ominus = -286\ \text{kJ/mol}$$
Standard enthalpy of formation($\Delta_\text{f}H_T^\ominus$) – Change in enthalpy during the formation of one mole of the compound from its constituent elements, with all substances in their standard states, and at a pressure of $100\ \mathrm{kPa}$ at $T^\circ K$.
It can be calculated using Hess's law if the reaction is hypothetical. An example is methane, $\ce{C}$ and $\ce{H2}$ will not normally react but the standard enthalpy of formation of methane is determined by Hess's law to be $-74.8\ \text{kJ/mol}$
$$\ce{\frac{1}{2}N2(g) +\frac{1}{2}O2(g) -> NO(g)}\qquad\Delta_\text{f}H^\ominus=90.25\ \text{kJ/mol}$$
Standard enthalpy of reaction($\Delta_\text{r}H^\ominus_T$) – Enthalpy change that when matter is transformed by a chemical reaction at $T^\circ K$ and $10\ \mathrm{kPa}$.
$$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)} +572\ \text{kJ}\qquad\Delta_\text{r}H^\ominus = −572\ \text{kJ/mol}$$
$$\Delta_\text{r}H^\ominus=\sum H^\ominus_\text{products}-\sum H^\ominus_\text{reactants}$$
$$\Delta_\text{r}H_\text{forward}=-\Delta_\text{r}H_\text{backwards}$$
Laws:
Le Chatelier's Principle – When an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change.
Yellow $\ce{Fe^3+}$ reacting with colorless thiocyanate ions $\ce{SCN-}$ to form deep red $\ce{[Fe(SCN)]^2+}$ ions:
$$\ce{Fe^3+(aq) + SCN-(aq) -> [Fe(SCN)]^2+(aq)}$$
When $\ce{NH4Cl}$ is added, $\ce{Cl-}$ reacts with $\ce{Fe^3+}$ ions to form $\ce{[FeCl4]-}$ ions. By the Le Chatelier's Principle, when $\ce{Cl-}$ is added into a solution of deep red $\ce{[Fe(SCN)]^2+(aq)}$ ions, the equilibrium will shift to $\ce{Fe^3+(aq) + SCN-(aq)}$, turning the solution pale red.
Kirchhoff's Law – Enthalpy of any substance increases with temperature.
Hess's Law – Total enthalpy change of a chemical reaction is independent of the number of steps the reaction takes.
Henry's Law – Amount of dissolved gas is proportional to its partial pressure in the gas phase.
Processes:
Constant Entropy – Isoentropic
Constant Pressure – Isobaric
Constant Volume – Isovolumetric
Side note: This is probably not a complete list as I may have missed some

Standard state does not include any stipulation of temperature.

@orthocresol Shouldn't it include, as due to Kirchhoff's law, the enthalpy of ___ will change with the temperature, so the temperature and pressure should be specified

Yes, you are correct that the enthalpy change will vary with temperature. Therefore, a temperature has to be specified. However, there is no condition that it has to be 273 K in particular (or 298 K or anything else).

@ArianaGrande I have edited your post using the \ce notation used for chemical equations. For example, write $\ce{A + B -> C}$ to get $\ce{A + B -> C}$. Furthermore, add \text{Input-text-here} on the units you write. Pretty good answer by the way.

@Ariana Standard conditions usually means a pressure of 1 atm or 1 bar ie 100 kPa (the value in the text is wrong).

Note I took the liberty of making a correction, the standard pressure: 1 bar = 100 kPa, not 10 kPa

Most of the chemistry happens in open containers. Open containers have constant pressure. At constant pressure work is done to carry out the reaction and also to "push" the air up (called as expansion work). This expansion work is $${\Delta PV}$$. Thus some part of the heat leaks as expansion work without being converted into Internal Energy.
So we define a new term $${\Delta H ~(Enthalpy) = \Delta U + \Delta PV}$$. It accounts for expansion work and internal energy. As the heat is used only for internal energy and expansion work, Enthalpy is equal to heat supplied.
There are various types of enthalpies depending upon Internal Energy is used to do what kind of work.
Phase is something that is uniform in composition and physical state. It is similar to state of matter (solid, liquid and gas), but these is only one state of matter (solid carbon) but there can be many phases (graphite, diamond)
Enthalpy of phase transition
It is the heat supplied at constant 1 atm pressure to bring about a change in phase. It can be $$\ce{H2O (liquid) <=> H2O (gas) -> +44KJ}$$ or $$\ce{C (graphite) <=> C (diamond) -> +4KJ}$$ Conversion of graphite into diamond is thermodynamically possible but kinetics does not allow it, as there are very strong bond to break and rearrange. This is true for every thermodynamical reaction.
Enthalpy of vapourisation and Enthalpy of Condensation
Enthalpy of vapourisation is the heat required to vaporise 1 mole of liquid into gas. (a phase change) Energy needs to be supplied and hence enthalpy is positive.
As Enthalpy is a state function, the reaction can proceed from the final state (gas) to initial state (liquid) by absorbing the same amount of energy. This heat absorbed at constant pressure is called as enthalpy of condensation and is numerically equal to negative of enthalpy of vapourisation.
Enthalpy of Fusion and Enthalpy of freezing
The heat required to melt 1 gm of solid into liquid at a constant pressure of 1 atm is enthalpy of fusion. The enthalpy required for the reverse reaction is (liquid to solid) is called enthalpy of freezing. It follows the same lines as of above two enthalpies.
Enthalpy of sublimation and Enthalpy of vapour deposition
Some substances directly vaporise into gas, without forming liquid and liquid is not stable at any temperature and pressure. The enthalpy required to do so is enthalpy of sublimation. As enthalpy is a state function, the enthalpy required to deposit the gas into a solid is called enthalpy of vapour deposition.
Enthalpy of atomic and molecular changes
Enthalpy of ionisation and Enthalpy of electron capture
The energy required to remove the first electron of one mole of gas at constant pressure is called enthalpy of ionisation. The energy required to remove the second electron is called Second enthalpy of ionisation.
The enthalpy of ionisation of 1 mole of gas at 273.15 Kelvin and 1 atm pressure is called "Ionisation Energy".
The enthalpy of capturing an electron by a neutral atom (neutral, not positive ion) is called enthalpy of electron capture. At 273.15 K and 1 atm, for 1 mole this energy is also called "Standard electron gain Enthalpy"
Enthalpy of disassociation and Enthalpy of bond formation
The heat required to disassociate a molecule into 2 entities is called enthalpy of bond dissociation Eg $$\ce{H2O -> H2 + O ~~~~ \Delta H = +927 KJ}$$
the opposite of it is enthalpy of bond formation Eg. $$\ce{H2 + O2 -> H2O}$$ Here the enthalpy of (H-O-H) is not equal to 2 times enthalpy of (O-H)
for this particular reaction, $$\ce{H2O(gas) -> H + HO^. ~~~~\Delta H = +428KJ }$$ and $$\ce{OH^. -> H + O^. ~~~~ \Delta H = +428 KJ}$$. Therefore total enthalpy is +927KJ. As Enthalpy of disassociation is dependent upon reaction itself $$\ce{\Delta H(O-H)~is~the~average~value~of~\Delta H(O-H)~of~H-OH~,~\Delta H(O-H)~of~CH3OH~etc.}$$


text:
Why won't my propane combust when I spark it?
I need to run an experiment and have some pretty basic questions. I'm trying to get combustion of propane in a controlled environment. I understand that that looks like this:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$
Meaning I need 5 times more $\ce{O2}$ molecules than $\ce{C3H8}$. But, I want to do this based on the volumes of the gases at STP. Do I just go with 5 times the volume of $\ce{O2}$ to the volume of $\ce{C3H8}$? Do the gas laws require that I take into account the size of a molecule?
And seeing that $\ce{O2}$ comprises about $\pu{21\%}$ of the volume of air, I would need to multiply the volume of oxygen by roughly 5 to get the correct volume of air needed...yes?
That gets me a 25 to 1 ratio between air and propane by volume. So, I tried it, but I couldn't get it to burn.
Doing further research, I see something called the Limiting Oxygen Concentration, but I think either I don't understand it, or this isn't the problem.
I also found that the lower and upper limits of flammability for propane are $\pu{2.15\%}$ and $\pu{9.60\%}$ respectively. My concentration is roughly $\pu{4\%}$, so this shouldn't be the issue either.
The only other thing I can think of is that my spark isn't "good enough". Is there a minimum amount of heat required from my spark to ignite the mixture?
Anyone have any ideas as to what I am missing here? Any help would be much appreciated!

Hello and welcome to Chemistry.SE. If you have any questions about the site please feel free to take the short [tour] or visit the [help]. Best of luck with your question!

Do you mean that you actually mix some volumes of air and propane, and then try to ignite it? Then it is probably a good thing that some of your components was faulty, otherwise you could blow your head off and get arrested for that.

@IvanNeretin Necroing this, but your order of actions is amusing.  Would they arrest the severed head or the headless body?

@Tristan Depends on the jurisdiction, I guess.


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What is the meaning of n-Pr and i-Pr?
Apologies in advance, I have a background in physics, so this question might seem awfully simple.
Anyway, I am interested in the chiral spin selectivity effect in organic molecules, and have read a paper in which they synthesize helicene molecules with side groups R = n-Pr, i-Pr with absolutely no reference only a slight hint as to what it is an abbreviation of in either the paper or supplementary information.  I thus assume that it must be some very common notation, and my guess is that Pr refers to some kind of propane/propene/propanol, and n and i refers to some normal and altered structure, respectively.  But I don't know.
I would be grateful if someone could shed some light on it for me.

Propyl is a simple hydrocarbon with three carbons in it. Propane is $\ce{CH3CH2CH3}$.
Propyl is the unit formed by attaching that chain to something else (which implies replacing one of the carbons with another bond). There are two ways to do this: one attaches the propyl unit using the end carbon on the chain (giving $\ce{X-CH2CH2CH3}$); the other involves attaching the middle carbon to something else (giving $\ce{CH3CHXCH3}$). The first is called (at least in older terminology) normal-propyl, n-propyl or $\ce{n-Pr}$; the second is called iso-propyl, i-propyl or $\ce{i-Pr}$. These reduce the amount of space required to write the structure.
In your compound helicene I'd guess that the propyls hanging off the rings are there to make the compounds far more soluble in organic solvents as flat aromatic structures tend to be fairly insoluble due to the way the pack in crystals.


text:
Is it possible to determine the age of a building by just looking at its window?
I've recently come across the fact that glass is an amorphous solid and is known as a pseudo solid or a super cooled liquid. Our teacher told us that buildings that if we were to closely observe the glass in the window panes, we would find that it's thicker at the bottom. I'm now wondering if we can calculate the difference between the width of the top and bottom of the window, would it be possible to determine an approximate age of a building?

It's possible, but you must know the fluidity of glass. Besides, it must be sure that the glass was initially made perfectly with even thickness.

@PrittBalagopal ...and that would be a problem if we are to deal with Drawn Sheet glass.

This is but an urban legend. Glass does not get thicker at the bottom. (Sure, glass is fluid to some tiny extent, and it can be demonstrated, but not in this way.) It is just that the glass panes were initially manufactured with uneven thickness. Then the people who stuck the glass into the frames would orient the thicker side down, maybe because it is more stable this way, or because it feels right, or just to mess with us. Sometimes they didn't pay attention and turned it randomly.
Supporting links: 1, 2.


text:
Can the hydride ion act as a ligand?
Hydride ion, as I know is a pretty powerful base, much stronger than hydroxide ion and cannot exist in an aqueous phase.
Can it act as a ligand in coordination compounds? Hydride ion has its electronic configuration as $\mathrm{1s^2}$ in a symmetrical s orbital so the tendency to donate a lone pair into a metal atom would be less. But can its high nucleophilicity allow it to bind to the metal atom despite the symmetry of its orbital?

Yes, there are many examples. For instance

The common reducing agents LithAl, $\ce{LiAlH4}$, and sodium borohydride, $\ce{NaBH4}$ (last one you might argue about)
$\ce{[ReH9]^{2-}}$, see https://en.wikipedia.org/wiki/Potassium_nonahydridorhenate
Metal carbonyl hydrides, see https://en.wikipedia.org/wiki/Metal_carbonyl_hydride
More generally many organometallic compunds, e.g. $\ce{(C5H5)_2ReH}$

A quick google also shows up http://www.ilpi.com/organomet/hydride.html

Are these coordination compounds? I thought in LiAlH4 and others, the hydrogen was covalently bonded with the metal atom. Is it a dative bond that exists between them?

Does the model of the bonding matter? In $\ce{MnF6^{2-}}$ the best model is probably ionic while in $\ce{PtI6^{2-}}$ covalent is probably nearer to the truth, and then you have $\ce{{Cu(NH3)_4}^{2+}}$ where some kind of dative bond is the most intuitive model. But I think we'd all be happy to call them metal complexes, which are examples of coordination compounds.

I guess thats true

There’s also the nice case of pentacarbonyl iron which, when hydroxide ions are added, reacts to give $\ce{[Fe(CO)4(COOH)]-}$ which then rearranges and liberates $\ce{CO2}$ to give $\ce{[FeH(CO)4]-}$. This compound is Brønsted amphotheric: it can be protonated to give neutral $\ce{[FeH2(CO4)]}$ or deprotonated to give $\ce{[Fe(CO)4]^2-}$. I think it’s fascinating!


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Why does nitrous oxide have 300 times the global warming potential of CO2?
Both nitrous oxide and carbon dioxide have roughly the same atmospheric lifetime (nitrous oxide slightly longer), so I thought that it would have to do with the infrared absorbance associated with each of them. So I looked at their spectrum.
Perhaps this is misleading because it is my understanding that not much infrared light at around the 2300 $\pu{cm^{-1}}$ region exists. But even so, that means that the two peaks that matter are $\ce{CO2}$'s 700 peak, and $\ce{N2O}$'s 1300 peak. It is my understanding that the blackbody radiation of earth peaks at around 700, and though sizable, is much less at 1300. So, I don't see why nitrous oxide would have a greater global warming potential than carbon dioxide.

On a molecule to molecule basis besides any basic transition intensity effect  perhaps it is due to the fact that  $\ce{CO2}$ belongs to the $D_{\infty h}$ point group and has zero nuclear spin which means that odd J rotational levels are absent.

I'm not sure whether global warming potential takes this into account, but $\ce{N2O}$ may oxidize/decay into other compounds which have far greater GWP, so "by proxy" $\ce{N2O}$ could end up having a significant impact.

From Wikipedia "Even if a gas absorbs radiation efficiently at a certain wavelength, this may not affect its GWP (Global Warming Potential) much if the atmosphere already absorbs most radiation at that wavelength. A gas has the most effect if it absorbs in a "window" of wavelengths where the atmosphere is fairly transparent." $\ce{N2O}$ happens to lie within one of these "windows" where nothing else is absorbing much.

@NicolauSakerNeto , methane is a good example of what you are talking about. It is a strong greenhouse gas with a short atmospheric lifetime (~12yr) whose primary sink is oxidation to other greenhouse gases, $\ce{CO2}$ and water.

@NicolauSakerNeto It is a good point to raise, but not applicable to the GWP of N2O as a discrete molecule. As said, it is important to raise, as N2O is involved in ozone depletion. It's not as effective at O3 depletion as CFCs, but the increasing concentrations of N2O (coupled with decreases in CFCs) make it a major player.

There are two really important properties of N2O that make it such an important greenhouse gas, both of which you have raised in your question but need some clarification.

Infrared absorption - it's good to think in terms of gaps here. There's a lot of overlap between CO2, CH4 and (a frequently ignored GHG, but very important) water vapour in the wavelengths which they absorb. At the 4-5 micron and 7-8 micron range, N2O is extremely efficient at absorbing infrared radiation. These are important regions, precisely because there is no overlap with other GHGs in these regions. N2O is absorbing infrared very efficiently, and is doing so without "competition".
Lifetime - you noted that the lifetime of N2O and CO2 are comparable, which is sort of true. N2O, as a stable, inert molecule that is well mixed in the atmosphere, has a very long lifetime of 120 years compared to some other GHGs (CH4, for example is 8 years, based on the fact that its removed by the hydroxyl OH radical). It's possible to determine this fairly accurately as the only sinks for N2O are its stratospheric photo-dissociation and reaction with O(1D) radical. CO2 by comparison has a lifetime that varies anywhere between 5 and 200 years. The comparable part comes from the fact that for calculations for global warming potential, CO2 is given an "averaged" lifetime.

With N2O, therefore, we have a very efficient infrared absorber at two "windows" at which there is no overlap with other GHGs, and a long lifetime. The Global Warming Potential is the time integrated radiative forcing for a 1 kg pulse emission of the compound, where the upper limit of the integration is time horizon (usually set at 100 yr), relative to the same quantity of the reference compound CO2. It becomes clear from this definition why a stable, inert molecule with a long lifetime and which is very efficient at absorbing at particular infrared wavelengths would have a greater global warming potential than CO2 (shorter lifetime, absorbing in "windows" shared by other compounds) on a kg to kg comparison.

Another facet to the window factor mentioned in many other answers is the fact that the absorption of any species is blocked by itself - ie, the concentration of any gas already in the atmosphere is already absorbing its wavelengths, so the marginal addition of 1 more kg of that gas will have less IR to absorb. Because the concentration of CO2 is over 400ppm while the concentration of N2O is more than 1000x less at around 329ppb, a new kg of N2O has a lot less "competition" to absorb its IR than a new kg of CO2. For the same reason, the GWP of CO2 is decreasing over time as more CO2 builds up in the atmosphere.


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Calculating the entropy of vaporization at temperatures other than the normal boiling point
I would appreciate if someone could explain why this is so:
To find the entropy of transition at another temperature, we have to break the calculation down into three steps . For example, to find the entropy of vaporization of water at 25  C and 1 bar, imagine that we
1. heat the liquid to its normal boiling point, 100 C
2. allow it to vaporize,
3. then cool the vapor back to 25  C
Why do you do these three processes? I don't understand why this would give you the entropy of transition at another temperature...

How would you propose doing it? The answer lies at the heart of thermodynamics - it doesn't matter how something gets from state A to state B, the Gibbs free energy difference is the same. So, you pick the easiest path to perform the calculation, knowing that the answer is the answer.

@jonCuster That's was my conclusion after thinking and discussing it with my friends! Thanks!


text:
Which analytical reagent can be used to detect Vinca alkaloids?
I am commencing a project on increasing the sensitivity of screening of some bioactive compounds produced as secondary metabolites from the plants! All of its chemical struture is known which is a dimer of vindoline ring connected to a catantine ring through carbon-carbon analysis! However there are no reagents specified in any research papers which could specifically bind with this compound to help in its quantitative analysis! Do I have to predict the reagent according to its chemical structure or is there some specific analytical reagent which can provide a detectable reaction with it?


text:
How can I dilute 13% Hydrochloric Acid to 5%?
This is likely too simple of a question, but I just want to make sure I'm understanding things correctly.
I have 1 liter of 13% Hydrochloric Acid solution. I need 5% solution. Is it correct that I can just add the proportional amount of distilled (?) water to end up with a 5% solution? (add 1.6 liters of distilled water?)

In order not to waste your resources I recommend that you specify the quantity that you need for the upcoming reaction.
To make a diluted quantity of $x$ liters you may use the equation $$c_1*v_1 = c_2*v_2 \\ v_2 = \frac{c_1*v_1}{c_2} $$
Where $v_2$ is the volume that you want to use for the reaction, $v_1$ is the volume from the concentrated solution that you need to take, $c_1$ is the concentration of the concentrated solution, and $c_2$ is the concentration of the solution that you want to use.
For your solution you may use the weight percentages or convert them to molarities if you want to use volumetric lab glassware.
To convert the weight percentages to molarities, use the equation $$c = \frac{x*\rho}{m*(100-x)}$$
Where x is the weight percentage of the solution so for 5% you use the number 5 in the equation, m is the molar mass of the hydrochloric acid. To facilitate calculating the molar masses of different compounds I use the on-line tool Molar Mass Calculator.
To get the best results I recommend using a volumetric flask and substituting the volume for the volumetric flask instead of the volume of the acid that you want to use. So the equation that want to use is $$v_{needed} = \frac{c_{5\%}*v_{volumetric}}{c_{13\%}} $$
You then put the volume that you obtained in the volumetric flask and fill the rest with distilled water till the mark and you have thus obtained your solution.


text:
How did hypophosphoric acid get its name?
Why does hypophosphoric acid has a strange name? My knowledge of inorganic nomenclature tells me that the prefix "per-" should be used with "-ic" acids and the prefix "hypo-" goes with "-ous" acids, which isn't the case here. Why is this so?
Just as part of an additional observation, two other things which I find strange about $\ce{H4P2O6}$ are:

I think this is the only oxy-acid of phosphorus having a $\ce{P-P}$ bond.
It has phosphorus in $+4$ oxidation state.

Don't look for logic in chemical naming, it is not always there. Maybe it's just because all shorter names were taken.

@IvanNeretin Thanks, I'm always at a hunt for logic, I think I must curb this habit of mine atleast in nomenclature

Chemists use whatever fits.  Nitric acid = oxidation state of nitrogen is +5, nitric oxide = +2, nitrous acid = +3, nitrous oxide = +1.  Perchlirate and permanganate each = +7 for the root element, but chlorate = +5 and manganate = +6.  And so on.

@IvanNeretin That is bad advice. There is pretty much always some logic behind the names, even if we know today it's wrong. And it always helps to know what the error was, if only to not fall into the same trap.

You are most of the way to the reason yourself:
$$\begin{array}{|c|c|c|}\hline
\text{name} & \text{formula} & \text{oxidation number of P}\\ \hline
\text{phosphoric} & \ce{H3PO4} & +5\\
\text{phosphorous} & \ce{H3PO3} & +3\\
\text{hypophosphorous} & \ce{H3PO2} & +1\\ \hline
\end{array}$$
The "hypo-" prefix means "beneath" or "less than". Hypophosphoric acid, $\ce{H4P2O6}$, has oxidation number of phosphorous as +4, which is in between phosphoric acid and phosphorous acid. Thus, "hypophosphoric" refers to "beneath" phosphoric (but above phosphorous) acid.


text:
Why not simulate every particle properties and interactions?
Today we have huge computational power (which is even significantly larger with supercomputers). I know that computational chemistry is used sometimes to predict particle properties. As I read on Wikipedia:

Present algorithms in computational chemistry can routinely calculate the properties of molecules that contain up to about 40 electrons with sufficient accuracy.

If that's so, why bother to try to find chemical interactions and properties experimentally, at least up to 40 electrons? For example, every year new drugs are being discovered. Wouldn't be it easier at least to find new chemical compounds, if not their properties, simply by computer simulation? What are the constraints and where do they come from? (I know that such constraints exist, but I'd like to know why).

By the time our computational power grew to the present state, all molecules with up to 40 electrons (which, mind you, constitute a tiny minority of all compounds) were already studied experimentally. As for the new drugs of today, they contain much more than 40 or even 400 electrons, so you can't calculate them reliably. Then again, computational chemistry is an increasingly valuable tool, as long as you know its limitations. Also, welcome to Chem.SE.

related https://chemistry.stackexchange.com/questions/4190/is-it-possible-to-compute-the-colour-of-any-molecule

As R.M. describes in the answer, they do do it, to some extent. https://en.wikipedia.org/wiki/Docking_(molecular)

It might be worth it to take a quick look at the Travelling Salesman Problem. While it's probably pretty easy to study a molecule and assume it's properties, there are a lot of possible molecules with up to 40 electrons -- The time it would take to go through all of them and list out their properties is presumably pretty difficult. It gets even harder above that. I don't believe the problem would be determining the behavior of an individual molecule type, but searching through them all.

Realistic systems have more than 40 electrons, more along the 10^22-23 or more atoms/electrons are typical

Drugs don't act by themselves, they bind to other molecules like enzymes and change their behavior. How many enzymes have electron count below 40?

@Sidney - 40 electrons is pretty small, so we could easily enumerate all molecules with up to 40e. People have already enumerated (and calculated with QM) essentially all molecules up to 9 heavy atoms: http://dx.doi.org/10.1038/sdata.2014.22

@GeoffHutchison Even so, the rules of the TSM Problem apply. If proper calculations were to require each electron be checked against all other electrons in the system, the calculation grows factorially. Assuming other comments are correct on the number of electrons in larger systems is correct, simulating the results would be 40! vs (10^23)!

@Sidney this is called "Full CI" for quantum calculations and I stick to my comment - 40 electrons is manageable (e.g. Benzene = 42e)

Forty electrons is tiny. Even if we limit ourselves to just the valence electons, cyclohexane already has 36 electrons. Anything drug-like has way more electrons that 40. For example, viagra has 178 valence electrons, and that's not necessarily a "large" drug. (Compare with vancomycin, for example.)
Even if you're dealing with things inorganic compounds, where the total number of atoms in the formula unit is small, the properties of the material don't come from a single formula unit, but come from the interaction of a large number of atoms. -- That's an example of a more general principle. The important properties for most materials you use (including drugs) don't come the molecule in isolation, but come from the interactions of the molecule with other molecules, either of the same chemical or of different chemicals. To be accurate, all of those interactions need a system with much more than 40 electrons.
The 40 electron limit comes from the implict assumption here that you're talking about quantum mechanical calculations. QM calculations are rather computationally expensive, as you have to account for all the interactions of all the electrons with each other at all positions in their delocalized superposition. There's various tricks (like DFT) which make the calculations for large numbers of electrons easier, but note that "easier" doesn't mean "easy". Even with DFT and other approaches, large systems take a lot of computer time to calculate accurately.
There are other approaches which don't suffer from the same limit as QM does, but they are able to make their gains in efficiency because they make approximations. For example, molecular mechanics approaches are able to simulate systems in the hundreds of thousands of atoms region. But they're able to do so because they don't actually calculate the position of electrons. Instead they treat the system "classically", and experimentally fit interaction potentials which approximate the underlying quantum effects. (For example, they don't exactly calculate the bond stretching potential, but instead approximate it as a harmonic one. That's "close enough" to the true bond stretching potential for the range of bond lengths typically seen in such simulations, but not 100% quantum mechanically accurate.)
There's many groups and companies which do use molecular mechanics and other similar approaches to inform their drug and material development process. The issue is that because the energetic potentials being used are only approximate the results from the simulation are also only approximate. Depending on what you're trying to simulate, the results of the simulation may or may not be accurate. As such, these simulations are treated mostly as a first step, to find potential leads/hypotheses, and then the scientists actually have to go into the lab and test the results to confirm.

and those are just the drugs, if you count the drug reseptor sites as well, 40 electrons is like a fart in a desert. Not likely to matter.

I think I'd particularly emphasize the intermolecular interactions and environmental interactions. While QM methods are quite good at treating individual molecules, I think beyond the "40 electrons" question, we're really just starting to tackle interactions.

Aside from the computational power needed to simulate larger molecules, there is also a lack of knowledge about the exact mechanisms that some drugs could potentially use. Think for example of experiments in yeast or Escherichia coli cells, which are used to find new biochemical mechanisms that could be exploited for new drugs. Even though we already know a lot about those cells, it would be computationally very demanding to include all known proteins and mechanisms into any kind of simulation. Furthermore, even if we could do such a simulation, there would still be a whole lot of other proteins, genes and mechanisms which we don't really understand yet but which could very well provide new mechanisms that could be used for new drugs. For this reason we would still need (biological) experiments even if we had much greater computational power than we have today.

Attempt exists, lots of drugs are first generated randomly, those most likely to have useful effects given structural category and simulation (simulation of molecular interactions and of known cell molecular pathways and physiological interactions) are tested, in cell lines, animals, then human.
All that has costs and risks.
References about cell simulation:

Wikipedia: Metabolic network modelling
Skip the Humans: Drug Discovery by Simulating Cells by Adrienne Lafrance, the Atlantic, May 30, 2014
R. Weinberg, M. Berkus, Int. J. Biomed. Comput. 1971, 2(2), 95-120. DOI: 10.1016/0020-7101(71)90025-0; Mirrored at deepblue.lib.umich.edu
(pdf).


text:
Acid-catalyzed epoxide opening
For the acid catalyzed opening of epoxides, the epoxide is first protonated to form the oxonium ion. I read in some textbooks that the $\ce{H2O}$ molecule, which acts as a nucleophile attacks in a $\mathrm{S_N2}$ concerted mechanism. Given that $\ce{H2O}$ is a weak nucleophile, how can a $\mathrm{S_N2}$ concerted mechanism be possible,  8 instead of $\mathrm{S_N1}$ which is expected of $\ce{H2O}$?
Also, for an unsymmetrical epoxide, why will the nucleophile attack at the more substituted $\ce{C}$?

Take a look in Organic Chemistry by Clayden & Warren, the whole thing is fully explained (as it probably is in any good organic textbook).

I can only find the base catalyzed epoxide opening, but not the acid catalyzed reaction.

https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map%3A_Organic_Chemistry_With_a_Biological_Emphasis_(Soderberg)/Chapter_08%3A_Nucleophilic_substitution_reactions_I/8.6%3A_Epoxides_as_electrophiles_in_nucleophilic_substitution_reactions This link might help.


text:
Goldschmidt alumino-thermic process
I recently came across this question in a book.

Among the metals $\ce{Cr}$, $\ce{Fe}$, $\ce{Mn}$, $\ce{Ti}$, $\ce{Ba}$ and $\ce{Mg}$, the one that cannot be obtained by reduction of its metal oxide by aluminum is:-
  (a) $\ce{Cr}$
  (b) $\ce{Fe}$
  (c) $\ce{Mn}$
  (d) $\ce{Ba}$
  (e) $\ce{Mg}$

So, one method to answer this question is by Ellingham Diagram and to remember the values of $\Delta G^0$(Gibbs free energy change) for oxidation of each metal into its oxide and it's variation with temperature.

So, we can see that $\Delta G^0$ for oxidation of $\ce{Al}$ is much lower than other metals except $\ce{Mg}$ at lower temperatures. Thus, $\ce{Al}$ can reduce other metals except $\ce{Mg}$ at lower temperatures.
Thus, the answer is (e) $\ce{Mg}$.
So, this was the answer based on the experimental data for metals.
Couldn't we determine this answer theoretically as we can't remember all these data?

The metal reactivity series is pretty fundamental to remember. The series would be (K,Na,Mg,Al,Zn,Fe,Cu,Ag,Au,Pt) to name a few

But, there aren't all the metals in your mentioned reactivity series like Cr, Mn, Ca, Ba, Ti etc. Can you please mention the whole reactivity series.

Wikipedia has an entire page about it: https://en.m.wikipedia.org/wiki/Reactivity_series

Actually, activity series works well in aqueous medium, while these reactions happening at high temperature will leave all waters long gone in vapours.
Well, here we go by (delta)G for the coupled reactions.
We know that a negative (delta)G amounts to SPONTANEOUS process.
The 2 reactions which need to be coupled must be such that sum of (delta)G1 for the reaction-1 and (delta)G2 for the reaction-2 is NEGATIVE.
Seeing the curve, we can conclude, Al can reduce all the metals listed except Mg, which of course could be above ~1600 degree celcius.
The conclusion: upto 1600 degrees Mg can reduce Al2O3, while above 1600 degrees Al can reduce MgO [Unlike what happens in aq. medium!]

But, my question was that how would you determine it theoretically.

The metal or C or CO curve lower in the Ellingham diagram for a given temperature will reduce the metal oxide above it...
Similar to what we see in activity series, albeit, it changes with temperature


text:
What will Happen if p-nitrophenol is heated with HCl and then cooled?
When working in my Lab with p-nitrophenol ,I added some concentrated HCl and heated until  it was fully dissolved. Then I cooled it in ice bath and found a white salt like compound which is not probably p-nitrophenol itself. I am totally puzzled about the reaction involved in that condition , what the product really is . Does anyone have any idea about the reaction occurred in that system ?

Why do you think a reaction occurred?

How do you know your white solid is not p-nitrophenol?

Because the color and state of two solids were different.. p-nitrophenol that was given to me was crystalline and greenish. But the product was powder like crystal and white.  That's why I was confused.

No reaction has occurred, you have recrystallised p-nitrophenol!
The white solid you have recovered is pure p-nitrophenol.
P-nitrophenol is surprisingly soluble is hot water (33g/L at 40C) and crystallises out on cooling. There is no reaction that can take place with cHCl.


text:
Method to know the number of isomers of metal complexes
Whenever I try to find the number of geometrical isomers (including optical isomers) of coordination compound, I got confused, and mostly miss few isomers. Is there any standard method to know the number of isomers of compounds such as $\ce{Ma2b2c2}$ or $\ce{[M(A-A)a2b2]}$.
where $\ce{M}$ is metal and $\ce{a,b}$ are monodentate ligands while $\ce{A-A}$ is a bidentate ligand.

related: https://chemistry.stackexchange.com/q/10620/4945

Might be helpful: How can I find the number of isomers of coordination compounds?- Quora

Do it many times over, so as not to get confused.
Alternatively, there is a thing called Polya's formula, but you won't be able to use it anyway. In trivial cases like this, the said formula is about 100 times more complicated than counting isomers by hand. It is not before polysubstituted fullerenes that its use in chemistry starts making any sense.

For an easier method without going much into mathematics you can use simple logic to do find the number of geometrical isomers of [Ma2b2c2]. Though such logics are very much situation based,  and I believe there's no such generic one.
Here you can start by choosing no. of pairs of same ligands to be in trans (i.e. 180 degrees to each other). After some thought, its clear that one can choose 0,1 or 3 such pairs, because choosing 2 will automatically force the 3rd pair to be in trans. For choosing 1 such pair there are 3 ways, namely (a,a),(b,b),(c,c); and only 1 way each for choosing none (0) or all (3) such pairs. Thus accounting for all 5 isomers. Take the following example of $\ce{[Co(NH3)2Cl2(NO2)2]-}$.

For further references you can visit this site.

For Ma2b2c2, you could go by:

All cis (aCa) (bCb) (cCc)
All trans (aTa) (bTb) (cTc)
Only 1 trans (aTa)
Only 1 trans (bTb)
Only 1 trans (cTc)

Similar cases could be done for M(AA)b2c2, except of the fact that, (AA) will remain always cis [all cases EXCEPT where the 2 donor atoms are connected by NOT-A-VERY-LONG-CHAIN]

And how to know about optical activity


text:
Mechanism of rearrangement reaction found in synthesis of retinol
Today I was solving a problem that was about the synthesis of retinol. This is a step I didn't understand:

In the solution, it's written that the mechanism goes through this intermediate:

I have not seen anything like this yet.

Check out the Claisen rearrangement: http://www.organic-chemistry.org/namedreactions/claisen-rearrangement.shtm

This reaction is the Carroll variant of the Claisen rearrangement. The reactant is prepared from 2-methylbut-3-en-2-ol and diketene.

When trying to propose a mechanism, especially one where the bond changes are not obvious, I like to start by numbering the atoms of the product and trying to map them to the reactants. Here, I've taken the numbering given by ChemDraw for the product and translated it to the reactant.

Carbons 1-3 are pretty logical. In the product, carbon 4 is alkyl with 2 hydrogens and between to other carbons. In the starting material, the carbon connected to C3 is a carbonyl with zero hydrogens. It seems unlikely to me that this would be C4.
So I look for another, more logical mapping. For example, carbons 7 and 9 are methyl groups in the product, and there are two methyl groups in the starting material, which I label as 7 and 9. In both structures, they are bridged by a single carbon, which I call 6.
In the product C5, is part of an alkene and so is the carbon connected to C6 in the reactant, so that seems logical. And the next carbon in the starting material has two hydrogens, so that seems a better mapping as C4. The last carbon of the reactant must be lost as carbon dioxide, so I've labeled it C10.
With a hypothetical atom mapping, now I can identify bonds broken and formed. C10 and its adjacent oxygens are being lost as carbon dioxide, which makes it easy to identify the bonds broken as C3-C10 and C6-O, with a shift of the pi-bond from C4-C5 to C5-C6.

Without being aware of [3,3]-sigmatropic rearrangements, or the Claisen rearrangement specifically, you might be stuck. The Claisen rearrangement is a cyclic movement of electrons converting an allyl vinyl ether into a gamma,delta-unsaturated carbonyl compound. This is driven by the formation of a carbonyl at the expense of an alkene.

The given starting material isn't quite an allyl vinyl ether, but as a 1,3-dicarbonyl, it is equilibrium with one by tautomerization. From the tautomer shown, the Claisen rearrangement is possible, which forms the C3-C4 bond we noted earlier and shifts the pi bond into the desired position. Also note that the resulting intermediate is the one shown in the solution manual. Decarboxylation through a similar cyclic movement of electrons gives the tautomer of the product along with carbon dioxide.


text:
In rdkit, how to preserve stereochemistry in a SMARTS subgraph?
In rdkit, when doing a MCS search for molecules bearing a chirality center, (how) is it possible to preserve the stereochemical information when exporting the subgraph to a SMARTS string?
Consider the following three molecules:
>>> mol_ccw = Chem.MolFromSmiles('C1=C[C@H](Cl)CCC1')
>>> mol_cw  = Chem.MolFromSmiles('C1=C[C@@H](Cl)CCC1')
>>> mol_lin = Chem.MolFromSmiles('C=C[C@H](Cl)CCC')

Doing a chirality-sensitive subgraph search leads to the somewhat expected result:
>>> rdFMCS.FindMCS([mol_ccw, mol_cw], matchChiralTag=True).smartsString
'[#6](=[#6])-[#6]-[#6]-[#6]-[#6]-[#17]'
>>> rdFMCS.FindMCS([mol_ccw, mol_lin], matchChiralTag=True).smartsString
'[#6]=[#6]-[#6](-[#17])-[#6]-[#6]-[#6]'
>>> rdFMCS.FindMCS([mol_cw, mol_lin], matchChiralTag=True).smartsString
'[#6]-[#6]-[#6]-[#6]-[#6]'

The subgraph over mol_cw and mol_lin includes the stereocenter. But unfortunately, the chirality information is not stored in the SMARTS string, and using [#6]=[#6]-[#6](-[#17])-[#6]-[#6]-[#6] for a chirality-sensitive substructure match leads to the expected result of the pattern made from that SMARTS string matching all three molecules:
>>> patt = Chem.MolFromSmarts('[#6]=[#6]-[#6](-[#17])-[#6]-[#6]-[#6]')
>>> len({mol_cw, mol_ccw, mol_lin}.GetSubstructMatches(patt, useChirality=True))
1

Manually inserting a lazy @H or a &*&H1 at the stereocenter leads -of course- to the desired result:
>>> patt = Chem.MolFromSmarts('[#6]=[#6]-[#6@H](-[#17])-[#6]-[#6]-[#6]')
>>> len({mol_ccw, mol_lin}.GetSubstructMatches(patt, useChirality=True))
1
>>> len(mol_cw.GetSubstructMatches(patt, useChirality=True))
0

Now, when matching mol_ccw and mol_lin, how do I get the stereochemistry-aware SMARTS string [#6]=[#6]-[#6&*&H1](-[#17])-[#6]-[#6]-[#6] as the substructure in the first place?

For anyone interested in this topic: I just learned that this is a bug that should already be filed on GitHub, but apparently the developers themselves do not find it anymore - probably that's why I didn't find it. Nonetheless, the bug is known and will be fixed in the next release.


text:
Hydrolysis of β-1,2 glycosidic bond
Are there any easy, not enzymatic, ways to hydrolyse $\beta$-1,2 glycosidic bonds like they occur in steviosids?
I thought of something like vinegar, which is commonly used to cleave the $\alpha$-1,2 glycosidic bond of saccharose when producing inverted sugar $-$ could that also work for the $\beta$-type?
Thank you in advance.

By easy do you mean involving less energy, or less work to do? Or any other thing?

I mean less work to do, so one could do it at home.

There are non-enzymatic methods for removing the three glucose moieties of stevioside 1. Unfortunately, dilute mineral acid hydrolysis not only removes the glucose entities but also effects Wagner-Meerwein rearrangement of the aglycone steviol 2 to isosteviol 3.1 While vinegar (5% aqueous acetic acid) is a weaker acid than hydrochloric acid, I am certain that it would have been tried to avoid the rearrangement of steviol 2. Exposure of stevioside 1 with aqueous KOH effects saponification of the ester group leading to steviobioside 4.---continued---

A more fruitful procedure that avoids rearrangement involves periodic acid cleavage of the five vicinal diols of the glucose units (marked in red in structure 1 below).2 This procedure liberates hexa-aldehyde 5 in addition to two equivalents of formic acid.3  Vigorous base treatment is required to saponify the hindered ester while base also liberates tetra-aldehyde entity by β-elimination forming steviol 2 in 75% yield. And NO, you can't do this at home.

1) Avent, a. G., Hanson, J. R., de Oliveira, Bras H., Phytochemistry, 1990, 29, 2712.
2) Ogawa, T., Nozaki, M., Matsui, M., Tetrahedron, 1980, 36, 2641.
3) On periodic acid cleavage of vicinal diols, see: Periodic cleavage of carbohydrates