text:
IUPAC nomenclature query
The given compound is: $\ce{CH3-CH2-CH2-CH(CH3)-CH(CH2CH3)-CH2-CH2-CH3}$
The IUPAC name I came up with was 4-ethyl-5-methyloctane but the given answer states 5-ethyl-4-methyloctane.
Are both acceptable? Is there any rule I'm missing out on? I went through the rules but couldn't find the appropriate rule.

Your name is correct, while the given one is not. When the set of locants for subsituents would be the same no matter what, you would assign the lowest locant to the substituent that appears first in the name, which, by alphabetical order, it would be the ethyl group in this case. Hence 4-ethyl-5-methyloctane is the right name.

"no matter what" did you mean to say "no matter what direction"?

I meant it as this is the last rule in priority to follow.


text:
Redox reaction involving a polyatomic ion
Does it count as an autoredox reaction if two different elements from the same polyatomic ion or molecule change their oxidation numbers?
For example, for a reaction like this:
$$\ce{2KNO3(s) -> 2KNO2(s) + O2(g)}$$
Here, the nitrogen and oxygen in the nitrate ion undergo oxidation and reduction to from nitrite and diatomic oxygen, respectively. Does this count as a redox reaction since the same polyatomic ion undergoes both?
In my text book, it says that when an element in a molecule or polyatomic ion undergoes reduction or oxidation, we refer to the entire molecule or ion as being reduced or oxidized. I don’t know what effect this has on this reaction though.

Yes, this is a particular type of redox reactions known as disproportionation reactions.

Hi dahen! Please use mathjax to format questions and answers! Thanks.

@MollyCooL sorry about that, I'm on phone right now so it was kind of inconvenient to do it

Any reaction in which one atom/ion loses some electrons and any other atom/ion gains those electrons is a redox reaction.
In your particular example, the oxygen atom is losing its electrons and the nitrogen atoms are gaining those electrons. So, it is definitely a redox reaction.
The reaction in which atoms from the same species are oxidised and reduced are called disproportionation reaction. So, the same reaction is also a disproportionation reaction.
Look at these examples too,

In both of these examples, the same molecule/ion/atom undergoes an oxidation as well as a reduction, so these are also disproportionation reaction.
Note that every disproportionation reaction is a redox reaction.

You should cite figures that you didn't personally make.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Chemiluminescence of luminol
Recently I have tried to perform an experiment with luminol and hydrogen peroxide (with potassium ferricyanide as the catalyst). Upon googling, I found two ways to conduct this experiment and was wondering which way is more suitable (because I have limited amounts of luminol and do not want to fail the experiment and waste luminol).
1st way: Preparing an alkaline stock solution of luminol and potassium ferricyanide and another stock solution consisting of hydrogen peroxide only.
2nd way: Preparing an alkaline stock solution of luminol  and another stock solution consisting of hydrogen peroxide and potassium ferricyanide.

Here they replace the ferricyanide for bleach, but the description does say that the luminol solution is stable in base for up to three weeks. So I would follow your 2nd way.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why don't we explode after drinking water?
It is known that acid should be added to water and not the opposite because it results in an exothermic reaction.
Our stomach contains HCl, so why don't we explode when we drink water?

It is surely better to dilute acid by adding water and not viceversa.  But it is not that all acids are H2SO4, as far "explosion" is of concern. Plus see answer

The point of the acid to water rule is to spread the acid around: a given mass of a given type of acid will make a certain amount of heat, and by having all the water already there, you give plenty of heat capacity to spread this heat around. But if there's not too much acid in the first place, then it doesn't matter.

Right @Ian.  I have experience with hydrochloric acid at  work, for that one the big issue is fuming.  I have found that if I add acid to enough water to drop the acid concentration to about 10%,  then I can safely make up additional volume adding water.  But upon water-rinsing the cylinder that contained 37% stock solution I see fumes from the traces of that acid.

There is an interesting video on YouTube exploring why it is recommended to add acid to water instead of vice versa: https://www.youtube.com/watch?v=fMYju7hMdqA.

Possible duplicate of Why first water and then acid?

@Karl I study in India. The science book of mine (and of 70% other students in my age group) read, "You should add acid to water and not vice versa, because it results in an exothermic reaction and the contents of the beaker might splash out. The beaker might break." Only this much. Though not exact but nothing more. I am 15, and in class 10th.

@Dmitry it doesn't seem to be. It seems like this one is building up on that one.

Yikes!  That YouTube video is bad!  Not proper prtoection (exposed skin)!  No fume hood!  Please do not try it this way at home!

@RamKeswani That sounds like safety advice for doing experiments. Safety advice is often much more cautious than is technically necessary.

@RamKeswani I'm sorry, it is a shame how lousy textbooks (and teachers!) are especially in developing countries. You should never accept anything written in a book or said by a teacher if it is not satisfactorily, logically explained why and under what circumstances it is the case.

The "other answer" would be...evolution.
A species with a high enough concentration of acid in its body to explode when water is added would not survive. As mankind is not yet extinct[citation needed], we can assume that evolution has dealt with this and obviously found a way to mitigate this issue.
Unfortunately, this answer belongs into Biology, not Chemistry.

This is also a good watch: https://www.youtube.com/watch?v=_f0vK8Wffgg

Totally unrelated except for "explode" and "water": A basic understanding of the Navier-Stokes, or Terry Tao's "exploding water" problem

The hydrochloric acid in the stomach is already quite dilute; its pH is in fact no less than 1.5 so that at the extreme maximum there is only 0.03 molar hydrochloric acid.  And even that small amount is, of course, stabilized by being dissociated into solvated ions.  There is just not enough stuff to react violently.

Is there a further stabilising effect from a buffer solution?

Not really.  Once ypu have a set pH the effective concentration is also set,  regardless of what else is in there.  I am unaware that any buffering would be used because the acid needs unbuffered solvated protons (in moderatoon) to work.

A basic understanding of the Navier-Stokes, or Terry Tao's “exploding water” problem ;-)

@OscarLanzi What about patients suffering from extreme hyperacidity ?

The rule that acid should be added to water and not the opposite only applies to the dilution of some concentrated acids that results in a strongly exothermic reaction, in particular to the dilution of concentrated sulfuric acid.
The dilution of concentrated sulfuric acid with a large amount of water at a temperature of 20 °C releases 95.33 kJ per mol $\ce{H2SO4}$. That means that, ignoring the heat capacity of the sulfuric acid itself, dilution of 1 mol of $\ce{H2SO4}$ releases enough heat to increase the temperature of 2 mol of water from 20 °C to the boiling point and evaporate this amount to steam at a pressure of 1 bar (ΔH = 46.678 kJ/mol).
Therefore, diluting concentrated sulfuric acid with small amounts of water (i.e. when adding water to the acid) can result in evaporation of the mixture of water and acid at the point where the water is dropping into the acid.
Gastric acid is not a concentrated acid. The free concentration of $\ce{HCl}$ is between 0.5 % and 1 % (which is about 0.17 mol/l–0.33 mol/l). Diluting an aqueous solution of $\ce{HCl}$ with a typical molality of 0.2 mol/kg to infinite dilution yields only 0.761 kJ/mol at 25 °C. This is not enough for a dramatic increase in temperature of the mixture.

Minor point.  My source gave a pH range suggesting a lower acid concentration in gastric juice, ad I answered accordingly.


text:
Removing chloride ions from humic acid purification
I used HF and HCl in humic acid purification, and now I need to remove the Cl ions from solution. What should I do to remove the Cl ions? Do I just wash it with distilled water, or is there another procedure for this?


text:
Total spin number 0 together with $B_2$ state - confusion
I'm trying to understand molecular symmetry and I got confused reading the Molpro documentation.
There is an example of a system with $B_2$ symmetry written like this:
$$
wf,10,3,0,
$$
which means, that the system has 10 electrons, $B_2$ symmetry and the total spin number is 0.
If I understard the total spin number well, it is 0 only when all electrons are paired in the orbitals (i.e. the system is in singlet  state).
But, on the other hand, I thought, that when the electrons are paired, then the orbital is "totally symmetric", i.e. $(\cdot)^2 = A / A_g / A_1$.
So, as far as I understand it, $B_2$ symmetry and total spin number 0 is mutually exclusive.
Is it true or do I understand it wrong? If I do, could you, please, explain my mistake and provide some simple example of the abovementioned configuration?

It might not be the solution to this particular case, but you can have an open shell singlet, where you have two unpaired electrons with different spins.

@Tyberius And otherwise is my understanding correct? And in open-shell systems it is possible to have an orbital with single electron, which has -1/2 spin?

I think otherwise you have the right idea, a fully paired system has to be totally symmetric. My understanding is that the total spin plugged in to determine whether it is a singlet, doublet, etc is taken to be the absolute value of the total spin.

@Tyberius Could you, please, describe your idea in a little more detail and write the answer, so I could accept it?

yeah I'll write something more detailed up when I get the chance.

We can see what a $^1B_2$ will look like by consider the MO diagram for water, a $C_{2v}$ molecule with $10$ electrons.
$\hspace{9ex}$
This is the configuration that we might expect for the ground state, which has $^1A_1$ symmetry. We could get a configuration with $^1B_2$ in a couple ways, namely by exciting form $1b_2$ into $4a_1$ or by exciting from either $2a_1$ or $3a_1$ into $2b_2$ without changing the spin of whatever electron we move. These new configurations would still be singlets because we haven't changed any of the spins and would have $B_2$ symmetry since clearly $B_2\otimes A_1=B_2$.
As to your question about spin multiplicity, if we consider a system with a single electron, we account for both the spin up and spin down configuration by determining the spin multiplicity. The spin multiplicity for $S=\frac{1}{2}$ is $2S+1=2$, a doublet, which gives us exactly the number of spins states we would expect.

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


text:
What is the meaning of the word root "aceto"?
What is the meaning of the word root "aceto"?
I believed that "aceto" meant a methyl attached to a carbonyl group. For example, as in acetamide. Then, why do we name $\ce{CH3C\bond{3}N}$ as acetonitrile, even though it does not have any carbonyl group?

related https://chemistry.stackexchange.com/questions/12250/why-is-it-acetone

The root "acet-", according to wiki, is derived from latin acētum, meaning vinegar, i.e. diluted acetic acid. Consequently, direct derivatives of acetic acid, like acetamide and acetonitrile are called so. Also, acetone called so because it can be produced from acetic acid (directly or by dry distillation of calcium acetate)
What isn't obvious is why acetonitrile  is considered a derivative of acetic acid and why nitriles in general are considered derivatives of carboxilic acids. Partially it's tradition and partially it's because compounds with carbon skeleton and oxidation state of carbon atoms unchanged are often rather easy to convert into each other.

I suppose that back in those days, one could have obtained acetonitrile by dehydration of acetamide, although I don't know if that was indeed the motivation for the name.

A common misconception is that a nitrile is the group CN (triple bond) when in fact the nitrile is the triple bonded nitrogen. CH3CN is acetonitrile and "cyclohexyl nitrile" is cyclohexylCARBOnitrile.

Acetamide (ethanamide) is an organic compound with the formula CH3-CO-NH2. It is the simplest amide derived from acetic acid.
Acetonitrile is the chemical compound with the formula CH
3-CN. This colorless liquid is the simplest organic nitrile
In organic chemistry, acetyl is a moiety, with chemical formula CH3-CO. It is sometimes represented by the symbol Ac.


text:
Clarification on IUPAC nomencalture

This is 1,6-dimethyl-cyclohexene. Why cannot this same structure have the IUPAC name 1,2-dimethyl-cyclohexene. In the given picture the locant sum comes to 7 but in my case it comes to 3, implying that my answer is in accordance with the rules. Aren't the alkyl groups supposed to be given the lowest number as per lowest locant sum rule?
Please do clarify. Thanks!

Look at this maybe it can help :)

How does the double bond influence the numbering?

Note that the widely-cited "sum of locants" rule is incorrect. See: IUPAC nomenclature: “Smallest sum of locants”? and the answers therein.

Related but not duplicate: First point of difference rule 1,6-dimethylcyclohex-1-ene, 1,6-dimethylcyclohexene and 2,3-dimethylcyclohexene, Which is the correct name: 2,3-dimethyl cyclopentene or 1,5-dimethyl cyclopentene? and What is the correct structure for trans-3,4-dichlorocyclopentene?.

With your alternative name, how should I know where  the double bond is?!? It's incomplete and ambigous.

Could you clarify how the first point difference rule works out here?

Between alkenyl and alkyl groups, alkenyl groups (double bonds) have seniority over simple alkyl substituents groups such as methyl and ethyl groups (See the new version of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Prepared by Henri A Favre & Warren H Powell); Royal Society of Chemistry, Cambridge, UK, December 2013, ISBN: 978-0-85404-182-4:
Briefly, it suggests "When several structural features appear in cyclic and acyclic compounds,"  low locants (numbering) are assigned to saturated/unsaturated groups in the following decreasing order of seniority: (a) low locants are given to ‘ene’ and ‘yne’ endings; (b) low locants are given first to multiple bonds as a set and then to double bonds; and (c) detachable alphabetized prefixes, all considered together in a series of increasing numerical order.
Thus, ene-group on cyclohexene get the highest priority such as cyclohex-1-ene (or, 1-cyclohexene). Since the double bond has the highest priority, the substituents on  it (if any) should get the next priority such that it should have been 1-methylcyclohex-1-ene (or, 1-methyl-1-cyclohexene), if there are no other substituents. But in your molecule, there is an another methyl group present. Therefore that has the lowest priority. Since 1,2-numbers belongs to the double bond as mentioned above, that methyl can have been only 6 with numbering. Thus, the compound is:  1,6-dimethylcyclohex-1-ene (or, 1,6-dimethyl-1-cyclohexene).
Note that 1-methylcyclohex-1-ene has higher priority over 2-methylcyclohex-1-ene (lowest locant rule within double bond). Therefore, naming the molecule as 2,3-dimethylcyclohex-1-ene (or, 2,3-dimethyl-1-cyclohexene) is also not possible within IUPAC rules.

If an alkyl group is there attached to a cycloalkenyl group then the double bond always gets the least possible locant numbering.Right?

Yes, but lowest number goes to the most substituted $\ce {sp^2}$ carbon (if double bond has the highest priority).


text:
Geometry optimization: energy not lowered in every cycle
I am performing geometry optimizations of an enzyme active site, covalently bonded to a substrate. I have a total of 122 atoms, of which 7 are fixed in space during the optimizations. I am using ORCA to perform D3-B3LYP/6-31+g(d,p) optimizations with the conductor-like polarizable continuum model (epsilon=4).
For certain starting geometries, I notice the energy fluctuates a lot during the geometry cycles, decreasing and increasing. With good geometry convergence I should expect the energy to be lowered in each cycle, but I do not see this. My model should be very flexible, with a lot of shallow energy minima. Could too big step sizes accidentally lead my optimization out of one minimum region and into another, making it difficult to converge the calculation?
Or, more generally, which factors could lead to such a convergence behavior, and are there any common strategies to improve geometry convergence?

This depends a lot on the algorithm you use for the geometry optimization. For instance, using a quasi-Newton scheme, one is not guaranteed smooth convergence, but it will usually get to the desired minimum. Following both gradients or eigenvectors can also lead to this oscillation because you may be following the gradient or eigenvector in one direction, when the ordering of these switches, so one starts making quite different changes to the geometry that sometime leads to increased energy. It is certainly possible though that too large of step sizes can be a problem with very shallow minima.


text:
What's the difference between aqueous hydrogen chloride and hydrochloric acid?
From my understanding, both aqueous hydrogen chloride and hydrochloric acid are written as $\ce{HCl(aq)}$, is there an actual difference between them?

There is no difference.
Actually, it would be more accurate to refer to gaseous (and dry) $\ce{HCl}$ as "hydrogen chloride", and to aqueous $\ce{HCl}$ as "hydrochloric acid", but often the terms are used interchangeably.
The difference is due to the fact that, as you might guess, an acid is a proton donor (it means that it tends to give off $\ce{H+}$ in a solution). On the other side, gaseous and dry $\ce{HCl}$ is simply a diatomic molecule. The dissociation in its ions (and the showdown of its acidic properties) only happen in a polar medium (typically water).
In the paper - Zotikov, V.S. Russ J Appl Chem (2009) 82: 1733. https://doi.org/10.1134/S1070427209090389 - you can find a cool example of the difference between a dry hydrogen halide and a wet hydrogen halide (here, the "terrible" HF is involved):

In  dry  hydrogen fluoride,  carbon  steel  is  subjected to  minor  corrosion  up  to  $\pu{250^\circ C}$.  In  aqueous  solutions the  carbon  steel  is  stable  only  at  the  concentration  of hydrogen fluoride  above  75  wt  %  and  temperature not exceeding $\pu{50^\circ C}$


text:
Can Hydronium Act as a Ligand?
Can hydronium function as a ligand? For instance could there exist $\ce{[Al(H3O)4]^{+7}}$ or $\ce{[Fe(H3O)6]^{+9}}$ cations? I asked my chemistry teacher and he said that he's never encountered a coordination compound with cationic ligands (however I know that there are because $\ce{NO+}$ can act as a ligand). What are the properties of these cations, if they exist? How are they synthesized? And finally, how would you name these complexes?

If this is to work at all, I would suspect we would need a 4d or 5d transition metal in a low oxidation state.

@OscarLanzi why 4d or 5d? Shouldn't larger atomic radii be more stable due to increased distance of the positive ligands from the nucleus?

I think a better pi donor metal would help, and 4d/5d with a low oxidation state is where I would look for that.

This is kinda like follow-up of https://chemistry.stackexchange.com/questions/93809/is-hydronium-an-electrophile-nucleophile-both-or-neither Hydronium can be protonated by fluoroantimonic superacid, but only transiently, but even accepting h-bond makes it a ligand.

Complexes you mentioned are much to strongly protonated,  related https://chemistry.stackexchange.com/questions/74528/formation-of-mno2-vs-mnh2o64-in-aqueous-solution

@OscarLanzi what is a pi donor metal? Trying to read  wikipedia I got fairly confused? Does it bond to pi acceptors? Is it like platinum in zeise's salt and iron in ferrocene?

Something like Ir in Vaska's complex where the CO ligands accept the pi electrons.

This is not the answer OP would expect because I couldn't find metal complexes of type $\ce{[M(H3O)_x]^{y+}}$ (the reason for non-existence(?) is mentioned in the comments) but I did find a complex of an organic compound where hydronium ion act as a ligand i.e a complex of 18-crown-6 and hydronium ion(emphasis mine):

The observation that cyclic crown ethers can bind cations with high
affinity and selectivity has had a tremendous impact in chemistry. The
complex between 18-crown-6 (18c6) and the hydronium ion, $\ce{H3O+}$,
is a prototypical representative of this important host−guest paradigm
and has been the subject of numerous experimental and theoretical
studies. The computational evidence presented so far has favored a
structure with three linear $\ce{O−H···O}$ hydrogen bridges as the
lowest, stable minimum, whereas a bifurcated arrangement has turned
out to be a transition state for $\ce{H3O+}$ rotation 4−5 kcal/mol
higher in energy.

The closest metal complex I could find where hydronium ion is present as an entity is hydronium zinc(II) pyrazine-2,3-dicarboxylate(again not the type of complex OP is looking) where 2,3-PZDC (μ-pyrazine-2,3-dicarboxylato-N,O,O′) act as ligand(emphasis mine):

The structure of dihydronium
[catena-bis(μ-pyrazine-2,3-dicarboxylato-N,O,O′)zinc(II)],
$\ce{(H3O)2[Zn(2,3PZDC)2]}$, is composed of polyanionic ribbons of
zinc(II) ions linked by double bridging 2,3-PZDC ligand molecules.
Each ligand uses an N,O bonding moiety formed by one carboxylic group
[$\ce{Zn–O}$ 2.071(2) Å; $\ce{Zn–N}$ 2.184(2) Å] and a monodentate oxygen atom of
the other carboxylate group [$\ce{Zn–O}$ 2.092(2) Å]. Coordination around the
zinc(II) ion is strongly distorted octahedral. Hydronium cations
$\ce{H3O+}$ link the ribbons by hydrogen bonds.

References

Hydronium Ion Complex of 18-Crown-6:  Theory Confirms Three “Normal” Linear Hydrogen Bonds
Michael Bühl, Ralf Ludwig, Rachel Schurhammer, and, and Georges Wipff,
The Journal of Physical Chemistry A 2004 108 (51), 11463-11468
DOI: 10.1021/jp045879+
Michal Gryz, Wojciech Starosta & Janusz Leciejewicz (2005) "Doubly bridged molecular ribbons in the structure of an ionic complex, hydronium zinc(II) pyrazine-2,3-dicarboxylate", Journal of Coordination Chemistry, 58:11, 931-935, DOI: 10.1080/00958970500055492


text:
When is it necessary to consider size consistency?
For what kinds of reactions is it essential to account for size consistency? To my understanding for dissociation reactions $(\ce{AB -> A + B})$ it is important to consider. However, is it also necessary to consider for isodesmic reactions? How about for other addition or insertion reactions?

Your terminology is, to some extent, rather obscure https://en.wikipedia.org/wiki/Isodesmic_reaction Dunno what "integration" is supposed to mean, insertion maybe?

Yes, my apologies. I meant insertion reaction.

@Mithoron Isodesmic is a fairly often-used term in my reading of the computational literature.  Isodesmic reactions are commonly used as exemplars of systems where the electronic character of the products and reactants are similar, allowing benchmarking/comparison of methods while introducing the minimum of effects from variations in gross electronic structure.

@hBy2Py Well, more importantly what is "size consistency" supposed to mean here? Lack of significant difference in size between atoms?

@Mithoron It's a specific term of art for a particular kind of error inherent to certain kinds of calculations. See here.

The obvious answer is that you should always worry about size consistency, however, the only method that truly suffers from it is truncated configuration interaction.


text:
Does MgCl2 sol coagulate gold sol faster than Fe(OH)3 sol?
Here is a question from the former AIEEE 2005

The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged respectively. Which of the following statements is not correct?
(a) Mixing the sols has no effect
(b) Coagulation in both sols can be brought about by electrophoresis.
(c) Magnesium chloride solution coagulates the gold sol. more readily than the iron (III) hydroxide sol.
(d) Sodium sulphate solution causes coagulation in both sols.

It is a single answer correct, and it is obvious that (a) is false - hence is the answer. But the statement (c) also seems to be false (and therefore should also be an answer).
According to Hardy Schulze law,

Greater the valency of the flocculating ion, greater will be its coagulation power.

Doesn't that mean $\ce{Fe(OH)3}$ sol, having $\ce{Fe^{3+}}$ ions, will coagulate the gold sol more readily than $\ce{MgCl2}$ sol, which contains $\ce{Mg^{2+}}$ ions? Or is there any specific reason why the $\ce{MgCl2}$ sol coagulates the gold sol more readily?

Option (c) is actually correct. Magnesium chloride does coagulate gold sol faster than ferric hydroxide, because ferric hydroxide does not coagulate in the first place.
Wikipedia mentions that ferric hydroxide is insoluble at $\ce{pH}=7$, so there aren't any $\ce{Fe^{3+}}$ ions to coagulate the gold sol anyway.


text:
The decomposition of NaCN in water
One method for making NaCN is the reaction of $\ce {2K4FeCN6 + 7H2SO4 = 12HCN + 2FeSO4 +2KHSO4 +K2SO4}$. The $\ce {HCN}$ is distilled into 50% $\ce {NaOH}$...after awhile the solution turns dark brown and you can smell ammonia.The reaction is likely $\ce {8NaCN +3H2CO3}$ (from water and $\ce {CO2}$)$\ce {=2NH3 + 6NaOCN + Na2CO3 + 2C}$.
What can be done to reverse this reaction? Is there any way to get rid of the dark brown color? The Merck Index says water solutions of $\ce {NaCN}$ rapidly decompose...if that is true then how are any $\ce {S_N2}$ reactions done with $\ce {NaCN}$ in water--ethanol solution? The internet has next to nothing on this problem.

Sodium cyanide is slowly decomposed by water, but the implication is that it can still be used over short time periods provided that acids are kept away and proper safety practices are followed.  See here.


text:
Excitation and de excitation...
I had learn that whenever there is an excitation  in an atom, it is single step that is excitation takes place in only one step.
On ther other hand whenwhere there is de excitation  it is multiple stepped that is de excitation takes place in multiple steps.
If this is right,i wants to why is this.?

Could you expand a little bit? The typos make it specially hard to understand what you're saying here.

The excitation typically is due to an absorbed photon. This is a one-step process. The de-excitation will usually proceed by several one-step processes, some of which will emit a photon and some will be "internal conversion" and thus without emission.
Note that an excited state can very well absorb another photon to reach an even higher state - but this is unlikely, because you typically have many atoms in the ground state and few in the excited state.


text:
Effect of stereoisomerism and racemization on solubility
I was on the Wikipedia page of Tartaric Acid, and I noticed that its solubility in water varies depending on whether the solution is racemic, meso, or optically active:
$$\begin{array}{|c|c|} \hline
 & \text{solubility (kg/L)} \\ \hline
\text{D or L (active)} & 1.33 \\ \hline
\text{DL (racemic)} & 0.21 \\ \hline
\text{meso} & 1.25 \\ \hline
\end{array}$$
After some pondering, I can see why the D-solution and the meso solution may have different solubilities; I would guess it is because of varying degrees of intramolecular H-bonding that can occur between the -OH and the -COOH groups. I would be grateful if anyone could formally justify this.

What surprised me, however, was the difference in solubility of the D-solution versus the racemic mixture. I would imagine that due to symmetry, every individual molecule of the acid in a racemic mixture would have an equal chance of being ionized/solvated as any other, as well as equal to the chances of molecules in a pure D-solution. Ergo, their solubilities should be identical. However, it appears my intuition is wrong here.
What is the reason, then, for these differences in solubility between the three types of solutions?

Solubility is represented by something like A(s) <=> A(aq). You’re essentially only looking at the stability of A(aq), measured by inter/intramolecular interactions, but ignoring that of A(s). I don’t know what the exact interactions in the solid state are but this is probably the fundamental issue

I understand that, but the solid state aside, why would the A(-)(s) <=> A(-)(aq) vary depending on whether there are more A(-) or A(+) molecules in the solution? unless some of the solvation energy is due to other molecules of A rather than only those of water...

I was ready to comment by using @orthocresol argument as well as your last lines in your comment.  The first is correct because is the general way to think. The second makes sense as for just based on similes cum similicum, a solvent containing a certain amount of X can be a better solvent for a compound Y as compared to itself alone. For the interaction between between X and Y we are back to how they pack in solid state. In the Tartaric ac. case, there is a stronger interaction between a D and a L molecule.  When perfectly balanced in rac. mix. this lead to a higher melting point. This. ...

.... means that the behaviour in your question is not as surprising as it is at a first glance.  I have no time to make a nice looking answer out of this. Perhaps @orthocresol can or I will later.  But this is the answer.

Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.
Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.
As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.
However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.
In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.
However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.
This is right the case for tartaric acid,
Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.
Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower
according to the experimental data that you have posted.
(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).

Interesting. I realize that until now, for some reason, I'd always assumed that the only manner in which these solutions differ is their rotation of plane polarized light. Am I correct in imagining, then, that the rates of different reactions that any given compound participates in would differ slightly, depending on what kind of mixture (D or racemic or meso) it is, because the activation energies for some of the steps would differ slightly?

It is  ast  really. Because the reaction will be already between solved

It is as you said concerning the attained solution optical activity. No you cannot expect different reaction activation energies unless the reaction is taking place in the solid state . Again, is rather the solid side of the story that matters.  Think of it as a "third compound" unless dissolved, In the latter case just as a mixture in solution. For sake of completeness:  not all 1:1 mixtures of enantiomrrs give a raceme with higher melting point.


text:
Best drying agent for ozone?
For cleaning purposes I do bubble ozone trough $\ce{H2O}$.
However the ozone gas is moist then, but I need it as dry as possible.
I intended to use silica gel as drying agent.
However the orange silica gel that I usually use has an organic moisture indicator included.
I think that $\ce{SiO2}$ does not react with $\ce{O3}$ itself, but I worry that the organic moisture indicator will react with the $\ce{O3}$ and form compounds that influence the following processes.
Does anyone have experience of what drying agent to use on ozone?
Is there a reason to worry about the organic moisture indicator in the $\ce{SiO2}$?

Do you really need to pass the ozone through water? You could try dry alumina?

Yes I need water for a simple reason. The feed gas of the O3 generator contains a little N2, so subsequently some NO2 is formed in the corona discharge. Water seems the best way to get rid of the NO2 while the O3 can pass on once the H2O is saturated.


text:
Are bridge bonds in diborane longer than the terminal H bonds?
The compound diborane on alkylation loses the terminal $\ce{H}$-bonds rather than the bridge bonds indicating that the bridge bond is stronger but on the other hand it's length is found to be longer than terminal hydrogen bonds. But usually bond length is inversely proportional to bond strength.
What is the reason for this?

I don’t really see how you can conclude anything about bond strengths from one reaction.

@ortho Why not? Does this method have known exceptions? Or is it just illogical?

Premise is false, also it's not first case with this weird idea about strength of those bridging bonds. No idea where it's stemming form.

Is there an answer to this question with a proper reference? It still seems unsolved almost 2 years later

The answer by Oscar Lanzi Here is a good one, but the reference seems a bit shady (it has Wikipedia as its citations). I’m looking for a definitive answer on whether the bridge bond is stronger or weaker (In terms of bond enthalpy) than the terminal bonds.

As ocresol said, its not wise to conclude the bond strength from one reaction. Rather, considering the structure of the molecule is a more viable option.
Actually the bridged bond is weaker than the terminal $\ce{B-H}$ bonds. This is because of the presence of multielectron density region. Since it is a pair of shared electrons spread between two regions, so more volume and less the electron density.

Image Source: Quora.
Also, the bond order of terminal $\ce{B-H}$ bonds is 1, whereas the bond order of the bridged bond is half of it ie. 0.5. And it is intuitive that bond strength is directly related to bond order.

Also, a source on quora claims that on alkylation, whichever H atom is released, the B-C bond will remain/transfer to a terminal position because that would be energetically more favorable. I’m not sure about it though.

So can anyone explain why the terminal bonds rather than the bridge bonds are lost during alkylation?

@M.S.L Thats what I told you, its apparently not that the terminal bonds are broken. Even if the bridged hydrogen is released, the B-C bond formed because of alkylation, returns to the terminal position because it is energitically favourable.

Yes i read that..but i am sorry, i didn't quite get it why its more energetically favorable .Correct me if i am wrong but if the bridge is reformed again like you said that means the bridge renders some stability to the molecule and that would mean the bridge bond is more stronger than the terminal ones..but then again u said the bridge bonds are weaker ......I am sorry but what am i missing?

@M.S.L No, it actually means the opposite I would say. Terminal position is energitically we say, meaning if the B-C bond stays in the bridge, its actually unstable there, which implies the bridge is weaker, and that is why it re-arranges to terminal position. I hope that makes sense!

Your last comment is not 100% rigorous either. The B–C bond does not move in isolation: moving a bridging methyl group to a terminal position means that you have to move a terminal hydrogen into the bridge. If the rearrangement is thermodynamically favoured, it only implies that the bridging B–C–B bond plus the terminal B–H bond, combined, are weaker than a terminal B–C bond plus a bridging B–H–B bond.

@orthocresol Oh, yes I get that, only the combined stabilty is more, so we cannot compare the individual stability directly. So is there anyother reason that I missed?

Diborane is supposed to be a dimer of BH3, which is an electron-deficient species so it needs to look for a way to gain some respite from the drought of electron density. So it adjusts for this 3c-2e bonded system where it shares the B-H σ-bond electron density and shares it to the Boron atom of another BH3 unit, forming a dimer.

The 'bridging' bond has a bond order of 0.5 as @MollyCooL stated, which means it is certainly a weaker bond than the terminal ones, which have 'unity' bond order. However, in the case of alkylation reaction, the reacting unit being the actual species BH3, the dimer has to break its bridges, let BH3 react, and then when the product is formed, look for a way again to stabilize itself by forming that bridge bond again.
When we look at the final product, it seems like the bridge hasn't broken at all. But it has only reformed.
Here, the bond length does run in an inverse relation with the bond strength as per the image data (Wikipedia). The bridge is indeed weaker than the terminal bond.
However it may not be true in all cases. (Thanks Martin for correcting)
P.S. I am a new contributor and not very experienced with the subject so please correct me if I am wrong.

Welcome to Chemistry.se! We use MathJax for mathematical and chemical formulae, if you want to know more, please have a look here and here. While it often appears that a longer bond is weaker, this is strictly speaking not true, and also only an observation which cannot be proved.

Thanks Martin. Noted, will edit the answer. BTW do you have any theoretical examples to back that up?

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


text:
Which is the strongest base?
When I was in lower grades, I was taught $\ce{NaOH and KOH}$ are the strongest bases. But I realised later that there can be more stronger ones. After a bit of research on the internet, I found these three to be spoken more about.

Lithium diisopropylamide (LDA)
Methyllithium (MeLi)
Butyllithium (BuLi)

Wikipedia on LDA says that the $\mathrm{p}K_\mathrm{a}$ of its conjugate acid is $36$, which means the $\mathrm{p}K_\mathrm{b}$ value at $\pu{298K}$ is $-22$. (Since $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$).
Likewise, for BuLi and MeLi I found the $\mathrm{p}K_\mathrm{b}$ value to be $-36$ and $-34$ respectively from the $\mathrm{p}K_\mathrm{a}$ values of their conjugate acids. So, can we conclude that BuLi is the strongest base? Is this absolute basic strength defined?

Sources for $\mathrm{p}K_\mathrm{a}$ values:

Wikipedia-LDA
pKa of MeLi and BuLi

Depends on the solvent system to some extent

pKb as measure of basic strength takes into account the solvent and the proton afinity of the species. pKb is effectively a free energy of reaction. So it takes into account the whole system: what the species is, what conentration, temperature, the solvent. Proton affinity by itself is a measure of the energy associated with being a base -- i.e. extracting protons, it's somewhat more universal. But then again the environment of a species effects it ability to extract those protons so if you wanted to make a basic solution you'd need to take into account the pKb too.

Looks like it's neither of the three; ortho-diethynylbenzene dianion has the highest calculated proton affinity.
From Wikipedia

according to calculations, ortho-diethynylbenzene dianion is the strongest base, with a proton affinity of $\pu{1843 kJ \cdot mol^{−1}}$.
This base claimed the title of the strongest base 2016 ish, another base may have grabbed the spot but I haven't been able to find it.
B L J Poad et al, Chem. Sci., 2016, DOI: 10.1039/c6sc01726f

Any mention on the pKb value? Also I asked another question reg. the absolute basic strength.

Ofc no absolute basic strength is not defined . No idea regarding the pKb values

Oh! Why is it not defined when we can say a single base is the strongest as of now? And is proton affinity the only factor?

@Molly the doi link I provided deals with all that

Seems a little unfair to bring in di-anions, but whatever.

How can you get the di-anion to start with?  Neutral compounds or even mono-anions are more accessible.

@OscarLanzi Ah! Alright. Was thinking about the same.  Do you find any of that kind stronger than what I have mentioned? Thanks.

@AvnishKabaj I read the link twice/thrice. But I couldn’t find an explanation for “why proton affinity is the decisive factor?” when it comes to base strength. Thoughts?

@Molly More energy released = more stable the protonated product

@AvnishKabaj was asking more like “ Why in comparison to pKb”? I get the stability of the di-anion will obvi be more when it gets protonated.

I have added my queries into the question. Sorry if I’m asking too many questions. Thanks!


text:
Are my intuitions about how batteries work right?
Being really fascinated by how electricity actually works, I started studying batteries.
I initilally understimated the complexity of this apparently-simple objects, and thorugh studies and researches I have understood how they are the result of complex chemical and physical discoveries, blendend together in a triumph of engineering.
Anyway there are  some things that I still have to figure out completely, I made some pictures that aim to clarify my doubts and the question objects.
This is the picture of a zinc-copper cell with a non-specificated electrolyte:

It represents essentially how I imagine a cell of this kind.
Zinc and Copper have both reached a dynamic equilibirum with the electrolyte, so charges leave electrodes at the same rate as charges enter.
This creates an electrode's electric field, and the elctrode potential at the same time.
The net electric field represents the net force driving the elctrons from the zinc to the copper, resulting from the superoposition of the elctrodes fields.
Given that this analysis is right ( If it's not, please help me understand where is flawed), I don't understand how this electric field ( and so the potential) changes as time passes.
If electrons are leaving zinc to reach copper (and eventually react with some positive ions in the electrolyte,right?), how the electric field is mantained so that the flow can go on?
As electrons leave, how the zinc electrode field is affected?
Are positive ions leaving the zinc at a faster rate in order to compensate for th electrons deficency? (That would explain the zinc corrosion)
Is there an electrode's size limit (corrosion limit) where the electrons  that make up the elctrode aren't enough to keep the same voltage(the chemistry's votlage)?
Is it at the point that we start to see a decrease in the voltage of the cell?
Of course, all the considerations presume the precence of a salt bridge capable of mantaing the elctrolyte charge stable.

Hi, if  anyone who put the question on hold has any suggestion on how to make it more concise I'll be happy to follow it. I think it would be bad to close this quesiton since it touches some useful point that may be unclear  to a lot of people. Moreover I spent a lot of time, trying to make it good and clear :)

I'll assume you are someone less than 16 years of age, since usually it's around 16-17 years (11th or 12th grade) that you'll probably learn how batteries work.
You're assumptions about how electrons are taken from the zinc cell and moved to the copper cell are correct. But considering your setup, as the charge moves, an electric field and a potential difference would build up, won't it? In such case a battery would eventually stop functioning.
While batteries do stop functioning after a while, it's due to the reaction being completed (all of the zinc converted to $\ce{Zn^{2+}}$ and/or all of the $\ce{Cu^{2+}}$ deposited as copper onto the copper electrode. Charge separation does not occur. Then how do the charges remain balanced, you might ask. And that's where the salt bridge comes into the picture.
A Salt Bridge is a tube that connects the two half-cells to maintain electrical neutrality. This tube makes sure that the ionic solutions on both sides are electrically connected, but at the same time prevents diffusion of both solutions to become one (see the link to know exactly how this works).
Feel free to comment if you need any help or clarifications.

Thank you for your help Pritt! :) I'm 20 unfortunately but still very curious even if I don't have a strong chemistry background! I will try to explain my understanding with a simplified version :)   Let'say  the dynamic equilibrium  in the zinc cell it's reached  when there are always 7 "unpaired" electrons on the surface and 7 positive  ions in the electrolyte, while in the copper cell it's reached when there are 3 electrons on the surface and 3 positive ions in the electroyte.
The net electric field favours the copper cell.

Two electrons will then travel from zinc to copper in order to balance charge, they will reach the electrode and join the positive ions in the copper cell. The salt bridge will then release 2 positive ions in the solution in order to keep the ions concetration unaffected.In the meantime, in the zinc cell, as soon as the 2 electrons leave, 2 zinc ions will be released by the zinc in the electrolyte in order to mantain the dynamic chemical equilibrium, so that there are always 7 "unpaired" electrons on the surface.

The salt bridge will then release 2 negative ions in order to mantain the electrolyte ions concentration unaffcted.
This cycle repeats.
My guess was that voltage start to decrease when Zinc electrode's size start to be so small that the chemical equilibirm can't be mantained anymore. I hoep my analysis is clearer and it's now easier for you to address any misconception.

@GabrieleScarlatti To an extent yes, but it's still an approximation. Let's consider the "unpaired electrons" in terms of a more easily measurable term called "emf", which is simply the work done in moving a unit charge across the electrodes. Emf is also the potential difference across the electrodes. The emf is not always constant (for routine purposes it appears to be) and actually depends on the concentration of the ions in the solution. For your copper-zinc case, the reaction is $$\ce{Zn + Cu^{2+} <=> Cu + Zn^{2+}}$$ The equation for emf is infact, continued in next comment...

$$E = E^{o}-\frac{RT}{nF}{\log{\frac{\ce{[Zn^{2+}]}}{\ce{[Cu^{2+}]}}}}$$

Thank you for all your time and patience, I think that this formulas are a little advanced for my current level. I hope that my understanding even if a little naive, it's not flawed and reflects the general process even if it abstracts from the formulas. If there is something you wish to correct that you omitted, I would be glad to hear it, or if you have some suggestions. Thank you very much, I'm in debt.


text:
IUPAC naming of quinol and hydroquinone

This is benzene-1,4-diol. Its common name is quinol and hydroquinone. How is hydroquinone an accepted name since the functional group here is an alcohol and not a ketone? Is this the only such case or does it apply elsewhere too?

This is benzene-1,2,4-triol also known as hydroxyquinol. How does the position of alcohol influence the name of the compound here?

Pay attention to the prefix "hydro-", it can definitely change the meaning of the name.
Quinone is the common name given to cyclohexa-2,5-diene-1,4-dione:

However, hydroquinone is not a ketone. Hydroquinone is basically hydrogen added to ketone. As a result, it's not a ketone, but rather it's a phenol.

I must agree, the naming does cause confusion, as this isn't the systematic nomenclature. The common nomenclature doesn't follow any strict rules, and are often named after the compound they were formed from (such as this case). Don't rely on such names too much.

So a hyrdro-(ketone) is always an alcohol?

In a way, yes, alcohols are formed by adding $\ce{H2}$ to a ketone.

First of all, the functional groups are not alcohols. They are OH (pronounced oh--aitch) or hydroxyl groups. A 2-electron oxidation of hydroquinone gives quinone, a "ketone". A 2-electron reduction gives back HYDROquinone. Therefore, benzene-1,2,4-triol is hydroxyquinol (there is only one position on quinol for the additional OH). No number required. Phenols may look like alcohols; they are not.


text:
Will distilled water dissolve aluminum
I have an aluminum drinking-water container. If I fill with distilled water will the aluminum eventually be attacked by the water?

No, unless you destroy the oxide layer.

Don't repost bad questions https://chemistry.stackexchange.com/questions/94045/aluminum-in-contact-with-distilled-water

Sure aluminum can be attacked by water.  Happens all the time when it rains on aluminum siding, as we can plainly see.  Why do people outside of deserts buy that stuff?  Uh ... not so, as you might have guessed.

In a clean controlled environment Al is resistant ,but in the real world of CO2, traces of salts , etc, it will pit. Al foil wrap put over food will often corrode through overnight, if not protected by oil and grease from the food.

No. Your aluminum drinking-water container has a thin coating of aluminum oxide which is unreactive to water, whether or not it is distilled.


text:
Protonating/basic site and Brønsted basicity orders of beta, gamma, and delta lactams
Consider the three cyclic amides:

$\beta$-lactam
$\gamma$-lactam
$\delta$-lactam

My question
What is the basic site (site of protonation) for these three lactams? And what should be their Brønsted basicity order?
My approach
If you consider $\ce{O}$ as the protonating site for the lactams ($\ce{O}$ can act as the basic site as the lone pair on $\ce{N}$ is delocalising with the carbonyl carbon, which increases electron density on $\ce{O}$, which helps to donate its lone pair to $\ce{H+}$, and after forming the conjugate acid, its resonance stabilisation is still there, which can be lost if it gets protonated through $\ce{N}$ (due to non availablity of lone pair). Delocalisation is not at all possible in $\beta$-lactams, due to huge angle strain. But $\delta$-lactam is the ideal situation for delocalisation which will increase the electron density on $\ce{O}$ atom, and hence will be the most basic due to the most availability of electron density of $\ce{O}$.
So, basicity order will be $$\ce{\delta-lactam > \gamma-lactam > \beta-lactam (considering O as the protonating site)}$$
But if you consider N as the protonating site for these lactams, the delocalisation actually decreases the basicity, as delocalisation induces non-availability of electrons on $\ce{N}$ atom.
Going by this logic , basicity order should be, $$\ce{\beta-lactam > \gamma-lactam > \delta-lactam (considering N as protonating site)}$$
So, these two logics contradict each other and hence I am unable to decide the protonating sites for amides, as in some cases $\ce{N}$ is seen to be protonating site, and in some cases $\ce{O}$. Can anyone explain what should be the proper logic to deduce the Brønsted basicity of these lactams?

Small tip: Before focusing on cyclic amides, focus on a simple amide. We are asking ourselves which site is basic and which one is acidic. Try writing the resonance structure of a simple R-CO-NR2 system. We know that an amine nitrogen is basic since it has a readily available electron pair to share. In the resonance structure that you wrote, what happened to the electron pair of that amide nitrogen?
Now, try writing the two hypotheses (protonated O vs protonated N) and their resonance structures. What do you see?

@The_Vinz I think his question is fair. He is aware of the resonance, but he is looking at the cases where the resonance is not possible due to torsional impediments of the ring strain. $\beta$-lactams are not planar, thus there is little, if any, delocalization of the nitrogen lone pair into the carbonyl (which is another reason why they are susceptible to hydrolysis).

As the significance of the two contrasting factors cannot be easily evaluated, let us first look at some evidence. Wan, Modro & Yates (1980) reported the $\ce {pK_a}$ values of the conjugate acids of the various lactams, as shown in the table below:
\begin{array}{|c|c|c|c|} \hline
\text{Lactam} & pK_a\\ \hline
\ce{$\beta$, 4} &  -0.96\\ \hline
\ce{$\gamma$, 5} & -0.65 \\ \hline
\ce{$\delta$, 6} & -0.18 \\ \hline
\ce{$\epsilon$, 7} & -0.46 \\ \hline
\ce{$8$} & -0.38 \\ \hline
\ce{$9$} & -0.18 \\ \hline
\end{array}
As the ring size increases, from the 4-membered $\beta$-lactam ring to the 9-membered ring, the basicity of the lactam increases. This seems to suggest that the oxygen atom may indeed be the protonating site for these lactams, even for those with the smaller ring sizes. However, it seems that the 6-membered $\delta$-lactam ring exhibits rather anomalous basicity.
The authors explain the general trend in basicity as follows, using the two resonance structures shown:

If the two canonical forms of the $\ce {O}$-protonated lactam are considered, the basicity will depend strongly on the ability of nitrogen to donate its pair of electrons
  to give a delocalized cation. Hence structure II will be more important in the conjugate acid than the analogous contributing structure in the neutral lactam. This will cause more serious ring strain the smaller the size of the lactam ring, and hence tend
  to reduce the basicity. As the rings become very large, such strain factors will become less important and the basicity would be expected to level off at a pK value similar to that of a typical secondary amide. Only the 6-lactam is unusual in that its base strength is much higher than expected. It is commonly observed (27) that for five-membered rings exo double bonds are preferred, while for six-membered rings endo double bonds are more stable, although these preferences are not easily rationalized. Therefore contributions from canonical form II in the protonated 6-lactam should be more important than in the $\gamma$-lactam case (and possibly the other lactams as well), leading to a higher basicity for the 6-lactam than its ring size would lead one to expect from the trends in Fig. 2.

Reference
Wan, P.; Madro, T. A. ; Yates, K. The kinetics and mechanism of acid catalysed hydrolysis of lactams. Can. J. Chem. 1980, 58, p. 2423-2432.


text:
Have I identified the IUPAC name of this ketone correctly?
$\ce{(CH3)2-CH-CO-CH2-CH2-OH}$
The IUPAC name I came up with was 1-hydroxy-4-methyl-3-pentanone because:

Ketone has higher priority over alcohol
The lowest locant set is achieved by numbering the C bonded to -OH group as 1.

But the answer given states it is 4-methyl-3-oxo-1-pentanol which violates the first rule mentioned above.
Please clarify the discrepancy between my reasoning and the textbook answer.

I think you book has a few nomenclature mistakes.

Your assessment is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

(…)
  16. Ketones
  17. Hydroxy compounds (includes alcohols)
  (…)

However, locants are placed immediately in front of the part of the name to which they refer. Therefore, the correct name for the compound that is given in the question is 1-hydroxy-4-methylpentan-3-one.


text:
What is V times dP?
Work is defined when there is a change of volume. Mathematically, $W=-\int P\, dV$. When pressure is constant $W= P\, \Delta V$. For an ideal gas expansion we have $W= nRT\ln\frac{V_2}{V_1}$. If its constant volume then $W=0$ in both cases.
Enthalpy is defined as $H=E+P\, V$, where $E$ is the internal energy. If it is a change with constant volume, $Q=\Delta E$ and $\Delta H=Q+ V\, P$. In this case, what is $V\,\Delta P$? It cant be defined as work since no volume changes. Maybe a form of energy transfer that is not work but similar to it? Does it have particular name? Or can it be included in  a kind of works?

The scenario you are describing is known as an isochoric process. Because there is no change in volume, the thermodynamic work done is zero. Some refer to this 'energy transfer that is not work but similar to it' as isochoric work.
The utility of isochoric processes is that all energy added to the system result in a one to one increase of internal energy.
$\Delta U = Q$
Here is a quick review of classical thermodynamics by Dr. Zhigilei to put it.

Thank you for your answer. What I wonder now is that, if I can say -VΔP is isochoric work, when in isochoric process, does Q have to be ΔU + VΔP? Why we don't consider VΔP energy when we calcualte Q? My physical chemistry textbook also says dQ=dU+PdV. (well it says a little about non-expension work but not enough for me to comprehend why no VdP)  It was not a big question until yesterday because I thought W is only defined when V is changed. But when considering ΔH in isochoric process, ΔH = Qv + VΔP (which is equal to nCpΔT consequently). ΔH consider VΔP. Why Q excluded this energy? help me..

The only explanation I can give to myself is that.... it's because just a definition. Q is developed in physics with piston engine. in this situation, maybe VdP was not a point of consideration because expasion work was much important. so Q is just defined with expansion work. In chemistry however, non expansion work probably is important factor. that's why they define the function H which can include all the work from Δ(PV). That's why I have to consider VΔP in ΔH. This is really my imagination, maybe misconception but I just wrote it anyway.

One last thing. I searched it on youtube channel and found some of lecture regard to the subject. Many of them explain that W=0 in isochoric system. Yes, that's what I've know so far. and I found another lecture, which explains W=-VdP. Drawing a P-V graph and a constant V line on it, others said it has no PΔV because of vertical line to V axis, and the other guy said VΔP, making an area not by V but by P axis.

The only differences between the lecture was a definition of system. One with W=0 mentioned closed system, while the other with W=-VdP mentioned open system. but.... constant volume process with open system?? ahh..... something is missing.....  but I don't think that's not my missing part. or.... is it?? enlighten me please....

Enthalpy H and internal energy U are both state functions
Irrespective of the pressure and volume changes from initial to final state, you can consider H or U to describe processes.
Enthalpy H is most useful for processes at constant pressure
If you are interested in the amount of heat a reaction gives off, you have to consider that a closed system can exchange energy with the surrounding via heat or work. In the special case of a reaction (or process) where the only work is P-V-work, and where the pressure is constant, the heat transferred is equal to the change in enthalpy. That is why reactions with negative change in enthalpy are called exothermic. If their is non-P-V work, however, an exothermic reaction can happen without giving off heat (e.g. reaction in a battery).
Internal energy is closely related to heat when the volume is constant
In the special case of a reaction (or process) where there is no work, the heat transferred is equal to the change in internal energy. This happens, for example, when the volume is constant and there is no non-P-V work (a constant volume implies that there is no P-V-work).
What about enthalpy changes when the pressure does change?
In that case, there is no easy relationship between heat and enthalpy. At constant volume, VΔP is the difference between enthalpy and internal energy. There is no P-V-work, and it makes sense to work with the internal energy U instead.

In this case, what is VΔP?

It is the difference between H and U in the case of constant volume, like stated in the question. It does not have any physical meaning. The way H is defined makes it easy to interpret for constant pressure processes. The flip-side is that it is difficult to interpret for all other processes, including those happening at constant volume.

Maybe a form of energy transfer that is not work but similar to it? Does it have particular name? Or can it be included in a kind of works?

No, there is no work associated with it. It is just what is mathematical left over from the difference between U and H when the volume is constant.


text:
Why does acetonitrile have a larger dipole moment and boiling point than acetaldehyde?
Experimentally, acetonitrile has a larger dipole moment than acetaldehyde, but I've never understood why.
I always thought that the charge separation between carbon/oxygen is larger than that of carbon/nitrogen, and since the $\ce{C=O}$ bond should be longer than the $\ce{C#N}$ bond, the dipole moment $\mu = \delta d$ would be larger with both larger charge separation and larger distance.
I assume geometry might have something to do with it but I can't really explain such a large boiling point difference.

Data: Dipole moment values:

acetonitrile: $\approx \pu{3.4D}-\pu{3.5D}$ (source)
acetaldehyde: $\approx \pu{2.7D}$ (sources: 1, 2)

Comments are not for extended discussion; this conversation has been moved to chat.

Using molcalc.org to get a qualitative idea about what is going on, here are the charge distributions and the partial charge assigned to the electronegative N (-0.19) or O (-0.40):

Even with the quick cheap calculation done in the educational software, the calculated dipole moment of acetonitrile (4.05 D) is larger than that or acetaldehyde (3.16 D).
How do we rationalize a smaller dipole for acetaldehyde given the larger negative partial charge on the oxygen atom? You could consider the dipole of the methyl group separately from the dipole of the two functional groups. In acetonitrile, these dipoles line up, so they are additive. In acetaldehyde, they are at an angle, so the (vector) sum will be less.
If you look at all the calculated partial charges in acetonitrile, the hydrogen atoms are assigned most of the positive charge (3 x 0.08), and the nitrogen atom most of the negative charge. This results in a large charge separation, which influences the dipole moment (charge difference x distance). Surprisingly (to me), the carbonyl carbon is assigned a charge of -0.06.
If you look at all the calculated partial charges in acetaldehyde, the largest positive partial charge is assigned to the carbonyl carbon (+0.33), making for a shorter distance of the charge separation.
To get a more quantitative picture, one would have to perform a more serious quantum-chemical calculation. This would also be an opportunity to look at the orbital shapes and their relation to the charge distribution. This quick calculation, however, already points out some of the salient differences.


text:
What do the six polymorphs of chocolate look like?
Today I learnt that chocolate tempering works by controlling the temperature to ensure the crystals set to one (type V) of the six polymorphs.

What do these polymorphs actually look like? By that I mean: what is the crystal structure? Also, what gives type V its properties?

To tackle the question of why form V is shiny, here a physicist refers to the shininess being attributed to

long, skinny molecules all stacked nicely in line—which produces the clean, reflective surface

Digging deeper than this article, I found a publication which analyzes the effect of smaller crystalline structures in form V and refers to the smaller structures allowing for softer form V chocolate, possibly implying that large crystalline structures in form V cause its rigid structure.
Reference
New Textures of Chocolate Are Formed by Polymorphic Crystallization and Template Effects: Velvet Chocolate
Laura Bayés-García, Teresa Calvet, Miquel Àngel Cuevas-Diarte, Enric Rovira, Satoru Ueno, and Kiyotaka Sato
Crystal Growth & Design 2015 15 (8), 4045-4054
DOI: 10.1021/acs.cgd.5b00660 /acs.cgd.5b00660

The actual polymorphic compound is the fat (cocoa butter). They pure phases will all look like white powders and the 'bloom' referred to in your diagram is white cocoa butter extruded by the solid chocolate, and probably the most obvious  visual difference.
The melt-in-your mouth aspect of Form V is going to be because of its appropriate melting point w.r.t your mouth temperature. The decrease in mechanical stability as the melting point decreases is fairly typical of solids (think of plastics softening as you heat them up).
Afraid I don't know why Form V is shinier.
C.f. https://www.sciencedirect.com/science/article/pii/S0969806X04003019

Please not from the tag, that this is a question about crystal structure. I have edited the question for clarity?

Ah ok – the linked paper contains some of the structures, and references to the rest.


text:
Is water with oxygen-17 safe to drink?
Is it safe to drink and what else can it be used for?

Since that isotope is stable, there would be no radiation danger. As @The_Vinz states, there would be no other practical biological difference.

Depends how much, I'd guess. Just like drinking $\ce{D2O}$, replacement of oxygen atoms in your body for heavier isotopes may lead to some metabolic reactions running slower than others, thus affecting metabolism. Of course, you'd need to replace most of the oxygen atoms contained in your body, which would require way too much $\ce{^{17}O}$ to see any noticeable effect.

It can be used to drain your bank account quickly...

The kinetic effect change would also not be as severe as in the case of $\ce{D2O}$.

The isotope 17 of oxygen is NMR-Active, so it can be useful if isotopic labelling (eg: in the experimental study of a reaction mechanism) is needed.
Aside from that, since Oxygen-17 exists in nature, you are currently breathing it, and you do drink it.
Now, the question is: could we survive by breathing and drinking only Oxygen-17?
I don't think that anyone has ever answered that question, for economical reasons (from a fast search, a gram can be bought for around 500$!).
But, if I have to take a guess, I would say yes, since I don't think that the nuclear and mass properties of such an "heavy" atom would interfere with biochemical processes (unlike hydrogen's isotopes, the mass of oxygen and its isotopes is so large that they are not involved in quantum tunneling mechanism, and from this point of view their behaviour should be quite similar).


text:
Can 99% isopropyl alcohol be stored in a clear container?
I'm planning to get hold of some 99% isopropyl alcohol (aka rubbing alcohol) for the purpose of cleaning electronics, and am going to transfer it from the plastic bottle it comes in to a smaller, clear dropper bottle or similar. However, I vaguely remember being told that isopropyl alcohol shouldn't be stored in a clear container - possibly something to do with speedier evaporation, but I'm really not sure.
Is this true? Are there any downsides to storing 99% isopropanol in a clear container? Will storing the solution in direct sunlight change this at all?

Generally more transfer means purity goes down. However this can be negligible.  No a clean container does nothing bad. Shadow and caps are generally better. Storing a solvent with no cap is not safe or likely no storing at all :)

Ps isopropyl alcohol for cleaning is justified in industrial environments. For most cases ethanol etc work as well

@Alchimista What do you mean by shadow?

I mean common sense.  Do not store anything on a beach

@Alchimista Definitely don't. Pirates are everywhere!

I would not expect that isopropanol would evaporate appreciably from a properly sealed container. Rather the main purpose of storing isopropanol out of light is to prevent peroxide formation.  Isopropanol can form peroxides and when distilled they can concentrate and explode. A clear container allows more light to pass and thus more peroxides to form.

I am no expert, but I have some experience dealing with solvents such as isoproponal in a research setting. Normally in a lab solvents are not stored in the open, exposed to light, for the exact reason that you mentioned- speedier evaporation. Hence, even if you were to store it in a clear container, I would make sure to cap it and stow it away in a dark place as soon as you are finished with it.

Alcohols exposed to light form peroxides. The guy responsible for lab safety in your place should get a hard kick in the back if he didn't explain why exactly you have to keep most solvents not only cool but also in the dark.

I think that what you’re talking about with the clear container problem is hydrogen peroxide, which is stored in brown containers because light causes oxidation to occur which causes decomposition of the hydrogen peroxide into water. Rubbing alcohol does not have this problem with the light.


text:
Henderson-Hasselbalch approximation
The Henderson-Hasselbalch approximation gives us a method to approximate the pH of a buffer solution. The basic equation is as follows:
$$\mathrm{pH} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[A^-]}}{\ce{[HA]}}$$

Note that HA is a weak acid. By definition, "HA does not dissociate completely and we can say $\ce{[HA]}\approx\ce{[HA]_i}$ and $\ce{[A−]}\approx\ce{[A−]_i}$. Hence, we can use the initial concentrations because
$\mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[A^-]_i}}{\ce{[HA]_i}} \approx \mathrm{p}K_\mathrm{a} + \log\dfrac{\ce{[A^-]}}{\ce{[H\!A]}} = \mathrm{pH}$

Source: Chem.Libretexts
Now they have given a question and it's solution as:

Find $\ce{[H+]}$ in a solution $\pu{1.0 M}$ $\ce{HNO2}$ and $\pu{0.225 M}$ $\ce{NaNO2}$ The $K_\mathrm{a}$ for $\ce{HNO2}$ is $5.6\times10^{−4}$.
$\mathrm{pH} = 3.14 + \log \left( \dfrac{1}{0.225} \right)$

So my question is why did they neglect the concenteration of conjugate base from the dissociation of the weak acid and included concentration of the base from the salt?

Your last equation is wrong. 2) The actual error is small (2.60 vs 2.62). 3) At these concentrations, you would need to account for ionic strength. Depending of what you want to do, it depends of how accurate you want to be or not.


text:
What is a good solvent for pentacene?
I am doing an experiment with pentacene molecule ($\ce{C22H14}$). I want to clean my sample reservoir after using it.
But I show in some literature that it is slightly soluble in organic solvent. Could you please suggest me a solvent for pentacene ?

Something high boiling.  Try toluene, chlorobenzene etc and scratching.  Otherwise chromic mixture if allowed, or "piranha". Skill needed for the last two.

@Alchimista Why not write an answer? ;)

It does not qualify as an answer in my opinion. Pentacene is a hard beast. The OP should try aromatic high boiling and rubbing. If he/she would have ask how to prepare a solution sensibly containing Pentacene , I would have commented "forget". @Gaurang Tandon

If you can heat your container strongly, it should burn off easily in air above ~500 °C, or even down to about ~350 °C, though in the latter case it may take a whole day.


text:
What is the necessity of re-hybridisation of carbon in the transition state of SN2 reaction?
In March's Advanced Organic Chemistry (The SN2 Mecahnism, page 374), it's stated:

When the transition state is reached the central carbon atom has gone from its initial sp3 hybridisation to essentially sp2 with an approximately perpendicular p orbital.

I would like to know why and how this happens. Is there an underlying mechanism? How is this rehybridisation feasible?
What makes it difficult to understand is that SN2 reactions are concerted because bond formation and breaking occur simultaneously.

Check out this animation (for reference, this is material which accompanies the beautiful "Clayden - Organic Chemistry" book).
As you can see, there is indeed an intermediate step (the "transition state"), when the nucleophile and the leaving group are in a intermediate (and unstable, being highly energetic) state (nor attached or detached). Look closely at the geometry of the carbon and the hydrogens: that's the sp2 hybridization of the transition state carbon.
The mechanism (also shown on the page linked) involves an electron pair transfer from the nucleophile (here, bromide ion) to the carbon and eventually to the leaving group (chlorine).
The curved arrow formalism, of course, is just a representation of what goes on in the real scenario: as you can guess from the animation, the charged bromide transfers part of its electron density to the carbon (which carries a partial positive charge!). The chlorine atom withdraws the electron density as it is transferred from the bromide to the carbon, and eventually, when a carbon-bromine bond is formed (full orbital overlap between Br and C), the chlorine atom has gained a full electron pair, which allows it to leave as a chloride ion.
This is the how.
As for the why, it's a matter of energy. The reaction conditions are energetic enough to push the system towards the unstable transition state, and the equilibrium is shifted towards the products because the energetic state of the products is lower than that of the reactants (that is: the products are more stable than the reactants)


text:
Why does increasing the pressure cause ice to melt more?
This question was given under chemical equilibrium.

Why does increasing the pressure on this system cause ice to melt more?
  $$\ce{H2O(s) <=> H2O(l)}$$

I'm pretty sure this has to be solved using Le Chatelier's principle; the hint given is that The molar volume of water is more that of ice.
My understanding of Le Chatelier's principle is that a system tries to oppose any change.
So the equilibrium must shift towards the side where the pressure decreases.
Given this hint, I fail to see the correlation between molar volumes and pressure.

I'm pretty sure this has to be solved using Le Chatelier's principle the hint given is that The molar volume of water is more that of ice

The hint is exactly backwards.
Per here, at $\pu{0 ^\circ C}$:
$$
\begin{align}
v_\ce{H2O(\ell)} &= \pu{18.0182 cm3/mol} \\
v_\ce{H2O\!(s)}  &= \pu{19.66 cm3/mol}
\end{align}
$$
In this situation, the pressure change is the Le Chatelier stimulus, and a reduction in total system volume is the response, in a fashion loosely analogous to the compression of, say, a block of rubber when pressure is applied. Since liquid water has a smaller molar volume than ice, one of the 'options' the system has for decreasing its total volume in response to applied pressure is to undergo the phase change to liquid water.

This is analogous to the situation where a change in the number of moles between reactants and products leads to a Le Chatelier response to a pressure change. Consider the following reaction:
$$
\ce{N2 + 3H2 <=> 2 NH3}
$$
Here, increasing the pressure will also drive the equilibrium to the right. In this case, the main 'option' the system has for responding to the increase in pressure is to react four moles of reactants to form two moles of products. As above, the Le Chatelier response is in the direction that works to mitigate the applied pressure increase.

But to counter the increase in pressure i.e. the decrease in volume, isn't the equilibrium supposed to shift in a direction with a larger molar volume?

The liquid and solid, to a reasonable approximation, can be considered incompressible (i.e., dV=0)

@GaurangTandon Let's continue in chat

@GaurangTandon No. The decrease in the total volume of the system is the Le Chatelier response to the increased pressure, and the melting phase change is one of the mechanisms available to the system to decrease its volume.

Le Chataliers principle can get you into trouble when reactions/products are 50/50 in this reaction...

@hBy2Py Going by the ice-water analogy, on increasing the external pressure, the nitrogen-hydrogen-ammonia system will try to resist the change by increasing the internal pressure and which can be done by increasing number of moles of the system. This will be achieved by shifting the equilibrium to left.

At equilibrium, the change in Gibbs free energy is zero.  So, to maintain equilibrium, if the pressure increases, the temperature must change in tandem in such a way that the change in free energy of the solid is the same as that of the liquid.  We have, for either phase, $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$So, $$\mathrm{d}(\Delta G)=-\Delta S \mathrm{d}T+\Delta V\mathrm{d}P=0$$where the deltas refer to the changes between liquid and solid.  We also have that, at equilibrium, $$\Delta S=\frac{\Delta H}{T}$$So, combining these equations gives $$\mathrm{d}T=\frac{T(V_\mathrm{l}-V_\mathrm{s})}{(H_\mathrm{l}-H_\mathrm{s})}\mathrm{d}P$$Since $V_\mathrm{l}<V_\mathrm{s}$, this equation indicates that the equilibrium temperature for the solid-liquid transition decreases as the pressure increases.
This shows that the melting temperature decreases. So if the pressure is increased, and the actual temperature is not changed, the temperature of the ice will be higher than the melting temperature, and the ice will melt.

Interesting. But I was taught that $\mathrm{d}G=\mathrm{d}H-T\mathrm{d}S$. From your first equation, it seems that you've taken $\mathrm{d}H=V\mathrm{d}P$. How did you derive that? I thought that this relation only holds in adiabatic conditions. But aren't we heating ice to change it into solid? (so it isn't an adiabatic process?) Also, shouldn't it be $T\mathrm{d}S$ instead of $S\mathrm{d}T$?

Actually, the correct equation is $dG=dH-d(TS)=dH-TdS-SdT$.  When combined with $dH=TdS+VdP$, this becomes $dG=-SdT+VdP$.  This equation is in every thermo book.


text:
Why are the terminal Hydrogen bonds in diborane lost during alkylation?
Considering the fact that the bridge h bonds in diborane are weaker according to the answer  to the following question..>>.Why are bridge bonds in diborane longer than the terminal H bonds?
Why is then the terminal H bonds instead of the bridge H bond
lost during alkylation of diborane?
Is it energetically more favourable? If yes,Is there any reason why it is so?


text:
Why don't we consider VΔP when we define Q?
We only define $Q = \Delta U + W_\text{exp}$ (expansion work = $-P\Delta V$).
If heat can cause $\Delta U$ and work, why work is defined only as expansion work in the first place where there are other forms of works, such as isochoric work ($=V\Delta P$)?
Let's say, there is a container with ideal gas $T, P, V$ and $V$ can't be changed by this container. I give it heat therefore $T, P$ changes to $2T$, $2P$. In this constant volume process, expansion work, Wexp is 0. That's why $Q_v = \Delta U$. In the mean time (in this constant volume process), $\Delta H = Q_v + V\Delta P$ and $V\Delta P$ is work done in constant volume process. My question is that why can't we just say this heat $Q = \Delta + V\Delta P$? Do we just ignore the non-expansion work? or is it included in $\Delta U$? or because the definition of $Q$ is $\Delta U+P\Delta V$ originally?
One more thing. I googled it all day and someone wrote that $V\Delta P$ is work in flow process. I can't imagine how matter can flow in the constant volume container, and even though it can flow, what is the relationship between $V\Delta P$? They didn't explain it why so my question get bigger and bigger.

Do you mean to ask why not $\Delta U = q+ P\Delta V + V\Delta P$ ?

What the heck is isochoric work?  Work is force times displacement (over which the force acts), not a change in force times no displacement.

@ChesterMiller While I agree with you, do note that in the OP's previous question, the +2 score answer told the OP that $V\Delta P$ is "isochoric work". That answer got no downvotes or comments saying it is wrong. So, probably that's where the OP got this notion from.

@MollyCooL In a sense, yes. Could you explain why this ignore VΔP in that?

For what kind of process? Isochoric , isothermal or anything else?

@ChesterMiller Thanks to Gaurang Tandon, I don't need to explain why I mentioned isochoric work. and I know Work (expansion) is force times displacement surely but I guess I want to know more than that haha. https://youtu.be/m3raInLl9nA , in this video by the way, he explain works the way I want to be understood, which means work done is not only by force. I don't know why this concept of work is not applied to the other equations. That's what I wonder.

@MollyCooL Isochoric process. Like you said, if ΔU=q+PΔV+VΔP, why most books explain  ΔU=q in isochoric process?

Most books explain that $\Delta H=q$ for an isobaric process, not $\Delta U$

OP maybe meant $\Delta U = q_v $ for isochoric @ChesterMiller

@ChesterMiller isobaric process, ΔH=q. I was talking about Isochoric process, ΔU=q.

Let's start by the fundamental ideas. Assume, you have a container with volume $V$ with a piston on it of  cross sectional area $A$ and container is filled with gas.
Suppose, you push the container by $\mathrm dx$ amount. So, the gas is compressed and work is done on the gas.
By the definition of work classically,
$$\mathrm dW=F\,\mathrm dx$$
Now if we use the fact that $P=\frac FA$, and $\mathrm dV=A\,\mathrm dx$, we will have $$\mathrm dW=\frac FAA\,\mathrm dx= P\,\mathrm dV$$ So, the classical definition of work only tells us the that work done on the gas should be $$\int_{V_1}^{V_2}P\,\mathrm dV=P\,\Delta V$$ if $P$ is constant.
So, the definition tells us there is no $V\,\Delta P$ term in the expresion for work. And, therfore, in the first law of thermodynamics, i.e. $\Delta Q = \Delta U - W_\text{on the system}$, also, that expression doesn't occur from work, and it is not also included in $\Delta U$ either.
But if you consider enthalpy change for a system, there $V\,\mathrm dP$ term will occur as $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$

Actually, that last equation is not mathematically correct, unless it involves differentials rather than deltas.

After edit: Change “ VdeltaP term will occur” to VdP...

Thank you for your answers. But maybe my writing was not clear. I meant like a steel container of which volume can't be changed. Let's say I fixed your container's piston so that it can't be moved and put heat in it. volume can't be changed so only pressure get higher, followed by temperture higher. In this case, no expansion, no Work, -PdV=0 AND WHAT ABOUT -VdP? shouldn't we consider it because even if it's not work like others say, it's definitely other form of energy. enthalpy consider it. so why not Q? this is why I suggested this form of energy might be included in  ΔU.

Enthalpy and $Q$ are different quantities.  Enthalpy is a state function and it is defined as $H = U + PV$. But heat is not a state function, we can only measure its change i.e. $\Delta$Q, but can't measure it in a particular state of a system. On the other hand, We can measure enthalpy of a system at a particular state. So, better not to compare enthalpy with heat.

Isn’t it like we cannot measure enthalpy at a particular state and hence only the change?

If you know the internal energy of a system ( for example : for monoatomic gases, $U =\frac {3}{2} nRT$) and Temperature of the system, you actually know $PV ( = nRT) $ and hence you know  $H $.

But, Wikipedia says “The total enthalpy, H, of a system cannot be measured directly. The same situation exists in classical mechanics: only a change or difference in energy carries physical meaning. Enthalpy itself is a thermodynamic potential, so in order to measure the enthalpy of a system, we must refer to a defined reference point; therefore what we measure is the change in enthalpy, ΔH.” Am I missing something trivial?

As an answer to your comment, it is not that we ignore the P$\Delta V$ term. Let us consider the first law statement,
$$\mathrm{d}U = \mathrm{d}Q+ \mathrm{d}W$$
And we have, $\mathrm{d}W = P\mathrm{d}V$. Notice that I use differentials instead of delta terms.
Considering any process, Intgerating the expression and setting appropriate limits, we actually don’t get $\Delta W = P\Delta V$. This is because the pressure is not constant and cannot be taken out of the integral. Hence the $V \Delta P$ is technically not ignored.(This was also explained in Soumik’s answer). Only for an isobaric process $V\Delta P$ term can be ignored.
For all other processes involving an ideal gas, if we work with $PV = nRT$, differentiating we get,
$$P\Delta V + V\Delta P = nR\Delta T$$
And incase of an isochoric process, we get the following relation,
$$V\Delta P = nR\Delta T$$


text:
Mixing hydrochloric acid and sulphuric acid, both concentrated
In the chemistry lab, I stupidly decided to mix hydrochloric acid and sulphuric acid, both concentrated, in a random ratio, which was about $\pu{50 ml}$.
When I poured them in one test tube, the result was a vigorous reaction, with the green coloured liquid mixture "jumping" in the test tube and a white gas being released.
Could someone tell me what the resultant chemicals were?

Are you sure nothing else was in the test tube?

Did you add the sulfuric acid to the hydrochloric acid, or the reverse?  Hydrochloric acid stock solution is already about 63% water, so I would expect fewer side effects from adding sulfuric acid to hydrochloric.

https://www.quora.com/What-happens-when-we-add-HCl-to-H2SO4 might help but I need to get more sources

Basically, if you pour $\ce{HCl}$ onto $\ce{H2SO4}$, hydrogen chloride gas will be emitted simply because $\ce{H2SO4}$ has a strong affinity for the water the hydrogen chloride is dissolved in.

@Molly The testube was clean.Well ,I added H2SO4 first and then added the HCl.The resultant liquid was slighly lighter than the HCl colour and loads of the white fumes was emmited,though the reactants were small in quantity.Even with the ventilator fans on,the gas pretty much covered the room.

Could someone tell the resultant solution's name ?

Three things occurred. First, sulfuric acid as an high affinity for water (it is indeed used as a drying agent in some cases. For instance, if you wanted pure and anhydrous bromine, you could bubble wet bromine gas in concentrated sulfuric acid). So, one thing is the "stealing" of water by sulfuric acid, which took the solution one step forward the saturation point (with the release of gas).
The release of gas was aided by the exothermic reaction of sulfuric acid with water: (self) heating of the solution helped releasing the gas.
Edit: the following is an hypothesis, made by assuming that the "extreme" reaction conditions could have aided an unlikely redox reaction, but I am not sure about the oxidation of HCl:
And finally, sulfuric acid is a oxidizing acid: it oxidized part of the hydrochloric acid to chlorine, which made the solution green (which probably was released as a gas, together with HCl).

Does the release of Chlorine gas makes the solution green? I know the colour of it, is green.

Not the release of chlorine, but the presence of chlorine in solution!
When I was talking about the release of chlorine I was describing the composition of the gas, which probably contained part of the chlorine generated.
Oh, and it contained water vapor, probably (if the temperature was high enough, as I suspect)

The resultant was medium light green in colur,so the Clorine gas if formed,could'nt have dissolved in the solution.

What colour do you see when you put a droplet of red wine in a large glass of water?

But what you are talking about is diffusion,which strays us from my next sub-question:Did the acids mix to form an Aqua Regia type of solution , and if theexperiment was done in a closed environment ,could it have killed me ?

Yes, the acid did mix. It was not acqua regia, but yet it's a cute mix of acids.
And yes, in a closed enviroment breathing hydrochloric acid and chlorine gas is not too good for your body.
In fact, chlorine was used as an irritating (and toxic) war gas.
And acid... well, it's acid.
But don't worry: luckily nowadays we have aspiration apparatus, and a short exposure, like in your case, is not a terrible thing.
You just learned a lesson, I guess, which is what the laboratory is for.

Yeah,but I want to know the chemical and commercial name for the acid and the gas,as I could not have been the only guy ever to do so.

Are you sure sulfuric acid is oxidizing enough to oxidize chloride ions? I can't find the redox potential, but I do know that in the lab we prepare gaseous hydrogen chloride by dripping concentrated sulfuric acid into sodium chloride, and I've never seen any oxidation of our reagents from any chlorine gas that might form.

I confess that I am puzzled and that I was puzzled, but I saw no other explanation.
What puzzles me is that every redox chart says it doesn't happen.
But if you look at Pourbaix diagrams of chlorine and sulfur, you see that at pH=0 Cl2 has Eh of around 1.4 (decreasing with acidity), and hydrogen sulphate of around 1.25 (increasing with acidity). Now, there are two problems: 1.the pH is probably "negative". 2.Is our sistem aqueous, with such an amount of concentrated acid? If it's not, how does it behave?
My answer derived from the only explanation I could find for the green.

Also, your method is indeed used, and it works (it's a typical reaction employed in a Kipp generator). But I wonder if the extreme reaction conditions (high temperature, high concentration of acids in the contact area) could be an answer.

Could oxygen be involved? It's reduction potential turns more positive at lower pH too. And HCl bottles turn yellow over time too.

Sulfuric acid occupies several categories besides corrosive, one of which is as a dehydrant. The water removed from hydrochloric acid produce a vapour of hydrogen chloride. The solution is effervescent. There should not be a green colour unless the test tube was contaminated. I was hoping that you had done this in a fume hood.
I did this using 10 ml of each acid knowing what to expect. The fumes actually escaped as I shut the sash quickly causing turbulence. The fumes were quickly detected. If this were done in a prep room without a fume hood, with larger quantities you would be lucky to walk away from this! Despite their being stored together, they are not compatible.


text:
Comparison of chemical shifts among alkanes in NMR spectroscopy

I was reading through the factors influencing chemical shifts in proton NMR. I found the order of chemical shifts in the image above. Why is this order supposed to be as it is? I thought it would be the reverse as the inductive effect is the maximum in tertiary carbon, so the hydrogen attached to that carbon is shielded the most and thus has the least chemical shift. But why isn't that true?

A carbon atom is slightly more electronegative than a hydrogen atom. For tertiary hydrogens, the extra bonded carbons will inductively withdraw more electron density from the hydrogen. This means the hydrogen will be deshielded slightly more, and resonate at a higher ppm.
I would classify cyclopropane in its own class, as its molecular orbitals make it somewhat unique. The C-C bonds do not exhibit exclusive sigma bond characteristics, and due to that phenomenon the hydrogens will hold slightly more electron density.


text:
How is sulphur trioxide formed and why is it stable?
Structure of $\ce{SO3}$ (sulfur trioxide):

In the molecule, if each oxygen atom shares two electrons with sulfur atom then how does the sulfur atom remain stable? It already has 6 valence electrons and needs only 2 more to become stable but the oxygen atoms share total of 6 electrons with sulfur atom. So how is the molecule stable?

@JavaScriptCoder The linked question doesn't seem to explain the stability of hypervalent sulphur trioxide, which seems to be what the OP is asking.

At least from how I read it, @GaurangTandon the first answer talks about hypervalency and sulfur trioxide.

@JavaScriptCoder Sorry, you are able to infer the answer from the linked post, but I am unable to. But, nevermind.

For many hypervalent molecules such as SO3, stability is derived from firstly, the electronegative ligands (in this case, oxygen atoms) being able to reduce electron density accumulating in the central atom, thus minimising interelectronic repulsion within the valence shell of the central atom. Secondly, due to electronegativity differences, significant partial charges which build up on the central atom and surrounding ligands contribute an ionic character to the intramolecular bonds. This strengthens the intramolecular bonds, granting the molecule stability.

@JavaScriptCoder https://chemistry.stackexchange.com/questions/29101/why-is-the-bond-order-in-the-so%e2%82%83-molecule-1-33-and-not-2

Synthesis
Synthesis of $\ce{SO3}$ can be done by heating $\ce{CaSO4}$ with $\ce{SiO2}$, which produces $\ce{CaSiO3}$ and $\ce{SO3}$. Another way of synthesis is the heating of $\ce{NaHSO4}$ to $\ce{Na2SO4}$, $\ce{SO3}$, and water. (source)
As to why $\ce{SO3}$ is stable...
Sulfur forms an expanded octet. That means that it doesn't really obey the octet rule, allowing it to take on extra electrons. Sulfur is a 3rd-period element; hence it can use its 3d orbitals to make more than 4 bonds. Wikipedia has a page on hypervalency, and from an answer here:

Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: C,N,O, and F.

EDIT:
Users such as @Mithoron have pointed out that this analysis is not correct. Papers seem to show that there is no d-orbital utilization. According to this comment:

...the doubly bonded picture is a very crude oversimplification, though. -@Martin

This enlightening image
shows that sulfur trioxide is actually a resonance of several different structures, and should be more accurately expressed as several single bonds with a dashed line to show that the bond order is not actually two. This also solves the question as to why it is stable with apparently six bonds, as in reality it only has four bonds.

In the molecule, if each oxygen atom shares two electrons with sulfur atom then how does the sulfur atom remain stable?

Look at the structure of Xenon tetrafluoride $\ce{XeF4}$ and Xenon hexafluoride  $\ce{XeF6}$. Do the Fluorine donate any electron to the Xenon? Of course not. It is the other way, the Fluorine with 7 electrons in the second shell needs one more electron to fulfill tho noble gas configuration and the Fluorine takes this electron from the Xenon with its electrons in the fifth shell. You know, the ionisation energy is higher in low electron shells, this allows to compose these compounds.

The same happens to the sulphur atom with its 6 electrons in the third shell. The three oxygen atoms fulfill their shells at the expense of sulphur. And sulphur gets closer to the level of the noble gas Neon. Of course this is a oversimplification because the atoms share the electrons.

The sulfur element has D-shell available for bonding. That is why it can form six bonds and have an expanded octet. Oxygen on other hand uses p shell for bonding which is completely filled by sharing 2 electrons. As to why sulfur expands, sulfur has vacant d orbitals lying with comparable energy to the bonded ones hence it is energetically favorable to form more bonds. Also, note that typically atoms with large sizes form expanded octet with small surrounding atoms as there is less inter-electronic repulsion.