text:
Comparison of acidic strength between nitric acid and nitrous acid
These are the structures of nitric acid and nitrous acid respectively:

To compare acid strength, we compare the stability of the corresponding conjugate base. Here conjugate bases are nitrate and nitrite anions respectively.

Now, my point is that in $\ce {NO2-}$, the resonance effect is pronounced, but in the case of $\ce {NO3-}$, we have a case of cross resonance. Put simply, in  $\ce {NO2-}$ the negative charge de-localizes to double bonded oxygen atom. In the case of $\ce {NO3-}$, same thing happens, but we have another $\ce {O-}$ here due to which there is a cross of 2 resonance effects (If I am using the correct terminology). So, resonance effect of nitrate ion should be weaker than nitrite ion and therefore nitrite ion should be more stable.
If I am true then nitrous acid must be a stronger acid, but all my sources suggest otherwise. They insist that in $\ce {NO3-}$, the negative charge is distributed over 3 oxygen atoms, and hence its resonance is more powerful. My confusion is that there are also 2 negatives that need to be stabilized as opposed to one negative in $\ce {NO2-}$.
So what concept am I missing here?

Well, you seem to not get the difference between "formal charge" and real charge. Would be good if you wouldn't even denote them in the same way.

The correct terminology is "cross conjugation" and "extended conjugation" (but the latter makes more sense in the context of more than two pi bonds, as in penta-1,3,5-triene).
Now, you have correctly identified that the nitrate anion has cross conjugation while the other has simple conjugation. However, you've incorrectly concluded one of them to be more stable. The conclusion is incorrect because there is a much dominant factor at play instead.
The three canonical resonating structures of the nitrate anion are "equivalent" i.e. wherever the negative charge may delocalise, the new resonance structure is exactly the same as the original one. Therefore, all the resonance structures are "equivalent" in their potential energy. This leads to a much rapid delocalisation of the lone pair, between the three oxygen atoms, leading to a rapid charge dispersal.
Both nitrate and nitrite anions have equivalent resonating structures. But, in the former, the negative charge is dispersed over three oxygen atoms (instead of only two in nitrite). From this, we conclude that the nitrate anion is much more stable than the nitrite anion.
Therefore, the next time you apply the fact that cross conjugation imparts lesser stability than extended conjugation, do look out for this additional dominant factor of "equivalent resonance"!

I did not overlook that, but I dismissed that because both had equivalent structures. One thing though, you are absolutely correct that in nitrate the negative charge is dispersed over three atoms whereas in nitrite the negative charge is dispersed over 2 atoms only, but I again insist that nitrate ion also has 2 negative charges that need to be dispersed as opposed to the only negative charge in nitrite anion. So, why are we treating the presence of 2 negative charges as equivalent to presence of 1 negative charge?

@Sarthak123 That is a good point, but note that the negative charge that you see on the other oxygen atom is not a real negative charge but actually a "formal" negative charge. The bond between that $\ce{O}$ and $\ce{N}$ is a coordinate covalent bond - a lone pair is donated by $\ce{N}$ to $\ce{O}$. Hence, we depict a +ve sign on $\ce{N}$ and a -ve sign on the corresponding $\ce{O}$. These are "formal" charges only.

oh, well I think that his has to do with my poor understanding of formal charges. I never considered the +/- on nitrate ion as formal charge, my understanding was that the hypothetical charges (much like oxidation states) are formal charges whereas these actual +/- are not formal charges. I would love to be enlightened on this issue, can you elaborate please?

@Sarthak123 Oxidation states are a different thing altogether. The only use I have ever figured out for formal charges is that they are just another way to depict coordinate covalent bonds. In the structure you posted for nitric acid above, instead of showing a line with an arrow-head (indicating coordinate lone pair donation), the author has simply drawn a solid line and denoted the associated formal charges on both the atoms involved. You see a similar formal charge in the ammonium ion. Notice that in ammonium ion, the formal charge of N is only +1, whereas it's O.N. is -4. Understood?

but doesn't it make those +/- as real as they seem? what's the difference between putting a coordinate bond or putting +/- when they physically convey the same thing?

@Sarthak123 That formal -ve sign just indicates that neither of the electrons in the bond is from oxygen itself (in a normal covalent bond, one electron in the bond comes from each atom involved); similarly, the formal +ve sign indicates that both the electrons of the bond were given by this atom; your question basically boils down to "why is the lone pair on a real negatively charged O atom only considered while that of the formal negative charged O atom ignored"? (when comparing stability of nitrate and nitrite anions)

(cont) I'm sorry but I don't have a response to this question; you might wish to add this question also to your original question, so that future answer posters can accommodate that point.

So I thought about this concept and I seem to have grasp the difference that you are trying to tell me. I think I now understand the difference between formal chargees and actual charges, though not sure, I am like 80% satisfied. Your answer also makes more sense to me now and hopefully I will soon accept it. I just need to think about it a little more. Thankyou.


text:
Why is hydrobromic acid a stronger acid than hypobromous acid?
If my understanding is correct, bromous acid ($\ce{HBrO2}$) is a stronger acid than hypobromous acid ($\ce{HBrO}$) because the additional electronegative oxygen atom draws the electron away from the hydrogen atom, making dissociation for $\ce{H+}$ easier.
Then why is hydrobromic acid stronger than hypobromous acid, despite the latter having one more electronegative atom to pull the electron? Are other factors involved here? And how do we decide which factor weighs more?

$\mathrm{p}K\mathrm{a}$ data:

$\ce{HBr}=-9.00$  (source)
$\ce{HBrO}=8.65$ (source)

The structure of $\ce{HBr}$ and $\ce{HBrO}$ is different, and also the place from where the proton leaves the molecule is also diferent.So, we have to take account of this and also the stability of the conjugate base formed after deprotonation.In $\ce{HBrO, }$ the proton is attached to the $\ce{O}$ atom of the molecule.The molecule can lose proton as, $$\ce{H -O -Br <=> H^+ + (O -Br)^-}$$ The negative charge here mainly resides on the Oxygen, which makes $3$ lone pairs on $\ce{O}$ atom, adjacent to that, there is also electron rich $\ce{Br}$ atom, which also has $3$ lone pairs on it, so there are inter electronic repulsions. For the conjugate base to get stabilisation, the weak $\ce{-I}$ effect of $\ce{Br}$ atom is not at all sufficient. Instead, if the lone pairs can somehow delocalise that will contribute better to stability of the conjugate base. But the lone pair resides on the $\ce{2p}$ orbital of Oxygen, which can't at all effectively overlap with filled$\ce{4p}$ orbital of $\ce{Br}$, and  $\ce{4d}$ orbitals are hugely high in energy. So, due to high electronic repulsion, no chances of delocalisation, and very weak $-I$ effect of $\ce{Br}$ can't make the conjugate base stable at all. So, $\ce{HOBr}$ has high $\ce{pK_a}$.But if you come to $\ce{HBr}$, the bond between $\ce{H}$ and $\ce{Br}$ is very weak. More over if you dissolve it in a sloution, the weak bond is easily broken, and the individual cations and anions can get extensively solvated which makes the  whole process of dissociation thermodynamically very much favourable.That's why, the $\ce{pK_a}$ of $\ce{HBr}$ is very low which means it can dissociate  very easily in a solvent.

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text:
Why does neopentane have a higher melting point than n-pentane?

Why does neopentane have a higher melting point than n-pentane?

I know that we can say that, due to stacking, it has a higher melting point as it freezes easily. But then, why doesn't this happen in the case of n-hexane too?
n-Hexane has a melting point of −95 °C whereas 2,2-dimethylbutane has a melting point of −98 °C. Instead, if we go by stacking, 2,2-dimethylbutane should have a higher melting point.

image source

No time to go through your Q but it can be related to this Q and my A at it. https://chemistry.stackexchange.com/q/89689/48509

Neopentane is completely symmetrical, whereas neohexane loses this symmetry. Neohexane thus has to be "flipped around a certain way" to fit into the crystal compared to neopentane where all the methyls are equivalent.

Many symmetrical molecules undergo a phase change before they melt.  This is usually interpreted as the ability to rotate without disrupting the lattice structure.  This has a much greater effect on the entropy of the solid than on its energy.  This reduces the entropy of fusion more than it reduces the enthalpy of fusion, thereby raising the melting point which equals delta-H/delta-S.  This is very dramatic in the comparison of melting points of octane, cyclo-octane, and bicyclo-octane.

Is there a name for this phase change? Or can you give a source describing it? It sounds reasonable, but I've never heard of it.

Some call it "Order-disorder phase transitions", and yes, they are pretty much a thing.

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text:
Why doesn't the bromide anion attack during electrophilic addition of bromine in methanol solvent?
It is known that addition of bromine (in methanol) to ethene will yield 1-bromo-2-methoxyethane. The reaction proceeds with the formation of a bromonium intermediate in the first step. In the second step, methanol attacks either carbon through the lone pair on its oxygen atom. In the last step, the $\ce{CH3OH+}$ loses a proton, which combines with the bromide anion obtained in the first step, to release $\ce{HBr}$ as a by-product.
My question: Why doesn't the $\ce{Br-}$ (obtained in the first step) attack as a nucleophile in the second step?
We know that nucleophilicity is directly proportional to the atom's charge and size. Bromide anion has a unit negative charge as opposed to the neutral oxygen atom in methanol. Bromide anion is also much larger in size. Both these factors favor bromide anion being a much better nucleophile than the neutral oxygen atom in methanol. Yet, the latter only attacks in the second step. Why?

There's a very simple explanation for this. Your reasoning regarding $\ce{Br-}$ being a better nucleophile is absolutely correct. The reason as to why methanol attacks instead of $\ce{Br-}$ is because it is the $\mathsf{\color{red}{solvent}}$. The solvent concentration is simply too high. Only one $\ce{Br-}$ is generated.
More or less the answer is this paragraph from J. Clayden - section "Unsymmetrical bromonium ions open regioselectively" in chapter 20 "Electrophilic addition to alkenes" :

...when a bromination is done in a nucleophilic solvent—water or methanol, for example—solvent molecules compete with the bromide to open the bromonium ion. As
  you know, alcohols are much worse nucleophiles than bromide but, because the concentration of solvent is so high (remember—the concentration of water in water is $\pu{55M}$), the solvent gets there first most of the time. This is what happens when isobutene is treated with bromine in methanol. An ether is formed by attack of methanol only at the more substituted end of the bromonium ion.

Note: Concentration of pure methanol is around $\pu{24.75 M}$, calculated from the density given on ChemSpider


text:
Oxidation of conjugated dienes into carboxylic acids by potassium permanganate
In the book -  MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed, Chapter Alkenes Q32 - the following reaction is given:

The given solution says that:

Conjugated dienes on oxidation by $\ce{KMnO4/\Delta}$ give oxalic acid

However, the mechanism for this reaction is not given. Also, the formation of a dicarboxylic acid on the parent ring is not explained. I am also unsure what would the products be in a simpler case like buta-1,3-diene. What products will it give?
I checked the internet but could not find a mechanism for the same. The related question of oxidation of arene subsituents by $\ce{KMnO4}$ is not a duplicate.
What is the mechanism for this oxidation?

related https://chemistry.stackexchange.com/questions/88668/oxidation-of-naphthalene-with-kmno4 https://chemistry.stackexchange.com/questions/54369/oxidation-of-cycloalkenes

Here is a relevant extract from a PDF (Dr. P.G. Hultin, 28 July 2016) depicting oxidative cleavage.  For the compound in hand (your question) the process will be done twice. $\ce{H2O2}$ will serve as the source of $\ce{H3O+}$.

When alkenes are treated with $\ce{KMnO4}$ in acidic solutions, the diol is not formed. Instead, the alkene is cleaved. The reaction proceeds by the same mechanism at the start, forming a cyclic manganate ester (although since the reaction is under acidic conditions the structure is protonated). In the schemes below, the alkene carbons are highlighted throughout, so you can see where they end up in the product.

Now, under these conditions the manganate ester is not very stable and it undergoes a cyclic fragmentation process, which results in breaking the C-C bond between the two oxygens. Notice that in this case, since there was a hydrogen atom attached to each of the alkene carbons in the starting material, there is a hydrogen attached to the carbonyl carbon in the product and therefore the product that is initially formed is an aldehyde

Aldehydes are very easily oxidized to carboxylic acids, and thus the aldehydes formed in the cleavage reaction do not survive. They are rapidly transformed into carboxylic acid groups, by a complex reaction whose mechanism you need not worry about.

Now, if the alkene had not had any hydrogens attached, the product in that case would have been a ketone
  rather than an aldehyde. Ketones are not easily oxidized further, and the reaction would have stopped at that
  stage.

If one of the alkene carbons had a hydrogen substituent, while the other did not, then we would get both acid
  and ketone groups in our product, as shown below.

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text:
Why are H+ and OH- ions preferentially discharged in the electrolysis of NaCl(aq)?
There are 4 types of ions in an $\ce{NaCl(aq)}$ solution: $\ce{H+}$, $\ce{OH-}$, $\ce{Na+}$ and $\ce{Cl-}$. According to the E.C.S. table, $\ce{H+}$ and $\ce{OH-}$ should preferentially discharge at cathode and anode, respectively.
However, according to the Wikipedia page on self-ionization of water, water only dissociates slightly into $\ce{H+}$ and $\ce{OH-}$ in neutral solution. (This is the reason why pure water conducts electricity poorly). Therefore, in the $\ce{NaCl(aq)}$ solution, the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely small, and by comparison those of $\ce{Na+}$ and $\ce{Cl-}$ are much higher. As a result, it seems to me that $\ce{Na+}$ and $\ce{Cl-}$ should be preferentially discharged because of the concentration effect.
However, my chemistry textbook says that in fact $\ce{H+}$ and $\ce{OH-}$ are preferentially discharged, and hence the main products of the electrolysis are hydrogen and oxygen. So apparently my understanding is wrong. But where?

Possible duplicate of While electrolyzing concentrated aqueous sodium chloride, why is it that chlorine is discharged but not sodium?

@Mithoron Related, yes, but not duplicate.

Water does not need to be dissociated into ions in order to participate in electrochemical reactions. While it is common to refer to electrode reactions as "discharging" this or that species, there is no requirement that the reactant of an electrode reaction must always be an ion.
In neutral solution, it's quite common for the water electrolysis reactions to be written as:
$$
\begin{align}
\text{Cathode:} &\quad \ce{4H2O + 4e- -> 2H2 + 2OH-} \\
\text{Anode:} &\quad \ce{2H2O -> O2 + 4H+ + 4e-}
\end{align}
$$
It's absolutely the case that the electrochemistry written using the ions as reactants in many cases is likely to be more facile:
$$
\begin{align}
\text{Cathode:} &\quad \ce{4H+ + 4e- -> 2H2} \\
\text{Anode:} &\quad \ce{4OH- -> O2 + 2H2O + 4e-}
\end{align}
$$
But it's still far (far) more favorable for $\ce{H2O}$ to be reduced than $\ce{Na+}$, and in many cases more favorable for $\ce{H2O}$ to be oxidized than $\ce{Cl-}$. You're quite right that the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely low in neutral salt solutions, but that fact is mostly irrelevant to the electrochemistry.

So, water is actually discharged in the form of complete molecules, rather than in the form of H+ and OH- ions? Also, according to your first two equations, water is simultaneously reduced and oxidized, so this is actually an example of disproportionation?

@user161070 Water can be "discharged" in the form of a complete molecule, yes, at both the cathode and the anode.

@user161070 Given that the oxidation and reduction reactions are happening at electrodes separated in space, and thus it is never the same $\ce{H2O}$ molecule undergoing both oxidation and reduction, I would argue that this is not an example of disproportionation.

It isn't true.  In "chlor-alkali" electrolytic processes it is chloride ions that get discharged at the anode, not water.  See here for an introductory look.

Is it still false if the solution is in very dilute condition? Also, how about sodium ions?

Well ... there is even a setup where sodium is reduced instead of hydrogen, although this cell, where they form an amalgam, is on its way out.  Industrially the brine has to be concentrated so nothing is said in the article about very dilute solutions; but there is such a thingcas running out of chloride ions thus forcing water oxidation.

Thanks, but actually my doubt is about the "claim" of the question. Maybe I should ask the question in this way:
Is the following claim true: Water only dissociate very slightly into H+ and OH− by itself (which is the same reason why pure water almost doesn't conduct electricity). Therefore, in any aqueous solution, the concentration of other ions will be a lot higher than that of ions of water. If we take this aqueous solution into an electrolysis process, the ions of water will never get preferential discharged, because of the way higher concentration of other ions.

According to the principle of electrolysis, for different cations (in this case, Na+ and H+), the H+ is discharged since it has a low reduction potential (i.e. has low reactivity) as compared to the Na+. Thus, H+ is prefered.
For anion (in this case, Cl- and OH-), two quantities are used. 1. The one with high concentration will discharge over the anion with low concentration. We know that water is in dynamic equilibrium, so the concentration of OH- is obviously lower than that of Cl-. Thus, Cl- is discharged.
This leaves the Na+ and OH- in the solution.


text:
Why do more substituted alkenes undergo epoxidation, but the less substituted alkenes undergo hydrogenation?
We know that, in case of a choice, catalytic hydrogenation will hydrogenate the less  substituted alkene, because the less substituted alkene is less stable. But, we also know that, in case of a choice, peroxyacids perform epoxidation of the more substituted alkenes, because the latter are more electron-rich (i.e. more nucleophilic) (details)
In one case, we are favoring the stability of the pi bond. But in another case, we are prioritizing the nucleophilicity of the pi bond. Why is this so?
I believe this definitely has something to do with their respective mechanisms, which I've studied from Libretexts (hydrogenation and epoxidation). The only difference I see is that in hydrogenation, the pi bond first attaches to the catalyst surface, and then a hydrogen atom (already attached to that surface) is transferred to the pi bond. But, in epoxidation, the alkene directly attacks the electrophilic $\ce{-OH}$ group of the peroxy acid. So, this explains the need for more nucleophilicity in the latter case, but how does this explain the need for less stability in the former case? Or is this direction of reasoning entirely wrong?

As you already know, the alkene has to approach the catalyst surface in the first step. For optimum adsorption, the molecule must be able to orient itself parallel to the surface of the metal. Less number of substituents promote this.

The activated hydrogen is in the form of $\ce{Pt-H}$ or $\ce{Pd-H}$ bonds on the surface of metal particles, and hindered alkenes can't approach the $\ce{M-H}$ bonds easily.

Source
Epoxidation on the other hand is free from any such need for to attach to the surface. So we simply consider nucleophilicity.


text:
Why is the effect of beta-branching neglected in this comparison of SN2 reaction rates?

Which compound would give a faster SN2 reaction?

2-bromobutane
1-bromo-2-methylpropane

The answer is given (2).
While I agree that the primary position is preferred as compared to the secondary position (for an SN2 attack), but won't beta branching hinder the nucleophile's attack, hence, reducing the reactivity for SN2?

Yes, the beta branching will hinder the nucleophile a bit, but the substrate is still a primary halide with 2 hydrogens on the carbon that is attacked. The steric effect of having a methyl group and a hydrogen on that carbon vs two hydrogens is very much larger.


text:
Ethanol in gasoline — will it evaporate?
If gasoline with 10% ethanol is stored for a few weeks, will the ethanol evaporate?
I have read that to check for ethanol in gas is to add a little water to a container; mark the level and add gas. If the water lever stays the same, it is ethanol-free. I purchased ethanol free-gas and it passed the test. I then tested some gas with 10% ethanol that had been stored for approximately 1 month in a plastic gas can and got the same results. According to my source, the water/ethanol level should have gone up.

Whether or not the ethanol will evaporate depends on your container. If the container isn't air-tight then both your gas and the ethanol will leak into the atmosphere.
With regards to the 'level test'. Generally, ethanol mixes well with both water and organic solvents. If this is the case with fuel (and I suspect it is - it is mainly hydrocarbons), you will not see appreciable changes in the level of the 'water'.
However, for this type of qualitative test, you should keep a few things in mind.
Depending on the diameter of the container you used and the amounts of water and fuel, you will get varying amounts of level change, if any. You could try a very thin tube and mix the components well. After mixing leave the phases to separate for a good while before checking your level.


text:
Which of - carboxylic group or carbaldehyde group - has more negative inductive effect?

Which has more -I effect $\ce{-COOH}$ or $\ce{-CHO}$, and why?

I tried to analyze it on my own but I was confused on what to do. My teacher today taught it by shifting of electrons from $\ce{C=O}$ to the formation of a charge separated structure. Still, I was confused about it.
Mostly, I'm confused on why we are even allowed to create a charge separated structure. Even with it, it seems $\ce{-COOH}$ should have more -I effect. But I am not sure. I need help here.

Are you sure your teacher explained the -I effect by "the formation of a charge separated structure" and the -M effect instead?

Yes, he made the charged structure and I don't know why it was feasible. I'm not sure under what conditions we are allowed to do so. Sometimes these things seem to be bit mechanical and goes over my head.


text:
How to calculate the molarity of a gas?
If I have $X$ moles of a gas and I put them in a container at constant volume $V$, will the molarity of the gas then be $X/V$?

related https://chemistry.stackexchange.com/questions/19702/what-is-the-difference-between-molarity-and-concentration

Molarity is defined as "moles of solute per volume of solution", which implies that the system is in the liquid state.
As such, molarity is undefined for anything in a gaseous state -- there is no solute/solvent distinction in the gas phase.
You can define the concentration of a gas, though, and that calculation would be exactly the one you've described: placing $X$ moles of a gas in a container of volume $V$ yields a concentration $C=X/V$, with units of, e.g., $\pu{mol\over dm^3}$.
Note that while these units might look the same as the units of molarity, they are different. In this concentration value, the volume is the volume of the container; in a molarity value, the volume is the volume of the solution.

Ok! If the value of the volume is not specified, but if I am told that it is constant, can I say that the concentration of the gas is, for example, $X$ mol / l?

@GuillemusCallelus No... if the volume is not specified, even as a variable, then it's impossible to write an expression for the concentration.

@GuillemusCallelus Huh. Reading this again, three years later, I think I might have misread your question -- you can definitely write an expression for the concentration using an unknown variable, but it would make the most sense to me to write that expressing using a variable in place of the volume (the specific unknown thing), and not the concentration, which you could write as $n/V$, where $n$ is the amount you have of the gas of interest.

The answer given by hBy2Py is correct - "only solutions have molarity" is likely the right thing to say on the quiz.  It will help you remember that in calculations of equilibrium constants and Nernst potentials, gases are referenced to a standard pressure rather than a concentration, and that pressure corresponds to 1 bar at 0 C = 1 mol / 22.7 L, not 1 mol/L.
But ... that answer is also wrong, and here's why.

Many modern texts define "solutions" to include mixtures of gases.  By that semantics, you correctly described the molarity of a solution in your question.

It is commonly said and often useful to know that water at STP is 55.5 mol/L.  Though converting this to a vapor normally involves boiling and a visible change of state, we also have the option to pressurize it to 300 atm, heat to 700 K, then lower the pressure.  This goes around the critical point on the phase diagram.  Although subtle and interesting boundaries between supercritical liquid and supercritical gas can be drawn, there is no point on that path at which we would feel a sudden need to stop using mol/L to report the moles of water per liter.  In a study of hydrothermal water you might need to work in any region of the phase diagram along that path using consistent terms.

Most importantly, thinking about the concentration of gases can help you to understand the colligative properties of solutions.  For example, the ideal gas law PV = nRT can be rewritten as P = cRT, where c = n/V is the concentration of the gas as you defined it.  That is the same formula as we use (at first approximation, and remembering some substances dissociate into multiple moles of particles) to determine the osmotic pressure of a solution!  It takes some doing to understand how these relate to the thermal energy of individual particles, but by regarding the situations side by side you will have an easier time of it.

Mike Serfas raised a very good point and it exposes the fallacy behind the concept of up vote, down vote or accepted answers. How does the OP know that the answer is "right"? There is no guarantee that the accepted answer is the right one and the down voted one is the wrong one. Since when science started relying on populism?
How can we say that the molarity of a gaseous mixture is undefined? Who said so?
If we look up IUPAC's molarity, they suggest the term "amount concentration"

Amount of a constituent divided by the volume of the mixture. Also
called amount-of-substance concentration, substance concentration (in
clinical chemistry) and in older literature molarity.

https://goldbook.iupac.org/terms/view/A00295
There is no such requirement that the solute must have some interaction with the solvent for molarity to be well-defined.


text:
Nomenclature of complex benzene-based substituents
For a complex alkyl substituent, (at least the ones I have come into contact with thus far) you name it as a chain numbering from the first non-parent carbon and placing the name in square brackets: e.g. isopentyl can also be named [3-methylbutyl]. What is the convention for more complex benzene-based substituents? Specifically, can you still use the more common names like phenol and toluene as a parent for your name if say you bond C5 of 2,3-dimethylphenol to your parent and what suffix will you use?

“For a complex alkyl substituent, you name it as chain numbering from the first non-parent carbon and placing the name in square brackets.” This is not generally true.

Note that your example 3-hydroxy-4,5-dimethylphenyl is a compound substituent group and not a complex substituent group.

But when adding it to your parent name won't you use [3-hydroxy-4,5-dimethylphenyl]?

Intro
Your statement, that complex substituent names are placed in square brackets, is incorrect. These are to be placed in parentheses (round brackets). Correct preferred IUPAC name (PIN)[1] example:

  2-(3-methylbutyl)phenol (PIN)

Square brackets are used when the substituent name already contains parentheses, e.g.:

  2-[(2R)-2-methylbutyl]phenol (PIN)

Retained aromatic compounds allowed for substitution
You can see that the names above are based on phenol, which is one of retained aromatic functional parent compounds, with allowed substitution, in preferred names.
The other ones are aniline and benzoic acid (see IUPAC rule P-34.1.2.[1]). E.g.:

  2-hydroxybenzoic acid (PIN)
  (not 2-carboxyphenol, as $\ce{-COOH}$ group is preferred over $\ce{-OH}$)

Retained alkyl-substituted aromatic hydrocarbons (not allowed for substitution in PINs)
There are also retained alkyl substituted aromatic parent hydrocarbons (P-22.1.3):

toluene (PIN)
  methylbenzene

and

  1,2-xylene (PIN)
o-xylene
  1,2-dimethylbenzene

(and the two other xylenes).
But none of them is allowed for substitution in preferred names, e.g.:

  1-chloro-4-(chloromethyl)benzene (PIN)
  α,4-dichlorotoluene

Xylenes are not allowed for substitution even in the general non-preferred names:

  1,2-bis(bromomethyl)benzene (PIN)
  α-bromo-2-(bromomethyl)toluene
  (not α,α′-dibromo-o-xylene)

(Yet another retained name mesitylene for 1,3,5-trimethylbenzene (PIN) is allowed only for general names, not allowed for substitution.)

Complex aromatic substituents
Now if a complex substituent is made from substituted phenol, aniline or benzoic acid, the substituent name is no longer based on these names. They can be thought as suffixes, but now we have another higher-priority “suffix”, -…yl. E.g.:

  3-hydroxy-4,5-dimethylphenyl acetate (PIN)
  (not 2,3-dimethylphenol-5-yl acetate)

(In case you didn't hear ‘phenyl’ yet – it's not to be confused with ‘phenol’, it's the only, notorious, name for “benzenyl”, $\ce{-C6H5}$)

References:

Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)

Suppose the carboxylic acid was not attached directly attached to the benzene ring and it were an ethanoic acid derivative attached to ring, how would you name it?


text:
Need help with an IR spectrum (polymer identification)
I have been given an unknow polymer and I have to identify it.
DSC measurements showed no glass transition (which means it's below room temperature) and a melting temperature at around 100-120°C. So, I thougth it was a polyethylene. But then, the IR spectrum showed some bands that I didn't expect.

The only thing I'm sure about are the peaks at around 2900 being C-H streching.
Maybe the large band between 3200-3400 is an O-H bonded? At 3500 cm-1 a N-H ? 1370 I really don't know...
I don't find any corresponding polymer. Could it be a copolymer ? Or an impurity ?
Could you help me please ?


text:
Atomic structure
We've got a homework and I know this question may seem so primitive, but I have already read whole textbook and I've been looking also on the Internet, but I still cannot find the right answer. I'd be grateful if you could help me.
The question is what is the basic structural unit of an atom.
I'm considering whether to take into account only proton since "a basic structural unit" of something is usually only one thing and also atom of hydrogen consists only of a single electron and proton. But on the other hand, neutrons make up quite a considerable part of weigh of other atoms and are a part of the nucleus.
The answer options are the following:
a) atom
b) proton
c) neutron
d) electron
e) positronium

It's rather structural unit of element in this question of yours? Otherwise there wouldn't  be option "atom" which is correct for such question.

No, it's not element. They are asking about atom. It is a worksheet with some options that are wittingly incorrect.

sigh Well then I guess b) c) and  d) would be kinda OK but calling them  "basic structural units" isn't best wording.

If the answer is single choice, and you understand the atom's structure, then just skip about the question. It is indeed poorly worded.

I quote Gaurang Tandon. As it is the question is garbage. It happens.

An atom, I'd say, consists of a central nucleus with orbiting electrons.
Take away those neutrons from the nucleus, and you have just protons. A proton is a part of an atom, but it doesn't have to be the basic 'structural unit' of the whole thing - there are two other things that can contribute.
One thing I don't understand are the possible answers. 'Atom' is the basic structural unit of the atom? They sound like poor answers. If you can choose more than one, choose 'proton', 'neutron' and 'electron'. I'd say that's the textbook composition of the atom (although the hydrogen atom has no neutrons...)

It is a worksheet with some answers that are illogical or trying to confuse students, but that wasn't the thing I was worried about :) I would tell the same as you, that the atom consists of these three parts and due to it I can't decide which of these should be considered as a basic one. Anyway, thank you for your answer :)

The guy was confused and asked which options he should choose from a poorly worded question... I gave my answer and the guy agreed with what I said... How this doesn't provide an answer is strange.


text:
How to determine the angle between non-bonding electron pairs?
Is there a way to determine the angle between non-bonding electron pairs that aren't obvious? Take $\ce{H2S}$ for example:

The two non-bonding electron pairs would be at the yellow end of this molecule, but I am curious as to what the angle is between them? Is there even a way to tell?

It's quite difficult to tell https://chemistry.stackexchange.com/questions/50906/are-the-lone-pairs-in-water-equivalent

There are some ways one could claim to calculate such angles, but any calculations of that sort would be ... highly controversial. Meaning, naturally, I will happily perform such a calculation (when I get a chance) to stoke the fires of such controversy. :-D

Yes, there are ways one could claim to calculate an angle between two non-bonding electron pairs.
BUT: As Mithoron points out, this Chem.SE question illustrates how quantum chemical calculations and photoelectron spectroscopy both demonstrate the non-equivalence of the lone pairs of $\ce{H2O}$, an analysis which presumably applies equally well to the analogous $\ce{H2S}$. Thus, the methods used for calculating such an angle will be controversial, and the results may or may not be of any particular practical value given that they're at odds with PES data. That being said, I'll show here one way that a putative 'non-bonding electron pair angle' can be calculated.
The calculation below is based on application of the quantum theory of atoms in molecules (QTAIM) to the quantity called the electron localization function (ELF), which is a scalar function in $\mathbb R^3$.  QTAIM is useful for identifying intrinsic features of various three-dimensional fields that arise in quantum chemical calculations, with a significant focus placed on the 'critical points,' where the field gradient is zero. It was originally developed for analysis of the electron density distribution, but has been extended to the ELF and other quantities. One of my favorite papers illustrating coupled QTAIM/ELF analysis is a 2009 review by Matito and Solà (Coord Chem Rev 253: 647); it also discusses the localization and delocalization indices (see here and here), which I won't go into in any detail here.

I'm going to start out by emphasizing the distinction between electron density and electron localization. The electron density at a point is fairly straightforward: it's a measure of, on average, how many electrons (or fraction thereof) are located at that point, per unit volume. Due to its waveform nature, each electron's position is a continuous distribution function, and the electron density of the system represents the collective distribution function of all the electrons present.
The electron localization at a point can be harder to get a grasp of: it's a measure of how "spread out" the "location distribution" is, of the electrons that contribute to the density at that point. Regardless of the magnitude of the electron density, the electron localization at a point is high when the electrons contributing to the density at that point overall contribute relatively little to the electron density in other parts of the system.  That is to say, these electrons associated with the point of interest don't "travel around very far" from that point, in their quantum-mechanical meanderings about the system.  Conversely, the electron localization is low when the electrons associated with the point do contribute significantly ("wander") to "distant" regions of the system.
My answer here provides a more concrete example of how the electron localization can differ significantly among systems whose electron density distributions are otherwise relatively similar.  It also anticipates the discussion below, noting features of the three-dimensional ELF field that suggest the locations of electron lone pairs on N, O and F atoms. It is this ELF field that I will be focusing on in the following analysis.

I performed DFT calculations of $\ce{H2O}$, $\ce{H2S}$, $\ce{H2Se}$, $\ce{H2Te}$ and $\ce{H2Po}$ for this analysis using ORCA v3.0.3, and I carried out the subsequent QTAIM/ELF analysis in Multiwfn v3.3.7.  The ORCA input for $\ce{H2S}$ was
! RKS PBE0 def2-TZVP def2-TZVP/J RIJCOSX
! OPT GRID4 GRIDX5 PRINTBASIS

* xyzfile 0 1 H2S.xyz

H2S.xyz was an initial geometry file generated by Avogadro, overwritten with the optimized geometry at the end of the ORCA run.  The inputs for $\ce{H2O}$ and $\ce{H2Se}$ were identical except for the identity of the central atom; I added relativistic effects for $\ce{H2Te}$ via the ! ZORA simple keyword, and for $\ce{H2Po}$ via the use of a 60-electron effective core potential by adding ! ECP{def2-TZVP,def2-TZVP/J}. (While there is an argument to be made that I perhaps should've included relativistic effects for $\ce{H2Se}$ also, I expect them to minimally affect the geometry and overall valence-electronic structure of that system, which is what is of interest here.)
I generated a MOLDEN wavefunction file for each computation, to serve as inputs to Multiwfn. Using $\ce{H2S}$ as an example, the shell command was:
$ orca_2mkl H2S -molden

I then renamed the resulting H2S.molden.input to H2S.molden, which is necessary for Multiwfn to correctly identify it as an ORCA-generated MOLDEN file.
After loading H2S.molden into Multiwfn, I used the following series of commands to locate the 3-D maxima ("attractors") of the ELF distribution:

Main menu:  [17] Basin analysis
[1] Generate basins
[9] ELF
[3] High quality grid

Multiwfn's search routine found six ELF attractors, one of which was identified as a cluster of 'degenerate attractors'. The figure below plots these attractors (via Multiwfn sub-command [0] and some MS Paint manipulation) atop two different views of the $\ce{H2S}$ molecule (click to enlarge):

As can be seen, there are two attractors located right where chemical intuition would expect lone pairs to sit, along with two more attractors located at the hydrogen atoms.  The final non-degenerate attractor is right in the center of the sulfur atom, and could be interpreted as representing the $n=1$ core electron shell. The degenerate attractor is distributed around the $n=1$ attractor, and could similarly be interpreted as the $n=2$ core electron shell. Interestingly, the two $n=3$ valence electrons of the sulfur that are involved in bonding to the $\ce{H}$ atoms do not exhibit independent attractors---each shares an attractor with the electron originating from its respective hydrogen atom.
The QTAIM method provides a way to subdivide the space occupied by a molecule based solely on the properties of the ELF distribution, and associate portions of the electron density to each of these attractors. Integrating these subdivisions of the electron density then provides the number of electrons associated with each attractor:
$$
\begin{array}{cccc}
\hline
\text{lp} & \text{H} & \text{S 1s} & \text{S 2sp} \\
2.114 & 1.855 & 2.143 & 7.865 \\
\hline
\end{array}
$$
All of these values seem pretty reasonable---the attractors corresponding to single orbitals $(\ce{lp}$, $\ce{H}$, $\ce{S 1s})$ have about $\ce{2 e-}$ associated with them, and the $\ce{S 2sp}$ attractor has about $\ce{8 e-}$. In my experience, the signs and magnitudes of these deviations from $2.0$ and $8.0$ are typical for ELF basins.
So, leaving the question of actual physical significance aside, I would argue that these ELF attractors provide a reasonable representation of the (non-)bonding structure of the molecule. Thus, to the main question: what is the structure?  Conveniently, angles and distances among the attractors and atoms can be calculated via Multiwfn sub-command [-2]:
$$
\begin{array}{cccc}
\hline
\angle\,\ce{H-S-H} & 92.3^\circ & r_\ce{S-H} & 2.535\,\mathrm{Bohr} \\
\angle\,\ce{lp-S-lp} & 127.8^\circ & r_\ce{S-lp} & 1.838\,\mathrm{Bohr} \\
\angle\,\ce{H-S-lp} & 108.5^\circ \\
\hline
\end{array}
$$
(In the above, the actual nuclear positions of the atoms were used where relevant, not the positions of the associated attractors. While each such attractor generally falls close to its associated nucleus, it's usually displaced by a tiny amount, $<0.1\,\mathrm{Bohr}$. I've always attributed this to numerical precision limits of the calculation methods, but I don't know for sure if that's what causes it.)
Therefore: By this QTAIM/ELF method, the angle between the two lone pairs of $\ce{H2S}$ is calculated to be $127.8^\circ$. This is considerably greater than the $\ce{H-S-H}$ angle, consistent with the intro-chem conception of the increased 'steric bulk' of a lone pair.
For comparison, I used the same procedure to generate results for $\ce{H2O}$, $\ce{H2Se}$, $\ce{H2Te}$ and $\ce{H2Po}$:
$$
\begin{array}{c|c|ccccc}
\text{Quantity} & \text{Units}  & \ce{H2O} & \ce{H2S} & \ce{H2Se} & \ce{H2Te} & \ce{H2Po} \\
\hline
\int_\text{lp}{\rho} & \ce{e-} & 2.265 & 2.114 & 2.206 & 2.265 & 2.390 \\
\int_{\ce{x}\ce{-H}}{\rho} & \ce{e-} & 1.667 & 1.855 & 1.861 & 1.807 & 1.806 \\
\int_\text{core}{\rho} & \ce{e-} & 2.129 & 10.008 & 27.273 & 44.567 & 77.609 \\
\hline
r_{\ce{x}\ce{-H}} & \mathrm{Bohr} & 1.813 & 2.535 & 2.772 & 3.129 & 3.289 \\
r_{\ce{x}\ce{-lp}} & \mathrm{Bohr} & 1.103 & 1.838 & 2.526 & 3.216 & 3.399 \\
\hline
\angle \ce{H-x}\ce{-H} & ^\circ & 105.1 & 92.3 & 91.2 & 90.9 & 89.6 \\
\angle \ce{lp-x}\ce{-lp} & ^\circ & 114.9 & 127.8 & 139.5 & 156.8 & 156.8 \\
\angle \ce{H-x}\ce{-lp} & ^\circ & 109.1 & 107.8 & 104.0 & 98.1 & 98.2 \\
\hline
\end{array}
$$
Due (presumably) to numerical precision limitations, the calculated values of the four different $\ce{H-x}\ce{-lp}$ angles for each system varied by $1^\circ$ or so; I've reported the means of the obtained values above.
Those at home who are doing their math carefully will note that there is some electron density missing in the systems with the heavier central atoms.  This is because at the 'high' grid quality used, the deep core electrons are poorly captured by the numerical integration involved.  (Indeed, the deep core electrons may not be accurately captured at just about any computationally reasonable grid size!) I think the $\int_\text{lp}{\rho}$ and $\int_{\ce{x}\ce{-H}}{\rho}$ values should be pretty reliable, though.
Naturally, $\ce{H2O}$ does not exhibit a degenerate attractor for the $n=2$ valence electron shell, though it does have a non-degenerate attractor for the $n=1$ core shell.  The two heaviest chalcogenides examined possess degenerate attractors representing their $n\geq 2$ core electrons, as with sulfur.
In general, the $\ce{x}\ce{-H}$ and "$\ce{x}\ce{-lp}$" bond lengths increase as the central atom is varied down the group, which is unsurprising ("bigger atoms are bigger"). Interestingly, while the $\ce{x}\ce{-lp}$ distance starts out considerably shorter than the $\ce{x}\ce{-H}$ bond length for $\ce{x}=\ce{O}$, by the time one reaches $\ce{x}=\ce{Te}$ the relative magnitudes are reversed.  I don't have a particularly good explanation for this trend; presumably it's due to some sort of counterbalance between the nucleus-nucleus repulsion and nucleus-electron attraction forces.
Where these results are particularly dramatic is in the clear trends in the bond angles: as the central atom is made heavier, the $\ce{lp-x}\ce{-lp}$ bond angle increases markedly while the $\ce{H-x}\ce{-H}$ and $\ce{lp-x}\ce{-H}$ angles steadily decrease. The lone pair 'steric bulk' apparently increases uniformly down the group for this series of analogous systems, at least as determined by this QTAIM/ELF approach.  The answers to this question provide more detail into the physical reasons for this decreasing trend in $\ce{H-x}\ce{-H}$ bond angles.


text:
Need help explaining OH-peaks in IR
I have synthesized (1S, 2S)-1,2-diphenylethene-1,2-diol, and I'm wondering why the $\ce{OH}$-peaks in the IR-spectrum looks the way it does. I would expect a broad signal (because the solution wasn't diluted, and because free $\ce{OH}$-stretch happens at $\pu{3580-3700 cm^{-1}}$). So why these two separate, distinct, signals at $\pu{3394}$ and $\pu{ 3499 cm^{-1}}$?
(the IR-spectrum)

Could you describe the preparation method (solvent? water?) and IR technique (i.e., thin film on salt plates?)?

Crystalline product dissolved in diethyl ether (didn't measure the amount of product dissolved). I'm not really sure what type of IR-instrument i was using, but I placed some of the solution on the instrument and let the ether evaporate, then i ran the test.

Would be see any vicinal coupling effects? Or are only geminal ones seen in IR?

Off top of my head, I would think this is due intramolecular hydrogen bonding (a 5 membered ring), giving rise to a sharp peak at $\pu{3499.14 cm^-1}$, and intermolecular bonding, giving a slightly broader peak at $\pu{3394.45 cm^-1}$. But I don't have any references, this is just a hypothesis.

1,2-Ethanediol, 1,2-diphenyl (also known as hydrobenzoin) can exist in three crystalline forms. One is S,S (formerly known as d or -), and its twin, R,R (formerly known as l or +), with a mp of 148-149 C; second is the meso form R,S (same as S,R), mp 139-140; and a third form, easily indicated as dl (or S,S:R,R), mp 122-123. It would be a eutectic.
Tartaric acid has the same kind of enantiomeric situation, but the racemic form (dl) is the highest melting form and the meso form is the lowest.
The formation of the low-melting dl form of diphenylethanediol suggests an intermolecular bonding which would likely be thru the hydroxide groups. So intramolecular bonding could give rise to one type of hydrogen bond while intermolecular bonding could lead to another type, with a different stretching frequency.
This is merely an expansion of the answer by ralk912.


text:
Cooling Curves differences between Solution and Pure Solvent
To put it short, why is the cooling curve for a solution slanted during the phase change? The pure solvent plateaus at a fixed temperature, but for a solution, the phase change occurs over a range of temperatures. What explains this phenomenon?

Well, it would be kinda surprising in freezing/melting of mixture had sharp peaks, even for pure compounds they can be quite broad, or even variable.

During the phase change, the composition of the solution phase is changing from that of the original solution as the solidifying component is removed.  This doesn't happen for a pure liquid.  In a simple system, the slant continues until both components begin to solidify simultaneously - the eutectic point.  The temperature then flattens out because the composition of the liquid phase doesn't change.  It stays flat until no liquid remains.  This can be complicated if there are other solid phases or formation of new chemical components.


text:
Why is ordinary soap and water adequate to decontaminate possible nerve agents?
In the news it was announced that anyone who might have been exposed in the recent Salisbury nerve agent incident should wash any potentially contaminated clothing as normal using regular detergent and water; other personal items can be cleaned with wipes.
The agent in question (identified only as Novichok, a family of chemicals) has been described as the most potent ever made.
How is it that ordinary cleaning can suffice as a method of decontamination? Is it a very unstable molecule? Does water (or detergent, or detergent solution, or alcohol solution) act strongly to decompose such chemicals? If unstable in water, how can it remain stable in the body to act? Could any viable agent remain in the wash waste water (going down the drain) after washing and pose a hazard downstream?
Edited:
The focus of my question is entirely about the apparent paradox of a chemical able to remain stable in an everyday environment, yet readily destroyed/decomposed by everyday means. In the specific case, a chemical which happens to be lethal in trace quantity that can persist, yet simple washing apparently destroys all trace, removing any potential for harm.

Not sure why the close votes. A perfectly reasonable chemistry question.

@ToddMinehardt Then the fact that a wide range, if not most, "modern" nerve agents are based on a common fluorophosphonate core which is the target for the hydrolysis to less harmful substances should alleviate the worry. Every question relates to this single chemical feature.

Nerve agents are usually susceptible to hydrolysis leading to non-toxic degradation products
Most modern nerve agents are based on fluorophosphonate cores. That is they contain a $\ce{PFO(OR)(X)}$ core where R is some organic radical (-iPr in sarin) and X can be a range of radicals (Me- in sarin, OR or N=R in novichok agents).
Most are not very stable in storage as they are subject to degradation from hydrolysis so have to made not long before use or have to be made ultra pure to avoid any contaminants catalysing their degradation by acid or base hydrolysis. As a result any were deployed as binary agents (where two tables components are mixed inside the weapon to create the active ingredient).
The key to their reactivity is the P-F bond which is easily hydrolysed to give much less harmful products. This hydrolysis is not fast enough to degrade the compounds inside the body before they kill you but is fast enough that they don't hang about for long periods in the environment after use (it is worth bearing in mind that they are extremely powerful inhibitors of acetylcholinesterase, without which your nervous system can't function: it doesn't take a big dose or a long time for them to work). It is also something that can be accelerated by washing. It seems likely, for example, that mildly contaminated clothing can be decontaminated by a standard domestic clothes washing cycle as this would dramatically accelerate any hydrolysis reactions.
It is also worth noting that, if the contamination were large enough, you would notice it very rapidly. So any residual contamination in people not already ill would be of a very low level and easily rendered insignificant by washing.-

Side question: does hydrolysis follow the half-life pattern? In other words, even if the majority of such an agent were quickly destroyed, could some quantity of the original agent survive "decontamination" (i.e. remain un-hydrolyzed, even in water)? Is the washing approach viable only because of the assumption of "mild" contamination and that half-life is short enough that assumed residual quantities are brought down to no more than a molecule here and a molecule there?


text:
Buffer solution calculation using pka

A solution is obtained when $\pu{10.0 cm3}$ of $\pu{0.100 moldm−3}$ sodium hydroxide is added to $\pu{20.0 cm3}$ of $\pu{0.100 mol dm−3}$ ethanoic acid ($\ce{\mathrm{p}K_\mathrm{a}}$ = 4.76). Find its $\mathrm{pH}$.

I got $\mathrm{pH} = 4.76$, but the answer key is $4.46$. I think the answer is wrong. What do you guys think?
Here is how I solve it:
mole of $\ce{OH-} = 0.1*10*10^{-3} = 10^{-3}$ moles
mole of Acid =$\mathrm{ 0.1*20*10^{-3} = 2*10^{-3}}$moles
$$\ce{Acid + OH- -> Base + Water}$$
Start    $\mathrm{ 2*10^{-3}}$ ,   $\mathrm{10^{-3}}$ ,      $\mathrm{ 0}$
Change   $\mathrm{ -10^{-3}}$ , $\mathrm{-10^{-3}}$ , $\mathrm{10^{-3}}$
End     $\mathrm{10^{-3}}$    ,  $\mathrm{0}$  ,  $\mathrm{10^{-3}}$
$\ce{[Acid] = \frac{mole}{volume} = \frac{10^{-3}}{30*10^{-3}}}$
$\ce{[Base] = \frac{mole}{volume} = \frac{10^{-3}}{30*10^{-3}}}$
$$\mathrm{pH = pKa + log \frac{[Base]}{[acid]}}$$
$$\mathrm{  = 4.76 + log \frac{1}{1} }$$
$$\mathrm{= 4.76}$$

$K_a = 10^{-pK_a} = 10^{-4.76}$ = 1.74x10-5 mol dm-3
Initial moles $\ce{CH3COOH}$ = 2x10-3 mol and  initial moles $\ce{NaOH}$ = 1x10-3 mol
$\ce{CH3COOH}$ is in excess $\therefore$ final moles $\ce{CH3COOH}$ = 1x10-3 mol, final moles $\ce{CH3COONa}$ = 1x10-3 mol
New volume = 0.03 dm3 $\therefore$

$\ce{[CH3COOH]}$ = $\frac {1x10^{-3}} {0.03}$ = 0.0333 mol dm-3
$\ce{[CH3COONa]}$ = 0.0333 mol dm-3

$K_a = \frac {\ce{[H+]} {\ce{[CH3COO^{-}]}}} {\ce{[CH3COOH]}} \therefore \ce{[H+]} = \frac{{K_a} {\ce{[CH3COOH]}}} {\ce{[CH3COO-]}} = \frac{{1.74x10^{-5}} * {0.333}} {0.333}$ = 1.74x10-5 mol dm-3
$pH = -log([H^{+}]) = -log(1.74x10^{-5})$ = 4.76

I would imagine that the answer is merely a typo, unless we're both making mistakes.

The final volume is 30 ml as the question states that 10 ml is added to 20 ml, so the concentrations of acid and base are not the same but in the ratio 2/3 to 1/3 of 0.1 mol /dm$^3$ and this produces pH= 4.46. The calculation given only works when the acid /base concentrations are large compared to $\ce{H^+}$ and the equilibrium $K_w=\ce{[H+][OH-]}$ can be ignored.

@porphyrin Hi mate, what you said sounds logical. However, I get taught that a concentration of base and a new concentration of acid must be calculated from the new volume, using the final moles. This makes me unsure.

I agree that a $2:1$ proportion gives the answer, but I don't see why though. The answer proposed calculates final concentrations of acid and base (i.e. it takes into consideration volume change). The amount of moles of base added is half the amount of moles of initial acid, so final moles (thus concentration because they're in the same vessel) will be the same.

Also, as I mentioned in the old post, the concentration of acid and base is a thousand fold the concentration of protons, which I think should be close enough to ignore water dissociation.

@ralk912 It's making me wonder whether the answer doesn't take into consideration new concentrations, due to new moles and a new volume. I wouldn't imagine that the initial moles and concentrations will remain the same even after a reaction.

If that's the case, I would argue the answer is wrong. This is a typical titration of a weak acid with a weak base, and this is the half equivalence point, which corresponds to the $\text{p}K_\text{a}$ of the acid. Exceptions are, of course, deviations of Henderson-Hasselbach's equation, as mentioned by @porphyrin (more here: https://pubs.acs.org/doi/pdf/10.1021/ed078p1499)

It is the concentrations initially present that mater not the number of moles. The total amount of acid is $c_a=\ce{[AH] + [H+] - [OH-]} $ and base $c_b=\ce{[B]  - [H+]  +[OH-]} $ where $c_a$ and $c_b$ are the initial amounts and $\ce{[AH], ,[B]}$ the equilibrium amounts that you put to the equation for $K_A=\ce{[H+][B]/[AH]}$. As the initial concentrations are so large vs $\ce{[H+]}$ or $K_w/\ce{[H+]}$ these can be ignored giving  $\ce{[H+]}=K_Ac_a/c_b$

You do the reaction and end up with a new volume and new moles, so new concentrations have to be calculated. The initial moles/concentrations/volumes can't be used. The 20cm3 of acid and 10cm3 of base are now together at 30cm3. The acid and base have reacted to give new moles of each, so this means new concentrations.

I agree with @KianStevens on this one, the concentrations involved in the $K_\text{a}$ expression are AFTER equilibrium is stablished, not before.


text:
Action of soaps on dirt
In my book,  It is given that the action of Bile salts (a part of Bile juice) is similar to the emulsifying action of soaps on dirt. The role of bile salts is to break down the fat into smaller globules.
I don't understand this and cannot find it on the internet. Can someone help me with this?

The bile salts have an ionic (hydrophilic), region and a hydrophobic region. The hydrophobic part can interact with the hydrophobic fats, whereas the hydrophilic part can interact with water, making the fats soluble in water (they form micelles, with the hydrophobic groups in the middle and the polar groups out.
Soap does exactly the same thing with fats.

What happens in the case of fat and bile salts?

What I just explained. Bile salts have a hydrophobic region (formed by the ring parts of cholic acid, for examole) that interacts with the fats you eat (cause fats have a large hydrophonic region), and a highly hydrophilic region (for example, the sulfonic acid part of taurine, which can be conjugated to cholic acid to form taurocholic acid, this exists as a sodium or potassium salt, i.e. it's ionic and very hydrophilic), that interacts with water.

Can you make it simpler.  I still don't get it.

What part don't you understand?

Correct me if I am wrong.  In the case of bile salts,  the fat is the dirt and salts are the soap molecules

Correct me if I am wrong.  In the case of bile salts,  the fat is the dirt and salts are the soap molecules

Yes. I think the problem might be in what are you thinking about dirt. Soaps clean hydrophobic compounds (usually oils, fats), which we call dirt.


text:
Chemical reaction to create a delayed mark on skin
I'm looking for a dye or chemical reaction that can create a mark on the hand.
Substance A is placed on hand in advance.  When substance B is added, the reaction creates a black mark or stain on the hand.
Ideally, after a delay of a few seconds, a stain or mark appears on the hand. It could possibly be two clear substances that when mixed together create the reaction.
Does anyone have experience with anything that might achieve this effect?  I'm working on this for a visual experiment.

I have a feeling that people would object in strongest terms against you leaving marks on them.


text:
Why does automotive glass shatter into tiny pieces but glass used in our houses into shards?

Image Source : Bigstockphoto

Image source : pinterest
The first picture depicts the glass that is found from an accident site, while the second picture depicts the glass that we use in our home.
What is the difference between these two glasses? Why doesn't the first glass, which is used as a wind shield in vehicles, breaks like the glass that we use in our home?

The first is prestressed, or tempered glass. It is designed just for that purpose: to break in a relatively safe way.

Related: https://chemistry.stackexchange.com/questions/38922/why-laboratory-glassware-shatter-in-big-pieces

BTW, in most countries windshields (windscreens) may not be made of tempered glass, though side and rear windows, because even minor damage causes the glass to suddenly disintegrate.

In the US, glass in exterior doors and showers and some others are  also tempered glass. ( I have swept up 2  windows from doors.)

Let's start out by covering the
Why?
Glasses used in automobiles are designed to break in tiny pieces.
Imagine you're driving down a dark highway hitting dangerously high speeds and then suddenly you lose control and crash.
If the glass used was not to shatter in tiny pieces rest assured you would not remain in one piece. Essentially it shatters for safety reasons.
Household glass on the is generally not involved in such drastic accidents so there is no need for it to shatter into tiny pieces .
How is it done?
From How Stuff Works

Tempered glass is created by heating and then rapidly cooling the glass to room temperature by ushering it through a system of blowers.
The surface of the glass cools much faster than the center of the glass and contracts, causing compressive stresses, while the center of the glass expands because of its temperature, producing tensile stresses. What does that mean? Imagine a piece of glass that could be pulled or stretched to a certain length (tensile stress), while being pushed down and compressed (compressive stress) simultaneously. Both the pulling and pushing stresses achieved through the heating and cooling process give tempered glass its tensile and compressive strength. The differences between these two give the glass 5 to 10 times the amount of strength it originally had.

This system holds a lot of energy due to the stress so when it breaks all of it is released causing the glass to shatter into tiny pieces
Tl;Dr
Automotive glass contains more energy than normal household glass. When it breaks all that energy is released which causes the glass to break into tiny pieces.

Mostly right, but an automotive windshield is also laminated to reduce the risk of injury.


text:
NGP mechanism vs the simple carbocation mechanism
Question:

Predict the product of: ($\ce{C^*}$ is a $\ce{C^14}$ carbon)

My attempt:
I simply assumed the super-leaving group $\ce{-OTs^-}$ would leave, forming a primary ethylbenzene carbocation. That would rearrange via a hydride shift to form a secondary carbocation on $\ce{C^*}$ atom. The nucleophilic $\ce{-OH}$ group of the $\ce{-COOH}$ group will then attack and lose a proton to form 1-phenylethylethanoate (with the ester linkage at the $\ce{C^*}$ carbon):

However, my book has instead taken a different route. In the very first step, after $\ce{OTs-}$ leaves, they have taken the NGP mechanism and formed this transition state:

and said that now the $\ce{OAc-}$ will attack either of the two carbons (in that trigonal ring). Hence, according to their logic, both types of 1-phenylethylethanoate will be formed - one with the ester linkage at the $\ce{C^*}$ carbon, and the other at the $\ce{C}$ carbon.
My question:
I have confirmed from Clayden (Chapter 37 Rearrangements page 976) that the product ratio is indeed 50:50. So:

Why is the NGP mechanism being favored here over the other mechanism I described, even though the NGP mechanism involves a loss of aromaticity?
Are there other cases where the NGP mechanism dominates over the simple carbocation mechanisms, or is this the only such case?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q16 in  Alkyl Halides (Substitution)

At some level, the mechanism must be similar to that described (losing aromaticity etc) as that predicts the observed results. Not the other way around, surely.

Your second question doesn't make much sense. If there was only a unique case there wouldn't be even such a name as NGP. Every time when neighbouring group can delocalise carbocation it does that.

The experimental results dictate the choice of mechanism. The spiro cation is a stabilized cation, even if aromaticity is interrupted and it is less stable than its precursor. Is there an electrophilic aromatic substitution that doesn't interrupt aromaticity?

Hydride shift? Do you really want to form a primary carbocation?

@ApoorvPotnis I mentioned that "That would rearrange via a hydride shift to form a secondary carbocation on C* atom", so I am not really sure why you think I'm forming a primary carbocation instead :/

But you are forming a primary carbocation as an intermediate which is not stable. Or are you proposing the hydride shift and OTs removal in a concerted step?

@ApoorvPotnis Well, obviously, in both of the mechanisms I showed above, $\ce{-OTs-}$ will be removed and a primary carbocation will be formed. This would then rearrange to a secondary carbocation in my mechanism, and form a spiro carbocation in the second mechanism. So, I still don't get what you're saying? (and what is a "concerted step"?)

From Clayden (Chapter 37 Rearrangements page 976):

Intramolecular reactions (including participation of a neighbouring group) that give three-,
  five-, or six-membered rings are usually faster than intermolecular reactions.

It also gives the exact same example as your question:

Notice that both products are given to be exactly 50% each.
NGP is very often faster than bimolecular reactions as for the latter to take place two molecules have to collide first.
1:1 ratio can be attributed to the fact that in NGP the transition state involves loss of aromaticity BUT NGP supports the ionization of the starting material.
There are a lot of examples  this one reaction isn't a black swan. Interested readers may see chapter 36: Participation, rearrangement and fragmentation in Clayden to find more details and examples.

I added a few things. You may wanna delete your 2 bullet points. The data is exactly 50:50 given in book.

I don't understand why NGP is favored even over losing aromaticity. And both of your reasons don't make sense. Can you please give a page number where I can read about "favored by high entropy changes"? And what exactly do you mean by "The ionization of the starting material?

@Gaurang See clayden CH31 pg 806-807 for the ring thing . Ionization of the starting material means ngp doesn't depend upon a collision so it ionizes faster.

Pg 806-807 Ch31 doesn't mention "entropy" anywhere. And $\ce{OTs-}$ is a good leaving group, I doubt if it even needs any collisions to leave.

@Gaurang my south asian edition does

Please quote some relevant details from it into your answer.

@Gaurang That's a pretty big topic

What do you mean by it doesn't need any collisions to leave you gotta break the bond it's easy for it to leave having an NGP speeds it up faster than a collision

Let us continue this discussion in chat.


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How can I attempt to inhibit the hydrolysis/degeneration of aspartame?
I've surfed the internet, looking for some possible inhibitors for the hydrolysis process of aspartame, but couldn't find any. I'm wondering if such a thing is possible and if yes, how should I proceed?
I'm writing an essay about how the rate of degradation of aspartame could be slowed down and wanted to attempt using some compounds to inhibit the reaction.

Hydrolysis under what conditions? In vivo? In aqueous solution?

aqueous solution

Aspartame is hydrolysed in both acid and alkaline solution with the methyl ester being the first site hydrolysed, so control of the pH of the aqueous solution is important. Addition of buffer solutions would enable control of the pH, keeping it close to neutrality.
https://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html


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Minimal basis for Li
Currently, I'm trying to get into the field of quantum chemistry and was reading about minimal basis sets. As I understood this is the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons.
For example, H has one electron, so a minimal basis set would be $\mathrm{1s}$ (STO, STO-nG, or something else).
Li has 3 electrons, so my guess would be that we need 1s and 2s. But in the literature, also the $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$ orbitals are included in the minimal basis. Why is this the case? Together with the spin degrees of freedom, the three electrons already fit in the 1s and $\mathrm{2s}$ orbitals?

A minimal basis set is a set that employs just one function to describe each orbital. For $\ce{H}$ you take into account one orbital ($\mathrm{1s}$), so you use one function, and for $\ce{Li}$ you consider five orbitals ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$, $\mathrm{2p_z}$) and you use five functions, one for each.
What you described ("the minimal number of orbital/spatial wavefunctions needed for a specific number of electrons") would use a function for each occupied orbital - leaving unoccupied orbitals, even in a partially occupied subshell, out of the picture. This is a very bad idea - shells are a set of degenerate solutions for the hydrogen atom, so reducing the number of functions to describe them will seriously distort the depiction of the atom.
The usual approach is, instead, to always provide functions at least for the entire valence shell of the atom (and its inner shells). So for all 2nd period elements, you would use all 2nd shell orbitals ($\mathrm{2s}$, $\mathrm{2p_x}$, $\mathrm{2p_y}$ and $\mathrm{2p_z}$), even in a minimal basis set, no matter how many electrons actually occupy that shell. What still makes that basis set a minimal one is that you are only using one function for each orbital.
Similarly, a minimal basis set calculation for $\ce{Na}$ would include one function for each orbital in the 1st, 2nd and 3rd shells, and so on.

Hi, thanks for this answer! :-) So, a minimal basis set include one function for all possible atomic orbitals for a given principal quantum number, right? The definition would be clear for me, but I find it a bit surprising, that the minimal basis of Li is the same as for Ne. Intuitively, I would guess that Li is much better approximated by this minimal basis then Ne is.

Take into account that the functions will still represent the same orbitals - so both your basis set and the number of orbitals in your output will be the same for $\ce{Li}$ and for $\ce{Ne}$ - what will change is the number of occupied vs. unoccupied orbitals.

Thanks again! But having more unoccupied orbitals increases the "goodness" of the result, or? So, the ratio of unoccupied vs. occoupied orbitals is better for Li than for Ne, meaning that the minimal basis set is a better approximation for Li than for Ne?

Having a larger basis set definitely improves the variational approximation of the results - as more functions can be combined to find a better solution within the functional space defined by the basis set - that's the rationale behind multi-zeta (i.e. non-minimal) basis sets, diffusion functions, etc. So a calculation for $\ce{Li}$ including third shell orbitals would definitely be more accurate than a calculation not involving them. However, it's not immediately obvious to me than the same reasoning can be applied to two calculations involving the same functions but different occupancies.


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What are good naming conventions for chemical properties?
I've noticed (as I'm not a chemist) that some terminology in chemistry seems very confusing to me compared to how I'm used to view symbols and relations in mathematics.
For example, today I was reading about α acids. I have no idea why has someone chosen to name them α acids. But I presume that α is merely "first" in some other collection of acids. However, since acids are not particular to beers, then this seems like an overly general name to give for some beer-specific acidic properties.
So what are good naming conventions in chemistry and what do you think of the α acids for example as a name?

You might want to [edit] your question to focus on a particular set of compounds whose naming conventions you may wish to understand. Otherwise, "what are good naming conventions in chemistry?" on its own is a broad question. (the other part about alpha acids seems fine though)

Nothing's preventing from giving "not that opinionated" answers. That is, if there's a particularly "logical" naming convention, then present it.

There are naming conventions in chemistry. Specifically the IUPAC (International Union of Pure and Applied Chemistry) has defined intricate rules of how to systematically name molecules -- see the Wikipedia article on naming organic compounds. Often, these names can become very long, though, so chemists often use shorter "trivial names". Also, if you want to sum up compound classes, you have to find a way of describing what is common to the class.
I am not sure where the term alpha acids comes from, but I guess the following: It is due to the fact that these acids do not have the normal carboxylic acid group, where the double bonded O and the OH would be attached to the same carbon. Instead, they have an OH-group attached to a carbon that is standing "alpha" to the CO-group. See Wikipedia for an explanation of the term alpha carbon.


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Static electricity with calcium carbonate particles
We are working with small calcium carbonate particles (microfossils) but one of the problems is they stick to the plastic or metal tools we are using, presumably due to static charge. For example, small particles on plastic tray will stick to the surface even if it is upside down.

I'm guessing the static potential is related to how much calcium carbonate wants to gain or give away electronics?
What is this called in this context?
How do I choose a material that has a low potential for static electricity (i.e. sticking) with calcium carbonate particles?

Some links of a commercial equipment to avoid problems of static electricity in powders during weighing:
https://www.mt.com/us/en/home/products/Laboratory_Weighing_Solutions/Accessories/anti_static_devices.html
http://lab.mt.com/antistatic/
https://www.mt.com/dam/mt_ext_files/Editorial/Generic/6/Anti-static_0x000246700004f0770006d93b_files/ion_e.pdf

Hi, answers that contain barely more than links are discouraged, and might get removed.

Markos, can you explain how these products will help and simply use the links as a way to access further information.  Also, if you are affiliated with the company, please disclose that.

My intention was to help by giving some reference.                                  I received a flyer of this company some time ago and I remembered about this product when I read the question.
I also had this type of problem in laboratory when weighing organic fine powders (pigments).
I don't work for this company so I don't know the working principle of this device.
I suggest to ask the manufacturer or other supplier about about the operating principle of the equipment.


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Can inductive effect be the sole driving force for carbocation rearrangement?
I've been taught that a carbocation mainly rearranges because of:

increasing degree (1 to 2, 1 to 3, or 2 to 3)
+M stabilization
ring expansion (in an exceptional case, ring contraction as well).

However, I've never been taught whether the following carbocation A would rearrange:

My professor would say that this carbocation would not rearrange, since at the alpha position we only have two hydrogen atoms, and either of the hydride shifts would still yield a two degree carbocation B:

However I believe that B is stabilized by the strong inductive effect of three extra methyl groups (at the alpha position) that were absent in A. I believe this is a strong enough driving force for rearrangement.
So, can we say A will rearrange into B? Is there experimental evidence to support the fact that A rearranges/does not rearrange into B? Finally, based on these, can inductive effect be the sole driving force for carbocation rearrangement?

What's the source

@AvnishKabaj Source for what? I never claimed A definitely rearranges into B?

The answer to this question is quite difficult to deduce logically. In the first place, the carbocation that is formed is secondary. Rearranging will only change the position of the carbocation, but it will be still secondary. But then there is also a stabilising +I effect.
Instead, if we take into account that that these carbocations are intermediates to form alkenes, a conclusion can be drawn. So, the best way is to consider the thermodynamic favour of the rearrangements.
Prior to rearrangement, the carbocation is secondary. If you consider that this carbocation is going to form an alkene, the major product from it (Saytzeff product) will be  4-methyl pent-2-ene, which is of the form $\ce{RCH=CHR}$. After rearrangement, this will be still a secondary carbocation, so there is a significant energy required for this hydride shift. This energy is not really compensated by the +I effect of the isopropyl group. But if we consider that this carbocation is an intermediate for the alkene formation, the major alkene (Saytzeff product) will be 2-methyl pent-2-ene, which is of the form $\ce{R2C=CHR}$, which is thermodynamically more favourable.
So it is seen that in the rearrangement, the energy of the intermediate is increased, but the overall energy of the product is decreased. So, I think that, at higher temperatures, this rearrangement can be possible, because the ultimate product (if we think this as an alkene) is thermodynamically favourable, while at normal temperature this rearrangement is energetically restricted a bit.

"the energy of the intermediate is increased" but isn't the carbocation B more stable? So, its energy should be lower?

The energy should be higher because the strength of $+I$ effect in this case is not so strong as you are thinking. The Hydride shift requires much more energy, than $+I$ can stabilise the system.

I'm not going in the direction of whether the hydride shift requires higher energy or lower energy to occur. Instead, please think of it in this way: whether a carbocation has higher or lower energy is solely decided by its stability. Do you agree? if yes, then why do you say that "the energy of the intermediate is increased"? Do you disagree that the carbocation B is more stable?

How do you define stability ? Do you have any other way to define stability rather than thermodynamic stability ?

I am defining stability in carbocation to be due to aromaticity, resonance, or inductive effects. The first two are missing in both A and B, so I am forced to check inductive effects.

Do u agree that if all of your described effects are present, the potential energy of the compound will be lowered ?

Do u also agree that carbocation B is formed from A ?

Yes, I agree with both your comments.

Then do u agree that change in the potential energy from A to B  is combinational effect of HYDRIDE SHIFT  and inductive stability , because conversion from A to B includes these two effects  ?

I get it, I accept it, going by your logic, we may find A to have a lower potential energy. Fine. However, suppose that you were given these two carbocations separately and asked to judge their relative stability (aka relative potential energy)? Wouldn't you then say that B is more stable and consequently has a lower PE?

Yes, then B is. But your question was also considering that rearrangement. Isn't it ? . We then should consider the overall effect.

Exactly. And now we have arrived at a deadlock. Going by one route (hydride shift+inductive effect) we claim that B has higher energy. But, going by the simple route of checking only inductive effects (like we would usually do if rearrangement wasn't the case), B has actually lower energy.

I think that without any experimental data, it is no possible to decide. Personally, I don't think the inductive effect differences between the two ($\ce{CH3}$ and $\text{iBu}$ vs. $\text{Et}$ and $\text{iPr}$) and are large enough so that the energy gain from the stabilization of the carbocation is enough to overcome the energy barrier of the hydride shift.
From Hammett parameters , the $\ce{CH3}$ would actually be more electron donating than the $\ce{iPr}$ group, and since $\text{Pr}$ and $\text{Et}$ have the same $\sigma_\text{p}$ value, I wouldn't expect a difference between $\text{iBu}$ and $\text{Et}$ either. Of course, Hammett parameters aren't necessary a direct measurement of inductive effect, and these are not aromatic rings, but I think it paints a picture of how the inductive effect difference isn't that large.
Overall, I wouldn't say that carbocation B will never form from carbocation A, but it certainly will depend on conditions. My answer in an academic situation would be that it will not form, since the stabilization effect, if any, is minimal.
Edit: Upon more thinking, the Taft polar parameter, $\sigma^*$, might be a better representation of this case. These actually show a higher electron donating ability for both $\text{iPr}$ vs. $\text{iBu}$, and $\text{Et}$ vs. $\text{Me}$, respectively. These are also empirical values, and not necessarily the same as we would observe, but again, they can give an idea. The differences are so small that I would stand on arguing towards the side that the energy barrier for the shift is large enough that it would be minimal for stabilization by shifting to be favored. Again, it would definitely depend on conditions.

Thanks for your interesting answer! :) Actually, I was basing my assumption that longer alkyl chains have a larger +I effect based on this question. Could you review it and see if it's something what you agree/disagree with?


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Why is methamphetamine less polar then amphetamine?
The question is as stated in the title. Is there a general way to determine relative polarity of organic compounds?
This question is for use in the context of the manner in which polarity of the components of the mobile phase in high-performance liquid chromatography determine the polarity of the stationary phase.

Hydrocarbons are not polar. Some other organic compounds are, though.

Oh is there a method to determine which organic compounds are going to be more or less polar? I'm trying to reconcile an answer in the back of my workbook with my pretty limited understanding at the moment.

Yes, anything from quantum chemical calculations (which you might not understand) to pretty simplistic rules which I would argue are sufficient in this case ( something something electronegativity). You should maybe consider starting at simpler molecules, you know, at the beginning.

Yea, I'm asking about this as a year 12 chemistry student and am fairly certain we have not encountered quantum chemical calculations yet (but they do sound fascinating). Do you have any examples of simpler molecules for which I could apply a fairly defined process to, in order to determine their relative polarities?

@Kcronix I don't think the comment section below your question is the place to discuss this. Have you considered a google search?

Yes I have, I was wondering if anyone could summarise the process for the specific case mentioned

You took a pair in which is particularly difficult to offer a simple rule as for the two are relatively similar and not particularly polar. Not a classic example...


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What is the difference between ψ, |ψ|², radial probability, and radial distribution of electrons?
I am very confused regarding these four terms- ψ,|ψ|², radial probability and radial distribution
I know that ψ is called the wave function but is it the same as radial probability?
I also know that |ψ|² is called the probability density of finding an electron inside an atom but is it the same as radial distribution—which is the probability of finding an electron in a volume $\mathrm dV$ inside a nucleus?

You ask way too many questions at once. Radial probability is not a thing at all, and the rest are all different thing.

All the four terms are different and they represent different concepts in quantum mechanics.Firstly, the term $\Psi$ represents wave function of a particle which is distributed in a three dimensional space. This wave function is a function of four coordinates ($x$, $y$, $z$, and $t$), and it gives the values which are in complex-space. For a typical example, $$\Psi(x,y,z,t) = \sqrt{\frac{8}{abc}} \sin\left(\frac{n_x\pi x}{a}\right) \sin\left(\frac{n_y\pi y}{b}\right) \sin\left(\frac{n_z\pi z}{c}\right) e^{-2\pi iEt/h}$$ is an example of a wave-function.
But the $|\Psi|^2$ is mathematically defined as $\Psi\cdot\Psi^*$. Max Born interpreted the value of this real valued function as the probability of finding the particle in 3 dimensional space. If you consider the previous example then $$|\Psi(x,y,z,t)|^2 = \frac{8}{abc} \sin^2\left(\frac{n_x\pi x}{a}\right) \sin^2\left(\frac{n_y\pi y}{b}\right) \sin^2\left(\frac{n_z\pi z}{c}\right)$$ Here the complex part will not appear as in the previous example because $\Psi$ is multiplied with its complex conjugate.
The probability of finding the particle in a unit volume element $\mathrm dV$ is $|\Psi|^2 \mathrm dV$. In spherical polar coordinates, it is $|\Psi|^2 r^2 \, \mathrm dr \, \sin\theta \, \mathrm d\theta \, \mathrm d\phi$. When you are only concerned about the radial part, the polar angular integral and azimuthal angular integral are replaced by $4\pi$ as, $\int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta \, \mathrm{d}\theta \, \mathrm{d}\phi =4\pi$. thus we are left with only the radial part which is your radial probability distribution function. $$\Pr(r) = |\Psi|^2 4 \pi r^2 \, \mathrm dr$$ The radial probability can be thought as the probability of finding the particle within an interval of length $\text{d}r$ at $r=r_0$. So, the radial distribution is a function but the radial probability as described can be calculated by integrating that function from $0$ to $r_0$.

Nice answer. Just would change 'When you are only concerned with radial part' by 'if the radial part of the wave function is separable'

According to the Copenhagen interpretation of quantum mechanics,
$|\Psi|^2$ is the "probability density" (the probability per volume of finding a particle, such as an electron, in a given volume, in the limit the volume approaches zero).
If $\Psi(r,\theta,\phi)$ is separable as $R(r)Y(\theta,\phi)$, such as in the hydrogen atom, then:
"Radial Probability Density" is $R^2(r)$ (still proportional to probability per volume for s-states)
"Radial Probability Distribution" is  $4\pi r^2R^2(r)$ (probability per incremental distance from origin, as the incremental distance approaches zero)

Actually Born's rule doesn't depend on Copenhagen interpretation. In MWI, Bohm-de Broglie theory and any other interpretation it's still valid, because it is a testable rule, unlike internals of the interpretations.

@Ruslan I don't think there is complete agreement on that, according to this: https://en.wikipedia.org/wiki/Quantum_non-equilibrium

I'd say the wave function is your $ψ$. This wavefunction describes your system. When you want to determine something like its energy or other operations you need to describe your system like your atom using a wave function. $|ψ|²$ is as you already said the probability density.
Your wavefunction for example in solving the hydrogen atom using the Schrödinger-equation has this step where you transform to spherical coordinates and then seperate into a radial part and an angular part. From this radial part you can get the radial probability, so to find an electron for example at the distance x. And the radial distribution is this function times the area of a sphere having the size of the radius for that distance you are looking at the moment. So this will give you the probability for the electron to be located somewhere on a sphere at the distance x.
But I can't really tell you the actual difference between taking only the radial part or the wave function itself to describe probability.


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Estimation of change in enthalpy of a gas phase reaction using bond enthalpies
Estimate ∆H for the following gas phase reaction:
$$\ce{C2H4 + 3 O2 ⟶ 2 CO2 + 2H2O}$$
I have been having trouble on this problem for a long time, and I was hoping someone would help me with it. I tried using the average bond enthalpies, but that always got me the wrong answer. Below is a table of average bond enthalpies in kJ/mol.

Hi Smith!  welcome to Chem.SE! We require you to show your efforts on this problem. What formulae/steps did you try? Please add this to your question. Thanks!

Hi John, you said that "I tried using the average bond enthalpies, but that always got me the wrong answer." Do note that telling us what
exactly you did with the bond enthalpies will help us help you much better. Thanks!

How do you know the result is wrong and not just imprecise? See, the word "estimate" (rather than "calculate") is there for a reason.

Use Hess' Law:  $\Delta$H = $\Sigma$ Enthalpy of Bonds Broken - $\Sigma$ Enthalpy of Bonds Formed
Reactants

Bond enthalpy of C2H4 = 602 + (413 * 4) = 2254 kJ mol-1
Bond enthalpy of O2 = 3 * 498 = 1494 kJ mol-1
$\therefore$ $\Sigma$ Enthalpy of Bonds Broken= 2254 + 1494 = 3748 kJ mol-1

Products

Bond enthalpy of CO2 = (799 * 2) * 2 = 3196 kJ mol-1
Bond enthalpy of H2O = (463 * 2) * 2 = 1852 kJ mol-1
$\therefore \Sigma$ Enthalpy of Bonds Formed = 3196 + 1852 = 5048 kJ mol-1

Overall

$\Delta$H = 3748 - 5048 = -1300 kJ mol-1

After having a quick Google search for the enthalpy change of combustion for ethene, it came up as -1411 kJ mol-1. Our calculated answer isn't far off. I'm hoping we got the same answer.


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Finding molar mass of naphthalene using the freezing point of a solution?

Naphthalene is the main ingredient in some mothballs. The freezing
  point of a solution made by dissolving $\pu{7.01 g}$ of naphthalene in $\pu{200 g}$
  of benzene is $4.20 \pu{^\circ C}$ . What is the molar mass of naphthalene?

I started by using $T_f= K_f\times m $ (we're given the $K_f$ of benzene is $5.12 \space{} \pu{  ^\circ C/m}$).
$4.20~ \pu{^\circ C}=(5.12 \pu{^\circ C/m})(\pu{m})$
$m=0.8203125$ molal.
molality = moles of solute per kg of solution
$\implies 0.8203125$ molal= (unknown moles)/(kg solvent)
Am I supposed to use kg of total solvent or kg of naphthalene? I can't seem to figure out where to go next.

molality = moles of solute per kg of solution

This is incorrect. Molality instead is moles of solute per kg of solvent. In your case, do you know which one of nepthalene or benzene is the solvent? Whichever it is, use its mass in the formula's denominator.
Moreover, your usage of the formula, along the line "I started by using", is incorrect. The LHS should be $\Delta T_\mathrm f~(=T_\mathrm {f2}-T_\mathrm {f1})$ and not $T_\mathrm f$. The question is incomplete because it should have also mentioned the initial freezing point of pure benzene solvent.

Should not the boiling point of benzene (solvent) be affected differently since naphthalene is also volatile?


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Difference between basicity and reducing character
My understanding of basicity and reducing character:

Reducing character is the ability of a substance to reduce something else. So it itself must get oxidized. Since $\ce{BiH3}$ has a large radius, it will loose hydrogen atoms easily. Hence I conclude $\ce{BiH3}$ has more reducing character than $\ce{NH3}$.
Basicity is how easily can the compound give up hydrogen atoms/electrons.

Now I end up thinking basicity and reducing character are the same thing. And that $\ce{BiH3}$ should have more basicity than $\ce{NH3}$. But, this is not the corrrect answer.
So what's the difference between reducing character and basicity?
On mostly all Google searches I find that reducing character is directly proportional to basicity. But that doesn't say why they are different in the first place.

They have, like, nothing in common, unless you wanna talk about obscure Usanovich's theory.

@Mithoron That's the thing I don't understand. Why aren't they the same. I write why I think they are the same in the question.

Well, then it sounds like Usanovich, and also why his theory wasn't successful. Reasoning in vein of BiH3 barely holds to its protons and looses them in oxidation would be much better on your part.

Basicity is how easily can the compound give up hydrogen atoms.

Isn't this the definition for acidity instead? (assuming your "hydrogen atoms" = proton) Basicity instead refers to how easily a compound can accept protons on its lone pair, not give them up.
$\ce{BiH3}$ is a weak base, because the lone pair on bismuth is dispersed over a large surface area, as compared to that on the nitrogen in ammonia. (details here) Being diffused, the lone pair is unable to attack a proton, and hence, is a weak base.
$\ce{BiH3}$ is a better reducing agent, because on getting oxidised (losing a hydrogen), the negative charge formed on the bismuth atom is dispersed over a much larger surface area. Hence, the negative charge is stabilized much more on the bismuth atom than it would be stabilized on the nitrogen atom in ammonia.
Notice that the dispersion of the lone pair makes it more diffused, hence, less basic. While dispersion of the excess negative charge stabilizes the specifies formed after loss of a hydrogen, hence, a better reducing agent.

+1. I think the key word is polarizability (when you talk about surface area)..

@santimirandarp Thanks for your comment, but I don't get it exactly. Are you talking about the polarizability of the bismuth and the nitrogen atoms? How exactly does that fit in here? (I am more used to seeing polarizability in applications of Fajan's rule. So I may be missing something)

if the negative charge lies on the bismuth, then how come the hydrogen is responsible for the reducing character (if it doesnt even have the electrons)?

Your confusion seem to stem from the fact that getting hydrogen is reduction (so losing hydrogen is oxidation) and getting electrons is reduction (so losing electrons being oxidation) as well. So you must have concluded that greater reducing character means greater basicity as basicity is ability to donate electrons.
But here we are considering hydrides of pnictogens that involves lone pair of electrons as well as hydrogens.
$\ce {NH3}$ is most basic as it's lone pair of electrons is fairly well localised in the 2s orbital which is very small in size and thus could concentrate the charge density into a tiny volume. Such a high charge density of lone pair makes it quite easy to be donated and hence making it most basic. But on the other hand, $\ce {NH3}$ has the least reducing character in group 15 because nitrogen's p orbitals are also compact due to it's small size. This compact size means p orbitals are localised fairly well too and the H orbitals will very well overlap with that of N to form a very strong bond in $\ce {NH3}$. This high bond strength makes it difficult to lose H and hence $\ce {NH3}$ being least reductive.
Going to bismuth, as elucidated in the other answer, it has such a large size that it's lone pair is too diffused to be concentrated in a suitable volume for donation which makes it least basic. But the same factor of size and diffused orbitals make the overlap of its orbitals with that of H orbitals quite poor. So $\ce {BiH3}$ can easily lose hydrogen and hence has the greatest reducing character.


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Crystal structure database search for space groups
I'm usually using databases like the Springer Materials site or the free Crystallography Open Database to get crystal structure data. But none of these offers the possibility to search for all entries with a defined space group. I recently prepared a new compound and solved and refined the crystal structure. I would like to compare it to compounds known by literature to give the structure a name. It's quite common to compare structures to already existing ones. But I can't find any website that would give me the possibility to look for all entries with the same space group without knowing the elements.
Does anyone know a search engine for this specific problem?

A good librarian at a university library would be a good start.

You could use COD SQL interface. For example this little Python script returns all entries in space group F-43m that contain Si:
#!/usr/bin/env python
from mysql import connector
con = connector.connect(user='cod_reader', host='sql.crystallography.net',
                        database='cod')
cursor = con.cursor()
cursor.execute("""select file, formula from data
                  where formula LIKE '% Si%' AND sg = 'F -4 3 m'""")
for n, formula in cursor:
    print n, formula

The script prints entry number and formula for each matching COD entry:
$ python cod.py
1010995 - Si C -
1011031 - Si C -
1011254 - Al13 Cl F9 H9 O29 Si5 -
9000281 - Al13 Cl F3.84 H14.16 O34.16 Si5 -
...

Oh wow, thank you for that script. I will try it once I'm home today.