text:
Why is the melting point of PCl3 less than that of PCl5?
From just the IB chemistry bonding chapter information, you would normally deduce that $\ce{PCl3}$ and $\ce{PCl5}$ are both covalent molecules.
However, as $\ce{PCl3}$ has a dipole moment, it has dipole-dipole intermolecular attractions, implying that  $\ce{PCl3}$ has a higher melting point than $\ce{PCl5}$, which only has London dispersion forces.
So why is it that $\ce{PCl5}$ has a higher melting point than $\ce{PCl3}$ (I saw online that  $\ce{PCl5}$ is an ionic solid)?

We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details.

Maybe it is related to energy needed to break the bonds in case of both PCl3 and PCl5. See: https://chemistry.stackexchange.com/questions/40702/is-the-p-cl-bond-energy-same-in-pcl5-and-pcl3

What is "IB"? International Baccalaureate?

Yes @PeterMortensen.

Because $\ce{PCl5}$ does something which is not immediately obvious from its molecular formula: it autoionizes and becomes an ionic solid $\ce{PCl4+PCl6-}$. As such, it has much stronger interactions than $\ce{PCl3}$ with its mere dipole-dipole attractions, hence the higher melting point.
If not for that fact, you deduction should have worked just fine.

Is there any mathematical way or just method to deduce that $\ce {PCl5}$ is an ionic solid? Or is that from observation.

No, I don't think this can be deduced via logic alone. As for more mathematical way, in chemistry this usually implies that you are going to end up in a nocturnal abyss of quantum chemistry, which is not what you want, believe me.


text:
Acidity of ethers vs alkanes
I wonder if ethers are more or less acidic at the alpha position than alkanes ($\mathrm{p}K_\mathrm{a} \approx 50$). I'm not sure if the stabilising inductive effect from oxygen or the alpha effect (the repulsion between lone pairs on the carbanion and the adjacent oxygen) will dominate.
Based on $\ce{H2O}$ ($\mathrm{p}K_\mathrm{a} \approx 16$) vs $\ce{H2O2}$ ($\mathrm{p}K_\mathrm{a} \approx 11.5$), I would expect the inductive effect to be stronger, but I would like to check.

A good rule of thumb is: everything is more acidic than alkanes.

@IvanNeretin Thanks! So you would say that the inductive effect dominates over the alpha effect in this case? Is this a general case?

Yes, I think so.

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Deprotection (debenzoylation) of glucose (sacharide) using NaOMe leads to hydrolysis glucose
I have a problem removing the benzoyl groups of my glucose moiety using sodium methylate (standard procedure). Indeed, the benzoyl groups are removed. However, a significant amount of glucose gets hydrolysed leaving an alcohol group. I have tried different equivalents of sodium methylate and also other bases like potassium carbonate and ammonia. However, in all cases hydrolysis takes place.

Does anyone have an idea why hydrolysis takes place? Could it have something to do with the TMSOTf (strong acid) in the previous step? The only workup I did for compound 1 was quenching the reaction using the same molar amount of TEA as TMSOTf, then evaporating $\ce{CHCl3}$, then silica column chromatography ($\ce{CHCl3 : MeOH}$ in $10 : 1$, see picture). Is it possible that TMSOTf is still present in my compound?
Column of compound 1:

The mildest conditions I know of for ester hydrolysis is LiOH in 19/1 THF/water at room temp. Could be worth trying.

I very much doubt TMSOTf would survive the presence of the MeOH in your eluent

What's the water content of the methanol?

How do you know that the first step was complete. Was the benzoylglucoside pure?

MeOH is of AR quality.

The benzoylglucoside was pure as shown by MALDI and NMR.

Et3NTMS+ OTf- may still be a silylating agent that removes your fatty alohol in the CHCl3 evaporation (moist) or on silica gel.

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text:
Entropy at zero kelvin for pure atom crystal
The third law of thermodynamics describes that the entropy of pure ideal crystaline substance is zero.
Now we know that in practice this could be wrong for water (for example) because of its different crystalization states with nearly the same energy.
(Book: Atkins, Physical Chemistry 4th, German Translation, page 99)
But: Is the entropy always zero at zero kelvin for ideal crystalised and pure elements? Because there shouldn't be a difference in position of atoms.

Maybe someone can give the literature notice to the answer, cause I couldn't find any.

Are you referring to different phases of ice?https://en.wikipedia.org/wiki/Ice#Phases?

I'm refering the an atkins text parts (physical chemistry book), which describes that water can have different crystalization states cause of different bond length configuration at zero kelvin. And therefore an entropy greater than zero.

Could you please add that to the question too? Would clarify things a bit...

For a perfect sample of  crystalline Argon, for example,  one would expect entropy to be zero at zero, but not for CO where the orientation can be CO-CO-CO or CO-OC-CO etc so the experimental entropy is not zero for easily explained reasons. Water will be even more complicated.

That is what I thought of, but just don't know. :-) Do you have a literature source for this statement? Althought it seem to be right!

If you want to have it "entropy-free", you need isotopically pure crystals. Although the difference is hard to find, chemically. The size of the crystal could still be regarded as entropic, at least if you have more than one.

You seem to underline some perceived inherent difference between water and elemental compounds. There is none. A crystal of water ice, so to put it, is as crystalline as any other crystal.
There are quite a few crystals (indeed, certain elemental compounds are among them) that exhibit some kind of disorder; as you lower the temperature, their molecules (or atoms, or ions) freeze in somewhat different positions having nearly the same energy, so if you extrapolate to zero, you'll see that they retain some significant entropy even then. This has nothing to do with the Third law. These crystals are not ideal by a wide margin, and they are not going to become more ideal anytime soon, especially if you keep them near the absolute zero.
On a side note, the words "practice" and "at 0 K" do not sit well together, but that's another story.

So for a conclusion: Ideal crystal always has an entropy of zero at zero kelvin. Cause when they do not have a value of zero, these crystals are far away from being ideal.
But what about pure element crystals now?

You are still relying on a distinction which is simply not there. Pure element crystals are no different from any other crystals.


text:
How do we calculate the partition coefficient (including titration)
In a separating funnel we add $\pu{25.0 mL}$ of $\pu{8.5 M}$ $\ce{HCl}$ solution. We also add $\pu{25.0 mL}$ of a tributyl phosphate solution. We shake the funnel for 1 minute and then after the two phases have separated we remove the lower phase.
We add in the same separating funnel $\pu{23.0 mL}$ of the same $\pu{8.5 M}$ $\ce{HCl}$ solution and we also add $\pu{2.00 mL}$ of $\pu{0.5 M}$ $\ce{ZnCl2}$ solution. We shake the funnel again and we remove the lower phase. Now the biggest concentration of zinc is in the organic phase.
We add in the separating funnel $\pu{20.0 mL}$ of deionized water. We shake the funnel and now the zinc comes back again the aqueuous phase. After the two phases have separated, we transfer the aqueous phase to a volumetric flask. We repeat the previous step 2 more times and we collect the aqueuous phase in the same volumetric flask. In a conical flask, we add $\pu{25.00 mL}$ of the volumetric flask, we add $\pu{75 mL}$ of deionized water and $\pu{3 mL}$ of a buffer and also drops of an indicator. We titrate this with EDTA solution.
Now we want to calculate the partition coefficient using these formulas;
$D_\ce{Zn}$ is the distribution coefficient.
$C_\ce{{Zn}_{org}}$ and $C_\ce{{Zn}_{aq}}$ are the concentrations of zinc in the organic and aqueous phases respectively.
$D_\ce{Zn}=\dfrac{C_\ce{{Zn}_{org}}}{C_\ce{{Zn}_{aq}}}$
If $n$ denotes amount, then
$n_\ce{{Zn}_{org}}=\pu{0.0100 mmol//mL}\times V_\mathrm{EDTA}\times 4$
$n_\ce{{Zn}_{total}}=\pu{0.0100 mmol//mL}\times V_\mathrm{{EDTA}_{total}}\times 4$
$n_\ce{{Zn}_{aq}}=n_\ce{{Zn}_{total}}-n_\ce{{Zn}_{org}}$
I get confused about what the difference between the $V_\mathrm{EDTA}$ and the $V_\mathrm{{EDTA}_{total}}$ is.

Some formatting would certainly help.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

I briefly edited the experimental part, but I'm having a hard time trying to parse the formulas. If you could rewrite them using generally accepted notations and, ideally, MathJax, that would help a lot.

What a hopelessly convoluted way to determine a $D$! Good luck with that...


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Stability of canonical forms

Which of the following ions is more stable? Use resonance to explain your answer.

When we have to compare stability of resonance structures we generally check for the following features in the resonance forms (in sequence):

Neutral molecule;
More covalent bonds;
Complete octet of all the atoms;
Less separation of opposite charges;
Positive charge (if any) on more electro positive atom and negative charge (if any) on more electronegative atom.

In the above problem:

Both the molecules are charged;
Both have same nuber of covalent bonds;
Both have the octet of one carbon atom incomplete (the positively charged one);
Both have positive charges on carbon atoms.

From all of these considerations the molecules should be equally stable. But the answer given in my book is (A).
What is that I can't figure out?

For the ions to be equally stable, they would have to be identical. Here, you have a primary allyl carbocation vs. a secondary allyl carbocation. I also didn't get how (A) is more stable though...

One way of reasoning that would lead to determining A as more stable is to ignore the charge and take the cue from the double bond, which is more substituted in A. I also seem to recall that an endo double bond (as in ring system) is more stable than an exo double bond. However, these lines of reasoning are flawed since the two structures are in resonance anyway and, as pointed out by Eashaan Godbole, the primary allyl carbocation of A is less stable than B.

The reason given in the solutions of my assignment is planarity. It says "structure B is not planar and hence conjugation is not possible in it".

But still I'm not having a clear cut picture of what actually is happening, How exactly the ring B isn't planar?

"Use resonance "in the question does not make sense

Throw away the book, if it doesn't get this right, there are more distorted and wrong views about chemistry in it.

Throw away the book, if it doesn't get this right, there are more distorted and wrong views about chemistry in it.

The 2 structures are completely identical. You should burn the book.

Does this answer your question? How to determine the relative contribution of resonance structures when different rules give contradictory outcomes?

It has been noted in the comments and I shall repeat that the premise of the question is already bogus. The question does not consider two different ions but two mesomeric depiction of the same ion. This is also evident by the mesomeric arrow between the two forms. Since mesomery strictly means that no geometrical change takes place, both depictions are identical in all aspects and have the same (resonance-influenced) energy. Thus, any book claiming that either of the two have a lower energy is wrong.
However, what is commonly done is to determine a more likely resonance structure. This is, for example, commonly used to explain the strongly stabilised nature of certain aromatic systems, e.g. when DMAP (4-N,N′-dimethylaminopyridine) is used as a carboxylic activating agent. We can attempt to do that for your structure, too. If we compare the two we see — of course — that a lot of features are identical. The principal differences are:

A has a trisubstituted double bond while B only as a disubstituted (and terminal) double bond
B has a secondary carbocation while A has a primary one (both, however, are allylic).

Since allylic stabilisation for cations is much stronger than hyperconjugation, we can discard the second bullet point as being essentially equal for both resonance structures. Thus, we should argue with the location of the double bond. Both double bonds can benefit from hyperconjugation twice (the primary cation of A cannot hyperconjugate due to geometry), but in A the two hyperconjugations stabilise different carbons. Along with the general double bond stability given as:
$$\text{monosubstituted} < \text{terminal disubstituted} < \textit{cis} < \textit{trans} < \text{trisubstituted} < \text{tetrasubstituted}$$
we can identify A as being slightly better than B and therefore contributing slightly more towards the overall structure.

You mentioned in the comments that your book states:

structure B is not planar and hence conjugation is not possible in it

Your book probably wants you to assume that the following depiction in figure 1 is adequate for the two resonance forms:

Figure 1: structures B and A under the assumption of a chair-configured cyclohexyl structure for B.
Under the underlying assumption of a different geometry for both structures, this may be valid. However, there is no way that cation B will adopt a chair cyclohexane configuration. Instead, as I mentioned above, the following structures in figure 2 are much more likely and even show that we are dealing with resonance:

Figure 2: depiction of the mesomeric forms without geometry change.
You can easily see that the argument of non-conjugation is bogus, since there very well is a way to optimise geometry so that the orbitals may form an allyl system.

The question got mangled when it was copied and posted on the internet. It should read (I recreated the missing image):

Which structure, I or II, is more stable? Use resonance in your answer.

And of course it is the allylic carbocation, I, as explained in the answer key that is available on the internet (see e.g. Comparing stability of two canonical structures).
Here is an image of the original question (part d):

And here another one:

Makes more sense now! Where did you find them?

Google “for the following pair of ions which one is more stable”. @user55119

Or “Which of the following ions is more stable? Use resonance to explain your answer.”

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IUPAC name for [CoSO4(NH3)4]NO3
What is the correct IUPAC nomenclature for the coordination compound $\ce{[CoSO4(NH3)4]NO3}$?

I significantly edited your question and removed unnecessary parts (I hope this is OK); still, this looks like a homework-type question. Please include your ideas or suggestions to avoid closing.

Thank you so much. This wasn't a homework question. I was stuck on  something, thus I thought of getting it clarified, but soon I figured out my mistake. So I AM actually trying to remove this question. Apologies for that, as I am a newbie here. So it's taking a bit time to figure out how does this work :)

Sure, no prob. It's just the homework policy which is applied to the questions where no attempt has been shown. Others are not supposed to answer these questions, and they are typically end up closed. If you want, you can [edit] your thoughts in (or just the name of the compound you think is correct), so that others can actually help you with this:)

@Advdtii and if you have figured out the issue, you can include an answer to your own post to help future readers.

The answer I came up with is tetraamminesulphatocobalt(III) nitrate .

It's "sulfato" and not "sulphato".

@Loong Actually "sulfate" is preferred by IUPAC and "sulphate" is from British English. (Courtesy: wikipedia)


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Reason for Optical Isomerism (at electronic level)
I have been studying Optical Isomerism recently, and I have got one question, answer to which was not in the books that I have.
I have understood what the phenomenon is, and that, how one can determine whether the plane of polarised light is rotated, and how to tell from the structure whether the molecule will rotate the plane of polarised light or not.
However why does the phenomenon occur? If I go to the atomic level or the quantum mechanical level, it is surely a result of interaction of the polarised light with the electrons in the atomic orbitals and the molecular orbitals. Can anyone explain how this interaction occurs? That is, how is the plane of polarised light bent by the electron - photon interaction (if any)?

https://physics.stackexchange.com/questions/15503/how-do-optically-active-compounds-rotate-plane-polarized-light might be useful

or https://chemistry.stackexchange.com/questions/6271/molecular-chirality-and-optical-rotation


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Why the colour of KMnO4 at the end point of titration may disappear after some time?
In the redox titration of $\ce{FeSO4}$ with $\ce{KMnO4}$, the colour change of the solution at the end point is colourless to light pink. Most laboratory manuals and books advice us to check whether the light pink colour stays or not for 30 seconds. If the colour stays after 30 seconds, then end point is reached. If the colour disappears before 30 seconds, then the titration is not finished, so more $\ce{KMnO4}$ should be added.
As far as I understand, at the end point, all the reducing agent is depleted. So, the colour of $\ce{KMnO4}$ stays. And, if I have interpreted my book correctly, the colour of $\ce{KMnO4}$ is to stay for more than 30 seconds so that we can say that the end point is reached.
This means, that the colour of $\ce{KMnO4}$ might disappear after 30 seconds even though the end point is reached. But why? All reducing agents are depleted. Aren't they?
Now, I have found some reactions which might possibly cause the discharge of colour after the end point:
\begin{align}
\ce{3 MnSO4 + 2 KMnO4 + 2 H2O &-> 5 MnO2 + K2SO4 + 2 H2SO4}\\
\ce{3 MnSO4 + 2 KMnO4 + 8 H2SO4 &-> 5 Mn(SO4)2 + K2SO4 + 8 H2O}\\
\ce{4 KMnO4 + 2 H2O &-> 4 MnO2 + 4 KOH + 3 O2}
\end{align}
I am not sure which of them occur in reality.
So, which of the reactions are the reasons for the 30 second rule?
And why is it 30 second anyway? Why not 40 second or 1 minute?

I think the authors just wanted to provide a reasonable time span for the equilibrium to be established. The exact number has been most likely cherry-picked, or averaged from several experiments at best.

@andselisk I thought that as well.

The most likely reason for the colour of permanganate to vanish after a few seconds even though you thought you have the end point is: you have not reached the end point.
Acid/base reactions are typically extremely fast reactions; their rate is not even limited by the speed of diffusion since protons can be shuffled along via the Grotthuss mechanism much more rapidly. Thus, you can assume each acid/base reaction to instantly reach completion.
With practically all other reactions that is not the case. They are typically slower and can be much slower. The reacting species must meet each other in solution which can take time. At the beginning of the titration, there are still many iron(II) ions floating around ready to be oxidised by any incoming permanganate. Towards the end, iron(II) is getting more scarce. So a permanganate ion may linger around much longer before it meets an iron ion. You should give it a certain time in case it does find one. The value of $\pu{30s}$ is arbitrary. Ten seconds or two minutes would probably do equally well (except that ten seconds may or may not be slightly on the short side). The book/method just needs to present any ballpark value.
So the reaction occuring that causes the discolouring of the solution is indeed exactly the reaction you are observing: the oxidation of iron(II). It is merely slowed down so it can take a few seconds to proceed to completion when close to the equivalence point.

In redox titrations, we titrate a substance with a reducing nature against a substance with an oxidising nature.

$\ce{KMnO_4}$ is an oxidant. It reacts and reduces the other substance.
The colour of manganese compounds largely depends on the oxidation state. For instance, $\ce{K\overset{+VII}{Mn}O4}$ is intensely purple while $\ce{\overset{+II}{Mn}SO4}$ is pale pink.

The reactions mentioned in the question involve reaction of products of $\ce{KMnO4}$ with $\ce{KMnO4}$ and the third one is the decomposition(?) of $\ce{KMnO4}$ in water, which looks absurd.
While titrating with oxalic acid or $\ce{FeSO4}$, we typically dilute the samples, add some dil. $\ce{H2SO4}$ and heat the mixture.
The reaction which most likely takes place in acidic media involves the conversion of $\ce{\overset{+VII}{Mn}O4^-}$ to $\ce{\overset{+II}{Mn}^{2+}}$. The half-reaction for this is:
$$\ce{MnO_4^- +8H^+ +5e^-\to Mn^{2+} +4H2O}$$
Oxalic acid is oxidised to carbon dioxide as follows:
$$\ce{C2H2O4 \to 2CO2 + 2H^+ + 2e^-}$$
$\ce{FeSO4}$, i.e. $\ce{Fe^{2+}}$ is oxidised to $\ce{Fe^{3+}}$ as:
$$\ce{Fe^{2+} \to Fe^{3+} + e^-}$$
Combining these ionic equations and adding the right ions, we get the complete balanced equations.
We can now see that -

Dilution - So that the "pinkness" of $\ce{MnSO4}$, though only slight, does not interfere with the endpoint.
Addition of acid - To provide the $\ce{H^+}$ and the $\ce{SO4^{2-}}$ as counterion.
Heating - To accelerate the reaction.

The fact that heating is advised shows that the colour change is not as "rapid and conclusive" as in the case of normal acid-base titrations - we need to wait a bit.

This does not answer my question at all. I have asked why the colour of KMnO4 after the end point may fade. I already know that MnO4 is converted to $\ce{Mn2+}$ in presence of acid and reducing agent. My question is what reaction might occur after the reducing agent is completely oxidized.

When $\ce{K\overset{+VII}{Mn}O4^-}$ is reduced, the purple colour disappears. Why would a further reaction take place after all the reducing substances have been oxidised?

And that is the exact question I have asked.

By the way, I have modified my question, as it appears that it wasn't clear before. I hope you can understand it now. And I hope that you will modify your answer accordingly.

@permeakra Yes, this might be a cause. Can you make an answer so that I can accept it?

Permanganate can oxidize water if left untouched for several days, this is why it should not be stored in a water solution too long even if sealed.
However, in this case you should consider faster reactions and setup. Typically redox titration is performed in a lab, that has quite a bit of organic dust and organic solvents in the air. Both might react with permanganate. Permangate solutions used for redox titration are quite diluted, so the reaction might affect the color rather fast. At the same time, redox reaction typically have limited speed and during titration color might be preserved just because permanganate didn't have time to react with reducer that is in the solution.
So, we need to draw a line when we expect permanganate to be completely reduced by reducer present in the solution and not diffused from air. The exact timing is a subject of trial and error.


text:
How to remove formaldehyde (smell) from clothes/jeans?
Bought several jeans, then to discover they have a really nauseating smell, that is not removed by washing. Reckon it's formaldehyde, from what I can read. Anyway, assuming you have jeans with formaldehyde in the fabric, what may effectively remove/neutralize this?
I ordered $\pu{10 kg}$ sodium bicarbonate for a soak, or I plan to heat them for a long time.  But advice from chemists is perhaps more productive. So, any ideas?

So did the ammonia work? Which brand did you use? I bought jeans that has their weird funky smell to them, I also have soaked them in vinegar and baking soda and washed on cold and air dried for days, and just let them sit outside and it still smells bad. So I’m wondering does the ammonia really work? How long do I let the ammonia sit for before washing?

@Dee See my comments on the answer below. I never tried ammonia.

Formaldehyde is indeed used for disinfection, but usually there is a subsequent procedure involving formaldehyde neutralization with $10\%$ to $25\%$ aqueous ammonia solution introduced (as a spray) at elevated temperatures. Ammonia reacts with formaldehyde producing water-soluble hexamine which, even though also has unpleasant smell of rotten fish, is then easily washed out from the fabric:
$$\ce{6 CH2O + 4 NH3 -> C6H12N4 + 6 H2O}$$
I'd suggest to wash the clothes at higher temperature ($\pu{40 .. 60 ^\circ C}$) as you would normally do, but adding some ammonia; afterwards, soak the clothes in cold water for a while (solubility of hexamine increases in cold water), wring out and air dry.
In some countries there are also ozone-chambers (basically a sealed wardrobe with an $\ce{O3}$ generator inside, offered by dry cleaning companies) where organic pollutants are oxidized; formaldehyde is no exception:
$$\ce{CH2O + 2 O3 -> CO2 + H2O + 2 O2}$$
Both vinegar and soda can also be used, though I'd think their effectiveness is going to be lower.

I will test it and report the results (..in some time).

@bretddog - I would love to know how it went!  Specifically, how offensive was the ammonia smell?

@Tracy The smell was strong enough that I would certainly have thrown them away or returned them. Forgot exactly what I did with them, but I believe I soaked them in baking soda, or a mix of baking soda and vinegar perhaps. Then I think I froze them. And hang them outside for several days. Maybe it got like at least 50% better, but still the smell was present. So they were just laying around for some time. Then after a while there was surprisingly no smell at all, so I started using them. Perhaps also laying them out in the sun may help.

@Tracy It's possible I may have heated them in the oven too, for a few hours, on 75-100C maybe. Thinking chemicals evaporate at different temperatures.


text:
Interpreting from phase diagram whether O2 will float or sink

The normal melting and boiling points of $\ce{O2}$ are $\pu{-218 ^\circ C}$ and $\pu{-183 ^\circ C}$, respectively. Its triple point is at $\pu{-219 ^\circ C}$ and $\pu{1.14 Torr}$, and it's critical point is $\pu{-119 ^\circ C}$ and $\pu{49.8 atm}$. Will $\ce{O2 (s)}$  float or sink on $\ce{O2 (l)}$?

I quickly sketched the phase diagram of oxygen:

Pressure of the molecules will affect how compact the molecules are in solid and liquid, so I guess we can explain whether it will float or sink using the slope?
How does the slope tell us about the densities of $\ce{O2}$ solid and liquid ?
And in addition, how do I use intermolecular forces to explain this?

Floating or sinking is a matter of comparing the densities, or molar volumes of two phases; even though one can easily find the density of an ideal gas ($V_m = RT/p$), I'm unaware of how a phase diagram can be used for it. from what I remember, solid oxygen is heavier and will sink.

Yes, we can do that.
In fact, it is only the sign of the slope that matters. Say, you stand right below the gas/liquid curve; you apply a little more pressure, and you step from gas to liquid. How does that correspond to Le Chatelier's principle? Well, just like that: the system responds to increased pressure by decreasing its own volume, hence the liquid must be denser that the gas (which we of course already knew from independent sources).
Guess what?
The same reasoning applies verbatim to the liquid/solid curve, since it is sloped in the same direction.
As to the intermolecular forces, I think the usual hand-waving along the lines of "less freedom $\to$ less motion $\to$ lower volume" will do, and even that is redundant. Having solid denser than liquid is normal. If anything, it is the opposite situation that would require an explanation.


text:
Why is the electronic wavefunction of helium a product of the two 1s wavefunctions when ignoring electron-electron repulsion?

This question has also been asked on physics stack exchange:
Why is the helium atom wavefunction a product of the two 1s wavefunctions?

From [1, p. 116]:

In seeking an approximation to the ground state, we might first work out the solution in the absence of the $1/r_{12}$ term. In the Schrodinger equation thus simplified, we can separate the variables $r_{1}$ and $r_{2}$ to reduce the equation to two independent hydrogen like problems. The ground state wavefunction (not normalized) for this hypothetical helium atom would be:
$$\psi(r_1, r_2) = \psi_{1s}(r_1)\psi_{1s}(r_2) = e^{−Z(r_1 + r_2)}$$

Why is it only the product and not some linear combination of the two wavefunctions? I heard somewhere that it has something to do with "tensor product". Can someone provide a detailed explanation about this?
Reference:

Blinder, S. M. Introduction to Quantum Mechanics: in Chemistry, Materials Science, and Biology; Elsevier, 2012,. ISBN 978-0-08-048928-5.

I did a quick search and found the source for the quoted text and I edited it in. In the future, please do cite the source you are quoting.

The exact same question was posed by the same OP, here: https://physics.stackexchange.com/questions/373985/why-is-the-helium-atom-wavefunction-a-product-of-the-two-1s-wavefunctions Quite different answers too!

@Gert Exactly! The answers here didn't quite give any information about the tensor product which is why I had to go to physics SE.

@qforquantum Cross-posting is usually frowned upon, especially if undisclosed. You could have commented on the answers here for the missing information, or flagged a moderator for migration to physics.SE instead

@TobiasKienzler Why is there so much rigidity and "frowning upon" on SE? It makes it tedious to use. Half the time I am hesistant to do anything here. :/

@qforquantum You should at least mention the cross-post, otherwise users may waste their time writing an answer almost identical to one on the other site. It also helps others getting all answers. See https://meta.stackexchange.com/q/4708/146482. But actually I agree with you, there's a nice feature request: Build and strengthen the Stack Exchange community with “crossover questions” between sites

@TobiasKienzler Okay, I agree with that.

The same mechanics as for the separation of angular and radial parts of the hydrogen electronic wavefunction are in place here: You have a Schrödinger equation that is separable into two (i.e. one for each electron). You solve them and in order to combine them, they must be multiplied.
It so happens in this case that the solution for each equation is known and recognizable: the electronic wavefunction of $\ce{He+}$. The fact that the two wavefunctions are multiplied is not related at all to electronic structure theory per se (no involvement of LCAO etc.), but merely a result of the separability of the Schrödinger equation.

If an atom is represented by a wavefunction does this mean that they are also waves ?

@adosar: This question goes deeper than you may think. Some would say that atom (i.e. nucleus and electrons) are represented by one wave-function. Some would separete nucleus and electrons into two wave-functions to make the problem easier and justify that by the large mass difference between the particles. Some go further than that in their handling. (For more discussion, consider a chat)

As TAR86 answered, this is the separation of variables. This is only possible due to the neglected $1/r_{12}$ term, which would otherwise make separation impossible. Therefore, this multiplication is actually merely an approximation.
For completenesses sake, just like with the hydrogen orbitals (or any solution to a linear PDE obtained via separation of variables) the full solution to this approximated $He$ is a superposition of all multiplicative orbitals, i.e.
$$\Psi(r_1,r_2) = \sum_{n,l,n',l'}c_{nln'l'}\psi_{nl}(r_1)\psi_{n'l'}(r_2)$$
with adequate normalization.


text:
What is the IUPAC name of fullerene like molecules?
I wonder what the IUPAC names of fullerene-like molecules, I mean all those hydrocarbons with 3-D structure. Do we have another name of these compounds or are they classified as aromatic?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please see our guidelines on how to ask nomenclature questions.

From Does IUPAC nomenclature have the ability to name all organic compounds? (Chemistry.se) we find Pure & Appl. Chem. 1997, 69 (7), pp. 1411-1434 pdf mirrored on iupac.org. And obviously the blue book 2013: http://dx.doi.org/10.1039/9781849733069

It seems that you are asking two different Qs. It is possible to name each fullerene.. The synthesis guys do it so I have no refs at hand. Is better to consider them as a class, as for aromaticity is present or no depending on which aspect is accounted for.

I think I've found relevent sources here:
https://old.iupac.org/publications/pac/1997/pdf/6907x1411.pdf
http://www.sbcs.qmul.ac.uk/iupac/fullerene/
However, I still wonder if we have another name for all those hydrocarbons with 3-dimensional structures. Should they be regarded as a subclass of "aromatic" or do we have another name for them?

This is not really an answer and in my opinion should be edited in your question.

I am considering converting that to an edit. Right now it poses more questions than it answers. In addition, whilst the links may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link only for reference.

Polycyclic cage? http://www.ebi.ac.uk/chebi/searchId.do?chebiId=CHEBI%3A33640 - Agreed that this question could do with a cleanup


text:
Do bidendates occupy cis position always?
https://chem.libretexts.org/Core/Inorganic_Chemistry/Coordination_Chemistry/Properties_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_Transition_Metal_Complexes
In example 1.1, it is written that en which occupy only adjacent positions.....is it true for any bidendate for octahedral isomer?

It surely depends on the size of the specific ligand. Big floppy ligands might well be able to occupy opposite positions.

Its not impossible to have a trans-spanning ligand, but they seem to be rarer than cis-ligands due to their increased length and greater potential for steric strain. While en is too small of a ligand to reach trans-positions, it is possible with significantly longer ligands.
It seems octahedral ones are even rarer than their square planar counterparts, but there has been at least one reported synthesis of a trans-spanned octahedral$^1$ that I could find. In the linked paper, they report the synthesis of trans-$\ce{[Ru(Cl)(trpy)(C3SPAN)]}$ where trpy = 2,2:6,2-terpyridine and C3SPAN=$\ce{Ph2PC6H4CH2O(CO)(CH2)3(CO)OC6H4PPh2}$.
$\hspace{25ex}$

C. Bessel, et al. Chem. Soc., Dalton Trans., 1999, 2281–2292

Interesting, but isn't that for square-planar? Or does it apply to octahedral as well?

@gilleain good catch, I jumped on that too quickly. It seems to also be possible for octahedral as well. I will try to link to a paper.

So the question was solved on the basis of assumption or taking the case which is more bound to happen?

@HydrousCaperilla for the example you gave, it could only be cis because the en ligand is too small to be trans-spanning.

can en liagand be trans spanning in any other case?

@HydrousCaperilla I don't believe so. The ligand itself just doesn't reach far enough to access sites that aren't right next to each other (cis).

But in my book ,for [Cocl2(en)2] complex,the en liagand configures itself to show both cis and trans isomers.Is it bc it is square planar?

@HydrousCaperilla thats the difference between trans and trans-spanning. $\ce{CoCl2(en)2}$ can be cis or trans because the two en groups can be separated by 90 or 180 degrees, but neither of the en groups binds to sites 180 degrees apart, which is trans-spanning.

So square planar can have cis-trans while it is difficult for octahedral...am i correct

@HydrousCaperilla I think you still have trans and trans-spanning confused. In the picture on my post, the long ligand is trans-spanning because one of its ends is bound on the opposite side of the metal from the other. When you have two of the same ligand bound on opposites sides of the metal, we call that trans. CoCl2(en)2 is actually octahedral because en is a bidentate ligand.


text:
Neutralization of alkali solution by a mix of strong acids

$\pu{0.1 M}$ $\ce{HCl}$ and $\pu{0.2 M}$ $\ce{H2SO4}$ solutions are mixed in equal volume. This solution is diluted to double the volume. Find the volume of $\pu{0.1 M}$ $\ce{NaOH}$ which can be neutralized with $\pu{20 mL}$ of the former solution.

The answer is $\pu{50 mL}$.
I have tried a bit, but the answer I get is wrong. Let volume be $V$ at the beginning, therefore, no. of moles of $\ce{HCl}$ is $0.1V~\pu{mol}$, no. of moles of $\ce{H2SO4}$ is $0.2V~\pu{mol}$.
Total volume of solution after dilution is $4V$, therefore, no. of moles of $\ce{H+}$ in the solution is $(0.1V + 0.4V)/4V$.
No. of moles of $\ce{Cl-}$ in the solution is $0.1V/4V = 1/40$.
No. of moles of $\ce{SO4^2-}$ in the solution is $1/20$.
Since $\ce{NaOH}$ reacts only with $\ce{Cl-}$ and $\ce{SO4^2-}$, therefore total moles of $\ce{NaOH}$ needed is $(1/20 + 1/40)(20/1000) = 1/100~\pu{mol}$
Therefore, $\text{volume needed} = \text{molarity/no. of moles} = 0.1/0.01 = \pu{10 L} = \pu{10000 mL}$.
I Don't Know what I am doing wrong.
I tried again using some of the suggestions given: for the first step of mixing H2SO4 and HCl, M1V+M2V=M3(2V),
which comes as M3=0.5/2.
Then for dilution,
(0.5/2)*2V=M4*4V
thus M4=0.5/4.
Taking 20ml, (0.5/4)*20=0.1*x,     [0.1M of NaOH]
thus x=25ml.
Why doesn't this work??

Sodium hydroxide does not react with either chloride or sulphate, only with hydronium ions/protons.

I think you over-complicated things a bit by focusing on the formation of sodium salts, when in reality all you need to take into account is the net neutralization reaction:
$$\ce{H3O+ + OH- -> 2 H2O}$$
From the equivalence point condition of any neutralization reaction one can find volume of the base $V_\mathrm{b}$ expressed via the volume of acid(s) $V_\mathrm{a}$:
$$V_\mathrm{b} = \frac{C(\ce{H3O+})}{C(\ce{OH-})} \cdot V_\mathrm{a} \tag{1}$$
Since there are two moderately diluted strong acids and a base, one can conclude that the dissociation of each acid and a base is independent and complete:
\begin{align}
\ce{HCl + H2O &<=>> H3O+ + Cl-} \\
\ce{H2SO4 + 2H2O &<=>> 2H3O+ + SO4^2-} \\
& \\
\ce{NaOH &<=>> Na+ + OH-}
\end{align}
Note that concentration of protons is doubled in case of sulfuric acid. Total concentration of protons in the solution is
$$C(\ce{H3O+}) = \frac{n(\ce{H3O+})}{V(\ce{H3O+})} = \frac{n(\ce{HCl}) + 2 \cdot n(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \tag{2}$$
Taking into account that $n = CV$ and that two identical volumes of acid were mixed ($V(\ce{HCl}) = V(\ce{H2SO4}) = V(\text{acid})$) (2) can be transferred into:
$$
\require{cancel}
\begin{align}
C(\ce{H3O+}) &= \frac{C(\ce{HCl}) \cdot V(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}) \cdot V(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \\
&= \frac{\cancel{V(\text{acid})} \cdot (C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}))}{2 \cdot \cancel{V(\text{acid})}} \\
&= \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2} \tag{3}
\end{align}
$$
Now let's plug (3) in (1) and finally find the volume of alkali:
$$V_\mathrm{b} = \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2 \cdot C(\ce{OH-})}  \cdot V_\mathrm{a} = \frac{\pu{0.1 M} + 2 \cdot \pu{0.2 M}}{2 \cdot \pu{0.1 M}} \cdot \pu{20 mL} = \pu{50 mL}$$

please help me with the equation(2). why are we dividing the summation of volumes??

@Sri To find the molar concentration of any compound ($\ce{H3O+}$ in this case), we need to divide amount by volume. Since we mix two acids, the resulting volume is the sum of their volumes.

And where have we used the fact that the volume has been doubled??

@Sri In divisor  in eq. (2) we add two equivalent volumes of both acids, isn't that it?

@Sri I expanded the expression for the concentration of protons so that now it should be completely transparent what is what.

yeah! but the solution is diluted, doubling its volume. that is, 2*[ V(HCl) + V(H2SO4) ]. i don't understand it.

@Sri What exactly you don't understand? Why volumes are additive, or anything else?

This solution is diluted to double the volume. The question says this.

@Sri When you mix two identical volumes, the volume will be doubled. If you still have idea why is it so, take two same glasses of water half-full, and pour the water from one to another.

The question says, after mixing the solutions of H2SO4 and HCl, the solution thus obtained is again diluted with water. This water has volume equal to the initial volume of the solution. Thus the solution thus obtained has volume 4 times the volume of H2SO4 that was taken at the very beginning

I tried again using some of the suggestions given: for the first step of mixing H2SO4 and HCl,   M1V+M2V=M3(2V), which comes as  M3=0.5/2. Then for dilution,  (0.5/2)2V=M44V thus M4=0.5/4. Taking 20ml,  (0.5/4)20=(0.1x), [0.1M of NaOH] thus x=25ml. Why doesn't this work??

First, thank you for showing work for a homework problem. I think it is instructive in this case, not only to provide the correct solution, but also to point out where your problem solving went wrong. The exact solution to this problem may not be extensible to other problems, but the general guidelines for solving problems will, so if there's a take-home message, those are it.

You're talking about dilution. Why does that matter here? The crux to any problem is to figure out what exactly are the key features you need to extract. This problem requires understanding stoichiometry because you need to match amounts of a substance according to a reaction and concentration because you need to understanding the relationship between amount of substance, volume, and concentration.
Units. You have units in some places but not others. Units should be carried through the entire calculation. What is 100? It's a number. It's not a measurement or even a quantity because it has no units. Having units is a check on performing the correct calculations because units will cancel when dividing/multiplying and only two values with the same units can be added.
Sanity checking results. 10 mL is roughly a small mouthful of water. 10000 mL is the volume of a large pot. For two solutions of roughly the same concentration, there should not be a 1000x different in the amounts.
Make sure you have what you need before you start solving. You correctly tagged this as a stoichiometry problem. The correct stoichiometric ratios are $\ce{HCl}:\ce{NaOH} = 1 : 1$ and $\ce{H2SO4}:\ce{NaOH} = 1 : 2$, but even if you had said that, you don't have the means to justify it. Why? Because stoichiometry starts with balanced equations which you don't have. Once you determine the type of problem, make sure you have the appropriate ground work for solving problems of that type.

OK. So onto solving your problem. For point 4, you should write out balanced equations:
$$\ce{HCl + NaOH -> H2O + NaCl}$$
$$\ce{H2SO4 + 2NaOH -> 2H2O + Na2SO4}$$
At this point, you immediately know that you can decompose the problem into two separate problems. Why? Because hydrochloric acid and sulfuric acid don't interact with each other according to the equations written.
At this point, we should compute the total moles of sodium hydroxide required since this is the whole point of stoichiometry.
Note that since the initial solutions are mixed in equal volume, the concentration of each is half what is listed. This is definitely a dilution problem, but one that is simple enough not to require any kind of dilution math or ratios aside from dividing by 2.
$$0.02\ \mathrm{L}\cdot\frac{0.05\ \mathrm{mol\ \ce{HCl}}}{L}\cdot\frac{\mathrm{mol\ \ce{NaOH}}}{\mathrm{mol\ \ce{HCl}}}= 0.001\ \mathrm{mol\ \ce{NaOH}}$$
$$0.02\ \mathrm{L}\cdot\frac{0.1\ \mathrm{mol\ \ce{HCl}}}{L}\cdot\frac{2\ \mathrm{mol\ \ce{NaOH}}}{\mathrm{mol\ \ce{HCl}}}= 0.004\ \mathrm{mol\ \ce{NaOH}}$$
The total amount of sodium hydroxide is:
$$0.001\ \mathrm{mol\ \ce{NaOH}} + 0.004\ \mathrm{mol\ \ce{NaOH}} = 0.005\ \mathrm{mol\ \ce{NaOH}}$$
Now, concentration is the per unit volume amount of substance:
$$c = \frac{n}{V}$$
or
$$V = \frac{n}{c}$$
Plugging in:
$$V = 0.005\ \mathrm{mol\ \ce{NaOH}}\cdot\frac{L}{0.1\ \mathrm{mol\ \ce{NaOH}}} = 0.05\ \mathrm{L}$$
Now. Sanity check: 50 mL is much closer to 20 mL and makes more sense.

Thanks, I don't quite get  the first equation can you please explain it. It would be nice

@Sri If you don't understand the first chemical equation, that's why you have problems. Please review neutralization. If you don't understand the first mathematical expression, that also would indicate why you have a problem. Please review the concept of concentration.


text:
Why do the d suborbitals start at 3d instead of 4d?
Why do the suborbitals for d start at 3 instead of four? It is in the fourth row on the periodic table and all other suborbitals start according to the row on which they are located except for d. Why is that?

Related: https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium

Actually $d$ orbitals are not exceptional.  $4f$ orbitals come in row 6.

This is due to the underlying quantum physics. Quantum physics teaches us that orbitals are ultimately a result of so called quantum numbers that appear during the mathematical description of these quantum mechanic systems.
If you do the math you'll see that in the 3rd period there's the main quantum number $n=3$ which allows the secondary quantum number $l$ to take the values 0, 1 or 2. As $l$ determines the kind of the orbital you get a $s$, $p$ or $d$ orbital.


text:
How to calculate the enthalpy change from the reaction of silver nitrate and sodium iodide?

When $\pu{20.00 mL}$ of $\pu{1.00 M}$ $\ce{AgNO3}$ solution is added to $\pu{20.00 mL}$ of $\pu{1.00 M}$ of $\ce{NaI}$ at $\pu{25 ^\circ C}$ in a calorimeter, a white precipitate of $\ce{AgI}$ is formed. The temperature of the aqueous mixture increases to $\pu{40 ^\circ C}$.

I am trying to calculate the $\Delta H$ for the reaction per mole of $\ce{AgI}$.
The specific heat of the aqueous mixture is $\pu{4.184 J//g K}$, the density of the mixture $\rho = \pu{1.00 g//ml}$. And I assume that the calorimeter absorbs a negligible amount of heat.
How am I suppose to go about finding the $\Delta H$ per mole of $\ce{AgI}$?
I don't really know where to start here as I am unclear about the whole problem here. In order to use the formula
$$Q = m C_p \Delta T$$
I need to find the mass. But which mass do I take? $\ce{AgI}$, or $\ce{NaNO3}$,  or is it the mass of the entire product of the reaction?
This is the balanced equation
$$\ce{AgNO3 + NaI -> AgI + NaNO3}$$

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

When the reaction takes place and releases its heat, everything in the calorimeter  heats up: the water, the $\ce{AgI}$ and the $\ce{NaNO3}$. But of these the water is by far the most massive and for that reason, as a reasonable approximation, we consider that mass that absorbs the heat to be the water, i.e. $40.00\ \mathrm{g}$.
Use this to calculate $Q$.
Then determine how many moles of $\ce{AgI}$  was formed, by stoichiometry. The quotient of these two numbers gives you the reaction enthalpy per mole of $\ce{AgI}$

So, for Q I got 2510.4J (using mass as 40g as we assume no changes in volume. By mole ratio,  I got 0.02 Moles AgI . So the question wants per mole of AgI . So am I right to say that 0.02 moles of AgI makes 2510.4 J ? And now I just need to find 1 mole of AgI make much heat in joules ?

Yes. Note that "we assume no changes in volume" is irrelevant: this is about mass, not volume.


text:
What are the basic conditions for a carbocation to rearrange into a stable carbocation?
While rearranging,every rearrangement has to increase the stability of carbocation, other than this is there any other thing that we have to keep in mind while rearrangements?
Is there any limit to the number of rearrangement to the carbocation given in the first step?

https://chemistry.stackexchange.com/questions/27835/can-hydride-shift-and-methyl-shift-happen-one-after-another-in-a-carbocation https://chemistry.stackexchange.com/questions/59010/is-a-sequence-of-two-wagner-meerwein-rearrangements-possible

This seems too broad. Also not "every" rearrangement has to give lower energy intermediate.


text:
Rotovibrational selection rules for symmetric tops
The selection rules for symmetric top molecules are
$\Delta J = 0, \pm1$ and $\Delta K = 0$ for $K \neq 0$ (parallel transition)
$\Delta J = \pm1$ and $\Delta K = 0$ for $K = 0$ (parallel transition)
$\Delta J = 0, \pm1$ and $\Delta K = \pm1$ for $K = 0$ (perpendical transition)
My question concerns the possibility of a $\Delta J=0$ transition. For example, the only way how $\Delta J =0$ is possible for linear molecules is with degenerate normal modes (e.g. $\ce{CO2}$ bend), as these can "evolve" into one another and give rise to vibrational angular momentum, which ensures that conservation of angular momentum can still hold regardless of $\Delta J$.
The fact that $\Delta J =0$ is allowed for all transitions of symmetric tops means that there is vibrational angular momentum associated with all of them. My questions are:

What is the origin of vibrational angular momentum if the normal modes involved are not degenerate? Can you help me visualise it, for example for the $\ce{C-Cl}$ stretch of $\ce{CH3Cl}$ (a parallel non-degenerate transition)?
Why is the selection rule different for $K=0$ and $K\neq0$?

Some of the statements you make above are not quite correct, rather that correct individual points I think that it is simpler to start from scratch, as it were, and try to explain what happens.
Two quantum numbers are needed to define the rotational energy of a symmetric top molecule, such as CH$_3$I or benzene. The total angular momentum is characterised by $J$ and this generally does not lie along the 'figure' axis of the molecule which is the C-I bond axis in CH$_3$I for example. This quantum number has integer values starting at zero, i.e. $0,\, 1,\,2\cdots$  The total angular momentum is $ \hbar\sqrt{J(J+1)}$. The second quantum number $K$ is the projection of $J$ along the figure axis and so has $2J+1$ values ranging from $-J$ to $+J$ and therefore has values $0,\,\pm 1,\, \pm 2, \cdots \pm J$. The size of the projection quantum number is $K\hbar$ and clearly as it is the projection along the figure axis $K$ cannot exceed $J$.
The energy levels of any vibrational level form stacks of $J$ levels rather like those of a rigid rotor but where each $K$ value has its own stack beginning with $J=K$. Each type of vibration has sets of $K$ and associated $J$ levels for each vibrational quantum number.
We are considering rotation-vibrational transitions. The selection rule for parallel transitions, i.e. those with a dipole moment along the figure axis always have $\Delta K=0$ and then $\Delta J = 0, \pm 1$ unless $K=0$ then transitions with $\Delta J=0 $ are forbidden. The reason that $\Delta K = 0$ is that radiation cannot induce transitions between different $K$ values because rotation about the figure axis induces no rotating dipole moment. (The dipole lies along the axis in this type of transition). When $K=0$ there is no angular momentum component along the figure axis and so if the total angular momentum is also zero there is no rotating dipole to couple to the radiation.
In the case of perpendicular transitions the dipole has a component perpendicular to the figure axis. Bending vibrations can induce perpendicular transitions. The selection rules are $\Delta K=\pm 1$ and $\Delta J = 0,\pm1$. In this case changing $K$ does induce a change in the dipole and so transitions can occur.


text:
Molar absorption coefficient of H2SO4(aq) for UV-vis spectrophotometry
I am wondering if there is anything in the literature that can tell me what is the molar absorptivity (i.e. molar extinction coefficient) of aqueous sulfuric acid at a given wavelength (in my case, I am interested in $\lambda_\mathrm{max}(\mathrm{ferroin}) = \pu{508 nm}$).
In addition, is there a good resource in the literature to look up these values for common substances? I haven't yet been able to find anything of the sort.

I never ran into uv absorption data for (aqueous) sulfuric acid and I doubt that there is any absorption at $\lambda$ = 508 nm.
There is an old paper by K. F. Palmer and D. Williams in Appl. Optics, 1975,14, 208-219 (DOI) on the optical properties of sulfuric acid droplets with respects to the composition of clouds on Venus and a more recent one from the group of J. B. Burkholder in Phys. Chem. A, 2003, 107, 1112-1118
(DOI) on Vibrational and Electronic Spectroscopy of Sulfuric Acid Vapor. Both, while interesting and therefore mentioned here, are not relevant for your experiment.
My suggestion is simple:
Measure your ferroin sample in aqueous sulfuric acid (or whatever solvent is required) and use the same solvent as a blank. Most spectrometers will automatically perform the subtraction of the blank from your sample!
Concerning a (free) source of UV-vis absorption data, there is the NIST Chemistry WebBook.


text:
Why is ethanol used in catalytic reduction of nitro group to an amine?
$$\ce{R-NO2 +C2H5OH + H2 (g)->[Pt/Pd/Ni] R-NH_2}$$
The above equation represents catalytic reduction of nitro group to amine. What is the role of ethanol in it?
I understand that $\ce{Pt/Pd/Ni}$ plays an important role as reducing agent and hydrogen gas replaces oxygen atoms. But, I'm not able to understand the role of ethanol. Can someone help me with this?

Solvent? You're missing the actual reductant, which is presumably hydrogen gas.

Provide conditions. Ethanol may be the reductant.

The above equation does not represent the catalytic reduction of a nitro compound to an amine! As mentioned by @orthocresol, you're missing the hydrogen gas! I've done the reaction very often with nitro arenes. Ethanol as a solvent is great because it dissolves both starting materials and products (only the catalyst, e.g. Pd/C is suspended and can be filtered off later). It is cheap, can be removed easily on a rotavap and, in contrast to methanol, isn't overly harmful.

I'm terribly sorry for not mentioning Hydrogen. I did miss a really important reactant there. And thank you @Klaus-DieterWarzecha for making me understand the role of ethanol as a solvent.

Note that Pt/Pd/Ni is definitely a catalyst here and should be denoted differently (not as reactant). I suppose the same goes for the ethanol (solvent), but since it's a part of a question, I left it as it is, but still the reaction is incomplete. Is there water among the products?


text:
Why does CO2 only exist in gaseous and solid state under STP/SATP?
Why does carbon dioxide only exist in gas and solid state under STP/SATP conditions?
$\ce{CO2}$ pressurized to $\pu{5.1 atm}$ can exist in all 3 states: solid, gas, and liquid.
However, why couldn't the same happen in 1atm pressure? As a solid chunk of dry ice is the coldest you could get $\ce{CO2}$, why cant you warm it up slightly but still have it cold enough to turn into a liquid, instead of directly to a gas?

Because its pressure at triple point is higher then atmospheric.


text:
Wavelength-dependence of refractive index in internal reflection spectroscopy?
In many literature sources (web example), only a single value for the refractive index is assumed for the infrared element, and another is typically assumed for the sample substance. However, should not the refractive index be wavelength-dependent? So the wavelength-dependence of the penetration depth,
\begin{equation}
d_p = \frac{\lambda}{2\pi n_1 \sqrt{\sin^2\theta - (n_2/n_1)^2}}
\end{equation}
would not only be in the numerator, but also in the denominator: $n_1 = n_1(\lambda)$ and $n_2 = n_2(\lambda)$? Here $d_p$ is the penetration depth, $\lambda$ is wavelength, $\theta$ is the incident angle, and $n_1$ and $n_2$ are the refractive indices of the infrared element and sample, respectively.

Yes it should, but it doesn't matter much.
In physics, roughly speaking, everything depends on everything. We wouldn't be able to do a thing if we couldn't tell what is important and what isn't. Now let's see how much the refractive index actually changes as $\lambda$ goes, say, from 400 to 800nm (thus covering the entire visible range).

(source)
Looks like $n$ of a heavy flint glass goes from 1.85 to 1.75 (that's a change of 5%, in most other materials even less), while $\lambda$ changes by... how many percent?


text:
How to know whether to add latent heat when calculating final temperature?
From the examples in my textbook, it is usually very easy to see whether you need to account for the latent heat when calculating the final temperature. For example, if you add 2 x 13 gram ice blocks to a 250 ml cup of water, it is obvious that the ice will melt, and you will have to account for the latent heat of fusion for the ice.
However, what if it isn’t so obvious whether one of the substances will undergo a phase change? For example, what if we added 15 grams if ice to 20 ml of oil? If the final temperature is below 0, then we don’t need to account for the latent heat for ice, otherwise we will have to. How do we determine whether the ice will experience a phase change, and more generally, how do we determine whether one of the substance will experience a phase change?

It is less obvious than you seem to assume (say, if water cup in your example were already at $0^\circ$C, then the ice would not melt), but still pretty obvious.
The rule of thumb is:

Assume there is no phase change.
Calculate the final temperature.
Check whether there is a phase change in any of the substances involved as they go from starting to final temperature.
If no, we are done. If yes, assume there is a phase change and go back to step 2.


text:
What exactly does setting the net charge in a DFT calculation do?
In the implementations of DFT using Gaussian basis sets, it is common to set the net charge. What exactly does this do? The Gaussian (program) manual, for instance, says that it introduces a background charge distribution made of point particles. This raises more questions than it answers for me.
1) If it's made of point particles (presumably at some internally defined coordinates), how is it a background charge distribution?
2) What exactly does setting the charge do? Is it just adjusting the electronic energy? Does it impact the geometry optimizations or the spin states given a pre-specified spin multiplciity?
I'm having trouble understanding what the charge is actually doing in terms of the DFT calculation itself. It's not a setting in plane-wave calculations, so the parallels are not clear to me.

The Gaussian manual (or whatever that is) cites 2 articles. Did you read them?

Wow, I didn't see that. Apologies. I'll read them now that I know where to look.

The keyword has absolutely nothing to do with DFT, that is just what they use in the example.

Thanks, Martin. I'll check out the papers to learn more about what's going on. Good to know it's not related to the DFT implementation.

You can use it to add point charges to your calculation and see how that effects the energy, etc. I have never used it before, and that's only what I take home from the explanation in the manual. I would say, for everyday use it is not worth knowing.


text:
Relationship between pH and volume
I had prepared two different volume of solution: $\pu{500 mL}$ and $\pu{100 mL}$ with same concentration ($1\%$), but I obtain different $\mathrm{pH}$. Will volume affect $\mathrm{pH}$?

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Best of luck!

The volume will not affect the pH.
However, any change in the concentration will

If correctly prepared, the pH should have been the same. Logically, consider if the pH would be different if measured in 5 separate 100 ml samples poured from the 500 ml preparation.
Why might the pH differ in your tests? Some possibilities:

One container was not completely rinsed after cleaning and had some (often alkaline) detergent clinging to it.
The preparation absorbs $\ce{CO2}$ from the air, so the layer on top is more acidic.
Error in measurement.

Unless you have a buffered system, pH may vary abruptly with a slight change in ingredients.

No, $\mathrm{pH}$ is not volume-dependent as it's an intensive property derived from another intensive property (concentration):
$$\mathrm{pH} = -\log{[\ce{H+}]}$$
The reasons for the different $\mathrm{pH}$ might be different, among them there are the following:

Contaminated beaker;
Solution absorbs carbon dioxide from the air differently due to different surface area of the liquid exposed to air;
$\mathrm{pH}$-electrode is contaminated or/and improperly calibrated;
$\mathrm{pH}$ values were measured at different temperatures.


text:
How to perform a relaxed scan in GAMESS?
I have the following input Gaussian input for my system. The coordinates are given in Cartesian format and it is scanning through 8 steps, with a bond distance between 1 and 5 having a difference of 0.1 Å in each step, and the bond distance between 1 and 2 is fixed.
#hf/6-31g** opt=modredundant

OPT

1 1
 O     0.086655     0.133386    -0.150088
 O    -1.636963     0.213966    -1.809353
 H    -0.149749     0.260762     0.763793
 H     0.802607    -0.484925    -0.261350
 H    -0.423645     0.140377    -0.629197
 H    -2.313930    -0.430164    -1.970372
 H    -1.720592     0.938467    -2.416119

B 1 5 S 8 0.1
1 2 F

What is the analogue of this in GAMESS?

I am inclined to close that as homework. Reading and understanding the manuals of the software you are using is a prerequisite of operating these. Hence your own research how to tackle this problem should be included in the question body. || At the very least I fail to understand what z-matrix has to do with it.

Unfortunately, MathJax here has no support for \AA; the best workaround is to use a unicode symbol Å.

Just write a small script that does this for you or do it manually? My guess is starting this thread here takes about as much time as producing the 8 desired coordinate files. The thing is: while it is good to use functionalities that  are implemented in the programs you are using and not constantly reinvent everything, asking such a question kind of implies that the program is doing most of the thinking for you.

@Martin-マーチン I would like to let you know that I have done an extensive research on Google before posting this question. ZMatrix is the one which contains the parameters of $3N-6$ coordinates and is independent of the Cartesian coordinates.

@Raditz_35 I have thought of that, but how do I constrain the bond length, angle, dih, etc. during the optimisation at each step?

If you have done extensive research, then please include your findings - even the dead ends - into the question body, so that doubling of effort will be avoided. I know what a z-matrix is, however, you do neither use it in your Gaussian input (you use redundant internal coordinates) nor specify that you would want to use it in the GAMESS input, hence my confusion.

@MitradipDas I don't know because I don't use GAMESS. But that means you actually want to know how to keep coordinates fixed during a GAMESS geometry optimization? I'm pretty sure you can google that one

@Raditz_35 The option is available for Cartesian, but I could not find any good clue for internal coordinates

Unfortunately a feature like ModRedundant in Gaussian, which one can use to perform relaxed scans, is not available in GAMESS. Even the possibility to perform a rigid scan seems a bit complicated. You can request a calculation with RUNTYP=surface and you have to specify a $SURF group. See example 29 of the tests, which you should find in path/to/gamess/tests/standard/exam29.inp.
A relaxed scan is a bit harder to perform, and only straight forward for varying distances. You have to perform a series of single point calculations, with partially frozen coordinates. I describe the general procedure of doing that in How do I perform a partial optimisation in GAMESS?
Here I will outline it for your problem, based on what you can read there.
Before we do that however, you should be aware, that there is a python script by Andrew Hoos on github. I did not have the time to test it. It is quite old, and I cannot predict how the development of GAMESS will influence the functioning of rscan.
I assume you have read and understood the linked post, so in an effort to safe some space, I use a more automated approach here.
First, set up the general file, which you want to use for a check run. I have used your geometry.

 $CONTRL
   EXETYP=CHECK
   SCFTYP=RHF RUNTYP=ENERGY
   COORD=UNIQUE   ICHARG=1   MULT=1   NZVAR=15   $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   $END
 $GUESS
   GUESS=HUCKEL    $END
 $DATA
   H5O2
   C1
   OXYGEN      8.0      0.086655000         0.133386000        -0.150088000
   OXYGEN      8.0     -1.636963000         0.213966000        -1.809353000
   HYDROGEN    1.0     -0.149749000         0.260762000         0.763793000
   HYDROGEN    1.0      0.802607000        -0.484925000        -0.261350000
   HYDROGEN    1.0     -0.423645000         0.140377000        -0.629197000
   HYDROGEN    1.0     -2.313930000        -0.430164000        -1.970372000
   HYDROGEN    1.0     -1.720592000         0.938467000        -2.416119000
   $END

It is again essential that you include NONVDW descriptions. Since the proton will move in between the two water molecules, I have included both pairs. Note that the 1-2 distance will not be included in the list of internal coordinates (even if you specify it directly). You can use NONVDW(1)=1,2 instead though.
Now that you have checked everything you can run the first optimisation:

 $CONTRL
   SCFTYP=RHF   RUNTYP=OPTIMIZE    COORD=UNIQUE
   ICHARG=1   MULT=1   NZVAR=15     $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   IFZMAT(1)=1,1,2,     1,1,5
   FValue(1)=2.3938489, 0.700
   $END
 $GUESS
   GUESS=HUCKEL   $END
 $DATA
   H5O2
   C1
   OXYGEN      8.0      0.086655000         0.133386000        -0.150088000
   OXYGEN      8.0     -1.636963000         0.213966000        -1.809353000
   HYDROGEN    1.0     -0.149749000         0.260762000         0.763793000
   HYDROGEN    1.0      0.802607000        -0.484925000        -0.261350000
   HYDROGEN    1.0     -0.423645000         0.140377000        -0.629197000
   HYDROGEN    1.0     -2.313930000        -0.430164000        -1.970372000
   HYDROGEN    1.0     -1.720592000         0.938467000        -2.416119000
   $END

Note that I am keeping the automagic creation of the coordinates this time. To be on the safer side, you should probably follow more closely the other post. In any case, we are now ready to continue. Well use the optimised geometry of the last run, as well as the orbitals, which can be found in the auxiliary output.

 $CONTRL
   SCFTYP=RHF    RUNTYP=OPTIMIZE    COORD=UNIQUE
   ICHARG=1   MULT=1   NZVAR=15     $END
 $BASIS
   GBASIS=STO    NGAUSS=3   $END
 $ZMAT
   AUTO=.TRUE.   DLC=.TRUE.
   NONVDW(1)=1,5, 2,5
   IFZMAT(1)=1,1,2,     1,1,5
   FValue(1)=2.3938489, 0.800
   $END
 $GUESS
   GUESS=MOREAD
   NORB=15
   $END

! ----- RESULTS FROM SUCCESSFUL RHF      GEOMETRY SEARCH -----
! ----- COORDS, ORBS, GRADIENT, AND APPROX. HESSIAN -----
!  COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
!      ATOM  CHARGE       X              Y              Z
! ------------------------------------------------------------
 $DATA
   O2H5
   C1
   OXYGEN      8.0   0.0758752612   0.1029703596  -0.1380388056
   OXYGEN      8.0  -1.6329486914   0.2211188064  -1.8103049937
   HYDROGEN    1.0  -0.0725810430   0.4747088980   0.7504261439
   HYDROGEN    1.0   0.9330501410  -0.3174832331  -0.3330916306
   HYDROGEN    1.0  -0.4237766906   0.1376065810  -0.6270666776
   HYDROGEN    1.0  -2.3424944360  -0.4268887357  -2.0060894609
   HYDROGEN    1.0  -1.7921347121   0.9257714517  -2.4735439582
   $END

! --- OPTIMIZED RHF      MO-S --- GENERATED AT Fri Dec 15 00:48:25 2017
! E=     -150.1620919294, E(NUC)=   47.9559336673
 $VEC
 1  1 9.93613099E-01 3.20702947E-02-2.24634863E-03 1.56904217E-04-2.19957003E-03
 1  2 3.13313084E-04 9.04941985E-04 6.59436960E-04-4.52587984E-05 6.45901704E-04
 1  3-4.70970670E-03-4.70987324E-03-9.86284464E-03 1.88174472E-04 1.88109591E-04
 2  1-5.80880573E-04 1.71794788E-03-8.47977388E-04 5.87852394E-05-8.29926100E-04
 2  2 9.94111746E-01 2.64448946E-02-2.47981701E-03 1.59721737E-04-2.45660777E-03
 2  3 1.64987868E-04 1.65066081E-04-3.05915593E-03-5.57763329E-03-5.57725393E-03
 3  1 2.41238581E-01-8.54557950E-01 8.35552605E-02-5.83631489E-03 8.18285646E-02
 3  2 7.39936297E-03-2.26953665E-02-1.91551065E-03 1.44103019E-04-1.84324895E-03
 3  3-7.30179599E-02-7.30300125E-02-1.26560023E-01-2.80500756E-05-2.84225764E-05
 4  1-1.47611593E-02 6.57491207E-02 2.15954260E-02-1.48570875E-03 2.11365513E-02
 4  2 2.35829657E-01-8.53953087E-01 6.86900515E-02-4.36681579E-03 6.81986435E-02
 4  3 1.56173985E-02 1.56167812E-02-3.51168589E-02-1.45207104E-01-1.45193326E-01
 5  1 2.65621341E-02-1.38559271E-01-5.11829955E-01 3.54136523E-02-5.00838599E-01
 5  2 1.64839489E-02-6.35695788E-02 4.44897012E-02-3.02940337E-03 4.36454714E-02
 5  3-2.12387862E-01-2.12451288E-01 3.07944883E-01-3.22077160E-02-3.22120362E-02
 6  1-6.42384605E-06 3.19756766E-05-4.30678730E-01 3.39316521E-01 4.64142719E-01
 6  2-8.23546629E-06 3.50155053E-05-4.64034770E-03 3.65246668E-03 4.98286518E-03
 6  3 3.26546905E-01-3.26522265E-01-2.41992633E-05-2.15894395E-05 1.43171984E-05
 7  1 3.03726648E-05-1.46010812E-04 3.61316021E-01 8.78725612E-01-3.07129752E-01
 7  2-9.63734293E-07 4.12738444E-06 1.19803044E-02 2.92946887E-02-1.01831260E-02
 7  3 3.50564742E-05 4.37481046E-05 1.83831688E-05-2.02321039E-02 2.02282616E-02
 8  1 3.91710932E-06-1.95703405E-05 1.97311245E-02 4.80455996E-02-1.67388085E-02
 8  2 5.95251643E-06-3.08145320E-05-2.27533068E-01-5.59223458E-01 1.93232191E-01
 8  3 3.64914685E-05-2.06253221E-05 8.95712681E-08 4.24180458E-01-4.24177941E-01
 9  1-2.23062153E-02 1.54500963E-01-1.24786038E-01 8.66527729E-03-1.22090919E-01
 9  2 8.76656449E-02-4.57576664E-01-5.50893665E-01 3.63552632E-02-5.43466463E-01
 9  3-3.91196174E-02-3.91748912E-02-1.19065907E-01 2.64674392E-01 2.64677791E-01
10  1 1.09047481E-04-7.57403559E-04-4.23084857E-03 3.61227555E-03 5.66486586E-03
10  2 1.69891575E-04-8.14724604E-04 6.04758690E-01-4.68585751E-01-6.43938341E-01
10  3 5.24314248E-03-4.87589514E-03 5.88487922E-04 2.55282495E-04 2.68854270E-04
11  1 1.54833774E-01-1.06847947E+00-2.73466865E-01 1.87755601E-02-2.67446840E-01
11  2 2.46273283E-02-1.43176172E-01-9.48992300E-03 8.45045780E-04-8.70879842E-03
11  3 7.99564096E-01 7.99664119E-01 2.53819180E-01 7.71090350E-02 7.71448689E-02
12  1 5.72763567E-06-2.78066490E-05 5.58361863E-01-4.39877926E-01-6.01681240E-01
12  2-9.45283650E-06 6.04268124E-05-9.09590832E-05 2.17244079E-05 3.25012306E-05
12  3 8.69427097E-01-8.69318789E-01-3.23352070E-05-6.93390818E-05-3.27231202E-05
13  1 4.13030403E-03-7.72647532E-02 1.79424923E-01-1.24368684E-02 1.75491985E-01
13  2 1.34789464E-01-8.74151452E-01 4.71611018E-01-3.07542286E-02 4.66703957E-01
13  3-1.13602815E-01-1.13734341E-01 2.65565976E-01 7.69534468E-01 7.69759188E-01
14  1-1.26314109E-06 1.52361151E-05-7.19364333E-04-1.76212662E-03 5.75477468E-04
14  2 2.07150162E-05-1.49811355E-04 3.43071888E-01 8.42940764E-01-2.91210891E-01
14  3 2.75400846E-05-1.38879856E-05-3.40876355E-05 8.31340062E-01-8.31078482E-01
15  1-8.63113810E-02 9.44284502E-01-7.40065924E-01 5.12577143E-02-7.24284153E-01
15  2-1.65007257E-02 6.66544153E-02 2.89063586E-01-1.96228027E-02 2.83830925E-01
15  3 2.15524124E-01 2.15552519E-01-1.69321183E+00 1.62283414E-01 1.62340378E-01
 $END

And you'll have to do that for every step of you scan. If you are a bit talented with scripting, then you can surely automate the process.
I did not run the whole scan, but here are the energies of the ones I completed:

OH = 0.7   #1       NSERCH= 13     ENERGY=    -150.1620919
OH = 0.8   #2       NSERCH=  4     ENERGY=    -150.2905386
OH = 0.9   #3       NSERCH=  4     ENERGY=    -150.3504096
OH = 1.0   #4       NSERCH=  5     ENERGY=    -150.3755229
OH = 1.1   #5       NSERCH= 60     ENERGY=    -150.3875894
OH = 1.2   #6       NSERCH=  9     ENERGY=    -150.3898837
OH = 1.3   #7       NSERCH= 20     ENERGY=    -150.3872606
OH = 1.4   #8       NSERCH= 36     ENERGY=    -150.3774125


text:
Is this a viable method for graphene production?
In a cup of warm water add sugar until it no longer dissolves then add a few spoonfuls more of sugar.
Drop in a small amount of graphite powder into the sugar water mixture. Stir it up well. There should be a thin black film on top of the liquid.
Let it set for a few days.
The sugar crystals will grow on the graphene surface similar to how rock candy grows.
The sugar crystals grow on the bottom of the floating film of graphite and once heavy enough the graphene and sugar crystal wafer will sink. And then the process repeats.
Is this possible with graphite powder in water as a film or is a thin solid piece of graphite necessary for sugar crystal growth?

It is likely that what you get are graphite-coated sugar crystals. Why do you think this might produce graphene?

I know that if you add Graphite powder to water a thin film of Graphite appears on surface of water. It likes to group itself centrally in the liquid and on top. Is it necessary to chill the sugar water solution to prevent (or attempt to prevent) shearing ? Maybe instead of simply dumping the Graphite powder into the saturated sugar water you first mix up a lot of Graphite powder with a liquid (not necessarily water) and use an eye dropper to add the Graphite mix to the sugar water?

But where do you go beyond a thin film of graphite flakes? Graphene is not the same as graphite flakes.

When making rock candy the sugar in the saturated sugar water wants to latch onto things submerged in the solution and forms crystals on the object. Since the Graphite sits as a film on top of the liquid the crystals will grow on the nearest surface which is the bottom of the Graphite film. Will the accumulated crystals on the Graphite film cleave the underside and by its weight sink?

Also Hydrogen Peroxide 3% (do not use stronger concentrations!) could be later used (or warm water) to dissolve the sugar from the Graphene and Sugar wafer that was produced.

OK I get the idea. But most likely the answer is no since the weight of a sugar crystal is enough to sink a graphite flake and there is nothing strong enough in the surface tension holding it on the surface that would cause the flakes to cleave. This is hard to prove, but easy to test if you have a test for graphene, that is.

Thanks for the reply. I wonder if use of different mediums to suspend the Graphite mixture in before adding to sugar water or if different temperatures of the sugar water may be helpful. So what if you froze the top of the sugar water and Graphite film would that provide a viable anchor point from which cleaving may be achieved by sugar crystal growth?

Excellent and many thanks. I am sure of it that if the surface is frozen or semi frozen then it would provide a point from which graphene will cleave off. If this is indeed true then they will have to call this the White and Black method or the Black and White method lol :P


text:
Melting ice with salt - how does it start?
I know that salt reduces the melting point of ice. I can see how a little bit of salty water can melt ice to water which can dissolve even more salt. But how does it even start? There is ice, which is hard. There are grains of salt, which are hard. So how does the salt enter the ice, to begin with?

Well, there must be points where one hard body touches the other. That's where it all starts.

Related: https://chemistry.stackexchange.com/questions/43786/how-can-sodium-chloride-melt-ice-or-keep-it-frozen

Well, a surface of for example ice is not ice|air its more like ice|lessdenseice|chaoticice|water|air if you are in a normal household setting.

@IvanNeretin The way you formulated it that's how pregnancy starts, too:) To the topic, there is solid-state chemistry existing for a good reason; also, ice is considered to have a pronounced "surface melting" effect.

@IvanNeretin Can you explain why touching should be enough? This is not a chemical reaction. If we have two crystals (with no moving molecules) touch each other, how could one dissolve the other if there is no room for movement?

To have crystals with no moving molecules, you must freeze them all the way down to absolute zero. Otherwise they move. Admittedly, that's more of a wiggling than a free motion, but still they move. Also, how is that different from a chemical reaction?

And why is the wiggling in crystals enough to dissolve the salt? I created another question whether dissolving of salt is a chemical reaction.

Well, surface molecules wiggle harder than those in the bulk. Sometimes they jump into the air and return in a short while, or in a long while, or never. In effect, a brick of ice is covered by a few molecular layers (or more) of confined water which is neither quite liquid nor quite solid.

I believe that Ivan's comments are on the right track. Generally any melting phenomenon to do with ice begins with the fact that the surface of ice is disordered. By this I mean that ice on the whole is a periodic crystal like other solids, but at the surface there are imperfections, primarily due to the extra conformational freedom afforded to the surface molecules. This distortion then propagates down quite a few layers [1]. Ref. [1] discusses the properties of this surface layer which is liquid-like, and points out that this layer exists even down to quite low temperatures.
If these surface molecules "see" something charged (each ion that is) like $\ce{NaCl}$, then the surface molecules already have the conformational freedom to solvate the ions. Additionally, there is an energetic benefit to this because the hydrogen bonds between these ions and water are stronger than the water-water hydrogen bonds.
In general, melting phenomena are extremely complicated and the question of how melting "begins" is has not really been answered completely or satisfactorily (can provide refs. later if you want). This case is a bit different though because it is really a solvation phenomenon, which is adequately explained by the liquid-like layer at the surface.

[1]: Beaglehole, D., & Nason, D. (1980). Transition layer on the surface on ice. Surface science, 96(1-3), 357-363.

One possible mechanism:  water evaporates from the ice, condenses onto salt pellets, forms salt solution which can then directly take up more ice and salt.  Next thing you know, the road is covered in a corrosive solution that's getting all over the underside of your car.

Plausible as it sounds, I don't believe this is the main mechanism.

Hence, "one possible mechanism".  Would like to see, and probably +1, your answer.


text:
Is dissolving salt a chemical reaction or a physical change?

Physical changes are limited to changes that result in a difference in display without changing the composition.
  (source)

e.g. melting water or dissolving sugar in water

Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. (source)

e.g. burning $\ce{H2}$ to $\ce{H2O}$
But what is dissolving salt in water? Is that a physical change (like dissolving sugar) or is it a chemical reaction?

The short answer is: yes.

The distinction exists only in our heads, so you might just as well stop relying on it.

So, you would allow me to call the melting of water a chemical reaction? Or putting paint on a canvas?

Related: https://chemistry.stackexchange.com/questions/33278/how-is-dissolving-a-physical-change-and-not-a-chemical-change/33281#33281. I would go with Ivan's take on this one. Something being a physical vs a chemical change is relatively loose distinction and you don't gain much descriptive value from classifying things as one or the other.

I would think that dissolving a non-reactive salt, like $\ce{NaCl}$ in water can be considered as a physical change. This is because, you can recover the salt by evaporating the water away.
From the Wikipedia page of physical change

In general a physical change is reversible using physical means. For example, salt dissolved in water can be recovered by allowing the water to evaporate.

But the distinction between physical and chemical change can be misleading and confusing at times. In general, anything that does not change the chemical constitution of the substance, is a physical change.
When you dissolve $\ce{NH3}$ in water, the following equilibrium is set up:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
But if you heat the water up, you will get back the original ammonia without any chemical change in its composition. However, the ammonia does change its chemical composition in water, as $\ce{NH4+}$ ions are formed. By that argument, you can say that the dissolution of an ionizable salt is a chemical change.
In some cases, dissolution of a compound can be called a true chemical reaction. For example, when you dissolve Calcium oxide ($\ce{CaO}$) in water, a vigorous chemical reaction takes place, forming $\ce{Ca(OH)2}$. You cannot get back the original $\ce{CaO}$ by just evaporating the water. In case of aqueous solution of $\ce{AlCl3}$, if you try to evaporate the water by heating the solution, you will get back only $\ce{Al(OH)3}$ as anhydrous $\ce{AlCl3}$ reacts with water as soon as it is in contact with water, to form $\ce{Al[H2O]6Cl3}$.
What I am trying to say is that some changes can be classified into one of these types, while some other changes can not be given a clear classification. Melting of water is clearly a physical change, while reaction of $\ce{H2}$ and $\ce{O2}$ is clearly a chemical change. One another fact, putting paint on a canvas is not always a physical change. In fact some paints are manufactured in such a way that during drying, some chemical reactions occur, due to which the paint fixes firmly on the surface. This is the reason many water thinnable paints can not be dissolved in water once they become dry.
To conclude, you should not hold hard classifications like that in your mind. However, my final verdict: Dissolving a salt is physical change (mostly).

But what about dissociation? Is it physical or a chemical change?

@BoredComedy Dissociation of what? A salt? Then its difficult to say. For example dissolving NaCl is probably a physical change as you can recover the salt by boiling off the water. But dissolving $\ce{AlCl3}$ is technically a hydrolysis reaction, as it is irreversibly changed to hydrated alumina. The line between them is blurry.

By dissolving a salt you are splitting the salt in its components (the ions) which have different properties than the salt as a whole.
Therefore I'd say dissolving salt is as much a chemical reaction as splitting a molecule in its components is.


text:
Why are some materials more flammable than others?
Wood and oil are flammable materials, but water isn't. Moreover, flames spread a lot faster in oil than in a piece of paper. What is the reason that it is so?
I also don't know why a small flame can spread out to a very big flame; what is the physical reason for it? I would like to know the molecular reasons in detail what makes fire so powerful.

Have you tried the wikipedia article on fire? You are asking a lot of questions, but at least some of them should be answered there

Firstly, for a material to be flammable it needs to contain something that is oxidisable by air oxygen, typically carbon. This explain why hydrocarbons and other carbon bearing materials typically burn well but minerals, rock, glass, water etc do not. In the latter materials the elements are already in a high Oxidation State and cannot be further oxidised.
For a generic hydrocarbon the oxidation reaction by air oxygen is given by the following reaction equation:
$$\ce{C_nH_{n+2}(g,s,l) +\frac{3n +1}{2}O_2(g)-> nCO_2(g) +\frac{n +2}{2}H_2O(l)}$$
The second major factor that affects flammability is volatility. The more volatile a material is, the more flammable it will be. The hydrocarbons illustrate that well. Methane is the most volatile and flammable of all of them and by the time we get to the solid ones like paraffin wax, flammability has been greatly reduced because these aren't volatile at all.


text:
What are the factors that affect the redox reactions?
In my work, I used $\ce{Co(OH)2}$ and $\ce{Ni(OH)2}$ as an active materials to fabricate as symmetric supercapacitors in which I used 1-butyl-1-methylpyrrolidinium dicyanamide as an electrolyte.
From my CVs curve, I found that $\ce{Co(OH)2}$ have a high redox peaks while for $\ce{Ni(OH)2}$, it shows no redox reaction peaks. What might be the factor that affects this? As I've known both active materials have similar layered liked-structure. Is it the electronegativity differences between an electrolyte and the materials?
What could possibly affect the redox reactions and how could I study more on the reaction mechanism?

I don't know much about it, but I believe the simplest theory, and apparently a very good one, for describing these kinds of processes is Marcus Theory

I assume that you are using cyclic voltammetry, if this is the case then  electrode preparation and/ or surface properties are usually important. If the reaction is not reversible, due to side reactions/decomposition  then very odd traces can be obtained.

@Jheindel You are correct in noting that Marcus Theory is the one to use. It has been confirmed, mainly by photo-induced reactions, where the two reacting species are fixed in space relative to one another.  see Harrison, et al., Chem. Phys. 116, 429 (1987), & Reid et al. J. Am. Chem. Soc., 2002, 124 (19), pp 5518–5527.


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Order of reaction greater than molecularity?
We were taught that the order of a reaction is either less than or equal to it's molecularity. But then I came across a mechanism called E1CB, where the fact was totally reversed. Here molecularity is Unimolecular while the order of reaction is 2.
So, is it any kind of exception or am I missing something?

This is because the E1CB reaction is a two step process. The elimination itself is unimolecular, but the concentrations of the intermediate depends on the concentration of the initial substrate and the base.