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Differentiating between Prins reaction / ene reaction experimentally
The Prins reaction – stepwise addition of an alkene to a carbonyl compound – leads to the same product as a concerted ene reaction. This is illustrated here with tin tetrachloride as a Lewis acid, which (as far as I know) lowers the activation energy of both pathways:

How can these two mechanistic pathways be differentiated experimentally?
One thing that I can think of right now would be the stereochemistry of the alkene formed - presumably in a concerted pathway, the cyclic transition state (envelope-like, if I remember correctly?) the substituent R' would be positioned equatorial or pseudo-equatorial. However, I'm not confident enough to undertake an analysis of the stepwise reaction.
I'm open to hearing about other possibilities, e.g. kinetic studies or spectroscopic studies. I suppose the free carbocation in the Prins reaction could lead to a more negative coefficient $\rho$ in a Hammett analysis, compared with a stepwise mechanism?
I'd also be glad if somebody could provide an example in the literature, if it exists. (I confess to not having looked this up - been very busy.)

Determine the kinetic isotope effect for the $\ce{C-H}$ bond that is broken.

The traditional Prins reaction was run in aqueous acid (H2SO4, HCl) with formaldehyde. The carbocation formed by addition of the alkene was trapped by a nucleophile, or if E1 elimination occurred, the most stable alkene formed. The Lewis acid-catalyzed ene reaction is intramolecular without nucleophile capture and the less stable alkene is formed, at least when an isopropenyl group is present in the reactant. Two examples follow:

Malonaldehyde 1, which has two enantiotopic aldehyde groups, reacts via transition state 2 to produce alcohol 3 as the major product. The cyclopentadiene ring is in effect a t-butyl group that prefers being equatorial in the transition state. Note the exocyclic double bond as opposed to a more stable endocyclic position. The hydroxyl group is axial. Aldehyde 4 follows the same type of pattern forming axial alcohol 5. If an open transition state were operable, the equatorial alcohol would be anticipated.
[PS: You can Google much of this information.]
1) J. Am. Chem. Soc., 1990, 112, 2149-2158
2) J. Am. Chem. Soc. 1984, 106, 718-721

I probably think it does answer the question. The main point is that the transition state for a cyclic mechanism leads to different stereocontrol.


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The impossibility of 100% ionic bond
Recently, I read the definition of  oxidation state on Wikipedia. It read that a 100% ionic bond is impossible. So what does a 75% ionic and 25% covalent bond mean at all?

I don't think this is impossible. I can charge a capacitor and cut it in half.

@Joshua Yes, it's not really a problem to make two charged things be it anion and cation or two charge spheres, it's just that there's simply no chemical bond in there. "100% ionic bond" is not a chemical bond at all.

This is because when a force is supplied to the ionic compound it leads to dislocation of ions by merging the like charged ions together which causes great repulsion force that leads to break down of compound.

A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion.  That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation.  That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract.
Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds.  It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold.

Not only some attraction of cation to electrons of anion, but quantum theory has probability distribution of each orbital extending to infinity. There isn't any absolute cutoff distance from the nucleus.

Actually bonds that are $100\%$ ionic, or close to it, are possible.  But they require a little ingenuity.  Instead of combining atoms of very different electronegativity, we can build multiatomic ions in which the charge is (formally) buried or highly delocalized making covalent bonding unfavorable.  We use molecular orbital structures rather than electronegativity to drive the charge separation.
One example of this is the cyclopropenyl salt $\ce{[C3H3]^+[SbCl6]^-}$ produced by Breslow and Groves in 1970 (J. Am. Chem. Soc. 1970, 92 (4), 984–987).  To form a covalent bond between the ions would require either occupying a highly destabilized antibonding orbital in the cyclopropenyl ring, or an interaction between the ring and poorly overlapping orbitals in the bulky anion.  So we have very little covalent bonding; the molecular orbital structure does not favor it.  We could get covalent compounds by transferring a chloride ion, but this reaction is energetically unfavorable.  The salt, with its ionic bonding between the aromatic cation and complex anion, is surprisingly stable for a species with a three-membered carbon ring.  Further stabilization may be achieved and ionicity reinforced by placing substituent on the cyclopropenyl ring.
Another example is provided by compounds in which alkali metals are disproportionated into cations and anion (see Alkalide).  Because the ionization energy of alkali metals (at least from sodium on down) are so small, the cation may be stabilized by forming a complex where the (formal) positive charge is deeply buried and again, no obvious means exists for covalent bonding.  The alkalide anion must be bonded by essentially ionic attraction to the cationic complex.

Reminds me a lot about ionic liquids.

In a "100 %" covalent bond, like the $\sigma_{ss}$ bond in the dihydrogen molecule, the electron probability density is perfectly symmetrically divided between the two bonded nuclei because both have the same electronegativity.
Below: the schematised electron density $\psi^2$, for a 100 % covalent bond:

But when the two atoms have different electronegativities, the electron probability density will be higher towards the more electronegative atom, see e.g. $\ce{HCl}$. We say the bond is polarised. See below, schematised, the permanent dipole of $\ce{XY}$, with partial charges $\delta +$ and $\delta -$:

Below: the schematised electron density $\psi^2$, for a bond between atoms with differing electronegativity (highest electronegativity to the right):

If the electronegativity difference is really high, see e.g. $\ce{NaCl}$, the bond starts being so strongly polarised (electron probability density strongly skewed to towards the $\ce{Cl}$ atom) that the bond starts taking on an ionic character.
Ionic and covalent must be considered relative to each other: in a "75% ionic" bond the electron distribution is strongly skewed toward the electronegative element.
There is however no meaningful way to measure this "percentage".

There is however no meaningful way to measure this "percentage". ||  NBO analysis?

https://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound https://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist

There are "meaningful ways" and even for symmetric bonds ionic component can be assigned ^.

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Reactions in the presence of sunlight
Some reactions in my textbook specify the need for the presence of sunlight,
eg-photolytic decomposition of silver bromide and substitution reaction of methane with chlorine.
What is going on in these reactions?
is there something special in light or can these reactions take place if provided with another source of energy like heat?
Also, can thermal reactions(reactions requiring heat) be done in the presence of sunlight?

https://chemistry.stackexchange.com/questions/46567/difference-between-thermal-reactions-and-photochemical-reactions

There is nothing magical about light; it just brings in some energy. What's important is that it brings a huge lump of energy to one point, at once. If you'll heat the system in the hope that the heat will provide the necessary energy for the reaction, then you might or might not succeed. In the worst case, before initiating your reaction, heat will initiate some other reactions you don't want.
I expect the radical bromination of some huge branched alkane to be a particularly bad case. Put it under sunlight, and it will go smoothly; heat it, and you'll have all sorts of side reactions, fragmentations, isomerizations going on at once. Of your examples, decomposition of silver bromide is relatively "well-behaved" (after all, $\ce{AgBr}$ must decompose since it doesn't have many other things to do), and $\ce{CH4 + Cl2}$ is somewhere in between.
If you try it the other way around, it is the same: some thermal reactions can be started by using sunlight instead of heat, some can't.

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Real gas not to behave ideal
One of my teacher said that the one with more molecular mass will deviate more from behaving as an ideal gas and he showed a graph given below.

Is his statement is correct? If no, then what's the reason?
In th graph $\ce{CO2}$ have higher molecular mass than other gases and its deviation is more than others, so from his point of view his statement is OK?
If there are any other reason please give me with easy language to understand.
N. B. If the question is not clear, please comment.

(everyone: please be nice)

One of my teacher said that the one with more Molecular mass will
  deviate more from behaveing as ideal gas and he showed a graph given
  below

Yes, that's basically correct. But it has to be understood as a general trend, to which many exceptions can be found (especially when the Molecular Masses are close).
Ideal gases would be made up of point masses (taking up no volume at all) that collide perfectly elastically.
Real gases are made up of real masses that take up some volume (just condense them to see that) and substances with higher Molecular Mass (MM) tend to have larger molecules that take up more volume.
Secondly, higher MM usually means larger electron clouds surrounding the nucei. During collisions, due to transient polarisation (the electron clouds repell each other, remember?), there's more interaction (Van der Waals forces) between larger molecules. Larger molecules tend to be more polarisable.
So by and large, smaller molecules/atoms behave more ideally.

I think you should explicitly state that the teacher made a sweeping generalization for which numerous exceptions could be found. So generalization holds well if $m_1 >> m_2$ but not if $m_1 \approx m_2$

@MaxW: I think the question is about trends, at least that's how I read it.

@Gert yeah i remember about polarization

Added comment about trend.

@Gert - I understand.


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How to get force constants in internal coordinates in Gaussian?
I have tried quite a few different ways to get Gaussian to print the force constants in internal coordinates. If the size of the molecule is small then it prints both the internal hessian and Cartesian hessian in the log file on using Iop(7/32=5), it outputs only the Cartesian hessian without the Iop command. But if the size of the molecule is large then it stops printing the hessian in the .log file. Though the Cartesian hessian is always printed in the checkpoint file. How do I get the internal hessian? I am also using #p to get Gaussian to write everything in the log file.

I haven't attempted to validate this post yet, but it is suggested in the following link that if you include #p freq=intmodes, it should return the internal hessian under "Internal force constants" in the log file. https://www.researchgate.net/post/How_do_I_extract_the_Hessian_from_Gaussian

I have tried that, you do get the internal modes in the log file if the molecule is small. But for bigger molecules it stops printing either the cartesian or the internal hessian in the log file. Though the cartesian hessian can be found in the checkpoint file.

You have to use freq=intmodes and use the command
formchk -3 xxxx.chk xxxx.fchk

to convert the binary checkpoint file to a formatted checkpoint file, which consists of text. The -3 requests the version 3 of formatted checkpoint files, which support all features of Gaussian 09 and 16.
Source: Ruixing Wang's answer on Researchgate, where I asked the same question.


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What is exactly the pH value of distilled water?
This was asked by one of my students in my class. I know it is the pH of 7, but others say it can be 6.55 pH

And I say it's 6.2. What others say is not an argument. Atmospheric gases are dissolved in distilled water (mainly oxygen, nitrogen, argon, carbon dioxide). Due to dissolved carbon dioxide, distilled water has a weakly acid medium and its pH is typically between 6 and 7. It vastly depends on how it's stored and what the ambient conditions are.

Does it depend on the temperature?

Of course it does https://chemistry.stackexchange.com/questions/39608/effect-of-temperature-on-ph-of-water

Yep, it does, and quite significantly. The solubility of $\ce{CO2}$ decreases with the increase of the water temperature (A page ripped off from CRC Handbook which you can use to support your explanation to the students).

So for pure distilled water at $25\ \mathrm{^\circ C}$ the pH would be 7.00. But the $K_\mathrm w$ is a function of temperature and hence the pH of pure water will vary with temperature too. The problem is that ordinary distilled water will contain some dissolved $\ce{CO2}$ which will lower the pH. You have to take extra precautions not to get $\ce{CO2}$ dissolved in the distilled water.

I just measured battery water for pH 5.5 and EC 1 µS/cm at 21 °C, using calibrated lab equipment. It was quite hard to get a stable pH reading because distilled water doesn't have enough ions for the probe to work reliably. As others have stated, the acidity is caused by CO2 absorbed from air.

keep in mind that water (distilled, deionized, or tap) is NOT "pure" (i.e., pH equal to 7). The moment it comes in contact with air, CO2 gas begins dissolving into it, forming carbonic acid. The actual pH, therefore, will often be slightly less than 7.

The pH of water in equilibrium with aerial CO2 is reportedly 5.65.( or less due raising CO2 level )


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What happens when you mix AgClO4 and CaBr2?

Predict if a reaction will occur when silver perchlorate and calcium bromide are mixed. If a precipitation reaction occurs, complete the reaction and include the physical states of the products, and balance the equation. If no reaction occurs, explain why.

I think a reaction does occur, but I'm not positive. Silver perchlorate and calcium bromide would give silver bromide and calcium perchlorate:
$$\ce{2 AgClO4 + CaBr2 = 2 AgBr + Ca(ClO4)2}$$
I'm unsure about the physical state of the products.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

You propose a double displacement reaction.  In water solution you can get that if there is something insoluble.  Is there such a thing here?  What is it?

Thank you for editing the question. Assuming the reaction takes place in aqueous media, you should've checked the solubility of the reactants in water first, and if they do dissolve in water, then the reaction takes place. Same for the products to determine their physical states. I retracted my close-vote and upvoted the question.

You have already figured out the common concept:

in water salts may dissolve
if they do, they will dissociate into their ions
the ions may recombine and form insoluble products which then precipitate from the solution

Good!
How to figure out whats happens for a particular combination of salts (such as silver chlorate and calcium bromide)?

Look up the solubilities of the starting materials and the possible products. If you don't have solubility tables in your study materials, wikipedia is ok.
If you run into values for a solubility product ($K_{sp}$) rather than a solubility given in grammes per 100 mL, you should be on red alert!

In the case of $\ce{AgBr}$, you should find a value around $5\cdot10^{-13}$. Please read about the solubility product to figure out what it means!
If you figure out that it indicates remarkably low remaining concentrations for a particular combination of ions in solutions (while the majority of your material has precipitated) you're on the right track!
In such cases, you might then go into details in your reaction equation:
$$\ce{2AX(s) + BY2(s) ->[H2O] B^{2+}(aq) + 2 X-(aq) + 2AY v}$$


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Why does solvent particles move towards solution when semi permeable membrane is added?

When a SPM is added in between solute and solution the water molecules move towards salt water even though the concentration is same both sides. Why does this happen? According to SPM's perspective, concentration (of water) at salt water side is lower than that at fresh water side but how does that change the fact that the overall concentration is same both sides? My book says it depends on vapour pressure. If that's the case then osmosis should take place without SPM because it depends only on vapour pressure?

Osmosis surely does take place without a membrane. Related: https://chemistry.stackexchange.com/questions/49435/how-can-you-explain-the-energy-change-in-this-phenomenon

Here what happens is that the fresh water is a pure solvent and a pure solvent always has a greater vapor pressure than that of a solution. So due to the greater vapor pressure on water's side, the molecules of water will experience a force in salt water's direction as the pressure is less in salt water side. Therefore the water will flow towards the salt water.


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Can Gas diffuse according to its partial pressure, not according to its concentration in gas mixture?
by Raoult's law,
Partial pressure  =  total pressure  X  concentration(mole fraction)
In gas exchanging situation of 2 gas mixtures which are different in total pressure from each other(like higher temperature?) and include gas A of same concentration, each gas A in 2 mixtures will be the same in its concentration but different in partial pressure. Then, gas A will diffuse from high partial pressure to low partial pressure even though this diffusion is from low concentration to high concentration. Is this right??
Sorry for any misconceptions.

The problem is basically expansion of gas. You assumed in Q that gas A has the same molar fraction in both halves of the container. ...

Strictly speaking, diffusion occurs due to gradients in chemical potential.
One implication of this is that sometimes diffusion can occur from lower to higher concentration.
See, for example, this nice class summary
If the total pressure is different, the transport will also occur by fast flow convection instead of diffusion until total (mechanical) pressures are equal (unless thermal transpiration effects are significant).


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What happens when an alkylborane is treated with acetic acid?
In the hydroboration reactions of alkenes I've seen so far, they were always followed by oxidation with $\ce{H2O2/NaOH}$ (to yield alcohols).
Today I found a peculiar reaction in which acetic acid-d1 was added after hydroboration to give a deuterated product:

I don't quite understand what exactly happens after the hydroboration part. Can someone enlighten me here?

Acetic acid effects protonolysis of the borane:

If AcOD is used then RD (instead of RH) is formed.

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Why is thiourea a monodentate ligand?
I was going through a list of monodentate ligands, and I came across examples like $\ce{H2O}$ (which has 2 lone pairs), $\ce{Br-}$, $\ce{I-}$ (both of them have more than one lone pair). Similarly, thiourea has 4 lone pairs in all (2 from nitrogen and sulphur each).
When I searched about it, I realized that the oxygen in water can't simultaneously form bonds using both the lone pairs (due to some obvious geometrical reasons, it said).
However, I couldn't figure out any such (obvious) reasons in thiourea.
Can anyone please help?

There are plenty $\mu_2$-thiourea complexes with Cd, Pb, Ag, Cu. Just saying.

He said that thiourea can act as a bridging ligand. It is also wrong to say that water can only form one co-ordinate bond, the reasons why it often doesn't are not at all obvious. Could you comment, at which level you are studying and which source said that it is obvious.

This is not really a standalone answer, rather a pointer or an exeption. There are few complexes where deprotonated thiourea $\ce{S=C(NH2)NH}^-$ acts as a bidentate ligand. One example is (aminomethanethioamido)-hydrido-tetrakis(trimethylphosphine)-technetium(III) $\ce{[Tc(P(CH3)3)4(SCN2H3)]+}$ complex[1]:

The fact that thiourea ligand is deprotonated and is coordinated asymmetrically within equatorial plane in $\eta^2-\ce{N,S}$ bonding mode has been proven by NMR and SCXRD studies. From the article [1, p. 4642] (emphasis mine; $\ce{tu}$ = thiourea $\ce{S=C(NH2)2}$):

The $\ce{Tc-N}$ distance, 2.190(13) Å, is long when compared to
known $\ce{Tc(III)}$ complexes. The range for $\ce{Tc-N}$ bonds was found
to be 2.041 Å ... to 2.124 Å ... . This reflects delocalization of negative charge within the bidentate deprotonated thiourea ligand. This elongation is also seen in the $\ce{Tc-S}$ bond, 2.543(4) Å, which is significantly longer than $\ce{Tc-S}$ bonds in the parent compound $\ce{[Tc(\textit{S}-tu)6][Cl]3}$ ... .

As Martin pointed out, the reasons for the observed denticity are not always transparent. Simply put, it seems like in this case in order to be a bidentate ligand, thiourea molecule needs to increase its nucleophilicity and remove steric hindrance caused by $\ce{H}$-atoms lying within the $\sigma_\mathrm{h}$ plane.   Both can be achieved by eliminating a single proton from the thiourea molecule.
Reference

Watson, P. L.; Albanese, J. A.; Calabrese, J. C.; Ovenall, D. W.; Smith, R. G. Inorg. Chem. 1991, 30 (24), 4638–4643. DOI: 10.1021/ic00024a035.

Okay. So as far as it has all the H atoms, it acts as a monodentate ligand. If we somehow remove an H atom, it can act as a bidentate ligand, right? Is there any possibility that it can act as a tridentate ligand (because it has 4 lone pairs) ?

@user166465 Well, I haven't discovered any examples of existing compounds (certainly not in crystal structure database), plus how you would present it geometrically? Thiourea ligand is basically a triangle which you are trying to attach to a single point by all three vertices, which leaves two nucleophilic centers (metal and central carbon – refer to the picture) in the vicinity, which would greatly destabilize the complex. To answer quantitatively, one need to do quantum chemistry calculations, which I' unfortunately not capable of.


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Marker dissipates on White board paint
What causes white board markers to dissipate when writing on dry erase painted surfaces. It almost immediately start to pull together and seems as if it disappears

"Dissipate" is quite the opposite of what you seem to mean.

@FellowUsers: This is not a duplicate (though the poor wording makes it superficially similar). It's a phenomenon I experience quite often myself: Using markers to write on white-boards, result in smooth, continuous lines. However, if I use the same marker to write over another surface (like a plastic), there are times when I notice the lines of wet ink begin to "pull together" and shrink into shallow puddles. Zhe's answer correctly addresses this question. I vote to reopen.

The issue at hand is competition between the intermolecular forces that pull the solvent together and the forces that cause the solvent to adhere to the surface. In this case, the solvent's interactions with itself are greater, causing the ink to bead. Since the amount of ink that is actually drawn out when writing is very small, the beads of ink are tiny and very hard to see.


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What are the percentages of isotopes?
The average atomic mass of a sample of an element  'X' is 16.2u. What are the percentages of isotopes $\ce{^16_8X}$(atomic number = 8, atomic mass = 16) and  $\ce{^18_8X}$(atomic number = 8, atomic mass = 18) in the sample?

The formula for finding the percentage of isotopes in an element's atomic weight? Also related: https://chemistry.stackexchange.com/q/10206/4945

Hey Moderators, I tried something new on this obvious candidate for closed as OT, homework.

Hey @TAR86, that's nice of you.

We have two isotopes, $A,B$ with atomic masses $m_A, m_B$. We thus have two unknowns $x_A, x_B$ representing the percentage amounts of each isotope (i.e. the mole fraction), which are trivially connected as

 $$ x_A + x_B = 1 $$

We also know something about the average atomic mass $\overline{m}$, namely

 $$ x_A \cdot m_A + x_B \cdot m_B = \overline{m} $$

So we now have a system of two linear equations to plug in values and solve. Steps:

 $$ x_A = 1 - x_B $$

.

 $$ (1 - x_B ) \cdot m_A + x_B \cdot m_B = \overline{m} $$

.

 $$ x_B \cdot (m_B - m_A) = \overline{m} - m_A $$

.

 $$ x_B = \frac{\overline{m} - m_A}{m_B - m_A} $$

And plug back in the first equation.

$x_A$ representing the percentage amounts of each isotope = mole fraction?


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Will sodium sulfite effect the oxidization of toluene to benzoic acid with potassium permanganate?
I am trying to oxidize toluene to a mixed salt $\ce{K/Na}$ benzoate/ benzoic acid using the potassium permanganate method, but I am using OTC sourced $\ce{KMnO4}$ which has a significant amount of sodium sulfite contamination.
Will that hinder the reaction?  And is there a practical method for filtering said contaminants?

I don't believe KMnO4 can have sodium sulfite in it. Sulfate, maybe.

As @Ivan Neretin said. $\ce{KMnO4}$ is such a strong oxidiser that no sulphite could co-exist with it. It would be oxidised to sulphate.

How do you know your permanganate contains sodium sulphite?
It seems highly unlikely because $\ce{KMnO4}$ is a very strong oxidiser and sulphite is very easily oxidised to sulphate:
$$\ce{SO3^{2-}\to SO4^{2-} +2e-}$$
Also, it seems improbable to find a sodium salt in a potassium salt.
Even if the $\ce{KMnO4}$ contained some "surviving" $\ce{Na2SO3}$, during the oxidation of the toluene, it would quickly be oxidised to sulphate. At worst your benzoic acid yield would suffer slightly.

Sorry I wrote the wrong one.  The KMnO4 is a filtering aid for home green sand water filtration systems and on the container it states that there is sodium sulfate at roughly 20-40% with KMnO4 at 60-100%

You can test any significant amount of sulfite contamination in $\ce{KMnO4}$ soln. using anion analysis. You can add dil. $\ce{HCl}$ to the soln. to evolve any sulfur dioxide gas or add any barium salt to produce $\ce{BaSO3}$(white precipitate).
But I don't think  there would be any amount of sulfite left in the soln. as sulfite would be oxidized to sulfate as $\ce{KMnO4}$ is strong oxidizer.
$$\ce{3Na2SO3 + 2KMnO4 + H2O → 3Na2SO4 + 2MnO2 + 2KOH}$$
$\ce{MnO2}$ is a black insoluble salt that would turn the product solution black. It can be decanted/filtered off. Both sodium sulfate and potassium hydroxide are soluble and it is hard to distill off. But no need to worry, amount of sulfite contamination is so little that it is hardly taken into consideration. And what worse could happen? During oxidation of the toluene, sulfite would quickly oxidised to sulphate and your benzoic acid yield would suffer slightly by a little amount and that amount is insignificant.


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What all computing knowledge should a computational chemist traditionally have?
If I want to develop algorithms or computer programs for studying chemical systems, where should I start? Is learning one or two programming languages like FORTRAN enough? And if I do it, then how do I proceed further?
I would be grateful if someone can tell me about a book or any other learning resource.
Edit:
I am 3rd year undergraduate in chemistry. I am interested in pursuing computational chemistry for my master's as well as PhD. I am mostly interested in electronic structure theory and ab initio calculations.
Regarding the theory, I think I have got that part covered as I am taking courses and learning stuff on my own as well (QM, Linear Algebra, Quantum Chemistry etc.). What I want to know is the fields in computer science that I should have a knowledge of so that I can start learning it and be prepared with it if I encounter it in my grad school.

The Szabo and Ostlund book is probably the best for a beginner. https://www.amazon.de/Modern-Quantum-Chemistry-Introduction-Electronic/dp/0486691861

I guess you might want to specify what chemical systems and what properties you want to study (and add corresponding tags to attract the specialists). I doubt you start from scratch with Fortran like in 1970s, these days there are many ready-to-go solutions implemented in various libraries. Availability of those libraries often dictates the choice of programming language. E.g. I've heard that in chemical engineering python is quite popular, whereas large farmacompanies tend to hire Java-programists. I'm not a programmer BTW, just an observer, so take this with a bit of a skepticism:)

I would recommend to look for open jobs in the area you are interested in. Read the required skills carefully, word-by-word. This should give you a deep insight into the expected skills/knowledge.

The traditional computational chemist should have a much stronger background in chemistry than in computing. There is a long list of books in our resources post.

If you are looking to develop production level code that would be implemented in say Gaussian or QChem, I would recommend trying to learn Fortran and C++. It's not a requirement and I know people who do method development who didn't know these languages beforehand, but it would certainly give you a leg up. Beyond that, you will want to find a masters/PhD research group that focuses on methods development. From just a general background perspective, it also might be useful to find a course/book on numerical methods of integration and optimization.

Don't worry about the programming so much, rather think about chemical theories or problems you find interesting, and then see what methods are available. Mathematical maturity is much harder to come by than learning how to program. Although implementations are almost always based on linear algebra, method derivations require calculus and differential equations. Understanding batches of results requires statistics. All programming that you would plausibly do requires discrete math.

If "develop algorithms or computer programs" is crucial in your post then it's not what "traditional computational chemistry" is. It strongly depends on programming skills - you'd need to know many languages and be good in programming to significantly contribute in development.

@Mithoron sorry about the late response: this is partially a myth (no pun intended). Everyone is correct that it is not traditional computational chemistry, which is just applications with maybe some scripting. However, "you'd need to know many languages and be good in programming to significantly contribute", ignoring some definition of "significantly contribute", is not true. Anecdotally, most of the people I know are only remotely familiar with 1-2 languages and are nothing close to software developers.


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Is 2a¹H-cyclopenta[cd]indene aromatic?
In this question, 2a1H-cyclopenta[cd]indene is said to be aromatic.
My doubt is that the central C atom (connecting the three cyclic rings) is sp³ hybridized. So I think it should be non planar thus making compound non aromatic
Can someone please explain this compound’s aromaticity?

sp³ carbon can't and doesn't have to be planar. We just don't care about it at all. Otherwise that isopropyl tail would make cumene non-aromatic.

@orthocresol Do you know whether this compound by any chance has a short traditional name which can be included in title?:)

@andselisk, that crossed my mind, but I don't know of anything. A bit of research suggests that 7aH-cyclopent[cd]indene might be suitable, but I know nothing about this kind of nomenclature, so I opted to leave the title as is.

@orthocresol 7bH-cyclopenta[cd]indene corresponds to the former (1998) numbering, which is no longer recommended. The name 2a1H-cyclopenta[cd]indene is in accordance with current recommendations.

It's a modified form of [10] annulene (https://en.wikipedia.org/wiki/Cyclodecapentaene) where internal bridges are used to hold the outer, conjugated ring in a planar geometry and thus with the $4n+2$ rule that outer ring is aromatic.
You may be more familiar with the methylene bridged version of [10] annulene that works the same way, but with a simpler bridging structure.  See the above mentioned Wikipedia link.
Addendum:
As mentioned in comments, it is overly simplistic to use the $4n+2$ rule.  IUPAC justifies it only for simple annulene systems.  When you have internal bonds in the ring like the compound here, you no longer have the symmetry and degeneracy structure on which the rule is based.
When you have saturated interior atoms surrounded by an outer conjugated ring, the $\pi$ orbitals are only slightly altered and that is why we are tempted to still apply the $4n+2$ rule.  But beware: if we unsaturate any internal atoms, in this case by removing the hydrogen from the central carbon, the $\pi$ orbitals are altered more considerably and the $4n+2$ rule is well past its use-by date.
Maybe in the end we can use the rule as a guide, but ultimately we need to back our conclusions up with hard evidence.  In this case the conclusion of aromaticity is supported by Gilchrist et al.[1], in which the authors studied substituted versions of the compound involved here.  They found proton nmr shifts consistent with an aromatic outer ring.
Reference

Thomas L. Gilchrist,   Charles W. Rees,   David Tuddenham  and  David J. Williams (1980). "7b-Methyl-7bH-cyclopent[cd]indene-1,2-dicarboxylic acid, a new 10 π-electron aromatic system; X-ray crystal structure". J. Chem. Soc., Chem. Commun., 1980, 691-692.

That Wikipedia page also gives this nice reference: doi:10.1039/C39800000691 where the authors made substituted versions of the compound in question. In their analogues, the central "tetrahedral" carbon bears a methyl group which comes in at –1.34 or –1.25 ppm in the proton NMR, indicating aromaticity. The peripheral hydrogens also appear between 7.7 and 8.5 ppm. This would probably be a helpful thing to include in the answer.

Hückel's rules cannot be employed for this compound!

@Martin, I suspect they are so applied in the OP's textbook.  I rolled with the punch.

I know - and I am very sad about that - that Hückel's rules are often misrepresented in (especially) undergraduate studies, and consequently taught wrong. It is one of these common problems that leads to wrong conclusions, and a completely distorted understanding of chemistry, especially aromaticity. And it is so hard to get rid of it...


text:
Auto-catalysis Mechanism
I just read about the theory of auto-catalysis, and here's one thing which is sort of unclear to me:

In this type of catalysis, one of the reaction product catalyses the reaction. For example, in the oxidation of oxalic acid by potassium permanganate, Mn(II) cation is formed which is known to accelerate the reaction. So when potassium permanganate is added to a warm solution of oxalic acid in presence of dilute sulphuric acid (acid medium catalysis), initially there is a time lag before decolourisation occurs. As more permanganate is added, the decolourisation is observed to be nearly instantaneous.

Points I didn't understand:

How does the formation of Mn(II) cation accelerate the reaction? Could someone please help me with understanding the mechanism?
What is the reason for time lag in decolourisation of the solution? Also, what exactly do they mean by decolourisation here? My guess is that it's referring to the bright purple coloured permanganate solution losing its colour after getting reduced to Mn(II) cation.
How does the time lag reduce? Are there ways to estimate or determine the same? Of course, it'd depend on the reaction kinetics; but I'm looking for a mathematical viewpoint on how the catalyst influences reaction kinetics, except for decreasing the activation energy.

Thank you.

(1) There are several web sites giving details of this reaction and so this need not be copied out here.
(2,3) Rather than describe a complex reaction scheme it is easier to understand a generic autocatalytic reaction. The general autocatalytic reaction of species A with catalyst B has the form $\ce{A + B $\to$ P + 2B}$ and the rate equation is deceptively simple and is $da/dt =- kab$ if $a$ and $b$ are the concentration A and B. The important point here is that species B is both reactant and product. As more B is produced than that there initially, the reaction accelerates as it proceeds. This is the reason for the time lag: the reaction rate is initially small but as reaction proceeds the extra B produced speeds it up. Eventually A runs out and so the reaction stops. (The colour change you mention is just a measure of the concentration of permanganate. Note also that from the Beer-Lambert law describing transmission of light the absorbing species concentration is related exponentially to light transmission (and so absorption) and this probably complicates what you observe by eye).
If $x$ is species A concentration, the rate equation is, $\displaystyle \frac{dx}{dt}=-kxb=-kx(a_0+b_0-x)$ or $\displaystyle \int\frac{dx}{(a_0+b_0-x)x}=-kt+c$ which can be integrated using partial fractions;
$$\frac{1}{(a_0+b_0-x)x}=\frac{1}{(a_0+b_0)}\left[\frac{1}{x}+\frac{1}{a_0+b_0-x} \right]$$
which produces two log functions when integrated. The initial conditions are $x = a_0$ when $t$ = 0 then the integration constant is $\ln(a_0/b_0)$ and the rate equation,
$$ \ln \left(\left| \frac{xb_0}{a_0(a_0+b_0-x)}     \right|  \right) =-(a_0+b_0)kt$$
and the absolute value is added because the log cannot be negative. Rearranging produces $\displaystyle x = a_0\frac{a_0+b_0}{a_0+b_0e^{-(a_0+b_0)kt}} $.   This and also the increase of $B =a_0+b_0-x$ are plotted.

Figure Autocatalysis $\ce{A + B $\to$ P + 2B}$. Initially species A decreases slowly as the concentration of B is small. As the reaction proceeds, more B is produced and although A is reduced overall, the rate increases and A is consumed even more rapidly. At longer times, the concentration of A becomes so small that even though that of B is large the reaction rate is slow.

Thank you. Could you specifically share certain links which could answer the first part (1) of my question?

I searched for 'oxidation of oxalic acid  by permanganate' , there are several pages with details including this one  https://chemistry.stackexchange.com/questions/4310/what-is-the-reaction-between-oxalic-acid-and-potassium-permanganate


text:
Reaction Kinetics and Catalyst Addition
The addition of a catalyst is known to increase the rate constant of a reaction by providing an easier pathway for the reaction to occur, one with lesser activation energy.
My doubt is:
Is it possible that the addition of a catalyst, alters the order of a reaction?
I haven't come across any such reaction so far, so it'd be great if someone could provide an example as well; if the reaction order can be changed upon addition of catalyst
My thoughts:
Sometimes, catalysts participate in the reaction; but are regenerated in the end without change in mass and chemical composition.
The very fact that a catalyst can participate in a reaction, should mean that it can influence the reaction's order, right?

It certainly can. In most situations, the resting state has free catalyst, and if the catalyst is involved in the transition state of the rate determining step, then the catalyst concentration enters into the rate law for the catalyzed reaction.

Ohh! Could you please provide a more specific example, in which the catalyst concentration influences the rate law?

You can pick like any example you can think of.


text:
Do atoms in a molecule actually keep rotating freely in case they have sigma bond?
While reading about cis-trans isomers I found the concept of free rotation of atoms.it says 2 atoms connected by sigma bond do not have trans sis isomer because of free rotation.while in case of pi bond free rotation is not possible. My question is : do we mean actual rotation of atoms in molecule when we say free rotation? Does the rotation happen spontaneously? What things are responsible for the rotation? Or is it just a concept to understand the concept of cis-trans isomers?

Are you referring to molecular rotation? http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/rotrig.html

No I am not referring to molecular rotation

Oh, I see what you mean now. Will answer.

related https://chemistry.stackexchange.com/questions/64219/when-is-rotation-around-double-bonds-to-form-cis-trans-isomers-allowed

With a molecule the whole thing can rotate both in the gas phase and in solution. Additionally groups of atoms within a molecule can rotate provided there is enough thermal (or light) energy to overcome any steric interaction from bulky groups or bond energy that will otherwise prevent this. Thus C-C bonds rotate relatively easily at room temperature but C=C do not as thermal energy is not enough to overcome bond energy.

You really are referring to molecular conformations:

Conformation, any one of the infinite number of possible spatial arrangements of atoms in a molecule that result from rotation of its
  constituent groups of atoms about single bonds.

Below are given two possible (out of many!) conformations of the ethane $\ce{C2H6}$ molecule:

The single $\sigma$ bond allows free rotation around the inter-nuclear axis of the two $\ce{C}$ atoms. One consequence is that around the single bond no cis-trans isomers can exist.
But if we add a $\pi$ bond, so we have a double bond and ethene $\ce{C2H4}$, then the methylene groups can no longer freely rotate about the bond.
That's because the electron probability density added by the $\pi$ bond doesn't lie on the inter-nuclear axis, but above and below it. Double bonds, and by extension also triple bonds, are rigid, with respect to rotation of the atoms they bond. This can give rise to cis-trans isomers.
Below: highly schematised rendition of the electron density of a $\pi$ bond:

It doesn't exactly answer question, only elaborates on what OP told. Actual rotation depends on temperature.

Sorry to say that I still have questions.my question is that the "free rotation" is it like happening all the time? Do the c atoms keep rotating always? Is it spontaneous ?

@user136782: Hi. Did you read the link I provided? The term 'free rotation' is a little misleading, as different parts of the molecule will still electronically repel each other. But the impossibility of cis-trans isomers about a single bond does stem from the possibility of rotation about the inter-nuclear axis.


text:
What happens to the colour change of an equilibrium reaction when one of the reactants begins to form a complex? (specific problem included)
I'm working on an equilibrium lab/quiz for my chemistry class and I came across this reaction:
$\ce{Fe^{+3}\ \text{(pale yellow)} + SCN- <=> FeSCN^{+2}}\ \mathrm{(red)}$
Then the lab said a stressor was added: $\ce{Na2HPO4}$ was added to the equilibrium reaction -- which formed a complex with some of the $\ce{Fe^{3+}}$ ions.
This stressor changed the solution's color from an orange to a pale-yellow as seen in the image below:
(left: before stressor ($\ce{Na2HPO4}$) was added; Right: after stressor ($\ce{Na2HPO4}$) was added)

I was asked to explain why that after $\ce{Na2HPO4}$ is added (which actively forms a complex with the iron ion ($\ce{Fe^{3+}}$)) the system/solution turns increasingly pale-yellow.
Basically I want to know what is happening chemically when $\ce{Na2HPO4}$ is added, and more specifically why their is the resultant color change from orange to pale-yellow.
My basic confusion: wouldn't the formation of a complex between $\ce{HPO4^{2-}}$ and $\ce{Na2HPO4}$ actively remove the $\ce{Fe^{3+}}$ ions responsible for the pale-yellow color of the solution? Making the solution less rather than more yellow?
Thanks!

Going by the right hand picture, what was formed there is the very poorly soluble and lightly yellow $\ce{FePO4.2H2O}$. A precipitate and clear supernatant liquid can clearly be seen.
If so, the reaction is:
$$\ce{FeSCN^{2+}(aq) + HPO4^{-}(aq) + 3 H2O(l) \to FePO4.2H2O(s) + SCN-(aq) + H3O+(aq)}$$
This explains why on adding the $\ce{Na2HPO4}$ the red colour of the $\ce{FeSCN^{2+}}$ complex disappears and a yellowish, flocculant precipitate appears. This is not however, as the OP suspected, a complex. It's merely an almost insoluble salt.
Some info on $\ce{FePO4}$.

A true complex would have been formed by the addition of $\ce{CN-}$, which would have led to the formation of $\ce{Fe(CN)6^{3-}}$, which has a much higher formation constant ($K_f$) than the thiocyanate complex. It's also weakly coloured and would not form a precipiate. The reaction would have been:
$$\ce{FeSCN^{2+}(aq) + 6 CN-(aq) \to Fe(CN)6^{3-}(aq) + SCN-(aq)}$$
The solution would have evolved from bordeaux to weakly amber/red.

Gert this might very well be the right answer (and thank you a ton for the quick reply), but quick clarification: I didn't suspect that the reaction between Na2HPO4 with Fe+3 forms a complex, my lab question informed me so "the HPO4-2 ion forms a complex with the Fe+3 ion." So maybe both reactions happened?.... (the insoluble salt you spoke of aswell as the ironIII complex the lab spoke of)

@Charly = You can clearly see that something, the precipitate, is scattering light in the "yellow" test tube. A complex would have a "clear" (not colorless) solution as the red solution does. I'll point out that the lab almost certainly had a centrifuge. Centrifuging the yellow test tube would have separated the ppt nicely from the solution.

@Charly: sorry but your lab question was wrong. Wouldn't be the first time or the last.

Please note that the iron(III) thiocyanate complex is better described as the following: $\ce{[Fe(SCN){$x$}(H2O){$6-x$}]^{3-x}}; x \in {1,2,3}$. It is also with itself in equilibrium. || Most of the time there is no need for bracketing subscripts within mhchem constructs, i.e. \ce{Fe^3+} is actually preferred over \ce{Fe^{3+}}. There is also a wide variety of arrows available, that better typeset than \to, e.g. ->, <=>, <-->, etc. Just for further reference.


text:
Calculating weight of soluble salt in a solution
I got a solution which density is $\pu{1.035 g mL-1}$ and it contains soluble salt $\ce{Ba(NO3)2}$ at concentration of $\pu{0.250 M}$. The teacher is asking from me to "clear" the solution from the $\ce{Ba^2+}$ ions by using a sedimentation with $\ce{PO4^3-}$ ions that the concentration of $\ce{Ba^2+}$ would be less than $\pu{7.00 ppb}$.
Then she is asking if its possible (well yes it is, but I don't know how to explain this though)? And, if it is possible, what would be the weight of the soluble salt $\ce{Na3PO4}$ that we need to put into a $\pu{50.0 mL}$ of the solution?
And I got the $K_\mathrm{sp}(\ce{Ba3(PO4)2}) = \pu{4.03e-25}$.
Now I probably started by making the right ions equation, which is:
$$\ce{2 Na3PO4 + 3 Ba(NO3)2 -> 6 NaNO3 + Ba3(PO4)2}$$
Then, I calculated the net mass of the solution by multiplying the given density  $\pu{1035 g L-1}$ by given Volume $\pu{0.05 L}$.
And finally I calculated the amount of $\ce{Ba(NO3)2}$ and of $\ce{Na3PO4}$ which are $\pu{0.0125 mol}$ and $\pu{0.0083 mol}$, respectively.
That is the point where I don't know how to mix all this information and to get the right answer.
The final answer: It's possible. If we will put more than $\pu{1.80 g}$ of $\ce{Na3PO4}$ to $\pu{50.0 mL}$ of the solution, the ion concentration of $\ce{Ba^2+}$ will be less than $\pu{7.0 ppb}$.

Hello and welcome to Chemistry.SE. Please visit this page, this page and this one on how to [edit] this post or format your future posts better with MathJax and Markdown. Also, note that "I calculated the moles of" should read "I calculated the amount of" instead. Just like you would say "I calculated the mass of" not "I calculated the kg of". Anyway, best of luck with your question!

Thank you very much for your comment. Ill remember that for the next time :) @airhuff

@Chemgoat regarding your edit: that's not how you use mhchem. Please do refer to How can I format math/chemistry expressions here?. \pu{...} is used for physical units, \ce{...} for chemical formulas.

Ok... It is hard to read you post since the details are scattered around.
Problem Statement
Given: 50 ml of a 0.250 molar solution of barium nitrate with a density of 1.035 g/ml  and that barium phosphate has a $K_\mathrm{sp} = 4.03 \times 10^{-25}$.
Question: How much $\ce{Na3PO4}$ must be added to reduce the concentration of barium to less than 7.00 ppb.
Complications with problem
Initial density

The density of the initial solution, 1.035 g/ml, is useless.

Final density

We need a density for the final solution to convert 7.00 ppb $\ce{Ba^{2+}}$ to a concentration of grams per liter. There is no density for the final solution, so the assumption will be that the density is 1.000 g/ml which means that a liter of the final solution will weigh 1000. grams. Since each $\ce{Ba^{2+}}$ cation will be replaced by 2 $\ce{Na^+}$ cations, the final solution will be at least 0.50 molar in $\ce{NaNO3}$. Thus the assumed density of 1.000 g/ml will be too low.

A pragmatic point here about the experimental procedure.

You typically wouldn't want to add solid $\ce{Na3PO4}$ to a solution containing $\ce{Ba^{2+}}$ since $\ce{Ba3(PO4)2}$ would precipitate on the particles of $\ce{Na3PO4}$ sealing them from the solution and thus preventing complete dissolving of the $\ce{Na3PO4}$ particles. So experimentally you first need to dissolve the $\ce{Na3PO4}$ in a small amount of water and then add the solution of $\ce{Na3PO4}$ to the solution of $\ce{Ba(NO3)2})$. Thus the initial volume would really be over 50 ml.

Rounding errors

The calculations will be carried to 5 significant figures to try to avoid rounding errors due to intermediate results. The final result will be properly rounded to 3 significant figures

Calculations
Moles of barium in the initial solution
In the initial solution $\ce{[Ba]} = 0.250\ \mathrm{m}$ so 50 ml contains
$\mathrm{m}\ \ce{Ba^{2+}} = 0.250\ \mathrm{m/l}\times 0.050\ \mathrm{l} = 0.0125\ \ \mathrm{m}$
Moles of barium in the final solution
In the final solution for Ba:
$\ce{Ba^{2+}} = 7.00\ \mathrm{ppb}$
$\quad\quad = (7.00\times 10^{-9})\times 1000\ \mathrm{g/l}=7.00\times 10^{-6}\ \mathrm{g/l}$
$\quad\quad  = \dfrac{7.00\times 10^{-6}\ \mathrm{g/l}}{137.33\ \mathrm{ g/m}} = 5.0972\times 10^{-8}\ \mathrm{m/l} $
For 50 ml of solution this is:
$\mathrm{m}\ \ce{Ba^{2+}} = 5.0972\times 10^{-8}\ \mathrm{m/l}\times 0.050\
\mathrm{l} = 2.5486\times10^{-9}\ \ \mathrm{m}$
$2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ so the residual amount of barium in solution can be ignored.

(1) *In determining the amount of barium phosphate that will precipitate, the amount of barium left in solution is so small that it doesn't matter if the solution has a density of 1.000 g/ml or 1.05 g/ml. So the density of the sodium nitrate solution doesn't really matter for that part of the problem.
(2) $2.5486\times10^{-9}\ \ \mathrm{m} \ll 0.0125\ \ \mathrm{m}$ This is a significant figures argument. In other words for intermediate calculations I'm using only 5 significant figures.
    $$\begin{align}
0&.012\,500\,000\,000\,0 \\
-0&.000\,000\,002\,548\,6\\
\hline
0&.012\,499\,998\,551\,4
\end{align}$$ Now when trimmed to 5 significant figures the intermediate result is 0.012500. But the 0.0125 is exact, so I don't need the two extra zeros at the end.

Residual phosphate in the final solution
For the residual phosphate, in excess of that needed to precipitate the stoichometric amount of barium is:
$K_\mathrm{sp} = \ce{[Ba^{2+}]^3[PO4^{3-}]^2}$ or $\ce{[PO4^{3-}]} = \sqrt{\dfrac{K_\mathrm{sp}}
{[Ba^{2+}]^3}}$
$\ce{[PO4^{3-}]} = \sqrt{\dfrac{4.03\times10^{-25}}{(5.0972\times 10^{-8})^3}} = 5.5164\times10^{-2}\ \mathrm{m/l}$
For 50 ml this would be:
$\mathrm{g}\ \ce{PO4^{-3}}\ \mathrm{as }\ \ce{Na3PO4} = (5.5164\times10^{-2}\ \mathrm{m/l})\times 0.050\ \mathrm{l}\times 163.94\ \mathrm{g/m} = 0.45218\ \mathrm{g}$

This is too great of an amount of residual phosphate in solution to be ignored. Thus we need to add this to how many grams of trisodium phosphate are needed to precipitate 0.0125 moles of $\ce{Ba^{2+}}$ as $\ce{Ba3(PO4)2}$.

Stoichiometric amount of phosphate to precipitate all the barium
Each mole of $\ce{Ba3(PO4)2}$ requires 3 moles of $\ce{Ba^{2+}}$. So for 0.0125 moles of $\ce{Ba^{2+}}$ there would be
$\mathrm{moles}\ \ce{Ba3(PO4)2} = \dfrac{\mathrm{moles}\ \ce{Ba^2+}}{3}= \dfrac{0.0125}{3} = 4.1667\times10^{-3}$
But each mole of $\ce{Ba3(PO4)2}$ requires two moles of $\ce{PO4^{3-}}$
$\mathrm{moles}\ \ce{PO4^{3-}} = 2\times(\mathrm{moles}\ \ce{Ba3(PO4)2})= 2\times( 4.1667\times10^{-3}) = 8.3333\times10^{-3}$
$\ce{Na_3PO4} = \mathrm{MW}\ 163.94\ \mathrm{g/m}\\
\quad\quad\quad\quad= 163.94\times8.3333\times10^{-3} = 1.3662\ \mathrm{g}$
Total amount of trisodium phosphate needed
$\mathrm{Total}\ \ce{Na_3PO4} = 1.3662 + 0.4522 = 1.8184\ \mathrm{g}$
Now rounding the final result to 3 significant figures we get 1.82 grams.
Now the other pertinent fact is how much $\ce{Na3PO4}$ can be dissolved in 50 ml at 25 °C? Wikipedia shows 14.5 g/100 mL (25 °C), so 7.25 g/50 mL (25 °C). Thus it should be possible to dissolve the $\ce{Na3PO4}$.

!@#$%^&* Think that is 4 edits to get the calculations right. I really should do the problems on paper first.  Doing the markup is so distracting to me that I lose track of the problem. :-(

I see. Another happy $\mathrm{\LaTeX}$ user :D

Thank you very much MaxW, you are the best. It helped me a lot! I got a few more questions maybe you can answer me: 1. I didnt understand why you assumed that the final solution would have no density? 2. I know that ppb is 1 g particle X for Y g of a total solution, and i know that density is the mass per volume ratio, but how did you multiplied those together? what is the relationship between those two? 3. When can i say that the amount is too small and i can ignore it. I mean, your relation was that 0.0125mol is the max limit point, why cant it be higher? are there any rules? @MaxW

(1.) I didn't assume the final solution has no density, I assumed that the density was 1.000 g/ml. The density of NaNO3 solutions is probably in a table somewhere and could be looked up. But it would make the calculations a mess. I'll add that as an explicit assumption. (2.) 1 ppb is 1 part in a billion by weight. Since 1 ml of solution weighs 1.000 gram, then $1\ \mathrm{ppb} = 1\times10^{-9}\times 1.000\ \mathrm{g} = 1\times10^{-6}\ \mathrm{g} $ (3) I'll add an explanation in the solution.


text:
What's the chemical formula of "Everitt's salt"?
According to my school textbook (which is notorious, for being in possession of several errors), a certain "Everitt's salt" is a complex that results from the reduction of Prussian Blue.
Everitt's salt (according to our inglorious textbook)  has the chemical formula $\ce{K2[Fe(CN)6]}$... which I straightaway dismissed as nonsense, because it would mean it's got $\ce{Fe^{4+}}$ (ferryl ion). From my (very limited) knowledge of the subject, the ferryl ion doesn't look like something that would occur in any reasonably stable compound; since Everitt's salt is, well, a "salt" (and since salts can usually be crystallized into stable solids).
Google wasn't much help in this regard,

I've no idea how to interpret the formula that is supposedly that of Everitt's salt

Questions:
1) Is there really something called "Everitt's salt"? Is it produced by the reduction of Prussian Blue?
2) What's its chemical formula?

Everitt's salt appears to be of mixed valence in iron: $\ce{K2Fe[Fe(CN)6]}$ http://www.molbase.com/en/cas-15362-86-4.html

@Gert Hmm... so Everitt's salt is a ferrous ferrocyanide salt? :O

acc. to that source, yes. Never before heard of it.

@Gert I'm no chemist (so that site doesn't impress upon me as strongly as chemspider or sigma aldrich), but would you want to make that an answer? O:)

no because I don't know whether it is produced by the reduction of Prussian Blue.

@Gert  This Wiki section mentions that Prussian Blue can be created by the oxidation of the compound you mentioned; it doesn't call it "Everitt's salt" though :/ But it fits the bill (I think), since the ferrous ferrocyanide is a "reduced" form of Prussian Blue. :-)

http://pubs.acs.org/doi/abs/10.1021/j100323a055?journalCode=jpchax Yes it's made by reduction

@Mith Aha! Good find! Okay, so now there's a (fairly) reliable source that mentions Everitt's salt... but there's no mention of the formula. So I can't tell if what Gert found is the same thing :(

file:///C:/Users/User/Downloads/BHC39p18.pdf This source references the same formula for ES as my Molbase reference. Find the pdf here: https://www.researchgate.net/publication/281176633_On_the_History_of_Prussian_Blue_Thomas_Everitt_1805-1845_and_Everitts_Salt

@Gert Looks like that's settled then! (How about an answer now?) :D

@Mith Well, Gert seems to have nailed it. Consider posting an answer of your own too (referencing the document you linked), since you've confirmed that Everitt's salt is made by the reduction of PB; it'll  be a legitimate (albeit, not exactly complete) answer and will get my upvote O:)

There is an obvious typo in the image. Notice that there is a right bracket ] but no left bracket [. So the author made some kind of mistake.

Update: I have asked "What is the chemical composition of Ferrous Ferrocyanide?"

According to various sources Everitt's Salt has the formula $\ce{K2Fe[Fe(CN)6]}$, where both $\ce{Fe}$ are in the $+2$ Oxidation State. Everitt described it as a yellow precipitate, resulting from the reaction:
$$\ce{2 K4Fe(CN)6(aq) + 6 H2SO4(aq) \to 6 HCN(g) + 6 KHSO4(aq) + K2Fe[Fe(CN)6](s)}$$
Acc. Everitt, the wet salt can be oxidised to Prussian Blue by the action of air (oxygen).
Source.


text:
Obtaining 30% sulfuric acid

What volume of $10\%$ sulfuric acid solution ($\rho = \pu{1.07 g cm-3}$) do you need to put into $\pu{300 mL}$ $50\%$ of the same acid solution ($\rho = \pu{1.4 g cm-3}$) to get $30\%$ solution?

I've been searching for a solution for the past few hours, I've seeken out the same kind of problem and found only one which had water in it, not the same solution. So if anyone was kind enough to show a solution I'd be enlightened.
I tried "Pearsons letter", which gives me that it has a ratio of $1 : 1$ for making 30% solutions for some reason (doesn't make sense, because then it'd be 300+300 when the answer is 392.5), I also tried making an algebraic formula, which gave me an answer that was about $60000$ units above the actual answer. I only need the solution, so I could solve the same kind of problems.
P. S. English isn't my first language, if you need me to specify something on the problem, just say. Thank you in advance.

Hello and welcome to Chemistry.SE. This is a homework type of question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this. Best of luck...

Hello, I don't need the answer or the solution using the same exact units, the problem lies in me not finding an example problem which has the 2nd solution having units and the finished one not having units (by units I mean volume, mass etc.)

Let's discuss basics. What is a 10% solution of sulfuric acid?

10% solution means that it has a procentile mass of 10% sulfuric acid mixed with water, the same being with 50%, so by finding the mass of the whole solution 300 and multiplying it by ρ of the selected solution you get the whole mass of the solution which is 420 and that multiplying by 0.5(50%) you get 210grams of H2SO4

First off, solving these kind of problems we are assuming that

the resulting mixture is homogeneous;
the mixing of various solutions takes place instantaneously;
the volume of the mixture is equal to the sum of the volumes of the miscible solutions.

The approach is very similar, there slight deviations based of how the concentrations are expressed. I suggest to start by writing down the equation for the given fraction, and then expand it using the provided physical quantities, finally expressing the volume. A couple of examples:

Mass fractions
This addresses the original question. By definition mass fraction $\omega_i$ or $i$-component with mass $m_i$ in solution is determined as follows:
$$\omega_i = \frac{m_i}{\sum_i{m_i}}$$
At the same time mass is expressed via density $\rho$ and volume $V$:
$$m = \rho V$$
Now, the resulting mass fraction $\omega_3$ is:
$$\omega_3 = \frac{\omega_1 m_1 + \omega_2 m_2}{m_1 + m_2} = \frac{\omega_1 \rho_1 V_1 + \omega_2 \rho_2 V_2}{\rho_1 V_1 + \rho_2 V_2}$$
All you have to do now is solve this equation for $V_1$:
$$V_1 = \frac{\rho_2 (\omega_2 - \omega_3)}{\rho_1 (\omega_3 - \omega_1)} \cdot V_2 = \frac{\pu{1.40 g cm-3} \cdot (0.5 - 0.3)}{\pu{1.07 g cm-3} \cdot (0.3 - 0.1)} \cdot \pu{300 mL} = \pu{392.5 mL}$$

Molar and equivalent concentrations
In the comments @user55119 suggested a similar problem and expressed a doubt about the necessity of using density in the OP's original question:

How much $\pu{2 N}$ $\ce{H2SO4}$ must be added to $\pu{300 mL}$ of $\pu{10 N}$ $\ce{H2SO4}$ to prepare a $\pu{6 N}$ $\ce{H2SO4}$ solution?

By definition molar and equivalent concentrations are tied via the number of equivalents $z$:
$$C = \frac{C_\mathrm{eq}}{z} = \frac{n}{V}$$
The resulting molar concentration is:
$$C_3 = \frac{n_1 + n_2}{V_1 + V_2} = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2}$$
Solving the equation for $V_1$ and taking into account that the equivalence factor $z^{-1}$ is a proportionality coefficient:
$$V_1 = \frac{C_2 - C_3}{C_3 - C_1} \cdot V_2 = \frac{C_\mathrm{eq2} - C_\mathrm{eq3}}{C_\mathrm{eq3} - C_\mathrm{eq1}} \cdot V_2 = \frac{\pu{10 N} - \pu{6 N}}{\pu{6 N} - \pu{2 N}} \cdot \pu{300 mL} = \pu{300 mL}$$

Methinks this might goes slightly over the OP's head. But it is correct of course. Nicely formatted too.

@Gert Thank you, I've thought of graphical method, but in the end there is still some math to handle, so I decided to go with this one. Do you think I should've included intermediate steps when expressing $V_1$ for clarity?

(Even though it says to avoid using comments to thank people, I just had to do so(don't have 15 rep. points)) Thank you, a lot. My problem was that the formula I knew couldn't be used like that, so basically I had to figure out the solution by guessing. Is there a place where I could learn to extrapolate formulas out of other formulas that big? Thank you, once again for editing my post and also answering my question.

@Kurai You are very welcome:) I'm not quite sure I understood what you mean by "extrapolate", you basically take the basic formula for the corresponding concentration (mass % in this case) and try to express the unknown parameters using the equations for basic physical quantities, such as density or molecular weight. And yes, when given "$x%$ solution of compound" without further notes, it's usually safe to assume it's mass fraction $\omega$.

What if the question had been formulated in the following way. How much 2N H2SO4 must be added to 300mL of 10N H2SO4 to prepare a 6N H2SO4 solution?

@user55119 Same approach, replace the equation for mass fraction with the one for normality.

It was a rhetorical question. No densities in the question and the normalities are virtually the same as the percents provided. now what answer do you get?

@user55119 How about you post it as a separate question and include your attempt on solving this?

I posted my comment to stimulate some thought about the answer 392.5 mL. Feel free to answer it. n = meq./mL; V = mL  and n1V1 + n2V2 = n3V3. Is the density a red herring?

@user55119 What's to stimulate here? Am I supposed to do the same procedure for every possible way the concentration can be expressed? This is the question regarding mass fractions. If you have similar question, go ahead and post it, I'll come by and answer it.

@user55119 I've got some spare time and edited the answer to address your comments. But usually it's not how SE works, and it's always better to ask a standalone question if you expect a reasonable answer.


text:
Why does vibrational cooling occur much faster than sponteneous emission in the condensed phase?
Following the absorption of light by an arbitrary small molecule, the molecule is found to be excited both vibrationally and electronically. The excess vibrational energy of the molecule will usually be dissipated into the solvent in less than ~10 ps, and fluorescence emission will occur in about ~10 ns.
Since there are many transitions available to accept the excess energy form the vibrations in the surrounding solvent it makes sense that the relaxation occurs very rapidly. On the other hand, electrons can move much faster than nuclei can. What must happen for an electronically excited molecule to relax through the emission of a photon and why does it take so much longer than the vibrational relaxation?

As you point out, there are many places for this relatively small amount of vibrational energy to go. There are both radiative and non-radiative transitions which can take place to vibrationally quench the system in the electronic excited state. The explanation for both parts of your question (the vibrational and electronic parts) can be understood by Fermi's Golden Rule. I won't go into a lot of detail of deriving this or anything else because you can find anything you wanna know about it by searching online.
Fermi's golden rule says that the transition rate, $\Gamma_{i\to f}$, from some initial state, $|i\rangle$, to a set of final states, $|f\rangle$, is given by,
$$
\Gamma_{i\to f}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho
$$
where $\rho$ is the density of final states of the system and $H'$ is a perturbing Hamiltonian of some kind. For instance, the electric field of a photon. This is what it would usually be for absorption. In the case of decay, it may also be the field of a photon, but I am not as familiar with using Fermi's golden rule in this case.
Notice that this rule has dimensions of inverse time. It can be shown that the inverse of this rate is equal to the mean lifetime of the excited state (see this page).
Basically all of the physics you are asking about is contained within the fact that this transition rate is proportional to the density of possible states to which the system will decay. To put it simply, there are many many more intramolecular and intermolecular transitions which can take place to redistribute this energy. The vibrational density of states for this problem could probably be approximated quite well as being continuous because there will be many vibrational levels which are slightly perturbed due to distortions from equilibrium geometries. Think of the width of vibrational bands in IR.
On the other hand, a very large amount of energy must be dissipated for electronic relaxation to take place, and the density of these high-energy states is very low. Especially if one excites to the lowest electronic state. If, however, excitation occurs to a higher electronic energy level, decay should take place faster because the density of electronic states becomes much larger.
These processes have almost nothing to do with the speed of nuclear motion compared to electronic motion, as the condition which must be satisfied is really conservation of energy, and this is not always easy to do when the energy levels are quantized. From this perspective, the result is not surprising because vibrational energy levels are certainly closer to being continuous than electronic energy levels. Also, the spacing of vibrational state is generally smaller on an excited state than the ground state, so this adds to the quickness of the decay to the ground vibrational state.

As an aside, this paper is of some interest as it shows that vibrational relaxation times in the gas phase (for this one molecule) are slower than vibrational relaxation in solution. This can also be understood by the density of states argument discussed above.

Thank you very much for the clear explanation and the referenced paper! I had also not considered the case of higher electronically excited states decaying faster,  which helps explain Kasha's rule, and that is definitely something interesting to think about.

The processes available to the excited molecule are vibrational relaxation/redistribution, emission (fluorescence , phosphorescence), inter-system crossing and internal conversion. We assume no chemistry occurs. These processes occur in competition with one another. Often it is observed in solution that vibrational relaxation is far faster than other processes so that emission occurs only from the lowest vibrational level of the excited state.
When a photon is absorbed forming an excited state only one type of vibrational mode is excited, and this is because only certain vibrations have a large Franck-Condon factor and so increase the transition dipole which couples the two electronic states. Because of the relative disposition of the electronic states the excited state is usually formed with more vibrational energy than has the ground state, i.e. the excited state is vibrationally hot. This proceeds to cool by transferring (redistributing)  energy to other vibrational modes in the molecule. (The fact that electrons move far faster than nuclei do is why we can draw potential energy as a parabola with vibrational levels. This is often called Born-Oppenheimer approximation. The nuclei are effectively stationary while the electron 'moves' from ground to excited state. )
Now normal modes are supposed to be orthogonal but in reality all vibrations are anharmonic and this allows coupling between vibrational modes and so the energy flows out of the initial excited mode into others. The larger the number of vibrational quanta excited the more this occurs and this is also aided by the increasing density of states in the accepting modes: these can be many millions /cm$^{-1}$. Sometimes this is called the Franck-Condon weighted density of states ($\rho$) and the product of this with the operator $H$ coupling the levels forms the Fermi Golden rule expression ($k\approx |H|^2\rho)$ for the rate constant of the process.
All this happens in the isolated molecule (as in the gas phase) but the effect is small compared to that in solution where the continuous collisions with solvent cause far larger perturbations and larger coupling together of levels and so energy is lost faster out of the initial vibrational levels, generally of the order of a few  picoseconds. (In passing we note that there are exceptions that have been observed in some rigid aromatic molecules,  such as Nile-Blue, where excitation of the Franck-Condon mode in the singlet excited state lasts for hundreds of picoseconds. This is observed as fast periodic oscillations in the transient absorption).
Once in the vibrationless singlet state inter-system crossing to the triplet can occur, where just as in the singlet, vibrational energy can be lost before phosphorescence occurs. Also from the excited singlet, internal conversion to the ground state occurs. Internal conversion directly populates the ground state and so produces a vibrationally very hot ground state (as all the electronic energy is dumped into the ground state) and this looses energy rapidly to the solvent. You can observe this in transient absorption, but only for short lived excited states where this is the dominant pathway and lasts about 10 ps before the energy diffuses away in the solvent.
Internal conversion does not have to be fast, if it were always of picosecond duration then fluorescence would be almost unknown as the fluorescence yield is $\displaystyle \varphi=\frac{k_f}{(k_f+k_{ic}+k_{isc})}$ where subscripts mean fluorescence, internal conversion and intersystem crossing. The rate constant fro fluorescence is typically $10^8$ s$^{-1}$ and so if $k_{ic}=10^{12}$ the yield would be approx $10^{-4}$ whereas very many types of molecules have yields > 10%.
In competition to these processes fluorescence can occur, the rate constant for this is governed by the transition dipole moment between the ground and excited state $\bar \mu=\int \psi_i\mu\psi_fdRdr$ where $i$ and $f$ are label the total wavefunctions (electronic + vibrational) for the initial and final states and $\mu$ is the sum of electronic and nuclear dipoles.

Whoops, I missed the notification and did not see your answer until now.
The mention of anharmonicity allowing different 'orthogonal' modes to couple has given me some more valuable insight, and I will look further into this and into the FC weighted density of states. Thanks a lot!

There is more detail in my answer here https://chemistry.stackexchange.com/questions/28883/non-radiative-transitions-or-what-features-of-rhodamine-result-in-it-being-so-hi. The theory papers were done in the late 1960's and early 70's. There are numerous papers. The names to look for are papers by W. Siebrand, by S.H Lin,  by J. Jortner and  by K. Freed and their collaborators. There are also many paper on experiments done mainly on isolated molecules in the vapour phase.


text:
How to calculate concentration from reaction quotient and initial concentration?
Say I know that the reaction quotient $\left(Q = \dfrac{[B]}{[A]}\right)$ for a simple reaction $\ce{A\to B}$, at $t = 0$, $[A] = 1$ and $[B] = 0$, at some time in the future is $Q = 1$ (or any number really).
Is there a way to calculate the concentrations of A and B, or is more information required?

Concentration when?

just for a general case - I have used the Nernst equation to calculate Q based on the potential of an actual cell (OCV measurement over time) - but am unsure how to convert Q to actual concentrations. Thanks.

Not really. For the general reaction, you would definitely need a reaction rate coefficient. Then the reaction could be zero order (say if a catalyst was being used and was saturated) or first order. Also is the backward reaction possible? If so then you'd need a backward rate coefficient too. Again the reaction could be zero or first order. Welcome to kinetics... ;-)

You have everything you need.
$$\ce{A -> B}$$
From the initial condition, stoichiometry and the conservation of mass, always:
$$\ce{[A] + [B]=1}$$
At time $t$:
$$\ce{\frac{[B]}{[A]}=1\to [B]=[A]}$$
Substitute into the first equation:
$$\ce{\to [A] + [A]=1\to 2\times [A]=1\to [A]=\frac12=[B]}$$

For a more general case:
$$\ce{[A] + [B]=n}$$
At time $t$:
$$\ce{\frac{[B]}{[A]}=Q\to [B]=Q\times [A]}$$
Substituting as above, we get:
$$\ce{[A]=\frac{n}{1+Q}}$$
$$\ce{[B]=\frac{n\times Q}{1+Q}}$$


text:
How can the concentration of a solid be constant during a reaction?
The molar concentration is given by the ratio between density and molar mass. I understand the proof.
But how does this remain true during a reaction?
For example consider the reaction,
$$\ce{4Al(s) + 3O2 (g) -> 2Al2O3(s)}$$
If the amount of Al and O2 moles are equal, according to stoichiometry Al ought to be the reactant to end first. But, being a solid, Al’s concentration must be constant.
How can this be? What is wrong in my reasoning?
**Please note that I mistakenly assumed that the amount of Al and O2 are equal, without stating it, leading to the confusion that Al ends first at all times, I have edited the question appropriately

If Al ends first, then there is no more Al, hence its concentration is zero. What's unclear about that?

Yes, but they say that the concentration of a solid remains constant throughout. How is that possible?

Well, just like that. A huge piece of Al has exactly the same properties as small piece of Al.

RE: According to stoichiometry Al ought to be the reactant to end first. Stoichiometry doesn't really determine which is the limiting reagent. The limiting reagent is based on the relative number of moles of the reactants. So if there are 10 moles of Al and 1 mole of O2 then O2 is limiting. If 1 mole Al and 10 moles O2 then Al is limiting.

I don't understand why Al ought to be the reactant to end first (I assume by end, you mean run out?). Surely I can give you a sealed room with limited oxygen and a honking giant chunk of aluminum?

@chipbuster It seems I implicitly assumed that the amount of Al and O2 moles are equal. A slight mistake. I’ll edit the question as appropriate.

So any changes you may observe in the reaction rate depending on the size of the aluminium block/foil/whatever, would not be accounted for by the law of mass action but principles of transport phenomena.

In physical chemistry one does not use the concentration of a chemical but its activity (see https://en.m.wikipedia.org/wiki/Thermodynamic_activity)
In a solid the standard condition is the pure chemical. In case of the solid block of aluminum it is therefore in its standard condition which leads to the fact that the activity is 1 (the chemical potential is equal to the chemical potential in the standard condition as the block of aluminium is in its standard condition - see the link I provided for the definition of the activity).
Therfore it doesn't need to be considered in calculations like the law of mass action as multiplying or diving by 1 does not affect the result.
However in "real chemistry" one does not calculate with activities too often and uses concentrations instead and in order to apply the same trick one simply says that the concentration is constant.
So in the end it's just a result of sloppiness because one doesn't want to work with activities.


text:
Identifying the D/L form of any aldose or ketose in cyclic form

I understand that the compound above is  a ketose because it has a furanose ring. But, how do we know that it is a "D" form?

Is there any general rule by which I can identify if any given aldose/ketose is in D/L form?

Please see this related question's answer: Monosaccharides configurations (alpha beta, D L) identification

Your structure is a ketose because it is a hemiketal of a ketone. It is a furanose because it is a 5-membered (furan-oid) ring. Fructose, which your structure is not, is usually drawn as a furanose as in sucrose but fructose itself exists principally in the pyranose (6-membered ring) form.
How do you recognize the D-configuration? Your furanose 1a is in the D-series. Reverse your furanose to the keto form 1b by following the red bonds. Rotate about the $\ce{C4-C5}$ bond by $120^\circ$ as illustrated by the blue arrow in 1b. This rotation produces the Fischer projection 1c having the $\ce{C5}$-hydroxyl on the right. Thus, furanose 1a is of the D-series.
There is another method that can be used with qualification. At any carbon that determines D vs. L in a carbohydrate, as long as the substituents are $\ce{O, CH2OH, CH(OH)C}$ and $\ce{H}$, the D-series will be of the R-configuration. Conversely, the L-series will be of the S-configuration. In furanose 1a, the priorities are $\ce{O>C4>C6>H}$. This determination is easy using this method.
Structures 2a, 2b, and 2c are different representations of $\alpha$-D-glucose. All of the $\ce{C5}$-carbons are, necessarily, of the R-configuration and therefore in the D-series.

Structures 3a and 3b are Haworth renderings of sucrose. Both the glucose (pyranose) and fructose (furanose) moieties are of the R-configuration and in the D-series.


text:
Second order reaction with two reactants
I'm interested in the second order reaction $\ce{A + B -> P}$. Mass action kinetics is assumed.

Let $[A]_0 = a$ and $[B]_0 = b$, then $[A] = a - x$ and $[B] = b - x$. The rate law becomes
$$-\frac{\mathrm dx}{\mathrm dt} = -k([A]_0 - x)([B]_0 - x)$$

Why is the right-hand side of the differential equation negative? Shouldn't it just be the left side when we look at the decrease of the reactant $\ce{A}$? Or is it because $x$ is equal to the concentration of $\ce{P}$?

The right side can be negative because someone has placed a minus sign before it. Ditto for the left side. To me, both sides would look better without it.

But why does it make sense? Usually there is a minus sign only on the left part when you look at the decrease of a reactant..

But you don't look at the decrease of a reactant. x is the concentration of P, and it is increasing.

I recommend this page: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions

For the problem $\ce{a + B -> P}$ the mathematical equation could use a bit better notation to see what is going on. $$-\dfrac{d\ce{[P]}}{dt} = \ce{-k([A]_o - [P]_t)([B]_o - [P]_t)}$$

Assuming mass action kinetics, we have the following differential equation
$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = -\kappa \, [\ce{A}] [\ce{B}]$$
where $\kappa > 0$ is the rate constant. If we denote
$$x (t) := [\ce{A}]_0 - [\ce{A}] = [\ce{B}]_0 - [\ce{B}]$$
then
$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = \frac{\mathrm d}{\mathrm d t} \left( [\ce{A}]_0 - x \right) = 0 - \dot x = - \dot x = -\kappa \left( [\ce{A}]_0 - x \right) \left( [\ce{B}]_0 - x \right)$$

While I personally understand your answer (and up'd it), I would assume, that the last step is a bit too contracted for everyone to follow. (But in its brevity it's beautiful.)

The main issue I see here is that there's too much fixation on how the equation normally looks. Instead, you would find much more insight into looking at the physical meaning conveyed by the equation.
For example, you can plug in initial and final values to see if the function describes what we expect.
At the beginning of the reaction, $x=0$. Your equation gives $\frac{\mathrm{d}x}{\mathrm{d}t} = k\ce{[A]}_{0}\ce{[B]}_{0}$. This is exactly what's expected, as this is the maximum value of the rate of product formation.
Let's arbitrarily assume that $\ce{[A]}\geq \ce{[B]}$. Then the reaction is done when $x = \ce{[A]}_{0}$. If you plug this value into your equation, you find that $\frac{\mathrm{d}x}{\mathrm{d}t} = 0$ which is also correct.
And as Rodrigo pointed out, the main reason you have sign confusion is that you've chosen to represent the concentration of product as opposed to the concentration of reactant, and since one is produced while the other is consumed, you get an extra minus sign.


text:
Charge-transfer absorption complex - solvent sensitivity
This is a quote from Anslyn's Physical Organic Chemistry:

[...] One way to think of these systems is that the ground state is a weak,
  intermolecular complex, (D • A). Often such molecules are polar and /
  or quite polarizable, facilitating complex formation. Because one
  partner is intrinsically a good electron donor (low ionization
  potential) and the other is a good acceptor (high electron affinity),
  the excited state will have a large contribution from states such as
  (D+• A-) that involve electron transfer, a configuration not possible
  for either isolated molecule. This preferentially stabilizes the
  excited state, leading to a lower energy (longer wavelength)
  absorption.  Because of this highly polar character, charge-transfer
  absorptions are extremely sensitive to solvent polarity, moving to
  longer wavelengths as the solvent polarity increases.

Earlier I read about Franck-Condon principle but I lack QM-mathematical background behind, so that's why this principle still leave me at unease, and hence my question. How is it possible that absorption wavelength is so sensitive to solvent polarity, when significant charge develops not until the complex is excited? I mean how the excited state of the molecule "knows" there is solvent ready to solvate it, since according to Frank-Condon principle solvent rearrangement is much slower than absorption.

Without the complex A, the excitation of D would move the electron to a higher orbital of D. With the complex A, complex A has an empty orbital lower in energy than the higher orbital of D. So it takes less energy, a longer wavelength, to excite the D-A complex as opposed to the D atom.

In a charge transfer DA complex the lowest excited state is the charge separated species $D^+ A^-$. The 'normal' lowest excited singlet states of D and A are each at higher energy. As the molecules D and A come together the interaction between them, although small, (the absorption spectrum is only marginally altered) is nevertheless enough to cause electron transfer upon photon absorption.
The dipolar excited state possesses its own electrostatic self-energy E, as does any dipole. This is the sum of the Born charges $\pm q$ at infinity less the Coulomb energy of bringing the charges together to from the dipole. This works out as $E=  q^2/(8\pi\epsilon_0 \epsilon a)$ where $\epsilon_0$ is the permitivitty of free space and $\epsilon$ the dielectric constant of the solvent. The charges have a combined radius of $2a$. The equation can be re-written as $E=\mu^2/(4\pi\epsilon_0\; \epsilon\; L^3)$ if the dipole is defined as $\mu=qL$. Thus you can understand why the energy is lowered, and hence CT transition red shifted in a polar solvent, e.g acetoninitrile $\epsilon =37$ vs hexane $\epsilon \approx 2$.
Notes: Electron transfer upon absorption is in contrast to photoexcited electron transfer where the excited state of the donor or acceptor, whichever is excited, undergoes a thermal reaction in the excited state to form $D^+ A^-$. In the CT complex the charge is transferred on absorption between one molecule and the other, rather than from one part of the molecule to another part as in a normal absorption process. The absorption process is so rapid that, there is plenty of time for this to occur whilst the D and A molecules are close to one another and in the correct relative orientation because molecular diffusion is a relatively slow process.


text:
How to assign priorities around the asymmetric carbon with its substituents having multiple bonds?
How do I assign priorities to substituents with triple bonds?

If we take a look at the molecule above, the atom directly bonded to the asymmetric center is carbon for all three substituent.
Moving on to the atoms connected to those carbons, benzene side has carbons attached by a double bond and a single bond. The double bond is counted twice and the atoms count as $\ce{C, C, C}$.
For the sp carbon on the lower right, it is triple bonded to its neighboring sp carbon, so it counts as $\ce{C, C, C}$ as well.
The tert-butyl group is bonded to three carbons from methyl groups, so this gives another $\ce{C, C, C}$ making it a tie again.
If we look at the next set of atoms, the tert-butyl group has methyl groups so it's $\ce{H, H, H}$, giving it the lowest priority.
However, I'm not sure how to count the next set of atoms for benzene and the triple bond containing group. Judging from the R, S configuration the benzene group should have a higher priority, but I'm not sure how to figure that out. What should be my next step?

I would like to add a diagram to supplement @Raven's response and add a caveat about flipping structures to make R,S-assignments. The C's are the phantom carbons from the double and triple bonds. Working outward from the center of chirality, the first shell (in red) does not allow a decision to be made. At the second shell (in blue) the tertiary-butyl group drops out of the race and becomes the third priority group ahead of the hydrogen (4th priority). The third shell (in green) assigns the benzene ring as the top priority because the acetylene group has "run out of room" and must settle for second priority.
Flipping structures often result in drawing the enantiomer when one tries to place the lowest priority group in the rear. Leave the structure where it is and use your hands to determine chirality. Point your thumb from the chiral center to the lowest priority group. Curve your fingers so that they point in the direction 1 --> 2 --> 3. If you come up with 3 --> 2 --> 1, you are using the wrong hand! In this case your right hand (Rectus) will do the job.

Hydrogen: H; priority 4. Tert-butyl: C; [C,C,C]; 3 (H,H,H); priority 3. Alkyne: C; [C,c,c]; (c,c,H); priority 2. Phenyl: C; [C,C,c]; 2 (C,c,H); priority 1. You don't need the third shell.

The third shell was for the sake of comparison.

that is technically not the right way of looking at it. The first point of difference is not where you imply it is.

The benzene ring has the highest priority as it has another C connection after the C, C, C tie whereas the others only have hydrogen attached to them.
Therefore the order of priorities is cycling counter clockwise which would mean that it is a S isomer. However for the R, S nomenclature the residue with the lowest priority has to be facing backwards. But in this case the single hydrogen is sticking out the front so that the molecule has to be flipped which then leads to it being am R isomer.


text:
Chiral Centers and Stereochemistry
I was just wondering how many chiral centers this molecule has. This question might be a bit simple but I was especially confused by the bottom left "chiral center" that is bonded to a wedged ch3 and parts of a ring. Initially I thought it was a chiral center but upon further introspection it seems like two of its groups (ones next leading to ring) are exactly the same. What do you guys think about this molecule? Again I know that it is an sp3 carbon I am just confused about the terminology of its chirality. Thank you so much for your help

All chiral centers seem to be marked as it is drawn with stereo information... Therefore each carbon having a residue connected via a "triangle" is a chiral center...

Hey thank you for the answer! Would you be able to reword this more simply ? Would the bottom left carbon bonded to a methyl group count as a chiral center?

The methyl group is connected via a shim to the rest of the structure indicating that the carbon it is connected to is a chiral center. So yes the mentioned carbon is a chiral center.

The number of chiral centres are 6.And crabon with 4 different groups is assymmetric and achiral.Hope this should help you.


text:
Can hydroxides ionize alcohols?
For example, can $\ce{NaOH}$ ionize ethanol into ethoxide?
My textbook (Organic Chemistry, 6th edition, by M. Loudon and J. Parise, Roberts & Company 2016) suggests yes — it uses it in its mechanism for the intramolecular substitution of 4-bromo-1-butanol in the presence of $\ce{NaOH}$.
But I was taught that $\ce{^-OH}$ could only ionize thiols, not alcohols. After all, if simple $\ce{NaOH}$ can create alkoxides, why would chemists bother to use $\ce{NaH}$?

If you're going to use NaOH to ionize ethanol then you'd need anhydrous ethanol. To get and keep a solvent anhydrous requires a bit of care.

In a polar non-protic solvent such as 1,2-dimethoxyethane, it may be possible to get complete deprotonation of the alcohol by a hydroxide base. The hydroxide ion is much more basic than it may appear at first, because its reactivity is strongly levelled in water and other protic solvents. See this past answer of mine for background

Yes, and no.
Acid and base chemistry is a neat balance between equilibria. As you well know, water dissociates readily by itself. A tendency that can be observed in any compound with $\ce{-OH}$ groups, yet not to a similar extend.
In other words, a hydroxyl group tends to be slightly acidic. Given a base, i.e. sodium hydroxide in this particular case, the following equilibrium will develop:
$$\ce{EtOH + NaOH <<=> EtO- + Na+ + H2O}$$
While the equilibrium will mostly be on the left side, depending on what other substrates are available, a tiny amount of alkoxide might already be enough to drive the reaction forward. In this case it is important to look at what actually drives the reaction.
In the example of 4-bromo-1-butanol that your book uses I am assuming there is a cyclisation happening.
$$\ce{BrH2C(CH2)2CH2OH + NaOH <=>> THF + H2O + NaBr}$$
Unfortunately I cannot quickly find a reference, but sodium bromide is not very soluble in diethylether, so I would assume it's not very soluble in THF either. The driving force here would be having a well enough leaving group and the continuous removal of product, basically making it a one-way reaction.
In some cases you need a higher concentration of alkoxide, and that's where hydrides come into play.
$$\ce{EtOH + NaH <=>> EtO^-\bond{~}Na+ + H2 ^}$$
In this case dihydrogen, as it is a gas, will be removed from the reaction mixture, basically ensuring full conversion to the alkoxide.

You are asking for a clear yes/no answer in a situation where there is none. See, it is all about equilibrium. True, $\ce{OH-}$ is a weaker base than $\ce{RO-}$ (in other words, alcohols are weaker acids than water), but not by much, hence the reaction is still possible, if only a little. That is, an equilibrium is much closer to the starting compounds, but still it contains some products.
So the full answer to your question goes along the following lines: Yes, if ionizing one molecule in a hundred is good enough for you; otherwise no.


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Can I supercool and solidify a liquid multiple times?
Let's say that I supercool a bottle of beer and then tap it so it freezes and solidifies. If I were to allow the frozen beer to melt completely, and then freeze the same bottle again, would it still be able to supercool and not solidify?
I am guessing that it will solidify while freezing because it is no longer pressurized. But I don't know and don;t completely understand this concept.

You never said you opened your bottle, which implies that it must still be pressurized. Then again, it's hardly related to supercooling.

Actually you can supercool water as well. And actually the cleanest is tge system easier is the supercool it. So in principle you can repeat the cycle.

One can supercool very pure water, I'm afraid suppercooling beer would be at least very difficult, if not impossible.

@Mithoron many an collegiate "amateur scientist" has done this exact thing when they were in a rush on a weekend night and got distracted, myself included (on multiple occasions). Take the bottle out, bump the bottom, and watch the crystals travel up the bottle. It may be harder with unfiltered beer, though, and even more so for bottle-conditioned beer. My point here is unless this is a case where there is an general definition used by the public and stricter definition used by experts, I think your comment is incorrect.

If, as noted in the comments, you do not open the bottle, the process should basically be repeatable, as longer as the container does not crack. Additionally, the freezing may destroy certain larger things that do not tolerate ice crystals, such as yeast cells, if any are present.


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Hydroboration on Limonene
I saw that the hydroboration of Limonene (1-Methyl-4-(1-methylethenyl)-cyclohexene) does only happen on the methylethenyl group. Up to now, I wasn't able however to find out why. Generally speaking, a hydroboration can happen on any double bond, so why doesn't it react – at least a bit – with the cyclohexene?

I could argue, that the stereochemistry of the cyclohexene is not as favorable as the one of the methylethenyl group, but I don't see this being the case. On the cyclohexene the methyl group is in the way, where as on the methylethenyl group it is the cyclohexene.

The borane reagent in question, disiamylborane, is one of the many bulky hydroborating agents which preferentially add to less hindered double bonds.
In this case, the 1,1-disubstituted double bond outside the ring is more prone to hydroboration than the trisubstituted double bond within the ring, simply due to sterics.
There are many more examples of hydroborations with this reagent which are dictated by sterics: for further examples see its corresponding entry in the Encyclopedia of Reagents for Organic Synthesis. In all cases, disiamylborane is selective for terminal alkenes (i.e. either monosubstituted alkenes, or 1,1-disubstituted alkenes).


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Near melting thermoplastic material accumulates in the sharper corners. What is this phenomenon?
I have a part made of a thermoplastic material. When I heat it up and cool it down, the dimension of the part changes non-uniformly. Somehow the melted material tries to fill the sharper corners. So for example if there is a corner with a fillet of $r_1$ and wall thickness of $t_1$, after a high temperature experiment the part deforms to $r_2$ and $t_2$ where
$$\left(t_2-t_1\right)<\left(r_1-r_2\right)$$

I know it has to do with adhesion and capillary, but not knowing that much about chemistry jargon, I'm not able to find the right term for this phenomena. I would appreciate if you could help me know if this phenomena has a name and what parameters control this behaviour?

It can be probably explained as a tip effect in term of chemical potential. I think.

@Alchimista any links to know more about this tip effect?

@andselisk I'm sure I saw a comment about  anisotropic thermoelasticity Hill yield criterion but it just disappeared!

I think it is just capillary pressure, as simple as that.

Some things in this direction: http://chem.atmos.colostate.edu/AT620/Sonia_uploads/ATS620_F11_Lecture8/Lecture8_AT620_090911.ppt

@Foad Sorry, I read your question again and  realized what I suggested is irrelevant here. One can delete own comments within 5 min:) I agree with Ivan and what you suggested in the question: capillary effect, it's just that simple.

@andselisk One can delete own comments until forever. 5 min is just the limit for edits.

@andselisk NP. Well, I know it has to do with capillary and rheology of the melted plastics, but was wondering if this specific phenomenon has a name.  Maybe it has been studied before and I can just use the knowledge. instead of throwing experiments.

Yes of course has to be "capillarity". Though I want say a drop is spherical because of capillarity.  Another link in the same direction http://www.eng.utah.edu/~lzang/images/lecture-8.pdf