Why doesn't tetrachloridotitanate(II) obey the 18-electron rule?

There is more to complex chemistry than the 18-electron rule. It states one condition when complexes may be particularly favourable, namely when all metal d-orbitals are fully populated and the ligands favourably interact with all metal s- and p-orbitals leading to all paired spins and hopefully a large energy difference between populated and unpopulated orbitals. The list of complexes that do not obey the 18-electron rule may well be larger than that of those that obey it. In no particular preference that includes $\ce{[MnO4]-}, \ce{[CrO4]^2-}, \ce{[Fe(CN)6]^3-}, \ce{[Mn(H2O)6]^2+}$ and many more. (A few of those are not typically described as complexes but they are nonetheless.)

Further constraints that you need to be aware of include, but are not limited to:

steric stress, i.e. too many large ligands around too small a metal centre,
charge density, i.e. surrounding a weakly charged cation with too many anions,
the actual electronic configuration, which can render octahedral, Jahn–Teller-distorted octahedral or square planar complexes more favourable.

In the case of a chloridotitanate(II), we first need to realise that we are surrounding a moderately small cation (titanium(II)) with rather large anions (chloride) which may generate steric stress. We also need to observe that titanium(II) is only weakly charged and attempting to generate an octahedral configuration would give us an anion with $-4$ charge — not exactly favourable (exceptions may apply). Also, chlorido ligands are known to be weak-field ligands, so one would not expect an overly large crystal field split.

The two greatest factors are probably sterics and electronics. Consider tetrachloridocobaltate(III) $\ce{[CoCl4]-}$: while potentially six chlorides may fit around the rather small cation, even though it is $+3$ the electronic situation favours only tetracoordination. This results in a 14-electron complex rather than an 18-electron one. In analogy, I am tempted to say that the even lesser charged titanium(II) ion cannot accommodate for a total of six chlorides, allowing only four of them in its vicinity. Taking electronics into account, square planar coordination is out of the question due to lack of stabilising interactions and the complex would be tetrahedral.

In fact, the tetrahedral geometry allows for a mildly stabilised triplet state: both $\mathrm{e_g}$ orbitals are semipopulated which is generally a tad better than two of three $\mathrm{t_{2g}}$ orbitals as it would be in the octahedral case.


What is the difference between hydroxyl, hydroxide and hydroxy?

Hydroxide - Hydroxide is a diatomic anion with chemical formula $\ce{OH-}$. It consists of a hydrogen and an oxygen atom held together by a covalent bond, and carries a negative electric charge.

Hydroxyl radical- {hydro- + oxy(gen) + -yl + radical} - The hydroxyl radical, •OH, is the neutral form of the hydroxide ion (OH-1). It is the monovalent group -OH which contains hydrogen and oxygen. The suffix -yl indicates it's a radical just like carbonyl.

Hydroxy or hydroxy group: A hydroxy or hydroxy group is the entity with the formula $\ce{-OH}$. It contains oxygen bonded to hydrogen. Hydroxy is also prefix which means containing hydroxyl group.


How do copper sulphate and the like change crystal structure on adding water of hydration?

[Manishearth, I'm not sure I'm even answering the right question below. After reading your question more carefully, I think you were more interested in how water molecules help crystals form, rather than whether or not the anhydrous forms can form crystals also (versus being strictly amorphous). So, oops. Maybe some one else has a good answer on the water molecules part?]

I suspect the answer is an issue of process rather than of fundamentals.

That is, if you could find a non-aqueous medium that readily dissolves the anhydrous form of some salt, my very strong suspicion is that you could grow excellent clear crystals of it that would look almost nothing like the hydrous crystalline forms.

The problem for metallic-anion salts in particular is that finding a liquid that dissolves the salt without simultaneously forming strong coordination complexes with it is going to be tricky, since I'm guessing it's usually the coordination effect that help ionize the units of the material to make it soluble!

Here's a wild guess: Someone, somewhere, has likely done research on how to grow crystals of substances such as anhydrous $\ce{CaSO4}$ (hmm, why does this phrase "dead burnt plaster" come to mind? :) or anhydrous $\ce{CuSO4}$ by using low-temperature molten solutions of... something or other? Not sure what!

And what a delightful crystal that would be to have! It would be way out of the ordinary despite being so very... ordinary?... in one sense in its composition.

Update: Apparently, clear crystals of anhydrous $\ce{CuSO4}$ do indeed exist, and are easier to create than I thought. You just add concentrated sulfuric acid [a] to ordinary copper sulphate pentahydrate solution and let it evaporate slowly. Since $\ce{H2SO4}$ is one of the most aggressive drying agents in existence, my guess is that the $\ce{CuSO4}$ just can't compete for the remaining water molecules and is forced to crystallize without hydration.

[a] Please, never ever attempt to use concentrated $\ce{H2SO4}$anywhere except in a real lab with protections and procedures fully in place. This is a compound that wants water so badly that it willy synthesize water right out of the hydrogen and oxygen bonded into proteins and carbohydrates, leaving only charred carbon. Ironically, the actual acidity of concentrated (vs hydrated) $\ce{H2SO4}$ is quite moderate, comparable to that of vinegar.

Addenda:

-- Another common-exotic crystal may be lithium fluoride, $\ce{LiF}$, although the accuracy of that seems to depend on who you read. Unique how, you say? Even though lithium was an early result of the Big Bang, lithium and fluorine are both consumed rapidly by large hot stars, especially fluorine. Their existence on earth thus appears to depend on processes such as proton irradiation and possibly even intensive neutrino irradiation tht only occur during the explosion of a supernova star, rather than being inherited from before the cataclysmic event. Pretty much all of the elements (except $\ce{H}$) here on Earth are star-forged of course, but the elements in $\ce{LiF}$ crystals thus may be remnants of the supernova event itself. That would... cool!)

-- And speaking of unique, wearable crystals: I've always wondered if there's a market out there for diamond crystals synthesized using carbon that was captured as carbon dioxide during the cremation of a loved one. If there is, feel free -- that is definitely not my bailiwick!


How to calculate m/e value for the base peak in mass spectroscopy?

I would say that the question is not quite well formulated and that the answer that you quote is even further misleading in giving the proper answer.

Finding the answer

To start, let's consider fragmentation of primary alcohols, which is the question at hands here. First of all, can we rule out any of these answers?

  Propan-1-ol undergoes complete fragmentation in a mass spectrometer.

This sentence means that we can rule out the molecular ion $M^{\bullet +}$ as a base peak in the spectrum. Propan-1-ol has a monoisotopic peak at $m/z$ 60 so it rules out answer D.

Now we apply the standard fragmentation rules for alcohols: consider ionization on the oxygen lone pair, we have two possible $\alpha$-cleavage leading to $m/z$ 31 and $m/z$ 59. An additionnal side reaction could also be hydrogen transfer to the oxygen atom, followed by water loss. This would lead to an ion at $m/z$ 42.

In the options above, should we thus completely rule out $m/z$ 29? I think that in the spirit of those who asked the exam question, the answer is yes, and we should rule out answer A. But see below...

We are now left with two options which can be explained by the fragmentation chemistry. Which should we choose? Technically, the problem here is that now we rely on abundances and not on the simple occurence of these fragments anymore. This is much more tricky, as there are no absolute rules. If we have a look at good textbooks, we find that loss of the largest radical is favored. Therefore, we have here to compare loss of $H^{\bullet}$ vs loss of $CH_{3}CH_{2}^{\bullet}$. Thus the answer is B.

Is the above answer complete?

The problem in such a question (and in the answer that is given below) is that it assumes that in mass spectrometry one could correctly predict fragment ion abundances based on very simple rules. And, as I tell my students, currently, even if you were able to map out the complete potential energy surface of the fragmentation pathway and do some RRKM calculations on the transition states, you would probably be off by a factor of 2 from the experimental results in terms of abundances. So asking for predicting the base peak (i.e. the highest abundance peak in a spectrum) can be quite wrong when two fragmentation channels are quite close to each other.

Let's have a look at the experimental electron ionization spectrum. Well OK, the base peak is indeed $m/z$ 31, so we are not too bad.

But let's have a deeper look. Remember we crossed-out $m/z$ 29 as a possible base peak, and it happens to be the second most abundant peak in the spectrum, ahead of $m/z$ 42 and $m/z$ 59 which followed the basic rules!

$m/z$ 29 could be either $HCO^{+}$ or $C_{2}H_{5}^{+}$. The former could be a secondary fragmentation of $m/z$ 31 after $H_{2}$ loss, whereas the latter could be produced as a heterolytic bond cleavage of the C-C bond. (In fact it is the same bond elongation leading to $m/z$ 31, except that they electron remains on the opposite side.) Energetically, these two channels differ by the difference in ionization energy between the $C_{2}H_{5}^{\bullet}$ and $^{\bullet}CH_{2}OH$ radicals, i.e. 8.11 eV vs 7.56 eV. Although it amounts to almost 100 kJ/mol, with the fairly large energy distribution in electron ionization, this is still possible.

And for the second rule that we applied, is it always true that the largest radical if formed? This empirical rule works out quite well, even more so when the comparison is between a hydrogen radical (which is quite difficult to produce) and a carbon atom centered radical.

And how about the exam paper solution

Here I am very highly surprised by what they seem to affirm. OK as we just saw above, $m/z$ 31 is a quite characteristic peak of primary alcohols. And loss of water is also characteristic of alcohols having side chains longer than 3 carbons. But does it mean that these should be the base peaks?

Let's have a look at the series of small primary alcohols:

Alcohol              m/z 31        Water loss           Other base peak
Ethanol              100            10
Propan-1-ol          100            15
Butan-1-ol            95           100
2-methylpropan-1-ol   38            10                  m/z 43
Pentan-1-ol           38            50                  m/z 42

So I do not see the rational behind this answer. It starts failing as soon as the small alcohol sizes reaches four carbon atoms. Of course there is a rational behind each of these base peak (for instance, in the case of 2-methylpropan-1-ol, the ionization energy of the isopropyl radical is of 7.37 eV, i.e. lower than that of $^{\bullet}CH_{2}OH$ and thus it becomes the most abundant ion, with exactly the same bond cleaved), but you have to go through the mechanisms, the thermochemistry and sometimes even the transition states to get to the global picture.


Under what conditions does MnO2 acts as an oxidizing agent?

It certainly is  powerful oxidizer when heated! Thermite can be made with powdered aluminum and $\ce{MnO2}$, and is at least as spectacular as with $\ce{Fe2O3}$ as the oxidant.


Special electron configurations in p block on periodic table?

The simplest answer to this problem is related to relative energy differences between the Nd and (N+1)s orbitals in transition metals and Ns and Np orbitals. The Nd orbitals in transition metals are at a slightly lower energy than the (N+1)s orbitals, so for certain atoms there are some unusual changes in the electron configurations, e.g. Cr and Cu. However, the energy differences between the Ns and Np orbitals are quite large, with Np being higher in energy. In these cases, there is a substantial energetic penalty in moving an electron from the Ns to Np orbital.

The electron configuration problem for transition metals is complex, but this J Chem Ed article has a nice description of the theory, although it is somewhat technical.


Oxidation reaction of dimethylselenide / Selenourea with hydrogen peroxide

I'm not familiar with selenium chemistry but am trying to help with the oxidation reaction of selenourea with hydrogen peroxide part. Here what I found on that reaction in literature (Ref.1):

  Abstract: Reactions of biological oxidizing agents, such as hydroxyl radicals ($\ce{^•OH}$), singlet oxygen ($\ce{^1O2}$), hydrogen peroxide ($\ce{H2O2}$), and peroxynitrite ($\ce{ONOO-}$) with selenourea were studied. The kinetics of the reactions was followed using time-resolved techniques, and the bimolecular rate constants were determined. In all these reactions, under aerated conditions, elemental red selenium was produced as one of the reaction products. The average size of the selenium particles could be controlled and stabilized in the range of $\pu{20−100 nm}$ with the addition of bovine serum albumin (BSA) or sodium dodecyl sulfate (SDS). The particles were characterized by dynamic light scattering studies (DLS), which revealed that the size and distribution of the particles depended mainly on the amount of selenourea undergoing oxidation. Other factors such as the nature of the oxidant and the concentration of the stabilizer also are important in stabilizing the particles. Nanoselenium-reduced $\ce{ABTS^{•-}}$ to colorless $\ce{ABTS^2-}$ ($\ce{ABTS} =$ 2,2‘-azinobis(3-ethylbenzothiazoline-6-sulfonate) and oxidized dichlorodihydrofluoresecein diacetate (DCFA) to fluorescent dichlorofluorescein (DCF) indicating its ability to participate in redox and free radical reactions. The reactivity of selenium nanoparticles with these systems varied linearly with the surface area of the particles. The studies demonstrate that selenourea undergoes oxidation with both one-electron and two-electron oxidants to produce elemental selenium, which, on stabilization to nanometer size, exhibits size-dependent redox activity.

The reaction mechanism is depicted in following Scheme, which is directly from Ref.1:

Hope this help.

References:

B. Mishra, P. A. Hassan, K. I. Priyadarsini, Hari Mohan, “Reactions of Biological Oxidants with Selenourea:  Formation of Redox Active Nanoselenium,” J. Phys. Chem. B 2005, 109(26), 12718–12723 (DOI: 10.1021/jp051328n).


What chemicals are often confused for different chemicals?

One especially serious example is confusing methanol with ethanol. Methanol is highly toxic and sometimes is deliberately added to or inadvertently present in bootleg alcoholic beverages, causing many deaths and blindness in survivors every year around the world. It would be a life saver for someone to develop a cheap dipstick test for methanol.


What's the exact difference between (-)-BTX and (+)-BTX

From a cursory glance, batrachotoxin has 8 stereocenters. No single Cahn-Ingold-Prelog R or S designation is going to be enough. As a result, we simply identify which enantiomer we're talking about by using the standard optical rotation of the substance. The (+) and (-) forms are enantiomers. They rotate plane polarized light in equal and opposite directions and are non-superimposable mirror images of each other.


Why do we use the Kováts index?

The Kováts index is used to normalize GC data. This Wikipedia page(though not very long), sums up why you would want to do so.

Retention times of the same compound on even two different versions of identical instrument with the same temperature and pressure program using the same column from the same manufacturer might not be the same! Retention time is dependent on so many things:

The identity of the analyte
The identity of the stationary phase
The length of the column (did you know that columns sometimes get trimmed as they age to remove deposits?)
The diameter of the column
The thickness of the stationary phase film
The flow rate of the carrier gas
The temperature or temperature program of the oven

With so many variables, gas chromatography (indeed all chromatography) can seem like a forbidding black art in which one has no hope of being able to compare results from one instrument to the next or one lab to the next.

Fortunately, we're smarter than that. We can always compare our analyte's retention time to some retention times of well behaved standards. If we know the elution order (will not change) of a series of well-behaved standards on several instruments, and we know when our analyte elutes in relation to those standards on one instrument, we have a good starting point for predicting the retention time of that analyte on another instrument.

Ervin Kováts took this idea a step farther and created a quantitative index relating the retention time of an unknown analyte to the retention times of known standards in a way that produced system-independent constants describing the elution behavior of compounds. The system-independent part is important, given all of the factors listed above that can influence retention time. Every analyte should have (nearly) the same Kováts index value on any GC system under any set of parameters. Thus, if we determine the Kováts index of analyte $\ce{A}$ on one GC, we can know something about how $\ce{A}$ will behave on other instruments or with a different temperature program, etc.

  Why the linear alkanes?

The choice of the n-alkanes was probably not arbitrary, but I cannot say for certain. Of the two links from the Wikipedia article that might contain a rationale, one is dead and the other is in German behind a paywall. However, any well-understood series of standards would have worked. My guess would be that Kováts chose the linear alkanes for the following reasons:

They form a homologous series increasing one carbon at a time.
A great many of them are (were - Kováts developed this index in the 1950s) available commercially in high purity (high enough purity that they can be easily purified further). At Sigma Aldrich, you can buy all of the n-alkanes at least up to eicosane $\ce{C20H42}$.
The first 20 linear alkanes cover a wide range of boiling points (and thus retention times): Methane $\text{BP}=-161.5\ ^\circ \text{C}$, pentane (the first liquid at room temperature) $\text{BP}=35.9\ ^\circ \text{C}$, decane $\text{BP}=173.8\ ^\circ \text{C}$, and eicosane $\text{BP}=343.1\ ^\circ \text{C}$.
There is a relatively reliable linear relationship between $\log(t_r)$ and the number of carbon atoms in the chain.

  How do you determine and use the Kováts index?

Lets say we have an analyte $\ce{A}$ that has a retention time of 193 seconds on some system. First, we identify the linear alkanes that have retention times on either side of the analyte. Lets say that heptane has a retention time of 170 seconds and octane has a retention time of 200 seconds in this system.

The formula for the index looks like this (for non-isothermal programs):

$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$

$I$ is the Kováts index
$n$ is the number of carbon atoms in the alkane with the lower retention time
$N$ is the number of carbon atoms in the alkane with the higher retention time
$a$ is the analyte
$t_r$ is retention time, so $t_{r,a}$ is the retention time of the analyte

For our example:

$$I=100\times \left[7+\left(8-7\right)\dfrac{193-170}{200-170} \right]$$
$$I=100\times \left[7+\dfrac{23}{30}\right]=777$$

Now, we want to analyze $\ce{A}$ on a different instrument, where the retention times of heptane and octane are 217 and 289 seconds, respectively. We can plug values back into the equation for the index to predict the retention time of $\ce{A}$ on this system:

$$I=100\times \left[n+\left(N-n\right)\dfrac{t_{r,a}-t_{r,n}}{t_{r,N}-t_{r,n}} \right]$$
$$777=100\times \left[7+\left(8-7\right)\dfrac{t_{r,a}-217}{289-217} \right]$$
$$7.77=\left[7+\dfrac{t_{r,a}-217}{72} \right]$$
$$0.77=\dfrac{t_{r,a}-217}{72}$$
$$55.2={t_{r,a}-217}$$
$$t_{r,a}=272$$


Predict XPS Spectrum of Organic Monolayers on a Semiconductor

Some density functional theory packages, such as Quantum Espresso, can calculate XPS core level shifts using the pseudopotential formalism. See Phys. Rev. B 74, 045430 (2006).


At the triple point the water system is assumed to be adiabatic. The system has 3 phases, ice from pure water, liquid water, and pure water vapor. The various equilibria  are shown below.

Note that the ice floats on the water. So the ice is in contact with both the liquid phase and the gas phase. Furthermore The ice can't cover the whole surface like a frozen over pond. There must be some liquid water in contact with the gas phase too.
A subtlety here... You can't make a solid phase of ice that is homogeneously say 1 molar NaCl. If you take a 1 molar NaCl and try to freeze it you get relatively pure water first, then saltier and saltier water freezing as the solid phase gets colder and colder. Since the impure ice has doesn't have a homogeneous composition, it doesn't have a specific melting point.
So for the system you proposed there is pure water ice,  a 1 molar NaCl solution, and a gas phase of water vapor. Unfortunately humans can't make a perfectly adiabatic container, so as the ice melts, the 1 molar NaCl solution becomes more dilute. Thus there is no specific triple point.
In other words... if you just throw some ice into the 1 molar NaCl solution, the final temperature of the theoretical system will depend on the relative amounts of the three phases - How much 1 molar NaCl solution, how much ice, and the volume of the gas phase.


What causes the iridescent colour in laboratory grown bismuth?

The colour is from a thin film of bismuth(III) oxide that forms on the surface if the crystals are formed in air. At the elevated temperatures used to melt bismuth, the oxide forms quite quickly. The iridescence is a result of thin film interference—light waves constructively or destructively interfere as they bounce off the bismuth-oxide and oxide-air interfaces. The interference is dependant on the wavelength and incidence angle of light as well as the film thickness, which results in different wavelengths of light being attenuated by different amounts, making the surface appear coloured and the colour changing with viewing angle.


Why must a buffer solution contain both a weak acid and a salt solution of its conjugate base?

You are correct that a weak acid will only partially dissociate in water. Taking for example the weakly acidic salt $\ce{NH4Cl}$ of which the ammonium cation is acidic, in water we get reaction (1).

$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$

To analyse this reaction we need the equilibrium constant which in the case of acids is defined as an acidity constant $K_\mathrm a$ as shown in (2). (Wether we use $\ce{H+}$ or $\ce{H3O+}$ is irrelevant.)

$$K_\mathrm a = \frac{[\ce{NH4+}][\ce{H+}]}{[\ce{NH3}]}\tag{2}$$

Using the tabulated $\mathrm pK_\mathrm{a}$ value of 9.25, we can calculate the amount of $\ce{NH3}$ in a $\pu{1M}$ solution of $\ce{NH4Cl}$:

$$\begin{align}K_\mathrm a &= \frac{[\ce{NH4+}]^2}{[\ce{NH3}]}\\[0.3em]
K_\mathrm a (c_0 - [\ce{NH4+}]) &= [\ce{NH4+}]^2\\[0.5em]
[\ce{NH4+}]^2 + K_\mathrm a [\ce{NH4+}] - K_\mathrm a c_0 &= 0\\
[\ce{NH4+}] &= \frac{-K_\mathrm a \pm \sqrt{K_\mathrm a^2 + 4 K_\mathrm a c_0}}{2}\\
[\ce{NH4+}] &= \pu{2.37e-5 M}\tag{3}\end{align}$$

Whoops. We may have a good concentration of ammmonium chloride but only a very minor amount of it is still ammonium, $99.998~\%$ are conjugate base. That doesn’t look like a good buffer system. If you were to add an acid, that would be fine since the excess base can capture that. But if you add more base the pH will go up rapidly, because there are no ammonium ions in solution to buffer it.

Only if you add both a significant amount of an acid and its conjugate base will you have significant amounts of both compounds in solution which are then able to buffer by capturing excess protons or hydroxides.


Making Ferric Chloride using HCl and Ferrum

You can find a glass beaker somewhere and simply drop in iron and diluted HCl. That's enough to make $\ce{FeCl2}$.

$\ce{Fe + 2HCl -> FeCl2 + H2}$

That forms hydrogen gas, which is very flammable, so it would be wise to do this outside. You might have to heat or reflux (heat the hell out of it while condensing whatever tries to evaporate out) depending on what concentration you use. Anyway, the resulting solution is gold-colored.

You'd then need to oxidize the Fe from a +2 state to a +3 state. I found this guide that recommends hydrogen peroxide (probably as a safer alternative to potentially stronger oxidizing agents.) It says it can take several days, so that's something to watch for.

Good luck! Again, I'd definitely do this outside or in a lab.

http://www.instructables.com/id/Make-Ferric-Chloride-for-etching-printed-circuit-/

Also, I've heard that not all nails are made of iron, so a density test could help ensure that.


Here's a device that's been around for at least several decades, as far as I know:

(source: teploterm.ru)
I've been using something similar for around 15 years. The idea is you put firewood into the oven you can see at the bottom, and the chimney goes through a 90 liter water tank. The bottom is something like .5 - 1 cm thick cast iron, the rest is made from stainless steel sheet metal.
There are variants to the construction, but I think here are the two main points, coming back to your question:

This thing needs replacement around once a decade, being in use around 500-1000 hours a year.

If the stainless steel is thick, then rusting gradually becomes irrelevant. Does make the thing heavy, though.

Yes, there is corrosion, especially around the fiery hot parts.

Also, note that the plumbing within this thing is kind of special, as it needs to refill cold water while simultaneously mixing your hot and cold water. I remember that it was very damaging to run it empty.
And yeah, this is more engineery than chemicky...


In terms of remediation, what is sequestration?

Sequestration is any way of preventing the heavy metals from being able to harmfully react with biologically important molecules. In the case of zero-valent iron nanoparticles, the heavy metals are sequestered by sorption as well as reduction. However, there are other methods of sequestration, such as chelation.


Did my teacher specialist about sulfur compounds made an error in part of Ramberg-Bäcklund rearragement mechanism?

As discussed above, the pathway shown on the left is incorrect because it essentially shows insertion into a $\ce{C-H}$ bond by an anion with loss of $\ce{X^-}$, which is not the easiest path for this reaction to take by a long shot.

The actual mechanism for this reaction is as shown below.

First, the non-halogenated alpha-carbon is deprotonated.
The carbanion formed arranges itself such that the it can attack the halogenated carbon and in such a way that the two R-groups are cis to each other.
Attack occurs, with loss of $\ce{X^-}$. Product is cis.
In water, due to lack of solubility, decomposition is faster than redeprotonation and the (Z)-alkene forms.
In organic solvent, base deprotonates an alpha-carbon, which can then scramble.
The main reprotonation product is the trans product.
The trans product decomposes, and the (E)-alkene is the major product.

A few notes:

The red path makes almost completely (Z) product. The blue path makes a mix of the two with the (E) product as the main product.
Purple star: the reason the anion attacks to form the cis ring is unknown, but there are several theories, apparently.
Green stars: cheletropic ring decomposition is symmetry forbidden, so this is theorized to be a two-step decomposition through either diradical bond fissions or a dipolar mechanism.


Can an amide nitrogen be a hydrogen bond acceptor?

If I understood you correctly, you are talking about the peptide bond nitrogens ($\ce{R-C(=O)-\color{red}{N}H-R}$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide.
The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. However, this lone pair is actually significantly delocalised towards the carbonyl bond. If we ignore the two $\ce{R}$ residues, we could write this the following way:
$$\ce{O=C-N-H <-> ^{-}O-C=N^{+}-H}$$
The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors.
Further, the fact that a Grignard reagent cannot react with an amide gives great support for the fact that nitrogen’s electrons are delocalized onto the carboxylate.

Finally, thank you Orthocresol for supplying me with the paper given at the end of this post. The authors performed some quantum chemical calculations to extract the energy difference between free formamide and water molecules and various conceivable hydrogen bonds between them. The strongest stabilisation can be found in an amide–amide hydrogen bond of the following type:
$$\ce{H2N-CH=O\bond{...}H-NH-CO-H}$$
At around $-40~\mathrm{kJ \cdot mol^{-1}}$ for a $\ce{H\bond{...}O}$ distance of $285~\mathrm{pm}$ and a $\ce{C=O\bond{...}H}$ angle of $60^\circ$.
$$\ce{H2N-CH=O\bond{...}H-OH}$$
For the interaction of a water molecule donor and the carbonyl oxygen as depicted just above, the stabilisation is given as up to $-27~\mathrm{kJ \cdot mol^{-1}}$ for a distance of $280~\mathrm{pm}$ and a $60^\circ$ $\ce{C=O\bond{...}H}$ angle.
$$\ce{OHC-HN-H\bond{...}OH2}$$
For this interaction where the amide nitrogen is donor and water is acceptor, the stabilisation was calculated to $\Delta E \approx - 30~\mathrm{kJ \cdot mol^{-1}}$ — achieved twice. Once for the geometry $d(\ce{H\bond{...}O}) = 280~\mathrm{pm}$ and $\angle(\ce{H\bond{...}O-H2}) = 30^\circ$ (the angle is measured between the $\ce{H-O-H}$ plane and the amide plane), and once for $d = 280~\mathrm{pm}$ and $\angle = 15^\circ$.
$$\ce{OCH-H2N\bond{...}H-OH}$$
This is the one you are looking for. These energy differences are all given as positive. The lowest is $\Delta E \approx +0.8~\mathrm{kJ \cdot mol^{-1}}$ for a distance $d(\ce{H\bond{...}N}) = 360~\mathrm{pm}$ — $1.3$ times the previous distances. (The angle is not given but from the picture we can assume a $90^\circ$ angle between the amide’s plane and the $\ce{N\bond{...}H}$ axis.) For the $280~\mathrm{pm}$ which brought the most stabilisation in the other cases and which thus seems to be the ideal hydrogen bond distance, they report $\Delta E \approx + 30~\mathrm{kJ \cdot mol^{-1}}$ — this is about the dimension of the stabilisation of the previous hydrogen bond only in a destabilising manner!
These studies let us conclude that it is safe to say the amide nitrogen will never be a hydrogen bond acceptor.
Reference:
A. Johansson, P. Kollman, S. Rothenberg, J. McKelvey J. Am. Chem. Soc. 1974, 96, 3794. DOI: 10.1021/ja00819a013.


To answer (2) first, propionaldehyde is the common (Trivial) name and hence is derived and held-onto only due to historical reasons and now obsolete nomenclature reasons. The 3-C aliphatic saturated qroup was historically called Propion- and hence propanoic acid corresponds to propionic acid and propanal corresponds to propionaldehyde, and propanoate(ester) corresponds to propionate in trivial nomenclature. It is a consequence of the history of the original naming of the 3-C aliphatic group at the time of its discovery.
As for (1), it is a consequence of pure convention; just as water is not called hydrogen hydroxide!


Why are tertiary amines less basic than secondary and primary amines in water than when vapourised?

A possible answer from what I've searched on the internet so far..

The strength of basicity of amines depends on two factors: 1) the availability of the lone pair for donation to a proton, 2) the stability of the conjugate acid of the amine.

In methylamine, the conjugate acid, $CH_3NH_3^+$ will attract water molecules and ion-dipole interactions will form between the cation (the conjugate acid of methylamine) and the partial negative charge on O in the water molecule. This ion-dipole interaction or attractive force helps to stabilize the positive charge on the cation.

So, the more water molecules you have solvating your cation, the more stable the conjugate base. However, the more alkyl groups you have, the less stable the conjugate base, because these alkyl groups are hindering water molecules from coming and stabilizing the conjugate base.

[The N atom is "shielded" by the bulky alkyl groups which take up a lot of space and prevent any water molecules from stabilizing the conjugate acid. Water molecules can only come in from where the H atom is where the shape of the conjugate acid is that of a pyramid. Note that there is a positive charge on N.]

As methylamine has the least number of alkyl groups as compared to dimethylamine and trimethylamine, at this point, it would be the strongest base. However, basicity also depends on the availability of the lone pair of electrons on the N atom of the amine itself. As methylamine has the least number of alkyl groups, the inductive effect by the alkyl group is the least out of methylamine, dimethylamine, and trimethylamine.

So, it really is a measure of which factor is more dominant or which factor outweighs the other. In this case, although trimethylamine has the most number of alkyl groups, the three bulky alkyl groups prevent a lot of water molecules from stabilizing the conjugate acid. And, for dimethylamine, although dimethylamine has a fewer number of alkyl groups than trimethylamine, the inductive effect of the alkyl groups outweighs the steric hindrance from the 2 alkyl groups as compared to the three in trimethylamine.

Hence, in an aqueous solution, $(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$

However, in the gaseous state, there are no water molecules to stabilize the conjugate acid and basicity depends on the inductive effect of alkyl groups; there are no water molecules to hinder. As trimethylamine is trigonal bipyramidal in shape, the factor of steric hindrance does not apply here. Hence, as trimethylamine has the most number of alkyl groups, the lone pair of electrons on trimethylamine is most available for donation.


what happens to the intramolecular bonds within water molecule during phase changes?

The only intramolecular bonds in water are the O-H bonds. They are not modified during phase changes. The hydrogen bond is not an intramolecular bond. It is an intermolecular bond. These hydrogen bonds are modified during fusion or ebullition.


Are there chemicals that absorb and release upon heating oxygen more efficiently than potassium oxide?

If you are looking for something that can be made using elemental  oxygen as an oxidizer and reversibly release the excess oxygen below 1000 °C, sodium peroxide beats potassium peroxide by using lighter metal atoms.  It can be prepared by reacting sodium with oxygen in two stages (forming sodium oxide in the intermediate stage) at 130–200 °C and releases the excess oxygen at 657 °C (Wikipedia).

Lithium, magnesium and of course hydrogen form peroxides with even higher excess oxygen percentages, but these are not made via reversible reactions with elemental oxygen.


Why is CO2 more likely to dissolve in a basic solution than an acidic solution?

A simple way to think about this is as follows (assuming we are talking about an aqueous solution).

The dissolution of $\ce{CO2}$ in water is followed by three chemical reactions:

$$
\begin{align}
\ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\
\ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\
\ce{HCO3- &<=> CO3^2- + H+}\label{rxn:R3}\tag{R3}
\end{align}
$$

Now the more basic the solution becomes, the further reactions \eqref{rxn:R2} and \eqref{rxn:R3} (\eqref{rxn:R3} becomes significant at high pH) lie to the right as hydroxide consumes $\ce{H+}$. As reaction \eqref{rxn:R2} depletes $\ce{H2CO3}$, reaction \eqref{rxn:R1}  is also forced to the right. The net result is more $\ce{CO2}$ residing in solution (dissolved) as the solution becomes more basic.


What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

It is commonly said, that a cyclopropane fragment behaves somewhat like a double bond. It can conjugate, and pass mesomeric effect similar to a double bond, but the donor orbital is $\sigma_{\ce{C-C}}$ instead of $\pi_{\ce{C=C}}$. Cyclopropane can be considered as a complex of a carbene and an alkene, where the carbene $\mathrm{p}$ orbital interacts with the $\pi^*_{\ce{C=C}}$ orbital while the carbene $\mathrm{sp^2}$ orbital interacts with the $\pi_{\ce{C=C}}$ orbital, so this 'virtual' double bond behaves somewhat like a normal double bond.

On the other hand, the structure of the cyclopropylmethyl cation is downright weird. It is well known that both cyclopropylmethyl and cyclobutyl derivatives give very similar product mixture under $\mathrm{S_N1}$ hydrolysis conditions, resulting in both cyclopropylmethyl and cyclobutyl derivatives (see, for example, J. Am. Chem. Soc. 1951, 73 (6), 2509–2520). This is commonly described by the conjugation in following manner (here, the 1-cyclopropylethyl cation is depicted):

Here bonding of $\ce{C-4}$ with $\ce{C-1}$ and $\ce{C-2}$ can be roughly described as an interaction of the vacant $\mathrm{p}$-orbital of $\ce{C-4}$ with filled orbital of $\pi$-bond between $\ce{C-1}$ and $\ce{C-2}$. It does not matter much, where the original leaving group was - at  $\ce{C-2}$ or $\ce{C-1}$. Since the positive charge is more or less symmetrically distributed between three atoms and the small ring is somewhat relieved of its geometrical strain (both cyclopropane and cyclobutane are very strained molecules, not only due to angle strain, but also considerable steric interactions between hydrogens), the cation has remarkable stability.

Similar effects are common in the chemistry of small bicyclic systems, with norbornane derivitives being the chosen test subjects for decades with the 2-norbornyl cation being probably the most well-known example. March's Advanced Organic Chemistry, 7th ed., Section 10.C.i discusses such nonclassical carbocations in great detail, with the cyclopropylmethyl system being described on pp 404–406.

With further addition of multiple cyclopropyl groups, however, the full conjugation becomes sterically hindered, so extra groups beyond the first have less of an effect. Of course, the stability of these cations is far below that of the tropylium cation, which has very little strain and also possesses aromatic character, distributing the positive charge over seven(!) carbon atoms.  In fact, the stability of the tropylium system is so high, that even the cyclooctatrienyl cation (also known as the homotropylium cation) adopts a similar structure.


number of optical isomers of a cyclic molecule

1-Bromo-3-chloro-5-fluoro-7-iodoadamantane would be an interesting example of central tetrahedral chirality without the actual (physical) center. Exchanging halogen atoms will result in only two stereoisomers.

Due to its rigidity, its stereochemistry is related to tetrasubstituted tetrahedrane (tricyclo[1.1.0.02,4]butane; or rather its imaginary geometrical model, because it would be very strained and unstable), whose stereochemistry is related to bromo(chloro)fluoro(iodo)methane.

To your additional question: It depends how (far) you'd like to go with removal of the rotation restriction, anyway the resulting compounds will already not be original 1,2-dichlorocyclohexane.

I mean yes, (1R,2R)-1,2-dichlorocyclohexane and (1R,2S)-1,2-dichlorocyclohexane can probably be considered geometrical isomers (in addition to optical), they can even be named trans-1,2-dichlorocyclohexane (which has two enantiomers, the other one being (1S,2S)) and cis-1,2-dichlorocyclohexane; but this isomerism is not caused by rotation restriction, but by different arrangements on chiral tetrahedral carbons.

But note that geometrical isomerism is usually used in relation with double bond cis/trans isomerism; and together with optical isomerism are both deprecated terms.


What happens to the mobility of Group 1 elements going down the periodic table?

When you have charged particles in a liquid acted upon by a electric field, the experience a force, and said force accelerates them to a terminal, drift speed.

Note: The physical treatment of the problem, and the notion of terminal speed is quite similar to say a ball falling from the top of a sky scraper through air (a fluid medium)

Now, let's go back to our charged particles in fluids. Let's say we have applied a voltage $\Delta\phi$ between two large, planar electrodes, immersed in the liquid and separated by a distance $l$

The  magnitude of the electric field, and the electric force experienced by an ion of charge $ze$ are given by the following relations:

$$ E = \frac{\Delta\phi}{l}$$
$$F = zeE = (ze).\frac{\Delta\phi}{l}$$

A cation responds by accelerating towards the negative electrode, and an anion responds by accelerating towards the positive electrode. This acceleration, however, is short lived because the fluid medium offers a retarding force $F_{\text{friction}}$.

$$F_{\text{friction}}\propto s\text{, where s := speed} $$

Assuming we have spherical particles of radius $a$, let's say our retarding force is given by Stokes' Law

$$F_{\text{friction}} = (6\pi\eta a)s$$ $\eta$) is the viscosity of the medium.

When the electric force is exactly balaced by the viscous drag, the acceleration drops to zero and this final speed of the particle is called the drift speed ($s_d$).
$$F_{\text{friction}} = F_{\text{electric}}$$ and solving for the drift speed we get,
$$s_d = \frac{zeE}{6\pi\eta a}$$

At this point we note that the drift speed is proportional to the applied electric field and introduce the following definition:

$$s_d = \mu E$$ where, $$\mu = \frac{ze}{6\pi\eta a} $$

and we call $\mu$ the mobility.

From the aforementioned equations/definitions one can deduce that $\mu \propto \frac{1}{a}$. Thus, bigger/bulky ions ought to have lower mobilities, and lower drift speeds in solution.

However, $a$ is not the ionic radius, but the Stokes radius (or the hydrodyanmic radius), i.e the "effective radius" of an ion in solution taking into account its solvation shell (i.e the solvent molecules surrounding the ion). I believe the following illustration is instructive,

The size of this solvation shell depends on how "attractive" an ion appears to a solvent molecule, and that in turn depends on the electric field produced by the ions.

For spherical ions, we consider the electric field on the surface of a charged sphere: $$ E _\text{surface of a charged sphere} \propto \frac{ze}{r^2} $$

here $r$ is the ionic radius. So smaller ions have a larger solvation shell than larger ions (assuming they carry the same charge).

This is the reason why mobilities of alkali metal ions increase from $\ce{Li^+}$ to $\ce{Cs^+}$ (and not the other way around); the lithium ions are effectively dragging a large amount of solvent molecules with them and thus, move slower.


Useful Definitions (from the IUPAC Gold Book)

chirality plane - "A planar unit connected to an adjacent part of the structure by a bond which results in restricted torsion so that the plane cannot lie in a symmetry plane. For example with (E)-cyclooctene the chiral plane includes the double bond carbon atoms and all four atoms attached to the double bond; with a monosubstituted paracyclophane the chiral plane includes the monosubstituted benzene ring with its three hydrogen atoms and the three other atoms linked to the ring (i.e. from the substituent and the two chains linking the two benzene rings)."
planar chirality - chirality resulting from a chirality plane.
chirality axis - "An axis about which a set of ligands is held so that it results in a spatial arrangement which is not superposable on its mirror image. For example with an allene abC=C=Ccd the chiral axis is defined by the C=C=C bonds; and with an ortho-substituted biphenyl the atoms C-1, C-1', C-4 and C-4' lie on the chiral axis."
axial chirality - chirality resulting from a chirality axis

Interpretation

According to the above definition, a chirality plane must involve restricted torsion (restricted rotation) about a bond.
A molecule can contain both a chirality plane and a chirality axis; an example would be an ortho-substituted biphenyl where the restricted rotation is responsible for chirality (atropisomers).

Conclusion

  Axial or planar chirality in spiro[3.3]hepta-1,5-diene

There is no restricted rotation that brings about chirality in this molecule.  Unlike trans-cyclooctene or a biphenyl, this molecule cannot be made to lie in a plane by torsional motion (at least not as long as we maintain the tetrahedral nature about the spiro carbon). Therefore, this molecule does not possess a chirality plane.

Rather, this molecule is analogous to the allenes. It does have a chirality axis passing through the $\ce{CH}$ group in each ring furthest from the spiro carbon along with the spiro carbon. The molecule can be said to possess axial chirality.


What are some uses for an O2/N2 mixture that is more N2-rich than air?

Some laboratories do use, and possibly some processes use "synthetic air", which contains 5% oxygen. This is to carry out oxidation reactions or processes but reduces the amount of excess oxygen to prevent unwanted combustion e.g. if flammable solvents are also being used. Many university research labs will use this if carrying out pressurised reactions.


Are there contradictions about intermolecular forces in the postulates of kinetic theory?

The kinetic theory of gases is, empirically, a good approximation to how most gases behave most of the time. Its key assumptions are:

there are no attractive forces between the molecules or atoms that make up the gas
the molecules or atoms interact only by elastic collision
the molecules or atoms have negligible size

The two statements you have are (especially the second) are poorly worded versions of these assumptions. For the collisions to be elastic there have to be repulsive forces between the entities that make up the gas.

But don't forget that the assumptions are not strictly true, they are just good approximations under most conditions. Molecules are not infinitely small and attractive forces do exist. But, at least when most gases are far from the point where they liquefy, the approximation is good and it explains the way gases behave. For example, a given quantity of a gas (specifically the number of molecules or moles) always takes up the same volume at the same temperature and external pressure however big or heavy the component molecule or atom is. We observe this to be true (approximately).

The apparent contradiction you worry about is due to a careless way of stating the key assumptions and is not a problem with the (well validated) theory.


What happens to the NMR of an alcohol when D2O is added?

Thank you for the question. This is a very basic physical chemistry question but the way it has been asked is interesting.

In order to understand the probability of displacement of a H from any X-H bond (X is any atom), you need to know its pKa. The pKa values are as follows (ref.):

pKa of water = 14
pKa of OH (phenol) = 9.95
pKa of CH (benzene) = 43

If the pKa of a bond is $x$, it means that for one $\ce{X^- + H^+}$, there has to be $10^x$ no of $\ce{X-H}$ compound. Thus, it clear that for one dissociation of OH bond of phenol, ~$10^{10}$ phenol molecules are required, while for one dissociated CH bond of benzene, $10^{43}$ molecules of benzene is reuqired. To put it in perspective, in 1 mole of phenol, there will be ~$10^{13}$ dissociated OH bonds but only ~$10^{-20}$ dissociated CH bonds.

This hereby implies that dissociation of CH bond and exchanging with $\ce{D_2O}$ is highly improbable, but dissociation of OH bond and its exchange with $\ce{D_2O}$ is highly probable. That's why you get the OH peak to be lowered due to exchange but no similar phenomenon occurs for CH bonds in the system.

Please reply in comments if you have doubts.

Greetings from India. जय हिन्द. Jai Hind.


Yes, see Structural and Stablity of Nitric Oxide Dimer.

Just don't call it "hyponitrite molecule", it might get offended.


It provides limited corrosion protection. I would suggest just putting oil on it; more corrosion protection and it will not obscure the surface pattern which is  most all that separates Damascus steel from ordinary steels. As I remember regular "bluing" is a complex of ferocyanides ; Oxides of iron are red and black , brown rusts are hydrated oxides. I have now looked at Wiki and see some of the problem.  Molten cyanide salts are use to carbonitride steels;I believe these leave a blue surface- you can't do that. Another source of blue ( and straw and purple) color is different thicknesses of very thin oxide layers. I have a chart showing one hour at 540 F (in air) , or 10 minutes at 590 F, or 20 hours at 485 F will produce a blue. Higher temperature and longer time give thicker oxide. The color of the oxide is determined by the thickness. Assuming the blade is hard like Rockwell C 60, these temperatures will cause some softening. The other methods sound like they will etch the steel . Same conclusion, oil it.


Iron exists in the +2 oxidation state (iron(II)), also called ferrous iron, as well as the +3 oxidation state (iron(III)), also called ferric iron.  There are sever other oxidation states of iron that are not pertinent to your reactions.

  In your first reaction, the strongly oxidizing $\ce{Cl}$2
  oxidizes the neutral, metallic iron to it's highest stable oxidation
  state, iron(III).

  In your second reaction, $\ce{HCl}$ oxidizes iron to the lower stable
  oxidation state, iron(II).


what are possible isomers of (meso-stilbenediamine)(isobutylenediamine) palladium (II)?

The use of the term "meso" in the compound name indicates that both phenyl groups are on the same side of the molecule, "cis" to one another, not "trans".

Here are the tetrahedral and square planar representations.

Note: I used $\ce{Pt}$ instead of $\ce{Pd}$ as the central metal, but it makes no difference in terms of isomers.

The tetrahedral molecule has a plane of symmetry containing the "top" 5-membered ring, bisecting the 2 methyl groups on that ring and then bisecting the "bottom" 5-membered ring with the 2 "meso" phenyl groups. Since the tetrahedral molecule contains a plane of symmetry it is achiral and has no isomers.

In the square planar form, there is no symmetry element.  Therefore the square planar molecule is chiral and exists as an enantiomeric pair of isomers.


What does "at%" refer to in sol-gel method?

Atomic percentage. It can be converted to weight percentage (wt%) too.


Toothpaste is what is called a non-newtonian fluid, more specifically toothpaste is a Bingham plastic.  This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous alone is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon).

What the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy).

Now if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid.

Other examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty.


Determining a molecular formula of the catalyst for polymerisation of butadiene

Despite the absurd lack of data, this problem attracted my attention, probably because it closer resembles a real-life challenge rather than a textbook problem.
It looks like it's an adaptation of the problem from Schaum’s Outline of Theory and Problems of College Chemistry [1, p. 39]:

3.37 What is the empirical formula of a catalyst that can be used in the polymerization of butadiene if its composition is $23.3\%$ $\ce{Co},$ $25.3\%$ $\ce{Mo},$ and $51.4\%$ $\ce{Cl}.$
Ans. $\ce{Co3Mo2Cl11}$

Here we are dealing with zero knowledge as to what the remaining element(s) are, but with a couple of assumptions we actually can come up with an answer from the textbook, and an answer on top of that with a justification.
Let's dive in and assume we are dealing with an unknown compound with the following formula:
$$\ce{Co_xCl_y}\sum_{i=1}^N \ce{El}_{i,z_i}$$
where we account for $i$ other elements $\ce{El_$i$}$ with the respective coefficients $z_i.$
The first assumption required to advance with the solution is to treat the compound as stoichimetric. e.g. $x, y, z_i \in\mathbb{N}.$
This would immediately allow us to find the exact values for both $x$ and $y$:
$$y : x = \frac{m(\ce{Cl})}{M(\ce{Cl})} : \frac{m(\ce{Co})}{M(\ce{Co})} = \frac{\pu{25.7 g}}{\pu{35.45 g mol-1}} : \frac{\pu{11.65 g}}{\pu{58.93 g mol-1}} = 3.67\tag{1}$$
To satisfy $x, y \in\mathbb{N},$ and taking into account the "hint" decimal part $.67$ (which implies a triple factor to an integer):
$$x = 3, 6, 9, \ldots, 3n~(n\in\mathbb{N})\tag{2}$$
To pinpoint the exact allowed value of $x$ (and $y,$ as $y = 3.67x),$ we can use the total molar mass:
$$M = x × M(\ce{Co}) + 3.67x  × M(\ce{Co}) + \sum_{i=1}^N z_iM_i\tag{3}$$
$$\pu{759 g mol-1} = x × \pu{58.93 g mol-1} + 3.67x × \pu{35.45 g mol-1} + \sum_{i=1}^N z_iM_i\tag{3a}$$
$$\sum_{i=1}^N z_iM_i = (759 - 189x)~\pu{g mol-1}\tag{3b}$$
Since molar mass is a positive number, $759 - 189x > 0$ and the only $x$ to satisfy this criteria would be $x = 3.$
Accordingly, $y = 3.67 × 3 = 11,$ and at this point we are left with the following formula:
$$\ce{Co3Cl11}\sum_{i=1}^N \ce{El}_{i,z_i}$$
and the remaining sum
$$\sum_{i=1}^N z_iM_i = (759 - 189 × 3)~\pu{g mol-1} = \pu{192 g mol-1} \tag{3c}$$
The second assumption is that we are dealing with the common oxidation numbers of the elements, no exotic stuff.
With this in mind, and knowing that cobalt catalyst used in polymerization are vastly only $\ce{Co^{II}}$ and $\ce{Co^{III}}$ species, we can deduce the possible charge of the remaining sum:
$$(\ce{Co3Cl11})^{q-}\left(\sum_{i=1}^N \ce{El}_{i,z_i}\right)^{q+}$$
Since it's likely only $\ce{Cl-},$ the $q$ can adopt only the values defined by $\ce{Co^{II}}:\ce{Co^{III}}$ ratio.
For the total of three cobalt(II,III) atoms $q = 2,3,4,5.$
This will assist us in proposing missing element(s), and one can start to iterate over numbers.
The third assumption is that $i = 1$ and there is only one extra element with single coefficient $z_1$.
The table below summarizes possible outcomes:
$$
\begin{array}{ccccc}
\hline
z_1 & M_1/\pu{g mol-1} & \ce{El_1} & M(\ce{El_1})/\pu{g mol-1} & \text{Formula Example} \\
\hline
1 & 192 & \ce{Ir} & 192.22 & \ce{Co^{II}2Co^{III}Ir^{IV}Cl11} \\
  & & & & \ce{Co^{II}Co^{III}2Ir^{III}Cl11} \\
2 & 96 & \ce{Mo} & 95.95 & \ce{Co^{II}3Mo^{II}Mo^{III}Cl11} \\
  & & & & \ce{Co^{II}2Co^{III}Mo^{II}2Cl11} \\
3 & 64 & \ce{Cu} & 63.55 & \ce{Co^{II}Co^{III}2Cu^I3Cl11} \\
  & & & & \ce{Co^{II}3Cu^{I}Cu^{II}2Cl11} \\
4 & 48 & \ce{Ti} & 47.87 & \ce{Co^{?}3Ti^{?}4Cl11} \\
\hline
\end{array}
$$
Starting with $z = 4,$ there doesn't seem to be an appropriate set of oxidation numbers for the element to comply with the charge balance, so I'd say the candidates for the third elements are only iridium, molybdenum and copper.
Judging from the application (catalysis) as well as from the typical oxidation states, I'd actually propose iridium cobalt chloride $\ce{Co3IrCl11}$ as the answer, yet $\ce{Co3Mo2Cl11}$ would also be possible (AFAIK, molybdenum(II,III) isn't a common composition for halide salts), and $\ce{Co3Cu3Cl11}$ would be a stretch both due to higher deviation from the declared molar mass and presence of copper(I).
References

Rosenberg, J. L.; Epstein, L. M. Schaum’s Outline of Theory and Problems of College Chemistry, 8th ed.; Schaum’s outline series; McGraw-Hill: New York, 1997. ISBN 978-0-07-053709-5.


I haven't used MarvinSketch in a while, but isn't it a 2D editor?

Generation of 3D coordinates from 2D files often means trouble. I tried that recently with a 2D BKChem drawing of a trypticene-based ligand for another answer and the result was useless.

I definitely recommend the use of Avogadro as a free 3D editor. Please have a look at their website or the article in Journal of Cheminformatics.

My typical workflow here is:

switch off auto-adding of hydrogen
draw the skeleton with the right multiplicity of bonds, but using carbon only
change the "wrong" carbons to heteroatoms
let the program add the hydogen atoms
run a force-field optimization
manually adjust some dihedral angles
run the force-field optimization again
(optional) align a bond with one of the principal axis
save as CML or MOL

I put a MOL file of etravirine created that way on pastebin. Please give QuteMol a try at it.

I generated a PDB from the molfile mentioned above and ran it through QuteMol. The image below shows that the $\ce{C-Br}$ bond does exist in the PDB file.

Admittedly, checking the PDB file for the respective CONNECT x y line would have been a (less flamboyant) proof too ;-)


Manganese(II) sulphate is not colourless. It is a faint pink. Compare this image taken from Wikipedia, where a full list of authors is available:

However, it is true that the colour is rather faint, and you need a white background to properly see it. That’s because manganese(II) is a $\mathrm{d^5}$ system and a high-spin one of those meaning that all five d-orbitals are occupied by a single electron:

Exciting an electron directly is forbidden due to the spin-retaining rule: The excited electron must occupy an already populated orbital, which is only possible if it flips its spin, which is forbidden by selection rules. Special mechanisms are needed to account for the (obviously possible but faint) transition here; the page Orthocresol linked in the comments names spin-orbit coupling.


Thermodynamics - Series Reaction - Estimating mols of reactants based on Free Energy of Rxn


Stable alkynes are normally linear, that is, they have bond bond angles of 180°. That is necessitated by the fact that alkynes are formed by two π-bonds and one σ-bond, which must be mutually perpendicular. If cyclopentyne were to exist, it would be forced to adopt bond angles of roughly 108°, equal to those of the interior angles of a regular pentagon. This deviation from ideal geometry causes extreme instability by virtue of poor orbital overlap. The rigidity of the alkyne moiety would also exacerbate the inherent conformational inflexibility of small rings, wherein the inability of bonds to freely rotate results in eclipsing between atoms.

The only cyclic alkyne I'm aware of is benzyne. However, it's far from stable, existing only as a fleeting intermediate in specific nucleophilic aromatic substitution and pericyclic cycloaddition reactions.

Edit: After a quick search, I found the cycloalkyne Wikipedia page, which claims that cyclooctyne is the smallest cycloalkyne stable enough to isolate and observe. I don't know whether this information is completely current, however.


How does temperature affect equilibrium, according to Le Chatelier's principle??

In a simple equilibrium we have either $$\ce{A + heat <=> B}$$ where the forward reaction is endothermic and the back reaction exothermic, or $$\ce{A <=> B + heat}$$ where the forward reaction is exothermic and the back reaction endothermic.  So in a simple equilibrium, the reaction in one direction will be exothermic and the reaction in the other direction will be endothermic.

When the temperature of an equilibrium is increased, the rate of both the forward and back reaction is increased.   However the rate of the endothermic reaction (the reaction that absorbs the added energy) is increased more.  So in the first reaction shown above, adding heat will shift the reaction to the right; in the second reaction heat will push it to the left.

The reason behind this behavior is that both the forward and back reaction can be described by the Arrhenius equation $$\ce{rate=Ae^{-E_{act}/RT}}$$  and the rate with the larger activation energy (the endothermic reaction) will always be more sensitive to changes in temperature. Therefor, the endothermic reaction will increase faster with increasing temperature than the exothermic reaction will.  Put some numbers into the equation and convince yourself.


Does anything like this covalent concerted bond exchange reaction exist?

$$\ce{W=X + Y=Z ->[\mathrm{cat}] W=Y + X=Z}$$ is known as olefin metathesis, achieved using Grubbs/Hoyveda/Fürstner catalysts.

Water-soluble Ru catalysts have been prepared in the Grubbs group (DOI).


Your arguments do not contradict each other. You are trying to separate a process into two processes, which are coupled of course, but in first approximation this is a good way at looking it.

Let's state our model system again: A balloon with the pressure $p_1$, the volume $V_1$ is heated from temperature $T_1$. As a result we see an inflated balloon. Let's call those variables with $2$ as an index. Of course we treat everything as ideal as possible.

Now we look at the coupled process as two separate processes:

Isochoric heating, i.e. $V=\mathrm{const}$:
\begin{align}
&&p_1V&=n\mathcal{R}T_1\\
&&p_2V&=n\mathcal{R}T_2\\
\therefore
&&\frac{p_1}{p_2}&=\frac{T_1}{T_2}\\
\therefore
&&p&\propto T\\
\end{align}
That is Gay-Lussac's law.
Isothermal expansion, ie. $T=\mathrm{const}$
\begin{align}
&&p_1V_1&=n\mathcal{R}T\\
&&p_2V_2&=n\mathcal{R}T\\
\therefore
&&\frac{p_1}{p_2}&=\frac{V_1}{V_2}\\
\therefore
&&p&\propto \frac{1}{V}\\
\end{align}
That is Boyle's law.

You can also look at the whole process as the pressure stays the same while the balloon is heated. That then is an isobaric process, i.e. $p=\mathrm{const}$:
\begin{align}
&&pV_1&=n\mathcal{R}T_1\\
&&pV_2&=n\mathcal{R}T_2\\
\therefore
&&\frac{V_1}{V_2}&=\frac{T_1}{T_2}\\
\therefore
&&V&\propto T\\
\end{align}
That is known as Charles' law

Of course in reality there will be a mixture happening as the rubber of the balloon will expand, but the pressure that is possible is dependent on the material of it. In combination you will arrive at:
\begin{align}
&&p_1V_1&=n\mathcal{R}T_1\\
&&p_2V_2&=n\mathcal{R}T_2\\
\therefore
&&\frac{p_1V_1}{T_1}&=\frac{p_2V_2}{T_2}\\
\therefore
&&\frac{pV}{T}&\propto \mathrm{const}\\
\end{align}


Why is sodium sulfate neutral while its conjugate acid is weak?

Given $\text{pK}_{a_1} = -3$ and $\text{pK}_{a_2} = 1.9$ for sulphuric acid, the pH for a 1M solution is indeed expected to be $7.9$, $7.4$ for a 0.1M, and $8.0$ for a saturated solution. So it is slightly basic.


Can polyatomic ions (CO₃, PO₄, SO₄, NO₃) be considered conjugated systems?

"Conjugated" implies a 1,3-shift to move an electron or hole.  It is a remnant of LCAO modeling that is obviously wrong but fantastically useful short of the Woodward-Hoffmann rules.  I'm not sure inorganikers would like the name as such. MO modeling is accurate but unwieldly.  "Delocalized" or "resonance hybrid" is good, certainly for inorganic systems that may have (virtual) d-orbital participation.

Consider nitrate.  Inorganikers would say $\ce{N^{5+}}$, $\ce{^{-}O-N(=O)2}$, with five bonds to the nitrogen.  The negative charge 1,3-shifts around all three oxygens.  Organikers would see it as $\ce{N^{3+}}$, $\ce{[^{-}O{}-]_2N^{+}=O}%edit$, with four bonds to the nitrogen.  1,3 shifts, etc.  Is nitrate ever a bidentate ligand?  YES!  But it is just another resonance structure.

Inorg. Chem. 35(24) 6964 (1996)
http://pubs.acs.org/doi/abs/10.1021/ic960587b


This is a highly concentrated base, which is very corrosive. I would assume, that its concentration is quite unstable.

It will, for example, quite rapidly react with carbon dioxide from the air:
$$\ce{NaOH + CO2 -> NaHCO3}\tag{1}$$

Also it will corrode glass containers, dissolving minerals like magnesium and calcium from it:
$$\ce{Mg^{2+} + 2NaOH -> Mg(OH)2 v + 2Na+ }$$

A PE bottle should be used to avoid that.

Even if you worked in a very clean and low humidity environment, the surface of your pallets will be corroded since sodium hydroxide is hygroscopic and will further react according to $(1)$, even when stored carefully.

Therefore the concentration you obtain by weighing the pallets can only be quite approximate. If you need to know the exact concentration of your base, you have to calibrate it versus a primary standard, like dried succinic acid, before and after the intended use.
In your case standardisation will indeed be not easy to perform, since your concentration is very high. Most likely there are uncertainties to the procedure that will likely leave you with similar error bars than not doing the calibration.


How does sodium bicarbonate act as a deodorizing agent?

The key factor is that baking soda is amphoteric. This means that it can react both with strong acids and with strong bases which are typically smelly compounds (sour milk, rotten fish, fatty acids in sweat).

The explanation on the third page of article by Lamb [1] gives a nice summary:

  The chemical explanation of why bicarbonate of soda should act as a deodorant
  is of interest. Sweat (3) has an acid reaction, $\ce{pH}$ of $5.2\text{–}6.75$ which is due to the fatty acids or acid phosphates of sodium and potassium which it contains. It also contains $\ce{NaCl}$, $\ce{KCl}$, alkaline salts, organic acids and urea. Minute amounts of unstable fats and oils are secreted in sweat and hydrolysed to their corresponding fatty acids and glycerin. Free fatty acids may be secreted. Many of these acids are volatile and have a characteristic disagreeable odor. These include butyric acid, formic acid, caproic acid and valerie acid. The following are among the possible explanations of the manner in which sodium bicarbonate acts as a deodorant in the axilla.

  It forms a sodium salt with butyric, caproic and valerie acids. These sodium salts are comparatively mild in odor in comparison to the volatile "rancid-smelling" fatty acids.
  By changing the $\ce{pH}$ of the axillae, the bacteria which decompose the fats are unable to act. However, as a result of this the staphylococcic flora which prefer higher alkalinity might thrive and axillary furunculosis might result.

To give an example. The reaction of baking soda with butyric acid looks like this:

$$\ce{C3H7COOH + NaHCO3 -> C3H7COONa +H2O + CO2} $$

while butyric acid is extremely smelly (an intense vomit smell), the salt hardly smells at all.

Given this explanation, I think $\ce{NaOH}$ should obviously also work well for deodorizing in case the smell is caused by acids. The problem would be that smells caused by basic compounds are left untouched.

References

Lamb, J. H. Sodium Bicarbonate: An Excellent Deodorant. Journal of Investigative Dermatology 1946, 7 (3), 131–133. https://doi.org/10.1038/jid.1946.13


Why do higher orbitals have more energy?

potential energy is negative between infinity to nucleus

It varies with .Thus, U∝(−1)1/r so it means the father you go from nucleus the value become less negative that is higher in energy than equilibrium energy.


Validation of _atom_site_frac_*: Range outside of [0,1)?

I completely agree with @porphyrin comment. This is an example of crystal data which formatting does not entirely meet IUCr requirements, that's all. This happens a lot, and in most cases it's no big deal. This particular set of coordinates is just not uniform within a single cell. Different crystallographic software uses different means to "reassemble" atoms "correctly" and write changes to *.cif, e.g. Olex2 uses Standardise function for that:


Are there any datasets containing molecules with more than 38 heavy atoms?

This sounds like you were exploring work at least related to the work by the Lilienfeld group equally hosting a dedicated site here about data sets already used in their earlier and ongoing exploration of chemical space, programs used to work with the data, and publications.

To go considerably higher in molecule count than QM9, you could either go for

GDB-11 about small organic molecules up to 11 atoms of C, N, O and F which «contains 26.4 million molecules (110.9 million stereoisomers), including three- and four-membered rings and triple bonds», described in J. Chem. Inf. Model. 2007, 47, 342-353 (doi.org/10.1021/ci600423u), or
GDB-13, about «small organic molecules up to 13 atoms of C, N, O, S and Cl following simple chemical stability and synthetic feasibility rules. With 977 468 314 structures, GDB-13 is the largest publicly available small organic molecule database to date».  This one was described in J. Am. Chem. Soc. 2009, 131, 8732-8733 (doi.org/10.1021/ja902302h)

Convienently, you can download both -- including sub-sets like «containing only carbon and nitrogen», or «chlorine and sulfur», or «fragrance like» in case you don't want to fetch 2GB of already compressed data -- from the Reymond group. To quote: «All the molecules are stored in dearomatized, canonized SMILES format.»

The even larger GDB-17 («of up to 17 atoms of C, N, O, S, and halogens» with an universe of 166 billion entries, described in J. Chem. Inf. Model. 2012, 52, 2864-2875, [doi.org/10.1021/ci300415d, open access]) is accessible to the public on this site as a 50 million random subset only, partly because the gzipped archive is about 400GByte.  Among the publications citing this work is for example the Lilienfeld group again for machine learning (J. Chem. Phys. 143, 084111 (2015), doi.org/10.1063/1.4928757).

Initially, I misinterpreted the question but think the answer may be more rounded by addition of the following complementary publication:  «Chemical diversity in molecular orbital energy predictions with kernel ridge regression» (J. Chem. Phys. 150, 204121 (2019), doi.org/10.1063/1.5086105, preprint available here).  Aiming for a machine-learning analysis, the authors first compared QM9, 44k conformers of proteinogenic amino acids (AA), and a 64k set of organic molecules extracted from the CCDC potentially suitable for organic electronics (OE) for the content of atoms per molecule and found the following distribution:

To shed some light on them:

QM9 represents 133,814 small organic molecules with up to 9 heavy atoms (C, N, O and F)
AA is about «44,004 isolated and cation-coordinated conformers of 20 proteinogenic amino acids and their amino-methylated and acetylated (capped)
dipeptides. The molecular structures are made of up to 39 atoms including H, C, N, O, S, Ca, Sr, Cd, Ba, Hg and Pb.»
OE is about «64,710 large organic molecules with up to 174 atoms extracted from organic crystals in the Cambridge Structural Database (CSD). [...] The OE dataset is not yet publicly available. OE offers the largest chemical diversity among the sets in this work both in terms of size as well as number of different elements (Fig. 2). It contains the 16 different element types H, Li, B, C, N, O,  F, Si, P, S, Cl, As, Se, Br, Te and I.»

(The mentioned restriction sharing the original data relates to the user agreement with the CCDC.)

Further DFT-based property computations with these OE extracted molecular geometries lead to an ensemble of equilibrium molecular structures, and these derived geometries are accessible within a public Jupyter notebook.  Shared with the public here, the deposit comes with a guiding tutorial.ipynb, including an example how to retrieve these optimized geometries and display them with Jmol.


Could a strong acid become a strong base?

Take nitric acid for instance. Reduce the nitrate until ammonia forms. Then using a stronger base than amide ion, you can generate amide base.

Therefore, it is required to make a number of reactions to get a strong base from a strong acid. If you meant that could a strong acid be changed to a strong base by one step reaction, it might be possible with complex big organic or organoboron molecules by replacement of acid functional group with a basic one, but I am unsure about this part.


What happens when 3-(2-chlorophenyl)propanenitrile is treated with sodium amide?

You are correct that a benzyne is formed as an intermediate in this reaction. However, the nitrile is not an innocent bystander. Recall that protons α to a nitrile group are acidic: this is because the negative charge can be delocalised onto nitrogen. The pKa of a typical nitrile is around 30, lower than that of ammonia, which is roughly 40 (see: Evans pKa table). So, sodium amide is capable of deprotonating α to the nitrile.

This anion can now act as a nucleophile in the benzyne mechanism which you are no doubt familiar with. Note that the intramolecular attack of this nucleophile is likely to outcompete intermolecular attack of the amide ion. Also, since the nucleophile must attack intramolecularly, there is no question about which triply bonded carbon it attacks; it can only reach the position ortho to itself:

This specific reaction to form a benzocyclobutene is described in, for example, J. Med. Chem. 1966, 9 (5), 656–659.


Iodine is having 53 electrons and cesium is having 55 electrons in general. Iodine ion can be made by adding or by removing electrons from it. Here you added two electrons to iodine and made it an iodine ion in which number of electrons of iodine ion became equal to the number of electrons of cesium but the composition of nucleus(number of protons) is different in both iodine ion and cesium ion. Obviously cesium is having more number of protons than iodine, so cesium nucleus can hold it's electrons more closer in cesium than iodine nucleus in an iodine ion which results in larger radius of iodine ion than cesium..
Same number of electrons doesn't mean that their radii are same. Same logic is applicable for lithium ion and lithium.. Lithium ion is having less electrons and more protons compared to lithium.. So radius of lithium is larger than radius of lithium ion


What are the dangers of handling 0.12N sulfuric acid?

0.12 Normal = 0.06 Molar.

For comparision, concentrated sulphuric acid is 18 Molar.

So the 0.12 N sulphuric acid is over 99% water.

  Would it be safe to handle this without a respirator?

Yes, the vapor pressure of sulphuric acid is low, and it has already been diluted with considerable water, so there is no need for a respirator.

  My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes.

Goggles are the most important safety equipment.  An eye wash station and shower are also standard lab safety equipment.

The gloves are good because liquids will not pass through them.  Ordinary clothing isn't really helpful because if you did spill something on that part of your body, the substance would absorb into the cloth and you would just be delayed in washing that part of your body by having to take off the clothes.

  Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity?

Pronged contact may cause some damage, but nothing comparable to the concentrated acid.  If you spill some on yourself, stay calm, and wash with soap and water. If you spill on your clothes, take the wet clothes off first, then wash with soap and water.

  Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa?

This is always a good rule to keep in mind.  0.12N is dilute enough that little heat will be released from 0.5mL, but just follow the rule.

  However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. Will it still be dangerous even spraying a low concentration in the air?

Now you are diluting another 1000 fold.  The solution will be around pH 4, similar to some rain water.  I would still want you to wear goggles and not spray in the direction of anyone.


Molar mass of a protein from osmotic pressure

Osmotic pressure of a solute in water is given by the formula

$$\Pi = i\rm{MRT}\tag{1}$$

Where:
$\Pi$ is the osmotic pressure
$i$ is the Van't Hoff factor
$\rm{M}$ is the molarity of the solute (mol/L)
$\rm{R}$ is the universal gas constant 62.36 ($\rm{L}\space \rm{Torr}\space$ $\rm{K}^{−1}$ $\rm{mol}^{−1}$)
$\rm{T}$ is the absolute temperature ($^\circ\rm{K}$)

We'll assume $i =1$.

Let's use $\rm{M_W}$ for the molecular weight of the protein. So the moles of the protein is equal to $\rm{g/M_W}$.

$$\dfrac{0.010\space\rm{g}/\rm{M_W}}{0.010\space\rm{L}}\times 62.36\space\rm{L}\space \rm{Torr}\space\rm{K}^{−1}\space\rm{mol}^{−1}\times293\space\rm{K}  = 52.2 \space \rm{Torr}\tag{2}$$

$$\rm{M_W} = \dfrac{0.010\times62.36\times293}{0.010\times52.2}\space\rm{g/mol}= 350\space\rm{g/mol}\tag{3} $$

Your mistake was believing the book answer.

  If you need to check this sort of answer, then choose R with convenient units. It makes solving the problem much easier.


How to rationalise the protonation of a compound with multiple double bonds?

The key is delocalization of the positive charge in carbocations.

Most stable carbocation that you can get from the protonation of the compound above. Reasons for it being the most stable:

The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)
In one of the resonance forms, the positive charge is at a tertiary carbon atom

Second most stable carbocation.

The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)

When protonation happens on the third pi bond. No delocalization - not as stable as the carbocations shown above.

There are other possible ways in which the compound could have been protonated (below), however none of them are as stable as (1) or (2), because there is no delocalization.


What does it mean for a higher glass transition temperature in terms of crystallinity?

A higher glass transition indicates less energy is required for the bonds of the glass to relax their strain. This is usually due to a slower cooling rate. In a slower cooling rate regime, more time is allowed for crystallization to occur.  So while you are correct to observe a correlation, one does not cause the other but rather they have a common cause of the slower cooling rate.


What is the number of unpaired electrons in cobalt(II) tetrathiocyanate?

According to the tetrahedral splitting in crystal field theory, the filling of the orbitals is just opposite to the filling in case of octahedral splitting. Thus filling up the elecrons according to it and obeying the hunds rule (since it is a weak ligand) the number of unpaired electrons are 0 in $d_{x^2−y^2}$ and $d_{z^2}$ whose lobes are directed along the axis.


Which carbons in pyrrole have the highest electron density?

Interesting question, more subtle than I realized!

Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must also consider the inductive effect of the nitrogen atom, which draws some electron density away from C2 and C5 through the sigma bonds. This would imply that the two proximal carbons (closest to N) are more electron-deficient, and the distal (more distant) pair of carbons are more electron-rich.

Experimental considerations: Orthocresol has a good point that pyrroles are observed to form bonds with electrophiles at the proximal carbons (2/5). However, I think this trend may be due to the superior stability of the intermediate produced by that pathway (3 significant resonance forms) versus a distal (3/4) attack (2 significant forms). The cation from a proximal attack is better stabilized by resonance than the distal-attack cation. I expect that pyrrole forming C2 bonds with electrophiles has more to do with the stability of the intermediate of that pathway than necessarily greater electron density on C2. What I did not expect, but read on Wikipedia, is this:

  Unlike furan and thiophene, [pyrrole] has a dipole in which the positive end lies on the side of the heteroatom, with a dipole moment of $1.58\ \mathrm{D}$.

This is experimental evidence demonstrating that the electron density is lower on the proximal side (C2,5) and greater on the distal side (C3,4). Besides molecular modeling of the electron density clouds, I think the molecular dipole moment is about the most conclusive evidence one can hope for.

It really comes down to the context: in an undergrad first-semester course, you might be expected to say the distal carbons have more density because the nitrogen withdraws from the proximal carbons. In more advanced courses, you might take orthocresol's perspective and base your answer in reactivity, concluding the proximal have more electron density. But I expect a doctorate researcher would probably conduct molecular modeling and compute the electron distributions using quantum mechanics, and perhaps conduct a few experiments to measure the dipole. And at that level, I think C3 and C4 have the honey.


Reversible and irreversible processes of thermodynamics

For both reversible and irreversible processes, the pressure of the gas at the interface with the surroundings exactly matches the external pressure imposed by the surroundings $P_{ext}$, and the work done on the surroundings is always equal to $\int{P_{ext}}dV$.  (The equation you wrote for an irreversible process applies only if $P_{ext}$ is held constant).  However, for an irreversible process, the pressure of the gas at the interface is not given by the ideal gas law, because (a) there are viscous (dissipative) stresses that contribute to the pressure at the interface and (b) the pressure and temperature of the gas is not uniform within the cylinder.  So we need to specify how $P_{ext}$ varies during the process (say, as a function of the volume V) manually in order to analyze an irreversible process.

For more details on this, see the following link:  Calculate Work Done for Reversible and Irreversible Adiabatic process


Both pyrrole and furan have a lone pair of electrons in a p-orbital, this lone pair is extensively delocalized into the conjugated pi framework to create an aromatic 6 pi electron system.

Where pyrrole and furan significantly differ is that,

in pyrrole there is an $\ce{N-H}$ bond lying in the plane of the ring and directed away from the ring
whereas in furan, there is a full lone pair of electrons in roughly the same position.

The localized lone pair of electrons pointing away from the ring has a very significant effect on the dipole vector and is enough to cause the observed reversal in dipole moment direction between furan and pyrrole.


Relationship between freezing point depression and hydrogen bonds

The addition of one molecule (or salt) of solute produces $i$ effective solute particles. This defines the van't Hoff factor $i$, which encodes the strength of the solute with regards to colligative properties.

As simple (idealized) examples in aqueous solution, the van't Hoff factor of $\ce{NaCl}$ is 2, as $\ce{NaCl}$ dissociates into the two solute particles $\ce{Na+}$ and $\ce{Cl-}$ in solution, whereas the van't Hoff factor of glucose is 1, as glucose doesn't dissociate in solution.

These examples are idealized because they do not account for intermolecular or interionic interactions, and are therefore most valid in dilute solutions. In concentrated solutions, intermolecular and interionic effects are much more important: the presence of hydrogen bonding, for example, or ion pairs in solution. Attractive interactions "clump" the solute particles together and reduce the van't Hoff factor.

Returning to your example, if we assume that acetic acid does not dissociate in solution, and we're under conditions in which acetic acid dimer predominates, then indeed we will have a van't Hoff factor $i = 0.5$. It's important to note, however, that the hydrogen-bonding interaction in acetic acid is particularly strong, and in general intermolecular and interionic interactions will not change the van't Hoff factor so drastically.

It remains to consider how to correct for these interactions. Intuitively, it seems to me that the activity coefficient should be the desired correction factor, considering that the activity coefficient is a measure of how ideal (interaction-less) a solution is, but I do not have a more quantitative argument along these lines.


How was it determined that the electron observed in the cathode ray experiments was the same particle that gave an atom its balancing negative charge?

The idea that the atom contains small indivisible quantities of electric charge was developed in the 19th century by various contributors (e.g. Faraday, Stoney, Laming, Weber, and Helmholtz).

When Thomson (Nobel Prize in physics 1906) discovered the electron in 1897 and estimated its mass and charge, he used the free electrons of cathode rays.

The important experiment that showed that atoms contain electrons was the discovery of the Zeeman effect in 1896, i.e. the splitting of spectral lines in the presence of a magnetic field. Lorentz explained the phenomena with his electron theory. In 1902, Lorentz and Zeeman jointly received the Nobel Prize in physics.


Brillouin's Theorem: How do other determinants mix with the ground state?

Brillouin's theorem is the direct consequence of the so-called Slater–Condon rules which for the case of the HF reference determinant $\Phi_{0}$ and a singly excited determinant $\Phi_{a}^{r}$ show that the Hamiltonian matrix element between these determinants reduces to the Fock matrix element between the corresponding orbitals
$$
  \newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
  \newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
  \newcommand{\op}[1]{\hat{#1}}
  \newcommand{\el}{_{\mathrm{e}}}
  \newcommand{\core}{^{\mathrm{core}}}
  \newcommand{\oneeintphys}[3]{\bracket{#1}{#2}{#3}}
  \newcommand{\twoeintphysrantisymm}[2]{\langle{#1}\,||\,{#2}\rangle}
  \bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }
  =
  \oneeintphys{a}{\op{H}\core}{r} + \sum\limits_{j=1}^{n} \twoeintphysrantisymm{aj}{rj}
  =
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} } \, .
$$
Now taking the HF equations and orthogonality of spin orbitals into account we indeed get zero for $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ since
$$
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} }
  =
  \varepsilon_{r} \braket{ \psi_{a} }{ \psi_{r} }
  =
  0 \, .
$$
Now try to use the Slater–Condon rules for the case of the HF reference determinant $\Phi_{0}$ and a double excited determinant $\Phi_{ab}^{rs}$ to convince yourself that $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{ab}^{rs} }$ unlike $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ won't vanish. This is a very well-known result which manifests itself in many different ways. For instance, this is the reason why in order to improve the HF energies in configuration interaction scheme we include doubly excited determinants first (CID) rather than singly excited once (CIS): the later won't have any effect on electronic energy.


How does one apply the knowledge of the direction of a molecular dipole moment?

Take a look at chemlibretexts. I'll just write an overview.

A dipole is a vector representing the charge distribution in a molecule. It's drawn from the positive-density charge side towards the negative.

Some examples

Phase change

How do we explain $\ce{CO2}$ fusion temperature is $-78\,^\circ$C while $\ce{SO2}$ is $-10\,^\circ$C? Because of its dipole magnitude. In the first one, there's no net dipole, but there is for the second one.

The bigger molecular interactions the more they oppose to temperature-increasings.

Solubility is higher for alcohols than lipids* in water. Again dipole-dipole interaction is stronger (energetically more favourable) than dipole-induced dipole interaction.
CO$_2$ has no electric dipole in the symmetric vibration. According to quantum mechanics, if electric dipole does not change when it absorbes light of a specific energy, it will not be susceptible of a infrared measure. This particular vibrational mode will be (and it is) infrared inactive.


How to calculate the yield for an incomplete reaction?

Well the problem doesn't say anything about how to get the % yield. If you start with 0.527 grams of platinum, but have 0.155 grams left, then 0.372 grams reacted.

$\frac{0.372}{0.527} * 100$% = $70.6$%

As for the yield

$\frac{0.372}{0.527} * 0.732 = 0.517$ grams


How to calculate the solubility of AgCl in a MgSO₄ solution?

In pure water, as you calculated, for the reaction
$\ce{AgCl -> Ag+ + Cl-}$ with $K_\mathrm{sp} = 1.8\cdot10^{-10}$
 the concentration of $\ce{Ag+}$, i.e. $\ce{[Ag+]}$ will be $1.34\cdot10^{-5}~\mathrm{M}$.

In a solution of $0.25~\mathrm{M}$ $\ce{MgSO4}$ (which dissociates almost completely), there's an additional equilibrium here:
$\ce{Ag2SO4 -> 2Ag+ + SO4^2-}$ with $K_\mathrm{sp} = 1.4\cdot10^{-5}$

Looking at the $K_\mathrm{sp}$ of both the reactions, we can see that $\ce{AgCl}$, having the lesser solubility, decides the $\ce{[Ag+]}$. Here we can even check if the concentration of $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$ and $\ce{[SO4^{2-}]} = 0.25~\mathrm{M}$ exceeds the solubility limit of $\ce{Ag2SO4}$. It doesn't.
$(1.34\cdot10^{-5})^2 \cdot 0.25 = 4.5\cdot10^{-11} < (K_\mathrm{sp})_{\ce{Ag2SO4}}$

The final composition of the solution will be $\ce{[Ag+]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Cl- ]} = 1.34\cdot10^{-5}~\mathrm{M}$, $\ce{[Mg^2+]} = 0.25~\mathrm{M}$ and $\ce{[SO4^2- ]} = 0.25~\mathrm{M}$.

In conclusion, the $\ce{[Ag+]}$ is the same for both the cases.


The $\ce{Cu^{2+}}$ ion is capable of coordinating to the lone pairs of the nitrogens of the amide bonds.

$\hspace{1.4cm}$

The formation of this complex is associated with a certain energy change in the orbitals of the $\ce{Cu^{2+}}$ ion, and hence a color change.


Step 1 is a Claisen ester condensation with loss of ethoxide from the ester acting as the carbonyl part. The product is the $\beta$-ketoester (2).

Upon acidification and heating, the ester is cleaved, the resulting $\beta$-ketoacid decarboxylates to yield the symmetrical ketone.


Organic chemistry textbook for self-learning?

I'll provide a rundown of the general organic chemistry textbooks I've studied in detail:

Organic Chemistry, David R. Klein: A very solid introductory text. Emphasis is on expediency and development of intuition over rigorous theoretical correctness. The coverage of molecular orbital theory is extremely scant, and classic valence bond theory (with associated concept of orbital hybridization) is the major explanatory paradigm. This is probably the most accessible text for students with a minimal background, but it's also somewhat superficial.
Organic Chemistry, Jonathan Clayden, et al.: This is a very thorough text, with a slightly wider array of reactions discussed than Klein's textbook (most notably in the coverage of organometallic reagents). Again, the MO treatment is lacking, but it's invoked more frequently than in Klein. This book tends to delve quite deeply into the mechanisms underlying specific reactions, making it rather lengthy. There's also some superficial discussion of HSAB theory, which is mostly (if not completely) absent in Klein's book.
Advanced Organic Chemistry, Carey & Sundberg. This is something of a seminal classic, but it's not for the faint of heart or for those uninitiated in the fundamentals. Mechanisms are elucidated in great detail (primarily via MO theory, with results from both ab initio and semi-empirical methods), and the exposition of pericyclic reactions is notably excellent. It also contains a great deal of illuminating content on conformational analysis. This is definitely not appropriate for a first textbook, but I would consider it essential reading for advanced undergrads and above.

Ultimately, either of Klein's or Clayden's textbooks should satisfy all of your criteria, but I strongly recommend Carey & Sundberg for subsequent study, if possible.

Beyond those, I've found Fleming's Molecular Orbitals and Organic Chemical Reactions to be a very good introduction to qualitative MO theory, but it's not a general-purpose text. March's Advanced Organic Chemistry is the ultimate, all-encompassing reference manual, and I'd consider it the sine qua non of any good organic chemistry library. As an aside, McQuarrie and Atkins have both published very good physical chemistry textbooks, which in many ways have been more enlightening than any textbook specific to organic chemistry. I'm also a fan of Oxtoby's Principles of Modern Chemistry as a general chemistry textbook; its mathematical and physical rigor is incredibly admirable, even if it does sometimes lack severely in concision. I've also heard unanimous praise for Housecroft & Sharpe's Inorganic Chemistry, but I've yet to read more than a few sections of it.


There are several calcium phosphates.  I'll assume you mean monocalcium posphate, $\ce{Ca(H2PO4)2}$, because the others are not nearly soluble enough in water to allow even a 1.5 mmol/L solution.  (I assume the toothbase's most abundant ingredient is water.)

The formula weight is 234.  (I won't bother with many decimals because your range of molarities, 1.5 to 18 mmol/L, is so wide.)  I got this number from
http://en.wikipedia.org/wiki/Monocalcium_phosphate although it is easy to compute from scratch by adding the atomic weights for 1 Ca atom, 4 H atoms etc.

To calculate how much  of this compound you need given the molarity, ideally you should be looking at a known volume of paste.  Instead you have a known weight, 1g.  If you know the density of the paste, you can compute the volume by dividing 1g by the density.  Or you can just use an approximation.  I'll assume that 1g = 0.9mL (this will be a good approximation).  From now on I will assume volume = 0.9mL.

Multiplying 1.5mmol/L by 0.9mL gives you 1.35 micro moles (1.35$\times 10^{-6}$ moles).  Since 1 mole is 234 grams (formula weight), this means 316 micrograms or 0.315 mg.  The same calculation with 18mmol/L gives 3.79 mg.

So your answer (discarding  unwarranted decimals) is: between 0.31 and 3.8 mg .

I hope if your proprietary formulation is commercially successful you will make a generous contribution to the organization behind stackexchange!


Will "soft nucleophiles" undergo 1,4-conjugate addition even if the β-carbon is sterically hindered?

Your line of thought possibly is influenced by the Walden inversion which is characteristic for the nucleophilic substitution mechanism of $\mathrm{S_N2}$ where steric hindrance affects the rate when this umbrella-like structure flips upon entry of the nuclophile and departure of the nucleofuge around a formally $\mathrm{sp^3}$ hybridized center:

(credit)
The reaction shown by you however is an addition on to a carbon with molecular orbitals formally hybridized as $\mathrm{sp^2}$.  Similar to the Bürgi–Dunitz trajectory for the entry of a nucleophile on to a $\ce{C=O}$ bond a) it is plausible that the angle of attack for the 1,4-addition b) equally is tilted, rather than an attack from the back.

(composed with a picture from here)


Best solvent for crystallization + mass recovered from crystallization + purity of the product

Expanding on LDC3's comment a little.

The information in the table (irrelevant of how good or bad the method proposed is) tells you what the saturation points are for solvent mixtures at given temperatures for each compound.

Fundamentally, you want to separate your desired product from your undesired product. In this case, you are trying to recrystallise the 3-phenylacrylaldehyde, leaving the benzaldehyde in solution. You know this because for all temperatures and solvent mixtures it has lower solubility (this cannot be assumed for a more complex solvent array).

You need the minimum of hot solvent, here 100 degrees, to dissolve all of your crude compound, and you want to maximise the return of pure compound on cooling. This explains why you want the biggest ratio of hot to cold solubility as LDC3 indicated.

You also want to minimise the amount of your starting material that crashes out with your product. So, calculate the same ratios for the starting material, but here you want to minimise the return on cooling, so you want the smallest ratio.

You now have a table of solubility ratios, columns being 'solvent mix', 'rP' (ratio of product solubilities) and 'rSM' (ratio of starting material solubilities). The largest value of rP/rSM gives you the best solvent mixture.


How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock?

What you're describing is essentially the field of "qualitative inorganic analysis". That's a very broad subject, but a short description is that prior to the advent of advanced instrumentation, chemists developed extensive series of tests specifically to determine the elemental composition of samples.

For example, one would start with basic physical properties like color, density, and hardness, then look at reactions with heat, acids, bases, oxidants, etc. If the compound could be solubilized in any way, then one can look at how the solubility/color/etc is affected by addition of known cations and anions. If it was a geological specimen, then the location and the types of surrounding minerals would help inform the analysis as well.

A well trained inorganic chemist would have standard flow-charts of tests and could fairly confidently identify the elemental composition of common minerals and metals.