What is the molecular structure of the fragment ion m/z 161 of rosmarinic acid?

The source you link to says that the spectrum was acquired in negative mode.

  The CID experiment on $\ce{[M - H]-}$ ion at m/z 359.08 identified as rosmarinic acid gave the two main constituents of rosmarinic acid, namely caffeic acid at m/z, 179.0 and the 2-hydroxy derivative of hydrocaffeic acid at m/z
  197.0 as illustrated in Figure 3. Similar pattern of fragmentation of rosmarinic acid during CID analysis has been reported by several authors (17, 25, 26) in analyzing extracts of Lamiaceae spices.

This means that the parent ion is $\ce{(C18H16O8 - H)-}$, and thus the (hydrated) fragment in question is $\ce{C9H7O4-}$.

The first question in any CID mechanism is whether the fragment arises via charge migration or via charge retention.  I think in this case a charge retention mechanism is pretty likely, because phenolic hydroxyl groups are acidic enough to ionize readily via proton loss.

The next question is where the loss of water comes from.  It's worth noting that the mass spectrum of even just plain old catechol, $\ce{C6H6O2}$, shows a loss of water in negative-mode CID.  That is, the parent ion $\ce{(C6H6O2 - H)-}$ loses water somehow to form $\ce{(C6H3O)-}$.  I see no reason why the same thing couldn't happen to the rosmarinic acid fragment in question.

It's impossible to say for sure without doing a lot of esoteric spectroscopic characterization of gas-phase ions, or least the synthesis and fragmentation of a series of partially deuterated or $\ce{^{18}O}$-labeled rosmarinic acid isotopologues.  Neither of those is an easy task (far from it), so it's nearly impossible to say for sure what the structure of the fragment is.

But think a perfectly reasonable proposition is that the ion is the same as whatever the structure of the catechol dehydration fragment is, perhaps a deprotonated hydroxybenzyne or maybe an epoxybenzene anion?  Loss of aromaticity in CID fragmentation not unusual.  In fact, a previous Q&A here on Chemistry StackExchange provides yet another example of this.

High-energy gas-phase ion chemistry is weird.


Do lithium and beryllium want to lose electrons to be like helium?

Yes, lithium wants to lose electrons to be like helium because full valence shells are more stable states and all noble gases have full valence shells. So alkali metals lose one electron to achieve nearest noble gas configuration.

But beryllium is exceptional because of its small atomic and ionic radii. Due to this, its ionization potential is very high. So it mostly forms covalent bonds.


Comparing strength of the C-H, N-H, and O-H bonds

The answer should be C.
One way would be to write the thermochemical equations, however since we just want to compare, we can do it without going into that.
The formation of the H-Cl bond, (like all bonds) is exothermic. And the breaking of the C-H, N-H and O-H bonds is endothermic (since their formation is exothermic too).
Since the contribution of the H-Cl bond to the enthalpy is the same for all three, we can focus on the C-H, N-H and O-H bonds alone.

Now, the stronger the bond, the more energy it would need, and hence make a positive contribution to the enthalpy change. Hence we can see that since the total enthalpy of the reaction is highest(most positive) for the reaction involving O-H bond, the O-H bond is the strongest. (Infact, it is stronger than the H-Cl bond too as the enthalpy change is positive).

Note that this question also demonstrates the order of electronegativity, which is
O>Cl >~ N> C>H. Since the bond with most electronegativity difference would be the strongest, it would be the O-H bond.


How to remove a stuck stopper from a flask?

Is it a hollow glas stopper?

Sometimes, it helps to hold the flask at the body and hit the corner of the lab bench with the stopper. If that doesn't help, heat the neck of the flask carefully with a heat gun.


How to find the pH of acetic acid after adding sodium hydroxide?

If you add $3 \cdot 10^{-2}~\mathrm{mol}$ $\ce{NaOH}$ to $1~\mathrm{L}$ of $2 \cdot 10^{-2}~\mathrm{M}$ $\ce{AcOH}$ you end up with a solution that contains $2 \cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{AcO-}$ and $1\cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{OH-}$. So the pH is 12.


How can water with a very low hardness have substantial buffer capacity?

In an aquarium, there could be several non-mineral buffers. The most likely one is a phosphate buffer. There would also be carboxylic acid (amino acids) and probably urea, as well as carbon dioxide.


What can be done to increase the yield of a product in a chemical reaction that has reached equilibrium?

I agree with your initial starting point, that the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide may be treated as an example of an heterogeneous equilibrium

$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$

that still is of relevance today

(source, Hughes et al. in Ind. Eng. Chem. Res. 2004, 43, 5529-5539).

and actually at around $\ce{900 ^\circ{}C}$, the equilibrium pressure of $\ce{CO2}$ over $\ce{CaCO3}$ and $\ce{CaO}$ would equal to $\pu{1 atm}$.

Your first reasoning, while keeping the other parameters of this equilibrium fixed ($T, p$), a removal of $\ce{CO2}$ may increase the yield of $\ce{CaO}$ since this eventually (macroscopically) leads to an exhaustive decomposition of $\ce{CaCO3}$ is correct.

For the second part, you have to balance the general idea about chemical equilibria (which in this case indeed means, keeping other parameters fixed, adding more $\ce{CaCO3}$ as starting material yields more $\ce{CO2}$) with the fact that this particular reaction is heterogeneous.  Looking up the data for molar mass and density, (1, 2, and 3) allows you to estimate the volume one mole of each pure compound under standard conditions occupies:

| compound | molar mass |  density | molar volume |
|          |    [g/mol] | [g/cm^3] |   [cm^3/mol] |
|----------+------------+----------+--------------|
| CaCO3    |     100.09 |    2.711 |        36.92 |
| CaO      |      56.08 |     3.37 |        16.64 |
| CO2      |      44.01 | 0.001977 |     22261.00 |
|----------+------------+----------+--------------|

The value of the molar volume of $\ce{CO2}$ may remind you to the value of the volume occupied by the Ideal Gas at standard conditions ($\pu{22.7 L/mol}$) and demonstrates that the influence of $\ce{CO2}$ on the position of the equilibrium is much more important than the two other of the solids with their -- in comparison -- often negligible molar volumes. Hence, to underline that this equilibrium is this much pressure dependent, it is described by $K_{\text{p}}$, rather than by a normal $K$ value.

The general idea of influencing a chemical equilibrium along "an increase of  partial volume (or partial pressure) of starting material(s) will increase the yield of the product(s)" better full-filled in the case of homogeneous reactions, like the synthesis of ammonia

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}. $$


Entropy changes during transfer of heat between two bodies

This answer is correct.  After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature reservoirs running from its initial temperature to its final temperature.  At least, that is what your equation implies.  This is definitely the correct way to determine the entropy change.


How does turbidity affect dissolved oxygen?

According to this document, turbidity and oxygen content can be correlated because the more microbes in the water, the more turbid the water is, and the less oxygen is left (because they breathe it).

But a correlation is no cause-effect relationship. Both are linked in the particular case of microbes because they affect both by their presence.

Not a reliable method at all. More oxygen in the water does not mean it is potable. If you let sit a bucket of water contaminated with microbes, the oxygen levels will drop over time, but if you stir the water, you re-dissolve some up to saturation levels, but you still have the microbes present.

You may want to evaluate the oxygen demand, or the biomass (absorption at 600nm), centrigugate and evaluate visually, etc.


The first thing to realize is that "mole" is not a mass unit. It is simply a quantity - a number - like dozen or gross or score. Just as a dozen eggs is 12 eggs, a mole of glucose is $6.02 \times 10^{23}$ glucose molecules, and a mole of carbon atoms is $6.02 \times 10^{23}$ carbon atoms. "Moles" are only associated with mass because individual objects have mass, and thus a mole of objects also has a certain mass.

So why $6.02 \times 10^{23}$? What's so special about Avogadro's number? Well, nothing, really, it just makes calculations work out nicely. Avogadro's number is defined as the number of $\ce{^{12}C}$ atoms which weigh 12 g. So it's effectively a ratio: how many times larger is a gram than an atomic mass unit? If you have one atom of $\ce{^{12}C}$, it weighs 12 amu. If you have $6.02 \times 10^{23}$ of them, they weigh 12 g. If you have one molecule of glucose, it weighs 180 amu (or thereabouts). If you have $6.02 \times 10^{23}$ of them, they weigh 180 g (or thereabouts). -- This is just like if one egg weighs 60 g (on average), then a dozen of them weigh 720 g (on average), and if one cup of flour weighs 120 g (on average), then a dozen cups of flour weigh 1440 g (on average). The only difference is that dozen is defined forwards ("a dozen is twelve"), whereas a mole is defined "backwards" ("a dozen is the number of 60 g eggs that are in a collection of eggs that weighs 720 g.")

This convenience definition - pegging the value of Avogadro's number directly to the difference in scale between the amu and the gram - is what is probably throwing you. Moles are not a mass unit, but the definition is intimately tied to mass units. The equivalence in numbers at the atomic scale (amu) and at the macroscopic scale (grams) can also result in chemists playing fast and loose with terminology, quickly working back and forth from atomic to macroscopic scale, without a necessarily clear distinction between the two.


Why is the dz2 orbital so different from the rest?

The wikipedia is helpful in explaining why radial variations should arise in the density of non-s orbitals:

  The non radial-symmetry properties of non-s orbitals are necessary to localize a particle with angular momentum and a wave nature in an orbital where it must tend to stay away from the central attraction force, since any particle localized at the point of central attraction could have no angular momentum.

What's unique about the $d_{z^2}$ orbital (see the table above, from the wikipedia) compared to the other $l=2$ angular momentum wavefunctions is that the z-component is zero ($m=0$). This further constrains the geometry of the wavefunction.

The functions describing the angular dependence of the hydrogenic wavefunctions are Legendre polynomials $Y_{lm}(\theta,\phi)$, solutions of Legendre's differential equation. In the case of d-orbitals, they satisfy

$$ \hat{L}^2Y_{lm}(\theta,\phi)=\hbar^2l(l+1)Y_{lm}(\theta,\phi)$$

with $l=2$, where $ \hat{L}$ is the angular momentum operator. Since the z-component of the angular momentum is also quantized, the following eigenequation also holds:

$$ \hat{L}_zY_{lm}(\theta,\phi)=\hbar mY_{lm}(\theta,\phi)$$

with $m=0$ in the case of the $d_{z^2}$ orbital, and this last equation leads to the following condition:

$$ \frac{\partial\psi}{\partial y^2} = \frac{\partial\psi}{\partial x^2}$$

which implies that the solutions must be cylindrically symmetric about z. However, the condition $l\neq 0$ implies that the solution is not spherically symmetric. The result is the unexpected shape of the $d_{z^2}$ orbital.


How to calculate the number of moles present in a litre of saturated solution?

For your salt, calcium phosphate, or $\ce{Ca_3(PO_4)_2}$, this is the correct dissolution equilibrium:

$\ce{Ca_3(PO_4)_2 \leftrightharpoons 3Ca^{2+} + 2PO_{4}^{3-}}$

Upon dissolution of one unit of calcium phosphate, we get five units of ions total, true, but the solubility product is defined as follows (from the IUPAC Gold Book):

The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.

For our purposes, we will use concentrations instead of activities. Concentrations are a rough gauge of activities in very dilute solutions (such as this one involving the semisoluble salt, calcium phosphate).

Also since you mentioned this was a multiple-choice test, I highly doubt that the following hydrolysis reaction will be relevant, even though at such low concentrations of phosphate anion, the phosphate anion should be nearly completely hydrolyzed.

$\ce{PO_{4}^{3-} + H_2O ->HOPO_3^{2-} + HO^-}$

Back to the problem. Upon dissolution of one unit of calcium phosphate, we should get 3 units of calcium ion and 2 units of phosphate ion.

Therefore, based on the IUPAC definition:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}$

Taking $\ce{x}$ to be one unit of calcium ion:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}=[3x]^3[2x]^2$

Solving the above yields $x = 10^{-6}$.

Therefore, with respect to your original question, the number of moles of calcium phosphate present initially can be number greater than 0 moles. For the amount of calcium phosphate present at equilibrium, then yes, $5x$ appears to be the right answer.


It's relatively easy to add an electrophilic species, such as X-X, to an alkene. As @Philipp points out, the alkene is operating through the (filled) pi orbital.

It's relatively easy to add a nucleophilic species to a carbonyl. In that case, the carbonyl operates through its (unfilled) pi* orbital.


Product of intramolecular aldol condensation of 3-methyloctane-2,7-dione

For the enolates that can be formed, the others have already given comprehensive answers: All four enolates are possible and since we are under thermodynamic control (weak base), we expect o observe them all. Numbering the chain carbons according to IUPAC, and noting the non-carbonyl end of the enolate, I expect the enolate ratio to be $3>6>1\approx 8$.

But your question asked for the reaction’s major product. First of all, the two terminal enolates (carbons 1 and 8) are to be dismissed: The seven-membered ring is much less stable. Then, the reaction mechanism shows us which enolate leads to a thermodynamically much more stable product than the other:

The more stable enolate 2 unfortunately cyclises to give a β-hydroxyketone 5 which has a quarternary carbon between the ketone and the newly-formed hydroxy group. No elimination is possible here. The other enolate aldol product 3, however, can eliminate to the corresponding cyclopentene 4. This should likely be given as the reaction’s major product since all steps up to the elimination are reversible. (The elimination is, too, but it is still most stable.)

Indeed, Stork and Williard report that the cyclopentene 4 is obtained when 3-methyloctane-2,7-dione is refluxed with ethanolic KOH for 2 hours, although no yields are given as it is only mentioned in a footnote.[1]

Stork, G.; Williard, P. G. Five- and six-membered-ring formation from olefinic α,β-epoxy ketones and hydrazine. J. Am. Chem. Soc. 1977, 99 (21), 7067–7068. DOI: 10.1021/ja00463a053.


can metamers contain N as differing functional group?

This term is so old, I'd never even heard of it, had to look it up. Hooray for wikipedia

https://en.wikipedia.org/wiki/Metamerism

. As i understand it, this applies to the two compounds you give.

The proper modern term is "isomer", a substance having the same molecular formula as a second one, but in different structural arrangement. (And yes, the two are isomers, although you would only rarely come in a situation where you want to discuss them as such, because they are chemically very different.)


How to show that when the equilibrium constant is larger than one, some real-life equilibria are still reactant favoured?

The terms reactant-favored and product-favored
As discussed in the comments and in YB609's answer, these are not technical terms with precise definitions.
First possible definition
One straight-forward definition would be to call reactant-favored reactions those for which $K$ is smaller than one, and product-favored those for which $K$ is larger than one. You would have to specify the reaction (with phase information), the temperature and the standard state you are using. With this definition, you would be correct to say that such a reaction, if initiated at standard state, would go in the net forward direction to reach equilibrium if $K$ is greater one (and therefor, according to this definition, it is product-favored). Of course, if you take this definition, the question does not make any sense.
Second possible definition You could make a definition based on whether there is "more" reactant or product at equilibrium. This is problematic for multiple reasons. First, you could have a lot of one product, little of another product, and something in between for the reactant. So is there more reactant or more product? Second, if you use amounts of substance (moles) as the criteria for "more" and have a heterogeneous reaction, the relative amounts depend on the volumes (compare oxygen in equilibrium between hemoglobin and the dioxygen in a test tube vs in the entire atmosphere).
Third possible definition If you look at the concentrations to define "more", it still is confusing under some circumstances, even for a homogeneous reaction (everything in the same phase). If the sum of coefficients of reactants is not equal to that of products, diluting the system will disturb an equilibrium, so reactant: product ratios will depend on the dilution of the system.
Answering the question

How would I go about answering this question?

You would start by asking for the definition of reactant-favored and product-favored.

Do I approach it with regards to initial concentrations in the system, their initial states (mixed vs. unmixed), pressure and volume changes?

Once the terms are clearly defined, it would be sufficient to look at equilibrium states (and you need to know the definition of the standard state, otherwise the value of $K$ has no meaning).


Why can't Schiff's reagent distinguish between glucose and fructose?

Schiff test is a colorimetric method for determination of aldehyde groups. Both glucose and fructose are monosaccharides and exhibit oxo-cyclo tautomerism. Glucose is an aldose and has aldehyde group in linear form. Whereas fructose is a ketose and doesn't give a positive Schiff test.


Why does magnesium oxide not react with water?

Like a lot of things in chemistry, "it depends".

Magnesium oxide is commonly produced by calcination of magnesium hydroxide or carbonate. The magnesium hydroxide or carbonate decomposes at a fairly low temperature whereas magnesium oxide is itself stable to both decomposition and melting  beyond 2000°C. So, we can choose from a wide range of calcining temperature, and as explained here we can use this to dial in the reactivity we want.

At a low calcining temperature the magnesia is quite reactive, even hygroscopic (the "magnesium oxide" film formed on magnesium in ambient air is thus usually magnesium hydroxide); this "light-burned magnesia" is used where easy reactivity is required as in water or soil treatment.  And, yes, the reaction with water gives magnesium hydroxide.

Calcination at higher temperature gives less reactive "hard-burned" magnesia, or in the extreme case the totally unreactive "dead-burned magnesia" used in furnace refractories.


How to find the formula of an binary oxide based on the ionisation energies?

The point of this problem is to test if you can see where there is a jump in ionization energies. The third ionization energy is considerably larger than the others. Although ionization energies will naturally increase in succession, this is the most profound increase, which would suggest that this atom will tend to lose 2 electrons before obtaining a stable octet structure.

This would mean that its ion would be $\ce{A^{2+}}$. Since the oxide ion is $\ce{O^{2-}}$, the charges will be balanced if the ratio of the ions is 1:1, meaning the answer is $\ce{AO}$.


Is there a way to Determine if an Element can undergo Disproportionation?

Expanding on my original comment, I disagree with the OP's statement that knowing the possible oxidation states of an element will easily allow you determine if a disproportionate reaction will occur.

To predict such a reaction, thermodynamic data such as the $E_\mathrm{cell}^\circ$ of the two reactions must be known. Latimer diagrams are a really simple and compact way of showing the $E_\mathrm{cell}^\circ$ values for a variety of reduction and oxidation half-equations between different oxidation states of the same element. Below is an example of a Latimer diagram for manganese in acidic medium:

Just from this I can tell the $E_\mathrm{cell}^\circ$ potential for any oxidation or reduction reaction between any pair of manganese species (in fact there is extra information that isn't even required). For example by looking at this diagram I can tell what $E_\mathrm{cell}^\circ$ will be for the reduction of $\ce{Mn^7+}$ to $\ce{Mn^3+}$ (I won't go into how you would do that in this post, I suggest that if you want to know how to, have a look at the website that I have linked).

But wait, there is more! Now the neat thing about Latimer diagrams is that I can easily tell if a particular oxidation state of manganese is stable or not (in other words, will it undergo disproportionate reaction?). All you have to do is follow one very simple rule:

'if the number on the left is lower than the number in the right, the species is unstable to disproportionation'

For example, let's consider $\ce{Mn^3+}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO2}$ to $\ce{Mn^3+}$ is $0.95~\mathrm{V}$ (that is the number on the 'left') while the $E_\mathrm{cell}^\circ$ for the reduction of $\ce{Mn^3+}$ to $\ce{Mn^2+}$ is $1.51~\mathrm{V}$ (this is the number on the 'right'). Since the number on the left is smaller than the number on the right, $\ce{Mn^3+}$ will spontaneously disproportionate to form $\ce{MnO2}$ and $\ce{Mn^2+}$. The reaction is below:

$$\begin{align}
\ce{2Mn^3+ + 2H2O} &\ce{-> MnO2 + Mn^2+ + 4H+} & E_\mathrm{cell}^\circ = +0.56~\mathrm{V}
\end{align}$$

Note how for this reaction $E_\mathrm{cell}^\circ > 0$, hence it is spontaneous.

However these 3 species that are involved don't have to be right next to each other. For example, let's consider $\ce{MnO3-}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO4-}$ to $\ce{MnO3-}$ is $0.419~\mathrm{V}$ while the E cell for the reduction of $\ce{MnO3-}$ to $\ce{Mn^3+}$ is $2.61~\mathrm{V}$. Again, the number on the left is smaller than the number on the right, hence $\ce{MnO3-}$ will disproportionate spontaneously.

This may not be the answer that you were expecting originally, but hopefully this will help you with your original questions.


Why is there an increasing solubility in water for chlorides, chlorates, perchlorates in that order?

Science is all about discovering hidden regularities, patterns, and laws. In that sense your bold generalization is good, but will it hold if we check just one more data point? Let's see: $\ce{KCl}$ has solubility of 25g/100ml, $\ce{KClO3}$ has 8, and $\ce{KClO4}$ has 1.5. Er, well...

What we have here is an interplay between solvatation energy and lattice energy, the latter being determined by the crystal structure, which is a tricky thing, and hence so is solubility. Don't expect it to be predictable from qualitative considerations. You don't know it until you measure it. Chemistry is an experimental science, after all.


Energy-efficiency aside, what are the chemical constraints on CO₂ capture and methanation?

I'm only directly attacking your direct question, "what sort of catalysts might be used, and what are the performance / lifetime issues with them?" because the rest is ill-defined. I'll mention some electrochemical conversion aspects. I hope to illustrate why your question is problematic.

There are several different products one could imagine reducing carbon dioxide to:

methane
methanol
formaldehyde
carbon monoxide
formic acid
ethylene

Each will require different materials and conditions, but they all require energy for this to occur. Not to mention, these are all difficult reactions. Thermodynamically speaking carbon dioxide is at the bottom of this energy well and to get out of it requires spending energy.

One approach involves directly electrolyzing on a metal. Cu in aqueous solutions can give methane as a major product, but it also evolves hydrogen and many of the others I listed above. There are a couple of problems with this but the major problem is that it often requires large overpotentials at rates of interest for industrial purposes and this means more cost in terms of electrical input. Another major problem surrounding nearly all of these studies is the selectivity of a catalyst: in water at pH 7 evolving dihydrogen is thermodynamically more favorable than reducing carbon dioxide to everything except methane and methanol. Unfortunately these two are the most complex, as far as chemical transformations are concerned, to make. So the other products are much more likely to be produced along the way. This is why people have focused on making simpler things such as carbon monoxide or formic acid.

Where does this energy come from? Right now coal. So we would end up creating more carbon dioxide than we reduced to other species. It reminds me of how diesel-electric transmission works in freight trains: a dirty engine turns an electric generator that generates electricity that is used on motors instead of the motor directly driving the wheels; I think we can call this diesel-electric transmission an indirect drive whereas the latter is a direct drive. It turns out that making the former is far easier in engineering rather than the latter. The same is true here. It's easier to generate electricity somewhere else then reduce CO2 rather than have energy directly fed to reduce carbon dioxide. You can get efficiencies better by going the latter, but engineering/technical complexities increase immensely.

As a means to address this electrical burden via the direct-drive analogy, people have looked at these artificial photosynthesis approaches, i.e. using light to  drive these reactions or at least help ease the energy burden. This requires photocatalysts and often with these come semiconductors. Some semiconductors are naturally good at reducing carbon dioxide, such as CdTe, but this better performance for generating methanol is in turn related to an actual corrosion process of the semiconductor. Semiconductors are very complicated and expensive to manufacture well. They tend to corrode easily when they work well with sunlight. They require catalysts and this introduces huge compatibility issues and manufacturing complexities. We'll just leave the indirect approach at the state  the field is, complicated and incomplete.

As for the direct approach:

it's complicated by the fact that carbon dioxide is fundamentally limited in solubility at room temperature. Hence why other people prefer non-water systems. These unfortunately add other complexities...

Actually while typing this I came to the conclusion that I could write a much more succinct response to you rather than continuing to babble on: The best candidate will involve a transition metal, there is no doubt about that. It may end up being a molecular species, but there are so many possible structures that it is hard to say if we won't find a simpler heterogenous catalyst that is more capable first. By Murphy's law, it will be the most expensive material imaginable. But right now, considering everything, you can always look at all extensive work that has been done on copper, but realize it isn't very efficient or selective.


To understand why the exclusion principle isn't violated in this system, you really need to shift from valence bond theory to molecular orbital theory. The $p_{z}$ orbitals in benzene combine according to the $D_{6h}$ symmetry of the molecule to generate a set of bonding molecular orbitals (MOs) (which are lower in energy than the isolated $p_{z}$ orbitals) and a set of antibonding molecular orbitals (which are higher in energy). Bonding orbitals are generated from sets of $p_{z}$ orbitals which are mostly or totally in phase, whereas the highest energy antibonding orbital of the set is composed of $p_{z}$ orbitals that are totally out of phase with respect to their adjacent neighbors.

Here, for instance, is the MO that is generated from the totally in-phase set (I calculated this at a spin-restricted RI-BP86/6-311G* level of theory in ORCA(1) and visualised the isosurfaces in VMD(2)):

Now, this MO is shared equally across all of the carbons, i.e. it is not localised. However it only contains 2 electrons and thus satisfies the exclusion principle. (F'x mentions spin-orbitals - this is where the orbitals are split on the basis of their spin, which entails a single electron per orbital. Benzene is a closed-shell singlet so we would expect the $\alpha$ and $\beta$ spin orbitals to be spatially indistinguishable.)

Not to sound like a walking advertisment for Housecroft and Sharpe, but H&S has a great visual introduction to MO theory.

P.S.

It was suggested in chat that I expand upon the generated orbitals - here's one of the two highest occupied molecular orbitals, from the same calculation.

As you can see, there's a big node along one of the mirror planes. This orbital is higher in energy than the earlier example because it's composed of two antiphase sets of 3 $p_{z}$ orbitals, so this orbital has antibonding character for two of the carbon-carbon pairs. It should be easy to extrapolate to the highest lying MO of this type - antibonding all the way around.

(1) Neese, F. ORCA – an ab initio, Density Functional and Semiempirical program package, Version 2.6. University of Bonn, 2008

(2) Humphrey, W., Dalke, A. and Schulten, K., 'VMD - Visual Molecular Dynamics', J. Molec. Graphics 1996, 14.1, 33-38.


There is nothing wrong with the ±0.004 uncertainty you obtained. The relative uncertainty of the difference between close results is supposed to be far larger than that of the original measurements. One way of seeing why is stretching your scenario a bit: suppose you got 0.048 rather than 0.044 as the result of the second measurement. The difference, 0.002, would be small enough to be meaningless, as the second result would encroach into the uncertainty range of the first measurement. Your example is not so extreme; still, the large relative uncertainty is justified.

(Aside: an analogous effect can affect computer calculations done with limited precision. Suggestively enough, it is referred to as catastrophic cancellation).


Calculating the physical properties of isotopes

The Debye frequency of a crystal is a theoretical maximum frequency of vibration for the atoms that make up the crystal.

It seems safe to assume melting will be occurring at the Debye frequency.

It is quite possible that the authors of your paper used the Lindemann model, which was published in 1910, in conjunction with Debye frequency, which is unique to each topoisomer, to predict the melting point using the following equation of Lindemann:

$$ T_m = \cfrac{4\pi^2 m \nu^2 c^2 a^2}{k_B} $$

$$m = \mathrm{atomic\ mass}$$ $$ v = \mathrm{frequency} $$   $$ k_B=\mathrm{Boltzmann\ constant}$$ $$T=\mathrm{temperature}\ \left(^\circ \mathrm K\right)$$ $$c = \mathrm{Lindemann\ constant}$$   $$a = \mathrm{atomic\ spacing}$$

Different diastereomers have different resonant frequencies which can be found through either Debye calculations or IR and Raman spectroscopy.

Substituting in the Debye expression for frequency (I copied this from Wikipedia; I didn't actually derive it myself) gives:

$$ T_m = \cfrac{2\pi m (\theta^2_D k_B)c^2 a^2}{\hbar^2} $$

Where $\theta^2_D$ is the Debye temperature and $\hbar$ is Planck's constant.

Another way of doing it is by using a hard sphere model in combination with Debye/Lindemann (PDF link).


" What is Carbon tri chloride purpose?" is not answered in the link provided but the answer starts with a carbene attack on furan .

Trichloromethane in presence of a base undergoes 1,1-elimination (or an alpha elimination) because both groups are removed from the same carbon.The electron-withdrawing effect of the halogens makes the hydrogen acidic enough to be removed by a strong base such as hydroxide or alkoxide ion.

A halide ion then acts as a leaving group from the conjugate base, producing the
 carbene.1

Carbenes find a pair of electrons  to complete their valence shell of electrons.carbenes attack a lone pair, a C=C double bond (electron-rich or -poor), or even a C–H bond.In its reactions ,

It is an electrophilic reagent and is known to give ring expansion of aromatic rings.
Electrophillic attack takes place on pyrrole where the intermediate is most stable.

This leeds to normal Reimer–Tiemann reaction product pyrrole 2-aldehyde.

Carbene attack of a C=C double bond leads to an ring expansion leading to 3-chloropyridine.

The reaction of bicyclic compounf acts on similar lines of dichloro carbene attack.2

References

1 ORGANIC CHEMISTRY ,JOSEPH M. HORNBACK ,UNIVERSITY OF DENVER

2 MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE ,SIXTH EDITION


How does life choose it's nominal temperature preferences?

Here's a link to a brief Scientific American article that presents the results from a study of this question.  Basically, for every $\ce{1^{o}C}$ increase in body temperature, 6% fewer fungal species can survive - so the hotter, the better (the longer you're likely to survive).  The downside of "hot" is the increased metabolic rate required, which translates into more time spent consuming food.  If you were to plot the effect of these two competing effects on longevity versus body temperature you would get something like an inverted (upside-down) parabola.  The study found a maximum longevity at 98.1 F, not far off from the observed average body temperature of 98.6 F.


Can we use peaks in any chromatogram to identify a compound

Compounds have specific retention times that are used to identify their respective peaks in chromatograms. Usually these times are dependent on many factors (temperature, column type, mobile phase...), and as such some kind of pre-calibration of the machine with respect to compounds in the mixture has to be done. The area of the peak is then indicative of the amount of substance, also subject to calibration (or comparison with reference substances with known concentrations).

For structure elucidation of previously unknown compounds in mixtures other methods such as GC-MS are much better suited.


Why there is an Upper Explosive concentration Limit?

In simple terms, an explosion takes place when the reaction front propagates faster than the speed of sound. If it is slower, the expanding gas cloud can push away the unreacted mixture before it reaches ignition temperature.

With too much methane, the combustion reaction slows down, because the additional material takes up energy for heating, but cannot react.

A very simplistic microscopic picture would be that a hot methane molecule too often does not hit a cold oxygen molecule, with wich it could react to produce more heat, but a cold methane molecule, leaving two lukewarm methanes, than won't react any further.


Is it possible to have anions of metals?

Yes.  And, ironically, the most clear-cut examples involve alkali metals.  In these compounds you have the cation, often an alkali metal itself, stabilized in a complex so that electrons can be transferred to the recipient alkali metal (Na or heavier, based on today's known compounds), and the latter is a full-fledged alkali metal anion.


Strange "far out" orbitals in HF and DFT calculations. Why do they occur?

As already mentioned in the comments you have to be careful when interpreting the meaning of the virtual orbitals in a ab-initio or DFT calculation. In the calculation the only orbitals that are optimized are the occupied orbitals while the virtual (unoccupied) orbitals give you a guess of the shape you expect from them if they were occupied. To look at it from a simple perspective, somehow you are using the LCAO method, you obtain as many MO as AO you are using in your basis set, but you only optimize the MO that are occupied.

There is a second point that may be relevant to your question. In many programs the default orbitals you get in your calculation are the canonical orbitals (delocalized orbitals) that even if occupied may not have an evident chemical interpretation on its own. For those porpouses it may be better to use localized orbitals (e.g. NBO) that may be easier to interpret.

Two book recommendations:

C. J. Cramer, Essentials of Computational Chemistry. Theories and
Models, Wiley (2004).
Frank Weinhold, Clark R. LandisDiscovering chemistry with natural
bond orbitals, Wiley, 2012


Greatest Energy Output Reaction to Atmosphere Air

this is a way too generalized question. you certainly need experimental data or extensive screening to make sure you are on the top of the volcano plot. have you heard the term adiabatic flame temperature? The ratio of inert to gasoline/fuel is a determining factor to get high combustion energy. For example the adiabatic flame temperature for methane is higher for a mixture of methane and air with 5-15 mol% methane. 5-15 mol% methane can ignite without any external source of fire.

Please see the attached link to compare fuels according to their energy density:  http://en.wikipedia.org/wiki/Energy_density

And for your question you can't have two variables to optimized with one equation, either you have to specify the chemical X or you have to specify what is the output energy of this process. If you know the chemical X with it's thermodynamic property than using basic heat balance equation you can calculate the output energy.


Estimating stability and mean life of anomalous tautomers

TLDR: With a quantum/semi-classical transition state theory.
I order to determine the lifetime of the tautomer, you need to know the rate constant for the decay from the tautomer to the conventional dimer.  This is a kinetics question rather than a thermodynamic question so you won't be able to answer this by comparing single point DFT energies.  The rate will also vary with temperature.
A complicating factor is that the nuclei moving are protons in hydrogen bonds.  Hydrogen bonds are frequently double well potentials with two minima, one near each end of the bond - in this case the tautomer and conventional dimer.  In these case the physical distance between the minima is small and the proton may tunnel through the potential barrier between the two minima - hence nuclear quantum effects are likely to be important, at least at low temperatures.
Methods that can calculate nuclear quantum effects can be very computational expensive and may require you to work in a reduced dimensional represention - only considering the two protons quantum mechanically and potentially simplifying the surrounding structure.
Two potential methods that could calculate a rate taking account of nuclear quantum effects are  semiclassical transtition state theory, or ring-polymer molecular dynamics which has been applied within pockets of larger systems such as proteins.


There is no set answer to your problem, it depends on the geometry of the diffusional process. I give two examples.

(a) Suppose there is a volatile solute in a solvent placed in a beaker and the solute evaporates from the surface, then there is a concentration gradient of solute away from the surface as the solute is lost into the air. Similarly, if there is an electrode at which ions are being deposited as a metal, or a precipitate is being formed, then a concentration gradient is set up, but as material is removed no steady state can be set up as there is a finite amount of substance present.

Solving Fick’s 2nd law produces an equation describing the concentration vs. time which is $c_{x,t}=c_0erf(x/(2\sqrt{Dt}) $ where erf is the error function, D the diffusion coefficient, t time and $C_0$ the initial concentration. (The initial conditions are $t=0, x\ge 0, c=c_0$;  $ t>0,x=0, c=0$;$ x \rightarrow \infty, t=0, c=c_0$ ).

The depletion distance, that is the distance from the sink that the concentration is just becoming less than than $c_0$ is given by $\delta = \sqrt { \pi Dt}$ at time t.

(b) In a closed system, for example, a cell with half of it containing a solution and the other half solvent, equilibrium will become established once diffusion into all the cell is allowed. Similarly an insulated bar half of which is initially hot will effectively come to an equilibrium at the average temperature.

Consider a tube filled with solvent and then closed and some solute injected at some place, diffusion will ensure that equilibrium will eventually be reached. Because the initial distribution of concentration  must level out to a constant value, the solution to Fick’s equation must have both a time dependent and a constant part.

At long times the temperature or concentration will become uniform and if $f(x)$ is the amount initially added at position x and $0 < x < L$, the long time value is the average value;  $\frac{1}{L}\int _0^Lf(x)dx$ . The initial condition is given by the shape of the concentration profile $f(x)$.

The concentration of added solute is $c(x, t)$  at position x and time t. Because no material (or heat)  leaves the concentration gradients at the ends of the tube are zero at all times, or
$$\frac{dc(0,t)}{dx} = \frac{dC(L,t)}{dx} =0, ~~~t>0$$
Starting with the diffusion equation
$$\frac{\partial c}{dt} = \frac{\partial ^2c}{dx^2} $$
and separating variables, gives
$$\frac{\partial c_t}{\partial t} = -Dk^2c_t$$
 and
$$\frac{\partial ^2 c_x}{\partial x^2} =-c_xk^2 $$
where k is a separation parameter to be determined.
The solution is $c=c_tc_x$.

The time dependent equation integrates to  $c_t=e^{-Dk^2t}$ but the spatial part needs special attention to incorporate the boundary conditions.

The separation parameter k can have values of $0$, or it can be positive or positive and imaginary. Only when k is positive are the solutions non-trivial and are $c_x=a\sin (kx)+b\cos (kx)$.

The gradient boundary condition ensures that the solution is the cosine term because the derivative, $kb\sin (kx)$, is zero both at $x = 0$ and at L provided that $k=n\pi/L$, where n is an integer.

Next we use k as if it were an integer and only substitute its actual value at the end. The solution is
$$c_k=e^{-k^2Dt}cos(kx)  ~~~~~~~~  k=0,1,2…$$
and the final solution is a linear combination of the k terms including $k = 0$ and this will give a term that is independent of time. The initial condition ensures that
$$ f(x)=\Sigma _{n=0}^{\infty}b_k\cos (kx)$$
giving
$$ b_k=\frac{2}{L}\int _0^Lf(x)\cos (kx)dx$$
where $k=n\pi /L$. The complete solution is
$$ c = \frac{b_0}{2}+\Sigma _{n=1}^\infty b_ke^{-(n\pi /L)^2Dt}\cos (n\pi x/L )  $$
The first term $b_0/2$ is divided by 2 to make it equal to the average concentration at long times, $\frac{1}{L}\int _0^Lf(x)dx$ because by its defining equation $b_0$ is otherwise 2 times too large.
The results below show how the initial profile eventually reaches a constant, time-independent value, the total amount of material, or heat being conserved. Uniform concentrations are found when $ t > \approx  5/ (n\pi /L)^2D $, n is found to be in the range $10$ to $20$.

figure: One dimensional diffusion in a closed tube. The summation was taken to $n=10 $ terms (left) and $20$ (right).  (Left) The solute is injected in the centre and has a gaussian profile ($\exp (-5000(x/L-0.5)^2)$ at $t=0$  Right. Case where the solution is half the cell. The initial profile is a sigmoid curve $(1+\exp (-100(x/L-0.5))^{-1}$. The diffusion coefficient $D = 1$ and $L=1$. The profile at different times is shown.


What is the mechanism for the oxidation of a lactol to a lactone?

Methyl(trifluoromethyl)dioxirane (TFDO) in 1,1,1‐trifluoro‐2‐propanone (TFP)may be able to do your sought oxidation (see image [1] below): In there, cyclic ethers (shown example is tetrahydropyran) yield lactones by subsequent oxidation of a stable cyclic hemiacetal intermediate such as lactol in your case.

It is considered as selective and reactive oxidizing agent, capable of epoxidation of unreactive alkenes and arenes, oxyfunctionalization of alkanes, and oxidation of alcohols and ethers in following reference: Encyclopedia of Reagents for Organic Synthesis (1; March 2016); 11 pages. (https://doi.org/10.1002/047084289X.rm267.pub3)

You may also read: Oxidations by methyl(trifluoromethyl)dioxirane. 5. Conversion of alcohols into carbonyl compounds: J. Am. Chem. Soc., 1991, 113(6), 2205–2208 (DOI: 10.1021/ja00006a044).


Cutting a molecule (expressed with SMILES strings) into fragments yields sub-structures.  Typically these fragments can't be considered as molecules for their open, non-bonding valencies.  Derived from SMILES, it is then the SMARTS notation which is used to represent them, or to search with this notation of sub-structures in databases, e.g. in bioinformatics.
Open Babel supports the display of SMARTS-encoded fragments with branching points marked by an asterisk. The phenyl group, [16*]c1ccccc1 thus is shown as:

Consequently, bidental fragments are represented with two asterisks, e.g., [4*]CC[4*]:


What makes diazo compounds so unstable and explosive?

Well, your question is equivalent to “what is it about α-diazoketones that makes them so much more stable?”, which is easier to see. Compared to an alkyldiazo, the α-diazoketone has a resonance structure in which the negative charge goes to the ketone’s oxygen (and far away from the positively-charged nitrogen atom):

Because the oxygen is a quite electronegative element, the resonance form is quite stable and explains the extra stability of α-diazoketones. It is for the same reason that the protons in position α to the ketones are always more acidic than alky chain protons.

Coming back to alkyldiazo compounds, you have to realize that merely being able to write a resonance structure does not intrinsically imply stabilization: the resonance structure has to have some intrinsic stability factor. In the alkyldiazo, the resonance form you wrote is a carbanion, which is considered quite unfavourable unless it has a further stabilizing factor. Moreover, the most common reactions gives N2, which is a very stable compound… the reaction is thermodynamically very favourable.


How to find the mass of a solute given a mass fraction?

In a $20\%$ solution of $\ce{CuSO4}$ we have $\pu{20 g}$ of $\ce{Cu}$ and $\pu{80 g}$ of $\ce{H2O}$. Then

in every $\pu{259.6 g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{159.6 g}$ $\ce{CuSO4}$,

so for every $x~\pu{g}$ $\ce{CuSO4 * 5 H2O}$ we have $\pu{20 g}$ $\ce{CuSO4}$

$$x = \pu{31.87 g}$$


Differences between alkyl halides and alkyl tosylates

Both can be leaving groups. In general, TsO- are better leaving groups, thats why they are often specifically used as so. Halides can be used in many more reactions, like preparation of organometallic compunds.
TsO- more common function is to transform an alcohol group in a good leaving group. Formation of an halide, usually is not the way to go in this case.

But their basic reactivity is very similar.
If you dive into more complex reactivity, you can complicate things to a very very large extent...


Copied table below from "Quantum chemistry", Donald A. McQuarrie, University Science Books,  2008, page 472

This illustrates the notion that $p^2$ and $p^4$ are equivalent.

I think the 2,4 superscript in Cowan's book for $^{2,4}S^\circ P^\circ D^\circ$ is a peculiar notation of his. Most other books seem to presume the reader knows the equivalency.


CuI, CuSCN Solubility in Ammonia Solution

The reason these salts are soluble in ammonia (including aqueous ammonia -- so why buy ammonia in EtOH?) is that they form complexes with ammonia:
$$
\ce{CuI}\text{(s)}+2\ce{NH3}\text{(aq)}\to\ce{Cu(NH3)_2^+}\text{(aq)}+\ce{I-}\text{(aq)}.$$
Have fun!


Is an alcohol-based spray cleaner ok for cleaning an aluminum heat exchanger?

When aluminum metal is exposed to the oxygen of the air at room temperture, it forms a protective air-tight layer of aluminum oxide. So as long as nothing is present that will disrupt the oxide layer on aluminium it is OK for aluminium at 20 oC to be in contact with air.

If an aluminium alkoxide is exposed to water then the aluminium alkoxide will react with the water to form aluminium oxide and the alcohol. So in alcohol which contains some water it is not possible for the aluminium to dissolve in the alcohol.

But if the alcohol is anhydrous then it will be able to enter a scratch in the aluminium and start to react with the metal thus causing corrosion. For titanium that anhydrous methanol is very corrosive.

So as long as some water is present and air is present then at room temperture


There exists a range of ω-3 acids.

Fatty acids are made of a chain of carbon atoms (generally linear), with a carboxylic acid at one end. Usually the carbon atoms are linked with a single-bond, but in unsaturated acids one or more of these are double-bonds.
In order to describe the positioning of these double bonds, the carbon-atoms are numbered, starting either from the carboxylic acid end, also called the α-end,  or from the methyl-end (the other end), which is called the ω-end. Therefore, an ω-3 acid is a fatty acid with a double-bond from carbon number 3 to 4, counted from the methyl/ω-end.

Fatty acids can have several double-bonds in its chain, making it a polyunsaturated acid. An example is Eicosapentaenoic acid, which has 5 double bonds. Edit: As pointed out in comments, it is the position of the first double-bond that determines if the acid is called an ω-3 acid. Likewise: if the first double bond is from carbon number 6 to 7, the acid is an ω-6 acid.

Wikipedia has some very informative articles about both ω-3 acids and fatty acids  in general.


I don't remember to have seen the term tensid outside the context of the German language.

The term is typically applied to agents that decrease the surface tension (hence tensid) of water. Note that the logic behind the English equivalent surfactant is just the same.

So, in summary tensid and surfactant mean the same.


Standard curve of absorption and concentration

Supposed that Yellow Dye #5 refers to the FD&C numbering, this is a food colorant more known as tartrazine outside the U.S.

A standard curve correlates absorbance at a particular wavelength with the concentration of the dye.

Supposed that no distortion (e.g. due to aggregation effects) is present, the relation is given by the Lambert-Beer law: $$E_\lambda = \epsilon_\lambda \cdot c \cdot d$$

where $E_\lambda$ is the absorbance, $\epsilon_\lambda$ is the molar absorption coefficient (specific for a particular dye at a particular wavelength $\lambda$ in a certain solvent), $c$ is the concentration of the dye and $d$ is the path length of the cuvette. Please pay attention to the units in your calculations!

In order to create a standard curve,

Record the uv spectrum of the stock solution. If you have two-channel spectrometer, put a second cuvette with the neat solvent in that channel.
Get a couple of volumetric flasks and dilute the stock solution.
Record the uv spectra for each of these solutions.
Plot absorbance (probably at around 420 nm) vs concentration.


Froth floatation process - Use of stabilizers in froth floatation method

Bubbles tend to collapse when the liquid that makes up the walls drains away and the walls weaken. The walls can be strengthened by choice of surfactant (and addition rate) and by adding thickeners like water-soluble gums (e.g., xanthan) or polymers or glycerol (also for children's bubbles).

The small particles being floated may act as imperfections that break the walls more easily, so coating the particles with appropriate polymeric dispersants can make the (coated) particles more compatible with the (thickened) wall.

On the other hand, many industrial operations sometimes develop undesirable bubbles/foam which be eliminated by spraying a small amount of low surface tension silicone or non-viscous solvent like isopropanol on the top of the foam. Avoid these if you need to stabilize the foam.


Process of transferring kinetic energy in regard to photons.

You may be experiencing a very common misconception, which I blame on Star Trek and Star Wars and all their talk of pure energy that always comes as bright lights, with the implicit assumption that energy is some kind of fiery bright substance, like a glowing fluid or something.  It isn't.  Energy is a property of matter, and of the rather peculiar substance we call electromagnetic radiation, chunks of which are called photons.  Think of energy as kind of like an adjective that describes a noun (some kind of matter, or radiation).  An atom can be energetic or not energetic, and so can a photon, just as a cat can be furry or not furry.  But you would  never mistake "furriness" as some kind of object itself.  In the same sense, you should not mistake energy as some kind of substance itself.  It's a property of substances, just like furriness is a property of cats.

So energy has to be transferred from a hot substance to a cold one, and indeed it could go by way of photons.  One particle could emit a photon, transferring part of its energy to the photon, and maybe slow done, and then this photon could be absorbed by another particle, which would receive the transferred energy, and maybe accelerate (go faster).  That's definitely one way it could happen.  But it doesn't have to happen that way, because energy can be transferred other ways.  For example, the first particle could simply run into the second, bang, and the impact itself could transfer the energy directly, no photons involved.  Or the first particle could run into a third particle, or the wall of a container, and the second particle could pick up the energy from the third particle, or the wall of a container.  And so on.


Technically, alcohol is the name of a class of organic compounds containing one or several hydroxyl groups.

Colloquially, the term "alcohol" is understood as you have described: A solution (of varying degree of purity) of ethanol and water. Pure ethanol is impossible to create via the traditional method of purification (atmospheric distillation) as the water-ethanol mixture is azeotropic:¹

  An azeotrope [...] is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This happens because, when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture.

  Source: Wikipedia

This is why a chemist would never think "100% pure ethanol" when you mention "alcohol".

If you check for example this vendor page for the purest ethanol (>99.5%) I could find, you see various substances listed as impurities:

<0.2% Water
<0.1% Methanol
Other organic substances, such as isopropanol and acetone

Concerning your second question, I would argue that many distilled spirits (vodka, whisky, and cognac just to name a few) are solutions of ethanol, water, and various flavour compounds.

Some alcoholic beverages, such as unfiltered beer and wine, usually also contain solid particles and would as such be correctly termed "suspensions".

¹ I wonder now whether it would be possible to produce 100% ethanol from another mixture somehow. The problem is probably that every common starting mixture already has some amount of water. But synthesizing ethanol in, say, an ionic liquid under an inert atmosphere and then distilling it away might be an interesting approach.


Is [Co(NH₃)₄Cl₂]Cl paramagnetic or diamagnetic?

The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $\ce{Co^3+}$ is $\mathrm{d^6}$, and virtually all "octahedral" $\mathrm{d^6}$ complexes are low spin - essentially some complexes of $\ce{Fe^2+}$  and a very small number of fluoro complexes of $\ce{Co^3+}$ are the only exceptions to the rule that all $\mathrm{d^6}$ octahedral complexes are low spin.

Why is this? Well let's use a bit of Crystal Field Theory (much of below stolen from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy). One can consider the crystal field stabilization energy which is how stabilized an ion is compared to the free ion when in a given electronic state electronic and geometry. Here we are interested in high and low spin, and octahedral geometry. Taking a table of these values from the above link:

In the above $\Delta_\circ$ is the splitting energy of the orbitals. We can see that low spin $\mathrm{d^6}$ has the largest possible stabilization energy of any electronic configuration compared to the high spin case ($2\Delta_\circ$-P). Thus unless the splitting is very small octahedral $\mathrm{d^6}$ prefers low spin. So how do we get small splitting to see high spin $\mathrm{d^6}$? Well small splitting is favoured by

Low charge first row transition metals - for instance $\ce{Fe^{2+}}$
Very weak field ligands

And what were our exceptions above?


Where can I find a complete and accurate table of CPK colours?

Indeed, this is annoying. Because of this problem, several years ago, one of the Jmol developers and I sat down and worked out a color scheme. It was similar to Accelrys, although we tried to make certain "known colors" match (e.g., rust-like for Fe, golden for Au, etc.)

Certainly, there's been an effort to keep color consistency between Jmol and Avogadro

These colors are open source and available, e.g., through Open Babel

If there are differences between Jmol and Avogadro, I'm sure both the Jmol developers and the Avogadro developers (including myself) would be happy to reconcile them.


Why does pouring soda slowly and smoothly seem to keep more of the carbonation?

Because your soda is a supersaturated solution of $\ce{CO}_2$, and such solutions are non-equilibrium and hence unstable. Any physical disturbance (shaking, touching a rough surface, etc.) causes it to emit the gas and return to equilibrium state. Throw in some salt, and you will see a reaction even more vigorous.

Supersaturated solutions of solid compounds exhibit similar behavior, but they are not that easy to come by (I mean, starting from the typical household items). Sugar solution is too viscous, and table salt is no good because its solubility does not depends on temperature all that much, so you can't really supersaturate it.

There is, however, another pretty common example of non-equlibrium state, and that is superheated water. If you pour some bottled water into a clean cup, heat it in a microwave for a while, and then touch it not too gently, it may almost literally explode in your face, so beware.


TL;DR IUPAC hasn’t made up their mind, but plain old water appears to be an appropriate name. However, chemical derivatives of water may not be named using water.

In Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 it is stated (P-21.1.1.2) that

  The common names water, ammonia, [...] are used in these recommendations but their use as preferred IUPAC names is deferred pending publication of recommendations for the selection of preferred inorganic names; thus, no PIN label will be assigned in names including them.

I’m not sure when the next edition of the Red Book (i.e. inorganic chemistry nomenclature recommendations) is coming out, but I can quote the relevant section from the most recent version.

Table IR-6.1 of Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005 lists "oxidane" as the parent hydride name for $\ce{H2O}$. However, it also adds the caveat that

  The names ‘azane’ and ‘oxidane’ are only intended for use in naming derivatives of ammonia and water, respectively, by substitutive nomenclature, and they form the basis for naming polynuclear entities (e.g. triazane, dioxidane). Examples of such use may be found in Section IR-6.4 and Table IX.

Therefore the compound $\ce{ONONO} = \ce{(ON)2O}$, a nitrosylated derivative of water, is named "dinitrosooxidane" and not "dinitrosowater". More examples may be found in Table IX of the same publication.

Water itself is still called water. For example, tritiated water $\ce{H^3HO}$ is named (3H1)water (Section IR-2.2.3.2).


How does amine inversion affect basicity?

Amine inversion is also known as nitrogen inversion. You can find more here: http://en.wikipedia.org/wiki/Nitrogen_inversion

It has nothing to do with relative basicity of amines, that is defined by 1) energy of particular N-H  bonding (increases with +I effect) and 2) energy of ammonium cation hydration. (decreases with alkyl group number and their size). The relative balance of said factors is fragile and cannot be easily predicted without reference to experimental data except cases when amine has lone pair involved into bonding interactions (aniline, pyrrole, amides) or connected with extremely strong electron acceptors.


What does chemical bonding look like at the quantum level?

I believe that generally in chemistry the wave characteristics of the electron are far more prevalent/relevant that the particle characteristics. So it's usually easiest just to think of it as a wave (or I think "region of space" is easier).

A chemical bond looks like the electron wave or area spread out between two or more atoms. This shape of the wave has lower energy than any of the orbitals it could occupy on just one atom. Because it has lower energy, that means it needs energy from somewhere for the electrons & atoms to change back to a different configuration. So unless it gets some more energy it cannot change away from the bonded state. That explains why chemical bonds are stable and hold atoms together.


Why does nucleophilic substitution of a haloalkane with ammonia produce ammonium, not a hydrogen halide?

The answer to this is quite similar to one I gave recently for another set of reactions with hydrogen halides as possible products.

Your proposed last step is:

$$\ce{CH3CH(NH3^+)CH3 +Br- \rightarrow CH3CH(NH2)CH3 +HBr}$$

Here, you have a weak acid $\ce{(CH3CH(NH3^+)CH3)}$ and a very weak base $\ce{(Br^{-})}$ reacting with each other to turn into a weak base $\ce{(CH3CH(NH2)CH3)}$ and a very strong acid $\ce{(HBr^)}$. This is highly unfavourable from a thermodynamic sense, because you're trying to inject a lot of chemical potential into the molecules at no cost. Though your teacher suggests it might happen, I wouldn't be very comfortable saying that, if only for the fact that there's another, easy option.

It is possible that the previous argument is slightly circular, so it's best to explain why $\ce{HBr}$ is such a strong acid / $\ce{Br^-}$ is such a weak base. As you mention, $\ce{Br^-}$ has a negative charge, and one might expect that to mean it is a better base. However, that affirmation is only useful as a rule of thumb when comparing substances of different structure, being strictly valid solely when looking at a single sequences of related structures (such as $\ce{PO_4^{3-}}$/$\ce{HPO_4^{2-}}$/$\ce{H_2PO_4^{-}}$). The strength of a base is determined by several factors, such as charge localization, bond strength between base and a proton, steric hindrance, and strength of interaction of the protonated/deprotonated species with the solvent (if it exists). There are very strong neutral bases and very weak negatively charged bases ($\ce{SbF_6^-}$) known.

$\ce{Br^-}$ is a poor base because, while it has a net negative charge, that extra electron is bound to a somewhat electronegative atom, thereby stabilizing the charge. Not only that, but the bromide anion is quite large, so the charge is "spread out" over a significant volume and would attract a proton less, in some sense. Meanwhile, though $\ce{NH_3}$ has no extra charge, and even though the central nitrogen atom is more electronegative than bromine, the non-bonding pair of valence electrons in $\ce{NH_3}$ is highly localized, especially due to the fact that second period p-block elements are anomalously small. This makes $\ce{NH_3}$ an attractive target for a proton.


Optimization completed on the basis of negligible forces

According to the Gaussian Frequently Asked Questions:

In general, a stationary point is found when one of two criteria is met for a point in an optimization calculation:

All four values listed in the output are smaller than the indicated thresholds.
The Maximum Force and RMS Force are two orders of magnitude smaller than the thresholds shown, regardless of the values of the displacements.

You indicate three molecules where the second criteria activate - the forces are really, really small. So the optimization gives up.
In your final case, you can see that the displacements are also small, so the first criteria activate.
You then want to know how "reliable" the geometries are when the forces are small, but the displacements are not. That depends a lot on your needs. The optimization quits because such small forces imply a very flat potential energy surface and it may take a long time before "complete" convergence. My guess is that you still have some imaginary frequencies, but that might not matter depending on your needs.
It's worth pointing out that many quantum chemical programs have different criteria for declaring an optimization "finished." (In other words, your calculation in a different program might be declared done because of negligible energy and force/gradient change.)
If you want to push for a true minima, you should take that optimized geometry, run a force / frequency calculation to get an accurate Hessian, and save the result to a checkpoint, then re-start the optimization using the saved Hessian:
%Chk=myfile
# B3LYP/6-31G(d,p) OPT=ReadFC Freq Geom=AllCheck Guess=Read

Of course, that means probably breaking your relaxed scan calculation into a set of separate steps. (Honestly, this is how I usually do relaxed scans anyway - a set of separate jobs with each stepped geometry.)
One other very important point. If you're doing a scan of intermolecular interactions with B3LYP, you really should be using a dispersion correction such as EmpiricalDispersion=GD3BJ. I often see incorrect repulsive potential energy curves from B3LYP without dispersion corrections.


Why is so different the temperatures of a compressed gas compared to a liquid/gas mixture of the same fluid and they are both at the same pressure?

Generally, compression of any gas transforms mechanical energy to thermal energy and the gas warms up, until heat dissipates.

If you have ever inflated bike tyres, you would notice the air pump gets warm ( MTB ) up to hot ( high pressure road bikes ).

When air drops down by 1500 m ( from 0.85 to 1 bar ) after passing over mountain range, it warms up by adiabatic compression by 15 deg C. Warming up to 90 Deg C to get pressure 16 bar does not seems to be much.

See Adiabatic_process

A gas, with or without its liquid phase, has 3 variables: temperature, pressure and volume. You adjust 2 of them and the gas adapt to it by adjusting the 3rd one according laws of gas behaviour.

When you keep gas/liquid mixture at the given temperature and volume, it adjusts to it by reaching its saturated vapour  pressure for given temperature  by either evaporation either condensation.

When you set a gas to have a particular volume applying a particular pressure ( by pumping it in or out ), it adjusts to it reaching a particular temperature.

For propane gas and adiabatic compression ( without heat exchange with surrounding ):

$$p_1^{1-\gamma} T_1^\gamma=p_2^{1-\gamma} T_2^\gamma $$

where $\gamma=1.13$

If we consider as initial conditions 1 bar and 20 Deg C:

$$293.15^{1.13}=16^{-0.13} \cdot  T_2^{1.13 }$$

$$ T_2^{1.13 }=   293.15^{1.13} \cdot  16^{0.13} $$

$$ t =  T_2 - 273.15 =  {( 293.15^{1.13} \cdot  16^{0.13} )}^{\frac1{1.13}} - 273.15$$

For given data it leads to about 130 deg C.

If you suddenly compress propane to 16 bar, it would get hot. If you compressed it enough and there was some oxygen, it would  self ignite. This effect  is used in diesel engines.

But if the compression of propane is gradual and gas is in contact with surrounding,  heat gets time to partially dissipate.

If the hot compressed propane is kept at constant external pressure 16 bar, it gradually contracts while cooling down.

When it's temperature reaches the temperature, at which saturated propane vapour pressure, it becomes gradually condensation.

Gradual condensation at constant pressure maintains constant temperature, as dissipative heat loses are compensated by releasing the condensation heat.


Constrain bond angle In Gaussian molecular structure optimization

Fixing the bond angle to a set value seems like an extreme solution to this problem.

It is. To be honest, fixing parameters in a calculation is almost always no solution at all. The resulting geometry has little to no meaning and all derived properties are for a purely hypothetical case. If the located geometry is no stationary point, what does that even mean? A wrong solution is worse than no solution at all. You have to be very careful with what you fix and when you fix, and you should always confirm what you have found with a full optimisation and a frequency calculation. (On various levels of theory.)

I think Greg's advice is also very important. Symmetry can be a very powerful tool if utilised correctly, but you should make sure that the implied symmetry is correct. You also need to check whether your methodology matches the system you are computing.

All that being said, there are certainly a few very productive uses for partial optimisations:

finding higher or unusual local minima
scanning the potential energy surface
modifying converged geometries
fixing transitional modes
etc.

There are a few possibilities how to make partial optimisations in gaussian. All have some caveats and all may lead to undesired results. There are certainly restrictions to all of them. I'll give a short overview the most common ones with a few practical examples.

You have already noticed, that fixing a bond angle to 180 degrees is a challenge. This is due to the fact, that it is a requirement to have linearly independent variables, which is simply not possible with threes point on a line. However, there are certain ways to work around the problem. Spoiler alert: It is only possible with z-matrices or very rigid symmetry (where you fix the symmetry, not the bond angle).

Partial optimisations with z-matrices

Let's start with something very simple. A full optimisation of methanol without any symmetry restrictions.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt
int(ultrafinegrid)
sym(none)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc4         2 hch4          3 dih4
 o    1 oc5         2 och5          3 dih5
 h    5 ho6         1 hoc6          2 dih6

hc2           1.1
hc3           1.1
hch3        109.5
hc4           1.1
hch4        109.5
dih4        120.0
oc5           1.4
och5        109.5
dih5       -120.0
ho6           0.9
hoc6        109.5
dih6        180.0

This will result in the following optimised geometry:

 c
 o    1 oc2
 h    1 hc3         2 hco3
 h    1 hc4         2 hco4          3 dih4
 h    1 hc5         2 hco5          4 dih5
 h    2 ho6         1 hoc6          3 dih6

oc2           1.414851
hc3           1.109697
hco3        107.476
hc4           1.118853
hco4        113.586
dih4       -118.329
hc5           1.118853
hco5        113.586
dih5       -123.342
ho6           0.974680
hoc6        107.408
dih6        179.999

Fixing an angle in a z-matrix is straight forward. You need to specify that you are doing a partial optimisation (popt) and move the fixed parameter to the constants section. As an example, I am fixing the COH angle, and also invoke symmetry.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc3         2 hch3          3 dih4
 o    1 oc5         2 och5          3 dih5
 h    5 ho6         1 coh           2 dih6
  Variables:
hc2           1.100000
hc3           1.100000
hch3        109.5
dih4        120.000
oc5           1.400000
och5        109.5
dih5       -120.000
ho6           0.900000
dih6        180.000
  Constants:
coh          90.0

This will give you the following list of optimised parameters:

       hc2         1.1104
       hc3         1.1173
      hch3       107.0973
      dih4       116.0195
       oc5         1.4471
      och5       108.92
      dih5      -121.9694
       ho6         0.9796
      dih6       180.0092
       coh        90.0

Now we will try the same, but we will make the bond angle 180 degrees. Here we need to introduce dummy atoms:

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc3         2 hch3          3 dih4
 o    1 oc5         2 och5          3 dih5
xx    5 xxo6        1 xxoc6         2 dih6
 h    5 ho7         6 xxoc6         1 dih7
  Variables
hc2           1.100000
hc3           1.100000
hch3        109.500
dih4        120.000
oc5           1.400000
och5        109.500
dih5       -120.000
dih6        180.000
ho7           0.950000
  Constants
xxo6          1.000000
xxoc6        90.000
dih7        180.000

Which will give you once again the following set of parameters:

       hc2         1.1209
       hc3         1.1211
      hch3       106.3981
      dih4       113.09
       oc5         1.3644
      och5       112.3772
      dih5      -123.4351
      dih6       180.0
       ho7         0.9465
      xxo6         1.0
      xxoc6       90.0
      dih7       180.0

And the following Cartesian coordinates:

    6
scf done:  -115.581115
 C     0.000000     0.000000     0.000000
 H     0.000000     0.000000     1.120932
 H     1.075500     0.000000    -0.316498
 H    -0.421786    -0.989342    -0.316498
 O    -0.695163     1.052866    -0.519431
 H    -1.177401     1.783245    -0.879763

Quick and Dirty fixing with Cartesian coordinates

Appending -1 to an element in a row in the Cartesian coordinates block will fix it. This is a very quick and dirty approach, which may lead to a brute force computation. You are basically restricting the algorithm to rotate the molecule, which leads to less effective calculations. Fixing Cartesian coordinates also fixes the bond length, but it is the only way to force a 180 degree angle with this choice of input.
Here is an example input:

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)
symm(loose)

title card required

0 1
 C  -1    0.000000     0.000000     0.000000
 H        0.000000     0.000000     1.300000
 H        1.221600     0.000000    -0.444626
 H       -0.479083    -1.123738    -0.444626
 O  -1   -0.695162     1.052869    -0.519426
 H  -1   -1.177400     1.783250    -0.879755

This will result in the following optimised geometry:

    6
scf done:  -115.581115
 C    -0.663620    -0.000005     0.000000
 H    -1.090591     0.003267    -1.036538
 H    -1.090556     0.896080     0.521076
 H    -1.090648    -0.899281     0.515456
 O     0.700780    -0.000004     0.000001
 H     1.647272    -0.000004     0.000001

Note that this calculation takes about five times longer.

Fixing parameters with modredundant

This option is described quite straight forward in the Gaussian online manual. But let's make an example where we try to restrict the COH angle to 90 degrees.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt(modredundant)
int(ultrafinegrid)

title card required

0 1
 C     0.000000     0.000000     0.000000
 O    -0.659966     1.143095    -0.466667
 H    -0.000000     0.000000     1.089000
 H     1.026720     0.000002    -0.362997
 H    -0.513360    -0.889165    -0.363000
 H    -0.502133     0.869719    -1.359507

A 5 6 1 F

With this option it is impossible to fix the angle to be 180 degrees, since this is not an allowed value. You also cannot use dummy atoms, as these do not exist in the redundant coordinate system. It might be a good idea, to kill this variable completely, to ensure that the  program will not fail trying to construct it, or when an optimisation is getting close to that value.

Alternatively the angle can be restricted via a linear bend. This is the best option, if you know that the angle is 180.0 degrees.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt(Modredundant)
int(ultrafinegrid)
sym(none)

title card required

0 1
C        0.000000000      0.000000000      0.736983516
H        1.037151577      0.000000000      1.103476471
H       -0.518575789     -0.898199613      1.103476471
H       -0.518575789      0.898199613      1.103476471
O        0.000000000      0.000000000     -0.663016475
H        0.000000000      0.000000000     -1.563016262

L 1 5 6 2 F

The last line freezes the linear angle between atoms 1, 5, and 6. Atom 2 needs to be specified to determine the plane in which the linear bend lies, or is orthogonal to.

Fixing linear angles with general internal coordinates (GIC)

A new way of handling the coordinate system was introduced with Gaussian 16: generalised internal coordinates. It is in many ways superior to earlier versions, however, also a little bit harder to learn. Nevertheless, it is absolutely worthwhile trying.

Here is a sample input for oxoketene $\ce{O=C=C=O}$. In this case I specified the complete set of coordinates necessary.

#p PM6
opt geom(AddGIC)
scf(xqc,MaxCycle=250)
int(ultrafinegrid)
symmetry(none)

Oxoketene (constrained)

0 1
O        0.000000000      0.000000000     -2.210000000
C        0.000000000      0.000000000     -0.770000000
C        0.000000000      0.000000000      0.770000000
O        0.000000000      0.000000000      2.210000000

BondC1O1(Active)=R(2,1)
BondC2O2(Active)=R(3,4)
BondC1C2(Active)=R(2,3)
LinBend1O1C1C2(Freeze)=L(1,2,3,0,-1)
LinBend2O1C1C2(Freeze)=L(1,2,3,0,-2)
LinBend1C1C2O2(Freeze)=L(2,3,4,0,-1)
LinBend2C1C2O2(Freeze)=L(2,3,4,0,-2)

The format is

L(     i,      j,      k,      l,                                 M)
L(Atom 1, Atom 2, Atom 3, <Atom 4 | Orthogonal to Plane>, component)

The linear bend is therefore defined between i, j, k as Atom 1, Atom 2, Atom 3 and must not be dummy atoms (and with GIC these are hardly necessary anymore). The fourth value l, can be another atom to determine the orthogonality/ plane to, or it can be defined as a plane where l = 0, -1, -2, -3 standing for automatic, $yz$, $xz$, $xy$, respectively. The last value M gives the component, of which two are necessary, and must be M = -1, -2.

The methanol example is as follows:

#p PM6
DenFit
opt(AddGIC)
int(ultrafinegrid)
sym(none)

title card required

0 1
C        0.000000000      0.000000000      0.736983516
H        1.037151577      0.000000000      1.103476471
H       -0.518575789     -0.898199613      1.103476471
H       -0.518575789      0.898199613      1.103476471
O        0.000000000      0.000000000     -0.663016475
H        0.000000000      0.000000000     -1.563016262

L(1, 5, 6, 2, -1) Freeze
L(1, 5, 6, 2, -2) Freeze


Why are protein hydrophobic cores denatured by heat when entropic forces should be favoured at high temperature

You are correct that the hydrophobic interactions are stabilized at higher temperatures. However, I believe the addition of heat mainly disrupts protein structure by breaking hydrogen bonds through increased thermal motion.

Once the secondary structure starts to go, any tertiary interactions will get scrambled, likely resulting in the separation of hydrophobic residues. The refolding of the polypeptide backbone must be stronger than the enhanced hydrophobic interactions.


Is the amount of substance of acid and base always equal at the equivalence point?

We speak in this case of multiple equivalence points.

As a concrete example, if we are titrating $\ce{H2SO4}$ with $\ce{NaOH}$, then the first equivalence point occurs when all the $\ce{H2SO4}$ has reacted to form $\ce{HSO4-}$, and the second equivalence point when all the $\ce{HSO4-}$ has further reacted to form $\ce{SO4^2-}$. If you started with 1 mole of acid in solution, then adding 1 mole of base gets you to the first equivalence point, and another mole (2 moles total) gets you to the second equivalence point. In the same vein, one might consider triprotic acids, which have three equivalence points, and so on.

Incidentally, an interesting case to think about might be that of a titration between a diprotic acid and a dibasic base.


Which chlorine species can act as an oxidizing agent but not as a reducing agent?

For a species to act as a reducing agent, it should have the ability to get oxidised. Any element in its highest oxidation state, cannot oxidise anymore, and can hence not act as a reducing agent. $\ce{ClO4-}$ has $\ce{Cl}$ in $+7$ oxidation state, and cannot have a higher oxidation state, and hence cannot act as a reducing agent.


Is there an upper limit on the coordination number of a coordination compound?

I don't see why not. It is a matter of having

enough of an interaction between the center and the ligands, usually in the form of donor/acceptor orbital pairs, and
enough space around the coordination center.

These go hand-in-hand, as any atom that will have enough electrons to donate to ligands will also most likely have a larger atomic radius than a first-row transition metal like iron.

For example, here is a thorium complex that has CN = 15 in its crystal structure and CN = 16 in an isolated gas-phase density functional calculation:

Daly, Scott R.; Piccoli, Paula M. B.; Schultz, Arthur J.; Todorova, Tanya K.; Gagliardi, Laura; Girolami, Gregory S. Synthesis and Properties of a Fifteen-Coordinate Complex: The Thorium Aminodiboronate $\ce{[Th(H3BNMe2BH3)4]}$. Angew. Chem. Int. Ed. 2010, 49, 3379-3381, DOI: 10.1002/anie.200905797


How do you determine a chemical formula given the molecule's crystal structure?

Knowledge of the crystal structure implies that you are also able to describe the unit cell of the compound. To determine its formula, count all sorts of atoms inside the unit cell, also considering fractional atoms which belong to more than one unit cell. For example, an atom at the corner of the unit cell counts as $\frac{1}{8}$ atom because it is shared between 8 adjacent unit cells. Likewise, an atom at the edge of the unit cell counts as $\frac{1}{4}$ as it belongs to 4 unit cells etc. When you obtain fractional values for the atom numbers, multiply the values of all atoms with a certain number so that all values are integers.

Let us take sodium chloride ($\ce{NaCl}$) as a simple example. Its unit cell, which is shown below, contains $\ce{Na+}$ (violet) and $\ce{Cl-}$ ions (green).

Counting the number of chloride ions in the unit cell gives

$$12 \cdot \frac{1}{4}+1=4 \ \ce{Cl-}$$

because there are 12 $\ce{Cl-}$ on the edges and 1 $\ce{Cl-}$ in the center of the unit cell. For the sodium ions, which are situated at the corners and on the faces (count as $\frac{1}{2}$ because they are shared between 2 cells) of the unit cell, we obtain

$$8 \cdot \frac{1}{8}+6 \cdot \frac{1}{2}=4 \ \ce{Na+}$$

The composition of the unit cell is therefore $\ce{Na4Cl4}$. Since the number of formula units per unit cell is 4, the formula of the compound is $\ce{NaCl}$.

For compounds in which the building blocks of the crystal structure are molecules, the procedure is essentially the same.


What will methyl (2-oxocyclohexyl)acetate produce when reacted with sodium borohydride?

Both carbonyl carbons (and everything else in the molecule) are in aliphatic environments.  The six-membered ring you drew is not aromatic, it is saturated like a cycloalkane.

The carbonyl group in this ring is a ketone function, bonded only to other carbon atoms.  The carbonyl group in the chain is bonded to oxygen (did you miss that?) in an ester function, which is a less active electrophile than the ketone function.  Yes the additional oxygen atom is electronegative, but with its octet satisfied it can (to a degree) share a nonbonding pair with the carbonyl carbon, and that effect dominates making the ester carbon less electron deficient. The sodium borohydride reacts accordingly.


Lithium ion batteries are pretty poorly recycled as is and so when I was looking through literature reviews I found only one mention of lithium ion polymer batteries being recycled. The paper that was using this method was also recycling lithium ion batteries the same way though so it wasn't clear if anything different had to be done.
The main thing that seems to be different with lithium ion polymer batteries is just the electrolyte. The electrolyte is often not recycled anyway since the main focus is typically on recycling the cathode because it contains cobalt and/or nickel and is profitable to recycle.
If you want a more detailed overview of the different processes lithium ion batteries are recycled this review is open access and explains the ways different companies currently recycle lithium ion batteries:
https://www.mdpi.com/2313-0105/5/4/68
There's no mention of lithium polymer batteries in the review but this patent mentions recycling lithium ion and lithium polymer batteries together so I assume some of the same methods can be used:
https://patents.google.com/patent/US8979006B2/en
There's no standard method for recycling lithium ion batteries but usually methods involve mechanical, pyrometallurgical and/or hydrometallurgical processing. Usually the batteries are first discharged and then shredded or dismantled.
If mechanical processing is carried out it usually relies on different physical properties in the materials that allow it to separate out components. For example, the electrodes are a fine powder so if you put the shredded battery through a sieve you can separate the electrodes from the other materials. The cathode and anode can also be separated using froth filtration since they have different hydrophobicities. This can allow the cathode to be isolated and further processed to obtain metals like cobalt if the battery has a lithium cobalt oxide cathode. The review goes into more detail about the other mechanical steps that can be carried out but there's quite a lot of them.
Pyrometallurgical processes involve putting the battery in a furnace and you end up with an alloy of metals (containing cobalt, nickel and iron usually) and a slag (containing aluminium, lithium, manganese and some other metals). The metals in the alloy can then be extracted.
Hydrometallurgical processes involve using solvents to leach metals. This is generally combined with other methods like mechanical processing so the battery components you have are slightly more homogenous and easier to process. Sulfuric acid is the most common acid used to leach cobalt from lithium cobalt oxide. The leached metals can be precipitated out and recovered.
It's a bit difficult to clearly explain how lithium ion batteries are recycled since it does seem like every company does it differently. This just mentions some of the processes that are being used at the moment but there's lots more as well that are being researched. I know this doesn't clearly mention how lithium polymer batteries are recycled but I hope it's of some help anyway!


Energy comparison between an atom and its ion

Ionization energy is, exactly, a comparison of the energy of (i) a given species and (ii) that same species with one electron removed.

Ionization energies can be thought of as a reaction like the following, using the oxygen atom as an example:

$$
\ce{O -> O+} + e^-
$$

When evaluating quantities like the ionization energy, "external" energies (this is my term, not an official/formal one!) like kinetic and gravitational potential energy are disregarded. Further, the reference energy is generally defined to be zero for an isolated electron at "infinite separation".

Thus, the energy of the above reaction is:

$$
E_\mathrm{rxn} = \left(E_{\ce{O+}} + E_{e^-}\right) - E_{\ce O}
$$

But, since the electron is taken as being in its reference state, this is simply:

$$
E_\mathrm{rxn} = E_{\ce{O+}} - E_{\ce O} \equiv IE
$$

PLEASE NOTE that the above energies are only the internal/electronic energies of the $\ce O$ and $\ce{O+}$ species, and not their enthalpies $\left(H_{\ce{O+}},\,H_\ce{O}\right)$, free energies $\left(G_{\ce{O+}},\,G_\ce{O}\right)$, etc.


Why is xenon hexafluoride not stored in glass?

$\ce{XeF_6}$ can't be stored in glass as it reacts with the silica that makes up the glass

$\ce{2XeF_6 + SiO_2 -> 2XeOF_4 + SiF_4}$

This can proceed further to eventfully make $\ce{XeO_3}$:

$\ce{2XeOF_4 + SiO_2 -> 2XeO_2F_2 + SiF_4}$

$\ce{2XeO_2F_2 + SiO_2 -> 2XeO_3 + SiF_4}$

Dry $\ce{XeO_3}$ is highly explosive ...

That is why both $\ce{XeOF_4}$ and $\ce{XeF_6}$ are stored in nickel containers at room temperature.

These are generally considered as acid-base reactions using the Lux-Flood definition, see https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Lux.E2.80.93Flood_definition

I don't have them to hand but any decent University level Inorganic Chemistry text book such as Housecroft and Sharpe or Greenwood and Earnshaw should cover this.


Horizons of a bachelor degree in chemistry

You would need certain prerequisite courses to get the master's degree in a different field.  For example, Boston University explains:

The Master of Science (MS) in Computer Science program is intended for computer professionals and for people who intend to move into the computer field from other areas of study (emphasis added).

So yes, you can come from a different background.

Applicants to the program are required to have a baccalaureate degree from a regionally accredited institution and proficiency in the areas covered by the following courses:
◾MET CS 201 Introduction to Programming
◾MET CS 231 Programming with C++ or MET CS 232 Programming with Java
◾MET CS 248 Discrete Mathematics
◾MET CS 341 Data Structures with C++ or MET CS 342 Data Structures with Java
◾MET CS 472 Computer Architecture

So it would be fine to switch to physics, biology, biophysics, or many other programs as long as you have the right prerequisite courses.
However, some schools only accept people wanting a PhD, and give a master's degree along the way, which you get to keep if you don't finish the PhD.


Hybridization of sulfur in sulfur dioxide

The structure of sulfur dioxide ($\ce{SO2}$) is quite complicated.
The image from socratic.org* explains it quite well, please find the $\ce{SO2}$ hybridization diagram via the Internet Archive.
As seen, all the atoms have $\ce{sp^2}$ hybridization.
I'll only focus on the central sulfur atom.

Two $\ce{sp^2}$ orbitals form $\ce{\sigma}$-bonds with the two oxygens.
The other $\ce{sp^2}$ orbital is where the lone pair lives in.

Now, we have dealt with 4 electrons, and only have 2 electrons to deal with.
The 2 remaining electrons actually live in the unhybridized $\ce{p}$ orbital.
At this point, the two oxygen atoms will also have 2 electrons left to pair with the sulfur atom.

The two oxygen atoms get to keep their own electron.
Sulfur shares two electrons among itself and the other two oxygen atoms.

Thus, no electron lives in the $\ce{d}$ orbital.

$\color{Red}{\mbox{red}}$ represents number of electrons (I am too lazy to draw the fish-hooks).
$\color{Green}{\mbox{green}}$ represents $\ce{sp^2}$ orbital.
$\color{blue}{\mbox{Blue}}$ represents $\ce{p}$ orbital.

Note that the two $\ce{sp^2}$ orbitals between sulfur and the two oxygen atoms are in $\ce{\sigma}$-bond.

* The website is no longer available; there is no version in the Internet Archive.


Where do the electrons not in the S1 orbital get this "extra energy" from?

In simple terms: The further apart the negatively charged electrons are from the positively charged nucleus, the higher is the potential energy of said electrons. When increasing the quantum number $n$, you automatically get further away from the nucleus. That is why the $2s$ electrons have a higher energy than $1s$ electrons. (That, and adding that the $1s$ electrons shield some of the nuclear charge.)


Addition of oxygen molecule to given aromatic compound

The three hydrogen bond at -OCH3 would be reactive and
-CH3 at attached directly to ring would react so 6 products possible.


When discussing this question, one needs to think about all the processes involved (or at least those that account for most of what happens). In general, any process will occur spontaneously, if $\Delta G$ is negative. To determine $\Delta G$, equation (1) is used.

$$\Delta G = \Delta H - T \Delta S\tag{1}$$

There is an enthalpic term related to the heat a reaction produces or requires and an entropic term that is temperature-dependent. We cannot say a lot about the entropic term a priori, but we can about the enthalpic term — which is luckily what the question is about. For that, we first need to write a reaction equation (2):

$$\ce{NH4Cl (s) + n H2O <=>> NH4+ (aq) + Cl- (aq) + m H2O}\tag{2}$$

Dissolving of the ammonium chloride can also be thought as two separate processes, see equation $(2')$.

$$\ce{NH4Cl (s) -> [NH4+ (g)] + [Cl- (g)] -> NH4+ (aq) + Cl- (aq)}\tag{2'}$$

Meaning that first we break up the ammonium chloride crystal structure and second we dissolve the bare ions. If we want to write that in single enthalpy terms, we can do that as shown in equation (3).

$$\Delta H_\mathrm{tot} = -\Delta H_\mathrm{lattice}(\ce{NH4Cl}) + \Delta H_\mathrm{solv}(\ce{NH4+}) + \Delta H_\mathrm{solv}(\ce{Cl-})\tag{3}$$

All of these values can be looked up. I have given those as calculated by Jenkins and Morris in table 1.[1]

$$\textbf{Table 1:}\text{ Values of enthalpies used in this answer as}\\\text{quoted from Jenkins and Morris (reference 1).}\\\begin{array}{cccc}\hline
\text{compound} & \Delta H_\mathrm{lattice} [\mathrm{kJ/mol}] & \Delta H_\mathrm{solv} [\mathrm{kJ/mol}] & \Delta H_\mathrm{tot} [\mathrm{kJ/mol}] \\\hline
\ce{NH4Cl} & -709.1 & - 694.7 & 14.4 \\ \hline\end{array}$$

So we need energy (a lot of it!) to break up the crystal lattice of $\ce{NH4Cl}$. We then re-gain energy by creating the hydrated, i.e. dissolved, ions in solution. If energy is required, it is typically (excluding photochemical reactions — not the case here) heat energy which is simply drawn from the surroundings. If energy is released, it is typically (same caveat) released as heat into the surroundings.

Thus, the answer to your question is:

The heat is absorbed by the solid ammonium chloride. It is used to break up the salt crystal according to equation $(2'.1)$. Part of it is re-released (unnoticed) by forming solvent–ion hydrogen bonds (the solvation mechanism of $\ce{NH4Cl}$ — equation $(2'.2)$).
This heat is drained from the surroundings, which in this case is primarily the water in which you wanted to dissolve the ammonium chloride. (Of course, the solution will then further exchange heat with whatever is around it, meaning that either the air around a bottle/flask or an outer flask as in your described experimental setup will lose heat at the benefit of the $\ce{NH4Cl}$-solution.)

Reference:

[1]: H. D. B. Jenkins, D. F. C. Morris, Mol. Phys. 1976, 32, 231. DOI: 10.1080/00268977600101741.


Here

B denotes Brønsted base
and HA denotes acid


Antibonding, Nonbonding and Bonding orbitals

In basic MO theory, when two atomic orbitals mix, they form an in-phase (bonding) and out-of-phase (antibonding) molecular orbital. The primary criteria for this mixing are having the same symmetry and being near in energy. If you mix your atomic orbitals together, but find that some of the atomic orbitals have no symmetry match with a similar enough energy, then these unmatched atomic orbitals do not mix and become non-bonding MOs.

Consider hydrogen fluoride. The 1s orbital on hydrogen has the same symmetry as the $p_z$ orbital on fluorine. These mix into a bonding and antibonding orbital. Hydrogen however does not offer any other occupied atomic orbitals with matching symmetry for the $p_x$ and $p_y$ atomic orbitals on fluorine. Therefore, these orbitals remain as non-bonding. Furthermore, the 2s fluorine atomic orbitals are much lower in energy than hydrogen's 1s orbital, so this does not mix appreciably either, and can therefore also be treated as non-bonding.

For elementary concepts, I think you just need to know about how these contribute to bond order. The occupancy of non-bonding orbitals does not have any effect on the bond order and are typically considered lone-pairs.


Why not use all the colours in colorimetry?

Why not send in white light (with all the wavelengths at once)?

This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.

Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!

In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.

Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.