Percolation network analysis with 1H MAS solid state NMR

Solid-state (MAS) NMR can most likely help you understand more about the interaction between the percolating additive and the resin. If you have not yet carried out solid-state NMR analysis, it can be a useful complement to information that you have determined by other methods. However, it is hard to be 100% sure that it would answer your exact question in terms of guiding you on how to modify the percolation threshold.

Should you do $\ce{^1H}$ MAS solid-state NMR?

Unlike in solution-state NMR, $\ce{^1H}$ MAS solid-state NMR is not yet a routine procedure and has much fewer applications (compared to the widespread use of $\ce{^1H}$ in solution-state NMR) until fairly recently (within the last 5-10 years).

$\ce{^1H}$ has a high natural abundance and high sensitivity. However, unlike in solution-state NMR where the free tumbling leads to averaging of such anisotropic interactions, in solid-state, anisotropic interactions are retained. Thus, the high abundance and sensitivity of $\ce{^1H}$ lead to a dense network of anisotropic dipolar interactions in the solid-state, meaning that a "straight-forward" $\ce{^1H}$ spectrum would yield a broad unresolved hump rather than resolved, informative (sharp) signals.

Thus, the approach taken more commonly is to detect heavy atoms with better inherent chemical shift resolution, such as $\ce{^13C}$. Techniques such as cross polarisation (CP) to tackle the problems of lower sensitivity with heavy atoms have been around for decades and can now be considered routine in solid-state NMR. So, if you may want to consider $\ce{^13C}$ CP-MAS solid-state NMR instead of $\ce{^1H}$ MAS solid-state NMR.

What access do you have to $\ce{^1H}$ MAS solid-state NMR facility and expertise?

$\ce{^1H}$ MAS solid-state NMR has undergone some dramatic technological improvements in the past few years. While it is still not routine, it is now possible to acquire (more) resolved $\ce{^1H}$ solid-state NMR spectra. However, highly specialised hardware is required.

The hardware required is a high magnetic field (700 MHz to 1 GHz) and very high MAS rates (at least 40 kHz but ideally 60-100+ kHz). Both of these types of hardware have been available commercially since about 2009. However, there are not many labs that I know of that have both of these pieces of hardware, and nearly all are national or pan-EU facilities (and unfortunately the ones I can think of nearly are nearly all focussed on structural biology). It is also not straightforward to operate these hardware, so you will need to interest these labs in a collaboration.

There are some labs that are trying to do $\ce{^1H}$ MAS solid-state NMR with lower magnetic fields and/or MAS rates by developing specialist pulse sequences to carry out the $\ce{^1H}$ homonuclear decoupling required. It is not an area of ssNMR that I am very familiar with though I believe that there are people doing this in Israel and India.

Generally, the groups doing $\ce{^1H}$ MAS ssNMR are not going to be doing it like a "service". You cannot just give someone a sample and expect them to give you a spectrum. If you can find a group that specialises in $\ce{^1H}$ MAS ssNMR to collaborate with, then it is worth doing $\ce{^1H}$. Otherwise, if you find a ssNMR group with other interests, you may be able to ask them to carry out a (relatively routine) $\ce{^13C}$ CP-MAS spectrum for you.

Sample amounts and types

I know you did not directly ask about this, but I feel it may be helpful to give an idea of how much and what types of sample you will likely need.

For very fast MAS spinning and $\ce{^1H}$ detection, you need only a small amount of sample. Depending on the density of your polymer it is likely that no more than 10 mg is required. (My estimates are based on biopolymers - mostly carbons and protons, so adjust as appropriate for your polymer.)

For heavy atom detection, more sample is usually advised because that will help with the lower sensitivity. Around 50-100 mg would be a good amount.

It will also be helpful to you (and your collaborator) if you can provide the monomer of the resin, and the pure additive (as monomer if necessary). It can be helpful if you have several rubbers that are more simple "models" of what you are trying to investigate (known to have one particular type of polymer links, etc).

...and what information can you expect to obtain?

In the best case scenario, you will have a fairly resolved spectrum of your polymer sample, and by comparing the spectrum of your monomers and your polymer you can assign nearly all the signals. By looking at which signals have shifted/broadened the most, you can determine the atoms which are interacting (or have become bonded). Knowing which atoms are interacting, you can then have some guidance on what functional groups may allow you on to adjust the percolation point as you wish.

But! If you are used to looking at solution-state NMR spectra, solid-state NMR spectra can look incredibly broad. For polymers, the resolution is often poor and signals are fairly broad (even by solid-state NMR standards). In my experience (detecting $\ce{^13C}$) by CP on a 400 MHz spectrometer with a MAS rate of 10 kHz), the linewidths of polymers I have seen are around 150-300 Hz.

The resolution is a very sample-dependent parameter, and is one of the main reasons why I cannot answer definitely whether ssNMR will help you or not. If, as you said, the additive and the polymer are very similar, their signals may well overlap or be insufficiently resolved for you to determine if there has been any changes.

Moreover, polymers are often heterogeneous. This means that one signal in the monomer may broaden and split into multiple signals in the polymer. If this happens at too many atoms, it will make it difficult to assign your polymer spectrum.

There are ways, again, to overcome these problems. It is possible to employ multidimensional ssNMR spectroscopy detecting $\ce{^1H}$ and/or heavy atoms, leading to heteronuclear correlations or homonuclear double-quantum colrreations. The additional dimension often improves resolution and therefore helps with assignment. However, sensitivity will most likely suffer. Alternatively, the molecular dynamics within the polymer can give insight on where the interaction may be occurring in cases where you have sufficient resolution to assign but it is unclear where the interaction has occurred from observing the chemical shift alone. These approaches are again going into non-routine territory where you will need to find a collaborator rather than having it as a "service".

Here is a review if you are interested in some examples of recent multidimensional $\ce{^1H}$ MAS ssNMR work on polymers. Here is a more dated paper which shows how dynamics can help in identifying the nature of interactions at a rubber-filler interface.


Thermite reaction beween zinc and iron(II, III) oxide

I'd check the Galvanic series.

$\ce{Fe(II)}$ and $\ce{Fe(III)}$ have an higher standard potential than $\ce{Zn}$ and therefore your reaction should take place.

I don't know how reactive it is but I had good experiences with also using $\ce{HCl}$. Use $\ce{Zn + HCl}$ to produce $\ce{H2}$ and heat $\ce{Fe2O3}$ in the presence of $\ce{H2}$ to produce $\ce{Fe}$ and $\ce{H2O}$.


I think this should address your query:

  The separation of macromolecules such as polymers and DNA by means of electrophoresis, gel permeation chromatography or filtration exploits size-dependent differences in the time it takes for the molecules to migrate through a random porous network. Transport through the gel matrices, which usually consist of full swollen crosslinked polymers depends on the relative size of the macromolecule compared with the pore radius. Sufficiently small molecules are thought to adopt an approximately spherical conformation when diffusing through the gel matrix, whereas larger ones are forced to migrate in a snake-like fashion. Molecules of intermediate size, however, can get temporarily trapped in the largest pores of the matrix, where the molecule can extend and thus maximize its conformational entropy. This 'entropic trapping' is thought to increase the dependence of diffusion rate on molecular size. Here we report the direct experimental verification of this phenomenon. Bragg diffraction from a hydrogel containing a periodic array of monodisperse water voids confirms that polymers of different weights partition between the hydrogel matrix and the water voids according to the predictions of the entropic trapping theory. Our approach might also lead to the design of improved separation media based on entropic trapping. – Source

I don't know if you needed a textbook definition, but thought you'll get a more elaborate view from this.

This is a schematics of an entropic trap device(ref: Anal. Chem. 2002, 74, 394-401)


How is dioxygen stabilised by coordination to haemoglobin?

I think you have a misconception. The binding of oxygen to haem and its release are processes that are entirely equilibrium-controlled by the equilibrium shown in $(1)$.

$$\ce{haem + O2 <=> haem{-}O2}\tag{1}$$

Meaning that wherever high partial oxygen pressures are observed, oxygen preferentially binds to haem but likewise it is released where low partial oxygen pressures are observed. Thereby, Le Chatelier’s principle nicely controls where oxygen is released from the red blood cells’ haem groups — it will always be in those areas where oxygen is required. Haem itself has evolutionally acquired the role of binding oxygen rather well but only to make sure that it actually reaches the distant parts of the body.

Oxygen in its $\ce{O2}$ modification is also not as bad as your initial post seems to assume. Yes it can react with quite a few substances and many of its reactions have a low activation barrier — but the concentrations in which it would be released into the blood stream ‘undesiredly’ are concentrations the body has learnt to cope with over millenia of evolution. So while minor damage may occur, most damage is quickly repaired by dedicated enzymes. Most importantly, any liberated oxygen will always be $\ce{O2}$ and never superoxide or peroxide.

Only while being bound to haem do oxygen’s (and iron’s) electronic properties change. On a microscopic level, the oxygen molecule causes a one-electron oxidation of iron converting itself to superoxide. This also causes a spin-flip in the iron centre which goes from high to low spin. The combined $\ce{\overset{\mathrm{+III}}{Fe}-O2}$ system can finally undergo antiferromagnetic coupling to give an overall observed diamagnetic ground state. Details can be found in this answer. However, while it is superoxide the $\ce{O2}$ moiety will never leave the iron centre — or, more precisely described, as soon as $\ce{O2}$ leaves, the entire process is reversed again. Thus, even though the bound state includes a superoxide ion no superoxide can ever be directly released into the surrounding blood stream.

$$\ce{[\overset{\mathrm{+II}}{Fe}(por)] + ^3O2 <=> [\overset{\mathrm{+III}}{Fe}(O2^{.-})(por)]}\tag{2}$$


Do all adiabatic processes involve a change in temperature?

The first law of thermodynamics says,
$$\Delta U = Q - W$$

By definition, no heat is exchanged between the system and the surroundings in an adiabatic process. Therefore, the 1st law reduces to,

$$\Delta U = -W$$

In an adiabatic process, $PV^\gamma =const. $ has to be followed. If you change pressure, the above equation implies that volume changes. If the volume changes, there will be some work done which in turn will change the internal energy of the system. Therefore, the temperature changes.

You can also come to the same conclusion using ideal gas equation.

We know the ideal gas equation,

$$PV = nRT$$

If there should be no change in temperature, $PV$ should remain constant.

But in case of an adiabatic process, we know that
$$PV^\gamma =constant$$

So if you change one of those parameters, the product $PV$ will obviously change (so as to keep $PV^\gamma$ constant). Hence, temperature always changes in an adiabatic process.

*I have ignored ridiculously insane cases where matter can exit/enter the system through an adiabatic wall (which isn't possible). If you could discover a wall that allows matter to move across and does not allow heat to conduct through it, you deserve a nobel prize.

*Free expansion and mixing gases are two processes where the process is adiabatic and there is no temperature change.


What is there in this 3D representation of organic molecules I fail to grasp?

You have drawn the compounds correctly, and yes

a solid wedge means the bond is coming out of the plane of the screen towards you
a dashed wedge means the bond is going behind the plane of the screen away from you
a solid line means the bond lies in the plane of the screen

Here is a Newman projection of your molecules, sometimes Newman projections can make it easier to "see" things.

Build a model (I did) or flip the drawings around, there is no way you can interconvert the isomers labeled "S,S' and "R,R"; they are chiral enantiomers.  The meso compound and its mirror image are easily converted into one another (try rotating the molecule about a $\ce{C_2}$ axis that lies in the plane of the screen, 2) is perpendicular to the mirror plane [the dashed line in my drawing] and 3) bisects the $\ce{C_2-C_3}$ bond in the molecule); they are indeed representations of the same molecule.


As stated in both the links Geoff and Philipp have kindly commented (1, 2) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are "allowed" to interact together in reactions.

In this case the $\mathrm{t_2}$ groups three of the metal atom's d-orbitals into a certain class while two of the orbitals belong to the $\mathrm{e}$ class. The $\mathrm{t}$ means triply degenerate while the $\mathrm{e}$ means doubly degenerate (degenerate means have the same energy).

The $\mathrm{g}$ is not about how many energy levels are degenerate rather it is an indication of the answer to a certain operation we can perform on an orbital. It instead relates to how the orbitals behave if we hypothetically were to put a line bisecting the orbitals or "invert" them around a centre point (crudely put "going from one corner to the other". If the phase of the wavefunction (or perhaps more pragmatic) the "colour" of the orbital lobes changes while doing this we label the orbital with a $\mathrm{u}$ if we have no phase change we label it with $\mathrm{g}$. They indicate the orbitals response to an inversion about the centre of symmetry.

We can do lots more of these operations to investigate the symmetry of the orbitals which together lead us to assigning the molecule to a symmetry group. Knowing this is very important for a plethora of reasons.


Why are tertiary alcohols more reactive towards Lucas reagent but show low rates of esterification?

Do you know the mechanism of esterification? The first step - which is also the rate determining step - involves the nucleophilic attack of the lone pair on the alcohol's oxygen atom on the electrophilic carbon atom of the carboxylic acid. Hence, the more nucleophilic the oxygen atom, more will the equilibrium shift to the right, favoring esterification.

Now, can you tell which degree alcohol is the most nucleophilic? And can you tell now which degree alcohol will favor esterification the most?

For Lucas reagent, the reaction involves the formation of a carbocation (at the position where the hydroxyl group originally was) by elimination of the hydroxyl group as an oxonium ion. Thus, the more stable the carbocation, the more the Lucas reagent favor the white precipitate.


I can't tell if the modifier "formal" applies only to "definition" or to "algorithm" also, in the first sentence of your question's body. In case any algorithm would do: this looks like a pretty standard instance of the cycle-detection problem that one learns in undergraduate computer science.

If you only want to detect the presence of cycles, then the algorithm is essentially just to attempt a topological sorting of the graph; if the attempt fails, you have a cycle, and I guess that would mean that you have some sort of catalysis going on. (I didn't major in chemistry; please yell at me if that's non sequitur.)

Okay, so mere cycles are not enough - it seems you want cycles that have "gain". How about a tortoise and hare and have these multiply a state variable by the ratio of outgoing to incoming edges to reaction nodes (denoted by squares in your diagram)? If the hare catches up to the tortoise, that implies a cycle, hence catalysis; then if you compare their state, if the hare has accumulated a larger "gain" than the tortoise, I would say that that shows the presence of autocatalysis?

I'm not sure how this works if there are multiple, linked cycles. You could have one cycle in your graph that's autocatalytic, and another cycle linked to it that "uses up" its catalyst (something of a contradiction). For example:

B + C -> D + E
A + D + H -> 2C + F
2F + G -> H

Is that still an autocatalytic system?


As Geoff already pointed out in the comments, the connection between a molecular orbital theory (MO) point of view and a Lewis picture (LT) is tenuous at best.

Maybe the best example for this is the mentioned $\ce{{}^{-}CN}$ and obviously the isoelectronic $\ce{CO}$. But there are many more molecules.

In the MO point of view the electron pair which can be donated to an acceptor orbital is of course the highest occupied molecular orbital (HOMO). Often this comes with the more trivial notation of "lone pair", based on the coefficients of a certain element in that orbital (percentage contribution). In the case of $\ce{CO}$ this becomes fairly obvious, but since this is not the question at hand I will only focus on the isoelectronic counterpart.

Please consider the above MO scheme only as a sketch, it is not to scale and contributions were estimated only be a quick browse of the output file. I also did not compensate the calculation for the negative charge, so even the bonding orbitals are antibonding with respect to fragmentation. But it shall still provide the point.
We can clearly see that the highest coefficient of the HOMO is at the carbon end, this is consistent with the fact, that it will most likely bond with this end to a more electropositive element. But this is, however, not a real lone pair since it is $\sigma$ bonding and also has quite a big coefficient at the nitrogen end. (In $\ce{CO}$ the difference is much bigger.)
Below that we see the two $\pi$ orbitals, followed again by two $\sigma$ orbitals. All five orbitals make up the bonding of the molecules and determine also its reactivity.

Now in the Lewis theory picture this is quite different. We traditionally describe it as a carbon oxygen $\sigma$ bond, two carbon oxygen $\pi$ bonds, a oxygen lone pair and a carbon lone pair. All made up from some kind of hybrid orbitals. In computational chemistry we can analyse the wave function and obtain localised orbitals. One of these methods is natural bond orbital theory (NBO). I have carried that out and sketched the common bonding picture along with it:

While this picture is also consistent with a valence bond theory (VB) approach, it explains bonding in a different way than MO theory. However, the conclusions we will ultimately get are essentially the same.

It shall be noted, that MO theory and VB theory are complementary theories, but the different explanations should not be mixed, even when the conclusions are essentially the same.


What toxic gases are emitted during the combustion of potassium ferrocyanide?

Potassium ferrocyanide cannot burn, because it is the product of a combustion. When meat or blood is overheated, carbonized and burnt, the remaining ashes contain potassium ferrocyanide. And this product was initially recovered by washing these ashes and evaporating the obtained solution. In German, this compound is called Blutlaugensalz, which means "salt from washing blood" This is even the way this product was originally discovered in the old times of beginning chemistry.
So potassium ferrocyanide resists high temperature, and will not produce any $\ce{HCN}$ by heating or by combustion. The only way of producing HCN out of it, is to heat it with diluted sulfuric acid, according to the equation : $$\ce{K4Fe(CN)6 + 3 H2SO4 -> 2 K2SO4 + FeSO4 + 6 HCN}$$This reaction is difficult to carry out properly, because, if the sulfuric acid is more concentrated, it will destroy $\ce{HCN}$ according to  $$\ce{H2SO4 + 2 HCN + 2 H2O -> (NH4)2SO4 + 2 CO}$$


What does washing and drying a solution mean?

This is standard for purifying substances.
To wash means to add your product solution to an aqueous solution (or just water, but frequently a saturated solution) to a separatory funnel. After shaking, you drain the lower layer (which is usually aqueous). This process removes water soluble impurities. This is frequently repeated.
Drying is accomplished by taking the organic solution of your product and mixing it with a desiccant (usually anhydrous sodium sulfate or anhydrous magnesium sulfate). This removes any water that was left from the washing in the above step. The desiccant may then be removed by filtration.


Partial Aromaticity and Anti-aromaticity

From a classification point of view,  there is no reason for not calling the molecule aromatic.

First, saying a molecule is aromatic doesn't imply that  aromaticity fully extends over it. A trivial example is, just to say, cyclohexylbenzene, or whatever aromatic molecule of your choice.

Now, let us assume the two rings being completely independent so to exclude any delocalisation. The phenyl moiety will be just aromatic while the Huckel antiaromatic cyclobutadiene ring will simply adopt a rectangular shape, so effectively avoiding the antiaromatic unstable configuration.

Again, there wouldn't be reason for not calling the molecule aromatic.

This is often a sort of misunderstanding when antiaromaticity is discussed. It is just a criterion to decide if full delocalisation within planar rings is attained. If the answer is no, because the system is in fact antiaromatic, then there is nothing actually specific to attain.

In short: while a ring that is Huckel-aromatic does actually tends to be aromatic, the Huckel-antiaromaticity is simply a relative high energy situation that a system will always try to avoid.*

Finally let us consider the actual case. The resonance structures you wrote indicate indeed a certain character of aromaticity in the cyclobutadiene moiety and a partial double character of the interring bond. This alone, let aside the above considerations, definitely tell us that the molecule is aromatic.

For more insights I've found the following papers, analyzing delocalisation in phenylcyclobutadiene experimentally (enthalpy of hydrogenation and EPR) and theoretically.

https://pubs.acs.org/doi/abs/10.1021/ja0525511

https://www.academia.edu/25340037/Aromatic_pathways_in_conjugated_rings_connected_by_single_bonds

As a very quick resume enthalpy of hydrogenation point to a destabilising effect of the Huckel-antiaromatic moiety, while electron spin resonance and calculation show overall aromatic rings current and an interring C-C bond length of about 1.44 Å, about midway from a single to a double C-C bond length.

*note that chemists are sometimes able to constrain structures in otherwise highly unfavourable systems. I won't be surprised if exotic beasts forcing an inner ring into antiaromaticity were synthesised. I might even have seen a related Q here in SE.


Why is carbon monoxide not a greenhouse gas?

$\ce{CO}$ is not considered a primary (or significant) greenhouse gas due to the weak (but non-zero) absorption of energy in the infrared. However, it does increase global warming by reacting with certain chemical species in the atmosphere which in turn lead to an increase in concentration of primary greenhouse gasses, most notably methane and ozone (see final paragraphs cited below). Thus, it is not completely accurate to dismiss it as a contributing factor.
From the National Oceanic and Atmospheric Administration's page on $\ce{CO}$ as a greenhouse gas:

Carbon monoxide (CO) is not considered a direct greenhouse gas, mostly because it does not absorb terrestrial thermal IR energy strongly enough. However, CO is able to modulate the production of methane and tropospheric ozone.

Carbon monoxide does contribute to global warming indirectly, as noted by the American Chemical Society's page entitled "Which Gases Are Greenhouse Gasses?":

Several other gases can have indirect effects on atmospheric warming. This occurs when chemical reactions in the atmosphere produce or destroy greenhouse gases, including tropospheric ozone. Indirect effects also occur when a gas influences atmospheric lifetimes of other gases or affects atmospheric processes like cloud formation that alter Earth’s radiative energy balance by increasing Earth’s albedo. Gases that can cause these indirect effects include carbon monoxide ($\ce{CO}$), oxides of nitrogen ($\ce{NO_{\rm x}}$) and non-methane volatile organic compounds (NMVOC).

The above is elaborated in the text Global Climate Change Linkages: Acid Rain, Air Quality, and Stratospheric Ozone (James C. White, Editor. Elsevier (1989), p. 106):

As has been noted, ozone also initiates reactions that generate the highly chemically active oxidants, including hydroxyl. The lifetime of radiatively active methane in the atmosphere is determined by hydroxyl abundance. Carbon monoxide and methane are both sinks for hydroxyl, $\ce{CO}$ being the more reactive of the two. Greater concentrations of carbon monoxide in the atmosphere lead to lesser concentrations of $\ce{HO\cdot}$, which in turn lead to higher concentrations of methane.
Carbon monoxide and ozone are linked in the non-urban atmosphere that is low in hydrocarbons through the interaction with $\ce{HO\cdot}$, as described above, and $\ce{NO2}$. Hydroxyl is destroyed by $\ce{CO}$, increasing the ambient concentration of $\ce{NO2}$. The resulting increased proportion of $\ce{NO2}$ to $\ce{NO}$ leads to a higher steady state concentration of ozone. Carbon monoxide, a non-radiatively active species, enhances the greenhouse warming by raising the levels of methane and ozone.

In summary: $\ce{CO}$ is not a direct greenhouse gas due to low absorption in the infrared but is an indirect greenhouse gas because it increases the concentration of direct greenhouse gases (methane and ozone most prominently) by reacting with atmospheric $\ce{HO\cdot}$, which otherwise would decrease the concentration of those gasses. The toxicity part is not germane to the greenhouse gas issue.


Measure low DOM absorbance values in a microplate reader. Is there a suitable deep well microplate with flat bottoms available?

I am not aware of suitable microplates for this application.  Here are some alternative approaches.

1. Low-volume quartz cuvettes.

There are quartz cuvettes in "micro" formats that may be suitable for you.  Mostly commonly they work with 1 cm pathlength.  The volume of the cells is controlled by the width of the non-optical dimension.  This can be made extremely small.

One example I found lists cuvettes that still have path lengths of 1 cm with total fill volumes as low as 15 μL.

2. Sample concentration

Details depend heavily on the details of your DOM and application, but you could concentrate your samples and then use existing microplates.

3. Longer measurement times.

Most plate readers (and cuvette-based spectrophotometers) let you control the number of "shots" or of the total amount of time a measurement of light intensity is integrated over.  You could crank this number up to get better sensitivity.


NIST suggests here and here (search for Ni) that $\mathrm{[Ar] (3d)^8 (4s)^2}$, specifically in the state $\mathrm{^3 F_4}$, is the fundamental electronic state for neutral nickel atoms in empty space. However, as Wikipedia describes, if you don't specify the term symbol for the configuration, then the average energy of all terms for the $\mathrm{[Ar] (3d)^9 (4s)^1}$ configuration is slightly lower than the average energy of all terms in the $\mathrm{[Ar] (3d)^8 (4s)^2}$ configuration.

Now, why one configuration has a lower energy than another is a much harder question to answer in close cases such as this, as it involves a precise quantitative analysis of electronic energies for every possible electronic configuration of a nickel atom (a multielectronic atom). The general idea is that even though, for $Z=28$, the configuration $\mathrm{[Ar] (3d)^8 (4s)^2}$ is higher in energy than $\mathrm{[Ar] (3d)^9 (4s)^1}$ if you look at the sum of occupied orbital energies, the latter case has more interelectronic repulsions, which causes its total energy to be slightly higher. A lot of information on this can be found in this previous Chem.SE question and the several nested questions within.

Interestingly, the gas-phase fundamental configuration of the monocation $\ce{Ni^+}$ is $\mathrm{[Ar] (3d)^9}$, which further shows how nickel's electronic configuration is teetering on the boundary between the options.


Does chelation take ions out of solution

No- still too small for normal filters. Those ions are in water complexes anyway. It doesn't stop (prevent) chemical reactions rather alter the thermodynamics. EDTA and other organic complexes are the common way to allow animals to absorb ions that would otherwise tend to form oxides when swallowed so its biologically (chemically) available.


Uv vis and fluorescence spectroscopy: sensitivity

Fluorescence is a 'zero background' or absolute type of measurement meaning that single photons can be measured against a 'dark' background so the sensitivity is huge, and limited by the fraction of light caught and efficiency of the detector.

Absorption is a relative measurement as it is the difference in two large numbers, one the light intensity with no sample $I_0$ and the other when the sample is present I. In terms of photons these intensities are huge, say $I_0=10^6$ and a 1% change would reduce the number of photons to $10^6-10^4$; still a large number and so the absorption measurement is looking for the small difference in two large numbers. This measurement is limited by noise in detectors and other electronics.   (Its a bit like trying to weigh the captain of a ship by weighing the ship with him standing on the deck and then when he is not there.)


How to find the concentrations of the acid-base equilibrium between sulfuric acid and sulfur dioxide for a known pH using Octave?

Your code seems to be calculating the differences between a guessed concentration vector x and a calculated concentration vector.  These differences in concentrations between guessed and calculated are y.  For example, the guessed pH is apparently x(1), and the calculated pH is 10^(-pH).  The difference between them is y(1).

Thus, its no surprise that the y values are sometimes negative.  To solve the problem you need to feed your function to a numerical solver such as fsolve.  That function will take your guessed solution x, other parameters such as pH, the various pKas, and M, and your function phCalc as input, and return the value for x which makes all the y variables zero.  That solution is the one you want.

On the chemistry side, you mention sulfur dioxide, but I don't see any equations for the solubility of sulfur dioxide gas in water.  Additionally, sulfur dioxide is reduced relative to sulfuric acid, so there would need to be an additional reaction, and parameter (extent of reaction or degree of reduction or something) that describes the relative amounts of reduced sulfur (as sulfur dioxide, bisulfite, or sulfite) vs. oxidized sulfur (as sulfate, bisulfate, sulfuric acid, or sulfur trioxide).

There is also the issue of charge balance: what concentration of inert cations (e.g. sodium or potassium or whatever) are you assuming?  Charge balance must hold in bulk aqueous solution.  If you don't have any sodium / potassium etc., then you should include the self ionization of water as an equation, hydroxide as a species, and y(7) = x(1) - x(2) - 2*x(3) - x(4) - x(OH) as an equation.  If you do have sodium/potassium/whatever at significant concentration, you can neglect hydroxide and just go with the equation y(7) = x(1) - x(2) - 2*x(3) - x(4) + x(Na) where x(Na) is the cation concentration, which I have assumed is sodium.

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How to predict the outcome of a crossed Cannizzaro reaction?

To answer this, first let us look at the mechanism of the Cannizzaro reaction in general. It is depicted in scheme 1 as a cyclic transition state. The cycle itself isn’t as important, the realisation that one aldehyde is present as a (deprotonated) hydrate after the action of $\ce{OH-}$ is.

Scheme 1: General mechanism of a Cannizzaro reaction.

In your reaction, you are reacting formaldehyde with a (substituted) benzaldehyde. The question boils down to ‘can we predict, which aldehyde will form a hydrate mor easily?’ And the answer is: we can!

$$\begin{array}{cc}\hline
\text{carbonyl} & \text{hydrate ratio} \\ \hline
\ce{H2CO} & 99~\% \\
\ce{MeCHO} & 60~\% \\
\ce{Me2CO} & < 1~\% \\ \hline\end{array}$$

We have a clear tendency for formaldehyde to almost exclusively form hydrates under standard conditions. Thus, we can conclude that formaldehyde will alwys be the hydrate partner, end up oxidised and reduce its reaction partner as shown in scheme 2.

Scheme 2: Cannizzaro reduction of an aldehyde with formaldehyde as sacrificial aldehyde.


Internal Energy Changes in Adiabatic Reactor

You omitted the energenics associated with breaking chemical bonds and making new chemical bonds.  If this contributes a negative change to the internal energy (exothermic reaction), then the temperature would have to rise to offset this effect so that the overall change in internal energy would be zero.  If it contributes a positive change to the internal energy (endothermic reaction), then the temperature would have to drop to offset this effect so that the overall change in internal energy would be zero.  Saying this in a different way, when you have a chemical reaction occurring, the internal energy can change even if the temperature is held constant.


Why does Ruthenium only have one electron in its 5s orbital in the neutral oxidation state?

From Oxtoby et al, Principles of Modern Chemistry, 7th edi p219 and Levine, Quantum Chemistry, 5th edi, p314, the orbital energies of Sc and the elements beyond, 3d<4s. The electrostatic repulsion makes energy of 3d$^3$ higher than 3d$^1$4s$^2$.

Just my pure guess, Ru is bigger than Fe, electrostatic repulsion is smaller, so 5s$^1$. However, the 6s$^2$ electrons in Os are extra stable due to relativistic effect, so again 6s$^2$.

Quantitative calculation will be highly appreciate for justifying this hand-waving argument


How do you calculate the delta H for multiple reactions?

Find the energy interval that produces atomic nitrogen in the flask but not any ions

This is the key. The question wants you to determine how much energy you need to supply just to only have atomic nitrogen, no ions.

If you supply $941~kJ/mol$, then you have supplied enough energy to move reaction 1 forward. In your flask, you would have $N(g)$ inside. Now, if you further supply energy up to $1402~kJ/mol$, then you have enough energy to make reaction 3 proceed, and that will convert your atomic nitrogen to nitrogen ions, so that's a no-no.

So you have to maintain $E$ that is in between $941~kJ/mol$ to $1402~kJ/mol$ to prevent reaction 3 from proceeding.

If you supply $1402~kJ/mol$ after breaking the bond, then you will definitely produce $N^+$ ions as you have met reaction 3's $E$ requirement.


Woodward Fieser rules for dienes - non conjugated double bond as exocyclic?

Woodward's rules are an empirical method for predicting the maximum absorption of a molecule based on the functional groups present (that is, conjugated functional groups which possess a distinctive maximum absorbance such as a diene or an unsaturated carbonyl).

  Should the marked double bond considered as an exocyclic double bond?

Exocyclic is defined as being outside of the ring system (or in more complex systems, outside of the ring system being considered), the bond highlighted is indeed inside the ring system and is as such endocyclic.

If we start with the alkene at the bottom left, then it is in conjugation with an endocyclic alkene (to the immediate right) and an exocyclic alkene (in the ring to the right hand side). These double bonds are all in conjugation with one another.

The bond you've pointed at isn't in conjugation with the 'triene'

The Woodward rules are specifically related to the effect of groups in conjugation, a simple test of which is to see if you can push curly arrows between the pi bonds. If you try this for your molecule you'll quickly see that the bond you've highlighted is too distant from the conjugated system to have any involvement.


Particle size vs dissolving rate of sucrose

I have found in Germany sugar crystals that are about 5 mm thick. If you are able to get some, you may drop one of these crystals in a beaker with 50 mL water and an stirring bar. If not take a standard sugar cube. Using a watch, measure how long it takes for the crystal to be dissolved. Now start again with another rotating speed. Now break a new crystal in two or three parts with a hammer. And start again with smaller crystals (to be weighed in advance). Repeat the measurements. Report these values on a graph vs rotating rate, or vs. weight of sugar, or vs. temperature. Now start again with powder, and even with the sugar powder used in bakery which is made of extremely fine grains. This is just ideas.


Compound with a bright green emission when electrically excited

While it is not a gas, copper and barium compounds come to mind. If you could introduce a volatile copper compound, you could get something remotely like a mercury vapor lamp, except with copper.


The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions.

Electrophilic nitration reactions like the production of nitrocellulose require sulfuric acid because of the dehydrative properties of sulfuric acid. Sulfuric acid is a non-nucleophilic acid - all it can do is mediate proton transfers and dehydrate things (sulfuric acid has a large negative enthalpy change of mixing with water). Thus,$\ce{NO2+}$ is free to react with whatever you want it to.

$$1)\ \ \  \ce{H2SO4 + HNO3 <=> HSO4- + H2NO3+} $$
$$2)\ \ \ \ce{H2NO3+ <=> H2O + NO2+} $$
$$\text{overall)    }\ \ \ce{HNO3 + H2SO4 <=> H2O + NO2+ +HSO4-}$$

Hydrochloric acid and nitric acid produce aqua regia. Hydrochloric acid is a nucleophilic acid, and thus the decomposition products after proton transfer to nitric acid are different. The chloride ion is a good nucleophile, and reacts with the nitronium ion, eventually producing $\ce{NO2}(g)$ and $\ce{Cl2}(g)$. There is a big difference between $\ce{NO2+}$ and $\ce{NO2}$.

$$1) \ \ \ \ce{HCl + HNO3 <=> Cl- + H2NO3+}$$
$$2) \ \ \ \ce{H2NO3+ <=> H2O + NO2+} $$
$$3) \ \ \ \ce{2NO2+ + 2Cl- <=> 2NO2 + Cl2}$$
$$\text{overall)    }\ \ \ce{2HCl + 2HNO3 <=> 2H2O + 2NO2 + Cl2}$$

  So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions.

Yes. However, starting with nitric acid shifts the equilibrium towards nitronium a little.

  Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway? Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time?

Maybe, but not enough to be useful. Nitric acid is not available pure. It is only stable to a maximum concentration of 70-75% by weight (remainder water). The other reason that sulfuric acid must be used is that it is available "pure" (or at least 98% by weight) and it is so hydrophilic so as to sequester the water. Hydrochloric acid has a max concentration of 35-40% by weight remainder water. All the extra water gets in the way and shifts the equilibrium in the opposite direction (the reaction produces water).


Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?

Here is a rather simple structure explanation presented on this site. To quote, where 'pieces' refer to an organic molecular with a non-symmetric branching structure:

You’ll notice something – the simpler the pieces are, the easier they are to stack together, which provides a tighter fit with fewer spaces. Here, by putting a kink in the block, we make them harder to stack.
What’s this got to do with chemistry?
When compounds freeze, the process is a lot like stacking bricks. The more symmetrical the molecules are, the easier it will be  and the fewer spaces there will be between the molecules. Fewer spaces = better stacking. Hence, when you compare hexane to its structural isomer, 2-methylpentane, hexane has a much higher melting point due to the regular arrangement of its structure.
Better stacking, higher melting point. Case closed. Right? Not quite.
It’s also about surface area [...]
It looks like as we increase branching, we’re increasing melting point and decreasing boiling point. What’s going on? [...]
Going from “branched” to “highly branched”  makes a molecule more compact and sphere-like. As the surface area of the molecule decreases (remember that spheres have the lowest surface area/volume ratio of any shape) they will become more compact and thus easier to pack. This explains the melting point phenomenon [...]
Boiling point is related to the forces between molecules, which in the case of hydrocarbons is Van Der Waals interactions. If you’ve ever seen microscope images of a gecko’s feet – which allow it to climb walls – you’ll see that there is no adhesive but the pads contain a tremendous amount of surface area. It’s all about the Van der Waals interactions.
As we decrease surface area, we are going to decrease intermolecular Van Der Waals interaction and therefore decrease boiling point.

The summary relationships are presented as follows:

linear versus branched —> higher melting/boiling points due to better stacking and surface area contact.
highly branched vs. branched —> more sphere-like –> better stacking –> higher melting point
highly branched vs. branched  —>more sphere-like – -> lower surface area —> lower boiling point.

The above connects boiling point to branching structure, and this paper "Hypothetical Thermodynamic Properties: The Boiling and Critical Temperatures of Polyethylene and Polytetrafluoroethylene" found here links BP to thermodynamics. Per the opening reference statement: "A lower heat of formation is equivalent to a higher thermodynamic stability".
This completes the argument as to why the heat of formation decreases for organic radicals with decreased size and degrees of branching.


How to interpret orbital transition in TDDFT?

To visualise the orbitals of your calculation, use a program of your choice. For Gamess, there are a few options availabile. I use ChemCraft and Molden, and they work quite well. Here is a compilation of some with the former mentioned:

You can further use the summary to identify the most interesting, strongest transitions:

                          SUMMARY OF TDDFT RESULTS

   STATE             ENERGY     EXCITATION  TRANSITION DIPOLE, A.U.  OSCILLATOR
                    HARTREE          EV         X       Y       Z     STRENGTH
   0  A         -193.0290234748    0.000
   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000
   2  A         -192.7831335626    6.691     0.4051 -0.0004  0.0000    0.027
   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002
   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042
   5  A         -192.7210047911    8.382    -0.0121  0.0000 -0.0006    0.000
   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001
   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000
   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022
   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290

For the calculated 9 states that is the last one, with an oscillator strength of 0.290. Now skip back one section to where you find:

      -------------------
      SINGLET EXCITATIONS
      -------------------

Look for excited state number 9:

 STATE #   9  ENERGY =    9.326387 EV
 OSCILLATOR STRENGTH =    0.290423
 LAMBDA DIAGNOSTIC   =    0.561 (RYDBERG/CHARGE TRANSFER CHARACTER)
 SYMMETRY OF STATE   =    A
                 EXCITATION  DE-EXCITATION
     OCC     VIR  AMPLITUDE      AMPLITUDE
      I       A     X(I->A)        Y(A->I)
     ---     ---   --------       --------
      9      17    0.137515       0.018404
     15      17    0.694034      -0.078753
     13      18   -0.085768      -0.006093
     14      19   -0.059549      -0.004510
     16      19   -0.289828       0.018031
     15      20    0.051487       0.010566
     11      21    0.053380       0.009416
     13      21   -0.060120      -0.009392
     16      22   -0.607830       0.005676
     11      24    0.052546       0.012226
     13      25   -0.039117      -0.008987
     16      26    0.112895       0.016712
     16      30   -0.041659      -0.014274
     16      33    0.052880       0.019973
     16      36   -0.031783      -0.011699
     13      40   -0.030352      -0.013004

There you find the orbital transitions. You can also have a look at the amplitudes to identify the most dominant one. In this case it is probably 15 to 17 and therefore corresponds to $\pi\to\pi^*$.

The more complicated the molecules get, the more confusing will this process be. Also the higher the level of theory, the more transitions you will need to consider. In this case Natural Transition Orbitals will certainly become very helpful, see Richard L. Martin, J. Chem. Phys., 2003, 118, 4775-4777.


Why do only ketones exhibit keto-enol tautomerism?

Esters, amides and acids do exhibit tautomerization to the enol form, but the enol content in these compounds is just much less than that found with aldehydes and ketones. Why?

The keto-enol process is an equilibrium; to understand where an equilibrium lies, we must compare the relative stabilities of the products (enol) and reactants (carbonyl compound).

Let's compare some resonance structures for an ester and a ketone and their corresponding enols.

The ester is stabilized by resonance structures such as A, whereas in the case of the ketone, the analogous resonsnce structure A' contributes, but not nearly as much; so ester carbonyls are stabilized compared to ketone carbonyls.

In the enols, the ester resonance structure B is not particularly favorable due to the negative charge on carbon and positive charge on oxygen, just the reverse of what we would want based on electronegativities.  At least in the ketone case resonance structure B' has a positive charge on hydrogen, so resonance structure B' contributes more than B (this is why carbon-carbon double bond stability increases with increasing alkylation).  As a result the enol resulting from the ketone is stabilized compared to the enol resulting from the ester.

Just where the carbonyl-enol equilibrium lies will depend on the relative stabilities of the carbonyl compound and the enol.  As we've just discussed, the ester carbonyl is stabilized relative to the ketone carbonyl, this will push the equilibrium further to the carbonyl side for the ester.  Also, the enol from the ketone is somewhat more stabilized than the enol from the ester, this will push the equilibrium to the enol side for the ketone.

So both factors, starting material (the carbonyl compound) stability and product (enol) stability will work to increase the enol content for ketones and reduce the enol content for esters.


Does glass adsorb methanol to an exceptional extent?

Exceptional is subjective term, and can't be responded to sensibly.  But more/less can be discussed.  This is somewhat complex.  The isotherm is an equilibrium between adsorption and desorption.

Methanol is small, and when adsorbing to glass, presumably through hydrogen bonding, a single molecule will do very little to block other adsorption sites.  Other, larger, molecules might 'lay down' on the surface or have branched structures that may inhibit their peers from adsorbing, giving smaller surface coverage.  However, long linear molecules will tend to orient, and there is an energetic favorability to forming a monolayer for long linear alcohols (and thiols, carboxylates, amines, silanes, etc).

In these highly ordered linear systems there is an attractive force of roughly 0.2 kcal/mol per methylene unit.  Since methanol has the fewest number possible, if a methanol molecule desorbs it can (almost) freely leave the molecular layer, as opposed to longer chain molecules.  When you begin to discuss molecules that are more topologically diverse than simple linear chains, this attraction is significantly weakened, and as such those molecules may also (almost) freely desorb.

In essence the methanol isotherm has maximum kinetic lability, and as such the coverage will happen reasonably quickly, and be fairly complete.  At equilibrium I would expect methanol to have apparently higher coverage than any branched molecule, and before equilibrium, I would expect methanol to have more coverage than long-chain analogs.

Please note this is on a per molecular basis, and the mass equivalents of methanol will be smaller than any other molecule to be compared with (save water).


How to explain disagreement between Le Châtelier's principle and the simplified Gibbs free energy equation?

The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ >0$ and $\Delta G^\circ<0$ must have a larger $K$ at a higher temperature because $\Delta G^\circ$ is more negative. If that were true, we would have a relationship of the form $\Delta G^\circ = -cK$, where $c$ is a constant. Instead, the key relationship is

$$\Delta G^\circ = -RT\ln K.$$

So $\Delta G^\circ$ is proportional to $-T\ln K$. In the case above where $\Delta G^\circ <0$ and $\Delta S^\circ>0$, as T increases, $\Delta G^\circ$ increases in magnitude (becomes more negative), but so does $-RT$, so we don't necessarily need to have a larger $K$ to satisfy the equation. To figure out the temperature dependence of $K$, we need to substitute $\Delta G^\circ$ with $\Delta H^\circ - T\Delta S^\circ$ and then rearrange things:

$$\Delta H^\circ - T\Delta S^\circ=-RT\ln K$$

$$\frac{\Delta H^\circ}{T}-\Delta S^\circ=-R\ln K$$

From that equation, hopefully it is clear that if $T$ increases (which reduces the magnitude of the $\frac{\Delta H^\circ}{T}$ term), $K$ will only increase if $\Delta H^\circ > 0$. If $\Delta H^\circ < 0$, K will have to decrease with increasing $T$ to maintain the equality. $\Delta S^\circ$ is a constant term that does not affect the change in $K$. Thus, our result is completely consistent with both Le Chatelier's principle and with the van't Hoff equation analysis.


Reaction of octahydroazecine with iodine crystals

Chlorination and bromination of alkenes are very general reactions, and mechanistic study of these reactions provides additional insight into electrophilic addition reactions of alkenes. Although much less detail is known about fluorination and iodination of alkenes, it is believed that iodination follows the similar mechanistic steps of much studied bromination (Ref.1) to give corresponding 1,2-diiodoalkane. In certain compounds, if neucleophilic center is available within vicinity of cyclic halonium center, the last step of nucleophilic attack of halide ion would be over written by much faster intramolecular cyclization as depicted by following mechanism:

This mechanism is well supported by the examples provided in Ref.1 and Ref.2.

References:

Yoshinao Tamaru, Masato Mizutani, Yutaka Furukawa, Shinichi Kawamura, Zenichi Yoshida, Kazunori Yanagi, Masao Minobe, “1,3-Asymmetric induction: highly stereoselective synthesis of 2,4-trans-disubstituted γ-butyrolactones and γ-butyrothiolactones,” J. Am. Chem. Soc. 1984, 106(4), 1079-1085 (https://doi.org/10.1021/ja00316a044).
Yvan Guindon, François Soucy, Christiane Yoakim, William W. Ogilvie, Louis Plamondon, “Diastereoselective Synthesis of 2,3,5-Trisubstituted Tetrahydrofurans via Cyclofunctionalization Reactions. Evidence of Stereoelectronic Effects,” J. Org. Chem. 2001, 66(26), 8992-8996 (https://doi.org/10.1021/jo010873r).


How do I find the oxidation number of a coordination complex

Think of it like this. Ammonia and Water are both neutral molecules, therefore they won't contribute to the charge of the two coordination spheres. There are also no subscripts, which indicates that iron and copper both have the same magnitude of charge. Now copper here could be +1 and iron could be in -1, or copper in +2 and iron in -2. (Of course, this isn't possible since this molecule doesn't exist.) I cannot conclusively say which though, without any further data, such as maybe their spin only angular momenta.

For other complex compounds, i suggest you practice, and eventually you'll get the hang of what oxidation states are common for different metals. Also use the concepts you use for normal covalent compounds. Trust me, it's very easy!


Some of the major components of tea are polyphenols. They are fairly acidic and may dominate the pH of tea.

For some examples of the types of polyphenol that are common see this site on tea chemistry:

  The major flavanols [a specific type of polyphenolic compounds] in tea are: catechin (C), epicatechin (EC), epicatechin gallate (ECG), gallocatechin (GC), epigallocatechin (EGC), and epigallocatechin gallate (EGCG).

EGCC has this structure:

from which some degree of acidity is likely to result from all those phenolic OH groups.


Endothermic reaction absorbs heat from environment

An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released
into the surroundings, resulting in an overall negative quantity for the heat of reaction (-ΔE ).
An endothermic reaction
occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat.
Endothermic reactions result in an overall positive heat of reaction (+ΔE ).
Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (ΔE ), the sum of
all potential and kinetic energies.
ΔH is determined by the system, not the surrounding environment in a reaction.
A system
that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the
products is lower than the enthalpy of the reactants of the system.
The enthalpies of these reactions are less than zero, and are therefore exothermic reactions.
A system of reactants that absorbs
heat from the surroundings in an endothermic reaction has a positive , because the enthalpy of the products is higher than
the enthalpy of the reactants of the system.
Because the enthalpies of these reactions are greater than zero, they are endothermic reactions.
EDIT ( See this )
I understand your confusion. See, in endothermic reaction the product will be "colder"....this might seem counterintuitive as in endothermic reaction heat is absorbed by the system form the environment. Therefore the heat absorbed must increase the temperature right?
No! See temperature shows the direction of heat flow. If the temperature of the system increased ( assuming that previously the system and environment were in thermal equilibrium ~ same temperature ) this would mean that heat should flow from system to the environment as the system has a higher temperature than environment.
The following example makes everything clear:
$$T_{(env)}= 25°$$
$$A + B _{(25°)}  = C _{(20°)}  ...... (1)$$
$$A' + B' _{(25°)}  = C' _{(30°)} ..... (2)$$
Compare the temperature of system and environment in the above rxn.
Therefore in rxn. 1 heat flows from environment to system ( from 25° to ~20°).
This heat energy is used to break bonds and stuff.( Endothermic )
And in rxn. 2 heat flows from system to environment ( from 30° to 25°).( Exothermic )
Also see this practical video:
https://youtu.be/NHOHKH3808c


Bond lengths: 1-D particle-in-a-box for 1,3-butadiene

As porphyrin said in his comment, the $\ce{C-H}$ bond is not part of the delocalized $\pi$ system that makes up the one dimensional box. This is partially because hydrogen does not have a p orbital to contribute to the system. The $\pi$ system that creates the box is the result of overlap between p orbitals found on adjacent carbon atoms, meaning that you would be able to approximate the length of the box by summing up the bond lengths between all adjacent carbons in the delocalized system. While resonance means that there are not actually alternating double and single bonds across the $\pi$ system, all of the $C-C$ bonds will be the same length, which should be close to the average of the bond lengths for the most stable resonance contributor.


Vapor Pressure, Equillibrium, and Increasing Pressure

There are a couple of gotchas' with the problem.

The volumes of the liquids and gases are specified. So 1 drop of water in a million gallon tank of dry air isn't going to be enough water to reach saturation.

You stated that "This container will maintain a temperature of 30 C after the start of this experiment." However the problem statement doesn't explicitly state that the contents of the container will be in thermal equilibrium with the container. That does seem to be a reasonable interpretation however. (If the evaporation had taken place adiabatically, then the liquid and gas would have to cool to account for the evaporation.)

There is another significant factor here. The water would absorb ethanol, and the ethanol would absorb water. You can't really figure this out without knowing all the volumes, or at least the relative volumes.

Now if you assume that there is enough liquid to saturate the gas phase, that the system (gas phase and liquid phase) absorbs heat from outside the system, then the vapor pressure of the liquid at $30\ ^\circ\mathrm{C}$  is added to the internal pressure of the gas phase before the evaporation takes place.


Why is nitration of veratrole specific in these positions?

When there are activating and deactivating groups on an aromatic ring, the direction of the electrophilic aromatic substitution is generally controlled by the activating groups.
Think that activating groups improve the substitution in certain positions by increasing its electron density. On the other hand, deactivating groups favor one position over another, not by activating it, but deactivating it less than the other positions. So, the directing effect of a deactivating group is masked by the directing effect of the activating ones, which is stronger.

You most likely don't need the sulphuric acid due to the presence of two activating groups. In contrast, as an example, last nitration step of trinitrotoluene is so disfavored that it has to be carried out in boiling sulphuric acid.


How to select a schematic of the coordinate vs energy representing a rearrangement reaction?

The answer is (b).  The initial step of the reaction should take energy.  Otherwise the reactant molecule itself is unstable.  The very initial portion of the curve in (c) is downward-sloped, which doesn't make sense.  If the reacting alkene could relax to a different conformation without any activation barrier, then it would do so.  But the fact that the reactant is a stable, isolable molecule with a well-defined chemical structure indicates that no such barrier-less relaxation is possible.

Thus, (b) is a better answer than (c).


Can buffer solutions be used to control the pH of ethanol?

Although pH scale is rather strictly used for aqueous solutions, the concept of pH in organic or non-aqueous solutions is not trivial. It is called apparent pH or operational pH when organic solvents are present. There is a practical problem here as well, do you think your buffers will dissolve in ethanol? Once you figure this out, you can read a section in  "pH measurements in Nonaqueous Solvents" by Laitinen & Harris, “Chemical Analysis ─ An Advanced Text and Reference". It is a very useful book.


How is sulfuric acid with a concentration of 12 M also 72% w/w?

Your problem is that you've assumed the density is 1 gram/mL.

Remember that a molar is defined as a mole of solvent per liter of solution, not solvent. Usually, in introductory chemistry classes, we skip over the fact that adding solute to a solution increases its density, because it makes life more complicated.

As you just found out though, sometimes you really need to account for the change in density that occurs as you add solute. This webpage claims that the density of $12~\mathrm{M}$ sulfuric acid is about $1.634~\mathrm{g/mL}$.

Using this updated calculation, we still have the original $1177~\mathrm{g}$ of sulfuric acid. However, the solution mass is now $1000~\mathrm{mL}\cdot 1.634~\mathrm{g/mL} = 1634~\mathrm{g}$. Finally, we obtain that

$$
\frac{ 1177~\mathrm{g} \, \ce{H2SO4}}{1634~\mathrm{g} \, \mathrm{solution}} = 72.03\% \; \mathrm{w/w} $$

A note about this calculation: it's possible we're begging the question, since I don't know which values on that linked page are measured and which are calculated. It's possible that the density was calculated from measured data. The important thing here isn't how the data was obtained though, it's that the density of solution changes as you add more solute, and so you have to change how you calculate things from the molarity.


You ask:

  Does a gold detector exist?

Absolutely. Here's a metal detector that is fine-tuned for gold detection, available ${\it via}$ Amazon.

That said, if you are looking for something that detects only gold, I'm not sure the best way to proceed in that case. This site appears to be a comprehensive review of some popular precious metal-tuned detectors on the market. The site is maintained by a person with an active interest in prospecting and metal detection. It is stated on the linked page that:

  A quick note to those who know nothing about these machines. These are metal detectors. There is no such thing as a "gold only" detector. These detectors will also find lead, copper, aluminum, and other metals. These units are best used to look for relatively larger pieces of gold at relatively shallow depths. Concentrations of gold dust are not detectable. Some of these units can hit gold that weighs as little as a grain (480 grains per ounce) or less but only at an inch or two. Only the larger nuggets can be found at depths exceeding a foot. Only world class nuggets weighing many ounces can be detected at over two feet. The vast majority of nuggets found are found at inches, not feet.

In any event, you can purchase a device which has been engineered to discriminate base metals from precious ones (such as the model I have linked here).

  Also, does a metal detector detect gold?

Yes - see the reviews for the product offered at Amazon (and many others like it) as well as the informational site I've linked (above). Also see the Wikipedia page on metal detectors for more information, including the issues relating to fine-tuning the devices to discriminate between metals with similar/different properties "detected" by the devices.


How to identify a pseudo-first-order reaction?

There are four steps:

Identify the mechanism.
Find the rate-determining step.
Write the corresponding rate equation according to the stoichiometric coefficients of that step.
Remove abundant reactants.

1. Identify the mechanism

Assuming that there is no acid to catalyze the reaction, it proceeds via a nucleophilic addition/elimination reaction.

An acetic anhydride molecule $(1)$ is attacked by the lone pairs on the water to form the complex $(2)$. This is the addition step.

The complex tautomerizes to give $(3)$, which is converted to our products, namely 2 acetic acid molecules $(4)$. This is the elimination step.

2. Find the rate-determining step.

Presumably, it is the first step.

3. Write the rate equation.

In the first step, one molecule of acetic anhydride reacts with one molecule of water. Therefore, it is a first-order reaction with respect to both of them:

$\text{Rate} = k [\ce{Ac2O}] [\ce{H2O}]$

4. Remove abundant reactants.

Water is abundant, so it is removed from the equation:

$\text{Rate} = k [\ce{Ac2O}]$

Conclusion

It is a pseudo-first-order reaction [A].


Conceptual understanding of solution / solute in a question

Mole fraction is the ratio of the number of mole of a component to that of the number of moles of the total mixture. In your case, 6.853g of NaCN equals 6.853/49.002 = 0.139 moles.

The rest is water i.e. 107.166 - 6.853 = 100.313g. this corresponds to (100.313/18.015) = 5.568 moles.

Now, total moles = 0.1398 + 5.568 = 5.708moles.

The mole fraction of NaCN = 0.139/5.708 = 0.024.


If dS ≥ dq/T for a spontaneous change is dS ≤ dq/T for a non-spontaneous change and is that even possible?

As you perfectly say, Clausius inequality states that $$\mathrm{d}s\geq \mathrm{d}q/T.$$ This has to be true for any process.

For reversible processes, what happens is that $\mathrm{d}s= \mathrm{d}q/T$, and for non-reversible processes, $\mathrm{d}s > \mathrm{d}q/T$. If for a given process $\mathrm{d}s< \mathrm{d}q/T$, it would not fulfill Clausius inequality.

Consider the situation where the system is isolated from its surroundings. In this case, $\mathrm{d}q_\mathrm{(sys)} = 0$ , and thus the Clausius relation implies $\mathrm{d}S\geq0$, so this tells us that the entropy of an isolated system cannot decrease in the course of a spontaneous change.

Since the universe is itself an isolated system, this result shows that we can use entropy as the signpost of spontaneous change. Processes are only spontaneous if they cause an increase in the total entropy of the universe.


Why is chemical accuracy defined as 1 kcal/mol?

Short answer: the goal of "thermochemical accuracy" for computational chemistry is to match or exceed experimental accuracy. Thus, ~1 kcal/mol comes from the typical error in thermochemical experiments.

The drive began with John Pople, who begin the modern effort to consider "Model Chemistries," comparing the accuracy of different methods across many molecules and often multiple properties. He realized that for thermodynamic properties, one could approach the accuracy of experiments. (See, for example his Nobel lecture).

  As the model becomes quantitative, the target should be that data is reproduced and predicted within experimental accuracy. For energies, such as heats of formation or ionization potentials, a global accuracy of 1 kcal/mole would be appropriate.

He then started work on composite methods like G1, G2, G3, etc. that could approach predicting many chemical properties to this accuracy.


Why can enolizable aldehydes undergo Cannizzaro reactions?

There is never an either–or in chemistry; it is typically a both instead. In the case of a hydroxide ion approaching an aldehyde with an α-hydrogen, it can both deprotonate the hydrogen creating the corresponding enolate and attack the carbonyl nucleophilicly to generate the required intermediate for a Cannizzaro reaction.

Most of the time, if there are two (or more) possible pathways and neither of them is intrinsically less hindered due to a strong reason, both pathways will be taken.


It's just adding up to the existing answer since I don't have enough rep to comment. 0.000609 is the no. of moles of limiting reagent. You don't multiply it by it's stoichiometric co-efficient because it is the no. of moles you were given. Stoichiometric equations just signify the proportions in which compounds react.


How does the production of pure, crystalline silicon via the Czochralsky or Siemens process?

How does trichlorosilane $\ce{HSiCl3}$ help in the purification of silicon?

Trichlorosilane is a highly volatile compound with a boiling point of 32 °C. This means that it can be destilled off the raw material and separated from impurities.

The purified trichlorosilane can either be decomposed thermally

$$\ce{4 HSiCl3 ->[\Delta] Si + 3 SiCl4 + 2 H2}$$ or reduced with hydrogen

$$\ce{HSiCl3 + H2-> Si + 3 HCl}$$

This does give you rather pure silicon, but it is, at best, polycrystalline.

What's the Czochralski process good for?

Note that this in't just a melting/cooling cycle!

The crucial point in the method is the use of a rotating, monocrystalline seeding crystal, dipped into (previously purified) molten silicon. The goal is to obtain huge, monocrystalline silicon rods of defined diameter, that can be used in the production of wafers.


In the Wikipedia article Phosphorescent paint,

  Phosphorescent paint is commonly called "glow-in-the-dark" paint. It is made from phosphors such as silver-activated zinc sulfide or doped strontium aluminate, and typically glows a pale green to greenish-blue color. The mechanism for producing light is similar to that of fluorescent paint

The process is summarised in the article Fluorescence vs. Phosphorescence,
The explanation from the article:

  Phosphorescent materials produce light in a similar way as does fluorescence materials. A visible difference between these two types of luminescence, the ability of phosphorescence materials to glow after the excitation energy source is removed.

The website has a diagram exhibiting the process (in a bit of a simplistic, but understandable way - just replace the 'UV' with 'car headlights etc'):

Essentially, photons are emitted from every step back from the excited state to the ground state, unlike fluorescence, where the photon is emitted when the electron goes straight to the ground state.


Why is HF considered a covalently bonded compound?

I don't think I've ever seen a reputable source claiming the bonds in HCl, HBr and HI are mainly ionic. The fact that they are all gasses at room temperature clearly suggests otherwise. HF has the most ionic character of all the hydrogen halides, but it too has a rather low boiling point (just below room temperature), which is uncharacteristic of ionic compounds.

It is best to think of all hydrogen halides as covalent polar molecules, with the polarity increasing in the order HI < HBr < HCl < HF, as suggested by the electronegativity differences.


Why is a set of p-functions added to the minimum basis sets of Li and Be?

By your definition, any minimal basis set would only include as many orbitals (including spin state) as there are electrons. This implies that each atom would be described as a single determinant and therefore the entire system would be described by a single determinant. Without unoccupied orbitals, there is no binding. Hence, a minimum basis set has to fulfill a few more constraints. A common requirement is the inclusion of the full valence shell and invariance under spin-flip. For second period elements, this includes p-orbitals.


How can I convert the number of particles per liter to gram per liter

$$x \frac{particles}{L} \times \frac {1 mol}{6.023 \times 10^{23}particles}\times \frac{MW gm}{mol}$$


Can we find 100% probability region in an orbital if the position of an electron is limited by speed of light?

The (Wave)functional forms for the surface which contains the 100% probability for finding the electron extends to the literal edge of the universe.  In the context of 'wave' it is nonsensical to measure an electron's position, and hence its speed may not be determined.

In the context of particle, an electron, having nonzero mass, may not be accelerated to the speed of light.


Calculation of fluorescence quantum yield

After some more reading, it turns out that the protocols suggesting to use the integrated intensity are wrong and it is necessary to correct for the variable wavelength across the emission peak. Würth et al. (2013) provide a description of the correct method. First, they state that

  all fluorescence spectra have to be converted to a (relative) number of photons per unit time

Second, they describe the conversion as discussed in the question:

  the radiometric quantity has to be multiplied with $\lambda/(hc_0)$


How to calculate the sufficient ratio of acid and corresponding base of the indicator that allows to clearly observe the endpoint of a titration?

If you assume that the $pK_{a}$ of the indicator is 7.0 and you assume that you're going to the end point at pH 8.5 then the Henderson Hasselbach equation

    ${pH = pK_{a} + log \frac{[In^-]}{[HIn]}}$

can be rearranged to:

    $\frac{[HIn]}{[In^-]} = 10^{(pH - pK_{In})} = 31.62+ \approx 32$

However:

(1) There is no real statement in the problem that $pK_{a}$ of the indicator is 7.0 which was assumed from:

    $\dfrac{1}{2}(5.5 + 8.5) = 7.0$

There is little real chemical validity behind this assumption.

(2) There isn't a real justification of why color at pH 8.5 is any better than color at pH 5.5. So the ratio is either 32 or 1/32. Flip a coin :-(


Look for superabsorbent balls whose expansion is reversible

Yes. The ones I have are known commercially as Jelly Marbles or Orbeez. They're sold at gardening stores, home decor, and online. They're used as toys but also as gardening products that absorb excess moisture in wet conditions and supply water in dry conditions. Mine grow up to about 1.5cm from a starting diameter of ~0.1cm. I've been told that some grow up to ~7.5cm; and they're advertised as increasing by 100-300 times in volume at full saturation. The process is reversible (e.g. by putting them in a cup with a wad of paper towels). I use them as a drying agent where powdered $\ce{CaCl2}$ etc. aren't ideal.

Note the actual identity of the polymer in the beads is a trade-secret, so I don't think you'll be able to find that information.


Condensation Rate is Increased with a Decrease in Temperature?

The problem in your question is the over-simplified statement that the rate of condensation is simply proportional to the concentration of the gas.  That is only part of the very complex picture here.

The primary key to condensation rate is the degree of supersaturation of the water vapor at the temperature of the object on which the water is condensing.  Under conditions of sub-saturation, i.e. the ambient water vapor concentration (vapor pressure) is less that the vapor pressure of water at the temperature of the object, no bulk (excluding molecular layers of chemisorbed water) condensation occurs.  This of course make your equation R = k[Gas] completely false under these conditions.

Once the partial pressure of water vapor exceeds the saturation vapor pressure at the temperature of the object, i.e. the relative humidity exceeds 100% (by decreasing the temperature), then bulk condensation of water can occur.  Because our atmosphere has an excess of water condensation nuclei, relative humidities of greater than 102% rarely occur.  So, at a given temperature, the condensation rate is roughly proportional to the partial pressure of water for relative humidities of 100-102% at the surface of the object.  Note that the relative humidity increases with decreased temperature. Under these specific conditions, your equation is approximately correct.  I say approximately because there are other factors like air movement (wind velocity), and kinetic limitations like the rate at which water diffuses across the nm or so layer of air at the surface of the object at which the wind velocity approaches zero.

Further complicating the issue is the temperature dependence of the water vapor concentration at 100% humidity.  In other words, there is much more water vapor in the air at $\mathrm{30^oC}$ and 100% humidity vs. $\mathrm{5^oC}$ and 100% humidity.  This means that the amount of water vapor available for condensation at any given degree of supersaturation, say 101% humidity, is a function of temperature.

Because of the complexity of this issue my answer has been largely qualitative in nature.  For a thorough discussion of this issue (albeit even more complex as it includes discussions of water vapor-to-ice condensation), see this article, "Supersaturation of Water Vapor in Clouds", Journal of Atmospheric Science, December 15 2003, p. 2957.


Is it possible to apply reaction progress kinetic analysis on non catalytic reactions?

Apparently the method works normally with non catalytic reactions according to the example presented in a Mettler Toledo webinar (Titled: Reaction Progress Kinetic Analysis, https://www.mt.com/us/en/home/library/on-demand-webinars/automated-reactors/kinetic.html) where a simple Diels-Alder reaction is analyzed by RPKA.


Reaction of a non-metal with alkali to produce hydrogen gas

This is the extension of Alaska Ron's answer. All reactions taken from this site.

Group 13 elements

All group 13 elements except indium and thallium reacts with aqueous alkali to produce a alkali salt and hydrogen. I'm taking the example of aluminium.

$$\ce{2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2}$$

$$\ce{2Al + 2(NaOH•H2O) ->[400-500°C] 2NaAlO2 + 3H2}$$

$$\ce{2Al + 6NaOH ->[450°C] 2NaAlO2 + 3H2 + 2Na2O}$$

$$\ce{2Al + 6H2O + 6NaOH → 2Na3[Al(OH)6] + 3H2}$$

Group 14 elements

Silicon react with sodium hydroxide to produce sodium orthosilicate and hydrogen.

$$\ce{Si + 4NaOH → Na4SiO4 + 2H2}$$

Sodium hydroxide react with silicon and water to produce sodium silicate and hydrogen.

$$\ce{2NaOH + Si + H2O → Na2SiO3 + 2H2}$$

Tin and lead however forms a different product but eventually gives hydrogen gas. Lead only gives the first reaction.

$$\ce{Sn + NaOH + 2H2O → Na[Sn(OH)3] + H2}$$

$$\ce{Sn + 2NaOH + 2H2O → Na2[Sn(OH)4] + H2}$$

$$\ce{Sn + 2NaOH + 4H2O ->[\Delta] Na2[Sn(OH)6] + 2H2}$$

Group 15 elements

Arsenic react with sodium hydroxide to produce sodium orthoarsenate and hydrogen.

$$\ce{As + 3NaOH ->[\Delta] Na3AsO3 + H2}$$


Side product obtained in reduction of cuprous oxide with carbon

@ Paras Khosla. Your first equation is correct, but not your second one : $\ce{Cu_2O}$ reacts with $\ce{C}$ without getting decomposed previously into $\ce{Cu + O_2}$. $\ce{Cu_2O}$ is more stable than $\ce{CuO}$ above $1000$°C
Furthermore, $\ce{Cu_2O}$ does not react easily with Carbon in the solid state. The contact between the reacting solids is not good. The contact is better if $\ce{Cu_2O}$ is liquid. But $\ce{Cu_2O}$ melts at $1026$°C. So, the reaction starts at this temperature, forming $\ce{CO}$ which is more stable than $\ce{CO_2}$ at such a high temperature.


No, pure malachite $\ce{Cu2CO3(OH)2}$ usually doesn't produce hydrogen because it is a copper carbonate. Copper is already in a higher oxidation state (+2) so it can't reduce hydronium to hydrogen. The right reaction is the following:
$$\ce{Cu2CO3(OH)2 +2H2SO4 -> 2CuSO4 + CO2 ^ + 3H2O}$$

However, if there are traces of metallic copper some hydrogen can be produced but not with sulfuric acid because:

$$\ce{Cu + 2H2SO4 -> CuSO4 + 2H2O + SO2 }$$

Nor with $HCl$ (copper doesn't react with $HCl$)  but only with nitric acid or aqua regia you can produce hydrogen from the metallic form. So, in fact, the gas you see is carbon dioxide.


Color imparted to flame and color of coordination compounds

You seem to be mixing up a term. Let me get that out of the way first.

Ionisation is the name for a process creating an ion from a neutral species. However, if you heat an ionic substance and break the ionic bond, you are not ionising but deionising (although that term is usually used for a different process). The reaction equation for sodium chloride would be the following:

$$\ce{NaCl ->[\Delta] Na + Cl}$$

Now to the actual question. The perception of colour is based on one of two principles (as you noted):

out of white light, certain wavelengths are absorbed leaving behind a remaining spectrum and causing the perceived colour to be complementary to that of the absorbed light
light of a certain wavelength is emitted, directly creating the impression of said colour.

As you also noted, this could mean that we see up to two colours in the same electronic system: One due to the absorption and the second, complementary one during emission. So what is it that we see?

For coloured compounds at room temperature — that includes, but is not limited to transition metal salts and coordination complexes — we typically see absorption colour. Take for example the hexachloridoferrate(III) anion $\ce{[FeCl6]^3-}$: It has an intense yellow colour which can be explained by the absorption of very short visible wavelengths. The shortest visible wavelength is a blue one, so the perceived colour is yellow. The colouring is due to ligand-to-metal charge transfer transitions. In the corresponding fluoridocomplex $\ce{[FeF6]^3-}$, the orbital energies of the fluoride ions is predicted to be lower, giving a higher transition energy and shifting that transition just into UV wavelengths — the hexafluorideferrate(III) complex is colourless.

Also in organic chemistry, there is a number of yellow-ish compounds which occur because they just absorb at high enough wavelengths to shift out of the UV range. (UV absorption is very common for organic compounds containing a phenyl ring.)

We only see one colour, however. This is because the absorbed energy — which immediately excites an electron into a higher orbital — is liberated into the surroundings as vibrations and ultimately heat. The electron does not fall back into the ground state in a single transition but via a number of intermediates loosing energy all the way; hence we do not see a corresponding emission colour.

In flame colouring, one typically takes white (or colourless) compounds and heats them in a flame (the flame ideally being colourless, too). Only in the heat of the flame do they show their colour — and in a typical ion-lottery (qualitative inorganic analysis) lab, this is done in a dark corner of the lab, because there is actual emitted light. What is happening here?

Through the heat of the flame we are transferring energy into the compound, first causing the bonds to break (liberating atoms) and then also causing these atoms to vapourise so that we end up with gaseous metal atoms and gaseous non-metal atoms. Further heat is introduced which — if it is enough — will finally cause the transition of an electron into a higher orbital. This is, of course, unstable, but here the relaxation is immediately back down into the ground state for a significant percentage of these atoms; each of them releasing a photon with exactly the wavelength of the corresponding transition. And what we see are those photons, corresponding exactly to the wavelength of the energy difference between two orbitals.

And finally, just to confuse you, I am going to quote an experiment I saw in my school’s biology course. We had isolated chlorophyll from leaves of a plant and had it in a solution in an Erlenmeyer flask. It appeared green. Our teacher held it into the strong light beam of an overhead projector and it appeared as red. The experiment was explained as follows:

Red light is absorbed by chlorophyll, causing a green impression under normal conditions.
in the strong light beam, so much red light was absorbed and so many chlorophyll molecules were excited that the energy had ‘nowhere to go’; thus, the same red light was emitted to all directions and we were able to see that red colour.

So to sum it up:

in coloured salts at room temperature, we see the complementary colour of an absorbed photon. We do not see the emission (and thus the actual colour) because it is too unfrequent; most energy absorbed is released as vibrations and heat.
in flame colouring, we have highly excited atoms and we do not actually see the absorption since the energy necessary is transferred by heat. However, we do see emitted photons of exactly the energy difference wavelength from a significant number of electrons falling back into a lower energy state. We see the actual colour but not the complementary one.


How to measure corrosion rate of an iron nail in salt water?

There is an ASTM standard(G01) that gives detailed descriptions of cleaning procedure for corrosion measurements.

For iron and iron oxide, the recommended way is to leave the sample in a cleaning solution(e.g a dilute acid solution), followed by a light brushing repeatedly until the weight stabilizes.

You could also try sonication in a solvent to see if that helps.

In any case, to make sure you are not losing significant amounts of the base metal, you should repeat the cleaning procedure on some uncorroded control sample. This will help you dial in the proper procedure for your samples.


Why are the sulfide ores generally converted to an oxide and then reduced?

Why, it's simple: you heat a metal oxide with carbon, and you get the metal you were after, plus a byproduct of $\ce{CO}$ or $\ce{CO2}$ which easily flies away. Coal is still abundant on our planet.

Try to pull that with a sulfide! I guess the reaction will not go in the first place, since $\ce{CS2}$ is a great deal less exothermic(*) than $\ce{CO2}$. In the unlikely case if it will, you'll have to deal with the byproduct. $\ce{CO2}$ can be simply dumped in the air (though this is no longer free). Dump $\ce{CS2}$, and all your neighbors will pay you an extremely unfriendly visit, possibly armed with pitchforks.

(*) So much so as to be endothermic, in fact.

Copper is less active, so instead of going via oxide, we have a chance to go straight to the metal in one step. Would be stupid not to use it.

Then there is aluminum which is not reduced like that, but then again, we don't convert it from sulfide to oxide either, primarily because we don't have much sulfide around. Should it be otherwise, it isn't immediately clear would we choose to convert it or not. Arguably, $\ce{Al}$ has no form which is suitable for reduction.

So it goes.


Calculating enthalpy change and energy required to convert lead sulphide to lead

Part 1:

It's easier to imagine \begin{equation}\ce{A + B -> C + D} \tag{1}
\end{equation}
\begin{equation}\ce{D + E -> F + G} \tag{2}\end{equation}
\begin{equation}\end{equation}
Now if we wanted to write a 3rd equation where both the reactions 1 and 2 happen, we'd get
\begin{equation}\ce{A + B + E -> C + F + G} \tag{3}\end{equation}
Why?

D is both a reactant and a product. So, it's a reaction intermediate. It's like it's produced in 1 and again consumed in 2. So, if both reactions take place, it's neither on the reactants' side nor the products.
The conventional way of writing reactions doesn't show the mechanism of the reaction. You have to image the reaction happening, to completely understand why D isn't written.

So, as the third reaction is reaction 1 and 2 happening at the same place, $\Delta H_{r_3}$ is the algebraic sum of $\Delta H_{r_2}$ and $\Delta H_{r_1}$. Thus, your final result is correct: $$\mathbf{-413.7 ~kJ \times mol^{−1} +106.8 ~kJ \times mol^{−1}=−306.9 ~kJ \times mol^{−1}}$$

Part 2:

We have to take a more careful look at the unit of $\Delta H$. What does it mean? It means that for each mole of the reactant ($\ce{PbS}$), one mole of product ($\ce{Pb}$) is formed. (With regards to their stoichiometric coefficients in reaction 3) So, we've got to simply 'punch' the numbers in the formula $n= \frac{m}{M}$ (Where $n$ is the number of moles, $m$ is the given mass and $M$ is the molar mass) And then multiply that number by the $\Delta H_{r_3}$. Thus, your final answer is correct too.


Reaction of magnesium with the surrounding air

I believe that magnesium exposed to air quickly ends up with a coating of magnesium oxide on its surface.  This oxide coating is quite uniform and does not allow any further oxygen to reach the surface.

Several metals, but NOT iron, have this property of a protective oxide coating.  In iron the coating forms a spongey surface easily penetrated by further oxygen.


Are there any sodium salts that are insoluble in water?

From here:

  Sodium tends to form water-soluble compounds, such as halides,
  sulfate, nitrate, carboxylates and carbonates. But examples of many
  insoluble sodium compounds such as sodium bismuthate ($\ce{NaBiO3}$),
  sodium octamolybdate ($\ce{Na2Mo8O25·4H2O}$), sodium thioplatinate
  ($\ce{Na4Pt3S6}$), and sodium uranate ($\ce{Na2UO4}$) is found in
  nature.

Other examples include:

Feldspars(aluminosilicates of sodium, potassium, calcium)
Sodium metaantimonate, $\ce{NaSbO3.7H2O}$ in the pyro form $\ce{Na2H2Sb2O7.H2O}$ has a solubility 0.7g/L.
Sodium metaphosphate, $\ce{NaPO3}$ has both soluble and insoluble form.


What is Sodium Chloride like in gas state?

No, in the gas phase sodium chloride exists as a monomer (the sodium chloride molecule) along with its dimer $\ce{Na2Cl2}$.  The dimer has a roughly rectangular shape and is quite floppy with chlorines located diagonally across from each other.  The dimer makes up about 27% of the mix.  All of the bond lengths, etc. can be found in this thesis.  Go to the end and you'll see the full paper that appeared in JACS with all of the data.


If I mix carbon dioxide and sodium hydroxide in the right amounts in water, do I get baking soda?

This is certainly one method by which sodium bicarbonate can be produced.  According to this Wikipedia article:

$\ce{NaHCO3}$ may be obtained by the reaction of carbon dioxide with an aqueous solution of sodium hydroxide. The initial reaction produces sodium carbonate:
$$\ce{CO2 + 2 NaOH → Na2CO3 + H2O}$$
Further addition of carbon dioxide produces sodium bicarbonate, which at sufficiently high concentration will precipitate out of solution:
$$\ce{Na2CO3 + CO2 + H2O → 2 NaHCO3}$$

The net reaction for this two-step process is then simply:
$$\ce{CO2 + NaOH -> NaHCO3}$$
Note that this is not the common method by which sodium bicarbonate is produced commercially, but it is a perfectly valid method of it's synthesis.


Details of Boltzmann distribution derivation

What you say is a good idea, but is not quite correct because we must maximize the distribute subject to two constraints. I will reiterate the derivation following along with McQuarrie's Statistical Mechanics.
Constraints in the Ensemble:
The constraints I mention are actually the properties that determine the ensemble we are working with:
$$
\sum_ja_j=A
$$
$$
\sum_j\epsilon_ja_j=E
$$
The first constraint says that sum over all states of the population numbers for each state $j$, given by $a_j$, must equal the total number of members in the system $A$. This is the number of molecules if you like.
The second constraint says that the energy of each state, $\epsilon_j$, multiplied by the population of that state, $a_j$, must equal the total energy, $E$, of the system when the sum is taken over all $j$. These constraints are so intuitive physically that it is easy to forget they even have to be enforced.
By the way, we are not actually constraining the total energy. We only constrain that whatever the energy is, it can be found from the second constraint. In fact, in McQuarrie, this derivation is undertaken for the canonical (NVT) ensemble.
Method of the Most Probable Distribution:
Now, as you say, the distribution which we primarily observe will be the one which maximizes the number of ways to arrange the members of the ensemble. Notice I say primarily observe. This is discussed below. This number of ways is given by the equation you write down,
$$
W(\mathbf{a})=\frac{A!}{\prod_ka_k!}
$$
There are many possible distributions that satisfy the constraints we have given. Given this indeterminate (but probably infinite) number of ensembles which satisfy our constraints, the probability of being found in state $j$ is given by,
$$
P_j=\frac{\bar{a}_j}{A}=\frac{1}{A}\frac{\sum_{\mathbf{a}}a_j(\mathbf{a})W(\mathbf{a})}{\sum_{\mathbf{a}}W(\mathbf{a})}
$$
Notice that we are now summing over the vector of all the quantum states for each ensemble, $\mathbf{a}$, because we want to find $\bar{a}_j$, the average population of quantum state $j$ in each different ensemble satisfying our constraints.
This is clearly not very useful to us because there are so many possible ensembles we will never be able to do something with them. We can, however, take advantage of the form of $W(\mathbf{a})$. Namely, by just playing around with the numbers, you will find that the multinomial coefficients are very strongly peaked at their maximum when all of the $a_j$'s are large. Fortunately, this is exactly the regime we live in because we are thinking about something on the order of $10^{23}$ particles, so the population of states will be very large.
This means we can make an approximation, which is formally exact in the limit as $A\rightarrow\infty$, that $\bar{a}_j/A=a_j^*/A$ where $a_j^*$ is the $a_j$ in the distribution which maximizes $W(\mathbf{a})$. So, the fact we are going to maximize this distribution comes about through an approximation, albeit one which becomes exact in the limit near where we are working.
So, we can simplify our equation for the probabilities to,
$$
P_j=\frac{1}{A}\frac{a_j^*W(\mathbf{a^*})}{W(\mathbf{a^*})}=\frac{a^*_j}{A}
$$
Method of Lagrange Multiplies:
We want to find $a_j^*$ which means we must perform a maximization subject to the two constraints we listed at the beginning. Maximizing something subject to constraints does not seem to be easy, but there is a very nice method for doing it called the method of Lagrange multipliers. Essentially, we introduce some undetermined coefficients to the maximization equation so that the constraints are included, but in reality we are just adding zero to the equation.
What I mean by this is that I can rewrite the constraints as follows:
$$
\alpha\left(\sum_ja_j-A\right)=0
$$
$$
\beta\left(\sum_j\epsilon_ja_j-E\right)=0
$$
where $\alpha$ and $\beta$ are the so-called Lagrange multipliers. Then, we can simply maximize our distribution as normal:
$$
\frac{\partial}{\partial a_k}\left(\ln W(\mathbf{a^*})-\alpha\left(\sum_ja_j-A\right)-\beta\left(\sum_j\epsilon_ja_j-E\right)\right)=0
$$
It really isn't obvious that this should maximize the system while satisfying the constraints, so you will have to read elsewhere why this works, but the fact is that it does work.
Notice, I have actually played a trick. Instead of maximizing $W(\mathbf{a^*})$, I am maximizing $\ln W(\mathbf{a^*})$. This is because $W(\mathbf{a^*})$ has factorials in it, so by taking the logarithm we can employ Stirling's approximation. This is also more or less exact in the limit of a large number of particles, which we always satisfy. The reason we can do this is because the logarithm of a function will have a maximum at the same coordinate as the function itself and the logarithm is smooth such that will not somehow create another maximum or something which would spoil the trick.
After taking the derivative, we lose the constant and every term in the sum where $k\ne j$. So, we have,
$$
-\ln a_j^*-\alpha-1-\beta E_j=0
$$
Or, after rearranging and solving for $a_j^*$:
$$
a_j^*=e^{-(\alpha+1)}e^{-\beta E_j}
$$
Putting it All Together:
So, all that remains is to solve for the undetermined mutlipliers and plug it all back into our probability statement, $P_j=a_J^*/A$.
This can be done very simply (and we actually don't care about the mutliplier $\alpha$). If we take the sum over all $j$ of the equation we wrote for $a_j^*$, we realize that the left side is equal to $A$ due to our first constraint and we can rearrange to write $\alpha$ in terms of $\beta$:
$$
e^{\alpha+1}=\frac{1}{A}\sum_je^{-\beta E_j}
$$
Then, we can find the final probabilities from,
$$
P_j=\frac{a_j^*}{A}=\frac{e^{-(\alpha+1)}e^{-\beta E_j}}{A}=\frac{A}{\sum_je^{-\beta E_j}}\frac{e^{-\beta E_j}}{A}
$$
We finally arrive at the final probability distribution which defines Boltzmann statistics:
$$
P_j=\frac{e^{-\beta E_j}}{\sum_je^{-\beta E_j}}
$$
Thus, the only one of these Lagrange multipliers we care about is $\beta$. We have not actually found what this multipliers is yet, but I will stop here as it should be no surprise that the multiplier is given by $\beta=1/k_bT$.

So, I believe the above answers your question 1. That is, you have to worry about the constraints associated with you ensemble so what you mention does not quite work.
As to your second question, it is not quite true that translational energy is continuous. Pretty much everyone pretends it is, but kinetic energy is quantized. It is, however, very rare that the boundaries a particle is confined to are sufficiently small (and impermeable) that quantization of the kinetic energy is noticeable. Also, because there are many of these states which are closely spaced, one must reach very low temperatures for the quantization to be important.
What I have described above is the origin of Boltzmann statistics (not Maxwell–Boltzmann, which is often confused with it). This is a common confusion because of the famous Maxwell–Boltzmann distribution of velocities. I only bring this up because the Maxwell–Boltzmann distribution of velocities makes exactly the approximation which you describe by pretending the kinetic energies are continuous.
That is, if we simply plug the kinetic energy of a particle $E=1/2mv_x^2$ into our final result (and note the deonominator is the partition function), we get,
$$
P_{v_x}=\frac{e^{-\frac{mv_x^2}{2kT}}}{Q}
$$
Then, the partition function, $Q$, is just a normalization constant. Because the states of different kinetic energies are so closely spaced, we can elevate $P_{v_x}$ to a function, $P(v_x)$, and integrate over all velocities to determine the probability of finding a particle with a velocity between $v_x$ and $v_x+dv_x$. I will leave this as an exercise. Notice this is just an integral over a gaussian function.


Why acyl halides are not named as 1-substituted aldehydes?

Well, the rest of the substituted ethane in the chlorethane is still like an alkane, but the rest of to be 1-chlorethanal is not an alkanal in the first place. An alkanal must have the aldehydic group.

If there would be 1-chlorethanal, why not 1-hydroxyethanal ( aka known as the acetic acid  ), or 1-methylethanal ( aka acetone )  ?  :-)


Determine which molecule has stronger intermolecular force?

Look at the boiling points.  Interaction is based on geometry (surface area) and polarizability, among other things - and the phase: solid, liquid, vapor.  Molecular weight is not diagnostic.  SiH4 is "fluffy" versus methane being "hard."  Compare the melting and boiling points of decane and isodecane  versus adamantane, all being $\ce{C_{10}}$.  Compare the melting and boiling points of n-octane versus 2,2,3,3-tetramethylbutane, both $\ce{C8H_{18}}$.  Complicated.

You might check your MW for methane.  Bells and whistles should have gone off, $\ce{CH4}$ versus $\ce{SiH4}$.