Is there a known chemical that makes it possible to control the opacity of a large surface?

There are

thermochromic materials that change color in response to heat
photochromic materials that change color in response to light
and electrochromic materials than change color in response to an applied electric field.

All of these materials are sub-categories of Smart Glass. I suspect that electric field sensitive Smart Glass materials are what you are looking for.  There are several different types of such Smart Glass materials.

Those with suspended particles that are normally randomly aligned, but when an electric field is applied they align an alter the light transmission properties of the material.  These materials have already found commercial application (airplane) and are moving down to lower cost applications (autos)
Electrochromic devices where a molecule undergoes a chemical change between colored and non-colored states when a field is applied.
Liquid crystal materials dissolved in a polymer matrix are normally randomly aligned, but, when a field is applied they align thus altering the light transmission properties of the material.

There are also some non-electrochromic smart glass technologies that are also discussed in the Smart Glass Wikipedia article linked above.  These are generally lower tech and therefore lower cost.  They work based on

Having several aligned windows, but with the alignments in different directions and then manually moving the windows into place to adjust the opacity.
Having two reflective windows with a space in between.  When the space is filled with a material having the same refractive index as the reflective windows, the assembly becomes transparent.
Using multiple polarized panels and moving them into or out of place to adjust opacity.

Which method you choose will depend upon how quick you need the materials to change color, the level of manual intervention you're willing to tolerate and how much your willing to spend.  If you google "Smart Glass" you'll find that this field is already well advanced.


How do you list all possible reactions and intermediates of a reaction?

You've discovered a secret. In undergraduate chemistry classes, the field of chemical kinetics is treated with a level of detail very far from reality. The truth is that virtually any reaction you can devise is vastly more complicated than we would like to think. This is especially true for gas phase reactions like the one you write down because it is possible for very unstable intermediates to exist for long enough to collide with lots of different species or undergo an internal conversion (when the molecule is large enough to do so) which introduces minute components that could nonetheless make a big difference in the kinetics.

Some people actually attempt to study these reactions without just ignoring the small sideways paths that lead to weird intermediates or weird products. Generally they do this by using what is called a Master Equation. I would encourage anyone to read about these and a very nice overview can be found here: Miller, J. A., & Klippenstein, S. J. (2006). Master equation methods in gas phase chemical kinetics. The Journal of Physical Chemistry A, 110(36), 10528-10544.

What you'll find when reading this is that they care a lot about what the potential energy surface looks like everywhere. This is because one formulation of the Master Equation is as follows:$$\frac{dn_i}{dt}=\sum_j(p_{ij}\cdot n_j(t)-p_{ji}\cdot n_i(t))$$This basically says that we are looking at the change in probability of finding the system in a state $n_i$ as a function of time and we are doing this by describing the probability per unit time $p_{ij}$ of moving from state $j$ to state $i$ as well as the process which goes the other way. Note that $n_i$ is essentially the same as a number density for a large number of molecules, but strictly this is a probability.

You can imagine then that in order to use this equation (which isn't very usable in its current form), one would need a detailed picture of what possible intermediates can form immediately from the state the system is in right now.

Thus, the answer to how you write all these down is that you look very carefully at the state of any system and ask yourself, what intermediates could I possibly imagine forming? Many of these intermediates are very unstable at low temperatures, but these kinds of models are frequently used when studying combustion chemistry or flames, so the number of intermediates can becomes very large and the number of reactions those intermediates can undergo (the number of states $j$ you can jump to from state $i$) is even larger. To operationalize this, you would basically write out all known intermediates, from experiment or theory, and imagine what reasonable combinations you could make of those that eventually lead to your product.

In the video you link, some unusual intermediates are listed, such as $\ce{HO2}$, and it is said that this is a simplified model. Basically what that means is that they had even more possible combinations of intermediates, but for computational simplicity or good experimental reasons, decided that many combinations of those intermediates contributed negligibly to the reaction of interest.

One final note is that it's not really as simple as combining everything you can imagine because you have to consider where a given intermediate came from. For instance if a single molecule is formed by means of a bimolecular reaction, the angular momentum must be conserved. This can lead to situation where certain states you could imagine hopping to are not actually possible because it would violate conservation of angular momentum. Indeed, one could imagine this leading to some reaction mechanisms which are very different from what intuition would dictate. The paper I linked above notes that when angular momentum becomes an important factor, the intermediate usually undergoes an internal conversion to basically store that angular momentum before reacting with other things... So, when you're writing out all of these intermediates don't forget that $\ce{HO2}$ might very well exist as $\ce{HOO}$ or $\ce{OHO}$ (though that seems unlikely I only meant it as an illustration).

I hope this addressed the spirit of what you were really asking.


No, it would not.

$\ce{Al}$ is normally covered by oxide layer which effectively renders it inert to most things. Your cell would stop dead, as if the electrodes were made out of wood.

Then again, you might want to know whether it is possible to remove the said layer. Yes, it can be done, but then $\ce{Al}$ turns out to be just too active. It reacts with water quite readily (albeit not as spectacularly as $\ce{Na}$). Merry little bubbles of $\ce{H2}$ will pop up, flakes of $\ce{Al(OH)3}$ will drop to the bottom, and in a matter of an hour or so your electrode will be gone. While it lasts, the cell will produce some voltage; as Dr.J. correctly noted, it will be even greater than that of the normal $\ce{Cu/Zn}$ cell.

So it goes.


The radius of a sulfate ion is  0.242 +/- 0.007 nm. (Yirhak Marcus. Chem. Rev. 1988, 88, 1475-1498.)
In order to know if it would displace something could you tell us what exactly you are trying to displace?


Will Sodium Amide react with Epoxides in the same manner as Grignard Reagents?

It actually is possible but it works well only with simple epoxides so it is not wildly used. $NaNH_2$ is extremely basic (it can deprotonate alkynes) and too many functional groups can react with it: alkyl halides (elimination), aldehydes, ketones...

To work around that basicity, sodium azide $NaN_3$ is commonly used instead, followed by the reduction of the azido group into an amine.

For example, see this article.


How can I tell if a liquid is (not) emitting light?

In a photograph, a light source is always brighter than any surface it is illuminating.

Therefore, if the liquid is the actual light source, the liquid should be significantly brighter than the rest of the bottle. See for example this picture by David Mülheims CC BY-SA 2.5, via Wikimedia Commons:

In the given picture, however, the parts above and below the water level have a similar brightness.


How do I arrive at the α-trimethylsilyloxy ketone from an unmodified ketone using TMSCl, triethylamine and mCPBA?

In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.

In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:

γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system

By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.

If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.

The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.

Scheme 1: mechanism of the Rubottom oxidation.[1]

Reference:

[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.


Why does chemisorption decrease with increasing temperatures?

Let's answer this question with the help of some thermodynamics and kinetics. Chemisorption is a chemical process with $\Delta S<0$ and $\Delta H<0$. Thus, in simpler words it is an enthalpy driven exothermic equilibrium
Every chemical process is an equilibrium with varying extent of completion. So your idea of chemisorption not being an equilibrium is wrong. Every chemical reaction is theoretically reversible in nature. What we call irreversible reactions in general are such reactions which are heavily favoured on one side of the equilibrium.
Now,
$\ln K = -\frac{\Delta H}{RT}  + \frac{\Delta S}{R}$
It is evident from this above expression that the $K$ will decrease when we increase $T$ when $\Delta H<0$ and thus the aforementioned conclusion.
But this is only half the story. Like all other chemical reactions, chemisorption also needs some activation energy to overcome the transition state. At lower temperatures there are very few molecules which have the kinetic energy enough to overcome the activation energy. The fraction of these molecules increases with increase in temperature and thus increasing the extent of chemisorption.
Conclusion
Due to the combination of thermodynamic and kinetic effects the rate of chemisorption first increases with the increase in temperature but after a certain saturation point the thermodynamic effects start dominating and thus the extent of chemisorption starts to decrease.
The graph below might help you understand how the extent of chemisorption changes with change in temperature.

Here,
$x$ = Mass of adsorbate
$m$ = Mass of adsorbent


Why the valence shell configuration of Lead is 6s2 6p2?

This is convention. To be perfectly honest, this order is not always kept up. But it makes sense to think of the orbitals by shell (first 1, then 2, …) and not by energy level.

For lead and all elements of the p block, whether the d and f orbitals are fully populated or not does not matter, they do not take part in chemistry (only the s and p orbitals of the outermost shell do). Considering that, we can write the d and f orbitals right at the beginning where it is ‘clear’ they are not needed.

For elements of the d block, we actually are doing chemistry with the d electrons. So it is maybe more customary to actually put 3d behind 4s, just to state that they are important.

There is no physical meaning behind the ordering in either example.


What side reactions are suppressed by the use of TMSCl in the acyloin condensation?

The acyloin condensation of diethyl adipate (1) on your Wikipedia citation generates ethoxide that can cause a Dieckmann condensation (intramolecular Claisen condensation, structure 6). TMSCl scavenges ethoxide and silylates the enediolate 3 formed in the acyloin condensation. The process is a 4-electron reduction (see details below). Mild hydrolysis of 4 gives the acyloin 5.


Does there need to be a fume hood with acetone?

Unless you're using very large quantities of it or in a confined space, usually not. I've spent a fair bit of time washing glassware out with acetone at the sink (using a bin to capture the runoff for transfer to a waste container) and I never had issues. Though I don't recommend doing this yourself, I've also had several errant splashes directly on my skin when working with stuff from the hardware store--again, no problems.

Acetone is staggeringly nontoxic: the NIH suggests an 8500 ppm EEL, meaning that in emergency situations, it's considered acceptable to breathe air that is nearly one full percent acetone for up to one hour. Hell, the air in my house contains less water than that sometimes.

  If I can get a ductless fume hood, what am I looking for to be sure that it can effectively take care of the acetone problem?

Look at this link. If the acetone is heated in the box, you should be able to meet all the criteria for being able to use a ductless hood:

A limited number of different chemicals should be used.

Acetone only. Check.
No extreme heating should be carried out in the hood, i.e. acid digestion applications.

You only need to boil acetone. That's 56C, not even close to the boiling point of water. Check.
Modest chemical volumes should be used, around 500 mls or less per chemical.

Even if you need more than this (you shouldn't, 500mL of acetone is quite a bit), it'll mostly be sealed in a box, so you shouldn't have nearly that much escaping into the hood. Check.
Moderate chemical exposure times should be maintained.

I don't know exactly how long it takes to smooth a print, but I don't think it can really be more than a few hours. You're probably okay on this one.
Will the chemicals involved in your application be effectively filtered with the available carbon filters?

Yes. Acetone is so common I can all but guarantee that every company will have a filter that can capture it (probably whatever is used for other general organic compounds as well). In Labconoco's case, that's the organic vapor filter.
How often will you have to replace your filters?

I don't know, that'll be your club's responsibility.

Finally, make sure someone doesn't get the idea that the hood will stop all fumes and try to do a perchloric acid experiment using that hood--that will go badly.

Now, all this being said...

What you really need isn't to make this safe, what you need is to convince administration that you're doing things safely. This may have nothing at all to do with whether things are actually safe or not.

At my old school (where I was briefly president of the local ACS chapter), I could have put an experiment into a fume hood with no filter, not plugged the thing in, and gotten an okay from the administrators because they believed that fume hood = safe and no fume hood = unsafe no matter what else was going on.

I say this not to encourage you to try to trick your way out of this, but to point out that, unless your administration is scientific (i.e. actually understands science), then it's fairly likely you're going to have to jump through a few silly hoops in order to satisfy them.

Best of luck!


What is the difference between "reaction in both directions" and "equilibrium"?

As far as I know, they are used indistinguishably and interchangeably.
One or another could be chosen to emphasize whether you are talking about chemical equilibrium or just trying to show that a reaction can take place in both directions depending on the conditions (which can be understood as completely the same...).

I saw the $⇄$ symbol used to specify some conditions for the reaction to take place in one direction and other kind of conditions for the reaction to occur in the other direction, and $⇌$ if conditions are the same for both reactions

For example:

First case is not an equilibrium, but it's true that the reaction can go both directions using the appropriated reagents in each case. But that could not be expressed using $⇌$.

The second reaction is an equilibrium, both reactions are happening continuously under the same conditions.
It also could be expressed using $⇄$.

That is the only difference I can tell. Like Martin pointed out, IUPAC does not give more hints about it, so the question cannot be answered for 100% sure.


How do I know what the end products of the reaction of glucose and fructose?

There are three possibilities. The first is knowing the reaction. I.e., you just know because you rote memorised it, that one molecule of water will be lost and then you apply that.

The second is knowing the mechanism. If you take a good look at it, a water molecule must be displaced for the mechanism to work out. Anything else would generate something highly unstable.

And finally, if you know the product structure, you can calculate double bond equivalents. Both reactants have a single double bond equivalent: $(2 \times 6 - 12)/2 + 1 = 1$. This corresponds to the ring in the reactants’ structures. The product has two rings, thus it requires two double bond equivalents. The number of carbons must stay the same, though. Thus: $(2 \times 12 - n)/2 + 1 = 2$; rearranging gives $n = 22$. Two hydrogens must be ejected and that biochemically typically happens with water.


What is meant when people say a molecule has high multi-reference character?

When doing a CI calculation based on Hartree-Fock orbitals, then usually the HF configuration has a very high weight in the CI vector (>90%). The next largest weight is then very small (for example only 1%), and there is a very large number of those very small configurations.
Such cases are called weakly or dynamically correlated. Because of the large weight, it is reasonable to optimize the orbitals for the leading configuration only and then run some truncated CI calculations (typically CCSD) on top.

If there are multiple configurations with similar (or same) weight, then it is no longer reasonable to optimize the orbitals for just a single configuration, because this will introduce a bias towards this configuration. The number of these configurations is rather small.
Additionally, those configurations are not necessarily related by Single or Double excitations, so CISD may not include all. However, they are commonly generated by an active space of just a few orbitals.
This is called strong or static correlation, and is typically found in systems with partially occupied close-degenerate orbitals, e.g. d orbitals in transition metals.

In FCI this orbital bias would not matter, because the basis is (numerically) complete. But in truncations like CISD or CCSD this leads to a qualitative error. Some important configurations are poorly represented (due to the bias), other are completely missing (due to the truncation).

A CASSCF calculation resolves the bias by considering all important configurations of similar weight (by chosing an appropriate active space) and balancing the orbitals to all of the considered configurations. Since only some configurations are required for static correlation, it is often feasible to optimize both (CI and orbital coefficients) simultaneously.

The term multi-reference then refers to a calculation where all (or most) of the CASSCF configurations are used to generate excitations from.


What does x mean in (1-x)A – xB composite materials, molar ratio, weight ratio or volume fraction?

Here, $x$ is a variable $(x\in[0;1])$ denoting mole fraction of a given constituent in a phase of variable composition (non-stoichiometric compound).
This is an algebraic way of denoting a given compound from the phase diagram for a solid solution: $x$ corresponds to its abscissa axis.

And no, volume fraction is not equal to mole fraction and is overall a very inconvenient variable to use, so it's usually being avoided.
For example, volume fraction would deviate for the different morphologies and allotropes of the same solid, which is not useful for the description of chemical composition, not to mention it's also temperature-dependent.


Nano zinc oxide is a real material used in sunscreens and looks like this.  It is just little crystals of zinc oxide which are about 30 nm across.

Chelate refers to something which is produced by a chemical process of chelation.  Chelation means that a metal atom has more than one chemical bond to surrounding organics.  (That is kind of a bad definition, but I'm trying to keep the lingo simple.)  Chelation can be used to produce zinc oxide, but I don't think this is the process used for manufacturing most industrial zinc oxide.


Why do some reactions need to be maintained at a particular temperature for prolonged periods of time?

Quite simply, reactions happen faster at higher temperatures (because who likes waiting), and some may not proceed at all until heated. From a statistical viewpoint, you are giving more reactant molecules more energy, which thereby increases the number of collisions, which in turn increases the number of successful collisions (those which produce product). A typical way to attain such conditions is to place a reaction under reflux.


Why draw "up" arrows first when filling orbital diagrams?

There is no real reason about that and drawing them "down" would be exactly the same as long as you draw them always in the same direction (it needs to be consistent).

The same applies to orbital lobes which can be hollow (white) or plain black depending on their relative sign. When two molecules' orbitals interact, you just consider they have the right sign for an interaction.


Reaction between 1,2-dicyanoethylene and 2-phenylbuta-1,3-diene in toluene and heat

The presence of an electron-donating group on the diene, or an electron-withdrawing group on the dienophile, is not an absolute requirement for the Diels-Alder reaction to occur. There are many Diels-Alder reactions that don't have such features.

In any case, the cyano groups on the dienophile are electron-withdrawing groups. So, it's still a pretty favourable scenario for the Diels-Alder to occur.

Since the dienophile is symmetrical there aren't any regioselectivity issues, and the relative stereochemistry of the two cyano groups in the product will be determined by the configuration of the double bond in the dienophile. A trans or (E) double bond will lead to the trans dinitrile (both enantiomers of it), and a cis or (Z) double bond will give rise to the cis dinitrile.

By the way, according to your title, there's no benzyl group on the diene, only a phenyl group. (Not that it makes much of a difference, though.)


Can the 5+ and 4+ oxidation state of Vanadium ion exist independently?

When an atom $M$ looses one or more electrons, it becomes positively charged and becomes the ion ${M^{z+}}$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $H_2O$, and it repells the proton of $H^+$.

If $Z$ = $1$, the ion $M^+$ attracts several molecules water around it, and it makes a Debye layer of oriented $H_2O$ molecules around the ion, and no chemical changes.

If $Z = 2$, the attraction is strong enough to repell one proton $H^+$ from one of the molecules adsorbed around the $Z^{2+}$ ion. The solution becomes a little bit acidic, and the $M^{2+}$ion is partially hydrolyzed into the ion [${M(OH)]^+}$. It is an equilibrium :$$M^{2+} + H_2O \ce{<=>}[M(OH)]^+  +  H^+$$

If $Z = 3$, the attraction between $Z^{3+}$ and water is strong enough to at least transform all $M^{3+}$ions into $[M(OH)]^{2+}$, and part of them to [$M(OH)_2]^+$ with the two reactions :
$$M^{3+} + H_2O\ce{->}   [M(OH)]^{2+}  +  H^+$$
$$[M(OH)]^{2+} + H_2O \ce{<=>} [M(OH)_2]^+  +  H^+$$
This is the case if $M^{3+}$ is $Al^{3+}$ or $Fe^{3+}$

If $Z = 4$, all ions $M^{4+}$ are at least hydrolyzed into $[M(OH)_3]^+$ and even into $M(OH)_4$, according to :  $$M^{4+} + 4 H_2O \ce{->}M(OH)_4  +  4 H^+$$It should be mentioned that $M(OH)_4$ has a tendency to loose water and produce a structure that has acidic properties. So it is often presented under the formula $H_4MO_4$ or $H_2MO_3$. That is the case with $M = Si$. Silicic acid is often presented as $H_4SiO_4$ or $H_2SiO_3$

With Vanadium $V^{4+}$, it goes another way. The molecule $V(OH)_4$ does not exist, as the ion $V^{4+}$ is too small to fix $4$ groups $OH^-$. As a consequence, $V^{4+}$ gives another positive ion $VO^{2+}$ according to :
$$V^{4+} +  H_2O \ce{->}  [VO]^{2+}+2H^+$$ Strangely enough, it looks as if the non-existent $V(OH)_4$ behaves like a basic hydroxide and gets dissociated according to : $$V(OH)_4\ce{->}[VO]^{2+}+ 2 OH^–$$
If $Z=5$, the ion $M^{5+}$ can be hydrolyzed according to $$M^{5+} + 3 H_2O \ce{->}  [MO_3]^{-}+6H^+$$$$M^{5+} + 4 H_2O \ce{->}  [MO_4]^{3-}+8H^+$$
But with Vanadium $V$, the ion $V^{5+}$ may be hydrolyzed  according to :$$V^{5+} + 2 H_2O \ce{->}  [VO_2]^{+}+4H^+$$$$V^{5+} + 3 H_2O \ce{->}  [VO_3]^{-}+6H^+$$The second reaction is favored at high pH.

This theoretical development is taken from General Chemistry, by Linus Pauling, W. F. Freemann and Co., San Francisco, 1953, Chapter XXI.


Why substitution and elimination reactions are favored respectively at lower and higher temperature?

When the question is asking about favorability, it is really asking about the magnitude of the Gibbs free energy. The more negative this value, the more favorable a reaction is. Gibbs depends on enthalpy and a temperature scaled entropy: $$\Delta G=\Delta H-T\Delta S$$

When we think of nucleophilic elimination, we know that a hydrogen atom is removed by a nucleophilic group resulting in our final product. Since a hydrogen is removed in elimination, while left intact during substitution, there are more final species formed during an elimination reaction. According to general chemistry principles, this means that there was a larger increase in entropy compared to the substitution. As temperature increases this larger entropy becomes more noticeable as it is now scaled by a larger $T$.

Therefore at low temperature, the enthalpy term dominates Gibbs and the substitution reaction prevails. However as $T$ grows, so does the entropy term and at high $T$ this overrides enthalpy, making the elimination reaction more favorable.


What happens to the complex part of the wave function when doing LCAO?

I understand that in general the wave function can take complex numbers [...]

It is not that it can, rather it should: by the very postulates of quantum mechanics wave function is a complex-valued function. So, both the spin orbitals (that are one-electron wave functions) and the Slater determinant built out of them are in principle complex-valued functions.

However, in solving the HF equations by the SCF procedure it is quite typical to impose some constraints on the spin orbitals, for instance, restrict them to be real-valued functions rather then complex-valued ones. As with any other constraint this restriction might lead to what is called SCF instabilities, i.e. situations when relaxing the constraints lead to a different variational solution of lower energy.

To elaborate a bit more let us mention few other well-known (as well as lesser-known) constraints for the SCF procedure:

The infamous restricted Hartree-Fock (RHF) model with the requirement that spin orbitals come in pairs: two spin orbitals corresponding to two different pure spin states are constructed from the same spatial orbital:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}(1) \beta(1)  \, .
\end{aligned}
$$
In the unrestricted Hartree-Fock (UHF) model the requirement above is relaxed and we exclusively use spatial orbitals from one set to construct $\alpha$ spin orbitals and spatial orbitals from another set to construct $\beta$ spin orbitals:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}^{\alpha}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}^{\beta}(1) \beta(1)   \, .
\end{aligned}
$$
A bit less known is the fact that the unrestricted Hartree-Fock method is not so unrestricted. In fact, there is a constraint here: each and every spin orbital describes an electron in a pure spin state, either $\alpha$ or $\beta$, while in general (in accordance with the postulates of quantum mechanics) electron can be in superposition of this states:
$$
  \psi_{i}(1) = \phi_{i}^{\alpha}(1) \alpha(1) + \phi_{i}^{\beta}(1) \beta(1)
$$
This most general setting is known as the general Hartree-Fock (GHF) method.

And all these three methods (RHF, UHF, GHF) exist in two variants: the real one, in which spin orbitals additionally required to be real-valued functions, and the more general complex one. All this give rise to six variants of the HF method with many possible instabilities between them discussed it some details in the seminal paper by Schlegel & McDouall1:

There is also a very similar earlier paper by Seeger & Pople.2 which suddenly is not freely available.

Its important to realize that any constraint of the mentioned above can only raise the electronic energy: with a great deal of certainty, we may expect that if any constraint is relaxed, the variational procedure will result in a lower energy due to greater variational freedom. Thus, for instance, for the most usual different real variants of the Hartree-Fock method we have
$$
    E_\mathrm{e}(\mathrm{RGHF}) \leq E_\mathrm{e}(\mathrm{RUHF}) \leq E_\mathrm{e}(\mathrm{RRHF}) \, .
$$
The same, of course, is true for the corresponding complex variants of these methods,
$$
    E_\mathrm{e}(\mathrm{CGHF}) \leq E_\mathrm{e}(\mathrm{CUHF}) \leq E_\mathrm{e}(\mathrm{CRHF}) \, .
$$
And for any of the three formalism
$$
    E_\mathrm{e}(\mathrm{C}x\mathrm{HF}) \leq E_\mathrm{e}(\mathrm{R}x\mathrm{HF}) \, , \quad \text{where} \quad x = \mathrm{G}, \mathrm{U}, \mathrm{R} \, .
$$

RHF/UHF/GHF and correct spin symmetry

The possible rise of electronic energy that accompanies introduction of more and more constraints in the GHF -> UHF -> RHF sequence seems to be quite contrary to the goal of the variation method. However, UHF and RHF constraints are nothing but symmetry constraints: they arise by requiring an approximate electronic wave function to have the same spin symmetry as the exact non-relativistic one, i.e. to be an eigenfunction of spin operators, the total spin-squared operator $\hat{S}^2$ and the $z$-component of the total spin operator $\hat{S}_{z}$.

With respect to $\hat{S}_{z}$ operator it is well known that any Slater determinant built out of spin orbitals corresponding to pure spin states is an eigenfunction of $\hat{S}_{z}$. Thus, RHF and UHF wave functions are eigenfunctions of $\hat{S}_{z}$, but GHF wave function is not.
The situation with $\hat{S}^{2}$ is a bit more complicated, but all in all only in RHF (but not in UHF or in GHF) formalism it is possible to construct an approximate electronic wave function which is an eigenfunction of $\hat{S}^{2}$.3

To conclude, as it was pointed out by Löwdin, we always face with a dilemma here: should we seek a solution that is a true variational minimum or should we seek a solution with the correct spin symmetry?

1) H. B. Schlegel and J. J. W. McDouall, Do You Have SCF Stability and Convergence Problems? in Computational Advances in Organic Chemistry: Molecular Structure and Reactivity, Springer Netherlands, 1991, pp. 167-185. DOI: 10.1007/978-94-011-3262-6_2. Free PDF from wayne.edu.

2) Seeger, R., & Pople, J. A., Self‐consistent molecular orbital methods. XVIII. Constraints and stability in Hartree–Fock theory. The Journal of Chemical Physics, 66(7), 1977, 3045-3050. DOI: 10.1063/1.434318.

3) This can be done even for an open-shell electron configuration, although it would require a linear combination of Slater determinants with constant coefficients, none of which alone is an eigenfunction of $\hat{S}^{2}$.


Are the energies of orbitals that are not filled in an atom, equal?

Sure, if they are unoccupied they contribute nothing to the total energy of the system. Remember that orbitals are mathematical constructs and only have meaning with regard to their energy when being physically occupied by an electron that adheres this mathematical description. The energy required to fill such an orbital however differs (vastly) from orbital to orbital and nucleus to nucleus (see for example http://www.colby.edu/chemistry/PChem/notes/AOIE.pdf).


Relationship between Gibbs free energy and potential energy (related to reaction coordinate diagrams)

Equation (1) is an empirical one that does not consider entropy. If you substitute it for the Eyring equation (considering free energy of activation $\Delta G^\ddagger$) you end up with equation (6) as expected, not equation (5). Wikipedia has a nice section on the relationship of activation energy and Gibbs energy of activation:

  Although the equations look similar, it is important to note that the Gibbs energy contains an entropic term in addition to the enthalpic one. In the Arrhenius equation, this entropic term is accounted for by the pre-exponential factor A. More specifically, we can write the Gibbs free energy of activation in terms of enthalpy and entropy of activation: $ΔG‡ = ΔH‡ – T ΔS‡$. Then, for a unimolecular, one-step reaction, the approximate relationships $E_a = ΔH‡ + R T$ and $A = (k_B T/h) \exp(1 + ΔS‡/R)$ hold. Note, however, that in Arrhenius theory proper, A is temperature independent, while here, there is a linear dependence on T. For a one-step unimolecular process whose half-life at room temperature is about 2 hours, ΔG‡ is approximately 23 kcal/mol. This is also the roughly the magnitude of Ea for a reaction that proceeds over several hours at room temperature. Due to the relatively small magnitude of TΔS‡ and RT at ordinary temperatures for most reactions, in sloppy discourse, $E_a$, ΔG‡, and ΔH‡ are often conflated and all referred to as the "activation energy".

So in a more rigorous treatment, the pre-exponential factor is temperature-dependent (this is almost impossible to measure because the exponential is "very" temperature-dependent) and has an entropy component. That is why equation (5) in the question is not accurate (equation (6) is correct, and can be derived in a rigorous manner from first principles and definitions).

Related question: Y-axis of the reaction co-ordinate graph


In freshman physics, we were taught that $q=mC\Delta T$, but in all the cases we considered, no work was being done.  When we got to thermodynamics, we learned that q represents energy in transit, and is a function of process path.  We also learned that, when work is being done, q is not equal to $mC\Delta T$.  Instead, we learned about the internal energy of a system U, which is expressed as $$\Delta U=q-w$$  When no work is being done, $$\Delta U=q=mC\Delta T$$All this misunderstanding is resolved and made consistent in thermodynamics when we change the definition of heat capacity slightly, defining it strictly in terms of U, rather than in terms of q:$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$Then, whether or not work is being done, we get the correct answer for the heat transferred and the change in internal energy.  The advantage of this definition is that C is properly recognized as a physical property of the material being processed rather than a function of the process path.


Why is trichloromethane more acidic than trifluoromethane?

We know that fluorine ($\ce{F}$) is more electronegative than chlorine ($\ce{Cl}$). Therefore, -I effect (Inductive Effect) of $\ce{F}$ would be more and trifluoromethane ($\ce{CHF3}$) would be more acidic.

But this is not the case. As -M (Mesomeric Effect) effect also plays an important role in determining the acidity. We also know that -M is considered over -I (Inductive Effect).

Now, If we compare the conjugate base of both the given molecules, $\ce{Cl3C-}$ and $\ce{F3C-}$, Chlorine has vacant $\mathrm{3d}$ orbitals whereas $\ce{F}$ does not have any vacant $\mathrm{d}$ orbitals (in fact, it does not even have $\mathrm{d}$-orbitals). Therefore, $\ce{Cl}$ can exert -M effect due to availability of vacant $\mathrm{d}$ orbital but $\ce{F}$ cannot exert the -M effect. Generally, -M effect stabilises the negative charge and increases acidic strength. Therefore, trichloromethane ($\ce{CHCl3}$) is more acidic than trifluoromethane ($\ce{CHF3}$).


Important naturally-occurring radioactive lead nuclides are:

$\ce{^{214}Pb}$ $\left(t_{1/2}=26.8\ \mathrm{min}\right)$ from $\ce{^{238}U}$

$\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ from $\ce{^{238}U}$

$\ce{^{211}Pb}$ $\left(t_{1/2}=36.1\ \mathrm{min}\right)$ from $\ce{^{235}U}$

$\ce{^{212}Pb}$ $\left(t_{1/2}=10.64\ \mathrm{h}\right)$ from $\ce{^{232}Th}$

On a geological time scale, their half-lives are very short. Nevertheless, these lead nuclides are members of the $\ce{^{238}U}$, $\ce{^{235}U}$, and $\ce{^{232}Th}$ decay series. Thus, they are constantly reproduced by decay of their respective mother nuclides. Therefore, all environmental samples (including soil, water, air, plants, and animals) naturally contain significant amounts of radioactive lead nuclides. However, in various materials (e.g. residues from mining or oil and gas production, fertilizers, building materials), the concentration of natural lead nuclides can be strongly enhanced by technology.

When lead is freshly refined (chemically purified), most of the mother nuclides are removed and the short-lived lead nuclides quickly decay. Nevertheless, the lead still contains significant amounts of $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$. The presence of $\ce{^{210}Pb}$ is a source of background mainly through the lead X-rays and bremsstrahlung caused by high-energy beta radiation emitted by its decay product $\ce{^{210}Bi}$. This background is a real problem when lead is used as a shielding material for low-background detectors. Therefore, some shields are made from selected lead with certified low $\ce{^{210}Pb}$ content. Preferably, very old lead is used. Since $\ce{^{210}Pb}$ decays with a half-life of 22.3 years, samples of lead that are many decades old are relatively free of this activity. In one extreme example, 2000-year-old lead was salvaged from a sunken Roman ship.


Finding membrane potential equation using the Nernst equation?

The Nernst Equation is $E = E^\circ - \frac{RT}{nF} \ln{Q_c}$. From that you can derive the approximation you gave. $Q_c$ is the product of the activities of the products over those of the reactants. This can be, for dilute solutions where activity approximates concentration, the typical [B]/[A] for the reaction A→B. So, the only thing you need to "explain" is why E° would be zero. Review the definition of E°.  hint:What is ln(1/1)? See the discussion on concentration cells in:  http://www.chem1.com/acad/webtext/elchem/ec4.html


When an iodide/iodate solution is acidified with H2SO4 instead of thiosulfate, why should it be titrated immediately?

After acidification with $\ce{H2SO4}$, starch starts to hydrolyze to its component sugars. Lugol's iodine does not react with those sugars to form the blue complex.

Our amylose is gone, today,
Its trace we'll see no more,
For what we thought was $\ce{H2O}$
Was $\ce{H2SO4}$.


If $\ce{CO2}$ is acceptable in the feed gas, along with $\ce{CO}$ and a bit of left-over oxygen, then carbon (coal, charcoal etc.) might do. It's ignition temperature is ~1,000 K.

Magnesium is more expensive, but leaves less oxygen and has no gaseous byproduct. It's self-ignition temperature is ~800 K.

Many other materials could be used; it really depends on how the gas is to be used.


Can a drink with such composition have 0 caloric value (whether manufacturer lied)?

You have to keep in mind the bane of all introductory chemistry students: excess precision and rounding. Every measurement has error bars, even if those error bars are implicit. It's senseless to report a value that has (much) more precision than your error bars.

A value that's measured as zero isn't necessarily exactly zero. ... it's only "mostly zero".

The particular rounding rules that are used in nutritional labels depend on the specific regulatory agency, but in the US, the FDA provides rounding rules for nutritional labels. It counts anything less than 5 (dietary) Calories per serving as being zero. Likewise, anything with less than 0.5 g of protein, fat or carbohydrate can be listed as having zero grams.

This is how Tic-Tac breath mints can be advertised as being "zero calorie" despite being made almost entirely of sucrose (table sugar): a serving contains less than 5 Calories, and so can be rounded to zero on the label.

In your case, there's a few things which could contain calories. Carmel color is made from sugar, and may contain residual saccarides. Aspartame is a methyl ester of a dipeptide, which is converted in the body to its constituent amino acids, providing a small amount of calories. (Gram for gram, aspartame would contain almost the same amount of calorific energy as sugar does. It's about 200 times sweeter than sugar, though, so on an equivalent sweetening basis it has much fewer calories.) Depending on what the "natural flavor enhancer" is, that might also contain carbohydrates, proteins or fats. (Natural flavors are often fats or fat-soluble, so flavorings often contain trace amounts of dietary fats.)

For a single serving (and for 100 ml) there's only a small amount of carbohydrate, only a scant amount of amino acids, a negligible amount of fat, and an insignificant number of calories. They're a rounding error. But it is true that if you had enough of the drink, you could indeed have an appreciable number of calories, carbohydrates, protein, etc. It's just that it would come along with enough water to make actually consuming that amount physically unfeasible.


Phenolic compounds, or in general aromatics actually, has electrons which many oxidants can easily abstract or bind to molecule.

http://www.springer.com/cda/content/document/cda_downloaddocument/9781441978257-c1.pdf?SGWID=0-0-45-1106848-p174056970

From the above link in page 17,

as you see the highest rates, order of magnitude or even more higher ones come from aromatics, or at least pi-bond between carbons. It can reasonably assumed that if you have these compounds in a mixture, the possibility of any oxidant introduced reacts with them is higher than the other ones. In short, they are good anti-oxidants in that way.


For somebody who is not well-versed in quantum mechanics, it might be natural to guess that this is the case. But if I were you, I wouldn't assume so.

Both electrons share the same p-orbital, and are therefore still paired. They both occupy the entire volume of the p-orbital. What they do not do, is to divide the p orbital up into two half-orbitals, with each electron taking one half.

There is a somewhat related question here, which might address some issues: How do Electrons Cross Nodes in Orbitals?


If you’re talking about a gentle placement of the drop on the paper, I’d guess it’s likely due to the fibrous structure of the paper providing “easier” paths for the ink to flow. If you’re talking about an impacting drop, the drop will vibrate when it impacts the paper until the forces of surface tension are exceeded and the drop “splatters”.


In an ideal world you would not have to do that. You'd simply wait until the reaction completes and then make a single measurement of temperature.

Too bad our world isn't ideal. The moment your reaction vessel becomes slightly hotter than the environment, it starts losing heat. If you wait for too long, it will cool all the way down to room temperature, so there will be nothing to measure in the first place. If you make your measurement in the first seconds so as not to let any heat escape, that's too early; the reaction has barely started. If you wait just until completion (which is a few minutes into the process, judging by your graph), that's too late, because by then some amount of heat is already lost.

So what do we do? We account for the loss. We assume that the temperature drop is linear (which of course is just an approximation, but still, it is much better than nothing), make a dozen or so measurements, build that trendline, and extend it to the origin. That's where your temperature would stop if there were no drop.

So it goes.


The activation energy of a single reaction does not change with time but there might be more than one reaction happening

An activation energy is always constant for a given single reaction. But that doesn't mean the reaction speed will be constant (it might depend on concentration or temperature).

But what about autocatalytic reactions? There are two possible sources of confusion here. A very simple autocatalytic reaction might involve a changing concentration of an intermediate. In this case the reaction might go faster because that concentration is changing as the intermediate builds up (but the activation energy stays the same). In more complex cases, the catalytic route to the product involves a different reaction. What you have is not a changing activation energy for a specific reaction but the addition of a different reaction pathway to give the same product. This reaction will have a different activation energy but the activation energy of the original pathway is not changed.


Why does ammonia act as an acid in only the first of these two reactions?

Ammonia has a $\mathrm{p}K_{\mathrm{a}}$ value (a description what that means can be found in this answer of mine) of 33 and is thus not acidic at all - you'd need a $\mathrm{pH}$ value around 30 to deprotonate $\ce{NH3}$ to any appreciable amount! This means that $\ce{NH3}$ very rarely acts as a proton donor itself.

So, what happens in your first reaction is the following: It is widely known that when you dissolve $\ce{Na}$ in liquid ammonia you get solvated electrons (in the form of electride complexes). But electrons have of course very strong reducing power, so after a while they attack $\ce{NH3}$ abstracting one of its $\ce{H}$ atoms under the liberation of hydrogen gas. The important point here is that there is no dissociation of $\ce{NH3}$ here, i.e. no reaction $\ce{NH3 <=> NH2^{-} + H^{+}}$, so $\ce{NH3}$ does not act as a proton donor in the same sense a Bronsted acid would.

Your second reaction is an $\mathrm{S}_{\mathrm{N}}2$ reaction where $\ce{NH3}$ acts as a nucleophile. So, it doesn't act as a proton donor in this reaction neither. What it does is that it replaces (substitutes) the $\ce{Cl}$ in $\ce{CH3Cl}$, thus forming a methylammonium salt, $\ce{CH3NH3^{+}Cl^{-}}$.

Now, the $\mathrm{p}K_{\mathrm{a}}$ of an alkylammonium cation lies somewhere around 10 (the $\mathrm{p}K_{\mathrm{a}}$ of $\ce{NH4^{+}}$ is 9.24). So, it is much more acidic than $\ce{NH3}$ and also more acidic than the surrounding alcohol (I assume that "(alc)" means that the reaction is done in alcohol), which has a $\mathrm{p}K_{\mathrm{a}}$ between 15 and 17, and thus it (the ammonium salt) acts as a proton donor.


Well, not quite. Here's the quick summary on subscripts in Mulliken notation for point group symmetry.

“g” stands for "gerade" and refers to the symmetry around an inversion center. (That is, the representation retains symmetry around the inversion center.) Yes, octahedral ($O_\mathrm{h}$) molecules have an inversion center, but so do others.

“u” stands for "ungerade" and refers to anti-symmetric around an inversion center.

“1” or “2” refer to symmetry (1) or anti-symmetry (2) around a secondary $C_2$ axis (e.g., in a $D_x$ point group like benzene).

The $'$ and $''$ assignments (if needed) are used for symmetry around the $\sigma_\mathrm{h}$ operation.

And, while you didn't ask, for “A” versus “B”, these are one-dimensional representations that differ with respect to the "main" rotational axis. “A” representations are symmetric “+1” and “B” are anti-symmetric “−1” for that axis.

You only need enough subscripts to uniquely identify a representation. For example, here's the character table for $D_{4h}$:

$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline
\end{array}$$

Note that all the “g” representations have +1 or +2 for the character of the “$i$” symmetry operation. We don't need “1” or “2” to distinguish the $\mathrm{E_g},$ but we do for the different $\mathrm{A}_{x\mathrm{g}}$ representations. For those, we go to the $C'_2$ axis, which separates $\mathrm{A_{1g}}$ from $\mathrm{A_{2g}}$. These are unique and we don't need to assign $'$ or $''$ for this point group.

(If you look up the $D_\mathrm{3h}$ character table, you find $'$ and $''$ needed.)


Raoults Law says that, for an ideal liquid mixture, the partial pressure of each component in the gas phase is equal to the vapor pressure of the pure component (at the temperature of the system) times the  mole fraction of that component in the liquid phase.  If the system is at equilibrium, the mole fraction to use in the equation is the current mole fraction.


Why does a lowered dielectric constant cause RNA/DNA precipitation in ethanol precipitation?

For ethanol precipitations [...] shouldn't the water-phosphate attraction be increased in magnitude by the same ratio as that by which the phosphate-cation in increased in magnitude?

The interaction between negatively charged nucleic acid and positively charged counter ions is an ion:ion interaction. The interaction between ions and water is an ion:dipole interaction.

  An argument based on the relative higher amounts of cations and ethanol, and lower in water, would make sense in increasing the likelihood of a cation-phosphate bond over a water-phosphate hydrogen bond, but then any mention of the dielectric constant in explaining ethanol precipitation seems entirely superfluous.

The presence of water between ions of opposite charge diminishes their long-range attraction (the dielectric constant is related to how well solvated ions are "shielded" by the oriented dipoles of the solvent).
Ion:dipole interactions are short range and rely on the dipole to be properly oriented. Adding ethanol to the aqueous DNA solution does not make the ion:dipole interactions of water with ions stronger. In contrast, it does make the ionic interactions stronger because ethanol is less effective in shielding charges than water.

It is important to compare dielectric constants of water and ethanol because both molecules are uncharged and polar (and capable of ion:dipole interactions), but water is the better solvent for ionic compounds for the reasons discussed above.

  [from comments] I don't understand why [we add] ethanol. Ion-dipole interactions and ion-ion interactions are both interactions based on charge and should thus both follow Coulomb's law. Ion-dipole interactions are certainly weaker, and hence shorter range, but that doesn't seem like it should exempt them from Coulomb's law. Dipole orientation is something I hadn't considered.

A dipole has a positive and a negative partial charge that add up to zero. Imagine you are the cation that interacts with a dipole. When the dipole is near you, the negative partial charge will be right next to you, and there will be a large attraction. The positive partial charge will be a bit further, so there is a smaller repulsion. Overall, it adds up to an attractive force. Now move the dipole away from you, maybe by a factor of 10. Now, the positive charge will attract much less, and be at approximately the same distance as the negative charge, so the repulsion and attraction will almost cancel out.

Take a look at pentavalentcarbon's answer to a question on intermolecular interactions. There is a nice table with the dependence of interaction strength on distance.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the first relevant criterion to be considered in choosing a principal chain is the length of the chain.
When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached:

P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)

Since both possible chains in this case have two multiple bonds, Rule (a) is not enough to make a choice.
According to Rule (b), the principal chain is the hepta-1,6-diene because it has the greater number of double bonds.
Therefore, the preferred IUPAC name (PIN) is 4-(prop-2-yn-1-yl)hepta-1,6-diene.


Difference between freezing point and melting point

The Bodner research group at Purdue University has posted this excellent description of the difference between melting point and freezing point:

Pure, crystalline solids have a characteristic melting point, the temperature at which the solid melts to become a liquid. The transition between the solid and the liquid is so sharp for small samples of a pure substance that melting points can be measured to $\pu{0.1^\circ C}$. The melting point of solid oxygen, for example, is $\pu{-218.4^\circ C}$.
Liquids have a characteristic temperature at which they turn into solids, known as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small differences between these quantities can be observed.
It is difficult, if not impossible, to heat a solid above its melting point because the heat that enters the solid at its melting point is used to convert the solid into a liquid. It is possible, however, to cool some liquids to temperatures below their freezing points without forming a solid. When this is done, the liquid is said to be supercooled.
An example of a supercooled liquid can be made by heating solid sodium acetate trihydrate ($\ce{NaCH3CO2 . 3H2O}$). When this solid melts, the sodium acetate dissolves in the water that was trapped in the crystal to form a solution. When the solution cools to room temperature, it should solidify. But it often doesn't. If a small crystal of sodium acetate trihydrate is added to the liquid, however, the contents of the flask solidify within seconds.
A liquid can become supercooled because the particles in a solid are packed in a regular structure that is characteristic of that particular substance. Some of these solids form very easily; others do not. Some need a particle of dust, or a seed crystal, to act as a site on which the crystal can grow. In order to form crystals of sodium acetate trihydrate, $\ce{Na+}$ ions, $\ce{CH3CO2-}$ ions, and water molecules must come together in the proper orientation. It is difficult for these particles to organize themselves, but a seed crystal can provide the framework on which the proper arrangement of ions and water molecules can grow.
Because it is difficult to heat solids to temperatures above their melting points, and because pure solids tend to melt over a very small temperature range, melting points are often used to help identify compounds. We can distinguish between the three sugars known as glucose ($\vartheta_\mathrm m = \pu{150^\circ C}$), fructose ($\vartheta_\mathrm m = 103{-}\pu{105^\circ C}$), and sucrose ($\vartheta_\mathrm m = 185{-}\pu{186^\circ C}$), for example, by determining the melting point of a small sample.
Measurements of the melting point of a solid can also provide information about the purity of the substance. Pure, crystalline solids melt over a very narrow range of temperatures, whereas mixtures melt over a broad temperature range. Mixtures also tend to melt at temperatures below the melting points of the pure solids.


"Encyclopedic" / semi-exhaustive reference work on Chemistry

One of the most comprehensive descriptive chemistry books is Holleman and Wiberg's Inorganic Chemistry. It fits your encyclopedic criterion with about 2000 pages with all major atomic structure concepts plus descriptive chemistry of the elements. It is being published for more than 100 years and originally it was in German.
Another classic, again of about >1000 pages was Glasstone's "A Textbook of Physical Chemistry". It is slightly dated (1940s) but very exhaustive in terms of covering all the classical concepts to be ever encountered in physical chemistry.


Stripping zinc electroplating with phosphoric acid - hydrogen danger?

Yes, mineral acids like $\ce{H3PO4}$ react with electropositive metals like zinc to produce dihydrogen gas ($\ce{H2}$).

The quantities generated of course depend on how much metal is being dissolved that way. About $65\ \mathrm{g}$ of zinc would produce about $44.8\ \mathrm{L}$ of hydrogen (in Standard conditions).

This may seem like a lot but your nuts and bolts are covered with only a very thin layer of zinc.

Hobby chemists dissolve metals in mineral acids all the time, with minimal danger. Just avoid open flames (use an electrical hot plate) and make sure your work place is well ventilated and you should be fine. Consult any safety advice supplied with the Ortho.


Is xenon electropositive in compounds of xenon?

How can a highly oxidized element be electropositive?

An element can be electropositive, the highly oxidized element less so.

There are a number of things that don't sound quite right in the passage you quote.  For example, the Pauling electronegativity for xenon has been assigned (2.6, if you'd like to see the methodology used to caculate the electronegativity of a noble gas, see this earlier answer). Also, depending on the compound, hypervalent (or hypercoordinate) bonding, rather than use of d orbitals, might be used to explain the bonding in some noble gas compounds


Percent degree of dissociation of gas by pressure

If $\alpha$ is the degree of dissociation, at equilibrium $1-\alpha$ of XY is present and $\alpha$ of X and $2\alpha$ of Y making a total of $1+2\alpha$. The partial pressure of XY is $\displaystyle p_{XY}=\frac{1-\alpha}{1+2\alpha}\frac{P}{P^\mathrm{o}}$. The $P^\mathrm{o}$ is the standard pressure used to make the equilibrium constant dimensionless. In our world this is 1 bar, in your new world it is 2 bar, thus $P/P^\mathrm{o}=0.1$. If you calculate the partial pressures for the other species in terms of $\alpha$ and form the equilibrium constant then $\displaystyle K=\frac{4\alpha^3}{(1+2\alpha)^2(1-\alpha)}\left(\frac{P}{P^\mathrm{o}}\right)^2$.(please check).  As $K$ is small we will assume that $\alpha$ is also then $\displaystyle K=4\alpha^3\left(\frac{P}{P^\mathrm{o}}\right)^2$ which produces $\alpha \approx 3.10^{-3}$.


TL;DR: The primary transition responsible for color is apparently the ${\pi{-}\pi}^*$ transition of the conjugated ketyl radical anion. This absorbs usually around the green to orange portions of the visible spectrum, leaving behind the purple-blue, and sometimes green color.

Original answer (background info):
According to Wikipedia/Benzophenone, the ketyl radical is formed by first, transition of the ketone from a singlet (double bond) to a triplet (diradical) state, and then abstraction of an electron from a donor to form the ketyl radical anion.
As I understand, only the radical anion is colorful. Both ref 1 and Wikipedia/Ketyl indicate that color disappears if the radical anion is allowed to oxidize (back to the ketone, which should be in equilibrium with the diradical). So my guess is, when an extra electron is added to the triplet state, this enables new electronic transitions that produce color.
These two sources also indicate both solvent and substituent effects: In the Wikipedia article, the solvent is toluene, and color is deep blue for benzophenone. In the paper, the solvent is isopropyl alcohol, and color is greenish-blue for benzophenone, and varies from blue to green depending on substituents. This also suggests that conjugation plays an effect, since the substituents can directly affect only the phenyls. But this, in turn, affects the energy levels of the double bond.

Edit:
A Google search for "benzophenone ketyl spectrum" (including the image search) gives several results of varying relevance, such as refs 2-4. In particular, from ref 2 (dealing with uranium ketyl radicals):

The electronic absorption spectrum of 2 in toluene features an intense absorption band ($\lambda_\text{max} = \pu{562 nm}$, $\epsilon = \pu{5680 M^{-1} cm^{-1}}$) in the visible region (Figure 8, top). This absorption gives rise to the ketyl complex’s distinctive and intense purple color and is assigned to the ${\pi{-}\pi}^*$ transition of a highly conjugated ketyl system. The transition energy of $\pu{562 nm}$ is blue-shifted from that of the sodium benzophenone ketyl radical, which exhibits the ${\pi{-}\pi}^*$ transition at $\pu{618 nm}$.

Ref 3 talks about $D_1 \leftarrow D_0$ and subsequent $D_n \leftarrow D_1$ transitions, and provides an energy level diagram. I assume $D_n$ refers to the doublet states that form upon abstraction of a hydrogen by (or addition of an electron to) the triplet diradical.
Finally, ref 4 adds:

Bzp is known to have a high ISC quantum yield following UV excitation, but until recently the mechanism of population of the lowest triplet excited state ($T_1$) after photoexciting to its $S_1$ (${nπ}^*$) state remained uncertain. ... These investigations indicate that the dominant pathway involves fast ISC from $S_1$ to $T_2$ (${\pi\pi}^*$) as an intermediate state, with the subsequent rate of population of the lowest vibrational levels of $T_1$ controlled by the time scale for dissipation of excess vibrational energy to the solvent bath after internal conversion (IC) from $T_2$.

(Bzp = benzophenone, ISC = intersystem crossing). It then proceeds to discuss these transitions in more detail.

J. Am. Chem. Soc. 1933, 55 (1), 391–395; doi:10.1021/ja01328a057
J. Am. Chem. Soc. 2008, 130 (20), 6567–6576; doi:10.1021/ja801007q
J. Phys. Chem. A 2004, 108 (40), 8147–8150; doi:10.1021/jp047058a
J. Am. Chem. Soc. 2019, 141 (38), 15222–15229; doi:10.1021/jacs.9b07047


Is there air in the volume of crystallized sodium hydroxide?

Imagine a container with sand; it may seem "full", but surely you may pour in some water, and it will fit in the cavities. Same thing here, except in case of $\ce{NaOH}$ I'd rather not check it directly with water, for the response may be quite intense.

$\ce{NaOH}$ commonly comes in granules; if you'd have one big solid brick inside your container, that would be another story.


What are some materials that expand a lot proportionally to heat?

You are looking for thermal expansion and its coefficient.  Of course such tables, and diagrams exist

(source)

as it is of relevance in daily live.  Thermometers based on liquid mercury, or alcohol, depend on this.  Bridges take this into account in expansion joints, too:

(source)

Joining two materials of different expansion coefficient equally is of relevance for the design of bimetallic stripes, found not only in thermometers and mechanical clocks:

(source)

So in reply to your "there is no source available" ...


Positive vs. negative heat of solution confusion

The heat of solution (aka enthalpy of the solution and $\Delta\mathrm{H}_{\text{sol}}$) is negative if heat is released. The reason is that energy is leaving the system. The reason you got two different answers is that the sources you referenced are describing two different things. The $\Delta\mathrm{H}_{\text{sol}}$ is positive (endothermic) when dissolved in water, but Erythritol is commonly combined with Inulin which produces an exothermic reaction (cancelling out the cooling effect of the Erythritol).


Describe the function of a salt bridge, What would happen if the half-cell compartments of a galvanic cell were not connected?

My electrochemical knowledge is a wee bit rusty, but I believe the salt bridge helps balance the charges and not mass.

What charge you ask? So, let imagine you start up your galvanic cell (below). Electrons from the Zn move to Cu electrode. You start to build up a negative charge on the Cu which would resist/repel further addition of electrons after a period of time, hence production of electricity ceases.

By having a salt bridge, the electrolyte ions can move from one side to the other to mitigate this issue.


How is the bonding in the [Au6C(PPh3)6]2+ cluster explained?

This is by no means a trivial case.
You are dealing with a so-called template or interstitial atom placement within a cluster shell.
There is a thorough review available [1, pp. 18–20] (reference numbers updated):

  An alternative way of satisfying the closed shell electronic requirements and retaining a positive charge on the cluster is possible if an interstitial atom is introduced.
  For an octahedral gold cluster an additional gold atom is too large to satisfy the steric requirements of the interstitial cavity, but a small main group atom, e.g. C, N or O, is able to satisfy these steric requirements.
  Figure 11 gives the relevant interaction diagram for octahedral $\ce{[Au6C(PPh3)6]^2+}$ and an interstitial carbon atom, which contributes just the right number of electrons to lead to a $\mathrm{[S^σ]^2[P^σ]^6}$ pseudo-spherical ground state conﬁguration.
  It is noteworthy that this simple theoretical analysis was ﬁrst published in 1976 [2], and the existence of $\ce{[Au6C(PPh3)6]^2+}$ was conﬁrmed by Schmidbaur and his coworkers in 1989 [3–5].
  This compound had been isolated and structurally characterised previously, but the interstitial carbon had not been identiﬁed.
  The introduction of the interstitial atom strengthens the radial interactions signiﬁcantly as a result of effective overlaps between the carbon $\mathrm{2s}$ and $\mathrm{2p}$ orbitals and the matching $\mathrm{S^σ}$ and $\mathrm{P^σ}$ cluster molecular orbitals.
  In broad brush terms the stabilisation of the valence orbitals of the central atom are stabilised by $\mathrm{nβ^σ}$/degeneracy of the molecular orbitals.
  If $\mathrm{β^σ(s)} = \mathrm{β^σ(p)} = \mathrm{β^σ(d)},$ the relative stabilisations are $\mathrm{6β^σ(s)},$ $\mathrm{2β^σ(p)},$ and $\mathrm{3β^σ(d)}.$
  Therefore, the greatest stabilisation involves the $\mathrm{s}$ orbitals of the central
  atom and increases as the number of metal atoms, $n,$ increases.
  For filled shells the stabilisation energies are independent of geometry as long as the ligand polyhedron approximates to a sphere.
  It follows that gold clusters with main group interstitial atoms are characterised by a pec of $12n + 8$ (sec 8) valence electrons, since each $\ce{AuPPh3}$ fragment is associated with a filled $\mathrm{d}$ shell containing $10$ electrons and a bonding $\ce{Au–P}$ bonding molecular orbital.

  Fig. 11 Molecular orbital interaction diagram for $\ce{[Au6C(PPh3)6]^2+}.$
  Similar analyses may be constructed for trigonal bipyramidal $\ce{[Au5N(PPh3)5]^2+}$ and tetrahedral $\ce{[Au4O(PPh3)4]^2+}$

References

Gold Clusters, Colloids and Nanoparticles II; Mingos, D. M. P., Ed.; Structure and Bonding; Springer International Publishing: Cham, 2014; Vol. 162. DOI: 10.1007/978-3-319-07845-8.
Mingos, D. M. P. Molecular-Orbital Calculations on Cluster Compounds of Gold. J. Chem. Soc., Dalton Trans. 1976, No. 13, 1163–1169. DOI: 10.1039/DT9760001163.
Schmidbaur, H., Grohmann, A., Olmos, M. E.; Organogold chemistry. In: Schmidbaur, H. (ed) Gold progress in chemistry, biochemistry and technology. Wiley, Chichester, 1999. pp 648–731.
Steigelmann, O.; Bissinger, P.; Schmidbaur, H. Assembly of the $\ce{[CAu6]^{2⊕}}$ Cluster with a Tailor-Made Diphosphane Spanning the Octahedral Edges. Angewandte Chemie International Edition in English 1990, 29 (12), 1399–1400. DOI: 10.1002/anie.199013991.
Schmidbaur, H.; Brachthäuser, B.; Steigelmann, O. Direct Observation of the Central Atom in $\ce{[C\{Au[(C6H5)2(PC6H4NMe2)]\}6](BF4)2}$ by $\ce{^{13}C}$ NMR Spectroscopy. Angewandte Chemie International Edition in English 1991, 30 (11), 1488–1490. DOI: 10.1002/anie.199114881.


How to determine the sign of the enthalpy of formation?

$\ce{Al2O3}$, upon the formation of its lattice, releases heat. This heats up water. The increase in the temperature is in water, not $\ce{Al2O3}$. We usually refer to the water as the surroundings and $\ce{Al2O3}$ as the thermodynamic system.

  A thermodynamic system is the content of a macroscopic volume in space, along with its walls and surroundings; it undergoes thermodynamic processes according to the principles of thermodynamics. Wikipedia

So, upon the formation of lattice (among with other, mostly, endothermic reactions) a lot of heat is released and that's what you have estimated. Since $\ce{Al2O3}$ is the system, it loses heat. Thus, $\Delta H_f$ is negative.


Why shouldn't Uranium come in contact with water?

Interesting question. The documentary was almost certainly referring to the extremely hot lava generated by the nuclear meltdown at Chernobyl coming into contact with water. This lava is made of a substance colloquially known as 'corium' and is a combination of molten fuel rods, moderator, reactor walls and whatever else is melted by the incredible temperature of the reactor failure. This is as far as I know why the failure mode is called a 'meltdown', because the corium melts and pools in the bottom of the reactor vessel. In a molten pool, the fissile material may achieve uncontrolled criticality and just get hotter and hotter.

If the temperature is high enough the corium may melt through the bottom of the reactor. This is really bad, because at least hypothetically the lava may melt down to the water table and cause a steam explosion. This is quite a bit like what happens if you try to extinguish an oil fire by throwing water on it (something you should never do) - the water violently boils and expands and sprays burning oil everywhere. However in this case the burning oil is radioactive lava.

This is not a chemical reaction, per se, but rather a case of extremly vigorous heating of water and associated violent expansion.

Additional

If you remember the Fukushima disaster, at least one of the buildings was destroyed in a hydrogen explosion. As I understand it, in the event of a meltdown, the fissile material can get hot enough to effect the thermolysis of water. (The wikipedia article of the event seems to confirm this) Inside a sealed building, the water will develop a headspace filled with hydrogen and oxygen gas. This mixture is very explosive.


It is one of the man-made tools to conceptualize the nature of a bond between two atoms, to know if there is a single / double / triple / quadruple *) bond between them, and to account for both electrons involved in bonding / antibonding molecular orbitals.  Because the nature of electrons is neither "true and only" particle, nor wave alone, hence sometimes escaping our day-to-day experience with macroscopic objects.

In addition, rarely you look at the bond order of two atoms in a molecule without keeping in mind the situation with other atoms present in the same molecule, too.

*)  Sorry, I forgot at least the three-centre-two electron bonds in diborane...


I have become familiar with many peroxide like Na2O2, BaO2, etc. But not able to recognise between 1 and 2. Is there any way for this type of question to recognise peroxide oxygen?

Yes. In the exam, if you do not know whether (1) or (2) contains peroxy, simply look at the charges of Chromium if all the oxygen is  in the -2 oxidation state.

For $\ce{CrO5}$, if all the oxygen have a charge of -2, then Cr would have to have a charge of +10, which is impossible, therefore you know that some of the oxygen must have a charge of -1(peroxy). On the other hand, for $\ce{Cr2O3}$, Cr has a charge of +3, which is possible (Cr(III) salts are plenty)


If we put an metal electrode in water with no metal ions present potential should tend to minus infinity?

Yes. The potential is in principle infinite if the water is pure. But in reality, when a metallic plate is dipped into pure water, you cannot prevent one or two atoms from being oxidized by air and pass into solution. On the other hand, the same thing can be said about $\ce{O2}$ producing one or two $\ce{OH-}$ ions. As a consequence the potential goes quickly from infinite to a finite value.


In phosphorus’ case, the connection between stability of the allotrope and its colour is only accidental.

White phosphorus, i.e. $\ce{P4}$ molecules, has very strained bonds because of the very small tetrahedral angles the atoms need to accomodate for. Going by the edges of a regular tetrahedron, the bond angles are $60^\circ$ which is very strained. This is the reason for its reactivity. Its colour is white, because there are no electron transitions excited by visible light, much like similar compounds of other elements.

Black phosphorus, on the other hand, forms large extensive networks which are well on the way to being metallic. Metallic bonding is typically such a widespread bonding network that every atom contributes to every molecular orbital with more or less the same contribution. You can compare this to large organic π systems. This allotrope is more stable because each phosphorus atom has bond angles that well suit the corresponding angles between atomic orbitals (somewhere just above $90^\circ$) and because that large network is hard to break. And since we have a lot of molecular orbitals with rather small differences in energy, light of the entire visible spectrum can be absorbed. There will always be a corresponding transition. Because of this, black phosphorus appears black.

Red phosphorus is somewhere in-between. The network is not as extensive, it is not as ordered and thus there are not as many molecular orbitals with good energy differences. Therefore, not all visible light is absorbed but green light is preferentially absorbed (giving a red colour). And also therefore, the reactivity is somewhere in-between white and black phosphorus.

(There is also violet phosphorus, which is somewhere between red and black.)


for calcitriol:

Assuming only Z and E isomer 64 isomers are possible .For rest of isomers (3 ,ZZ,EZ,EE)
each will have 64 isomers.

For Z and E isomer 64 isomers are :

enter image description here

Quoting
MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE SIXTH EDITION
Michael B. Smith ,Jerry March


How to calculate the Gibbs energy for the vaporisation of solid ammonia?

Actually for this reaction $\Delta G=\Delta G^\circ+RT\ln Q$ holds only when $Q$ is used in terms of partial pressures. This is because the activity of an ideal gas is measured by its pressure, not its concentration. This equation can be derived from the definition of the Gibbs energy.


Why are chelating ligands often used to treat poisoning by heavy metals (e.g. mercury, lead)?

Lead, mercury and other heavy metals form complexes with both monodentate and polydentate ligands. But the complexes made with polydentate ligands like EDTA are much more stable than with monodentates. Usually metallic complexes become toxic if the central metallic ion (non-complexed) is released or not entirely surrounded by ligands. Monodentate complexes do not surround completely the central metallic ion, and they may be decomposed by the acidity of the stomach. Polydentate complexes resist to this acidity, and often they surround the metallic ions much better than monodentate. So if a person has ingested one of these heavy metal ions, its toxicity will be hidden by a complexing agent like EDTA.


Are there any molecules with the same spectrum?

In addition to the R/S chiral molecules mentioned by @user26143. There are also molecules with a distinct atomic composition that have experimentally indistinguishable spectra, even though theoretically there will be minor differences. One example, according to Rabitz et al. 2009 PRL (freely available for viewing) are flavin mononucleotide and riboflavine and in fact many other flavins. Due to there very similar structure you need unconventional techniques to be able to distinguish them.


As you are dealing with Grignard reagents, the carbocation argument is inadequate since the reaction is in basic media (carbocations are brønsted acids, so they would be easily deprotonated to produce an alkene). In fact, this mechanism is concerted (everything happens simultaneously, not in 2 distinct steps such as in the formation of a carbocation followed by a nucleophilic attack).

The Grignard reagent could attack either the methylene $\ce{(CH_2)}$ carbon or the methylidyne $\ce{(CH(CH_3))}$ carbon. The methylene carbon is attacked preferentially because its substituents are smaller than the methylidyne ones (2 hydrogens against one hydrogen and a methyl group). So steric hindrance is the key factor for predicting the product correctly in this question, not carbocation stability.

ps.: I drew a specific enantiomer just for visualization, but in this case, the absolute configuration doesn't matter.


In short, yes. Single-element ions are common for both metals and non-metals. Some simple examples: $\ce{F^{-}, Cl^{-}, I^{-}, S^2-}$ Multi-element non-metal cations are a bit more common: $\ce{NH4+}$ and $\ce{H3O+}$ being the obvious examples and there are plenty of anions like $\ce{CN-}$, $\ce{SO4^{2-}}$, $\ce{NO3-}$, etc. Things like proteins and polypeptides are also frequently ions in solution and may be positively or negatively charged depending on the pH.

Simply due to their electronegativities, all the single element non-metal ions are anions in solution, but in the gas phase, all bets are off and you can knock off an electron to get a cation, (or have negatively charged metal ions) so long as it doesn't bump into anything else.


What chemical regularly transported by road in the UK would cause serious injury/loss of life if it escaped in a road accident?

Organophosphates in general would be a great idea as you can go from potent insecticides to chemical warfare as it fits the drama of your story.


How can you identify the anode and cathode of a car battery of indistinct shape by using a potassium iodide solution?

By using the battery as a source of electricity in an electrolytic cell that contains potassium iodide as an electrolyte.
Electrolysis takes place where iodide ion is oxidized at anode and brown colour of iodine solution appears at anode.
At cathode hydrogen gas evolves which is indicated by formation of bubbles at cathode.


Why do people add H(core)+F in Hartree-Fock?

Actually, $F$ is double-counting the electron-electron interaction when evaluating the total energey. By adding $H^{\text{core}}$ and using a factor of $\frac{1}{2}$, which I suspect is implicit in $D$, one can obtain the correct energy. This is basically an occurence of the HF energy not being the sum of orbital energies (which are a result of $F$).


Synthesis of a weinreb amide from an acid

Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here


Hückel theory would still work if there would be overlap.

In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system.

There is the opposite view in Extended Huckel Methods whereby the assumption is that the overlap entries depend upon the off-diagonal elements. In ordinary Huckel theory, the diagonal elements are represented as
\begin{align}
\alpha_{ii} &= \langle \psi_i | \hat{H} | \psi_i \rangle \\
\hat{H} &= -\frac{\hbar^2}{2}\nabla^2 + V
\end{align}

But in the extended thoery, the offdiagonal elements are not denoted by $\beta_{ij} = \delta_{ij}$ but more as

\begin{align}
\beta_{ij} &= - \frac{C}{2}\Big \langle \psi_i \Big |-\frac{\hbar^2}{2}\nabla^2 + V \Big | \psi_j \Big \rangle\\
&= - \frac{C}{2} \left(\alpha_{ii} + \alpha_{jj} \right)S_{ij}
\end{align}
Where $S_{ij}$ is the usual notation for the overlap between the corresponding atomic orbitals.

Here $C$ is a constant, usually taken equal to $7/4$ and has units of energy.


Possible Synthesis of MVK from 2-Butanone?

As pointed out in the comments, there is no way to make it safe, as methyl vinyl ketone itself is a very nasty compound.

I am also not an expert on organic synthesis, but I found some procedures, that might help you in adopting a synthetic route. The one you suggested might work, but be aware, that you are dealing with a methyl ketone, which can undergo the haloform reaction quite easily.

For example, in [1] they describe the bromination with bromine and absolute methanol at the methyl group for 3-methyl-2-butanone at a nice moderate temperature range 0-10 °C. The yield is 95%. But this is not what you want. Adding an acid as catalyst might work in your favour though, as after protonation the partial positive charge is better stabilised at the higher substituted carbon.

In [2] they use N-bromosuccinimide and aqueous hydrogen bromide to introduce the halogen in α position. They convert hexanoic acid with thionylchlorid to the acid chloride and subsequently to 2-bromohaxanoylchloride. The reaction takes place in the also quite nasty carbon tetrachloride at about 85 °C. This might also work for your system, it's a bit warmer than you expected though.

In [3] they use sulfuryl chloride in carbon tetrachloride to convert pantane-3-one to 2-chloropentane-3-one at about 45 °C with about 80% yield. How selective this will work with an asymmetric ketone is not clear.

Once you have obtained the 3-halogeno-butan-2-one, things will not get easier. I found a patent [4], where they convert 3-chlorobutan-2-one via pyrolysis with sodium acetate at 500-550 °C.

I could not find a reference for an elimination, but that should also work as long as you use a bulky base.

Wikipedia [5] has a different route for the synthesis, which might be a bit more economical and easier.

Well, good luck!

Org. Synth. 1976, 55, 24.
Org. Synth. 1976, 55, 27.
Org. Synth. 2002, 79, 251.
J. R. Long: Preparation of methyl vinyl ketone. US 2256149 A, 16. 09. 1941.
http://en.wikipedia.org/wiki/Methyl_vinyl_ketone#Production


As the user above states, if you mix concentrated solutions of $\ce{Cu(II)}$ salts and $\ce{NCS}$ you get $\ce{Cu(NCS)_2}$ which is a black solid. I'd disagree with them that it's not longer a complex because it 100% is - and you are correct to assume that the local coordination is $\ce{Cu(NCS)_4}$. To be more precise, each copper is coordinated by $\ce{Cu(NCS)_2(SCN)_2}$, and there's actually a large Jahn Teller distortion so it could also be described as $\ce{Cu(NCS)_2(SCN)_4}$. They're right that the colour can't be easily explained by $d-d$ transitions (i.e. the kind you're thinking of w.r.t. ligand field splitting) though.

$\ce{Cu(NCS)_2}$ is black probably because of ligand to metal charge transfer (the same reason $\ce{Fe(III)NCS}$ is blood red) - i.e. on absorption you transiently form $\ce{Cu(I)}$ and $\ce{NCS}$ from $\ce{Cu(II)}$ and $\ce{NCS–}$. It's more complex that that for sure, but I think it's fair to say no one knows yet, because we only worked out the structure two years ago.

If you want a lot more information about $\ce{Cu(NCS)_2}$ we published on it here https://journals.aps.org/prb/abstract/10.1103/PhysRevB.97.144421 (on the arxiv at https://arxiv.org/abs/1710.04889).


Byproducts of LiAlH4 reduction of amides

Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required:  e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:

In the workup conditions, $\ce{H3O+}$ is followed by $\ce{^–OH}$. An aqueous acidic solution is often used to carry out the protonolysis step that follows the $\ce{LiAlH4}$ reduction (as shown in the following mechanism). The excess of acid that is typically used converts the amine, which is a base, into its conjugate-acid ammonium ion. Hydroxide is then required to neutralize this ammonium salt and thus give the neutral amine.

Although water itself rather than acid can be used in the protonolysis step, for practical reasons the acidic workup is more convenient. Thus, the extra neutralization step is required.

The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
$$\ce{2RCONR'R'' + LiAlH4 -> 2RCH2NR'R'' + LiAlO2}$$
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:

When less than stoichiometric quantities of $\ce{LiAlH4}$ are used for the reduction of primary amides, the corresponding nitrile may be formed:

Moreover, high selectivity toward the nitrile is also observed with amides containing electron-rich quinoline systems. If traces of water are present,
partial reduction of amides to aldehydes over $\ce{LiAlH4}$ can also comprise an undesirable competing reaction. Hydrolysis to the alcohol product has also been reported for secondary amide substrates.

GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References

Loudon, G. M.; Parise, J. Organic chemistry, 6th Ed.; Roberts and Company Publishers: Greenwood Village, Colorado, 2016. ISBN 978-1-936221-34-9.
Nystrom, R. F.; Brown, W. G. Journal of the American Chemical Society 1948, 70 (11), 3738–3740. DOI 10.1021/ja01191a057.
Smith, A. M.; Whyman, R. Chemical Reviews 2014, 114 (10), 5477–5510. DOI 10.1021/cr400609m.
Thiedemann, B.; Schmitz, C. M. L.; Staubitz, A. The Journal of Organic Chemistry 2014, 79 (21), 10284–10295. DOI 10.1021/jo501907v.


Solvation is the process by which a species is dissolved in a solvent. The most classic example of this is when a metal ion is dissolved in water. The electronegative oxygens in the water molecules are attracted electrostatically to the positive charge on the metal ion. A solvation shell of water molecules result.

According to this article, which written about the hydration of alkali metal ions:

  In spite of many conducted studies the knowledge of the structures and bonding properties of the hydrated alkali metal ions in aqueous solution is scarce and deviating.

Also according to another letter, which reports research done on the stability of the different possible forms of the ion, there is not a consensus on only one form. This letter discusses three possible forms of the ion all in the form of $n_1 +n_2$, where $n_1$ is the number of molecules in the first shell of hydration and $n_2$ is the number of molecules in the second shell of hydration. The forms discussed are:

demonstrating that the water molecules do not necessarily have to be coordinately bonded to the metal itself, but in fact that the second shell of hydration could include water molecules interacting with the water molecules in the first shell of hydration.

In short the hydration of sodium (and the other alakali metals) is not well understood yet, though it is under study.


What happens to electrons in metal complexes after excitation by visible light?

Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity.

$$\ce{S_0 ->[$h\nu_\mathrm{ex}$] S_1}$$

In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the solvent is actually heated up, although it might not be measurable.

There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and then emits a photon to reach the ground state.

$$\ce{S_1 -> S_0 + $h\nu_\mathrm{f}$}$$

This deactivation via radiation is known as fluorescence. Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one. This means that the emission spectrum is shifted to the red end of the uv spectrum ($\lambda_\mathrm{f} > \lambda_\mathrm{ex}$). The effect is known as Stokes shift.

A classical case for a fluorescent transition metal complex is $\ce{Ru(bpy)3^{2+}}$.

In acetonitrile, it absorbs around $\lambda_\mathrm{ex} = 460~\mathrm{nm}$. The fluorescence is observed around $\lambda_\mathrm{em} = 620~\mathrm{nm}$.

Other bidentate ligands - more rigid than 2,2'-bipyridine (bpy) - often used are 1,10-phenanthroline (phen), dipyrido-[3,2-a:2',3'-c]phenazine (dppz), and benzo[i]dipyrido-[3,2-a:2',3'-c]phenazine (dppn).

In other cases, the excited singlet state undergoes a change in the spin multiplicity (Intersystem crossing) to a triplet state.

$$\ce{S_1 ->[ISC] T_1}$$

These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as phosphorescence.

$$\ce{T_1 -> S_0 + $h\nu_\mathrm{p}$}$$

Keeping all these processes in mind, one question arises - particular from younger students or pupils:

If our metal complex absorbs a photon and forms an excited state with such a short lifetime, why can we see a nevertheless observe permanent colour?

The answer is in the enormous numbers!

Continuous irradiation with white light means that we send a huge number of photons. Moreover, the solution of our transition metal complexes in a cuvette (or a test tube) also contains more than one molecule. In fact, much more!

As a result, we have a huge number of incidents

The complex absorbs a photon of a certain energy, which means that one colour is "swallowed". The complex forms a short-lived "excited state"
the solution reflects the remaining light
the excited state complex deactivates again

Since this happens all the time while we shine the light on our sample and for a lot of molecules, we observe a summary of the events: a coloured solution.


First solution is perfect.

The issue is in the second solution.

If we start with 1 mole of $\ce{NH3}$ then with 20% dissociation we are left with 0.8 moles as 0.2 moles react these 0.2 moles give o.1 mole of $\ce{N2}$ and 0.3 moles of $\ce{H2}$

Hence $$K_\mathrm p = \frac{[\ce{H2}]^3[\ce{N2}]}{[\ce{NH3}]^2} = \frac{[\ce{0.3}]^3[\ce{0.1}]}{[\ce{0.8}]^2}$$


"Middle row anomaly" of the periodic table

This is due to the transition metal contraction.

Bromine has the electron configuration $\ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. $The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation states of bromine (I, IV, VI for the examples you give) because the ionization energies are higher than you might expect from a simplistic approach to periodic trends.


What would the pH change be if I drop one gram of sodium hydroxide into a pH 8.1 solution

If we know that the pH is 13 (let's take it as 13 for the simplicity), then we can back-calculate the concentration of sodium hydroxide in this solution from $\ce{14=pH + pOH}$. This gives us pOH=1, thus the concentration of sodium hydroxide of 0.1M. Since 0.1M sodium hydroxide is hardly a very concentrated solution, we can safely assume that 1 gram of the solution is going to be roughly 1 mL. Thus, with each 1 gram of the lab solution you're adding about 0.0001 moles of $\ce{NaOH}$ into your 1 liter of sea water giving you $1*10^{-4}$M solution with the pH = 10.

Using the same trick to calculate the concentration of $\ce{OH-}$ in your sea water as we did for the stock sodium hydroxide solution, we get about $1.25*10^{-6}$M, which is negligibly small compared to $1*10^{-4}$M solution that you get after adding sodium hydroxide to it. Thus, after 1 gram of sodium hydroxide in a liter of sea water your pH should go from 8.1 to about 10. If you add 2 grams, you'll get a $2*10^{-4}$M solution, which yields pH of about 10.3.


Can a charcoal generating process be a low carbon footprint incinerator?

If kept at 200°C with water under pressure, any organic molecules break down after a while, that's called hydrothermal carbonisation, basically how coal was formed geologically
With pure sugar, the gross reaction is
$\ce{4CHO -> 3C + CO2 + 2H2O }$.
Lignin is approximately $\ce{C31H34O11}$ (varies a lot of course), so
$\ce{C31H34O11 -> 31C + 6H2 + 11 H2O}$
or
$\ce{C31H34O11 -> 28C + 3CH4 + 11 H2O}$,
and other hydrocarbons, depending on the exact conditions.
You can also throw in finely shredded plastic, works the same.
The process obviously produces heat, and some $\ce{CO2}$, because once formed, it only very slowly reacts with hydrocarbon to carbon and water (you need higher temperatures and a potent catalyst for that).
Sulfur containing products can be handled by the same processes as in regular coal power plants, burn and wash out and turn into elementar sulfur.


What is GMEC (Global Minimum Energy Conformation)?

Yes. It is the conformer with the lowest energy out of all possible conformers.

There are also local minima which have a lower energy than conformations either side of the them on the energy diagram but do not have the lowest possible energy.

In this diagram showing the conformers of butane we can see that all three stable conformers; the two gauche conformers and the anti conformer are at local energy minima but the anti conformer is the global minimum.


Relative basicity of fluorine and oxygen in silyl ether deprotection

Because oxygen has a lower electronegativity than fluorine, oxygen is less stable with a negative charge (and more stable with a positive charge). This is why $\ce{OH^-}$ is more basic than $\ce{F^-}$ and why $\ce{HF}$ in water is an acid (hydrofluoric acid), while $\ce{H2O}$ in water is neutral.


Karl Fischer Titration simple explanation

Attempting an almost non-chemical description, you need to know that the concentration of a sample, $c$, equates the number of moles $n$ per unit of volume $V$; expressed as $c = n/V$.
For once, a known volume $V$ of your liquid to be analyzed for its concentration $c$ of water into is injected into a chamber equipped with a stirrer and electrodes.  There is a stock of reagent solution for which you know ahead of the analysis its concentration. You equally know ahead how many molecules of the reagent will react with how many water molecules to yield a balanced reaction equation (stoichiometry).  The reagent solution shall react with water only.
The addition of the reagent solution to the analyte is monitored; for one, you know the volume of the reagent solution added.  Because of this known volume added, and the known concentration of the stock of reagent solution, you know how many molecules of the reagent were added.  The mixing and stirring ensures that each molecule of the reagent quickly encounters a molecule of water to react with.
For two, during the addition of the reagent solution to the analyte, the electric potential between the electrodes dipped into the analyte is monitored. Initially, the addition of a little volume of reagent solution changes this potential only a little.  However, continuing the addition of small volumes of reagent solution, there is one point where this change of potential suddenly is large.  This is the point when 1) basically all the water molecules reacted with the reagent, 2) the electrodes now monitor an excess of the reagent in respect of the water molecules, and 3) the titration may be stopped.
Because you know the volume of the reagent (of known concentration) added and the balanced reaction equations used in this analysis, you may calculate the number $n$ of water molecules consumed to reach this point of sudden change of the potential.  However, because you know both the number of water molecules consumed as well as the volume $V$ of analyte initially injected, you may compute the concentration of water molecules in the analyte, too ($c = n/V$).

The Karl-Fischer analysis offers a rapid characterization of organic liquids (minute scale), provided they do not react differently with the reagent.  Samples containing alcohol, for example, may differ in the underlying reaction equations and stoichiometries from samples containing water.  Generally speaking, to work well, typical samples submitted do not contain more than 1 mol/L of water (about 18 g/L).