Difference between oxidation and formal charge

This is best explained in an example. Let’s take methyl isocyanide:

$$\ce{\overset{-}{C}#\overset{+}{N}-CH3}$$

This structural formula already includes the formal charges. How do we arrive at these? First, we draw a Lewis structure with the correct, required connectivity.

In this case, we have to know that a $\ce{CN}$ fragment is connected to the methyl group via the nitrogen. We can also find out that we need to form a total of seven bonds, three of which must be $\ce{C-H}$ bonds in the methyl group. Meaning that from the intermediate of $\ce{C-N-CH3}$ we need to add two more bonds but the carbon on the right-hand side already has a completed octet so these can only be between nitrogen and the left-hand carbon.

We also know that we should have a single lone pair; the only atom that can accomodate this is the left-hand carbon. All atoms now have a completed octet, our new intermediate solution is $\ce{C#N-CH3}$.

To determine formal charges, each bond is split homolytically, meaning that each bonding electron pair is split apart giving each atom one electron. Lone pairs are left alone, since they are, well, lone. The total we get is compared to the number of electrons an atom should have according to its position in the periodic table. This gives us:

$$\begin{array}{lcccc}\hline \text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{charge}\\ \hline
\ce{C} & 5 & 4 & 1 & -1 \\
\ce{N} & 4 & 5 & -1 & +1 \\
\ce{C(H3)} & 4 & 4 & 0 & \pm0 \\ \hline\end{array}$$

This concludes the determination of formal charges; the left-hand carbon gets a single formal negative charge while the nitrogen gets a single formal positive charge.

For oxidation states, a different method is employed. Here, all bonds are split heterolytically in such a manner that the more electronegative partner gets both electrons. Nitrogen is more electronegative than carbon which in turn is more electronegative than hydrogen. The resulting number is again compared to the number an atom should have; it will likely be different than before. This method will give us the following:

$$\begin{array}{lcccc}\hline
\text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{oxidation state} \\ \hline
\ce{C} & 2 & 4 & -2 & \mathrm{+II} \\
\ce{N} & 8 & 5 & 3 & \mathrm{-III} \\
\ce{C(H3)} & 6 & 4 & 2 & \mathrm{-II} \\ \hline \end{array}$$

If we add the three $\mathrm{+I}$ of the three hydrogens, we arrive at a neutral $\pm0$ again.

It is important to note that both resuts are quite different and the only thing in common is that they both add up to zero for a neutral molecule or the ionic charge of a molecular ion (unless you did it wrong). The formal charge rarely has any non-formal meaning except in ammonium derivatives where it denotes the actual centre of positive charge. The oxidation states corresponds to the actually observed electron density in an atom somewhat.

Both however are formal concepts and you need to know how to cope with them to use them to your advantage.


The formation of the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ due to addition of ammonia has mainly three functions.

1.
The Tollens test takes place under alkaline aqueous conditions. The complexation of $\ce{Ag+}$ with $\ce{NH3}$ prevents precipitation of brown $\ce{Ag2O}$:

$$\ce{2Ag+ + 2OH- <=> 2 AgOH <=> Ag2O + H2O}$$

2.
The Tollens test shall be selective for aldehydes. $\ce{[Ag(NH3)2]+}$ is a milder oxidising agent than $\ce{Ag+}$.

$$\begin{alignat}{2}
\ce{Ag+ + e- \;&<=> Ag}\quad &&E^\circ = +0.799\ \mathrm{V}\\
\ce{[Ag(NH3)2]+ + e- \;&<=> Ag + 2 NH3}\quad &&E^\circ = +0.373\ \mathrm{V}
\end{alignat}$$

The difference in redox potential can be explained using the stability constant of $\ce{[Ag(NH3)2]+}$:

$$K_\text{B} = \frac{\left[\ce{[Ag(NH3)2]+}\right]}{\left[\ce{Ag+}\right]\left[\ce{NH3}\right]^2}$$

$$\begin{aligned}
E&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Ag+}\right]\\
&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{\left[\ce{[Ag(NH3)2]+}\right]}{K_\text{B}\cdot\left[\ce{NH3}\right]^2} \\
E_{\ce{[Ag(NH3)2]+}}^\circ&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{1}{K_\text{B}} \\
&\approx E_{\ce{Ag+}}^\circ-0.059\ \mathrm{V}\cdot\log K_\text{B}
\end{aligned} $$

3.
The complex formation equilibrium slows down the overall reaction. A slow, controlled reaction is important for creating the desired silver mirror. If $\ce{Ag+}$ is reduced too quickly, colloidal silver metal would appear, which would create a black cloudy liquid.


How is adenosine synthesized in biology?

Adenosine is synthesised from Inosine Monophosphate (IMP) as the nucleotide form (adenosine monophosphate).
IMP does not accumulate in the cell but is rapidly converted to AMP and GMP. AMP, which differs from IMP only in the replacement of its 6-keto group by an amino is synthesized in a two-reaction pathway.

  In 1948, John Buchanan obtained the first clues as to how nucleic
  acids and their components are synthesized by describing the synthesis
  of purine ribonucleotides.

Buchanan investigated how this process occurs de novo by feeding a variety of isotopically labelled compounds to pigeons and chemically determining the positions of the labelled atoms in their excreted uric acid (a purine).

The biosynthetic origins of purine ring atoms. Note that C4, C5, and N7 come from a single glycine molecule but each of the other atoms is derived from an independent precursor.

The actual pathway by which these precursors are incorporated into the purine ring, was elucidated in subsequent investigations performed largely by Buchanan and by G. Robert Greenberg. These investigations showed that the initially synthesized purine derivative is inosine monophosphate (IMP),

(Here I am not going to explain the biosynthesis of IMP but focus on the former)

IMP is converted to AMP (or GMP) in separate two-reaction pathways:

In the first reaction, aspartate’s amino group is linked to IMP in a reaction driven by the hydrolysis of GTP to GDP + Pi to yield adenylosuccinate.
In the second reaction, adenylosuccinate lyase eliminates fumarate from adenylosuccinate to form AMP (adenosine monophosphate).

It is important to note that free adenine (and guanine) can be reconverted to corresponding ribonucleotides through salvage pathways requiring the enzyme Adenine phosphoribosyltransferase (APRT) (mediates AMP formation)

References

Voet and Voet Biochemistry: Section 28-1. Synthesis of Purine Ribonucleotides


Why does the BET equation take into consideration a multimolecular-layer, even though it deals with chemisorption?

Brunauer–Emmett–Teller (BET) explains the physical adsorption of gas molecules on a solid surface, and doesn't address chemisorption. Even the wikipedia link that you provided, makes that quite explicit. However, you are correct to think that it is in a extension of the Langmuir adsorption model, which only accounts for monolayers.

The assumption that underscore this model are as follows:

  Adsorptions occur only on well-defined sites of the sample surface (one per molecule)
  The only molecular interaction considered is the following one: a molecule can act as a single adsorption site for a molecule of the upper layer.
  The uppermost molecule layer is in equilibrium with the gas phase, i.e. similar molecule adsorption and desorption rates.
  The desorption is a kinetically-limited process, i.e. a heat of adsorption must be provided:

  these phenomenon are homogeneous, i.e. same heat of adsorption for a given molecule layer.
  it is E1 for the first layer, i.e. the heat of adsorption at the solid sample surface
  the other layers are assumed similar and can be represented as condensed species, i.e. liquid state. Hence, the heat of adsorption is EL is equal to the heat of liquefaction.

  At the saturation pressure, the molecule layer number tends to infinity (i.e. equivalent to the sample being surrounded by a liquid phase)

References: The original article

P.S: Good reading material on BET theory can be found here


Do I really need to wear protective clothing while using isopropyl alcohol?

You'd be surprised at what things you might not think of as "chemicals" can do.  One of my co-workers got a spritz of  oil in his eye while peeling an orange, and now he is afraid when I tell him that I eat orange rinds because they are rich in vitamin C.  The moral of the story: your eyes are so sensitive that at the very least, eye protection is a must with just about anything in the lab, even oils.  I second Jon Custer's suggestion of reading the MSDS.


Finding partial derivatives for an ideal gas and van der Waals gas

Use the equation of state, express $P$ as a function of $T$ and $V$, and find the derivative with respect to $V$, assuming $T$ to be constant.


There is no order for filling atomic orbitals of the same azimuthal quantum number (l) such as px, py, and pz since they are degenerate ie. have the same energy (however they lose their degeneracy in a magnetic field, see Zeeman effect). Therefore, one electron in a p sub-cell can be in either 3 of them and the order of filling doenst matter. For when more electrons are added in degenerate orbitals look up Hund's rule and the Pauli principle.


How to switch Mercury-vapor lamp between UVA and UVC bands?

TL;DR - You need two different lamps, one that has a phosphor to absorb the 254 nm light, and one that doesn't. Alternatively, you can use one lamp, and use a filter to select the wavelengths.

Mercury vapour lamps emit a significant amount of both UV-C light (around 254 nm) and UV-A light (around 365 nm).  This includes fluorescent bulbs used for lighting a room. The ultraviolet emission is a fundamental property of the element, not some kind of engineering accident. Light at these wavelengths is harmful, so you don't want it flying all over the place. The solution? Coat the inside of the lamp with a substance that absorbs the UV light. In some cases (but not all), the glass of the lightbulb itself can absorb the UV light.

Except now you're turning lots of luminous energy into heat, which is wasteful. A smarter solution is to coat the lamp with a phosphor, which absorbs the UV-C and UV-A light and reemits the energy at longer wavelengths. Depending on the phosphor, the reemitted light will have a different spectrum.

If you look at a 254 nm UV-C bulb and a 365 nm UV-A bulb (while they're turned off..), you'll notice that the 254 nm lamp is completely see-through, whereas the 365 nm lamp often has a white powdery coating the inside. This powder is a phosphor, which is chosen to absorb the 254 nm UV-C light, while letting the 365 nm UV-A light through (as well as other lower energy visible wavelengths). Some 365 nm UV-A lamps also have a coating to absorb the any visible light generated, such that relativity pure invisible 365 nm light shines through. This is popularly called a "black light".

Note that this does NOT imply every mercury lamp with a phosphor coating blocks the UV-C light it generates.


How to write equations for dissociation of ionic compounds in water?

Your ionic charges are not correct for iodine. Looking at your attempt.

As $\ce{Be}$ is in group 2, the ionic charge for beryllium ion is fine, but iodine is in group 17, so its ion is $\ce{I-}$.  When the ions dissociate, they become aqueous or (aq) as the state of matter.

Then the ionic charges need to balance, thus:

$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$

To balance the ionic charges in this example, you need 2 $\ce{I-}$ (iodine ions) to balance the $\ce{Be^{2+}}$ (beryllium ion).

A similar example (and further explanations) are provided on the UC Davis ChemWiki page Unique Features of Aqueous Solutions (including an example of the dissolution of $\ce{MgCl2}$ - another compound with group 2 and 17 elements).

So,

determine the group, hence ionic charge of each dissociated ion
balance these charges
state that the dissociated ions are aqueous

Now, use the process to determine the dissociation of $\ce{LiI}$


What is maximum energy level of molecules?

In the painting "Dream Caused by the Flight of a Bee" by Salvador Dali, among other strange items, we may see a fish flying in midair. There is also one in the drawing "Big Fish Eat Little Fish" by Bruegel the Elder. One may conclude that there is a flying fish in every picture out there. One may even publish a paper to that effect. But I believe you see the shortcomings of this way of reasoning.

Same thing here. Molecules don't have the maximum energy level. You may think of the energy required to break a  molecule, but that's not a level; rather, it is a lower bound of the continuous spectrum. You may think of the energy levels of its constituent atoms, but those are quite irrelevant to the energy levels of the molecule. Moreover, atoms don't have the maximum energy level either.

At this point you may wonder: what were the authors thinking? Were they (together with the editors) just lunatics? Not at all! Then what did they have in mind? Why, they said it loud and clear: the spectra of graphs. How are those related to the spectra of molecules? Is a molecule the same as a graph? Of course it isn't... unless we are looking at it within certain simplistic approximation where it is.

Yes, this is the old good Hückel method once again. You take a conjugated $\pi$ system, draw a graph of the molecule, solve it for eigenvalues, and get the $\pi$ energy levels of the molecule (only upside down). Naturally, there is but a finite number of those, and there inevitably is a maximum. That's what they mean when talking about the "maximum energy level of a molecule".

Now, these $\pi$ levels typically include the frontier orbitals (HOMO and LUMO), and those next to them, and answer pretty much any chemical question about the molecule. All in all, it is quite natural and easy to forget altogether about all other orbitals and energy levels of the molecule ($\sigma$ bonds and stuff). But they are still there, and what I said before is still true.

So it goes.


Using RDKit to count "RR, the number of rigid single or fused ring systems in the molecule"

My group has thought about this as we pursue better methods for finding conformers which would ideally include flexing rings.
I'm not aware of a current descriptor for "rigid" rings because it's one of those concepts that chemists discuss but is more of a heuristic for "I can move some of the atoms." (Which would imply you need some sort of molecular dynamics to determine - solvable but not fast to compute.)
Let's try something like this:

To get the number of fused ring systems, you'd go through the SSSR (smallest set of smallest rings) and see if a ring bond is also a member of another ring.  If so, that's a fused ring system. You'd then need to do graph traversal to find all ring bonds in the fused ring system (i.e., in cholesterol there's only one fused ring system).

To get "rigid single rings" you'd need to go around the ring atoms and see if there is at least one "flexible" ring atom. The best heuristic I've derived for a potentially flexible ring atom seems to be "both neighboring atoms only have single bonds and are not connected to each other."

This last heuristic would ignore 3-membered rings (e.g., neighboring atoms are connected), benzene (i.e., atoms have double/aromatic bonds), .. seems to get most things right.
IMHO, even fused ring systems can flex, which is why my group has centered on item #2 - finding ring atoms that can move. But if you're implementing someone else's heuristic from a paper, you'll need to carefully do the ring traversal in #1 to accurately count the fused ring systems.


Is the condenser for a Soxhlet extraction left open on top

First thing, you should NEVER perform any heated reaction with a stoppered condenser. You shouldn't even be allowed to work independently in the lab if this rule is not coming automatic to you. I am not sure what you mean by a special condenser being used in Soxhlet extractions. You can use any condenser that fits. Since you don't do a distillation you can use a shorter condenser so your whole Soxhlet set-up has a reasonable height.


Struggling to understand wavevector K for deriving energy levels in reciprocal space

The Hamiltonian for a particle in a periodic array of atoms is invariant (doesn't change) under translation by any vector of the form $n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3$, where the vectors $\vec{a}_1,\vec{a}_2,\vec{a}_3$ for the unit cell and $n_1,n_2,n_3$ are any integers. The set of all such translations form a group.

When a Hamiltonian is invariant under all elements of some group, it follows that its eigenfunctions can be written as functions of definite symmetry with respect to that group. A function of definite symmetry with respect to the group described above turns out to be characterized by a "wavevector" $\vec{k}$. This is completely analogous to situation for the hydrogen atom, where spherical symmetry implies $[L^2,H]=0$ which in turn implies that total angular momentum is conserved and that our eigenfunctions possess a quantum number $l$ (along with $n$ and $m$) that quantifies the wavefunction's total angular momentum. To make the analogy clear:

Hamiltonian of hydrogen atom commutes with arbitrary rotations $\to$ eigenfunctions of Hamiltonian have quantum number $l$
Hamiltonian of particle in periodic potential commutes with arbitrary translations $n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3$ $\to$ eigenfunctions of Hamiltonian have wavevector $\vec{k}$.

A function $\phi_{\vec{k}}$ of wavevector $\vec{k}$ happens to have a special behavior under group operations:
$$
\phi_{\vec{k}}(\vec{x} + n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3) = e^{i\vec{k}\cdot(n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3)} \phi_{\vec{k}}(\vec{x})
$$
At this point we usually note that if two wavevectors $\vec{k}$ and $\vec{k}'$ differ by some $m_1 \vec{b_1} + m_2 \vec{b_2} + m_1 \vec{b_2}$, where
$$
\vec{b}_i = 2 \pi \frac{ \sum_{j,k=1}^3 \epsilon_{ijk} \vec{a}_j \times \vec{a}_k}{\sum_{m,n=1}^3 \epsilon_{imn} \vec{a}_i \cdot \left( \vec{a}_m \times \vec{a}_n \right)}
$$
and $m_1,m_2,m_3$ are integers, then
$$
\left( \vec{k'} - \vec{k} \right) \cdot(n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3) = 2 \pi \left( n_1 m_1 + n_2 m_2 + n_3 m_3 \right)
$$
so that the wave vectors have equivalent symmetry properties. This allows us to restrict our consideration of wavevectors to those of the form $c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3$, where $ 0 \leq c_j \lt 1 $. More commonly we equivalently restrict the possible wavevectors to lie in the Brillouin zone.

As to why the concept is introduced, the answer is that recognizing that our energy eigenfunctions can be classified by some symmetry often enormously simplifies problem solving. Using again the hydrogen atom as an example, recognizing the Hamiltonian's spherical symmetry allowed us to break up a 3-dimensional Schrodinger equation into three 1-dimensional differential equations, which is an enormous simplification. For a periodic pontential, knowledge that our eigenfunctions have a wavevector $\vec{k}$ allows us to find the eigenfunctions by solving a differential equation over a single unit cell (as opposed to all space).


How can change in entropy be the same for all processes if the entropy production $\sigma$ is present for irreversible processes?

When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function.
The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings.
Since the entropy change for the system is independent of the process, the difference in entropy production between rev. and irrev. processes is seen in the entropy change of the surroundings.
More broadly, since path differences (which in turn means differences in heat flow and/or work flow) have no effect on the final state of the system, they instead always manifest themselves as differences in the final state of the surroundings.


Are all NO3- salts soluble in water? If so, why?

No, the correct way of putting it is
$$\mathrm{Almost~all~of~the~\mathbf{inorganic}~nitrate~salts~are~soluble~in~water. }$$

The families of organic nitrate salts are typically nitrates of azoles and imidazoles. Some bright examples are (R) & (S)-miconazole nitrates, isoconazole nitrate and econazol nitrate. Econazol nitrate (Other names: Spectazole, EN) is the most common organic nitrate salt, and is believed to be very slightly soluble in water, less than $0.1 \mathrm{\frac{g}{100~g}}$ (1, 2, 3)

In inorganic nitrate salts, with a bit of patience, I've noticed that barium nitrate, mercury(I) nitrate, and cobalt(III) nitrate are the least soluble of nitrate salts with a monoatomic cation. They are in the brink of insolubility, but I still wouldn't call them insoluble, since

Their solubility isn't that low at STP; somewhere around $\frac{5~\mathrm{g}}{100~\mathrm{g}}$. ($\ce{Hg2(NO3)2.2H2O}$ decomposes in water, so as RSC says)
Their solubility increases greatly with temperature.

The reason, you say?

Well, let's think of why one compound is soluble in water while the other isn't. When an ionic compound is formed, energy is released. This is called Lattice enthalpy in constant pressure. The process of formation of lattice from separate ions is usually highly exothermic. Therefore, when an ionic compound dissolves in water, and when this lattice "breaks up", the reaction would be highly endothermic.

So, to dissolve in water, the ions have to 'overcome' the lattice energy. How? Of course, the net resulting attractions between water molecules and the ions have to be stronger. This attraction is kind of a formation of a very very weak bond (and its weakness is the reason it's not categorized as such) and thus, is exothermic. This energy release when the dissolution occurs is called enthalpy of hydration, provided that the ions be in the gaseous state.

So, nitrate is a really big anion, with a single charge. The lesser concentration of negative charge results in relatively lesser lattice enthalpies. Also, the possibility of the creation of hydrogen bonds with water enhances solubility by increasing enthalpy of hydration.

That's why almost all of nitrates are soluble.


Although there are several sources of weighing error that would be minimized by using the weighing by difference method (as discussed in the comments), I want to illustrate what I see as the limiting problem, that of the relative uncertainty of the measurement, which roughly scales (no pun intended, really) with the mass being measured.

Lets use a bit of an exaggerated situation for illustration. In particular, we'll say that we have one balance that can measure down to $\pu{0.001g}$, and can handle up to $\pu{100g}$ total mass. In reality you would likely have a high-mass balance for weighing directly into the beaker, or a low-mass balance for using the weighing boat. We are just using one really good balance to simplify this illustration, but the concept is the same either way.

Lets say our beaker has a mass of $\pu{50g}$, our weighing boat is $\pu{1g}$, and we are to measure out $\pu{1g}$ of material.

The beaker plus the material to be measured will then have a measured mass of ~$\pu{51.00 +/- 0.05g}$. This means that the relative uncertainty in the measured mass of the material is no better than $\pu{5\%}$($\pu{0.05g}$ out of $\pu{1g}$).

The weighing boat plus the material to be measured out will have a measured mass of ~$\pu{2.000 +/- 0.002g}$. This means that the relative uncertainty in the measured mass of the material is $\pu{0.2\%}$ ($\pu{0.002g}$ out of $\pu{1g}$).

So, even with two measurements rather than one, using the weighing boat method results in much less total uncertainty in the measured mass of your material as compared to weighing a small quantity of material directly into a large beaker.


Why do some solutions eventually stop boiling and start "popping" as they get more concentrated?

What you witnessed was probably something similar to superheating/bumping. Essentially, when a solution is heated too rapidly, large pockets of vapor can form beneath the surface of the liquid that have an internal temperature higher than the expected boiling temperature of the liquid. In those situations, even small physical disturbances and/or further addition of heat can provoke the bubbles to burst explosively as the vapor is released. Usually, boiling chips (made of inert, inorganic materials like aluminum oxide, silicates, carbonates, etc.) are added to the reaction vessel in order to prevent that occurrence. The boiling chips provide an uneven and porous surface around which smaller and more numerous vapor nuclei can form, rather than the large bubbles that can form in smooth containers which have relatively few nucleation sites. Furthermore, the chips themselves provide additional surface area for the radiation of heat, allowing more uniform heat distribution. The likelihood of bumping can also be reduced by using wider reaction vessels, which allow superheated bubbles to escape without disturbing all of the surrounding liquid above them.


Hypervalency in elements in the second period

Normally when we talk about a single covalent bond, we are referring to a 2-centre 2-electron bond, which means that there are two electrons holding two atoms together.

Carbon never forms 5 bonds. The only exception that I know of is the $\ce{CH5+}$ methanium cation, the bonding in which can be explained by a 3-centre-2-electron bond. The same kind of bond appears in diborane ($\ce{B2H6}$). In both cases, the octet rule (or duplet rule in the case of the bridging hydrogens in diborane) is not violated. It is just that those 2 electrons are shared amongst 3 different atoms, so each "bond" is effectively half a bond (in MO theory parlance we say that the bond order is 0.5). You could think of it as three of the C-H bonds being normal 2-electron bonds, and two of the C-H bonds being half-bonds (having one electron each). The total number of electrons around carbon is therefore $3 \times 2 + 1 + 1 = 8$.

The neutral species $\ce{CH5}$ does not exist, because it has one more electron than the $\ce{CH5+}$ cation. That would mean that you either have to put 9 electrons around carbon, or put 3 electrons around hydrogen, both of which are of course not allowed.

The hydronium ion $\ce{H3O+}$ is not actually hypervalent. It is similar to the ammonium ion $\ce{NH4+}$ in that a dative bond is formed from the lone pair on O to a $\ce{H+}$ ion.


Mesylate the alcohol, react that with the enolate, epoxidise the double bond then reform the enolate.


Most electron components of computer chips and boards are semiconductor devices like transistors and diodes.
Silicon is, from the Wikipedia article on semiconductor devices:

  By far, silicon (Si) is the most widely used material in semiconductor devices. Its combination of low raw material cost, relatively simple processing, and a useful temperature range make it currently the best compromise among the various competing materials

Silicon is the second most common element in the earth's crust after oxygen, and present in a variety of silicate minerals.

Electronics grade silicon can be produced easily by molten salt electrolysis of these minerals followed by purification by zone refining.

Copper is a conductor, and therefor not suitable for semiconductor devices. Tellurium is one of the rarest elements in the earth's crust.


Why is distilled water such a poor conductor of electricity?

You are mistaken.

Water is not composed of these ions. There is ongoing autodissociation equilibrium reaction$$\ce{2 H2O <<=> H3O+ + OH- }$$ shifted strongly toward left, with concentration of ions  $\pu{1e-7 mol/L}$ at $\pu{25 ^{\circ}C}$ .

Therefore conductivity of water itself is very low.


Why is it that weak bases usually contain nitrogen?

The short answer is that yes, the fact that many weak bases contain nitrogen does have to do with nitrogen itself. However, it's not really accurate to say that weak bases "usually" contain nitrogen.

For simplicity, I'm limiting my explanation to aqueous solutions, the Brønsted-Lowry definition, and main-group compounds. A base must do one of two things: either remove a proton ($\ce{H+}$) from water or dissociate to produce something else (typically $\ce{OH-}$) that can remove a proton from water. Most compounds that produce $\ce{OH-}$ directly are ionic; even if they don't dissolve much (like alkaline earth hydroxides), these ionic compounds dissociate completely and are therefore strong bases.

A base that removes protons from water must have a free pair of electrons for the proton to bond to. In period 2, boron tends to form compounds that tie up all three of its electrons. Carbon compounds, likewise, tend to have no free electron pairs. Oxygen and fluorine compounds (e.g., water and hydrogen fluoride) have free electron pairs, but because of their high electronegativity these elements keep their electrons to themselves and are unlikely to grab protons from water if they already have a full octet. Nitrogen is the only element in this period that balances two factors: it has a free electron pair in many of its compounds (e.g., ammonia and organic amines) and has a low enough electronegativity that it can share those electrons with a proton to form a cation (e.g., ammonium, $\text{NH}_3$).

Covalent compounds of elements in the third period and below tend to be very weak bases, if at all. I'm not sure why, but I suspect that it's because bonds involving these elements tend to form much weaker dipoles than bonds involving nitrogen. The weaker dipole would limit their ability to forcibly pull protons off of water molecules.

All that said, there are many ionic compounds that form basic solutions due to containing the cation of a weak acid. In these cases the base is a cation, not a molecule. Certainly some of these cations contain nitrogen, but many do not. Examples without nitrogen include any soluble carbonates and bicarbonates and any salts of carboxylic acids.


It is not quite clear what do you mean by "in relation to a plane within itself or in relation to other molecules", but I think the answer is no, not all molecules are symmetric. Many of them belong to what is known as $\mathrm{C}_1$ point group, or, in other words they are are totally asymmetric. For more information on molecular symmetry and point groups read this Wiki article.


Nucleophiles in protic and aprotic solvent

From what I learned in organic I, their nucleophilicity does not change with the solvent because if the solvent is polar Protic, each nucleophile will be hindered in a similar manner (due to the solvent cage). If the solvent is polar aprotic, this hindrance does not matter. In both cases, nucleophilicity matches base strength.

I believe the solvent only affects nucleophile strength going down the periodic table. There are also polarizability effects for larger molecules but I’ve been told in general, a smaller molecule is a stronger nucleophile and larger molecules are weaker in aprotic solvents. This trend reverses for polar Protic solvents.

Also keep in mind, the strength of the nucleophile only matters for SN2 because of the backside attack being the slow step.


Why is the amount in a mole equal to the Avogadro's number?

Avogadro articulated the concept that is Avogadro's number, but didn't pick the number. From Wikipedia:

  Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

https://en.wikipedia.org/wiki/Avogadro%27s_law

The Number is a logical extension of this law: So what is that number in relation to the volume and what is the mass of that amount of material?  By agreement, a mole is defined as the mass, in grams, equivalent to the atomic weight of a substance. This was laid out before Avogadro's Number was actually determined.  Then it was simply a matter of measuring the mass of a single atom to determine how many atoms were required to have a Mole of a material.  Over time, Avogadro's Number has been refined to more and more accuracy.


What's the difference between enthalpy of combustion and heat of combustion?

Looking at Wikipedia for the definitions:

The standard enthalpy of combustion is the enthalpy change when one mole of a reactant completely burns in excess oxygen under standard thermodynamic conditions (although experimental values are usually obtained under different conditions and subsequently adjusted).

The heat of combustion $(\Delta H_c^\circ)$ is the energy released as heat when a compound undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon reacting with oxygen to form carbon dioxide, water and heat.

On first inspection, the two seem the same, however, there is a very slight different use in terminology and wording.

1.) What are standard conditions?

Referring back to Wikipedia:

Standard conditions for temperature and pressure are standard sets of conditions for experimental measurements established to allow comparisons to be made between different sets of data.

In chemistry, IUPAC established standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.987 atm, 1 bar)

As you know, the standard enthalpy of combustion can be referred to as as $\Delta H ^{\circ} _{\mathrm{total}}$ whose units are in kJ. The molar enthalpy of combustion is how much energy is released per kJ/mol. How is the standard enthalpy of combustion found?

Using the individual enthalpies of formation, and Hess's law, we can calculate the standard enthalpy of combustion, or $\Delta H ^{\circ} _{\mathrm{total}}$.

$\Delta H_{reaction}^\ominus = \sum \Delta H_{\mathrm f \,(products)}^{\ominus} - \sum \Delta H_{\mathrm f \,(reactants)}^{\ominus}$

The standard enthalpies of formation, or the negative values are located here https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table) and as well, generally in your textbook in an appendix along with the standard entropy of formation, and Gibbs free energy of formation, as all three are state functions.

2.) Why do the signs differ in the values from your textbook compared to the table in Wikipedia?

For the heat of combustion, it is generally expressed in units of higher heating value, lower heating value, and gross heating value. As the article states in Wikipedia, the heating value (or energy value or calorific value) of a substance, usually a fuel or food (see food energy), is the amount of heat released during the combustion of a specified amount of it.

In summary, the sign difference that I see is due to different terminology, they are specifically referring to units of HHV, LHV, and GHV and expressing those in the table below. Although, heat of combustion may also be calculated as the difference between the heat of formation $\big( \Delta H_f^\circ \big)$ of the products and reactants just as the standard enthalpy of combustion, it seems to me that the heat of combustion is specialized in a way to be applied to fuels, and as such, different from the standard enthalpy of combustion that we are accustomed to.

Perhaps someone could clarify a little more on this?


Biphynyls are a class of compounds with two phenyl rings attached by $\ce{C-C'}$ carbons in each phenyl ring (See Figure $\bf{\text{A}}$):

The phenyl rings can rotate around this $\ce{C-C'}$ bond (acts as an axis). Keep in mind that any molecule capable of rotating one part of it around fixed axis (a bond) is never planner. As a rule, if the rotational energy barrier is less than an activation energy barrier of $16$ to $\pu{19 kcal mol-1}$ at room temperature, the part of the molecule constantly rotate around the said axis. The experimental barriers for rotation about the $\ce{C-C'}$ bond of parent biphenyl (no substitution) are $\pu{6.0 \pm 2.1 kcal mol-1}$ at $\theta = 0^\circ$ and $\pu{6.5 \pm 2.1 kcal mol-1}$ at $\theta = 90^\circ$ with minimum energy barrier at $\theta \approx 44^\circ$ (Ref.1 & Ref.2; See Figure $\bf{\text{B}}$ for $\theta$). This values are less than required energy barrier to prevent rotation ($16$ to $\pu{19 kcal mol-1}$) at room temperature, thus, the biphenyl is rotate around $\ce{C-C'}$ axis (See Figure $\bf{\text{B}}$).

In addition, the rotation about $\ce{C-C’}$ can be restricted if each of the phenyl rings of a biphenyl has at least one same or different ortho substituent (e.g., 2,2'-dicarboxylic acid in $\bf{\text{C}}$). However, if these substituents are not large enough, it won’t prevent the rotation about $\ce{C-C’}$ and relevant conformers would not be able to isolate at room temperature (e.g., 2,2'-dicarboxylic acid in $\bf{\text{C}}$). If there are two different ortho substituents on each of two phenyl rings of a biphenyl (e.g., 6,6’-difluro-2,2'-dicarboxylic acid in $\bf{\text{C}}$), the rotation about $\ce{C-C’}$ would be not possible at lower temperatures since thermal (kinetic) energy is not sufficient to provide the necessary activation energy for racemization. As a result, the twisted $\theta = 90^\circ$ dihedral angle conformer becomes chiral. In order to inter-convert such conformers with their mirror image structures, a rotation through the higher energy coplanar form must be made. How easy this inter-conversion would be depends on the size of the ortho substituents. Unlike 2,2'-dicarboxylic acid, the two additional substituents in 6,6’-difluro-2,2'-dicarboxylic acid have a higher activation energy for racemization, and hence, can be resolved if care is taken to avoid heating them (Note: As a rule, an activation energy barrier of $16$ to $\pu{19 kcal mol-1}$ is required to prevent spontaneous room temperature racemization of substituted biphenyls). Since fluorine is smaller than a nitro group, 6,6’-difluro-2,2'-dicarboxylic acid racemizes more rapidly on heating than does the nitro compound (6,6’-dinitro-2,2'-dicarboxylic acid in $\bf{\text{C}}$). In fact, two conformers of 6,6’-dinitro-2,2'-dicarboxylic acid can be isolated at room temperature. Those conformational isomers that are isolable due to high energy barriers are called atropisomers.

For example, those conformational isomers have been studied comprehensively and computationally (Ref.3):

  Using B3LYP/6-311+G* and other methods, twist angles and torsional energies were obtained for biphenyl (BP), 2-halogen BPs, 2,2'-dihalogen BPs, and 3,3'-dihalogen BPs, the halogens ranging from $\ce{F}$ to $\ce{I}$. The results were compared with available gas phase and X-ray data. 2,2'-difluoro BP has a rotational double minimum, at $57.9^\circ$ and $128.9^\circ$, whereas the other 2,2'-dihalogen BPs have a single minimum at dihedral angles ranging from $84.9^\circ$ to $94.8^\circ$. All 3,3'-dihalogen BPs have a double minimum at about $45^\circ$ and $135^\circ$. Optimized twist angles and energy barriers were also calculated for 2,2'-dimethyl BP and for perfluoro as well as perchloro BP. Most structures are accounted for by steric effects. For 2,2'-dihalogen BPs, however, attractive forces also appear to play a role, as evidenced by the dihedral angle of 2,2'-dichloro BP lying well below $90^\circ$.

References:

Mercedes Rubio, Manuela Merchán, Enrique Ortí, “The internal rotational barrier of biphenyl studied with multiconfigurational second-order perturbation theory (CASPT2),” Theoretica chimica acta 1995, 91(1), 17–29 (https://doi.org/10.1007/BF01113859).
Otto Bastiansen, Svein Samdal, "Structure and barrier of internal rotation of biphenyl derivatives in the gaseous state: Part 4. Barrier of internal rotation in biphenyl, perdeuterated biphenyl and seven non-ortho-substituted halogen derivatives," J. Mol. Struct. 1985, 128(1-3), 115-125 (https://doi.org/10.1016/0022-2860(85)85044-4).
Friedrich Grein, “Twist Angles and Rotational Energy Barriers of Biphenyl and Substituted Biphenyls,” J. Phys. Chem. A 2002, 106(15), 3823–3827 (https://doi.org/10.1021/jp0122124).


What defines a proton as "labile" in NMR?

For your first question you may want to look up "proton exchange NMR", as partly addressed in for instance this post:

Proton attached to N or O in HNMR spectroscopy

$\ce{^1H/^2H}$ exchange is a solvent catalyzed process that results in rupture of the original O-H or N-H bond with formation of a new bond to previously H-bonded solvent $\ce{^2H/^1H}$. What keeps the exchange process from occuring is an activation energy barrier.  Available nonbonding electrons on O or N atoms in the solute or solvent which can interact with $\ce{^1H}$ or $\ce{^2H}$  affect the process, as does the pH (exchange can proceed through acid or based-catalyzed mechanisms). Catalysis of C-H bond rupture with formation of a new C-H bond has a larger activation energy barrier or may lead to chemically nonidentical products and reactants.

As regards the second question, if the (protonated) solute is dilute and the solvent is deuterated, there isn't a significant concentration of solvent $\ce{^1H}$ to drive reversal to the protonated state. For instance, if your molecule is present in water at 1 mM concentration and contains one exchangeable proton, and if all of the protons exchange and the solvent water was originally 100% deuterated, then the final concentration of protons in the solvent is $\pu{\rho = ( 0.001/110) mole/mole = 9 \mu mole/mole (or 9 ppm)}$ and the solute will exhibit a similar final isotopic concentration ratio (ignoring isotope effects).


How can I measure the amount of quinine in a home-made tonic water?

Without standard solutions and a fluorimeter, you can do a side-by-side comparison using a commercial tonic water and your solution. I just prevailed in getting a poor photograph of the beautiful blue fluorescence from a commercial tonic water (begins with an S), using 405 nm excitation (BluRay laser diode). The photo is below:

The two cuvettes contain tonic water (undiluted on the right) and diluted about 5 times (on the left). The laser entered from the right. The bubbles must be removed because they cause significant light scatter that gives red artifacts in the camera. By naked eye, it actually looks lots better: no red artifacts.

Removing the dissolved gas, to avoid bubble formation, is often done done by using an ultrasonic bath, sparging or by simply shaking the bottle repeatedly and waiting. Adding NaCl is a no-go because it quenches quinine fluorescence (a nice Stern-Volmer experiment).

Fluorescence is very sensitive, but the standard joke is that only 10% of all substances fluoresce ... and they are the wrong 10%. But this is a good case!

So, you would simply A-B the solutions: make dilutions of the commercial tonic water and see which one most closely matches your home-made tonic water. Cheers!

Note: you can use near-UV LEDs, i.e., 395 or so nm peak emission wavelength. The absorption max for quinine sulfate in dilute sulfuric acid is around 350 nm, and the peak emission wavelength is around 460 nm. But the photo shows that 405 nm works fine: it is not always necessary to be at the optimum excitation wavelength. Driving an LED is trivially simple: just a DC voltage supply, such as a battery or 'wall wart', and a current limiting resistor. Or use a constant current source that can output 20 or so mA.

One more thing: if you dilute the tonic water sufficiently, you will not be able to see the fluorescence with the naked eye. But a fluorimeter can detect it at even lower concentrations. Eventually, with sufficient dilution, you reach the limit of detection. But that is way below whatever you might want in usable tonic water.


In case your mercury didn't come with an assay of the supplier, or has been used before, you might want to remove oxidizable metals by repeatedly dripping the mercury though diluted nitric acid and distilled water.

Subsequent distillation of mercury is certainly possible (at 25 mm Hg or lower), and has been reported by E. H. Riesenfeld and W. Haase in Ber. Dtsch. Chem. Ges., 1925, 58, 2828-2834 (
DOI).

In their article Über die Destillation von gold-haltigem Quecksilber ( = On the distillation of gold-containing mercury) the authors examined the distillation of gold-saturated mercury and stated that after two distillations, the gold content in the purified mercury was lower than 1.5 ng $\ce{Au}$ in 1 g $\ce{Hg}$.

UPDATE

In Joel H. Hildebrand, PURIFICATION OF MERCURY, J. Am. Chem. Soc., 1909, 31, 933-935, (DOI), the author seems to discuss further different methods,

THE PURIFICATION OF MERCURY BY AN ELECTROLYTIC METHOD
Science, 1933,
78(2027), 414-415, (DOI) might be woth a look too.

However, please notice that these are rather old references. Please make sure, that the methods proposed are still in agreement with currents standards of lab safety and waste management!


The effect of temperature on the caffeine intensity

When coffee is being brewed in a moka pot, the water keeps evaporating until it reaches the coffee powder, then it condenses and falls down again, bringing some extracts from the coffee powder: it's just like a small reflux condenser.

When the pressure is high enough, the extract reaches the valve, and it escapes, giving the brewed coffee.

When temperature is too high, the pressure raises too quickly, and water fails to make a good extraction (just think about a reflux condenser which is kept running only for a few cycles).

If the temperature is appropriate, the water performs many evaporization/condensation cycles before it comes out of the valve, hence performing a better extraction (which, I guess, implies an higher caffeine level).

That's why usually a low temperature and a slow brewing are recommended.


What is the correct term of a chemical formula for very large molecules if not just "chemical formula"?

Chemical formulas of the type you describe are just as valid for large molecules as for small ones. The reason they are not commonly used is that the number of possible molecules described by a given formula goes up exponentially with the number of atoms.

Thus, it is much more common to describe larger molecules with more precise terms. In the case of polymers like proteins or nucleic acids, the primary sequence of the constituents (amino acids or bases) is often used if space allows. Alternatively, a protein or nucleic acid  might be referred to by a common name with which a sequence is associated.

Even for small molecules, IUPAC formal names or common names are used in preference to molecular formula in many cases, because the names provide information about the atom connectivity that is absent in the molecular formula. Other representations that include connectivity information are forms such as SMILES, which is a standardized system for representing connectivity in text.

In the scientific literature, names of molecules are often accompanied by pictorial representations that provide atom connectivity information in an easily understood format. For very large molecules like proteins whose connectivity cannot be easily represented in a small image, pictorial representations are more often used to show secondary and tertiary structural features.


Why is the reaction quotient Q always equal to one at standard state?

As already said in the comments, you cannot use any unit you want in the reaction quotient; it has to be formulated in terms of activity (for solutes) or fugacity (for gases).
The "standard state" of a chemical species is a reference state in which the properties of the chemical species are defined. This standard state is a theoretical state, it does not need to be practically reachable by the chemical species. By definition, solutes in standard state have an activity of 1, and gas have a fugacity of 1. Therefore, whatever you reaction, if all the species are assumed to be in standard state, the reaction quotient will be a product of 1s, that is, 1.


Well, that is just the way that it is...

You should also look at this a bit differently. Urea has a molecular weight of 60 and water has a molecular weight of 18.

$\dfrac{108}{60} = 1.8$ moles urea

$\dfrac{100}{18} = 5.6$ moles water

$\dfrac{5.6}{1.8} = 3.1$ molecules of water for every molecule of urea


When mixing lye and citric acid, which citrate will form?

Citric acid has three pKs: 3.1, 4.8 and 5.4.  If you add 3 equivalents of lye, you will get essentially only trisodium citrate.  If you add a lesser amount, you will have a mixture of the various citrates (at least from microscopic point of view), as well as some remaining citric acid if you add less than one equivalent.

One "equivalent" being one mole of lye per mole of citric acid.


Why is fabrication of ordered graphite substrates difficult?

Graphite is a semimetal, and no more mambojumbo than with metals, but it doesn't do anything with the problem.
Silicon/silica can be melted, purified, doped, can single crystals grown, can oxidize the surface, making the silicon semiconductor an silica insulator. By "can" I mean that engineeers know well tested methods, optimized for half a century.

Graphite cannot be reasonably melted, cannot grown to single crystals, cannot doped with similar techniques, highly anisotropic, so cutting maybe an issue, and when you make thin layers, it is hard to oxidize in a controlled way to make an insulator out of it. The resulted oxides are semiconductors, and there is a large variety of local structures that can appear with different electronic properties. By cannot I mean that no one really though about these problems until the last decade, so there might be equivalent solutions with e.g graphene, nanotube or other nanocarbon structures.


Why are real orbitals used to define the orbital of hydrogenic atoms instead of complex orbitals?

The main disadvantage of complex atomic orbitals (AOs) is that many of them are hard to visualize.  For example, try making a simple sketch explaining how $p_{-1}$ differs from $p_{1}$. Conversely the complex AOs have very few  advantages over real MOs - they are all eigenfunctions of $\hat{L}_z$ but that's not really important for molecular  and closed shell systems.

So with many key advantages and very few disadvantages real orbitals are understandably the most popular choice.


What chemicals are in tap water that gives it a slightly basic pH level

The tap water is likely "hard," i.e., contains some dissolved mineral salts, most likely (primarily) calcium and magnesium carbonates and bicarbonates. The anions of these salts are slightly basic, consuming $\ce{H^+}$ ions and thereby raising the pH. Carbonates (with the obvious exception of alkali metal salts) are only sparingly soluble in water, while bicarbonates are typically very soluble. A couple of other subtle factors may affect the pH:

Lower atmospheric pressure encourages the evaporation of dissolved carbon dioxide from water. Since dissolved carbon dioxide exists in equilibrium with carbonic acid, it would tend to lower water pH. If you live in an area with comparatively low atmospheric pressure, your tap water might have a lower concentration of $\ce{CO_2}$, which would favor higher pH.
At higher temperatures, bicarbonate decomposes to water, carbonate, and carbon dioxide. Carbonates are relatively insoluble, and the consumption of bicarbonate should drive forward the dissociation of any dissolved carbonic acid, so a higher temperature should have the effect of lowering pH.


Balancing the equation when MnO₂ is reduced to MnO by H₂(g)

This is balanced: I count the same number of elements on each side, which is the definition of balanced. What is not explicitly stated is that your hydrogen goes from $0$ to $+1$, and is oxidized, thus reducing your manganese from $+4$ to $+2$. But this all balances!


The answer given in the solution was wrong. It has been corrected, and it eventually proved my point.


Lab test to compare the reactivity of liquid 1-heptene and liquid heptane

According to this Wikipedia article:

  "In electrophilic halogenation the addition of elemental bromine or
  chlorine to alkenes yields vicinal dibromo- and dichloroalkanes
  (1,2-dihalides or ethylene dihalides), respectively. The decoloration
  of a solution of bromine in water is an analytical test for the
  presence of alkenes."

Alkanes tend to be much less reactive toward bromine, so yes, your bromine water test should allow you to distinguish between heptane and 1-heptene.


Iodine color change in different complexes

The formation of charge transfer complexes between iodine and aromatic hydrocarbons according to

\[\ce{Ar + I2 <=> Ar\cdot I2}\] is a long known phenomenon and has been described by Benesi and Hildebrand in A Spectrophotometric Investigation of the Interaction of Iodine with Aromatic Hydrocarbons. The following data for the absorption maxima of iodine in the VIS and UV range (wavelength in nm) in different solvents are taken from that article.

\begin{array}{lrrr}
\mathbf{solvent} & \lambda_{max}(\textrm{VIS}) & \lambda_{max}(\textrm{UV})\\
\hline
\textrm{benzene} & 500 & 297\\
\textrm{toluene} & 497 & 306\\
\textrm{p-xylene} & 495 & 315\\
\textrm{mesitylene} & 490 & 333\\
\hline
\textrm{trifluormethylbenzene} & 512\\
\textrm{n-heptane} & 520 & \\
\end{array}

Note that the effect is stronger in the UV than in the VIS range and the shift correlates with the (combined) $+I$ effects of the methyl groups, while a $\ce{CF3}$ substituent disfavours then formation of a charge transfer complex.

Ethanol, on the other hand, is a protic and polar solvent, nice to stabilize charged species.

One might argue that under these conditions, other species than in the case of aromatic solvents exist - the solvation of triiodide $\ce{I3-}$
 is certainly possible here.

In addition to the coordination of iodine by ethanol - we cannot exclude interaction with several molecules of ethanol:

\[\ce{n\ EtOH +I2 <=> I2\cdot(EtOH)_{n}}\]

it is conceivable that disproportionation of iodine and solvation of the resulting charged species might play a role here:

$$\ce{ I2\cdot(EtOH)_{n} + I2 <=>[\ce{EtOH}] I+\cdot(EtOH)_{n} + I3-} $$


Is the S–O bond strength in sulfur trioxide or sulfite anion larger?

Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-aromatic, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with respect to all $\ce{S-O}$ bonds.

The following valence molecular orbitals were derived at BP86/cc-pVTZ, Energy level decreases from top to bottom (enlarge).

The bonding picture is usually trivialised as each $\ce{S-O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=-\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three $\pi$ bonding orbitals, resulting in electron density deficiency.

However, the $\ce{S-O}$ bond in sulfurtrioxide is still stronger and shorter, than in the sulfite anion. (BO means bond order)

\begin{array}{lrr}
              & \mathbf{d}(\ce{S-O}) & \text{BO}(\ce{S-O})\\\hline
\ce{SO3}      & 1.458                & 1.4                \\
\ce{SO3^{2-}} & 1.578                & 1.1                \\\hline
\end{array}

The main reason for this is, that upon reduction of sulfurtrioxide to sulfite, $$\ce{SO3 + 2e- <=> SO3^{2-}},$$ you will fill the LUMO. As already stated, this orbital is antibonding with respect to the $\ce{S-O}$ bonds. This is energetically unfavourable for the molecule, therefore the molecule distorts to get rid of the symmetry constraints, i.e. it adapts the lower $C_\mathrm{3v}$ point group. This results in pyramidalization and a different bonding pattern.

You will now no longer find any $\pi$ type bonding orbitals. More accurately this molecule is best described consisting of three equal $\ce{S-O}$ covalent single bonds, and three lone pairs at each oxygen. This results in a charge at each oxygen of $q=-1$ and at the sulfur of $q=1$ (which does not fit the traditional Lewis type bonding description.)

This is also reflected in the molecular orbitals (BP86/cc-pVTZ, energy decreases from top to bottom, enlarge).

It is also very important to understand that the latter molecule would not be stable in gas phase, since it is too highly charged. Without compensation of solvent and/ or counter ions, this molecule will most certainly fragmentate.


Nobody has given an answer yet, so I'll give it a crack.
In electrolysis, as you probably know, there are two electrodes – the cathode (negative) and the anode (positive). We can deal with what happens at each separately, then tie them together.
At the cathode:
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
You're interested in the mechanism of these reactions. We know that water exists in solution in an equilibrium:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
At the cathode, electrons are supplied to the $\ce{H3O+}$ ions, reducing them:
$$\ce{H3O+ + e- -> H2O + H* }$$
With many reactions happening at once, the $\ce{H*}$ radicals react together, forming $\ce{H2}$. By removing the $\ce{H3O+}$ ions, it shifts the equilibrium of water to the right, to oppose the change – removing $\ce{H3O+}$ ions and increasing the concentration of $\ce{H2O}$:
$$\ce{H2O + H2O <=>> H3O+ + OH-}$$
Because the concentration of $\ce{OH-}$ ions doesn't change, the local solution around the cathode becomes basic – this can be done in a school lab. Add indicator. Electrolyse. Observe the color change near the cathode.
At the anode:
$$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Here I suspect the mechanism is a bit similar:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
We take the $\ce{OH-}$ and oxidise it:
$$\ce{OH- -> OH* + e-}$$
This results in a radical – the hydroyxl radical – being formed. The steps from here, to get to water being formed are probably quite complicated as a mechanism.


Well, the MSDS entry tells me that the solubility is 50%. Converting, we get 0.5 g per gram of water, or 0.0005 g per liter of water. This is $s=1.4\times 10^{-6}$ moles per liter of water.

$K_{sp}=(3s)^3\times(2s)^2=6.14\times 10^{-27} \rm M^5 $


Can artificial sea water be prepared by mixing sea salt with water?

Yes, it is possible though not easy to make artificial seawater by mixing sea salt and fresh water. An interesting example of this happens at the oceanarium in Lisbon, Portugal, which is one of the largest in Europe. After considerable investigation, it was found that the seawater present in massive quantities just outside the building could not be economically filtered, purified and adjusted to attend the surprisingly demanding and delicate needs of the marine organisms to be displayed. The chemical composition of seawater is actually quite complex, varying significantly with the location and time of collection. As with all life, marine life can be very sensitive to small changes in concentration of compounds, not to mention possible biological incompatibilities between the micro- and macrobiota present in seawater from different regions. The solution chosen for the Lisbon Oceanarium was to import some 16.5 metric tons of sea salt monthly from the Red Sea in Israel and mix it with local fresh water in order to replace the tank water. [1] [2] (Couldn't find a reference in English)

Having said that, you really should consult people experienced in setting up seawater aquariums before trying anything yourself. Seawater aquariums have significantly tougher upkeep and maintenance requirements compared to fresh water aquariums; you can't just mix any old cooking sea salt into water, it has to be in the right concentration, and replacement of the tank water has to be more gradual and frequent, not to mention that the salty water will corrode the actual aquarium more quickly. A good aquarium supplies store should be able to better inform you.

Edit: Now I realize you may actually be interested in cultivating shrimp rather than putting them on display. That's a whole different matter, many many times more difficult. You really need to contact some experts in marine cultivation.


I refer you to the second link in my answer for How does one recognized Td/Oh symmetry in molecules?, which (according to the abstract) describes some high symmetry species of unusual point groups $T$, $O$ and $I$. Additionally, the molecule [6.6]chiralane has $T$ symmetry.

Buckminsterfullerene is a well known $I_{h}$ symmetric molecule.

Don't know of any $T_h$ molecules but I'm confident they can be devised.


Why do ligands have such a small effect on overall absorption of a complexed ion?

This is absolutely not true. Many ligands can strongly dictate the color of transition metal solutions.

Your example picks three simple salts and then questions whether the color of the solutions (dictated largely by $\ce{[Cu(H2O)6]^{2+}}$) is very different. No, because the resulting majority complex in aqueous solution is likely identical.

Even taking a simple ammonia complex, e.g. $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ you can see a substantial color change.

Many copper acetonitrile compounds are weakly colored or colorless, with UV/Vis optical absorption occurring near the edge of the red to near-IR.

When you get into octahedral complexes, you can find substantial changes in color due to MLCT and LMCT absorptions, as well as significant modulation of the d-d transitions due to the ligand (i.e., high-field and low-field ligands).


Here is the reflectance spectrum for silver

It reflects the entire visible spectrum, absorbing light only in the ultra-violet region.  An ingot of elemental silver will have a silvery-white color.  In fact silver is said to be the "whitest" of all metals.  If you grind the silver up into smaller and smaller particles, the silver will appear progressively grayer or darker.  This is because smaller particles become less effective at reflecting light and produce more scattering.  Eventually when the particles are very small, even less light is reflected to the eye and the material appears black.  So depending upon the particle size, the color of silver can change from white to black.

Also, silver can react with chemicals present in air and tarnish.  It is the reaction of silver with sulfides in the air to produce silver sulfide that turns silver black, but this is through a chemical reaction.


Since PbCl2 is insoluble in water, how is it that lead leached from pipes due to Cl- reaction can become ingested?

Lead(II) chloride is not the sitting-at-the-bottom-like-a-brick type of insoluble. Rather it is the well-ma~ybe-I’ll-dissolve-a-little type of insoluble. So enough lead chloride will be dissolved — as Ivan mentioned, enough to poison somebody. Its solubility is given as $10.8~\mathrm{g/l}$ on Wikipedia. There are much less soluble salts like mercury(I) chloride ($2~\mathrm{g/l}$), barium sulfate ($2.4 \cdot 10^{-3}~\mathrm{g/l}$) or silver(I) iodide ($3 \cdot 10^{-6}~\mathrm{g/l}$). Note that for mercury(II) sulfide a solubility is not even given, it is merely termed insoluble (and is the sitting-at-the-bottom-like-a-brick type of an insoluble compound).

All values were taken from Wikipedia and are given at $20~\mathrm{^\circ C}$.


What would be the identification reactions of selenate(IV) ion?

You can use reducing agents to reduce the $\ce{SeO3^2-}$ salt to elemental selenium. This paper1 discusses the use of iron(II) salts in acidic medium (phosphoric - hydrochloric acid) as reducing agent. The reaction proceeds at r.t.
You can use other reducing agents like hydrochloric acid, sulfur dioxide, hydroxylamine hydrochloride, hydrazine hydrochloride. The reduced selenium has a red color which turns greyish-warm on warming, but be careful on overheating (boiling or evaporating turns to serious losses of selenium by forming volatile $\ce{SeCl4}$).
References

Rajkumar Kalaparthi , Srija Korapu , Padmarao Chekuri, 2020, New Spectrophotometric Determination of Sodium Selenate [Selenium (IV)] and Sodium Selenite[Selenium(VI)] with Iron(II), and Their Quantitative Analysis of Selenium (IV) – Selenium (VI) Present in A Binary Mixtures, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 09, Issue 05 (May 2020), DOI: 10.17577/IJERTV9IS050094
Vogel, A. I., & Svehla, G. (1996). Vogel's qualitative inorganic analysis


According to the IUPAC Gold book, these are known as azanes, from the information:

  Saturated acyclic nitrogen hydrides having the general formula $\ce{N_{n}H_{n + 2}}$

For the example you are looking at, according to the IUPAC document P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement
analogues as parents for naming organic compounds (p. 15) is named as 2-aminotriazane.


Mechanics behind auto-ignition temperature

To be clear, I'm no expert on the ASTM standards writers' thinking, and all I have to go on is the standard you've posted.

Yes, that's exactly right.  Some of the reaction's heat will be lost to the surroundings, but enough of it will go into the reaction mixture to raise its temperature and promote further, more rapid, combustion.
There are reactions that produce lots of heat but no flame, hydrogen peroxide decompostion, for example.  The standards writers had to come up with a clear definition for ignition, and as you note, they did: "the appearance of a flame accompanied by a sharp rise in the temperature of the gas mixture."  They even say that ignition is a subjective term, so they are careful to define it clearly.  There might be some chemical reaction that produces a large amount of heat in air with no visible flame (even in the dark, as the standard requires) and it might even qualify as an ignition event to you or I or any other sensible person watching, but it would not fit the standard.  They have to set some limit, after all, and that was what they set.
My original thinking on this was flawed; I misunderstood the air temperature reference.  The reason that they mention the air at all is that they are being specific that the reaction mixture should be an air/fuel mixture, rather than something else (oxygen/fuel, for instance.)  The predetermined temperature aspect of that point has to do with the AIT determination procedure itself.  They are setting the temperature of the system to some value T, then adding the fuel, then waiting for ignition.  So yes, the air in the flask is more or less in thermal equilibrium with the substance being tested.  It can, however, leak out and exchange with the air outside the flask. If the substance does not autoignite, the flask is purged and the temperature increased (including the air in the flask.)  I'm not sure whether they want a completely new sample of the substance for each temperature measurement.  It seems likely, since we're dealing with small amounts of generally volatile compounds.


In the ideal gas equation $$p_iV_i = nRT$$ $p_i$ is the pressure if there was no intermolecular forces between the molecules and $V_i$ is the free volume available for the molecules to do motion in, which is the volume of the container.
In real gases however, the pressure that you observe (real pressure) is less than the expected pressure. This is because of the intermolecular forces. So you convert the $p_r$ to a pressure which the gas would exert if there were no intermolecular forces.  $$p + a\frac{n^2}{V^2}$$
The free volume available to real gas is less than the volume considered in ideal gas equation. That is because Kinetic theory of gases made an assumption that the volume occupied by molecules of an ideal gas is negligible. But in real gases, the molecules are of considerable size. Thus, the full container will not be free for motion, some volume of the container will be occupied by the molecules. Let the volume occupied by one mole gas molecules be $b$, for n mole gas this is $nb$, so free volume available to gas is less than the volume of container $$V_i =  V_{\text container} - nb$$
Edit: In your question, the $V_r$ they're talking about is volume of the container. And $V_i$ is actual free volume available to the real gas.

Real Volume: Volume of the container
Ideal volume: Free space available
Real pressure: Observed pressure
Ideal pressure: Pressure if the gas didn't have any intermolecular forces.


Intensity of “surface plasmon resonance “ absorption peaks for a nano-rod

The intensity of a plasmon resonance is related to the magnitude of the transition dipole moment. Longitudinal resonances have a larger transition dipole moment compared to transverse resonances, and so have more intense absorption peaks. You can rationalize this by imagining the change in dipole moment from a longitudinal versus transverse electron density oscillation.


Safe and responsible disposal of aluminum chloride?

Sure, you can react the $\ce{AlCl3}$ with $\ce{NaHCO3}$ solution, and you'd also get $\ce{NaCl}$, in addition to $\ce{CO2}$.

Though $\ce{AlCl3}$ is toxic, it reacts readily with water, forming the less toxic chlorohydrate (used in antiperspirants!), and the baking soda should leave an innocuous sludge, safe to dispose of in a sink.

N.B.: The reaction is exothermic and bubbly, so use care when mixing in the $\ce{AlCl3}$ to avoids spatters.


Would this be a reasonable mechanism for the formation of a diazomethene?

The accepted mechanism is here: 1

KOH attacks the carbonyl of the urea, the tetrahedral intermediate breaks down with Me-N=N-O- leaving. This picks up a proton from solvent to give Me-N=N-OH. The second step is OH- removing a proton from the Me group of Me-N=N-OH which then loses OH- giving diazomethane.


How to find the balanced reaction equation for the decomposition of potassium dichromate in non aqueous conditions?

I think that part of your difficulty lies in incorrect naming of the products which are simply potassium chromate(VI) and chromium(III) oxide together with oxygen. The equation then becomes:

$$\ce {4K2Cr2O7 -> 4K2CrO4 + 2Cr2O3 + 3O2}$$

As for what is oxidized and reduced; the chromium is reduced from chromium(VI) to chromium(III) in going from $\ce{K2Cr2O7}$ to $\ce{Cr2O3}$ and some of the oxygen is oxidized from −2 in the dichromate to 0 in the oxygen gas.


What is the multiplicity of this signal in the ¹H NMR of glutamine in D₂O

You're thinking about deshielding the wrong way around, the most deshielded signal appears at the highers ppm value, so to the left. In this case it would be the signal at around 3.8 ppm, as the signal at 4.7 ppm is residual water.

The two signals at the right are multiplets, if you take a look they're obviously caused by rather complicated coupling patterns that you can't easily resolve. The reason for that is that the protons you see in this spectrum are not magnetically equivalent, even the CH2 protons are not magnetically equivalent. This leads to the complicated couplings you can observe here.


Stability of carbocation intermediates: secondary alkane versus primary allyl?

There are two competing effects at work here; hyperconjugation, and resonance involving the double bond.

According to this website, a secondary alkyl cation and a primary allyl cation have roughly the same stability.

  Carbocation Stability: $\ce{CH3+ < CH3CH2+ < (CH3)2CH+ ≈ CH2=CH-CH2+ < C6H5CH2+ ≈ (CH3)3C+}$

This is not really something you can predict from qualitative considerations - it is difficult to assess the strength of resonance versus hyperconjugation - but it requires experimental evidence, or detailed computer calculations to give a definitive answer.

In addition, the effect of adding more hyperconjugated groups is not linear. The difference in stability between tertiary and secondary alkyl cations is bigger than that between primary and secondary cations - tertiary alkyl cations are actually pretty stable.

Lastly, (thanks to @permeakra for pointing this out) the stability of carbocations in influenced by the environment which they are in. In particular, different solvents affect the stability of different cations to a varying degree and so the above order of stability should be taken as a rough guideline, not a hard and fast rule.


Electric Attraction Vs Resonance Energy in Electrophilic Aromatic Substitution Reaction

The positive charge, instead of being localized, is actually delocalized throughout the ring because of resonance, making the charge shared between three carbons, lessening the ability for the nucleophile to attack the positive charge.


Does diisobutylaluminium hydride (DIBAL-H) reduce only carboxylic acids? or  only carboxylic acid esters? or reduce both both types of compounds?

The answer is yes, it does reduce both. Not only them, it also reduces nitriles to aldehydes, and is a more selective reagent than lithium aluminum hydride (LAH) in the reduction of nitriles (Ref.1). About reduction of carbozylic acid, Miller et al. (Ref.1) found that DIBAL-H can be used to reduce benzoic acid to give 72% yields of benzyl alcohol in 1959. In same publication, they have also shown that esters can be reduced to corresponding alcohols as well.

In reduction of nitriles to aldehydes, following results were achieved and compared with the result obtained by Amundsen and Nelson (Ref.2) using $\ce{LiAlH4}$:

The summery of reduction of carboxylic acids and their esters are illustrated in following scheme, where two different groups (Ref.1 & Ref.3) achieved two different results for ester reduction under different conditions:

References:

A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).
L. H. Amundsen, L. S. Nelson, “Reduction of Nitriles to Primary Amines with Lithium Aluminum Hydride,” J. Am. Chem. Soc. 1951, 73(1), 242–244 (DOI: 10.1021/ja01145a082).
L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).


Why is buffer capacity at a maximum when the ratio of its components is 1?

With a buffer, you're trying to maximize your ability to neutralize acid or base. The log dependence of the pH on the ratio of interest in the Henderson-Hasselbalch equation is why the buffer works well:

$$\mathrm{pH} = \mathrm{pK}_{a} + \log\frac{\ce{[A-]}}{\ce{[HA]}}$$

Now, your ability to neutralize acid is determined by the amount of the conjugate base $\ce{A-}$. Your ability to neutralize base is determined by the amount of the acid $\ce{HA}$. When the ratio between these is 1, you have equal ability to neutralize either. If you're skewed too far one way or the other, then you will not have much of the respective species around to neutralize added acid or base.


Yes, the process is correct, but do you understand why?


Carbonic acid is not necessarily more stable than the carbonate ion.

Why carbonic acid ( p$\ce{Ka = 6.3}$ )  is not stable?

The reason for this is same as the one given for the instability of aluminium carbonate, iron (III) carbonate etc. That is, the carbonate ion has a highly polarisable electron cloud, which can be distorted by densely positively-charged species, such as the hydrogen ion, in this case. Thus, carbonic acid is not a very stable species and compared to the bicarbonate ion, which only has one densely-charged hydrogen ion distorting its electron cloud, it is definitely more unstable. Thus, the position of the 2nd equilibrium mentioned in the question should be to the left, favouring the more stable species.

Why is the bicarbonate ion ( p$\ce{Ka = 10.3}$ ) more stable than the carbonate ion  ?

This is due to there being a smaller negative charge on the bicarbonate ion and there being still 2 equivalent resonance structures to spread out the negative charge. Compared to the carbonate ion, which has twice the charge density but yet only 1 more resonance structure, the bicarbonate ion is more stable. Therefore, the position of equilibrium in the first reaction mentioned in the question should be to the left, not to the right.

Therefore, in solution, the dominant species among the three, should be the bicarbonate ion. This should be the correct line of thought.


Is KHF2 an ionic compound or a covalent compound?

Yes, $\ce{KHF2}$ is an ionic compound and a covalent compound. There is nothing exceptional about it; in fact, most compounds that we call ionic (not $\ce{NaCl}$, though) have covalent bonds in them. Consider $\ce{MgSO4}$ with ionic bonds between $\ce{Mg^2+}$ and $\ce{SO4^2-}$, and covalent $\ce{S-O}$ bonds within $\ce{SO4^2-}$. Consider $\ce{Na2CO3}$ with similar $\ce{C-O}$ bonds. These are the ways of the world.


Yes, your book is correct.
To rationalize it, throw away hydridization and electron exitation and work with concept of three-center four-electron bond. The idea is that in $\ce{XeF2}$ the real structure is a mix of $\ce{F-Xe+...F^-}$ and $\ce{F^-...Xe+-F}$, so the bonding is half-covalent and half-ionic. By packing more fluorine around the xenone atom, we force xenone into state with higher effective atomic charge, so the ionic part of the bonding becomes stronger as $\ce{Xe^{3+}}$ pulls fluorine anions much stronger.

same logic applies to many other cases, in particular sulfur oxides and sulfur fluorides.

For context, I quote experimental bond lengths from paper. Note, the paper itself quotes the values from other sources with slightly different values from different sources.

XeF2 ... 1.977

XeF4 ... 1.94

XeF6 ... 1.94 & 1.850 (two bond types)


$\ce{en}$ is bidenate so it has to take two neighbour corners from octahedron (let say equatorial position). The second $\ce{en}$ can bind with 1st nitrogen also on equator but then the 2nd has a choice: it can bind in equator or on axial position. So in theory it would be possible to see two isomers. The reality has however more to do with energy. $\ce{en}$ is very good ligand and $\ce{Ni++}$ makes very stable quadratic planar complexes. For that reason both $\ce{en}$ goes into equatorial position and the weaker ligands $\ce{Cl-}$ occupy axial positions.


How can silver work as a disinfectant in a bathtub if precipitates in the presence of chloride?

If you're referring to chlorination of public water supplies, keep in mind that this is likely not the main source of chloride ion. Elemental chlorine disproportionates via the following reaction:

$\ce{Cl2 + H2O <=> HCl + HOCl}$

Most chloride ion probably derives from various natural and industrial sources, as well as miscellaneous human activity. One source (somewhat outdated) I found indicates an average chloride ion concentration in U.S. public water of $11.5$ mg/L, which equates to a molarity of $3.24 \cdot 10^{-4}$. Using the $K_{sp}$ of $AgCl$ you can calculate the concentration of silver ion necessary to produce a precipitate, but given that it has a value of $1.77 \cdot 10^{-10}$ under standard conditions, you can assume that a fairly minimal amount of silver would be required, even with the large volume of a tub and increasing solubility in hot water. It's likely the case that any precipitate is just finely dispersed in solution and not clearly visible.


Aromatisation of a ring which includes nitrogen

This would be the mechanism I propose for the reaction:

I have taken the $\ce{pH}$ of the medium to be $4.5$. That should explain the protonation of the $\ce{-OH}$ groups.


Your error occurs when you write that

$$dS_{tot}=\frac{\delta Q}{T_{r}}-\frac{\delta Q}{T}$$

since this assumes that

$$\delta Q=-TdS$$

which implies reversibility and therefore that $dG=0$ at constant T and p.

The more general statement is that

$$dS_{tot}=\frac{\delta Q}{T_{r}}+dS$$

which means that

$$dS_{tot}=-\frac{dH}{T_{r}}+dS$$

(since positive heat here means exothermic)

Then

$$T_{r}dS_{tot}=-dG+(T_{r}-T)dS$$

or

$$dG=-T_{r}dS_{tot}+(T_{r}-T)dS$$

Now if you let $T=T_{r}$ (thermal equilibrium with surroundings) then

$$dG=-TdS_{tot}$$


Why is 2-fluorobenzoic acid more acidic than 4-fluorobenzoic acid despite its intramolecular hydrogen bonding?

You seem to be asking the question "Why does this intramolecular hydrogen bonding not cause a considerable effect to the $\mathrm pK_\mathrm a$ values of these compounds?"

TL;DR That is because this type of bonding is only present in a specific conformer only and this type of conformer of the compound 2-fluorobenzoic acid exists in low population levels, 2%$^{[1]}$ to not affect the ${\mathrm pK_\mathrm a}$ of the compound considerably.

Regarding the conformers of the 2-FluoroBenzoic Acid

A throughout conformational search on the B3LYP/6-311++G(d,p) potential energy surfaces of the
studied molecules (the mono- and di- ortho fluoro- or/and chloro- substituted benzoic acids, plus the
parent benzoic acid, for completeness) was undertaken. The identified conformers are presented$^{[2]}$

Here both the cis conformers are planar and low energy while only one trans conformer is planar while both of them are relatively high energy of which Trans II is of relatively lower energy.

The two cis conformers have rather similar energies, with cis-II (where
the fluoro-substituent stays the same side of the molecule as the acid oxygen atom and a C-H . . . O=
stabilizing interaction exists), being the most stable form. The cis-I conformer has a relative energy of
2.82 kJ·mol−1
. In this conformer, the C-H . . . O= interaction present in cis-II is replaced by a weaker
stabilizing C-H . . . OH interaction. The C-H . . . O= interaction is more efficient in stabilizing cis-II than
the C-H . . . OH interaction in stabilizing cis-I due to the more favorable localization of the interacting
carbonyl oxygen lone electron pair, which stays in the molecular plane, whereas in cis-I both lone
electron pairs of the acid oxygen atom are out of the molecular plane. On the other hand, the interactions
between the fluoro-substituent and the oxygen atoms, which are of repulsive nature, also favor a lower
energy for cis-II compared to cis-I. This can be rationalized in similar terms as for the interactions
involving the ortho hydrogen atom discussed above. In cis-I, the F . . . O= repulsion is stronger than the
F
. . . OH repulsion in cis-II because the interacting lone electron pair of the carbonyl oxygen atom is in
the same plane as the fluorine atom in the former conformer, while the lone electron pairs of the acid
oxygen atom in cis-II are both out of the plane in relation to the fluoro-substituent.$^{[2]}$

It is important to note that though the population level of Trans II are low, it is still a stable compound and nevertheless has been detected.$^{[3]}$

So.. Why is 2-fluorobenzoic acid more acidic than 4-fluorobenzoic acid?

I could not find any research for this but I could make an educated guess that the acidity of 2-fluorobenzoic acid comes from the similar stabilization as that in salicylic acid.$^{[4]}$ in the conformer cis II as shown below:-

References:

Kuş, N.; Fausto, R. Effects of the matrix and intramolecular interactions on the stability of the higher-energy conformers of 2-fluorobenzoic acid. J. Chem. Phys. 2017, 146 (12), 124305. [DOI: 10.1063/1.4978794]
(https://doi.org/10.1063/1.4978794).
Ogruc Ildiz, G.; Fausto, R. Structural Aspects of the Ortho Chloro- and Fluoro- Substituted Benzoic Acids: Implications on Chemical Properties. Molecules 2020, 25 (21), 4908. DOI: 10.3390/molecules25214908.
Daly, A. M.; Carey, S. J.; Pejlovas, A. M.; Li, K.; Kang, L.; Kukolich, S. G. Gas phase measurements of mono-fluoro-benzoic acids and the dimer of 3-fluoro-benzoic acid. J. Chem. Phys. 2015, 142 (14), 144303. DOI: 10.1063/1.4917031.
Dunn, G. E.; Penner, T. L. Effect of intramolecular hydrogen bonding on the relative acidities of substituted salicylic acids in benzene solution. Can. J. Chem. 1967, 45 (14), 1699–1706. DOI: 10.1139/v67-274.


Acetaldehyde with few drops of concentrated sulphuric acid

You're not wrong.

Your mechanism describes the classic acid catalyzed synthesis of 3-hydroxybutanal by dimerisation of acetaldehyde (Borodin 1869, Wurtz 1872).

Image: Whoopie23, Wikipedia

However, under different conditions acetaldehyde forms different addion and condensation products. E.g. with concentrated sulphuric acid at low temperatures metaldehyde is formed whereas at room temperature paraldehyde (the answer given in your book) is formed.


What would be the major products of this reaction?

I will give this my best shot. There doesn't appear that there is any conclusive evidence for this reaction since some sources say elimination is favoured while others say substitution is the major product. However, for this reaction, the difference likely resides in the fact that it is an allylic bromide. This is extremely favourable for substitution reactions because the intermediate will be very stable. Just because of that, substitution may be favoured over elimination and result in the major product being Br substituted for a deuterium atom, just as your book mentioned.


Does CHT=CHT have a geometric isomerism because of tritium atoms?

Yes isotopes can induce geometric isomerism.  You can induce chirality by using different isotopes of atoms; the molecule $\ce{CHDClF}$ is chiral with the deuterium atom (you do not need radioactivity) while ordinarily $\ce{CH_2ClF}$ is not.  Chirality won't work for ethylene (that molecule is still planar and thus mirror-symmetric about its own plane no matter what iostopes are used for any atoms), but cis-trans isomerism with two deuterium atoms still can work.


Reaction of cyclohexene with hydrochloric acid

Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:

After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:

A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?)
B: rearrangement of the carbocation (why did it happen?)
C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)

The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).


As can be seen from the periodic table below, oxygen, sulphur, and selenium belong to the same group, and they are all non-metals, so one should expect their chemical properties to be similar.

Having established their chemical similarity, one can transform the two ions in question into more familiar forms.

In thioformate anion ($\ce{HCSO-}$), we are not too familiar with sulphur $\ce{S}$, so let us transform it into the more familiar oxygen $\ce{O}$ whose chemical properties are similar. Then, it becomes formate ion ($\ce{HCOO-}$), whose resonance structure is well-known:

Just transform one oxygen back to sulphur and we are done:

In selenocyanate anion ($\ce{[SeCN]-}$), we are not too familiar with selenium $\ce{Se}$, so let us transform it into the more familiar oxygen $\ce{O}$ whose chemical properties are similar. Then, it becomes cyanate ion ($\ce{[OCN]-}$), whose resonance structure is well-known:

Just transform the oxygen back to selenium and we are done:

Copyright issues:

The first image is copyrighted by Michael Dayah.
The second image is from Wikipedia.
The third image is a derivative work of the second image.
The fourth image is taken by HGTCChem.
The fifth image is a derivative work of the fourth image.


Why is a LCAO necessary within Hartree Fock?

Even with single-particle systems, analytical solutions are only possible for a small subset of quantum mechanical problems: https://en.wikipedia.org/wiki/List_of_quantum-mechanical_systems_with_analytical_solutions For example, if you were to change the external potential of the particle in a box into something even a tad weirder, like a polynomial function, I would surmise that it is impossible to solve it analytically. The same goes with the one-electron operators in Hartree–Fock theory, because you have the mean-field potential to contend with.

Therefore, we need to bring in numerical methods. One way of doing that is to express the eigenstates as a linear combination of certain basis functions. If the set of basis functions – the basis set –  is complete, then in principle we can express any function we like as a linear combination of these basis functions. Mathematically, this is expressed by a "completeness relation":

$$\hat{1} = \sum_i |i\rangle\langle i| \quad \Longleftrightarrow \quad |f\rangle = \hat{1}|f\rangle = \sum_i |i\rangle\langle i | f \rangle = \sum_i c_i|i\rangle \text{ where }c_i = \langle i | f \rangle$$

Unfortunately, in the case of an atom or molecule, we need an infinite number of basis functions to have a complete set. This is obviously not possible, so we need a finite number of basis functions, judiciously chosen such that the result obtained using this limited basis is close enough to the result obtained with the infinite basis. The AOs (or approximations to them) are extremely convenient in this regard, and thus LCAO is born.


Reactivity of primary and tertiary alkyl halides

in this question, I think, we are not talking about priority of SN2 or E2, but according to the Clayden, (D) would go through E2 and (C) would choose SN2.

But we know that reactivity of tertiary halides are much more than primary ones. So, I think (D) is the right choice, either.


Elimination of bromine with sodium iodide and electrocyclization

Pöytä: Your conclusion that D contains four bromine atoms symmetrically distributed is sound. The treatment of D with NaI serves to substitute one, if not two bromine atoms with iodine to form "J". Iodide acts as a reducing agent in a vinylogous fashion. (Compare this reaction with the reduction of a vicinal dibromide with iodide to form an alkene.) The stereochemistry of E is likely to be E,E which arises from a transition state for elimination that avoids steric interactions involving bromine. I am assuming that the "electrocyclization" is thermal since there is no indication that light was involved. Achiral E will undergo a conrotatory electrocyclization in two directions to give racemic F and, eventually, diiodide G via SN2 displacement. (The electrocyclization may be rationalized as involving the HOMO of a butadiene or octatetraene. Look here for a discussion.) Reductive elimination of iodine via Zn-Cu couple affords benzocyclobutadiene H. (The elimination is akin to the formation of ethylene from 1,2-dibromoethane with magnesium.) The strained, reactive double bond of H undergoes a Diels-Alder reaction with cyclopentadiene to form adduct I. The stereochemistry shown arises from a so-called "endo" transition state.I found this question somewhat amusing in that information was provided for all species save the final product. ;)


Why are there 14 electrons in the third shell of iron atom?

Your doubt is - Total number of electrons in 3rd shell  (i.e. electrons from 3s + electrons from 3p + electrons from 3d) is not yet 18 but still how we filled up the 4th shell?

You have a general misconception than atomic orbitals always follow the 2,8,18... electron filling rule. According to this 2,8,18.. electron filling rule, 2 electrons should get filled up in 1st shell, then 8 electrons in second shell, then 18 electrons in 3rd shell........

Remember this rule mentioned above is completely elementary! Just remember 2,8,18... electron filling rule is actually a waste. Simply I would recommend forget about that 2,8,18... rule. It is very elementary rule... simply forget it. In 2,8,18... electron filling rule, "lower energy level orbitals gets filled up first"- this statement is not considered in that elementary 2,8,18..rule. But this statement which that elementary rule doesn't consider is very important statement to be followed while filling up of electrons. So remember 2,8,18... electron filling rule is an elementary rule for beginners who  just start to read about atoms and orbitals! Forget about that rule.

Its not necessary that 3rd shell has to have 18 electrons for 4th shell filling up to start.
The electron filling depends upon the energy levels of the orbitals. Lower energy orbitals get filled up first.

Iron's configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6. You can note from the configuration, 4s gets filled up before 3d . The reason behind this is lower energy orbitals are filled up first. Even though 4s belongs to 4th shell, still its energy is less than 3d which can be reasoned out by (n+l) rule. 4s energy is less than 3d, hence the two electrons which you expected to be going to 3d goes to 4s before 3d filling starts. Hence you end up having 14 electrons in 3rd shell and 2 electron in 4s. Remember 4s got filled up before 3d and the reason is 4s energy is less than that of 3d.


What's the convention for the symbols of enthalpies of formation, reaction, etc.?

I think you have already answered your question… IUPAC's usage should be a pretty good indicator of the proper notation.

There is, however, an even better reference than the Gold Book: IUPAC's Green Book, or Quantities, Units and Symbols in Physical Chemistry (full text here), on section 2.11.1 Other symbols and conventions in chemical thermodynamics (page 59 and following):

  (iii) Examples of the use of the symbol ∆
  The symbol ∆ denotes a change in an extensive thermodynamic quantity for a process. The addition of a subscript to the ∆ denotes a change in the property.

It says later:

  [∆vapH] can also be written ∆Hvap, but this usage is not recommended

The reason is that ∆Hvap could be understand as a change in Hvap, which it is not. It a change of H during a process, so the subscript should be on the ∆.


On the more improvised/DIY side, you may be able to get away with something like a gooseneck clamp of the sort commonly used for microphones; something like this:

(from here)

Obviously they're not really designed to be exposed to hazardous chemicals, though, and you'll want to make sure that it can handle the weight you need to put on it without bending further.


Why doesn't H₄CO₄ exist as a stable compound?

Wikipedia reports that calculations show orthocarbonic acid spontaneously decomposes to the ordinary (meta)carbonic acid $\ce{H2CO3}$ plus water.  However, orthocarbonate esters $\ce{C(OR)4}$ do exist.

The orthoacid may, however, become stable at high pressure [1, with a nontechnical summary given in 2] and possibly exist within Uranus and Neptune.  It is further posited that $\ce{H2CO3}$, ordinarily stable only in aqueous solution, becomes a stable solid in the interiors of icy moons such as Europa, possibly impacting interactions thought to be necessary for the development of subterranean life.

A final note: the swastika-like shape of the orthocarbonic acid molecule has led unfortunately to the name "Hitler's acid".  Actually the swastika long predates the Nazis and retains non-Nazi usages to this day.


Reaction of urea and thiourea with nitrous acid

As implied in your link, the reaction of urea and nitrous acid will follow the mechanism you posted, repeating a second time with the nitrogen on the remaining amide. As for why the nitrogen is more nucleophilic, I can speculate, but I don't want to be wrong.

The difference for thiourea is that sulfur can also act as a nucleophile instead of the nitrogen in the thioamide. The product is dependent on the pH. You can read more about it here. Here is the mechanism:

That's a thiourea that I squeezed in on the far right as a product from the strong acid. Sorry for the mess.


Lewis acids are by definition electrophiles. Electrophiles love electrons, or negative charge. Boron has an empty 2p orbital and there exists a strong partial positive charge on the boron due to the extremely electronegative fluorine atoms covalently bound to boron. This strong partial positive character, coupled with a vacant orbital, makes BF3 a potent Lewis acid and thus an electrophile.

That being said, note that acidity and basicity are both thermodynamic properties, while electrophilicity and nucleophilicity are both kinetic properties. That is, while you can label a Lewis acid an electrophile, it would be incorrect to say, for example, that because a certain Lewis acid is a "strong" Lewis acid, then it must be a strong electrophile.

Formation Constants; Lewis Acidity


Alcoholic KOH mechanism with alkyl halides

With a strong base $E2$ will always predominate over $E1$.

Both elimination mechanisms have a preference for tertiary substrates over secondary substrates over primary substrates. This behavior is on contrast with the substitution reactions, where steric factors are important for $S_N2$. Steric factors are less important for the proton transfer necessary for $E2$.

Since $E2$ and $E1$ have the same substrate preference order, you may feel that you cannot decide which is more likely. However, $E2$ is usually faster because $E2$ is bimolecular. Consider the rate laws:

$\mathrm{rate}_{E2}=k\mathrm{[substrate][base]}$
$\mathrm{rate}_{E1}=k\mathrm{[substrate]}$

If we make the assumption that the rate constants are of similar magnitude, then increasing the concentration of the base increases the rate of $E2$.


The Pitt Quantum Repository can get you what you want, at least for molecular species.

Go to the PQR at http://pqr.pitt.edu.
Search for a molecule, like benzaldehyde.
Click "Detailed" on the properties pane.
You can see HOMO and LUMO in the list.
Scroll down to the bottom of the page and click on the link to "Get JSON data".
The JSON file will open in your browser or you will download. You should be able to open it with a text editor. There will be a list of orbital energies for the set of molecular orbitals computed from the valence atomic orbitals. For example, for benzaldhyde.

"homo": -10.168,
"lumo": -0.712,
"orbitals": [
"-31.27648",
"-30.67710",
"-26.81219",
"-26.23865",
"-23.01445",
"-21.46816",
"-20.14594",
"-18.05465",
"-16.39792",
"-15.70143",
"-14.95762",
"-14.67105",
"-14.63486",
"-14.38581",
"-13.35746",
"-12.66484",
"-12.26937",
"-10.28746",
"-10.23037",
"-10.16786",
"-0.71215",
"-0.10196",
"1.01720",
"2.30473",
"3.35595",
"3.89337",
"4.34241",
"4.54954",
"4.71066",
"4.82113",
"4.99402",
"5.26380",
"5.29598",
"5.54486",
"5.70599",
"5.83103",
"5.89240",
"6.03437"
],


What are the best cleaning agents for carbon residues on glassware?

Are there any other cleaning agent that'll work better than aqeous sodium hydroxide?

Yes, but I would not recommend them to anybody running experiments at home!

Back in the days in the lab we used arm-deep tanks filled with a saturated solution of potassium hydroxide in isopropanol to soak dirty flasks. You need long (!) chemical-resistant gloves and once in a while, the whole mess has to be discarded and replaced. Down the sink is not an option ;-)

And there is piranha solution, a mixture of 30% hydrogen peroxide and concentrated sulfuric acid. While is is effective, it is infamous for going off with a bang, see this report.

To sum it up: Stay with the sodium hydroxide solution. Don't risk your health and don't screw up your own environment to rescue a spoiled flask.