Does a weak buffer solution still maintain a mostly steady pH if the acid or base load it has to buffer is small?

First, a disclaimer: This answer addresses only the question of the chemical buffering capacity of the questioner's formulation, given the information provided. Neither I nor Stack Exchange / Stack Overflow can take any responsibility for the safety or efficacy of the proposed formulation, for the effectiveness of the buffer concentration indicated below for maintaining the formulation $\mathrm{pH}$, or for any other hazard, harm, or adverse event, without limitation, arising from the use of the information provided below.

To answer the general question of the title: Yes.

You haven't given enough information to answer your question in the particulars, though.

For your particular application:

Worst-case, buffer-wise, as long as the following assumptions are accurate, the maximum amount of buffering capacity you'd need can be estimated from the total number of moles of active ingredient:

None of the other ingredients have appreciable acid-base character (appears to be the case here)
Degradation of the other ingredients into acidic/alkaline species can be neglected
No acidity or alkalinity is likely to enter from the environment

At $0.03\%$ w/w, assuming the overall density of the formulation is approximately $1\,\mathrm{g\over cm^3}$ (probably reasonable, with $67\%$ w/w composed of aqueous buffer) a species with a molecular weight of $420\,\mathrm{g\over mol}$ will have a mole concentration of $0.714\,\mathrm{mM}$.  In general, I would probably want the buffering species to have a concentration about five to ten times my expected net acid/base load. So, I would choose no less than about $3.6\,\mathrm{mM}$ for the total concentration of all forms of citrate, perhaps going as high as $7.2\,\mathrm{mM}$ to be on the safe side.

Note: The $\mathrm pK_\mathrm a$ of the ionizable group on the active ingredient is sufficiently far from your target $\mathrm{pH}$ that it will be fully protonated in your formulation.  Thus, if it is possible to add this ingredient in its protonated form, the required buffer concentration is probably somewhat lower than the above.

General Principles

As long as the number of moles of acid or base to be neutralized is small compared to the total number of moles of the relevant buffering species in the solution, and as long as the target $\mathrm{pH}$ is within about one unit of the nearest $\mathrm pK_\mathrm a$, the $\mathrm{pH}$ should hold steady.

As the number of moles of acid/base to be buffered against approaches the number of moles of the buffering species, the target $\mathrm{pH}$ needs to be closer and closer to the $\mathrm pK_\mathrm a$ in order to hold the $\mathrm{pH}$ steady.

If the number of moles of acid/base to be buffered against exceeds the number of moles of buffering species, the buffer will in general fail to hold the target $\mathrm{pH}$. The main exceptions would be polyprotic species with multiple $\mathrm pK_\mathrm a$ values in close proximity (polyphosphates come to mind).


$\ce{HSO4-}$ look at the valencies of the hydrogen and sulphate ion, because of the numbers don't match i.e. hydrogen has a +1 charge that means it is deficient by 1 electron while sulphate has a -2 charge which means it has two extra electrons (it can also be viewed that it wants to 'give away' these two electrons) these valence numbers basically show the amount of bonds the molecules can form (since at this level of chemistry bonds can be seen as the sharing of electrons) thus the sulphate can form two while the hydrogen can only form one and in this case since it forms $\ce{HSO4-}$ that means there is one 'unhappy electron' or one that is unbonded since the bond between the sulphate and hydrogen makes hydrogen's charge neutral and takes an electron from the sulphate ion leaving the unbonded electron thus giving the molecule an overall charge of -1. Note that here I am just explaining the bonding that is occurring between the molecules but $\ce{H2SO4}$ DOES form I am just explaining the charges and values involved and also note that bonding and chemical reactions usually involve the VALENCE electrons!.

However this molecule ($\ce{H2SO4}$) will disassociate in water or solvent of some sort, and form ions as the person before me pointed out. into $\ce{H+}$ ions and $\ce{HSO4^-}$ thus forms the conjugate base of the acid ($\ce{H+}$ is the basis of most acidic reactions)

If you're confused as to why $\ce{H2SO4}$ doesn't form, it does but $\ce{HSO4-1}$ is what happens after it disassociates in solution.

I don't know which part of the question or problem is where your confusion lies, hope I answered or helped!

(I am only a first year university student if my answer is wrong please notify me but it correlates with my notes, textbooks and what I was taught back in high school)


Is there a connection between M1V1=M2V2 and Boyle’s Law?

Yes, they are related. The first comes directly from the conservation of number of moles of the solute in a dilution,

$$n_1 = n_2 $$

Since $n = MV$,

$$M_1 V_1 = M_2 V_2$$

The second is related to the conservation of the total number of moles in an isothermal compression or expansion,

$$n_1 = n_2$$

Using the ideal gas law $n = pV/RT$,

$$\frac{P_1 V_1}{R T} = \frac{P_2 V_2}{R T}$$

Or $P_1 V_1 = P_2 V_2$.

Again, both are related to the conservation of moles (which by the way, isn't an universal law, since it's broken in trivial situations like the event of a chemical reaction), although the first one refers to solute only and the second to total number of moles.


Can someone explain the equation for entropy?

Your relation is not correct, so you must not understand something. If I were you I will asked the question in Physics SE but nevermind.

The good relation is : $$\mathrm{d}S=\frac{\delta Q}{T}$$

I let Richard Feynman explain it to you here (when you click on the link wait few seconds)

I hope it will help you, have fun, live long and do chemistry !


There are two distinct phenomena, which are somewhat related.

Dynamical effect of molecular vibration

Bonds do vibrate like springs, and as you said they can store energy by changing its vibrational state. Like a string, the more energy stored in the vibrational motion, the larger the magnitude of such vibration.  Different from a spring, such vibrations are associated with discrete quantum numbers, and even at absolute zero temperature, such vibration still possess a zero-point energy and can never stop.

Non-dynamical effect that change the statistical average of bond length

Molecular vibration often (but not always) occurs at an extremely short time frame, and the vibration itself is often (again not always) of a very small magnitude, so when we refer to bond length we often mean the average bond length over time.
    Normally a molecule stays near the bond length where the energy is the lowest.  There can be a few reasons for a the average bond length to change while keeping the connectivity of the molecule intact, with some examples:

Each conformational isomers will have different bond lengths.  Usually this is a small number.  Examples of large bond length change between conformational isomers with the same bond connectivity are those with Jahn-Teller effect.
By changing the vibrational state of the molecule, the average bond length can be different.  Usually, because of anharmonicity, the higher the vibrational state, the higher the average bond length.  Temperature change, pressure change, collision or photon excitation (especially infrared light) are the most common causes of a change in vibrational state.
Through inter-molecular interactions or external forces, the potential energy surface of the molecule can be changed so that the energy minimum happens at a different distance.  In this case, the bond length will change, but the energy stored in the bond will not change.  This can be achieved through external electronic field, or intermolecular forces such as electrostatic force from an ion, hydrogen bonds, charge transfer, electron donator/acceptor interaction, etc.


Help understanding external magnetic field sensitivity of radical pair reaction rate

(This is really a comment but  has become too long)
There are many examples of  triplet states in magnetic fields as studied by electron paramagnetic resonance (ESR or EPR) experiments. See chapter 8 'Introduction to Magnetic Resonance' by Carrington & McLachlan. It is an old book but does contain all you need to know. (A paper I happen to know showing magnetic field effects is Hyde et al. Chemical Physics v151,p239 1991 but there must be many others if you search)

As to the bird experiments, eqn 1 of the paper by Schulten et al. as described in your question describes the situation: scheme copied below.

$$\begin{align}
(D^*&+A)\leftrightharpoons \;^1(^2A^-+\;^2D^+) \stackrel{B}\leftrightharpoons \; ^3(^2A^-+\;^2D^+)\to \;^3T\\ &\downarrow\\ &\;^1X \end{align}$$

What it seems is necessary is (a) The reaction starts with a light induced electron transfer from excited donor $D^*$ to a nearby molecule that is the acceptor $A$. (b) The singlet state $^1(^2A^-+\;^2D^+) $ species produced has energy levels that cross those of one of the  of $m_I=\pm 1$ levels of the triplet states $^3(^2A^-+\;^2D^+) $ three levels at some value of the magnetic field $B$ and so singlet to triplet crossing occurs($^1$). (c) This enables the yield of products $^3T$ to $^1X$ to vary as the value of the magnetic field varies as the animal moves. (d) The protein has to be fixed relative to the bird otherwise this will not work. (This is the 'anisotropy' mentioned)  Being fixed could simply mean being in a cell membrane in one part of the retina. (e) Somehow the ratio of products has to be able to be distinguished.  Clearly this mechanism clearly will not work at night.

So all in all a complicated mechanism, involving subtle processes, and this makes it very interesting.

($^1$) In a molecule as the triplet often differs significantly in energy from the singlet a huge magnetic field is needed to reach the S-T crossing point, but in pairs of molecule the interaction is very small and so the energy difference is v small also. This is advantageous as then small magnetic fields can cause S-T crossing. (Normally the singlet is above the triplet in energy, unlike the in figure, but the crossing is clear nonetheless)

also see Can magnetic fields affect a chemical reaction?


What is the lowest mass measurable using a 4 decimal point analytical balance?

Keep in mind that for many analytical instruments, you cannot assume that the uncertainty is ± the last digit.
It's usually necessary to look at the specifications in the manual or on a sticker to find the measured uncertainty.
For instance, one of our balances, a Sartorius CPA225D (manual), is readable to 0.01 mg (i.e. the display can display increments of 0.01 mg), but the specifications list the repeatability as only ±0.05 mg (±1 s.d.).
This means that repeated measurements of the same mass will be (ideally) normally distributed with a standard deviation of 0.05 mg and that even though the instrument has the resolution to report in 0.01 mg increments, the uncertainty in a measurement is higher.


Why is degree of dissociation used to estimate strength of an acid?

I would answer only the constant of dissociation $K_a$, because it is a thermodynamical constant and you can calculate it from potentiometric measurements without dealing with activities and so on. If you have the $\Delta G$ of reaction, which is related with the difference of electric potential of the cell, you can calculate the constant as $\Delta G = -RT \ln K$. The pH is not an indicator of the acid strenght, but only its consequence. If you have a very diluite solution of sulfuric acid, its pH will be near to 7, but it does not mean that sulfuric acid is weak. Also the degree of dissociation can give you incorrect information if you are dealing with very diluite or very concentrate solutions.


What is the chemical formula(s) for liquid potassium-based soap?

You'll get an answer quickly if you search for "formulation liquid soap". There are for instance DIY guides on making liquid soap starting from natural oils (sunflower, coconut) by saponifying the oil with KOH. The fatty acid composition of sunflower oil can vary significantly depending on breed (see e.g. the Wikipedia) but generally contains mono- and polyunsaturated FAs in large proportion. Unsaturation contributes to the fluidity of the product. It is also worthwhile to check common nomenclature for FA chain length (e.g. see the Wikipedia). Sunflower oil contains palmitic (C16:0) and stearic (C18:0) acids as principal saturated FAs. Coconut oil contains shorter (lauric, C12:0) and mainly saturated FAs. Mainly because of the extent of saturation, coconut oil has a higher melting point and is used as the minor component.
In addition there are plenty of commercial vendors willing to provide recipes you can follow to make your own liquid soap. Note not all require a potassium salt. For instance, the company Stepan provides a recipe which consists of mixing the following (in decreasing order of weight %) to form one phase (which is then mixed with another lacking detergents):

Deionized Water
BIO-TERGE® AS-40 (or BIO-TERGE® AS-40 HP)  ==  SODIUM C14-16 OLEFIN SULFONATE
STEPANOL® AM  == AMMONIUM LAURYL SULFATE
STEPANOL® WAT == TEA LAURYL SULFATE
NINOL® M10  == COCAMIDE MIPA

Admittedly ammonium counterion (and TEA) is like potassium in that they are larger and softer counterions than sodium, so similar in their effect on the detergent.


Why is trimethylsilanol a stronger acid than tert-butanol?

The involvement of d-orbitals in the stabilisation of silanol anions is minimal. The ions $\ce{H3CO-}$ and $\ce{H3SiO-}$ were investigated by Hopkinson and Lien,1 who concluded that:

  Silicon is a more electropositive element than carbon and consequently the silyl group might be expected to be an electron donor. As such the silyl group should stabilise cations and destabilise anions. However, ab initio molecular orbital calculations show both the α-silyl and β-silyl to strongly stabilise anions and to weakly destabilise cations. Inclusion of d orbitals on  silicon does not significantly alter these conclusions. This contradictory behaviour can be largely understood in terms of n $\to$ π* delocalisation from the lone pair to π* of the silyl group.

These π* orbitals are antibonding orbitals derived from linear combination of Si 3p and H 1s. A partial MO diagram for the $\ce{H3Si}$ fragment is provided here (from the same paper):

Delocalisation of the oxygen lone pair into the $\pi_x^*$ and $\pi_y^*$ orbitals therefore leads to stabilisation. The authors write that the π* orbitals of the $\ce{H3Si}$ fragment are lower in energy than the corresponding π* orbitals of the $\ce{H3C}$ fragment, and hence the stabilisation is greater for silicon than it is for carbon.

A good test for this is to examine how the Si–H bond lengths change upon deprotonation. If the delocalisation is into Si–H antibonding orbitals, then the bond lengths should increase upon deprotonation. In contrast, if the delocalisation is into Si 3d orbitals, then there should be no change. Upon deprotonation, the Si–H bond length increases from 1.473 Å to 1.528 Å, which indicates it is the former.

However, there isn't a thorough explanation of why the energy levels are so. I would surmise that there are two factors: one is the fact that Si–H overlap is poor, compared to C–H overlap. Therefore, Si–H antibonding orbitals are lower in energy than C–H antibonding orbitals, and can act as better acceptor orbitals. Another factor (and one commonly taught in textbooks) is that since Si is more electropositive, the π* orbitals have a larger coefficient on Si; the π* orbitals of the $\ce{H3C}$ fragment have a smaller coefficient on C.

There is also an issue of why the stabilisation is larger in the anion than it is in the neutral molecule: if both the anion and the neutral molecule were equally stabilised, then there would not be any difference in the acidity, which is essentially a measure of the relative stability of the two species. It may be easily explained - at least partly - by the fact that the lone pairs in the anion are higher in energy than those in the neutral molecule, and consequently act as better donor orbitals.

Now, this explanation can be easily extended to the methylated compounds $\ce{Me3CO-}$ and $\ce{Me3SiO-}$. Due to the poorer acceptor capability of Si–C antibonding orbitals, we expect the factors above to play a smaller role in the preferential stabilisation of the silicon anion. As such, the difference in $\mathrm pK_\mathrm a$ between $\ce{X3COH}$ and $\ce{X3SiOH}$ should be smaller when $\ce{X = Me}$ compared to when $\ce{X = H}$.

Is there a way to test this? A crude analysis is as follows. I will take the value of $\pu{27.6 kcal/mol}$ given by Hopkinson and Lien1 for the reaction

$$\ce{H3SiO- + H3COH -> H3SiOH + H3CO-}$$

which corresponds to a $\mathrm pK_\mathrm a$ difference of 20 units when $\ce{X = H}$. This is obviously not rigorous, but serves as a good preliminary benchmark. Now, if Wikipedia's sources are to be believed, then when $\ce{X = Me}$, the $\mathrm pK_\mathrm a$ difference is decreased to a mere 8 units - which is in line with our prediction.

The topic of negative charge stabilisation by silicon has been studied much more thoroughly in the context of carbanions, because they are synthetically useful species (e.g. Peterson olefination). In this context, Schleyer et al.2 also conclude that d-orbital participation is negligible. The myth of d-orbital participation in main group bonding has been thoroughly debunked decades ago,3 so it is regrettable that it is still being taught nowadays.

References

Hopkinson, A. C.; Lien, M. H. A theoretical study of the effects of silyl and methyl groups on acid-base reactions at oxygen, nitrogen and carbon. J. Mol. Struct.: THEOCHEM 1983, 92 (1–2), 153–166. DOI: 10.1016/0166-1280(83)80065-7.
Schleyer, P. v. R.; Clark, T.; Kos, A. J.; Spitznagel, G. W.; Rohde, C.; Arad, D.; Houk, K. N.; Rondan, N. G. Structures and stabilities of α-hetero-substituted organolithium and organosodium compounds. Energetic unimportance of d-orbital effects. J. Am. Chem. Soc. 1984, 106 (22), 6467–6475. DOI: 10.1021/ja00334a001.
Reed, A. E.; Schleyer, P. v. R. Chemical bonding in hypervalent molecules. The dominance of ionic bonding and negative hyperconjugation over d-orbital participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. DOI: 10.1021/ja00160a022.


Proportion of gases filled in cylinder of sea-divers

Hint:
Try to use Henry's Law. According to which solubility of gases increases with increase in pressure. If we go deeper downwards, pressure increases and when we come to the surface, pressure decreases, so there might be that soem buubles are created in the blood which may be fatal. So air cylinders are made keeping this in mind.


How to calculate the enthalpy change for the formation of ethylene epoxide from ethylene?

When you use bond enthalpies to calculate $\Delta H$, you need to subtract the sum of the bond enthalpies for the products from the sum of the bond enthalpies for the reactants.  In other words "reactants - products".  This is because all bond enthalpies are positive, since energy must be put into the bond to break it.  Bond formation releases energy; the minus sign compensates for this.  In short, bonds of reactant molecule must break, and new are formed to form product molecules.  So, your answer should be negative.


The trouble with the question is that hydrophobicity is not a meaningful concept in a gas.

The concept is mostly useful when characterising how surfaces behave. Typical hydrophobic compounds don't interact strongly with water (e.g. non polar hydrocarbon waxes) and tend to disrupt water's hydrogen bond network. So water doesn't "wet" the surface and tends to form droplets that seem to be trying to minimise the contact area with the surface. The contact angle that a droplet makes with the surface is used as a measure of static hydrophobicity (if the droplet spreads out because the surface likes to interact with water, the angle is low, if the surface reels water the angle is high).

It is less clear that describing liquid phases as hydrophobic is meaningful, though it is not uncommon in casual usage. Compounds that would be hydrophobic as solids will probably exist as liquids that don't mix with water or can form micelles where they cluster together in very small droplets in the water phase. The bulk effect that matters is that hydrophobic compounds don't mix with water and form separate phases. The hydrophobicity of the molecules describes why the separation happens, but the effect is immiscibility.

None of this matters in the gas phase. The point of gases (compared to solids or liquids) is that the bulk properties are not determined by molecular interactions. The molecules of a gas don't interact much and are, to a good approximation, interchangeable whatever the individual molecules are made of. Gas equations describing gas behaviour can effectively ignore the molecular interactions.

What this means is that if you take a compound that would be hydrophobic as a solid and immiscible with water as a liquid, it will mix quite happily with water vapour. So a hydrophobic compound can be a gas, but a gas can't be hydrophobic.


Let establish some thermochemistry terminology first:

Work: the total energy transferred by a system to another. There are several types of work. In thermodynamics, the most common one is pressure volume work, given by: $w = -p_{ext}\Delta V$

Internal Energy: the total kinetic and potential energy of a system

Enthalpy ($H$): is the total energy of a system. It also can be thought as the amount of energy that is required to make all the bonds in the system plus all the energy that is required to make room for the system. Hence it should be clear that it is given by the formula: $\Delta H = \Delta U + \Delta {(pV)}$

Heat Transfer (q): the amount of heat energy (kinetic energy) that is transferred from a system given by the formula: $q = \Delta U -w = \Delta U + p_{ext}\Delta V ~~$ (when only pressure volume work is done)

In your question, you are actually referring to $q$ (heat transfer), not $\Delta H$. This is because $\Delta H$ also accounts for the change in pressure and volume.

However under certain conditions, $q$ and $\Delta H$ can be equal. If the external pressure is constant (which is a common case as most experiments are done in the open atmosphere which has a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p_{ext}\Delta V = q$$


Is there a list of must-have columns for LC-MS/MS used in biochemistry/clinical lab setting?

This answer will change fairly frequently because technologies are always improving.  These days, I think a bare minimum list might be something like:

a 2.1 mm × 50 mm C18 column for rapid reversed-phase methods.
a 2.1 mm × 150 mm C18 column for slower, broader-coverage reversed phase methods.
a 2.1 mm × 50 mm HILIC column for rapid methods to detect polar metabolites.
a 2.1 mm × 150 mm HILIC column for slower, broader-coverage methods to detect polar metabolites.

Some further notes:

If your lab is going to use ion-pairing based methods, you will probably want separate columns for ion-pairing methods and "clean" methods without ion-pairing reagent.  Ion-pairing agents like tributylamine and/or perfluoroheptanoic acid are very difficult/impossible to remove from columns once they have been used for this purpose.
Fairly soon the best answer might be to use 1.0 mm diameter columns instead of the "standard" 2.1 mm diameter that is most common nowadays.
The most widely used HILIC column seems to the "BEH Amide" column chemistry from  Waters.  Following close behind may be ZIC-HILIC from SeQuant and/or Agilent's HILIC-Z chemistry.
For C18 chromatography, performance differences between different vendors are comparatively minor (although real).  A wide variety of chemistries are used.


Effect of rain on water and soil samples contaminated by petroleum

It has much sense since the rain will modify the concentrations of both the heavy metals and the petroleum.
The initial quantities will be redistributed.
One part will penetrate deeper in the ground, another will be moved due to water runoff and what is left will be measured in your samples.

If possible, you should take the first samples as soon as possible and the the next ones towards the end of next season.
It is also important to get the value of the total precipitation during the rainy season.
This will serve to estimate how long it will take to eliminate the effects of the disaster.


LCAO (Linear Combination of Atomic Orbitals) and Phases

The concept of linear combination of atomic orbitals (LCAO) to form molecular orbitals (MO) is probably best understood, while digging a little deeper into quantum chemistry. The method is an approximation that was introduced for ab initio methods like Hartree Fock. I don not want to go into too much detail, but there a some points that need to be considered before understanding what LCAO actually does.

The time independent Schrödinger equation can was postulated as $$\mathbf{H}\Psi=E\Psi,$$
 with the Hamilton operator $\mathbf{H}$, the wave function $\Psi$ and the corresponding energy eigenvalue(s) $E$. I will just note that we are working in the framework of the Born-Oppenheimer approximation and refer to many textbooks for more details.

There is a set of rules, the wave function has to obey.

It is a scalar, that can be real or complex, but the product of itself with its complex conjugated version is always positive and real. $$0\leq \Psi^*\Psi=|\Psi|^2$$
The probability of finding all $N$ electrons in all space $\mathbb{V}$ is one, hence to function is normed. $$N=\int_\mathbb{V} |\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)|^2 \mathrm{d}(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)$$
The value of the wave function has to vanish at infinity. $$0 = \lim_{\mathbf{x}\to\infty}|\Psi(\mathbf{x})|$$
The wave function has to be continuous and continuously differentiable, due to the second order differential operator for the kinetic energy $\mathbf{T}_c$ included in $\mathbf{H}$.
The Pauli Principle has to be obeyed. $$\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N) = -\Psi(\mathbf{x}_2,\mathbf{x}_1,\dots,\mathbf{x}_N)$$

Also the variational principle should hold for this approximation, stating that the expectation value for the energy of any trial wave function is larger that the energy eigenvalue of the true ground state. One of the most basic methods to approximately solve this problem is Hartree Fock. In it the trial wave function $\Phi$ is set up as a Slater determinant.
$$
\Phi(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_N) =
\frac{1}{\sqrt{N!}}
\left|
   \begin{matrix}
      \phi_1(\mathbf{x}_1) & \phi_2(\mathbf{x}_1) & \cdots & \phi_N(\mathbf{x}_1) \\
      \phi_1(\mathbf{x}_2) & \phi_2(\mathbf{x}_2) & \cdots & \phi_N(\mathbf{x}_2) \\
      \vdots               & \vdots               & \ddots & \vdots \\
      \phi_1(\mathbf{x}_N) & \phi_2(\mathbf{x}_N) & \cdots & \phi_N(\mathbf{x}_N)
   \end{matrix} \right|
$$

The expectation value of the energy in the Hartree Fock formalism is set up, using bra-ket notation as $$E =\langle\Phi|\mathbf{H}|\Phi\rangle.$$
Skipping through some major parts of the deviation of HF, well end up at an expression for the energy.
$$E=\sum_i^N \langle\phi_i|\mathbf{H}^c|\phi_i\rangle +\frac12 \sum_i^N\sum_j^N \langle\phi_i|\mathbf{J}_j-\mathbf{K}_j|\phi_i\rangle$$

To find the best one-electron wave functions $\phi_i$ we introduce Lagrange multiplicators $\lambda$ minimising the energy with respect to our chosen conditions. These conditions include that the molecular orbitals are ortho normal. $$\langle\phi_i|\phi_j\rangle =\delta_{ij} =\left\{
\begin{matrix}  0 & , \text{for}~i \neq j\\ 1 & , \text{for}~i = j\\\end{matrix}\right.$$
I will again skip through most of it and just show you the end expression.
$$\sum_j\lambda_{ij}|\phi_j\rangle = \mathbf{F}_i|\phi_i\rangle$$
with the Fock operator set up as
$$\mathbf{F}_i = \mathbf{H}^c +\sum_j (\mathbf{J}_j − \mathbf{K}_j)$$
with $i\in1\cdots{}N$, the total number of electrons.
We can transform these trial wavefunctions $\phi_i$ to canonical orbitals $\phi_i'$ (molecular orbitals) and obtain the pseudo eigenwertproblem $$\varepsilon_i\phi_i'=\mathbf{F}_i\phi_i'.\tag{Fock}$$
This equation is actually only well defined for occupied orbitals and these are the orbitals that give the lowest energy. In practice this formalism can be extended to include virtual (unoccupied) molecular orbitals as well.

Until now we did not use any atomic orbitals at all. This is the next step to find an approximation to actually solve these still pretty complicated systems.
LCAO a superposition method. In this approach we map a finite set of $k$ atomic (spin) orbitals $\chi_a$ onto another finite set of $l$ molecular (spin) orbitals $\phi_i'$. They are related towards each other via the expression
\begin{align}
\phi_i'(\mathbf{x}) &= c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) + \cdots + c_{k,i}\chi_k(\mathbf{x})\\
\phi_i'(\mathbf{x}) &= \sum_a^k c_{a,i}\chi_a(\mathbf{x})\\\tag{LCAO}
\end{align}

From $\text{(Fock)}$ you can see, that there will be $l$ possible equations depending on the chosen set of orbitals, in the form of
$$\sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\mathbf{F}_i|\chi_b\rangle = \varepsilon_i \sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\chi_b\rangle$$
or for short there are $l$ equations $$C_{ab}F_{ab}=\varepsilon_i C_{ab}S_{ab}.$$ Since $a,b \in [1,k]$ you can gather $C_{ab}$ in a $k\times k$ matrix $\mathbb{C}$. Since $i\in[1,l]$, there will only be $l$ $\varepsilon_i$ and therefore the matrix of the Fock elements $F_{ab}$ has to be a $l\times l$ matrix $\mathbb{F}$. The whole problem reduces a matrix equation $$\mathbb{F}\mathbb{C}=\mathbb{S}\mathbb{C}\epsilon\!\!\varepsilon,\tag{work}$$
from which it is obvious, that the dimension of the involved matrices have to be the same. Hence $k=l$.

Too long, didn't read

The total number of elements in the finite set of atomic orbitals is equal to the total number of elements in the finite set of molecular orbitals. Any linear combination is possible, but only the orbitals that minimise the energy will be occupied.

The coefficients $c$ from $\text{(LCAO)}$ will be chosen to minimise the energy of the wave function. This will always be constructive interference. This is also independent of the "original" phase of the orbital that is combined with another orbital. In other words, from $\text{(LCAO)}$
$$\phi_i'(\mathbf{x}) = c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) \equiv c_{1,i}\chi_1(\mathbf{x}) - c_{2,i}[-\chi_2(\mathbf{x})].$$


Why don't the electrons move through the electrolyte (instead of the circuit) in a galvanic cell?

Not in water. Free electron in water is really unfavorable, so no significant concentration of them can be generated chemically, and it almost immediately reduces water itself to hydrogen (but I heard rumors about generation of solvated electrons in water in very special experiment with short half-life)

In liquid $\ce{NH_3}$, however, solvated electrons can occur, so self-discharge of galvanic cells with $\ce{NH_3}$-based electrolyte may occur through travel of electrons via electrolyte.


Do all decomposition reactions involve heat?

Plainly put: no, decomposition reaction does not always have to entail providing heat. A decomposition reaction (in simple language) is when one compound breaks into multiple other compounds, like so:

$$\ce{A -> B + C}$$

As a general rule, if the product side is relatively more stable, it will decompose without you having the to provide extra heat. In the case of the reaction you gave, $\ce{H2O}$ is more stable than $\ce{H2}$ and $\ce{O2}$, and thus you will have to provide heat for the breaking of bonds.


How are these molecules chiral and achiral?

Regarding the first molecule: I understand that the simplistic Fischer projection - with all its bond depicted by solid lines - can be very confusing at first glance in determining optical activity. You may believe it to be optically inactive, hoping for a center of symmetry to exist, but it is not so.
You must keep in mind how the molecule actually exists in 3d! Recall the rules of Fischer projection, the horizontal bonds project outward, while the vertical bonds extend inward. Hence, the molecule would actually look like this:

Now, I hope you can tell, there is neither any center of symmetry nor any plane of symmetry here. Hence, the molecule is chiral.
PS: Your second molecule is chiral by similar logic, as also mentioned in the comments, and as you yourself noted.


Why lone pair occupies more space than bond pair?

The basic premise of this statement is that the repulsion exerted by an electron system depends upon the area it occupies. More the area, more is the repulsion.

As the bond pair is formed from the sharing of electrons, it experiences attraction from not one, but two nucleii. On the other hand, the lone pair is attracted by only one nucleus.

So, the bond pair is more localised while the lone pair is spread out and occupies a larger area.

The 'size' here refers not to the size of electrons but to the size and nature of the orbitals these electrons occupy.


in order to calculate the heat of reaction, you need the bond energies (which are measured in $\mathrm{kJ/mol}$ and usually provided in the question or in an appendix in the back of the chemistry book). The heat of reaction is equal to the energy of bonds broken minus the energy of the bonds created. It would be equal:

$[4*(\ce{S-O} $ bond energy$) ~+ ~1*(\ce{O=O} $ bond energy$)] - [6*(\ce{S-O} $ bond energy$)]$.

This will give a negative answer if the reaction is exothermic and vice versa. Since it is measured in $\mathrm{kJ/mol}$, multiply by moles to get the total energy released.


On mathematics behind molecular orbital energies

An internal or interior eigenvalue problem seeks a solution somewhere in the middle of the spectrum of eigenvalues as opposed to trying to find the smallest or largest eigenvalue. Its generally trickier to solve for a particular eigenvalue than it is to solve for the smallest/largest one. The HOMO and LUMO are necessarily somewhere in the middle of the eigenspectrum.
The Kohn-Sham Hamiltonian is the orbital-DFT approximation of the molecular Hamiltonian. The Hamiltonian describes the interactions within the molecule and allows one to solve for the molecular wavefunction through the Schrödinger equation. Most of electronic structure is focused on approximations of the Hamiltonian or the form of the wave-function that make solving this problem tractable. There is a lot more that can be said about Hamiltonians than would ever fit in an answer here, so you should try to find good textbooks and papers that discuss them.


Orbital Electronegativity Considerations in Resonance Structures

As far as hybridization goes, the hybridization of all of the atoms must be the same in all resonance structures. In considering resonance structures, none of the nuclei should be moved in space. Since hybridization is designed to explain the arrangement of nuclei/electrons in space, hybridization cannot change between resonance structures, otherwise nuclei may be forced to occupy new positions.

In practice, this means that certain atoms may appear to be sp3 hybridized but with consideration of resonance structures actually should be treated as sp2. This often means that a lone pair is not in an sp3 orbital as expected, but an unhybridized p-orbital.


Strong Base combined with Sp2 carbons in Nucleophilic Substitution and Beta-Elimination Reactions

The strong base is capable of being a nucleophile in this sort of situation; deprotonating tert-butyl alcohol will increase its nucleophilicity. In an SN1 reaction the overall rate is only dependent on the loss of a leaving group, so a strong base/poor nucleophile will react via SN1 as long as there are not plausible E1 pathways (e.g. sp3 bonded hydrogen alpha to carbocation).


Assuming an adiabatic wall between gas and surroundings then the expansion of the gas will do work on the surroundings.  What is that work?  It is the pdV work using the external pressure acting against the piston from the surroundings.   You are asking why not use the pressure of the gas.  You could, but only if the process was reversible.  That is, the gas is at equilibrium throughout the expansion process.

But what if the gas is not at equilibrium throughout the expansion?   The pressure may not be uniform inside the gas throughout the expansion.  If the gas is expanding irreversibly then the pressure just behind the piston will be lower than the other parts of the gas.   It's easier to calculate the work on the surroundings since you know the external pressure is uniformly applied.


Inactivation of enzyme: Why inactivation constants do not follow the Arrhenius equation?

Without attempting to elaborate (this is not my area of expertise) I rely on the abstract to an article [1] reviewing the subject for an answer:

  The Arrhenius equation has been widely used as a model of the
  temperature effect on the rate of chemical reactions and biological
  processes in foods. Since the model requires that the rate increase
  monotonically with temperature, its applicability to enzymatic
  reactions and microbial growth, which have optimal temperature, is
  obviously limited. This is also true for microbial inactivation and
  chemical reactions that only start at an elevated temperature, and for
  complex processes and reactions that do not follow fixed order
  kinetics, that is, where the isothermal rate constant, however
  defined, is a function of both temperature and time.

Reference

[1] Peleg et al Critical reviews in food science and nutrition 52(9):830-51 (2012)


The most comprehensive one (so not "all" chemicals) that I've seen is the Aldrich website. By law they have to supply MSDS for everything they sell. However, they may ask for a lot number or the function may not work unless you're logged in as a customer.


What is the purpose of calculating other orbitals than HOMO and LUMO?

A practical answer: in most of molecular quantum chemistry, the cost of obtaining unoccupied ("virtual") orbitals is next to nothing because of the algorithms involved. In short, one diagonalizes some matrix (i.e. finding eigenvalues and eigenfunctions, which are the orbital energies and orbitals) and the virtual orbitals are usually delivered anyway.

A more theoretical answer: While a lot of chemistry involves only the ground state, there is UV/vis spectroscopy and photochemistry. For those fields, virtual orbitals are immensely useful, because they will be used for the description of the electronic transitions that happen in those fields. These transitions are not necessarily between HOMO and LUMO because of symmetry, energy scales involved etc.


Does RedP+HI reduce all carbon functional groups to alkane?

Nope, in general it doesn't. You are correct. See also for example:

A. Aramini, M. R. Sablone, G. Bianchini, A. Amore, M. Fanì, P. Perrone, A. Dolce, M. Allegretti, Tetrahedron 2009, 65 2015–2021. DOI: 10.1016/j.tet.2009.01.005:

M. Dobmeier, J. M. Herrmann, D. Lenoir, B. König, Beilstein J. Org. Chem. 2012, 8, 330–336. DOI: 10.3762/bjoc.8.36 Open access PDF. (This also includes your one as a reference, too.):

If it did, it would be a pretty poor reagent since it would have no functional group tolerance, so not very useful. However, it is not unlikely that under special conditions it might be able to reduce carboxylic groups as well.


Are Lewis Structures the same throughout solids, liquids, and gases?

Lewis structures are a representation of the atomic connectivities and electronic lone pairs present in an individual molecule of a substance. The molecule of the substance does not change in these states, so the Lewis structure would remain the same.

What does change in these states is the way individual molecules interact with each other. If you were to draw a diagram of how multiple Lewis dot structures of the molecule interact with each other, that would be different for these different states.


Generally a specification like 3% of potassium sulfate is wt%. Such a solution would have 30 g in a about a liter of water (weighing 1 kilogram). $$30/1030 = 2.91 \%$$ which should be close enough. You could use 970 g of water to be more exact. But just filling a beaker to the 1 liter mark would work fine.

I never though of cooking in Europe, but I'd guess that they do have measuring spoons in some $\mathrm{cm}^3$ (mL) units. Again good enough would almost certainly work. $$ 1 \text{ teaspoon} = 4.92892 \text{ mL}$$


Is it correct to include the concentration of water into the concentration equilibrium constant?

1. Reaction of Trichloroisocyanuric acid with water:

Water is omitted from the equilibrium expression only if it is a solvent in that reaction because it is a pure liquid. We can't increase the concentration of a pure liquid or pure solid and hence they are omitted from the expression.

However water in gaseous state cannot be ignored. It would be included in the equilibrium expression.

The above reaction is actually mixing of Trichloroisocyanuric acid in water to work as a disinfectant. So in the above reaction H2O is basically pure liquid (not gas).

Though in the expression water concentration is mentioned but in the later steps it would be substituted as 1. Actually some authors tend to write the formula of equilibrium constant including all the reactants & products concentrations (including pure solids and liquids) first and then in the next step they substitute 1 in the place of concentration of pure solids and liquids and other given numerical concentration values in place of other reactants and products.

So that expression is just a formula and in the next step when you would be substituting the values, 1 will be substituted inplace of concentration of water.

2. Coming on to the reaction you have sent in the picture i.e. reaction of aldehyde and alcohol:

Here water is not going to be omitted.

Okay water is always not neglected from the equilibrium expression. When I said water concentration is assumed to be 1, I had actually assumed water to be a solvent. Remember if water is a solvent in your reaction, then you can neglect the water concentration term but if water is not a solvent, then water term needs to be included. In the above question, water is being created in the process .i.e it is a product. If water had been a reactant as well as product (i.e. a solvent), then you could have neglected it.

Water in the above reaction is not present/taken from initial situation. It was eventually formed during the reaction and hence it is a product ( not a solvent). Hence you need to include it in the Kc expression.

Do you know what actually happens, when water is a solvent, its concentration term appears on reactant as well as on the product side. Hence while writing Kc expression it is cancelled out from the numerator and from the denominator. But if water is a product (i.e. not a solvent), then reactant part wont contain any concentration of water term and hence it can't be cancelled out in Kc expression .i.e. it will appear in the numerator of the Kc expression.

Hope it helps.


Which gas is a oxidizing agent as well as a reducing agent?

I'm guessing your teacher is looking for sulfur dioxide as the answer, but I don't see how or why you're supposed to be able to arrive to this answer logically. Either you'd need to read about it specifically, or maybe you're supposed to stare at a table of standard reduction potentials and notice that $\ce{SO2}$ appears as both a reagent and as a product with comparable voltages:

$$
\begin{array}{}
\ce{SO2(aq) + 4 H+(aq) + 4 e- &<=>& S(s) + 2 H2O(l)} &  E^\circ = \pu{+0.50 V}\\
\ce{SO2(aq) + 2 H2O(l) &<=>& SO4^2-(aq) + 4 H+(aq) + 2 e-} &  E^\circ = \pu{-0.17 V}
\end{array}
$$

(The bottom equation is the inverse of what you would find in a reduction potential table)

Extrapolating from these aqueous potentials, it appears $\ce{SO2}$ is a modest reductant and a weak oxidiser, but both reactions can be accessed in reasonable conditions.

I think you're technically correct (as Ivan mentions), but what makes hydrogen gas a "less correct" answer is that it only rarely acts an oxidiser (in the presence of very strong reducing agents, such as alkali metals), whereas it can act as a reductant much more commonly. Compare these equations with the ones above:

$$
\begin{array}{}
\ce{H2(g) + 2 e- &<=>& 2 H^-(aq, extrapolated)} &  E^\circ = \pu{-2.3 V}\\
\ce{H2(g) &<=>& 2 H+(aq) + 2 e-} &  E^\circ = \pu{+0.00 V}
\end{array}
$$

(Reference for the hydride potential)

Note the much larger gap in potential between the two reactions. In this sense, the reducing nature of $\ce{H2}$ vastly overwhelms its oxidising nature, so most people will think of it as a reducing gas.


Can acidic conditions break disulfide bonds

No, acidic conditions (i.e. low pH) shouldn't be enough by itself to break a disulfide bond. The main reaction that breaks a S–S bond is cleavage by reduction. This is in a way similar to peroxides, which are stable even at very low pH (see e.g. mixtures of sulfuric acid and hydrogen peroxide).


What are the methods of fluoride removal from drinking water?

There are many filters which do not remove fluoride from potable water. Boiling or freezing does not help either. Reverse osmosis filtration may be a solution. However they are expensive ($100+).

Activated alumina defluoridation filters are used in areas where contamination from fluoride is common. These can reduce fluoride levels from 0.5 ppm to 0.1 ppm, but these require frequent replacement and are not a solution for domestic treatment. Distillation filters can be purchased to remove fluoride.


IUPAC numbering of carbon atoms in a chain

Consider a n-hexane compound

$\ce{CH3-CH2-CH2-CHCl-CH2-CH3}$

The Cl is different than the H of n-hexane.

Going left to right the Cl is on the 4th carbon. Going right to left it is on the 3rd carbon. So you should use the right to left numbering in this case.


How can I find out a substance name based on its structural formula?

That's betanin, from which beets derive much of their deep purple color, and which may have neuro-protective (Ref. 1) features. The following is the structure retrieved from the wikipedia (on 2019-10-5), which closely matches that in the mosaic:

I used the editor available on the Sigma-Aldritch website to generate a structure and submit a search:

The program identified betanin as the top hit despite my sloppy structure.

The SMILES code is: C(OC(C1O)OC(C(=CC2[N+]3=CC=C(CC(N4)C(=O)O)C=C4C(=O)O)O)=CC=2CC3C([O-])=O)(C(C1O)O)CO

The IUPAC name (retrieved from the wikipedia) is (2S)-1-{2-[(2S)-2,6-dicarboxy-2,3-dihydropyridin-4(1H)-ylidene]ethylidene}-5-(β-d-glucopyranosyloxy)-6-hydroxy-2,3-dihydro-1H-indol-1-ium-2-carboxylate

As noted in a comment, the stereochemistry in the wikipedia image above and that in the mosaic does not match that of the structure in the Sigma-Aldrich catalogue. I used ACD/ChemSketch freeware to generate the following image that does match the one in the catalogue:

American Chemical Society. "Vegetable compound could have a key role in 'beeting' Alzheimer's disease." ScienceDaily, 20 March 2018.

As an aside, it's interesting to note that work on L-dopa, a biosynthetic precursor of betanin, led to a Nobel prize award for Swedish scientist Arvid Carlsson. Also, quoting from the wikipedia:

  The 2001 Nobel Prize in Chemistry was also related to l-DOPA: the Nobel Committee awarded one-quarter of the prize to William S. Knowles for his work on chirally catalysed hydrogenation reactions, the most noted example of which was used for the synthesis of l-DOPA.

The structure-drawing engine (and presumably also search engine) provided through the Sigma-Aldrich structure search web service is JSDraw2 (JSDraw V3.2.3) by Scilligence.


How to calculate the amount of water produced by the incomplete combustion of propane?

Yes the complete combustion equation is correct as you stated and the equation of the incomplete combustion is certainly one possibility. A more general formula is:
$$\begin{multline}\ce{C5H12 + $\left(\frac{x}{2}+y+3\right)$\,O2 -> $(5-x-y)$\,C + $x$\,CO +$y$\,CO2 + 6H2O}\\
\left\{x,y\in\mathbb{R}^{+}_0; (x+y)\leq5\right\}
\end{multline}$$
This is assuming all the hydrogen will be converted into water. In some cases you might have incomplete combustion, where ethene or ethyne (ar many others) are formed, but I think this is taking it a bit far at this point.
The above case reduces to your formula for $x=5$ and $y=0$.

For the second question, your equation has a small error, correct is:
$$\ce{C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (g)}$$
But this is not the equation you need, as the question explicitly states, that the combustion products are carbon and water. So you working equation becomes:
$$\ce{C3H8 (g) + 2O2 (g) -> C (s) + 4H2O (g)}$$
So you need twice the number of moles of oxygen than propane, to produce four times as many moles of water than you had propane.

You cannot convert a volume of gas into a mass by an easy conversion factor, you have to use the ideal gas law:
$$pV =n\mathcal{R}T$$
You can then use $m =n\cdot M$ to convert to grams, but you don't yet have to do that.

You have been given the temperature $T=25~^\circ\mathrm{C}=298.15~\mathrm{K}$ and pressure $p=100~\mathrm{kPa}$ and the volumes for the reactants, so you can calculate the amount of substance for propane and oxygen.

   $n(\ce{C3H8})\approx 0.7~\mathrm{mol}$ and $n(\ce{O2})\approx 1.4~\mathrm{mol}$

From the reaction equation you can now determine the amount of substance of water, that will be produced.

   $n(\ce{H2O})\approx 2.8~\mathrm{mol}$

You can convert that easily into a mass now.

   $m(\ce{H2O})\approx 50~\mathrm{mol}$

There are two notable changes you will observe. Since you are combusting a gas, obviously no bubbles can appear. If you do the combustion in a see-through container you will see black stains due to the solid carbon that is produced by the reaction and liquid water drops will form on colder surfaces.
You are of course right, that you will also have a rise in temperature.

Note: The here given values are crude approximations only, that might not satisfy the task at hand. You need to do the calculation yourself to find the correct answer.


Determining enthalpy of neutralization and heat capacity of calorimeter by overlying with straight lines or "eye-o-metric method"

Ah, I understand now. It's of analogous reason to the first case. By adding heat during a time interval we expect that some of it will be absorbed by the calorimeter.


Why do some ions form tetrahedral complexes more readily than others?

Ligands use lone pairs of electrons to form coordinate bonds with the metal ion of the complex. The coordination number is the number of coordinate bonds from the ligands to the central ion. For instance, if a central ion has two coordinate bonds formed to it, one from each ligand, then the coordination number would be 2.(1)

According to Lumen Learning, for certain metals the maximum coordination number is related to the "electronic configuration of the metal ion (specifically, the number of empty orbitals) and to the ratio of the size of the ligands and the metal ion". Furthermore, it states that large metals with small ligands result in high coordination numbers, presumably due to there being more attraction from the positively charged central ion. Conversely, small metals with large ligands will have low coordination numbers, due to there being less electrostatic attraction towards the lone pairs on the ligands as its surface area is smaller.(2) Extrapolating this knowledge towards your question, we can parse the question to look at its components:

Co2+ is has an ionic radius of 74 pm while Fe2+ has an ionic radius of 76 pm. If we examine the example complex you gave, [CoCl4]2-, Chlorine has a large atomic radius of 175 pm relative to Co2+ ionic radius.(3) Using what we learned from Lumen Learning, this is a small central ion with relatively large ligands. Hence, the coordinate number must be smaller, such that it is 4, meaning it forms a complex with a tetrahedral geometry. Fe2+ on the other hand has an ionic radius of 76 pm and for a complex with an octahedral geometry, let us say [Fe(H2O)6]2+, it has 6 water molecules that function as ligands. The hydrogen and oxygen atoms have atomic radii of 25 and 60 pm respectively.(4) Consequently, the size of the water atoms are considerably smaller than the Chlorine atoms and thus, according to Lumen Learning, since we have smaller ligands, more of them can form coordinate bonds with the Fe2+ central ion. If we are to look at a different example, e.g. [Fe(CO)4]2−, we can see that it has ligands with large atomic radii (i.e. Carbon and Oxygen) and thus fewer ligands can bond with the central ion, resulting in a tetrahedral geometry.

References:

Brown, Catrin. "Chapter 3 – Periodicity." Higher Level Chemistry. Ed. Mike Ford. 2nd ed. N.p.: n.p., n.d. 123-33. Print.
Mott, Vallerie. "Introduction to Chemistry." Lumen. N.p., n.d. Web. 16 Apr. 2019.
"Cobalt Chemistry." Cobalt Transition Metal Chemistry Cobalt(II)Co2+ Complex Ions Stabilised Ligand Substitution Cobalt(III) Co3+ Complexes Redox Chemical Reactions +2 +3 Principal Oxidation States GCE AS A2 IB A Level Inorganic Chemistry Revision Notes. N.p., n.d. Web. 16 Apr. 2019.
Kwilkis. "Helium Atom, Radon Atom and Water Molecule Sizes and Concrete?" Helium Atom, Radon Atom and Water Molecule Size... | ACS Network. N.p., 03 Oct. 2018. Web. 16 Apr. 2019.


Shouldn't weak acids act as strong ones in alkaline solutions?

While titrating weak acid with strong base, the titration does occur like a normal acid, because the base added quickly takes the dissociated $\ce{H^+}$ ion of the weak acid and thus the equilibrium of the acid is distorted. To maintain constant $\ce{K_c}$, the weak acid also produces $\ce{H^+}$ instantaneously, and thus its behaviour almost becomes like a strong acid if base is added fast(which is normal in titration) to maintain equilibium.But the $\ce{pH}$ at end point is not $7$ not due to the acid but the salt formed as a product of the titration.Suppose the weak acid($\ce{HA}$) is being titrated with a strong base(say for example,$\ce{NaOH}$), and both are taken in same equivalents. Then the reaction does go until all the acid (though weak) is exhausted.$$\ce{HA + NaOH -> NaA + H_2O}$$At t= o, say both $\ce{HA }$ and $\ce{NaOH }$ has $c$ equivalents each. So, at the end of the reaction, both $\ce{HA }$ and $\ce{NaOH}$ will be totally exhausted and $\ce{NaA }$ will be $c$ equivalents.Now, in the medium only $\ce{NaA }$ is present which will control the $\ce{pH}.$ But $\ce{NaA}$ is actually "SALT OF A WEAK ACID AND STRONG BASE". So it hydrolyses as,$$\ce{NaA + H_2O <=> HA + Na^+ + OH^-}$$So  you can see as $\ce{HA }$ is a weak acid ,it's conjugate base (stronger) $\ce{A^- }$ takes the proton of water and leaves only $\ce{OH^- }$, which in turn is the reason for incresing the alkalinity of the medium.That's why generally, $\ce{pH }$ is above 7 at the end point. By doing mathematical calculation, it can be easily shown that if the weak acid's $\ce{pK_a}$ is given and the final concentration of the salt is say, $\ce{ c' M}$. Then the $\ce{pH }$ of the medium can be calculated as ,$$\ce{pH = 1/2 pK_w + 1/2 pK_a + 1/2logc'}$$. Assuming the titration is being carried at $25$ degrees, then$$\ce{pH= 7 + 1/2pK_a +1/2logc' >  7 (generally) }$$ if your acid is tending to strong i.e. $\ce{pK_a}$ is very less positive, and $\ce{c'}$ is close to $1 $, the $\ce{pH}$ can be close to $7$, but $\ce{pK_a}$'s  are normally of the order of $3$ or $4$. So, generally $\ce{pH}$ exceeds $7$.


Reacting carbon dioxide with magnesium in ribbon form

The reaction you are working with is:

$$\ce{2 Mg(s) + CO2(g) -> 2 MgO(s) + C(s)}$$

where the carbon should be primarily graphitic.

Because you are producing two solids in this reaction, they will be physically intermixed and thus the purity of either solid will be poor, being contaminated with the other solid.

The separation of carbon from $\ce{MgO}$ is a fairly straight forward process, however. Upon contact with water, $\ce{MgO}$ becomes the basic $\ce{Mg(OH)2}$, which is readily soluble in dilute hydrochloric acid. Once the $\ce{MgO}$ has been dissolved in acid, you can filter off the acid-insoluble carbon. After the hydrochloric acid has been evaporated off, $\ce{MgO}$ will re-form from the $\ce{Mg(OH)2}$.


Enantiomorphs are non-superposable mirror images (all signs of one coordinate axis reversed).  What is the maximum enantiomeric divergence possible?  If you have a  quantitatively negligible chiral divergence, there is nearly no quantitative difference to be realized in calculation or measurement.  BTW, the maximum energy difference between enantiomers cannot be more than a part-per-trillion difference/average.  If it were larger, there would be easily detectable enthalpy of fusion anomalies for calorimetry secondary standard benzil.

Chirality is geometry.  Geometric chirality is easily observed (no ${S_n}$ symmetries[A]) and easily calculated[B].  alpha-Quartz and gamma-glycine single crystals have $\text{CHI} = 1 \text{, COR} = 1 \text{, DSI} = 0$, maximum mathematical chirality[C], but it cannot be measured.  Solution optical rotations ignore atomic mass distribution[D].  Silver thiogallate, $\ce{AgGaS2}$ in achiral space group ${I}$-${42d}$, rotates 522°/mm along [100] at 497.4 nm[E].  ${P3_121}$ quartz and ${P3_221}$ berlinite are both  levorotatory[F].  Achiral $\ce{PhCOCH3}$ impurity changes resolved $\ce{PhCH(OH)CH3}$ optical rotation[G].  CIP notation flips with composition at constant geometry (look at natural protein cysteine versus alanine, both L-amino acids but opposite CIP notations).

How do you obtain $100\%$ optically pure maximally divergent enantiomers to your specs?  No problem!  Benzil, $\ce{(C6H5CO)2}$, is achiral when molten, gas phase, or dissolved. Solid state crystal lattice forces twist and stack benzil molecules into homochiral helices, space groups ${P3_121}$ (right-handed)or ${P3_221}$ (left-handed)[H].

You obtain (racemic) benzil, modestly resolved (1 $\mathrm{S}$)-(-)-alpha pinene and (1 $\mathrm{R}$)-(+)-alpha pinene (both enantiomers), and mixed xylenes.  $20$ ml scintillation vial, foil-lined cap.  $100$ mg of benzil or so, $10$ ml resolved, alpha-pinene, heat to $100$ C. Drip in mixed xylenes until the molten benzil is just soluble, add another drop or two, very slowly cool undisturbed.  Only one space group crystallizes out.

Enantiomorphic space groups[I] are mathematically perfectly divergent geometric chiralities.  Here is the calculation of benzil crystal lattice's atomic mass distribution from Petitjean's software, CHI = $1$ (rapidly asymptotic) COR = $1$ DSI = $0$

If you want a discrete molecule as opposed to a crystal lattice, resolved camphor comes in around $\text{CHI} = 0.5$.  Resolved ${D_3}$-trishomocubane is a much better geometric example and is easily synthesized in large quantities starting with benzoquionone and cyclopentadiene (Diels-Alder).  If you have a methyl group sticking up from its ${D_3}$-axis, calculated $\text{CHI} = 0.996$.

Look at that methyl-bearing carbon!  It has no ${S_n}$ symmetries, for R must reflect or invert as S.  This chiral center's sense of chirality cannot be named.  Try doing it with CIP notation.

Uncle Al has been very naughty with geometric chirality, "[6.6]Chiralane: a remarkably symmetric chiral molecule," Symmetry: Culture and Science 19(4) 307-316 (2008).  Note that the core carbon atom has four rigorously identical substituents yet it is a maximally geometrically chiral center.  Chemical chiral nomenclature is fundamentally defective, for said chiral center (and the four to which it is attached) cannot be assigned a direction of handedness by any means, even in theory.

[A] http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
  pp. 11-13.
[B] J. Math. Phys. 40, 4587 (1999)
  http://petitjeanmichel.free.fr/itoweb.petitjean.freeware.html#QCM
[C] http://www.mazepath.com/uncleal/qzdense2.png for alpha-quartz
  http://www.mazepath.com/uncleal/glydense2.png for gamma-glycine
[D] http://www.mazepath.com/uncleal/norbors.gif
    http://www.mazepath.com/uncleal/norone.png
[E] Appl. Cryst. 33, 126 (2000)
[F] J. Appl. Crystallogr. 19, 108 (1986)
[G] J. Org Chem. 38(10), 1870 (1973)
[H] Acta Cryst. B43 398 (1987)
[I] http://elib.mi.sanu.ac.rs/files/journals/publ/69/7.pdf Section 2: eleven pairs of enantiomorphic space groups;
 http://www.math.ru.nl/~souvi/papers/acta03.pdf  Section 3ff.


London forces are the forces that you should always consider in the end. They're like the slow pawns on the chessboard, their effect is negligible when the fast rooks and the industrious queens are around!

In this case, our queens are the different structures of these elements. Let me first tell you the fact that even the melting point of boron ($\pu{2349K}$) is more than the boiling points of thallium and indium!

So, first off, why is the melting point of boron higher than that of all other group thirteen metals? From this Wikipedia page, its α-rhombohedral allotrope has B-12 icosahedral units arranged in a crystalline structure, and connected by strong covalent bonds[a]. As a comparison, the other metals in their solid state have unit cells whose atoms are bonded by only weak metallic bonding. The larger the atoms get, the weaker the metallic bonding gets, and hence, the melting point decreases down the group.[b]

Coming to the trend in boiling points now. Part of the difference in the trend has to arise from the excessively large melting point of boron. You see, thallium boils off even when boron has yet to melt, so that is a contributing factor in the excessively large boiling point of boron over the others.

According to this paper[1], for liquid thallium,

  The coordination number changes from 11.5 atoms at $\pu{350^\circ C}$ to 10.9 atoms at $\pu{700^\circ C}$

This is said to be due to:

  creation of vacant spaces about an atom. Although in liquid thallium the interatomic distances remain unchanged, some of the atoms are moving farther apart within the first coordination shell.

Another paper[2] however shows that, while CN of liquid indium decreases with increasing temperature, their inter-atomic separation also decreases. We can interpret this to be a "reinforcement" that helps increase the boiling point of indium over thallium.

With indium and thallium cleared, notice that the boiling points of indium, gallium, and aluminum are almost identical. Sadly, I've been unable to find any research papers dealing with these three elements in liquid state near their boiling point. For example, this paper[3] deals with liquid aluminium, but only upto $\pu{1273K}$.

We do have some data for boron. You'll be surprised how strong its covalent bonds are. In fact, this paper[4] states that, even at as absurdly high temperatures as $\pu{2600K}$, though the "icosahedra are destroyed", the "atoms still form an open packing with sixfold coordination". Moreover, this paper[5] observed that the density of liquid boron, at temperatures above $\pu{2360K}$, is higher than that of solid boron, which is unlike aluminum and indium (though for gallium the density at melting point is still slightly higher). Another paper[6] has measured the properties of liquid boron, but that sadly only at around $\pu{2400K}$.

I have checked a dozen best-matching research papers, and this was all the relevant data I could pull out. This isn't the complete answer, but again, without any proper data in the first place, there really cannot be a proper qualitative analysis either. So, this answer is really all there is.

References:

Temperature Dependence of the Structure and Transport Properties of Liquid Thallium, N. C. Halder and C. N. J. Wagner, The Journal of Chemical Physics 45, 482 (1966); https://doi.org/10.1063/1.1727593
Temperature Dependence of the Structure of Liquid Indium, H. Ocken and C. N. J. Wagner, Phys. Rev. 149, 122 – Published 9 September 1966
The structures of liquid mercury and liquid aluminum, P. J. Black and J. A. Cundall, Acta Cryst. (1965). 19, 807-814 https://doi.org/10.1107/S0365110X65004383
Structural and electronic properties of liquid boron from a molecular-dynamics simulation; N. Vast, S. Bernard, and G. Zerah; Phys. Rev. B 52, 4123 – Published 1 August 1995
Millot, F., Rifflet, J.C., Sarou-Kanian, V. et al. International Journal of Thermophysics (2002) 23: 1185. https://doi.org/10.1023/A:1019836102776
Structure of Liquid Boron; S. Krishnan, S. Ansell, J. J. Felten, K. J. Volin, and D. L. Price; Phys. Rev. Lett. 81, 586 – Published 20 July 1998; https://doi.org/10.1103/PhysRevLett.81.586

[a]: actually "the icosahedra are connected via a combination of two-center and three-center bonds" as claimed in [6], but suffice to say that they are almost as strong as the covalent bonds, if not more
[b]: except a few exceptions


At pH 7 lysine has a net charge of very close to +1.

The carboxylic acid group is fully deprotonated (-1 charge).

The alpha amino group is about 99% protonated (+1 charge).

The side chain amino group is fully protonated (+1 charge).

You need to look at the pK of each ionizable group, consider what fraction of the group is in each ionization state (for example using the Henderson Hasselbach equation), and add the charges of each group to get the net charge.

For isoleucin, at pH 7, the net charge will be almost neutral, because 7 is well above the carboxylic acid pK of 2.32 and significantly below the amino pK of 9.76.  There will be a small fractional negative charge because 7 is closer to the pK of 9.76 than to the pK of 2.32.  Note in the first linked table in the question, there is a column for "pI".  The pI will be the only pH where the amino acid has a net neutral charge.


Because on C; the negative charge should be on oxygen (the more electronegative element) instead of on carbon


Why do we get slightly more inversion product than retention product in SN1 reaction?

It is well known that SN1 reactions often give incomplete racemisation:

Although many first-order substitutions do give complete racemization, many others do not. Typically there is 5–20% inversion, although in a few cases, a small amount of retention of configuration has been found. These and other results have led to the conclusion that in many SN1 reactions at least some of the products are not formed from free carbocations but rather from ion pairs.

We can explain this by realising that the SN1 and SN2 mechanisms are not really discrete, as often taught, but rather two extremes of a 'mechanistic continuum'. This has implications for both the stereochemical outcome of many substitutions (i.e. proposed SN2 are not always as stereospecific as one might expect), and also the kinetics (i.e. measuring the rates of substitutions won't always give data consistent with purely SN2 or purely SN1 mechanisms).

At the SN1 extreme, there is no covalent interaction between the reactant and the incoming nucleophile during the transition state (TS) for the cleavage of the leaving group.
At the SN2 extreme, the TS has covalent interactions in which there is concerted formation of a new bond between the incoming nucleophile and the reactant , with cleavage of the bond to the leaving group.

This mechanistic continuum was first explained by Winstein, who studied both the kinetics and resulting stereochemistry of nucleophilic substitutions at saturated carbons.[1] This model is widely accepted, though rarely taught at the undergraduate level.
Winstein's theory proposes the presence of several additional intermediates along the reaction coordinate- these are known as 'ion pairs' and represent varying levels of association between the reagent and the leaving group. The energy levels between these ion pairs are small, and nucleophilic attack can occur at any stage (hence the continuum).

 Fig1: Winstein model for nucleophilic substitution. Taken from Carey's Advanced Organic Chemistry. Part A. /2
The Winstein model fits with what is commonly taught, namely that SN1 gives racemisation and SN2 gives inversion:

SN2  reactions occur when the nucleophile intercepts the contact ion pair and give stereospecific inversion of configuration since the leaving group is still associated with the carbocation, essentially shielding one face from attack
SN1 reactions occur when the nucleophile intercepts the fully dissociated free carbocation and gives complete racemisation as both faces of the planar carbocation are equally accessible

Which mechanism operates is largely determined by how stable the carbocation is. Primary carbocations are highly unstable and as such there is little (if no) dissociation. Tertiary cations are stable and as such complete dissociation can occur.
The interesting case is where nucleophilic attack occurs on a solvent separated ion pair - we see neither complete inversion nor complete racemisation, as some association still exists at this stage (partially, but not completely blocking one face of the cation).
Modern Physical Organic Chemistry[3] gives an overview of all of the possibilities, though the figure is a little hard to follow:

References

Winstein, S.; Clippinger, E.; Fainberg, A. H.; Heck, R.; Robinson, G. C. Salt Effects and Ion Pairs in Solvolysis and Related Reactions. III.1Common Ion Rate Depression and Exchange of Anions during Acetolysis2,3. J. Am. Chem. Soc. 1956, 78 (2), 328–335. DOI: 10.1021/ja01583a022.

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th ed.; Springer: New York, 2007.

Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science Books: Sausalito, CA, 2006.


Why is the water dissociation constant valid for acid/base solutions?

If Kw is written in terms of concentration (as opposed to activity), as [H+][OH-], it is only $10^{-14} M^2$ in the limit of zero ionic strength, but it is approximately correct in general.

image source


Aromaticity is not binary, but rather there are degrees of aromaticity.  The degree of aromaticity in benzene is large, whereas the spiro-aromaticity in [4.4]nonatetraene is relatively small. The aromaticity in naphthalene is not twice that of benzene.

Aromaticity has come to mean a stabilization resulting from p-orbital (although other orbitals can also be involved) overlap in a pi-type system.  As the examples above indicate, the stabilization can be large or small.

Let's consider $\ce{C_{60}}$:

Bond alternation is often taken as a sign of non-aromatic systems.  In $\ce{C_{60}}$ there are different bond lengths, ~1.4 and 1.45 angstroms. However, this variation is on the same order as that found in polycyclic aromatic hydrocarbons, and less than that observed in linear polyenes.

Conclusion: aromatic, but less so than benzene.

Magnetic properties are related to electron delocalization and are often used to assess aromaticity.  Both experiment and calculations suggest the existence of ring currents (diamagnetic and paramagnetic) in $\ce{C_{60}}$.

Conclusion: Although analysis is complex, analysis is consistent with at least some degree of aromaticity.

Reactivity - Substitution reactions are not possible as no hydrogens are present in $\ce{C_{60}}$.  When an anion or radical is added to $\ce{C_{60}}$ the electron(s) are not delocalized over the entire fullerene structure. However, most addition reactions are reversible suggesting that there is some extra stability or aromaticity associated with $\ce{C_{60}}$.

Conclusion: Not as aromatic as benzene

Resonance energy calculations have been performed and give conflicting results, although most suggest a small stabilization. Theoretical analysis of the following isodesmic reaction

$$\ce{C_{60} + 120 CH4 -> 30 C2H4 + 60 C2H6}$$

suggested that it only took half as much energy to break all of the bonds in $\ce{C60}$ compared to the same bond-breaking reaction with the appropriate number of benzenes.

Conclusion: Some aromatic stabilization, but significantly less than benzene.

This brief overview suggests that $\ce{C_{60}}$ does display properties that are consistent with some degree of aromatic stabilization, albeit less than that found with benzene.


Are there any hazardous products that would come from mixing glass cleaner with lubricant?

Nothing compromising safety will happen. Now, these are two completely different cleaning agents based on alcohol/water and organic-fuel respectively. Most likely will turn white  because these substances are immiscible. Therefore, it will not be a good idea to apply a mix for you cleaning tasks.  This will be  specially so  over wood or any rough porous surface likely as they will absorb the substances resulting in unappealing white or smear white spots.  Application over polished metal surfaces will be ok.


Difference between single point energy and dimerization energy

Definitions

Rather than use the phrase "dimerization energy", which is the energy difference specific to two chemically identical species coming together, such as $\ce{2 BH3 -> B2H6}$, it is better to use the phrase interaction energy. See here and here for a detailed distinction between interaction energies and binding energies. I will simplify a bit some of the text there and in this answer.

The single point energy (SP energy, SPE) is the total energy of a system at one and only one configuration of the nuclear positions. In the case of two monomers (sometimes called fragments), the interaction energy will be the energy difference between the complex (sometimes called the supersystem) and the sum of the individual fragments or monomers at the same nuclear positions as in the complex. This gives rise to the equation

$$
\Delta E_{\text{int}}(\text{AB}) = E(\text{AB}) - \left[ E(\text{A}) + E(\text{B}) \right]
$$

where there are two molecules, A and B, leading to three calculations, one for each of the monomers and one for the AB complex. In the dimerization of borane from above, or your specific example of the $\ce{HF}$ dimer, assuming that the internal coordinates (bond lengths, bond angles) of each monomer are identical, then the equation above can be simplified as A and B are identical:

$$
\Delta E_{\text{int}} = E_{\text{dimer}} - 2 E_{\text{monomer}}
$$

To take your specific example of the $\ce{HF}$ dimer, you mention the role of the "orientation of atoms" in both the single point and dimerization energies. For a single monomer, assuming again that internal coordinates are not changing, the relative positions of the nuclei are not changing, and the total energy is invariant to translations and rotations. However, in the dimer or complex geometry, the relative orientation of one monomer to the other will change the total energy of the dimer, leading to a change in the interaction energy. For example, in addition to the linear form $\ce{F-H}\cdots\ce{F-H}$, you might be interested in the "flipped" versions $\ce{F-H}\cdots\ce{H-F}$ or $\ce{H-F}\cdots\ce{F-H}$, or forming a T-shape, etc. The interaction energy of each will tell you about the relative stability of that nuclear configuration relative to the other configurations.

The most important variables then in an interaction energy calculation are the relative orientation of the interaction species and the distance between each of the fragments. How do you decide this? Usually some intermolecular coordinate is scanned, such as the distance between closest atoms on the monomers, still while leaving the monomer internal coordinates fixed to give an unrelaxed potential energy surface scan.

Calculation Examples

For example, here is the interaction energy for the $\ce{F-H}\cdots\ce{F-H}$ configuration with a variety of methods using the cc-pVTZ basis set and an intraatomic distance of 0.927 angstroms, scanning the $\ce{H}\cdots\ce{F}$ distance:

From a plot like this, we see that MP2 overestimates the interaction and CCSD underestimates it compared to CCSD(T); the shape of the B97-D potential is probably incorrect. HF is missing some repulsive interaction at long-range that the other methods include. It isn't dispersion, as the total dispersion interaction is always attractive and the minimum of the potential is qualitatively correct. You could certainly use any density functional or other method of your choice here (SCS-MP2, SOS-MP2, CISD, ...).

You could similar plots for other configurations, or choose a single method and place multiple configurations on the same plot (here, CCSD(T)/cc-pVTZ):

From a plot like this, it's pretty clear that only the $\ce{F-H}\cdots\ce{F-H}$ configuration is ever bound; it would be interesting to compare the T-shaped configurations to the linear ones.

Interaction Energy Decomposition

You mention "electrostatic monomeric interactions", and we also see some interesting long-range behavior in plot 1. One of the most powerful techniques for understanding intermolecular interactions is symmetry-adapted perturbation theory (SAPT), which gives a decomposition of interaction energies in terms of electrostatics, Pauli exchange, polarization (called induction), dispersion, and charge transfer (though the definition of the latter is often ambiguous).

Here are each of the interaction energy components (SAPT0/jun-cc-pVTZ, monomer-centered basis):

Combining all the electrostatic (1), exchange (1), induction(3), and dispersion (2) terms gives

This makes it pretty clear that all points, $\ce{F-H}\cdots\ce{F-H}$ is bound mostly by electrostatic attraction, and dispersion is negligible. Maybe you knew that, but I didn't. My guess is that the presence of dynamic correlation in all the wavefunction methods damps the long-range electrostatics compared to Hartree-Fock.

References

Interaction energies

J. Řezáč and P. Hobza, Chem. Rev., 2016, 116 (9), pp 5038–5071

SAPT

B. Jeziorski, R. Moszynski, and K. Szalewicz, Chem. Rev. 94, 1887 (1994)
E. G. Hohenstein and C. D. Sherrill, J. Chem. Phys. 132, 184111 (2010)
E. G. Hohenstein, R. M. Parrish, C. D. Sherrill, J. M. Turney, and H. F. Schaefer, J. Chem. Phys. 135, 174107 (2011)
T. M. Parker, L. A. Burns, R. M. Parrish, A. G. Ryno, and C. D. Sherrill, J. Chem. Phys. 140, 094106 (2014)

Technical notes

All of these calculations were run with Psi4. None of the interaction or SAPT energies are CP-corrected.


Supercritical fluid technology is used primarily to enhance the chemical extraction process.  The solubility of a compound in a supercritical solvent increases as the system pressure increases at constant temperature. This is due to the highly pressurized supercritical solvent disrupting the intermolecular attractive forces between the molecules of the compound to be extracted, and replacing those forces with increased attractive forces between the compound and the supercritical solvent.  Carbon dioxide is a very common material used for supercritical extraction.  It is cheap, not flammable, has low toxicity and, after the extraction is complete you simply return the system to STP, allow the carbon dioxide to evaporate and you are left with only the extracted material.  Some examples of commercial uses for supercritical carbon dioxide include caffeine extraction from coffee beans, sterilization of surgical equipment, fracking and dry cleaning.

Chemical reactions involving supercritical carbon dioxide are now being intensely studied.  One reaction that has been commercialized involves reaction of supercritical carbon dioxide with hardened, alkaline cements. A harder cement containing carbonate groups throughout the entire volume of cement is formed along with water (environmentally friendly process).


What's the reaction type of the Hydrogen shift?

This reaction can be viewed as a 1,2-sigmatropic hydrogen shift or as a 2+1 (if we are counting atoms, 2+0 if we are counting electrons) cycloaddition (adding the carbene's empty p-orbital across the $\ce{C-H}$ bond - analogous to its addition to an olefinic pi bond to produce a cyclopropane). However the reaction is most commonly viewed as and referred to as an "insertion" reaction because the carbenic carbon is inserting itself in between the carbon and hydrogen atoms that comprised the $\ce{C-H}$ bond.

(image source)

Here's what is known.

Thermal decomposition of most diazo compounds produce singlet carbenes.

Singlet carbenes add stereopecifically to olefins.  That is, addition of a singlet methylene carbene to trans-2-butene produces only the corresponding trans-dimethylcyclopropane

(image source)

Further, insertion of singlet carbenes into stereogenic centers results in retention of stereochemistry at the chiral center.  These facts suggest that both the cycloaddition and insertion reactions of simple singlet carbenes either proceed in a concerted fashion, or if any intermediate is formed it is extremely shorted-lived in order that stereochemical information be preserved.

Simple carbenes are typically mildly electrophilic.  The HOMO (the filled $\ce{sp^2}$ orbitals) and LUMO (the empty p-orbital) for a singlet carbene are shown on the left side of the following figure.

(image source)

Therefore, we would expect the interaction of the carbene LUMO (the empty p-orbital) and the $\ce{C-H}$ bond HOMO to be controlling, as the energy separation here is much less than the other possible HOMO-LUMO interaction between the carbene HOMO (the filled $\ce{sp^2}$ orbital) and the much higher in energy $\ce{C-H}$ antibonding orbital, the LUMO.

This controlling HOMO-LUMO interaction involving the empty carbenic p-orbital in the rate determing step is consistent with the observed electrophilic nature of the carbene.

All of this information taken together suggests that interaction of the empty carbenic p-orbital with the electrons in the $\ce{C-H}$ bond is the critical step.  A geometry like that shown below is likely preferred.  As the insertion proceeds, the trigonal carbenic carbon rotates.  The rotation starts to bring the carbenic $\ce{sp^2}$ orbital into alignment with the p-orbitals in the 7-membered ring.  Also during the rotation the $\ce{sp^2}$ orbital starts its rehybridization into a p-orbital. When completed we have created the final product, heptafulvene.  The reaction is likely concerted and certainly does not involve any long-lived intermediates. Note too that the heptafulvene is formed with increased electron density on the former carbenic carbon in accord with the dipolar nature (dipole moment ~0.5 D due to resonance stabilization) of the molecule.


Be careful with desiccant: silica may create tiny dust particles that could scratch the cornea, and hygroscopic salts may liquefy, possibly irritating.

A traditional anti-fog treatment is to rub the interior of the lens with bar soap (or a drop of mild detergent) and polish it. The soap decreases the surface tension of water, causing droplets to make a smoother, more transparent film. Wikipedia suggests some other anti-fog treatments.

There are also vented anti-fog goggles (the vents should have foam or other filter to make them splash resistant.

My personal choice for lab work or mechanical work is a full-face splash shield, which may offer more protection and does not fog.


Consumer choice; harmful environmental aspects of polyethylene vs polypropylene

Every material is terrible if not properly collected and processed after use.

Wether you re-use it (like glass bottles), or recycle the material, or burn it to make electricity and district heating, is small potatoes.

What material you use is also not very important, only the actual energy and environmental costs of "bio"materials like paper or cotton are very high. You have to re-use them very often before they make sense!


Can the reaction intermediate be lower in energy than the reactants or products?

The intermediate is a proper chemical species but a reactive one under 'normal' conditions and is not a transition state, i.e. it has a short, but measurable, lifetime. The intermediate can change only by crossing one of two transition states, one back to reactants and the other to another intermediate or to product.

Any intermediate must not be lowest energy species, including reactant and product, but can be lower or higher in energy than the reactant (but less than any transition state) and higher than product. The limiting activation energy will be the largest one in the whole scheme, so may be that between reactant & intermediate or between intermediate & product. Which it is depends on the exact species involved.


Which "exotic salt" can lower water's freezing point by 70 °C?

I recently got a chance to attend a talk by someone who was working on developing analytical instrumentation on Mars. The interesting story is that the initial results by ion-selective electrode was that Mars soil is full of nitrates. Nobody knew on Earth that the nitrate ion selective electrode is far more responsive to perchlorate than nitrate. After learning this, it was an eye opener for analytical chemists! Now they wish to use chromatography rather than electrochemistry. So this was a good lesson for us on Earth.

The perchlorate ion was discovered in 2008 by the nitrate selective electrode. No specific electrode was attached to detect perchlorate, it was rather an accidental discovery. The Science Report makes a footnote "Detection of Perchlorate and the Soluble Chemistry of Martian Soil at the Phoenix Lander Site" (paper: Science 2009, 325 (5936), 64–67)

A Hofmeister anion ISE was intended to monitor nitrate from a $\ce{LiNO3}$ reference electrolyte that was part of the leaching solution, but was ultimately used for perchlorate detection

  [Footnote] The relative sensitivity of the Hofmeister series ISE to perchlorate
  over nitrate is 1000:1, and substantial quantities of perchlorate will
  overwhelm any other signal. If, as was observed, >1 mM perchlorate
  accounts for the observed signal, it would require >1000 mM nitrate to
  produce the same response. This would correspond to more than the mass
  of the entire sample.

Now that they know it is a perchlorate ion, people did some studies on supercooled brines. See this paper: Toner, J.; Catling, D.; Light, B. The formation of supercooled brines, viscous liquids, and low-temperature perchlorate glasses in aqueous solutions relevant to Mars. Icarus 2014, 233, 36–47 (also available here). They clearly show that if calcium or magnesium perchlorates are slowly cooled, one can get supercooled brines up to -120 Celcius. This is a rather amazing finding. They call it a glassy state.


Gaussian Calculations - What is the Basis?

The basis set is used to describe the molecular orbital wave functions.

I suggest reading through a number of links:

Wikipedia
The Gaussian manual - Basis Sets - a nice description of the different basis sets, if a bit technical at times.
Naming of quantum chemistry basis sets
GAMESS Crash Course - applies to Gaussian or other packages


Possible to Calculate Electronic Charge Distributions/Densities for Organic Molecules?

If I understand correctly, you want the total electric field at some point $\mathbf{r}$ due to a molecule. This will have at least three components. The first due to the electric field from the nuclei. This can be calculated classically from the field due to point charges. The second due to the dipole moment of a molecule. This can also be calculated completely classically (though you will have to use some electronic structure calculation to get the dipole and direction). Finally, you will need to find the total charge density of the electrons.

The charge density can be found using either DFT or any post-Hartree Fock method such as MP2 or CCSD(T), etc. When using something like MP2, you will get the charge density from the molecular orbitals. You can do this manually, but one code I know of that will give you the relevant information from the HF orbitals is NWChem. You will want to use the dplot command. This allows you to print out electron densities for each orbital using a grid. You will want to use a fine enough grid to capture the actual spatial changes in density. Notice that you also provide limits i x, y, and z. This is because, in principle, the charge density extends to infinity in all directions. Thus, you will want to choose a cutoff where the density is sufficiently small as to be negligible. Then, you can again calculate the field due to the electron density from classical equations which can be found online or in an E&M textbook. You can either calculate the field discretely from the points on the grid, or fit it to some functional form and then do a true integral over the density.

Another way to do this is to use a code which will give you approximate charges per atom from some type of population analysis such as Mulliken charges. You can do this with Gaussian using population keyword. This will aggregate the nuclear charge and the electronic charge into one field. Fair warning, however, Mulliken charges (and all population analyses) are approximate and not even necessarily physical.

There may be other ways to do this from DFT, but I surprisingly cannot find any software packages that do this directly.


Do liquid O₂ and N₂ form ideal solutions?

It's not perfectly ideal.  Try to get access to Excess enthalpies of the liquid mixtures nitrogen + oxygen... for quantitative data.


How to determine the shape of hybridized atomic orbitals in VB theory?

What we call "Hybridization" is really just a mathematical transformation of an approximated wavefunction. Your basic summation formula describes how to build some sort of customized wavefunction by forming a linear combination of some set of "basis" orbitals; you will find this equation over and over if you look up "basis set expansion". Now, of course you can't just use any random coefficients $c_{ij}$ of your liking for this custom wavefunction; you have to adhere to some physical rules. Most importantly, the overall energy that the wavefunction represents must not change. I won't go into detail about the constraints, but if you are interested, google "unitary transformation".
For orbital hybridization, you first have to find a set of basis orbitals from a physically sound starting wavefunction. If you only look at an isolated atom, then atomic orbitals are obviously a useful choice. From there, you can form combinations of those orbitals (by some criteria that really aren't uniquely defined) while adhering to the rules of the unitary transformation. In the end, you will arrive at your preferred hybrid orbitals that fulfill your chosen criterion as best as they can while keeping the wavefunction energetically equivalent to the original one. Some of the coefficients $c_{ij}$ in your equations may come out to be 0, typically for reasons of symmetry. For example, an $sp^2$ hybrid that is oriented along the $y$ axis of your coordinate system would obviously only have contributions from the $p_y$ orbital, not from $p_x$ and $p_z$. But the exact values of the cofficients will generally depend on the specifics of your problem, most importantly the definition of your coordinate system. So you really can't say per se which coefficients will drop out to be 0 and which won't -- it's one of those "it depends" situations.
General Thoughts:
Unfortunately, hybridization is a concept that is so misleading that I would consider its general use in chemistry to be outright physically incorrect. People invoke it to suggest and visualize bonding situations in molecules and to explain the resulting molecular shapes, but unfortunately always seem to forget some very basic undergraduate quantum chemistry in this process.
Hybrid orbitals are linear combinations of eigenfunctions to different energy eigenvalues of the underlying one-electron Hamiltonian; think $s$ and $p$ orbitals. As such, they are not solutions to the time-independent Schrödinger Equation anymore, which in turn means that they are time-dependent functions. This implies two things:
1.) Hybrid orbitals do not have a fixed shape; what you see visualized is just a single snapshot in time. The orbital shapes periodically "morph" back and forth and can at other points in time look nothing like those suggestive localized "bonding" orbitals anymore. In fact, the hybrid orbitals can even change places among each other over time if the symmetry of the system permits.
To better drive this point home, here are some (semi-quantitative) visualizations of the time development of the bonding and non-bonding orbitals in water.
2.) Hybrid orbitals do not have defined energies, since they are not solutions to the time-independent Schrödinger Equation. If you were to do any sort of "energy measurement" on a hybrid orbital, you would end up with a Schrödinger's Cat situation where you observe the energies of the constituent atomic orbitals with certain probabilities that are connected to the expansion coefficients $c_{ij}$.
In addition, there is a third problem. When the pure integer mixtures of s and p orbitals (e.g. $\mathrm{sp}^2$, $\mathrm{sp}^3$) were not sufficient to explain molecular shapes, people started to come up with weird fractional mixtures like $\mathrm{sp}^{1.5}$ or some such. How are those numbers obtained? - As stated above, from the wavefunction side this comes down to a "localization" procedure of the molecular orbitals; however, the criteria by which to do this are not strictly defined, and there exist multiple different schemes for it. So this raises the third problem:
3.) Hybridization is not stringently mathematically defined if one regards it as an orbital localization procedure.
So, the next time someone presents you with hybrid orbitals to explain e.g. why methane is tetrahedral, ask them about the energy and the electronic density distributions of those orbitals, and how they would determine the hybrid orbitals for e.g. chloromethane where the bond angles and energies are clearly different than in methane.


Why does a carbocation undergo covalent bonding with, say, chloride, rather than ionic bonding, like what occurs between Na+ and Cl-?

Some carbocations do form true ionic compounds.  Many are known in which the positive charge is incorporated into an aromatic ring (e.g., cyclopropenyl, tropyl) or conjugated with multiple aromatic rings (e.g., triphenylmethyl).  But in most carbocatioms you have a compact, relatively low-energy vacant orbital that will overlap to form a bond with an electron pair from, say, a chloride ion.  Alkali metals don't have such compact valence orbitals and thus do not form a strong covalent bond even with the positive charge all on one atom.


Are benzyl alcohol and o-cresol functional isomers or position isomers?

The chloride example is interesting because it highlights a fine line separating both.

The distinction between functional and position isomers isn't as simple as the typical examples make it to be - sometimes, what a substituent is attached to is critical to its functional group, so the same sequence of atoms with the same connectivity can behave, chemically speaking, very differently depending on what they are bonded to. And, conversely, position isomers in general don't have exactly the same chemistry either - you know, for instance, that n-butanol, 2-butanol and t-butanol will behave similarly but not identically.

So where do we draw the line? Technically, it's true that both are position isomers. In both cases the same substituent is attached to different positions (unlike, say, acetone and propenol, where the connectivity is clearly different). But it's also true that their chemistry is distinct - aryl compounds, in general, have different reactivity than their equivalent alkyl compounds, which is why we talk about the reactivity of "phenols" and "alcohols" as separate ideas.

So, from your examples, I'd say benzyl alcohol and o-cresol are clearly functional isomers, as their chemistry is quite different. For instance, compare pKa values:

o-cresol: $\mathrm pK_\mathrm a=10.3$

benzyl alcohol: $\mathrm pK_\mathrm a=15.4$

2-ethylphenol (similar to o-cresol): $\mathrm pK_\mathrm a=10.2$

2-phenylethanol (similar to benzyl alcohol): $\mathrm pK_\mathrm a=15.9$

Chloride, on the other hand, is less sensitive to the substrate; but I'd still classify chlorotoluenes and benzyl chloride as functional isomers, since I think they are different enough to merit that distinction, but it's in the twilight zone of vague definitions.