Why are there as many molecular orbitals as atomic orbitals?

The orbitals, both atomic and molecular, must be orthogonal, that is, they have no net overlap or correlation:  equal constructive and destructive interference. You can think of them as forming the basis set of an n-dimensional vector space. If you think of the atomic orbitals as the standard basis...

Example for a 4-dimensional space:

$\langle 1,0,0,0 \rangle$,
$\langle 0,1,0,0 \rangle$,
$\langle 0,0,1,0 \rangle$,
$\langle 0,0,0,1 \rangle$

...Then the molecular orbitals are just a rotated basis. In other words, they take linear combinations of the atomic orbitals to span the same space.
Ex:
$\langle \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \rangle$,
$\langle \frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2} \rangle$,
$\langle \frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2} \rangle$, $\langle \frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2} \rangle$

The same n-dimensional space requires at least the same number (n) of vectors to span it. This is exactly the same phenomenon as rehybridising individual s and p orbitals to form sp$^3$ orbitals.

You can find explicit formulae for the orbital vectors with the actual numbers (one possible representation), along with fantastic illustrations here Darmstadt, LCAO Theory.

There are two normalisation conditions when combining atomic orbitals to make either hybridised orbitals or molecular orbitals. The first is that every atomic orbital must be used completely. That is, the sum of its contributions to all the molecular orbitals must be 1. The other is that each of the new orbitals cannot have more than 100% orbital. Although I think that it is this second part that you were questioning the reasoning for.


Validity of rationale for balancing chemical equations

When you write the equation

$$\ce{H2 + O2 -> H2O}$$

you are implicitly saying that the stoichiometric coefficient of every substance is $1$. It is like writing the equation $a + b = c$. By not having coefficients in front of $a$, $b$, or $c$, you are implying that $1a + 1b = 1c$ and not $2a + 3b = 4c$ or anything like that.

Obviously that cannot be true for the reaction above because if you react one $\ce{H2}$ molecule and one $\ce{O2}$ molecule, you get one $\ce{H2O}$ molecule and one leftover oxygen atom that doesn't know what to do with itself. The only balanced equation that does not lead to stray atoms is

$$\ce{2H2 + O2 -> 2H2O}$$

or any integer multiple of that, i.e.

$$\ce{$(2n)$H2 + $(n)$O2 -> $(2n)$H2O}$$

Humans prefer to work with small integers or half-integers, so $n$ is usually either chosen to be $1$ or $1/2$.


What is a barrier-less reaction in Quantum Chemistry?

Usually, in the reactions with a barrier, the saddle point with one imaginary frequency is taken as TS. However, many unimolecular dissociation reactions are barrierless: the reaction path on potential surface, connecting (minimum of) reactant with (minima of) products, is such that the energy along the path does not have a local maximum. Kinetic treatments of such reactions are more difficult: it is still possible to assign a TS, but it no longer a saddle point. In order to find it, one can use variational TST, but it is somewhat expensive.

A general recommendation after location of a saddle point: run reaction path calculation to see that this saddle point indeed connects the products and reactants that you expect.

About this small eigenvalue: this eigenvalue can easily pop up instead of zero eigenvalues of rotations and translations. Even if your calculation explicitly projects them out, numerical noise can be this order of magnitude (as pointed out above).


Reaction of carboxylic acid with phosphorus trichloride

A plausible mechamism of this reaction is discussed in a paper by Xiao et. al1 where they tried to react benzoic acid with $\ce{PCl3}$ to get acyl chloride. They noted the following mechanism:

Although the detailed mechanism was not clear, a possible mechanism is
proposed [...] the first step should be a reaction of the acid with
$\ce{PCl3}$, probably via a four-center transition state A, with
carbonyl oxygen attacks the electron-deficient $\ce{P(Cl)}$ while $\ce{Cl}$
attacking the electronically deficient carbonyl carbon. Formation of a
$\ce{Cl-C}$ bond and $\ce{P-O}$ bond (A’), further $\ce{C-O}$ bond cleavage and
hydrogen transfer took place to give the products B and C.
Product C then reacts further with the acid to give the
corresponding chlorides and $\ce{HP(O)(OH)2}$

This product C will again react with carboxylic acid two times so as to replace all the chlorine atoms and ultimately forming phosphorous acid (actually phosphorous acid technically exist as $\ce{HPO(OH)2}$ called phosphonic acid) So, the final reaction would be three moles of carboxylic acid reacting with $\ce{PCl3}$ to form three moles of acyl chloride and phosphonic acid thus fully utilizing the three chlorine atoms in $\ce{PCl3}$. The major advantage of this reaction is that there is only one side product i.e phosphonic acid which is non-toxic and easily removed by filtration whereas reaction with $\ce{PCl5/SOCl2}$ gives two side products making the reaction more volatile.
Reference

Atom-efficient chlorination of benzoic acids with PCl3 generating acyl chlorides, Jing Xiao and Li-Biao Han, Journal of Chemical Research, 2019, Vol. 43(5-6) 205–210 (PDF)


What is the reaction between nitric acid and fluorine?

It was first discovered by Cady that when fluorine is bubbled through $\pu{3 N}$ nitric acid, then a remarkable compound, fluorine nitrate ($\ce{FNO3}$) is formed. It boils at $\pu{-42 ^\circ C}$ and is a very strong oxidizing agent and is capable of liberating iodine from iodides and in gaseous states, it explodes around $\pu{200-300 ^\circ C}$.

References

The reaction of Fluorine with Nitric acid and with solid Potassium Nitrate to form NO3F by Don M. Yost and Alan Beerbower,  J. Am. Chem. Soc. April 1935, 57(4) 782. DOI: https://doi.org/10.1021/ja01307a520
Prof. Dr. Otto Ruff, Dr. Walter Kwasnikie,  Die Fluorierung der Salpetersäure. Das Nitroxyfluorid – NO3F, April 1935, DOI: https://doi.org/10.1002/ange.19350481604 [German paper]


The solubility of a solute in a solvent is determined by the chemical potentials of the solute for the solution and the pure solute. Substances move from high chemical potential to low chemical potential. At the concentration of solubility, these two chemical potentials are equal so there is no net transfer. Below this concentration, the chemical potential of the solute in the solution is lower than that of the pure solute so molecules will transfer from the pure solute to solvent until it is all dissolved or saturation is reached.

When you invoke the common ion effect, you are increasing the chemical potential of the common ion in solution which decreases the solubility. It is possible to have a nearly saturated solution and add enough common ions to reduce the solubility below the original concentration.


How strong are molecular bonds in layman's terms?

I will assume that by "a single sheet of graphite" you mean graphene.

The link in the comments to the wikipedia page on orders of magnitude of different forces, provided by @Alchimista in the comments, is helpful in getting a perspective on at least a few of your questions. Two useful reference values are the force exerted by a 1-kg weight due to gravity at sea level, ~10 N, and the force required to rupture a typical covalent bond, ~1 nN.

Perhaps the first question would seem to be the easiest to answer, but I found it is not: mechanoreception by sensors on skin is complicated. In the same way you can sense very gentle contact, you may be able to sense the graphite sheet touching your skin. It would probably be an odd feeling. Note also that the sheet would be just barely visible.

The strength of a material is described by more than one property. Perhaps the simplest property to describe is the tensile strength, the stress required to rupture the material by pulling it apart under tension. You can use estimates of the geometry of a sheet of graphite and the C-C bond enthalpy to arrive at the force per unit length of the material require to rupture it:

$$\pu{F_l(N/m)} \approx \frac{\pu{350 kJ/mol}}{(\pu{2.46 Å*1.42 Å})}$$

That's $F_l\approx\pu{16 N/m}$, so you might expect a single layer of graphite that is 1 meter long to be able to hold up a 1-kg weight suspended normal to this dimension, assuming the force is evenly distributed. If we assume the sheet is ~$\pu{2 Å}$ thick then we arrive at an estimate of the tensile strength of ~$\pu{10^11 Pa}$, making it the "strongest material ever tested."

Here I've estimated the number of bonds per unit length normal to the direction of tension as 1 per 2.46 Å based on this figure depicting the dimensions of a sheet of graphite. I also assumed the average force required to break one CC bond is $\Delta_{bond}H/r_{CC}$ and used a low estimate of $\Delta_{bond}H$. You could derive a similar value of $\Delta_{bond}H$ from the heat of sublimation of graphite.

Based on this information I would assume that it would be easy to tear the sheet, but that you would not necessarily tear the sheet by grasping it, provided you did this very carefully. On the other hand, you could certainly punch a hole with your finger. Ignoring subtle issues regarding the geometry at which forces are applied, we can use the previous estimate of the rupture force per unit length and an estimate of the radius of a finger (~1/2 cm) to arrive at the force required to punch the sheet, or $\approx\pu{0.5 N}$, or equivalently the pressure:

$$p=\frac{2F_l}{r_{finger}}$$

or $\approx\pu{7 kPa}$. A fat cat would only exert a pressure of about 300 Pa.

Could you fold it into a paper aeroplane? Apparently graphene is tough enough to be significantly bent out of shape without rupturing (Refs. 1 and 2), so I'd have to guess "yes".

References

Briggs et al., Electromechanical robustness of monolayer graphene with extreme bending. Appl. Phys. Lett. 97, 223102 (2010)
Edges of graphite and refs therein


The actual mechanism is believed to be strongly dependent on the nature of halide, metal (not only $\ce{Na}$ is used in this reaction), and solvent. The 'ionic' mechanism you have written here is the part of the most accepted mechanism which involves both, an 'ionic' and a radical reaction.
First step. Through the single-electron transfer from $\ce{Na}$ to the halogen atom, an alkyl radical $\ce{R^.}$ and sodium halide is formed:
$$\ce{Na + R-X -> Na+X^- + R^.}$$
Alkyl radical further acts as an electron acceptor from another $\ce{Na}$ atom to form highly nucleophilic intermediate (can be isolated in some cases):
$$\ce{R^. + Na -> R^-Na+}$$
This alkyl sodium intermediate reacts with another alkyl halide in the aliphatic nucleophilic substitution:
$$\ce{R^-Na+ + RX -> R-R + NaX}$$


Yes the equilibrium certainly exists. In fact, there exists a third equilibrium, the self ionisation of water. Considering any of the two equilibria gives the concentrations of all the species involved, there is no need to write down the third equilibrium as the third equation will be a combination of the first two. The problem arises only when you invoke stoichiometry. For example, if $0.1\ \mathrm{mol}$ of the acid dissociated, it would certainly give $0.1\ \mathrm{mol}\ \ce{A-}$, but $0.1\ \mathrm{mol}$ would not be the amount of $[\ce{H3O+}]$ as it is part of another equilibrium. However, if we do make this approximation, ignoring the hydrolysis of $\ce{A-} $, we can justify it in some cases. Say if $K_\mathrm{a}$ is around $10^{-5}$ then the equilibrium constant for the hydrolysis would be $K_\mathrm{w}/K_\mathrm{b}$ which would be around $10^{-9}$. Thus it would not be a bad approximation to ignore the second equilibrium for finding the pH.


Is there a geometrically intuitive explanation of the equivalence of the numbers of irreducible representations and symmetry classes?

SUMMARY
The geometrically intuitive justification for why the number of irreducible representations and classes are the same is because:

The irreducible representations of a molecule’s symmetry group correspond to a resolution of the molecule’s spatial structure (visualised even as a simple ball and stick model) into a variety of quite different, component geometrical objects – the electronic molecular-orbital components, each obeying the whole set of the symmetry group’s operations in a distinct (orthogonal) manner
As vectors in the Hilbert space of the molecule’s electronic structure, the molecular orbitals occupy (and are symmetry-bound by) the invariant subspaces of the Hilbert space that are reached by the operations of the corresponding classes of the symmetry operations of the group on any of the vectors.
Hence the number of irreducible representations of the molecule equals the number of symmetry classes of its point group.

NB A detailed reply to @Porphyrin's comment under the question, which is too long to insert as a comment in reply, is given at the end of this answer.
DETAIL
For those who, like me, think in more of a visual/intuitive than abstract algebraic manner:
In chemistry, we can easily visualise the groups of the physical symmetries of molecules as combinations of symmetry operations transforming an image of the molecule (for atoms, it is simply the group SO(3) of rotations in R3).
The irreducible representations of the group illustrate the transformation properties of the component molecular or atomic orbitals under the operations of the molecular group, and can also be visualised as corresponding directly to both geometrical objects in physical space (i.e. the spatial distributions that are modelled as the MOs or AOs themselves) and, a bit more schematically, as vectors in invariant subspaces of a Hilbert space of functions representing AOs and MOs.
I think of this in terms of the visualisable spatial structure of the molecule overall (even as a ball and stick model for the purposes of understanding its symmetries) being analysed by group theory (or, if you prefer, resolved into) a set of different component geometric/physical structures – the molecular or atomic orbitals (all having the correct symmetry properties, regardless of the approximation method used in their computation).
[Please note that I’m choosing my words very carefully as there’s much confusion in discussions of representation theory due to frequent abuses of language, especially by mathematicians and by others who are so familiar with the material that they cease to be aware of how easily students can be confused by such abuses.]
So the transformation properties of the component MOs under the overall molecule’s group operations are encoded in the irreducible representations of the group and, for most purposes, conveniently summarised in the associated character tables, listing the traces of matrix representations of the group operations on suitable basis functions).
Character tables can be considered as a shorthand version of the irreps since they contain most of the information of use to chemists.
Indeed, one can say that “the molecular orbitals are the irreducible representations” of the group, in much the same way that Wigner’s Classification leads to the rather more mysterious statement that “fundamental particles are the irreducible representations of the Poincare’ Group”; in that case, it’s the symmetries of transformations in spacetime, under the Lorentz Group plus translations, yielding the Poincare’ group).
Just as we can visualise the orbital geometries described by the irreducible representations, likewise the classes of a molecule’s symmetry operations are also easily visualised.
Therefore the answer to the original question comes by considering the other way of imagining the MOs, as vectors in invariant subspaces of a Hilbert space of functions.
The characters of the irreps will be the same for each symmetry operation of a given class; so for $C_\mathrm{3v}$, for example, all three mirror planes σ will have characters of 1 for the A1 rep, -1 for the A2 rep and 0 for the 2-dimensional E rep.

This says to me that the molecular-orbital wavefunctions (i.e. irreducible representations) each span a subspace of the Hilbert space of molecular orbitals defined or bounded by the classes. No symmetry operation will take an MO with characters = 1 for the mirror planes (i.e. those MOs unaffected by reflections in any of the three mirror planes) into any subspace in which reflections invert the sign of the MO function.
There are three such classes for $C_\mathrm{3v}$, so the MOs appear to be limited to, or bounded by, three subspaces of the Hilbert space.
I conclude therefore that the geometrically intuitive justification for why the number of irreps and classes are the same is that:

the irreps correspond to a resolution of the molecular structure into a variety of molecular orbital components, each obeying the whole set of the symmetry group’s operations in a distinct way
the same molecular orbitals are distributed across as many invariant subspaces of the Hilbert spaces as there are classes of the group.

APPENDED REPLY to @Porphyrin’s comment:
i)  I’m aware of group theory’s abstract nature, for example permutation groups; but even they can have a physical realisation. When group theory is applied to material systems, there should be some sort of physical interpretation.
ii) Some people are content to work with a purely abstract sense of groups, while others find it helpful to have a more intuitive, often geometric sense. You might be familiar with the famous mathematician Jacques Hadamard’s On the Psychology of Invention in the Mathematical Field in which his survey of leading mathematicians and theoretical physicists’ in the mid C20th found a propensity to think in visual imagery.
iii)    It’s reasonable to claim that mathematical models often usefully represent physical realities. Recall the title/quote from the 1960 lecture by one of the supreme masters of symmetry in physics, Eugene Wigner: The Unreasonable Effectiveness of Mathematics in the Natural Sciences


Reaction mechanism with benzylchloride (KMnO4 and Soda lime)

This is not a magical reaction. Yet, it is a two step reaction as depicted in diagram below:

As @Waylander's comment elsewhere, benzyl chloride might have converted to benzyl alcohol before get oxidized to benzoic acid in acidic $\ce{KMnO4}$ solution (the first step). This oxidation is well known (see here). The second step is heating (may be $\gt \pu{200 ^\circ C}$) of solid benzoic acid in the presence of solid soda lime to cause the decarboxylation. This reaction is well described by Dr. Jim Clark (here) and also here.


Why is the carbon-fluorine bond stronger than the other haloalkanes?

From wikipedia article of carbon-fluorine bond,

  The high electronegativity of fluorine (4.0 for F vs. 2.5 for carbon)
  gives the carbon–fluorine bond a significant polarity/dipole moment.
  The electron density is concentrated around the fluorine, leaving the
  carbon relatively electron poor. This introduces ionic character to
  the bond through partial charges ($\ce{C^{δ+}—F^{δ−}}$). The partial
  charges on the fluorine and carbon are attractive, contributing to the
  unusual bond strength of the carbon–fluorine bond. The bond is
  labeled as "the strongest in organic chemistry," because fluorine
  forms the strongest single bond to carbon. Carbon–fluorine bonds can
  have a bond dissociation energy (BDE) of up to 544 kJ/mol. (This is why) The BDE
  (strength of the bond) is higher than other carbon–halogen and
  carbon–hydrogen bonds. For example, the molecule represented by
  $\ce{CH3X}$ has a BDE of 115 kcal/mol for carbon–fluorine while values
  of 104.9, 83.7, 72.1, and 57.6 kcal/mol represent $\ce{C–X}$ bonds
  to hydrogen, chlorine, bromine, and iodine, respectively.


What is the use/importance of the equilibrium constant Kc

$K$ is the value of $Q$ at equilibrium. The expression for $Q$ and $K$ are the same however $Q$ describes the state of a system and $K$ describes the system at equilibrium. Le Chatelier's principle tells us that a system that has $Q\neq K$ will approach $K$. When $Q$ is not $K,$ the system is not at equilibrium.

K tells you where an equilibrium lays. For a given reaction, if $K$ is larger than 1 than the equilibrium is towards the right. If $K$ is less than one then the equilibrium is towards the left.
   For a simple equilibrium such as acid dissociation, the forward reaction's equilibrium constant is

$$K = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
so you can compare the strengths of acids by comparing the equilibrium constant. Since the $K$ of nitric acid is larger than the $K$ of hydrofluoric acid we know that it is a stronger acid.

Hope this helps!


Why does stale bread go soft in the microwave?

I believe this has to do with its moisture content. While certainly dry, it still has some moisture, and what little moisture is left gets excited by the microwaves, heats up (the heat helps it soften too) and steams the bread essentially. I think that MaxW's explanation of glutens for the question 'Where does the moisture come from?' is a likely explanation.

Note that if the bread is sufficiently dry enough, this little microwave trick wont help. It is usually advised that you add a few drops of water to the bread before heating up in the microwave in this way. This is to ensure the bread doesn't actually become harder in the microwave, as it boils the water and will drive it off if microwaved in excess.

Also, note that bread WILL become blackened, charred and start smoking if left in the microwave long enough. I did this by typing 30 minutes instead of 30 seconds. It started smoking around 5 minutes.


The acid in soft drinks dissolves your teeth. Which type of chemical could you use to stop the acid?

What you meant is known as erosion. Erosion is the chemical loss of enamel due to acid.When acid continuously attacks teeth, they cannot repair themselves and will gradually begin to turn fuzzy and dissolve. Most soft drinks contain acids such as citric, phosphoric and malic or tartaric acids. A study was done where extracted human teeth were placed in softdrinks. Within two days, the teeth became very soft and the enamel surface lost much of its calcium.

Brush your teeth at least twice a day with a fluoride-containing toothpaste.A mouth rinse containing fluoride can help prevent tooth decay, according to the American Dental Association. For children, sealants can be used.


Does UV water purification cause cancer?

Is it possible that this use of UV radiation may lead to cancer in humans?

No!

The lamps used for water treatment are usually low pressure mercury vapour lamps. These lamps emit in the UV-C range with a maximum at $\lambda = 254~\mathrm{nm}$.

No mercury can escape from an intact lamp.
In principle, UV-C radiation may cause severe sun burns and harm your eyes. However, the lamps are housed and the consumer is protected from the direct exposure to radiation.
There is no change to the water due to UV irradiation.
UV-C radiation is efficient in killing microorganisms in water.


can we use con HCl zn amalgam to reduce alcohol?

Here is the mechanism for reduction of aldehyde/ketone by clemmenson reduction.

Note that alcohol is formed as intermediate which further reduces. Considering that alcohol can be reduced by Zn/HCl


Having some level of experience in fire training in various scenarios over a number of years, I highly recommend that you take a commercial approach to generating your smoke, rather than a homemade recipe that compounds your liability. Any smoke you generate from a homemade smoke bomb is likely to be considerably toxic, and should be avoided. Fire fighters would not enter a smoke-filled boat to attack an active fire without the appropriate safety gear which would include correct breathing apparatus.

You can purchase smoke cartridges of various sizes, colours and burn times, and these are designed for this very purpose. Examples can be found using a quick search for fire training smoke cartridges: here, here and here are examples from companies that I in no way endorse. They are pretty cheap for the effects they produce. I use the 20m3 burn capsules for fire fighting training with vehicle fires - these have a burn time of about 5 minutes, and they are very authentic, apart from limited smell. The smell from a fire in a boat is likely to be an acrid toxic affair, quite unlike your typical woodfire smell, so you would be best to avoid this. For better effects, burn a couple together, and you can get burning cups with electronic ignition that allows you to remotely activate it whenever you are ready.

Another option for generating limited visibility that is frequently used is a smoke/fog machine that is used for theatrical smoke and at nightclubs. Search for fog machine or smoke machine. Here is an example - again, I have no affiliation with this company. I have used ones that have come off ebay, and they are very cheap, but do the job nicely. I have used these a lot simulating structural fires and vehicle fires, and they can produce great amounts of 'smoke'. You need to take care that before any water is applied that the smoke machine has been turned off and removed well aware from the area. These also work very well for using with ram fans to visualise wind flow patterns through buildings/structures to understand convection flows and the path of travel for fires.

Be sure to read and understand the SDS of all materials before you use them in an enclosed space such as a boat, and ensure you have a plan to sufficiently ventilate any space between fire drills and afterwards.


Which d orbitals can form sigma, pi, delta bonds?

Sigma, pi and delta denote how many planar nodes are in the bond. Sigma bonds have no node, pi bonds have one and delta bonds have two. You can tell what kind of bond forms by how the orbitals overlap. Two single lobes form a sigma bond, two pairs of lobes form a pi bond and two quartets form a delta bond.


Controlling the reaction of alcohols with acid to either ether synthesis or elimination

By modifying the conditions, in this case: the concentration.

The first and rapid step of the mechanism is the protonation of the alcohol:

$$\ce{HA + CH3CH2-OH <=> A- + CH3CH2-OH2+}$$

From then on, it depends what happens:

If the molecule is left alone, i.e. its concentration is low, something that has a just enough basicity might at the correct time be close enough to a methyl hydrogen to allow the elimination to happen. $\mathrm{E1}$ is not an option since the primary carbocation is extremely unstable and Wagner-Meerwein rearrangements for enhanced stability are not possible, thus the elimination must preceed by the $\mathrm{E2}$ mechanism. If $\ce{A-}$ is the only basic compound present, it must accept the proton.
$$\ce{A- + H-CH2-CH2-OH2+ -> A-H + H2C=CH2 + H2O}$$
If the molecule happens to find one if its kind before the former can take place, the two of them could engage in a substitution reaction by the $\mathrm{S_N2}$ mechanism. If the molecule’s concentration is high, this gets more likely. And if it happens, the second molecule’s hydroxy group is nucleophilic enough (even though alcohols are generally bad nucleophiles) to displace the extremely good leaving group $\ce{-OH2+}$ and liberate water. The extraneous proton can later be removed as the ether sees fit.
$$\ce{CH3CH2-OH2+ + HO-CH2CH3 -> CH3CH2-(OH+)-CH2CH3 + H2O}$$
$$\ce{CH3CH2-(OH+)-CH2CH3 + A- <=> CH3CH2-O-CH2CH3 + HA}$$

Thus, the first reaction will always happen but be rather slow. Its rate is close enough to being independent of the molecule’s concentration. The second reaction’s rate is directly dependent on the molecule’s concentration and will get faster the higher the concentration is. High concentrations (and low acid concentrations) can thus favour the substitution mechanism over the elimination one.

This will hold true more or less for all cases where two reactions compete against one another. It is always possible to modify the conditions to prefer one or the other. However, it is often not as simple as in this simple case.


SARS-CoV - relative size of the spike protein

Although this is a homework question, I also think OP deserves a help as orthocresol suggested. I'm not going to go into details but leave it at orthocresol's suggestion of this being a Western Blot of protein analysis, which is absolutely correct. Hence, OP should start reading of electrophoresis and western blots techniques since, I assume, s/he taking molecular biology class.
In brief, the given image is a X-ray photograph of the nitrocellulose membrane after transferring the proteins from the SDS-PAGE gel (after electrophoresis). I'm suggesting that because the image looks like visualized by radioactivity (I suggest OP to read about these techniques in details to understand). In SDS-PAGE gel electrophoresis, the proteins separated mainly by molar mass of each protein while heaviest protein go first since electric current run from top to bottom as sepicted in following diagram:

As marked in the diagram, SARS-CoV consists of SARS-M ($\pu{20 kD}$, SARS-N ($\pu{54 kD}$, and SARS-S (~$\pu{175 kD}$ proteins. I may give you your answer but I strongly suggest to read about the subject for better understanding. I do not support online courses, but there are no choice you can made during a bad pandemic era.


Tricky question, but it probably has a trivially simple answer if you have experience with this problem, such as the 'look' of the mixture in the tanker. Not quantitative of course but possibly accurate enough, thus it could be worth asking others with experience for their opinion.

Assuming that water is intimate with the organic, i.e. dissolved or as an emulsion, some water needs to be separated out then there is the Distillation method which uses an immiscible organic with a boiling point higher than water, say xylene. The water is quantitatively carried over with solvent and can then be analysed.

Conductivity would be one way but not conclusive, as both industrial water and sea water may conduct or industrial water may not conduct and so be similar to distilled water. However, the distilled water may also conduct as its not de-ionised water. A simple test unlikely to be definitive definitive, but if its sea water, then a simple flame test should show sodium, as may industrial water, but distilled water far less so. Industrial water may contain soluble organics as well as various inorganic ions so may be coloured, acidic or alkaline and have a distinct smell. The sea water and distilled water have known pH.

Ultimately you will want to do a proper analysis, say GCMS.
Finally, after separation, water could still be present in small amounts in the organic then there is the Turbidity method or the Karl-Fisher titration to determine the amount of water left in the organic layer. However, I suspect you are asked to to a test at the back of the tanker or at least with very simple apparatus.

Mitigation is decided by company policy and this probably depends on what the cost is to correct the problem.

I hope that someone with real experience of this can help, that is when they have stopped laughing at my amateurish ideas.


How does the choice of metal (oxide) catalyst affect the range of unsaturated compounds that can be hydrogenated?

The currently accepted hydrogenation mechanism for metal like palladium is the Horiuti-Polanyi mechanism. (image)  This mechanism works with good yield for alkenes and alkynes.  For an aromatic compound, the activation energy is much higher because the aromaticity of the target molecule is being broken by the first hydrogenation.  Indeed, 1,3-cyclohexadiene is very unstable and will revert to benzene and hydrogen instead of continuing to cyclohexane.

Addition of a Lewis Acid makes palladium an acceptable catalyst for the hydrogenation of benzene, however.  An intermediate compound with the benzene and the LA is first formed, and hydrogen is added to this intermediate.  This paper discusses the mechanism of hydrogenation in the presence of a Lewis Acid.

Rhodium catalysts work slightly differently.  They generally have ligands attached that help stabilize the intermediates.  You can see a simplified mechanism here and there are papers discussing specific catalysts here and here.  Basically, in the case of benzene, the relatively unstable 1,2-cyclohexadiene is held in place on the catalyst until it has been completely hydrogenated.


DIY Brass Etching methods (safer etchants, waste, and disposal of)

An easy step-up from vinegar would be hydrochloric acid, $\ce{HCl}$. It's known as muriatic acid and is sold in bulk at pool-supply and home-improvement stores. The concentration from such sources is typically ~8.9M, or 32% (by mass). Such is what I measured from home; it's typically advertised as 30%-32%. There are rumors (which I haven't been able to confirm but suspect are true) that $\ce{HCl}$ from these sources often has iron contamination, but that is generally a minor concern for etching processes.

$\ce{HCl}$ has been widely used for etching aluminum hardware and paint-thinning in construction etc. It is highly reactive with (unoxidized) aluminum but can still etch other metals with more effort / more concentrated solution. You probably already know, but it's best practice to clean and polish a surface before etching, to remove impurities like oil and dirt, but also to expose the pure metal by removing protective oxides on the surface. The rougher the better (you can always polish-up later).

Brass is a simple alloy of $\ce{Cu}$ and $\ce{Zn}$. The reaction enthalpy of $\ce{Al + HCl}$ is -1,062.74 kJ, whereas that with $\ce{Zn}$ is only -153.90 kJ. Copper is hardly reactive with $\ce{HCl}$ alone, so it follows that $\ce{HCl}$ is a much stronger etchant for aluminum than for brass, although it is still possible to use it. There are many types of brass -- generally, the more $\ce{Zn}$, the better $\ce{HCl}$ will etch the material:

It is fine to dispose of $\ce{HCl}$ down the drain etc., but only do so after diluting it (carefully, slowly and substantially with water).

You can boost the performance of $\ce{HCl}$ as an enchant by introducing hydrogen peroxide, $\ce{H2O2}$. This combination has very strong oxidative effect and can dissolve metals like copper with ease. If you want to be environmentally friendly, this would be a good stopping point. The most powerful etchant will result from higher concentration of $\ce{H2O2}$. It's typically sold at 3% in stores but food-grade 30% $\ce{H2O2}$ can be bought online or made at home via this method.

I have had great success etching copper (and nickel) using a home-made solution of copper II chloride $\ce{CuCl2}$. Copper (I/II) chloride is very hazardous to aquatic life and (dissolved in solution or not) human life, but with careful handling it's a fine thing to use. It is quite easy to produce and handle, and does not corrode other (more precious) metals, like gold, so you can get selective etching if you have a certain alloy or something like RAM cards from PC parts. (I have isolated gold metal from these using $\ce{CuCl2}$ solution). To make it, simply add copper metal to a solution of concentrated $\ce{HCl}$ and very slowly add concentrated $\ce{H2O2}$. The copper metal will dissolve/oxidize into $\ce{Cu^{+2}}$, and if you dry the resulting blue-green solution (you don't have to) completely, you'll be left with a brown $\ce{CuCl2}$ powder. It's hygroscopic so if you leave it out it will form a blue hydrate. So I keep mine in a tightly closed plastic bottle. To use it as an etchant, just take some of the solid and dissolve it in water (again). The chief downside, as you mentioned, is the toxicity of copper chloride -- I have never thrown mine away. You can re-use it many times.

If I ever had to dispose of it, to do so properly I would use electrolysis. Attaching a DC power supply with the (-) terminal on a piece of copper metal (in dissolved $\ce{CuCl2}$ solution) and the (+) terminal in the solution (or attached to an iron nail) would do the trick, though you'd have to wait some time for the copper in solution to reduce -- and to push equilibrium you'd need a moderate/high voltage. Once you've reduced most or all of the aqueous copper to solid metal, the solution leftover is relatively harmless.


Why is the melting point of lactic acid higher than pyruvic acid

Your analysis is practically correct but not quite — more due to technical reasons. You said:

  Is this because lactic acid has 2 $\ce{OH}$ groups, and pyruvate only has 1 […]?

You probably know the structures of both; they both include a $\ce{COOH}$ group which you apparantly counted as a hydroxy group. You should not do that; the carboxy group ($\ce{COOH}$) has different properties than a lone hydroxy group ($\ce{CR2OH}; \ce{R} \ne \ce{OX}$). The most obvious is the $\mathrm{p}K_\mathrm{a}$ value which differs by about $10$ logarithmic units.

So you should say that lactic acid has higher melting and boiling points becuase it contains a (or: one) hydroxy group while pyruvic acid has a carbonyl group in the same position.


Why is the sneaky method of smothering a grease fire more effective?

By smothering the fire your starving it of oxygen. Without oxygen it can't burn.

If you cover the fire and take the cover off quick it reignites because it quickly interacts with the oxygen again because the heat of the oil is still around the same temperature as when it caught on fire.

If you slowly slide of the lid your increasing the time you've starved the fire of oxygen and also allowed the grease to cool slightly so when the oxygen comes back in contact with the grease at a slow rate the grease won't reignite.

The best way to deal with a fire is to prevent it from happening in the first place , make sure you know the boiling point and smoke point of what your using and use a thermometer .


The problem with this question is that the exact answer is "it depends..."

First off, it depends on your definition of acidity and how you measure it. Everyone seems to be using Brønstead acids (i.e. $\ce{H+}$ donors). I see two different measures in other answers:

Proton affinity: This is a gas-phase measurement of $\ce{A^{−} + H^{+} -> HA}$
Hammett acidity ($H_0$): This is a solution measurement, given by $\mathrm pK_{\ce{BH^+}} - \log\left(\frac{[\ce{BH^+}]}{[\ce{B}]}\right)$

Secondly, it depends on medium as mentioned by LDC3.

Leveling effect: The solvent leveling effect reflects the lowest possible $\mathrm pK_\mathrm a$ in a particular solvent, based on the basicity of the conjugate base. So you need $\ce{HF}$ or fluorosulfuric acid to reach low $\mathrm pK_\mathrm a$.

So the problem in my mind is that carboranes and fluoroantimonic acid are solution measurements, but $\ce{HeH+}$ is a gas-phase measurement. It does have the highest gas-phase proton affinity. But I'd put my money on things I can use in lab.

Incidentally, the Reed group prepared the fluorinated carborane acid this year: Angew. Chem. Int. Ed. 2014, 53 (4), 1131–1134. So that compound, $\ce{H(CHB11F11)}$ wins the crown for strongest solution-phase Brønstead acid (for the moment, at least).

There's also a nice review article "Myths about the Proton. The Nature of H+ in Condensed Media" by the same group in Acc. Chem. Res. 2013, 46 (11), 2567–2575.


The two mentioned systematic names correspond to different types of nomenclature. The name “1-phenylethan-1-one” is formed according to substitutive nomenclature, and the name “methyl phenyl ketone” is formed according to functional class nomeclature. Both systematic names are in accordance with current IUPAC recommendations. However, the preferred IUPAC name (PIN) for such acyclic ketones is generated using substitutive nomenclature. Note that the omission of the locant “1” for the suffix “one” in “1-phenylethanone”, while permissible in general usage, is not allowed in PINs, thus the name “1-phenylethan-1-one” is the PIN. Furthermore, the name “acetophenone” is retained, but only for general nomenclature.

The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-64.2.2.1 Acyclic ketones

  Unsubstituted acyclic ketones are systematically named in two ways:

  (1) substitutively, using the suffix ‘one’ and the prefix ‘oxo’; the presence of several ‘one’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc.; the final letter ‘a’ of a numerical multiplying prefix is elided before the suffix ‘-one’, for example, ‘tetrone’;

  (2) by functional class nomenclature using the class names ‘ketone’, ‘diketone’ etc.; substituent groups are placed, as separate words, in alphanumerical order before the class name.

  Method (1) generates preferred IUPAC names.

Note that the rule for functional class nomeclature (method 2) stipulates that the groups shall be written as separate words.


Lab safety - Long vs. Short Sleeves on Lab Coats

The short sleeve has no place in a chemistry lab.

Considering the coats purpose as personal protective equipment (PPE) is to reduce or eliminate the potential or risk of exposure to materials then the long sleeve coat would be the correct PPE to wear in a chemistry laboratory along with extra PPE which should have been assessed prior to performing the procedure.

Even in a microbiology or molecular biology lab, where a lot of the solutions might not seem harmful, the cellular material in suspension is classed as a bio hazard, which the long sleeve will provide better protection.

Hazardous chemicals are still commonly used in these labs, too, and will most likely be present whether you using them or not. It's not just for chemical exposure either also for general protection.


How to predict SN2 reaction when comparing nucleophilicty and leaving group ability

This is very famous "Finkelstein Reaction" that involved the exchange of one halogen to another through SN2 mechanism. The driving force for the reaction is the solubility of salts (NaI, NaBr, KF) under equilibrium.
There are different factor that effect the equilibrium position of a reaction, firstly, nucleophilicity of anion, presence of  good leaving group, and stabilization of anion in a given solvent.

first part of your question, I don't think with the absence of sodium cation the reaction would take place, because the driving force is the precipitation of salts of NaBr and NaCl in acetone.

In the second part of your question, you are right iodide is better leaving group compared to Bromide because iodide is weaker base compared to bromoide. Apparently, iodide can't replace bromine but here the reaction is driven toward products by "mass action" due to the precipitation of the insoluble salt as mentioned above.


Why is time-dependent density-functional theory (TD-DFT) used to describe excited states?

TD-DFT is used and developed because there is an excited state theorem, the Runge-Gross theorem, that is analogous to the Hohenberg-Kohn theorem for ground states.

(...) the Runge-Gross theorem shows that for a many-body system evolving from a given initial wavefunction, there exists a one-to-one mapping between the potential (or potentials) in which the system evolves and the density (or densities) of the system.
The Runge–Gross theorem provides the formal foundation of time-dependent density functional theory. It shows that the density can be used as the fundamental variable in describing quantum many-body systems in place of the wavefunction, and that all properties of the system are functionals of the density.

This means that any state of the system can be described by an electron density, not just the ground state. If I understand this correctly, the R-G theorem is more general than the H-K theorem, which can be derived from the R-G theorem by removing the explicit time dependence.
Here is a paywalled link to their original paper.
An interesting comment from the Wikipedia proof is

The proof relies heavily on the assumption that the external potential can be expanded in a Taylor series about the initial time. The proof also assumes that the density vanishes at infinity, making it valid only for finite systems.

In the case of applied fields, such as a slowly varying or oscillating electric field, these may be treated with perturbation theory rather than self-consistently; this is the same thing as the Taylor expansion. A consequence is that TD-DFT may not be valid in the presence of strong applied fields or a strong external potential.


As you say, it appears to be a kind of paint? So perhaps if you get a razor or something sharp you could try scraping and it will probably come off eventually, that would be one physical way. On the other hand if you want to go crazy try putting some paint thinner on there and see what happens, but don't blame me if you ruin your shirt experimenting on it, I have never used paint thinner before but I am guessing it consists of a non-polar compound and dissolves similar compounds (paint) so that it might be washed off with water. This is all hypothesis of mine, I have no idea what will happen except for what a razor can do.

Edit: I believe that Mr. Bean may have already demonstrated this experiment for us.


Why do the first and second laws of thermodynamics not contradict each other?

It is something of a historical accident that entropy has units of J/K.  It came out of the fact that the connection between heat, temperature, and energy was not obvious to early scientists, and so they effectively picked different units for measuring temperature and for measuring energy.

In the more modern statistical interpretation of entropy, the entropy of a system is simply a number.  Specifically, if the number of microstates associated with a given macrostate is $\Omega$, then $S = k \ln \Omega$.  The number of microstates ($\Omega$) is just a number, without any units, and therefore so is $\ln \Omega$.  You can see that we actually have to insert Boltzmann's constant, with its units of J/K, to make the units "come out right".

Arguably, a more natural way to define entropy would be to just make it it a dimensionless quantity:  $S = \ln \Omega$, without the factor of $k$.1  This would be equivalent to measuring temperature in units where $k$ is equal to 1 exactly, rather than defining our unit of temperature such that $k= 1.380649 \times 10^{-23}$ J/K exactly.
If we did this, we'd be effectively measuring temperature in units of energy as well;  for example, in an ideal monatomic gas with a "temperature of 1 J", the average KE of each molecule would be $\frac{3}{2}$ J.  Quantities such as Helmholtz free energy would still have units of energy, since we'd still define $F = U - TS$, with $T$ having units of energy and $S$ being dimensionless.

Of course, in this parallel universe where entropy is defined as a dimensionless number, another gen. chem. student would be asking why temperature is not the same thing as energy, even though they're measured in the same units.  But that's another question and another answer.

1 In fact, entropy is defined exactly this way in information theory, since there's not really a notion of energy (or temperature) to speak of in such contexts.


There may be better answers forthcoming, but a chemical does not need to melt (i.e. be in the liquid phase) to be reactive. For example, your food very likely has some liquid component, and that could be enough to react with some solid. Now polyethylene is used all over the place and is very nonreactive in many non-rigorous situations (like food service), so you are probably just fine. But I wanted to make the more general point that you don't have to be a liquid or gas to be reactive.


Why do we need to add hydrogen to amorphous silicon?

Well, one either adds it on purpose: In order to reduce the defect volume - by passivating the dangling bonds. Unsaturated silicon atoms in a-Si is a serial killer of charge carriers. But then the issue is the SWE - you have to handle it somehow. Options are few but research is being done. The common way of handling it right now is to be honest about it. This pv cell will lose some efficiency the first months, but then stabilize. Not a good selling point, but that is where it is at.

Or one adds it accidentally / coincidentally by using a deposition technique or raw material that contains hydrogen. ($\ce{SiH4}$ or $\ce{SiH_xCl_{4-x}}$). But you can usually thermocycle these problems away, as long as your substrate can handle it. If you did it, accidentally, on a plastic surface - it is usually referred to as "scrap" and recycled...


Is it safe to store muriatic acid, hydrogen peroxide and acetone?

You don't need to take any special precautions when storing these chemicals together compared to how you would store them individually. They should all preferably be stored in their original, properly labeled containers with properly fitting lids.

Regarding your concern for hydrogen gas forming from the HCl fumes, that is simply not a reaction that will take place from only the compounds you've listed, or in combination with anything else you would typically find in a utility or storage shed. You will want to avoid breathing HCl fumes while working with it of course, but as far as a storage concern even this will not be an issue for you.

I commend your extreme caution here in general, although your fears of the acetone igniting from sitting in a hot shed are unwarranted. The flash point of a compound is the lowest temperature at which it can be ignited with an ignition source, like a spark. This is commonly confused with the autoignition temperature, which is the lowest temperature at which a compound can spontaneously combust without an ignition source. So, while you really don't want your acetone near a flame or spark source at any temperature, it's autoignition temperature is $\pu{465^oC}$ and so you don't have to worry about it igniting just from sitting in a hot shed.


dependence of Intensity on penetration of light

In general, when light penetrates a material (water for example) its intensity decreases exponentially with increasing depth. This process is known as attenuation or absorption. The severity of attenuation differs with the wavelength of electromagnetic radiation. In the region of visible light, the red part of the spectrum attenuates more rapidly than the shorter-wavelength blue part.

But what you are asking about is a non-linear optical property known under the name: Saturable absorption It is a property of materials where the absorption of light decreases with increasing light intensity. At sufficiently high incident light intensity, atoms in the ground state of a saturable absorber material become excited into an upper energy state at such a rate that there is insufficient time for them to decay back to the ground state before the ground state becomes depleted, and the absorption subsequently saturates. The material becomes transparent at that wavelength. Saturable absorbers are useful in laser cavities.

Another kind  of non-linear materials is known under the name "optical power limiting materials". In these materials the absorption of light increases with intensity such that beyond a certain input intensity the output intensity approaches a constant value. Such a material can be used to limit the amount of optical power entering a system. This can be used to protect expensive or sensitive equipment such as sensors, can be used in protective goggles, or can be used to control noise in laser beams.

For more details, please see: http://en.wikipedia.org/wiki/Saturable_absorption
and http://en.wikipedia.org/wiki/Two-photon_absorption


How do I know if concentration of Ammonia affected my absorbency measurements?

The concentration of ammonia will matter, but only up to a certain point.

When you dissolve copper in nitric acid, you form copper nitrate, which is very soluble, but as you say, not very absorptive.

$$\ce{Cu(s) +2HNO3(aq) -> Cu(NO3)2(aq)+H2(g)}$$

One of the problems with copper nitrate is that it dissociates readily in water.

$$\ce{Cu(NO3)2(aq) -> Cu^{2+}(aq) + 2NO3^{-}(aq)}$$

The copper(II) ions then coordinate with varying amounts of water molecules. Each balue of $\ce{x}$ in the equation below is a slightly different aqua complex with slightly different optical properties.

$$\ce{Cu^{2+}(aq) +xH2O-> Cu(H2O)_{x}^{2+}(aq)}$$

The addition of ammonia changes the situation. Ammonia binds strongly to copper to preferentially form the tetraamminecopper(II) complex, technically $\ce{Cu(NH3)4(H2O)2^{2+}}$, which is highly absorptive. Since this formation is highly favored, you only need to add enough ammonia to provide at least 4 stoichiometric equivalents.

If the ammonia solution you added was labeled "Ammonium Hydroxide" (as you say in your question), then you know how much ammonia you added. Your instructor just might not have told you how to do it.

Ammonium hydroxide exists only by the ionization of ammonia in water (which is very small except at high concentrations). Labeling a solution as "Ammonium Hydroxide" has come to mean a "saturated solution of ammonia in water", which is 35% ammonia by mass (approx. 18 molar). Now you can calculate the amount of ammonia that you added (as long as you know the volume of solution that you added).

To determine the effect of amount of ammonia, you can do the following:

  Dilute the original ammonium hydroxide solution. For example, dilute 25 mL of the solution to 50 mL with water to get an ammonia solution of about 9 molar. Then repeat the penny experiment and add exactly the same volume of the new ammonia solution. This guarantees that the overall copper concentration remains constant even when the ammonia concentration changes. Then measure the absorbance. Then dilute the ammonia solution again (either dilute the new solution or dilute the original solution by a different amount). Rinse and repeat.


stability of a negative charge in halogens and that of orbitals

When we talk about hybrid orbitals and stability of charge on them, the electronegativity factor plays a more prominent role. The sp hybridised orbital is more electronegative than the others, owing to Bent’s rule which states - Greater the s character of a hybrid orbital, greater is its electronegativity.
Therefore, species having the negative charge on an sp hybridised atom tend to be less basic.
(You might be interested in knowing that, an sp hybridised carbon atom is actually more electronegative than a nitrogen atom.)


Benzaldehyde and Cyanide - if you can smell one can you smell the other?

Answer is, they are most probably not correlated. Smell requires molecules to bind to receptors, which they do by structure and functional group. CN is a very small linear molecule that does not even share its heteroatoms with benzaldehyde, an aromatic compound several times cyanide's size. Ergo, it's improbable they share the same scent receptors.


What is the product of the titration of potassium permanganate and iron(II) sulfate?

From your description I'd say you were titrating ferrous sulphate, $\ce{FeSO4}$ solution (the analyte), with potassium permanganate, $\ce{KMnO4}$ solution (the titrant), in acid conditions (dilute $\ce{H2SO4}$ present).

The permanganate ion $\ce{MnO4^{-}}$ is a strong oxidiser and oxidises the ferrous ion $\ce{Fe^{2+}}$ to the ferric ion $\ce{Fe^{3+}}$ very easily. In dilute solution both ferrous and ferric ions are more or less colourless.

In this process the very strongly coloured permanganate is reduced to the manganous $\ce{Mn^{2+}}$ ion, which is also colourless in dilute solution.

When all the ferrous ions have been oxidised, one more drop of the strongly coloured permanganate solution then turns the analyte solution pink and this is your end point.

  I recently did an experiment at school, where I had to titrate $\ce{KMnO4}$
  with $\ce{FeSO4}$.

In that case you'd have started from a purple solution (analyte) which would gradually, during the titration, have turned colourless. This method is also used.

As this is a redox reaction, I suggest you write out (or look up) the reduction (permanganate to manganous) half reaction and the oxidation (ferrous to ferric) half reaction, to complete your knowledge of this titration.


In terms of Bohr model ionization potential $E_\mathrm{i}$ is the work $A_\mathrm{i}$ on eliminating an electron in vacuum from its current non-excited orbital level to infinity:

$$E_\mathrm{i} = \frac{A_\mathrm{i}}{e}$$
$$A_\mathrm{i} = h\nu = \frac{hc}{\lambda}$$

Unknown wavelength $\lambda$ can be determined from the Rydberg formula:

$$\frac{1}{\lambda} = R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

so that final equation for ionization energy looks like this:

$$E_\mathrm{i} = \frac{hc}{e}R_\infty Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Since we are determining 4th ionization energy of beryllium ($Z = 4$), $n_1 = 1$ and $n_2 = \infty$:

$$E_\mathrm{i}^\mathrm{IV} = \frac{\pu{6.63e-34 m^2 kg s-1}\cdot\pu{3e8 m s-1}}{\pu{1.602e-19 C}}\cdot\pu{10973732 m-1}\cdot 4^2\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = \pu{217.86 eV}$$

This value is in a good agreement with the one listed in CRC Handbook of Chemistry and Physics [1, p. 10-204]: $E_\mathrm{i}^\mathrm{IV}(\ce{Be}) = \pu{217.71865 eV}$.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; CRC Press, 2016. ISBN 978-1-4987-5429-3.


Hill function and stoichiometric matrix: confusion about reaction rates

If the reaction is $\ce{aA + bB = pP + qQ} $ then we define the rates as

$$-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{p}\frac{d[P]}{dt}=+\frac{1}{q}\frac{d[Q]}{dt}$$

As to the Hill function it seems that the complex nSX is just one species.


Why can DAP be used to elute a protein in affinity chromatography?

In this report, diaminopropane is used to elute a lectin from an affinity column. They mention the pH, which is the key to the question.

  The  bound  lectin was  eluted  with  20  mM diaminopropane  containing  0.15M NaCl,  pH ~11.

The diaminopropane is unbuffered, and the high pH weakens the interactions between lectin and carbohydrates.


Is the Gibbs standard free energy always constant?

In short, no, the standard Gibbs free energy change is not constant; it is a function of temperature. The same is true for practically all other standard-state quantities.

This gets a little confusing because of how standard-state properties are often explained in lower-level (high-school and college-freshman level) textbooks. The standard Gibbs free energy change, $\Delta G^\circ$, is the value of $\Delta G$ at a certain constant pressure (usually 100 kPa or 1 atm) and constant concentration of solutions (usually 1 molal or 1 molar) and a specified temperature (usually $0~^\circ \text{C}$ or $25~^\circ \text{C}$). What is intended but not immediately obvious is that the pressure and concentration are constants but temperature is a variable; since the temperature is a variable, $\Delta G^\circ$ is a function of temperature. When the value of $\Delta G^\circ$ is given, the temperature must also be specified.

In your equation, then, $\Delta G^\circ$ is the standard Gibbs free energy change of the reaction at the specified temperature ($T$ in the equation). Change $T$, and the value of $\Delta G^\circ$ also changes; but it is still the standard Gibbs free energy change at the (new) specified temperature. (This equation cannot be used to calculate $\Delta G^\circ$ at a given temperature because  $K_\text{eq}$ is also a function of temperature.)

The IUPAC "Green Book" (ed. 3) has a good explanation of what is meant by the term "standard state"; the discussion starts on p. 61 (as printed on the page; it's p. 76 in the pdf file). It explicitly expresses the standard chemical potential ($\mu^\circ$), which is very closely related to $\Delta G^\circ$, as a function of temperature.


Does all fire only emit light on its outermost shell?

The answer is NO.
The article you quote makes a completely unwarranted generalisation that "all flames are hollow". This is true of some flames but only because the fuel that is burning is only able to burn when mixed with oxygen from the air. In those flames, the flame is "hollow" because only in regions where air can mix with the fuel can the fuel burn and the rate of mixing is dependent on the flow of fuel and the rate at which air can mix or diffuse into the fuel rich region. Bunsen burner flames are like this when the air is not premixed with the fuel gas (because the air intake on the bae is closed) and candle flames are like this because the wax has to evaporate to mix with air.
A bunsen flame when the air intake is open is not like that as the fuel gas and air are pre-mixed and can burn throughout most of the region of the flame.
Another factor worth considering is what causes the emission of light. In candles and many other poorly controlled flames, there is a lot of light from incandescence not from the underlying burning reaction. Incandescence involves black-body emissions from hot solid particles. Those particles are usually small particles of solid carbon produced from partial, incomplete burning of hydrocarbons. They emit light because they are hot. Candles are tuned to create a lot of these particles as this gives far more light than complete combustion. In contrast, a well-controlled bunsen flame emits very little light (the light is not black-body emission but comes from specific excited states in the reactants).
As to how this applies to your specific examples here are some analyses.
1. Campfires
Campfires are a mixture of various phenomena. They have flames  a little like candle flames. But they also have very hot solids which emit a lot of black-body radiation because they are hot. Some of the emitted gasses have little emission of light as they are like a bunsen flame, but the overall effect is dominated by the hot solids. And, of course, some camp fires have far better mixing of air and fuel than others so it is hard to generalise.
2. Rocket exhaust
Most rockets need to be as efficient as possible. So they tend to have very carefully designed mixing of fuel and oxidiser to achieve maximal energy output. They are not hollow and tend not to emit huge amounts of light (compared to the equivalent candle flame anyway).
3. Oil platform explosion
Impossible to generalise. But unintended explosions don't tend to be well controlled so will have some areas where mixing is good and others where it is not so some areas are like a candle and others like a bunsen. Incidentally, movie explosions emit a lot more light than real explosions because, well, hollywood likes visual spectacle not realism (see this youtube explanation and illustration of how movies deliberately add fuel to create poorly mixed candle like flames for more visually interesting explosions).
4. The sun
The sun does not emit light because of burning (at least not in the normal chemical sense). The sun emits light because the surface is very hot. And this happens because of nuclear fusion deep in its core (not what a chemist would call burning, though a nuclear physicist might, but what do they know?). The light we see is mostly black body emissions because the surface is hot. It is certainly not hollow either.


Can carbon be the central atom of a complex? Can it form a quadruple bond with itself?

Asking multiple unrelated questions at once is considered a bad practice, but let's forget that for a moment.

Your first molecule has more or less satisfactory electronic configuration, but that's not enough for the long-term survival. N would be more than willing to relinquish a proton, and C to accept it, and thus in no time you will have $\ce{H2N-CH2-NH2}$, which then may or may not transform into something else.

Your second molecule has been discussed here before on multiple occasions. Long story short, it does exist, but doesn't have a quadruple bond. (Such bonds are known, but not in $\ce{C2}$; they have a delta component, and that requires d orbitals). The problem with your description is that the sigma-oriented 2p orbital of one carbon will interact not just with the similar 2p, but also with the 2s orbital of the other. If not for this fact, it would work out pretty much as you expected.


Does the reaction of sulfur and potassium nitrate involve production of sulfur trioxide?

Generally a mixture of sulfur, potassium nitrate with charcoal or an organic carbon compound is used in gunpowder mixture.

$$\ce{3C + S + 2KNO3 → 3CO2 + N2 + K2S}$$

(Source)

There are many questions in chem.SE related to gunpowder and explosives which are all based on this reaction. You can find them of help.

What is the oxidation mechanism of gunpowder?
What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder?
Why can sulfur act like a catalyst?
Write the chemical equation of this reaction:


Recall that, for an ideal monatomic gas, $$U = \frac{3}{2}nRT.$$ (Diatomic and polyatomic gases have a larger coefficient.) For an isothermal expansion, reversible or otherwise, we therefore have $$\Delta H = \Delta(U+PV) = \Delta(U+nRT) = \frac{5}{2}nR\Delta T = 0,$$ since $\Delta T = 0$.


Does increasing electronegativity imply more covalent character?

The main force responsible for holding together covalent bonds  and ionic bonds is different. Ionic bonds are held together through electrostatics. A transfer of electrons causes one atom to be positively charged and another one to be negatively charged. In other words, electrons are given away or taken away by atoms. Covalent bonds are held together due to the delocalization of electrons, i.e. electrons are shared by the atoms. For more detail look at What makes the difference between ionic and covalent bonds? (physics.se)

There is no meaning to the statement that an element is covalent. The reason is that the nature of a bond (metallic, covalent, ionic) is determined by the elements participating in the bond. Sodium may form a metallic bond with itself, but it will form an ionic bond with halogens. Halogens will form covalent bonds with themselves but will form ionic bonds with alkali metals. Furthermore, the conditions can determine the nature of a bond. For example, hydrogen is metallic under high temperature and pressure.
Just a side note, compounds can be ionic. I recommend that you read this thread explaining the difference between compound, element, mixture, etc: What is the definition of of 'compound', 'mixture', 'element' and 'molecule'?

Tldr: Wikipedia is not being exact. Saying that an element is covalent is inaccurate because the terms covalent and ionic refer to bonds not elements.


Determine the order of the reaction, and the value of the rate constant. Predict the concentration of AB at 25s

If n is the order of this reaction, then the rate equation is given by
$$-\frac{d[AB]}{dt}=k[AB]^n$$
If we take the log of both sides of this equation, we obtain:
$$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$
This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$.  So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk.  So we could determine both the reaction order and the rate constant all in one shot from this single graph.

But how would we reduce the given data to plot such a graph?  Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval.  If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as:
$$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$
The average concentration over this same time interval would be:
$$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$
So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table.  The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.


why does milk of magnesia have an expiration date?

Many countries require anything marketed as edible to have an expiry date. You can speculate about what might spoil your Magnesia (see comments above), but likely it will be good four or fourty times as long as indicated.

Of course the lid might not be perfectly tight, and it could dry up or get contaminated by bacteria, but this can happen anytime. And certainly after some time it will separate into clear water and a thick suspension of magnesia particles and you'll have to shake or stir it very thoroughly before consummation.

Anyway these expiry dates are 99% baloney. Very few things really get bad before you see the mould grow on them.

Fun fact and (german) exemption to that rule: After six months at the latest, the taste of beer (the non-pasteurised stuff) starts to deteriorate. Yet, according to the date on the bottle, you have 12 more weeks to drink the remainder of that crate with increasing disgust before it "expires".


Affinity to mobile or stationary phase during column chromatography

I don’t know in which universe you would consider a dinitrobenzene less polar than an ethyl ester but this universe is not the correct one.

As your experimental data (the $R_\mathrm f$ values of the two eluted compounds) clearly show, the dinitrobenzene is more polar than ethyl benzoate.


Why is PLGA so expensive in comparison to PLA?

Poly(lactic-co-glycolic acid) or PLGA is a biodegradable copolymer of lactic acid ($\ce{CH3CH(OH)CO2H}$) and glycolic acid ($\ce{HOCH2CO2H}$). Poly(lactic acid) or PLA is, on the other hand, a polymer of lactic acid, which is also biodegradable. Since both polymers are linked together by ester linkages of their monomers, the resulting polymers can be categorized as linear, aliphatic polyesters:

PLGA is one of the most successfully biodegradable polymers, and hence it attracts considerable attention in the development of drug delivery systems due to its desirable properties and to its Food and Drug Administration (FDA) and European Medicine Agency (EMA) approval for parenteral administration. PLGA encapsulated drug delivery enjoys drug protection from degradation. In addition, PLGA also provides prolonged controlled release of encapsulated drugs. Therefore, as you have suspected, these medical uses (current and developing applications) needed GMP manufacturing of the product for the purpose of the use in Humans. This GMP manufacturing of drugs is always costly. The price of the compound also depend on the other factors such as the size of the polymer (molecular weight), percent enantiomeric ratio (Lactic acid has two enanthiomers), end groups, etc., which are critical in human use. Also remember, synthesis of copolymer with desired ratio is not a easy task.
PLA has become a popular material due to it being economically produced from renewable resources. In 2010, PLA had the second highest consumption volume of any bioplastic of the world. For example, PLA is the most widely used plastic filament material in 3D printing. This commercial success of PLA makes its bulk production (no need of GMP manufacturing processes). Thus, it is much cheaper than demanding PLGA. Yet, PLA can be expensive as well, depending on its use, which require careful synthetic processing. For example, D- and L-lactide copolymer is used in bone engineering (Ref.1).
Sigma-Aldrich listed series of PLGA and PLA with some different properties:
$$
\begin{array}{c|ccc}
\hline
\text{EXPANSORB$^®$} & \text{PLGA or PLA} & \text{L/DL Ratio} & \text{End group} & \text{Molecular weight} & \text{Price}/\pu{10 g}\\
\hline
\text{DL 100-2A} & \text{PDLLA} & 0/100 & \ce{COOH} &  \pu{10-25 kD} & \pu{501.00 USD}\\
\text{DLG 50-2A} & \text{PLGA} & 50/50^a & \ce{COOH} &  \pu{5-20 kD} & \pu{446.00 USD} \\
\text{DLG 75-9E} & \text{PLGA} & 75/25^b & \text{Ester} &  \pu{100-150 kD} & \pu{458.00 USD} \\
\text{DLG 75-2A} & \text{PLGA} & 75/25^b & \ce{COOH} &  \pu{5-20 kD} & \pu{446.00 USD} \\
\hline
\end{array}\\
^a: \text{D/L-Lactic : Glycolic} = 50:50; \ ^b: \text{D/L-Lactic : Glycolic} = 75:25
$$
The list was posted to show when synthesized for biological usage, either polymer can be expensive. For example, the entry 1 of the above table is essentially PLA (made by 100% D/L-lactic acid), but it is the most expensive one on the list.
References:

C. X. F. Lam, R. Olkowski, W. Swieszkowski, K. C. Tan, I. Gibson, D. W. Hutmacher, "Mechanical and in vitro evaluations of composite PLDLLA/TCP scaffolds for bone engineering," Virtual and Physical Prototyping 2008, 2(4), 193-197 (https://doi.org/10.1080/17452750802551298).


How to calculate the electrical potential and mols of electrons transferred from combined half cell equations?

TL;DR

This basically boils down to the fact that $\mathrm{E_{cell}}$ is an intensive property, not an extensive property.

Long Version

Lets define some terms first:

Intensive Property:
A property that doesn't change when the size of the sample changes. For example, density and concentration are examples of intensive properties

Extensive Property:
A property that changes when the size of the sample changes. For example, mass and volume are examples of extensive properties

$\mathrm{E_{cell}}$ is the energy in joules of 1 coulomb of charge (1 volt = 1 joule per coulomb). Hence it is an intensive property. Therefore the number of coulombs of charge or number of electrons doesn't effect the value of $\mathrm{E_{cell}}$.

Another way of thinking of it is that by doubling the amount of moles, the  amount of coulombs and joules also double, hence $\mathrm{E_{cell}}$ remains the same.


Why does Pb normally have an oxidation number of +2?

Why can't lead gain four electrons? The lower you go on the periodic table, the lesser you will find elements that have a more non-metallic character (so with this I mean elements that will have a lower tendency to gain electrons). Just visualize a lead atom with a very small nucleus and a large area surrounding the nucleus, filled with electrons. Now, the more you go away from the nucleus, the weaker the positively charged nucleus can 'pull' on the outer , negatively charged, valence electrons. Elements that haven't got a very tight grip on these valence electrons, will lose their valence electrons. However, elements with a small atomic radius (such as Fluorine) will have such a tight grip on its valence electrons, that it will even be able to snatch electrons from other atoms.

Of course the electron configuration of Radon is stable, because it has fully filled orbitals and it has a noble gas configuration, but you need to keep the nuclear forces in mind. This you can also see by looking at the electronegative value of lead : it is 1.87 (Pauling scale). Now compare it to carbon, which has 2.5. The lower you go in the same group, the lower the electronegativity goes and the more metallic character you will get. Elements with more metallic properties (such as lead) will mostly form ionic bonds and elements with more non-metallic properties (such as carbon) will mostly form covalent bonds. There exists a rule to predict (however it isn't always correct) if a compound will be covalently/ionically bonded : if the difference in electronegative value is higher than 1.66, then you have an ionic bond and if it is lower than 1.66 then you have a covalent bond.

Why does lead have an oxidation number of 2+ while carbon and other elements in the same group have an oxidation number of 4+? Just to correct you on this : lead does have an oxidation state of 4+ and elements above such as carbon and tin also have 2+ as an oxidation state. Just look at the electron configuration : the outer $\ce{s^2 p^2}$ orbitals will be able to lose 2 or even 4 electrons, or even gain 4 electrons. In fact these are all possible oxidation states of the elements in the carbon-group:

   Carbon 4, 3, 2, 1, 0, −1, −2, −3, −4
   Silicon 4, 3, 2, 1 -1, -2, -3, -4
   Germanium 4, 3, 2, 1, 0, -1, -2, -3, -4
   Tin  4, 3, 2, 1, -4
   Lead   4, 3, 2, 1
   (source : wikipedia)

Please do note the tendency to have more positive oxidation states, the lower you go in the group. (the reason why? see the beginning of the answer) Note : not all of these oxidation states are (very) stable.

Also have a look at this picture:

Notice the metallic (shown by red) character increases the lower you go on the periodic table.


Why do you need excess LAH to reduce esters?

Lithium aluminium hydride or LAH is the more everyday form of $\ce{LiAlH4}$. When originally produced, most of it is indeed $\ce{LiAlH4}$ but that quickly partially hydrolyses giving the grey powder you may or may not know from the lab. As such, $\ce{LiAlH4}$ is practically never entered into reactions in pure form. Since it is impure, excess reagent is added to make sure the reduction job is performed as desired.

Also, remember to consider the reaction meachnism:

$$\ce{R-COOR' ->[LAH][- HOR'] R-CHO ->[LAH] R-CH2OH}\tag{1}$$

Note that the second step, the reduction of an aldehyde, is more rapid than the first as aldehydes are more reactive towards nucleophiles than esters. Thus, if you attempted to bargain, hoping to arrive at the aldehyde selectively, you would end up with under half of your reactant reduced to the alcohol and half of it still as an ester. Another reason to always add excess just to be sure.


When naming alcohols based on alkanes, when do you keep the "e" at the end of the alkane's name?

The names pentane-1,2-diol and pentan-2-ol are correct.
Systematic elision of the terminal letter ‘e’ or other vowels appears on several occasions in the IUPAC nomenclature.
A brief rule for simple alcohols is mentioned early in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-15.1.7.2.1 Alcohols are named by attaching the suffix ‘ol’ to the name of the parent hydride, with elision of the final letter ‘e’ in the parent hydride, if present. When alone in the structure, the characteristic group(s) must receive the lowest locant(s) possible, which is (are) cited immediately in front of the suffix (see P-14.3.2).

For example: pentane → pentan-2-ol
Extensive general rules for the elision of vowels can be found in Subsection P-16.7:

P-16.7 ELISION OF VOWELS
P-16.7.1 Vowels are systematically elided as follows:
(a) the terminal letter ‘e’ in names of parent hydrides or endings ‘ene’ and ‘yne’ when followed by a suffix or ‘en’ ending beginning with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’, or ‘y’;
(…)

Clearly, this rule also applies to the suffix ‘ol’ for alcohols.
A similar rule is mentioned later in Chapter P-6 concerning the applications of general principles to specific classes of compounds:

P-60.2 PRESENTATION OF NAMES.
(…)
(a) names formed substitutively using suffixes follow the general method described in P-15.1. Substitutive names are formed by adding a suffix such as ‘al’, ‘ol’, ‘yl’, ‘carbaldehyde’, ‘carboxylic acid’, etc., to the name of a parent hydride, with elision of the final letter ‘e’ of the parent hydride, if any, before ‘a’, ‘i’, ‘o’, ‘u’, and ‘y’;

Clarification concerning multiplying prefixes and numerals can be found in Section P-59 concerning name construction:

P-59.1.9 Characteristic groups
(…)
If, and only if, the complete suffix (that is, the suffix plus its multiplying prefixes, if any, as described in Appendix 1) begins with a vowel, a terminal letter ‘e’ (if any) of the parent hydride name is elided. For example, ethanol (not ethaneol). Elision or retention of the terminal letter ‘e’ is independent of the presence of numerals between it and the following letter, for example, propan-2-ol (not propane-2-ol).
(…)

Since the complete suffix (the suffix plus its multiplying prefixes) of pentane-1,2-diol is ‘diol’ and thus begins with ‘d’ and not with a vowel, the terminal letter ‘e’ of the parent hydride pentane name is not elided.


How to calculate molecular emission spectra?

As you mentioned, Gaussian can do it but is proprietary.

I believe ezSpectrum (https://github.com/iopenshell/ezSpectrum) might be able to do what you are looking at but I have not used it myself so far.

There used to be a program available called FCClasses which would run the Franck-Condon computations for you based on (I think) Gaussian output, but it seems like the author took it down, I can only find a front-end to it right now: https://github.com/jcerezochem/fcc_tools The FCClasses homepage is mentioned as http://www.pi.iccom.cnr.it/fcclasses on that page but, as mentioned above, no longer works.


Electrolysis of PbCl2 and use of Nernst Equation

I've finally read this question. I don't know what the source is, but I disagree with both the answer and the explanation.

Consider first the reduction half reaction for lead:

$$\ce{Pb^{2+} + 2e- -> Pb}, E^{\varnothing}=-0.13\ \mathrm{V}$$

Now, this is for the reduction at standard conditions, which means that the concentration of lead ion is $1\ \mathrm{M}$. If you decrease the concentration of the reactant ions (and you are, if you consider that the concentration of the dissolved lead chloride is less than 1), then the reaction should be less favorable. This means that the potential should be more negative, not less negative.

This is easy to check based on the form of the Nernst equation as the lead ion concentration enters into the denominator of $Q$. Since $\Delta Q<1$, $\Delta \left(\frac{RT}{nF}\ln Q\right) > 0$.


If volume is understood classically, i.e. as the quantity of three-dimensional space enclosed by a closed surface, then quantum systems in general do not have a certain volume. One could, in principle, redefine the notion of volume for a quantum system to be, say, an area chosen so that there is a certain probability (for example, 90%) of finding the system within it, but that would be a completely different quantity, of course. It should also be noted that in quantum theory statements about a system having or possessing certain values of physical quantities has a lot of issues, even when the quantities are well defined. For instance, does a quantum system have (possess) certain values for physical quantities before the corresponding measurements are made?

  On other hand In general Chemistry books often spin quantum no. is
  referred as electron spinning on its axis.

You already know that such picture is an absurd, and you are perfectly right here: electrons are not spinning. So there is no any contradiction there: they don't have a classical volume and they are not spinning.


Oxygen is relatively insoluble in water, its solubility being only 264 µM at 25 °C. That explains in part why you (and fish) require dedicated oxygen carriers in your blood to transfer sufficient oxygen around your body.
That oxygen is weakly soluble can be explained, as you point out, on the basis of the hydrophobic effect, which amounts to saying that water likes to hydrogen bond, and an oxygen molecule is not very good at that compared to another water molecule.
That oxygen is soluble at all can be explained on two grounds. First, statistics, just the random mixing of matter. Some oxygen molecules will happen to find their way into water and not find their way out for a while. The other explanation is that oxygen is not entirely nonpolar: it lacks a permanent dipole moment, but it is polarizable. Even molecules that we think of as highly water repellent can exist as liquids (oxygen is a gas at room temperature but can be cooled into a liquid state), which means something must be keeping them together, and that is dispersion forces due to the polarizability of all atoms and molecules.


Is there an algorithm that takes atom coordinates and produces VSEPR geometries?

Quantum chemistry software usually exploits molecular symmetries to speed up computations. To do so they need some way to determine the point group for a given set of atoms and their coordinates. So this is basically what you are asking about, just that get something like $D_{4h}$ instead of "square planar", or $T_d$ instead of "tetrahedral".

I am not sure whether this is actually used by the mentioned software, but it is quite simple and efficient if you only want to test for certain point groups:

Each symmetry operation can be represented as a $3\times 3$ matrix, for example mirroring in the $xy$ plane is

\begin{equation}\sigma_{xy} = \begin{pmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & -1
\end{pmatrix}\end{equation}

We can represent the atom coordinates in a $3\times N$ matrix, where $N$ is the number of atoms. The molecule needs to be properly oriented (the center of mass should be in the origin of the coordinate system and the principle axes of inertia should coincide with the $x$, $y$ and $z$ axes).
For example $\ce{H2}$ (or any other homonuclear diatomic):

\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}

To get the transformed coordinates according to a certain symmetry operation, we just apply the matrix of the symmetry operation to the coordinates matrix.

\begin{equation}\begin{pmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & -1
\end{pmatrix}\times
\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}=
\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}
\end{equation}

We can then compare the resulting set of coordinates with the original one. Since for $\ce{H2}$ both atoms are the same, we can interchange the 2 columns and see, that we have identical matrices. Thus, $\sigma_{xy}$ is a valid symmetry element of our input coordinates. At this point you may implement a tolerance threshold, by allowing the norm (distance) between two "same" points being larger than 0.
The set of found symmetry operations will then directly give us the point group of that molecule (this information is available in character tables). So we can check for a certain point group by checking whether the symmetry operations of that point group can be found in our input geometry. This however means we need to define which point groups we want to test for.

There are also flow charts (available in books about molecular symmetry or via your favorite search engine), telling you which symmetries elements to check for and then guide you to the point group.

I am not aware of a more general approach. If someone knows more, please extend.


how to destroy excess iodomethane in reaction mixture?

Yes, you can destroy small amounts of methyl iodide by hydrolysis with an aqueous solution of sodium hydroxide. You can just add some sodium hydroxide solution to your reaction mixture, if your molecule is stable to basic conditions.

The reaction is: $\ce{NaOH + MeI -> MeOH + NaI}$

From: "A Comprehensive Guide to the Hazardous Properties of Chemical Substances" by Pradyot Patnaik, 2007


Is tea powder better dissolved when put in hot water?

I would say that your method is actually quicker (if that's what you mean by 'better'). My reasoning is as follows: the solubility of the powder in water increases with increasing temperature, as does the dissolution rate but the dissolution rate remains finite all the time.

1) If you add the powder and start heating at $t=0$ the concentration of dissolved powder will start to increase from $c=0$ to $c>0$ immediately. Once the water reaches the boiling point at $t=t_b$ already a certain amount (maybe even all) of the powder has dissolved i.e. some concentration of dissolved powder is present $c=c_1$.

2) If you boil the water first and then add the powder you start with $c=0$ at $t=t_b$, which is a lower concentration than in the powder-first case.

Because there is already more powder dissolved at the boiling point ($c=c_1$ vs. $c=0$ at $t=t_b$), the moment at which all powder is dissolved comes sooner for the powder-first case (situation 1).

Note that I approach this question purely from the physical point of view, I have no clue whether the two methods might result in differently tasting tea.

P.S. I ignored any possible changes in the boiling point due to the already dissolved powder. For low concentrations, as in the case of tea, this is no problem. If the concentrations become much higher then the change in boiling point might become dominant and the story changes


What is the maximum amount of medicine that could accumulate in the body if the compound has a half-life of 24 hours?

1st day: 25 mg
2nd day: (12.5 + 25) mg
3rd day: (6.25 + 12.5 + 25) mg
4th day: (3.125 + 6.25 + 12.5 + 25) mg
The amount get closer and closer to 50 mg but never exceeds it.


Why can't an electrochemical cell be a single cell?

Yes, if you dip two different metals in an aqueous solution containing ions, you will have an electrochemical cell. This is how all lemon batteries work, which consist of zinc or copper inserted into lemons (approximately an aqueous citric acid solution). You can see a potential difference between zinc and copper. The key point is that one cannot determine Nernstian potential in such cases.
Half cell is just a mental construct as it cannot exist alone. If there is a reduction, something must be oxidized. Electrochemists usually separate the "half-cells" in order to conveniently calculate Nernst potential.

Imagine a hypothetical cell, which is partitioned into half cells with a plantinum sheet. Each compartment contains an electrode Cu and Zn, dipped in their respective salt solutions. What will happen in that case?


If you have seen pink salt crystals from Khewra mines in Pakistan (the so-called Himalayan pink salt) sold at exorbitant rates in grocery stores, you would find out it is nothing but NaCl with a small amount of iron oxides as stated above. If you dissolve large tablespoons of such salt crystals in water, you will see a brownish precipitate sitting at the bottom on the container.

The word muriate reminded me of ancient chemistry books of the 1940s. It is a very very old name for a salt of hydrochloric acid.


Mineral oil serially penetrates the wood to render its surface hydrophobic.  Nothing eats mineral oil and it does not easily oxidize (go rancid).  It is also the solvent vehicle to transport beeswax (plasticizer, compatibilizer, internal soft sealant) and the hard wax surface sealant. Nothing polymerizes (re linseed or tung oil) or chemically bonds to the wood.  All three are FDA GRAS.

Like dissolves like.  While the surface will be sealed against polar materials and provide no food for bacteria and fungi of itself (and absence of water is a good biostat), it can imbibe fats and oils.  Butcher block surfaces and such are best made of dense oily woods that are naturally sealed against water.  Top them off.  Lignin is a fair antibacterial and antifungal agent.  Polyethylene cutting boards have higher sustained levels of viable bugs than do wood cutting boards.

http://faculty.vetmed.ucdavis.edu/faculty/docliver/Research/cuttingboard.htm


Why does a gas' ability to absorb infrared radiation affect its potency as a greenhouse gas?

There is only one difference between a greenhouse gas and a non-greenhouse gas: a greenhouse gas has a greater ability to absorb radiation in the infrared (IR) region of the electromagnetic spectrum than does a non-greenhouse gas.

Once a greenhouse gas absorbs IR radiation, that energy must be converted to another type of energy such that the the energy is conserved. The only thing that can happen at IR energy levels is that the absorbed IR energy gets transformed into thermal energy, resulting in an increase in heat in the atmosphere.

From there it follows that the more strongly a particular gas absorbs IR radiation, the more heat is retained in the atmosphere, which might have otherwise just radiated off into space as IR radiation. This means a gas absorbing strongly in the IR is a more "potent" greenhouse gas than one that only absorbs weakly or not at all in the IR.


Should I include ions from the solvent in solubility product constant calculations?

Yes, absolutely, all nitrate counts in solubility product regardless of source. The system cannot distinguish what a nitrate ion (or any dissociated species) was bonded to before it dissociated.


If you put two blocks of an element together, why don't they bond?

Why, when you bring the two bars together so that they touch each other, do they not instantly bond with each other forming one larger bar or block? ... Why do we need to 'weld' two bars together - why don't they just bond on their own?

The problem is generally one of two things: gases (air) or metal oxides get in the way.

You can actually bond two pieces of metal this way: it's called cold welding. But, in order to get it to work with large pieces of metal, you have to (1) get all the air out of the way, and (2) you have to clean both surfaces very thoroughly in order to remove all traces of surface oxides, and (3) you have to make sure both surfaces are perfectly matched, either precisely flat or with precisely the same curvature.

Once you've taken care of those preconditions, my expectation is that all metals should cold weld. No guarantees as to how hard it might be to actually satisfy these conditions, though.

  We can weld elements together so they 'stick' to each other, but what is the process that actually causes two like elements to bond together?

In the most common welding methods, both of the surface to be welded are actually melted in the area right around the weld, which takes care of (1) and (3) above.  Requirement (2) is dealt with in a variety of ways, such as by surrounding the weld area with an inert gas blanket or by using a 'flux' material that either chemically reacts with any metal oxides to turn them back into metal, and/or that "floats" on top of the weld puddle to protect it from oxidation.

ADDENDUM: After reading DarioOO's answer, I realized that I should note that cold welding does not satisfy the "forming one larger bar or block" aspect of your question. An assembly of cold-welded parts is not attached as strongly as it would be if it were machined from a single piece.


Why doesn't CaO decompose into Ca and O2?

Stability, the thermodynamic property, is defined as whether a decomposition reaction is exergonic or endergonic at a certain temperature. The reaction $\ce{CaO -> Ca + 1/2 O2}$ is endergonic across a very large temperature range. But that doesn’t necessarily help you.

In that list of four, we have to clearly molecular compounds and two mostly ionic ones. We need different discussions for both. Basically, the molecules are unstable towards decomposition into $\ce{O2}$ and something ($\ce{H2O}$/$\ce{O2}$) because they contain energetic bonds whose replacement by an $\ce{O=O}$ bond is favourable.

For the ionic ones, we have a number of different discussions we can do. For one, the metallic partner is calcium versus mercury; while the latter is known since ancient times and pretty stable in liquid form the former is not: It will readily react to form salts. For another, the $\ce{Ca^2+}$ ion is rather small and thus a hard Lewis acid according to Pearson’s concept. $\ce{Hg^2+}$ is much larger and thus a much softer acid. Therefore, the adduct of $\ce{Ca^2+}$ and $\ce{O^2-}$ is predicted to be more stable (hard–hard interaction) than that of $\ce{Hg^2+}$ with $\ce{O^2-}$. Finally, we can also consider heats of formation which should (I unfortunately don’t have values) be much more negative in the case of $\ce{CaO}$ than for $\ce{HgO}$ — therefore, the latter can be decomposed with enough heat due to $\Delta G = \Delta H - T \Delta S$ while the former cannot.


Original concentration calculation homework

The mass of $\ce{FeCO3}$ is given as $m_{\ce{FeCO3}}=23.758\ \mathrm g$.

You already found that the molar mass of $\ce{FeCO3}$ is $M_{\ce{FeCO3}}=115.85\ \mathrm{g\ mol^{-1}}$.

For a pure sample, molar mass $M$ is defined as
$$M=\frac mn$$
where $m$ is mass and $n$ is amount of substance.
Thus, the amount of $\ce{FeCO3}$ can be calculated as follows:
$$\begin{align}n_{\ce{FeCO3}}&=\frac {m_{\ce{FeCO3}}}{M_{\ce{FeCO3}}}\\[6pt]
&=\frac {23.758\ \mathrm g}{115.85\ \mathrm{g\ mol^{-1}}}
\end{align}$$

The chemical equation
$$\ce{Fe^2+ + CO3^2- <=> FeCO3}$$
indicates that $1\ \mathrm{mol}$ of $\ce{Fe^2+}$ yields $1\ \mathrm{mol}$ of $\ce{FeCO3}$. Therefore,
$$n_{\ce{Fe^2+}}=n_{\ce{FeCO3}}$$

The volume of the solution is given as $V=100.00\ \mathrm{ml}=0.10000\ \mathrm l$.

The concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.
Thus, the concentration of $\ce{Fe^2+}$ can be calculated as follows:
$$\begin{align}c_{\ce{Fe^2+}}&=\frac {n_{\ce{Fe^2+}}}V\\[6pt]
&=\frac {n_{\ce{Fe^2+}}}{0.10000\ \mathrm l}\end{align}$$


Why are solids and liquids ignored in the equilibrium constant?

A solid or liquid that should take part in a solution or gas phase reaction has to dissolve or evaporate before it can react in any meaningful way.

If you add more solid solute to a saturated solution, its concentration does not change, and so doesn't the equillibrium constant of a reaction of the dissolved solute.

(The solute might also react directly out of the solid. Theoretically. But this different reaction needs much more energy than the one from solution, and it will only add a minuscule shift to the total equillibrium, even if your solid has a huge surface. Surely neglectable, most likely not even measurable.)


Why Inorganic chemistry reaction mechanisms are hard to find?

Reactions can be of order $0, 1 , 2$ or $3$ , but never more. No collision may happen between more than three particules simultaneously. If an equation is written with more than $4$ substances on the lefthand side, it means that the corresponding reaction is a sum of several elementary reactions.
Mechanisms in inorganic reactions are hard to establish because they are made of a lot of successive reactions, in cascade. And each step is so quick that it cannot be studied. For example your oxidation of $\ce{Fe^{2+}}$ ion by $\ce{KClO3}$ is probably a series of at least $6$ steps which are all nearly instantaneous : To be able to study such a mechanism, the $6$ intermediate states must be observed for at least some nanoseconds. If the intermediate states are too short living, all we can do then is to make hypothesis without experimental arguments.