Why are weak acids not considered good leaving groups?

Leaving group ability is tied to the stability of the leaving group. Acidity is directly related to the stability of the conjugate base. If you're a strong acid, the conjugate base is relatively stable in solution. Less so, if you're a weak acid. So the stability of the conjugate base is directly proportional to both acidity and leaving group ability.

Edit:

$$\ce{HA->H+ + A-}$$

$\ce{A-}$ is the species we're evaluating as the leaving group. Basically, it's a good leaving group if it's stabilized. If that's the case, the acid dissociation equilibrium also shifts toward products, making $\ce{HA}$ a stronger acid. You should be careful to differentiate between $\ce{HA}$ and $\ce{A-}$ when you're thinking about these concepts.


How to compare dipole moment in thiophene, furan, and pyrrole?

As another (downvoted) answer points out, the key to this comparison involves the heteroatom donating an electron pair to the ring in the aromatic compound.  This shifts the negative charge away from the heteroatom's corner into the ring.  Thus we either do not get as much of a negative end at the electronegative heteroatom's corner, or with a stronger interaction we can reverse the dipole.  In the examples shown, pay attention to the directions as well as the numbers of the given dipole moments!

Clearly nitrogen has this stronger interaction, while oxygen and sulfur give the weaker effect, with a reduced but not reversed dipole moment.  Oxygen is so electronegative that even in forming the aromatic ring it donates an electron pair only weakly.  The large sulfur atom has lower electronegativity, but overlaps poorly with the smaller carbon atoms. (In the nonaromatic tetrahydrothiophene, the larger size of the sulfur atom makes the dipole longer and thus may give a larger dipole moment with less charge separation.)  Nitrogen in pyrrole has the moderate electronegativity and strong overlap which, together, enable the stronger interaction.


Why does this compound only have one pair of enantiomers?

Usually it doesn't as rotation around single bonds is generally very quick in room temperature. However, sometimes it matters because rotation can be blocked by bulky substituents, or other parts of the molecule. Even double bonds usually connected with E/Z isomerism, can in certain situations cause optical activity.

These are all examples of so called axial chirality. In the first case (rotation blocked by substituents) it's called atropoisomerism. Second (blocking by other parts of molecule) is present for example in helicenes. Third case (cumulated double bonds) occurs in allens.


Why is the atomic mass unit less than the mass of both a neutron and a proton?

The source of the discrepancy is the mass defect, which is described in the article below as :
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation 4, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The equation four referenced is just E=mc^2.  When these nucleons form carbon-12, which is the reference isotope for the definition of the amu, the mass defect appears, as you mentioned, by conversion to binding energy.

https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry/2%3A_Atomic_Structure/2.07_Mass_Defect_-_The_Source_of_Nuclear_Energy


A free neutron is not considered to be an element and there is no element that does not have at least one proton.

Free neutrons could be included on the graph as they are unstable and decay with a half life of around 10 minutes ($611.0\pm1.0~\mathrm{s}$) although most versions I have seen do not usually include them as they aren't really nuclei which is what the graph is about.


Could a non-spontaneous reaction occur "on its own"?

An analogy may help.  Imagine a rock perched at the top of a hill.  If given a small push, the rock will roll all the way down the hill without any further input of energy (in fact, releasing energy as it goes).  That's what we mean by a "spontaneous" reaction.  The reverse (pushing the rock up the hill) is a "nonspontaneous" reaction into which you have to continuously put energy for it to proceed.

But now let's consider that small initial push.  Just how small is it?  If the rock is small and perched on a very sharp hilltop, maybe a strong breeze or a squirrel jumping on it is enough.  But if the rock is big and perched on a rather broad hilltop, it may need an earthquake or big tree falling on it to get moving.  Of course, earthquakes and big trees falling are much rarer events than a squirrel jumping around or high wind.  So if the rock needs a tiny push, we expect it to come flying down the hill quite soon after it's put up there, and if it needs a big push, it will be quite a long time before it gets rolling.

This is what we mean by the kinetics of the reaction -- how fast it goes.  A fast reaction needs hardly any push to get started, a slow reaction needs a big push, and small pushes are much more common than big pushes.  (The "push" for a real chemical reaction is some fluctuation in energy in the environment, like the coincidence of a bunch of simultaneous high-speed collisions of molecules.)

Hopefully the analogy makes clear that the thermodynamics of a reaction (whether it is spontaneous or not) is completely separate from the kinetics (whether it is fast or not).

Finally, use caution when everyday words (like "spontaneous") are used in scientific language.  The implications may not be the same!  In everyday speech "spontaneous" often has the additional implication of "quick."  If I said "I spontaneously break into song during meetings" you would find it odd if I only did it every 25 years.  There's an implication, in everyday speech, of speed as well as no need for a driving force. But that implication does not carry over to the scientific use of the word!


What is the chemical structure of ceramic nonstick coating on pans?

There is a near continuum (and thus a potential infinitude) of relative concentrations of the different components of ceramics and their means of fabrication, but US Patent US7879449 B2 (PDF) gives a thorough description of their own non-stick ceramic invention as well as descriptions and references to others.  Below are several excerpts from the cited patent:

The novel coating includes a $\ce{SiO2}$ ceramic matrix that includes the condensation product of hydrolyzed alkoxysilane and a colloidal silica sol. A series of new abrasion resistant and high hardness non-stick coating materials have been developed with ceramic matrix and organic compounds by sol-gel process.
This non-stick material has two main component of silicon dioxide. One is silica sol which is pure $\ce{SiO2}$, another is an organoalkoxysilane which is an organic-inorganic hybrid material. Both materials can make strong chemical bond from chemical reaction, condensation. $\ce{(OR')_{4-x}}$ groups are hydrolyzed with water and then make condensation with silica sol and each other. The ceramic non-stick hybrid coating is applied by a sol-gel process. Two kinds of non-stick organic compounds make a chemical bond with silica ceramic network or matrix and it can exist in the ceramic structure in a stable state.
The main matrix is composed of a network of silica formed by the chemical reaction or condensation of colloidal silica sol (water base) and an organoalkoxysilane such as MTMS (methyltrimethoxysilane). The colloidal silica sol is a very stable state because of the electrostatic repulsion of silica particle surface.
The organoalkoxysilane preferably has the general formula $\ce{R_xSi(OR')_{4-x}}$. Wherein $\ce{R}$ is one or more organic radicals chosen independently from among methyl, ethyl, and propyl, $\ce{R'}$ is a low molecular weight alkyl radical, and x is at least one and less than four. Commonly x is one, so that the organoalkoxysilane has three hydrolysable sites. This increases cohesion and bonding of the components and also provides a thickened layer and increases the links with the inorganic silica surface.
The surface of colloidal silica is coated by organoalkoxysilanes, in other words, hydrolyzed organoalkoxysilane make a thin layer or coating on the silica surface as a result of the condensation reaction with the hydroxyl groups of the silica surface.

Again, this is just an overview of a single process. The patent is well written and diagrammed and I suggest reading in detail to best answer your question (Images > Full Pages).


On biological detergent breaking down organic material

In the household, hair and fat residues in the pipes are typically dissolved using strongly alkaine drain cleaners.

In addition to the hints and directions given in the commentaries, namely this and that you might want to consider, at least in part and for the dramatic effect, a biological solution:

In the preparation of skeletons for museum displays, adult insects and larvae of the bacon beetle (Dermestes lardarius) are frequently used. After gutting the carcass (I'll leave the gory details to the novellist), plan a couple of weeks for them.


What is the relationship between the constant of equilibrium (Kc) and the position of equilibrium?

What you are probably referring to is the degree of dissociation $\alpha$ when you ask about the position of equilibrium. At a given temperature the equilibrium constant is just that, constant, but as the pressure changes, say for a gaseous reaction then $\alpha$ changes.An example shows this.

In the gas phase reaction $\ce{A2 <=> 2A}$ with partial pressures $p_{A_2}$ and $p_{A}$ respectively, the  equilibrium constant is $\displaystyle K_p=\frac{p_A^2}{P_{A_2}}$ and is defined at standard conditions of 1 atm.  If $P$ is the total pressure then the two partial pressures are $\displaystyle p_A=\frac{2\alpha}{1+\alpha}P$ and $\displaystyle p_{A_2}=\frac{1-\alpha}{1+\alpha}P$ where and  so after some simplification $\displaystyle K_p=\frac{4\alpha^2}{1-\alpha^2}P$.

Thus, as the equilibrium constant is constant at any total pressure $P$, the degree of dissociation must change as the pressure changes to compensate, thus the 'position of equilibrium' changes.

[Notes: The amount of A$_2$ reacted is $1-\alpha$ and the amount of A produced is $2\alpha$, thus the mole fractions of each are $(1-\alpha)/(1+\alpha)$ and  $(2\alpha)/(1+\alpha)$ and so the partial pressure are as given above. The total amount needed for the mole fraction is $1-\alpha+2\alpha=1+\alpha$. You can derive related expressions in terms of concentrations.]


Isn't the change in entropy zero for a reversible process?

As mentioned in the comments, there is an entropy change that occurs in the system, however, when combined with the entropy change of the surroundings, is zero.
When solving for the change of entropy in the system, you can use the equation below,
$ \Delta S = R \ln(\frac{P_1}{P_2})$
converting MPa to atm and plugging in, we get
$ \Delta S = 8.314 \ln(\frac{49.3}{9.87})$
which yields an entropy change that is equal to the answer you provide.
The above equation can be rearranged to have different variables instead of pressure. Here is a link for a ChemLibretexts article on entropy and reversible processes if you're interested.


What is the minimum number of given sets of information required to complete an ICE chart for an equilibrium reaction?

Alright I think you have the basics down, first of all I have a couple pointers for you since I have done many (many) RICE charts. What I would do if I were you? Since I have some experience in programming I would just use molarities of the initial and final.
You must make the user input the data, say if the user was given moles then make him/her convert it to molarity.
I think you have it pretty much down, you need either the initial concentrations and K or find the K through using the initial concentrations of the products and reactants or find the initial using the product and K.
I mean then again I would work through making this first part down because those are the essentials (atleast what I've learned) and then work on making updates to that program.
Since I do have programming experience I would just use if statements to do the entire thing. If this is available then solve for this.
Don't worry about the entire thing yet. Otherwise I would have to say you have pretty much everything down.
NOTE: This is not about Ka, Ksp, Kb, just for Keq. Because then you would have to work with the pH and pOH more equations.


The delocalised nature of hyperconjugation

I'll try to answer your question through the exemples you give us.
Where are the electrons ?
If you look at the hybridized orbitals making the C-H and the C-C bonds, you can see that the more the bond is polarized, the worse electron donor you have. It's quite understandable since if the bond is polarized, the electron density will be closer to the heteroatom but further to the carbocation that needs stabilization.

Usually the donor ability goes in this order increasingly:
C-F < C-O < C-N < C-H < C-C
Hyperconjugation from C-C to carbocation
Following the "rule" stated before, the stabilization of a carbocation will be best with an adjacent C-C bond. And this is in accordence with what we learn in school since the more alkyl groups nearby, the more stable is the carbacation. Not only because of the increased overall electron density (donor inductive effect) but also because of the hyperconjugation. This is also the base of some reactivity (1,2-C-C bond migration)
Hyperconjugation in the bromonium case
We can easely extend the thing we just saw for the exemple of a bromonium intermediate. In fact it is important to dissociate the cases of the bromonium and the transition state for the bromonium openning.
The bromonium has two bonds with a carbon atom. One could expect the opening reaction to occur in competition between these two sites. We can see that the C-Br bond (Scheme 2, left) can be aligned with a C-C bond, so if we take a look at the molecular orbitals, we will see that this filled sigma C-C bond can interact with the empty C-Br sigma*. Thus creating a destabilization of the C-Br bond (filling a anti-bonding molecular orbital destabilizes the bond) trough an hyperconjugation intercation. But if we look at the second C-Br bond (right) we see that there are no such interaction since there are no filled molecular orbital which can interact with the sigma* of the C-Br bond.

In term of reactivity, the destabilization of the left-hand C-Br bond will favor the breaking of this  bond instead of the other one.


Why a higher s character increases a carbon atom's electronegativity?

Electronegativity is the power of an atom to attract bonding pairs of electrons to itself. It clearly depends on the nuclear charge: the larger it is, the more strongly the nucleus attracts electrons and the closer to the nucleus they stay, so the larger is the electronegativity. Besides, electronegativity also depends on the number of other electrons present in the atomic shells ahead of valence electrons participating in chemical bonding. These electrons shield the valence electrons from the positively charged nucleus decreasing its effective charge and consequently lowering the electronegativity of the atom.

Now, back to the question. $\mathrm{2s}$ orbitals are lower in energy than $\mathrm{2p}$ ones, so that electrons occupying the former are (at average) closer to the nucleus than that occupying the later. Thus, the more $\mathrm{s}$ character a hybrid orbital has, the lower its energy is and the closer (at average) electrons occupying it are to the nucleus. The lower the energy of a hybrid orbital of an $\ce{X}$ atom is, the lower is the energy of the molecular orbital in formation of which it participates, and consequently, the closer (at average) the bonding pair of electrons stay to that $\ce{X}$ atom.

So, with respect to a bonding pair of electrons that occupy a molecular orbital to which a hybrid orbital of an $\ce{X}$ atom contributes, the $\ce{X}$ atom indeed appears to be more electronegative as the $\mathrm{s}$ character increases. Specifically for $\ce{C}$ atoms, the larger the $\mathrm{s}$ character of a hybrid orbital is, the lower in energy is a bonding $\sigma_{\ce{C-C}}$ orbital formed from the hybrid ones: $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$. So that electrons occupying the $\sigma_{\ce{C-C}}$ orbital stay closer to $\ce{C}$ atoms (at average) as the $\mathrm{s}$ character of hybrid orbitals increases.

With respect to the question OP asked in comments,

  But if we compare between $\ce{C}$ and $\ce{H}$, is there some other factor to be accounted for because clearly H atom due to its very lower potential energy electrons should attract the shared electrons more.

the factor OP is looking for is the nuclear charge, which is 6(!) times larger for $\ce{C}$ atom comparing to the $\ce{H}$ one.


Is salt an electrolyte?
Common table salt (NaCl) is an electrolyte, and when this is dissolved in water to form salt water, it becomes sodium ions (Na+) and chloride ions (Cl-), each of which is a corpuscle that conducts electricity.
Let's go back to conductivity. Conductivity is an index of how easy it is for electricity to flow. In water, it is the ions that pass electricity from one to the next. This means that the more Na+ and Cl- contained in water the more electricity is carried, and the higher the conductivity.
Here's a chart:

(Don't mind the sugar water side)
Source linked here.


Detail from Integral Evaluation in Szabo (Appendix)

As commented, do note that the integral (A.20) is taken over all $\mathbf{k}$-space, whereas the integral (A.21) is one-dimensional. The explicit calculation of this integral follows as

$$\begin{align} \int\text{d}\mathbf{k}\,\Psi(k)e^{-i\mathbf{k\cdot r}} &=\int_0^\infty\text{d}k\,k^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\varphi\,\sin\theta\Psi(k)e^{-ikr\cos\theta}\\
&= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\int_0^\pi\text{d}\theta\,\sin\theta e^{-ikr\cos\theta}\\
&= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\left[\frac{e^{ikru}}{ikr}\right]_{u=-1}^1\\
&= \frac{4\pi}{r}\int_0^\infty\text{d}k\,k\Psi(k)\sin(kr).\end{align}$$

In the first equality we convert this integral from Cartesian to spherical coordinates, in the second equality we evaluate the integral over $\varphi$, in the third equality we substitute $u = -\cos\theta$, and in the fourth equality we make use of Euler's formula to get the exponential representation of $\sin(kr)$.


Why are protic acids not used for the formation of thioacetals?

The pKa/leaving group ability between alcohols and thiols may have something to do with it.

  Alcohol pKa around 17, thiol around 13.

As you pointed out the whole reaction is in equilibrium. But in the case of the thioacetal, after the initial addition theres a strong driving force to kick back out the sulfur to reform the carbonyl rather than committing to the thioacetyl formation since the sulfur is a relatively stable leaving group.

Protic acids are almost always wet, and as such the water would prevent the reaction, whereas Lewis acids are generally either neat, or in ethereal solutions.


One way to do it would be do first do a hydroboration-oxidation reaction, which leads to the anti-Markovnikov addition of $\ce{H2O}$. So you will get an $\ce{-OH}$ group instead of $\ce{-Cl}$. Then, we can replace the $\ce{-OH}$ group with $\ce{-Cl}$ by treating it with $\ce{PCl3}$ or $\ce{PCl5}$.


Grignard reagent reaction mechanism: Which bond breaks during the decomposition of the adduct?

As can be seen above, the end product is the same whether we split the $\ce{Mg-O}$ bond or the $\ce{C-O}$ bond.

The problem is, then, which pathway is more favourable?

This, I have no idea. Maybe both pathways are favoured. It doesn't matter because they give the same products.


How to prepare cyclohexanone without starting with cyclohexanol?

Do you have access to phenol? If you have, then use that to make cyclohexanol (hydrogenation of phenol using nickel catalyst):

Then treat cyclohexanol with copper(II) and chromium(III) oxides to convert it to corresponding ketone (Dehydrogenation):

Benzene could work as a starting material but the reaction might be costly and hectic as the catalyst is not a household item$\ce{^{[2]}}$. Secondly, it is used as a starting material in the synthesis of nylon and the reaction intermediate is a mixture of cyclohexanol and cyclohexanone(mixed oil)$\ce{^{[3]}}$ which add to your task of separating them. Better convert it to phenol$\ce{^{[4]}}$.

Notes and References

http://www.leck.co.uk/demo/eci3/polymers/polyamides.html (above reaction synthesis source). It is a part of polyamide synthesis
https://link.springer.com/article/10.1023/B:CATL.0000011080.03035.eb
http://www.greener-industry.org.uk/pages/nylon/4nylonpopup.htm
https://www.quora.com/What-is-the-method-to-convert-benzene-to-phenol
https://russianpatents.com/patent/220/2205819.html


What are some good research topics for a high school student?

Green chemistry is a highly advanced topic. It ain't for the weak of heart, so it is probably not for below university level.

I personally favor pyrotechnics as relatively safe iff every bit of safety rules is strictly followed and stunningly spectacular topic that is pretty easy and cheap to do by own hands (most materials are inexpensive as well as most tools). However, it is extremely time-consuming and has questionable legal status in some countries (and for a good reason: safety may be easily compromised and potential for disaster is high).

Another interesting topic is fluorescence. Many compounds shine under UV light, and there are families of such compounds that are not exhaustively studied and yet are relatively easy to make. However, a deep study is required, but it is probably advanced school level.

With access to advanced materials/tools/labs more topics become available, but a full list of such tools is required to recommend anything.


Mechanism for bromoetherification of olefinic alcohol

As you already know, treatment of an alkene with aqueous bromine leads to halohydrin formation. This results when the intermediate bromonium ion is trapped by the solvent, water:

Generally, water is a lousy nucleophile, but this process can happen because there is simply so much water around. In our case, the alcohol in the same molecule acts as the nucleophile, instead of water which may or may not be present. Alcohols aren't great nucleophiles, but what this has going for it is that the process (called bromoetherification, because you form a bromo+ether) is intramolecular.

As you can see in the scheme above, there are two possibilities in the formation of the bromonium ion: either on the top face of the alkene, or on the bottom face. There is probably a slight bias because of the presence of the substituent on the ring, but it turns out that this probably does not matter.

When the alcohol attacks the bromonium, it is an $\mathrm{S_N2}$ process, which requires the oxygen to attack carbon from behind the C–Br bond. Now, if the bromonium ion forms on the top face, this can't happen, because it's tough for the oxygen to loop all the way to below the plane of the alkene. Only if the bromonium ion is formed on the bottom face can the oxygen attack. Thankfully, bromonium formation is pretty much reversible, so even if the bromonium ion is initially formed on the top face, it can revert to starting material.

Extra

This bromoetherification reaction doesn't seem to have been carried out on this particular substrate - perhaps it's too simple - but it is a thing. Here's a more impressive example (Org. Lett. 2004, 6, 2961):

It also has a somewhat more popular cousin, called bromolactonisation, where the nucleophile is a carboxylate instead of an alcohol. Here's an example, taken from E. J. Corey's prostaglandin synthesis. The halogen used is different ($\ce{KI3}$ acts as a source of $\ce{I2}$, so this is an iodolactonisation) and the nucleophile is different, but the mechanism is generally the same - except that $\ce{NaHCO3}$ deprotonates the carboxylic acid before it attacks the bromonium. Can you work it out?

See also: Facial selectivity of iodonium formation in an iodolactonisation for a related question on iodolactonisation.


Can resonance take place in between biphenyls?

I thought since both phenyl rings are not coplanar, the resonance between them is not possible.

This is not correct. If the rings are coplanar you get the best orbital overlap and the strongest resonance effect, while if you twist it the effect will be smaller but still present until you are at an dihedral angle of 90°, where no overlap can take place.

Let us look at phenyl-tetrazine. The phenyl substituent loweres the LUMO of the tetrazine through conjugation and raises the HOMO. The most favored confomer is the planar one, so the ideal system to study this. If it's correct that resonance only takes place at a planar conformation we should see a sudden jump in LUMO energy between planar conformation and any other and for all angles not equal to 0° the LUMO level should be the same.

In addition resonance is stabilizing, so the energy should drop rapidly between non-planar and planar.

Calculated energies (M06-2X/6-311+(d,p)) while twisting the bond looks like this:

So the energy goes up continuously if we twist the phenyl ring, there's no jump in energy. Since there's basically no steric effect, unlike in biphenyl, the energy is most likely only influenced by resonance.

Let us look at the HOMO level. I just picked some angles:

As expected the HOMO is raised in case of planar conformation in contrast to 90°, where we expect no overlap but we can clearly see that there's an influence on the HOMO at nearly every angle getting much stronger if we get near planarity.

Since the HOMO is completely localized at the tetrazine even in the planar form let's look at the HOMO-3 as an example. This orbital is part of the conjugated aromatic system and looks like this if the angle is 0°. Of course orbital energies will get shifted if we twist the rings and finding "the same" orbital is hard, but I'll show the orbital with the same symmetry at the tetrazine.

Here's the picture for 0°, a nice conjugated system:

Here's the picture for 90°, there's no conjugation with the pi-system, little bit with some, I assume, sigma* orbitals.

Now here for 15,30 and 60°:

We can see quite nicely that the orbital in case of 15 and 30° still looks quite similar to the one at 0°, while the one at 60° already looks quite like the one at 90°, with a little bit of orbital in para position at the phenyl ring. Showing that there's a gradual loss of conjugation.

In addition, if the conjugation would be possible only at exact 0° dihedral angle then two things could happen: Either we won't see the effect of conjugation ever, since the molecules move all the time and if rotation around that bond is possible then the chance that the molecule is at exact 0° is basically zero. But we know that conjugation exists. Or the planar conformation would be much more stable due to the conjugation and it would be exact 0° all the time, but from experiments we actually know that those bonds rotate quite well.

  coplanarity the topmost condition for resonance –

The most important condition is an overlap between the orbitals, this usually works best if planar, but it most certainly doesn't have to be planar.


Ca and Mg hydrated radius in clay flocculation

Charge density is a measure of electrical charge per volume. It is independent from electronegativity, and increases with the number of charges of an ion and with decreasing radius, because a smaller ion has also a smaller volume. The number of charges for $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ is the same (+2), but $\ce{Mg^{2+}}$ has the smaller atomic radius. Its charge is therefore distributed over a smaller volume, and the charge density is higher.

The hydrated radius of an ion can somewhat differ from its atomic radius, because it is influenced by the charge density. If the ion charge is concentrated in a smaller volume, it exerts stronger attractive forces on the dipoles of water molecules, which results in the formation of a larger hydration shell. The smaller $\ce{Mg^{2+}}$ attracts more water dipoles and forms a larger hydration shell. Hence, its hydrated radius can be bigger than that of $\ce{Ca^{2+}}$, which has a larger atomic radius but a smaller hydrate shell.


What will be the major product of the reaction between (2R,3R)-2,3-dimethyloxirane and triphenylphosphine?

First, note that with the mechanism you have drawn you should expect to get a mixture of cis and trans isomers. In the intermediate you drew, rotation about the C–C bond will be fast and will lead to loss of any stereochemical information that was previously there.

The only way to get exclusively trans-2-butene is for the reaction to occur in a single step without a discrete intermediate:

However, that's not the case. The mechanism of epoxide deoxygenation by triphenylphosphine isn't a concerted pathway, but is stepwise.

Triphenylphosphine opens the epoxide ring first;  because this epoxide is symmetrical, it doesn't matter which carbon you attack. This gets you to a charge-separated intermediate 1a (a "betaine"). These species tend to combine their positive and negative charges if they can, and here they certainly can, but first the C–C bond in the middle has to rotate to get the negatively charged oxygen and the positively charged phosphorus on the same side (1b). Note that this bond rotation brings the right-hand side methyl group to the front of the molecule.

Cyclisation to the oxaphosphetane 2 then occurs, and collapse of this to give triphenylphosphine oxide and cis-2-butene is concerted and stereospecific, exactly like the final step of a Wittig reaction.

See also the Barton–Kellogg reaction for a similar mechanism, just with sulfur instead of oxygen.


In the periodic table, why doesn't the 2nd row have exactly 2 elements?

The pattern is better expressed this way:

Row 1: 2 elements
Row 2: 2+6 elements
Row 3: 2+6 elements
Row 4: 2+6+10 elements
Row 5: 2+6+10 elements
Row 6: 2+6+10+14 elements
Row 7: 2+6+10+14 elements

The reason comes down to how the electrons fill the available energy levels.  The thing that differentiates one element from another is the proton number, and each time a proton is added and a new element defined it requires one more electron to neutralise the charge.  That electron naturally occupies the lowest available energy level in the atom.

The energy levels available are defined by the quantum state, including the quantum numbers $n, l,$ and $m_l$.  $n$ is the principle quantum number and relates to the period the element is in, or the shell.  $l$ is the angular momentum quantum number which defines the sub shell s, p, d, f, of which there are $n$ subshells whose values are $l=0,\dots {n-1}$.  The magnetic quantum number $m_l$ further subdivides the subshell into orbitals, of which there are $2l+1$ orbitals whose values are $m_l=-l,\dots {+l}$.

subshell                 number of orbitals    subshell
label     l value       (number of ml values)  electron capacity
 s         0               1                   2
 p         1               3                   6
 d         2               5                   10
 f         3               7                   14

Each orbital (i.e. each value of $m_l$) may contain 2 electrons  The available quantum numbers are:

n    l values   (subshells)   ml values        total shell electron capacity
1    0           (s)           0                 2
2    0,1         (s,p)        -1,0,1             2+6
3    0,1,2       (s,p,d)      -2,-1,0,1,2        2+6+10
4    0,1,2,3     (s,p,d,f)    -3,-2,-1,0,1,2,3   2+6+10+14

The pattern is there, but it doesn't seem to match up because by row 7 you would have 98 electrons in the shell and might expect the row to contain 98 elements.  This is not the case!  The energy levels for the $l$ values with large deviations from $0$ (i.e. orbitals with high angular momentum) become increasingly far apart, so even though 3d orbitals exist in the third row, they are so much higher in energy than the 3p orbitals that they are higher even than the 4s orbitals.  This happens again in row 4 where 4f orbitals are so much higher in energy than 4d orbitals, they are even higher than 5s.

The different in energy is not always so large in specific cases, so there are examples where bonding or ligands or symmetry cause the energy levels to switch around, but otherwise the trend is true. If they can produce (or discover) some elements in the next row, we might expect to see the 5g shell finally start to get filled and have a new block in the table, but realistically it doesn't look like they will be stable atoms.

I think I'm missing the reason why we arrange the elements into this table in the first place.  The periodic table is arranged so that all elements in the same column have the same number of outer electrons.  This is useful because elements with the same outer shell configuration of electrons react in similar ways, so they are grouped together in columns. This has the effect of creating 'periods' that begin with a reactive metal and end with an inert gas. Row 1 has two elements, and it's special in a way because it's the first in the series.  Hydrogen has 1 electron, and a maximum capacity of 2 electrons.  This means it behaves in a similar way as other elements that only have 1 electron in their outer shell (group 1), and also in a similar way to elements that only need 1 more electron to complete their outer shell (group 7).  Where do we position it then?  Often it is placed somewhere in the middle.  Helium then has a full outer shell, which means it has many similarities to other elements with a full outer shell and goes in group 8 (or group 0) on the right.

Edit:
The alternative period table that you mention, the left-hand version, is laid out so as you read it from left to right you fill up the orbitals.  This takes in to account that situation I describe where 3d orbitals are higher than 4s, and using this left-hand table you clearly see that. However this alternative table does not allow you to easily read the valance shell configuration like the Mendelev one does.  Other types of table highlight different kinds of properties or relationships, which can be more useful in certain situations.  The labels on the left (1s; 2s; 2p 3s; etc) relate to the outer subshell in that row, so Carbon is in the "2p" section of that chart while Magnesium is in the "3s" section on the same row.


What is the role of pH in azo coupling reaction of diazonium with phenol and aniline?

Deprotonation of the phenol and protonation of aniline result in species that easily react with a diazonium cation in the intended manner.

Let's have a look at the species involved. On the one hand, there is the diazonium cation:

Deprotonation of phenol yields phenolate, for which a resonance structure with a negative charge in para position to the substituent can be written:

We conclude: Deprotonation of the phenol means activation!

In the case of aniline, the lone pair on the nitrogen atom is the preferred centre for the reaction with the diazonium cation:

This isn't what we want! Protonation of aniline avoids this side reaction.

We conclude: Protonation of aniline means "masking" the wrong centre!


Why and how does carbon hybridize itself according to its need?

Hybridization is part of the language chemists use to speak to one another.  When someone says that a carbon atom is $\ce{sp^2}$ hybridized, it's a shorthand way of telling others

that there is a p-orbital on that carbon
that the angles around the carbon are roughly 120°
that the bonding orbitals (the $\ce{s^1p^2}$ orbitals) have 1 part in 3, or 33%, s-character and 2 parts in 3, or 67%, p-character
that the bonds around that carbon are likely to be a bit shorter than the bonds around an $\ce{sp^3}$ hybridized carbon (because the $\ce{sp^2}$ bonds have more s-character and are therefore closer to the nucleus, e.g. shorter).

So when someone says that a carbon atom is $\ce{sp^2}$ hybridized, they're really saying a lot more.

Using hybridization to describe an atom is similar to using VSEPR or MO theory to describe a molecule.  Hybridization is just another way to try and describe reality; just like the other methods, often it works well, but sometimes it doesn't.

Also keep in mind that there is a continuum of hybridizations available to a molecule all the way from pure s to pure p with everything else in between.

$\ce{s^1p^0...s^1p^1...s^1p^2...s^1p^{2.2}...s^1p^3...s^1p^4...s^0p^1}$

Of course, we normally drop the "1" and the "0" superscripts and write those hybridizations as

$\ce{s...sp...sp^2...sp^{2.2}...sp^3...sp^4...p}$

As to how carbon (or any other element) "decides" which hybrization to use, it adopts the hybridization that produces the overall lowest energy for the molecule. Let's take the case of a carbocation.  Suppose we start with an $\ce{sp^3}$ hybridized carbon in a molecule like $\ce{R_3C-I}$ and subject it to conditions where it ionizes to an $\ce{sp^2}$ hybridized carbocation and $\ce{I-}$.

(image source)

Your question is basically, why did the central carbon rehybridize from $\ce{sp^3}$ to $\ce{sp^2}$? The answer is, the ionized molecule (the carbocation) will reoptimize its geometry (rehybridize) in order to attain the lowest energy possible. In this example note that the carbocation p-orbital is empty - it contains no electrons. So why keep that orbital $\ce{sp^3}$ hybridized, why waste a part of a valuable, low-energy s-orbital in a bond with no electrons in it to stabilize? Instead, why not remove the s-character from the $\ce{sp^3}$ bond making it a pure p-orbital and mix that s-character into the other 3 orbitals (or bonds) to convert them from $\ce{sp^3}$ to $\ce{sp^2}$ thereby lowering the energies of the electrons that are in those 3 bonds. In fact, this rehybridization, or geometry reoptimization, is exactly what does occur in this reaction in order to stabilize the electrons as much as possible and produce a molecule (or ion) with the lowest possible energy.


I'm getting basically the same answer when I work out the math. I got $T=3398.45\ K$, but we probably just rounded the constants differently.

This is a good lesson in how much thermal energy is carried around at any given time by molecules in the air, as well as the importance of distinguishing between thermal processes and photo-processes.

That is, if we wish to sustain an ensemble of excited particles, there must be sufficient energy such that when we spread the energy out evenly, which is one of the things that this equation implies given that it claims all possible microstates are equally likely, some of the energy must be stored in the form of electronic excitations.

The reason this probably seems like a very high temperature is because we use equipment all the time that relies on electronic excitations and we never have problems with the equipment melting due to it being at thousands of degrees Celsius. That is simply because in these photo-process, meaning the energy is carried in the form of photons, the temperature never is that hot.

That is, if I shoot a photon in, or even a stream of photons at a sample with the transition you are describing, each photon carries a very small amount of energy, and even though there may be quite a lot of them, the energy is quickly dispersed because the excited state is very short-lived, so the photon gets re-emitted, and likely just makes it way out into an infinite reservoir known as the surroundings.

Thus, it is very difficult to maintain a population of excited states thermally, because the system needs to basically be a closed system or else it will eventually just radiate all the energy away.

Another way of thinking about this is that the value of $k_bT$ at room-ish temperature of $298\ K$ (in a familiar unit) is $207\ cm^{-1}$. This makes it clear then why rotationally excited states are frequently occupied at room temperature, because these transitions are on the order of tens of wavenumbers. On the other hand, we often get off free assuming everything is in the vibrational ground state because these transitions are on the order of thousands of wavenumbers.

This is also where the idea of rotational temperature and vibrational temperature come from. These can be understood physically as,

  an estimate of the temperature at which thermal energy (of the order of $k_bT$) is comparable to the spacing between rotational [or vibrational] energy levels.


Migratory aptitude in pinacol-pinacolone rearrangement

In the pinacol rearrangement, a 1,2-diol is treated with acid and rearranges to a carbonyl compound. Here is a reaction scheme showing a mechanism for the rearrangement.

In this case the molecule is symmetric and methyl migration is the only reaction pathway available.  Methyl migration can 1) help stabilize the developing carbocation center and 2) once fully migrated, produces a resonance stabilized hydroxyl-carbocation.

If one or more of the methyl groups is replaced with some other substituent (for example, alkyl, hydrogen, aryl), then our reaction possibilities become more complex; which hydroxyl group will protonate, which group will migrate?  Generally, both hydroxyls will protonate and lead to separate products.  In terms of which group will migrate, the only thing agreed upon is that a tertiary carbon will migrate in preference to a secondary carbon which will migrate in preference to a primary carbon.  Clearly, the ability of the migrating group to stabilize positive charge plays a role.

The reaction is reversible and product distribution can change as reaction conditions are changed (thermodynamic control vs. kinetic control).  All of these complexities make it difficult to sort things out in detail, thereby making it difficult to determine the true relative migratory preference of other groups like H and phenyl.  This is why the migratory aptitudes of these groups (H and phenyl) can change from textbook to textbook.

See section 2. Pinacol Rearrangement in this reference for a very nice, straightforward discussion of this reaction mechanism, and all its subtleties.


Mechanism for the formation of butanoic acid from ethylacetoacetate

I thought about adding solid $\ce{NaOH}$ to ethanol and therefore used $\ce{EtO-}$ ;) :


Regularity of electron configuration for heavier atoms

Yes, the nucleus has a complex structure, too.  The most basic evidence for this comes in the existence of patterns in which nuclei are most stable, which manifests itself in the decay lifetime and to some extent in the cosmic abundance of isotopes.  I don't think there is any reflection of it in the electronic structure: from the point of view of the electrons, the nucleus is pretty much just a point positive charge.  The patterns in the electron configuration you are noticing are due to the Pauli exclusion principle and the various ways quantum mechanical particles can fit into a spherical potential well.   You can find a description in any basic general chemistry textbook, and the math will be developed in any undergraduate physical chemistry textbook.

You can get started on nuclear structure by googling "nuclear shell structure."  That will lead you to very basic introductions, e.g. on Wikipedia or Hyperphysics, and you can decide where to go from there.  I don't know any textbooks that treat it simply, because the assumption is that if you are interested in the nuclear physics of nuclei you are already very familiar with the quantum mechanics of electrons.


How to calculate saturation point of a solution?

Yes. But it would probably be faster to simply find some empirical data.

You would need to:

approximate the hydrophobicity of your vitamins (for rather big molecules like vitamins, this would need some mathematical modeling, but there are some computer programs that can do this rather accurate.
this data could be used to approximate the solubility in water and probably also olive oil. But milk is quite complex, as there will be a solubility equilibrium in/between both the fat phase and the water phase. And all the salts and other compounds in the milk (and to a small degree in olive oil) will affect the solubility.

If you know A-level chemistry and know someone to help you irl (or some fab ppl here :-) ), I'd guess there's some days of work in it!


Carbon Free energy vs Temperature Diagram

Here is the link to a set of Ellingham Diagrams

At high temperatures, smelting and carbon reductions have several practical challenges some of which are energy efficiency, formation of metal carbides, equipment concerns etc. Therefore, to assess the reducing ability of any reducing agent, we must work within practical limits. Let us select a temperature of $1600\ \mathrm{K}$ as our upper bound beyond (which is still quite high) which, practical concerns outweigh any benefit of having a good reductant. At this temperature $\Delta_\text fG(\ce{CO2})$ is approx $-394.39\ \mathrm{kJ/mol}$ and $\Delta_\text fG(\ce{CO})$ is $-504\ \mathrm{kJ/mol}$. These values are high enough and can already reduce several oxides of metals like $\ce{Co, Fe, Cu, Pb, Ni, Bi}$ and $\ce{MnO2, CrO2}$ etc (subject to practical constraints). This gives it a good reducing character, better than $\ce{H2}$ or ammonia.

Another thing to notice, is that, although the variation of $\Delta_\text fG$ for carbon dioxide is flat, for carbon monoxide it slopes downward and hence with increasing temperature, it can reduce many more oxides further expanding its capacity to reduce metal oxides. $\ce{V2O5, VO2, SnO2, WO3, Mn3O4, MoO3, GeO2}$ can be reduced by carbon. (All this just implies thermodynamic favourability; it can still be unfavourable kinetically, or due to practical restraints.) Hence, the negative sloping $\Delta_\text fG$ for $\ce{CO}$ makes carbon a very good reducing agent, and it can reduce a huge variety of metal oxides, within a considerable temperature limit.


Because the second pKa of sulfuric acid is weak, sulfate withdraws protons from water.  Therefore, the contribution of sulfate alone is to make the solution slightly more basic than pH 7.

However, as quantified in CE4501 Environmental Engineering Chemical Processes, Cu will form species such as CuOH+, which is the reason the solution will be acidic overall.  This could be approximated by thinking of $\ce{Cu(H2O)6^{2+}}$ as having a pKa of ~6.3.

Because  $\ce{HSO4-}$ is a stronger acid than CuOH+ is a base, the solution is acidic.


Due to Chappuis absorption, ozone does have a bluish color. To determine exactly what kind of blue it is, let's first look at the spectrum of absorption in the Chappuis band. The following plot was done using these data for 293K.

This is spectral cross-section of absorption. To determine color from this spectrum, we need to choose some parameters:

Number density of ozone molecules,
Thickness of ozone layer we're trying to visualize,
Illuminant.

If we denote spectral radiance of our illuminant as $L(\lambda)$, thickness of ozone layer as $d$, ozone molecule number density as $\rho$, and absorption cross-section as $\sigma(\lambda)$, then we'll get the following expression for spectral radiance transmitted through the layer:

$$L_T(\lambda)=L(\lambda)\exp\big(-\sigma(\lambda)\rho d\big).$$

The most sensible illuminant to choose for showing color of a material on the web is the CIE illuminant D65, whose color is the white point of the sRGB color space. Its spectrum can be found e.g. here.

We can find the color in XYZ space using CIE 1931 color matching functions (can be found e.g. here). The expression is

$$c_X=\int_{300}^{830}L_T(\lambda)\bar x(\lambda)\,\mathrm d\lambda,$$

and similarly for $Y$ and $Z$ coordinates. Then these can be transformed to sRGB using linear transformation matrix $\mathrm{XYZ}\to\mathrm{sRGB}$ given e.g. here and gamma-correcting to $\gamma=1/2.2$ to yield final sRGB values.

Then, for ozone molecule number density $\rho=10^{25} \frac{\mathrm{molecule}}{\mathrm{m}^3}$ we'll get the following colors for different layer thicknesses:

For comparison, typical ozone column in the atmospheric ozone layer is about 300 Dobson units, which is equivalent to $2.687\times10^{20}\frac{\mathrm{molecule}}{\mathrm m^2}$; with our $\rho$ chosen above this corresponds to $d=8\,\mathrm{mm}$. So for daylight, ozone column has negligible effect on the sky color (unlike the evening — see the history of Chappuis absorption!).

As can be seen in the above plot, hue does change with increasing layer thickness. If we normalize the RGB values to see the hues of the thick layers (this would correspond to increasing illuminant power to compensate for absorption), we'll get the following hues:

Note that the violet hues in the above plot aren't accurate: they can't be accurately represented on sRGB monitors, so the plot only approximates them. They should be more saturated. Here's how the chromaticity changes from the white point to the most violet with increasing layer thickness (dashed triangle denotes the sRGB gamut):


Is the refractive index of a solution of four substances the median of the four substances' refractive index?

There is no single equation that accurately predicts the refractive index of a mixture.  This reference discusses 9 approaches to estimating the refractive index of mixtures and provides the formula for each method. The simplest method is the Arago-Biot approach.  It is reasonably reliable and is described by the following formula

$$n = n_1 \phi_1 + n_2\phi_2$$

where $n$ is the predicted refractive index of the mixture, $n_x$ is the refractive index of the $x^{th}$ component, and $\phi_x$ is the volume fraction of the $x^{th}$ component.  The volume fraction of the first component is given by

$$\phi_1 = x_1 V_1 / \Sigma~ x_i V_i$$
Here $x_i$ is the mole fraction of the $i^{th}$ component and $V_i$ is the molar volume (the molecular weight divided by the density) of the $i^{th}$ component. The more ideal the liquids, the more accurate the predictions.


Cracking of dicyclopentadiene - was my simple Liebig condenser distillation setup adequate?

The Friedrichs condenser is for refluxing dicyclopentadiene (bp ~170 C). It's kept at 50 C to let the cyclopentadiene (bp ~40 C) pass without condensing.


What causes the stairstep structure in bismuth crystals?

Hopper crystals, that is, the "staircase" form, develop when a crystal forms so rapidly that material is used up before the interior finishes.

Table salt, $\ce{NaCl}$, also forms nice hopper crystals. If you want to try this, use "kosher" salt or CP $\ce{NaCl}$, which doesn't have anti-caking and iodide additives that would spoil the crystal structure.


Creating 1L of 1000 ppb stock solution of Pb2+

Making the solution in a single step is inaccurate up to impossible.

Try to find a crystal of lead nitrate with the mass close enough to $\pu{1 mg}$
Try to weight up $\pu{1 mg}$ on scales with good enough accuracy.
Try to make such $\pu{1 ppm}$ solution stable enough.

The usual thing is to create a concentrated stock solution ( e.g $\pu{1000 ppm}$ and creating the working solution in 1 (1000 times) or 2 step ( 20x50 times) dilution whenever needed.

There is need to consider stabilizing by an acid, to prevent hydrolysis and adsorption. Very diluted solutions, like for AAS, are often stored in PE instead of laboratory glass to prevent ion Vs glass surface interactions.


Difference between biosynthesis in organism versus lab

More often than not, the synthesis in organisms involves the action of enzymes. In the lab, you often avoid these for various reasons. The enzymes may not be available, or you're interested in a more general procedure, that allows to synthesize a number of derivatives of the target structure or just a part of it.

When the synthesis aims to obtain a particular molecule, just as found in nature by any means, you'll often find a term like natural product synthesis.

When a procedure aims to yield a product by a reaction similar to processes in living  organism, the term biomimetic is frequently used.


Gibbs Free Energy Spontaneity Criterion Derivation

For a system in contact with a constant temperature reservoir at the same temperature as the system in its initial state, and also in contact with a constant pressure surrounding at the same pressure as the system in its initial state, $$\Delta U=Q-W=Q-P\Delta V-W_{npv}$$where $W_{npv}$ is the amount of non-PV work done by the system.  In addition, from the 2nd law of thermodynamics, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\frac{Q}{T}$ is the amount of entropy transferred to the system from the surroundings across the interface between the system and surroundings (recall that the latter is always at the initial temperature of the material comprising the system).  Note that this does not say anything about how the temperature may be varying with location and time within the internals of the system during the process.  In addition, the parameter $\sigma$ represents the amount of entropy generated within the system due to irreversibilities during the process. Note that $\sigma$ can only be positive (irreversible process) or zero (reversible process)
If we combine these two equations, we obtain:
$$\Delta U=T\Delta S-T\sigma-P\Delta V-W_{npv}$$or$$\Delta G=-W_{npv}-T\sigma$$ According to this, if no non-PV is done by the system on the surroundings during the process, $\Delta G$ must be negative (irreversible) or zero (reversible).


What happens if hydrogen peroxide is added to silver oxides?

AgO (silver(I, III) oxide) is unstable and decomposes to produce $\ce{O2}$ in aqueous solutions. Hydrogen peroxide is thermodynamically unstable too and slowly decomposes to form water and oxygen. Decomposition of hydrogen peroxide can be catalyzed by different compounds, including transition metals (such as Ag) and their compounds. Probably, the silver (I) oxide will directly/inderectly catalyze the decomposition of hydrogen peroxide according to the reaction:$$\ce{Ag2O + H2O2 → 2Ag + H2O + O2}$$

Given that the majority of silver compounds/salts are photosensitive, they will be decomposed when exposed to regular light, especially to the ultraviolet light. Even if you have other substances/compounds as impurities in hydrogen peroxide solution which can react with silver oxide you still will probably end up with elemental silver ($\ce{Ag}$).

  and some water?

Sounds like a joke. The available hydrogen peroxide solutions are approx. 3% (drug store) or 30% (most chemical labs), thus, the solutions already have a plenty of water, at least 70%. Do you really need more water?


Why is the rate of a reaction proportional to the concentrations of reactants raised to their stoichiometric coefficients?

A late answer...

In a molecular approch to chemical reaction rates we asume that the molecules move randomly in solution and have a velocity randomly assigned according to the Boltzmann distribution and that the chance of being in one position in solution is the same as being at any other, i.e. the molecules are in thermal equilibrium at some temperature T.

The reaction rate for the reaction $\mathrm {A+B \rightarrow}$ can be found by considering the number of collisions per second and then this number reduced by the chance that a reaction occurs on collision. To arrive at a rate the chance that a randomly chosen combination of molecules of type A and B at time t will react in the next infinitesimal time interval $t+\delta t$ and has not reacted up to t has to be calculated. As a result the total number of collisions /second between $N_A$ molecules of type A and $N_B$ of type B is found to be $\sim N_AN_B\sigma_{AB}$, where $\sigma_{AB}$ is the cross sectional area for a collision, and this expression produces the rate  when  multiplied by the probability $p_{AB}$ of A and B reacting, i.e. rate $\sim N_AN_B\sigma_{AB}p_{AB}$. The probability of reaction $p_{AB}$ depends critically on the activation energy.

The important point is, however, that the  rate is proportional to the product of the numbers of molecules of type A and B which in constant volume is the same as being proportional to the product of concentrations. That it also is proportional to the collision cross section and the chance of reacting is also physically realistic. The rate constant is proportional just to the probability of reaction and the cross section for collision.

Thus the origin of the product of concentrations in the rate expression is found in the chance that a molecule of type A will collide with that of type B in a small time interval $t, t +\delta t$.  Thus if we ignore the fact that there is variation due to changes in experimental conditions, such as temperature, the electronics of the detectors used to measure the reaction etc. perfect kinetics would be measured, i.e $dA/dt$ or $ dB/dt$ vs. time would be noise free.

If we now consider the random nature of the collisions between molecules a slightly different approach has to be taken. If a small volume of the solution is considered and the rate calculated in that region, because the number of species entering and leaving in any given time period varies, the rate expression is also potentially subject to variation. This happens simply because of the random nature of the motion of molecules in a solution or gas. In this case the rate of change of a species X should be written in terms of its average (or mean) value represented as $\langle X(t)\rangle$;

$$ \frac{d\langle X(t)\rangle}{dt} = \langle f\times X(t)\rangle $$

where $\langle f\times X(t)\rangle$ represents some rate expression such as $k_1\langle X(t)\rangle + k_2\langle X(t)\rangle^2$ etc. except that the number of  $X$ are represented as their average values. Clearly the $X$ are always positive values. These  values have a variance (square of standard deviation) and so we would expect that different experiments give slightly different results because of this. (This is the approach taken by Gillespie, see 'Markov Processes' publ. Academic Press 1992)

In a bimolecular reaction such as $X + X \rightarrow $ with rate constant k the rate expression is

$$ \frac{d\langle X(t)\rangle}{dt} = k\langle \;X(t)(X(t)-1)\;\rangle/2 $$

The term $\langle X(t)(X(t)-1)\rangle /2$ arises becaus of the number of distinct X-X pairs in the reaction volume at time t. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that X is large compared to $1$ and so $\langle X(t)(X(t)-1)\rangle \rightarrow \langle X^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle X^2(t)\rangle \sim \langle X(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $X(t)=\langle X(t)\rangle$ which means that $X$ is a deterministic process, so that the randon nature of the numbers of $X$ entering into the reaction volume is always the same, which is not what was assumed at the outset and seems to fly in the face of what we know about how molecules move in solution, i.e. their random diffusional nature.

However, in any normal kinetic experiment the number of molecules we measure is vast, $\langle X(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the  number of molecules is tiny and the approximation made above is valid.

$$ \frac{\sigma_X}{\langle X \rangle} \sim \frac{\sqrt{\langle X \rangle}}{\langle X \rangle} \sim \frac{1}{\sqrt{\langle X \rangle}}\sim 10^{-9}$$

Even if there are as few as 1000 molecules the realtive fluctuation  is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.

Notes :

(A)  To see that the variance is zero use the definition $\sigma^2_X = \langle X^2 \rangle -\langle X \rangle^2$ where $\sigma$ is the standard deviation)

(B) There are other situations where the rate equation approach breaks down in macroscopic systems and that is when they are sufficiently non-linear, one example is the Schlogl system of equations, which for certain values of concentrations and rate constants becomes bi-stable;

$$B_1+2X \rightleftharpoons 3X; \;\; B_2 \rightleftharpoons X$$


Mixing Supplements - Will the Distributions Remain?

It's hard to maintain homogeneity for the mix of powders. The bigger the batch, the less homogeneous it's going to be. Ideally you want to disperse solid ingredients; there are several techniques to do that. Two of the most popular ones are:

dissolving all the components in a non-reactive solvent and they dry out or precipitate the powder;
using a ball mill.

With supplements, however, it's somewhat trickier, as they may react with each other over time, with the solvent, or be contaminated in grinder. Also they are often granulated for a good reason: this reduces the contact with atmosphere and prolongs shelf life.

If you decide to avoid the dispersion part, then due to Brazil nut effect the largest powder particles will remain on top of the batch, the smallest – on the bottom. This is especially noticeable when the batch has been shaken, e.g. after being transported in a car trunk, or just being kicked a few times lying on the kitchen's floor.

All in all, I wouldn't recommend doing this. The probability of suffering from an imbalanced diet is too high. There are some multivitamin and mineral complexes designed for humans, maybe look for something similar for your pet.


Constitutional Isomers or Stereoisomers?

Correct, no constitutional isomers, but there are stereoisomers (different substituents in the apical and equatorial positions).  Just for the record, what is your fifth substituent?


How can an inexperienced chemist determine the chemical structure of a molecule?

So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure?

Yes.

I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars.

Then probably no (at least not with the same accuracy you would if there weren't these constraints).

Mass spectrometry and nuclear magnetic resonance are the two most powerful analytical tools to identify completely unknown substances. This doesn't mean it's as trivial as putting your purified substance in the equipment and finding its structure. For reliable results you need a purified substance, an optimized experimental routine and of course the instrument properly set up.

In fact, before trying to determine the structure of the substance, you must purify it, and that alone is a challenge. As I'm sure you are aware, the plant contains thousands, millions of different molecules. How are you going to extract something you haven't defined? The first step would have to be defining what is it that you're looking for.

Let's consider you're looking for whatever makes plants green:

You could macerate the plant, add some water, filter the mixture, check the colors of the filtrate and of the residue, and you would notice most of the green wasn't extracted. You would find different results with different solvents though, and a better option would be ethanol. Even then, it wouldn't mean you have a pure chlorophyll solution, so several other steps would probably be required to purify the substance*. Generally speaking, you would use some kind of chromatography.

For your project, I suggest you try to first precisely define what is it you're looking for. The substance that gives the plant its color? Its smell? Its pharmacological effects? You must use this property to analyze your experimental results. You can visually inspect the color, you can smell extracts, and perhaps test physiological effects (this is obviously dangerous and shouldn't be done unless you really know what you are dealing with!). Use the target substance properties' to decide what's the best way of extracting and purifying it.

Then, considering you did purify something, without the high-tech analytical tools mentioned, there are some experiments which can indicate what you have, but they require a reference substance and usually can't be used to say that a substance is the same as another, but rather that a substance isn't the same as the reference. For example, you can compare boiling/melting points and/or thin layer chromatography retardation factors, both which can be done at home home (to some extent), and also UV/Visible light absorption/emission, which might not be trivial to do at home but require instruments that are a lot easier to find in chemistry teaching labs.

For precise tips on the extraction and identifications, it would be necessary to know what you're looking for.

It's also important to note that there's a lot you can do besides determining a chemical structure. There are several properties of a substance you can study in order to characterize it. Some examples are solubility (in different solvents, at different conditions (temperature, pH, salinity)), density, reactivity (in face of pretty much anything you find at home) and physical properties such as texture, color, smell, etc.

Finally, do not think I'm saying you should give up, your wish to discover what something is made of is what made chemistry a science, but you need to have a clearer problem to be solved. Decide precisely what it is you're looking for (in terms of macroscopic properties, chemical group or something like that), and then research (and ask!) what could you do about that without a full lab.

*Chlorophyl is actually a group of substances which have a critical role in photosynthesis, and are, of course, green.

In summary:

You already have your sample. Now choose your analyte, the molecule(s) you want to study.
Do research. Look for information on how people usually study this (group of) substances. You'll have better luck looking in education-related sources, which usually have simpler experiments.
Once you've found something that fits your interests and available material, do experiments. Observe the results. Change parameters, do them again, try to understand what's happening.
If you can't explain the results, do research, check your references, and of course, ask here. We'll be glad to help.


On what principle do electric alcohol detectors work?

I believe you are referring to breathalyzers. When you breathe into a breathalyzer, any ethanol present in your breath is oxidized to acetic acid (the main component of vinegar apart from water) at the anode (there is a diagram below).  At the cathode atmospheric oxygen is reduced. The overall reaction is the oxidation of ethanol to acetic acid and water. The electric current produced by this reaction is measured by a microprocessor and displayed as an approximation of overall blood alcohol content (BAC).

This website can give you more information.

Hope this helps!


Reaction of alkyl bromide in presence of piperidine

@User6376297 raised a question:

  "Could the cyclopropane ring behave like a double bond, making this the analogue of an allyl bromide, and yielding $\ce{R2NCH2CH2CH=CH-tBu}$ by attack on a cyclopropyl $\ce{CH2}$ and ring opening?" Especially given the steric hindrance near the carbon bearing the leaving group.

Indeed, that is the process going on in this case when the reaction was taken place in piperidine either at refluxing conditions or at room temperature. When (1-bromo-2,2-dimethylpropyl)cyclopropane (the alkyl bromide in question) was reacted in refluxing piperidine, it has yielded $88\%$ of (E)-1-(5,5-dimethylhex-3-en-1-yl)piperidine as a major product, which corresponds to the homoallylic analogue of the well-known SN2 process observed with allylic halide.

In addition, $8\%$ of 1-(1-cyclopropyl-2,2-dimethylpropyl)piperidine was also yielded as a minor product (Ref. 1 & 2). The authors claimed the formation of the minor product was not compatible with simple SN2 displacement due to the great steric hindrance imposed by the neighboring $\ce{t-Bu}$ and cyclopropyl groups, but the tight ion-pair mechanism (not separated by solvent) may explain the presence of SN2 product very well.

The percent yields of SN2' and SN2 products for the reaction at $\pu{25^\circ C}$ had been $73\%$ and $8\%$, respectively.

References:

Reaction of amines with cyclopropylcarbinyl halides: SN2' or
solvolysis?: J. Org. Chem., 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).
Homoallylic substitution reaction of piperidine with 1-bromo-1-cyclopropylalkanes: Tetrahedron Lett., 1983, 24(46), 5031–5034 (https://doi.org/10.1016/S0040-4039(00)94033-4).


Product operator analysis for CHn groups in HSQC and HMQC

Your analysis is generally correct. In the case where all the protons are equivalent, you don't need to worry about proton–proton coupling. The main thing you need to worry about is that when you have a term such as $2I_{1z}S_x$ you will get evolution of different proton–carbon couplings, represented by (e.g.) the Hamiltonian $2\pi J_{IS} I_{2z}S_z$. This could, in theory, generate more complicated product operators such as $4I_{1z}I_{2z}S_y$. You probably already know that

$$S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta) \, S_x + (\sin \theta)\, 2I_{2z}S_y$$

where $\theta = \pi J_{IS}\tau$. A term on a different nucleus, such as $I_{1z}$, does not evolve under this coupling, so you can "multiply" by $2I_{1z}$ on both sides* to get

$$2I_{1z}S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta) \, 2I_{1z}S_y + (\sin \theta)\, 4I_{1z}I_{2z}S_y$$

Thankfully you do not need to worry about this, because the only times that these can evolve are during the $t_1$ periods of both experiments, when there is magnetisation present on carbon. Conveniently, both experiments have a proton 180° pulse in the middle of $t_1$ which refocuses all C–H couplings anyway. Outside of the $t_1$ periods the magnetisation of interest is single-quantum on proton, so the only coupling that evolves is that one proton–carbon coupling, which behaves in exactly the same way as for a CH system.

For a more detailed treatment I suggest consulting Chapter 12 of James Keeler's Understanding NMR Spectroscopy, 2nd ed. (2010).

However, if the protons are not equivalent, then you will start to get evolution of H–H homonuclear couplings, which can't simply be refocused by a 180° pulse. For the HSQC this is less important as it just leads to a reduction in intensity from magnetisation that goes down non-useful coherence pathways. Because the magnetisation during $t_1$ is single-quantum carbon, it does not evolve under homonuclear couplings. However, for the HMQC, the multiple-quantum coherences present during $t_1$ will evolve under homonuclear couplings, which means that you will get multiplet shapes in the indirect dimension after Fourier transformation. In practice it is difficult to resolve these splittings and so it is just manifested as line broadening.

There are also considerations to be made when using a sensitivity-enhanced HSQC sequence, but I assume those are not relevant for your current situation. The interested reader is directed to J. Biomol. NMR 1994, 4, 301–306.

* To be more precise, this is because $I_{1z}$ commutes with the Hamiltonian. So, it also commutes with the unitary propagator $U = \exp(-\mathrm i H\tau)$, and we can write

$$2I_{1z}S_x \xrightarrow{H\tau} U(2I_{1z}S_x)U^\dagger = 2I_{1z} \cdot US_xU^\dagger$$


What exactly happens to mass during deuterium fusion?

Fusion of H-2 does NOT give He-4.  It gives ~50:50 H-3 (1.01 MeV) + p (3.02 MeV) or He-3 (0.82 meV) + n (2.05 MeV).  Either way, the missing "mass" is the mass equivalent of binding energy, E/c^2.  Binding energy is negative.  A bound system masses less than its separated constituent parts.  You must input energy (that being mass-equivalent) to unbind it.

Fusion converts weakly bound nuclei into more strongly bound nuclei.  The net difference in binding energies comes out as the kinetic energies of the product nuclei and particles.  Count neutrons and protons going in versus neutrons and protons coming out.  If one convetrs to the other, there must be elecron neutrinos or electron antineutrinos emitted to balance the books for energy, momentum and lepton generation number.

http://en.wikipedia.org/wiki/Nuclear_fusion
   Middle


Better functionals to work with transition metals

This question is like the snake that eats its own tail. There is no functional good enough for metals. To my limited knowledge, fitting too many parameters -which happens with metals- feels like DFT is not first principle anymore and could be called "empirical" (which is also not correct). You allow a number of parameters to to be fit(1,2,more!!) to a specific system. Those dozens of parameters -at the end- may give you desirable results but you "trained" the set of energies according to your will which is not that realistic. You train your dragon not in the jungle but in a cage. The result is never the same. This "fixing" not only contradicts the spirit of DFT but is also a weak form of empirism. It is like adding more "dark force" to the community. Imagine in what extend many of those "good", modern functionals affect research daily. To end, maybe other comments will come up with accurate functionals about Ti or Zn with papers in good journals etc but without being against DFT, I would just keep a small basket.


How did scientists discover the concepts of anode and cathode without knowing about electrons?

Electricity is a macroscopic phenomenon. It does not depend on the notion of an electron.

In 1733 Charles François de Cisternay du Fay discovered the existence of positive and negative charge. In 1800, after the invention of the voltaic pile, the first electrolysis experiments that showed the effect of electric current on water have been made.

In 1833 Michael Faraday defined the terms cathode and anode:

  The anode is therefore that surface at which the electric current, according to our present expression, enters : it is the negative extremity of the decomposing body; is where oxygen, chlorine, acids, &e., are evolved; and is against or opposite the positive electrode. The cathode is that surface at which the current leaves the decomposing body, and is its positive extremity ; the combustible bodies, metals, alkalies, and bases, are evolved there, and it is in contact with the negative electrode.

His definition does not involve particles of any type. In fact Richard Laming firstly hypothesized the concept of an indivisible quantity of electric charge in 1838.

You may want to study https://en.wikipedia.org/wiki/History_of_electrochemistry for further information.


What are common procedures for lab absorption spectroscopy of gases?

Non-dispersive infrared absorption spectroscopy is commonly used to measure $\ce{CO}$ and $\ce{CO2}$ (and other gases, like methane and $\ce{NO2}$). The absorption overlap from water interferes with $\ce{CO2}$ measurements in particular but various algorithms can be used to mitigate this.

$\ce{NO2}$ is more commonly measured by chemiluminescence because of its high sensitivity, but FTIR and other techniques are used for higher levels (i.e. exhaust pipe and stack emissions).


Is water committed to vapor state at boiling point or after?

At equilibrium, not only are molecular bonds of the liquid breaking to form vapor.  There are also molecular bonds of the vapor reforming to produce liquid.  At equilibrium, the rates of the two processes are equal.  But, if you add heat to the liquid, the rate of breaking bonds exceeds to rate of reforming bonds (i.e., the temperature and vapor pressure of the liquid increases slightly), and some of the liquid changes to vapor.  But once you stop adding heat, equilibrium is re-established.


How could alcohol be made into a powder?

Cyclodextrins, which comprise a class of cyclic carbohydrates, use their properties of being hydrophobic within their ring(s), yet hydrophilic outside. Alcohol can be put inside these rings. Then, the entire molecule is still capable of dissolving in water, releasing the encapsulated alcohol molecules. (src)

This popsci link has one way of making edible powdered alcohol. A nice casual step-by-step tutorial, actually...

Still you want to maintain a balance between maximizing alcohol (or do you?) and having the powder "feeling wet" due to high absorption of alcohol.

Also this may be a relevant Wikipedia article.


Corey-Schmidt oxidation of primary alcohols

As stated in the previously posted answer:

  This oxidation into
  carboxylic acids succeeds when the intermediate aldehyde equilibrates with a liberal proportion of hydrate; that is, when the intermediate aldehyde belongs to the aliphatic kind and is not highly hindered. The water necessary for the formation of the intermediate aldehyde hydrate may proceed from the decomposition of PDC.$^{[1]}$

Though the text does state that water may form from decomposition of the dichromate anion, since DMF is so hygroscopic as a result of its high polarity ($\epsilon_\mathrm{r}=38^{[2]}$), I would not be surprised to learn that little to no over-oxidation occurs in completely dry DMF under an inert atmosphere.

This is only speculation though, and the dichromate anion may very well be decomposing to produce water in the solution. As such, I will propose a reaction mechanism for this decomposition. The first step can only occur when pyridinium dichromate is well solvated and protons are free to be transferred between the species, which occurs in DMF but not in DCM:

$\hspace{.6cm}$

This produces chromic acid, another chromium(VI) oxidizing agent. Unlike dichromate oxidation, however, chromic acid oxidation actually produces water as a byproduct, which is then free to hydrate the resulting aldehydes:

$\hspace{.8cm}$

$[1]$ Tojo, G.; Fernández Marcos. Oxidation of Alcohols to Aldehydes and Ketones: A Guide to Current Common Practice; Springer: New York, NY, 2006; p 28.
$[2]$ Wikipedia: Solvent


Does the Hydrogen/Chlorine reaction in light proceed instantaneously?

It depends on how much UV is used, and how it's distributed... from a point source, the flame front would spread as if from an automotive spark plug. In fact, lasers are being tested as a means of ignition in Otto-cycle engines because the flame front can be better controlled than from a spark.

If evenly spread over the surface, as in laser fusion attempts, then the wave-front would propagate from the exterior inwards. Theoretically, using fiber optics inside the container to distribute and delay the beam equally, the explosion could propagate as fast as desired, with the caveat that the delay would still be half the largest space between tips.

This means, for example, that a one-cubic-meter container could be made to explode in far less than 1/3*108 seconds, which would be the minimum if it were to be ignited at the edge. Though not truly simultaneous throughout the container, one can approach any finite speed, even, apparently, greater than light. This is similar to other phenomena that exceed C, such as the crossing point of scissors or the phase velocity of a wave.


The first step is to figure out a reaction equation.

[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.

$\ce{Na + H2O -> NaOH + H2}$

We need to balance this.

$\ce{2Na + 2H2O -> 2NaOH + H2}$

$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).

$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.

--

Mathematically...

$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$

and

$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$

Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.

For every 2 moles of water used, 2 moles of sodium hydroxide are formed.

$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$


Is it correct to write Na2B2O4 as 2NaBO2?

No, $\ce{Na2B2O4}$ can not be written as $\ce{2NaBO2}$. $\ce{Na2B2O4}$ means 1 mole of $\ce{Na2B2O4}$ and $\ce{2NaBO2}$ means there are 2 moles of $\ce{NaBO2}$. Both are different compounds and have different molar mass.

Actually in real $\ce{2NaBO2}$ should be written as $\ce{(NaBO2)2}$. Now, both $\ce{Na2B2O4}$ and $\ce{(NaBO2)2}$ are same.


Visible Light Spectrophotometer Recommendations

Disclaimer: I'm not affiliated with any company

However,
Ocean Optics cuvette spectrophotometers are a good choice

This model is used in some undergraduate labs and is good because the cuvette holder is built in (i.e. it doesn't require a fiber optic cable).

As long as the optics don’t get jumbled, the prism detector mechanism work well and reliably for many years, although it says to get it checked each year.

If you're on a tight budget I'd recommend looking at a DIY spectrophotometer. In particular the RasberryPI designs seem relatively simple

The Hackaday website has a blog dedicated to DIY spectrometers.


Identification: was this gallium or mercury from broken thermometer?

Gallium, and Galinstan have the ability to wet glass, while mercury does not.

So it's most probably gallium (or Galinstan).


How to determine Delta H from the heat of combustion?

Are you supposed to determine $\Delta_rH$ for the reaction (which is $\Delta_cH$ for the combustion of 2-propanol) or are you supposed to determine $\Delta_fH$ for the formation of 2-propanol?

  If the first case, it might be hard.

The change in internal energy $U$ is equal to the sum of the heat transferred $q$ and the work done $w$. If we assume that only pressure volume work is done, then:

$$\Delta U=q+w = q-p\Delta V $$

The negative sign in front of $p\Delta V$ is to ensure that a decrease in volume is work being done on the system (system gains energy).

As a thermodynamic potential, internal energy is only easy to use if the pressure is constant (and $\Delta V$ is measurable), which is not true in a bomb calorimeter, or the volume is constant (so $\Delta V =0$), which is true in a bomb calorimeter.

At constant pressure: $\Delta U=q$.

Enthalpy is defined such that
$$\Delta H = \Delta U + \Delta(pV) = q - p\Delta V + p\Delta V + V\Delta p $$
$$\Delta H = q + V\Delta p$$

Thus, enthalpy change is only easy to use if the volume is constant (true in a bomb calorimeter) and the pressure change is knowable (you were not given that data) or if the pressure does not change (so that $\Delta p=0$).

  The second case is simpler.

If you were asked to determine $\Delta_f H$ for the formation of 2-propanol, you probably are encouraged to assume that for the combustion reaction $\Delta_rH\approx \Delta_cU$ and go from there. Then, you will need a balanced equation (2-propanol + oxygen forms what?) and some standard enthalpies of formation for the products.

$$\Delta_rH \approx \sum_{products}{\Delta_fH}-\sum_{reactants}{\Delta_fH}$$


Why is AlF3 soluble in anhydrous HF only in the presence of KF?

$\ce{AlF3}$ is insoluble in the anhydrous $\ce{HF}$ because the $\ce{F-}$ ions are not available in intermolecular hydrogen bonded $\ce{HF}$, but it becomes soluble in the presence of $\ce{KF}$ due to the formation of soluble complex, $\ce{K3[AlF6]}$.
$$\ce{AlF3 + 3KF -> K3[AlF6]}$$


Firstly, you need to understand what it means when a system is in equilibrium. Equilibrium is the state of a system where the rate of the forward and reverse reaction are equal.

Note: this doesn't mean that the reaction completely stops, all this means that the concentration of the reactants and products remain constant.

A simple example of this can be shown in an experiment where a cube of sugar is placed in a saturated solution of sugar. If you leave it for a few days and come back, you will find that the shape of the sugar cube has changed. Instead of being a smooth, perfect cube, it is a irregular shape. However when if weigh the solid sugar, you find that it will weight exactly the same as the original sugar cube. Now this can be explained by the fact that the solid sugar is in equilibrium with the dissolved sugar in the saturated solution. Despite the forward and reverse reaction being equal (as no more sugar will dissolve) the reaction is still occurring. At a microscopic level, glucose molecules in the sugar cube are dissolving into the solution while dissolved sugar molecules precipitate out onto the sugar cube. However where they precipitate on the cube is random, hence they don't form a perfect cube. This is why the sugar cube changes shape but has the same original mass.

To do this question, it is important that you understand the concept that the reaction still occurs when a system reaches equilibrium. Now, lets look at each graph separately (remember, when we are looking at these graphs, we are only interested at what is happening at the end, what happens at the start is irrelevant):

1: Here the forward and reverse reaction are equal, so from our definition of equilibrium, this graphs shows a system in equilibrium. You should note that despite the system being at equilibrium, the reaction is still going

2,3: Here in both graphs the forward and reverse reaction are not equal. Therefore the system can't be in equilibrium. This is because in graph 2, the rate of the backward reaction is greater than the rate of the forward reaction. Hence the concentration of the products will be decreasing while the concentration of the reactants will increase. Vice verse for graph 3

4,5,6: Here, since the concentration of the reactants and the products are not changing over time, this means that the forward and reverse reaction is the same. Therefore, the system is in equilibrium

7: In this graph, the concentration of the products is increasing while the concentration of the reactants is decreasing. Hence the forward reaction must be greater than the backward reaction. Therefore, the system isn't in equilibrium.

So to recap the ideas that we heave just learnt, here is the  2 main things to check to work out if a system is in equilibrium or not:

Is the forward and backward reaction equal? if they are equal, this means that the system is in equilibrium. If they aren't equal, the system isn't in equilibrium
Is the concentration of the products and the reactants constant over time? if they are constant, the system is equilibrium. Note, the concentration of the products and reactants don't have to be equal. If concentration of the product and reactants are aren't constant, the system isn't in equilibrium.


Laboratory setup for the distillation of sulfuric acid

Just on the purification of sulphuric acid, well the one he gets has a dark green tinge, and what is in the sulphuric acid is anticorrosive inhibitors, etc, and they are all organic in nature, and the simple heating up a beaker of sulphuric acid, and before it gets to hot, slowly add 100 ml of 3% grocery store bought food grade hydrogen peroxide, and stir whilst slowly heating, well before long the acid will be crystal clear, any fumes are the skeleton structures of the organic additives, and concentrating it back to 98% sulphuric acid is either density determined, temperature, to insure water has been removed hence both will get a much purer, crystal clear sulphuric acid. Anything organic will be consumed by the peroxide and what doesn't will turn into a gaseous product that evaporates of. The product described has served his associates and himself, for many years.


Questions about combustion and isotopic abundance


What chemical equation proves that KHCO3 can extinguish a fire?

According to this Wikipedia entry:

  Decomposition of the bicarbonate occurs between 100 and 120 °C (212
  and 248 °F):

  $$\ce{2 KHCO3 → K2CO3 + CO2 + H2O}, \Delta H>0$$

So, you produce a nonflammable powder and two nonflammable gases, and you absorb heat from the fire in an endothermic reaction.

The picture here is more complex than just a chemical equation however. The $\ce{KHCO3}$ also acts as an energy-absorbing material and provides a solid surface on which free radicals can be destroyed.


They say that enthalpy of reaction is the heat when at constant temperature and pressure.

They should not say that. At least not without saying in the absence of non-PV work. And in a closed system.

  So if [...] heat is TdS [...]

You should not say that without specifying a reversible (near equilibrium) process in a closed system where dS refers to the surrounding:

$$q_\mathrm{rev} = T dS_\mathrm{sur}$$

The sign of $q_\mathrm{rev}$ is from the perspective of the system, i.e. if there is heat transfer from system to surrounding (positive value), the entropy of the surrounding will increase (we don't know what happens to the entropy of the system because there is more going on in the system whereas the only change in the surrounding is that it got heat transferred to it).

On the other hand, all the thermodynamic quantities in your formula refer to the system. By the way, the enthalpy and the entropy of a reaction are not linked, there is no general formula that lets you infer reaction enthalpy from reaction entropy without knowing other quantities.


When are the prefixes bis, tris etc. used in nomenclature of complexs?

For naming coordination compounds, the multiplicative prefixes ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., are used in the case of simple ligands, for example:
$\ce{[PtCl4]^2-}$  tetrachloridoplatinate(2−)
The alternative multiplicative prefixes ‘bis()’, ‘tris()’, ‘tetrakis()’, ‘pentakis()’, etc. are not only used for bidentate ligands. They are generally used with composite ligand names or in order to avoid ambiguity, for example:
$\ce{[Pt(PPh3)4]}$  tetrakis(triphenylphosphane)platinum(0)
The corresponding section in the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

IR-9.2.2.2 Number of ligands in a coordination entity
Two kinds of multiplicative prefix are available for indicating the number of each type of ligand within the name of the coordination entity (…).
(i) Prefixes di, tri, etc. are generally used with the names of simple ligands. Enclosing marks are not required.
(ii) Prefixes bis, tris, tetrakis, etc. are used with complex ligand names and in order to avoid ambiguity. Enclosing marks (…) must be placed around the multiplicand.

The use of multiplicative prefixes is also explained in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
Composite ligand names include ligand names which are substituted or which themselves contain multiplicative prefixes or locants.
For example, the normal prefix is used in ‘diammine’ $(\ce{(NH3)2})$, but the alternative multiplicative prefix is used in ‘bis(methylamine)’ $(\ce{(CH3NH2)2})$ to make a distinction from ‘dimethylamine’ $(\ce{(CH3)2NH})$.
Note that the use of the alternative multiplicative prefixes is not limited to coordination compounds. They are generally used for compositional names in inorganic chemistry when a simple multiplicative prefix may be misinterpreted. For example,
‘tris(iodide)’ is used for $\ce{(I- )3}$ (rather than ‘triiodide’, which is used for $\ce{I3-}$);
‘bis(phosphate)’ is used for $\ce{(PO4^3- )2}$ (rather than ‘diphosphate’, which is used for $\ce{P2O7^4-}$.


What is the volume occupied by the atoms and the edge length in these cubic lattices?

This is the case of CCP (simple cubic closed packing) and FCC (face centered cubic closed packing)
For CCP
Imagine sitting in a cubical room. Now place some big spheres on each of the vertices of the room such that the centre of sphere and the vertices coincide.
You will see that some part of sphere goes outside the room itself. In chemistry, this is called sharing.
If your room is surrounded in all directions by other rooms, you share one sphere with 7 other rooms. (top 4 and bottom 4)
Thus a single particle gets shared by 8 rooms. The part in your room is 1/8
And the number of such particles are 8 (as cubic room has 8 vertices)
Net particles in your room $= \frac{1}{8} \cdot 8 = 1$
So in CCP the space occupied by all eight particles is basically equal to that of a single particle.
Now as assumed in the question, the particles are tightly packed, the
Radius of sphere equals half of the edge length. In other words edge length is twice the radius of the sphere.
$$\pu{Packing efficency = \displaystyle \frac{\text{Volume of sphere}}{\text{Volume of cube}}}$$
$$= z \cdot \frac{\frac{4}{3} \pi r^3}{(2r)^3} $$
(Here $z$ is net particle contribution; for the current example it is equal to 1)
Calculating this you'll get packing efficiency $\approx 52.4%$

For FCC
Now imagine putting one more sphere in the room, but this time, at the centers of the face. (Do this for each face). Now the particle in center gets shared only by 2 rooms. Yours and the room which is opposite to you.
Therefore, each particle has net contribution of half.
Since a cube has six faces, the net contribution of all spheres at the center the of sphere are
$$\frac{1}{2} \cdot 6 = 3$$
Add the already existent corner particles to it (whose net contribution is 1 as we know)
Thus the total number of particles in FCC are $3 + 1 = 4$

If you're good at imagining things you might have observed that now spheres won't fit in until you reduce their sizes. So we must falsify the assumption that twice the radius of sphere is the edge length. Spheres have shrunk.
I want you to imagine the face of the room now.
There is little space between two spheres on corner but not between the center sphere and corner spheres. Draw a diagonal line in the sphere passing through the diameter of central sphere. Now using Pythagorean theorem:
$$ \pu{(\text{Edge of cube})^2 + (\text{Edge of cube})^2 = (\text{Diagonal of cube})^2}$$
Let edge be 'A':
$$(A)^2 + (A)^2 = D^2$$
$$2(A)^2 = D^2$$
Since the center spheres have no space with corners ones. Therefore
4(radius of sphere) = Diagonal length
$$\sqrt{2}A = D$$
$$\sqrt{2}A = 4r$$
$$\sqrt{2}A / 4 = r$$

There you go. You've found the radius of sphere. Now put it in packing efficiency formula and take $z$ = 4
You will find it somewhere near 72%
On a sidenote: Observe that simple cubic has only 1 sphere which takes up nearly 50% of space of cube. While face centred has 6 + 8 = 14 smaller spheres which take up nearly 70 % of all space in cube. It's like comparing apples and bananas.
You might think that FCC is better than CCP as it uses space judiciously. But since in CCP, the particle size is large therefore CCP is best suited for particles of that size and FCC for smaller sizes.
Although we can discuss hexagonal packing form with CCP.


What's the mechanism for this isotopic substitution reaction?

To answer this question you should understand the kinetic isotope effect.

Since you deliver no real data about reaction rate constants I assume that the dissociation of $\ce{HF}$ is faster (has a larger $k_\text{dis}$) than the dissociation of $\ce{DF}$. I postulate the following mechanism:

$\ce{HF + D+ <=>[k_\text{dis}^\ce{HF}] H+ + F- + D+ <=>[][k_{\text{dis}}^\ce{DF}] DF + H+}$

Essentially it is a pretty simple reaction mechanism, but it tends to forming $\ce{DF}$ over $\ce{HF}$ because the deuterium dissociates more slowly.


It seems as though your issue with the formula arises due to the fact that there is a feeling of not knowing what is considered to be factors that are internal or if they are external factors. To prove the formula, the method would be as follows.
The formula for the given reaction would be:
$$\ce{C12H22O11 + H2O ->  C6H12O6 (glucose) +C6H12O6 (fructose)}$$
Drawing a modified version of a RICE table with optical rotation, and the case at t=$\infty$, we get:
\begin{array}{|l|c|} \hline
\text{R} &\ce{C12H22O11   &C6H12O6(glucose)  &C6H12O6 (fructose)} &\text{Optical Rotation} \\ \hline
\text{I} & n & 0& 0 & \alpha_{s}n+\mathrm E \\
\text{C} & -x &  x&  x & - \\
\text{E} & n-x &  x & x & \alpha_{s}(n-x)+ \alpha_{g}(x) + \alpha_{f}(x)+ \mathrm E \\
\infty & 0 & n & n & (\alpha_{g}+\alpha_{f})(n) +\mathrm E \\ \hline
\end{array}
Now, for a general first order reaction, the formula would be:
$$k=\dfrac{2.303}{t}\log \frac{n}{n-x}$$
Therefore, we need to find the value of $\dfrac{n}{n-x}$. Using the given data, we get:
$$\frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}=\frac{n}{n-x}$$
Therefore the formula given is true, even considering an external factor. You consider glucose and fructose to be internal.
$$k=\dfrac{2.303}{t}\log \frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}$$


Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it?

Honey is indeed a complex mixture containing more than hundred compounds.

As for Wikipedia and depending on the point of view it is

a supersatured liquid solution
a viscous supercooled liquid (in the sense that it can get so viscous as to appear solid, without affecting its status of being a supersatured solution, and undergoes glass transition).

https://en.wikipedia.org/wiki/Honey#Physical_and_chemical_properties

The facts that you describe in the question are less surprising if we consider that:

with respect to sugars crystallization, of which at least  there are two different ones, glucose and fructose, the rest of the non-sugar components must be considered to be impurities
impurities, even in traces, often hamper the crystallisation of a compound, even in simple mixture of a single compound and the above traces. This is common after organic synthesis, in which often a viscous "oil" is attained that might crystallize only upon prolonged storage or a careful removal of the disturbing trace compound(s)
finally and most important, the solubility of sugars in water is very high, and very sensitive to temperature. For instance, at room temperature glucose is already soluble in the reason of 90 g per 100 ml of water, that means a saturated solution already contains about fifty percent weight per weight of sugar

A table is here (I didn't cross check the values):

http://www.mpcfaculty.net/mark_bishop/supersaturated.htm

All this, viscosity included, makes the attainment of a supersatured solution particularly easy, as in the kitchen in the case of sucrose:

https://sciencing.com/make-supersaturated-solution-sugar-6199355.html

As such, heating crystallized honey does indeed dissolve the sugar, and a supersaturated solution is attained upon subsequent cooling.