Estimation of ground state energy from wave function

The variational method uses a linear combination of functions as an estimate of the true wavefunction. It works by adjusting coefficients in the linear combination to minimize the energy of the trial wavefunction. This is the best estimate of the true wavefunction using the restricted basis set. The variational principle states that adding more (new) basis functions to the linear combination always improves the estimate.

In your problem (or according to your instructions), you don't really optimize anything (except for the last comparison), because the normalization condition fixes the value of the $c_i$ coefficients, but you can estimate which of the two functions appears most like the true wavefunction, as this should return a minimum energy.

The ground state solution has the form $$\psi \propto sin(\pi x)$$
when x is in units of the box length, with bounds at $x=0$ and $x=1$ (and null wavefunction outside these bounds). This function has nodes at $x=0$ and $x=1$ and is concave with a single maximum at $x=\frac12$. By evaluating the functions at the bounds, in the midpoint and at one or more other points you should get an idea about the shape of each function and thereby which are better. It would of course be easier to plot them. Consider to the extent possible symmetry and whether bounds are satisfied (which they must for the solution to be at all physically reasonable - remember that the energy blows up if the wavefunction samples points out of bounds.

The last comparison is a "trick question". There you have to take into account the variational principle which is mentioned in the first paragraph of this answer.


Determining the reaction time of water synthesis by use of differential equations

How do I determine the reaction speed?

The rate of reaction can be determined experimentally or theoretically, the latter case requiring that a reaction mechanism be known. Proposing a reaction mechanism requires knowledge of the reaction at multiple regimes and some guesswork. Evidence for a certain reaction mechanism is usually indirect.

Theoretically speaking, once the reaction mechanism and rate constants are known, we can write down a coupled system of kinetic rate equations and solve them simultaneously to find the rate of reaction.

  I know you need a value, k, and order, which is only gained through experimentation, but how do the differential equations play in? How do I plot the result?

The differential equations govern the reacting system, and their solution completely specifies the system, enabling you to find the rate of reaction. You would then simply plot, say, the rate of reaction vs. time.

  How do I proceed? I now have this function but I am not sure what to do with it. I do not have the means to observe the reaction and figure out its k-value or order.

I don't believe you can proceed, because your initial assumption was wrong. The combustion of hydrogen is far more complicated than a first-order reaction, which you have assumed; you should not, therefore, expect that your first-order analysis should describe your system well.

  Is this reaction a known one, though? Is the order and k-value something you can find on the Internet?

It has certainly been studied. I will refer you to Reaction mechanism of combustion of H2 and the links therein.


Why can oxygen have a coordination number of 3 in a complex between silver and SiO2?

High coordination numbers are really not uncommon in solid-state compounds. For example, consider the structure of $\ce{Li2O}$, an anti-fluorite structure. In fluorite proper, calcium ions form a cubic closest packing structure and the fluoride ions occupy the tetrahedral voids. In $\ce{Li2O}$, it is reversed: the oxide ions form a cubic closest packing and lithium occupies the tetrahedral voids. That gives each lithium a coordination number of 4 and by maths and symmetry oxygen must have a coordination number of 8.

Tricoordinated oxide anions are also well-known in solid-state structures. The prime example would be the rutile structure of $\ce{TiO2}$ with hexacoordinated titanium and tricoordinated oxide ions.

In your specific case, the silver cations are acting as Lewis acids while the oxide anions (which have to lone pairs not directed towards neighbouring silicium atoms) can act as Lewis bases. A coordination number of three points to a somewhat covalent (and not so ionic) structure — which makes sense considering significant fragments of $\ce{SiO2}$ are still there.


Hydrogentaion of alkenes in absence of catalysts

The main point of these exercises is that not all C=C bonds are created equal. Their reactivity towards hydrogenation can be influenced by steric effects and specifically applying here, the degree of substitution of the double bond. So for the bottom example, the exocyclic disubstituted double bond will react but not the endocyclic one which is much more stable. Now here there is some uncertainty in this exercise since the actual result is dependent on the exact conditions which are not provided. For example hydrogenations are well known to be very capricious on the selection of metal catalyst and $\ce{Pd}$  can give a different result than using $\ce{Pt}$ or $\ce{Rh}$. And this only an example as there are so many different set of conditions possible with variable results.

The top one is an even worse exercise in the sense that the outcome is very specific on the exact conditions. Now, knowing the exact outcome we can rationalise as follows: Here the electronics also play a role and this diene because of conjugation will be more stable and thus more resistant towards hydrogenation. When one of the exocyclic bonds gets reduced (even if it only partially does so) then in the intermediate formed the other unreacted double bond will isomerise (much much faster than the reduction of the exocyclic conjugated bond in the starting material)  to the more stable product with the  endocyclic double bond i.e. Structure a. This structure will be unreactive towards further reduction and eventually will be the sole product of the reaction.


Why is DNA negatively charged and what makes it so?

As cibrail said, DNA is a polymer of nucleotides. They join themselves through phosphodiester bonds (a specific kind of covalent bond) that can grow to as much as millions of nucleotides.

What part in the strand contributes to the overall non neutral charge?

The reason why DNA is negatively charged is the phosphate group that makes up every nucleotide (pentose + nitrogenous base + phosphate). When forming part of the phosphodiester bond, they retain 1 of 2 negative charges (the other being lost to form the other ester bond to a new pentose, that's why the bond is called "phospho-di-ester").

Lehninger 5th edition. Part of figure 25-5

DNA is not isolated in the body, so what keeps it stable while being charged?

The main molecules in charge of that are proteins called histones. They have alkali features and positive charge, and so, they neutralize these negative charges. On the other hand, that's not the main point of histones. Histones play a really important key role in the regulation and compaction of DNA within the nucleus of the cell, and are important targets of diverse molecules that want to alter DNA structure and function:

Lehninger 5th edition.Figure 24-26

Why is it important for DNA to be charged?

It has great importance because cell machinery is adapted to that property. If DNA wasn't negatively charged, the structural fundamentals of interactions between DNA and transcription factors would be completely different.
Thanks to the fact that it is negatively charged, it becomes spontaneously pretty straight when denaturalised, because negative charges repel themselves. It forms very wide curves.
And of course, histones wouldn't be able to join DNA and exert their key regulating functions.


How to calculate the heat released when sodium hydroxide is dissolved in hydrochloric acid solution?

If that's all the information you have (no concentration or mass or density for the HCl solution, no mass for the final solution), and you weren't instructed to obtain any of that information, then what you did (assuming 50g and heat capacity of water) is ok for an introductory lab class.

There will be many sources of error:

The density of HCl solution is greater than water, so the initial solution will be more than 50g.

The heat capacity of HCl solution is less than that of water.

Without knowing the concentration of the initial solution, it is unknown if the final solution is NaCl solution, a solution of HCl and NaCl, or a solution of NaCl and NaOH.

The heat capacity of the final solution will be less than that of water.

The final solution will be somewhat more than 52g.

Just dissolving NaOH in water is exothermic. It can't be assumed that all the NaOH reacts without knowing the concentration of HCl.


Why do pregnant women turn toilet seats blue?

The question cannot be answered without a chemical analysis.

However, given that there exist toilet seats with silver based anti-microbial coatings, the bluish grey color may come from metallic silver particles produced by a reduction process similar to the photographic development process.

You could proof the presence of silver by applying Caffenol, an alkaline aqueous solution of instant coffee, washing soda, and vitamin C (ascorbic acid) to the toilet seat. I suggest to use Caffenol-C-M whose preparation is described here. Let the Caffenol react on the toilet seat as long as possible. If grey or black stains are formed that persist after cleaning, your toilet seat contains silver with high probability. Note that the stains cannot be removed anymore.

If the test is positive, the next step would be to investigate why your pregnant partner has reducing powers.


What material is synthesized in the rubber egg demo?

The shell dissolution reaction is the calcium carbonate in the egg being dissolved by the acetic acid, resulting in release of $\ce{CO2}$ and formation of calcium acetate.

$$\ce{CaCO3 + 2 CH3COOH -> Ca(CH3COO)2 + CO2 + H2O}$$

The rubberization of the egg white is primarily a protein denaturation reaction.  Albumin in egg whites is a globular protein, meaning each molecule is a single chain that is tightly bundled into a ball-like ("globular") form.  In this form the molecules can stay nicely solvated and the egg white is liquid.  However, the low pH from the acid (and also possibly the high concentration of calcium acetate) causes each chain to unfurl and then to bind with other neighboring chains, forming a complicated cross-linked tangle of molecules.  This tangle is no longer a liquid but is a rubbery solid-like material.

So the products are primarily calcium acetate, $\ce{CO2}$, and denatured albumin.


Why are higher-energy bonds preferred over lower-energy ones?

When a mole of $\ce{S-O}$ bond is formed, $340$ kJ of heat is released.

This $340$ is not how much energy is needed to start a reaction. That energy (activation energy) cannot be derived from the bond dissociation energy.


Until there's water present (I'm assuming that it's not the solvent), the proton can't come from water. The proposed intramolecular proton transfer would go through a 5-membered transition state, which is reasonable. If you don't like that mechanism, then two of the intermediates could transfer protons to each other. Depending on the concentration, one or the other or both could be occuring.

The exact nature of these proton transfers are difficult to prove and typically inconsequential, so you will often see "shortcut" mechanisms drawn which imply intramolecular proton transfer even when it is more unlikely than the one you show here.


This is a very broad question, so below I mention just the very key points.

Big Bang

According to the prevailing cosmological model for the early development of the Universe, known as the Big Bang theory, the Universe begun at about 14 billion years from a singularity that can be thought of a state of an infinite density and temperature, and since its creation the Universe began to expand and cool.
Right after the Big Bang the Universe was immensely hot, the energy density was incredibly high and no particles existed, but the Universe was extremely rapidly expanding and cooling, so that soon enough energy was allowed to be converted into various elementary particles: quarks, leptons, bosons.

Baryogenesis

It is generally assumed that initially there was an equal numbers of particles and corresponding antiparticles, but at some point an unknown physical process referred to as baryogenesis lead to a very small excess of quarks and leptons over antiquarks and antileptons.
As the Universe continued to decrease in density and fall in temperature quarks and gluons combined to form baryons such as protons and neutrons, and the small excess of quarks over antiquarks led to a small excess of baryons over antibaryons, which in turn after mass annihilation resulted in the predominance of matter over antimatter in the Universe.

Big Bang nucleosynthesis

After annihilation it was just the matter of time for the Universe to cool down to a point where remaining protons, neutrons and electrons could combine to form atoms.
Already a few minutes after the Big Bang neutrons combined with protons to form the lightest nuclei (H-1, H-2 or D, H-3, He-3, He-4, Li-6, Li-7, Be-7, and Be-8) in a process called the Big Bang nucleosynthesis, and roughly 378,000 years after the Big Bang the Universe cooled to the point when the formation of light neutral atoms (mostly H and He with trace amounts of Li) became energetically favored.
Eventually, most of the protons in the Universe were bound up in these neutral lightest atoms and matter was uniformly distributed in the Universe.

Stellar nucleosynthesis

But the gravitational attraction of the nearby matter by slightly denser regions and the subsequent growth of such regions resulted in formation of gas clouds, stars, galaxies, and other large structures.
The formation of stars meant that the extreme conditions (temperatures and pressures) for the nuclear fusion were recreated and light elements (H, He and Li) from which stars were formed started to turn into heavier one by a process referred to as the stellar nucleosynthesis. Stellar nucleosynthesis has its endpoint, namely, when nuclear fusion reaches nickel-56 which then undergoes radioactive decay into iron-56 the process stops. The reason is that iron-56 has one of the highest binding energies of all of the isotopes, and thus, no more energy can be extracted from an iron-56 nucleus, either by fusion or by fission. But it is not the end of the story.

Supernova nucleosynthesis

When the core of a star converts to iron-56, the nuclear reactions stop. And that is a big problem, since so far the star balanced the forces of gravity pulling its materials inward by nuclear reactions pushing them outward, but now all the reactions pushing outward stop. What happens then depends on how massive the star is. Supergiants, that have masses from 8 to 12 times the Sun, gravitationally collapse in supernova explosions accompanied by a large burst of neutrons. These neutrons are captured by heavy nuclei, such as iron-56, and the resulting neutron-saturated isotopes then undergo beta decays producing stable isotopes of elements heavier than iron.


First of all, let's make something crystal clear: The more acidic is a compound, they smaller will be its $pK_a$.

To evaluate the acidity of these compounds you need to analyze the stability of the corresponding conjugate base of each one, in effect, the stability of the same compound but without a proton and a negative charge on an oxygen atom.

First of all, you have two kinds of compounds:

Two alcohols: V and II
Two carboxylic acids: I, III and IV

From those compounds, alcohols will always be less acidic than carboxylic acid. Why? Because the negative charge left by a proton leaving the molecule can be delocalizaed by resonance troughout the carbonyl, but in the case of the alcohols it will be localized on the only oxygen atom, which makes the conjugate base less stable, and therefore, the compound less acidic.

That being said, from the two alcohols, V will be more acidic, because the presence of the chlorine atoms substract charge from the carbon they are bond to (chlorine is more electronegative than carbon), and helps stabilizing the negative charge that is left in the molecule after the proton is substracted.

In regard to the carboxylic acids, you can use a similar reasoning for the carboxylic acid.
The less acidic is the one in which R is just a hydrocarbon chain, which has no effect in contrast to I, which has 1 chlorine atom that stabilizes the negative charge, and to III, which has 3 chlorine atoms that stabilize the negative charge of the conjugate base even more.

Therefore, the order presented in option E is justified.


Has anyone even taken a picture of a molecule to confirm the geometry predicted by theory?

Yes. Researchers have been using atomic force microscopy (AFM) and scanning tunneling microscopy (STM) for some time for this purpose. Do note that these images are not photographs in the sense that we usually think of "pictures" and are indirect measurements of constituents of the molecule. However, they do yield "pictures" that show the geometry of the molecule(s) in question.

See, for instance, this piece from UC Berkeley News published in 2013 which describes the techniques in a manner that is appropriate for a non-chemist. The same article includes images of molecules before and after a chemical reaction occurs.

I also encourage you (and others) to have a look at images in the IBM STM Image Gallery, which are quite spectacular.

Finally, as others have commented, X-ray diffraction merits mention from both a historical and scientific perspective. As one person notes, the helical secondary structure of DNA was determined by Watson and Crick after examining the fiber X-ray diffraction plates obtained by Franklin's group.

The importance of Franklin's contribution and that of Watson and Crick are monumental achievements. That said, the structure that was elucidated was the secondary structure of a large biomolecule and as such might not fit your criteria for how "complete" the "picture" needs to be to satisfy your question.

Here is the famous Photo 51 result obtained by Rosalind Franklin's doctoral student Raymond Gosling that was subsequently used by Watson and Crick.

Recently, researchers have imaged single atoms using a combination of STM and MRI - see Nature Physics. P. Willke et al. Magnetic resonance imaging of single atoms on a surface. 1 July 2019.


Breaking of carbon chains due to annihilation of electrons using positron beams

This a problem of radiation chemistry. It has been thoroughly studied in the years 1950 - 1960 - 1970 with gamma rays. When intense source of gamma rays strikes water, electrons are extracted from the water molecule. And some of these electrons are coming from the valence bond O-H. The bond is then weakened, and may even be broken. Usually the electron soon comes back to its original atom. This produces individual H atoms and OH radicals. Usually H and OH soon recombine and produce again H2O a few nanoseconds later. But if the irradiation is intense, they may recombine with similar neighbors. And this produce dissolved H2 and H2O2 molecules. So the water sample contains more and more dissolved H2 and H2O2. Titration of H2O2 in solution is possible, and may be used to determine the doses and the intensity of the radiation absorbed by the sample.

The same experiments have been done with hydrocarbons. After irradiation, hexane, for example, is partly converted into dodecane, with a lot of different isomers. The distribution of these isomers allows to determine where the initial breaking in hexane happened, in other words which C-H or C-C bond has been initially broken, and in which proportion. Why don't you read the publications by Ausloos or Gäumann or others in this old literature. Today nearly nobody is interested any more in this chemistry, which is considered to be "finished".


The $\ce{K_a}$ expression for acetic acid, in aqueous solution, is given by:

$\ce{K_a} =\dfrac{\ce{[CH3COO-][H+]}}{\ce{[CH3COOH]}}$

The expression can be rearranged to give

$\ce{[CH3COOH]} = \dfrac{\ce{[CH3COO-][H+]}}{\ce{K_a}}$

so no matter what the pH, all the acetic acid will never be reacted.

So in a titration let's assume that we start out with 0.1000 moles of whatever acid $\ce{[HA]_{initial}}$.

After base has added and $0.1000 \gg \ce{[HA]_{final}}$ and thus $0.0001 \gt \ce{[HA]_{final}}$, then the reaction is  quantitatively complete.

This is strictly a significant figures argument. You can't do a titration to 10 significant figures. 3 is decent, if you get 4 you're a magician.
When 0.1000 moles of base have been added to the solution the equivalent point has been reached.

The key point is that the moles of base is equal to moles of acid, regardless of the pH.
When ever you get the "signal" that the titration is over you have reached the end point.

The signal could be, for example, a color change of an indicator, the swing of the needle on a pH meter, or the formation of a precipitate.


Why don't elements having the same valency as Carbon bond so readily?

It's not that they don't bond readily, it's that they can't form as stable of networks as carbon. Carbon has many different allotropes, while silicon has only a few. As you can see, carbon forms uniform hexagonal sheets with massive $\pi$ systems as in graphene and many of its other allotropes, while silicene is incapable of remaining flat and is bumpy. This is a result of the bond distances and p-orbital overlap.

$$
\begin{array}{|c|c|}\hline
\text{Bond} &\ \Delta H ^\circ_f \text{ (kcal mol}^{-1}) &\ \text{Bond Length (pm)} \\
\hline
\ce{C-C} &\ 82.70 &\ 154 \\
\hline
\ce{Si-Si} &\ 53.06 &\ 233 \\
\hline
\ce{Ge-Ge} &\ 44.93 &\ 241 ^{[1]} \\
\hline
\end{array}
$$

As you can see, as the atoms get larger, the bond length increases. $\pi$ bonding is completely reliant on the overlap of adjacent p-orbitals, so the farther away these orbitals are, the weaker the bonds.

Bond dissociation enthalpy also increases with atom size. This leads to an overall decrease in the stability of larger silicon or germanium compounds. Tin and lead are metals, which have much different chemistry altogether on account of their electronegativity and occupied d-orbitals, in addition to their size.

${[1]}$ Bond Enthalpy Table


What is the phase transformation temperature anatase to rutile phase in tin dioxide(SnO2)?

Anatase and rutile phases form in $\ce{TiO2},$ the presence of cationic dopants or impurities significantly affect the transformation kinetics and phase stability [1].
It should be noted that anatase is metastable at all temperatures and pressures.

References

Hanaor, D. A. H.; Sorrell, C. C. Review of the Anatase to Rutile Phase Transformation. J Mater Sci 2011, 46 (4), 855–874. https://doi.org/10/ct79p8.


Difference between 18 electron rule and effective atomic number rule

Both of these rules are equivalent and are related to transition metals attaining a noble gas configuration of electrons in coordination complexes. The difference between them is the number of electrons given to the central metal.

The effective atomic number uses all of the metals electrons. For example, in an $Fe^{2+}$ complex, the central iron atom has 24 electrons, and you would want the ligands coordinated to the metal to contribute another 12 electrons to reach 36 electrons, the next noble gas configuration.

The 18-electron rule is a simplification of the EAN. There is always a noble gas configuration of electrons for the metals core orbitals (e.g. 1s2s2p3s3p for $Fe^{2+}$), so you can ignore these electrons in the counting scheme. For transition metals, this means that the only important orbitals are the valence nd(n+1)s(n+1)p orbitals, which can hold a total of 18 electrons. For $Fe^{2+}$, there are 6 electrons in these orbitals. Again, the ligands would need to donate 12 electrons to reach the 18 electron limit.


Does hydrogen peroxide exhibit intramolecular hydrogen bonding interactions?

Preamble:

I will conveniently avoid entering into a debate as to what a hydrogen bond is, given the numerous theories which attempt to do so from various perspectives.
I will, however, attempt to use the relatively cheap and easy natural-bond orbital (NBO) analysis method to examine the possible existence of an intramolecular hydrogen bond (HBond) in the ground-state, global minimum configuration of H$_2$O$_2$.  In doing so, we will assume that the mere existence of an $n\rightarrow\sigma^{*}$ overlap indicates that an intramolecular HBond is indeed present (a commonly accepted metric).  For comparison, I will look at the planar form of H$_2$O$_2$ (containing a trans-type bond configuration) in order to minimize any overlap between the oxygen lone pair and the corresponding $\sigma^{*}$ orbital.  I note here that the NBO analysis is purely a qualitative method, and caution should be used when attempting to make quantitative comparisons with this method.

Methods:

The H$_2$O$_2$ molecule was constructed in $C_2$ symmetry (Structure A). A full optimization and corresponding harmonic vibrational frequency computation was employed at the MP2/heavy-aug-cc-pVTZ level of theory, where non-hydrogen (i.e. heavy) atoms are augmented with diffuse functions.  The basis set is denoted henceforth as haTZ.  The magnitudes of the residual Cartesian gradient components of the optimized structure were less than $4.0\times10^{-5}$ $E_{\textrm{h}}$ $a_0^{-1}$.  The frozen-core approximation was invoked for all computations.
A subsequent NBO analysis of the fully optimized H$_2$O$_2$ molecule was performed at the MP2/haTZ level of theory using the SCF density.
This procedure was also applied to a H$_2$O$_2$ molecule constructed in $C_{2h}$ symmetry (Structure B).
All computations were performed with the G09 (Rev. D01) software package.

Results:

Structure A is a minimum on the MP2 potential energy surface with zero imaginary modes of vibration whereas structure B is a transition state with one imaginary mode of vibration.

The NBO analysis of structure A reveals two $n\rightarrow\sigma^{*}$ interactions (where $n$ is an electron lone pair on oxygen and $\sigma^{*}$ is the $\sigma$ antibonding orbital of the neighboring O$-$H entity.  The existence of exactly two of these interactions is due to the symmetry of the molecule, and only one will be reported here for simplicity.

The following figure illustrates the overlap of a lone electron pair on O with the adjacent $\sigma^{*}$ orbital in structure A.  The occupation number for this antibond, 0.00351, indicates a very small overlap.  The corresponding second-order perturbation energy E$^{(2)}$(donor$\rightarrow$acceptor) is a modest 1.74 kcal mol$^{-1}$.  We note that this is merely an estimated measure of the strength of the interaction.

Interestingly, upon rotation of the torsion angle, $\tau$(H-O-O-H), to 180$^\circ$ (structure B), the $n\rightarrow\sigma^{*}$ interaction disappears.  The intramolecular hydrogen bonding  may play an important role in stabilizing the lowest-energy conformer in H$_2$O$_2$ (structure A) as the transition state, structure B, does not contain this interaction.

Comparing the strength of an intramolecular HBond to an intermolecular HBond is not straightforward.  Intermolecular interactions are frequenly and efficiently computed by using the supermolecular approach by taking difference of the electronic energy of a complex and the energies of each non-interacting body which make up the complex.  Intramolecular interactions cannot be determined in this way without breaking bonds, thereby skewing the resulting number.  To facilitate comparison to the intramolecular HBond in H$_2$O$_2$, an NBO analysis was performed on the water dimer (H$_2$O)$_2$ (in $C_S$ symmetry) due to the presence of a strong, intermolecular hydrogen bond.   The corresponding E$^{(2)}$ for an $n\rightarrow\sigma^{*}$ interaction is a much larger 6.82 kcal mol$^{-1}$, nearly 5 kcal mol$^{-1}$ stronger than that seen in H$_2$O$_2$

Conclusions:

An NBO analysis of the lowest energy conformer H$_2$O$_2$ indicates the possibility of the presence of weak, intramolecular hydrogen bonds.  These intramolecular hydrogen bond-like interactions could play a role in stabilizing the H$_2$O$_2$ complex.
The strength of the interaction is much smaller than that computed for the water dimer by about 5 kcal mol$^{-1}$.

(Note: I strongly recommend performing a Bader analysis on H$_2$O$_2$ to compare with the NBO analysis before reaching any solid conclusions.)

ADDENDUM:

I have tried to avoid calling this intramolecular interaction a hydrogen bond.  Based on the metric imposed in the preamble, it has a property of a hydrogen bond, however, the strength of this intramolecular interaction is so weak, it really does not reflect a true hydrogen bond (e.g. A typical hydrogen bond is about 10 kcal mol$^{-1}$ and a weak hydrogen bond is usually considered to be on the order of 2-3 kcal mol$^{-1}$.  If we take the NBO results at face value, the interaction is just too weak to be considered as a hydrogen bond.  Coupled with the fact that the geometry itself does not facilitate reasonable HBonding configurations.

Upon reflection of these issues, I conclude that there is, in fact, no intramolecular hydrogen bonding in H$_2$O$_2$.


First, a point that is more like a comment: I'm not sure that just because electrons get put in to d-orbitals (and thinking of crystal band structures in terms of atomic orbitals is of questionable value anyway) that it would not be a metal. Many elements (one could even say most elements) with d-electrons are quite happy as metals.

You might want to start with L.F. Lundegaard et al., Phys Rev B80 020101(R) (2009), where they do DFT calculations on potassium at different pressures. I will quote from their conclusions section:

  The observation of the oP8 structure in K reveals it to
  have a pressure-induced structural behavior like that of Na,
  as well as having behavior like Rb and Cs as indicated by the
  observation of the tI4 and oC16 phases. The similarity to Na
  is unexpected in relation to the recent electronic structure
  calculations of Ma et al.$^{19}$ As noted, they found K, Rb, and Cs to have the same high-pressure structural sequence, and
  they attributed this to these heavier alkalis’ being dominated
  by d electrons arising from s-d transfer, whereas the ultrahigh
  pressure behavior of Na arises, they suggest, from p-d
  hybridization. Thus our results question this picture of a
  simple dependence of structural stability on d-electron dominance
  or p-d hybridization.


GGA functionals requiring second derivatives?

It turns out that my original reference is now somewhat out of date, and it is quite possible to compute the relevant properties without resorting to second spatial derivatives. [Johnson1993] has a nice overview of the relevant math, including clean derivations of both YLP and B88 functional terms, and states:

Note that Eq. (7) does not require evaluation of the spin density second de- rivatives as is necessary in many other formulations, a major computational advantage [48]. The fact that these can be avoided has been previously noted by Kobayashi et al.

Johnson, B. G., Gill, P. M. W., & Pople, J. A. (1993). The performance of a family of density functional methods. The Journal of Chemical Physics, 98(7), 5612–5626. https://doi.org/10.1063/1.464906
and
Kobayashi, K., Kurita, N., Kumahora, H., & Tago, K. (1991). Bond-Energy calculations of Cu2, Ag2, and CuAg with the generalized gradient approximation. Physical Review A, 43(11), 5810–5813. https://doi.org/10.1103/PhysRevA.43.5810


Why does covalent bonding not break down if observer effect can be applied to atomic electrons?

I think part of your confusion stems from neglecting the fact that the electrons in a single molecule are all entangled with each other, so while an individual electron can occupy many states simultaneously, in each of those states it is entangled with all the other electrons such that the sum of states remains the same.
To start with a simple example, consider an atom of helium in its singlet ground state. Prior to measurement, neither electron has a defined spin. Both are a superposition of $+1/2$ and $-1/2$ states, but that does not mean that the helium atom can have $+1$ or $-1$ spin. The net spin is still 0. If we measure one electron's spin to be $+1/2$, the other will necessarily have $-1/2$, even though neither had defined spin prior to measurement.
We can extend this to multi-atomic molecules. Consider ground state methane. We can calculate delocalized molecular orbitals that collectively describe the charge density at any point in the molecule, with each of those orbitals being composed of a standard hydrogenic orbital of carbon (eg 2s or 2p) and athe 1s orbitals of all four hydrogen atoms. Alternatively, we can describe the same charge density distribution with four localized orbitals, each comprised of a single hydrogen 1s orbital and a hybrid sp3 orbital on carbon. Prior to making an observation, neither of these descriptions is correct, but neither is incorrect. Both correctly describe the charge density of the molecule, but both incorrectly assign electrons to specific states. Instead, the electrons are collectively in both states (and an infinite number of others that describe the same density). Because every contributing state correctly describes the charge density, there is no change in the covalent structure resulting from the superposition.


Groundwater contain besides other things $\ce{Fe^2+}$ a lot of organic material Lenntech, Natural Variations in the Composition of Groundwater. Oxygen released during electrolysis denatures organic proteins and the $\ce{Fe^2+}$ oxidize to $\ce{Fe^3+}$ that resulting to the reddish colorization.


Moles of chlorine gas needed to make up an equivalent amount of saturated salt solution

a) molar concentration
There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$.
I'd prefer to write the molar mass as "$\pu{58.44 g mol^-1 }\ \ce{NaCl}$" or "$\pu{58.44 g/mol}\ \ce{NaCl}$" to keep it all together.
b) molal concentration
You got the right value, but you did it the wrong way. For the solvent:
$$\dfrac{\pu{980 kg/m^3}}{\pu{1000 L/m^3}} = \pu{0.980 kg/L}$$
c) moles of $\ce{Cl2}$ gas
The book is correct. There are 6.14 moles of $\ce{NaCl}$. Each molecule of chlorine gas contains two atoms of chlorine, so you need half as many moles to get the same number of atoms of Cl. So 6.14/2 = 3.07
This part of the question has poor phrasing. You can't dissolve 3.07 moles of $\ce{Cl2}$ in water and get a solution of $\ce{NaCl}$.
d) density of the mixture
Your answer is correct.


Product of reaction between aluminium chloride and trimethylamine

Here are a few tips to get you started:

Lewis acids are electron pair acceptors; Lewis bases are electron pair donors.
This looks like a Lewis acid-base reaction. You mention that this is an acid-base reaction but there are no easily accessible protons in either reactant. Aluminum trichloride has no protons; trimethylamine has does have protons but these are $\ce{sp^3~C-H}$ type protons - which are extremely weak. Also, there is no path to stability; deprotonation of one of the methyl groups leaves a localized negative charge on a carbon.

There are other heuristics one can use. For example, note that aluminum trichloride has an electron-deficient aluminum atom (the Al doesn't have an octet). Acceptance of an electron pair will provide it with an octet. This heuristic works fairly well; consider boron trichloride for example. Boron is similarly electron-deficient. And boron trichloride is also a Lewis acid.

Also note that the amine has a lone pair. Keeping in mind that Lewis acid-base reactions are about the movement of electrons (rather than protons) then we might reasonably suspect this to be a Lewis acid-base reaction. Lone pairs generally indicate Lewis basicity, as these are generally available for donation to a Lewis acid. Delocalization of lone pairs can weaken the lone pair's Lewis basicity. Thus, the presence of a lone pair only serves as an indicator of Lewis basicity rather than the strength of Lewis basicity. And of course you can also donate bonding electrons rather than lone pairs (halogenation of ethene would be an example; the $\ce{C=C}$ double bond is broken and the electrons previously involved in that double bond are put into the antibonding orbital of some diatomic halogen)


Sum of a rate constant gives the total decay rate

I am simply confusing two different variables here, rate and rate constants.

In this situation, the sum of the rate constants is the sum of the individual rate constants of the processes occuring in the reaction i.e.

$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$

This is consistent with my current knowledge.

It should be referred to as "total decay rate constant"

Of course, the rate equation can still be written here as:

$$\ce{v_{rate} = k[A]}$$


"Mass" in this context refers to the large number of particles and consequent behavior in a stochastic manner, as opposed to a small set of particles, where Maxwell's demon might have been able to influence outcomes.Of course, Maxwell didn't know know of Heisenberg's uncertainty principle.


What is the precision of intensity data in XRD measurement?

In short: what does the manual of your diffractometer claim?  Were you able to replicate its standards in first place?
The peak area depends both on the peak's height and width.  But this is difficult to answer because your question does not detail how data were recorded in first place.  Beside content (elements) and quality of your sample (remaining mechanical stress, orientation / size / shape of the crystallites, sample thickness), temperature and pressure while recording the data, the peak area will depend on the source (e.g., wavelength monochromators, slits widths on source and detector), type of detector, i.e. if a point detector was used moving around the sample, or a fixed image plate / area detector and data processing (e.g., how do you account for background intensity, (how) do you deconvolute peaks in question).

(modified from Stoe's IP-PSD)
If you opted for an area detector recording all intensities along $2\theta$ simultaneously, was it the same as for your reference sample?  You are set by the geometry and the inner design of this detector (pixel size and response to the radiation used) on $2\theta$-resolution and may improve intensity resolution only by change of the time of the exposure.
With a point detector, you are less limited for angular resolution because the detector may travel as far around the capillary or rotating dish as mechanics permit in tiny steps.  Large machines

(credit)
claim increments down to steps of 0.0001° (reference) which may equate to only one experiment per day but likely better peak resolution (think peak deconvolution), potentially with a much better signal/noise ratio than the area detector, especially if the sweep is recorded more than once.


Unexpected reaction during acid hydrolysis of cellulose

We miss some data to make a good hypothesis.

The concentration of the $\ce{H_2SO_4}$ solution.
The temperatures.

These are the main two factors that affect cellulose hydrolysis. Usually are used temperature above 200 °C and pressure above 25 MPa. For me its strange that the exothermic reaction between SA and $\ce{NaOH}$ could lead to an increase of temperature such that.  However some studies show that the reaction can occur at lower temperature with concentrated acid.

Another possible solution is the formation of alkali cellulose.


How to identify an emerald green solution with possibly two anions?

You can have two different anions if your solution is a mixture of two different metal salts, a chloride and a sulfate. The color change from green to blue upon dilution with water fits well with $\ce{Cu^2+}$ being the corresponding cation. Concentrated solutions of $\ce{CuCl2}$ are green due to the presence of chlorocuprates, like $\ce{[CuCl3]^{-}}$ and $\ce{[CuCl4]^{2-}}$. When water is added, a stepwise ligand exchange takes place, and the pale blue color of $\ce{[Cu(H2O)6]^{2+}}$ starts to appear. The image below shows copper(II) chloride solutions of different concentrations: Green when the chloride concentration is high, and blue when it is lower (Wikipedia: copper chloride).

In summary, the observations (color change, precipitates) indicate that you have an aqueous solution of $\ce{CuCl2}$ and $\ce{CuSO4}$.


The IUPAC rules use "aminoethyl" for such a group. This is copied from the IUPAC "blue book" which can be found online here

Rule 811.3 - An -NH2 group, when not the principal group, is named by the prefix "amino-".

Your original name is very close to what would be used if the N were attached to the ring and the ethyl attached to the N. That would be "3-ethylamino-2-methyl. . ." according to this rule:

811.4 - Radicals RNH-, R2N-, and are named as substituted amino groups by changing "ine'' of the amine to "ino".


How to make approximation of rotational partition function of diatomic linear molecules?

Sorry for the late answer. I just discovered this question.

Preludium

Firstly, I think you might have an error or some non-standard notation in your formula for $Q_{\mathrm{rot}}$. The rotational energy levels are given by

\begin{align}
  E_{\mathrm{rot}} = \frac{\hbar^2}{2 I} J ( J + 1 ) \ ,
\end{align}

where $I$ is the moment of inertia.
Then the rotational constant $B_{0} = \frac{\hbar^2}{2 I}$ (in units of energy) is introduced, so

\begin{align}
  E_{\mathrm{rot}} = B_{0} J ( J + 1 ) \ .
\end{align}

Thus

\begin{align}
  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{B_0 J(J+1)}{kT}\right)} \ ,
\end{align}

whereas your exponent contains another $h$ which should not be there. At least that's the standard notation I am used to. For convenience I'll also introduce the rotational temperature $\Theta_{\mathrm{rot}} = \frac{B_{0}}{k}$, so that

\begin{align}
  Q_{\mathrm{rot}} = \sum_{J=0}^{\infty}{(2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right)} \ .
\end{align}

Main Answer

Ok, now that that's sorted out, let's start the real work. It is actually a pretty standard approximation that is used here: They use the Euler-Maclaurin formula

\begin{align}
  \sum_{n = a}^{b} f(n) &= \int_{a}^{b} f(n) \, \mathrm{d} n + \frac{1}{2} \bigl( f(b) + f(a) \bigr) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} \Bigl( f^{(2i - 1)}(a) - f^{(2i - 1)}(b) \Bigr)
\end{align}

where $f^{(k)}(a)$ is the $k$th derivative of $f$ evaluated at $a$. The $b_{2i}$s are the absolute values of the even Bernoulli numbers, $b_{2} = \frac{1}{6}$, $b_{4} = \frac{1}{30}$, $b_{6} = \frac{1}{42}$, $...$.
It provides a way to approximate the summation of a function over a discrete variable by an analogous integration over a continuous variable.

In the special case at hand,

\begin{align}
  f(J) = (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \ ,
\end{align}

where the summation limits are $a=0$ and $b=\infty$ and where $f(\infty) = f^{\prime}(\infty) = \ldots = 0$ this formula boils down to

\begin{align}
  Q_{\mathrm{rot}} = \sum_{J = 0}^{\infty} f(J) &= \int_{0}^{\infty} f(J) \, \mathrm{d} J + \frac{1}{2} f(0) + \sum_{i=1}^{\infty} (-1)^{i} \frac{b_{2i}}{(2i)!} f^{(2i - 1)}(0)
\end{align}

The first term is what is usually called the classical or the high-temperature limit. It can be easily calculated by making the substitution $K = J(J + 1)$ $\Rightarrow$ $\mathrm{d}J = \frac{\mathrm{d}K}{2J + 1}$. So

\begin{align}
  \int_{0}^{\infty} f(J) \, \mathrm{d} J &= \int_{0}^{\infty} (2J+1)\exp\left(-\frac{\Theta_{\mathrm{rot}} J(J+1)}{T}\right) \, \mathrm{d} J \\
  &= \int_{0}^{\infty} \exp\left(-\frac{\Theta_{\mathrm{rot}} K}{T}\right) \, \mathrm{d} K \\
  &= \frac{T}{\Theta_{\mathrm{rot}}}
\end{align}

Now, only the derivatives $f^{(k)}(0)$ need to be sorted out. Those that contain terms below the third order in $\frac{\Theta_{\mathrm{rot}}}{T}$ can be calculated to be:

\begin{align}
  f(0) &= 1 \\
  f^{(1)}(0) &= 2 - \frac{\Theta_{\mathrm{rot}}}{T} \\
  f^{(3)}(0) &= - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \\
  f^{(5)}(0) &= 120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \ .
\end{align}

Doing this by hand is very tedious. Best you use something like Mathematica. Plugging all this into the Euler-MacLaurin formula and dropping all terms above second order in $\frac{\Theta_{\mathrm{rot}}}{T}$ you get exactly the result of Gordy & Cook

\begin{align}
  Q_{\mathrm{rot}} &\approx \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} \cdot 1 + (-1) \frac{1}{6 \cdot 2!} \left( 2 - \frac{\Theta_{\mathrm{rot}}}{T} \right) \\
 &\quad + (-1)^{2} \frac{1}{30 \cdot 4!} \left( - 12 \frac{\Theta_{\mathrm{rot}}}{T} + 12 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\
  &\quad + (-1)^{3} \frac{1}{42 \cdot 6!} \left(120 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - 180 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + 30 \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} - \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{5} \right) \\
  &= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{2} - \frac{1}{6} + \frac{1}{12} \frac{\Theta_{\mathrm{rot}}}{T} - \frac{1}{60} \frac{\Theta_{\mathrm{rot}}}{T} \\
  &\quad + \frac{1}{60} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} - \frac{1}{252} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \\
  &= \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right)
\end{align}

As a side note: If you have a homonuclear diatomic molecule you need to introduce the so called symmetry number $\sigma$ which represents the number of indistinguishable orientations a molecule can have. It originates from symmetry constrains: for homonuclear diatomic molecules there is a two-fold axis of symmetry perpendicular to the internuclear axis which gives rise to two indistinguishable orientations, so the above formula for $Q_{\mathrm{rot}}$ overcounts the number of available quantum states
by a factor of two because of the inherent indistinguishability of the nuclear pair. Thus, $Q_{\mathrm{rot}}$ needs to be divided by $\sigma = 2$. For heteronuclear diatomic molecules you simply have $\sigma = 1$.
With the introduction of $\sigma$ the equation for $Q_{\mathrm{rot}}$ becomes:

\begin{align}
  Q_{\mathrm{rot}} &\approx \frac{1}{\sigma} \left( \frac{T}{\Theta_{\mathrm{rot}}} + \frac{1}{3} + \frac{1}{15} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} \right) \right) \\
  &= \frac{T}{\sigma \Theta_{\mathrm{rot}}} \left(1  + \frac{1}{3} \frac{\Theta_{\mathrm{rot}}}{T} + \frac{1}{15} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{2} + \frac{4}{315} \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{3} + \mathcal{O}\left( \left(\frac{\Theta_{\mathrm{rot}}}{T}\right)^{4} \right) \right)
\end{align}


To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:

$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V_1 + V_2 + V_x} \tag{1}$$

where $C$ – concentration, $n$ – amount, $V_1$ – volume of water remaining from permanganate solution, $V_2$ – volume of water synthesized during the reaction, $V_x$ – additional volume of water to be found.

From the balanced redox reaction, which it seems you have established correctly:

$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$

one can express both $V_1$ and $V_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:

$$V_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$

$$V_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$

$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$

From (1):

$$V_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V_1 - V_2 \tag{5}$$

Combining (2–4) into (5), and taking into account that

$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$

one can obtain the following expression for the volume of water to be added:

$$V_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$

The problem is, plugging in the current values returns negative volume, which means the solution is already too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:

$$V_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$


Does the octet rule only apply to the second and third shell or does it apply further?

The shell concept is useful rather for 1-electron hydrogen-like atoms, where energies of ns, np, nd, nf orbitals are all the same ( well, not fully true due the very tine QED effect of interaction of electrons with quantum vacuum ).
For multielectron atoms, energy wise, $\mathrm{nd}$ orbitals from originally nth shell belong now rather to the (n+1)th shell. Similarly, $\mathrm{nf}$ orbitals from originally nth shell belong now rather to the (n+2)th shell.
So the octet rule for the 4th and higher periods have limited applicability for the $\mathrm{s}$ and $\mathrm{p}$ orbitals only for groups 1-2 and 13-18. It is not applicable for transition metals involving partially occupied $\mathrm{d}$ or $\mathrm{f}$ orbitals.


How to calculate and denote small uncertainties?

When a number is given without any further information, it is generally interpreted so that the last digit is rounded.

Therefore, the first given value $a=1.0$ is generally assumed to represent a value between $0.95$ and $1.05$, or $a=\left(1.00\pm0.05\right)$. Thus, the assumed uncertainty of $a$ is $u(a)=0.05$.

Conversely, the second given value $b=\left(0.1000\pm0.0005\right)$, where the assumed uncertainty of $b$ is $u(b)=0.0005$, might simply be written as $b=0.100$.

When multiplying numerical values, according to the simplified rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. Therefore, when you multiply $a=1.0$ (two significant digits) by $b=0.100$ (three significant digits), the result is $\boldsymbol{y=a\cdot b=0.10}$ (rounded to two significant digits).
This result may be estimated to represent a value between $0.095$ and $0.105$, or $y=\left(0.100\pm0.005\right)$; i.e. the assumed uncertainty of $y$ is $u(y)=0.005$.

By way of comparison, a generally better result for the uncertainty $u(y)$ may be obtained by using the rules of propagation of uncertainty (or propagation of error) as follows.

When calculating the product of $a$ and $b$

$$y=a\cdot b$$

and if we assume that the uncertainties of $a$ and $b$ are not correlated, the relative uncertainty of the result $y$ may be estimated as

$$\begin{align}
\left(\frac{u(y)}{y}\right)^2 &= \left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2 \\[6pt]
\frac{u(y)}{y} &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2}
\end{align}$$

Thus, the absolute uncertainty of $y$ is

$$\begin{align}
u(y) &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot y \\[6pt]
&= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot a \cdot b\\[6pt]
&= \sqrt{\left(\frac{0.05}{1.00}\right)^2 + \left(\frac{0.0005}{0.1000}\right)^2} \times 1.00 \times 0.1000\\[6pt]
&= 0.005024938\\[6pt]
&\approx 0.005
\end{align}$$

The same result may be obtained using the general formula

$$\begin{align}
u(y) &= \sqrt{\left(\frac{\partial y}{\partial a}\right)^2 u^2(a)+\left(\frac{\partial y}{\partial b}\right)^2 u^2(b)} \\[6pt]
&= \sqrt{b^2\cdot u^2(a)+ a^2\cdot u^2(b)} \\[6pt]
&= \sqrt{(0.1000)^2\times (0.05)^2+ (1.00)^2\times (0.0005)^2} \\[6pt]
&= 0.005024938\\[6pt]
&\approx 0.005
\end{align}$$

Thus, in this case, the result that is obtained by using the simplified rules of significance arithmetic is a good approximation of the result that is obtained by using the rules of propagation of uncertainty.

(Note that the numerical values of the estimate $y$ and its uncertainty $u(y)$ should not be given with an excessive number of digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.)


Estimation of change in enthalpy of a gas phase reaction using bond enthalpies

Use Hess' Law:  $\Delta$H = $\Sigma$ Enthalpy of Bonds Broken - $\Sigma$ Enthalpy of Bonds Formed

Reactants

Bond enthalpy of C2H4 = 602 + (413 * 4) = 2254 kJ mol-1
Bond enthalpy of O2 = 3 * 498 = 1494 kJ mol-1
$\therefore$ $\Sigma$ Enthalpy of Bonds Broken= 2254 + 1494 = 3748 kJ mol-1

Products

Bond enthalpy of CO2 = (799 * 2) * 2 = 3196 kJ mol-1
Bond enthalpy of H2O = (463 * 2) * 2 = 1852 kJ mol-1
$\therefore \Sigma$ Enthalpy of Bonds Formed = 3196 + 1852 = 5048 kJ mol-1

Overall

$\Delta$H = 3748 - 5048 = -1300 kJ mol-1

After having a quick Google search for the enthalpy change of combustion for ethene, it came up as -1411 kJ mol-1. Our calculated answer isn't far off. I'm hoping we got the same answer.


A way to do it would be:

\begin{align}
\left( \frac{\partial H}{\partial U} \right)_P \left( \frac{\partial U}{\partial V} \right)_P &= \left( \frac{\partial H}{\partial V} \right)_P = \left( \frac{\partial U}{\partial V} \right)_P + P\\
\left( \frac{\partial H}{\partial U} \right)_P &= \frac{\left( \frac{\partial U}{\partial V} \right)_P + P}{\left(\frac{\partial U}{\partial V} \right)_P}\\
\left(\frac{\partial U}{\partial V} \right)_P &= \frac{P}{\left( \frac{\partial H}{\partial U} \right)_P - 1}\\
\end{align}


Stability of α, β, γ allotropes at different temperatures

I think most of the time, they sort according to the order which temperature the allotrope/polymorph is more stable. Let's say element Z has two allotropes A and B. A is stable below 100 °C and B is stable above 100 °C. Thus allotrope A would be denoted alpha-Z and allotrope B would be denoted beta-Z.
Gray tin (alpha-tin) and white tin (beta-tin) is an example. Gray tin is stable only below 13.2 °C and white tin is stable above this point. So that's why gray tin is called alpha-tin because it is stable at a lower temperature, and ditto with white tin. As you can see in this case the beta allotrope would be the one stable at room temperature one.
Quartz (a crystalline form of silicon dioxide) is another example. See this from Wikipedia:

Quartz exists in two forms, the normal α-quartz and the high-temperature β->quartz, both of which are chiral. The transformation from α-quartz to β-quartz >takes place abruptly at 573 °C (846 K; 1,063 °F).


What properties of an element determines the maximum number of bonds it can make?

Considering the number of other atoms a central atom can be bonded to, "hybridization [is] irrelevant" and "the coordination number ... is not restricted by the number of
hybrid orbitals that the atom can form", quoting On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.

Examples such as $\ce{CLi6}$ and $\ce{TeF8^{2-}}$ are cited and emphasis is placed upon the ionic character of the bonding by the cited reference.

In my opinion, the main limiting factor is how many atoms can simultaneously be close to the central atom, this increasing with increasing radius of the central atom and with decreasing radius of the surrounding atoms.


How did chemists in the 18th century learn that magnesium gain mass when it's burned?

The phlogiston hypothesis is about a substance, phlogiston, which was supposed to exist within matter that be consumed or released when it is combusted.  If this were true, then when magnesium is combusted the mass would decrease.

To someone who believed this to be true, it would come as a great surprise to find that if they combusted 100g of magnesium the mass would increase by about 65%.  This would be easy to measure with a set of balances, even with much smaller masses of a few grams.

A layman might not come across many day-to-day examples of a combustion reaction with a mass increase, since most things that burn release a lot of mass as gases, and that effect is what phlogiston was intended to explain.  One possibility, aside from burning metal, is the oxidation of metal in air.  Someone could weigh an iron nail or other object, allow it to rust, and weigh again to show it has gained mass.  This could easily be demonstrated by any merchant with a set of scales, but it's not obvious the rust is an oxide related to the combustion products.


The base, as you pointed out is nearly always the salt of the acid corresponding to the annhydride.

This is because the base which differs in structure will react with the given carbonyl compound and yield some other compound rather that abstracting a proton.

Eg.
Let's say you have tertbutyl salt there instead of what we are having right now.The tert butyl could have directly attacked on the Carbon having the double bonded oxygen as it has a partial positive charge due to polarization.Which will result in to eliminating the other symmetric half of the annhydride and replacing it by tertbutyl.This changed your basic reacting molecule.And hence the
 product you obtained is definitely not the right one. Whereas using the salt of the acid that formed the annhydride will not change the annhydride even if it reacts with the annhydride the way I stated above.

Using the salt of the acid they produced the annhydride does nothing but abstracts the proton or simply regenerate the original carbonyl compound/annhydride if it reacts rather than abstracting a proton.


What hinders SN2 more: geminal alkyl group or alkyl group on the aromatic ring?

The correct answer is X because the 2 methyl groups on the Y are bulky enough to cause steric hindrance. Surprisingly, from what I have seen, if you only had one methyl group on Y, the SN2 reaction would proceed.


In a mixture of NaOH and Na2CO3, why doesn't phenolpthalein change its color until Na2CO3 is half neutralised?

a) That's right, $\ce{NaOH}$ reacts first. But you are wrong in thinking that $\ce{NaCl}$ has a pH of 7. In a way, $\ce{NaCl}$ has "no pH at all". It just sits there and does nothing. It is other component(s) of the solution that determine its pH. At that point, the other component is $\ce{Na2CO3}$, which is alkaline due to hydrolysis. As the titration progresses, it turns to $\ce{NaHCO3}$, hence the change in pH.

b) Well, $\ce{H2CO3}$ would indeed make the solution slightly acidic, but not enough so to trigger the color change in methyl orange. It is $\ce{HCl}$ that does the job. After the neutralization is completed, one tiny extra drop is enough.


The unit of energy, 1 dm³ atm, is equal to how many calories?


Why does ice cream not go back to its original form upon refreezing?

Ice cream is a frozen emulsion mixed with air. When the ice cream melted, the air escaped. When you re-froze it, it formed a solid mass.


Can rate of reaction depend on both the reactants and the product?

There exist autocatalytic reactions than can show such a behaviour. E.g. a reaction $\ce{A + B \rightarrow 2\ B}$ might show the rate $k[\ce{A}] [\ce{B}]$.

The reaction of permanganate with oxalic acid is a well known autocatalytic reaction.
$$
\mathrm{ 5\ (COOH)_2 + 2\ MnO_4^- + 6\ H^+ \rightarrow 10\ CO_2 + 2\ Mn^{2+} + 8\ H_2O}
$$
Here the generated manganese(II) ions catalyze the reaction. The reaction starts slowly if no initial manganese(II) ions are present, and then speeds up.


Do all polymers contain either carbon or silicon?

No, neither C nor Si have to be present in a polymer.
IUPAC defines polymer simply as

  A substance composed of macromolecules.

In turn, there is no limitations on the elements the macromolecules may consist of.
In fact, there is an extensive class of inorganic polymers free of carbon and silicon.


Creating Boric Acid from Borax and Hydrochloric Acid

Assuming that boric acid is completely soluble in water, you did everything correctly so far, you simply missed the last step: dilution.

First you calculated the amount of substance of boric acid you would need to produce a two molar solution. With that you basically assume, that you dissolve solid boric acid in water.
You then calculated the amount of borax you need for complete reaction with hydrochloric acid. Since this is already hydrogen chloride dissolved in water, you bring some of the water you assumed before. But this is as your calculation says only about 40 mL.
Still assuming that it is completely soluble, you would have produced a solution that is about ten molar. You still need to dilute it to the final volume, to have the desired concentration.

Now let's go one step further. Boric acid is not very soluble in water under standard conditions, see Wikipedia. As A.K. pointed out, solubility is 47.2 g/L at 20 °C. That translates to 0.76 mol/L. As you can see solubility increases with temperature, in boiling water the maximum concentration is about 4.5 mol/L.
So what you are trying to attempt is not possible, at least not in pure water at room temperature. Try heating it up a bit, somewhere around 70 °C should be enough.

Alternatively PubChem has kindly provided us with a little more statistics about solubility.
In methanol you can obtain the desired two molar solution at room temperature, as solubility is 173.9 g/L at 25 °C.


This is perhaps not the best example, since hydroxide ion will react with bicarbonate ion to form carbonate ion, and also as you titrate carbonate ion note that it will form bicarbonate again. But, generally, yes, the strongest base will be consumed first as you titrate. The titration curve will tell you the amounts of the bases present and the pKas of their conjugate acids.


Differences between chemical physics and physical chemistry?

You are of course right, there is a lot of overlap. There are also the scientific publication series ChemPhysChem and Physical Chemistry Chemical Physics amongst others, that deal mainly with subjects from the both areas.

In principle they are most likely dealing with almost the same subjects. And you often find physicists working in chemistry and vice versa. However, there is a little difference in the general approach to problems.
Physical chemistry usually approaches the problems at hand from a chemist's point of view, focussing (initially) mainly on macroscopic properties of an ensemble and use physical laws to determine them. It is often considered the parent field for many other subjects, like physical organic chemistry, chemical kinetics, and spectroscopy, among others. Most common publication titles are Zeitschrift für physikalische Chemie and The Journal of Physical Chemistry A, B, C, and letters. And many others...
Chemical physics on the other hand studies chemical ensembles and reactions from a physicist's point of view. They focus more or less, like you assumed, on ions, atoms, clusters, surfaces, free radicals, and molecules interacting with each other. This does include, but is not limited to, the study of solvation effects and single entities like quantum dots. Here are fewer specialist titles, like The Journal of Chemical Physics and Chemical Physics Letters. Of course publications are often widely accepted also in other journals of neighbouring areas. (Sometimes it is also treated as a subfield of physical chemistry.)

In modern times the various fields are not separated by a sharp line, they most likely blur into each other. You could probably ask the same question for Biochemistry and Chemical Biology, or Computational Chemistry and Quantum Chemistry, or Polymer Chemistry and Material Sciences. And there are more. Interdisciplinary cooperation is encouraged by many researchers and institutions, it is important not to care too much about the labels anymore.


Can you find the mass of solvent with mass of solute, volume of solution, and solution density?

That works!  In most chemistry (but not nuclear reactions!),  mass is always conserved, so

$$m_\text{solution} = m_\text{solvent} + m_\text{solutes}$$

Mass is also equal to density times volume, so

$$m_\text{solution} = \rho_\text{solution} V_\text{solution} = m_\text{solvent} + m_\text{solutes}$$

Solving for $m_\text{solvent}$:

$$m_\text{solvent} = \rho_\text{solution} V_\text{solution} - m_\text{solutes}$$

Which is exactly the calculation you did.


First some basics....
Lithium has three electrons.

$\ce{Li}$ electronic configuration: $1s^22s^1$
$\ce{Li -> Li+ + e-\quad\quad\mathrm{Ionization\ Energy= 0.520\ MJ/mol}}$
$\ce{Li+}$ electronic configuration: $1s^2$
$\ce{Li+ -> Li^2+ + e-\quad\quad\mathrm{Ionization\ Energy= 7.298\ MJ/mol}}$
$\ce{Li^2+}$ electronic configuration: $1s^1$
$\ce{Li^2+ -> Li^3+ + e-\quad\quad\mathrm{Ionization\ Energy=11.815\ MJ/mol}}$
$\ce{Li^3+}$ electronic configuration: $1s^0$, i.e. no electrons
Ionization energies from Wikipedia

Thus the different ionization reactions remove the electrons from the outer most to the inner most electrons.
Below is an another image of the idealized spectra of lithium that has the binding energies stated. (from https://www.andrew.cmu.edu/course/09-105/S98Slides9.html)

$\mathrm{Binding\ Energy= 0.52\ MJ/mol}$ corresponds to $\ce{Li -> Li+ + e-}$
or $1s^2 2s^1\ce{->} 1s^2$, hence the first ionization energy of a Li atom.
However $\mathrm{Binding\ Energy= 6.26\ MJ/mol}$ corresponds to $\ce{Li -> Li+ + e-}$
or $1s^2 2s^1\ce{->} 1s^12s^1$, which is not the first or second ionization energy of a Li atom. Rather it is the binding energy of the 1s orbital in a neutral lithium atom.
Hence the photoelectron spectra is called the binding energy spectra because that is what it measures. Note that the sweep of the spectra would not have to cover the ionization energy. For example the spectra could have been taken from 5 MJ/mol to 100 MJ/mol.


Predicting sigma bond overlap strengths of s-s, p-p, s-p, sp-sp etc

I think the issue might be what orbitals are being referred to. Both books could be correct if they are referring to different $\mathrm{p-p}$ overlap. For example, suppose two atoms are bonding along the z-axis. $\mathrm{p_z}-\mathrm{p_z}$ overlap would be greater than $\mathrm{s-s}$ because the lobes of the p orbitals extend out to meet each other between the two molecules. If however we looked at $\mathrm{p_x}-\mathrm{p_x}$ overlap, the $\mathrm{s-s}$ overlap would be greater because the p lobes don't extend toward each other. Increasing hybridization increases overlap because it combines the directionality of p orbitals (that they extend out away from the atom) and the electronic density of s orbitals (larger space for overlap to occur).


Does hyperconjugation into sigma-star explain methylamine's increased basicity relative to ammonia?

Yes, I once thought about this interesting phenomenon also. There is no good reason why hyperconjugative interactions cannot take place between the $\ce {N}$ lone pair orbital and the $\ce {C-H}$ $\sigma$. However, the interaction may be less favoured if you consider the geometry, compared to the the usual hyperconjugation to alkene $\pi$ systems or to carbocation $\ce {p}$ orbitals. Although geometry does not help to favour this interaction as much, the change of atom from $\ce {C}$ in the usual case to $\ce {N}$ does help to decrease the energy gap for the interaction since $\ce {N}$ is more electronegative than $\ce {C}$ and thus, its atomic orbitals would all be of a lower energy compared to $\ce {C}$.

Raising the energy of the HOMO of a general nucleophile would increase its nucleophilicity by decreasing the energy difference between its HOMO and the LUMO of the electrophile. In the same way, by this filled orbital-filled orbital interaction, the HOMO of the methylamine is now raised, allowing it to more easily interact with acids. Thus, $\ce {MO}$ theory can be used to explain the increasing basicity of amines with increasing alkyl substitution.


Does (chloromethoxy)ethane react in a bimolecular or monomolecular mechanism?

Probably both mechanisms can happen and it would require experimental data from reaction kinetics and computational calculations to determine which one of the mechanism is domintant.

One example of determined mechanism about neighboring group effect, also called anchimeric assistance is mustard gas (chloromethine) alkylation reaction with DNA.

Effect of chloromethine is based on its ability to block DNA replication by alkylating guanine, and after the war it has been also used as cancer drug.


Help understanding how "steric effects" are distinct from "electronic effects"?

The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds
All the normal physical interactions we experience are arguably electronic. When you touch your desk, you feel force because of interactions between the molecules of the desk and the molecules of your hand because the molecules of one interact "electrostatically" with the molecules of the other.
But your hand and your desk are not bonded together.
When a noble gas condenses into a liquid, it has a finite volume. The atoms become liquid because there are weak attractive forces (not usually described as "bonds") but have a fixed volume because the strong forces between their filled electron shells repelling each other prevent the atoms packing together any more closely. This second effect is very similar to the force stopping you hand penetrating your desk.
Arguably this is all electronic interactions.
But, if you are a chemist, it is worth drawing a line because it makes it much easier to talk about the properties of molecules. Electronic interactions to a chemist are usually interactions transmitted via bonds in a molecule.
So a molecule like fluorescein (shown below) has an extended pi-system so the oxygens at either end of the three fused rings can interact via bonds (and the overall molecule has electronic transitions that make it a useful fluorescent dye).

But in a molecule like this substituted biphenyl the electronic interactions might drive the molecule to be planar (so the two pi-systems can communicate) but the steric size of the 4 methyl groups prevent that because the methyl groups would come too close in 3D space to allow a planar configuration.

So the molecule will not have two aromatic rings in the same plane but will have significant angles between the rings. A "space filling" view (below) shows the methyl groups take up too much space to allow the rings to be coplanar.

Chemists think of the forces trying to make this molecule planar as "electronic" as they are transmitted via the bonds but the forces preventing this as being "steric" interactions. They are more like the forces preventing your hand from penetrating your desk than they are alike the forces holding the molecule together.
It is a useful distinction even when the forces are internal to a molecule.


See H++ if you want a rough prediction of $\mathrm{p}K_\mathrm{a}$.

If you want to more accurately calculate the $\mathrm{p}K_\mathrm{a}$ of your molecules, you should run a quantum mechanics calculation with a continuum solvent model (example calculation: [1]).

The general outline is as follows. Run a geometry optimization and vibrational mode analysis on your system in basic and conjugate acidic forms. So, $\ce{BH+}$ and $\ce{B}$, or $\ce{BH}$ and $\ce{B-}$, depending on what your system looks like. This will allow you to extract a free energy of reaction for each of your lipid molecules.

Importantly, these calculations will not give accurate single point free energies, but the relative values ($∆G$) will be. As such, you will need a reference $∆G$ from literature of a similar reaction to "set the 0" of your values. Finally, you can get $\mathrm{p}K_\mathrm{a}$ from $∆G$ using

$$\mathrm{p}K_\mathrm{a} = \frac{∆G^\circ}{2.303RT}$$

Refrences

Liptak, M. D.; Shields, G. C. Accurate $\mathrm{p}K_\mathrm{a}$ Calculations for Carboxylic Acids Using Complete Basis Set and Gaussian-n Models Combined with CPCM Continuum Solvation Methods. Journal of the American Chemical Society 2001, 123 (30), 7314–7319. https://doi.org/10.1021/ja010534f.


Empirical and Molecular formula: Homework

The molecular formula and empirical formula are right. As we known, empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

If the compound molecular formula is C2H4Cl2, so we divide it with 2, 4, and 2 GCD / greatest common divisor. The GCD is 2, so the empirical formula is CH2Cl.


How would you find the mass in grams, if you only know the number of particles of a portion of the formula?

The formula $\ce{Fe2(SO4)3}$ tells us that 3 molecules of $\ce{SO4^{2-}}$ exist for every $\ce{Fe2(SO4)3}$ molecule.

$\frac{\ce{1 Fe2(SO4)3 molecule}}{\ce{3 SO4^{2-} molecule}} * 3.59*10^{23} \ce{SO4^{2-} molecule}$ can be used to find the number of molecules of ferric sulfate.

You can then use $\frac{1 \ce{mole}}{6.022*10^{23} \ce{molecule}}$ to find the moles of ferric sulfate.

You can then use the given information to solve for grams of ferric sulfate.

Write a comment if there's any ambiguity.


My suggestion is to stop trying to predict the product by rules like “rich becomes richer” or “proton goes to carbon having greater number of hydrogens attached to it”, etc. for Markovnikov addition and opposite for anti-Markovnikov addition.
These are subject to many exceptions.
Instead, draw the direction of dipole in the molecule at the site of the double bond, (order of precedence: +R effect, +H effect, +I effect for considering direction). After doing this, $\ce{H+}$ will attach the carbon with δ− charge (as electrophile will attack first, and carbocation formed on δ+ carbon which is more stable), and nucleophile going to other carbon atom in case of Markovnikov addition.
The opposite holds for anti-Markovnikov addition, i.e. $\ce{H+}$ will attack carbon with δ+ charge.
Now coming to your question. After you figure out on which carbon $\ce{H+}$ will attack, the carbocation can further be stabilized by Hydride shift (always check for rearrangement possibilities like hydride, methyl, ethyl shift, ring expansion, etc.). After rearrangement only does the nucleophile $(\ce{Br-}$ here) attack.


What is an intuitive explanation for Electron Configurations

Electron configurations are a bunch of numbers and letters, and they also follow a pattern. But there is also a logic behind the whole thing.

Basically, chemists use electron configurations to see in which atomic orbitals the electrons are found. From this several things can be derived, most of all the reactivity of said electrons (which is where the chemistry happens).

The intuitive answer is this: Electron configurations are a way of telling in which atomic orbitals the electrons are put in, according to the Aufbau principle and the Pauli exclusion principle. Decoding it is pretty easy, as it always follows this pattern:
$n$$(s|p|d|f)^x$ where $n$ is the main quantum number (or shell number), $(s|p|d|f)$ denotes the atomic orbitals with the quantum number and $x$ is the number of electrons found in the orbitals.

The orbital "groups" ($p$, $d$, $f$) contain several "orbitals" (side note: an orbital is strictly seen a one-electron function) where two electrons with opposite spin fit in. $p$ has 3 ($p_x,\; p_y,\; p_z$), $d$ has 5 ($d_{xy},\; d_{xz},\; d_{yz},\; d_{x^2-y^2},\; d_{z^2}$) and $f$ even more. This means that the maximal number of electrons in the $p$ group is $6$, and in the $d$ group $10$. So when you fill in the electrons according to the aufbau principle, you have to go to the next orbital group when you reach this number.

Bottom line: Every electron in an atom is defined by the four quantum numbers $n$ (main or principal quantum number), $l$ (azimuthal), $m_l$ (magnetic) and $m_s$ (spin). Using the electron configuration we can derive the first three of these four numbers.


Run a command at every step of a relaxed scan

Guess what: It is possible (in your specific case).

I was wrong. At least a little bit. Most of what is described below is true for the general purpose, and I will not change that part of the answer.

However, in the specific case where you only want to compute the volume, you are able to do that because I think it is part of the population analysis module. You can actually request a population analysis at every step of the optimisation, but be careful what you wish for (running NBO will take forever). That way the optimisation will be a little slower, but that is the most convenient way. Therefore the solution to your very specific problem is adding the following line to the route section:

pop=always volume

Be aware, that the output will be an even bigger nightmare than before.

Here is a complete input of the $\ce{H3B-NH3}$ I have used previously.

%chk=df-bp86-d3.def2svp.chk
%nproc=4
%mem=16000MB
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
pop=always                         ! Perform a population analysis at every step
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Scan

0 1
  N       0.00000        0.00000       -0.80852
  H       0.92634        0.00000       -1.36025
  H      -0.46317       -0.80223       -1.36025
  H      -0.46317        0.80223       -1.36025
  B       0.00000        0.00000        0.19148
  H       0.52348        0.90670        0.85928
  H      -1.04696        0.00000        0.85928
  H       0.52348       -0.90670        0.85928

B 1 5 S 5 0.2

For comparison (I'm not 100% sure I used the right values):

Energies
Step1 SCF Done:  E(RB-P86) =  -82.7531561505     A.U. after    8 cycles
Step2 SCF Done:  E(RB-P86) =  -83.0511089272     A.U. after    7 cycles
Step3 SCF Done:  E(RB-P86) =  -83.1459883448     A.U. after    7 cycles
Step4 SCF Done:  E(RB-P86) =  -83.1661412429     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1606918387     A.U. after    8 cycles
Step6 SCF Done:  E(RB-P86) =  -83.1485268431     A.U. after    8 cycles

Volumes
Step1 Molar volume =  491.032 bohr**3/mol ( 43.819 cm**3/mol)
Step2 Molar volume =  415.148 bohr**3/mol ( 37.047 cm**3/mol)
Step3 Molar volume =  401.060 bohr**3/mol ( 35.790 cm**3/mol)
Step4 Molar volume =  406.785 bohr**3/mol ( 36.301 cm**3/mol)
Step5 Molar volume =  514.428 bohr**3/mol ( 45.907 cm**3/mol)
Step6 Molar volume =  453.278 bohr**3/mol ( 40.450 cm**3/mol)

The energies differ very slightly only (from the below approach), but the volumes do differ at some points. Now that could be, that I just used the wrong population analysis, or something else. I don't have the time to investigate. (Leave a comment or edit if you find out.)

I've again looked into that, but without some serious hacking via the external keyword interface, I don't think it is possible. I also couldn't find a suitable iOP to control what gets punched to checkpoint file and when, and then use the result from that for a property/analysis run.
The punch keyword is equally disappointing, only outputting the last state.

While the manual suggests that you can access a given step of a scan in the checkpoint file, I found this to be not true. The documentation is fuzzy on that part and it also warns, that not everything is stored. Since it is a binary file, there is also no real way of checking what is contained.
I guess the real reason for the existence of that option is to restart a calculation that went rogue with different parameters from an intermediate calculation.

I was equally unable to find a way to set a given redundant coordinate to a specific value and then freezing it in order to perform a partial optimisation at that point, and then manipulate that coordinate while reading in the guess and perform another calculation.
Even with the new and generally awesome way to define general internal coordinates in Gaussian 16 (GIC) I was unsuccessful.

From all that trying, I came up with the probably easiest (=lazy) way (not computationally most efficient) to do what you want is to perform the scan, extract the (p'optimised) coordinates and run a series of single point calculations.

A slightly less (probably) computationally heavy approach would be to perform an initial scan on a very low level (like pm6, if your molecule is well-behaved), and the do p'optimisations on the resulting steps. Usually pm6 is able to offer reasonable starting geometries, which in many other cases may also help you reduce computational effort. So this is probably easy and quite efficient.
There are various options to extract geometries from a scan, the following question deals exactly with that: Extract all structures of Gaussian 09 molecular dynamics calculation using babel?

There certainly are more efficient ways to do this kind of task, but in the end, most of them require manual labour.

N.B.: I cannot believe it, but the pm6/popt approach is actually a halfway decent application of compound jobs. Although I have to admit I have not tried it yet.
It works as the following example demonstrates. (I used Gaussian 16 Rev. A.03, but it should work with Gaussian 09 Rev D.01, too. Maybe even earlier.)

I set up an initial scan on the pm6 level of theory for $\ce{H3B-NH3}$ and scanned the $\ce{B-N}$ bond:

%chk=pm6.scan.chk
%nproc=2
%mem=8000MB
#P PM6
OPT(MaxCycle=100)
SYMMETRY(loose)
GEOM(ModRedundant)
Volume

title

0 1
 N     0.000000     0.000000    -0.764994
 H     1.019690     0.000000    -1.164250
 H    -0.509845    -0.883078    -1.164250
 H    -0.509845     0.883078    -1.164250
 B    -0.000000     0.000000     0.235006
 H     0.509845     0.883078     0.634262
 H    -1.019691     0.000000     0.634262
 H     0.509845    -0.883078     0.634262

B 1 5 S 5 0.2

I have extracted the pre-optimised geometries with Chemcraft and added a header section and made use of the --Link1-- separator. After the first optimisation I read the MO for a slight speedup. I'm sure some of this can be further automatised, but for the sake of demonstrating, I did it by hand. The commented input:

%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 1

0 1
  N       0.00000        0.00000       -0.80852
  H       0.92634        0.00000       -1.36025
  H      -0.46317       -0.80223       -1.36025
  H      -0.46317        0.80223       -1.36025
  B       0.00000        0.00000        0.19148
  H       0.52348        0.90670        0.85928
  H      -1.04696        0.00000        0.85928
  H       0.52348       -0.90670        0.85928

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 2

0 1
  N       0.00000        0.00000       -0.90852
  H       0.94043        0.00000       -1.36414
  H      -0.47021       -0.81444       -1.36414
  H      -0.47021        0.81444       -1.36414
  B       0.00000        0.00000        0.29148
  H       0.53853        0.93276        0.86317
  H      -1.07706        0.00000        0.86317
  H       0.53853       -0.93276        0.86317

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 3

0 1
  N       0.00000        0.00000       -0.99853
  H       0.95023        0.00000       -1.38525
  H      -0.47511       -0.82292       -1.38525
  H      -0.47512        0.82292       -1.38525
  B       0.00000        0.00000        0.40147
  H       0.55372        0.95907        0.87762
  H      -1.10743        0.00000        0.87762
  H       0.55372       -0.95907        0.87762

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 4

0 1
  N       0.00000        0.00000       -1.07752
  H       0.95590        0.00000       -1.42049
  H      -0.47795       -0.82783       -1.42049
  H      -0.47795        0.82783       -1.42049
  B       0.00000        0.00000        0.52248
  H       0.56746        0.98288        0.89885
  H      -1.13493        0.00000        0.89885
  H       0.56746       -0.98288        0.89885

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 5

0 1
  N       0.00000        0.00000       -1.14730
  H       0.95649        0.00000       -1.46941
  H      -0.47824       -0.82834       -1.46941
  H      -0.47824        0.82834       -1.46941
  B       0.00000        0.00000        0.65270
  H       0.57849        1.00197        0.92763
  H      -1.15698        0.00000        0.92763
  H       0.57849       -1.00197        0.92763

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 6

0 1
  N       0.00000        0.00000       -1.21413
  H       0.95402        0.00000       -1.53258
  H      -0.47701       -0.82621       -1.53258
  H      -0.47701        0.82621       -1.53258
  B       0.00000        0.00000        0.78587
  H       0.58528        1.01374        0.96868
  H      -1.17057        0.00000        0.96868
  H       0.58528       -1.01374        0.96868

B 1 5 F

For comparison:
Energies

Step1 SCF Done:  E(RB-P86) =  -82.7531561507     A.U. after    8 cycles
Step2 SCF Done:  E(RB-P86) =  -83.0511087360     A.U. after    8 cycles
Step3 SCF Done:  E(RB-P86) =  -83.1459860378     A.U. after    7 cycles
Step4 SCF Done:  E(RB-P86) =  -83.1661471840     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1606709617     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1484921220     A.U. after    8 cycles

Volumes

Step1 Molar volume =  459.690 bohr**3/mol ( 41.022 cm**3/mol)
Step2 Molar volume =  479.017 bohr**3/mol ( 42.747 cm**3/mol)
Step3 Molar volume =  401.060 bohr**3/mol ( 35.790 cm**3/mol)
Step4 Molar volume =  406.785 bohr**3/mol ( 36.301 cm**3/mol)
Step5 Molar volume =  570.344 bohr**3/mol ( 50.897 cm**3/mol)
Step6 Molar volume =  498.606 bohr**3/mol ( 44.495 cm**3/mol)

Animation


Reaction between halogens and concentrated nitric acid

This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:

$$
\begin{align}
\ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\
\ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &= \pu{0.934 V} \tag2\\
\ce{2NO3- + 4H+ + 2e- &<=> N2O4 + 2H2O} &\quad E^\circ &= \pu{0.803 V} \tag3
\end{align}
$$

Now, let's look at reduction potential of halides:

$$
\begin{align}
\ce{F2 + 2e- &<=> 2F-} &\quad E^\circ &= \pu{2.866 V} \tag4\\
\ce{Cl2 + 2e- &<=> 2Cl-} &\quad E^\circ &= \pu{1.3583 V} \tag5\\
\ce{Br2 + 2e- &<=> 2Br-} &\quad E^\circ &= \pu{1.066 V} \tag6\\
\ce{I2 + 2e- &<=> 2I-} &\quad E^\circ &= \pu{0.5355 V} \tag7\\
\end{align}
$$
Accordingly, we can list oxidizing power at standard conditions in decreasing order:

$$\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{NO3-} > \ce{I2}$$

Therefore, at standard condition, concentrated $\ce{HNO3}$ acid (meaning $\ce{NO3-}$ in acidic medium) can easily oxidize only $\ce{I2}.$ One such example is documented in Ref.1. According to the authors, the oxidation of iodine by hyperazeotropic nitric acid ($\pu{16–21 M}$) was shown to proceed in at least two steps:

$$
\begin{align}
\ce{I2 + 3HNO3 &<=>[$k_1$][$k_2$] 2I+ + 2NO3- + 2HNO2 + H2O}\\
\ce{I+ + 6HNO3 &<=>[$k_3$][$k_1$] I^5+ + 4NO3- + 2HNO2 + 2H2O}
\end{align}
$$

Since reduction potentials of $\ce{HNO3}$ and $\ce{Br2}$ are in such close range that each can oxidize the other. In fact, the kinetic studies of this phenomenon has been performed in literature (Ref.2).

In different aspect, there is a publication (Ref.3) claiming that the presence of bromine in fuming nitric acid makes it a brominating reagent (electrophilic aromatic substitution):

This bromination has worked well with deactivated aromatic nucleus. In absence of bromine, the reagent nitrate the ring. Thus, it is safe to suggest that $\ce{HNO3}$ might promote formation of '$\ce{Br+}$', meaning oxidation.

References:

J. C. Mailen, T. O. Tiffany, "The reaction of iodine with concentrated nitric acid," Journal of Inorganic and Nuclear Chemistry 1975, 37(1), 127-132 (https://doi.org/10.1016/0022-1902(75)80139-4).
Istvan Lengyel, Istvan Nagy, Gyorgy Bazsa, "Kinetic study of the autocatalytic nitric acid-bromide reaction and its reverse, the nitrous acid-bromine reaction," J. Phys. Chem. 1989, 93(7), 2801–2807 (https://doi.org/10.1021/j100344a021).
Alexander M. Andrievsky, Vera I. Lomzakova, Mikhail K. Grachev, Mikhail V. Gorelik, "Aromatic Bromination in Concentrated Nitric Acid," Open Journal of Synthesis Theory and Applications 2014, 3(2), Article ID:44941, 5 pages (doi: 10.4236/ojsta.2014.32003).


Does a messier room actually have higher entropy?

A room can be a complicated system, (at least mine can be) so I shall assume a simplified version: an enclosed space bound by 4 walls, a ceiling and a floor and the only piece of furniture inside the room is a desk, and  two books on it -- book 1 and book 2.

In the this scenario, I would define a clean room as one with both the books neatly stacked on the desk, and in a messy room you can find one of the books on the floor.

Now, if I were to try enumerate the number of ways of organising the books in a clean and a messy room, I would find that there is exactly one way in which I can achieve a "clean room configuration", and that is by having both books on the desk. On the other hand if you wish to achieve a "messy room configuration", you can do so in two ways-- either have book 1 on the floor and book 2 on the desk, or have book 2 on the floor and book 1 on the desk.

The no. of ways in which you can achieve a configuration is called its weight and is defined as:

$$W = \frac{N!}{N_0!N_1!N_2!....}$$

In the first case, $W=1$ (clean) and in the second $W=2$ (messy)

The boltzmann formula for entropy is: $ S = k_b\ln W$

Plugging in $W$, we get $$S_\text{clean} = 0$$ and $$S_\text{messy} = k_b\ln2$$

Thus, the messier room has a higher entropy. This somewhat "silly" argument can be reproduced for a room with more furniture, and more items in it. Feel free to toss in some clothes, papers, chairs, book cases, toys, shoes, pets...go nuts.

So, I believe the analogy holds good. But again, what I did here is a silly exercise to illustrate a point, and it isn't meaningful as such.

If you wish to learn more, watch this lecture by Leonard Susskind