In the case of $\ce{HNO3}$, according the abstract of this paper$^\dagger$, nitric acid vapors act as nucleation sites for ambient water, causing the mists/fumes that are observed above concentrated nitric acid solutions:

  The gas phase reaction [forming nitric acid from other reagents] supplies $\ce{HNO3}$ nuclei on which the water vapor readily condenses to form strong nitric acid mist droplets.

According to this book$^\ddagger$:

  Hydrochloric acid has thermodynamic properties rather similar to those of nitric acid.

So, I would hypothesize that the physical chemistry phenomena of evaporation and hygroscopic nucleation are responsible for the fuming of $\ce{HCl}$.

$^\dagger$ Goyer, J Colloid Sci 18(7), 616 (1963)

$^\ddagger$ Kulmala and Wagner, "Nucleation and Atmospheric Aerosols," 1996, p591.


Whats the effect of pressure on water electrolysis of sea water with increase in depth?

The higher pressure itsels doent make it much more difficult. But...

For water electrolysis you need high overpotential, preferably large electrodes with small electrode gap between them. You want clear, filtered water and no fishes electrocuted, burning down the whole submarine due to an accidental short.

And you will need plenty of gas for buoyancy control, especially at $200{-}1000\ \mathrm{m}$ depths, therefore you need a lot of electricity:

According to Faraday's law $$m=(Q/F)(M/z)$$ where $F = 96485\ \mathrm{C/mol}$ you can make like $4.5\ \mathrm{l}$ of hydrogen gas (1 atm, 25 °C) with a $10\ \mathrm{A}$ current in 1 hour! That is like $0.04\ \mathrm{l}$ at $1000\ \mathrm{m}$ ($100\ \mathrm{atm}$) of a potentially explosive gas.


What solute should I use with dialysis tubing if I want a 100% retention rate?

Polyethylene oxide, AKA polyethylene glycol or PEG has MW > 100,000. You can find higher-MW PEG in disposable diapers and similar absorbent products.

You could also use agar, pectin or gelatin, available in groceries and from lab supply companies.

One of my favorite demos was to make an agar solution with a bit of phenolphthalein added and put it in sausage casing (AKA dialysis tubing). Holding the hotdog over an open bottle of ammonium hydroxide turned it red, and over hydrochloric acid, back to a clear frank.


Explanation of Lemon Juice-Invisible Ink

Lemon juice works as a sympathetic ink partially due to its caramelization, as @Mithoron profoundly noticed -- especially upon extreme heating.
Recent review of the degradation process of lemon juice [1] suggests that browning of lemon juice is associated with 3 processes:

ascorbic acid degradation;
Maillard reaction;
caramelization.

One of the first in-depth investigations revealed the major role of ascorbic acid degradation referring to the browning phenomena [2]:

Compared with other fruit products, lemon juice contains few natural pigments to mask.
...
Because of the high acidity (pH 2.5) of this product, it was unlikely that browning was due to sugar-amine condensation and the results showed that ascorbic acid was the main precursor. Browning of lemon juice and model systems was proportional to the level of ascorbic acid; the presence of amino-acids in model systems increased the intensity of browning.
The principal role of amino-acids in the non-enzymic browning of fruit products appears not to be in the initial reaction leading to the formation of reactive compounds, but to increase the browning potential after the oxidation of the ascorbic acid to reactive carbonyl compounds. Thus, the pathway for the build-up of carbonyls depends on the pH and nature of the product, but in all cases the formation of melanoidin complexes originates from the polymerisation of carbonyl and $\alpha$-amino-groups.

It was also confirmed that the $\alpha/\beta$-unsaturated carbonyls are potent browning agents and also that dicarbonyls of the glyoxal type make a contribution to browning in the early stages.
More recent investigation at elevated temperatures (up to $\pu{120^\circ C}$) also supports the major role of ascorbic acid degradation among other factors of browning [3]:

Browning of clariﬁed cashew apple juice was caused by the degradation of ascorbic acid. Changes in the absorbance at $\pu{420 nm}$ and the ascorbic acid levels during thermal treatment of clariﬁed cashew apple juice were
described by ﬁrst order kinetics, showing the correlation between ascorbic acid loss and colour formation (browning).

It is concluded in all mentioned sources that formation of intermediate undesirable carbonyl compounds (mostly furfural and 5-hydroxymethylfurfural (5-HMF)) and various products of their polymerization and combination with amino acids (furfural is an aldehyde) upon heating is one of the major factors of browning.

Maillard reaction is a chemical reaction between an amino acid and a reducing sugar, usually requiring the addition of heat. The reaction is initiated by a condensation between the free amino group of an amino acid, peptide, or protein and the carbonyl group of a reducing sugar, leading to the formation of Amadori compounds (various furoylmethyl derivatives in case of fruit juices)[4].

Caramelization generally occurs at high temperatures (above $\pu{140^\circ C}$), when the juice is dehydrated and has high sugar content. Basically it is a pyrolysis and a final step of decomposition.

To sum it up, these are the products of first 2 processes (ascorbic acid degradation and reaction with amino acids (Maillard process)) (from [3], referred to as browning markers):

Bibliography

Bharate, S. S.; Bharate, S. B. J Food Sci Technol 2014, 51 (10), 2271–2288. DOI 10.1007/s13197-012-0718-8.
Clegg, K. M. Journal of the Science of Food and Agriculture 1964, 15 (12), 878–885. DOI 10.1002/jsfa.2740151212.
Damasceno, L. F.; Fernandes, F. A. N.; Magalhães, M. M. A.; Brito, E. S. Food Chemistry 2008, 106 (1), 172–179. DOI 10.1016/j.foodchem.2007.05.063.
del Castillo, M. D.; Corzo, N.; Olano, A. Journal of Agricultural and Food Chemistry 1999, 47 (10), 4388–4390. DOI 10.1021/jf990150x.


Why doesn't baking soda react with the solid form of an acid?

How do you know a reaction is not occurring when you mix the two solids? The only cue we have that a reaction occurs in the solution state is the formation of bubbles of escaping carbon dioxide. If there is no liquid to displace, would we be able to tell if carbon dioxide was being produced? Yes, but not with the naked eye.

  The real issue is that the solid state reaction is much, much slower.

To produce carbon dioxide, the bicarbonate ion needs to come in contact with an acidic proton on citric acid. This is unlikely in the solid state since there is very little kinetic energy.

In solution, there is more kinetic energy, so the molecules are more likely to encounter each other. Additionally, the proton transfer can propagate quickly through the hydrogen bond network in water.

$$\ce{HCO3- \bond{...} H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-Cit}$$


Sintered vs perforated plate Büchner funnels

The thing on the left is not usually a called a Büchner funnel, but a frit. Frits are generally meant to be used for stuff that can be redissolved afterwards, because they tend to clogg permanently, if very small particles get stuck inside.
You do not normally put a filter paper on it, because that would leak on the edge. The filter grade should be etched onto the glass somewhere. https://en.wikipedia.org/wiki/Fritted_glass Sometimes a frit can be used to hold another, powdery filtering substance, if that has a coarse enough grain size.
Advantage of the frit is the well defined pore size, you can heat it for drying, it can be mounted in a glass tube with two ground glass connectors (for evacuating), it can be used with concentrated and/or oxidising acids or other solutions, ...
Advantage of the Büchner funnel is that it does not get clogged with insoluble stuff, ease of use, cheap, available in large diameters, very mechanically solid, tolerates alkalic solutions (glass frits do too, but only briefly), ...


The rate of reaction is defined in IUPAC's Gold Book as:

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

where $[\ce{X}]$ denotes the concentration of the substance $\ce{X}$, and the lower case letters represent the coefficients of those substances in the balanced equation. The rates for $\ce{P}$ and $\ce{Q}$ are positive, as these are products in the reaction.

You are right in that you can determine the reaction rate by studying the change in concentration of a particular reactant (or product) over time. But, to determine the rate r for the reaction, not just for a particular species, you must take into consideration the coefficients in the balanced equation. The stoichiometric numbers are included in the definition of the rate, $r$, so that it is independent of which reaction species you study. For example if $a = 1$ and $b = 3$ then $\ce{B}$ is consumed three times more rapidly than $\ce{A}$, but $r = -\mathrm d[\ce{A}]/\mathrm dt = -(1/3)\mathrm d[\ce{B}]/\mathrm dt$ is uniquely defined.

Source: https://en.wikipedia.org/wiki/Reaction_rate


Your question is not very specific. So the answer is it depends. Think of water as an analog.

Methane "snow" would be white. (Due to scattering of light).
Solid methane under sufficient pressure would form a clear colorless solid.
Methane "snow" (which would be on surface" of some celestial body) could be any color if contaminated by even small amounts of other hydrocarbons. A reddish color seems to be typical. See ref: http://www.caltech.edu/news/astronomers-find-ice-and-possibly-methane-snow-white-distant-dwarf-planet-1714


I might suggest that you write a short program to simulate the spectra.  The problem with using real data is the amount of points (upwards of 32k) that make publishing CSV files of raw spectra unreasonable.  Below is a quick attempt to give you what you are looking for.  I apologize for those who do not use Mathematica; however I tried an Excel version and it became unwieldy very quickly.

I have three functions:

Clear[feature];
feature[shift_, n_, coupling_, intensity_] := Module[{},
  Plus @@
   Table[Binomial[n, k]*intensity/(n + 1)*
     Cos[(x)*(shift + coupling (k - n/2))], {k, 0, n}]]
Clear[fid]
makefid[feature_, broadening_] := Module[{},
  Exp[-broadening x] Plus @@ {feature}]
Clear[transform]
transform[fid_] := Module[{resolution = 4096, $y, $x},
  $y = Re[Fourier[Table[fid, {x, 0, resolution, 0.1}]]];
  $y = Take[$y, Round[Length[$y]/2]];
  $x = Range[Length[$y]]/(resolution/(2 Pi));
  Transpose@{$x, $y}
  ]

The function feature creates a peak with a chemical shift as the first argument, the splitting as the second, the coupling constant (use 0.1 for starters) and the intensity (its integral).  makefid creates a simulated FID based on the feature that it is passed and transform makes the spectrum.  It can be used like this:

a = feature[1.2, 2, 0.1, 3];
b = feature[2, 0, 0.1, 3];
c = feature[4.1, 3, 0.1, 2];
fid = makefid[Plus @@ {a, b, c}, 0.01];
ListLinePlot[transform[fid], PlotRange -> {{0, 5}, All}]

The features {a,b,c} are simulated ethyl acetate yielding the following spectrum:

The data you want is in transform[fid] which can be exported as a CSV file:

Export["myfile.csv",transform[fid]]

Which will make it accessible to other software packages and you can perform your math lesson plan.  To make sure this is working properly in Mathematica, we can make a function out of our data and numerically integrate the three peaks:

data = Interpolation[transform[fid]];
NIntegrate[data[x], {x, 0.9, 1.5}]
NIntegrate[data[x], {x, 1.5, 2.2}]
NIntegrate[data[x], {x, 3.7, 4.5}]

This produces areas of 0.32, 0.25, and 0.33, respectively, which is what we expected.


Comparing the Hückel and extended Hückel methods

In the simple Hückel method (SHM) the basis set is limited to p orbitals. This set is limited in a great extent to pz orbitals which constraints the molecular plane to be the xy plane. Basically, you are limited to planar molecules.
  The inclusion of all valence s and p orbitals in the extended Hückel method (EHM) naturally lifts the spacial constraints and you can work with non-planar molecules.

These are the two first differences between the two methods. The other points are:

SHM:

Orbital energies are limited to same-atom interactions, adjacent-atom interactions while all other interactions are 0.
Fock matrix elements are not actually calculated.
Overlap integrals are limited to 1 or 0.

EHM:

Orbital energies are calculated and vary smoothly with geometry.
Fock matrix elements are actually calculated.
Overlap integrals are actually calculated.

You can look up the derivation and steps for the implementation of these two methods in this book that I used as a reference:

Errol G. Lewars; Computational Chemistry, Introduction to the Theory and Applications of Molecular and Quantum Mechanics, Second Edition;
Springer: 2011. DOI: 10.1007/978-90-481-3862-3


2-step reaction mechanism for the decomposition of NO2Br

It is more probable  like
$$\begin{align}
\ce{NO2Br &-> NO2 + Br} \\
\ce{NO2Br + Br &-> NO2 + Br2} \\
\ce{2 Br &->  Br2} \\
\end{align}$$

The last reaction is a minor one in case concentration of $\ce{Br}$ is low.

The reaction rate order can be concentration dependent and need not be the integer.

In fact, it is rather mathematical parameter, related to solution of differential equations for a complex reaction system.

If the 2nd reaction is fast enough, the overall reaction rate is given by the slow rate of generation  of $\ce{Br}$, which fast reacts to form $\ce{Br2}$

If the 2nd reaction is slow enough,
it's rate $$k_{\rm 2}\cdot [\ce{NO2Br}][\ce{Br}]$$ can be written as $$k_{\rm 2a}\cdot [\ce{NO2Br}]^2$$

The exact solution is to solve system of differential equations for the rates of the concentration changes.

$$\frac{\mathrm{d}[\ce{Br}]}{\mathrm{d}t}=k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}] - k_3 [\ce{Br}]^2$$

For the dynamic equilibrium of the steady concentration of $\ce{Br}$:

$$\begin{align}
0&=-k_1.[\ce{NO2Br}] + k_2.[\ce{NO2Br}][\ce{Br}] + k_3 [\ce{Br}]^2 \\
[\ce{Br}]&=[ -k_2.[\ce{NO2Br}]+\sqrt((k_2.[\ce{NO2Br}])^2+4.k_3.k_1.[\ce{NO2Br}])]/(2.k_3) \\
\frac{\mathrm{d}[\ce{NO2Br}]}{\mathrm{d}t}&=-k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}]\\
\end{align}$$


What happens with the boiling point of a liquid in a vacuum

"Sealed in a vacuum" is an oxymoron, a contradiction in terms, much like "frozen with fire". If you seal a liquid in a flask containing nothing else but vacuum, then a part of the liquid will quickly evaporate and fill the flask with vapor, so it would no longer be a vacuum.

The said vapor will exert some pressure, depending on the temperature and the nature of the liquid. For example, with water at room temperature, it will be a lot lower than the normal atmospheric pressure. Indeed, the boiling point of water at that pressure will be a lot lower than the usual $100\;^\circ\rm C$. Guess what? It will be the same as ambient temperature. That's how equilibrium works.

Now what if we just keep the liquid at vacuum by pumping out the vapor as soon as it forms? Well, the simplified† answer is: the liquid will evaporate gradually, bit by bit, until nothing is left. There will be no more liquid, and consequently no more boiling point. That's what happens with the boiling point in a vacuum.

† In fact, it is more complicated. As the liquid evaporates, the rest of it gets progressively colder. At some point, it will get cold enough to freeze, and then you'll be dealing with ice. The molecules will still be leaving its surface, but slower and slower, and the complete sublimation might well take forever, or even longer. That's why the space rocks are still there after all these years.


How are pi-orbitals delocalized in a porphyrin chain?

What you are seeing are molecular orbitals which, by definition, extend across the entire molecule, restricted only by inherent symmetry. It is important to note and is not clearly visible in your image that you are looking at a $\ce{H2-Fe-Zn-H2}$ system, i.e. from one side to the other the centres of the porphyrin system contain two hydrogens, an iron ion, a zinc ion and another two hydrogens. Replace the zinc with iron or vice-versa and you would get an axis of symmetry with molecular orbitals that are either perfectly symmetric or perfectly antisymmetric to this axis of symmetry.

As it is drawn, the molecule is $C_\sigma$, i.e. there is a single plain of symmetry along the plain of the molecule. All orbitals must be symmetric or antisymmetric to this plain, and since we are looking at a $\unicode[Times]{x3C0}$-system, it will be antisymmetric.

All the rest is the result of a calculation! So we cannot really answer why, all we can say is ‘because that is what maths gives us’. But we can slightly rationalise it. First, there is a difference between the top and the bottom, it is just not as easily visible. This is because the only difference between the top and the bottom is ‘where exactly are the two hydrogens and the double bonds?’ Second, the two ions are clearly different ones with different electronegativities that create entirely different environments. Therefore, it is understandable that one orbital is centred more around one of the two ions and the other around the other. And third, the LUMO$+1$ is clearly delocalised across the entire molecule.


The driving force behind the hydrophobic effect is entropy. Suppose I put several hydrophobic molecules in water. These molecules have two choices. (1) The molecules can stay separate from each other and randomly move around in solution. (2) They can group together and form a cluster. It turns out that it is entropically favorable to cluster. The reason is that it takes more water to individually solvate the separate molecules. It takes less water to solvate a cluster of the molecules. When water solvates these molecules, it is ordering itself. In fact, it is believed that water forms cages around hydrophobic molecules in solution. Therefore, if the molecules cluster, less water will have to order itself. Less ordering means greater entropy (randomness).

It also true that London dispersion forces will occur in these clusters. However, this is not the driving force for the hydrophobic effect. The driving force is entropy.


Displacement of leaving groups in organic reactions

In that kind of reactions, the full explanation would require advanced quantic calculation to determine the energy of all the present entities.

In chemistry, we usually prefer giving a half-quantitative explanation to explain the formation of the observed products.

In your example A —> B + C + D, A is a secondary alcohol and it CAN be transformed to a carbocation leading to B... but you can see it is far from perfect since B represents only 50% of the yield. The result would be much more clear-cut with a primary alcohol (unstable carbocation) or a tertiary one (very stable carbocation). I think you are expected to explain all that.


Is there a point where further dilution of ethanol is not exothermic?

Mixing ethanol and water is exothermic all over the range at about room temperature. Data are available, e.g., here excess enthalpy ethanol-water

However, at relatively high temperature and at least around azeotropic composition,  the mixing should become endothermic otherwise the ethanol-water azeotrope should boil at higher temperature instead of having Bp of 78.1 °C.


Chemicals in a Rubber Compound - NSR 20, Accel CS, Accel DPG, Anitiox Q, Antioz PMB

NSR 20 and SMR 20 are natural rubber (Nigerian Standard Rubber and Standard Malaysian Rubber)

"Accel" and "Orac" mean accelerator.

Accel DPG is diphenyl guanidine.

"Antiox" and "orox" mean antioxidant

"Antioz" means antiozonant

"Orac CS" is likely N-cyclohexyl-2-benzothiazole sulphenamide

"Orac TMTD" is likely Tetramethylthiuram disulfide

"PVI" may refer to pre-vulcanization inhibitor


Heterolytic vs homolytic cleavage of the C−H bond in cyclopentadiene

The cleavage of the C-H bond could be envisioned to take place in two steps:

Homolytic cleavage to hydrogen atom plus cyclopentadienyl radical
Ionization of the hydrogen atom to $\ce{H+}$ and addition of the electron to the cyclopentadienyl radical

Ionization of the hydrogen atom requires 313.8 kcal/mol. Aromatic stabilization, calculated from unexpected differences between calculated and experimental heats of hydrogenation, yield an "extraordinary 36 kcal/mol" greater stability than expected. Well, 36 kcal/mol could be extraordinary under certain circumstances, but compared to an ionization energy, it's not very large.

The molecular orbital situation for the cyclopentadienyl radical isn't exactly non-aromatic, it's just not as fully aromatic as it could be. The five pi electrons of the radical are all bonding, but there's room for one more bonding electron. The question is whether the extra stabilization of that electron is sufficient to cause the ionization of the hydrogen atom. A typical carbon-carbon single bond has a dissociation energy of about 100 kcal/mol and a pi bond would be a bit less. A bond would involve two electrons, so one electron would be responsible for about half that, or less than 50 kcal/mol.

So between the "extraordinary" stabilization of benzene (36 kcal/mol) and the extra bond energy of the sixth electron (50 kcal/mol), I don't think there is enough energy available to ionize the hydrogen atom. Homolytic cleavage would be easier than heterolytic. However, if you provide an ionic solvent that stabilizes the proton and cyclopentadienyl anion, circumstances could change completely.


Which is the most acidic hydrogen in vitamin C

The correct anion is stabilized by the mighty resonance.

That's about it.

Hydrogen bonding is irrelevant, since it affects both variants the same way.


Yes, the vapor pressure is the point of equilibrium between the liquid and the vapor. But air moves about so the liquid never saturates the air and an equilibrium is never reached. So more and more water evaporates until it is all gone.


I'm not sure what you mean by 'regular' NMR experiments. NMR experiments which observe multi quantum signal are still 'regular' experiments, so I'll assume you asking why you wouldn't see DQ signal in a basic 1D 1H spectrum, for instance. I will attempt to explain without the use of product operator formalism, and scary looking maths.

In its simplest terms, the frequency of a double quantum coherence is about twice the normal observation frequency, and so cannot be observed without transformation back to single quantum coherence. Similarly, zero quantum coherence is typically less than a few kilohertz, and so is also not observable. If you set your receiver to the correct frequency, this transition would be observable, although significantly lower in intensity due to the lower probability of this transition path.

In the case of an isolated, uncoupled spin, there are only two energy states, and the transition between these two states (the single quantum transition) is the only possible transition. For the case of coupled nuclei, the state of spin I is affected by the spins state of nucleus S. The diagram above shows the typical spin transitions for a spin system, IS. The most common transition is a single quantum transition, where only a single spin changes state, and this generates magnetisation in the transverse plane. When both spins change state simultaneously, you can get either zero quantum (the spins flip to opposite states) or double quantum (spins flip the same way) transitions. The frequency of the double quantum transition is roughly twice that of a single quantum transition. In the vector model, double quantum transitions have no magnetisation in the transverse plane (only transverse magnetisation can be observed), but can be transferred to single quantum transition through the process of generation of SQC(first pulse), evolution (linked to couplings) and generation of MQC (second pulse) in a multi quantum experiment. This is typical of a DQF/TDF COSY type experiment, but equally utilised in studies of dipolar couplings of small organics in liquid crystal media (where quantum orders of >5 are typically observed)


Understanding the increase in pH of a buffer solution upon incremental additions of NaOH analytically

I just answered what I thought was this question but realized that it was essentially the same question asked 5 years ago. So here's that answer modified for the slight difference in the way it is asked here:

We need to know the the pH of the buffer i.e. what does it measure before any NaOH is added. Later we will assume that it is 4.76, the pK of acetic acid.
$[\ce{HAc}] = [\ce{Ac^-}]$ ([·] symbolizes the molar concentration of ·). In order to solve the general problem we need to be able to calculate the fraction of the total Ac that is dissociated and the fraction that isn't.

The fraction that is dissociated comes right out of the Henderson - Hasselbalch equation and is $$f_1 = 1/(1 + r_1)$$ where $$r_1 = 10^{(pH - pK_1)}= [\ce{Ac^-}]/[\ce{HAc}]$$ as is clear from inspection of the Henderson - Hasselbalch equation. The subscript 1 indicates that $r_1$ is the ratio of the number of acid ions that have lost 1 proton to the number that have lost $ 1 - 1 = 0$. In $f_1$ the subscript is indicative that $f_1$ is the fraction of the total Ac molecules that has become singly charged by loss of a single proton. When dealing with monoprotic acetic acid the subscripts aren't that important as there is only one proton to loose. But if the acid is polyprotic we have unionized, once ionized, twice ionized etc. ions to consider. The fractions of those ions are $f_0$, $f_1$, $f_2$... and we have other ratios as well

$$r_j = 10^{(pH - pK_j)} = [H_{n-j-1}Ac^{−j}]/[H_{n-j}Ac^{-(j-1}]$$ Here "Ac" stands for the acid anion and n is the number of protons it can yield when fully dissociated. For acetic acid. $\ce{CH_3COOH}$, and Ac is $\ce{CH_3COO}$, n = 1 and j only has values of 0 or 1. With phosphoric acid, $\ce{H_3(PO_4)}$, Ac is $\ce{(PO_4)}$, n = 3 and j = 0,1,2 or 3.

So suppose now we have a polyprotic acid with values for $r_1$, $r_2$, $r_3$... and that there are x moles of the acid in a solution. Then there would be $xr_1$ moles of the singly deprotonated species, $xr_1r_2$ moles of the doubly deprotonated, $xr_1r_2r_3$ moles of the triply deprotonated and so on.Then the total number of moles of Ac would be the sum of the number of moles of each$$C_{Ac} = x + xr_1 +xr_1r_2 +xr_1r_2r_3...$$ The fraction of the total that is undissociated is $$f_0 = x/(x + xr_1 +xr_1r_2 +xr_1r_2r_3...) = 1/(1 + r_1 +r_1r_2 + r_1r_2r_3...)$$ The fraction that is singly dissociated is $r_1$ times this $$f_1 = r_1f_0$$ and the fraction that is doubly dissociated is $r_2$ times that $$f_2 = r_2f_1$$ and, in general $$f_j = r_jf_{(j-1)}$$

At this point let's remember that $f_j$ is a function of the solution pH and all the pK's of the acid in question and that it is the fraction of the anions of that acid that carry charge -j. Thus we can write an expression for the total charge on all species of Ac at a given pH. This is $$Q_{Ac} = -C_{Ac}(0f_0 + 1f_1 + 2f_2 + ...)$$

Before going on to show you how Q solves buffering problems let me stop to suggest that the simplest way to work with it is to make an Excel (or other) spread sheet. Designate a column for pKs and a cell into which the pH goes. For example, put pH into cell A1 and start the pKs list in A2, Then in B2 put =10^($A$1 - A2).  Using $A$1 lets you copy and paste B2 into as many cells as you have pKs. The B column now contains the r corresponding to the pK in the cell to the left of it. Now enter the formula for $f_0$ in a cell and make another column with the $f_j = r_jf_{(j-1)}$ formula in it. So how many pK's. I say make the spread sheet for 5 or 6. Why? Well lets go back to $\ce{CH_3COOH}$ for a minute. It doesn't have 1 proton to give, as we have been assuming. It actually has 4. Are the other three ever coming off? Not with any base in my lab but we can model those other protons simply by assigning pKs that are so high (say 50) that the f values for anything other than $f_0$ or $f_1$ are 0. The point being that if you are going to go to the trouble to make the spreadsheet you might as well make it big enough to handle any acid you may ever encounter as it easily handles anything up to its maximum size using this trick.

Now how to use $Q(pH)$. If there are a total of $C_{Ac}$ moles of Ac in a solution the negative charge on them at $pH_0$ is $C_{Ac}Q(pH_0)$. At $pH_1$ it is $C_{Ac}Q(pH_1)$. Thus to move the solution from $pH_0$ to $pH_1$ you must supply or remove charge of $$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = C_{Ac}Q(pH_1) - C_{Ac}Q(pH_0)$$ by adding or absorbing protons. If $ pH_0  > pH_1$ then $Q(pH_1) > Q(pH_0)$ (less negative) and so the difference will be positive indicating that protons (acid) will need to be added to effect this pH shift.

The acid species in the solution are not the only thing that emits or absorbs protons when pH changes. The solvent does too.

$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-pH_0} + (10^{(pH_0 - pK_w)} - 10^{(pH_1 - pK_w)})$$ represents the number of protons that must be supplied (or absorbed) to change the pH of water from $pH_0$ to $pH_1$.

Now let's use this to solve the original question. We have 300 mL of 4.76 buffer with $C_{Ac}$ = 0.2 mol ( 0.1 mol of Ac and 0.1 mol of $Ac^{-1}$. This implies that $f_0 = f_1 = 0.5$ Also
$$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = 0.02(Q(pH_1) - Q(4.76))$$
$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-4.76} + (10^{(4.76 - pK_w)} - 10^{(pH_1 - pK_w)})$$

Aparently we want to know how much NaOH we would need to raise the pH to 5.76.  With our formulas in a simple Excel spread sheet and using 5.76 for $pH_1$ we quickly find that $\Delta Q_{Ac}(4.76\ce{->}5.76) = -0.0818 mole. IOW protons must be absorbed. At the same time we find $\Delta Q_{W}(4.76\ce{->}5.76) = -1.56E-5 mole per liter. We only have 0.3 L so the protons to be removed to adjust the water are insignificant. We apparently need 81.8 mL of 1 M NaOH. Note that your digits are pretty similar so one of us is off by a power of 10. As 0.8 mol OH- to shift an 0.2 mol buffer is a lot, I think it's you. As we both used Henderson Hasselbalch, we had better get the same answer.

So why would you do what I'm suggesting instead of what you are doing (assuming you find the factor of 10). If you do a lot of problems like this it makes cranking them out a snap.

Now this post has gone on long enough so I won't have space to tell you about extension of this technique to much more complicated systems. You can use it to find the pH of complicated mixes of weak and strong acids and bases or any materials which have buffering capacity. I ginned it up (this is just using the 'proton condition' to solve a complex system - I didn't invent it) to predict the pH of brewer's mash which is a mixture of weak acid (bicarbonate) and several malts (each of which has its own buffering properties).


Calculation of the specific rate constant (k)

We assume 1st order kinetics in hypochlorite and 1st order kinetics in the color. Putting this in the equation we get the following:
$$\frac{\text{d}[\text{Red}]}{\text{d}t} = -k\, [\text{Red}]^1 \, [\text{bleach}]^1 $$

Because you keep the concentration of bleach the same throughout the trial, you can write
$$k_\text{obs} = k\times [\text{bleach}]$$

Now, we can continue onwards as follows:
$$ k_\text{obs} = k \times [\text{bleach}] \Leftrightarrow k = \frac{k_\text{obs}}{[\text{bleach}]} = 0.0299~\mathrm{s^{-1}\; M^{-1}}$$


How do I create a concentrated solution of singlet oxygen (in the dark)?

Potassium ferricyanide (and other ion complexes, like hemoglobin) will decompose hydrogen peroxide into water and (presumably) singlet oxygen. This process is necessary for the luminol reaction.

$$\ce{2H2O2 \overset{Fe}{->} 2H2O + ^1O2}$$


XeF4 - why are there only 2 dihedral mirror planes?

This is a situation of coordinate systems. In the $D_\mathrm{4h}$ point group, as you've noticed, there are 4 $C_{2}$ rotation axes in the plane. The easiest way to set the coordinate system is to have one pair be coincident with the $x$ and $y$ axes and the other pair coincident with the functions $x=y$ and $x=-y$.

Herein lies the issue: which pair is which? I think most people would agree that a reasonable choice is that the primary set should be coincident with the Cartesian axes. Indeed, an example character table for this group makes this choice as well (these are the $C_{2}'$ elements). This then defines which pair of mirror planes are $\sigma_\mathrm{d}$ and which pair are $\sigma_\mathrm{v}$.

This covers $\ce{XeF4}$ only, not ethene, because ethene belongs to the $D_\mathrm{2h}$ point group.

Here's the character table for that group. There's no dihedral mirror plane because the mirror planes perpendicular to the plane of molecule are coincident with the rotation axes.


Does bleach (without rinsing) leave residues and can vinegar, baking soda neutralize them?

After a thorough wash, rinse and dry cycle, little or no bleach should be left in the fabric, but there is possible concern about the effect of chlorine bleach on organic compounds. However, if disinfection is needed, then the infinitesimally small risk of VOC* is outweighed by the elimination of microorganisms.
If any remaining smell of chlorine bothers you, an extra water rinse should be helpful. @Brian suggests a reducing agent, and $\ce{Na2S2O3}$, sodium thiosulfate, is inexpensive and is commonly used to remove residual chlorine for aquaria. However, leaving garments to air-dry or airing them for a day would be as effective.
(*) "Volatile organic compounds"


Percent composition by volume of the reaction H2 + Br2  2HBr

Your equation and your math appear correct to me: I also get $x=0.808$ from the quadratic equation, and calculate the same percentages.

As long as you are supposed to assume ideal gas behavior in the question, it appears to me that the answer key is incorrect.


Formation of azobenzene from nitrosobenzene and N-hydroxyaniline

Below is my proposed mechanism. You'll see that we go through some of the same key intermediates, so I think you are on the right track. In your second step, it is much more likely that the proton is removed by acid base reaction instead of through the electron transfer process you've shown. Since the initial intermediate (my second structure) is zwitterionic, we can just say that it will neutralize itself. If the reaction conditions are not acidic (from what I see, it's just methanol and magnesium), then there will not be an appreciable amount of the protonated hydroxylamine. The mechanism doesn't require it, as nitrogen is strong enough to "kick out" hydroxide. The result is essentially an azoxybenzene, so we just need to figure out how to lose the other hydroxyl. As the cationic form, it will want to accept electrons, such as from the magnesium as a free electron. Back to a neutral (radical) compound, the other hydroxyl can be kicked out, giving a radical cation on nitrogen. Filling in a second electron from magnesium results in the azobenzene product.


Conductivity as a function of acid concentration

At low concentration, conductivity is proportional to concentration (a linear relationship).

Each ion will have its own unique mobility, as discovered by Kohlrausch. $\ce{H+}$ has the highest mobility. As you can see in your graph the acids have higher conductivities than the salts. $\ce{OH-}$ is also highly mobile. As concentration increases, the linear relationship breaks down for two reasons.

Firstly, in infinitely dilute solution, for the strong electrolytes in the graph, there is complete dissociation into separately solvated ions. However, as concentration increases, a portion of the electrolyte exists as ion pairs.  See for example Equations for Densities and Dissociation Constant of NaCl(aq) at 25°C from “Zero to Saturation” Based on Partial Dissociation J. Electrochem. Soc. 1997 vol. 144, pp. 2380-2384.

Secondly, the mobility of the ions that are solvated is decreased by the fact that they are no longer moving through water, but are instead moving past other ions as well.

To have a maximum in the curves of the question, and to account for the above factors, it is necessary to subtract a term from the linear term.

You need a function of the form:

$$\text{Conductivity} = Ac - Bf(c),$$

where $A$ and $B$ are constants, $c$ is concentration, and $f(c)$ is some function of concentration. Historically, the function $Ac - Bc\sqrt{c}$ was the first and most simple to use.

I would start by trying to fit your data to that function.

More advanced treatments replace $\sqrt{c}$ with $\sqrt{I}$, where I is ionic strength. Further advances involve higher order $I$ in addition to $\sqrt{I}$, such as $I \ln (I)$, $I$, and $I^{3/2}$.

The $I \ln (I)$ and $I$ terms arise upon considering the ions as charged spheres of finite diameter, rather than a simple point charges, as explained in Electrolytic Conductance and Conductances of the Halogen Acids in Water and references cited therein. This is the reference to look at if you want the best function to fit the HCl curve in your graph as it includes numerical coefficients.

For more information see The Conductivity of Liquids by Olin Freeman Tower, which although not the most recent work, has the advantage of being understandable, or Calculating the conductivity of natural waters.

For a very advanced consideration, see "Electrical conductance of electrolyte mixtures of any type" Journal of Solution Chemistry, 1978, vol. 7, pp. 533-548.


Calculating Molar Volume from crystal volume

well figured out the answer, I had to take the correct value of Z from the database. The difference in values was because of the incorrect value of Z (which was actually 8) . When you use Z =8 , the two solution yield 'approximately' equal results.


In desperation, for some unknown reason, I need 15% aqueous $\ce{HF}$. What do I really need? A weak solution of $\ce{H+}$, say about pH 2. And a solution of fluoride ions to wreak havoc on some unsuspecting metal or glass(Ooops, I may have given away the secret).

My subconscious suggests that I could get them from two different sources, like $\ce{NaF}$ and $\ce{HCl}$. Yeah, that'll work.

Maybe. Maybe even with $\ce{CaF2}$(The use of $\ce{H2SO4}$ with the calcium salt suggests the reason was to precipitate the $\ce{CaSO4}$ and flash off the $\ce{HF}$, thereby separating, not just generating, the $\ce{HF}$).


What is the empirical formula of the polyethylene?


The octet rule has no theoretical basis. It has been observed to work quite often. But that is all we could say. It cannot be proven. The octet rule works nearly perfectly in organic chemistry,

There are many examples where this rule is not verified. For example the Boron hydrides, the simplest being $B_2H_6$. This compound is commercial, but it does not follow the octet rule. Another example is the familiar molecule of Oxygen $O_2$, which has no double bond as often given in elementary chemistry courses. Other examples : $PCl_5, SF_6$. Another simple example is Nitrogen oxide $NO$. And a lot of compounds made with metals from the transition range.


Your best bet is to publish the procedure. Procedures from pier reviewed journals are generally reproducible (there are exceptions). So if they say "dissolve x grams in y mL" just do it.

NaOH is poorly soluble in dimethoxyethane. This might be sufficient if base is used as a catalyst or if reaction is slowly shifting to product.

Devil is in details. Heat, presence of water and other factors might impact soluility greatly.


Why do vinegar and other acids remove rust?

The acid can dissolve rust (e.g. $\ce{FeOOH}$). The chemical reaction is:

$\ce{3CH_3COOH + FeOOH -> Fe(CH_3COO)_3 + 2H_2O}$

EDIT: ($\ce{Fe(OAc)_3}$is NOT soluble in water) The acid forms a water-insoluble salt with the iron oxide, which then probably just crumbles from the rust layer and precipitates in the container.

Mind that rust also contains for example $\ce{FeO(OH)_3}$ and $\ce{Fe_2O_3·nH_2O}$. The mechanism is the same.

Also, the acid oxidizes the upper layer of the iron under the rust to form a rust-resisting oxide-layer.


Abundance of sulfur containing compounds in the air

The tarnishing of silver by sulfur is primarily due to hydrogen sulfide. Hydrogen sulfide is found throughout the Earth's lower atmosphere, and has many sources including the human body$\pu{^1}$.

The amount of hydrogen sulfide naturally found in the air has been estimated as $\pu{0.11-0.33 ppb}$ ($\pu{0.15-0.46 ug/m3}$). Lower levels ($\pu{0.02-0.07 ppb}$; $\pu{0.03-0.1ug/m3}$) have been observed in some remote areas$\pu{^2}$.

Although these are relatively low concentrations compared to many other trace gases found in our atmosphere, silver is highly reactive towards hydrogen sulfide and thus tarnishing of silver is a ubiquitous issue. Additionally, some tarnish results from the oxidation of silver by oxygen.

Regarding your question about silver tarnishing from carbon, it's a matter of reactivity. The overwhelming source of carbon in the air is carbon dioxide, and methane to a much lesser extent. Neither of these gases are reactive towards silver nearly to the degree of hydrogen sulfide and oxygen.

1) Toxilogical Profile for Hydrogen Sulfide and Carbonyl Sulfide, U.S. Dept. of Health and Human Services, November, 2016
2) Ambient Air Guidelines for Hydrogen Sulfide, CAS Registry Number: 7783-06-4, March 27, 2006


Not necessarily.  If you get the pH right and the stoichiometry absolutetly 1:1 then you might form $\ce{Cl2}$.  However your reaction will compete with other reaction and you will more likely oxidize the chlorine into chlorate ions ($\ce{ClO-}$, $\ce{ ClO2-}$, $\ce{ClO3-}$, or $\ce{ClO4-}$).


Effect on kinetic energy of a body with change in temperature

You might find it interesting to read up on the differences between the different temperature scales. The kelvin scale effectively measures the vibration (average kinetic energy of molecules) so it makes sense that a doubling in the kelvin means a doubling in kinetic energy. On the other hand the Celsius scale is defined by the melting and boiling points of water (quite arbitrary).


What is the state of the oxygen free radical in the following reaction?

According to the reaction $$\ce{O2\,(g) ->[h\nu][\nu\,<\,240~\mathrm{nm}] 2O}$$
photolytic cleavage of oxygen yields two radicals. These will most likely be in the high spin triplet state. And yes these are free radicals according to the definition of the IUPAC Goldbook, as they have two unpaired electrons.

I assume you are talking about state of matter and then of course, you are right. Oxygen radicals can exist neither in aqueous solution, nor liquid, nor solid. Even in gas phase the concentration of oxygen radicals will be very low as they are an incredibly short lived species. They might exist in plasma, too. Keeping that in mind it is probably unnecessary to assign a state to the oxygen radical.


How were old style stereographic structures produced?

Created in FORTRAN by Carroll K. Johnson, of the Oak Ridge National Laboratory (ORNL), and first released in 1965, ORTEP (Oak Ridge Thermal-Ellipsoid Plot Program) rapidly became a favorite of crystallographers and protein crystallographers to produce illustrations of structures for conference presentations and publications. A key strength of ORTEP was its capacity to generate stereoscopic images automatically.

For source of above and more information see: History of Visualization of Biological Macromolecules

For the ORTEP program itself, check ONRL Report #3794: https://web.archive.org/web/20170210061629/http://web.ornl.gov/info/reports/1965/3445600599779.pdf (big file warning)


What is the lattice structure of manganese?

Ah, good old Mn. This is one of the uglier ones in terms of phases.
$\alpha$-Mn, the room temperature phase, is CBCC (A12 family), fairly unusual. Near 1000K, Mn transitions to $\beta$-Mn, with a simple cubic (A13 family) crystal structure. $\gamma$-Mn is FCC, face-centered cubic (A1), and $\delta$-Mn is BCC, body-centered cubic (A2).

One excellent reference is A.T. Dinsdale, SGTE Data for Pure Elements, CALPHAD 15(4) 317-425 (1991). There is a readily available PDF of an updated version of that paper that can be found (Dinsdale). This paper gives the accepted Gibbs free energy functions for the elements in their various stable, and some unstable, forms.

Editing to add additional information on the $\alpha$- and $\beta$-Mn crystal phases. Both of these phases are pretty weird, and each is the first observed crystal with their structure (hence they are the 'prototype' in crystallography terms).

$\alpha$-Mn was described by A.J. Bradley and J. Thewlis in 'The Crystal Structure of $\alpha$-Manganese', Proc. Royal Society 115, 456-471 (1927). The unit cell is based on body-centered cubic, but contains 58 atoms representing 4 distinct positions. It can be thought of as a bcc Bravais lattice with a 29 atom basis. Mn is the only element that exhibits this crystal structure.

$\beta$-Mn was described by G.D. Preston in Phil. Mag. 5(33) 1207-1212 (1928). The unit cell is simple cubic, containing 20 atoms in 2 groups. Preston's unit cell drawing is significantly worse than that of Bradley and Thewlis, although representations of the unit cell can be easily found on the web these days. Again, Mn is the only element that exhibits this crystal structure.

The $\delta$-Mn and $\gamma$-Mn are normal, boring bcc and fcc crystals, so no need to point to references for them.

I would provide links to the papers if I could - my institute does not have direct access to them.

As yet further info on the crystal structure references.  The A12, A13, ... notation is the Strukturebericht notation used by crystallographers. Unfortunately, there are a number of other notations used as well, including Schonenflies and Hermann-Mauguin. The Strukturebericht is pretty basic, mostly a catalog of crystal types, while the others convey the actual symmetries of the crystal structure. Some of this confusion of notations seems to date from the early days of crystallography when it was done in many places and languages. There are seven crystal systems, 14 Bravais lattices, 32 crystallographic point groups, and 230 space groups to be described to some level by the various notations. For this answer I stuck to the Strukturebericht because that is the notation used in the Dinsdale reference on Gibbs free energies of the elemental phases.


Fireplace window: Can one influence how fast soot is building up?

I don't have a fireplace or wood stove, so I don't have a practical technique that I know works to prevent soot buildup on glass, but I can answer some of your questions about glass and soot. To start, I should add a caveat here: while many fireplaces use tempered glass windows, wood stoves generally have glass-ceramic windows to withstand the increased heat. As far as surface properties go for this, glass-ceramic isn't dramatically different from normal glass.

  does each glass (especially my one) have pores with a size that starch
  particles can fit in? I read somewhere that starch has a diameter
  between 2-170 µm depending on the type.

While special techniques exist for making porous glasses, normal glass is not porous and float glass (the standard technique for fabricating flat glass panes) produces quite smooth pieces even without polishing (hundreds of nm peak-to-peak), so probably whole starch particles are not getting stuck in the surface of the glass and the surface roughness probably has little impact on soot adhesion in this case.

  I'm making a fire where I burn wood! Why wouldn't the starch simply
  burn and create itself the soot?

Unless the flame is right up to the glass, it probably doesn't get hot enough to really burn starch (if any is even there). That site didn't compare it directly with not applying starch so it's not clear that it makes any difference.

  How does soot building work? I mean, why does the soot stick so hard
  at the window? After all it's just little particles flowing around and
  landing on the window. (I know it's called burning in, but how does
  the heat work on a already burned particle?)

Soot is basically the deposition of incomplete combustion products from a flame. A wood flame appears yellow or orange because tiny particles of mostly carbon are incandescing. There is insufficient oxygen to fully combust the carbon in the fuel to $\ce{CO2}$, so the combustion results in carbon molecules in various states of oxidation from small polymers to elemental carbon. These species are really only stable as gases within the heat of the flame itself and will thus readily deposit onto cold surfaces, like the window glass. Many modern wood stoves seem to have "air-wash" systems that flow air past the inside of the window to reduce how much soot can actually reach the surface.

What is actually deposited as soot is a complex mixture, but what is emitted from the flame is related to how much air is available and the temperature of the flame (the two are also connected with each other). If the oxygen supply is increased, more complete combustion is favoured so less carbon species that can deposit as soot are formed. The temperature that is attained influences the composition of soot particles: higher temperatures produce more aliphatic and less aromatic compounds. The practical significance is that reducing the flame intensity by limiting oxygen can greatly increase the production of soot. (and may have some effect on how well it adheres to the glass, though I wasn't able to find anything concrete to that effect)

From the non-scientific sources I glanced through, it seems to be well known that excessive soot buildup can be due to wet fuel or low airflow and that burning a very hot fire can help make soot easier to remove. As for the corn starch, it may indeed be leaving some deposit that hinders soot adhesion (there are commercial cleaners that claim to leave a film behind that makes soot easier to remove), the act of applying it may be removing existing soot that's too small to see but new soot adheres readily too, or it may do nothing. Perhaps you might want to try applying corn starch to half of a freshly-cleaned window to see if there is a noticeable difference.


Is the reaction of hydronium and thisulfate an acid-base reaction?

It starts with the protonation of the anion, so that part is acid-base, but then it continues as a redox reaction.


Is there a relationship between acidity of the water solution and oxidation/reduction reactions?

Fenton chemistry based on iron is definitely more reactive at lower $\mathrm{pH}$. This is due, in part, to speciation issues and associated solubility. Source: 'Effect of $\mathrm{pH}$ on Fenton and Fenton-like oxidation', by
Jung YS, et al., at https://www.ncbi.nlm.nih.gov/pubmed/19278159.

However, the use of chelates (like EDTA) can expand the $\mathrm{pH}$ range of reactivity (see full paper at https://www.jstage.jst.go.jp/article/jwet/11/1/11_21/_pdf ).

There are other transition metals, like Copper, which appear to be active in a much wider $\mathrm{pH}$ range where the chemistry is referred to as Fenton-type. There is an old french patent on water purification (from microbes,...) based on copper ions ($\ce{CuSO4}$ in small amounts) and natural waters containing dissolved oxygen. Interestingly, the addition of EDTA is still recommended, but now as a means to TERMINATE the redox reaction at near-neutral $\mathrm{pH}$.

There is a similar property for Cobalt ions, however, its toxicity has generally precluded it from commercial applications (like remediation of polluted waters through advanced oxidation processes referred to as AOP). Here is a reference on non-iron fenton processes at https://www.sciencedirect.com/science/article/abs/pii/S0304389414003239 and a more general review here, click on pdf download feature at https://www.researchgate.net/publication/322020986_A_Review_on_Advanced_Oxidation_Processes_for_Effective_Water_Treatment/link/5a4b0cd5458515f6b05b5221/download).


The polarity of a single bond might indeed be the same for some different molecules. When speaking of polarity of molecules however, the whole molecule should be taken into account, since the macro properties are dictated by all the parts of a molecule.

Regarding the fatty acids: The 'head' of such a molecule is indeed polar, and in specific cases this means that the interactions and reactivity will be like polar molecules. This effect however is only contained to the head of the molecule. Fatty acids as a whole are still considered non-polar because this type of molecule will not be dissolved in water. Note that the main reason for this is not due to the interactions of water with the fatty acid. The fatty acid is just as 'happy' interacting with water as it is with other fatty acid chains. Water, on the other end, does not 'want' to interact with fatty acids, since it will have to rearrange around the fatty acid molecule to ensure the same number of hydrogen bonds are made. This is an entropic cost, that makes the total Gibbs Free Energy change unfavorable.

The result is that water molecules do not rearrange the hydrogen bonds with other water molecules and the fatty acids interact with each other. Due to this effect, water not being willing to rearrange hydrogen bonds, fatty acids are considered non-polar.

Hope this has helped, and if I used to much terminology or jargon, please let me know!


What to do when there is an odd number of valence electrons left on the the central atom?

There are some molecules where the full octet isn't possible (like you've found with $\text{NO}$ and $\text{ClO}_2$). In these cases, the important thing to consider is trying to keep the formal charge (Wikipedia) on each of the atoms as close to zero as possible.

For example, for the $\text{NO}$, we know that $\text{N}$ contributes 5 electrons and $\text{O}$ contributes 6 electrons for a total of 11 electrons. We essentially have to decide between the following Lewis structures:

$$
(1) \qquad :\ddot{\text{N}}::\ddot{\text{O}}. \quad \text{Single electron on O} \\
\qquad \text{vs.} \\
(2) \qquad .\ddot{\text{N}}::\ddot{\text{O}}: \quad \text{Single electron on N}
$$

Here is where calculating the formal charge is important. For $(1)$ above the formal charge on the N is (5 - 4 - (4$\div$2) = -1) and for O is (6 - 3 - (4$\div$2) = +1, whereas for $(2)$ the formal charge on the N is (5 - 3 - (4$\div$2) = 0) and for O is (6 - 4 - (4$\div$2) = 0.

Therefore the second Lewis structure is more favorable.


Why do dihaloalkanes lose both halogen atoms during mass spectrometry?

It isn't just both bromine atoms that are lost, although that might be expected in mass spec given the relative weakness of the carbon-bromine bond.  You also lose a hydrogen atom.  And that is key.
We might suppose that a single bromide ion (or bromine atom plus electron) comes off first, forming a $\ce{C4H8Br^+}$ ion.  That is the peaks at 135 and 137, recalling that in mass spec the mass of a bromine atom is both 79 and 81 which are the most common isotopes.  There is nothing spectacular about such an ion per se, but now it can eliminate $\ce{HBr}$, or $\ce{H + Br}$ to give a resonance-stabilized cation:
$\ce{CH2=CH - \overset{+}{C}H - CH3}\leftrightarrow\ce{\overset{+}{C}H2 - CH=CH - CH3}$
Thus the formation of this ion is energetically favored, and it would account for $\ce{C4H7^+}$.
In the case of 1,4-dibromobutane you should expect the initial (primary) $\ce{C4H8Br^+}$ carbocation to rearrange with a hydride transfer to the end carbon that first lost the bromine, making the ion secondary.  Then the ion will match that obtained with the other isomers.


From (1):

  Salicylic acid decarboxylated to give phenol and carbon dioxide, at a convenient rate for kinetic measurements, at $200-230^\circ$, in a homogeneous solution of benzoic acid and certain other high-boiling solvents.

  ... there was a considerable amount of interest in the possibility of a bimolecular decarboxylation mechanism involving an attack by a proton on the free acid or its anions. A variety of compounds such as anthracene-9-carboxylic acid, substituted cinnamic mesitoic acid, aminosalicylic acid, and salicylic acid, appeared to represent examples of this type of decarboxylation mechanism. In most cases, pseudo-first-order rate constants were
  observed experimentally.

  ... mechanism is also proposed for the decarboxylation of salicylic acid in benzoic acid solution. A competition between an oxygen atom and the terminal ring carbon atom for the proton to give either the undissociated acid or a $\sigma$-complex leading to products is shown by eq. 3.

  The effect of substituents on the rate of decarboxylation of salicylic acid has been used to support a bimolecular mechanism. Electron-donating groups in the ortho or para positions have led to an increase in rate. The increased electron density on the ring carbon atom bonded to the carboxyl group would presumably facilitate an attack by a proton at this position.

  ... An attack by a proton on the terminal ring carbon atom of the salicylate ion would appear, at first glance, to be extremely unfavorable because of the close proximity of a formal negative charge. Any interaction should merely lead to undissociated acid. The hydroxyl group in the ortho position appears to influence the distribution of the charge because of the steric configuration.
  Resonance and tautomeric forms, eq. 5, which do not involve an extensive shift of atoms, can be drawn which would distribute the charge in a manner favorable
  to the proposed mechanism, eq. 3.

  This unique steric arrangement is not possible with m- or p-hydroxybenzoic acid, the effect of which is borne out by the fact that these two acids did not decarboxylate at a measurable rate in benzoic acid solution at $212^\circ$. Since inductive and resonance effects would be expected to be similar for the ortho and para isomers, the steric effect appears to exert a critical influence.

(1) Kaeding, W. W. Oxidation of Aromatic Acids. IV. Decarboxylation of Salicylic Acids J. Org. Chem. 1964, 29 (9), 2556–2559. DOI: 10.1021/jo01032a016


In the definition of an s shell, you will find that its $\ell$ number is zero. In classical terms, that corresponds to an orbit with zero orbital or angular momentum — which for a large object is a clear impossibility. For an electron, it gives the peculiar result that any electron in any s shell is, classically speaking, moving back and forth through the nucleus, rather than around it. So, in one curious sense, what you just asked is exactly what happens: the classical analogy is that the electrons do go through the nucleus, which is why they have such a lovely spherical symmetry.

The second part of the answer, however, is that electrons can't go through the nucleus unless they are hugely more energetic than the ones found in a typical small-nucleus atom. Clearly there's a bit of a paradox going on there!

The resolution of the paradox is that very-low-mass charged particles must be treated by quantum rules. So, for example, rather than the electron behaving like a well-defined particle, it behaves like a standing wave. That standing wave can in turn be thought of as two simultaneous versions of the electron, one going (for example) clockwise and the other counterclockwise around. (The real situation has an infinite number of such components; I'm just picking out a single pair of them that demonstrate the principle.)

Each of these components can furthermore be thought of as being refracted by the powerful spherical charge field of the nucleus, curving around it without every striking it. This refraction is not the same as an attraction. In fact, it is this refraction effect that prevents the density of the electron cloud from reaching infinity at the nucleus — that is, from striking the nucleus. If you think of how a tank of water can cause a beam of light to bounce off the surface instead of entering the tank — and that's a terrible analogy, I know, I know — you can at least get some idea of how an increasing "optical density" towards a central point might keep light away rather than bringing it closer.

So for an electron that behaves "like" two waves going both clockwise and counterclockwise, the combined waves curve around the nucleus rather than striking it. This is a very quantum sort of event, since for a classical object such "splitting" of the object is simply not possible and the object simply dives straight into the source of attraction. But if objects are light enough, that kind of particle-like behavior simply ceases to be available to the object. Instead, you get waves that neatly and with perfect spherical symmetry curve around the nucleus, never obtaining enough energy (which makes it more particle-like) to connect directly with that nucleus.

Finally, notice that electrons in s (and other) shells necessarily combine multiple paths all at the same time. For each "image" of the electron that is traveling clockwise, there must also be an exactly balancing "image" of the same electron traveling counterclockwise, so that the two images always balance out to zero orbital momentum. What an amazing thing that is! And an important thing, too, since it's what makes chemistry possible.

So, good question, even if it really is more of a physics question per se than a chemistry question. But it's such an important chemistry question! It's like asking how the engine that powers the car works. You can accept it as a given that all cars and vehicles have engines, and that they all work in a certain way. However, sometimes it's nice to dive in a little deeper and try to understand why these peculiar things do the things that make chemistry possible — that is, how the engine really works.


This page suggests that throwing power is the ability to coat a surface evenly regardless of current density differences, and thus to produce an even plating thickness across the whole of a surface.

A solution with high throwing power would thus have the ability to produce a good, close-to-uniform coating on more difficult surface shapes.


Is a transition metal with 7 d electons ever diamagnetic?

As I noted in my comment above, I think the question is whether there are 7d electrons in the atom or in the ion.

For example, if you take 3 electrons from 7, you get 4d electrons in the $\ce{M^{3+}}$ ion which obviously could (and should) be diamagnetic.

This is all, of course, a guess given what little you posted.

Let's say I had a similar question in my class. Students might then assume this $\ce{M^{3+}}$ is octahedral. (Personally, I'd make it a bit more obvious in the question.)

Given that the ligand is high-field, you get an initial MO diagram like this:

But... we're not done. This electron configuration is subject to a weak Jahn-Teller distortion. So you'd end up with an MO diagram like this:

That, of course, is diamagnetic.


Compound with pressure sensitive melting point at moderate temperature and pressure

I'm not going to do your research for you (at least not for now—perhaps if the mood takes me I'll do the Mathematica programming at some later point), but I will give you guidance that will help you find the answer:

It's hard to find data on melting points at different pressures for a wide range of compounds (and you want to search through a wide range of compounds).  What is much more comprehensively available are data that can be used to calculate such pressure dependency, namely the specific densities of substances in their solid and liquid phases, and their enthalpies of fusion.  So:

According the Clapeyron equation,
$$\frac{dT}{dp}=\frac{\Delta V_m}{\Delta S_m}$$
where $T$ is the phase transition temperature, $\Delta S_m$ is the molar entropy change for the phase transition, and $\Delta V_m$ is the molar volume change for the phase transition.
For a phase transition, $\Delta S_m$ = $\frac{\Delta H_m}{T}$, where $T$ is in Kelvin.  And you can determine $V_m$ (in $\pu{\frac{{cm}^3}{mol}}$) from the molecular mass, $M_r$ (in $\pu{\frac{g}{mol}}$), divided by the specific density, $\rho$ (in $\pu{\frac{g}{cm^3}}$):
$$V_m = \frac{M_r}{\rho}\equiv \frac{\pu{\frac{g}{mol}}}{\pu{\frac{g}{cm^3}}} = \pu{\frac{{cm}^3}{mol}}$$
Thus:
$$\Delta V_{m, fus} = \frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}$$
The subscript "fus" means "fusion", indicating this is a solid-liquid phase transition.
Substituting, we obtain:
$$\frac{\Delta T_{fus}}{\Delta p} \approx \frac{dT_{fus}}{dp}=\frac{\Delta V_{m, fus}}{\Delta S_{m, fus}} =\frac{\frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}}{\Delta S_{m, fus}}=T_{fus}\frac{\frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}}{\Delta H_{m,fus}} $$
What you want is to identify the substances that have the largest values for the above expression, and then filter according to your other criteria.
Note that this expression only gives an exact value for $\frac{\Delta T_{fus}}{\Delta p}$ if you integrate with respect to $T_{fus}$ using $\Delta H_{fus} (T), \rho_{liq}(T) \text{ and } \rho_{solid}(T)$—i.e., if you account for the temperature-dependencies of each of these. However, since you are determining $\frac{\Delta T_{fus}}{\Delta p}$ over a range not far from $T_{fus}$ at 1 atm, and since the densities for the phases that are stable at room temperature are probably at $25 ^\circ C = 298 K$, which is close to your desired $T_{fus}$, and since they will not change much over your pressure range, this should give you a very good approximation. Once you narrow it down to the actual substance, you can apply corrections to get a more precise estimate.

You could do this search manually by plugging in values into an Excel spreadsheet, or semi-manually by finding downloadable data tables with these values and importing them into Excel, but the best approach would be to use a program like Mathematica*, which has access to large chemical databases, and ask it to search for all substances for which it has density data in the solid and liquid phases, and enthalpies of fusion, calcualate the values for the above expression, and sort by magnitude.  You can then apply additional filters (e.g., melting point and enthalpy of fusion), as desired.

[*This of course requires knowing how to use Mathematica, or learning to use it.  It's possible someone on MathematicaSE could help you write such a query. Alternately, it's possible the online CRC is also searchable in this way (certainly seems it should be), though I don't know myself. One way to get help might be to check the reference desk of your local library.  They might have the capability to query chemical databases in this way.]
Once you have this, you can then identify the best candidate compounds, and further filter by cost and stability.

Note: It may not be possible to find a compound whose melting point changes enough (a comprehensive chemical database search would help you determine this).  Finding a large pressure-induced change in melting point is challenging, because substances don't change volume that much in going from a solid to a liquid. Where you do get a large pressure-induced change in a phase transition temperature is when going from liquid to gas because, under normal conditions (room temp and 1 atm), gases have about 1000x the molar volume of liquids. That's why water boils at a notably lower temperature when you go into the mountains, even though the change in pressure there is a mere fraction of an atmosphere. [By "notably lower" I mean enough of a change to affect cooking times.]

Plus you make your search even more challenging by requiring a high $\Delta H_{fus}$, since a high $\Delta H_{fus}$ lowers $\frac{\Delta T_{fus}}{\Delta p}$ (everything else being equal).


What is the correct pH value of weak and strong alkalis?

There might be confusion as below:

strong/weak "Alkali"(1) - or rather an alkaline solution, which is related to the $\mathrm{pH}$ value. The conductivity and the strength increases with $\ce{pH}$, but conductivity can be obviously affected by presence of salts.
There is not a strict threshold between strong or weak alkaline $\ce{pH}$. It is often considered relatively and/or it depends on context. As a guide the threshold can be between 10-12, context dependent.

strong/weak "Alkali"(2) - or rather a base ( as an opposite to an acid ), which is related to the base dissociation constant. IF a base has too low dissociation constant, so there is significant fraction of not dissociated base ( like ammonia ), the base is considered a weak base. There is ongoing equilibrium $\ce{ BOH <=> B+ + OH-}$, resp. $\ce{ B + H2O <=> BH+ + OH-}$

E.g.
$\ce{NaOH}$ solution with $\mathrm{pH} = 10$ is weak "alkali"(1), but strong "alkali"(2).

$\ce{NH3}$ solution with $\mathrm{pH} = 11$ is stronger "alkali"(1) then above, but weak "alkali"(2).

Note: For historical reasons, "alkali" is reserved to hydroxides of alkalic metals, so it should not be used for ammonia. As the general term, a base is preferred, with alkalic/alkaline ( along with basic ) as adjective being accepted.


Why would a polar protic solvent favor the addition of HBr to an alkene?

why a protic solvent?

Protic solvents tend to have higher dielectric constants than typical, unreactive, non-protic solvents.  Look at the following table.
\begin{array}{|c|c|c|}\hline
\text{Solvent} & \text{Dielectric constant}    \\ \hline
\text{methanol} & 32.6    \\ \hline
\text{ethanol} & 24.6   \\ \hline
\text{methylene chloride} & 9.08 \\ \hline
\text{chlorobenzene} & 5.69\\ \hline
\text{chloroform} & 4.81    \\ \hline
\text{diethyl ether} & 4.27    \\ \hline
\end{array}
The higher dielectric constant stabilizes ions which means that 1) $\ce{HBr}$ will tend to be more dissociated and 2) (as Dissenter noted) the resultant carbocation intermediate will be more stabilized thereby lowering the energy of the transition state leading to it.


Relation between Temperature and viscosity of Linseed Oil?

There are many formulas!  This usually means that some of them are OK in general for estimation purposes, but none of them will be especially accurate.  Not easy to choose from the many formulas.

One of the most simple and popular seems to be the Wright equation,
$$\log_{10}\left(\log_{10}\left(\nu + \lambda \right)\right)=A-B\log_{10}(T),$$
where $\nu$ is the kinematic viscosity, $\lambda$ is a parameter, usually taken to be $0.7$, and $A$ and $B$ are compound-specific parameters.  With knowledge of linseed oil viscosity at least one other temperature than the one you provided in your question, you could fit $A$ and $B$ parameters and apply the equation.

However, the best answer to your question is to obtain some empirical data.  A 1977 paper[1] reports on linseed oil viscosity as a function of temperature, for example.  They found that a different equation (not the Wright) gave a reasonable fit to the data.  Specifically, they found
$$\ln(\eta) = -5.73 + \frac{2820}{T}$$
for viscosity $\eta$ in centipoises and temperature $T$ in K.

Another thing to keep in mind is that linseed oil is prone to oxidation.  Oxidized linseed oil may have different properties than undergraded, "pure" linseed oil.

[1]: T. Gallagher; D. E. Kline; J. Am. Oil Chem. Soc. 1977, 54 (2), 68-70. DOI: 10.1007/BF02912392


Why there were changes in molar mass of lithium?

This doesn't necessarily completely answer the question, but it at least moves closer to the root of the problem. From the Wikipedia article that MaxW included in his comment:

  Brande also described some pure salts of lithium, such as the chloride, and, estimating that lithia (lithium oxide) contained about 55% metal, estimated the atomic weight of lithium to be around 9.8 g/mol (modern value ~6.94 g/mol)

So it seems to stem from an erroneous estimation of the percent composition of lithium oxide. If we go to the source (Brande's manual of chemistry:https://books.google.com/books/reader?id=zkIAAAAAYAAJ&printsec=frontcover&output=reader&pg=GBS.PA190), we find that this error likely stems from the fact that lithium oxide was difficult to obtain and so very limited experimentation had been done to characterize it.


Analytical solution of the Schrödinger Equation for AB^+ systems

First, one has to be really careful with the words: when it is said that the Schrödinger equation can be solved the dihydrogen cation completely analytically, several remarks has to be made.

There are few different Schrödinger equations out there, so it is better to explicitly mention that one which can be solved analytically for $\ce{H2+}$ is the electronic Schrödinger equation. So, first of all, we are talking about the time-independent Schrödinger equation here, and secondly, we employ the Born-Oppenheimer approximation and solve only the electronic problem. No claim is maid with respect to analytical solutions of the whole time-independent Schrödinger equation without any approximation introduced, as well as regarding analytical solutions of its time-dependent counterpart.
The second remark is about the meaning of the word "analytical", or "analytic". The definitions taken from the Wikipedia are as follows:

  An equation is said to have an analytic solution if and only if at
  least one solution can be expressed as an analytic expression.

  An analytic expression is a mathematical expression constructed using well-known operations that lend themselves readily to calculation.

These definition seems to be rather vague for me, especially taking into account that,

  [..] the set of well-known functions allowed can vary according to context [...]

so we have to be careful relying on some intuitive notions of "analytic expression" and "analytic solution".

Having said so, and in hope that my intuitive notions of "analytic expression" and "analytic solution" are not miles away from the truth,  it is important to note that $\ce{H2+}$, for which analytic solutions are known, is (a special case of) a three-body system. And since there is no claim about analytic solutions for a four-body system, a five-body one, etc., from the beginning we have to restrict ourselves for three-body systems.

From that viewpoint the notation $\ce{AB+}$ used by OP is misleading, because the only three-body chemical system of $\ce{AB+}$ kind is $\ce{H2+}$. Already for the system composed of $\ce{H}$ and $\ce{He}$ the charge has to be +2 since we have to remove 2 electrons so that we end up with just one of them which together with 2 nuclei constitutes a three-body system, and not a four-body one.

Now, what about analytical solutions for $\ce{AB^{n+}}$ systems where $n$ is chosen appropriately so that the system is a three-body one? Well, I think, as for $\ce{H2+}$, there are analytic solutions for electronic Schrödinger equation in the Born-Oppenheimer approximation. The reason is pretty simple in fact: the clamped nuclei model reduces the three-body  problem to a single-body problem of the motion of a single electron within the fixed nuclear potential.

Why there are no related references (at least I could not find one)? It is very likely because chemist has little or no interest in three-body chemical systems of $\ce{AB^{n+}}$ type in general, and specifically, little or no interest in analytic solutions for these systems. Except for the simplest $\ce{H2+}$ system which serves as a primary model for the concept of the one-electron bond.


How does formylation of a nitrogen change the hydrophobicity of a target protein?

Formylmethionine is used as the start amino acid for a protein chain in certain organisms (the prokaryotes.)

As the alpha nitrogen is formylated, it cannot be in the middle of the chain. This is never seen in real proteins, partly because there would be two carbonyl groups attached to the nitrogen, which would have a very strong electron withdrawing effect and be quite unstable.

http://en.wikipedia.org/wiki/N-Formylmethionine

Other organisms (the eukaryotes, including the animals and humans) use methionine as the start amino acid of the chain.

http://en.wikipedia.org/wiki/Genetic_code

As there can only ever be one N-formylated methionine in a real protein (and it is often removed after protein synthesis in a step called post-translational modification) the effect on hydrophobicity will not be very much.

If on the other hand, you were to take an existing protein and formylate the -(CH2)4-NH2 side chains on all the lysine amino acid residues, I would expect a slight increase in hydrophobicity in neutral solution. In acid solution I would expect the difference to be more marked, as the formylated NH2 groups would lose their ability to become ionised by accepting a proton from the acid.


How are lowercase m and uppercase M used in general chemistry courses?

Lowercase m

$m$ is the official symbol for the quantity mass.
m is an SI-prefix that stands for milli ($\pu{e-3}$), for example in mg (milligram).
$m$ is sometimes uses as the symbol for molality. This is not IUPAC recommended (they recommend $b$) and leads to the confusing definition $$m_\text{solute} = n_\text{solute} / m_\text{solvent}$$ where the first $m$ is for molality and the second one for mass.

Uppercase M

$M$ is used as symbol for the quantity molar mass (dimensions mass per amount of substance).
M is an SI-prefix that stands for mega ($\pu{e6}$), for example in MW (megawatts).
M is often used as an abbreviation for mol/L, and mM for mmol/L.
$M$ is sometimes used as a symbol for the quantity molarity (dimensions are amount of substance per volume, official IUPAC name is amount of substance concentration, and official IUPAC symbol is $c$). Textbooks that follow this convention sometimes use MM or ℳ as an alternate symbol for molar mass to avoid name conflicts. Combining the quantity $M$ with the unit M gives silly statements such as: $$M = 5~\text{M}$$

Other useful information

Make sure your lowercase m and uppercase M look different when you write or type, otherwise you are confusing your readers. If you type, quantities should be italicized and units should not. That also helps to distinguish the different meanings of m.
You can't cancel m with m if they don't have the same meaning. It is fine to cancel the milli in mg/mL to obtain g/L, but not cancel $m$/mg to obtain 1/g.


How to know if something will be a good oxidizing/ reducing agent? (Besides the activity of the metal)

The relative strengths of oxidising and reducing agents can be inferred from their standard electrode potentials. The strongest oxidizing agents are shown from the standard electrode table.


Intermolecular forces - Covalent bonding

Permanent dipoles are the result of charge separation within a bond, and normally results from bonds where the two elements have a large difference in electronegativity. Dipole moment is a measurement of the polarity of a compound. For example, the carbon-oxygen bond in acetone gives the compound a permanent dipole, explaining its large dipole moment. A temporary dipole occurs when electrons randomly group together in one part of a bond. These dipoles result in London Dispersion Forces, the weakest king of Van der Waals force. Van der Waals forces are a collective term for all residual attractive and repulsive forces between compounds that are not the result of ionic or covalent interactions. Induced dipoles are dipoles that result from a non polar compound becoming polarized as a result of interactions with an ion or polar compound.


This is an example of terminology which should be taken with a grain of salt.
The term "effective nuclear charge" is often casually symbolized with $Z_\mathrm{eff}.$ This is a universally accepted simplification, but you should keep in mind that the effective nuclear charge is, strictly speaking, $Z_\mathrm{eff}e,$ where $e$ is the elementary charge.
Similarly, the full nuclear charge is $Ze,$ where $Z$ is the atomic number.
So, it would be more appropriate to refer to $Z_\mathrm{eff}$ as to effective atomic number since $Z_\mathrm{eff}$ is defined as a function of $Z$ $(Z_\mathrm{eff} = f(Z_i)).$
For example, the common form provided in every introductory physical chemistry textbook is
$$Z_\mathrm{eff} = Z − σ,$$
where $σ$ is the shielding constant of the nucleus (empirical dimensionless parameter).
Since $Z$ is dimensionless, so is $Z_\mathrm{eff}.$
However, effective nuclear charge $Z_\mathrm{eff}e$ is not, and can, for instance, be expressed in coulombs (in the SI system $e = \pu{1.602176634E−19 C}).$


Which has more inductive effect: an aldehyde or a carboxylic acid?

I believe the Wiki page is more accurate, because the Hammett Equation substituent constants (para) indicate -CHO should be more "electron withdrawing" than -COOH, but there is little reason to use such lists as "absolute" - I think they are simply good guidelines for reactivity.

I looked at the data in this link: https://www.bluffton.edu/homepages/facstaff/bergerd/classes/CEM311/handouts/sigmas.pdf

However, substituent constants are not absolute proof of "inductive effect" and it is difficult to conclude from such data that a particular substituent is more "inductively" electron withdrawing than another. To do that, we would need a model (or a set of compounds, equivalently) where we can achieve separability of sigma and pi electrons.


What is the preferred IUPAC name of hexamethyldisilazane?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the name ‘disilazane’ is no longer recommended for $\ce{H3Si-NH-SiH3}$. Such structures are now named as amines. (Also note that the name ‘hexamethyldisilazane’ is ambiguous and therefore never was in accordance with IUPAC nomenclature.)

P-21.2.3.1 Heterogeneous parent hydrides composed of alternating heteroatoms, i.e., [a(ba)n hydrides], excluding carbon atoms and the halogen atoms.
Compounds containing an unbranched chain of alternating atoms terminated by two identical atoms of the element coming later in the seniority order O > S > Se > Te > N > P > As > Sb > Bi > Si > Ge > Sn > Pb > B > Al > Ga > In > Tl may be named by citing successively a multiplying prefix denoting the number of atoms of the terminal element followed by the ‘a’ term for that element, then the ‘a’ term of the other element in the chain and the ending ‘ane’. The terminal letter ‘a’ of an ‘a’ term is elided when followed by a vowel; the terminal vowel of a numerical prefix is not elided even when the ‘a’ term begins with the same vowel. When nitrogen atoms are present, amine names (see P-62) are preferred because of the higher functionality of amines. This concept does not extend to other elements. These parent hydrides are preselected parent hydrides (see P-12.2) and have priority to receive preferred IUPAC names, as long as they are used to name compounds containing carbon.

Symmetrical secondary amines are named according to Subsection P-62.2.2.1, which describes three different methods.

P-62.2.2.1 Symmetrical and unsymmetrical secondary and tertiary amines are named only by the same methods described in P-62.2.1.2.
(1) substitutively using the retained name ‘aniline’ or the suffix ‘amine’ and the name of a parent hydride with further N-substitution;
(2) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘azane’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
(3) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘amine’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
Method (1) generates preferred IUPAC names. Functional parent names like diethylamine and triethylamine are deprecated. The prefixes in names of such secondary and tertiary amines formed by method (2)[sic: read (3)] are set off by parentheses to distinguish them from these deprecated names.

The corresponding names for $\ce{H3Si-NH-SiH3}$ are actually given as examples in the Blue Book:

(1) N-silylsilanamine
(2) disilylazane
(3) disilylamine
(not disilazane; see P-21.2.3.1)

Note that, since $\ce{H3Si-NH-SiH3}$ does not contain any carbon, no preferred IUPAC name (PIN) is indicated in the Blue Book for this compound.
The corresponding names for the complete compound that is given in the question are:
(1) 1,1,1-trimethyl-N-(trimethylsilyl)silanamine
(2) bis(trimethylsilyl)azane
(3) bis(trimethylsilyl)amine
Since P-62.2.2.1 explicitly states that Method (1) generates preferred IUPAC names even for symmetrical amines, the preferred name for the complete compound that is given in the question is 1,1,1-trimethyl-N-(trimethylsilyl)silanamine.


Why is heat capacity defined as a partial derivative?

I think you are starting on the wrong foot.

$U$ and $H$ are state functions. They both depend on several variables. $U$ is a function of entropy $S$ and volume $V$ and composition. For a pure fluid there is no dependence on composition, so we can just say $U = f(S,V)$. Similarly $H$ is a function of entropy and pressure $p$ i.e., $H=f(S,p)$. You go from U to H by using a Legendre Transform (but ignore Legendre Transforms for now... I will give you a good reference to read for thermodynamics at the end, it will make your life better).

Important

Reality has a way of changing things. For instance experiments are done almost always at a constant $T$, and either constant $p$ or $V$. Therefore, we like thinking of things in terms of these variables. Lets face it... entropy is tricky.

Assumption

Okay, lets now say that $U=f(T,V)$ and $H=f(T,p)$ this is alot easier to handle. In thermodynamics we are always after changes in things. Therefore, the differential forms of $U$ and $H$ are desirable.

Also, this is not a haphazard switch, entropy and temperature are intricately related to each other $T \equiv \left( \frac{\partial U}{\partial S} \right)_{x's}$ where $x's$ are all other independent variables being held constant.

The differentials for both $U$ and $H$ are as follows (with respect to our new independent variables).

$$U(T,V)$$
$$dU = \left( \frac{\partial U}{\partial T} \right)_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$
and
$$H(T,p)$$
$$dH = \left( \frac{\partial H}{\partial T} \right)_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$

Endgame

It is here that we now define what $C_V$ and $C_p$ are

$$C_V \equiv \left( \frac{\partial U}{\partial T} \right)_V$$

$$C_p \equiv \left( \frac{\partial H}{\partial T} \right)_p$$

These are fundamental definitions and can't be tampered with. In introductory thermodynamics teachers play games and try to frankly dumb things down, and this leads to, in my opinion, improper definitions and understanding. This link provides a definition (without much context) that ties $Q$ into the mix.

If we insert these definitions into our differential equations we have

$$dU = C_V dT + \left( \frac{\partial U}{\partial V} \right)_T dV$$
and
$$dH = C_p dT + \left( \frac{\partial H}{\partial p} \right)_T dp$$

I leave it as an exercise, if you wish, to find the definitions for $\left( \frac{\partial U}{\partial V} \right)_T$ and $\left( \frac{\partial H}{\partial P} \right)_T$

Final Resource

Herbert B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd Edition

The man called Callen

The Textbook


PIB wave functions and the effects of e- delocalization on UV absorption in conjugated systems

You can use the PIB model as a very approximate way of thinking about energy levels in conjugated ($\pi$ electron) molecules. Absorption in triatomics is much more tricky, however, and usually a Walsh diagram is necessary.

The general idea is that as the conjugation gets larger the absorption wavelength for electronic transitions gets smaller and so shifted to the red part of the visible spectrum. (A similar argument applies to particles on a ring for aromatics)

The energy levels of a PIB vary as $E_n \approx (n/L)^2$ where n is the quantum number and L the length of the box, or conjugation. Take this to be sum of all conjugated bond lengths. Thus as L gets larger the energy levels become closer together. Secondly as the number of conjugated bonds increases the number of quantum levels filled also increases, as there are more electrons. The optical transition is from the highest filled level to the next one up (HOMO-LUMO), the energy difference is $E_{n+1}-E_n$ at each value of L. If you take the length of the box to be the number of bonds m of length a, then $L=ma$ it is easy to show that the absorption spectrum shifts to the red as the conjugation increases, although you should definitely not expect accurate values in terms of energies.


Aldehyde and ketone nomenclature when aldehyde/ketone group is bonded to an alkyl group

The characteristic group $\ce{=O}$ corresponds to the substitutive prefix ‘oxo’.
Thus, the substitutive names that you are looking for are ‘5-(2-oxoethyl)decane’ and ‘5-(2-oxopropyl)decane’.

However, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred names for aldehydes and ketones derived from alkanes are constructed using the suffixes ‘al’ and ‘one’, respectively.

Furthermore, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes and then to parent hydrides when different classes are present.

Therefore, the preferred IUPAC names for the given compounds are ‘3-butyloctanal’ and ‘4-butylnonan-2-one’.


How is it possible to increase mercury's specific weight?

Specific weight is a property of the element. It is product of the atomic mass and the volume. You can't easily alter either.

You could mix an element with another element of a different density and the density of the mixture will change. But you won't be changing the specific mass of the elements, just creating a mixture with a different density.

You can't readily change the mass of an element. But an element might be made up of multiple isotopes (mercury has 4 significant ones with masses ranging from 199 to 202 atomic mass units). If you enrich the mercury for one of those isotopes, you can change the mass, very slightly (for other elements this can be important; fissile isotopes of uranium are concentrated by doing it when making nuclear fuels).

In practice there is nothing practical you can do to alter the density (specific mass) of an element though you can create mixtures with different densities (but then you still haven't altered the properties of the element itself).


How to calculate Hirshfeld charges from Gaussian?

%nproc=8
%mem=1600mb
# mPW1PW91/6-311+G(2df,2p) pop=Hirshfeld scf=tight

Methanol

0 1
8 -0.046153000 -0.751257000 0.000000000
6 -0.046153000 0.658469000 0.000000000
1 0.857779000 -1.062372000 0.000000000
1 -1.086394000 0.976180000 0.000000000
1 0.437376000 1.072716000 0.889654000
1 0.437376000 1.072716000 -0.889654000

will do. All I did was to introduce blank lines after Route section (the one starting with #) and Title section (Methanol). Also, make sure that there is a blank line after the Molecule specification part. Gaussian is usually a convenient entry point to computational chemistry and you can find a lot of advice for it. The Gaussian manual is not bad, either. Have a look at this to get an impression about the syntax of input files - this is something Gaussian isn't forgiving about.


Confusion about the SN1 and SN2 Mechanism

Your reasoning is correct. It's the relative rates for a given alkyl halide that matter, and it's difficult to draw conclusions about those, because the relative rates are not fixed, as the $\mathrm{S_N2}$ rate is dependent also on the nucleophile concentration.

We can write the rate of the $\mathrm{S_N1}$ as $k_1[\text{alkyl halide}]$ and the rate of the $\mathrm{S_N2}$ as $k_2[\text{alkyl halide}][\text{nucleophile}]$, so the relative rate at a fixed concentration of the alkyl halide is given by

$$\frac{\mathrm{S_N2}}{\mathrm{S_N1}} = \frac{k_2[\text{nucleophile}]}{k_1}.$$

The identity of the nucleophile will be a factor as well, since that will affect $k_2$.

So all in all, you can't make a general conclusion that one reaction type will always go faster than the other, although it is likely that there are alkyl halides for which one reaction type is faster in a large majority of cases.


According to the Wikipedia page on the Kipp's apparatus, if the reactant is a powder, and soluble in water, you can soak pieces of pumice in the solution, dry the pumice, and use it in a Kipp's apparatus. There is also a variant of the apparatus that allows two liquids to be used together rather than a liquid and a solid reactant. In that case, you would pre-dissolve the sodium sulfite in water.