How to create different levels of hydrophobicity on glass beads surface?

Lets assume first that the surface $\ce{Si-O}$ bonds of teh glass are terminated by $\ce{Si-OH}$ groups. Hydrophilicity will depend greatly on the ration of $\ce{Si-OH}$ to $\ce{Si-R}$ groups on the surface of the glass. Chlorotrimethylsilane is great for a completely hydrophobic surface as it will cap all of the $\ce{Si-OH}$ groups to $\ce{Si-O-Si(CH3)3}$ groups allowing no polar interaction of the surface with the water. dichlorodimethylsilane is not as good leaving you with $\ce{-Si-O-(Si(CH3)2)-O-Si{-}}$ and $\ce{-Si-O-Si(CH3)2OH}$ groups at the surface which affords a bit more interaction.  Octadecyltrichlorosilane does provide even more polar groups than the last two, but because of the large octadecyl group this will likely make a more hydrophobic surface than the chlortrimethylsilane.  The tetraethyl orthosilicate or rather tetraethoxy silane (TEOS) will hydrolze to $\ce{Si(OH)4}$ in water which you can use to help make the surface more hydrophilic.

If I had to do this I would make various binary mixtures of TEOS, chlorotrimetyhlsilane, and dichlorodimethylsilane and treat the surface with them under constant conditions and see what spectrum of hydrophilicity you achieve. For the most hydrophobic surfaces I would use mixtures of octadecyltrichlorosilane and chlorotrimetyhlsilane.


Is common ion effect applicable on two strong electrolytes

Certainly not.

Suppression of dissociation of a strong electrolyte is not a very rational thing to discuss, owing to the unmeasurably large value of $K$.

To see what I mean, try looking up the $K_{a1}$ of $\ce{H2SO4}$.
(At least I couldn't find it!)

Your textbook's given answer is correct.

The equilibrium suppressed by the introduction of $\ce{KHSO4}$ is not the dissociation of $\ce{H2SO4}$, but rather this one:

$$
\ce{HNO3} \ + \  \ce{H2SO4} \rightleftharpoons \ce{NO2+} \ + \ \ce{HSO4-} \ + \ \ce{H2O}
$$

Nothing too auspicious: all that remains is the run-of-the-mill common ion effect we've known from Kindergarten days!

PS.
Conclusion: $\ce{KHSO4}$ sucessfully decreases the concentration of $\ce{NO2+}$, which happens to be the electrophile and a key participant in the rds(rate determining step) of Nitration of benzene.

$$
Rate=k[\ce{NO2+}][\ce{C6H6}]
$$


Converting 1,4-benzenedimethanol to benzene and methanol?

Oxidize to the dialdehyde (Jones reagent), decarbonylate with rhodium (do it at the bench before building the pilot plant), reduce the CO to methanol.  Why would you want to do that?


White phosphorus can spontaneously ignite in air, and produces white smoke.

Some oils, like linseed oil, on bunched-up cloth can also spontaneously combust.


How to obtain the radial probability distribution function from a quantum chemical calculation?

This is gladly not a big problem as MultiWFN (open source) can do that for you. I'll guide you through the process from start to finish. My current setup is Gaussian 09 Rev D.01 and MultiWFN 3.3.8; there are newer versions out there with which this should work, too.

Run the calculation, for example ar.com at the HF/STO-3G level of theory, but anything else will also be fine:

%chk=ar.chk
#P HF/STO-3G
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo6

title card

0 1
Ar    0.0  0.0  0.0

This will generate a checkpoint file ar.chk which needs to be formatted:

formchk -3 ar.chk ar.fchk

Now fire up your MultiWFN program, you should get a prompt similar to the following:

 Multiwfn -- A Multifunctional Wavefunction Analyzer (for Linux)
 Version 3.3.8, release date: 2015-Dec-1
 Project leader: Tian Lu  (Beijing Kein Research Center for Natural Sciences)
 Citation of Multiwfn: Tian Lu, Feiwu Chen, J. Comp. Chem. 33, 580-592 (2012)
 Multiwfn official website: http://Multiwfn.codeplex.com
 Multiwfn official forum (in Chinese): http://bbs.keinsci.com
 Bug reporting, question and suggestion, please contact: Sobereva@sina.com
 ( The number of threads:  2   Current date: 2017-03-09   Time: 14:46:57 )

 Input file path, for example E:\Haiyore!Nyaruko-san\Nyaruko.wfn
 (Supported types: .wfn/.wfx/.fch/.31/.chg/.pdb/.xyz/.cub/.grd/.molden, etc.)
 Hint: To reload the file last time used, simply input the letter "o". Input such as ?miku.fch can open miku.fch in the same folder of the file last time used

If you loaded MultiWFN from the directory you're files are located, simply put in ar.fchk. It will load the file and output some basic information and take you to the "Main function menu":

 Please wait...
 Loading various information of the wavefunction
 Loading basis-set definition...
 Loading orbitals...
 Converting basis function information to GTF information...
 Generating overlap matrix...
 Generating SCF density matrix...

 Total/Alpha/Beta electrons:     18.0000      9.0000      9.0000
 Net charge:     0.00000      Expected multiplicity:    1
 Atoms:      1,  Basis functions:      9,  GTFs:     27
 Total energy:  -521.222880802727 Hartree,   Virial ratio:  2.01704078
 This is a restricted single-determinant wavefunction
 Orbitals from 1 to     9 are occupied
 Title line of this file: title card
 Formula: Ar1
 Molecule weight:        39.94800

 Loaded ar.fchk successfully!

                    ------------ Main function menu ------------

For the sake of sanity I will not copy every menu screen here, just simply what you have to put in. On this screen the radial distribution function is hidden in the "Other functions (Part 2)" and there it is option 5, also see the manual chapter 3.200.5 on page 175 (v.3.3.8). So choose

200 // Other functions (2)

and then

5 // RDF

You should get the following menu:

   ====== Plot radial distribution function for a real space function ======
 -1 Exit
 0 Calculate radial distribution function and its integration curve
 1 Select real space function, current:   1
 2 Set sphere center, current    0.0000    0.0000    0.0000 Ang
 3 Set lower and upper limit of radial distance, current:   0.0000   5.0000 Ang
 4 Set the number of integration point in each shell, current:  2030
 5 Set the number of radial points, current:   500

Here you can select a couple of options. The electron density is already preselected, but you can change it in option 1. You'll get a table with all available functions.
The only thing you might want to adjust are the limits, so choose

3 // Limits
0.0,2.0

and calculate the RDF:

0 // Calculate RDF

Choose 1 to plot and view the function. Or choose

3 // Save graph

or choose

5 // Export to RDF.txt

for post-procesing (with gnuplot).

You can shorten the journey if you prepare a file like this mwfn.in

ar.fchk
200 // Other functions p2
5   // RDF menu
1   // Select real space fucntion
1   // Electron density
2   // Set sphere centre
0.0,0.0,0.0
3   // Set limits
0.0,2.0
0   // Calculate RDF
3   // Save graph
5   // export to text file

and pipe it into the program:

multiwfn < mwfn.in

The native graphing capabilities of MultiWFN don't allow for customisation, so you might want to run it through gnuplot. Here is a file which gives you the graph below, I am using gnuplot 4.4, but later versions are more feature rich.

# General Settings
set terminal svg enhanced font "Arial,12" size 800,600
set tics font "Arial, 9"
set output "ar-rdf.svg"
unset key
#or set key top right Left # etc
set title 'Argon Radial Distribution Function HF/STO-3G'
#
# X-Axis: radius / angstrom
set xlabel 'r / Å'
set xtics in scale 0.6, 0.4 nomirror
set xtics 0,1
set mxtics 2
set xrange [0:2]
# Y-Axis: RDF/a.u.
set ylabel "RDF / a.u." rotate left
set ytics  in  scale 0.6, 0.4 nomirror
set ytics 0,10
set mytics 2
set yrange [0:25]
#
#
# PLOT!
plot \
     "RDF.txt"   using ($2):($3) title "HF/STO-3G"   with lines lt 1 axes x1y1

This is the resulting graph:


Reorganization during ionisation for d block elements

As your textbook notes, the relative energy of atomic orbitals is not fixed as you start ionizing your atom. As you ionize, all orbitals drop in energy. However, higher angular momentum orbitals drop more than lower ones, and specifically in your case, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ orbitals. So consider vanadium, the atom has a $\mathrm{[Ar]4s^23d^3}$ configuration, but after you ionize it, the $\mathrm{3d}$ orbitals drop below the $\mathrm{4s}$ ones and you get $\mathrm{[Ar]3d^4}$ as the electron configuration of the $\ce{V+}$ ion. So your measured ionization energy includes both the energy it takes to remove an electron from $\ce{V}$ and also some reorganization energy. You know there is some reorganization energy since two electrons (both $\mathrm{4s}$ ones) have "moved" but you only removed one.

Now, what about $\ce{Cr}$? Recall that $\ce{Cr}$ is a bit special and has a $\mathrm{[Ar]3d^54s^1}$ configuration when neutral. After we remove an electron (in this case, a $\mathrm{4s}$ one), we have an $\mathrm{[Ar]3d^5}$ configuration for $\ce{Cr+}$. Note that no electrons had to move around besides the one we removed. So there is no reorganization energy in the case of $\ce{Cr}$, as your textbook notes.


Sodium metal will react with iron (III) oxide to produce sodium oxide and iron.

$\ce{6Na +Fe2O3 -> 3Na2O + 2Fe}$

This is also like a thermite reaction, so try to use the soam\e method as the reaction.


Second-step activation energy in SN1 reactions

The $\mathrm{S_N1}$ reaction has two steps i) the formation of carbocation and 2) the formation of product.

(image taken from libretexts)
In the second step an unstable carbocation reacts with the nucleophile. You are asking why this step has to have an activation energy. You can rationalize that by considering the principle of microscopic reversibility. Think about what would happen if the $\mathrm{S_N1}$ reaction went in reverse—first the nucleophile leaves the product, forming a carbocation, and then the carbocation joins with the leaving group to give the reactant. Obviously when the nucleophile leaves the prdouct, and carbocation forms, there is an activation barrier for that.
So regardless of the themodynamics of the total reaction, whenever you detach a functional group and form a
(carbocation + anion), there has to be an activation barrier i.e. a transition state that is higher in energy than both (carbocation + anion) and the initial state. Which means that if you wish to reattach an anion to the carbocation, there will also be an activation energy.


Why do the peaks in a radial distribution function graph have a probability >1

The radial distribution function isn't a probability or a probability density. The usual definition is $$g(r)=\frac{\rho(r)}{\rho_0}$$ where $\rho(r)$ is the number density at a particular distance $r$ away from some specified atom and $\rho_0$ is the bulk density of the material.
It can be larger than one if the local density is greater than the bulk (which you see in the first peak of your graph), so it can't be a probability. It also can't be a probability density, as the local density should eventually converge to bulk at long distance, so $g(r)$ converges to 1 (as you start to see on the right-hand side of your graph) so the integral would continue to increase as you increase the range of integration.
So what does the RDF tell you? Well if you integrate it over a spherical shell and multiply by the bulk density, you obtain the number of atoms in that shell. One particular shell we are often interested in is the first coordination sphere. Using the RDF, we can define the number of particles in the first coordination sphere as $$n(r')=4\pi\rho_0\int_0^{r'}g(r)r^2dr$$ where $r'$ is the location of the first minimum of the RDF. We can similarly determine the number of atoms in the $n^{\text{th}}$ coordination sphere by integrating from the$(n-1)^{\text{th}}$ minimum to the $n^{\text{th}}$ minimum.


Question regarding the mechanism of oxymercuration-demercuration of alkenes

This reaction, in my opinion is an extension of Markovnikov’s rule

To understand why Markovnikov’s rule works, consider the
structure and stability of carbocations and about the general nature of reactions
and transition states.

The first point to explore involves structure:

  A great deal of experimental evidence has shown that carbocations are
  planar. The trivalent carbon is sp2-hybridized, and the three
  substituents are oriented toward the corners of an equilateral
  triangle.

Because there are only six valence electrons on carbon and all six are used in the three s bonds, the p orbital extending above and below the plane is
unoccupied.

The second point to explore involves carbocation stability. Thermodynamic measurements show that,
indeed, the stability of carbocations increases with increasing substitution so that
the stability order is tertiary > secondary > primary >methyl.

Now here comes the question…

Why are more highly substituted carbocations more stable than less highly substituted
ones?

There are at least two reasons.

Part of the answer has to do with inductive effects (as you noted in question) , and part has to do with hyperconjugation.

  Inductive effects, result from the shifting of electrons in a s bond
  in response to the electronegativity of nearby atoms. (electrons from
  a relatively larger and more polarizable alkyl group can shift toward
  a neighboring positive charge more easily than the electron from a
  hydrogen

. Thus, the more alkyl groups there are attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs)

Hyperconjugation, is the stabilizing interaction between a p orbital and properly oriented C- H sigma bonds on neighboring carbons that are roughly parallel to the p orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation


Reaction (or lack thereof) between oxygen and halogen gas

Your on the right track: in order to form a new compound, old bonds must be broken. This requires energy, even though the reaction may, overall , be exothermic. This is called the activation energy.
Think of it as pushing a car a short distance up one side of a hill before it reaches the peak and then rolls a long distance down the other side.
For example, in the dark, at room temperature, $\ce{Cl2}$ and $\ce{H2}$ coexist quietly in a flask... until light activates the chlorine (wait for the bang).
On the other hand, add enough energy, and even endothermic reactions take place, and, if the product is comparatively stable, or metastable, some can be left afterwards. For example $\ce{O2}$ is fairly stable, but given some energy input, e.g. from an electric discharge or from shortwave ultraviolet radiation, $\ce{O3}$, ozone, is produced, though that reaction is not thermodynamically favorable. The ozone is waiting to decompose... once it's activation energy is achieved.


At what temperature (in kelvin) are most of the elements on the periodic table liquids?

Now that's a great question indeed! Evidently, at 0K all elements except helium are solids, at 10000K they are all gases, so someplace in between the number of liquids must reach a maximum; what and where might that be?

Well, there is no formula or theorem that says liquid hydrogen must boil at 20K, nor is there such a thing for any other element, so this line of thinking hardly provides any insight. Let's try another way.

Let's take a look at all elements' melting and boiling points. (I used the great site referenced in a comment by ringo, where all data are conveniently stored on line 47 in the file http://www.ptable.com/Static/interactivity-2fcd37b.js). Let's simply make a joint list of all melting and boiling points, sort it (keeping track of which is which), and then walk through it from bottom to top, calculating the number of liquids on each interval. If you know some kind of programming, you can do it with a script.

It turns out that the maximum number of simultaneously liquid elements is 49, and it is reached several times between 2183K (melting of vanadium) and 2792K (boiling of aluminium). By dragging the slider you may actually see which elements are included in these 49. Now, the data are not terribly reliable (for example, in higher actinides they claim some melting points obtained by God-knows-what extrapolation, but no boiling points at all), yet still this is something to think about.


Syn-addition or anti-addition to fumaric acid?

I don’t know what molecule you ended up with, but the way I draw it, I end up with (2S,3S)-2,3-dihydroxysuccinic acid as shown below.

The syn-addition step would technically be the addition of the electrophile permanganate to the double bond. From there, the oxygens are in place and are not moved. If you reorder the structure to the identical zig-zag representation, you should arrive at a clear syn-orientation.

Note that there are two rotations in this: the first step includes a rotation of fumaric acid by $90^\circ$ so that I can draw the five-membered ring in a plain. After cleaving off $\ce{MnO4^3-}$ (which further reacts to $\ce{MnO4^2-}$ in basic conditions) I arrive at the syn-product.

With maleic acid, everything would be the same, except for the orientation of the carboxylic groups in the second and third structures. That would lead to the hydroxyl groups pointing in different directions even though we are still dealing with a syn-addition.


How to find the angle of emission of an X-ray during impact of accelerated cathode rays?

The angle of resulting X-ray radiation, product of the impact, is related to the amount of energy or momentum of the cathode ray. Bremsstrahlung relation will amount how much energy the produced photon has.
$$hf = \frac{hc}{\lambda} = K - K'\tag{1}$$
$$\lambda_\mathrm{min} = \frac{hc}{K} = \frac{hc}{e\,\Delta V}\tag{2}$$
Back scatter the possible photon angles with the Compton effect equation,
$$\lambda' - \lambda = \frac{h}{m_\mathrm{e}c}(1 - \cos\theta)\tag{3}$$
A difference in kinetic energy is as difference in wavelength. Precise results should consider a vacuum and use imaginary numbers; also probability distribution on superposition.
For more details refer to the chapter 3 of Krane's Modern Physics [1].
Reference

Krane, K. S. Modern Physics, 3rd ed.; Wiley: Hoboken, NJ, 2012. ISBN 978-1-118-06114-5.


Infrared spectroscopy of substances on a OHP film

Your are going to analyze both, the OHP foil alone, and the part of the OHP foil that is covered by the coating and calculate the difference of the later to get the data of your coating material.
If you do want to do the classical way (transmission IR), just cut-off a little of the sample (a hole puncher as in the office may do the job, too), attempt to grind it with KBr and press the pellet as normal.

If you want to stick to ATR, place the foil on the dish without prior sample preparation.  This is quicker, and to some degree allows a space-resolved analysis, too -- the window (center dish of the table) of a single reflection geometry instrument (like the one below, Nicolet iS10 with smart-ATR-IR module) is of about 3 mm in diameter).  The window material of the ATR module may be Germanium (as here), yet there are instruments with a window of Diamond, too.


In order to test whether there may be $\ce{s}$-$\ce{p}$ mixing and also to check whether there may be actually a bound state, I optimised this molecule (at PBE0-D3/def2-TZVPD) using NWChem 6.6 and analysed it using natural bonding orbitals (NBO 5.9).
I found following minimum geometry:

Figure 1: Minimum geometry found for $\ce{N2^{-2}}$ (distance in angstrom)

So there seems to be a bound state.

Natural charge on each $\ce{N}$ is -1.0 as would be expected (natural electron configuration was found to be $\ce{1s^{2.00} 2s^{1.65} 2p^{4.17} 3s^{0.06} 3p^{0.05} 3d^{0.07} 4d^{0.01}}$).
Wiberg bond order in the NAO basis for the $\ce{N-N}$ bond was found to be 2.2511.

Two bonding NBOs were found:

    (Occupancy)   Bond orbital/ Coefficients/ Hybrids
-------------------------------------------------------------------------------
  1. (2.00000) BD ( 1) N  1- N  2
              ( 50.00%)   0.7071* N  1 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
              ( 50.00%)   0.7071* N  2 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
  2. (2.00000) BD ( 2) N  1- N  2
              ( 50.00%)   0.7071* N  1 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)
              ( 50.00%)   0.7071* N  2 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)

That is,

$$\pi_{\ce{N1-N2}} = 0.7071 \ce{p}_{\ce{N1}} + 0.7071 \ce{p}_{\ce{N2}}$$

Figure 2: $\pi_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$

$$\sigma_{\ce{N1-N2}} = 0.7071 \ce{sp^{1.70}}_{\ce{N1}} + 0.7071 \ce{sp^{1.70}}_{\ce{N2}}$$

Figure 3: $\sigma_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$

Furthermore,

NATURAL BOND ORBITALS (Summary):

                                                    Principal Delocalizations
          NBO                 Occupancy    Energy   (geminal,vicinal,remote)
===============================================================================
Molecular unit  1  (N2)
  1. BD ( 1) N  1- N  2        2.00000     0.18488
  2. BD ( 2) N  1- N  2        2.00000    -0.43475  10(g),42(g)

Thus, according to the simple Lewis picture provided by NBO,

$\ce{s}$-$\ce{p}$ mixing ($\ce{sp^{1.70}-sp^{1.70}}$) happens for the formation of $\sigma_{\ce{N1-N2}}$, as expected;
the formation of $\pi_{\ce{N1-N2}}$ happens through pure $p$ hybrids; and
$\pi_{\ce{N1-N2}}$ lies in energy above $\sigma_{\ce{N1-N2}}$ by 0.61963 a.u..


Do solid insoluble ions react with soluble ions in aqueous solution?

What you seem to be asking about is a two phase system, a solid phase in equilibrium with ions in solution. For such a system there are two reactions occurring at the same rate. Thus some of the solid is constantly dissolving and some of the ions from solution depositing at the same rate.
There is a named process, Oswald ripening, that describes the interaction in such a solution. The overall effect is that smaller particles are more likely to dissolve so the particle size tends to grow as the process continues since larger particles have a smaller surface area to volume ratio.


Determining water of crystallisation in a hydrate from the mass percentages

In this case you only have one variable to solve for ($x$) and you have one piece of information (Zn being 21.5% by mass) so you have enough information to solve the question.

The molecular mass of the hydrate can be expressed in terms of $x$:

$$\begin{align}
M(\ce{Zn(ClO3)2.xH2O}) &= M(\ce{Zn}) + 2\cdot M(\ce{Cl}) + (6+x)\cdot M(\ce{O}) + 2x\cdot M(\ce{H}) \\
&= \pu{65.38 g mol-1} + 2(\pu{35.45 g mol-1}) + (6+x)(\pu{16.00 g mol-1}) + \,\\
&\qquad 2x(\pu{1.01 g mol-1}) \\
&= (232.28 + 18.02x)~\pu{g mol-1}
\end{align}$$

Now

$$\begin{align}
\frac{M(\ce{Zn})}{M(\ce{Zn(ClO3)2.xH2O})} = \frac{\pu{65.38 g mol-1}}{(232.28 + 18.02x)~\pu{g mol-1}} &= 0.215 \\
65.38 &= 0.215(232.28 + 18.02x) \\
x &= 3.99 \approx 4
\end{align}$$


Ammonium bicarbonate, "baker's ammonia", is sometimes used as a leavening agent for baking (e.g. cookies) as it decomposes to gaseous ammonia, water, and carbon dioxide which are driven off in the oven heat. You can get it in some food stores. Hope this is helpful.


What is the structure of [Co(NH3)5SO4]+?

There's a certain amount of misinformation here:

  sulfate is a bidentate ligand

Sulfate is actually a flexidentate ligand, which is a term used to describe a polydentate ligand which is found to show different denticity in different complexes. Sulfate can act both as mono- or a bidentate ligand. Remember, just because it has two donor atoms doesn't make it a bidentate ligand.

The occurrence of complexes where sulfate acts as a bidentate ring would also be rare owing to the fact that four-membered rings are strained. This page lists $\ce{[Pt(SO4)(P(C6H5)3)2]}$ as one example of such an occurrence.

  coordination number of cobalt would become 7. How is this possible?

If this implies that coordination numbers of 7 aren't possible, it isn't true. Coordination numbers of 3, 5, 7, 8-12, 14 are possible, albeit such compounds are quite rare. See this Wikipedia page for some examples.

Thus, the structure of the complex $\ce{[Co(NH3)5SO4]+}$ should be octahedral with cobalt showing a coordination number of 6.


Why does a mixture of nitrous acid with concentrated nitric acid cause nitration of benzene instead of nitrosation?

You were absolutely correct when you said that nitrosation takes place.

What actually happens is that after nitrosation occurs, concentrated $\ce{HNO3}$ oxidises the nitroso group to form nitrobenzene.

Source
A Guidebook To Mechanism In Organic Chemistry; Chapter 6, section 6.2;Peter Sykes


Which salt precipitates from a solution of ammonium oxalate and silver nitrate?

Your current solution is wrong and you don't seem to be familiar with the concepts involved.

Let's walk through this together and change that!

You take solid ammonium oxalate $\ce{(NH4)2C2O4}$ and silver nitrate $\ce{AgNO3}$ and dissolve both salts in water.
The salts dissociate into their ions. Let's do that separately for both salts first.

$$\ce{ (NH4)2C2O4 ->[\ce{H2O}] 2 NH4+_{(aq)} + C2O4^{2-}_{(aq)}}$$

$$\ce{AgNO3 ->[\ce{H2O}] Ag+_{(aq)} +  NO3^{-}_{(aq)}}$$

Since you are dissolving both salts in the the same beaker with water, your solution now contains all the ions of both equations.

Forget about the spectator ion fanciness.

It is much more important to identify the chemical units (= the ions from which the salts are composed). Leave these ions intact and don't rip them apart into weird combinations of elements.
Precipitation will occur when the combination of these ions (chose a cation and an anion) yields a salt with a very low solubility in water. There are only two cations and anions here, so the number of possible combinations isn't too big. Look up the solubilities of the possible products to figure out which one will precipitate.
Do not forget to balance the reaction!


Does the reaction of 1-bromobutane with cyanide ion follow an SN1 or SN2 mechanism?

Agreed, $ \ce{Br-} $ is a good leaving group, so one would expect that an SN1 mechanism is not too difficult to have in a polar protic solvent. But do not forget that the mechanism of any reaction is determined by not only the nature of the leaving group, but also the carbon skeleton. A primary carbocation is extremely difficult to obtain, and will not, in general, be formed unless the corresponding anionic complex is extremely stable. For example, the equilibrium $$ \ce{R-CH2-F + SbF5 <=> R-CH2+ + SbF6-    }$$ lies predominantly towards the products side because the anion $\ce{SbF6-}$ is an extremely stable system (which is why the mixture $\ce{HF + SbF5}$ is considered a super acid.). Had $\ce{AlBr3}$ been used before adding $\ce{KCN}$ to the solution, I would've gone for SN1 mechanism.

So, firstly, the primary carbocation is a difficulty if you consider an SN1 mechanism.

Moreover, the formation of a carbocation would immediately facilitate rearrangements. Since a primary carbocation is unstable, Wagner-Meerwein rearrangements will take place and your end product would have been a substitution at the second carbon instead of the first (terminal) one.

As for the solvent, it is a mixture of $\ce{Et-OH}$ and $\ce{H2O}$ to facilitate the dissolution of both reactants. Because one of them is an ionic compound, and the other an organic compound, getting them both in the same state can only be achieved by the mixture of an organic solvent and water; the organic solvent being such that it is, itself, miscible with water. A polar protic solvent weakens the ionic bonds by screening, and favours ionisation. This doesn't imply that a polar protic solvent will always completely ionise your solute.


In principle, you can use the integrated Clapeyron and Clausius–Clapeyron equations to calculate $P$ vs. $T$ co-existence lines and find the crossing point which is at the triple point. You need the $\Delta H$ for fusion and evaporation and the densities of the solid and liquid phases.
The equations are $$ p_\text{liq-vap}=\exp\left(\frac{\Delta H_\text{vap}}R\left(1/T_1-1/T\right)\right)$$ and for the solid vapour line use $\Delta H_\text{fus} +\Delta H_\text{vap}$

For the solid-liquid line use $$ p_\text{sol-liq} = P_1 + \frac{\Delta H_{\text{fus}}}{\delta V}\left(\ln\left(T\right)-\ln\left(T_1\right)\right)$$

where $\delta V = M\left(1/d_\mathrm l-1/d_\mathrm s\right)$ where $M$ is the molecular weight and $d$ the densities of liquid and solid. $T_1$ and $P_1$ are temperature and pressure at the triple point. Thus you have to iterate $T_1$ and $P_1$ by fitting (say by least squares) to the calculated curves to whatever data you have to find the triple point. Tricky!

Naively, I am surprised that using Lennard-Jones parameters would work as they describe interaction potential between two atoms and not in a solid or liquid.


Mechanism for the addition of hydrogen iodide to 3,3‐dimethylbut‐1‐yne

There seems to be an issue with both mechanisms on the front that an SN1 reaction would not take place since the carbocation formed is a vinylic carbocation which is highly unstable.
The actual reaction follows a termolecular mechanism where the rate of reaction is given to be:
$$\text{Rate}=[\ce{HX}]^2[\text{alkyne}]$$
Now, according to Advanced Organic Chemistry by Francis A. Carey, the reaction mechanism would be as follows:$^1$

Step $1$:
A concerted termolecular reaction...
This involves an acid/base reaction, protonation of the alkyne developing positive charge on the more substituted carbon. The π electrons act pairs as a Lewis base.
The other part is attack of the nucleophilic bromide ion on the more electrophilic carbocation creates the alkenyl bromide.

Step $2$:
In the presence of excess reagent, a second protonation occurs to generate the more stable carbocation.$^2$

Step $3$:
Attack of the nucleophilic bromide ion on the electrophilic carbocation creates the geminal dibromide.

[$1$]: The reaction mechanism stated above uses $\ce{HBr}$ and not $\ce{HI}$ instead. However this reaction takes place for $\ce{HX}$.
[$2$]: The same reaction using $\ce{HI}$ would have a comparatively lower yield of the geminal product since geminal diiodides are unstable due to the steric hinderance posed by the large size of the iodine atoms.


Hydrogenation side product: potential mechanisms for C-C bond making

First of all, I would check for impurities in styreneoxide. There is a good chance it was prepared from styrene having some 2-phenylpropylene in it, and this could account for your impurity.

Other than that, I would think among lines that hydrogenation is usually reversible, so above strong hydrogenation catalys phenylethanol is in equilibrium with corresponding aldehyde and methanol is in equilibrium with formaldehyde. Crotone condensation of said carbonyl compounds with subsequent hydrogenation may account for the impurity you see here.


Calculation of free energy from potential and kinetic energy

The sum of the potential and kinetic energies is the internal energy, E (or U).
However, you don't necessarily need to know the entropy to calculate the free energy.  If you perform an equilibrium molecular dynamics simulation, you can derive the free energy from the equilibrium constant.
See, for instance:
https://onlinelibrary.wiley.com/doi/full/10.1002/jcc.21776
If you are modeling a constant T and V system, where there is no work done, the equilibrium constant would give you the Helmholtz free energy:
$$\Delta A = - R T ln K^{'}_{eq}$$
where A = E - TS
If you are modeling a constant T and p system, where you have no non-pV-work, the equilibrium constant would give you the Gibbs free energy:
$$\Delta G = - R T ln K^{''}_{eq}$$
where G = H - TS = E + pV - TS
The reason for this is that, at constant T and V, with no work term, minimizing A_sys corresponds to a maximization of S_univ, and thus determines the equilibrium condition. By contrast, at constant T and p, with pV-work only, minimizing G_sys corresponds to a maximization of S_univ, and thus determines the equilibrium condition.
I've marked the equilibrium constants as prime and double-prime to indicate that these would be numerically different.


How can I balance this reaction using the algebraic method?

Now all you have to do is variable substitution, which is kinda tedious since it's such a large equation. Remember that solutions are defined up to multiplying everything by a constant factor, so we might as well pick

$a = 2$ (this guess is inspired by the equation for $\ce{Fe}$)
$\ce{Fe}{:~} e = 1$
$\ce{C}{:~} h = 12$
$\ce{N}{:~} g = 12$
Because $\ce{Mn}{:~} b = f$, we can rewrite $\ce{O}$ as $4c = 4d + 12e + 3g + 2h + i$
Subtract 4 times $\ce{S}$, and we get $0 = -4f + 3g + 2h + i$, so $4f = 60 + i$ (call this $\ce{O}'$)
Now rewrite everything in terms of $b$:

$\ce{K}{:~} 8 + b = d$, so $d = b + 8$
$\ce{Mn}{:~} f = b$
$\ce{O}'{:~} 4b = 60 + i$, so $i = 4b - 60$
$\ce{S}{:~} c = (b + 8) + 3 + b = 2b + 11$
$\ce{H}{:~} 2(2b + 11) = b + 8 + 12 + 2(4b - 60)$, so $4b + 22 = 9b - 100$, so $5b = 122$ and $b = 122/5$. This is a bit unfortunate since later on we have to multiply all variables by 5.

We now have $a=2$, $b=122/5$, $c=299/5$, $d=162/5$, $e=1$, $f=122/5$, $g=12$, $h=12$ and $i=188/5$.
By multiplying everything by 5 we get the final solution:

$a = 10$
$b = 122$
$c = 299$
$d = 162$
$e = 5$
$f = 122$
$g = 60$
$h = 60$
$i = 188$

You can check that these figures will make the equation balanced.


What concentration of sulfur hexafluoride is safe to inhale

I'm afraid this answer doesn't address where you can get $\ce{SF6}$, as I don't know any suppliers myself (it would probably help to specify where you are, though). I just want to straighten out the safety details and make sure you (and anyone else who reads this in the future) know what you're doing. All chemistry experiments can take a turn for the worse if done without adequate preparation.

Pure sulfur hexafluoride is a rather chemically inert gas (unlike the related but very toxic sulfur tetrafluoride $\ce{SF4}$ and disulfur decafluoride $\ce{S2F10}$), and it can be safely handled in regular situations. Decomposition of $\ce{SF6}$ usually requires aggressive conditions, such as exposure to molten sodium metal at 200°C. There are some situations where $\ce{SF6}$ reacts at low temperatures, but that requires irradiation with deep-UV photons or direct contact with solvated electrons from alkali metals dissolved in ammonia. I don't suppose you'll have to worry about such situations. It also seems that $\ce{SF6}$ can decompose into a small amount of the other aforementioned toxic sulfur fluorides if it is exposed to electrical arcs, but since it is an electrically insulating gas, the electric fields required for arcing are even higher than in air, so this also shouldn't be a problem.

All in all, both pure $\ce{SF6}$ and helium are very unlikely to cause chemical damage to the body, even indirectly. However, that doesn't imply they are completely safe. Of course, they can still act as an asphyxiant simply by displacement of oxygen in the lungs (make sure to read this!). Be aware that the brain is incapable of directly measuring the oxygen content in your blood, and instead measures oxygenation rather indirectly, by the amount of carbon dioxide. It is possible to faint and even die from hypoxia without any warning. Additionally, since $\ce{SF6}$ is relatively heavy, it requires more effort from your diaphragm muscles to expel from your lungs, making $\ce{SF6}$ inhalation slightly more dangerous than helium.

As a safety measure, do not take consecutive breaths in the non-oxygenated gas and do not spend more than 5 seconds without inhaling air. While your lungs are devoid of oxygen, not only are they unable to transport oxygen into your blood, the lungs actually work in reverse and help expel oxygen from your blood. After performing the trick, take deep breaths in regular air for a minute or two before trying it again, to be sure all the previously inhaled $\ce{SF6}$ or helium has exited your lungs and your oxygen supply has returned to normal. Hyperventilating before or after inhaling $\ce{SF6}$/helium is likely a bad idea, as it further messes with the balance of oxygen and carbon dioxide in the blood. You should avoid any possibility of becoming light-headed and fainting, especially if you decide to do the trick multiple times. Also be mindful that the gasses can accumulate in poorly ventilated spaces, with $\ce{SF6}$ being a particular problem as it will tend to stay close to the ground and disperse with more difficulty.

If you're going through the trouble to get $\ce{SF6}$, you should do the floating aluminium boat display, too. It's pretty cool! Here's another video showing you can even pour the gas into the boat and make it sink.


No, stricly speaking metallic sodium is not a strong base. Strictly speaking, metallic sodium is not a base at all. (Sodium amide clearly is one, though: due to the amide part.)

Not everything that reacts with an acid is a base. If an acid reacts to form hydrogen and there are no hydrides present, that is a very strong indicator that the reaction was not an acid-base reaction but a redox reaction. And that is what metallic sodium is: a strong reducing agent.

Just because reducing agents can react with acids (proton donors) to generate hydrogen radicals and thus nascent hydrogen gas does not make them bases. For the reaction partner of an acid to be a base, it must accept (and keep!, at least to a certain extent) the proton donated by the acid — or, in the Lewis theory, it must donate an electron pair.


Why did my deep eutectic solvent solidify?

You have to keep mixing the components until the liquid forms. If you have a sludge, you will have mixed something, but not everything intimately, so you won't have the correct component ratio. The DES is not going to form spontaneously,  so leaving it will not have helped.
We have had DES's remain as liquids for 6 months and counting.


Is argon preferred over nitrogen for maintaining an inert atmosphere?

Speaking purely from an organic point of view, there isn't really a huge difference, labs tend to stick to one or the other, largely depending upon the availability (some universities use the boil off from the liquid nitrogen cylinders to furnish labs with nitrogen gas, whilst others have no such facilities meaning labs must buy cylinders of either nitrogen or argon).

For serious organic groups (organometallics, total synthesis) doing a lot of air-sensitive chemistry, there can be some advantage to argon due to its higher density [argon (1.714 g/L) nitrogen (1.251 g/L)], meaning it tends to blanket the reaction, and if you need to quickly open the flask to put a reagent in then the argon doesn't escape so quickly.

For inorganic chemists, there are situations where nitrogen will react with certain chemicals, and in these cases argon must be used.


Determination of +/- reflections in sigma molecular term symbols

The $ \pm $ superscripts refer to symmetry operations defined in the $D_{\infty h}$ point group.  The operation on the molecular orbital is reflection in a plane containing the nuclei. If the orbital in unchanged then the sign is $+$, if changed then $-$.

The electron configuration for ground state oxygen is
$(1\sigma _g)^2(1\sigma _u)^2(2\sigma _g)^2(2 \sigma _u)^2(3\sigma _g)^2(1\pi _u)^2(1\pi _g)^2(1\pi _g)^1(1\pi _g)^1$

However, the only electrons we need to consider are the two in the $\pi_g$ orbitals. Adding up the individual orbital and then spin angular momenta separately (equivalent to Russell-Saunders coupling in atoms) then combining both gives the term symbols $^3\Sigma, ^1\Delta ,^1\Sigma $. The calculation is done in a similar way to that for atoms and is shown below in the notes.

The subscript g is found because there is an even number of $\pi_g$ electrons.
The superscript $ \pm $ is because any state ( singlet or triplet) must be asymmetrical overall to exchange and as the triplet has a symmetrical spin wavefunction the spatial part is asymmetrical, and vice-versa for the singlet, the superscript for the triplet is $-$. Thus the superscript gives the term symbol $^3\Sigma_g^-$. Hund’s rule dictate that this is the lowest energy level, followed by $^1\Delta$ at $7918 \pu{cm^{-1}}$ and $^1\Sigma_g^+$ at $13195.22 \pu{cm^{-1}}$.

The $\ce{O_2^-}$ has one unpaired electron in a $\pi$ orbital (as does $\ce{O_2^+}$) so the term symbol is $^2\Pi_g$.

As to your third question the overall symmetry should not change when you go from atomic to MO's.

Notes

The individual $m_l$ and $m_s$ are listed below. The values restricted by the Pauli principle are removed.
The $m_l$ values are $\pm 1$ and the $m_s \pm 1/2$. The the table the spins are labelled as $\pm$ only. The total angular momentum is given as $\Omega = \Lambda + \Sigma$ by analogy with $J=L+S$ in atoms.

$$\begin{matrix}
m_{l1} & m_{l2} & m_{s1} & m_{s2} & \Lambda & \Sigma \\
1 & 1 &  + & + &\text{etc.  Pauli forbid'n}\\
\\
1 & 1 &  + & - & 2 & 0 & ^1\Delta\\
1 & 1 &  - & + & 2 & 0 &  .. \\
-1 & -1 &  + & - & -2 & 0 & ..\\
-1 & -1 &  - & + & -2 & 0 & ..\\
\\
1 & -1 &  + & +  & 0 & 1 & ^3\Sigma\\
-1 & 1 &  + & +  & 0 & 1 & ..\\
1 & -1 &  - & -  & 0 & -1 & ..\\
-1 & 1 &  - & -  & 0 & -1 & ..\\
\\
1 & -1 &  + & - & 0 & 0 & ^1\Sigma\\
1 & -1 &  - & + & 0 & 0 & ..\\
\end{matrix}$$


Which is the specific heat of human body?

A specific heat for humans typically quoted in the literature is 3.5 kJ/(kg K) (for example, pg 16 of Herman's "Physics of the human body", 2nd ed).   This is below the value of water (4.2 kJ/(kg K)) primarily due to body fat, which has a lower specific heat than water.

Despite its title, the article:

Faber and Garby, "Fat content affects heat capacity: a study in mice", Acta. Physiol. Scand., 1995, 153, 185-187, https://doi.org/10.1111/j.1748-1716.1995.tb09850.x

contains a brief summary of the literature concerning the average specific heat of humans.  The authors find that the specific heats of lean and fat mice agree reasonably well with a relationship originally proposed for humans:

$$ c_P = (3.72 - 1.84 \: m_f ) \text{ kJ/(kg K)}$$

where $m_f$ is the fraction of body fat by weight.  Wikipedia  ( https://en.wikipedia.org/wiki/Body_fat_percentage ) indicates that body fat percentage (by mass) is 17-21% and 25-30% for "average" men and women respectively (both neither "fit" or "overweight").  Substituting $m_f = 0.22 $ into the formula above gives 3.3 kJ/(kg K), in rough agreement with typical value of 3.5 kJ/(kg K) quoted in the literature (the literature value corresponding to $m_f = 0.1$, a value typical for a male athlete).

There is some uncertainty in my mind about the accuracy of the formula above, as it is based on a heat capacity of fat significantly smaller than recently reported (see database reference below).  However, I have trouble imagining a situation where the average specific heat capacity of a human would need to be known to great accuracy (it is needed for the artificial, but instructive Problem 1.54 in Schroeder's Introduction to Thermal Physics).  On the other hand, the heat capacity of various components of the human body could be of significant utility (the influence of localized heating etc...).  These specific heats (blood, bone, etc...) are tabulated (with citations) in the "Tissue Properties Database"  http://dx.doi.org/10.13099/VIP21000-03-0 (V3.0).


I'm trying to make an optical dissolved oxygen sensor. I need help finding an appropriate oxygen sensitive dye

This review article might be useful for your purpose:

Title: Optical methods for sensing and imaging oxygen: materials, spectroscopies and applications
Chem. Soc. Rev., 2014, 43, 3666-3761 (DOI: 10.1039/C4CS00039K)

!http://pubs.rsc.org/en/content/articlehtml/2014/cs/c4cs00039k

This is a review article with 694 references (96 pages). The discussion focused on main sections such as direct spectroscopic sensing of oxygen, absorptiometric and luminescent probes, polymeric matrices and supports, additives and related materials, spectroscopic schemes for read-out and imaging, and sensing formats (e.g., waveguide sensing, sensor arrays, multiple sensors and nanosensors). They also discuss future trends and applications, and concluded it with summary of the properties of the most often used indicator probes and polymers.

There are several probes has been discussed. Some of them may be useful to develop oxygen sensors: Those include: a) Hemoglobin and myoglobin as optical probes for oxygen (the affinity of hemoglobin for oxygen is a function of pH and ionic strength); b)  Molecular absorptiometric probes for oxygen (e.g., several cobalt–organic compounds are capable of reversibly binding molecular oxygen); c) Methods based on the oxidative power of oxygen (the reduced or "leuco" form of Methylene Blue (MB) is colorless and quickly converted back
to blue MB by oxygen); etc.

PS: Selected irreversibly acting chromogenic probes for oxygen along with respective polymeric supports and color transitions, and Organic fluorescent probes for oxygen were also given in Table 2 & 3, respectively.


Why does water cause a jet flame when poured over a bunch of tealights?

The wax doesn't need to become gaseous to cause the column of flame (if that's what you were thinking). Water, when in contact with the liquid wax at a couple of hundred degrees above its boiling point, will flash-boil in a very chaotic manner as if it were "exploding" ("transition boiling" in this graph). The water closest to the wax very quickly vaporises and tries to escape, but is temporarily blocked by the layer of liquid water above it, so the system ends up being vigorously churned to make way for the vapour.

The problem is that the turmoil will make tiny liquid wax droplets airborne, as if wax were being shot out of an atomizer, creating a wax aerosol. Suddenly, the same amount of wax has its surface area increased tremendously and is exposed to much more oxygen. Even though the wax in the aerosol quickly loses heat and may become solid again in a fraction of a second, if even a small region of the wax aerosol reaches ignition temperature, the flame quickly propagates as a huge amount of heat is dumped at once in the vicinity and triggers the ignition of all the airborne wax.

This trick is effectively the same as what happens if you try to put out a grease fire with water.


Flexible & thermally conductive materials?

There is a reasonable answer that does not involve aluminium. Thin silicone sheets are extremely available, but an even better solution to your problem would be a polyurethane elastomer with a thermal conductive filler however these are less readily available. Obviously they are not going to be as thermally conductive as metals but that is the sacrifice you make (currently) for flexibility.

Hopefully in the future graphene will be cheap and readily available but with the green energy crisis, I doubt it!


In the gas phase I would suspect that the entropies would be equal as both systems have the same number of translational, rotational and vibrational modes (18).

However you specifically ask about the liquid phase system, since you show that both "a" and "b" are solvated in aqueous media. In this case, the ionized molecule (b) has charges on both ions and this will cause increased ordering of the solvent molecules surrounding the ions, more so than neutral molecule "a".  This increased ordering of the system in case "b" will reduce the entropy of the system.  Consequently system "a" will have greater entropy.


Does double replacement reaction only happen to aqueous reactants?

Generally, reactions occur with liquids and gases (which encompasses solutions). This is because there is the fundamental problem of mass transport. When reactants react, they leave product in the wake of the reaction, which isolates the reactants from each other. For fully solid reactants, it means that more reactants have trouble finding each other. When reactions take place in liquid or a gas, there is the prospect of diffusion, which allows more reactant to find each other.

In the reaction you point out, hydrogen sulfide is a gas so it can easily diffuse to find more zinc oxide. The product zinc oxide might not form a perfect coating (i.e., it has holes in it) that allow the hydrogen sulfide to diffuse through this product to find more zinc oxide.

Other examples: when you burn a piece of wood, this is a solid reacting with a gas. But the combustion reaction is actually between vaporized wood and oxygen. The presence of a gaseous fuel is what allows good mixing between it and oxygen to produce a flame.

Finally, an example from a chemistry demonstration: solid barium hydroxide and solid ammonium chloride are mixed together and shaken. They reaction is quite endothermic and can freeze water on the outside of the reaction vessel. The trick here is that the barium hydroxide is usually the octahydrate. The hydrate provides enough water to create small amounts of solution for the reaction.


Can we determine the amplitude of vibration of O-H bonds in water?

My question is: is it possible to know the amplitude of
  oscillation/stiffness constant of a O−H bond?

Yes.

Your approach is along the right lines but -

The Hooke's Law formula you used

$$E = \frac{1}{2}kA^2$$

is for the stretching of a spring.  Hooke's law changes if the spring has mas(ses) attached at the end(s), now it is a harmonic oscillator.  In this case Hooke's Law becomes

$$E = \frac {h}{2 \pi} \sqrt{\frac{k}{\mu}}$$

where k is the force constant you are interested in and $\mathrm{\mu}$ is the reduced mass of the system.

Since

$$E=h\nu$$

The frequency, which can be experimentally measured, is given as

$$\mathrm{frequency} = \frac {1}{2 \pi} \sqrt{\frac{k}{\mu}}$$

The frequency is independent of the amplitude, affected only by the masses and the stiffness of the spring.

If the masses of the atoms at the ends of the spring are given as A and B, then

$$\mu =\frac{A \cdot B}{A+B}$$

So for an $\ce{O-H}$ bond $\mathrm{\mu=\frac{1 \cdot 16}{1+16}}\ \mathrm{u}$. If we measure the frequency of the $\ce{O-H}$ stretch absorption, then we can calculate the force constant.

Here is a link to a one-page example that shows the actual computation and explains the specific units to be used very nicely.

You also ask about the amplitude of oscillation.  If, by this you mean the bond length, then yes this can also be calculated from spectra (relatively easy in simple diatiomic cases, but more difficult in complex polyatomic cases.  The methodology involves assuming a rigid rotor and then using the Schrodinger equation to solve for the energy states and extracting the bond length.  A nice example using $\ce{HCl}$ can be found here, look down the page for the section entitled "Bond Length of HCl".


If I mix an aqueous solution of hydrogen chloride with an organic solvent, will the organic solvent contain hydrogen ions?

$\ce {HCl}$ is so strong a proton donor that bases weaker than water could be protonated.  Alcohols and etherss, if they form a separate phase from water, are likely to be protonated.  Going the other direction, some organic compounds with little solubility in water might become more soluble in a hydrochloric acid solution if they are sufficiently basic.  The basicity of some cyclopropenones can be measured that way.  See (behind a paywall) http://pubs.acs.org/doi/abs/10.1021/cr60288a003?journalCode=chreay.


Gallium makes aluminum soft like a wet tissue. Any other metals or alloys that are vulnerable to similar damage?

Gallium and Mercury, the latter in particular are absolute pests and poisons to most metals. There are too many to mention that will suffer the same consequences. I will mention the few that don't. Iron, Platinum, Tungsten and Tantalum all tolerate them fairly well. There are certainly more examples, but the common rule is; most don't.


Difference between organic and inorganic chemistry

IUPAC is the International Union of Pure and Applied Chemistry, they make recommendations on the nomenclature. IUPAC mentions that the difference between organic and inorganic is not distinct. To quote "The boundaries between ‘organic’ and ‘inorganic’ compounds are blurred." in Brief Guide to the Nomenclature of Inorganic Chemistry R. M. Hartshorn, K.-H. Hellwich, A. Yerin.

Since the terminology of organic vs. inorganic is all human classification, it is not a binary system 0 or 1. What we can say now is that traditionally, all organic compounds do contain carbon. It can come from natural sources or purely synthetic. There is no such restriction. Plastic is an organic compound because it contains a lot of carbon chains.

Note that this word organic, as used in chemistry, has nothing to with the buzz word used in marketing of organic food, organic fruits, organically grown stuff. The word organic comes from French organique designating the jugular vein, hence related to organs or living beings.

It is a myth that Wohler started the distinction of organic vs. inorganic. There is a nice French article. The 16-paged paper is titled “Les origines de la chimie organique au-delà du mythe fondateur”  (The origins of organic chemistry beyond the founding myth), Comptes Rendus Chimie, 2012, 15, 553–568.

Read the table at the end, and the abstract.

It clearly shows that Berzelius coined the term organic chemistry between 1807-1813.

Use Google Translate or DeepL to translate it.


What is the effect of adding HCl on a certain solution at equilibrium?

You are absolutely correct. Adding $\ce{HCl}$ adds protons to the mix, which you can imagine as applying pressure on the right-hand side, which means that the reaction will shift backwards slightly and the acetate concentration will decrease. For $\ce{NaOH}$, you also need to consider the underlying equation:

$$\ce{H2O <=> H+ + OH-}$$

which will be shifted to the left by additional hydroxides. This means that the overall proton concentration decreases and you can imagine that as lifting a weight from the right-hand side: The reaction will shift forwards.


What wavelength of light is released when an electron drops an energy level in complexes?

Not necessarily. The drop may occur gradually so that the colour is "blinded out" by the nonadsorbed colours. The drop may also occur in stages, so that several low energy emissions occur. Some are outside visible range and others are too weak or go unnoticed.


how to obtain potential curves from database?

The RKR method determines the classical turning points $r_{\pm}$ of the potential energy based on knowing experimentally determined spectroscopic constants. ( Rydberg, 1931, Z. Physik. v73,376, Klein, 1932 Z. Physik. v76, 226, Rees Proc. Phys. Soc.(Lond.) 1947, v59, 998.)
The energy equation takes the usual form. Parameters all in cm$^{-1}$, distances in m, reduced mass $\mu$ in u, $h$ in cm$^{-1}$s.
$$E_v=\omega_e(v+1/2) - x_e\omega_e(v+1/2)^2 + y_e\omega_e(v+1/2)^3 + z_e\omega_e(v+1/2)^4 +\cdots$$
The vibrational quantum number $v$ should be considered as a variable that has values from $-1/2$ to whatever positive value is required.
The derivative of the rotational energy wrt. $J$ at $J=0$  is also needed,
$$B'(v)= B_e-\alpha_e(v+1/2)+\gamma_e(v+1/2)^2+\cdots$$
The function representing the turning points is
$$r_\pm= \frac{f(v)}{2}\left (\sqrt{1+\frac{4}{f(v)g(v)}}\pm 1 \right)$$
where functions $f, \, g$ are integrals. These functions 'blow up' at the upper integration limit $x \to v$ so a correction is added to prevent this. The basic equation is
$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\int_{-1/2}^v\frac{dx}{ \sqrt{E(v)-E(x)}}$$
and the modified one
$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\left(\int_{-1/2}^{v-\delta}\frac{dx}{ \sqrt{E(v)-E(x)} }+ 2\sqrt{\frac{\delta}{Q_v}}\right)$$
where $\delta$ is small $\approx 10^{-5}$ and
$$Q_v=\omega_e - 2x_e\omega_e(v+1/2) + 3y_e\omega_e(v+1/2)^2 + 4z_e\omega_e(v+1/2)^3 +\cdots$$
The function $g$ with correction is
$$g(v)=\frac{4\pi\sqrt{2\mu}}{h} \left( \int_{-1/2}^{v-\delta} \frac{B'(x)}{\sqrt{E(v)-E(x)} }dx+2B'(v)\sqrt{\frac{\delta}{Q_v} } \right) $$
A set of values for CO is
$\mu = 6.85841,\; \omega_e= 2169.8135,\; x_e\omega_e=13.2883,\; y_e\omega_e=0.010511,\; z_e\omega_e= 0.000057$
$ B_e= 1.9312808,\; \alpha_e= 0.0175044,\; \gamma_e= 0.000000548$.
Plot energy $E(x)$(cm$^{-1}) $ vs $r_\pm(x)$ (pm) is shown. The value at $x=-1/2$ should be  $r_e=103.14$ pm the equilibrium bond length with energy of zero. As values are based on cm$^{-1}$ a scaling of $10^{14}$ is needed to go to pm.

I used Jupyter notebooks (via Anaconda) and python 3 to do the calculation using built in integrator from numpy/scipy and matplotlib to draw the graph. All are free and very easy to use.


What's the difference between mass spectrometry, mass spectroscopy, and mass spectrography?

According to the Encyclopaedia Britannica, mass spectrographs use(d) photographic plates or films for the detection of ions, as in the apparatus thought out by F. W. Aston in 1919.


How to identify hydrogen bonds and other non-covalent interactions from structure considerations?

It is safe to say that there will always be intermolecular forces at play. At the time where you will consider these you should already have a good idea about the molecules involved in your system.
Based on the composition and molecular structures you can make certain assumptions. In a molecule it is straight forward to estimate (bond) polarities based on electronegativities, then infer from these how they might arrange. I will work an example based on the interactions between adenine and thymine later.[1]

With the advent of the information age, tools that every chemist has at her or his disposal have become more sophisticated. We have access to many digital resources like databases,[2] or publication servers[3] to retrieve a vast amount of information. Molecular modelling,[4] or even more sophisticated quantum chemical calculations,[5] have become more important; and free tools are available for everyone to use.

With that being said, here are some points that might help you identifying hydrogen bonds and other non-covalent interactions. For that purpose, let's have a look at our example molecules:

Immediately we can formulate a couple of assumptions based on the schematic representation. In adenine there are five nitrogen atoms, which have a higher electronegativity than carbon. Negative partial charges will therefore be located at mostly at these. Hydrogens in organic compounds usually carry positive partial charges; albeit $\ce{C-H}$ tend to be a lot less polarised than $\ce{N-H}$ bonds. Whenever a hydrogen is involved in a bond, that bond can potentially act as a hydrogen-bond donor (see below).
Similar observations can be made for thymine. Here we have two terminal oxygen atoms, which will carry a negative partial charge, since they have one of the highest electronegativities. These are often able to accept hydrogen bonds. On the other hand we also have $\ce{N-H}$ bonds, which can act as hydrogen bond donors.

Charges & Electrostatic Potential Surfaces

For many molecules structures are readily available. If not, some molecular editors give you the possibility to use implemented force fields to optimise built (guessed) structures. Based on those you can already do a few analyses. One tool that is quite powerful for various tasks is Avogadro, it let's you read crystal structures, perform basic calculations and much more. If you are just playing around, this is a really good choice.
For example, I have imported the crystal structure of adenine into Avogadro, optimised it, and calculated the electrostatic potential. Or after extracting Cartesian coordinates, Molden let's you easily calculate the charges.[6]

Hydrogen Bonds

Many molecular editors try to guess hydrogen bonds based on their implemented cutoff values. That certainly is very helpful, but not everything can be automatised in this way. And especially weaker interactions won't be found. One has to go a bit deeper then.
As a nice and concise example I have picked an intermolecular 2:1 complex between adenine and thymine, for which the crystal structure is available.[1]

There are two principle structural parameters to decide about hydrogen bonds:  (a) The distance of the hydrogen $\ce{H}$ and the hydrogen-bond acceptor $\ce{Y}$ is significantly shorter than the sum of their respective van-der-Waals radii, $d(\ce{XH\bond{...}Y})<r_\mathrm{vdW}(\ce{H})+r_\mathrm{vdW}(\ce{Y})$.[7]
(b) The angle around the hydrogen is nearly linear, $\angle(\ce{XH\bond{...}Y}) \approx 180^\circ$. For weakly polarised $\ce{XH}$ bonds, isotropic dispersion forces become more important (while the directional electrostatic and covalent contributions decrease), therefore the angle becomes more flexible.

We easily see that the bond angles are close to what we expect for hydrogen bonds. I have reproduced a few values from Batsanov's paper below, with the caveat that the value for hydrogen strongly varies depending on the chemical environment from $\pu{110 - 161 pm}$, so I used the classic from Bondi.[7c] Since all the distance are around $\pu{200 pm}$, they are well below the threshold we set earlier.

$$
\begin{array}{lr}
\text{Element }\ce{Y} & r_\mathrm{vdW}(\ce{Y})/\pu{pm}\\\hline
\ce{H}   & \approx 120\\
\ce{C}   &  196 \\
\ce{N}   &  179 \\
\ce{O}   &  171 \\\hline
\end{array}\hspace{2ex}
\begin{array}{lr}\\
\text{H-Bond }\ce{XH\bond{...}Y} & \sum r_\mathrm{vdW}(\ce{Y},\ce{H})/\pu{pm}\\\hline
\ce{CH}   &  316 \\
\ce{NH}   &  299 \\
\ce{OH}   &  291 \\\hline
\end{array}
$$

A quite interesting approach of revealing non-covalent interactions was presented by Johnson et. al., and the corresponding program is easy to use and only requires Cartesian coordinates.[8]

The surfaces between the molecules represent these interactions, where green represents weak interactions, typically found for dispersion. Blue represents stronger interactions, typically found for hydrogen-bonds. Red displays repulsive forces, typically found within ring or cage systems.

If you have access to quantum chemical software, then you can obtain this plot also for wave function files .wfn.

Another possibility is to analyse the electron density in terms of the quantum theory of atoms in molecules (QTAIM).[9] For this you do need a wave function file. The analysis, however, is straight forward and will yield a bond path or not. If there is a bond path, we can estimate the strengths of these bond with the methodology developed by Espinosa et. al.. According to this the bondstrength is approximately half the value of the potential energy density at the bond critical point.
$$E_\mathrm{H-Bond} = \frac{1}{2}V(r_{\mathrm{BCP}[\ce{XH\bond{...}Y}]})$$

I have performed such a calculation on the DF-B97D3/def2-TZVPP level of theory with Gaussian 09. The optimised geometry will be at the end.

\begin{array}{lr}
\text{H-Bond} & E_\mathrm{H-Bond}/\pu{kJ mol-1}\\\hline
\mathrm{N(37)H \cdots  O(1)}   &  46.6\\
\mathrm{N(5)H  \cdots  N(36)}  &  38.5\\
\mathrm{C(41)H \cdots  O(2)}   &   3.2\\
\mathrm{N(20)H \cdots  O(2)}   &  50.5\\
\mathrm{N(3)H  \cdots  N(24)}  &  29.9\\
\end{array}

A general warning shall be applied to the above. Absolute values of these are only approximate, but fall within the range of what is expected. A very nice side effect of this methodology is, that it can be applied to intramolecular hydrogen bonds, too.

Concluding remarks

Dispersive interactions and hydrogen bonds become more and more important in rational reaction design. Be it for understanding of molecular structure of biomolecules, or as a guiding principle for catalyst-substrate interactions.
With further development of computer technology, it should become more accessible to everyone. I hope this post demonstrates that gaining more insight can actually be quite easy (and free).

Notes and References

(a) Based on the structure from S. Chandrasekhar, T. R. Naik, S. K. Nayak, T. N. Row, Bioorg. Med. Chem. Lett. 2010, 20 (12), 3530-3533. DOI: 10.1016/j.bmcl.2010.04.131 PMID:
20493694 CSD: 739016
(b) Adenine, CSID: 185
(c) S. Mahapatra, S. K. Nayak, S. J. Prathapa, T. N. Guru Row, Cryst. Growth Des. 2008, 8 (4), 1223–1225. DOI: 10.1021/cg700743w, CSD: 652573
(d) Thymine, CSID: 1103
(e) G. Portalone, L. Bencivenni, M. Colapietro, A. Pieretti, F. Ramondo, Acta Chemica Scand. 1999, 53, 57-68. DOI: 10.3891/acta.chem.scand.53-0057, CSD: 136916
(a) The Cambridge Structural Database (CSD), https://www.ccdc.cam.ac.uk/
(b) Crystallography Open Database (COD), http://www.crystallography.net/cod/
(c) Computational Chemistry Comparison and Benchmark DataBase, http://cccbdb.nist.gov/ (Only for 1799 small molecules and atoms)
(d) Handbook of Chemistry and Physics, http://hbcponline.com/faces/contents/ContentsSearch.xhtml
(e) ...
(a) SciFinder, https://www.cas.org/products/scifinder
(b) Google Scholar, https://scholar.google.de/
(c) Web of Science, (formerly known as Web of Knowledge) http://www.webofknowledge.com/
(d) ...
(a) MolCalc, http://molcalc.org/
(b) Pitt Quantum Repository, https://pqr.pitt.edu/ (At the time of writing it was dead.) Github: pittquantum
(c) Many open source molecular editors include the possibility to use force field calculations. For example: Avogadro, molden
(d) For more on molecular modelling in the open source domain see S. Pirhadi, J. Sunseri, D. R. Koes, J. Mol. Graph. Model. 2016, 69, 127-143. An updated online version of this catalog can be found at https://opensourcemolecularmodeling.github.io.
(a) For an extensive, but not necessarily complete, list of quantum chemistry software see Wikipedia.
(b) For the purpose of this demonstration I will be using the proprietary software Gaussian.
(c) To view crystal structures, Mercury can be obtained (for free) from the Cambridge Crystallographic Data Centre (CCDC), which also hosts CSD.
https://www.ccdc.cam.ac.uk/solutions/csd-system/components/mercury/
(a) Tutorial for Avogadro (b) Tutorial for molden
(a) A concise (and as far as I can tell newest) list of van-der-Waals radii of many elements can be found in S. S, Batsanov, Inorg. Mat. 2001, 37, 871-885. DOI: 10.1023/A:1011625728803 (mirrored pdf)
(b) A list of van-der-Waals radii can also be found on Wikipedia
(c) A. Bondi, J. Phys. Chem. 1964, 68, 441-451. doi: 10.1021/j100785a001
(a) The original publication: E. R. Johnson, S Keinan, P. Mori-Sánche§, J. Contreras-García, A. J. Cohen, W. Yang, J. Am. Chem. Soc. 2010, 132, 6498-6506. DOI: 10.1021/ja100936w
(b) The presentation of the program: J. Contreras-Garcia, E. R. Johnson, S. Keinan, R. Chaudret, J-P. Piquemal, D. N. Beratan, W. Yang, J. Chem. Theory Comput. 2011, 7, 625-632. DOI: 10.1021/ct100641a
(c) Download the code: http://www.lct.jussieu.fr/pagesperso/contrera/nciplot.html
(d) You'll also need VMD (Visual Molecular Dynamics) from the University of Illinois
(a) A very brief introduction can be found on Wikipedia. The corresponding book:
Bader, Richard (1994). Atoms in Molecules: A Quantum Theory. USA: Oxford University Press. ISBN 978-0-19-855865-1. (publisher)
(b) Multiwfn - A Multifunctional Wavefunction Analyzer; http://sobereva.com/multiwfn/ corresponding paper: T. Lu, F. Chen, J. Comput. Chem. 2012, 33, 580-592. DOI: 10.1002/jcc.22885
(c) Startup script (and examples) for Linux version: https://github.com/polyluxus/runMultiwfn.bash (shameless self-plug)
(d) E. Espinosa, E. Molins and C. Lecomte, Chem. Phys. Lett., 1998, 285, 170–173.

Appendix

Optimised Structure of the adenine-thymine 2:1 complex calculated at DF-B97D3/def2-TZVPP in Gaussian 09 Rev. E.01

  45
E(RB97D3/def2TZVPP/W06) = -1388.51095169
O         19.03780       11.79565        1.63996
O         14.74303       13.36808        1.71115
N         15.07940       11.09762        1.64043
H         14.04669       10.93403        1.64128
N         16.88238       12.55083        1.67499
H         17.23771       13.53695        1.70257
C         17.83067       11.52964        1.63864
C         17.29332       10.17534        1.60048
C         15.51719       12.40225        1.67775
C         15.94232       10.03610        1.60357
H         15.46489        9.06076        1.57654
C         18.24710        9.02139        1.56030
H         18.89656        9.08263        0.68030
H         18.90669        9.02997        2.43483
H         17.71085        8.06866        1.53480
N          8.28060       10.09397        1.65692
H          7.68055        9.28386        1.63653
N          8.91986       12.24943        1.71837
N         10.48101        9.01368        1.60709
N         11.91467       12.94472        1.71757
H         12.92612       13.11883        1.71578
H         11.26613       13.71385        1.74609
C         10.03193       11.42408        1.68464
N         12.26612       10.63929        1.64378
C         11.41986       11.70069        1.68284
C          9.65939       10.07380        1.64587
C          7.89700       11.42267        1.70074
H          6.85459       11.71169        1.71748
C         11.75955        9.39245        1.60917
H         12.49648        8.59107        1.57896
N         21.22700       16.46826        1.76840
N         17.31372       17.43807        1.81488
H         16.80939       18.31038        1.84233
N         19.61652       18.27309        1.82815
C         18.69253       17.30751        1.80467
N         17.71219       15.24141        1.74956
N         20.64990       14.21957        1.70686
H         19.98577       13.44513        1.68592
H         21.63800       14.02512        1.69520
C         20.27365       15.50797        1.74532
C         16.77892       16.17095        1.78070
H         15.71578       15.97320        1.77996
C         18.91965       15.92447        1.76371
C         20.85479       17.75435        1.80728
H         21.67003       18.47621        1.82413


Living-cationic vs cationic polymerization

This difference between living and non-living polymerizations is small in words but large in effects. In principle cationic and anionic polymerizations can be living. In practice, it is not as easy as it looks on paper. See below.

A living polymerization is any chain (or addition) polymerization that is prevented from terminating, i.e. the ends of the chains are still reactive.  In a living polymerization, chain initiation has one initiation and propagation continues as long as there is monomer. Termination (almost) never happens.  If your polymerization stops because your monomer runs out, adding more monomer restarts the process and the existing chains get longer. The number of chains formed matches very closely with the number of equivalents of initiator, and the length of the chains depends on the ratio of monomer to initiator.

Most chain polymerizations (radical, anionic, cationic, transition-metal insertion, and ring-opening) do not behave this way. There is a mechanism for termination that becomes increasingly kinetically favored as the concentration of free monomer decreases. Sadly, the wikipedia article for chain termination is pretty sad. Most chain polymerizations propagate and terminate quickly, with termination usually initiating the next chain. Thus, in a non-living polymerization, new chains are constantly being initiated and old chains are constantly terminating until their is no more monomer.  In these cases, adding additional monomer later would not accomplish anything because the chain ends are "dead". Experimental evidence for this phenomenon includes the formations of more chains than there were equivalents of initiator. The length of the chain is more dependent on the relative rates of propagation and termination than on the initial ratio of monomer to initiator.

For radical polymerization, termination mechanisms include fragmentation by β-hydrogen abstraction or by recombination. Both of these mechanisms are inherent in the radical nature of the polymerization, and so are not preventable by exclusion of other species. Radical polymerizations are made living by the addition of species that "protect" the radical at the end of the chain to prevent it from terminating. Examples include ATRP and RAFT. The wikipedia articles for both are pretty good.

For cationic polymerization, termination mechanisms include β-hydrogen abstraction, chain transfer, and attack by nucleophiles like water. In principle, a cationic polymerization could be a living polymerization if great pains were taking to exclude all species which might induce termination. In practice this is doable, but difficult, as termination by chain transfer is always possible. See the wikipedia article on living cationic polymerization. The article is not great, but suggests that cationic polymerizations done in nonpolar solutions (not in bulk) with nonnucleophilic Brønsted or Lewis acids at low temperatures can become living polymerization.

For anionic polymerization, termination mechanisms include protonation by protic molecules, chain transfer, and reaction with electrophiles, which can include O2 in some reactions. Like cationic polymerizations, anionic polymerizations can be made living by rigorous exclusion of terminating species. Again, this is difficult, due to the possibility of chain transfer termination. However, with the correct choice of conditions, it is doable, as suggested by the wikipedia page for living anionic polymerizations. Another way to accomplish some anionic polymerizations in a living fashion is group transfer polymerization, even though it does not  follow a strictly anionic mechanism.

I know that Contemporary Polymer Chemistry by Allcock, Lampe, and Mark, Prentice Hall 2003 covers this topic well, but the book is a little intense to read. In particular, the book derives the kinetic expressions for various polymerization types in the living and non-living cases. I expect other textbooks focusing of the chemistry of polymers might also cover this topic in one fashion or another.


Is it possible to melt HDPE to a point where it has similar viscosity to engine oil?

You may see a small reduction in viscosity with increasing temperature, but at elevated temperature the polymer will start to oxidize and/or undergo thermolysis. If I were trying to reduce viscosity I'd try blending in some low molecular weight polyethylene (LDPE). Not sure if you have the capability of doing this or if you could tolerate the resulting changes in property (loss of tensile strength, etc.). This reference:

https://doi.org/10.1080/03602550600609663

suggests you may need to go as high as 40%+ to get a significant change in viscosity.


How mass spec formate and acetate adducts masses are calculated?

In your calculation, you are using the average molecular mass of the structure, where atomic masses are based on the natural abundance of all isotopes of the element.

H:  1.00794 u
C: 12.01070 u
O: 15.99940 u
O: 15.99940 u
─────────────
   45.01744 u

However, for the exact molecular mass (or monoisotopic mass) of the structure, the atomic masses of each atom are based on the most common isotope for the element.

H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
────────────────
      44.99765 u

Furthermore, you are missing the mass of the extra electron of the formate ion. The electron mass is $m_\mathrm e=0.000548579909070(16)\ \mathrm u$. Thus,

H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
e−:    0.00055 u
────────────────
      44.99820 u


Is there a better equation than the Nernst equation for low concentrations?

The Nernst equation does not need a replacement for low concentrations. Yes the limit is $\infty$, which essentially means you can't realistically have zero concentration of anything. We may think of absolutely pure water as an idealized situation. Well, as soon as it gets in touch with copper, some $\ce{Cu^2+}$ ions instantly jump into the solution, and from that point on, you have a certain non-zero concentration of them (though probably pretty low), and the equation works just fine.


What exactly is meant by 'multi-configurational' and 'multireference'?

Your problem seems to be with the terminology used in CI methods, so let me go through the different terms you mentioned:

A configuration is a certain occupation of (molecular) orbitals. Mathematically configurations can be represented in 2 ways. The first one is the Slater Determinant (SD), an anti-symmetrized product of spin-orbitals. Slater Determinants, however, are not eigenfunctions of the spin operator $\hat S^2$, but the electronic wave function needs to fulfill this requirement. Therefore one constructs spin-adapted Configuration State Functions (CSF) as certain linear-combinations of SDs. Using CSFs instead of SDs usually makes a calculation more stable. "Configuration" is a more general term, which does not explicitly say whether an SD or CSF is considered.
Multi-configurational just means the method considers more than one configuration.
Reference means we have a designated configuration from which the excitations are generated. Single-reference thus means we only have one such configurations (usually the HF configuration), e.g. in CISD or CCSD. Multi-reference means we have more than one configurations to generate excitations from.

So multi-configurational just means we have many configurations, while single-/multireference says something about how those configurations are selected/generated.

How does single-/multireference apply to the different methods?

In HF we do not generate any excitations, therefore this concept does not really apply here. But I think one could still call it single-reference and single-configurational.
CISD and CCSD are the typical examples of single-reference methods. Nothing special here.
In FCI and CASCI we don't restrict the excitations to certain degrees (Single, Double, etc.), instead, we just take all of them. The concept of having one (or more) reference configurations is a possible perspective, but it is not really necessary. FCI can be viewed as having one reference configuration and taking all excitation degrees (up to the number of electrons available), which would make it single-reference. But from another perspective, we can argue that every MRCI wave function is just a truncation of the corresponding FCI wave function. So if FCI covers everything and more than MRCI does, would it not be multi-reference as well? Again, I would just say the concept does not apply here.
MCSCF (CASSCF) is a combination of CI (CASCI) and SCF, where SCF means we optimize the orbitals as well. This is not done in CI(SD...), CC(SD...), MRCI, etc. Other than that, the above arguments apply in the same way to the CI space as before.

Multi-Reference CI and CASSCF

As argued above, personally I would not consider CASSCF to be multi-reference, as the concept does not really apply. But I can see why people would consider it as such.

One usually does CASSCF because single-reference methods fall short of describing the wave function correctly, in a qualitative way. Such systems are then called strongly correlated and CASSCF can treat that strong correlation (also called static correlation) which is missing in single-reference calculations. In turn, however, CASSCF is missing that kind of correlation single-reference method can treat well, called weak or dynamic correlation.

Multi-reference is now the approach to combine both, strong and weak correlation, by doing first a CASSCF, and then using those configurations as a reference space, to generate all the excitations from.

Orbital optimization, however, is usually not feasible for dynamic correlation, because dynamic correlation usually means to have a lot of configurations and therefore huge CI space. With static correlation there are much fewer configurations, so doing orbital optimization additional to CI optimization is feasible and yields qualitative improvements of the orbitals. Therefore this is done in CASSCF, but not in MRCI.

"Excited states" in MRCI

What is meant here by "excited states" are actually "excited configurations" (with respect to the HF configuration). If you, for example, choose a CAS as your reference space, then all configurations, except for the HF configuration itself, are excited configurations.

Please note, that "excited state" is not a good choice of words here since it is actually something different: a state is a result of a CI calculation and is physically observable. On the other hand, a configuration is a basis function you put in a CI calculation and therefore is more of an abstract mathematical tool. However, you can approximate a state as a configuration to some extent, this is for example done in HF.


How does partially thawing a frozen solution affect concentration?

In general, no, a frozen solution that partially thaws is not the same concentration as the original solution.

The reason for this is the same reason chemists can purify a compound by crystallizing it, and one of the reasons why north polar ice cap is freshwater, despite floating in salt water.

When something (e.g. water) crystallizes (e.g. forms ice) the crystal lattice tends to reject impurities (e.g. things dissolved in the water). This means that as the solution crystallizes/freezes, the solid portion becomes depleted in the solute, and the remaining liquid portion becomes enriched in the solute. Eventually, though, if things are cold enough, even the concentrated residual liquid will freeze. But this mean that you now no longer have a uniform concentration of solvent in your frozen sample.

The reverse happens when you thaw it. Since you don't have a uniform concentration of solute, not all portions of your solid sample have the same freezing point depression. So what tends to happen is that portions of the solid which are more concentrated in solute will thaw first (as they have a lower melting point), whereas portions which are depleted in solute will thaw later. -- This is all generalization, though. What actually thaws first depends a lot on the details of where these regions are in relation to the heat source, and how thermal energy is flowing through the sample.

The up shot of this is, though, that if you freeze a stock solution, you can't partially thaw it and expect to get the appropriate concentration. You have to completely unthaw the stock solution and mix it thoroughly to make sure you have a homogenous mixture of the correct concentration.

This is why it is typically recommended to make single-use aliquots of stock solutions like antibiotics. Not only do the smaller aliquots thaw faster, but you don't put your entire stock solution into multiple freeze/thaw cycles, which can degrade some of the more delicate compounds.


Why isn't calcium carbide more corrosive?

In these Safety Data Sheets for $\ce{Ca(OH)2)}$ and for $\ce{CaC2}$, both compounds are rated at "Causes serious eye damage – Category 1" and "Causes skin irritation – Category 2".
That said, $\ce{CaC2}$ is often sold as pellets or "stones". Because of large particle size, only a small amount of $\ce{Ca(OH)2)}$ is produced at a time from the action of water on $\ce{CaC2}$ since the release of acetylene bubbles slows penetration of water into the pellets.
$\ce{Ca(OH)2)}$, on the other hand, is more often sold as a finely divided powder. That dust is, in my experience more of a respiratory hazard. The "lime", as it's slaked in water, also evolves considerable heat, which exacerbates the caustic effect on skin.
Of course, $\ce{CaC2}$ has it's own hazards... it can cause considerable damage from an acetylene explosion.


At first thought, it might seem that rainwater should be clean like distilled water, as it evaporated, condensed, then fell back to earth. It sounds like a condenser or distillation apparatus in a laboratory. But in reality, the process is very different.

To start with, every raindrop must have a particle to condense upon, called a cloud condensation nucleus.  According to this Wikipedia article:

  Cloud condensation nuclei or CCNs (also known as cloud seeds) are
  small particles typically 0.2 µm, or 1/100th the size of a cloud
  droplet on which water vapor condenses. Water requires a
  non-gaseous surface to make the transition from a vapor to a liquid;
  this process is called condensation. In the atmosphere, this surface
  presents itself as tiny solid or liquid particles called CCNs.

Furthermore, the composition of CCN is frequently very acidic, having formed from sulfuric acid or sometimes weak organic acids. These types of CCN form in both polluted as well as pristine environments. Both natural and man-made sources of sulfur emissions result in the photochemical oxidation to sulfuric acid, which forms new CCN which are very hygroscopic (water absorbing).

The type of CCN described above are most commonly formed over land. In marine environments, there are natural sources of sulfur gases, but most CCN form from evaporated sea spray, which is of course much more pH neutral.

However, regardless of how the cloud drop was originally formed, they all have another source of acidity from the atmosphere; they absorb carbon dioxide, which is rapidly converted to carbonic acid when dissolved in a cloud droplet.  It is this absorption of naturally occurring, acidic carbon dioxide that is the primary reason that essentially all rain is somewhat acidic.


Hydrogenation stereochemistry-Pd,Pt, Ni catalysts

Using any transition metal catalyst (Pt, Pd, Ni) for hydrogenation of alkynes results in syn hydrogenation to not alkenes but alkanes. Using Lindlar's catalyst results in syn hydrogenation of an alkyne to an alkene.

In the case of sodium metal (not a transition metal) in ammonia, hydrogenation of alkynes results in a trans alkene.


Calculating empirical formula of vitamin C

You are using the wrong number for the grams of water, it should be 0.0819.  Go back and redo your calculations with that change.  Now you should get the correct ratio of the elements in the molecular formula for ascorbic acid.  If you don't, report back and show your math.


The electronic structure of NO2 leaves an unpaired electron on the nitrogen. This is one of the reasons it is quite reactive. But it is also the reason why it can easily dimerise to N2O4 as the two unpaired electrons in two molecules of the gas can pair up creating a weak bond. The dimer is favoured at low temperatures when the thermal energy isn't large enough to shake the dimer apart (I'm simplifying a bit). This is why the colour of the gas changes depending on temperature as the dimer is colourless but the NO2 is brown.


Why do we excite electrons before hybridisation?

Actually number of bonds formed by an atom is generally equal to number of unpaired electrons it has. Like oxygen and nitrogen make two and three bonds because they have these many unpaired electrons in their valence shell.
But, in its ground state, carbon got only two unpaired electrons, so it should make two bonds. But in reality carbon make four bonds. Two satisfy four bonds we need four unpaired electrons. This is possible to consider the excited state.
Hope its clear.


Can a lone pair be involved in cis/trans isomerism?

Yes, due to the lack of rotation about the C=N bond this will display the desired isomerism. For a much simpler example see $ \ce{N2F2} $

(taken from https://en.wikipedia.org/wiki/Dinitrogen_difluoride )


Boiling point elevation and liquid solutions

Use the ebullioscopic equation (the first equation) in this Wikipedia article,

$$\Delta T = K_\mathrm{b} m$$

First solve for $m_\mathrm{init}$, the initial molality of urea. Second figure out at what molality $m_\mathrm{fin}$ the boiling point is elevated by $\Delta T = \pu{1.5 °C}.$ Since the amount of urea is constant, the % change in the mass of water is then given by

$$\% \left(\frac{\Delta w_{H2O}}{w_{H2O,\mathrm{init}}}\right)= \frac{m_\mathrm{init}^{-1}-m_\mathrm{fin}^{-1}}{m_\mathrm{init}^{-1}}$$

where $m$ is the molality of urea, $w$ mass solvent.


Why are high pressures used in cracking of long-chain hydrocarbons?

First of all, cracking is not an equilibrium process, therefore La Chatelier-Brown principle doesn't apply! (irrelevant)
Also, the mechanism is much more complicated than this:

In gas phase, radical reactions generally have extremely complex reaction networks. E.g. burning of methane with oxygen contains thousands if not more elementary reactions. I am pretty sure that gas phase cracking is not a pure first order reaction.
In industry, cracking is generally done on heterogeneous catalyst, therefore not a gas phase reaction. Heterogeneous catalysis is pretty much controlled by the absorbed amount of molecules on the catalyst surface: high pressure, faster reactions.


Salt + Sodium Bicarbonate releasing water in carpet

I've never heard of sodium chloride/sodium bicarbonate being an effective dessicant. Pure sodium chloride is not hygroscopic at all.

However, in air sodium bicarbonate slowly decomposes to sodium carbonate which in fact is hygroscopic. If sodium carbonate, and hygroscopic impurities in table salt, such as magnesium chloride or calcium chloride, deliquesce in air of high humidity to give small droplets then these droplets start to grow due to a high concentration of dissolved salts (including sodium chloride and bicarbonate). The effect is reversible at low humidity or higher temperature.


Why does 2-chloropropane's mass spectrum have a peak at 27?

Here is a little mechanism I made up about it. Not quite sure if its correct, seems to fit the mass spectrum though.

Definitely have a look at McLafferty classical textbook. Maybe you will find the answer there.


In sourdough bread baking what do the wild yeast and bacteria not eat during dough fermentation and why?

Flours contain a diverse array of carbohydrates which vary depending on the specific type of flour (wheat, rye, teff, etc). There will usually be a lot of starch (itself having variable amounts of different types of branching), some small mono and oligosaccharides (again of varying types) and "fiber", which is a broad term for polysaccharides that are more resistant to digestion.
To add to the variables, the specific types of bacteria and yeast in a sourdough fermentation can vary in their ability to digest the different types of carbohydrates.
So one part of the answer to your question is that yes, there are carbohydrates that cannot be digested, but the specific molecules that are not digested will depend on the flour and the microbes. The cause of the undigestibility is either 1) the monosaccharides are linked in a way that they cannot be cut apart or 2) the microbes lack the metabolic capability to digest the monosaccharide.
But generally in a bread dough, the digestible carbohydrates have not been completely consumed, so some of the remaining carbohydrate is digestible. The reason it is not used up is that the baker stops the fermentation before it reaches that point so that the dough still has a good consistency.
It is also true that the fermentation activity slows down considerably before this point, even though it does not stop. There are two reasons for this. 1) the pH drops, which slows the metabolism of the bacteria and 2) the easily digestible carbs get used up first, so the later fermentation rate can be affected by the slow rate of breakdown of the more difficult to use carbohydrates. Ultimately, the dynamics of the fermentation are determined by the choice of flour, the microbial population and the conditions (temp, water content, whether and how often the baker works the dough, etc).


What is produced when citric acid (powder form) is burned, is it safe?

As @paracetamol mentioned, if you burn it completely and properly, you should get nothing other than $\ce{CO2}$ and $\ce{H2O}$.

But what if you don't? Then it's a bit complicated. You can't say exactly what will form. There are chances of carbon monoxide formation, which is quite toxic. By binding to the heme complex iron in a hemoglobin, it renders it incapable of carrying oxygen to the body, causing dizziness, nausea, and likely death. Consider that.

Second thing is, there may be unburnt carbon particles that are emitted if you don't burn it well. Particulate matter such as these is highly dangerous to the lungs.

Thirdly, considering that atmospheric combustion is a highly complicated process, you can't really say what you could have produced from that oxidation. It may be toxic and may have crazy properties that even the safety equipment you have cannot protect you. That's something you should consider as well.


Galvanic reaction between aluminum and carbon fiber composite

The reason is that aluminium carbide, $\ce{Al4C3}$ is formed. The reaction occurs at around 500°C on the interface. The problem with this substance is that it is brittle, and it also reacts with water:

$\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}$.

More info at this and this wikipedia page.


Is there a substance that would react with seawater to produce a dense, insoluble, nonporous precipitate?

A superabsorbent polymer sounds appropriate for this application, and the wikipedia article suggests they are used for "blocking water penetration in underground power or communications cable".

Will the water be removed from the pipe again after it has intruded, and if so do you need the seal to remain impermeable? I would expect this kind of polymer to dry out and shrink with time in the absence of water and it would not necessarily then be able to swell again with the same performance.


Friend threw mercury in a fire, is this dangerous?

Note that I generally do not use a lot of boldface in posts. Please recognise that it is required here.

Immediately get out of the room! Keep children and pregnant women away! Ventilate that room very well for a long amount of time! And never let your friend handle a mercury thermometer again!

Liquid mercury is not as dangerous as some people make it to be; a number of mercury compounds (e.g. dimethyl mercury) are much more toxic. However, that is because of the way most people would be exposed to mercury typically. If the thermometer breaks, there is a puddle of mercury on the floor and barring yourself doing anything stupid it will take a while for the mercury to fully evaporate. It’s inhaling mercury vapour that is most dangerous with respect to liquid mercury.

Mercury’s boiling point, according to Wikipedia, is some $350~\mathrm{^\circ C}$. A fire in a fireplace is typically hotter. Thus, by pouring mercury into the fire, your friend has chosen the most effective way to generate mercury vapour quickly from liquid mercury. Unless the fireplace is a closed oven, this vapour will now gladly move around the room with air currents, thus being easily breathed in and absorbed by the body.

Of all the ways to handle mercury, your friend chose the most dangerous and the most stupid!


Is this a correct way to write a chain reaction?

Your scheme is one of sequential reactions. The chain reaction has feedback so that one to two reactions occur repeatedly until the reactants are all consumed.
The general scheme is shown in the diagram where $R\cdot$ represents any radical species ( H or Cl in Maurice's answer) and $\alpha$ has to be at least 1 for species $R\cdot$ to continue the chain reaction.


Mechanism of radical cyclisation to form hydroazulene core

This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI:
10.1016/0040-4039(95)00502-4)

The heat (provided by refluxing) is required to initiate the reaction by the decomposition pathway you have drawn. The cyanopropyl radical thus formed abstracts a hydrogen atom from the tin hydride, a consequence of the weak Sn–H bond:

The triphenylstannyl radical is the one that actually does interesting chemistry. Generally, there are a few options for radicals at this stage. One common pathway is to abstract a halide or chalcogenide from the starting material, but there is no such atom here (C–F and C–O bonds are stronger and rarely broken, so the OTES group is out of bounds). The other common pathway is to add to a π-bond. In this case the best place to add is the terminal carbon of the alkyne, which is least sterically hindered.

After this, the remainder is a series of addition-elimination steps. Note that in each case the ring size formed is 3, 5, or 6: these are all kinetically fast cyclisations.

In particular, note that:

Intermediate 4 is stabilised by electron donation from adjacent oxygen (similar to carbocations, radicals - which are also electron-deficient species - can be stabilised by adjacent lone pairs).
You are correct that a ring expansion is needed – in fact, two are needed – but this does not require a Wagner–Meerwein shift in a carbocation. Radicals are perfectly capable of doing this job, usually by a sequence of cyclisation (5 to 6) and retro-cyclisation (6 to 7). See also J. Org. Chem. 2012, 77 (19), 8634–8647 for another example of this motif, which is very common in radical chemistry.
There isn't tin in the final product, so we will eventually have to kick it back out again: this can be achieved via a β-scission mechanism, so we have to work towards an intermediate which has a radical β to SnPh3, i.e. 7.

The original paper supports this mechanism (although they skipped some steps, as is usual in journals):


Naming substituted branched alkanes with different projections

You are right about the first one, it's called 2,3-dimethoxy-4-methylpentane. The $\ce{-OCH3}$ group is called methoxy. The second compound is a
newman projection
. It is basically the view that you will see when you are looking down the $\ce{C-C}$ bond axis. The second one should be called 2,3-difluoro-2,3-dimethylbutane.

Here's what it looks like (http://www.chemspider.com/Chemical-Structure.57499056.html?rid=29f2de9b-b762-46d4-a2ca-7452b8a5d6b6). In the newman projection you are looking down the the C2-C3 axis, and the arms show the substituents bonded to them.


Variation of cone angle with concentration of fuel-air mixture

The cone angle of the Bunsen flame is determined by the flame speed of the mixture and the average flow velocity. To first approximation one can write: $$ \sin \alpha = \frac{S}{U} $$
where $\alpha$ is the half cone angle, $S$ is flame speed and $U$ is flow velocity.

When the mixture is made fuel-lean, depending on how flame speed changes the cone angle will change accordingly.

Usually flame speed for fuel-lean mixture is less than that for stoichiometric mixture. Therefore we can expect the cone angle to decrease.