Your decomposition is for the hybrid orbitals that point from carbon to carbon, omitting the ones that point out towards the hydrogens, so you really want to project starting from one of the hybrid orbitals. First off we correct your decomposition:
$$\Gamma=A_{1g}\oplus A_{2g}\oplus B_{1u}\oplus B_{2u}\oplus2 E_{1u}\oplus 2E_{2g}$$
We only need half the operations of $D_{6h}$ because all the hybrid orbitals transform the same under $\sigma_h$, so we write out the matrices for the group operations of the subgroup $D_6$ and a set of matrix representations for each irreducible representation and the result of each group operation acting on $\phi_1$, first half:
$$\begin{array}{c|cccccc}R&E&C_6&C_6^2&C_6^3&C_6^4&C_6^5\\\hline
A_{1g}&1&1&1&1&1&1\\
A_{2g}&1&1&1&1&1&1\\
B_{1u}&1&-1&1&-1&1&-1\\
B_{2u}&1&-1&1&-1&1&-1\\
E_{1u}&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}
&\begin{bmatrix}-1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\\
E_{2g}&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}
&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}\\
R\phi_1&\phi_1&\phi_3&\phi_5&\phi_7&\phi_9&\phi_{11}\end{array}$$
Ummm... yeah, I assume the $x$-axis points right in your illustration and the $y$-axis points up. Also I number the hybrid orbitals $1$ to $12$ counterclockwise starting with the one just above the $x$-axis in the first quadrant. I will label the dihedral $C_2$ rotation axes $a$ to $f$ with $a$ parallel to the $x$-axis counterclockwise in $30°$ increments. So now we can produce the second half of our table:
$$\begin{array}{c|cccccc}R&C_{2a}&C_{2b}&C_{2c}&C_{2d}&C_{2e}&C_{2f}\\\hline
A_{1g}&1&1&1&1&1&1\\
A_{2g}&-1&-1&-1&-1&-1&-1\\
B_{1u}&1&-1&1&-1&1&-1\\
B_{2u}&-1&1&-1&1&-1&1\\
E_{1u}&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}
&\begin{bmatrix}-1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}\\
E_{2g}&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}
&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\\
R\phi_1&\phi_{12}&\phi_2&\phi_4&\phi_6&\phi_8&\phi_{10}\end{array}$$
Then it remains to apply projection operators:
$$\Gamma_j^{(\mu,i)}=\sum_{R\in G}D_{ij}^{(\mu)}\left(R^{-1}\right)R\phi_1$$
(Note the $R^{-1}$ in the above formula)
$$\Gamma_{A_{1g}}=\phi_1+\phi_3+\phi_5+\phi_7+\phi_9+\phi_{11}+\phi_{12}+\phi_2+\phi_4+\phi_6+\phi_8+\phi_{10}$$
$$\Gamma_{A_{2g}}=\phi_1+\phi_3+\phi_5+\phi_7+\phi_9+\phi_{11}-\phi_{12}-\phi_2-\phi_4-\phi_6-\phi_8-\phi_{10}$$
$$\Gamma_{B_{1u}}=\phi_1-\phi_3+\phi_5-\phi_7+\phi_9-\phi_{11}+\phi_{12}-\phi_2+\phi_4-\phi_6+\phi_8-\phi_{10}$$
$$\Gamma_{B_{2u}}=\phi_1-\phi_3+\phi_5-\phi_7+\phi_9-\phi_{11}-\phi_{12}+\phi_2-\phi_4+\phi_6-\phi_8+\phi_{10}$$
$$\Gamma_{E_{1u},x}^{(1)}=\phi_1+\frac12\phi_3-\frac12\phi_5-\phi_7-\frac12\phi_9+\frac12\phi_{11}+\phi_{12}+\frac12\phi_2-\frac12\phi_4-\phi_6-\frac12\phi_8+\frac12\phi_{10}$$
$$\Gamma_{E_{1u},y}^{(1)}=\frac{\sqrt3}2\phi_3+\frac{\sqrt3}2\phi_5-\frac{\sqrt3}2\phi_9-\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2+\frac{\sqrt3}2\phi_4-\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{1u},x}^{(2)}=-\frac{\sqrt3}2\phi_3-\frac{\sqrt3}2\phi_5+\frac{\sqrt3}2\phi_9+\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2+\frac{\sqrt3}2\phi_4-\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{1u},y}^{(2)}=\phi_1+\frac12\phi_3-\frac12\phi_5-\phi_7-\frac12\phi_9+\frac12\phi_{11}-\phi_{12}-\frac12\phi_2+\frac12\phi_4+\phi_6+\frac12\phi_8-\frac12\phi_{10}$$
$$\Gamma_{E_{2g},x^2-y^2}^{(1)}=\phi_1-\frac12\phi_3-\frac12\phi_5+\phi_7-\frac12\phi_9-\frac12\phi_{11}+\phi_{12}-\frac12\phi_2-\frac12\phi_4+\phi_6-\frac12\phi_8-\frac12\phi_{10}$$
$$\Gamma_{E_{1u},2xy}^{(1)}=\frac{\sqrt3}2\phi_3-\frac{\sqrt3}2\phi_5+\frac{\sqrt3}2\phi_9-\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2-\frac{\sqrt3}2\phi_4+\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{1u},x^2-y^2}^{(2)}=-\frac{\sqrt3}2\phi_3+\frac{\sqrt3}2\phi_5-\frac{\sqrt3}2\phi_9+\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2-\frac{\sqrt3}2\phi_4+\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{2g},2xy}^{(2)}=\phi_1-\frac12\phi_3-\frac12\phi_5+\phi_7-\frac12\phi_9-\frac12\phi_{11}-\phi_{12}+\frac12\phi_2+\frac12\phi_4-\phi_6+\frac12\phi_8+\frac12\phi_{10}$$


A phospholipid has a charged, hydrophilic phosphate head and a neutral, hydrocarbon hydrophobic tail.  In a phospholipid bilayer, two phospholipids combine tail-to-tail such that the two outer, or exposed, surfaces are comprised of the polar phosphate heads.  The bilayer can function as a cell membrane; it can keep what is in "in" and what is out "out".  It can also regulate critical cell functions such as pH and ion balance through the incorporation of proteins in the bilayer that can serve as ion pumps.  Clearly, any disruption to the bilayer could allow the cell contents to leak out and bad stuff to leak in, which would lead to either cell inactivation or cell death.

The mechanism of action of quaternary salts on lipid bilayers is still an active area of investigation, but here is a simplified overview of what is thought to be happening. .  The first step is thought to involve the electrostatic interaction of the positively charged ammonium ion with the exposed, negatively charged bilayer surface, followed by incorporation of the hydrocarbon chain of the ammonium ion into the bilayer. The electrostatic interaction affects membrane surface pressure which, in turn, disrupts cell function.  Shorter chain lengths (n=12-14) are more effective against gram-positive bacteria, while longer  chain lengths (n=14-16) are more effective against gram-negative bacteria.

Here's a reference to a recent thesis (2012) that studies the resistance of bacteria to quaternary ammonium compounds.  Section 1.2 (pages 2-5) provides a clear, concise summary of the latest thinking on the subject.


What is the Purpose of an Electrolyte in a Galvanic Cell?

Two issues are at play here.

1) You will measure a potential without any supporting electrolyte. As soon as you have everything connected, a potential will develop. Because a self-respecting voltmeter doesn't require much current (ideally zero), the potential will be measured.

2) As soon as you try to pass any current, as comments have mentioned, that current has to be carried from the electrode through the solutions and the salt bridge. Without any electrolyte in the solution, the solution resistance will be huge(as Geoff mentioned) and the amount of current passing will be cell voltage/cell resistance. This will essentially limit the current passing through the solution to very low values.


Is it possible to create a 4 inch by 4 inch channel through concrete using acid?

Offhand, this seems dangerous to the structural integrity of the slab, because the acid would follow the pore structure of the concrete and weaken the wrong places.

However, concrete cutting saw rental is available; e.g. United Rentals
and http://www6.homedepot.com/tool-truck-rental/Walk_Behind_Concrete_Saw_14/MK-CX3/index.html. Two 45-degree cuts ~6" or deeper should do the job.


Calculate $n(\ce{FeSO4})$ with $M(\ce{FeSO4})\approx 136\:\mathrm{g/mol}$

   $n(\ce{FeSO4})\approx0.08\:\mathrm{mol}$

Calculate $n(\ce{H2O})$

   $n(\ce{H2O})\approx0.5\:\mathrm{mol}$

Set them into relation $n(\ce{FeSO4}):n(\ce{H2O})= 1:x$

   $x\approx 6$


Petroleum cracking and incomplete combusion of organic compounds may give pyrene in a complicated mixture as discussed in else where. However, if you looking at a method to prepare pure pyrene or its derivative you may need to go through [2.2]metacyclophane procedure (For review: Ref.1):

As seen in the above image, [2,2]metacyclophane can undergo transannular ring closure to yield products with the 4,5,9,10-tetrahydropyrene (THPy) skeleton (e.g., Ref.2), which may then be dehydrogenated to pyrene (Ref.3).
The synthesis of [2,2]metacyclophane(s) can be achieved from m-xylene. This can be in two steps: First, the bromination of two methyl group using NBS to achieve 1,2-bis(bromomethyl)benzene and then subjected the brominated m-xylene to Wurtz reaction (e.g., Ref.4). Recent approach of some substituted [2,2]metacyclophanes is described in Reference 5 starting with substituted m-xylenes (The electronic supplementary material is attached for readers benefit).
The synthesis of highly substituted pyrenes via cyclisation of bis-alkynylbiphenyls has been also achieved (e.g., Refs.6-8). It is worth noting that these starting biphenyls were made using simple benzene derivatives (Ref.1):

References:

Juan M. Casas-Solvas, Joshua D. Howgego, Anthony P. Davis, “Synthesis of substituted pyrenes by indirect methods,” Org. Biomol. Chem. 2014, 12, 212-232 (DOI: 10.1039/C3OB41993B)(PDF).
Fritz Vögtle, Peter Neumann, “Stereochemistry of [2.2]Metacyclophanes,” Angew. Chem., Int. Ed. Engl. 1972, 11(2), 73–83 (https://doi.org/10.1002/anie.197200731).
Peter P. Fu, Ronald G. Harvey, “Dehydrogenation of polycyclic hydroaromatic compounds,” Chem. Rev. 1978, 78(4), 317–361 (https://doi.org/10.1021/cr60314a001).
Virgil Boekelheide, Rodger W. Griffin Jr., “Synthesis of [2.2]metacyclophanes,” J. Org. Chem. 1969, 34(6), 1960–1961 (https://doi.org/10.1021/jo01258a095).
Marco Blangetti, Helge Müller-Bunza, Donal F. O'Shea, “First asymmetric synthesis of planar chiral [2.2]metacyclophanes,” Chem. Commun. 2013, 49(55), 6125-6127 (https://doi.org/10.1039/C3CC42275E)(Electronic Supplementary Material).
D. Barney Walker, Joshua Howgego, Anthony P. Davis, “Synthesis of Regioselectively Functionalized Pyrenes via Transition-Metal-Catalyzed Electrocyclization,” Synthesis 2010, (21), 3686-3692 (DOI: 10.1055/s-0030-1258238).
Matsuda Takanori, Moriya Taisaku, Goya Tsuyoshi, Murakami Masahiro, “Synthesis of Pyrenes by Twofold Hydroarylation of 2,6-Dialkynylbiphenyls,” Chemistry Letters 2011, 40(1), 40-41 (https://doi.org/10.1246/cl.2011.40).
Mareike M. Machuy, Christian Würtele, Peter R. Schreiner, “2,6-Bis(phenylethynyl)biphenyls and Their Cyclization to Pyrenes,” Synthesis 2012, 44(9), 1405-1409 (DOI: 10.1055/s-0031-1290754).


What is the proof for saying all the compounds-of-an-element (except few), to have the same oxidation number?

This statement is a generalisation. As every generalisation it makes things simpler to remember but even here they give exceptions. Of course, chemists couldn't test all oxygen or hydrogen compounds. The assumption is based on electronegativity of atoms in the "usual" compounds. In OF2 oxygen is not in -2 oxydation state. Also in BH3 hydrogen is not +1. I would not even go as far as "In all the known hydrogen/oxygen compounds". "In most of known" would be enough.


How can I calculate the amount of HF in aquoues solution of NH4F and an aquoues solution of NH4HF2?

WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts.

Assuming:

(1) that concentrations can be used instead of activities (bad assumption...)
(2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations.
(3) Since HF is a relatively weak acid, little $\ce{HF}$ will dissociate into $\ce{H+}$ and $\ce{F-}$.

Using equilibrium:
$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4.7$$

From 5 molar $\ce{NH4F}$, $\ce{[F-] = \pu{5 molar}}$

From 0.18 molar $\ce{NH4HF2}$, $\ce{HF_2^- = \pu{0.18 molar}}$

So plugging the number in:

$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}} = \dfrac{0.18}{\ce{[HF]}\times 5}=4.7$$

$$\ce{[HF] = 0.008 molar}$$

For the final concentrations $\ce{[HF] \ll [H2F-]}$, so the assumption that the dissociation of $\ce{H2F-}$ can be ignored is valid. The $\ce{HF}$ will dissociate and acidify the solution, but still $\ce{[HF] + [H+] \ll [H2F-]}$


Disposal of Silver Halides dissolved in Ammonium Hydroxide

In the test for Cl- the chloride ions are converted into AgCl by the following reaction,
Cl- + AgNO3
→
NO3- + AgCl
which is then dissolved in ammonium hydroxide by the formation of diammine silver chloride by the following reaction:
AgCl + 2NH4OH
→
[Ag(NH3)2]Cl + 2H2O
On standing, the diammine silver complex subsequently breaks down to Ag3N.
The standard free energy of Ag3N is about +315 kJ/mol . Hence it is unstable and decomposes explosively to metallic silver and nitrogen gas .
2Ag3N → 6Ag + N2
So the diammine silver complex has to be destroyed after use so that it does not break down into Ag3N.
One way of destroying the complex is by adding nitric acid.
[Ag(NH3)2]Cl + 3HNO3
→
AgNO3 + 2NH4NO3 + H2O
So in all reactions involving the formation of diammine silver complex, the complex must be destroyed using acids after the purpose.
The complex is formed when AgCl, AgBr(sparingly soluble)and AgI(the complex will be formed only in trace amounts) are dissolved in NH4OH . The comlex is also formed in the well-known Tollen's test. Tollen's reagent is [Ag(NH3)2]OH. Ammoniacal Silver Nitrate is also used in Fontana–Masson Stain and also in silver mirroring(such as inside insulated vacuum flask).
The Wikipedia page on Silver Nitride
A reported accident


There are a variety of "open data" sources in chemistry. In your case, you probably want "open notebook science.".

There are a few practitioners in chemistry, although Jean-Claude Bradley sadly passed this last year.

JC Bradley
Andrew Lang
Open Source Malaria

I believe there will be a collection of articles published in the open access Journal of Cheminformatics as a memorial to Jean-Claude, but I don't think these have cleared review yet.


Mono-N-demethylation means the removal of a single (mono) methyl group from a nitrogen(N)-containing functional group. For example, a reaction which transforms the alkylated amine group $\ce{-NHCH3}$ into $\ce{-NH2}$.


Calculating the pH of a ammonium hydrogen sulfate solution

The hydrogen sulphate ion is a much stronger acid that the ammonium ion. Thus almost every hydronium ion in the solution comes from the dissociation of a hydrogen sulphate ion.

This said, the assumption

  $\ce{H3O+}$ has a concentration of $2x$ at equilibrium. $\ce{SO4^2-}$ and $\ce{NH3}$ each have a equilibrium concentrations of $x$.

is wrong.

When you solve the quadratic equation

$$K_a = \frac{[\ce{H3O+}] [\ce{SO4^{2-}}]}{[\ce{HSO4^{-}}]} = \frac{x^2}{c - x}$$

you will get a pH near your text book's value.


Entropy and enthalpy change on mixing liquids

This is difficult, because we do not know what is your current understanding and at what “level” of approximation you expect the question to be answered.

At a undergraduate level, or in the absence of any details (or thermochemical data or access to thermochemical tables), I would go with option (3).

Enthalpy: we are not given any indication at what is the enthalpy of mixing. In the simplest approximation, the enthalpy of mixing is zero, as in an ideal solution. It would correspond to a regime where the enthalpy of dissolution ($\Delta H_{\text{sol}}$) does not depend on the concentration of the solution. It's probably a good approximation, until you reach high concentrations.
Entropy: the entropy of mixture is positive, i.e. the process of mixing the two solutions is spontaneous. It can be explained by this reasoning.


Accelerated stability testing for chemicals: Shelf life / Expiry Date

Yes, there are guidelines. For example from ICH and WHO for pharma, and from CIPAC for pesticides. Some can be downloaded freely.

Yes, there are guidelines. For example, from ICH and WHO for Pharma, and from CIPAC for pesticides. Some can be downloaded freely.
Accelerated stability tests are generally not accepted as replacements for full-scale stability tests, but can provide an indication of the formation of undesirable by-products etc.

As a rule of thumb, a very general rule, chemical reaction rates in-/decrease by a factor of 2 for every 10C the temperature changes. According to this rule, if the test temperature is raised from 20C to 30C, the test period could theoretically be halved. However, if you look through the temperature/time combinations in the guidelines, you will find this relationship hard to find :-).


Products formed when solid CuCN is dissolved in aqueous KCN solution

No, both $\ce{[Cu(CN)3]^2-}$  and $\ce{[Cu(CN)4]^3-}$ species are formed. According to Wikipedia article of copper cyanide:

  ... but rapidly dissolves in solutions containing $\ce{CN−}$ to form
  $\ce{[Cu(CN)3]^2-}$ and $\ce{[Cu(CN)4]^3-}$, which exhibit trigonal
  planar and tetrahedral coordination geometry, respectively. These
  complexes contrast with those of silver and gold cyanides, which form
  $\ce{[M(CN)2]−}$ ions in solution.The coordination polymer
  $\ce{KCu(CN)2}$ contains $\ce{[Cu(CN)2]−}$ units, which link together
  forming helical anionic chains.

See this paper for more details.


Does it sometimes take energy to create a chemical bond?

"Molecules that require less energy to break than they give off when broken are called fuels." is the most imprecise sentence in the paragraph, because fuels don't give off energy when their bonds are broken, but when these broken bonds form new bonds (with oxygen) that are much more stable and release more energy than it took to break the original (carbon-hydrogen) bonds.

A candle is stable in air, even tho it does not have complete stability in an oxygen-containing atmosphere. But if you put sufficient energy into a few bonds, the excited atoms will combine with oxygen, which gives off enough energy to excite a few more bonds, which etc., etc. It's this new reaction, which is overlooked in the paragraph, which provides the increased temperature during the reaction.

The energy that "bonds require to form" is an activation energy which destabilizes a molecule enough so that it can reorganize with other reagents (like O2) to form more stable products. This activation energy may be supplied initially from outside sources (the match), but in a continuing reaction, the exotherm from the reaction provides the activation energy and then some.


Doesn't methyl amine give carbylamine test?

You are indeed correct.

$\ce{HOOC-CH2-NH2(X) ->[HNO2] HOOC-CH2-OH (Z) + N2(Y)}$

$\ce{HOOC-CH2-NH2(X) ->[soda lime] CH2-NH2(A)}$

Also, $ \ce{(A) + CHCl3 + 3KOH ->[Heat] CH3-NC + 3KCl + 3H2O}$.

Methylamine does give the carbylamine test.

So clearly, the answer given must be wrong.

The correct options are 1,2 and 4.


Sand may be dissolved in concentrated and hot solutions of NaOH. The dissolution is slow, and pressure may help this action. At the end of the reaction, a viscous solution of so-called sodium silicate is obtained. The formula of the solute is not well defined, and can be described as $(Na_2SiO_3)_n$. It is made of a long polymeric chain similar to the formula $HO-Si-O-(Si-O)_n-Si-O-Si-OH$, with supplementary negatively charged Oxygen atoms above and under each Silicon atoms. The whole chain is negatively charged. And of course, there are Sodium ions $Na^+$ in the vicinity of each negatively charged Oxygen atoms. To make the description more complete, one should mention that from time to time there are side branches of Si-O-Si-O chains, which get fixed on the Si atoms of the main chain. Furthermore there are no $Si=O$ bonds in this formula.

If you add some $HCl$ solution, you produce the precipitation of an insoluble gel whose formula has the same skeleton as the previous solution $HO-Si-O-(Si-O)_n-Si-O-Si-OH$. But the Oxygen atoms above and under the main chain are now holding a H atom, and the whole chain is neutral.

This gel can be filtrated and washed with large amounts of water. This removes the ions $Na^+$ and $Cl^-$. The result is a gel containing more water than silica. Usually this gel is then heated and the water is evaporated. An amorphous white substance is obtained with a formula between $H_2SiO_3$ and $SiO_2$. This is the common "silicagel" used in the labs for dehydrating purposes, because it has a tendency to adsorb water from the atmosphere.

The fresh gel can also be dried in a vacuum, instead of by heating. In this case, the silica gel is transformed into an extremely light substance, lighter than snow, which has a tendency to float in the air. It is a dangerous substance if you breeze it, because it will stay in the nose and the lung, without being able to get out. It may induce cancers, like asbestos.


Fluorine reacts with practically everything, especially at high temperatures, including zinc, wires, some ceramics, polymers, etc., so the system you describe may be prone to contamination. I don't see any obvious source of NO2, but it could easily form from nitrates, if present, in an acidic environment, which could result from fluorine attacking metals present.


How to calculate the entropy change of benzene when it changes from vapour into liquid?

You just used $\mathrm{d}S=\mathrm{d}H/T$. Also you didn’t told whether the heat capacities are $C_{\mathrm{m},p}$ or $C_{\mathrm{m},V}$. Anyways I found that you are using $C_{\mathrm{m},p}$ from the benzene data page on Wikipedia. You need to do something like this:
$$\Delta S=\left(1\;\mathrm{mol}\cdot81.7\;\mathrm{\frac{J}{mol\cdot K}}\cdot\ln\frac{373\;\mathrm{K}}{343\;\mathrm{K}}\right)+\left(\frac{30.8\;\mathrm{\frac{kJ}{mol}}}{373\;\mathrm{K}}\right)+\left(1\;\mathrm{mol}\cdot136.1{\;\mathrm{\frac{J}{mol\cdot K}}}\cdot\ln\frac{343\;\mathrm{K}}{373\;\mathrm{K}}\right)\\\approx78.012\;\mathrm{J/K}$$


Why is the electron-nucleus attraction modelled with only electrostatic interactions?

If I understand the question correctly, OP is somewhat surprised that Coulomb's law is used to describe the interaction between an electron and a nucleus, although it is usually pictured that electrons are moving and Coulomb's law describes interaction between static particles. Should not then the Lorentz law be used instead Coulomb's one?

First note, that electrons do not move around a nucleus in an atom. At least, they do not do so in the classical sense. Yet, a non-zero orbital angular momentum $\vec{L}$ of an electron in an atom gives rise to the orbital magnetic dipole moment $\vec{\mu}_{L}$ and so the magnetic field is indeed generated by the "moving" electron. Plus, an electron has an intrinsic magnetic moment (spin) which also contributes to this magnetic field. The nucleus possibly possess orbital magnetic momentum as well (if has non-zero angular momentum) that couples with its spin which interacts with that of electron. These interactions splits the energy levels of the atom and the resulting hyperfine structure can be measured using radio-frequency spectroscopy.

And in principle, for a complete description of an atom we could use a model in which we treat the interaction between electron and nucleus taking magnetic interactions into account, i.e. similarly to how it is done in classical electrodynamics with the Lorentz law. There is a problem with the Lorentz law though - it is formulated in terms of forces and velocities and this terms are not even present in the vocabulary of quantum mechanics. So, the magnetic field must appear somehow differently in the Hamiltonian, and to make a long story short the full (non-relativistic) Hamiltonian for one electron in an external electromagnetic field is:
$$
\hat{H} = \frac{1}{2m}(\boldsymbol{p} - e\boldsymbol{A})^2 + e\phi \, ,
$$
where $\phi$ is the scalar potential and $\boldsymbol{A}$ is the vector potential. And even in the absence of an external to the whole atom magnetic field, the nucleus itself gives rise to the vector potential $\boldsymbol{A}$.

So, this is certainly true that electrostatic interaction between electron and nucleus is not the whole story. However, electrostatic interaction dominates magnetic ones by far, so that usually the later are just ignored, unless a very high precision is required.


Brass Instrument: Dezincification or just scrubbed off?

    How do I identify dezincifiaction correctly? Can it be seen on those pictures? (most important question)

Instruments are made from solid brass, this is required as a stiffer metal like brass is required to have good sound quality. Your brass was not scraped off and the spots are indeed a result of dezincification. Water could get here for a number of reasons from fog mist to moist hands, it does not have to be saliva only.

  What do you think of chemical cleaning? It sounds dangerous to me. Isn’t it going to damage the instrument?

From what I can find chemical cleaning looks like they use a mild acid and chelatior (like citric acid or EDTA) to etch away some of the red rot. However the brass still has to be polished and the lacquer needs to be reapplied.

  What do you think of cleaning such an instrument soap? Is this going to damage the lacquer or material?

Soap won't damage the lacquer, but scrubbing might. That said your red rot shows that the lacquer is already damaged to you might as well scrub/polish the spots and reapply lacquer.


how to mathematically predict reaction results

There are two parts to quantification of reactions:

(1) How far it goes? (Thermodynamics, equilibrium). This all is dictated by ground state energy.

(2) How fast it goes? (Kinetics, reaction rates). This is dictated by  transition state energy vs ground state energy.

Equilibrium is much easier to understand. Let [Y] be concentration of reagent Y. You say that for 2A + B = C+3D +4E you have:

K = [C]$^1$[D]$^3$[E]$^4$/([A]$^2$* [B]$^1$)

Note how powers are equal to coefficients in the equation.

$\Delta$G= -RT*ln(K)

$\Delta$G is Gibbs energy (measured and tabulated for all of inorganic chemicals from school textbook).
R is a constant (Universal gas constant)
T is temperature.

How to use it? Well, imagine that you have

2*H2+ CO <-> MeOH

Lets say that K=10 under that conditions. I am sure I am off. Correcr way to find it is to use $\Delta$G= -RT*ln(K). You know T, R is constant. $\Delta$G is known for MeOH, H2 and CO. Calculate the difference to find $\Delta$G. You will need to find $\Delta$G for methanol vapor or take it for liquid MeOH and add $\Delta$G of evaporation.

You start with 1M of CO ( atm at 273C) and 1.5M of H2 (67.2 atm at 273C). Total pressure 112 atm. (Mol to pressure conversion using ideal gas law). What will be the equilibrium amount of MeOH if ?  You say that

Let x be the amount of product at equilibrium. H2 = 1.5 M - 2*x and CO = 1-x.

x/(1.5-2*x)$^2$*(1-x)=10=K

Now you have a cubic equation to solve (1.5-2x)$^2$*(1-x)*10 - x = 0;

With more reagents you have algebraic equation of higher power, but there are methods to solve those numerically and the idea is simple enough to explain.

This gets a bit more complicated when you have multiple reactions going. But This is the basis. When system gets complex you start ignoring some small members to simplify it. (1.0001)$^5$ ~ 1.0005 (reasonably accurate). YOu do the same in chemistry.

Kinetics is trickier. For A+ 2B -> C the first approximation is that

rate = k*[A]*[B]$^2$.

Also "k often increases by a factor of 2-4 per 10C of increase in temperature".

Now to treat it accurately you need to solve a system of equations. each equation is either

r$_i$ = $\delta$[A$_i$]/$\delta$t

or

r$_i$ = ($\delta$[A$_i$]*$\delta$[B$_i$])/$\delta$t

Rate if i$^{th}$ reaction = change of concentration of i$^{th}$ reagent per time period as the period goes to zero).

You end up with a system of ordinary differential equations. But don't get tricked by the word "ordinary". Most of them don't even have solutions in "elementary functions" (functions that you study in school or university as a chemist.

To understand what is a differential equation you would need to know what is an integral. We study differential equations in second year of university and that probably the most difficult subject in university.


Where I am struck is that I knew carbons occupy zinc sulfide like
  lattice so, how can this line be true.

Zinc sulfide can exist in two different crystalline forms, Zincblende (sphalerite) and wurtzite.  Zincblende is face-centered cubic (also known as diamond cubic), each ion is tetracoordinate and has local tetrahedral geometry - just like the carbons in diamond.  The wurtzite structure is hexagonal close-packed with interconnected 6-membered rings.  You may have been thinking of the wurtzite structure, while it is the zincblende structure that is related to diamond.  You can see pictures of the two structures here.


There was a method patented in 1896 whose claims would seem to meet your requirements.  Even with your restricted budget, used laboratory equipment companies can probably provide you what you need.

An excerpt from the patent states:

  "My invention relates to the production of graphite from any of the
  cheap and common forms of carbonaceous materials, such as mineral
  coal, coke, charcoal, gas-carbon, and carbids and the like; and it has
  for its object to provide a process or method of cheaply and
  practically producing pure graphite..."

The reagents used are common, cheap and readily available.  The patent states that:

  "[the mixture is] composed of powdered coke, sand, salt, and sawdust,
  the coke forming about fifty per cent., by weight, of the mixture, and
  these materials are thoroughly mixed and introduced into the electric
  furnace..."

The key issue for you will be designing the high-temperature inductively heated furnace, though the patent describes this thoroughly, including diagrams.

The availability of used equipment to fashion the device as stated is of course uncertain, and you will have to be creative with what is available.  I would start my search with LabX, with whom I have no affiliation whatsoever.  I hope this does "point you in the right direction", and Good luck!


Any differences in metal ion emission spectra?

$0.1~\mathrm{M}$ $\ce{KNO3}$ dissociates to give $0.1~\mathrm{M}$ $\ce{K+}$ ions.

$0.1~\mathrm{M}$ $\ce{K2SO4}$ dissociates to give $2 \times 0.1~\mathrm{M}$ $\ce{K+}$ ions.

Hence, in terms of emission intensity, $\ce{K2SO4}$ should give twice the signal if other parameters are same and if the detection response is linear.

Very sensitive instrument might detect the effect of counter-ion but that will be very difficult since the cation itself is always surrounded by water.


Acids / Bases - Need some clarity for final exam

1a) First use $pOH = 14 - pH$

Then use: $pOH = \log([OH^-])$

b) Simply use: $\dfrac{\text{dissociated}\ [OH^-]}{\text{original}\ [OH^-]}$

c) Use $K_b = \dfrac{[X^+][OH^-]}{[XOH]}$

2) Dissolving $KCN$ in water gives:
$\ce{KCN \rightarrow K^+ + CN^-}$

The following equilibrium reaction is thus formed

$\ce{CN^- + H2O \rightleftharpoons HCN + OH^-}$

Which is:

$K_b = \dfrac{[HCN][OH^-]}{[CN^-]}$

From the $K_a$, you can get the $K_b$ through $K_w\ =\ K_aK_b$ so that:

$\dfrac{K_w}{K_a} = \dfrac{[HCN][OH^-]}{[CN^-]}$

Let the amount of $CN^-$ that reacts to be $x\ M$, thus $[CN^-]$ becomes $0.1-x$ and you get $x\ M$ of both $[HCN]$ and $[OH^-]$. Solve for $x$ to get $[OH^-]$ and from there, $pH$ should be easy to calculate.

3) Use $Ksp = [Ba^{2+}][F^-]^2 = 1.7\times 10^{-6}$ and solve for $[Ba^{2+}]$.


Does CO2 dissolution in different salt concentrations affects the amount of carbonic acid formed?

100 mL of water and added 5 g of $\ce{CO2}$ and measured the time it took to dissolve completely

This statement doesn't make sense.  Firstly, if you know it dissolved completely, why try to determine how much dissolved?  Secondly, the only way that much CO2 would dissolve is if you did the experiment at high pressure.

If you want literature data to compare your experiment to, see Fig. 6c or equation 6 of Behaviour of the CO2-H2O system


Major product of the dehydration of alcohols

The mechanism of dehydration of 2° alcohol at high temperature (around 170°C-180°C) happens via $E_1$ mechanism. This means that the major product will be decided by the $\mathbf {Zaitsev}$ $\mathbf{rule}$, which says that major product will be the stables one. If you compare the three alkenes, you'll see that the middle compound you've drawn has the most hyperconjugative structures, and hence is the stablest. Therefore, that will be the major product.


No. Mathematically, that would not make sense.

Consider an imaginary element with only one known isotope: imaginarium-100.

The percent abundance of imaginarium-100 is $100\%$, so its relative atomic mass is: $$100 \cdot 1=100~\mathrm{amu}$$

Now imagine a group of scientists were to discover that in fact imaginarium-99 consists of $0.0001\%$ of all imaginarium samples.

The new relative atomic mass would be:

$$100 \cdot 0.999999+99 \cdot 0.000001=99.999999~\mathrm{amu}$$

The relative atomic mass goes down with the discovery of a less massive isotope. So the closest that the relative atomic mass of an element can be to the mass of its most massive isotope is if the most massive isotope is also the only known isotope, but the relative atomic mass can never go over this.


This paper actually explain the reason for which ruthenium is vulnerable to bleach. Bleach actually contains $\ce{OCl^-}$ ions which acts as an oxidising  agent and thus oxidises ruthenium. But the research on this phenomenon was done by using sodium hypochlorite, $\ce{NaOCl}$ in presence of a caustic alkali.

The experiment was done in two-fold process. First only using $\ce{NaOH}$ and the second using both $\ce{NaOCl}$ and $\ce{NaOH}$. In the first attempt, there was nothing occuring. The solutions with  $\ce{NaOH}$ were stable for a considerable time, but after several weeks began to break down, giving a deposit of the black oxide. In the second process, when a little of hypochlorite was given, it gave a orange-red solution as you can see in the above video.

Some more attempts were made by repeating the experiments and found that concentrated alkali hypochlorite more or less dissolved any ruthenium present. From these experiments, the conclusions were made that ruthenium, is rather soluble in concentrated alkaline solutions of hypochlorites and this might help to distinguish ruthenium from other metals. The overall reaction of this process is(from the video):-

$$\ce{Ru + 3NaOCl + 2NaOH -> Na2RuO4 + 3NaCl + H2O}$$

$$\ce{Ru + 7NaOCl + 2NaOH -> 2NaRuO4 + 7NaCl + H2O}$$

$\ce{NaOCl -> 2NaCl + H2O}$ (ruthenate catalyzed)

$$\ce{Na2RuO4 + NaOCl + H2O-> RuO4 + NaCl + 2NaOH}$$


In a neutralization reaction between a strong acid and weak base, can I assume the weak base is totally ionized?

You're never going to puzzle this out without some guidance so here goes...

  Why, if it is at all, correct to assume all of the base reacts?

All of the base will never react no matter how much acid you add. If you setup the equilibrium equation you can solve for $\ce{[NH3]}$ for any given pH. Thus there will always be some small amount of $\ce{NH3}$ in an acidic solution.

The gist is that there is $5.70\cdot10^{-4}$ moles of base. If you add "exactly" that much acid then the necessary assumption is that the acid balances the base and the solution is neutralized.

One other thing to consider in all of this is the significant figures. So +/- $1\cdot10^{-6}$ moles of acid doesn't make any real difference. That is enough to swing the final pH from 6 to 8.

Just like for the acid if $\ce{[NH3]} < 1\cdot10^{-6} $ then the free ammonia is negligible to the amount of $\ce{[NH4+]}$. Thus, considering the significant figures, the assumption that all of the ammonia was converted is reasonable, even though it is not actually absolutely true.


How can we determine the Acidity or Base only from information molecular formula provides?No concentration given

You cannot determine exact pH levels without knowing specific equilibrium constants (or equivalent values) and concentration of the substances.

Determining whether something is acidic is relatively straight forward. Inorganic compounds like the one you've presented can be broken down into cations and anions.

For the anions, you should consider:

Is this anion acidic ($\ce{H2PO4-}$)
Is this anion basic ($\ce{OH-}, \ce{HCO3-}$)

For the cations, you should consider:

Is this cation the hydrogen ion ($\ce{H+}$)
Is this cation a Brønsted-Lowry acid ($\ce{NH4+}$)
Is this cation Lewis acidic ($\ce{Al^{3+}}$)

For compounds where ions are both acidic and basic, the exact balance matters. Then, you're crossing into the complicated regime where things are interesting.


Coarse-grained force field for ion–water interactions

Forcefields for alkali ions in water have been developed for decades now, so the literature is rich with examples… and yet there still are significant developments (see this very new paper for only one such example).

If you want to have a forcefield for alkali ions that mix well with the TIP3P water model, you better look at what existing studies on such systems have done… this is an example, but you can find many others with searching through the literature.


The purpose of the vacuum jacket is to provide insulation of the column from cooling by ambient lab air.  You don't want to have to heat the "pot" (the flask containing the crude material you want to purify) any more than you have to, otherwise heat sensitive material in the pot may degrade and turn to tar.  Imagine if you cooled the exterior of your distillation column with cold water while performing the distillation.  You'd have to heat the pot a LOT more to get the material in the pot to finally reach the top of the column, then if you weren't cooling the column.  Air will have the same effect on your distillation, albeit to a smaller extent, than cold water.  So to avoid putting unnecessary extra heat into the pot, you vacuum jacket your column to preclude undesired cooling by ambient air.

Remember, a distillation column is different than a condensing column.  You don't want to cool a distillation column, you do want to cool a condensing column (or condenser).  Back in the day, very long columns were sometimes needed in the organic lab (5-10 feet) to separate closely boiling materials, they are still used in industrial applications.  If you had a 10-foot column that wasn't vacuum jacketed, you would turn most of your material to tar before it distilled over, because of the extra heat you had to supply to make up for the undesired cooling of the column by ambient air.  Eventually improvements like column packing materials (glass beads for example) and spinning bands (a metal helix rotating in the column) came along to increase column efficiency and allow chemists to be able to get good separation efficiencies (measured as "height equivalent theoretical plates", HETP) without having to resort to long columns.  Still a vacuum jacketed column is preferred to lessen unnecessary heating.


Thermal properties of aluminium alloy 6063

American Society for Metals has many publications. Temperatures below ambient are no problem for any aluminum alloys. They are commonly used for natural gas  separation liquefaction exchangers and piping down to −100 °C (guess at the temp.).
In the T6 condition (hardened) room temperature yield is 31000 psi; at 300 °F the short term yield is 20000 psi; 6500 psi at 400 °F, and falling fast as the temperature raises.
ASME Boiler Code will have allowable stresses which will be lower than these values. I looked in ASM Materials Handbook vol 1, 1961, it is probably not online. If this is long term, you will want creep or stress/rupture data which is not commonly available for aluminum; try in the aircraft industry.


My answer will unfortunately have almost nothing to do with chemistry.
The quoted statement is BS, there's no vitamin LE.
This claim appears to appear on several web pages, the common source seems to be a post from an nutrition/herbal oriented mailing list from 1998:[1]

…
OATSTRAW
Oatstraw as any livestock breeder will confirm, builds the strongest possible
body, with the greatest resilience. Used extensively in European cultures
throughout time for health and beauty, our country has let this valuable herb
slide into disuse.
Contains: calcium. high in Vitamins A, C, B complex, LE, G, phosphorus,
potassium, proteins.
Oatstraw contains Avenin, an amorphous alkaloid which is highly nutritive to
cells, improving normal cellular reproduction.
…

(also note e.g. that avenin is not an alkaloid but protein)
I have an evolution theory how the author could have made this typo, when (hopefully) compiling a literature:

… vitamins A, B complex, C, E, …
… vitamins A, B complex, C E, …
… vitamins A, B complex, CE, …
… vitamins A, B complex, LE, …
… vitamins A, C, B complex, LE, …

(Some intermediate forms might be purely hypothetical. An intermediate form “… vitamins A, B complex, C E, …” can be seen alive at the page mentioning oatstraw, at a different product though[2])
References:

http://www.geocities.ws/chdigest/herbs/pages/apr1998
https://www.healthylife-solutions.com/blog/2017/5/15/womens-herbal-tea-recipe


What's the relationship between supersaturation and temperature?

Supersaturation means simply that more of the substance is dissolved than should be. Supersaturation is an unstable condition. Given some "trigger" precipitation will occur and equilibrium conditions will then be restored.

The trigger could be a seed crystal, scratching a glass rod on a side wall or a piece of dust which acts as a nucleation point.


Setting up a water aspirator pump for vacuum distillation

Since you want the distillate, use the following setup.

This way if the water pressure drops then the water is sucked into the trap flask, not the distillate.


Why can't the energy of an electron exceed 0 eV?

The value of the energy in the Bohr model is zero when the quantum number is infinity  because that is the limiting value of the Coulombic potential at large distances, and because the electron is assumed bound to the nucleus (the atom is stable), which constrains the value of the total energy.
The energy of a stationary hydrogen-like atom is described by the equation
$$ U(Z,n) = R_\mathrm{E} \frac{Z^2}{n^2} $$
where $R_\mathrm{E} = \pu{-13.6 eV}$. It can equally well be written as
$$ U(Z,n) = \frac{-Zk_e e^2}{2r_n} \tag{1}$$
where $r_n$ is the allowed orbit radius in the Bohr model, which illustrates how the energy becomes more positive (goes to zero) as the average distance between the nucleus and the electron increases.
The Bohr model assumes the universe consists of just one electron and nucleus. It is implicit in the model that the electron is bound (the electron and nucleus can never get far enough apart to escape their mutual attraction). According to the virial theorem the total energy in the above equations is one half of the Coulombic potential energy and the negative of the kinetic energy. That means the expectation values of the potential and kinetic energy for a given quantum number are fixed, and that the value of the total energy is bounded by the value of the potential energy (see Eq. 1):
$$U(Z,n)=\frac{\left<U_{\mathrm{Coulomb}}\right>}{2}$$
The zero limiting value of the Coulombic potential can be interpreted as indicating that the interaction between the particles vanishes at large distances. On the other hand, the zero limiting value of the total energy can be interpreted as indicating that the reference frame of the atom is at rest. If the atom is assumed to have kinetic energy than this can be represented as a non-zero  offset of the total energy. In that case the energy converges to this threshold as the particles move farther apart.
In addition, in practice external fields might interact with the atom, contributing to the potential. It is possible for instance to superimpose a uniform linear potential on the Coulombic potential. The solutions (wavefunctions) for the new perturbed atom are not identical to those of the original isolated atom. Suffice it to say that if you now pull the particles apart far enough  they will eventually be free to follow trajectories that are no longer bound.
The following figures illustrate (roughly) the energies of an atom (for Z=1) before and after superimposing an external linear potential, plotted against the quantum number or the orbit radius in the original Bohr model. In the case of particles starting at rest, what determines whether particles will remain bound or fall apart is relative values of the potential. It is the slope of the potential that determines forces that drive particles together or apart. In the last figure, the (slight) downward slope at large r as the radius increases indicates that particles at rest starting at this internuclear distance are no longer bound.
Hydrogenic atom in Coulombic potential...

... and perturbed by linear potential

An electron can also be excited by a photon into the continuum of non-quantized states with energy greater than zero, but then the energy is no longer described by the Bohr model. The excess energy (greater than the ionization energy) amounts to the kinetic energy of the dissociated (nonbound) state (which is alluded to in the other answer).
You might also want to read these other posts.


I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation):

Let's consider the case of  the titration of strong acid (volume $V_a$ and concentration$C_a$: unknown) with a strong base (volume $V_b$ and concentration$C_b$), and we'll denote $V_b(eq)$ the volume of base at the equivalence:

The pH increases slowly at first because the pH scale is
logarithmic, which means that a pH of 1 will have 10 times the
hydronium ion concentration than a pH of 2. Thus, as the
hydronium ion is initally removed, it takes a lot of base to change
its concentration by a factor of 10, but as more and more
hydronium ion is removed, less base is required to change its
concentration by a factor of 10. Near the equivalence point, a
change of a factor of 10 occurs very quickly, which is why the
graph is extremely steep at this point. As the hydronium ion
concentration becomes very low, it will again take a lot of base to
increase the hydroxide ion concentration by 10 fold to change the
pH significantly.
Let's find the equation of the titration curve in the the region $0<V_b<V_b(eq)$:

The number of moles of $\ce{H3O+}$ before the titration is $C_aV_a$

The number of moles of $\ce{H3O+}$ after adding the volume   $V_b$ of the base is $C_aV_a-C_bV_b$

Then, the concentration of $\ce{H3O+}$ is $$[\ce{H3O+}]=\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
So, the titration curve in the region $0<V_b<V_b(eq)$  $$p\mathrm{H}=-\log\frac{C_aV_a-C_bV_b}{V_a+V_b}$$
If you trace this mathematical function $p\mathrm{H}=f(V_b)$, you will have a smooth slope at the beginning and as $V_b \rightarrow V_b(eq)$, the slope of the graph sharply increases as the function becomes not defined: $C_aV_a=C_bV_b(eq)$.


How would one favor the formation of p-TsCl over o-TsCl in the chlorosuflonation of toluene?

Chlorosulphonation using $\ce{ClSO3H}$ is an irreversible, electrophilic aromatic substitution reaction.
In the book Chlorosulfonic Acid, On taking Toluene as the aromatic substrate, it is seen to yield two products -

$para$-tosylchloride i.e. the thermodynamic product (because it is devoid of the steric hindrance offered by the methyl group)

$ortho$-tosylchloride i.e. the kinetic product (because of more activated ortho position due to proximity with methyl $(+I)$ group, which is favourable for the electrophile to attack on despite facing steric hindrance)

At low temperatures like $0-5°C$, the kinetic product is favoured and thus the ortho-isomer is formed. This reaction is exploited in the manufacture of the non-fattening sweetener $Saccharin$.
At higher temperatures like $75-80°C$, the thermodynamic product is favoured i.e. the para-isomer.
According to Chlorosulfonic Acid, The yield of para-tosyl chloride can be increased by

heating, or

addition of ammonium chloride or other catalysts

However, Clayden instead goes on to suggest on Pages 486 and 566, that the preference of ortho-isomer in chlorosulphonation and para-isomer in sulphonation is due to the fact that the latter is reversible unlike the former.

If the reaction is irreversible (as in chlorosulphonation), the alpha-product is usually formed. (with respect to naphthalene, here analogous to ortho position on toluene)
Sulphonation at low temperatures give the alpha-product by kinetic control and sulphonation at high temperatures gives the beta-product (corresponding to para-position on toluene) by thermodynamic control. The beta product is formed more slowly but it is more stable as there is less steric hindrance.... (as the case in toluene)

...which suggests that chlorosulphonation may not be temperature-dependent at all and shall always yield the ortho-isomer, which I am a bit doubtful about. (help needed)
However, the thermodynamic-kinetic explanation seems very satisfactory to me.


Does the carboxylate group show a positive or a negative inductive effect?

The carboxy group is electron-withdrawing both by inductive and by mesomeric mechanisms. I have drawn two depictions in the scheme below. The upper one is how the mesomeric effect of the carboxy group can be explained. This effect is only relevant if the $\ce{COOH}$ fragment is in the same plane as the ring.

In the case of 2,6-di(tert-butyl)benzoic acid, the bulky tert-butyl groups prevent the mesomerism depicted above. The carboxy group is perpendicular to the ring. Yet, this group still exercises an electron-withdrawing inductive effect. It can be explained by a mesomeric structure of the $\ce{C=O}$ double bond with charge separation:

$$\ce{C=O <-> C^+-O^-}$$

This positive charge is, naturally, electron-withdrawing and can be used to explain the $-I$ effect of carboxy groups. No matter which mesomeric depiction you choose, you can never get a negative charge on the carboxy carbon, thus it will always remain partially positive. That also makes sense considering it is bonded to two electron-withdrawing, strongly electronegative oxygen atoms.


The concept of normal coordinates of a molecule

Since I was reviewing this in some detail recently, I will try to supplement Prof. Hutchison's already good answer with some more detail as to why one would choose to work with normal coordinates, rather than the already quite simple Cartesian coordinates.

The short answer is that, under the assumption of small displacements, normal coordinates decouple the vibrational degrees of freedom from the rotational degrees of freedom and make all vibrations mutually orthogonal, such that the total energy of the nuclear motion of the molecule can be expressed as a sum of translational kinetic energy (3 dimensions), rigid rotation at the fixed equilibrium geometry (3 dimensions), and vibrations about the equilibrium geometry ($3N-6$ dimensions).

Now, it is not so difficult to show that if we allow a molecule to move in all $3N$ degrees of freedom, that the total energy of the system will have a term which couples the rotation of the molecule to the vibrations of the molecule, the so-called Coriolis energy. This is because if we work in strictly Cartesian coordinates, the axes are fixed, which means that as a molecule rotates, a simple displacement of two atoms (imagine a diatomic vibration) will not have the expected periodic behavior in Cartesian coordinates, but will appear to shift amplitude among the Cartesian coordinates.

So, we would like to work with a set of coordinates which circumvents this problem as we can see that it is an artifact of the coordinates we chose from the outset and does not have any useful physical meaning.

As it turns out, it is rather non-trivial to thoroughly show that there is a particular choice of coordinates which transforms the Cartesian coordinates such that the frame of molecular vibrations is fixed and hence eliminates the coupling between vibrational and rotational degrees of freedom. This is where I should mention that, if you like, the Bible of molecular vibrations is the book Molecular Vibrations by Wilson, Decius, and Cross[1]. Chapters 2 and 11 fully motivate and derive the transformation which completes this procedure. The coordinates turn out to be what we now call normal coordinates, and I will only point out that the transformation takes the following form,
$$
q_i=\sqrt{m_i}\Delta \alpha_i
$$
where $i$ is atom index which runs from 1 to $N$ and $\alpha=x,y,z$, so that we rescale the $3N$ Cartesian coordinates into $3N$ mass-weighted coordinates.

It shouldn't be too hard to see that the kinetic energy in these new coordinates takes the simple form,
$$
2T=\sum_{i=1}^{3N}\dot{q}_i^2
$$

Then, using a quadratic approximation to the potential energy, the potential energy takes the form,

$$
2V=\sum_{i,j=1}^{3N}f_{ij}q_iq_j
$$

where $f_{ij}=\left(\frac{\partial^2V}{\partial q_i\partial q_j}\right)$ is the harmonic force constant.

One can show that if you solve Newton's equations of motion using these expressions for the kinetic and potential energy, that one will have to solve an eigenvalue equation where the eigenvalues, $\lambda_k$, correspond to the frequencies of vibration and the eigenvectors correspond to the normal modes, i.e. the actual motion of atoms which vibrate at the the frequency $\lambda_k$.

Now, these $q$'s are not quite normal coordinates, which I will call $Q_i$, but they are very closely related to normal coordinates. Namely, what we call normal coordinates are the ones which have one additional nice property of having a potential energy of the form,

$$
2V=\sum_{k=1}^{3N}\lambda_k'\dot{Q}_i^2
$$

That is, we want the potential energy to not contain any terms involving more than one coordinate. Thus, we have to determine the coefficient $\lambda_k'$ which makes this transformation.

In Ref. [1], sec. 2-4, this derivation is performed and one finds the marvelous result that $\lambda_k'=\lambda_k$. Hence, one can make this representation of the potential energy diagonal by multiplying by the characteristic frequency which we got from each $q_k$.

Lastly, lest you convince yourself that vibration and rotation are not accounted for anywhere in this, when you solve the eigenvalue I mentioned above, you will find six eigenvalues which equal zero. This is only the case because we made this transformation to mass-weighted coordinates, and these six zeros tell us that six of our degrees of freedom appear to be completely stationary. These are translations and rotations.

So, people work with normal coordinates because they have these nice properties of providing a diagonal represent of the kinetic and potential energies when vibrations are harmonic, and giving a mutually orthogonal set of coordinates with which to work. This is nice because even when we wish to include anharmonicities and other corrections, we have a set of coordinates which have desirable properties and which are a usually a good approximation to the true "vibrational motions" of the system.

I put "vibrational motions" in quotes because it isn't clear that talking about the motions of atoms in a vibrational state is all that meaningful in quantum mechanics, and one can easily find a unitary transformation which causes the motion to look different in Cartesian space, but which leads to the same observable properties.

[1]: Wilson, E. B., Decius, J. C., & Cross, P. C. (1980). Molecular vibrations: the theory of infrared and Raman vibrational spectra. Courier Corporation.


To best understand, let us consider the reactions occurring.

First, we have $\ce{Cl-}$ ions in solution. Acidifying the solution serves to remove other anions that might give a false positive for the chloride test. Upon addition of $\ce{AgNO3}$, the following reaction occurs:

$$\ce{Ag+(aq) +Cl-(aq) -> AgCl (s)}$$

When enough ammonia is added to neutralize the $\ce{H3O+}$ in solution:

$$\ce{NH3(aq) +H3O+(aq) -> NH4+(aq) + H2O(l)}$$

The ammonia can then begin to form a metal complex with the $\ce{AgCl}$ that is soluble in water:

$$\ce{AgCl(s) +2NH3(aq)->Ag(NH3)2+(aq) +Cl-(aq)}$$

When $\ce{HNO3}$ is reintroduced, the ammonia ligand is protonated because it is more basic than water, and the metal complex is destroyed, reforming $\ce{AgCl}$:
$$\ce{Ag(NH3)2+(aq) +2H3O+(aq) +Cl-(aq) ->AgCl(s) +2NH4+(aq) +2H2O(l)}$$

Since you will likely be working with relatively small amounts, it is likely that you will be using concentrated acid in stoichiometric amounts to produce these products.


Why do electrons jump back after absorbing energy and moving to a higher energy level?

This is a very fundamental question and for really understanding the "why" some advanced physics is involved. I will describe the process rather superficially.
As you might know, the level energies of atoms and molecules can be calculated (in principle) using quantum mechanics. The simplest system is the hydrogen atom as it consists of a single proton and a single electron. Ignoring higher order effects (such as interactions of electron and nuclear spins and QED effects), the quantum mechanical calculation gives the same result as the Bohr model, that is, the level energies of hydrogen are given by the Balmer formula, which you probably know.
The calculation does not predict that the excited levels fall back to the ground state. An electron in an excited orbital will, according to this calculation, always stay in this orbital if nothing happens to the system.
Because we know that excited states decay, something must happen to the system to induce the decay. It turns out that in our calculation we have ignored the interaction of the atom with the photon field. Atoms can absorb light and emit light and we have completely neglected this. If you do take this coupling into account, you will find that excited states have a limited lifetime and that an excited electron will fall back to a lower level eventually.


Why does monoammonium phosphate (MAP) have 52% of phosphorus (P₂O₅)?

When it comes to fertilizers, usually what is advertised by general name monoammonium phosphate 10-52-0, in reality is ammophos, which is a mix of monoammonium and diammonium phosphates, $\ce{NH4H2PO4}$ and $\ce{(NH4)HPO4}$ with some water, so the brutto formula is

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}$$

where $0 ≤ x ≤ 1$.
To find out the range for $y$, let's find the average molecular mass from the given nitrogen ($\ce{N}$) and phosphorous ($\ce{P2O5}$) content $α_i$:

$$\bar{M} = \frac{1}{2}\cdot\left(\frac{M(\ce{N})}{α(\ce{N})} + \frac{M(\ce{P2O5})}{2α(\ce{P2O5})}\right) = \frac{1}{2}\cdot\left(\frac{\pu{14 g mol-1}}{0.10} + \frac{\pu{142 g mol-1}}{2\cdot 0.52}\right) \approx \pu{138 g mol-1}$$

Using the expression for molecular mass of brutto formula and its average value:

$$x = 1.35 - 1.06y$$

Since $0 ≤ x ≤ 1$, the following holds true for $y$:

$$0.33 ≤ y ≤ 1.27$$

resulting in formula with the following ranges for the variables:

$$\ce{(NH4)_{1 + x}H_{2 - x}PO4 * yH2O}, \quad x \in [0; 1], y\in [0.33; 1.27]$$

Since monoammonium phosphate usually prevails in this mix, there is a substantial amount of water; e.g. one of the possible brutto-formulas would be the following monohydrate:

$$\ce{(NH4)_{1.3}H_{1.7}PO4 * H2O}$$

Obviously, all the above is valid in the absence of third-party contaminations, e.g. if pure phosphoric acid and ammonia were used for the synthesis.


why does phosphorus makes 5 bonds with oxygen in P4O10 whereas 3 bonds in P4O6?

Phosphorus has electronic configuration of [Ne]3S2 3P3 As it is a 3rd period element it also have a vacant 3D orbital.

In case of P4O6 only the 3P3 electrons take part. Resulting in formation of 3 bonds with O. But in case of P4O10 one of the electron of 3S is excited to vacant 3D orbital. Now P has 5 unpaired electron and it can form 5 bonds. Therefore P forms 5 in P4O10.

Now, about the octet. Not every compound follow the octet rule. Just to understand The whole chemistry depends upon STABILITY i.e lesser energy of molecule. Therefore, if by forming more bonds (and does not obeying Octet rule) the molecule is gaining more stability, it will do so.


Evaporation is a surface phenomenon, boiling is a bulk phenomenon.  Evaporation will always be occurring for any liquid with an exposed surface.  Water molecules in a liquid sample have a distribution of kinetic energies.  It takes less work for a surface water molecule to break free from its neighbors, then for a water molecule in the bulk to escape, since the former has fewer neighbors (no molecules on top).  Those molecules with enough kinetic energy to break away from their neighbors will escape.

For water to boil its vapor pressure must equal the atmospheric pressure.

The only requirement for evaporation to occur is an exposed liquid surface.  I don't think relative humidity affects the boiling point, but I suspect it would affect the rate at which a liquid will evaporate from an open container.  Evaporation is always occurring, but to notice a change in the amount of material in an open system, more material must leave the surface then returns to the surface and the rate of return will be affected by the composition (water content) of the air above the open sample.


How would the "-COOCl" functional group be named?


A general method for estimating $\mathrm{p}K_\mathrm{b}$ values is to use the equation $\mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} = 14$ where $\mathrm{p}K_\mathrm{b}$ is that of the conjugate acid; in this case, applying it gives us $\mathrm{p}K_\mathrm{b}(\ce{CN-}) \approx 4.79$. We can generally assume that ‘strong’ and ‘weak’ in $\mathrm{p}K_\mathrm{b}$ terms are similarly to be defined as for $\mathrm{p}K_\mathrm{a}$ values. Therefore, cyanide is a weak base and the premise of this question is incorrect. This observation is consistent with the fact that neutral cyanide solutions are relatively safe as long as no acid is introduced to protonate $\ce{CN-}$.

Note that the general statement ‘the conjugate base of a weak acid is a strong base’ is incorrect. The opposite is often true, namely:

  The conjugate base of a strong acid is generally a weak base.

And vice-versa if you swap all occurences of acid and base. The word generally is required here, since compounds such as $\ce{H2Cl+}$ are very strong acids whose conjugate bases are still very strong — but these compounds are not typically easily accessable and certainly not in aquaeous solution. A weak acid needs to have $\mathrm{p}K_\mathrm{a} > 15$ for the conjugate base to be a strong one but those are seldomly still considered ‘acidic’. An example would be water ($\mathrm{p}K_\mathrm{a} = 15.7$ therefore $\mathrm{p}K_\mathrm{b}(\ce{OH-}) \approx -1.7$) which is a weak acid with a strong conjugate base.

Since it is such a weak base, it has no access to $\mathrm{E2}$ pathways and $\mathrm{E1}$ pathways are disfavoured against $\mathrm{S_N1}$ if either is applicable. We don’t even need to consider $\mathrm{E1_{cb}}$. Thus, cyanide will generally react via an $\mathrm{S_N2}$ pathway unless the conditions for $\mathrm{S_N1}$ are much better.


Is it a "Bary center" or a "Bari center"?

It is safe to say that Barycenter is the correct spelling for the sought word. The other versions are artifact of the original word as some authors preferred to name (especially in the internet).
Barycenter (or Barycentre in British English) is a commonly used term in astronomy and in chemistry (in Crystal Field Theory) as well. In both fields, it is used to explain center of mass balance and most of the literature has used the aforementioned spelling. For example, NASA explains the question "What Is a Barycenter?" as:

We say that planets orbit stars, but that’s not the whole truth. Planets and stars actually orbit around their common center of mass. This common center of mass is called the barycenter. Barycenters also help astronomers search for planets beyond our solar system!

In Crystal Field Theory (CFT), electrons in the $\mathrm{d}$-orbitals are repelled to different degrees by the ligands. The degeneracy of the $\mathrm{d}$-orbitals is lifted. The Barycenter is the average energy of the $\mathrm{d}$-orbitals (Coordination Chemistry II: Theories of Electronic Structure and Crystal Field Theory):


Why are overcoats used on aluminum if aluminum oxide forms on surface anyway?

Sapphire is an aluminum oxide which is transparent when in crystalline form. However when aluminum metal corrodes the aluminum oxide takes up more volume that the aluminum metal from which it was formed. So the aluminum oxide can't form the same crystalline structure as aluminum metal. Thus instead of one nice nice uniform crystalline sheet of aluminum oxide the oxide forms small crystals. Such an oxide coating reduces the reflectivity of the mirror. (Thinking of it like dropping table salt on a mirror.)

The overcoating materials keep oxygen away from the aluminum metal so that the oxide coating doesn't form.


The significant digits you use are the values of which you are confident, plus one more value which may be uncertain.

In this case, the balance is accurate to $\pm0.001$ mg, so you're confident about the first 2 decimal places, but the 3rd place is uncertain.

The measurement given is $9.00060$, but the decimal part is smaller than our uncertainty, so we can't count on it.  The most accurate we could record is $9.001$, because the first decimal we're uncertain about is the 3rd place, which leaves us 4 significant figures.

If the measurement had been $19.00060$ instead, then the same process would result in 5 significant figures.


Why does the Co³⁺/Co²⁺ couple have such a high reduction potential?

The electronic configuration has nothing to do with it. The reduction potentials of $\ce{Ni^3+}/\ce{Ni^2+}$, $\ce{Cu^3+}/\ce{Cu^2+}$ and $\ce{Zn^3+}/\ce{Zn^2+}$, if they have been/could be measured, would be even greater.

The reduction potential for $\ce{M^3+}/\ce{M^2+}$ is most dependent upon the third ionisation energy. If $I_3$ is large then it will be difficult to oxidise $\ce{M^2+}$ to $\ce{M^3+}$ (hence a high reduction potential). Likewise if $I_3$ is small then the reduction potential will be smaller.

$$\begin{array}{c|cc} \hline
\text{Metal} & I_3\text{ / eV} & E^\circ(\ce{M^3+}/\ce{M^2+})\text{ / V} \\ \hline
\ce{Sc} & 24.76 & (-2.6) \\
\ce{Ti} & 27.48 & (-1.1) \\
\ce{V}  & 29.31 & -0.26 \\
\ce{Cr} & 30.96 & -0.41 \\
\ce{Mn} & 33.67 & +1.60 \\
\ce{Fe} & 30.65 & +0.77 \\
\ce{Co} & 33.50 & +1.93 \\
\ce{Ni} & 35.16 & (+4.2) \\
\ce{Cu} & 36.94 & (+4.6) \\
\ce{Zn} & 39.72 & (+7.0) \\ \hline
\end{array}$$

Data from Weller et al. Inorganic Chemistry 6ed; Johnson Some Thermodynamic Aspects of Inorganic Chemistry. Numbers in parentheses indicate predicted values. Graph by me in Microsoft Office 365.

So your question is essentially: why is $I_3$ of $\ce{Co}$ so "large"?

Going across the 3d block, effective nuclear charge increases which leads to the general trend of IE3 and $E^\circ(\ce{M^3+}/\ce{M^2+})$ increasing.

Any stabilisation derived from "special" electronic configurations e.g. the half-filled subshell in $\ce{Fe^3+}$, or the ligand-field stabilisation energy of $\ce{Cr^2+}$, merely lead to relatively small deviations from the general trend.

Moral of the story is: look at the big picture first. Those $\mathrm{d^5}, \mathrm{d^{10}}, \cdots$ configurations can be relevant, but they aren’t responsible for the overall trend, and certainly shouldn’t be the first thing you consider when trying to rationalise this set of data.


According to VSEPR theory, each carbon has 2 electron "pairs".  These electron pairs are not what you would typically think of as electron pairs because VSEPR theory considers double bonds and triple bonds to still be a single "pair".

So considering each carbon atom, the shape is $AX_2E_0$ ($A$ is the carbon atom, $X$ is a bonding pair of electrons, and $E$ is a non-bonding pair (lone pair) of electrons).

$AX_2E_0$ corresponds directly to a linear molecule.

To directly answer your question, when trying to determine the shape of a molecule, consider double and triple bonds as only a single electron bonding pair.


Stability of 3d metal fluorides and iodides in different oxidation states

In general, iodides stabilise lower oxidation states and fluorides stabilise higher oxidation states, e.g. $\ce{CuF2}$ versus $\ce{CuI}$. This can be explained with some thermodynamics. Consider

$$\ce{M(s) + $\frac{n}{2}\,$X2 (g/l) -> MX_n (s)}$$

In order to form a halide salt with a higher oxidation state (i.e. larger $n$), you need to pay some energetic costs to get the metal to the higher oxidation state, namely:

more electrons are removed from the metal, so you need more ionisation energies
you need to generate more $\ce{X-}$ anions, so you need to break more $\ce{X-X}$ bonds (bond dissociation energies) and add more electrons to them (electron affinities).

This seems bad, so why would any higher oxidation state ever be formed? The answer is that when you pay these costs, you recoup some of the energy by forming more bonds between $\ce{M^n+}$ and $\ce{X-}$. If this bonding is strong enough, this can outweigh the costs described previously, and that makes the higher oxidation state more stable than the lower one.

For fluorine, the $\ce{M-F}$ bond is strong (regardless of whether you consider it to be ionic or covalent) and hence this energetic gain is large. On the other hand, the formation of fluoride ion is not particularly costly: the $\ce{F-F}$ bond is easy to break and fluorine loves picking up electrons, so the electron affinity works in your favour.

Overall, this means that there is a thermodynamic force for formation of fluorides in high oxidation states. Conversely, if you consider iodides instead, the formation of iodide ions is not as easy as the formation of fluoride. And the $\ce{M-I}$ bonding is weaker, too, so there isn't even much incentive to go to higher oxidation states.


Does it matter what electrolyte we use for a Galvanic Cell?

The electrolyte is very important for the functioning of the galvanic cell. The emf that the cell generates depends on the type and the concentration of the electrolyte used. In the Daniell cell, the reaction would be:

$$\ce{Zn + CuSO4 -> ZnSO4 + Cu}$$

As per the Nernst's equation, you get the emf of the cell to be:

$$\text{E}=\text{E}^\text{o}-\frac{2.303\text{RT}}{\text{nF}}\log_{10}{\left(\frac{\ce{[Zn^{2+}]}}{\ce{[Cu^{2+}]}}\right)}$$

Clearly, the emf of the cell depends on the concentrations of the cell. Dipping the rods in $\ce{NaCl}$ would not make the cell work.

And for why the salt bridge is needed, think about why you would build a cell in the first place. To generate power in a usable way right? If you let the solutions mix, they would react directly, and the energy would be released as heat instead, and not as electrical power. You want the ions to move towards the electrodes and get discharged there.


What is the difference between crystal structures of proteins, organic and inorganic materials?

Imagine an infantry unit of soldiers marching in file. Each of them may be quite irregular, but together they form a repeating pattern. And that's exactly what happens with protein molecules in a crystal.

When we say that a protein is irregular, we mean it on a different level. Indeed, one molecule of our protein is irregular if we are talking about the amino acid residues in it, of which there are probably a hundred: here goes Ala, then Gly, then all of a sudden Leu - what will come next? You can never tell. But wait, there is another molecule of the same protein sitting nearby, and it looks exactly the same as this one: Ala, Gly, Leu and so on. And there is yet another next to it, also similar, and many more of them, and together they form an ordered crystal structure which makes it possible for us to diffract X-rays on it and measure the intensities and then do some hardcore math and voila, here is your structure.


Calculating the partition function based on molar heat capacity

All you have to do is to use some of the well known results from Statistical Thermodynamics:
The molar heat capacity at constant volume is

\begin{equation}
  \bar{c}_V = \left( \frac{\partial \bar{U}}{ \partial T} \right)_V
\end{equation}

where $\bar{U}$ is the inner energy, $T$ is the temperature, and $V$ is the volume.
The molar inner energy in terms of the partition function $Z$ and the temperature is given by:

\begin{equation}
  \bar{U} = \frac{k_{\text{B}} T^{2}}{n} \frac{\partial \ln Z}{ \partial T} \ .
\end{equation}

where $n$ is the amount of substance.

Now, you simply have to plug the equation for $\bar{U}$ into the equation for $\bar{c}_V$ and after using the chain rule of differentiation you get the following differential equation:

\begin{equation}
  \bar{c}_V = 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} + \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} \ .
\end{equation}

Finally, you just have to use your formula for the molar heat capacity, $\bar{c}_V = \alpha T$, and solve the resulting differential equation

\begin{align}
  \frac{k_{\text{B}}}{n} T^{2} \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{k_{\text{B}}}{n}  T \frac{\partial \ln Z}{ \partial T} - \alpha T^2 &= 0  \\
 \frac{\partial^2 \ln Z}{ \partial T^2} + 2 \frac{1}{T} \frac{\partial \ln Z}{ \partial T} - \frac{n \alpha}{k_{\text{B}}} &= 0
\end{align}

for $\ln Z$.
If solving differential equations by hand is not your thing, you can simply charge Wolfram Alpha with the task and get:

\begin{align}
  \ln Z &= \frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \\
  Z &= \exp\left(\frac{n \alpha}{6 k_{\text{B}}} T^2 - C_1 \frac{1}{T} + C_2 \right) \ ,
\end{align}

where $C_1$ and $C_2$ are constants of integration which can be fixed by enforcing boundary conditions for the temperature.


Should carbon tetrachloride really be considered an organic compound?

Simple carbon compounds such as the tetrachloride blur the boundary between organic and inorganic. You should know that presence in natural systems is not an indicator of a compound being organic; many organic compounds are completely artificial, though they may be derived from natural feedstocks.

There is no strict criteria for organicity (is there a better word?). Consider perfluorinated carboxylic acids, they have no C-H bonds but are considered organic.


Mathematical proof of ΔQ=ΔH at constant external pressure

In an irreversible process, the total force per unit area exerted by the gas on the piston face is comprised of the local thermodynamic gas pressure at the piston face (as determined by the local gas density and temperature at the piston face) plus a viscous stress related to the rate of deformation of the gas in close proximity to the piston. So the thermodynamic pressure is not uniform throughout the gas and there are viscous stresses present.  By comparison, in a reversible process, the pressure is uniform throughout the gas and the viscous stresses are negligible.  In both cases, if the piston is massless, the force per unit area exerted by the gas on its side of the piston is exactly equal to $p_{ext}$, the force per unit area on the outside of the piston, and the work will be the same, as calculated from the force on either side of the piston.

If a frictionless piston has mass and the cylinder is horizontal, the piston will oscillate about its final position until the viscous stresses damp out its motion (i.e., its kinetic energy dissipates).  In the end, the final gas pressure will then be equal to $p_{ext}$, and, from a transient force and energy balance on the piston, one can show that the total work done by the gas will still integrate to $p_{ext}\Delta V$.

If the compression takes place irreveribly by, at time zero, suddenly increasing the external force per unit area from the initial gas pressure $p_0$ to the constant external force per unit area $p_{ext}$, the change in enthalpy of the gas from its initial state to its final state will not be equal to Q.  We will have $$W=p_{ext}(V_f-V_0)$$ and $$\Delta U = Q-W=Q-p_{ext}(V_f-V_o)$$so$$Q=\Delta U+p_{ext}(V_f-V_o)$$while$$\Delta H=\Delta U+(p_{ext}V_f-p_0V_0)$$


What are the effects of salt additives and impurities on the CMC?

Every chemical present can interact with the free lipid and/or with the micelles. Organic chemicals will tend to interact with the lipophilic portion of the lipids, while ionic ones will interact with the polar head.

Depending on their structure, all those species can have a stabilizing or destabilizing effect on the formation of micelles and so the CMC will change accordingly.

Of particular importance is the "ionic strength" of the solution (you should research that term).


What are these upside-down glass egg-shaped containers for?

This is called an inverted specimen jar, sometimes called a sample jar, for obvious reasons intentionally designed to stand on the stopper (it is not a standard one to close the other jars in the lab).  Their use was something more popular around 1900's, when showcasing dyes and other visual appealing chemicals, especially towards the general public. Your example invites to look at the granular character of the product.

A more traditional design is this one, more conveniently used in collections / museums.

(source)


The notation used does not necessarily represent a triple bond - for example, using one of the examples that you have provided:

  $\equiv\text{Si}-\text{OH}$

Which is for the functional group silanol, that is defined by the $\ce{Si-O-H}$ connectivity, it's chemical structure is

Image source

Using this example, the hydrogens represented by the  ≡ symbol can be replaced by other chemicals forming other members of the silanol functional group, for example, Trimethylsilanol:


Spontaneous inequilibrium in a gas in an isolated system?

The short answer to the question "Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?" is NO.  Once the system is in equilibrium it will remain there.

If it is possible, as is asked in the second part of the question, to have some reaction be triggered at a higher temperature reached by fluctuation, then as has been commented, the answer is that the system was never in equilibrium.

We should never forget that there is no "absolute" starting temperature for a reaction.  If the reaction is physically possible, it will take place, but the rate of that reaction can be horribly slow.


Why chlorine acts as a strong field ligand in tetrachloridoplatinate(II)?

I would like to add an answer stating a few exceptions which would come handy to you in future:

$\ce{Ni^{2+}}$ has a $3d^8$ configuration. In this case strong field ligands ($\ce{C}$ or $\ce{N}$ donors) will make the complex square planar. In all other cases, the complex will be tetrahedral.
In the case of $4d^8$ and $5d^8$ configuration and any ligand the hybridisation will be that of a square planar complex.
With $\ce{Co^{3+}}$, nearly all ligands behave as strong field ligands ($dsp^2$ hybridisation). Notable exceptions to this rule being $\ce{[CoF6]^{3-}}$ and $\ce{[Co(H2O)3F3]}$.
In case of bulky ligands (for instance $\ce{PPh3}$) present with $3d^8$ configuration, the ligands cannot get adjusted in the same plane leading to a tetrahedral shape ($sp^3$ configuration).


How expensive are Hartree Fock calculations?

I wouldn't call the Hartree-Fock methods 'very simple', I would consider it fairly complex both in terms of its derivation, interpretation, computational cost and implementation details. If Hartree-Fock calculations are now commonplace it is also because so many man-hours of study have been devoted to it.

In any case, as you say, its formal cost is $n^4$ ($n=$ number of basis functions): the number of two-electron integrals is $n^4$, and all these integral are necessary to build the Fock matrix, so the overall cost is also $n^4$. However, this is not the end of the story; as a system grows larger and larger in spatial extension many of the required integrals become vanishingly small and can be neglected, and other approximations become possible, at the cost of introducing a small and adjustable error. As long as the error introduced is (much) smaller than the intrinsic error of the Hartree-Fock method, it doesn't really matter. Some keyword often used in this context are 'linear-scaling' and 'density-fitted' methods.

The difficult, computationally expensive part of the Hartree-Fock method is dealing with the electron-exchange part of the Fock matrix, and most studies try to speed up this step. For example, a recent study of this type is:
Christoph Köppl and Hans-Joachim Werner, Parallel and Low-Order Scaling Implementation of Hartree−Fock Exchange Using Local Density Fitting, J. Chem. Theory Comput. 12, 3122−3134 (2016).

In this study the largest calculation they report on is for alpha-cyclodextrin, a polysaccharide with formula $\ce{C36H60O30}$ (126 atoms, 516 electrons); using the aug-cc-pVDZ basis set (2058 basis functions) the authors report that a density-fitted Hartree-Fock calculation took about 4 minutes on a beefy workstation with 20 cores (two Intel Xeon E5-2690 @ 2.8 GHz).


How to convert isopropanol to acetone and hydrogen

There are several ways to convert isopropanol to acetone.

Use of $\ce{Cr2O3}$ as catalyst[1]

  The conversion of isopropyl alcohol to acetone has been studied over $\ce{Cr2O3}$ powder at 473 K in a flowing microcatalytic reactor. Appreciable steady-state catalytic activity was achieved only when oxygen was present in the reactant gas flow. The effect of mild pretreatment of $\ce{Cr2O3}$ in $\ce{O2–N2}$ at 473 K on the initial activity, and the observed dependence of the process on $\ce{O2}$ pressure has been related to the oxidation–reduction cycle of the surface chromium species. $\ce{Cr^{(3+n)+}}$ on the surface of $\ce{Cr2O3}$ have been considered to be the active species, where n= 1–3.

2 . Catalytic dehydrogenation of isopropanol

a) Source 1.

  Catalytic dehydrogenation of isopropanol can be chosen as an
  alternative    synthetic route when high-purity acetone is required,
  such as in biomedical applications. Turton et.al. mentions that a
  single pass conversion of 85-92% with respect to isopropanol, with
  reactor conditions of 2 bar and 350° C, is generally achieved. A
  molten salt stream will be used a heat source for the endothermic
  reaction:

  $$\ce{(CH3)2CHOH -> (CH3)3CO + H2}$$

  The acetone produced in the reactor passes into a phase separator and
  then into a separation system that includes one stripping and two
  distillation columns. A recycle stream takes a mixture of unreacted
  isopropyl alcohol and water, with a trace amount of acetone, back into
  a mixer that feeds the reaction system. Using the catalyst which will
  be employed throughout this analysis, the reaction is first order with
  respect to the concentration of isopropanol and has an Arrhenius
  dependence on temperature with E=72.38 MJ/kmol and k=351,000 cubic m
  gas/cubic m reactor sec.

b) Source 2

  At the beginning of the process, feed including i-propyl alcohol and
  water, and recycle stream are mixed in feed drum. From here, this
  mixture is send to vaporizer to changestream’s phase as vapour. After
  vaporizer, mixture is heated to reaction temperature in theheater.
  Reactor used is a tubular flow reactor. Acetone, hydrogen gas (H2) are
  produced andwater and i-propyl-alcohol are discharged. The mixture
  with acetone, hydrogen, water, i- propyl-alcohol are sent to cooler
  and then to condenser. After condenser the mixture is sent toflash
  unit. Hydrogen, acetone, i-propyl-alcohol and water are obtained as
  top product. This top product is sent to scrubber to remove hydrogen.
  The bottom product of flash unit which isformed by acetone, water,
  i-propyl-alcohol are mixed with the bottom product of scrubber  before
  acetone column. In acetone column, acetone is obtained from top
  product with 99 wt%. i-propyl alcohol and water and also 0.1% of
  acetone is sent to i-propyl-alcohol column from bottom product. The
  top product of this column is sent to feed drum and bottom product is
  thrown away as waste water.

Using chromic acid [2]

  Oxidation at 25' to 40' is accordingly specified, requiring the
  mixture to be chilled in ice after each addition of oxidant. In 1926
  the writer worked out a student experiment which involves the same
  reaction, the oxidation being effected by gradual addition of chromic
  acid solution to the boiling solution of isopropyl alcohol in water.
  The reaction is carried out in a flask surmounted by a short
  fractionating column, through which the acetone is distilled during
  and after the oxidation. The crude acetone (collected below 65'), and
  several subsequent fractions, are redistilled through the column,
  permitting the easy isolation of a high yield (70% to 80% or more) of
  acetone boiling 57' to 60' and over 99% pure, and also the isolation
  of any unchanged isopropyl alcohol (76' to 82').

References:

Ilyas, M.; Shah, S.; Nigar, R.; Khan, H. Conversion of isopropyl alcohol to acetone catalysed by Cr2O3 at 473 K: role of molecular oxygen. Faraday Trans. 1994, 90 (16), 2413. DOI: 10.1039/FT9949002413.
Wagner, E. C. Oxidation of isopropyl alcohol to acetone. A student experiment. J. Chem. Educ. 1934, 11 (5), 309. DOI: 10.1021/ed011p309.


Choose the molecule with the strongest bond

The Bond Dissociation Energies of these molecules are in the order
Cl2 > Br2 > F2 > I2

All these molecules involve a single covalent bond. Covalent bonds rely on sharing of a pair of electrons between two atoms, an electron each from both atoms involved. The better the sharing, the stronger the covalent bond formed. This sharing is reflected in the valence shell orbital overlap that occurs between the two atoms, when a more effective overlap occurs, a stronger covalent bond is formed.

Now, the radii of the halogen atoms increases in the order of writing them down the group, F < Cl < Br < I < At. As the atoms get larger, so do the valence shell orbitals. The increase in size of orbitals carrying the same number of electrons would mean that the electron density would be lower in larger orbitals. So, for the same proportional overlap between two similar orbitals of similar atoms, the sharing of electrons would be poorer and would consequently result in a weaker covalent bond.

This should explain the strength of bonds decreasing as we go down the group.

However, in this orbital overlap consideration gets forgotten another factor that becomes important when atoms are small in size, and that is of electronic repulsions.

Fluorine is extremely small in size and so the electron density is pretty high as well. This results in an unstable arrangement if the atoms come too close, since like charges repel.

The interplay of the above mentioned effects results in Chlorine having the highest bond strength.

I may also add that the same is the case for (1) N-N bond energy as against P-P bond energy and (2) O-O bond energy as against S-S bond energy. But, do keep in mind this stands true only for single bonds.


Exceptions to the Madelung rule for electronic configurations

A lot of questions in here.

It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?

In the Schrödinger Equation, $\hat H\Psi = E\Psi$, the Hamilton operator $\hat H$ is generally considered to consist of a kinetic term $\hat T$ and a potential term $\hat V$. In general, $\hat T$ is defined via an electron’s impulse $\vec p$ rather than its velocity $\vec v$, meaning that we end up with a term $$\hat T = -\frac{\hbar^2}{2m}\nabla^2$$
Within this operator, the kinetic energy of an electron is considered and since $\hat T$ forms part of $\hat H$, we use an electron’s kinetic energy in the original Schrödinger equation and thus it is considered in the calculations.
Using relativistic quantum mechanics will complicate the calculations but the general idea will remain the same.

The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?

That’s … overly simplistic. In reality, it is a series of considerations we could make at the drawing board than when calculated leads to one electronic configuration being preferred over another. However, this question already covers that part.

Other exceptions to Madelung's rule:
For example, in the transition metals, the 4s orbital is of a higher energy than the 3d orbitals; and in the lanthanides, the 6s is higher than the 4f and 5d.

I don't understand the reason behind this.
Shouldn't it be that the 4s orbital is of a lower energy than the 3d orbitals (that's why it is filled first) and in the lanthanides, the 6s is lower than the 4f and 5d?

While your sentence is correct according to the aufbau principle or Madelung’s rule, this is no longer true when it gets to everything that’s not a neutral atom. Indeed, as soon as you turn scandium into a scandium(I) ion, the 4s orbital becomes less favourable than the 3d orbitals because the latter are still ‘closer to the nucleus’ and experience more stabilisation from an overall net positive charge. In general, once you go from neutral atoms to cations, the aufbau principle loses its relevance and a more ‘logical’ 3d-4s-4p order is favourable.

So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

Lanthanides tend to be rather subtle in their electronic effects if I recall my lecture correctly; as far as I am aware there is no explicit rule of any kind for them. But as for memorising: yes, you would have to do that.


Which proton in cyclohexene is the most acidic?

When comparing acidities of protons, we need to find out what the corresponding conjugate bases are.

The conjugate base upon removing $\ce{H_\mathrm{b}}$ is an allyl anion. There is (and should be) no question that it is the most stable carbanion of the three we are going to consider.

The conjugate base upon removing $\ce{H_\mathrm{a}}$ is an $\mathrm{sp^3}$ configured carbanion while the conjugate base upon removing $\ce{H_\mathrm{c}}$ is an $\mathrm{sp^2}$ configured carbanion. This is nice, because it allows us to directly compare the two using their hybridisation.

In general, lone pairs (and that explicitly includes those coming from anions) want to be in the lowest-lying atomic or molecular orbital possible. Since an s orbital is lower in energy than a p orbital from the same shell, lone pairs prefer s orbitals. If there is no pure s orbital available — as would be the case in practically all carbon-centred lone pairs — an orbital with the highest possible s character is preferred instead. If we look at the configuration $\mathrm{sp^3}$ we can say that it has $25~\%$ s character while $\mathrm{sp^2}$ has $33~\%$ s character. Since $33~\% > 25~\%$, the $\mathrm{sp^2}$ configured carbanion is more stable — which is also supported by corresponding $\mathrm{p}K_\mathrm{a}$ values.

The method outlined above is a very crude approximation building on a high amount of simplifications but gives us the correct result in practically all carbon-centred anion cases.

Thus, the acidity order is:
$$\ce{H_\mathrm{b}} > \ce{H_\mathrm{c}} > \ce{H_\mathrm{a}}$$

Note: Ron noted in a comment above that the acidity of an $\mathrm{sp^2}$ hydrogen almost parallels that of an allylic hydrogen. So while most courses may teach this simple ordering, you should keep that in mind as additional information.


Does formal charge affect bond polarity?

When you compute formal charges you split bonding electrons evenly between bonded atoms. This does not account for differences in electronegativity. It is similar to the difference between formal charge and oxidation state, as well explained in the Wikipedia.
A better picture (closer to the real electronic distribution) is, as Pauling would probably suggest, intermediate to the two, assuming a hybrid between possible Lewis structures (covalent and ionic).
In any case, when considering bond polarity, electronegativity overrules formal charge.    On the other hand inductive effects should not be dismissed.
The OP presented an excellent counterexample: CO. Here the carbon has a very slight negative charge relative to oxygen, as predicted based on the computation of formal charges.
I am not an expert on the development and application of electronegativity scales, but suffice it to say that these scales are (1) empirical and (2) provide useful guidelines. The original Pauling formulation of electronegativity cannot account for all possible bonding schemes. Case in point is the modification of the electronegativity scale to account for orbital hybridization, the following data via the Wikipedia:

Hybridization
χ (Pauling)1

C(sp3)
2.3

C(sp2)
2.6

C(sp)
3.1

'generic' C
2.5

Based on this modified scale the electronegativity of oxygen, 3.44, is no longer much larger than that of the carbon atom in carbon monoxide.
References

Data found in Wikipedia, original reference: Fleming, Ian (2009). Molecular orbitals and organic chemical reactions (Student ed.). Chichester, West Sussex, U.K.: Wiley.


Deduce the possible structure given spectroscopic data for the aromatic compound

It always helps to start with the ‘blooming obvious’: the signal at $\pu{9.60ppm}$. This is obviously an aldehyde. It also obviously couples ($^3J=\pu{9Hz}$) with the signal at $\pu{6.55 ppm}$. This proton is either in the range of a dihalogenated carbon, an electron-rich aromatic system or an alkene. We can rule out all possibilities except for the alkene: it can’t be aromatic because there would be no $^3J$ coupling to a non-aromatic proton (there has to be at least one quarternary carbon in-between).

Moving on, we have a very high coupling constant of $J = \pu{15.5 Hz}$. While it would, in theory, be possible that this arises from a geminal $^2J$ coupling of two diastereotopic protons (although those are typically on the order of $\pu{12Hz}$, so slightly less), the more reasonable explanation is an (E)-configured alkene. If it were a $\ce{CH2}$ group, we not only would get into trouble explaining the extremely high chemical shift but also why the aldehyde only couples to one of those protons. Thankfully, an alkene is what we already guessed from the chemical shift. So we now have three protons out of the way and a $\ce{?-CH=CH-CH=O}$ substructure.

Now we might choose to take double bond equivalents into play. The sum formula $\ce{C11H13NO}$ evaluates to 6 double bond equivalents. We have used two for the alkene and the aldehyde. The four remaning double bond equivalents together with the region the chemical shifts are in scream substituted benzene. We may now take a look at the actual integrals and couplings of the protons we deem can still be aromatic (all except for the 6 singlet protons at $\pu{3.00ppm}$). We have two sets of two and each are an apparant doublet. Therefore, we must have a disubstituted with a para-substitution pattern.

The two fragments we have so far, a $\ce{C6H4}$ and a $\ce{C3H3O}$ one, add up to $\ce{C9H7O}$, meaning that $\ce{C2H6N}$ is still missing. We should also note that all these six hydrogens are chemically equivalent. You can play around with this, but the only way to have six chemically equivalent hydrogens and a nitrogen in a group that connects only once is in a dimethylamino group. Thus, the final structure is:

Finally, allow me a few words on why your structure is incorrect apart from the missing proton corresponding to $\pu{9.60ppm}$:

The phenyl protons would show up as a set of three (so far so good). However, they would be a doublet and two triplets (one of the triplets corresponding to the single proton). Your spectrum does not contain any apparant triplets. Also, monosubstituted phenyl groups tend to have much more similar shifts to the point where they often only give a $\ce{5H}$ multiplet.
The six protons on the $\ce{NMe}$ groups must be equivalent. Since they are obviously on different carbons, there must be some rotation that maps one set onto the other. In my structure, you can easily rotate the $\ce{Ph-N}$ bond to transform one group onto the other. In your structure, rotation around the $\ce{C(O)-N}$ bond is restricted due to the partial double bond character.  Therefore, the two methyl groups would show up as two separate signals of three hydrogens each. Look up the spectrum of DMF (dimethylformamide) to confirm this.

Final note: I know nothing about IR and I do not consider it a helpful tool in structure determination. Thus, I cannot comment on the IR frequences and what they can, would or cannot correspond to.