Is it more reasonable to use stratification to separate similar compounds rather than using temperature?

The answer to your second question:

The halogenation of saturated hydrocarbons occurs by free radical substitution in the presence of UV light or heat and not by formation of carbocations. There's a very clear representation of the reaction mechanism on this Wikipedia page.

Since free radical intermediates are involved, this is a chain reaction and so what you said is correct, a large number of substituted products can be formed depending on the reaction conditions.

For example: If you consider the simple reaction between Methane and Chlorine in the presence of sunlight, Chloromethane, Dichloromethane, Chloroform, Carbon tetrachloride and a small amount of Ethane (by combination of Methyl radicals) are formed depending on the amount of chlorine available.

All these products will be formed even if the reactants are in a 1:1 ratio but they will be present in low concentration since there is less chlorine available.

In the case of higher alkanes like Butane and Pentane, the secondary carbons are substituted faster since the secondary free radical intermediates are more stable (reasons are the same as that for secondary carbocations).


Why heating up things before H2C2O4 &KMnO4 titration?

In general, the higher the temperature the faster the reaction. This is because increasing the temperature increases the number of molecules with more kinetic energy, and hence more molecules would have energy greater than the activation energy. If the reaction between oxalic acid and potassium permanganate wasn't heated then it would take extremely long to see the self-indicating titration occurring.


Maximum force that hydrogen-hydrogen bond can withstand

The force and the bond enthalpy are not necessarily directly related.

The bond dissociation energy is the amount of energy released when the bond is formed, or the energy required to break the bond.

The force constant of the bond derives from the curvature of the potential energy curve. For example, if we assume a harmonic oscillator (spring) for a bond, then:

$$ V(x) \approx \frac{1}{2}k_f \; x^2$$

Real bonds aren't purely parabolic, but clearly the second derivative of the potential energy curve near the equilibrium bond length will give the needed force constant.

Here I've plotted two fictional bonds with the same length. The blue curve gives a bond with a higher bond enthalpy and force constant (i.e., the curvature is higher than the red curve).

It should be clear that this isn't necessarily true. We can easily imagine some bond that has a shallow curvature but a deep enthalpy.

I should indicate that in many cases, these are related - stronger bonds typically have higher force constants and are harder to distort, but the point here is that the bond dissociation energy does not directly tell us anything about the mechanical strength of a bond.


Whilst not specifically an iodine clock reaction, Belousov-Zhabotinski reactions are oscillatory and can produce moving bands, spirals and concentric circles of colour. This works best in thin films (limiting convection) as the top commenter on the linked video points out.

There's a great thesis by Peter Hantz which looks at these kinds of reactions in detail (the first bit is in french, the rest is in english) and specifically covers Liesegang banding, where an oscillating chemical reaction in a mobile phase causes bands of precipitation in a gel column. As Hantz points out on page 22, the gel is required to prevent convection from destroying the bands, however the viscosity of the gel also freezes the precipitate in place as it forms. This would seem to be close to what you are interested in, however.

Here's a picture of Liesegang bands in test tubes from wikipedia:


What is the procedure of designing a salt bridge for a galvanic cell?

If you wish to do 25 trials with 5 trials per concentration, I would suggest the following:

You can use the same filter paper as long as the $\ce{Cu^{2+}}$ concentration remains constant. However, since you are changing the concentration 5 times, you will need 5 pieces of filter paper.
After every trial, thoroughly wash the filter paper with distilled water.

So I would say you need at least 5 pieces, but you would not need to go all the way to 25 pieces. Just make sure to follow step 2 after each trial.


Looking for linear non polar compound which is used heavily due to its boiling or melting temperature or its density

A possible answer is n-octane,

(image credit)

for what it

consists of linear organic molecules (vide infra)
has a boiling point of about $125\,\pu{^\circ{}C}$ / $257\,\pu{^\circ{}F}$
is a hydrocarbon you either find in, or generate from natural oil

A bit on the n in n-octane:  This is a shorthand for normal, because the carbon atoms are consecutively bound to each other in an open chain.  Which is not the same as if they would form branched molecules, then denoted iso-octane.  A prominent iso-octane (iso as in isomer, i.e. same number of atoms per molecule, but different spatial arrangment) is 2,2,4-trimethylpentane:

(image credit)

Blend in mixtures of the two, they serve as in octane rating of fuels.


Polymer properties in water often relate to the relative strength of polymer-polymer interactions versus polymer-water interactions. For example, the solubility of amylopectin > amylose > cellulose is attributed to how strongly molecules of each of these polymers interact with water molecules compared to other molecules of themselves.

In the paper making process, the cellulose fibers of paper dry out and interactions with water tend to get replaced by interactions with nearby polymer molecules. The resulting structure is strong because of polymer-polymer interactions as well as purely physical effects (e.g., "matting" of fibers, and surprisingly strong surface-surface interactions), all of which is enhanced by the compression and tension forces placed on the paper during its manufacture (google it to see photos).

When paper is later made wet, the entrance of water eventually disrupts these interactions, leading to local regions of what you might think of as micro solubility.  Then, when the paper dries out again as you describe, it probably does so without any compression/tension forces on the paper, thereby reducing the extent to which the matting of the fibers and chemical interactions between polymers takes place, and so it crumbles because of those micro regions of "solubility".

(Disclaimer: The above is just rationalizing off the top of my head. But since it is based on expertise as a PhD Chemist for nearly three decades, I'd give it a 70+% chance of being accurate ;)

The real science begins when you test the above ideas: For example, one test would be drying out paper under different degrees of compression and/or for different time periods and then testing the crumbling behavior.  Another might be to wet the paper while it's under different degrees of compression.  But the best test of the above ideas would be if there were some way to assay the extent of polymer-polymer versus polymer-water interactions -- no doubt such interactions result in very different vibrational modes. So there, I've outlined a Ph.D. thesis for you.


Why does only the cis-isomer of potassium dioxalatodiaquochromate(III) form?

I'm not an inorganic chemist, but to my understanding, both cis- and trans-dioxalatodiaquochromate(III) ions exist (Note that some also called it as Diaquabis(Oxalato)Chromate (III)). According to what I found online is that solid state reaction gives cis-form while aqueous reaction predominantly forms trans-isomer (it is mentioned in one of your attachments). However, it's still up to debate.

Hamm in 1952 had shown that trans—cis Isomerization using UV-vis spectroscopy ($trans \rightarrow cis$; Ref.1). According to Hamm, all previous workers have prepared trans-isomer because its lack of solubility. Yet, the existence of cis- and trans-isomers has been first recognized by Werner and has presented the preparation methods of two isomers individually (Ref.2), thus it is not true that the your concept of only the cis-stereoisomer forms.

Werner had mentioned that aqueous solution of trans-isomer is stable only in the cold temperatures. That may be the reason why only cis-isomer forms in solid state preparations because the preparation needs heating at high temperature.

Using the method given by Werner, trans-dioxalatodiaquochromate(III) isomer has been prepared and its crystal structure has been studied (Ref.3 & 4).

Later studies have showed this trans-isomer converted to cis-isomer by long standing in aqueous solutions (Ref.5). Based on their results, the authors have concluded that these color changes (dark red trans-isomer to blue cis-isomer) are due to trans–cis isomerization, uncomplicated by olation phenomena and that the transformation to the cis configuration proceeds to practical completion. They also stated that:

  Kinetically, the reaction (trans–cis isomerization) obeys a first-order law, and the dependence of the observed velocity constant upon temperature is found to be adequately described by the Arrhenius equation, where $A = 3 \times 10^9$ and $E = \pu{−17.9 kcal/mol}$.

Finally, in recent studies have been done on computational analysis of trans- and cis-Isomers of Potassium dioxalatodiaquochromate(III). The studies revealed that  cis-Isomer is slightly stable than trans-Isomer (Ref.6):

  The trans- and cis-isomers of potassium diaquabis(oxalato)chromate (III) were studied computationally and experimentally. The structures of trans- and cis-configurations of $\ce{[Cr(H2O)2(C2O4)2]-}$ were optimized by DFT methodology with various functionals namely: B3LYP, CAM-B3LYP, TPSS, PBE, M06-L and ωB97X-D along with the more sophisticated MP2 method. The calculations show that the most stable forms for both isomers are in quartet states. The results from all DFT methods reveal that the cis-isomer is literally more stable than the trans-isomer with the lower average relative energy of $\pu{2.1 kcal/mol}$. These are consistent with the results from MP2 calculation and experimental observation. The absorption wavelengths for the excited states of trans- and cis-structures were calculated by the time-dependent density functional theory (TDDFT) method. For the experiments, the trans- and cis-isomers of potassium diaquabis(oxalato)chromate (III) were synthesized and characterized by UV-Vis spectrophotometry. Both isomers have two maximum absorption wavelengths at $415$ and $\pu{560 nm}$.

References:

Randall E. Hamm, “Complex Ions of Chromium. II. trans—cis Isomerization of Potassium Dioxalatodiaquochromate(III),” J. Am. Chem. Soc. 1953, 75(3), 609-611 (https://doi.org/10.1021/ja01099a026).
A. Werner, W. J. Bowis, A. Hoblik, H. Schwarz, H. Surber, “Über Metallverbindungen mit komplex gebundener Oxalsäure. Zweite Mitteilung: Über Dioxaloverbindungen),” Justus Liebigs Annalen der Chemie 1914, 406(3), 261-331 (https://doi.org/10.1021/ja01099a026).
J. N. van Niekerk, F. R. L. Schoening, “Structure of Trans-Potassium Dioxalato-Diaquo-Chromiate,” Nature 1950, 166, 108 (https://doi.org/10.1038/166108a0).
J. N. van Niekerk, F. R. L. Schoening, “The crystal structure of trans potassium dioxalatodiaquochromiate, $\ce{K[Cr(C2O4)2(H2O)2].3H2O}$,” Acta Cryst. 1951, 4(1), 35-41 (https://doi.org/10.1107/S0365110X51000064).
G. E. Cunningham, R. W. Burley, M. T. Friend, “Potassium Dioxalato Diaquo Chromium (III) Complexes: Trans–Cis Isomerization,” Nature 1952, 169, 1103 (https://doi.org/10.1038/1691103a0).
Songtham Ruangchaithaweesuk, Juthathip Chorkate, Thana Maihom, Potjaman Poolmee, Piti Treesukol, Tipawan Rungsawang, Phornphimon Maitarad, Bundet Boekfa, "Combined Computational and Experimental Studies of Trans- and Cis-Isomers of Potassium Diaquabis(Oxalato)Chromate (III)," Key Engineering Materials 2017, 757, 103-107 (https://doi.org/10.4028/www.scientific.net/KEM.757.103).


Confusion about hybridization terminology

The number of hybridized orbitals is equal to the number of original orbitals. The name of the hybridized orbitals (e.g., $sp^{3}$) means just that the ratio of the contributions of $s$ to $p$ orbitals is $1:3$. We do not write $s^{\frac{1}{4}}p^{\frac{3}{4}}$.


Do bidendates occupy cis position always?

Its not impossible to have a trans-spanning ligand, but they seem to be rarer than cis-ligands due to their increased length and greater potential for steric strain. While en is too small of a ligand to reach trans-positions, it is possible with significantly longer ligands.

It seems octahedral ones are even rarer than their square planar counterparts, but there has been at least one reported synthesis of a trans-spanned octahedral$^1$ that I could find. In the linked paper, they report the synthesis of trans-$\ce{[Ru(Cl)(trpy)(C3SPAN)]}$ where trpy = 2,2:6,2-terpyridine and C3SPAN=$\ce{Ph2PC6H4CH2O(CO)(CH2)3(CO)OC6H4PPh2}$.

$\hspace{25ex}$

C. Bessel, et al. Chem. Soc., Dalton Trans., 1999, 2281–2292


Question related to Hydration of Alkynes

Due to -I effect from the carbonyl group and the halogen, not far enough for it to be negligible, the stability of the primary vinylic carbocation is counter-intuitively more than that of the secondary vinylic carbocation. Markonikov rule is applicable in absence of permanent electron shift effects, like resonance and inductive effects. The Markonikov rule, under these restrictions, translates into the stability of the carbocation, decided by the positive inductive effect of neighbouring carbons. As a clearer example, consider the hydration of acrolein,

The reaction predominantly yields the primary alcohol, as opposed to the product predicted by the markonikov rule, which again is not applicable in presence of permanent electronic effects. Hope that helps!


How to measure ozone output of an ozone generator using chemical methods

You can do the quantitative estimation of ozone ($\ce{O_3}$) by two reactions. First one is the reaction between $\ce{O_3}$ and potassium iodide ($\ce{KI}$), and the second one is an iodometric titration.

First, ozone is reacted with excess $\ce{KI}$ in presence of borate buffer ($\mathrm{pH} =9.2$), and due to this reaction, iodine ($\ce{I_2}$) and oxygen ($\ce{O_2}$) are liberated as follows, $$\ce{O_3 + 2KI + H_2O -> I_2(^) + O_2(^) + 2KOH}$$

Now this mixture of gases is passed through potassium pyrogallete solution, which absorbs $\ce{O_2}$, and we are left with $\ce{I_2}$. Next, this liberated iodine is dissolved in a solution to make a particular volume, and then titrated with sodium thiosulfate ($\ce{Na_2S_2O_3)}$ of known strength in the presence of starch indicator.

$$\ce{I_2 +_ 2Na_2S_2O_3 -> Na_2S_4O_6 (sodium tetrathionate) + 2NaI}$$

At the end point of this titration, colour of the solution turns colourless from blue ($\ce{I_2}$ + starch gives blue colour complex  which is lost at the end of this reaction).

So, thiosulfate reacts with iodine in a quantitative manner, and from there you can determine equivalents of $\ce{I_2}$ = equivalents of $\ce{Na_2S_2O_3}$. Thus, you can calculate what equivalents of ozone had produced that much of $\ce{I_2}$, and also the grams of ozone present.


Why are double bonds shorter than single bonds?

In general, in a single bond, there is involvement of sigma bonds only. In double bonds, there is also overlap between pi orbitals. More is the overlap shorter will be the distance between the atoms.


Similar to inox for inoxydable (inoxidable), it equally is a French coinage.  To tune mechanical properties of iron, mixing additives leads to numerous alloys of iron.  In case of rene or rené as contraction of résistant (durable) and on base of Nickel, and is example of the superalloys.

This source technologie des métaux et alliages - particulièrement en aéronautique (technology of metals and alloys, especially in aviation industry) by Dominique Ottello (2003), section 4.6.1.1 "Les alliages de nickel: / the alloys of nickel" on page 15 states:

  Ils  conviennent pour les trois types de sollicitations : MÉCANIQUES, THERMIQUES et CHIMIQUES (très grande résistance à la fatigue, au fluage et à  la corrosion à chaud). Il existe plusieurs familles d'alliages de NICKEL :

  Les alliages NICKEL + CUIVRE : alliages "MONEL" qui présentent une bonne résistance à la corrosion chimique, utilisables jusqu'à 500 °C.
  Les alliages NICKEL + CHROME + FER : alliages "INCONEL" qui sont très tenaces et utilisables jusqu'à 700 à 800 °C.
  Les alliages NICKEL + CHROME + COBALT : alliages "NIMONIC" et alliages
  "RENE". Ils sont très tenaces et présentent une bonne résistance au FLUAGE, utilisables jusqu'à 950 °C.

  NOTA : Dans les alliages de NICKEL à haute température, on rencontre :

  Des alliages plus spécialement réservés à la FONDERIE.
  Des alliages pour CORROYAGE.
  Des alliages à durcissement structural par traitements thermiques.

For our non-Francophon readers:

  These [Nickel-derived alloys] are suitable for three fields of application / strain: mechanical, thermic, and chemical (very high endurance against fatigue, compression, and corrosion at high temperatures).  There are several groups of nickel-based alloys

  alloys based on nickel and copper:  "monel" alloy, with a good resistance against chemical corrosion, serviceable up to $\pu{500 ^\circ{}C}$ [i.e., $\pu{932 F}$]
  alloys based on nickel, chrome, iron:  "inconel"alloy, tenacious and deployable up to $700$ or $\pu{800 ^\circ{}C}$ [i.e., $1292$ to $\pu{1472 F}$].
  alloys based on nickel, chrome, and cobalt: "nimonic" and "rene" alloys. These are very tenacious and exhibit a good resistance against traction, deployable up to $\pu{950 ^\circ{}C}$ [i.e., $\pu{1742 F}$].

  Note: One finds, among the nickel-based alloys with high melting point:

  alloys foreseen / dedicated for deployment in foundries / casting houses
  alloys for wrought (e.g., forging, rolling, etc.)
  alloys for hardening by thermal treatment

Voilà !


Why is the van der Waals coefficient b equal to four times the volume of the particle?

While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically.
In deriving the VDW equation, the particles are still assumed to be hard spheres, but this assumption is corrected for with the parameter $a$.
The hard sphere approximation forbids that two particles penetrate each other's radii. Thus, we find that two spheres in closest contact are surrounded by a sphere of radius $2r$ (or the diameter of one of the original spheres).

(source: nyu.edu)
Thus, the volume excluded by the particles from the larger sphere surrounding the two spheres shown is
$$b' = \frac{4}{3} \pi d^3 = 8 \cdot \frac{4}{3} \pi r^3$$
Thus, the excluded volume per particle $b$ is $b'/2$ or,
$$b=4 \cdot \frac{4}{3}\pi r^3$$
which, as you point out, is four times the volume of a single particle.
The interesting thing about this is that it does not represent the actual value of $b$ for any given atom, but represents the upper bound of $b$ for any given atom. What I mean by that is, $b$ could very well be correct by calculating four times the volume, but often experiment will show that it less than the calculated value of $b$ because atoms are not hard spheres.
For instance, using helium, which is the closest we'll get to a hard sphere:
$$b_{\ce{He},\mathrm{calc}} = 4 \cdot \frac{4}{3} \pi (140\ \mathrm{pm})^3 \cdot N_\mathrm{A} = 0.02767\ \mathrm{L\ mol^{-1}}$$
while,
$$b_{\ce{He},\mathrm{exp}} = 0.0238\ \mathrm{L\ mol^{-1}}$$
So, the experimental value of $b$ is indeed smaller, but the calculated value gives a rough idea.


There are 12 different constitutional isomers:

There is also the further possibility of stereoisomerism in some of the compounds above.

3: Two diastereomers (cis- and trans-) are possible.
7, 8, 9: Geometrical isomerism is possible in these alkenes.
11: C-3 (the carbon with the chlorine) is a chiral centre, which leads to two possible enantiomers.

That makes a total of 17 possible isomers.


Why does permanganate oxidation of 2-nitronaphthalene not form phthalic acid?

Looking at the substituents; NH2- activates the ring, making it more reactive than dibenzene ring system (easily oxidised), and the NO2- substituent is very deactivating - (strongly electron withdrawing), making it less reactive (not easily oxidised).

Although  a powerful oxidising, optimised oxidising agent (KMnO4) is being used, it is only able to oxidise the 2-naphthylamine compound to produce phthalic acid.

and 2-nitronaphthalene is very unlikely or at least takes longer to completion with very little yield.


Why is zinc sulfate used in daniell cell

Zinc sulfate is not needed. With any other neutral salt in its place, your cell will work just fine.

Without any salt at all, though, it will not work, since pure water is a poor conductor of electric current.


Here are some things to focus on:

How quickly\slowly did you add the diazonium salt to the NaI solution?
Did you maintain the reaction temperature below 10°C during the entire addition?
Did you also keep the diazonium salt cooled while adding it?
After addition was complete, did you let the reaction stir at 10°C until nitrogen evolution ceased and then let the reaction slowly warm to room temp?

Here is a tested lab procedure for the preparation of 4-iodonitrobenzene.  Compare it to your notes and see where you differed.

Keeping the reaction cold during the addition is key. If the diazonium salt warms up too much, the diazonium can react with itself (coupling) to give colored by-products.  The diazonium ion can also eliminate nitrogen to generate the extremely reactive phenyl cation which will react with any nucleophile around, water for example (see page 6, the Mechanism section in this reference).  Both pathways can  contaminate and reduce the yield of the desired product.


How many watts would I need to get CO2 from 30 °C to 4000 °C in 1 hour

I approached this in a similar way utilizing the heat capacity of $\ce{CO2}$ over a temperature range.

An estimation can be made by approximating the change in enthalpy of $\ce{CO2(g)}$ over a given temperature range. This can be accomplished through the integral form for changes in Enthalpy:

$$\Delta{H} = \bar{H_1} - \bar{H_2} = \int_{T_1}^{T_2}C_p(T)\,\mathrm dT$$

Which has units of [energy/mass].

Often times, the function $C_p(T)$ for certain gases and liquids will be tabulated in textbooks or chemical databases (such as NIST). In this case, I will utilize a function for the $C_p(T)$ of $\ce{CO2}$ in two temperature ranges (300–1200 K, 1200–6000 K)$^1$:

$$C_p(T)=A+BT+CT^2+DT^3+\frac{E}{T^2}$$

Range 1 (300–1200 K) coefficients:

$A=24.997,$ $B=55.187,$ $C=-33.691,$ $D=7.948,$ $E=-0.137$

Range 2 (1200–6000 K) coefficients:

$A=58.166,$ $B=2.720,$ $C=-0.492,$ $D=0.039,$ $E=-6.447$

Next, we can obtain the energy required for this temperature change by multiplying $\Delta{H}$ by the mass of $\ce{CO2}$ we're interested in, such that:

$$Q = m_\ce{CO2}\,\Delta{H} =  m_\ce{CO2}\int_{T_\text{initial}}^{T_\text{final}}C_p(T)\,\mathrm dT = m_\ce{CO2}\left[\int_{T_1}^{T_2}C_p(T)\,\mathrm dT + \int_{T_2}^{T_3}C_p(T)\,\mathrm dT\right]$$

Evaluating the integrals (keeping in mind that A, B, C, and D will change based on the temperature range in this case) yields:

$$m_\mathrm{CO2}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}}$$

$$\frac{100}{0.04401}\ \mathrm{\frac{kg}{\frac{kg}{mol}}}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}} = 1.651\times10^{16}\ \mathrm J$$

In watts, this value becomes:

$$\left(1.651\times10^{16}\ \mathrm J\right) \left(\frac{1}{3600}\ \mathrm{\frac{h}{s}}\right) = 4.560\times10^{12}\ \mathrm W$$

Calculating the solar-panned area required:

$$A_\text{panels} = \frac{4.560\times10^{12}}{200}\ \mathrm{\frac{W}{\frac{W}{m^2}}} = 2\times10^{10}\ \mathrm{m^2}$$

Source [1]: “Carbon Dioxide.” Carbon Dioxide, National Institute of Standards and Technology, webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Mask=1#Thermo-Gas.


Solutions of Group 1 and Group 2 metals in Ammonia

As per P. Bahadur's Objective Chemistry 1st Year Program for IIT-JEE and All Other Engineering Entrance Examinations, under the chapter "s-Block Elements - The Alkaline Earth Metals", under Solubility in Liquid Ammonia, it reads:

  Like Alkali Metals, alkaline earth metals also dissolve in liquid ammonia and form coloured solutions. When such a solution is evaporated, hexa-ammoniate, $\ce{M(NH3)6}$ is formed, which slowly decomposes to amides.
  $$\ce{M(NH3)6 -> M(NH2)2 + 4NH3 +H2}$$

(I noticed that it is almost exactly the same as JD Lee's Concise Inorganic Chemistry text.)

So far, it doesn't specify whether this metal $\ce{M}$ is only Beryllium or is the representation of the other alkaline earth metals as well... Since this is the case, I can safely assume that this applies to all the alkaline earth metals, and not just beryllium.

Another point mentioned in the book by Bahadur, under the chapter "s-Block Elements - The Alkali Metals", under Complex Ion Formation is that the Group 1 metals, i.e., the alkali earth metals have a larger size, low nuclear charge and hence don't tend to form complexes too easily. On the other hand, the Alkaline Earth metals are smaller, have higher nuclear charge, and hence, have a greater tendency to accept electrons, forming complexes. In Bahadur's Book, under "s-Block Elements - The Alkaline Earth Metals", under Formation of Complexes, he writes that:

  However, $\ce{Be^{2+}}$ on account of smaller size forms many complexes such as $\ce{BeF3-}$, $\ce{BeF4^{2-}}$ and $\ce{Be(H2O)4^{2+}}$. Of the others, only $\ce{Mg}$ and $\ce{Ca}$ show much tendency to form complexes in solution and these are usually with oxygen donor ligands.

$\ce{NH3}$ is a ligand, and hence acts as a powerful complexing agent. This is why the $\ce{NH3}$ coordinates with these metals, forming hexa-ammoniates.

Source:

Bahadur, P.; Objective Chemistry 1st Year Programme for IIT-JEE & All Other Engineering Entrance Examinations; GRB Books; 2016
Lee, J.D.; Concise Inorganic Chemistry; Wiley-Blackwell; Dec. 1998


Chemical equation for the reaction of Allura Red AC with NaClO

The two molecules created by breaking the $\ce{N=N}$ bond have something wrong. They both have a final bond $\ce{-N-OH}$. This is impossible. It may be at least $\ce{-NH-OH}$. Maybe the $\ce{N2}$ group is removed, and a phenol or naphthol group is formed.


As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The principal characteristic group is the carboxylic acid. Therefore, the lowest locant is assigned to the $\ce{-COOH}$ group in accordance with Rule (c).

Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without any further substituents is cyclopent-2-ene-1-carboxylic acid.

Finally, a low locant is assgined to the remaining substituent to Rule (f).

Thus, the correct name is 4-(acetyloxy)cyclopent-2-ene-1-carboxylic acid.


NaCl electrolysis with copper electrodes

Electrolysis of sodium chloride solution produces results at each electrode. At the cathode, water is reduced to $\ce{OH-}$ and makes the solution more alkaline. Copper hydroxide is a blue, bulky precipitate, feebly acidic, and soluble in concentrated alkali hydroxide solutions. So if you got copper ions into solution at the anode, you could form some cupric hydroxide which is reported to form blue polymeric species in solution [1].
Crystals from concentration of this solution will likely be $\ce{NaCl}$ contaminated with some $\ce{CuCl2}$. Examination with a microscope could show whether $\ce{CuCl2}$ has formed, as the blue-green rhombic dihydrate (because, after all, it is in water), or, if you actually did enough electrolysis to do more than color the water, whether you precipitated the yellow-green hexagonal basic salt $\ce{CuCl2.Cu(OH)2}$ or the hydroxide or some mixture [2].
Although cuprous chloride is highly insoluble in water, it does form complexes in concentrated chloride solutions: the main species in $1\rm{M}$ $\ce{KCl}$ is
$\ce{CuCl2^-}$ [1]. So you really have a complex situation with anode/cathode, $\ce{OH-}$/$\ce{Cl2}$, $\ce{Cu^0}$/$\ce{Cu+}$/$\ce{Cu^{+2}}$, $\ce{CuCl2}\cdot2\ce{H2O}$/$\ce{Cu(OH)2}$, and polymeric complexes with $\ce{OH-}$ and $\ce{Cl-}$, and various combinations of products depending on how long you do the electrolysis and, of course, whether you emit $\ce{O2}$ or $\ce{Cl2}$.
Of course, that's what makes chemistry so much fun. There are many variables, some of which you can control.
[1]: Advanced Inorganic Chemistry, Cotton and Wilkinson, 1966
[2]: CRC Handbook, 62$^{\rm nd}$ ed.


Formula of the compound based on occupancy of lattice voids

Since B forms an HCP lattice, it's z=6 (z is the number of atoms power unit cell of the lattice) then the ratio comes out to be 8:6 that gives us A4B3.


The only substance that absorbs $\ce{O2}$ to form an oxide at low temperature and release it at higher temperature is mercury. This experience was done by Lavoisier, Scheele and Priestley in the years around $1780$. They were able to produce pure oxygen $\ce{O2}$ by first heating pure mercury in air at about $300$°C to slowly produce mercury oxide $\ce{HgO}$, and then overheating this oxide $\ce{HgO}$ to about $400$°C. At this temperature, $\ce{HgO}$ is decomposed into mercury and pure oxygen $\ce{O2}$. This was the way oxygen was first discovered and produced.
Unfortunately, mercury ¡s unique in this respect, and cannot be replaced by another element. No metals have been found later on to have a similar property. It is a pity : presently this experiment cannot be repeated, because of the toxicity of mercury and its compounds.
This why your dream will remain at the state of dream.


Distinction between whole molecules and functional groups in IR spectra

Your question contains several points to address here.   And following Martin's suggestion, it is time to wrap up the points already mentioned in comments into a conclusive answer, too.

"IR spectrum graphs"-- there is nothing such.  The graph (or, the figure) represents the IR spectrum.  (And, for future reference, should there be need to deploy the plural form, it is "spectra".)
"Are IR spectra specific to compounds or to functional groups?"  If you record IR spectra in the typical interval of 4000 to 600 (or, depending on the window materials built in, about 500) 1/cm, then your IR spectrum will contain both: absorption bands that are characteristic for functional groups, like the classic ca. 1600 1/cm for carbonyls, or the ca. 2270 1/cm for the nitrile; AND between about 1400 to 500 1/cm what is called "fingerprint region" which contains absorption pattern that are characterisitic for the very molecule.
Attribution which absorption band in the fingerprint region belongs to what structural feature may be complicated enough that looking up a reference cataloge (by Sadtler, Aldrich, or other corporations) does not help, yet thankfully, modern computer chemical methods often allow a prediction of what vibrations may occur and where these should be observable in the IR spectrum. To render this fingerprint region more legible, often the $\tilde\nu$-scale is depicted with changes the scaling, like at 2200 and 1100 1/cm.
Which of the four compounds is more likely to be the one for the spectrum depicted?
 Well at first sight, 3 (1,4-dioxane) differs by the absence of a carbonyl group, typically yielding an intense absorption band around 1600 1/cm.  However, such an absorption band is recorded -- so it is either acetophenone (1), cyclohexanone (2), or $\delta$-valerolactone (4).
 There is not so much absorption in the region of 500--600 1/cm that were typical for aromatic rings, rendering 1 unlikely.
 Now, facing the distinction between either cyclohexanon 2, or the ester 4, it is indeed helpful that these compounds are literature known, already characterized by IR spectroscopy, and -- thankfully -- their data are made public; for example by AIST.  And as it turns out, it is likely cyclohexanone 2 which's spectrum was depicted here.

As a comment to the spectra by AIST:  They are searchable by chemical name, formula, or CAS number.  In the case of IR spectra, a little table ($\tilde\nu$), observed remnant transmittance) and a small image of the molecular structure is provided, too.

The excluded acetophenone 1:

The likely cyclohexanone 2:

The early exluded 1,4-dioxane 3:

The eventually excluded cyclic ester (lactone) 4:

So in turn, deciphering IR spectra is both learning / training to spot features (here: absorption bands), and comparison with reference data (here: spectra cataloges).  And typically supported by other means of analysis or characterization, like melting points / refractive indices compared with tables, or further NMR spectroscopy, for example.


Magnesium oxide dissolving in distilled water

All of that is true, these facts do not contradict each other.

The missed part is quantification. Salts and generally ion compounds are soluble, but solubility comes in range of many orders of magnitude.

Also, some compounds, even if largely ionic, react with water.

So magnesium oxide reacts with water forming magnesium hydroxide, both being almost insoluble in water.

Oxide anion  as a very strong Lewis base immediatelly reacts with water $\ce{O^2- + H2O -> 2 OH-}$. So the only chance for metal oxides to stay being oxides is to be insoluble, with their lattice energy too high to be broken by water hydration of ions.


Which mechanics or law is predominately seen on the nanoscale?

Its depence on the size of the system your are looking at. To understand spectroscopy of molecules, you usually use quantum mechanics. In molecular dynamics simulation, you'll need both: On the one hand, there is a potential needed based on QM-interactions, on the other hand, the classical equations of motion are good enought to move your atoms in a QM-force field.

Based on your problem, you have to decide how much classical mechanics you can justify. But in the nanoscale everything is made up with quantum mechanics and classical mechanics are used to approach a problem, which is not solvable in quantum mechanic way in an accetable periode of time.


How much water would one loose if one burns hydrogen and oxygen?

Theoretically, none.

Actually, since the gases are produced in a damp environment, there would be a lot of water vapor in the gases. You could easily remove most of the water vapor, but some (maybe as much as 1%) would still be in the gases.

I think it would be best to elaborate on what you mean as water being lost?


Would tin (II) sulfide be considered a covalent network solid?

There is no sharp border between ionic and covalent compounds, much like there is no sharp border between a boy and a man. Some compounds between a metal and a non-metal, like $\ce{LiF}$ or $\ce{NaCl}$, are predominantly ionic, some (think of $\ce{Mn2O7}$ or $\ce{TiCl4}$) have markedly covalent character, and the rest are somewhere in between. As you see yourself, $\ce{SnS}$ is ionic enough to have certain kinds of reactions typical of ionic compounds, but not so ionic in terms of electronegativity or crystal structure.

Now, people in some areas of chemistry would put forth some definitions to tell covalent from ionic. Your "difference of electronegativities of at least 1.7" is one of those. Another is based on $\ce{MgO}$ and would use it as a border. There are a lot of definitions, and it's not like they are all wrong or useless; in fact, they are necessary to keep our stuff in order. One has to remember, though, that they are all slightly different, and all inherently arbitrary (much like the legal age limitations for driving, voting, drinking, and other activities, if we'd talk once more in terms of the "boy or man" analogy).

This must have been said here like a dozen times before, but whatever. Repetition is the mother of learning, they say.


Which substance can be easily adsorbed by charcoal?

Gas molecules bind to activated carbon as a result of London dispersion forces. Therefore, the larger the molecules, as determined by the size and the number of atoms in it, the better they bind to the carbon.

If you were to rank these in order of highest dispersion forces to lowest, that list would read: $$\ce{SO2>CO2>H2>He}$$


There are a couple of ways to think about this. First, considering your atom economy equation. Typically a catalyst is used in place of a stoichiometric reagent. A catalyst is (strictly speaking) not a reagent, so its effect on the atom economy is neutral. The alternative stoichiometric reagent is a reactant, so it will reduce the atom economy. This leads to the conclusion that a catalytic method offers a better atom economy than an alternative, non-catalytic method.

From a slightly different point of view, atom economy is one aspect of measuring the amount of waste that comes out of a reaction. A more relevant quantity is really the ratio of the mass of the desired product(s) to the total mass of all of the reaction inputs (including reactants, catalysts, solvents, purification materials, etc). Since catalysts can be used in smaller amounts than stoichiometric reagents, there will physically be less mass put into the process, which increases that ratio.


So, why does simple distillation put emphasis on small amounts and
  fractional distillation, on large volumes?

It's not small and large volumes of material being distilled, but rather small and large volumes of impurities present in the material to be distilled.  Both simple and fractional distillations can be used to distill large quantities of material.

The efficiency of a distillation column is measured in "height-equivalent theoretical plates" (HETP), the more plates the more efficient the column, the more efficient the column the more likely that it will be able to separate two closely boiling materials.  If you have a small amount of a lower boiling impurity in your solution and you want to separate it from a large amount of the desired material, then you can probably get away with using a simple distillation column and discarding some of the forerun (the first drops of liquid that distill over).  On the other hand, if there is a large amount of impurity and it boils close to the desired product, then you will need a column with a higher efficiency (fractional distillation column) to effect a good separation between the two fractions (or "cuts" as they're sometimes called) and get pure product.

To sum up, perhaps the original question would be better written as

  So, why does simple distillation put emphasis on small amounts and
  fractional distillation, on large amounts of impurities?

and the answer would be that even an inefficient (fewer HETP) simple distillation column has enough power (enough theoretical plates) to separate a small amount of impurity from a large amount of product, especially if you are willing to discard the initial fore-run from the distillation.  If there is a larger amount of impurity, then we may need a more powerful (higher efficiency column) to produce the desired separation.

The higher the efficiency of a distillation column, the greater its ability to separate two things.  If we have to separate two materials that are both present in large quantity, or distill at boiling points close together, then we need a more efficient column than if they boiled further apart or one was present in only a small amount and we were willing to throw away the forerun (e.g. throw away some of the desired product in that early fraction that contained the impurity).  When we call a column a fractionation column, we're really saying that it has greater separating efficiency than a simple distillation column.  The more efficient the column, the better the job it will do on difficult separations.


Can solutions of polar covalent compounds conduct electricity?

Consider for example this lab where polar covalent bonds are said to be strong, weak or non-electrolytes : ELECTROLYTES AND NONELECTROLYTES CHEM 151

The part of the molecule which dissociates has strongly ionic character but the molecule itself is described as being polar covalent. Not not all bonds in the molecular have the same level of covalent character, but most are covalent and it is polar.

There are also other ways for molecules to allow conduction, conjugated systems can do it by a very different process see for example the wiki on Conductive Polymers


Your question has no sense – sorry.

You talk about a few small molecules (with coinage metals) but ask about lattice type for atoms. Atoms in molecular crystal have no their own lattice. Lattice describes how your molecule are aligned in 3D-space.

Assuming that you have a single crystal data then in the beginning of experiment your X-minion (that's the person that took X-ray data for you) has already found the cell parameters and then usually also Bravais cell. To be sure that selected lattice was right whole data set need to be collected, the data have to be integrated and then using Lattice exceptions (see International Tables for Crystalography Volume A,  2.2.13. Reflection conditions) you can tell what lattice type you have. If you do the same for glide planes and screw axis exceptions then eventually you will find space group.


How to calculate number of each type of atom in a residue in PyMOL?

Not super elegant, but this should work:

select resn dex and symbol C

Pymol will respond,

Selector: selection "sele" defined with 22 atoms.

Repeat for H, F, O.


There are a number of techniques but what you would use depends on how you plan to start the reaction. What use to be called Flash-Photolysis and now called Pump-Probe Spectroscopy is the most general and can be used from the femtosecond to multi-second time range. This is a general method as it measured the absorption of all species produced. It works by using a suitable duration and wavelength flash of light, usually from a laser, to initiate the reaction and then at various time delays after this,  the absorption spectrum of any species produced is measured. Absorption can be in the uv, visible or ir ranges. Sometimes Raman scattering or fluorescence can be used instead of absorption. What is used depends on equipment available and the species you need to measure.

If you cannot start the reaction with light then mixing methods can be used, but these are relatively slow, longer than microsecond time scale and more like milliseconds. In the Stopped-Flow method the reactant gases   are mixed as rapidly as possible in the chamber so as to start the reaction then this is monitored with absorption etc. as above. Additionally, there are flow methods whereby the gaseous reactants are mixed at the end of a long flow tube and are monitored spectroscopically at various points as they flow along a tube. The gases are pumped to cause the flow and so distance along the tube is an alias for reaction time.  Each of these methods can be used with gases or liquids.


Synthesing tetrahydropyran from ethylene oxide

Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM.

Make the Grignard of the THP-protected bromopropanol, react this with ethylene oxide. Work up and isolate to give the mono-THP-pentane-1,5-diol.

Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol.

Remove the THP group using cat. PTSA in MeOH

Form the alkoxide using LiHMDS in Et2O and cyclisation will occur to give the pyran. Isolate by careful distillation.


Electrolysis in aqueous solution – which equations to use to predict product at each electrode?

The given potentials $E = 0.00{\text{ V}}$ and $E = -0.83{\text{ V}}$ for the reduction of water apply to different $\text{pH}$.

The thermodynamic relation of the potential $E$ to the composition of the solution is generally known as the Nernst equation:

$$E = E^\circ  + \frac{{0.059{\text{ V}}}}{z}\log \frac{{{{\prod\nolimits_i {\left[ {{\text{ox}}} \right]} }^{{n_i}}}}}{{{{\prod\nolimits_j {\left[ {{\text{red}}} \right]} }^{{n_j}}}}}$$

For the given reaction

$$\ce{2 H+ + 2e- <=> H2}$$

the Nernst equation reads

$$E = 0.00{\text{ V}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{{{{\left[ {{{\text{H}}^ + }} \right]}^2}}}{{\left[{{{\text{H}}_2}}\right]}}$$

Since

$${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$$

the potential $E$ depends on $\text{pH}$:

$$E = 0.00{\text{ V}} - 0.059{\text{ V}} \times {\text{pH}} + \frac{{0.059{\text{ V}}}}{2}\log \frac{1}{{\left[ {{{\text{H}}_2}} \right]}}$$

Therefore, the potential is $E = 0.00{\text{ V}}$ at $\text{pH} = 0$ and $E = -0.83{\text{ V}}$ at $\text{pH} = 14$ respectively.


Why does color of halogen deepens and the boiling point decreases down the group?

As for the BP:
The attraction of covalent molecules is due 'dispersion' interaction. It is a quite complex effect, commonly described as interaction of 'instantaneous dipoles', i.e. interaction of dipoles formed by nucleus and electrons around it. Its efficiency generally increases with number of electrons (due to sheer numbers) and size of electron shell. Since outer electron shells vastly grow in size in fluorine-iodine row, the efficiency of dispersion interactions grow. In fact, close halogen-halogen contacts for heavier halogens (bromine and iodine) are very common in crystal structures of organic compounds.
As for color. As a guess, I'd say that the active transition is lone pair orbital $\rightarrow$ $\sigma$-antibonding orbital (see any introduction to MO theory for the terminology). The energy of this transition should lower from fluorine to iodine because of great decrease in bonding energy and increase of size of atoms. According to color wheel, the absorption in violet spectrum (with highest energy of photons in visible spectrum) should give yellowish apparent color (case of fluorine and chlorine), red apparent color to adsorption in lower-energy and violet color for adsorption at even lower-energy red spectrum.


Does the chemical in an alkaline battery make battery leaks unavoidable?

The reason for leaks in alkaline batteries is that as batteries discharge — either through usage or gradual self-discharge — the chemistry of the cells changes and some hydrogen gas is generated. This out-gassing increases pressure in the battery. Eventually, the excess pressure either ruptures the insulating seals at the end of the battery, or the outer metal canister, or both. In addition, as the battery ages, its steel outer canister may gradually corrode (in alkaline batteries there is some potassium hydroxide, a very strong base) or rust, which can further contribute to containment failure.
(see http://en.wikipedia.org/wiki/Alkaline_battery#Leaks )

I also found this online :

Heat is the biggest culprit. Storing your flashlights under the seat of your car, or in direct sunlight will cause the batteries to leak. The reason they leak is because they are filled with a chemical paste of which the liquid portion is composed of a base called Potassium Hydroxide. When the batteries get hot, the paste inside expands, and forces chemicals out of the seals in the bottom and top of the battery.
[...]
Regardless of the brand you buy,[...] all batteries will leak eventually, so if you don't use the flashlight for extended periods of 3 months or more, remove the batteries, or change them out every couple months.
(from here)


How can you find the concentration of copper(II) ions in a solution of 0.5 mole copper(II) chloride and 0.5 mole of copper(II) sulfate?

1 mole of copper(II) sulphate contains 1 mole of copper ($\ce{Cu^{2+}}$) and one mole of sulphate ($\ce{SO_4^{2-}}$). Or, if you want, 1 mole of copper, 1 mole sulphur, and 4 moles of oxygen.

This is by the definition of the mole as the unit of measurement for the amount of substance.

From this it follows that 0.5 moles of copper(II) sulphate contain 0.5 moles of copper. Similarly 0.5 moles of copper(II) chloride also contain 0.5 moles of copper. Together these are 1 mole of copper.

1 mole of copper ($\ce{Cu^{2+}}$) in $\mathrm{0.5\,l}$ water gives us a concentration of $\mathrm{2\, mol\,l^{-1}}$


Op's question is what is the correct $\mathrm{p}K_\mathrm{a}\ce{H}$ value of pyrrole. Thus, I'm not going to elaborate OP's findings, but would try to give a reasonable answer to the question. The most reasonable answer I found for $\mathrm{p}K_\mathrm{a}\ce{H}$ value for the pyrrole is $0.4$ (for the novices, this is the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of pyrrole).

It is well known that the best way to quantify the basicity of an amine is to examine the $\mathrm{p}K_\mathrm{a}$ of its conjugate acid (Master Organic Chemistry). The concept is the higher the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid, the stronger the base.

Let's compare two cyclic amine bases under this concept: Pyridine and Pyrrole. The following diagram illustrate the concept:

Accordingly, pyrrolium ion is about $10^5$ times stronger acid than pyridinium ion. Thus, pyrrole is about $10^5$ times weaker base than pyridine. Why? Read the note on the bottom of the box in the diagram. This is also well explained in chemistry.msu.edu with the diagram attached:

  In each case the heteroatom has at least one pair of non-bonding electrons that may combine with the four π-electrons of the double bonds to produce an annulene having an aromatic sextet of electrons. This is illustrated by the resonance description at the top of the following diagram. The heteroatom Y becomes $\mathrm{sp^2}$-hybridized and acquires a positive charge as its electron pair is delocalized around the ring. An easily observed consequence of this delocalization is a change in dipole moment compared with the analogous saturated heterocycles, which all have strong dipoles with the heteroatom at the negative end. As expected, the aromatic heterocycles have much smaller dipole moments, or in the case of pyrrole a large dipole in the opposite direction. An important characteristic of aromaticity is enhanced thermodynamic stability, and this is usually demonstrated by relative heats of hydrogenation or heats of combustion measurements. By this standard, the three aromatic heterocycles under examination are stabilized, but to a lesser degree than benzene. Additional evidence for the aromatic character of pyrrole is found in its exceptionally weak basicity ($\mathrm{p}K_\mathrm{a}$ ca. $0$) and strong acidity ($\mathrm{p}K_\mathrm{a} = 15$) for a $2^\circ$-amine. The corresponding values for the saturated amine pyrrolidine are: basicity $11.2$ and acidity $32$.

Furthermore, for everybody's convenience, I attached a orbital representation of pyrrole and furan to show their resistance to protonation.

Comment on different $\mathrm{p}K_\mathrm{a}$ for pyrrole in literature: The $\mathrm{p}K_\mathrm{a}$ value of $3.8$ (and $\approx 4$) is probably for the protonation of pyrrole at $\ce{C2}$ position. For protonation at $\ce{N}$, $\mathrm{p}K_\mathrm{a}$ is most probably $0.4$, as described above. We can trust the value given in McMurray's Organic Chemistry textbook, since it is widely used in sophomore organic chemistry university course in USA.


How to recycle olive oil manufacturing wastes?

The components of the 80% waste material could definitely be recycled using common procedures ( eg distillation, solvent extraction ) and could be used for many purposes like you mentioned or by converting the waste material to a more useful product.

I'm not sure what they make from the waste from olive oil production exactly but a good example of similar biotechnology is the use of waste cooking oils to synthesize biodiesel.


Your analysis of the expectation is correct, and we can look at it more quantitatively in this way:
Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of:

$$-\frac{d[\ce{CV}]}{dt}=k[\ce{HO-}][\ce{CV}]\approx k[\ce{HO-}]_{initial}[\ce{CV}]$$

For simplicity, we'll define a $k'$ as $k'=k[\ce{HO-}]_{initial}$ and treat the reaction as pseudo-first-order.

The half-life of a first-order reaction is given by $t_{1/2}\approx \frac{0.7}{k}$, so for the concentrated reaction we have $t_{1/2}=\frac{0.7}{k'}$.

For the 1:4 reaction, the initial hydroxide concentration is halved, so we have a pseudo-first-order rate constant of $0.5k'$ and $t_{1/2}=\frac{0.7}{0.5k'}$ or 2 times the half-life of the concentrated reaction.

Likewise, the half-life of the 1:8 reaction is twice that of the 1:4 reaction and four times that of the 1:2 reaction, so it's $t_{1/2}=\frac{0.7}{0.25k'}$.

Let's say that the concentrated reaction needs 6 half-lives for the concentration to get to the point where the color disappears. Since the 1:4 reaction starts at half the initial CV concentration, it only needs 5 half-lives. But the half-lives are each twice as long, so it should take a time equivalent to 10 half-lives of the concentrated reaction, so 10/6=1.7 times as long.

For the 1:8 reaction, the same analysis says that it will take 4 half-lives, but each is 4 times as long as that of the 1:2 reaction, so it takes 16 equivalent half lives, meaning 16/6=2.7 times as long.

The only instance in which the more dilute reaction will be faster is if the concentrated reaction requires less than two half-lives to be observed to be complete. That can't be the case here or there would be no color at the start of the 1:8 reaction.

So we must look for an alternative explanation for your observations. 0.5 M NaOH is a very concentrated solution, and it is possible that other reactions are occurring besides the intended one. To test that, I would follow Maurice's suggestion and run the reaction at a wider range of base concentrations. It would also be preferable to have the same concentration of CV in every reaction, so only one variable is involved. To do this, used a fixed volume of (NaOH + water) in each reaction.

UPDATE: This paper1 reports doing this reaction at much lower hydroxide ion concentrations and also notes that the reaction rate decreases with increasing ionic strength, which could be at least a partial explanation for why your observed rate at high concentrations is lower than expected.

1 Felix, L.D. Kinetic Study of the Discoloration of Crystal Violet Dye in Sodium Hydroxide Medium. J Chem Appl Chem Eng Vol: 2 Issue: 1


Electron tunneling
Tunneling refers to elementary particles moving "through" a potential barrier in the absence of sufficient energy to go "over the barrier". This is roughly equivalent to saying that a particle can occupy two locations and traverse between them even though all paths between those locations have a zero probability of the particle being there.
Some orbitals such as the p-orbital describe a probability density with areas of non-zero probability ("lobes") separated by space with zero probability ("nodes").
When electrons are transferred from one electron carrier to another, the transfer can occur through bonds or through space; the through-space transfer is either based on orbital overlap or on tunneling.
Proton tunneling
Proton tunneling is much slower because the proton is about 2000-times heavier than the electron. Tunneling has been used to explain reactions involving protons that are faster than expected from inspecting the potential energy landscape. This proton tunneling also explains proton transfer in hydrogen bonds.
For formic acid and acetic acid, proton tunneling explains cis-trans isomerism at ultra-low temperatures. Acetic acid, at room temperature, has a cis and a trans conformation.

There is a rotational barrier (which one may explain via resonance involving the carbonyl double bond) which is lower than thermal energy at room temperature [Ref 1]. For related systems like the peptide bond, cis and trans forms are isomers because the rotation barrier is higher.
At ultra-low temperature in a nobel gas matrix, the trans form is dominant and does not convert into the cis form unless you excited the molecule at the right wavelength. This means that the cis and the trans form are isomers with individual properties. Under these conditions, it is possible to enrich the cis form, and to observe the conversion to the trans form in the dark [Ref 2]. The temperature-dependence of the kinetics (and kinetic isotope effects) suggest that this cis-to-trans reaction occurs via tunneling.
Hydride tunneling in enzymes
Many redox reactions involve hydride transfer, e.g. from alcohol to NAD+ in the alcohol dehydrogenase catalyzed reaction. For enzymes, it is typically not possible to measure reactions at ultra-low temperatures; instead, a key experiment is to study the primary and secondary kinetic isotope effect [Ref 3]. From studies of multiple enzymes that catalyze hydride transfer it is evident that tunneling does occur, and the extent of tunneling is modulated by mutations in the active site. The review makes the argument that tunneling is easier to study for hydrogen (i.e. proton, hydrogen radical and hydride) transfer than for electron transfer because in the former case, isotopes (with a different deBroglie wavelength) are available as a probe. While the examples in the review are all redox reactions rather than E/Z isomerization reactions, one could envision the latter based on hydride tunneling as well.
"Double" tunneling
If there are two atoms connected to a double bond (such as in the molecule $\ce{H2C=CDH}$ mentioned in the comments), isomerization through tunneling would be much harder because the hydrogen and the deuterium have to somehow switch position. To avoid a clash, they would have to tunnel simultaneously, and could not both take a path along the saddle point of the potential energy landscape. Therefor, a cis-trans isomerization of $\ce{H2C=CDH}$ seems much less feasible, and there does not seem to be any experimental evidence for it.
References

Barriers to rotation adjacent to double bonds. 3. The carbon-oxygen barrier in formic acid, methyl formate, acetic acid, and methyl acetate. The origin of ester and amide resonance. Kenneth B. Wiberg and Keith E. Laidig. Journal of the American Chemical Society 1987 109 (20), 5935-5943. DOI: 10.1021/ja00254a006
Rotational isomerism of acetic acid isolated in rare-gas matrices: Effect of medium and isotopic substitution on IR-induced isomerization quantum yield and cis→trans tunneling rate. E. M. S. Macoas, L. Khriachtchev, M. Pettersson, R. Fausto, M. Rasanen
J. Chem. Phys. 121, 1331 (2004); DOI:10.1063/1.1760733
Tunneling and Dynamics in Enzymatic Hydride Transfer.
Zachary D. Nagel and Judith P. Klinman, Chem. Rev. 2006, 106, 3095-3118, DOI:10.1021/cr050301x


There are reactions similar to the one you show, which is straightforward catalysis, these are autocatalytic reactions such as $\ce{A + B\to 2B}$ or $\ce{A + 3B\to 3B}$ which lead to interesting non-linear behaviour. Of more importance is the Michaelis-Menten scheme  $\ce{E + S\leftrightharpoons ES \to P + E}$ where E is an enzyme , S a substrate, ES a complex of both, and P the product.


Relativistic effect: d-electrons in metallorganic complexes

First off, be aware that oxidative addition occurs by several mechanisms which are governed by different factors.[1]

As you move down a period, the outer electrons become more easily accessed, as the outer electrons are increasingly screened from the nuclear charge by the inner electrons.[2] This is why both ionization energy and electron affinity decreases down a period.

Classical organometallic chemistry uses metal carbonyl stretching frequencies as a direct measure of π back-bonding and electron density.[3] For a recent discussion, see: Chem. Sci. 2016, 7, 1174. Compare the stretching frequency of the group VI metal hexacarbonyls [4]: $\ce{Cr(CO)6}$: $\pu{2000 cm^-1}$ and $\ce{W(CO)6}$: $\pu{1987 cm^-1}$. Tungsten donates more electron density back into the carbonyl than does $\ce{Cr}$.

The importance of relativistic effects in structure calculations is frequently the subject of reviews.[5]

References:

Crabtree, R. H. The Organometallic Chemistry of the Transition Metals, Ch. 6
Waldron, K. A.; Fehringer, E. M.; Streeb, A. E.; Trosky, J. E.; Pearson, J. J. Screening Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends. J. Chem. Educ. 2001, 78 (5), 635. DOI: 10.1021/ed078p635.
Tolman, C. A. Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis. Chem. Rev. 1977, 77 (3), 313–348. DOI: 10.1021/cr60307a002.
Clark, R.; Crociani, B. Solvent effects on the infrared spectra of chromium, Molybdenum and tungsten hexacarbonyls. Inorganica Chimica Acta 1967, *1undefined 12–16. DOI: 10.1016/S0020-1693(00)93131-1.
Pyykko, P. Relativistic effects in structural chemistry. Chem. Rev. 1988, 88 (3), 563–594. DOI: 10.1021/cr00085a006.


Microscopic and mechanical approach to boiling point

Let’s ignore the air, first of all, and put it back later.

When you heat the bottom of the saucepan, the first thing that happens to the water at the top is nothing. It stays cold, because water is not very conductive.

The next thing that happens is that the water at the bottom becomes so hot that it is less dense than the water above it. The first thing that happens then is that nothing happens; the second thing is that the situation is unstable and the tiniest irregularity in the layers makes a blob of less dense water float upwards. The colder denser water sinks to the bottom to replace it, and then is heated in its turn.

(The long thin tube mentioned in the link finds it far harder to convect in this way, which is why the boiling, when it happens, is so explosive).

The rising blob of hotter water can cool by mixing with the water it meets on the way up. Mixing is more or less efficient depending on various things (such as how big the difference of temperature is, and so how buoyant the blob is), so the blob may still be travelling upwards when it hits the surface. Look sideways at a saucepan near to boiling and you will see this.

The boiling point of water is higher at the bottom than at the top, so at a certain moment you will have a blob that is below boiling point when it leaves the bottom finding itself above boiling point when it gets higher up. So boiling can start above the bottom of the vessel in that case; and indeed once a saucepan starts boiling, you see it boiling at the surface and only then does the boiling spread throughout the liquid.

Alternatively the blob may boil on the way up, and the bubble of steam then float up into the cooler layers and condense suddenly. This is the noise a kettle makes when it is near to boiling, and why it goes suddenly quieter when it really starts to boil.

An extra detail: water heated a little above boiling point does not boil. It stays a superheated until some irregularity or impurity acts as a nucleus for the change of state. This is why if you get a saucepan at the late “hot convection” stage and throw salt into it, it fizzes violently. The salt nucleates the superheated water and a bubble of steam forms round each grain.

Now air. Air is less soluble in hot water than in cold, so as the water is heated the air comes out of solution. First, bubbles form nucleated by any roughness of the surface of the base of the pan (if your pan is too shiny, scratch it). Eventually they grow big enough to detach themselves and float to the surface. The air bubbles stir up the liquid they float up through, so that convection happens quicker and in smaller blobs. That is why when you boil already-boiled water the second time, it makes such explosive noises: you get much bigger blobs (because no air bubbles intervene) making bigger steam bubbles which then implode more violently.

Finally, not heating from the bottom. If you grill water, it doesn’t boil at all, it evaporates in a quiet and genteel way. If you microwave water then there is little or no convection and most of it is superheated. This when you take your mug out of the microwave, it is still and innocent and unboiling. When you put a tea-bag into it then its rough surface nucleates the boiling everywhere (like the thrown salt) and you get an eyeful of boiling water.


How to calculate galvanic cell spontaneity?

To determine spontaneity of a reaction, one looks at the sign of $\Delta G$ for the process. And you have the following relation between the change in standard Gibbs free energy and the (standard)EMF of the cell:

$ \Delta G^0 = -n\,F\,E^0$

where, $n:=$ no. of electrons exchanged

$F:=$ Faraday constant (Charge on 1 mole of electrons. About $96,500\,\mathrm{C}$)

So, it is sufficient to compute the $E$ and look at it's sign. A positive sign for the EMF would ensure a negative value for $\Delta G^0$ and thus, you can conclude that your cell reaction is spontaneous.

The EMF of the cell is given as follows:

$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}} - E^0_{\mathrm{anode}} $$

Where, $E^0_{\mathrm{cathode}}$ and $E^0_{\mathrm{anode}}$ are the standard reduction potentials for your cathode and anode half reactions, and you can easily find tabulated values for most metals, ions.

An illustrative example:

Consider the following cell reaction:

$$\ce{Zn_{(s)} + Cu^2+_{(aq)} -> Zn^2+_{(aq)} + Cu_{(s)}}$$

Your cathode half reaction is:

$$\ce{Cu^2+_{(aq)} + 2e- -> Cu_{(s)}}$$

and your anode half reactions is:

$$\ce{Zn_{(s)} -> Zn^2+_{(aq)} + 2e-}$$

the tabulated reduction potential values for the two:

$$E^0_{\mathrm{anode}}= -0.763\,\mathrm{V} \qquad E^0_{\mathrm{cathode}}= +0.342\,\mathrm{V}$$

Plugging this into

$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}}-E^0_{\mathrm{anode}}$$

we obtain,

$$E^0_{\mathrm{cell}} = +1.103\,\mathrm{V}$$

Thus, this cell reaction is spontaneous.


What is the dipole moment direction in the nitrosonium ion?

In non-neutral species the dipole moment in calculations is dependent on the choice of the co-ordinate system. For this reason the origin is chosen to be the centre of mass. With $r=107.8~\pu{pm}$ on RMP2/def2-QZVPP previously determined by a geometry optimisation. The program used for the computations is Gaussian 09 Rev. D.01 and used input files are attached at the bottom of the post.

I am loosely following the same procedure/ conclusions as in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.) since $\ce{NO+}$ is isoelectronic with $\ce{CO}$.

According to the definition of the dipole moment
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
positive values in the following table indicate that the negative end is at the nitrogen atom.

\begin{array}{lr}\hline
\text{Orbital} & \text{Bond Moment}           \\\hline
\text{MO 1 } (\sigma)            &   -4.8336  \\
\text{MO 2 } (\sigma)            &    5.5232  \\
\text{MO 3 } (\sigma)            &   -0.7891  \\
\text{MO 4 } (\sigma)            &   -3.0061  \\
\text{MO 5 } (\pi   )            &   -1.1642  \\
\text{MO 6 } (\pi   )            &   -1.1642  \\
\text{MO 7 } (\sigma)            &    4.7383  \\\hline
\text{MO 8 } (\pi, \text{virt.}) &   (2.3831) \\
\text{MO 9 } (\pi, \text{virt.}) &   (2.3831) \\\hline
\text{Total } \sum_i^7 \text{MO }i               &   -0.6956  \\\hline
\end{array}

While we see the same polarisation of the molecular orbitals as in carbon monoxide, the magnitude differs and the HOMO of $\ce{NO+}$ cannot compensate for the polarisation of the electron density towards the oxygen.

Appendix

Base calculation for $\ce{NO+}$ after previous optimisation, E(RHF)=-128.972090453 input file:

%chk=mp2qzvpp.chk
#p MP2/def2QZVPP
scf(verytight)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

base

+1 1
N   0.0 0.0 -0.575312
O   0.0 0.0  0.503398

Analysis calculation input file for MO 1

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

analysis

+13 1

(Both empty lines are necessary.)

Analysis calculation input file for MO 2, etc

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
geom=check
guess=(read,only,alter)
symmetry(loose)

analysis

+13 1

2,1

The last line needs to contain the complete reordering procedure. The MO to be analysed needs to be in first place. Therefore for MO 3 it needs to read 3,1,2, for 4 4,1,2,3, etc.


Does melanostatin peptide whiten the skin in a similar manner as to how melanotan darkens it?

Melanotan

The "Melanotan" you're referring to is probably Melanotan I, also known as afamelanotide, which is an analogue of α-melanocyte stimulating hormone (α-MSH).  In other words, it activates (some of) the same receptors as α-MSH.

α-MSH has several functions, one of which is to stimulate the cells that darken the skin (melanocytes), so they produce more melatonin.  It activates these cells via the MC1 receptor.  Afamelanotide is more specific to that receptor, so it darkens the skin without some of the other effects.

It acts at the hormone level, so it won't be changing anyone's genetic makeup - stop using it and you'll go back to your normal colour.

Melanostatin

Confusingly, the name Melanostatin has been applied to more than one compound.  I've found two short peptides that might fit the bill.

The most likely candidate is described in this leaflet, which says it is used for skin lightening.  It's a receptor antagonist - it prevents α-MSH from binding to the MC1 receptor, so reducing melanin production.  So yes, it's working at the same site as Melanotan I but having the opposite effect.

The other possibility is also known as Melanocyte-inhibiting factor, which inhibits production of (natural) α-MSH, so it would be expected to lighten the skin.  However, since α-MSH has several other functions, reducing its production would also affect other systems - there would be side-effects, and skin lightening doesn't seem to be the main thing it's known for.  More here: Wikepedia


Corrosion of iron: no electron pushing mechanism?

Arrow pushing mechanism is just a formal way of thinking how the reaction might be occurring. Unfortunately, organic chemistry textbooks present it as if all of these reactions were occurring in real time, step by step, and we are able to watch it like a movie. The reality must be far more complex. If you were truly investigating a reaction mechanism, we have to design very clever experiments (there are books on this topic). It is experimental data that determines the true outcome of an organic reaction not the arrow pushing formalism. Usually, when chemists find out the experimental products, then they postulate arrow pushing mechanisms.
Coming to the rust question. Rust has puzzled chemists and engineers for decades because it causes millions of dollars of losses every year.  Your equation is a very simplistic way of thinking that rust forms in a single step.
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$
It does not proceed in a single step. Our atmospheric chemistry is very complex, you have carbon dioxide, you have sulfur dioxide, nitrogen oxides, ozone, water vapors, sunlight, plenty of free radicals and plenty of minor trace components. Then there are different phases of rust too. You can have a quick look at Google Scholar and one such representative abstract [1].
In short, people still do PhD in this field and one can only imagine how complex corrosion science is.
References

Misawa, T.; Hashimoto, K.; Shimodaira, S. On the Mechanism of Atmospheric Rusting of Iron and Protective Rust Layer on Low Alloy Steels. Corrosion Engineering 1974, 23 (1), 17–27. https://doi.org/10.3323/jcorr1974.23.1_17.


In this example, HA is indeed split by HD, but you have quite rightly identified that we don't see coupling to the alcohol proton, HC. This is pretty normal that we don't observe couplings to protons that are termed labile; -OH, -NH2, -SH, for example. Whenever samples contain trace amounts of an acid or base, there will be very fast exchange between the alcohol proton and that other species (water is often the culprit). In solvents like chloroform (which this spectrum was run in), the exchange is very fast, and you just observe a single peak. In other solvents, such as dmso, you might observe two broadened peaks. But, the exchange means that any coupling to that nucleus is lost

We do sometimes observe coupling to alcohol protons. If you can eliminate exchange, or slow it sufficiently, you will observe 3J couplings as you would for any other nuclear coupling pathways. 3J couplings to alcohol protons are in the order of 3-6 Hz, commonly.


Why are isotopes of hydrogen the only ones with special names?

Harold Urey and George Murphy used spectroscopy to identify deuterium late in 1931, announcing it at the 1931 Christmas meeting of the American Physical Society. Picking up out of 'From Nuclear Transmutation to Nuclear Fission, 1932-1939" by Per F. Dahl:

  If anything, the naming of the new isotope proved more problematic than its isolation.  At a special session on heavy hydrogen at the general June meeting in 1933 of the APS in Chicago, organized in conjunction with the Century of Progress Exposition, the ensuing discussion on its naming 'threatened to become acrimonious,' according to Francis Aston of the Cavendish Laboratory - the great authority on atomic weight measurements and a guest speaker at the discussion.  The argumentation had to do with whether to retain the name 'hydrogen' for the isotope, as Niels Bohr preferred; after all, it was not a new element and had the atomic number 1.  Both Gilbert Lewis and Ernest Lawrence opted for 'dygen' for the H$^{2}$ isotope and 'dyon' for its nucleus, wheras Rutherford preferred 'diplogen' and 'diplon' instead.  In the end, Urey had the last word, as he was entitled to, settling on 'deuterium' for the isotope and 'deuteron' for its nucleus.

Given the heavyweights in the field wanting a separate name for the isotope, even Neils Bohr could not hold back the tide.  Again, though, you have to remember that this was in the very early days of nuclear physics. While the existence of the nucleus dated back to Rutherford's experiments pre-WWI, they were still very unclear just what the nucleus was constructed out of.

This was all occurring just as Cockroft and Walton were using their new ion accelerator to perform the first human-induced nuclear fission ($^{7}$Li + p -> 2 $^{4}$He), published in the Proceedings of the Royal Society on July 1st, 1932. Alpha, beta, and gamma particles had been identified since the late 1800s, and natural sources were used for nuclear physics experiments. It was readily recognized that a source of energetic protons would be highly desired.

Also, remember that 1932 was the year that Chadwick discovered the neutron as a separate entity. Thus, also from Dahl:

  The discovery of the neutron would soon revise Urey's view of his deuteron as consisting of two protons and an electron.  However, as late as October 1932, it was still not obvious among physicists that the deuteron consists of 'one proton and one Chadwick neutron' instead of two protons and one electron.

Against this rapid-fire series of experiments on nuclear physics, it really isn't unusual that deuterium got a separate name - it was still not universally appreciated that it wasn't something weird. Than, once it was accepted, it was too useful to readily distinguish experiments with protons vs deuterons vs tritons to revert their names back to normal isotope nomenclature.


Why does dielectric constant of crystal fluctuates in case of Frenkel defect in a crystal?

The explanation provided in the text you quote is of the "hand waving" sort. Disrupting the crystal structure via Frenkel defects will have a rather complicated effect. That the effect is complicated is illustrated (in a rather rudimentary fashion) in the wikipedia, from which I derived the following figure:

Clearly the dislodged cations are shifted into closer proximity of other cations, but also of anions, and the orientational effects are not simple.

So the text you quote does not provide an entirely satisfactory explanation, I agree.

Now consider your line of argument:

  Now, if we think like that here, dipole moment is charge multiplied by
  distance, and distance has definitely reduced between a positive and
  negative ion here, so dipole moment should decrease and consequently
  dielectric constant should decrease but why is the opposite happening
  here.

The scenario you describe in the paragraph makes sense. You can write the dipole moment $p$ for two point charges of opposite polarity $q$ and separated by distance $d$ as

$$p = qd $$

Note that, as the wikipedia explains, the value of $p$ is independent of the choice of reference point, provided the overall charge of the system is zero.

Obviously decreasing $d$ decreases $p$.

The problem is that this does not describe the situation layed out in the text you quote:

  The closeness of like charges tends to increase the dielectric constant of the crystal.

Consider what it means to increase the closeness of like charges. Presumably if the crystal remains constant in volume and overall charge, it requires that opposite charges move apart.

But this is still a hand-waving explanation. You need to attempt some calculations with a model to get closer to a convincing explanation.


Conversion of 4-nitrotoluene to 2-bromobenzoic acid

Sorry but I would like to answer your questions from last to first,

(1)Yes,2-bromo-4-nitrotoluene is major product as -CH3 is ortho-para orienting group and preference of -CH3(activating) OVER -NO2(deactivating) will take place

(2) -COOH  is deactivating meta directing group (due to presence of  >C=O) so friedel-crafts bromination would take place at meta position.

(3)Yes,as -NH2 is better activator than -CH3 so bromination would take place ortho with respect to -NH2

(4)I think it is mistake of your book for not showing any catalysts like AlCl3 but It should be friedel-crafts bromination as -Br replaced  one of the -H at ortho (to -CH3) of benzene.

(5) As it is friedel-crafts bromination,Br+ ion will be easily satisfied(to be octet) by pi bond of benzene than sigma bond -C-H of-CH3.(since sigma bond is more stronger than pi bond so pi bond can be easily broken and form meta-stable sigma complex)


Why does aluminum chloride have a higher lattice energy than aluminum fluoride?

Lattice energy is a function of the radius ratio, which when tends to 1, improves the packing efficiency of the molecule. Here the radius ratio of the chloride ions and aluminium ions is much closer than the fluoride ions.
You could think of this from the born habers cycle also. Where the electron affinity of chlorine is higher, which leads to overall higher lattice energy. However, the radius ratio logic is much more convincing as the habers cycle logic would require the other parameters like the heat of formation also


Is there a free online resource where I can get the solubility of a solute in a solvent?

You need to narrow down your question.  Currently, the potential solutes include NaOH (inorganic salt), dextrose (organic neutral compound), and choline chloride (organic quaternary ammonium salt) on one hand, and on the other hand a wide range of potential solvents.  It is unlikely a single database will (ever) cover such a  broad range of possibilities.

Nevertheless, there are some resources that may be useful to consult.  For example solubility tables like this or the properties tables for some of the compounds in wikipedia, like here.  If you know both solvent, and solute, NIST-IUPAC's solubility database here may be worth to visit.

If your institution / library provides access to Reaxys/Beilstein (a commercial database by Elsevier), you may find solubility data in the category of physical properties. This database covers both organic and inorganic (including metal organic) compounds and will provide you both the numerical value, as well the primary reference.


Detection of NaF levels in aqueous solution also containing NaCl

Depending on the equipment you have available, you could buy some $\ce{CaCl2}$ de-icer, dissolve it in water, and add it to a known volume of your solution.  Highly insoluble $\ce{CaF2}$ will precipitate out; keep adding the $\ce{CaCl2}$ solution until no more precipitation is observed.

You can then filter the precipitated fluoride from solution, rinse and dry it, and weigh it.  The original fluoride concentration can then be back-calculated based upon the mass of obtained $\ce{CaF2}$, the percent by weight of fluoride in $\ce{CaF2}$, and the original solution volume.


How to report a triplet of septets in NMR tabulation?

I personally use a space, so "t sep" as long as both "t" and "sep" are defined. I think it's slightly clearer than "tsep", but I don't think there's any official adjudication on what's a good or bad acronym.

Also, of some interest to me is how you manage to distinguish 5.5 and 5.3 Hz. If it looks quite like an nonet, I would probably prefer reporting it as an apparent nonet, although different people may have different opinions... I know some groups will have "in-house" rules / guidelines for reporting NMR spectra, so it might be a good idea to check with your supervisor on this front.


Probability of measuring a particle in the ground state: having trouble with the integration

I think your tactics using partial integration from the start is the problem because $\lvert x \rvert$ is not differentiable at $x=0$ so working with its derivative is not quite kosher (and $v' = -1$ is certainly wrong since for $x < 0$ the gradient of $-\lvert x \rvert$ is $1$).

I suggest you go about it step by step:

\begin{align}
  c_1 &= \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right) dx \\
      &= \frac{2 \sqrt{6}}{a^2} \biggl( \frac{a}{2} \underbrace{\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{1} } - \underbrace{\int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{2} } \biggr)
\end{align}

Now, integral $I_1$ is easy:

\begin{align}
  I_{1} &= \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx \\
   &= \left[ \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right)  \right]_{-a/2}^{a/2} = \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) + \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) = \frac{2a}{\pi}
\end{align}

For integral $I_2$ I suggest that you use that

\begin{equation}
 \lvert x \rvert = \begin{cases} x &  x \geq 0 \\ -x & x \leq 0 \end{cases}
\end{equation}

and split up the integral into two integrals

\begin{align}
  I_{2} &= \int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx \\
   &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx
\end{align}

Now use integration by parts with $u'=\cos \left( \frac{\pi x}{a} \right)$ ($\Rightarrow u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $) and $v= x$ ($\Rightarrow v'= 1 $) to solve the integral

\begin{align}
  \int x \cos \left( \frac{\pi x}{a} \right) dx &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) dx \\
  &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \ ,
\end{align}

so

\begin{align}
  I_{2} &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx \\
        &= \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{0}^{a/2} - \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{-a/2}^{0} \\
        &= \frac{a^2}{2 \pi} - \frac{a^2}{\pi^2} - \frac{a^2}{\pi^2} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi} - \frac{2 a^2}{\pi^2}
\end{align}

Plugging this into the equation for $c_1$ gives

\begin{align}
  c_1 &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a}{2} I_{1} - I_{2} \right) \\
      &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a^{2}}{\pi} - \frac{a^2}{\pi} + \frac{2 a^2}{\pi^2} \right) \\
      &= \frac{4 \sqrt{6}}{\pi^2}
\end{align}

and that is indeed what Wolfram Alpha gives you.


No, bond angles are not constant. For example, the low-energy IR absorption of water (at $1595 \:\mathrm{cm^{-1}}$) is from the H-O-H bend

There are a few techniques to measure bond angles. For polar compounds, microwave rotational spectroscopy can often be used on small molecules in the gas phase.

I'd guess though, that most bond angles are measured from crystal structures using X-ray diffraction.

Keep in mind that spectroscopic techniques always have a timescale associated with them. So any experimental measurement is time-averaged. In the case of bond angles, while the angle will change, the average value is the equilibrium - at some points, the vibration will create a larger bond angle, and at some points in time, the bond angle will be smaller.


The paper by Order and Lindwall [1] agrees with you that the R-T reaction gives indole-3-carboxaldehyde.

  Since its isolation in 1903 by Hopkins and Cole (1), 3-indole aldehyde has been investigated very little. It was first prepared by Ellinger (2) from indole through the use of the Reimer-Tiemann reaction. This method was improved on later by Boyd and Robson (3).

References

Order, R. B. V.; Lindwall, H. G. 3-Indole Aldehyde and Certain of Its Condensation Products. J. Org. Chem. 1945, 10 (2), 128–133. DOI: 10.1021/jo01178a006.


How are pencil marks "adhered" to the paper?

The graphite itself can adhere to paper fibers with London forces; it's not a strong adhesion because the marks can be erased easily. Some of the graphite is just mechanically stuck between the paper fibers; rubbing a mark from a soft pencil can easily blur it. Enough graphite is deposited that a line drawn by a pencil can conduct electricity.

The clay gives the lead structural integrity; an "H" pencil has a harder lead, and deposits less graphite; a "B" pencil is softer, and an "F" pencil can be sharpened to a finer point.

In my time as an art major, I made a lot of sketches using vine charcoal on newsprint. Vine charcoal has no binding clay, and the marks were easily smeared. On paper without much "bite", it was easy to lift the marks off completely with a piece of kneaded rubber, and if you wanted to keep a charcoal drawing (or a soft pencil drawing) you really had to spray it with a fixative.

So I'd say that it's possible that the binding clay also plays some role in binding the graphite or charcoal to the paper.

  In the manufacture of graphite pencils, air-classified grades of Volclay sodium
  bentonite and micronized hectorite are regularly used to bind the graphite
  compound in pencil lead. --- American Colloid Company, Industrial and Household Applications

Bentonite and hectorite both have ionic surfaces that probably help bind it to OH groups on the surface of cellulose fibers.


How does a mostly non-polar molecule prevent a polar head from being pulled off by other polar molecules?

During the disolution of a compound C  in a solvent S, intermolecular forces between molecules of C and between molecules of S are broken (which requires energy), and intermolecular forces between each molecule of C and several molecules of S around it are formed (which releases energy). As such, dissolution of C into S is only favorable if the released energy is greater than the required energy.

Let's consider octanol $\ce{CH3(CH2)7OH}$ in water. We need to break the hydrogen bonds between octanol molecules and the London dispersion forces in them to separate the molecules, and we need to break the hydrogen bond network in water to accommodate each octanol molecule. Then, we surround each octanol molecule by water molecules. Only the polar head ($\ce{OH}$) will be able to form new hydrogen bonds (and account for the broken ones), but the interaction between water and the hydrophobic tail is merely permanent dipole-induced dipole, which are weak compared to all the hydrogen bonds we had to break between the water molecules to be able to accomodate the large hydrophobic portion, so we're losing energy rather than gaining it. Thus, this will be poorly soluble in water.


What kind of drugs can be absorbed through skin?

The skin is able to prevent some things from passing in and out of a person but it is not a perfect barrier.

The lipophilic things like nicotine, hydrogen flouride, VX, hexane and iodine vapour can pass through undamaged skin. These things are able to dissolve in the fatty parts of skin. Then they diffuse in through the skin into the person. Thus they enter the person.

Some water soluble ionic things can enter / leave a person through the skin, this can occur through the sweat pores.

For example cesium in humans does come out in sweat, in a 1980s case in Brazil where a lot of people were internally contaminated with cesium-137. It was found that cesium kept appearing on the surface of the skin of the contaminated people.


Can a magnesium atom combine with just one chlorine atom?

In solid state $\ce{MgCl2}$, can be approximately described assuming fully ionic bond, in which $\ce{Mg}$ atoms lose two electrons to form $\ce{Mg^2+}$, while a $\ce{Cl}$ atoms gain one electron to form $\ce{Cl-}$. To balance the two electrons released by $\ce{Mg}$, you need two chlorine atoms. The electrostatic force of attraction (aka the "ionic bond") binds them, and it has the formula $\ce{MgCl2}$.

Now, you wish to form $\ce{MgCl}$. To maintain charge neutrality, you'll want the $\ce{Mg}$ to lose only one electron. This means the valence shell of $\ce{Mg}$ would still have a single electron left in it. This is a very unstable configuration.

However, it is incorrect to say that $\ce{MgCl}$ cannot form. A research paper1 has explored the very $\ce{MgCl}$ molecule. Note that the conditions were very drastic, $\pu{500^\circ C}$ for pyrolysis and $\pu{2500^\circ C}$ for vaporization respectively. They ensured "the complete formation of $\ce{MgCl}$ molecular species, since $\ce{Cl}$ was the limiting reactant" as you had correctly identified yourself.

Another paper3 describes the formation of the $\ce{MgCl}$ molecule:

  Excited $\ce{MgCl}$ molecules were produced by two different techniques: (i) at Orsay, they were created by mixing Mg vapor with a flow of $\ce{He/Cl2}$ and excited in a “heated” Schuller-type discharge tube, and (ii) at Waterloo, they were generated by using a copper hollow cathode lamp loaded with $\ce{MgCl2}$ powder.

Note that these techniques are indeed quite "unnatural", in the sense that you might not find $\ce{MgCl}$ naturally, but you can certainly create it through artificial conditions. In fact, $\ce{MgCl}$ is fairly popular otherwise. Several papers2 have already discussed the emission spectrum of the $\ce{MgCl}$ molecule. Some4 also explore its laser cooling. Hence, by no means it is an unknown compound and in fact, can definitely be formed.

References:

Rina Lourena da S. Medeiros, Sidnei de Oliveira Souza, Rennan Geovanny Oliveira Araujo, Djalma Ribeiro da Silva and Tatiane de A. Maranhão, Chlorine determination via MgCl molecule in environmental samples
using high resolution continuum source graphite furnace molecular absorption
spectrometry, Talanta, http://dx.doi.org/10.1016/j.talanta.2017.08.026
Darji, A.B., Shah, N.R., Shah, P.M. et al. Pramana - J Phys (1985) 25: 571. https://doi.org/10.1007/BF02847235
Vervloet, M. Fourier Transform Spectroscopy 2001. https://doi.org/10.1006/jmsp.2002.8514
The Journal of Chemical Physics 143, 024302 (2015); https://doi.org/10.1063/1.4926389


Theses are potential energy diagrams, representing the 'stable' forms of both the reactant and the product of IVCT (and their immediate relationship with ligands and surrounding solvent). This is most easily seen in the class I case, where the curves are not merged at all, and instead represent the species as distinct (and the reaction unable to proceed). The reason the curves appear to merge is because the class II and class III cases are where electron transfer is able to occur, and thus the reactant has a reaction path with which to proceed to the product.
There are in fact two reaction paths that a class II mixed-valence species can proceed to the product, one being optical (which is the IVCT transition, caused by absorption of light), and the other being thermal. The optical pathway occurs upon absorption of light at the bottom of the leftmost potential well, where it is excited to the top left of the upmost curve. This corresponds to lambda, which confusingly represents the energy of this optical transition (E=hv=\lambda). The position that the ion is now at on the potential energy curve is representative of a state where the electron has been exchanged, but the molecule is yet to rearrange bond lengths, and the surrounding solvent has not rearranged to accomodate the new distribution of charge. The molecule then relaxes down the adiabatic potential energy curve (not shown here, some diagrams have the same structure as class I in dotted lines, these lines being the adiabatic energy curves), to the rightmost potential energy well, where the molecule has completely rearranged the surrounding solvent and changed bond lengths to accomodate the new charge distribution. The thermal process is less relevant, except to note that the hump in the class II diagram is the thermal activation barrier, where an activation energy equivalent to or greater than this barrier is required for the reaction to proceed via this pathway.

A class III ion is one where there is full mixing of the two potential energy curves, such that they become the structure you see above. Clearly here there is no thermal activation barrier, as there is no hump, and the optical transition energy is actually equivalent to or smaller than 2Hab (twice the coupling constant). The result of the now missing thermal activation barrier is that the ion can freely transition between both states, equilibrating between them, resulting in partially charged ions, and a very delocalised system. Note that optical transitions still occur in class III ions, and are typically more intense than transitions in class II ions. There is just no good way of displaying where the transition occurs as it is in the same position as the 2Hab gap.

I realise this thread was created long ago, so this response likely won't be useful to the original poster, but hopefully anyone else who is interested will have a greater appreciation of how these diagrams work.
I strongly recommend reading this paper https://onlinelibrary.wiley.com/doi/10.1002/9780470166314.ch1 if you can get access to it. It is by far the most understandable of all the papers on the subject, and very comprehensive.