Like you guessed, the pH should be around 7, since the concentration of HCl is very low. Obviously, it can't be 11.65, since it's still an acid. The reason for this discrepancy is that, for low concentrations, you must consider the self-ionization of water in order to get the correct concentration of $\ce{H3O+}$ ions.

Consider the equilibrium:

$$\ce{2H2O <=> H3O+ + OH-}$$
$$K_w = \ce{[H3O+][OH^{-}]} = 1 \times 10^{-14} \tag 1$$

and also the dissociation of the acid:

$$\ce{H2O + HCl -> H3O+ + Cl-}$$

Then
$$c_0 = \ce{[Cl^{-}]} = 2.26 \times 10^{-12} \tag 2 $$

and

$$\ce{[H3O+] = [Cl^{-}] + [OH^{-}]} \Rightarrow \ce{[OH^{-}] = [H3O+] - [Cl^{-}]} \tag 3$$

If we replace $(2)$ in $(3)$ and that in $(1)$, we end up with

$$ K_w = \ce{[H3O+]} (\ce{[H3O+]} -c_0) $$

$$ \ce{[H3O+]} (\ce{[H3O+]} -c_0) - K_w = 0$$

using $ x = \ce{[H3O+]}$

$$x^2 - c_0x - Kw = 0$$

This can be easily solved. Taking only the positive value of $x$:

$$ x = \frac{c_0 + \sqrt{c_o^2 + 4K_w}}{2}$$

Inserting the values of $x$ and $K_w$, the result is:

$$ x = 1.000023 \times 10^{-07} = \ce{[H3O+]}$$

So

$$pH = -\log\ce{[H3O+]} = -\log{1.000011 \times 10^{-07}} = 6.999995 $$

Which is not surprising at all, but is the calculated answer.

For a less diluted solution of $\ce{HCl}$ with concentration of $1 \times 10^{-8} \frac{mol}{L}$, which still can't be calulated directly, the result is $pH = 6.98$.


What is an example of chemical reaction that can be assisted by both an inorganic catalyst and an enzyme?

A good example is the conversion of hydrogen peroxide to oxygen and water
$$\ce{2H2O2 ->2H2O + O2}$$
It can be catalyzed by catalase, which has one of the highest known turnover numbers, or by an inorganic catalyst such as manganese(IV) oxide or lead(IV) oxide.
This has been suggested for instance as an educational project by the University of York (see references).
References

Decomposing hydrogen peroxide. Chemistry Review (1995), Volume 5, Number 1, p. 30. https://www.york.ac.uk/chemistry/schools/chemrev/


Electronic effects of substituents in the benzyne mechanism

the electron pair that is creating negative charge on ring is sp2 hybridized and its orbital direction is away from the pi orbital electronic cloud of benzene ring so not able to resonate along with pi electrons. OMe inductively stabilize this electron pair as it is electron with-drawing inductively. thats why NH2 group attaches at meta position rather than ortho position.


Partial Pressure and Partial Molar Volume

It doesn't work because partial molar volume of a species in a gas mixture is defined as the partial derivative of the total volume of the system with respect to the number of moles of that species at constant total pressure and temperature of the system.


Removal of octadecanoic acid from paraffin wax

It should be noted that typically, candles are about 10% stearic acid and 90% paraffin. Also, there are many different grades of paraffin waxes which are generally identified by their melting points (often given as ranges); candles are usually made from paraffin graded from 53.3 °C (128 °F) to 65.6 °C (150°F), while paraffin of 76.6 °C (170 °F) is more common for molding and sculpting.

So I'm pretty sure that you didn't know what grade of paraffin you are after, and I would guess that you are looking for something more like cheese wax. While cheese wax is significanly softer, it actually has a generally higher melting range of perhaps 72 - 82 °C (161 - 179 °F).

Cheese wax is a blend of paraffins:

<30%  paraffin wax fume (Cas# 8002-74-2)
<30%  paraffin waxes and hydrocarbon waxes, microcryst. (Cas#
63231-60-7)
<20%  petrolatum (Cas# 8009-03-8)
30% Slack wax, petroleum, hydrotreated (Cas# 92062-09-4)

However, there may be multiple grades of paraffin used for your candles or the grade used may be of a low enough melting point (MP) to be useful for making the product you desire (in conjunction with other materials to make it softer for sculpting or molding).

The "analysis" and the separation procedure would be based on the melting point. So, (I would) try to use the melting points to separate the stearic acid from the lower temperature fractions.

Stearic acid's MP is 69.3 °C (156.7 °F)

Paraffin's MP may be as low as 37 °C (98.6 °F), but finding that grade of paraffin in your candles is unlikely. With a bit of luck, a low temperature grade of paraffin (48.3 °C (119 °F)) might show up; but more likely would be grades having a melting point of 53.3 °C (128 °F) and/or 61.1 °C (142 °F), which are common for candle making.

Step 1. Melt the candles in a pot and remove the strings. If possible, you would want to use a hot plate with accurate, good-quality temperature control.

Step 2. Insert a (candy) thermometer and a spoon or ladle in the pot (maybe a screen-type of ladle, like what is used for noodles and dumplings). You need the ladle to be warm, so that the wax doesn't freeze and stick to it when you use it to scoop out the stearic acid.

Step 3. Reduce the heat until precipitates begin to form. This should begin to happen somewhere around 70 °C. With an open pot, the cooler surface will probably cause the top to skim over. The higher melting point fractions might be what solidify first on the top, like slag on molten steel. The first precipitates or slag should be higher in octadecanoic (stearic) acid concentration. Use the ladle to scoop/filter out the precipitates (put the stearic acid in a in another pot or a bowl). You should keep collecting precipitates until it reached about 55 °C or until it all becomes solid. Be sure to make note of the final temperature. If the temperature at which it all solidifies is at or above 60 °C, then the kind of paraffin that you have is probably not what you wanted; and it will be quite difficult to separate, and this procedure is not very useful for separating the stearic acid from the paraffin.

Step 4. Even if the paraffin has a low MP, this will not be a 100% perfect procedure (far from it); you will probably need to repeat the (enrichment) procedure a couple, or maybe several, times. Also, you could probably recover some (low MP) paraffin from the precipitated/enriched stearic acid filtride. It might be favorable to have a few bowls or pots which represent collections at different temperature ranges... like, for example, the first bowl would contain solids from 70 - 65 °C; the second bowl contains solids from 64 - 58 °C; and the third bowl contains solids from 57 - 52 °C.

Step 5. Measure the new MPs. A good anylysis of the cleanliness of this procedure will be to compare the MPs. A pure solution of stearic acid will melt/solidify at exactly 69.3 °C, but obtaining a pure solution in this manner, is a pretty lofty dream. So, after the stearic acid fraction has melted, let it cool and  mark the temperature at which it begins to solidify and the point at which it is completely solid. If the two fractions that have been separated have distinctly different melting points, that's pretty good. Specifically regarding the paraffin, if it doesn't begin to solidify until it's below 50 °C then, great job, it's pretty clean and you obtained the low temperature paraffin that you desired!

Step 6. Depending on the cleanliness (melting point) and amount of the stearic acid that you have, you may want to try to salvage some of the paraffin from the stearic acid (filtride). I believe that this will be more difficult, and I would advise using a third pot (or second bowl) to collect the salvaged paraffin. And then work on cleaning the stearic acid out of the salvage, checking the cleanliness as in step 5, before combining the salvaged paraffin with the first "batch".

Step 7. Further chemical reactions for cleaning up the stearic acid (or converting it to soap) could be performed with better results from the enriched fractions.


The various resonance structures that one can draw for an acid chloride are shown below.

Structure IV, is the "triple bond" structure that you mentioned.  The acylium ion ($\ce{R-C#O^{+}}$) is very stable, so structure IV does contribute to the overall description of an acid chloride.  Note that the $\ce{C#O}$ bond in IV is not a real triple bond as the carbon remains $\ce{sp^2}$ hybridized, with the attendant 120 degree (non-linear) bond angles.  I've also drawn a fifth resonance structure involving hypeconjugation.  This is probably a reasonable structure as well since it places negative charge on oxygen and positive charge on the hydrogen.  It also helps explain why hydrogens on the carbon alpha to a carbonyl have increased acidity.

Usually the term "backbonding" is used when electron density is transferred from one atom to another through the use of antibonding orbitals.  See here for a better description.  In resonance structure IV, the lone pair from oxygen is being shared with an empty carbon $\ce{sp^2}$ orbital, the one that had been used to bond to the chlorine.  I've never heard that referred to as backbonding.


When are transition state's energy barrier "reasonable" at a certain temperature?

Your question is really about what you consider to be a reasonable rate constant and so is somewhat subjective. The fastest a bimolecular reaction can be in solution is given by how fast the reactants can collide and then react on first collision. i.e reaction is diffusion controlled. In this case as reaction is at first meeting the activation energy must be small, $\le \pu{5  kJ mol-1}$. In water the diffusion controlled rate constant is in the region of a few times $\pu{10^10 dm3 mol-1 s-1}$. Most reactions are far slower than this.

The simplest, but very approximate, approach for other reactions is to use the Arrhenius equation to estimate the rate constant. The equation is
$$ k = A \exp(-E_a/RT),$$
where $E_a$ can be taken as your activation energy and you have then to estimate $A$. This can  be done for simple reactions using statistical mechanics and calculating the partition functions, but for anything complex you are left only with guessing values based on experimental data from similar reactions. However, because of the exponential dependence of the rate constant on activation energy even a small change in this can have a dramatic influence on the rate constant, far bigger than any similar change in $A$.

If you do have a potential energy surface then you can calculate trajectories along this surface by solving the (classical) equations of motion, repeatedly using energy sampled from a Boltzmann energy distribution,  and so work out a theoretical rate constant.


Is it possible to compute the colour of *any* molecule?

It is possible to compute electronic spectra with decent accuracy. Time dependent density functional theory (TDDFT) permits fairly rapid calculation of UV/vis spectra at a quantum mechanical level of theory. Whilst DFT is not truly a first principles method, it can furnish electronic spectra accurate enough to be useful, depending on care taken in selecting appropriate functionals, accounting for solvatochromic effects, etc. Obviously, DFT uses the Born-Oppenheimer approximation, but the beauty of the approximation is that it's actually very close to reality.

See, for instance, this paper by Jacquemin et al., where they calculate $\lambda_{max}$ for a wide range of dyes using TDDFT and get a MAE of 22 nm using CAM-B3LYP (a long-range hybrid functional), reducing that to 10 nm with an empirical calibration curve. The paper only really cares about accurately reproducing the location of $\lambda_{max}$, however this paper by Quartarolo and Russo computes complete spectra with fairly reasonable accuracy. Not perfect, but the general shape of the spectra are reproduced adequately. Part of the divergence between the experimental and computed spectra probably comes down to higher order broadening effects not reflected in the simple Gaussian convolution of the computed data, but I'm not a spectroscopist.

Additional: as for computing the colour of any molecule, well, that does depend heavily on the size of your molecule and any mitigating factors such as symmetry elements. Most QM techniques scale fairly viciously with the number of electrons in your molecule, and whilst you can always reduce the size of your basis set and heap on less and less conservative approximations, accuracy will suffer.


Yes, you can make aluminium oxide from aluminium chloride pretty much in your way but then again as @andselisk said, it is a pretty inefficient reaction and wastes a lot of fuel and energy and also produces many byproducts. Hence, the reaction is the other way round if considered on an industrial scale. The main preparation method is already described by @andselisk in the comment section. Only to verify its authenticity, here is a patent claim:

  A process for preparing alumina which comprises heating aluminum
  chloride hexahydrate at a temperature of from about 225 to 600 F. For
  a period of time such as to liberate only a portion of its chlorine
  content, commingling a basic reagent with the resultant
  chlorine-containing residue to form aluminum hydroxide, washing and
  filtering the latter and then calcining to produce the oxide.


Optical activity of cis coordination compounds

It is not always as easy to see from a standard octahedral depiction. Optical activity in octahedral complexes can be understood better, if you draw the octahedron as looking onto a triangular face rather than along an edge; this results in a hexagonal arrangement of ligands around the central atom with ligands alternately pointing into and out of the paper plane.

If we do that with $\ce{[Co(en)2Cl2]}$, we get the following two enantiomers which you cannot convert into each other:

The complex is chiral because the $\ce{en}$ ligands necessarily cross the paper plane and can do so either in a right-turn screw motion or in a left-turn screw motion.


Finding formula of hydrated potassium chromium(III) sulfate through experimentation

All the clues you need to solve the problem have already been given. A key clue is the second line; the word heating. What do you know about heating and how it affects water content? Use this as a starting point.

Thenceforth, the question’s solution should suggest itself.


I don't know what you are plating (nor do I care...) but here are some general observations.

(1) Plating is messy and exacting analytical measurements are overkill. Your calculation of 3728.79 ml water for instance is wild. I'm not sure what you intend to use for a container, but just diluting to 5 l +/- 100 ml would be more than enough precision.

(2) It is unclear if 200 g/l for copper sulfate is anhydrous copper sulfate or the pentahydrate. You're near the saturation limit for dissolving anhydrous copper sulfate at room temperatures.

(3) Your sulfuric acid calculation has the right number but wrong units. You need 271 ml per 5 liters.

SAFETY: You will require eye protection and appropriate clothing.  When diluting sulfuric acid always add the sulfuric acid to the water. The reaction is very exothermic. (If you try to add water to the acid it will boil violently and splatter sulfuric acid which is more than extremely dangerous.)

You should also not work alone, and you should have some sort of plan as to how you would handle an accident. What are you going to do if you get acid in your eye, or what happens if you drop the bottle of sulfuric acid? It turns out that concentrated sulfuric acid is very slippery...

(4) Your hydrochloric acid calculation is wrong. You have the right density, but HCl acid is only 37% HCl, the rest is water. Also HCl is only 97.3 % Cl, that pesky H is the rest. Your calculation is also again for 5 liters not 1 so the units that you show are wrong. I'm not sure about copper sulfate, but it wouldn't surprise me if there were some chloride in the sulfate.  That little hydrochloric acid isn't going to change the acidity of the solution. I'd be tempted to use copper (II) chloride as the source of chloride. It stores a lot easier, and would be easily measured. (Hydrochloric acid releases HCl fumes unless absolutely sealed - like in a glass ampule.)


Can I extract water from diethyl ether to get anhydrous diethyl ether?

No it would not work.  Ether boils at a lower temperature than water.  Plus, it forms an azeotrope with water:  a composition of 98.7 % (by weight) ether boils at 34.2 °C and the vapor phase has identical composition as the liquid, meaning that distillation cannot remove the final 1.3% of the water (at atmospheric pressure).

The standard way to remove water is (i) first to use drying agents such as magnesium sulfate to remove most of the water, and then (ii) if needed, eliminating the final traces of water with small amounts of sodium (or potassium) metal.


Why is acetic acid more acidic than phenol?

There are actually several factors that affect the stabilization of negative charge. The first is what kind of atom centers the charge can be delocalized on and the second factor is how many atom centers the charge can be delocalized on. The first factor is more important than the second. Just think it like quality is more important than quantity. Symmetry is another factor but not very important in your case.
Let's look at the ethanoic acid vs phenol case. In the acetate ion, the negative charge is delocalized on two oxygen atom centers, while in the phenoxide ion, the charge is delocalized on one oxygen and three carbon atom centers. Because oxygen is much more electronegative than carbon, the delocalization of negative charge over two oxygens is better than delocalization over one oxygen and three carbons. Quality beats quantity here.

If you have 2,4,6-trichlorophenol, because the strong election withdrawing chlorine atoms is helping carbon center to stabilize negative charge, it is a much stronger acid than phenol itself. This is an example of the inductive effect at work, as opposed to resonance.


How did hypophosphoric acid get its name?

You are most of the way to the reason yourself:

$$\begin{array}{|c|c|c|}\hline
\text{name} & \text{formula} & \text{oxidation number of P}\\ \hline
\text{phosphoric} & \ce{H3PO4} & +5\\
\text{phosphorous} & \ce{H3PO3} & +3\\
\text{hypophosphorous} & \ce{H3PO2} & +1\\ \hline
\end{array}$$

The "hypo-" prefix means "beneath" or "less than". Hypophosphoric acid, $\ce{H4P2O6}$, has oxidation number of phosphorous as +4, which is in between phosphoric acid and phosphorous acid. Thus, "hypophosphoric" refers to "beneath" phosphoric (but above phosphorous) acid.


By what reaction does (–)-sparteine decompose in ambient conditions?

Amines left out in the atmosphere will air-oxidize to give N-oxides, which typically have a yellow color. These N-oxides can decompose to give further byproducts. I would store your sparteine under nitrogen or argon atmosphere.

[edit to give examples and mechanism]
The reaction proceeds through a radical chain mechanism with molecular oxygen. Here is one study of the mechanism [from first page of Google search for amines air-oxidation] .

From wikipedia article for triethylenetetramine. "This oily liquid is colourless but, like many amines, assumes a yellowish color due to impurities resulting from air-oxidation."

"It can be observed that arylamines develop a yellow or black color if left exposed to air for prolonged periods". Ref Note - I realize that sparteine is not an aromatic amine, but the result is the same.

If you have access to triethylamine or other amines in your lab, do a control experiment: leave one vial open to air, and leave the other under identical conditions except sealed under an inert atmosphere. A few days worth of observation will convince you of the role in oxygen in the yellowing of amines.


Why was the definition of chiral carbon changed?

The IUPAC definition of asymmetric carbon is still:

  The traditional name (van't Hoff) for a carbon atom that is attached to four different entities (atoms or groups), e.g. Cabcd.

The definition of chirality centre is:

  An atom holding a set of ligands in a spatial arrangement which is not superposable
  on its mirror image. A chirality centre is thus a generalized extension of the concept
  of the asymmetric carbon atom to central atoms of any element, for example N+abcd,
  Pabc as well as Cabcd.


How does DeepHF (CRISPR sgRNA design tool) compute $\Delta G_{\text{binding}}$?

The simplest model for estimating binding energies (or melting temperatures) is to consider each base pair individually, without regard to sequence context. This is the basis of the 4 deg / 2 deg rule of thumb for DNA duplex. You can refine this model by considering the effect of ionic strength. More sophisticated models go beyond just considering the composition to include sequence information.
The nearest-neighbor model for binding is the simplest of these more sophisticated models, and it looks like that's what is used here. For DNA:DNA duplexes, you can find  parameters on Wikipedia. The DNA:DNA model has less parameters because of symmetry lacking for the RNA:DNA hybrids (i.e. AG/CT is the same as TC/GA for the former, but AG/CU is different from TC/GA).
The code does not indicate the units, which might be kJ/mol or kcal/mol. The parameters are usually derived from experimental data and some fitting algorithm, see e.g. https://pubs.acs.org/doi/abs/10.1021/bi00035a029
Missing in this implementation is the contribution of the first and the last base pair.


Is a negative pH level physically possible?

One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006, 83, 1465. Quoting the abstract in full:

  The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books.

The article text provides some counterexamples:

  For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$.


Densest gas
Densest element
Greatest range between melting and boiling temperature
Highest melting point
Lowest solubility product constant
Most dangerous: Is there any substance that's a 4-4-4 on the NFPA diamond?. But also, don't forget that the NPFA diamond doesn't even classify ClF3 as flammable (though everything else is flammable in it) and by any criterion it is far more dangerous than the other compounds in those answers)
Strongest acid
Strongest organic acid
Strongest reducing agent

Good external lists:

Most extreme compounds known to man.


TL;DR: When $\ce{PBr3/PCl3}$ is not available as reagent, red phosphorus is used which is reacted to bromine or chlorine to form the reagents in situ.
Long answer:
Hell–Volhard–Zelinsky halogenation reaction halogenates carboxylic acids at the α-carbon.  The reaction is initiated by addition of a catalytic amount of $\ce{PBr3}$, after which one molar equivalent of $\ce{Br2}$ is added. $\ce{PBr3}$ replaces the carboxylic -OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol, which will readily react with the $\ce{Br2}$ to brominate a second time at the α position.

Now, what happens if $\ce{PBr3}$ is not available?

You can use red phosphorus as a reagent. It is reasonably air stable (whereas white phosphorus combust in air almost spontaneously). Phosphorus reacts with bromine to give phosphorus tribromide which helps in the conversion of carboxylic acid into an acyl bromide.
$$\ce{2P_{(red)} + 3Br2 -> 2PBr3}$$

Why exactly $\ce{PBr3/PCl3}$ is needed and why an extra molar equivalent of
$\ce{Br2/Cl2}$ is needed?

When left out in the air, $\ce{PBr3}$ will readily combine with any water vapor present to form a new $\ce{P-O}$ bond along with $\ce{HBr}$. The strength of the $\ce{P-O}$ bond [about $\pu{140 kcal/mol}$] makes $\ce{PBr3}$ very useful for swapping out $\ce{C-O}$ bonds for $\ce{C-Br}$ bond and that's how carboxylic acids gets converted to acyl bromides.
However, $\ce{PBr3}$ by itself is not sufficient to perform the HVZ reaction. A molar excess of $\ce{Br2}$ is added to the already present $\ce{PBr3}$, which perform the halogenation of the alpha-carbon. That's why the reagent is always mentioned as ($\ce{PBr3 + Br2}$).
References

https://en.wikipedia.org/wiki/Hell%E2%80%93Volhard%E2%80%93Zelinsky_halogenation
https://www.organic-chemistry.org/namedreactions/hell-volhard-zelinsky-reaction.shtm
https://www.masterorganicchemistry.com/2020/08/26/the-hellvolhardzelinsky-reaction/

That generates another question: Can we use PI3 as reagent in HVZ reaction?


Pyrotechnic colorants- reactions & safety concerns regarding making charcoal discs

The best way for you to find out if a chemical is safe for your use case would be to look at the safety data sheets for your chemicals. For your convience, I have linked to relevant ones below.

Copper (II) Chloride
Calcium Chloride
Borax
Copper (II) Sulfate
Alum
Strontium Chloride
Potassium Chloride
Lithium Chloride
Sodium Chloride
Magnesium Chloride
Aluminium Powder

In general, you will find that most metal salts are acutely toxic. Take for instance, the Copper (II) Chloride, which is noted as having the following hazards, as noted in Section 2.1 Classification of the substance or mixture:

Acute toxicity, Oral (Category 4), H302
Acute toxicity, Dermal (Category 4), H312
Skin irritation (Category 2), H315
Serious eye damage (Category 1), H318
Acute aquatic toxicity (Category 1), H400
Chronic aquatic toxicity (Category 2), H411

From the same section in the SDS for Aluminum Powder, you will find that it has the following hazards:

Flammable solids (Category 2), H228
Acute aquatic toxicity (Category 1), H400
Chronic aquatic toxicity (Category 1), H410

You can find out more about these hazards from the EPA GHS Criteria Summary, and the Wikipedia GHS Hazard Statement List.

Overall, you should look over each SDS and make sure that the chemical you are proposing to use is safe for your purposes.

Cheers


Where do the percents in empirical formula questions come from?

How would those percentages be determined in real life? Spectroscopy?

You are right, this analysis comes in the domain of analytical chemistry. The fundamental requirement for formula determination is that your sample must be extremely pure.

There is a whole field. So there are two broad classes of approaches. One is called classical analysis where one would dissolve the sample in an "appropriate" acid or base mixture, and this is followed by target element specific titration or precipitation. This technique results in very reliable data but it is very time consuming

Today most people would use an instrument called the inductively coupled plasma optical emission spectroscopy (ICP-OES), you "srpay" the dissolve sample into a state of matter (plasma) which is as hot as the surface of the Sun. Under those those conditions, compounds break down into their constituent atoms and emit characteristic wavelengths.
For example, if we are interested in knowing how much aluminium is there, we would construct a calibration curve and interpolate the amount of Al in your unknown sample.
The same thing (although less effectively) can be done for sulfur.

Sometimes you need both classical or instrumental analysis.


In an O-H bond, will polarity increase/decrease if some of the electron density on the oxygen atom is reduced?

The polarity on the oxygen atom will decrease. This is because of the sp2 hybridised carbon atom which is highly electronegative. This will result in a decreased polarity in the OH bond and hence it will disassociate more easily in solvents such as water. One example of this effect is the increased acidity of phenol compared to most alcohols. This is because the OH is bonded to sp2 hybridised carbon in phenol and hence will disassociate in water easily when compared to normal alcohols.


There isn't really much sense in memorizing the periodic table. The elements you often use you will know them by heart after a while. And you can always use a table when you need it for the others.

That being said if you really want to do it, mnemonics are probably the best solution to memorizing the whole table. You can find some here :

https://www.mnemonic-device.com/chemistry/periodic-table/mnemonic-device-for-the-periodic-table/

or here :

http://en.wikipedia.org/wiki/List_of_chemistry_mnemonics#Periodic_table


Ethyl alcohols use on polycarbonate and acrylics

Not able to comment....

I think the main problem is propyl alcohol (not present in this cleaner). Propyl alcohol (propanol) is present in a lot of cleaning solutions and known to damage polycarbonate and acrylics.

Check this chemical resistance sheet:

Sheet

30% solution containing ethanol (ethyl alcohol) lightly attacks the plastic. However, a 10% (propanol) propyl alcohol already does the same damage. That is a factor 3!


Comparing the strength of metallic bonds

There are two main reasons:

Metallic bonding is more prevalent for transition metal elements because there are more electrons available for delocalization (from s and d orbitals), while Magnesium has only s electrons
Fe has a higher atomic mass than Mg --> van der Waals forces are stronger.


Scattering is the redirection of the 'beam' of light, as is illustrated in the diagram below:

Source: University of Wisconsin

The description from the document Fundamentals of modern
UV-visible spectroscopy (Figure 46) states that

  Scattering causes an apparent absorbance because less light
  reaches the detector

They also detail two main types of scattering in UV spectroscopy (Figure 47):

  • Rayleigh scattering: Particles small relative to wavelength
  • Tyndall scattering: Particles large relative to wavelength


Why does DFT-D finds different optimized geometries for the same structure when using different input geometry files?

Simply put, a geometry optimization finds a local minimum. Many such minima may exist for a given system and it is up to the user to ensure that a reasonable minimum is used for further calculations. Techniques may include using many handmade starting geometries, using a random mutation algorithm or starting from known, experimental structures.


This baseline roll seen in this spectrum is an example of acoustic ringing, and is quite normal for many low frequency NMR experiments that are recorded over wide sweep widths. It occurs as a consequence of slight breakthrough of the excitation pulse, as it rings down to zero. Ideally, you would like to start collecting signal at your receiver immediately after your excitation pulse, however it does take a small amount of time (microseconds) for the pulse to decay to zero. If your pre-scan delay (typically configured for your spectrometer during the configuration process) is too short, you can get this ringing artifact. It is more pronounced for weaker samples, but usually nothing to worry about.

To correct this as an experimental acquisition parameter, you would make your pre-scan delay longer; for Topspin, this is the parameter DE in your acquisition parameter list. It is normally defined during spectrometer configuration, and needs to be as short as practically possible to maximise your signal detection - typically less than 10 microseconds. For some nuclei, much of the signal has already decayed during this pre-scan period, and require some backward linear prediction to help. I'd speak to your local friendly NMR spectroscopist before mucking about with changing parameters like DE.

It is perfectly normal, and you should be able to get rid of it by applying a simple baseline correction algorithm - for Topspin you should use the commands abs or absd. This should work for your spectrum. As a word of caution when applying baseline corrections to your 1H spectrum, this will also apply an automatic integration for you, overwriting any integration you may have done previously. You can avoid this by using the optional n switch (abs n). For severe cases of baseline roll, apply a partial basline correction using absf between the limits of absf1 and absf2. Your Topspin manual can provide a fuller description, including other baseline correction modules.


Heat energy vs. electrical energy in a galvanic cell or in an electrochemical reaction

Operational electrochemical cells have $\Delta G\lt0$
And,
$$\Delta G=-nFE$$
$$\Delta G=\Delta H-T\,\Delta S$$
Knowing the entropy change, enthalpy change could be calculated
or
$$\Delta H=-nFE+nFT\frac{\mathrm d\Delta G}{\mathrm dT}\quad\text{at contant }p$$


I believe you are misreading the question. Rather than asking for the number of stereoisomers the product can have (this is where 2^n comes in), it is asking for the number of stereoisomeric products formed. Note that that the two products are diastereomers, which is a type of stereoisomeric relationship. It's not looking for any fancy calculation; just for the related products.

In a Gringard mechanims, the Gringard reagent can attack the carbonyl on either the re or si face, which in turn, can yield an R chiral center or an S center at the location of attack. This lends to the diastereotpic products.


Fiberfrax is the standard material for high temperatures in refinery, chem. plants, etc. There are also other brands , all readily available on the net. Ads do not give detailed composition other than "alumina and silica" .  Fiberglasses may not have  high melting /softening temperatures. There are higher temperature fibers like thoria, but are expensive and not easy to find.


Molecular Hamiltonian of Helium hydride ion

Your provided note is about the HF Hamiltonian in the one-electron basis set (atomic orbitals). Since a minimal STO-3G basis set is used here, the corresponding matrix representation of this Hamiltonian is a $2\times 2$ matrix.

The $3\times 3$ Hamiltonian from the SI is the FCI Hamiltonian in the many-electron basis set. These basis functions are Configuration State Functions, arising from linear combinations of (excited) Slater determinants (electron configurations), which in turn are built from the HF molecular orbitals.

Configuration Interaction calculations (like FCI) are typically done on top of a HF calculation, to correct for missing electron correlation effects. Your linked note is only about HF. For details about CI, I can recommend the book by Szabo and Ostlund which is also cited in your note.


E - Z configuration of cyclic substituents

Welcome to Chem SE, A Singh: You have done a nice job laying out the digraphs for the cyclopropene (A) and cyclohexene (B) rings. The double bond in A is accessed early in both directions but only in one direction in B does access to the double bond occur early. Which means that A>B and the double bond is of the (E)-configuration. But I offer you the more rigorous treatment in the diagram below.As you move outward from C1 and C4, sphere by sphere, compare sphere 2, C1 has [2,3,H] while C4 has [9,5,H]. This is a tie. Now move to sphere 3 (marked in red). The C1 path reveals [3,(3),H;2,(2),H] while the C4 path displays [6,(6),H;8,H,H]. Comparing atom to atom, C1 has one more carbon than C4 (See bold (2) and H above). Therefore, the cyclopropene ring (A) has priority over the cyclohexene ring B. The double bond is of  the (E)-configuration given that Cl>C. Recall that all "real" carbons are equal to duplicate carbons (in parentheses) but all duplicate atoms are attached to three atoms of atomic number zero. In some cases, one has to go this far to make a decision.


Space groups: Herman-Maugin notation, diagrams and Wyckoff positions

Well, knowing crystallography implies being able to deduce the picture above solely from the group's name $\rm Pmab$, and that's how it is done.

First you write down the symmetry elements mentioned explicitly in the group's name (that is, $\rm m,\;a$, and $\rm b$ in our case; in other examples there could have been axes among them, or even worse) and orient those accordingly, which you already know how to do. Then you combine these elements in all possible ways to find out what else is hidden beneath.

Essentially, any symmetry element is but a linear transformation that takes any vector $\bf\vec x$ to another vector $\bf A\vec x+\vec b$, where $\bf A$ is the $3\times3$ matrix of our transformation (rotation or reflection or otherwise), and $\bf\vec b$ is the shift, that is, a half-translation along the axis for a screw axis, or a half-translation in whatever direction for a glide plane, or $\bf\vec0$ in all other cases. The typical matrices are:
$$
\begin{array}{cc}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} &
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}&
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}\\
\text{Reflection $\perp$ X} & \text{Rotation about X}& \text{Center}
\end{array}
$$

Given two symmetry elements, you simply combine the two transformations by doing some matrix multiplication, which I sincerely recommend, if only to know how it feels. Alternatively, you may rely upon the wisdom of the predecessors, which says:

Two perpendicular planes give a twofold axis parallel to both.
A plane and a parallel twofold axis give another plane, perpendicular to the plane and parallel to the axis.
A plane and a perpendicular twofold axis give a center.
Two perpendicular twofold axes give a third axis, perpendicular to both.
All of the above remains true if you change

every "plane" to "plane or a glide plane" and
every "axis" to "axis or a screw axis".

When some of the combined elements have shift in them, the following happens:

If the shift is parallel to the resulting element, it becomes a part of the said element (that is, the result is going to be a screw axis or a glide plane).
If the shift is perpendicular to the resulting element, it moves the said element by half the value of the shift (that is, by $1\over4$ of a translation).

If you end up with a center of symmetry someplace other than at the origin, you relocate your origin there.

Now try to apply the rules 1-7 and see where this gets you.


If carbon dioxide is non polar, why does it react with water?

I can only repeat myself here: Polarity is an ill-defined concept that has a nice potential for confusion.
In most cases, when specifying a molecule as polar, one is colloquially referring to the presence of a dipole moment, i.e. one actually categorises the molecule as dipolar. As described by ron in "Why is carbon dioxide nonpolar?", $\ce{CO2}$ has no dipole moment, it is therefore not dipolar, or colloquially it is not polar.
However, $\ce{CO2}$ has two very dipolar bonds, and a significant quadrupole moment. If one were to extend the nomenclature, one would say the molecule is quadrupolar. However, this may lead to complications down the line.
On the other hand, as I have written in the linked question, toluene is often considered an unpolar/non-polar solvent, which is not really true considering it has a small dipole moment.
There are a couple of things one can predict with the concept of polarity, and fortunately, the more complex the molecules become, the better the approximation becomes. It is small highly symmetric molecules, which break these approximations.
Related reading:

Some more extended reading by others and myself can be found in this question: What are dipole moments in a molecule supposed to act upon?

And to further the confusion: Is the triiodide ion polar?

Where to draw the line? What is practical and what not? Why is CO practically nonpolar? What is the dipole moment direction in the nitrosonium ion?


Difference between microcanonical and canonical ensemble

Typically (at least how I learned this after three years of chemistry) the canonical NVT ensemble is created by considering many microcanonical NVE "ensembles" as microstates in this NVT ensemble. I quote from my lecture script for statistical thermodynamics:

  An ensemble with a constant number $N$ of particles in a constant volume $V$ and at thermal equilibrium with a heat bath at constant temperature $T$ can be considered as an ensemble of microcanonical subensembles with different energies $\epsilon_i$.
  The energy dependence of [the] probability density conforms to the Boltzmann distribution. Such an ensemble is called a canonical ensemble.

So your NVT ensemble is many NVE ensembles at different energies. Their statistical weights (the probability of finding a microstate in that particular NVE state) are Boltzmann distributed. The way this distribution of NVE subensembles looks like then defines the temperature of the overarching NVT ensemble.

To illustrate this I have written a small Python program that calculates this distribution for different temperatures of the NVT ensemble. The probability density as a function of the energy of the 48 possible (this number was arbitrarily defined, in principle it could be infinite) NVE subensembles is given below:

As you can see, the probability of finding NVE microstates with low energies drops drastically as the temperature increases. You can think of this as the thermostat taking a NVE microstate at higher energy, removing some of that energy, and then putting it in a lower energy state. Or the other way around if you want to "heat up" a system.


If $1$ liter $\ce{HCl} \pu{1 M}$ is gradually neutralized by adding $\pu{x mol }  \ce{NaOH}$, without change of volume, the pH of the obtained solution is given by : $$\mathrm{pH} = - \log(\frac{1-x}{2} + \frac{1}{2} \sqrt{(1-x)^2 + 4·10^{-14}})$$


What seems to be a set of simple questions leads one into the field of metallurgy, computational thermodynamics, mechanical properties, and goodness knows what else.

For example, steel has very different mechanical properties from pure iron.  There are about a bazillion different kinds of steel, each with different impurities mixed in to give specific mechanical properties, optimizing strength, fatigue life, corrosion resistance, etc. Much of the development has been through insight gained by experience.  More recently, the CALPHAD approach is giving more fundamental insight through thermodynamic modeling, but predicting the mechanical properties $\it a\ priori$ is still difficult to impossible.

As for mercury, there are a large number of binary phases of Hg with other elements.  These are readily found in binary alloy references, or online at the ASM Alloy Phase Diagram Database (assuming your library has access).  ASM has info on >60 Hg binary diagrams, and a number of ternary systems as well.


How do I write out the chemical reaction of ammonia with strong acid?

In the second part, for example, $\ce{Zn}$ is 2+ and $\ce{H}$ is -1. So using the criss-cross method, you get $\ce{Zn(NO3)2}$ and water. I understand this.

I don't agree on the -1 for hydrogen, let's look again at that example:

$$\ce{2HNO3(aq) + ZnO(s) -> Zn(NO3)2(aq) + H2O}$$

Metal oxides, such as zinc oxide, are basic anhydrides.

Imagine that

$\ce{ZnO}$ was made by heating $\ce{Zn(OH)2}$ until it loses water:

$$\ce{Zn(OH)2 ->[\Delta] ZnO(s) + H2O}$$

$\ce{ZnO}$ is a salt of $\ce{Zn^2+}$ and $\ce{O^2-}$

The reaction between nitric acid and zinc oxide and just a neutralization reaction between two protons and $\ce{O^2-}$ (which, as such, isn't present in solution).

Now for the reaction with ammonia: For this you may want to remember that there are different acid-base theories: The one that helps here is the Bronsted-Lowry theory:

Acids are proton donors

$\ce{HNO3}$ definitely is that.

Bases are proton acceptors

How can something else other than the $\ce{OH-}$ that you knew and the $\ce{O^2-}$ from $\ce{ZnO}$ be a proton acceptor = form a bond to a (proton)?

Is that possible if this acceptor doesn't even have a negative charge?

(Yes, otherwise I wouldn't buld up the tension like in a suspense story :D)

If the proton can't provide the electrons for the bond, the acceptor has to deliver them! And that can happen in the form of an electron pair, a double filled, non-bonding orbital.

Therefore a reaction, such as

$$\ce{H+ + NH3 <=> NH4+}$$

is perfectly valid. The rest of the example is just arranging the counter ions.


Water with Sodium VS Alcohols with Sodium

The dominant issue explaining the less vigorous reaction of long-chain (i.e. greater than about 3 carbons) alcohols with metallic sodium is simply a function of the molecular-mass of the alcohol. I will address this aspect last.

At the lower end of the molecular-mass for this series (i.e. water, methanol, ethanol, isopropanol, etc.) the destabilization of the hydroxyl intermediates by the electron-rich hydrocarbon groups (of course absent in water) also slows the progress of the reaction to some degree.

The reaction of sodium metal with water/alcohols is:

1)$$\ce{2Na(s) + 2ROH(l) -> 2NaOR v + H2 ^}$$
where R=H for water, and is the hydrocarbon chain for alcohols.

The inductive destabilization effect mentioned above also largely controls the relative acidities of the alcohols, which in turn are directly related to the reactivity of the alcohol with sodium.  The relative acidities of pure solutions of water and some selected $\ce{C1}$ to $\ce{C3}$ alcohols goes as:

$$\text{water > methanol > ethanol > isopropanol}$$

In the discussion above, we have considered the simple case in which the reactant of interest is also the solvent.  Any comparison of relative concentrations of the different reactants of course requires that we prepare some concentration of the water or alcohol in an appropriate solvent.  It is also desirable to dilute the water or alcohol because of the vigorous, highly exothermic nature of their reactions with sodium.

Because water is the most reactive of the compounds of interest, any sodium added to an aqueous solution of an alcohol would just preferentially react with the water solvent.  And of course you could not measure different concentrations of water in itself!  The solvent of choice in this case is DMSO.  DMSO is not reactive toward metallic sodium, and is miscible with  water, low molecular weight and even very high molecular weight alcohols.  In this system, the relative acidities of neat solutions of the $\ce{C1}$ to $\ce{C3}$ alcohols shown above holds, but with one exception, water:

$$\text{methanol > ethanol > isopropanol > water}$$

Although the acidity of water is low in DMSO due to the poor solvation of the hydroxyl anion (the alkoxy anions are solvated by DMSO in order of increasing hydrocarbon chain length), its reactivity toward sodium remains high.  This is because in the presence of sodium, the hydroxyl anion is the reactive species.  It reacts quickly and does not need to remain solvated for the reaction to proceed.

Of course the effects discussed thus far are also applicable toward higher molecular weight alcohols.  However, once you have a carbon chain of more than 3 or so, these  effects diminish with increasing molecular weight of the alcohol.  Here, the larger the molecular weight of the alcohol, the fewer molecules you have to react for a given weight or volume concentration of the alcohol.  Even if you have a pure alcohol solution, think about the "concentration" of the reactive hydroxyl functional group.  In the case of water, hydroxyl groups are everywhere.  In the case of say a $\ce{C20}$ alcohol, the sodium is mainly surrounded by hydrocarbons with a few reactive hydroxyl groups mixed in and available for reaction.


Is there an error in a Wikipedia article explaining the influence of oxidation states?

Please do not edit the Wikipedia article, as your wording is incorrect, and the existing wording, while awkward, is correct.
What the article is trying to say is that the electron density on each oxygen is less as the oxidation state of the Cl increases. We can understand this by imagining progressive oxidation starting with $\ce{HClO}$. Here, the Cl has oxidation state +1 and the O is -2. But of course the oxygen doesn't have a full negative 2 charge. Some of that charge is shared with the chlorine and the hydrogen, neither of which has a full charge of +1.
To get to $\ce{HClO2}$, we can imagine reacting $\ce{HClO}$ with a neutral oxygen atom (oxidation state 0). After the reaction, the oxygen will have oxidation state -2, and the Cl will now be +3. Again, these are not the formal charges, but give us a hint about where charge will be located. The oxygen will have a partial negative charge, which means the chlorine will have to lose some of its electron density, increasing its actual charge. The original oxygen from $\ce{HClO}$ is still there and feels a stronger "pull" on its electron density, so it gives more to the chlorine, and the oxygen's partial negative charge becomes less negative.
Completely making numbers up, we could imagine that in $\ce{HClO}$, the charge on oxygen is -1, on chlorine is +0.25 and on H is +0.75. After adding the second oxygen to make $\ce{HClO2}$, the charge on the new oxygen might be -0.5, on the already attached O might be -0.75, +0.75 on H and +0.5 on Cl. Do you see how the negative charge on each O gets smaller in magnitude, but the total negative charge on oxygen(s) gets larger in order to offset the increased positive charge of the more oxidized chlorine? The Wikipedia statement refers to the individual O's not the sum.


Do you know a simple fast front-end MD software for OS X

I'm biased, but I recommend Avogadro which can do simple force field optimization using MMFF94 or GAFF. It's not MD, but it would certainly work. It runs on Mac.

The water molecules shouldn't "fly away" during minimization, since there should be hydrogen-bond and Van der Waals terms attracting them to each other and to the protein.

I'll also point out that VMD runs on Mac and are quite easy to use. It can set up MD calculations with NAMD, GROMACS, etc.


Ligand exchange reactions... are they one way or reversible?

Let's take the binding of $\ce{O2}$ and $\ce{CO}$ to hemoglobin as example. $\ce{CO}$ does bind much better than $\ce{O2}$!

We can consider the reaction of oxygen-loaded hemoglobin with $\ce{CO}$ a ligand exchange reaction. If that would be completely irreversible, the logical consequences in the case of an intoxication would be: [...] six feet under ;)

Instead, intoxicated patients are treated with oxygen at higher pressure.

To me, that sounds like Le Chatelier in the ICU and it only makes sense in the case of equilibria.


Deciding which group twists out of the plane when ortho effect takes place

I am not sure the 3D conformation stills as reported on PubChem are all that accurate. Although crystal structures are sometimes referenced for specific molecules, some of the displays are in fact models, some of which do not make sense to me.
Here are some observations regarding the three molecules you noted.
2-Nitrobenzoic acid
The PubChem model shows both oxygen atoms of the nitro- group to be precisely coplanar with the phenyl ring. In the crystal structure, these oxygen atoms have a dihedral angle of $125.5^\circ$ and $129.1^\circ$ with respect to the aryl, i.e. away from coplanarity.
The PubChem model shows both oxygen atoms of the carboxylic acid to be precisely normal to the phenyl ring. In the crystal structure, these oxygen atoms have a dihedral angle of $152.9^\circ$ and $162.7^\circ$ with respect to the aryl, i.e. closer to coplanarity.

Pubchem
Crystal structure entry CCSJWHN

N,N-Dimethyl-2-nitroaniline
I could not find the crystal structure of this molecule. However, I’m not sure it makes sense that the N,N-dimethyl amine adopt an $\ce{sp^2}$ character at the nitrogen center, at least as far as the Pubchem model representation is concerned. In such a pose, there can’t be extended resonance from the phenyl to amine as obtained with unsubstituted aniline. Either the methyls are near coplanar with the aryl to allow the pi orbitals to enter resonance or the amine is in $\ce{sp^3}$.
2’-Aminoacetophenone
I could not find the crystal structure of this molecule. What I found was the 4,5-dimethyl variant as 1-(2-amino-4,5-dimethylphenyl)ethan-1-one.
The 2’-Aminoacetophenone PubChem model shows the acetone oxygen to be precisely $30.0^\circ$ out of the aniline plane. In contrast, the crystal structure of 1-(2-amino-4,5-dimethylphenyl)ethan-1-one indicates that oxygen to be precisely in plane with the aniline. This coplanarity may not be surprising in this case as there is an intramolecular hydrogen bond between the two substituents.

Pubchem
Crystal structure entry 1825318 (BEXTUW)

2'-Aminoacetophenone
1-(2-amino-4,5-dimethylphenyl)ethan-1-one

Original Post:

To my understanding, two things would effect this; one would be the extra stabilisation/destabilisation of a group due to not being in resonance any longer once it moves out of the plane of the benzene ring, and the second would be the "bulkiness" of the group itself, which would energetically be measured as the group's A-Value.

I think steric hindrance, you noted as “bulkiness” and estimated with an A-Values is the primary factor. There is little room for compromise on this one as two atoms can’t occupy the same space. The second factor is resonance as you noted. Another influencing factor is the presence of intramolecular interactions, such as a hydrogen bond. Solvation and pH are other influential factors to the ortho effect.


I did it accidentally, I had coins in the bottom of my school bag and a packet or paracetamol and ibuprofen leaked without me knowing, the tablets got crushed into a powder and were left in the bottom of my school bag with the coins for about 6 months. They're now green, and it doesn't wash off with water and it doesn't look like a coating.


Using table salt as a way to limit humidity in an unoccupied house

According to Transportation Information Service: Salt:

  At up to 74% relative humidity and 20°C, salt does not absorb any appreciable quantities of water vapor. The critical water content of sodium chloride (NaCl) is 0.5% at 74% relative humidity, which is the flow moisture point at which salt begins readily to absorb water vapor, and increases such that, at 75% relative humidity, the salt dissolves.

See this graph for better understanding.

So at high humidity, salt can absorb lots of moisture, enough to become liquid.

When it is liquid, it will creep up the walls of the container by adhesion.

Then, when humidity decreases, the salt will lose almost all its moist, and contract, but leave residue up to the height of the former meniscus.


Calculating approximate pH of polyprotic acids

For the first scenario, the Henderson-Hasselbalch equation is probably good enough. In this case, we assume the major species to be $\ce{HCO3-}$ and $\ce{CO3^{2-}}$ and we will use $K_2$ in the equation.

$$\text{pH}=\text{p}K_2 + \log_{10}{\frac{[\ce{CO3^{2-}}]}{[\ce{HCO3-}]}}$$

I was working on a more general approach using systematic treatment of equilibrium, and it was driving me crazy. I will try to come back to it for the second part of you question.


Why is bleaching with chlorine permanent but with sulphur dioxide temporary?

The bleaching action of Chlorine is permanent because it involves the process of oxidation. $\ce{Cl2}$ reacts with water to produce nascent oxygen which combines with the material in question (by oxidising it, as pointed out by @Mathew Mahindaratne
) and makes it colourless.
The bleaching action of $\ce{SO2}$ is temporary because it involves the process of reduction. $\ce{SO2}$ removes oxygen from the coloured substance to make it colourless. Atmospheric oxygen slowly takes place of the removed oxygen, because of which the material regains it's colour.


The column is filled with the carrier (liquid or gas) before the sample is injected. Thus if there is no interaction between the sample and the column, then the fastest that the sample can get to the detector is the dead time denoted by $t_M$ in the diagram. If the sample does interact with the column, then it is retarded by a time $t_S$  shown in the diagram. Thus the total retention time from injection to detection is $t_M + t_S$.


Resonance of lone pair of nitrogen in 4‐nitroaniline and piperidin‐2‐one

Considering that you're referring to the difference in the number of resonating structures, you're right in that aspect. 4‐Nitroaniline does have more resonating structures than piperidin‐2‐one. But in organic chemistry, you should be knowing that for a phenomenon to happen, there have to be several contributing factors.
While this includes the number of resonating structures as one, you have to take into consideration the other factors as well. And this includes the other factor you have mentioned in your answer — the position of the negative charge.
Since carbon is an atom which has an electronegativity of 2.55 (According to the Pauling scale), a negative charge on it would make it highly unstable, lowering the resonance energy, thus reducing the resonance stabilization in it. Nitrogen and oxygen have electronegativity of 3.04 and 3.44, respectively, thus a negative charge on those atoms would increase the resonance energy, thus increasing the resonance stabilization.


What is the partial molar volume of epsomite (magnesium sulfate heptahydrate)?

As you know, partial molar volume of given substance is the change in solution volume after one mole of this substance is added. When you add $\ce{MgSO4.7H2O}$ to water it dissolves to magnesium cation and sulfate anion but the hydration water simply mixes with the rest of water. So, I suppose that partial molar volume of the magnesium sulfate heptahydrate is the sum of partial molar volume of magnesium sulfate and seven times molar volume of water. If $\widehat{V}_i$ denotes partial molar volume of substance $i$, and $V_i$ denotes molar volume, then: $$\widehat{V}_{\ce{MgSO4.7H2O}}=\widehat{V}_{\ce{MgSO4}}+7V_{\ce{H2O}}.$$
Nevertheless, partial molar volume is function of composition and commonly quoted values are for very low concentration (adding substance to large quantity of water). If you want to obtain solution with higher concentration, finding its exact volume requires knowledge about $\widehat{V}_i$ as a function of composition and such function has to be integrated.


What fraction of air gets expelled upon heating an empty flask?

Your first equation should be:
$${V_1\over V_2} = {T_1\over T_2}$$

We want to know what $V_2$ is in terms of $V_1$. Here, I call $V_2$ final volume. Arranging the equation and using the values for temperature in kelvin:
$${\rm final\;volume} = {V_1 \times 750\over 300}= {5\over 2}V_1$$

To get the amount that was expelled, we subtract the initial volume ($V_1$) from the final volume:
$${\rm amount\;expelled} = {5\over 2}V_1 - V_1 = {3\over 2}V_1$$

The fraction expelled is then:
$${{\rm amount\;expelled}\over{\rm final\;volume}} = {3\over 5} $$


I agree with most of what the commenters and @jerepierre have written, with one or two exceptions.

  Is it possible to create a 100% identical copy of a chemical?

If your question were asked in an introductory chemistry class, I believe most instructors would answer with a simple "yes", for the reasons outlined above.  On the other hand, if your question were asked in an advanced chemistry class, or perhaps in a court of law, then I believe the answer would likely be "no".  A frustrating dichotomy, but such is life.

Let me explain the "no" answer.  To begin with, realize that simple molecules carry deuterium and $\ce{^13C}$ at levels found in natural abundance. Next, note that the way an organism would synthesize a vitamin, hormone, protein, etc., is likely different than the way a chemist might synthesize it in the laboratory. The differences would lie in the raw materials used in the process as well as the actual synthetic pathway.

Each step in a synthesis that requires breaking (or making) a carbon-hydrogen bond or a carbon-carbon bond will result in isotopic fractionation.  Simply put, $\ce{C-H}$ bonds and $\ce{^12C-^12C}$ bonds are easier to break than the corresponding $\ce{C-D}$ bonds and $\ce{^13C-^12C}$ bonds.  Therefore, if we repeatedly break some number of (they can be different or the same) C-H bonds and/or C-C bonds, then our final product will be enriched in molecules with lower deuterium and $\ce{^13C}$ content, the degree of enrichment being dependent upon the difference in the number of carbon-hydrogen and carbon-carbon bond breaking and bond making steps between the synthetic process as carried out within the organism and as carried out on the lab bench.  There may be further fractionation if the starting material used by the body has required C-H and/or C-C bond making and breaking and has, therefore, also suffered isotopic fractionation.  Such fractionation can easily be monitored through comparison of the mass spectra of natural and synthetic compounds.

In addition to this isotopic differentiation between synthetic and natural products, there will also exist a difference in reaction by-products and contaminants created in these two different synthetic processes.

Both methods (isotopic fractionation and by-product/impurity differences) could be used to determine if, for example, the testosterone in an athlete's body was produced naturally within the body or synthetically outside the body.  Also, we are probably all aware of various reports were a synthetic version of a drug (often illicit, but not always) naturally produced by a plant or metabolite has produced death upon administration due to toxic by-products produced in the lab synthetic route - clearly a different side-effect profile.

So in answer to your question, no, it is not possible to synthesize a natural product in the lab that will be 100% identical to the same material produced in a living organism.


Heaviest atom with which chemistry has been performed?

According to Adsorption behavior of super-heavy elements (Z
[greater than or equal to ] 112) on metal and inert surfaces :

  Nowadays the research focusses on the chemical properties of
  the element 114. So far, two gas chromatography experiments to study the interaction strength of
  element 114 with a gold surface have been performed with conflicting results. One experiment [reference 1]
  reported a weak interaction of element 114 with a gold surface, leading to adsorption only at very
  low temperatures of approximately –90°C, while in the second experiment [reference 2] adsorption on gold
  has been observed at the room temperature, indicating a much stronger bond between element 114
  and gold.

See also the earlier: Chemical characterization of element 112 :

  We therefore conclude that the stronger adsorption interaction of element 112 with gold involves formation of a metal bond, which is behaviour typical of group 12 elements


Why is the rate of disappearance negative?

Reaction rates are generally by convention given based on the formation of the product, and thus reaction rates are positive. So, for the reaction:

$$\ce{A->B}$$

$$\text{Rate} = \frac{\Delta[\ce{B}]}{\Delta t}$$

To ensure that you get a positive reaction rate, the rate of disappearance of reactant has a negative sign:

$$\text{Rate} = -\frac{\Delta[\ce{A}]}{\Delta t}=\frac{\Delta[\ce{B}]}{\Delta t}$$


When is equilibrium analysis appropriate?

It depends on the context.

The accepted threshold is "when everything else is slow by comparison." For example, if you have a reaction that takes place in two parts, a Jones oxidation taking a secondary alcohol to a ketone and Diels-Alder (which is far slower), you can assume for modeling purposes that the oxidation will be in equilibrium whenever you're looking at how the Diels-Alder will interact with the oxidation.

By contrast, if you're reacting something like a $\ce{CN7}$ anion (which will react very very quickly, detonating with no provocation), the Jones oxidation can't be considered in equilibrium by comparison.


Reaction quotient and Gibbs free energy at the start of a reaction

Your interpretation of the equation is correct. $\Delta G^o$ gives you thermodynamic favorability of a reaction under standard conditions (Q=1) and even reactions with positive values of $\Delta G^o$ (unfavored under standard conditions) can be driven to proceed if the concentrations of the reactants and products are made extreme enough.

The best way to think about the process is via a curve of G vs reaction extent $\xi$, this is related to Q and ranges from 0 (pure reactants, Q = 0, #1 on the graph) to 1 (pure products, Q = ∞, # 2 on the graph):

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/23.4%3A_Free_Energy_and_the_Gibbs_Function

Any reaction will have an equilibrium point on this curve, corresponding to a most stable composition (# 3 on the graph). The value of the equilibrium constant (K) tells us the x coordinate of the most stable position (i.e. how much reactant vs product we have in the equilibrium mixture).

No matter what the value of K, it will always be in the range $0<K<∞$. If we start the process with pure reactants, $Q = 0$ and we are to the left of the equilibrium point. The system will shift to the right and some of the reactants will convert into products.

This is what you mean by

  ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

you can calculate a negative $\Delta G{_r}$ for the reaction when you have no (or infinitesimal amounts of) products.

Many texts are moving away from the term "spontaneous" for this situation, because the system may not shift to the right (reactants convert to products) if there is a kinetic barrier to prevent this, so the forward reaction may not actually be observed. The term "thermodynamically favored" is often used instead, meaning having a negative $\Delta G{_r}$ under these conditions.

You can find lots more about these concepts at the following posts:

If change in free energy (G) is positive, how do those reactions still occur?

Calculating ΔG at the extremes of reaction extent


How to efficiently reduce of ilmenite to pure titanium?

Ilmenite mainly consists of $\ce{TiO2}$ and iron oxides. The idealized formula of ilmenite is approximately $\ce{FeTiO3}$.

Note that the aluminothermic process would yield ferrotitanium alloys, and the reduction with carbon would yield titanium carbide.

The titanium ore can be converted to titanium tetrachloride according to
$$\ce{TiO2 + 2Cl2 + 2C -> TiCl4 + 2CO}$$
or
$$\ce{2FeTiO3 + 7Cl2 + 6C -> 2TiCl4 + 2FeCl3 + 6CO}$$
(If a large content of iron oxides causes problem, it can be partially removed first by partial reduction with carbon.)

Titanium tetrachloride (boiling point: 136.5 °C) can be purified by fractional distillation.

Finally, titanium tetrachloride can be reduced to titanium sponge with magnesium (Kroll process):
$$\ce{TiCl4 + 2Mg -> Ti + 2MgCl2}$$

The remaining $\ce{Mg}$ and $\ce{MgCl2}$ can be removed with hydrochloric acid.


Determining standard Gibbs free energy for water?

Let's look at the reaction of the formation of water:

$\ce{2H2(g) + O2(g) -> 2H2O (l) }$

The Gibb's Free Energy relates the spontaneity of various reactions by looking at the change in enthalpy, temperature, and entropy.

We can relate those by:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

If we look at our reaction, it's clear that this is an exothermic reaction. Bonds are being formed as hydrogen gas and oxygen gas form water. Moles of gas are decreased and a liquid is formed so entropy is decreasing as well. We know that the formation of water is a spontaneous reaction, so $\Delta G^\circ$ must be negative. Therefore, this reaction must be run at temperatures where the difference of the enthalpy and the product of the entropy and temperature must be < 0.

The Gibbs free energy is also a state function, so looking at $\Delta G^\circ_f$, we can find $\Delta G^\circ_{rxn}$.

$$\Delta G^\circ_{rxn} = \Sigma \Delta G^\circ_{products} - \Sigma \Delta G^\circ_{reactants} $$

Both hydrogen gas and oxygen gas are in their standard states and have a Gibbs Free energy of 0. If you look at tabulated standard thermodynamic data at $25^\circ\ \text{C}$, $\ce{H2O (l)}$ has a Gibbs free energy of$ -237.13 \text{kJ}\cdot\text{mol}^{-1}$.

$$\Delta G^\circ_{rxn} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} - 0 \text{ kJ}\cdot\text{mol}^{-1} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} $$


You're right in that the SN2 Reaction Mechanism will invert the stereocenter. However, the naming convention is based on relative priority and the mechanism by which the product is formed has little to do in how it is named. This appears to be where your confusion lies since you're associating the mechanism and the named configuration when they are not related at all

In the case of an SN2 reaction:

If the attacking nucleophile had the same relative priority as the leaving group (in this case anything with an atomic number > F, since Br is the heaviest among H, F, and C), then you're right, S would turn into R.

However, in the case you quote the nucleophile is MeO-, and O has an atomic number lower than F. Following the CIP rules, we would have to assign the absolute configuration of 'S' even though the stereocenter has been inverted relative to the original compound.


How to remove nitrate from water sample?

Precipitation won't work

Nitrate removal is a very tough problem in wastewater treatment.  One of the reasons is that nitrate salts of nearly every common cation are highly water soluble.  Unfortunately, I can't think of a way to precipitate nitrate from your water sample.

How commercial nitrate removal processes work

Nitrate is removed from wastewater through a microbiological process called denitrification.  In this process, organic carbon substrates are used as food by special anaerobic bacteria.  In the absence of oxygen, the bacteria use nitrate as an electron acceptor so they can oxidize their food and obtain energy via respiration.  As an example organic food molecule, consider hexose sugar.  Then the reaction could be written as:

$$\ce{5 C6H12O6 + 24NO3- -> 30CO2 + 24 OH- + 18H2O + 12N2} $$

It will be difficult to purify a few mL of water in this way unless you know exactly the BOD of the water and are an experienced microbiologist with access to cultures of the appropriate bacteria.

Another not very good option

Another possibility would be electrochemical removal of the nitrate.  That is a topic which has been researched but which is not practices commercially to any significant extent.  Here is one paper where you can read about the technique.  I don't think it would be very easy for a home chemist either.

The tried and true

That leaves more tried and true methods.  You've simply got to distill your water!  Boil it and collect and condense the steam and you'll have substantially purified water to behold and enjoy!


What is the correct regular expression for InChI?

As long as you are not looking for something very specific, and you are not performing a sanity check of what you are matching against, the provided regular expression will match more often wrong than falsely not matching. It will even match more nonsense InChI than total number of existing ones. It is therefore not much better than the trivial catch them all:

^InChI\=1S?\/[^\s]+(\s|$)

The space (and some few others) is a character that will not be part of a correct InChI. The above will therefore match all correct InChI strings (and many more), and certainly more correct ones than the one given in the question (or in your answer).

The principal problem with the whole approach is basically given in the technical FAQ in section 5. Depending on what we are looking at, not all the layers will be present. There is not even a guarantee that a sum formula is present, as represented by the proton: InChI=1S/p+1. This one (and its relatives) is basically responsible why regex is doomed from the start, and that there is no correct regex in the first place.

If we wanted to implement only a tiny fraction of sanity control to reduce the number of mismatches, we will soon see that the whole string becomes about a page long.
Let's examine the given regex a little more to actually understand what it and why it's attempt is futile.

^InChI\=1S?\/[A-Za-z0-9\.]+(\+[0-9]+)?(\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/\?\;\.]+)*$

Beginning with

^

matches the start of the string, which is not really necessary

 InChI\=1S?

matches the prefix, while leaving it open if standard or not

           \/[A-Za-z0-9\.]+(\+[0-9]+)?

matches a sum formula (even the silly ones) or anything that doesn't have a main layer and goes straight to isotopes or charges, etc.. Better would here be to make the sum formula optional, i.e. (\/[A-Za-z0-9\.]+)?

                                      (\/[cnpqbtmsih][A-Za-z0-9\-\+\(\)\,\/\?\;\.]+)*$

this is pretty much a fancy way of saying match any kind of layer, as often as you want,( or not,) in what order is just convenient, with whatever content.
However, it does not match *, which it should: InChI=1S/2C2H5.Zn/c2*1-2;/h2*1H2,2H3; as given in the FAQ.

A much better approach would be to match the layers separately, to have a little control, because they have a predefined order. For example start with the connectivity 1.1 layer (\/c[0-9\-\*\;\(\)]+)?, move to the connectivity 1.2 or hydrogen layer (\/h[H0-9,\(\)\;]+)?. I am not too sure whether I caught all special characters that might be inserted. Then match on with charge (\/q(\+|\-)+[1-9]+[0-9]*)? and the proton balance \/p, the various stereochemistry layers, fixed layers, etc. There are plenty. I probably got the charge already halfway wrong, as I am not sure whether it needs ; or not.

Coming up with a correct and not too greedy regex for each layer is already a nice task, and if you can do something useful with it, good. For everything else, there is a reason that a software exists to create and parse those strings. For every other purpose the greedy match-it-all from the start should work perfectly fine. Now, InChIKeys are something very different ...


Mechanism for ring contraction from 2-methylenecyclobutanol to 1-methylcyclopropanecarbaldehyde

Mechanistically, both of those protonation options (at the alcohol or the alkene) are reasonable. Without any other information, I would probably try both and then see if either gets closer to the product.

However, two things suggest to me protonation of the alkene:

1) As suggested, protonating the alcohol generates a good leaving group (water). However, the product has a C-O double bond, so breaking the C-O single bond appears to be unproductive. That's not to say it's impossible, but it leads me to think about trying something else first.

2) The product has a methyl group, but the starting material does not. Somewhere in the mechanism, that methyl group has to be formed. If protonation of the alkene is the first step, then a methyl is formed, which I see immediately as productive. Again, that's not to say that this protonation is definitely correct, but at least something productive has happened.

One last tip before I show my suggested mechanism. For any mechanism that is difficult to draw, but especially rearrangements, it can be very helpful to draw in the hydrogens. In this case, drawing in the hydrogens just about gives away an exact mapping of the carbons from the starting material to product. With a mapping, it becomes easier to determine all of the bonds broken and formed, which is essentially a checklist of all of the goals of the mechanism.

I'll leave it to you to add the curved arrows, but the intermediates are shown below. Protonation of the alkene in 1 gives carbocation 2, which contains the methyl group seen in the product 4. Next is the ring contraction, sliding a C-C bond over the the carbocation and generating new carbocation 3. Compound 3 might appear to be less stable than 2, but it turns out to be quite stable due to resonance structure 3', which is fully octet. Deprotonation gives 4.


Put simply, there's no precedent for this unusual depiction of a singlet carbene.

Your comment about most carbene depictions, for both singlet and triplet carbenes is true, but a bit lazy. Yes, many chemists, frequently just draw two electrons as a lone pair on the carbon for both. In principle, with a moment's thought, you can often deduce the likely spin state.

As described, by the comment above, the standard for 2D chemical structure depiction is the IUPAC guide Graphical representation standards for chemical structure diagrams

In short, there's a difference in representing a triplet (two unpaired electrons), which should show preferably two separate electron "dots."

For a singlet, the lone pair is indicated, emphasizing the connection between the electron spins.


How to calculate the amount of aluminium that has been oxidised?

Initially, 1 kg of $\ce{Al}$ is present.
After some time, oxidation increases the weight by 1%, or 10 g (e.g. 10 g of oxygen has been added).
The atomic mass of oxygen is 16.0 g/mol.
The molar ratio of $\ce{Al}$ to $\ce{O}$, from the molecular formula $\ce{Al2O3}$, is $\mathrm{\frac{2~mol~\ce{Al}}{3~mol~\ce{O}}}$.

$$\mathrm{\frac{10~g~\ce{O}}{16.0~g~O/mol~O}=0.625~mol~O}$$

$$\mathrm{\frac{2~mol~\ce{Al}}{3~mol~O}*0.625~mol~O=0.417~mol~\ce{Al}}$$

$$\mathrm{\frac{27.0~g~\ce{Al}}{mol~\ce{Al}}*0.417~mol~\ce{Al}=11~g~\ce{Al}}$$

Note that even though only a single significant figure is given for each of steps 1 and 2 of the problem, I assumed that they should read "1.0 kg" and "1.0%" rather than just "1 kg" and "1%" for the purposes of illustrating the above calculation. Thereby, 0.42 mol or 11 g of aluminum was oxidized to $\ce{Al2O3}$.


Predict NMR spectra from the Mass Spectrometry

Sorry, but in my opinion this endeavour is unlikely to be successful.  The techniques are quite different by their underlying principles. This is, at least in part, why chemists working in (especially, but not limited to) organic sythesis use these and additional ones to get a sufficiently "rounded picture" about the substances analyzed, putting the information each technique provides together.

mass spectroscopy typically relies on fragmenting the molecules in question, and then separates these then charged fragments by their mass/charge ratio.  There are techniques keeping the molecules almost intact (softer ionization techniques, used for example in MS-mapping of tissues), and others (electon impact ionization) where the molecules are intentionally shattered into many differently sized fragments.  Depending on the ionization technique applied, the eventually intensity of these individual signals will vary. And if the molecules of interest are embedded in a matrix (e.g., a serum, a tissue), identification of this molecules by MS becomes even more complex.
NMR spectroscopy typically characterizes probes soluted, keeping the molecules of interest (smaller ones, or large proteins) basically intact, without fragmentation.  You basically probe your sample for specific atoms of a kind (isotopes) and their nearest chemical environment. Yet the NMR spectrum of a given sample -- say of an amino acid (building block for proteins) -- already looks differently for scrutinizing the either peripheral hydrogens, the backbone of carbon atoms, or nitrogens.

Addendum

I think a few pictures may provide a glimpse about the complexity here.  For a fairly simple molecule like acetophenone, given the molecular structure is known, a sketcher like Chemdoodle may compute within a blink of an eye an estimate how the NMR spectrum either about the hydrogens, or about the carbon atoms will look like.  This is possible because  the underlying physical principles are known.  In addition, there are very many similar compounds analyzed successfully by NMR, allowing (fine tuning) of details in the spectra simulated, e.g. where to expect the signals to appear.  Regarding MS, a conservative approach is to predict only the peak(s) that by theory should appear without change of the molecule in terms of its mass.  In other words, without fragmentation.  Hence Chemdoodle's estimation is this:

(accessed demo page)

There are equally projects to simulate MS spectra, like CFM-ID.  Under the menu Utilities -> Spectra Simulation, you have the possibilty to try out different modi of ionization (shortcut).  Acetophenone's structure, expressed in a machine-readable Smiles code is

C1=CC=C(C=C1)C(=O)C

Then, just choosing ESI mode about positive ions, without considering adducts, the three MS spectra simulated by CFM-ID of actephenone, in increasing hardness of ionization are these:

As you see, not only the presence / absence of the signals changes in function of the probability to record certain fragments of a specific mass/charge ratio (m/z), their relative intensity equally is affected by changing the parameters of ionization.

(Hovering the mouse over the streaks allows you to display the suggested fragments responsible for the signals in these simulations.  Because of the special conditions in a mass spectrometer, unstable specis you wouldn't be able to store in a shelf may be seen, too.)


Why were elements discovered "out of order" after 1950?

The synthetic trans-uranic elements (the "modern era" elements as you call them) are synthesized by bombarding a certain isotope of one element with a certain isotope of another element with a lot of energy in order to get nuclear fusion. The reason they are "out of order" is that the building blocks of these elements have to be very specific isotopes (extra rich in neutrons for example, somewhat abundant for another). This limits the possible sources considerably and so the heavy elements synthesized (and hence discovered) earlier were those where there was a convenient pair of smaller elements to combine to form the larger ones. You don't just bombard each successive element with protons to make the next one.

A lot of the early discoveries were made by bombarding sources with alpha particles, so there are several discoveries that are increases by 2 protons (e.g. Cm->Cf, Am->Bk). Indeed, the table you linked gives more examples of the pairs of elements used to make the heavy actinides and beyond.


Why do we use coefficients for RFM (relative formula mass) in atom economy?

Before we come to your actual question, we should clarify a few other problems related to quantities and units in your expression:

  moles = mass / RFM

It is essential to distinguish between quantities and units. For example, density is defined as “mass per volume” and not “mass per litre”. In your expression “moles” is a unit name; but it should actually be the quantity “amount of substance”:

  amount of substance =  mass / RFM

Names of quantities (such as “amount of substance” and “mass”) shall not be arranged in the form of an equation. You should use the corresponding quantity symbols:

  n = m / RFM

It is not permissible to use abbreviations (such as “RFM”) for unit symbols (“Mr”) or unit names (“relative molecular mass”).

  n = m / Mr

There shall be a space on both sides of most operators but not for the solidus “/”.

  n = m/Mr

And finally, your equation is wrong, since amount of substance is not given as mass per relative molecular mass. Remember that relative molecular mass is a dimensionless quantity. Actually, amount of substance is given as mass per molar mass:

  n = m/M

Now coming to your actual question.

Atom economy is defined as the quotient of relative molecular mass of the desired product by relative molecular mass of all reactants.

Thus, for a simple reaction

$$\ce{A + B -> Y + W}$$

where $\ce{Y}$ is the desired product and $\ce{W}$ is a by-product that becomes waste, the atom economy would be

$$x=\frac{M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{A})+M_\mathrm r(\ce{B})}$$

However, for a slightly different reaction

$$\ce{C + D -> 2 Y + V}$$

you would get two molecules of the desired product $\ce{Y}$ for each molecule of the reactant $\ce{C}$ and for each molecule of the reactant $\ce{D}$. The yield of product per used amount of reactants is twice as high; therefore, the atom economy is twice as high:

$$x=\frac{2M_\mathrm r(\ce{Y})}{M_\mathrm r(\ce{C})+M_\mathrm r(\ce{D})}$$

In the same way, any other coefficients in the chemical equation would have to be taken into account.


Is the negative excess volume of ethanol/water the cause of my math error?

When you mix ethanol and water they take up less space due to the way the molecules fit together, so you will have what is called a negative volume of mixing.  In this case, the 932.6 mL of ethanol and the 81.9 mL of water will give you a 100 mL solution of 90% ethanol.


Baking soda & magnesium chloride reactions

It is difficult to reconstruct what actually happened based on a single picture.

Nevertheless, the essential effects seem to be the hygroscopy and deliquescence of magnesium chloride. Magnesium chloride has a strong affinity for moisture. When exposed to the atmosphere, the salt will absorb relatively large amounts of water from the air. Magnesium chloride can form several crystalline hydrates: $\ce{MgCl2.2H2O}$, $\ce{MgCl2.4H2O}$, $\ce{MgCl2.6H2O}$, $\ce{MgCl2.8H2O}$, and $\ce{MgCl2.12H2O}$. Furthermore, the salt absorbs moisture from the atmosphere until it dissolves in the absorbed water and forms a concentrated liquid solution.

Apparently, the concentrated liquid solution of magnesium chloride ran out of the glass vessel and formed a puddle on the dish drainer.


I don't remember a direct palladation of arylboronic acids or their derivatives in any of the couplings. Typically, $\ce{Pd(0)}$ inserts into the $\ce{C-Hal}$ bond of arylhalides and is thus oxidized to $\ce{Pd(II)}$. Ouite often, $\ce{Hal}$ is then displaced with $\ce{OR}$ by the action of a base. Think in $\ce{KOt-Bu}$ often used in the Kumada coupling.

In the homo coupling that you described here, there is - for obvious reasons - no $\ce{Ar-Hal}$ for the oxiative addition step. Without $\ce{O2}$ for the oxidative activation of $\ce{Pd(Ph3P)2}$, the mixture will thus remain inactive.


Why would diclofenac have a higher skin permeability when formulated as diclofenac epolamine rather than diclofenac sodium?

Voltaren is a 1 -2% paste of the ammonium salt of diclofenac (epolamine is hydroxyethyl pyrrolidine) and is used a topical pain reliever. You can smell the ammonia.

It could be that the increased pH of the paste softens skin and makes it more permeable. High pH will peel your skin right off!


Is there a way to experimentally measure entropy?

The most common way of measuring $\Delta S^\circ$ for a chemical reaction is probably by making a van't Hoff plot. You measure the equilibrium constant $K$ at different temperatures and plot $\ln K$ vs $T^{-1}$.  The $y$-intercept = $R\Delta S^\circ$ and the slope = $-R\Delta H^\circ$.

Another option is to measure $\Delta H^\circ$ by calorimetry and measure $K$ by some other means. Then compute $\Delta G^\circ$ from $K$ and solve for $\Delta S^\circ$


Using a little extra phenolphthalein won't matter. You just need enough so that the color is visible once the pH becomes basic enough. If you were titrating a large volume, you might need more than a couple of drops, but 2 drops is enough to be sure that the color will be visible in a typical lab-scale titration vessel. One drop would probably even work. The reason they say "two drops" and not "at least two drops" is that they don't want you to be wasteful.


Equilibrium: Effect of change in composition and value of Kp, Kc on changing pressure

$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you  have more gas molecules in the products, so an increase in pressure will drive the equilibrium composition to the reactants.
As for why $K_p$ won't change with pressure:
$\ln K_p=-\frac{\Delta G^⦵}{RT}$ is at a specified $p^⦵$ (usually 1 bar), so $\left(\frac{\partial {\ln}K_p}{\partial P}\right)_T = 0 $.
And as for why $K_c$ won't change with pressure, see the answer here


Each of these materials is largely inert and there is no reason to expect that they would react with each other. Regarding supercritical $\ce{CO2}$, this Wikipedia article states the following:

  Supercritical $\ce{CO2}$ is becoming an important commercial and industrial solvent due to its role in chemical extraction in addition to its low toxicity and environmental impact. The relatively low temperature of the process and the stability of $\ce{CO2}$ also allows most compounds to be extracted with little damage or denaturing.

And regarding silicon nitride:

  Silicon nitride is a chemical compound of the elements silicon and nitrogen, with the formula $\ce{Si3N4}$. It is a white, high-melting-point solid that is relatively chemically inert, being attacked by dilute $\ce{HF}$ and hot $\ce{H2SO4}$. It is very hard (8.5 on the mohs scale). It is the most thermodynamically stable of the silicon nitrides. Hence, $\ce{Si3N4}$ is the most commercially important of the silicon nitrides and is generally understood as what is being referred to where the term "silicon nitride" is used.

Additionally, silicone nitride is used in the most critical, high temperature, high stress of applications, such as the bearings for NASA's (now defunct) Space Shuttle engine bearings.


Why is it necessary to measure the NMR frequency relative to a standard such as TMS?

The use of an standard makes it possible to compare spectra. The frequency itself is dependent on the used magnetic field, while the the chemical shift is independent of the magnetic field and thus comparable using different spectrometers.


Nuclear equations for the decay of radon-222 to lead-206

In any nuclear reactions the sum of the mass numbers $A_i$ (the sum of the indices on the top left) of the reagents and products is always the same. The same applies to nuclear charges $Z_i$ (indices in the lower left, which are often not indicated). Also, keep in mind the notation for metastable state:

$$
\begin{align}
\ce{^{222}_{86}Rn &→ ^{218}_{84}Po + ^{4}_{2}He}\tag{a} \\
\ce{^{218}_{84}Po &→ ^{214\mathrm{m}}_{82}Pb + ^{4}_{2}He}\tag{b} \\
\ce{^{214\mathrm{m}}_{82}Pb &→ ^{214\mathrm{m}}_{83}Bi + ^{0}_{-1}e + γ}\tag{c} \\
\ce{^{214\mathrm{m}}_{83}Bi &→ ^{214\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{d} \\
\ce{^{214\mathrm{m}}_{84}Po &→ ^{214}_{85}At + ^{0}_{-1}e}\tag{e} \\
\ce{^{214}_{85}At &→ ^{210\mathrm{m}}_{83}Bi + ^{4}_{2}He}\tag{f} \\
\ce{^{210\mathrm{m}}_{83}Bi &→ ^{210\mathrm{m}}_{84}Po + ^{0}_{-1}e + γ}\tag{g} \\
\ce{^{210\mathrm{m}}_{84}Po &→ ^{206}_{82}Pb + ^{4}_{2}He + γ}\tag{h} \\
\end{align}
$$


This is no different from what is observed in many binary systems. Here, the phenol and water exhibit a miscibility gap. As you noted, above point C, the liquid is one liquid with whatever composition it has. If one were to have a liquid at the composition corresponding to C, at a temperature above 66C. you would have one liquid. As you dropped the temperature to 66C, nothing would change.

Drop below 66C, and the liquid will phase separate in to two liquids of different compositions, given by the points on the miscibility curve. To figure out how much of each liquid you have, you do the appropriate weighted average as given by the book. That weighted average can be seen to be correct, since, say at 20C, sitting on point A you would be 100% liquid with the A composition, and sitting on point B you would be 100% liquid with the B composition. In between, you are balancing between the two liquids, which are in thermodynamic equilibrium.

If you have a book of phase diagrams handy, look at almost any phase diagram involving Pb, and you will likely find a liquid-liquid miscibility gap. For fun, look at Ag-Ni, two metals with fcc crystal structures, and you will see that the solids are immiscible, and there is even a liquid-liquid miscibility gap.


What's the mechanism of water catalyzing a reaction of hydrogen sulfide gas and sulfur(IV) oxide to yield sulfur?

Yes, hydrogen sulfide and sulfur dioxide reacts with each other in the presence of moisture. The reaction has been studied previously and was noted that the reaction did not proceed if there was no water vapor(dry gases). The mechanism of  reaction was determined and was adsorption. The gases adsorped on the surface of water and collided with each other to deposit sulfur and water. The formation of water further catalysed the reaction. This is a excerpt from the paper which extensively studies the mechanism of the reaction.$\ce{^{[1]}}$

  The results of experiment 3 are best explained by assuming that both
  the  $\ce{H2S}$ and $\ce{SO2}$ are adsorbed and that reaction takes
  place only between these  adsorbed molecules.

There is another paper which extensively discuss the mechanism of reaction. It says that both the gases react with a reaction to form a solution. This is called Wackenroeder solution which is a mixture of sulfur based acid(polythionic acids). These acids dissociate in aqueous solution and form and water. Have a good read at the paper:$\ce{^{[2]}}$

  Volynskii proposed a mechanism for the reaction of $\ce{H2S}$ and
  $\ce{SO2}$ in aqueous solutions. The formation of sulfur follows the
  steps below.

  $$\ce{H2S + SO2 -> HSSO2H }$$ $$\ce{HSSO2H -> S(OH)2 + S }$$
  $$\ce{S(OH)2 + H2S -> 2H2O + 2S (or~S2) }$$

  The overall equation for the three steps is reaction 2. The formation
  of polythionic acids results from the interaction of ionized $\ce{SO2}$ and
  monatomic or polyatomic sulfur, i.e.,

  $$\ce{SO2 + H2O -> HSO3- + H+ }$$
  $$\ce{HSO3- + S -> HS2O3- }$$
  $$\ce{S_{x} + 2HSO3- -> S_{(x +1)}O6- + H2S }$$ $$\ce{S_{x} + 2HS2O3- ->S_{(x +2)}O6- + H2S }$$
  $$\ce{S_{x} + HSO3- + HS2O3- -> S_{(x +3)}O6- + H2S }$$

  where, x can range from 1 to 7. Tiwari believed that the reaction of
  $\ce{H2S}$ and $\ce{SO2}$ in dilute sulfuric acid (below 10 wt %)
  follows the scheme from (7) to (9). His kinetic data show that the
  reaction is first order with respect to both $\ce{H2S}$ and
  $\ce{SO2}$, consistent with the scheme because he assumed reaction 7
  to be the rate-controlling

Reference

$\ce{^{[1]}}$ http://pubs.acs.org/doi/abs/10.1021/j150272a005

$\ce{^{[2]}}$ http://pubs.acs.org/doi/abs/10.1021/ie020275i


Electric field required to split water into oxygen and hydrogen

In part, there is no abrupt cutoff because the gradient of the field is not spread evenly across the electrolyte: the gradient is steep at the interface between electrodes and the electrolyte, where gas evolution actually takes place, not in the body of the electrolyte. Note where bubbles arise in this video.

However, if the conductivity of the electrolyte is low, whether due to electrode separation or low ionization, then electrolysis proceeds more slowly, perhaps imperceptibly, though there is no complete cutoff point.


Why is the vertical part of a titration curve longer for a strong acid by a strong base than for a weak?

Compare (1) the titration of $\ce{NaOH}$ with $\ce{HCl}$ and (2) that of $\ce{NaOH}$ with $\ce{AcOH}$.

At the equivalence point of each titration, solution (1) contains $\ce{Na+}$, $\ce{Cl-}$, and $\ce{H2O}$, with negligible concentration of $\ce{H+}$ and $\ce{OH-}$. In contrast, solution (2) contains $\ce{Na+}$, $\ce{AcO-}$, and $\ce{H2O}$, and in particular $\ce{AcO-}$ will react with water to form a buffer solution of $\ce{AcO-/AcOH}$.

This will be true in general for any weak acid/base titration. The resulting buffer resists changes in pH; hence a weak-acid titration curve is not as steep at the equivalence point as is a strong-acid titration curve.