Will heating alumina (Al2O3) to its melting point yield pure Al?

You've figured out one half of the Hall-Heroult process. Just melting alumina on its own won't liberate the oxygen. The reason is difficult to explain without getting into chemical thermodynamics - one way to think about it is that the lowest energy state is one where oxygen is bound to alumina - where the electrons from alumina have been "taken" by the oxygen. In diatomic oxygen, the electrons are "shared" (the bonds are covalent) and so the total energy is higher. This is a very brief and not-very-detailed explanation, but the end result is that oxygen and aluminum, if given the opportunity, would rather form a bond than not.

When you add thermal energy, unless it changes that balance (it doesn't, at least not until you get way past the melting point), all you are doing is making it more possible for the reaction to happen. If you kept going, you could eventually add enough thermal energy to completely dissociate the alumina, at which point (depending on how you did it) you might be able to remove the oxygen and recover the aluminum.

However - the amount of energy it would take to do that is extremely large. Instead, it turns out to be much easier to simply force oxygen to give its electrons back to aluminum using an electric potential. When you do that, you electrolyze the compound and create diatomic oxygen at the anode and aluminum at the cathode.

The Hall-Heroult process takes this another step to make this work at an even lower temperature (1000 C) by dissolving the alumina in molten cryolite. As a result, it is one of the most (maybe the most) efficient industrial process for producing aluminum, and is responsible for the widespread availability and low cost of aluminum in modern times (and also for the existence of Alcoa).

Your idea about using solar energy isn't bad - although you might be surprised at the amount of energy you need to produce aluminum - both to melt it, and to electrolyze it. One thing you might look into is using solar thermal heating (with a collector, for example) to melt the alumina/cryolite mixture, and then use photovoltaics just for the electrolysis step. That would be quite a bit cheaper than using photovoltaic electricity alone to run an electric heater.

Industrially, the cryolite mixture is kept molten by passing a large electric current through it. The electricity is usually produced by power plants, since the amount required is so large (15 kWh/kg according to Alcoa, and that's just for the electrolysis step!). Although coal and gas power isn't renewable, the advantages are a relatively low cost per unit of power (which might not stay true in the next few decades) and on-demand power generation (likely to always be a big advantage over solar).

There is also nuclear, hydroelectric, and geothermal power - all of which provide large amounts of on-demand power very inexpensively, once the plants are built.


What are the products of the reaction of methoxypropane with HBr?

The first step is protonation of the oxygen of the ether. This cationic species is then subject to nucleophilic attack by Br-. There are two potential sites of attack i.e. the carbons bonded to the protonated oxygen. Bromine is quite a bulky nucleophile, and of the two sites, the methyl group carbon is less hindered so this is attacked in preference, leading to bromomethane and propan-1-ol as your products.


Why do hairs (rubbed with plastic comb) fallen on water seems to be sticking?

It doesn't have to do as much with the combing as it does with surface tension. Water on the hair clumps them together, because smaller clumps have more surface area (net surface area) and thus more surface energy. Breaking these large clumps, therefore, requires energy input, which gives us the impression that they "stick" together.

As for sticking to the water individually, that's surface tension again. The surface is able to withstand a bit of weight before collapsing and letting the object drop through. Hairs are oily, and thus hydrophobic. The repulsion potential energy that would be formed / needed when it sinks outweighs the energy required to bend the surface (and get enough tension force to balance gravity). This is how water skeeters work.


For real gases there are some contradictions in the assumptions made while defining real gases.
Therefore some corrections were needed in the equation which were given by van der waals in his real gas equation
The van der Waals equation corrects for the volume of, and attractive forces between, gas molecules:

(P+a(n/v)^2)(V-nb) = nRT

There are two corrective factors in van der Waals equation. The first,   , alters the pressure in the ideal gas equation. It accounts for the intermolecular attractive forces between gas molecules. The magnitude of a is indicative of the strength of the intermolecular attractive force.

The factor - nb accounts for the volume occupied by the gas molecules. b has units of L/mol. Since b corresponds to the total volume per mole occupied by gas molecules, it closely corresponds to the volume per mole of the liquid state, whose molecules are closely layered. b is generally much smaller in magnitude than a . The values of a and b generally increase with the size and complexity of the molecule.


About purifying ethanol - What type of ethanol?

Forget about the denatured alcohol! Additives are typically chosen to render their removal difficult.

Supposed you have decent setup for distillation, analytics still might be a problem.

How will you determine whether your destillation was successful or whether it just was "thermal transfer" from one flask to another?

A hydrometer to measure the density of the liquid could be helpful.

Use some cheap vodka, mix with water and try to distill the vodka off. Measure the density before and after distillation. Note the temperatures, read about strange things that can happen: azeotropes. Cheers!

If you want it more simple and obvious, mix the vodka with some food colourant. Proceed as before and watch the effect. Cheers again.

Note that one usually would not digest anything that went through any lab equipment  for mixed used at a normal lab. But this is your home lab and you know what went through which labware. Anyway, keep your equipment clean and take care. Btw: Cheers!


In general nomenclature, the practice of omitting locants when there is no ambiguity is widespread. For preferred IUPAC names, however, the current recommendations given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) are prescriptive about when locants are omitted. In particular:

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(b) in monosubstituted homogeneous chains consisting of only two identical atoms;
(…)

Therefore, the preferred IUPAC name is indeed chloroethane and not the (correct, but not preferred) complete systematic name ‘1-chloroethane’.
(Nevertheless, note the correct capitalization and hyphenation of the names.)


What are the enthalpy and entropy contributions to carboxylic acids dissociating in water?

Electrostatic screening associated with solvation is obviously the key factor enabling dissociation in aqueous solution when compared to the gas phase. Without such dielectric screening, dissociation would be too costly ($\approx \pu{1 MJ/mol}$ or $\approx\pu{14 eV}$,  roughly the energy required to ionize a hydrogen atom) to occur to a significant extent at RT. The dielectric constant of water reduces the energy of the gas phase electrostatic interaction by a factor of $\approx 80$. This is an enthalpic effect. This energy is further reduced by water oxygen bonding to $\ce{H^+}$ to form $\ce{H3O+}$ (proton to hydronium).

Differences between dissociation constants of different carboxylic acid groups originate from enthalpy or entropy changes due to anion formation (since the formation of protium is something shared in common by all acids, its solvation cannot explain differences in behaviour of different acids).  The effects on the solvent are often described in terms of "structure making or breaking". As a rule there is an entropic penalty ($\Delta S^\circ<0$) with dissociation that can be attributed to solvent rearrangement around the ensuing charged species. The penalty is seen in thermodynamic data for many weak acids.  Trends are otherwise not quite as clearcut. Beyond the entropic penalty, other factors (enthalpic or entropic) may play a role in encouraging or discouraging dissociation, for instance charge delocalization (inductive, mesomeric effects) or intramolecular rearrangements.


How can metals like magnesium and sodium burn underwater?

Your last sentence was the key.  Metals like sodium and magesium react with water according to the following equations
$$\ce{2Na + 2H2O -> 2NaOH +H2 + heat}$$
$$\ce{Mg + 2H2O -> Mg(OH)2 +H2 + heat}$$
Both reactions produce hydrogen gas and a lot of heat.  As the hydrogen gas bubbles out of the water it mixes with atmospheric oxygen. The intense heat ignites the hydrogen-oxygen mixture and the mixture continues to burn (possibly explosively depending upon the quantities involved) as long as hydrogen continues to be generated.


What will form when Dihalides or Trihalides or Polyhalides react with Magnesium?

1,4-dibromobutane reacts with magnesium to form cyclobutane as main product. The  the side products are buta-1,3-diene, ethene, butane, (E) and (Z) but-2-ene. The reaction has been performed in xylene solvent to give a mixture of hydrocarbon(81%) and cyclobutane(13%). The best yield for cyclobutane from 1,4-dibromobutane can be obtained by using lithium amalgam in dioxane for 3 hours.

In general, for any dibromoalkane reacting with magnesium will always lead to formation of alkene or cycloalkane and magnesium bromide as side product. For e.g.- 1,2 dibromoethane reacts with magnesium to form ethene and magnesium bromide. In fact, it is being used as activating agents as its action can be monitored by the observation of bubbles of ethylene and also the side-products($\ce{MgBr2}$) is almost negligible and does not affect the reaction. (Wikipedia)
$$\ce{Mg + BrC2H4Br → C2H4 + MgBr2}$$
The reaction mechanism is that any dibromoalkane reacting with magnesium will first lead to magnesium bromine adduct of the alkane chain which will then decompose to form alkene or cycloalkane and magnesium bromide as side product.
$$\ce{Br-(CH2)_n-Br + 2 Mg ->[ether] BrMg-(CH2)_n-MgBr}$$
1,3-dibrompropane reacting with magnesium will form cyclopropane and magnesium bromide as side product.
$$\ce{BrCH2CH2CH2Br + Mg -> (\Delta) + MgBr2}$$

Scientists have discovered more exotic species called magnesacycloalkanes which is said to maintain a equilibrium between magnesium bromine adduct of the alkane chain and the final product. Some of the species include:

Now, the species formation depends on position of bromine atoms in alkane chain:

1,3 DiGrignard reagent species: formed by reacting 1,3-Dibromopropane and magnesium to form 3-Bis(bromomagnesio)propane (4a) and 1,6—bis(bromomagnesio)hexane (15%) as side product.

1,2-DiGrignard reagent species: Formed when aromatic 1,2 Dihalides are used

1,1-DiGrignard reagent species: Also called methylene—di—Grignard reagent It is prepared from dibromo- or diiodomethane.

$$\ce{CH2Br2 + Mg ->[(i-Pr)2O] CH2(MgBr)2 }$$
References

Bickelhaupt, F.. “Di-Grignard reagents and metallacycles.” Pure and Applied Chemistry 58 (1986): 537 -542, DOI: 10.1351/pac198658040537
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm
Science of Synthesis: Houben-Weyl Methods of Molecular Transformations Vol. 48: Alkanes, Georg Thieme Verlag, May 2014.


Explaning Cis & trans isomers through VSEPR

Answered my own bonus question (I believe): In case anybody was interested,

Trans seems to be favoured due to the methyl group on either side of a Cis bond repelling eachother (due to to both the methyl groups wanting to retain their tetrahedral structure, hydrogens are repelling one another and hence distorting things), and hence the trans is more energetically favourable.


Are there any poisons with an activation threshold?

Yes, there are certainly some chemicals that are harmless in low quantities but really dangerous in higher doses.

The common painkiller paracetamol (or acetaminophen) is one such candidate. In low doses it is a mild analgesic but in higher does it is deadly. The reason is that there is a metabolic side-reaction in the liver that produces a small amount of a toxic compound. Under normal circumstances this is mopped up by other pathways in the liver and causes no damage. But when large amounts are taken the liver's capacity to mop up the toxin is exhausted, the toxin builds up to dangerous levels and it causes severe liver damage often resulting in death.


Is it possible to use the school supplied algorithm to build the Lewis diagram of the iodate ion?

When I follow your school's algorithm for iodate I get a central iodine atom with three single bonds to oxygen.  I've shown my work below.  This is problematic, as you well know, but I can potentially see where your teacher is coming from.

When I teach formal charge and resonance structures and the like, I often have students use their intro chemistry rules to determine the structure of a polyatomic ion such as this.  If they follow the rules they get a pretty structure that looks proper - I then introduce the idea of formal charge by pointing out the wild imbalance of charges in the ion and how energetically unfavorable that would be, then we work out a way to rearrange the electrons to minimize that problem.

I also have a lot of students who immediately go to google to find Lewis structures rather than drawing them themselves, so when a student tells me that they think the structure should be something different, it takes an actual argument from them before I believe that they know why that should be.  I don't know that that is the case here; it also seems entirely possible that your instructor is just wrong.


The preferred IUPAC name is butan-2-one. This is explicitly mentioned in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the Blue Book:

P-14.3.4.1 terminal locants
P-14.3.4.2 the locant ‘1’
P-14.3.4.3 where there is only one kind of substitutable hydrogen
P-14.3.4.4 when no isomer can be generated by moving suffixes or prefixes from their position to another or by interchanging them between two different positions
P-14.3.4.5 when all substitutable positions are completely substituted
P-14.3.4.6 when all substitutable hydrogen atoms have the same locant.

None of these cases applies to butan-2-one. Therefore, the locant cannot be omitted in the preferred IUPAC name.


Cause of imaginary part of Warburg element in EIS

I am sorry, this is not an answer but a comment, too long to be edited in the comments section.

I don't agree with "straightforward" in the sentence :
" It is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. "

This would be an oversimplified view which implicitly considers that the phenomena involved are only on the energy dissipative kind (commonly modelized with resistors elements in the equivalent circuits). This cannot be so in case of Warburg impedance. Necessarily the phenomena involved includes electrical energy storage elements which are stocking energy during a part of the period and destocking during the other part of the period of the alternative current (commonly modelized with capacitor elements). One cannot forget the part of the alternative current due to the phenomena of energy storage kind. If we forget that, there is no phase shift and no complex impedance, thus no Warburg element.

In an intuitive way, but discutable because alternative models can be proposed, if we think that the more the diffusion layer grows, the shallower the concentration gradient gets also the shallower the electrical gradient of potential gets. So, the energy storage phenomena also decreases and the "out of phase" part of alternative current decreases as well. This is a loosely way for an intuitive approach.

The case of Warburg element (which is a particular case of CPE : http://www.consultrsr.net/resources/eis/cpe1.htm) can be related to dissipative energy phenomena and storage energy phenomena $\textbf{homogeneously}$ distributed. A possible equivalent circuit can be made of repeated identical pattern of resistor-capacitor. This is a particular case of network for example of the kind represented page 10, figure 7(a) in the paper : https://fr.scribd.com/doc/71923015/The-Phasance-Concept , only with the part noted $P_\varphi$, without $R_p$. Of course different interpretations and equivalent circuits can be proposed, depending on the kind of electrochemical system. This is briefly discussed page 26 in the above referenced paper.


K° here means elemental potassium (zero oxidation state). The author wanted to avoid the confusion of K with the equilibrium constant. There are more quantities than we have symbols, so letters and symbols get re-used. It is called a pigeon hole effect.
Now, that the OP has posted the reference, Solomons, Organic Chemistry, Ed 11, Problem 11.14 Solution manual, it is clear that the circle is a typo for zero. In the 11 and the 12th edition, the symbol stands corrected to proper numerical "zero" in the main textbook (pg 510). It is not unusual to find typos in solution manuals and textbooks. We are all humans-after all.


Why are wavefunctions with same principal quantum number and different azimuthal quantum number orthogonal

The mathematical answer to this specific question lies in the fact that s and p orbitals are simultaneous, non-degenerate eigenfunctions of the quantum mechanical $\hat{L}^2$ operator, which is hermitian:

$$\begin{align}
\hat{L}^2\left|\mathrm{s}\right> = [l(l+1)\hbar]^2 \left|\mathrm{s}\right> &= 0 \\
\hat{L}^2\left|\mathrm{p}\right> = [l(l+1)\hbar]^2 \left|\mathrm{p}\right> &= 2\hbar^2 \left|\mathrm{p}\right>
\end{align}$$

since $l = 0$ for s orbitals and $l = 1$ for p orbitals.

It is a simple proof that the eigenstates of hermitian operators that correspond to different eigenvalues are necessarily orthogonal. Suppose, for a hermitian operator $A$, we have $A\left|m\right> = a_m\left|m\right>$ and $A\left|n\right> = a_n\left|n\right>$ with $a_m \neq a_n$. Then:

$$\begin{align}
\left\langle m \middle| A \middle| n\right\rangle &= \left\langle n \middle| A \middle| m\right\rangle^* & &\text{(definition of hermiticity)} \\
a_n\left\langle m \middle| n\right\rangle &= (a_m\left\langle n \middle| m\right\rangle)^* & &\text{(extracting the eigenvalues)} \\
&= a_m\left\langle n \middle| m\right\rangle^* & &\text{(eigenvalues of hermitian operator are real)} \\
&= a_m\left\langle m \middle| n\right\rangle & &\text{(definition of inner product)} \\
(a_n - a_m)\left\langle m \middle| n\right\rangle &= 0 \\
\left\langle m \middle| n\right\rangle &= 0 & &(a_m \neq a_n)
\end{align}$$

You can likewise show that any pair of atomic orbitals in the hydrogen atom are orthogonal, using whichever of the three operators $\hat{H}$, $\hat{L}^2$, $\hat{L}_z$ that you need. All atomic orbitals are simultaneous eigenfunctions of these three operators, and have different eigenvalues which correspond to the quantum numbers $n$, $l$, and $m_l$:

$$\begin{align}
\hat{H}\left|n,l,m_l\right> &= E_n\left|n,l,m_l\right> = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 n^2}\left|n,l,m_l\right> \\
\hat{L}^2\left|n,l,m_l\right> &= L^2\left|n,l,m_l\right> = [l(l+1)\hbar]^2\left|n,l,m_l\right> \\
\hat{L}_z\left|n,l,m_l\right> &= L_z\left|n,l,m_l\right> = m_l\hbar\left|n,l,m_l\right>
\end{align}$$

Since no two orbitals have the same three quantum numbers, no two orbitals can have the same eigenvalues for all three operators. Consequently, there must be at least one operator which can be used to show that those two orbitals are orthogonal.


The molecule in black consists of a cyclic ketal and a hemiacetal. The hemiacetal is known to be in equilibrium with its aldehyde form, which is the the molecule in blue, in aqueous solutions even without $\ce{NaOH}$. Recall the reaction of glucose (a cyclic hemiacetal) with dilute sodium hydroxide, which is known as Lobry de Bruyn-van Ekentein rearrangement (Ref.1).
Nonetheless, a hemiacetals can be decomposed to the relevant carbonyl and the alcohol under basic conditions as explained here (a reliable mechanism is given). In your case both carbonyl and the alcohol groups are within the same molecule (the molecule in blue):

All acetals and ketals are stable to alkaline conditions since they don't carry free hydroxyl group.

References:

Roy L. Whistler, J. N. BeMiller, "Alkaline Degradation of Polysaccharides," Advances in Carbohydrate Chemistry 1958, 13, 289-329 (DOI: https://doi.org/10.1016/S0096-5332(08)60359-8).


Finding EMF of a galvanic cell without standard potentials

How to start solving the problem since the $E^\circ$ of individual half reactions is not given?

This is a concentration cell, i.e. the half reactions at the anode and at the cathode are the same (except for the direction).

AgCl(s) electrode
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
$$\ce{Ag+(aq) + e- -> Ag(s)} $$

AgBr(s) electrode
$$\ce{Ag(s) -> Ag+(aq) + e-}$$

$$\ce{Ag+(aq) + Br-(aq) <=> AgBr(s)}$$

The standard reduction potentials will cancel out, i.e. $E^\circ (\mathrm{cell}) = 0$.

Further thoughts

  [Comment by EdV] The Ag|AgCl electrodes I have are commercial, but making them in the lab is just a matter of oxidizing Ag wire in a chloride solution, so the electrode is Ag wire with an adherent coating of AgCl... I have never seen one of these made by just sticking an Ag wire in the Ag halide, but I guess it would work.

I added more to the answer prompted by that thoughtful comment.

  [...my own comment] I pictured the silver electrode submerged in the solution, with the solid halide on the bottom. I am puzzled now too. Does it make a difference if the electrode touches the solid, the liquid, or both?

What are the actual reduction potentials?

$$\ce{AgCl(s) + e− <=> Ag(s) + Cl−}\ \ \ \ E^\circ_\mathrm{red} = \pu{+0.22233 V}\tag{1}$$
$$\ce{AgBr(s) + e− <=> Ag(s) + Br−}\ \ \ \ E^\circ_\mathrm{red} = \pu{  +0.07133 V}\tag{2}$$
$$\ce{Ag+ +  e−  <=> Ag(s)}\ \ \ \ E^\circ_\mathrm{red} = \pu{  +0.7996 V}\tag{3}$$

Are they related?

If you subtract (3) from (1), you get the dissolution reaction of AgCl, if you subtract (3) from (2), you get the dissolution reaction of AgBr. Thus, standard reduction potentials for (1) and (3) should be different by

$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgCl})$$

and standard reduction potentials of (2) and (3) should be different by

$$ -\frac{RT}{zF} \ln K_\mathrm{sp}(\ce{AgBr})$$

Finally, standard reduction potentials (1) and (2) should be different by

$$ -\frac{RT}{zF} \ln \frac{K_\mathrm{sp}(\ce{AgBr})}{K_\mathrm{sp}(\ce{AgCl})}$$

Numerical answer using half reaction (1) and (2)

$$\ce{AgCl(s) + Br-(aq) <=> AgBr(s) + Cl-(aq)}$$

$$Q = \frac{[\ce{Cl-}]}{[\ce{Br-}]} = 200$$

$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$

$$= \pu{(0.22233 V− 0.07133 V) - 0.13612 V = 0.0149 V}$$

Numerical answer using half reaction (3) twice

$$\ce{Ag+(c) + Ag(b) <=> Ag(c) + Ag+(b)}$$

"c" stands for chloride side, and "b" stands for bromide side. For consistency, I am using the following values for the solubility products (derived from difference of standard reduction potentials of half reactions (1), (2) and (3)).

$$K_\mathrm{sp}(\ce{AgCl}) = \pu{1.74e−10}$$
$$K_\mathrm{sp}(\ce{AgBr}) = \pu{4.89e−13}$$

$$[\ce{Ag+}]_c = K_\mathrm{sp}(\ce{AgCl}) / [\ce{Cl-}] = \pu{8,27e−10}  $$
$$[\ce{Ag+}]_b = K_\mathrm{sp}(\ce{AgBr}) / [\ce{Br-}] = \pu{4.89e−10}  $$

$$ Q = \frac{[\ce{Ag+}]_c}{[\ce{Ag+}]_b} = 0.560 $$

$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{R T}{z F} \ln(Q)$$

$$\pu{= 0 - (-0.0149 V) = 0.0149 V}$$


What makes the enthalpy of hydration of calcium chloride dihydrate unique?

The heats of hydration of your salts, plus some others, can be obtained by subtracting the heat of formation of the anhydrous from the heat of formation of the hydrated salt, using the numerical values from the Handbook of Chemistry and Physics, 43rd Ed., 1961, pages 1807 - 1831. The so obtained values are rather different from yours, but the tendency is the same. Here are the obtained molar heats of hydration, divided by the number of water molecules. They are strangely similar.
$\ce{Na2CO3·H2O....... 295.3 kJ (mol H2O)^{-1}}$
$\ce{Na2CO3·10 H2O..... 295.3 kJ (mol H2O)^{-1}}$
$\ce{NaCH3COO·3H2O ... 297.6 kJ (mol H2O)^{-1}}$
$\ce{Na2SO4·10H2O ..... 294.3 kJ (mol H2O)^{-1}}$
$\ce{Na2S·4.5H2O ...... 305.0 kJ (mol H2O)^{-1}}$
$\ce{Na2S·5H2O....... 304.6 kJ (mol H2O)}^{-1}$
$\ce{Na2S·9H2O ....... 301.1 kJ (mol H2O)^{-1}}$
$\ce{Na2S2O3·5H2O..... 297.2 kJ (mol H2O)^{-1}}$
$\ce{Ca(NO3)2·4H2O.... 297.7 kJ (mol H2O)^{-1}}$
$\ce{Ca(CH3COO)2·H2O . 291.2 kJ (mole H2O)^{-1}}$
$\ce{CaCl2·2H2O ...... 179.6 kJ (mol H2O)^{-1}}$  ........ (!)
$\ce{CaCl2·6 H2O ......301.8 kJ (mol H2O)^{-1}}$
It is immediately clear that the heat of hydration of one molecule $\ce{H2O}$ in these compounds is $\ce{300 ± 10 kJ (mol H2O)^{-1}}$. But there is one odd exception : $\ce{CaCl2·2H2O}$. Has anybody an explanation ?
I would have been happy to show my previous numerical values in a table, or an array with lines and several columns, showing the different heats of formation. But I have not been able to write it correctly. There was always a message saying : \hline incorrect. Can somebody tell me how to proceed ? Thank you in advance.


Is CO2 produced when KHCO3 reacts with H2O?

Compiling the comments by Max and orthocresol:

  What is the reaction taking place to produce CO2?

$$\ce{HCO3− (aq) + H2O (l)
<=> H2CO3 (aq) + OH− (aq)
<=> H2O (l) + CO2 (aq) + OH−(aq)}$$

or
\begin{align}
\ce{HCO3− (aq) + H2O (l) &<=> H2CO3 (aq) + OH−}\\
\ce{H2CO3 (aq) &<=> H2O (l) + CO2 (aq)}
\end{align}

  $\ce{KHCO3}$ does not readily decompose into $\ce{CO2}$ when dissolved in water.
  It is just that the plant is using the $\ce{CO2}$, so that pulls the equilibrium over, Le Chatelier's principle.


Coagulation of milk is a result of casein micelles clumping together. This is favoured by higher temperatures and also by favour a pH of around 4.6 as this is the isoelectric point. At this isoelectric point, the amino acids within the casein micelle are electrically neutral. This means there is less electrostatic repulsion between the casein micelles, facilitating coagulation Furthermore, the addition of acid means that the kappa casein, i.e. hairs on the outside of the casein molecule which stabilise it, become protonated. This protonisation causes them to destabilise, making it easier for the micelle to fall apart and clump together with other micelles that have fallen apart. Relatively High temperatures also increase reaction rates, which overall speeds up the aforementioned processes as well.

Even if the coffee does not reduce the pH by much, any change in pH which leads to the pH being closer to the isoelectric point (4.6) of the casein micelle will increase the rate of the coagulation process.

Using old milk also increases coagulation, as milk naturally coagulates due to rennet, an enzyme, cleaving the kappa-casein, causing the casein molecules to destabilise.

To avoid coagulation during heating, perhaps try heating the milk gently. The ideal temperature at which milk coagulates is about 43ºC, so perhaps staying below that could work. This source also provides tips on how to avoid coagulation while heating milkAlternatively, you have yourself some cheese to enjoy :)


I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the preface to this method, including the potential concerns.
Essentially, the book talks about determining the type of reactivity of inorganic molecules by looking at their structures. While this may not reflect with 100% accuracy the actual reaction pathway, we can take it to be a pretty good approximation of what happens in reality. While the completely correct mechanism can only be found out experimentally, we can get a probable mechanism by employing an arrow-pushing approach, and that should at least mirror the actual pathway to a sufficient extent, if not entirely.
Now, if we look at the structures of Aluminium oxide and Gallium oxide ($\ce{Al2O3}$ and $\ce{Ga2O3}$), we can see that they possess a similar bridged dimer-kind of structure as shown below for aluminum oxide(the same structural skeleton is valid for gallium oxide as well, just by replacing aluminum with gallium below) :

This structure now possesses some unique properties. It contains an $\ce{Al-O-Al}$ bridge with an $\ce{Al=O}$ group on each side.Now,the $\ce{Al=O}$ group can be said to be structurally similar to the carbonyl group from organic chemistry.
Hence, like the carbonyl group, it can be said that there will be high partial positive charge on the $\ce{Al}$ atoms due to the electron-withdrawing effect of the $\ce{Al=O}$ groups. Hence, they can serve as the ideal electrophilic site for the attack of a nucleophile like $\ce{OH-}$ in basic medium, and lead to bond cleavage(and further reaction) in alkali as follows:

In a basic medium, the $\ce{AlO2H}$ moiety will also be deprotonated, and thereby you will end up with two moles of $\ce{AlO2^-}$ . If the base being used was $\ce{NaOH}$, then these products are better reported as $\ce{NaAlO2}$ for aluminum. In an analogous case, you would get $\ce{NaGaO2}$ if the we had started out with gallium oxide.
In excess of alkali, the reaction can proceed further, as further additions of hydroxide are still possible at the $\ce{Al=O}$ site as follows:

The reaction can either finish after the formation of the hydroxide, or a continued process may lead to the addition of yet another $\ce{OH^-}$ to form the complex ion $\ce{[Al(OH)4]^-}$ , which might be called the aluminate ion here.In another analogous case,we would have ended up with $\ce{[Ga(OH)4]^-}$ (called the gallate ion) if we had started out with gallium oxide.
Now, for the acidic medium, let us suppose that $\ce{Al2O3}$ reacts with $\ce{HCl}$. Here, the central oxygen of the $\ce{Al-O-Al}$ bridge represents a weak link that can be easily cleaved after protonation(because it has moieties that can function as good leaving groups on either side). Also,the counter-ion of the acid(that is, $\ce{Cl^-}$), can serve as a nucleophile to break down the dimer skeleton after protonation as follows:

The $\ce{AlO2H}$ can again be protonated to form an oxonium ion that can expel water and take up a $\ce{Cl^-}$ to form another mole of $\ce{AlOCl}$(shown in the next diagram).
Following that, the $\ce{AlOCl}$ can be engaged in a repetitive process of protonation, attack by chloride and removal of water to yield the final products as follows:

The products for these reactions can be verified for aluminium and gallium oxides from their Wikipedia pages respectively.
So we can see that one major reason for the amphoteric nature of oxides can be their bridged structures, which allows them to react with both acids and bases as illustrated above.
Now, coming to indium oxide, the formula of this compound as stated here is $\ce{In2O3}$, which bears a resemblance to the formula of $\ce{Al2O3}$ and $\ce{Ga2O3}$ which we stated above. So, I would expect this oxide to atleast have some amphoteric properties. This is indeed the case, as stated here for indium hydroxide(since hydroxides are simply hydrated forms of their oxides, we can expect a similar nature for indium oxide as well):

The result indicated conclusively that In(OH)3 is only slightly amphoteric, being much more basic than acidic.

Also, on the Wikipedia page for indium oxide it clearly states that this oxide is soluble in acids but not in alkali. This is pretty much expected because when you move down a group, the elements tend to become more metallic in nature, and hence their oxides tend to become more basic as well. Hence, an amphoteric oxide slightly tending towards the basic end seems reasonable for indium(III) oxide. The reaction between acid and $\ce{In2O3}$ should follow the general scheme that is illustrated above for $\ce{Al2O3}$ and $\ce{HCl}$, and should yield similar products. This indeed is the case as can be seen here
As for thallium, due to the inert pair effect, it generally forms an oxide in which thallium is in the +1 oxidation state, with the formula $\ce{Tl2O}$ as stated here. This doesn't fit in with the general formulation of $\ce{Al2O3}$ and $\ce{Ga2O3}$ that we discussed above. Hence, one would expect this oxide to simply become more basic than the other group members(as it is an oxide of the lower-most element of the group) and not show much signs of an amphoteric nature. This matches with what is stated here.
Before wrapping up the answer, I would like to re-iterate a point I made at the beginning of the answer:
All these pathways are simply possible mechanisms, and there is no way of knowing whether they are completely correct or not without performing experiments. However, they are highly probable propositions, and you can expect these pathways to represent the actual mechanisms to a good approximation


I dont know about HPLC column tolerance to various reducing agents, this would need to be checked. However, in biochemistry we routinely add reducing agents to avoid disulphide-bridge formation in the proteins.
Overall there are three main reducing agents used, to avoid this problem. The choice of which one to use depends on your needs. They include:

TCEP: Considered the golden standard, it is relatively odourless compared to the other reducing agents. Once you have dissolved it, it should also remain very stable over time. Typically used at a 0.4 mM concentration.

DTT: Cheaper than TCEP, although I have heard that it is not as stable over time

2-Mercaptoethanol: The absolute cheapest, stinks like crazy. Typically to a final of 5 mM.


Cyanide is generally better than iodide in transition metal complexes because it acts as a pi-acceptor. It has empty orbitals that overlap with the partially filled d-orbitals of the metal, lowering the energy of these electrons.

Iodide is a sigma-donor, meaning that it donates electron density to empty metallic orbitals. The effect is small compared to a pi-acceptor in a transition metal, because the major contribution to stability is the energy of d-orbital electrons.

However in Pb(II) complexes, the d-orbitals are full already, so here the sigma-bonding mechanism for stabilisation is key. As cyanide isn't as good of a donor as iodide, then iodide forms a complex whereas cyanide doesn't.


Is (2R,3S)-2,3-dichloropentane optically inactive?

To have an optically inactive meso compound with two chiral centers:

1)  the chiral centers must have opposite configuration

2)  the chiral centers must have the same set of fragments bonded to them

Among other things the 2 carbon has a methyl group bonded to it versus the 3 carbon having an ethyl group instead.  You do not satisfy (2), so you do not have the symmetry required for a meso compound, and the molecule is still chiral and optically active.


Is there a real detection limit in NMR spectroscopy?

This is an interesting problem which also appears in gamma spectroscopy. Both 1D NMR and normal gamma spectroscopy are related. A sample emits photons, the photons are then captured by a detector.

In the case of a modern FT NMR machine the nuclei are excited by a radiofrequency pulse, the nuclei in the sample then re-emit radiowaves and the sample relaxs back to the way it was before the pulse in the free induction decay. After capture of the signal a analouge-digitial converter (ADC) supplies it to a computer which then uses the fourier transform to convert it into a histrogram of intensity against frequency.

Gamma counting uses something similar a energy dispersive detector records a series of events, these events are then fed into a ADC whose output is recorded by the computer. Normally again as a histrogram of events at different energies.

In both NMR and gamma spectroscopy there are two sources of noise, I define noise as unwanted signals in a system or spectra.

If we use gamma spectrscopy as an example for a moment, we have random noise due to cosmic rays (things like muons) and effects like Compton scattering. This will cause the random appearance of events in the detector with random energies. This will have a similar effect on the appearance of the spectrum to the random noise which exists in the electronics of a NMR machine. If we were to lower the noise level of the amplifers in the signal path before the ADC in a FTNMR machine we would be able to record spectra faster.

The more times we run the experiment the smaller the effect of the random noise will be on the final spectrum. I once ran an experiment for three weeks to get a spectrum from a lump of trinitite I bought off ebay. It turned out to be genuine.

The question is about the measurement of a stable sample. When I was a PhD student I worked on some samples which were unstable. I suspect to this day that I managed to dissolve for a short time somethings in my NMR samples which then later formed a superinsoluble solid. Once I saw a peak appear in the FT-NMR spectrum at the expected chemical shift which then disappeared again. We can consider this side question also.

Now one problem I see in both systems is if some non random unwanted signal enters the spectrometer before the ADC.

Consider for a moment if we were to do NMR spectroscopy in a 1:1 mixture of CHCl3 and CDCl3. We would have a very strong chloroform peak, this would limit the sensitivity of the amplifiers which can be used before the ADC, by forcing the system to reduce the senstivity of the amplifiers before the ADC what will happen is that the effect of random noise in the ADC will become greater. This would make it harder to make the measurement.

Also with a NMR spectrometer we have a limit as to the number of bits in the computer memory which handles the output from the ADC. If we had eight bit memory. then when one of the points in the free induction decay was to reach 11111111 then the system should not attempt to pulse the sample to record any transient.

If we were to pulse the sample again and then try to increase the count in this point (in time) in the FID, then we would have a nonsensical situation which would cause mayhem with the fourier transform. I think it could then fail to work as intended

Also in gamma spectroscopy we would have the problem if we were to overflow the memory storing the output of the ADC we would start to get a distortion of the spectrum. I suspect that in gamma spectra the effect would be less. We would merely get the highest peak to become zero again. But I suspect that the rest of the spectrum would still be OK.


Mesh is a unit representing the grain-size distribution of a material. So it represents either the opening of the mesh, or the number of grain per surface unit depending on what you are considering.

You can find here the conversion in grain size and surface unit related to the "Mesh".


Are there square planar complexes with sp2d hybridization?

0.1) Hybridization is to be used with caution in inorganic chemistry above school level. It is proven not working for one-electron properties at the very least.

0.2) Depending on the details, you may or may not be taught about hypervalent compounds using d-orbitals of outer shell. While this concept fell out of favor, it still is taught here and there.

1) ignoring 0.*, $\ce{PnX5}$ family where $\ce{Pn=P,As,Sb}$ and X is a halogen (typically $\ce{F}$ or $\ce{Cl}$) adopts trigonal-bipiramidal shape and was viewed as an example of $sp^3d$ hybridisation. Square planar compounds for p-elements are much rarer, but $\ce{XeF4}$ adopt such structure.

2) A rare anion $\ce{[Ni(CN)5]^{3-}}$ may adopt such structure, specifically in $\ce{ [Cr(NH3)6][Ni(CN)5]\cdot 2 H2O}$ Actually, it is often said that square planar complexes may coordinate weakly an additional ion to form a square pyramid and this is why they are typically much more reactive, than octahedral complexes.


Balancing a reaction where dichromate ion reacts with ethanol to give Cr(III) and carbon dioxide

Let's get the simple question out of the way first:

How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.

You can not balance it then. Also, you wouldn't be given such a reaction and is always a typo when they've said that it is a redox reaction.

Now, onto the bit more complicated question.

Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:
$$\require{cancel}\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
In a basic solution:
$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$
[sic] It seems like in the answer for the reaction in acidic medium, the left side of the reaction has a charge of +14 and the right side has a charge of +12.
However,  in the basic medium the answer has the left side with a total charge of -2 and the right side -4. How can these answers be correct if the charge isn't balanced?

A simple explanation would be that the excess charges that you feel are there are actually not. Try breaking the given reaction in the answer into each atom with their own oxidation state. I'll do the one for acidic medium, and the other will be up to the reader.
$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
Here, we start from the reactant side, we assume the following for simplicity:

$\ce{C-H}$ bond leads to positive charge on $\ce{H}$ and a negative charge on $\ce{C}$
$\ce{Cr-O}$ bond leads to positive charge on $\ce{Cr}$ and a negative charge on $\ce{O}$
$\ce{C-O}$ bond leads to positive charge on $\ce{C}$ and a negative charge on $\ce{O}$

\begin{array}{lc} \hline
&\text{Reactants} &\text{Products} \\ \hline
\ce{Cr} & \ce{Cr^6+} & \ce{Cr^3+} \\
\ce{O} & \ce{O^2-} & \ce{O^2-} \\
\ce{H} & \ce{H+} & \ce{H+} \\
\ce{C} & \ce{C-,C^3-}  & \ce{C^4+} \\ \hline
\end{array}
Now, we split the actual equation into this form:
$$\ce{4Cr^6+ + \cancel{22\mathrm H^+} + C- + C^3- +\cancel{15\mathrm O^{2-}} -> 4Cr^3+ + 2C^4+ + \cancel{15\mathrm O^{2-}} + \cancel{22\mathrm H^+}}$$
Which simplifies the reaction to:
$$\ce{4Cr^6+ +C- +C^3- -> 4Cr^3+ +2C^4+}$$
Adding the charges on both sides we get:
\begin{align}
\text{Charge on reactant}=(4\times 6 + (-1) + (-3)) = 20 \\
\text{Charge on products}=(4\times 3 + 2\times4)= 20
\end{align}
Therefore the charge on both sides are equal ...
Note: The reaction in basic medium uses $\ce{CrO4^2-}$ and not $\ce{Cr2O7^2-}$
$$\ce{4CrO4^2- + 7H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 20OH-}$$


Why does permanent marker writing come off a white board when coated with white board marker?

The solvent of the marker dissolves the permanent ink and mixes it with the whiteboard ink, leaving a dark dust that doesn't stick to the board.

The link in Wikipedia doesn't name common ethyl alcohol (denatured or not) as being a solvent for old marks. That is what I have used.


Sudden pressure rising in a pipe of an old plant transporting 33% ammonia solution

Ammonia is more soluble at lower temperatures so the temperature change between the outside of the factory and the inside could raise the pressure inside the pipes.

Gas dissolved in water are metastable systems most of the times. Nucleation points in many cases suffice to promote the release of the gas from the solution (have you ever tried to put some salt in your beer?). It is possible that somewhere in the old pipes there are some areas with higher roughness (e.g. rusty connections) where some nucleation points promote bubble formation and so the ammonia can be released.


No strong nuclear force is involved. Instead, tt's the same force as always: electromagnetic interaction. By staying in the same low orbital as the other electron, the electron in question will continue to be subject to a stronger attractive force from the nucleus. Note that the MO diagrams (see e.g. wikipedia) typically already account for the intra-orbital electron-electron repulsion, and you still have HOMO-LUMO gap.

It is also helpful to remember that electrons are "spread out" over the orbital, so once you do the math, you realize that the average distance between electrons does not have to be small compared to the average electron-nucleus distance.

Let's get semi-quantitative: $\ce{H_2}$, 1.4 bohr, HF/STO-6G($\zeta = 1.20$). After the SCF is completed, I transformed the integrals from the AO basis to the MO basis, which yields the following, relevant integrals in Mulliken notation:

(1|h|1) -1.24964
(2|h|2) -0.49918
(11|11)  0.65720
(22|22)  0.68091
(11|22)  0.64686
(12|12)  0.17440

Nuclear repulsion is 0.71429, all values in hartree. We can then calculate the low-spin, ground-state energy as:
$$
E_\text{gs}^\text{RHF} = E_{nn} + 2\cdot(1|h|1) + (11|11) = -1.12778,
$$
wherein we add the nuclear repulsion energy, the one-electron energy of the lower MO (once for each electron) (one-electron meaning that this is the sum of the electron's kinetic energy and its nuclear attraction) and the relevant two-electron integral for the Coulomb repulsion. Since the electrons are of opposite spin, no exchange term (as is responsible for the spin-pairing energy, cf. Hund's rule) is needed.

The energy of the high-spin, "excited" state can be calculated according to:
$$
E_\text{triplet}^\text{RHF} = E_{nn} + (1|h|1) + (2|h|2) + (11|22) - (12|12) = -0.56207,
$$
where an additional exchange term enters the equation to account for partly-correlated electrons due to Pauli-repulsion.

This calculation is semi-quantitative because RHF in such a small basis set does not yield quantitative results. One would have to go to natural orbitals obtained through full CI in a larger basis set for fully quantitative results. However, for illustrating the relative size of the quantities involved, this level of calculation is sufficient.


Does MgCl2 sol coagulate gold sol faster than Fe(OH)3 sol?

Option (c) is actually correct. Magnesium chloride does coagulate gold sol faster than ferric hydroxide, because ferric hydroxide does not coagulate in the first place.

Wikipedia mentions that ferric hydroxide is insoluble at $\ce{pH}=7$, so there aren't any $\ce{Fe^{3+}}$ ions to coagulate the gold sol anyway.


Is it true that organic chemistry as an independent research field is no longer developing?

The closest thing to organic chemistry would now be the shape and sequencing of protiens and or DNA. That field is best left alone as it takes too much time to pay anyone to get any applicable results.
Advances in things like machine learning are promising to the unraveling of these complex structures. Humams are better at recognising the patterns hence the "fold it" game.


Reduction of butadiene with Na/NH3 - stereoselectivity

At first ... I don't like books that don't give proper solutions to their questions$^{*0}$. This doesn't help anyone ... or at least not me.

Old sources

I started my search on Reaxys and there are two old papers that I found when searching for $$\text{buta-1,3-diene} \xrightarrow{\ce{Na}} \text{but-2-ene}$$
They are:

K. Ziegler, F. Häffner, H. Grimm, Liebigs Ann. Chem. 1937, 528, 105
W. Hückel, H. Bretscheinder, Liebigs Ann. Chem. 1939, 540, 157

In both (german$^{*1}$) papers they work at very low temperatures around $\mathrm{-70\ldots-60~^\circ C}$. Both say that they get mainly the Z-isomer and discuss somehow why this happens. They think about their reaction as follows: $$\ce{dien + 2 M + 2R2NH -> dien.H2 + 2R2N.M}$$
They think that the butadiene is somehow pre-oriented towards the sodium surface from which the preference for the cis-isomer follows.

Old but not soooo old sources

Through the paper that M.A.Ramezani mentioned (W. Reeve, D. R. Kuroda, J. Org. Chem. 1980, 45 (12), 2305–2307), I found another paper (N. L. Bauld, J. Am. Chem. Soc. 1962, 84 (22), 4347–4348) that clarifies some of the mysteries. It is about the stereochemistry of metal addition to conjugated systems, esp. dienes. One of the first things that is shown is the following table.

  $$\small\begin{array}{ccc}
\hline
\text{Diene} & \text{Temp., }\mathrm{^\circ C} & \text{% cis-Olefin}\\
\hline
\text{Butadiene} & -33 & 13\\
\text{Butadiene} & -78 & 50\\
\text{1,3-Pentadiene} & -33,\,-78 & 68\\
\hline
\end{array}$$

It contains the cis-yields from v.p.c. analyses for two different temperatures for the reduction of butadiene with sodium in liquid ammonia. As one can see, there is less Z-but-2-ene produced when the temperature is higher$^{*2}$ (13 %) and more when the temperature is very low (50 %).

What the author also mentions is, that there is in every case more Z-product produced than one would expect, because there exists only about 3-7 % of Z-butadiene at room temperature (which must be even lower at lower temperatures). So somehow the metal is included, wich is described by Bauld with a cyclic complex of sodium and the radical anion that is supported by the experiments of Reeve and Kuroda.

As there are many possibilities how the sodium ion "binds" to E- and Z-form$^{*3}$ and there is also the possibility that there are two (or maybe none?) sodium ions involved, this is far off from the scope of this answer.

Besides that, I roughly calculated the structures from the mechanism that is schown in Scheme 8 from the other link by M.A.Ramezani --which, as I found out in this moment, is also from Prof. N. L. Bauld--. Below are the corresponding $\pi$-MO's$^{*4}$ -- you can imagine the mechanism with involved sodium ion(s) to be similar but somehow with the above cyclic intermediate.

TL,DR

The question in Flemings book lacks some informations. The reaction in liquid ammonia does not only depend on temperature (which was not stated!) but also on the rests that are applied to the diene. For Buta-1,3-diene in $\ce{Na/NH3}$, the Z-isomer is favored when cool temperatures are applied and therefore yield the kinetic product of the reaction. This follows from a cyclic radical anion intermediate with the sodium ion. But when the reaction temperature is rised, the thermodynamically preferred E-product is favored, as is shown by Bauld (see the table).

To get to know more about what happens when different methyl and ethyl groups are added, read the paper by Reeve & Kuroda.

$^{*0}$ Dare you, Szabo and Ostlund ಠ_ಠ
$^{*1}$ Old german papers are a joy to read ... usually the older the better.
$^{*2}$ $\mathrm{-33~^\circ C}$ is the boiling point of ammonia
$^{*3}$ In my case it seems to be purely vdW-interactions but it is gas-phase, this is a small basis set and even if the functional fits there was no dispersion correction.
$^{*4}$ be aware ... those are Kohn-Sham orbitals


Hybridisation of terminal nitrogen in diazomethane

As usual with these small molecules, the bonding situation is a lot more complicated than expected. In this very molecule this is to the extend that not only the Lewis formalism, but also the conventional resonance formalism breaks down.
However, Jan's description is sound, and provides a good starting point for further analysis.
Unfortunately formulating a clear cut valence bond description is difficult, because it involves strongly coupled electrons. This is in part due to the high symmetry of the molecule, $C_\mathrm{2v}$, and the even higher local symmetry of the $\ce{CNN}$ moiety, $C_\infty$. Because of the nature of molecular orbital theory, i.e. being better at describing delocalisation, it is actually easier to understand.

Here are the molecular orbitals computed at the DF-M06L/def2-SVP level of theory:

And here is the same thing in numbers

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-14.323 is N2-s=1.00
 Alpha occ 2 OE=-14.252 is N1-s=1.00
 Alpha occ 3 OE=-10.099 is C3-s=1.00
 Alpha occ 4 OE=-1.055 is N2-s=0.51 N1-s=0.29 N1-p=0.10
 Alpha occ 5 OE=-0.802 is N2-p=0.31 C3-s=0.31 N1-s=0.14 N2-s=0.12 C3-p=0.07
 Alpha occ 6 OE=-0.518 is C3-p=0.26 C3-s=0.24 H5-s=0.16 H4-s=0.16 N1-s=0.07 N2-s=0.06
 Alpha occ 7 OE=-0.487 is N2-p=0.45 C3-p=0.24 N1-p=0.14 H5-s=0.08 H4-s=0.08
 Alpha occ 8 OE=-0.450 is N2-p=0.61 N1-p=0.29 C3-p=0.09
 Alpha occ 9 OE=-0.407 is N1-s=0.44 N1-p=0.41 N2-p=0.08
 Alpha occ 10 OE=-0.366 is N1-p=0.28 C3-p=0.24 N2-p=0.16 H5-s=0.15 H4-s=0.15
 Alpha occ 11 OE=-0.206 is C3-p=0.57 N1-p=0.38
 Alpha vir 12 OE=-0.082 is N1-p=0.55 N2-p=0.32 H5-s=0.06 H4-s=0.06
 Alpha vir 13 OE=-0.005 is C3-p=0.33 N2-p=0.33 N1-p=0.32

From the visual inspection we can see that the terminal nitrogen has only one orbital that could be classified as a lone pair, i.e mo 9; the hybridisation according to this calculation is approximately sp. There are three bonding π orbitals, i.e. mo 7, 8, 10. And there is also one anti-bonding π orbital, i.e. mo 11. The sigma bond is formed by two bonding (mo 4, 5) and one anti-bonding orbital (mo 6).
This is no coincidence, but pretty much always the case for terminal atoms.

Due to symmetry constraints, the hybridisation pattern at terminal atoms is usually:

for the orbital forming the single bond $\mathrm{sp}^{\geq1}$
two orthogonal (π) lone pairs $2\times\mathrm{p}$
one (σ) lone pair $\mathrm{sp}^{\leq1}$

Localisation of the orbitals is not easy for this compound. I performed a natural bond orbital analysis, which transforms the (delocalised) canonical orbitals into hybridorbital like ones, but it results in a low occupied carbon lone pair orbital and a partially occupied anti-bonding nitrogen-nitrogen π orbital.

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. - 3. core orbitals (omitted)
   4. (1.97583) LP ( 1) N  1            s( 67.37%)p 0.48( 32.60%)d 0.00(  0.03%)
   5. (1.33940) LP ( 1) C  3            s(  0.00%)p 1.00( 99.92%)d 0.00(  0.08%)
   6. (1.99655) BD ( 1) N  1- N  2
               ( 43.82%)   0.6620* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 56.18%)   0.7495* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
   7. (1.99587) BD ( 2) N  1- N  2
               ( 51.93%)   0.7206* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 48.07%)   0.6933* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
   8. (1.98025) BD ( 3) N  1- N  2
               ( 39.11%)   0.6254* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 60.89%)   0.7803* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)
   9. (1.99580) BD ( 1) N  2- C  3
               ( 65.65%)   0.8103* N  2 s( 53.76%)p 0.86( 46.18%)d 0.00(  0.05%)
               ( 34.35%)   0.5861* C  3 s( 30.11%)p 2.32( 69.75%)d 0.00(  0.14%)
  10. (1.95419) BD ( 1) C  3- H  4
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  4 s( 99.91%)p 0.00(  0.09%)
  11. (1.95419) BD ( 1) C  3- H  5
               ( 60.92%)   0.7805* C  3 s( 34.95%)p 1.86( 65.03%)d 0.00(  0.02%)
               ( 39.08%)   0.6251* H  5 s( 99.91%)p 0.00(  0.09%)
 ---------------- non-Lewis ----------------------------------------------------
  12. (0.01076) BD*( 1) N  1- N  2
               ( 56.18%)   0.7495* N  1 s( 33.02%)p 2.02( 66.72%)d 0.01(  0.26%)
               ( 43.82%)  -0.6620* N  2 s( 46.00%)p 1.17( 53.92%)d 0.00(  0.08%)
  13. (0.64416) BD*( 2) N  1- N  2
               ( 48.07%)   0.6933* N  1 s(  0.00%)p 1.00( 99.63%)d 0.00(  0.37%)
               ( 51.93%)  -0.7206* N  2 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  14. (0.07754) BD*( 3) N  1- N  2
               ( 60.89%)   0.7803* N  1 s(  0.00%)p 1.00( 99.54%)d 0.00(  0.46%)
               ( 39.11%)  -0.6254* N  2 s(  0.00%)p 1.00( 99.89%)d 0.00(  0.11%)

This is usually an indication of a strongly delocalised system such as a three-centre bond. Investigating the molecule with natural resonance theory gives the rather odd contribution of long range N1 - C3 bond.

 TOPO matrix for the leading resonance structure:

     Atom  1   2   3   4   5
     ---- --- --- --- --- ---
   1.  N   2   2   0   0   0
   2.  N   2   0   2   0   0
   3.  C   0   2   0   1   1
   4.  H   0   0   1   0   0
   5.  H   0   0   1   0   0

         Resonance
    RS   Weight(%)                  Added(Removed)
 ---------------------------------------------------------------------------
    1*     33.02
    2*(2)  32.31    N  1- N  2, ( N  2- C  3), ( N  1),  C  3
    3*(2)  28.81    N  1- C  3, ( N  2- C  3), ( N  1),  N  2
    minor contributions omitted
 ---------------------------------------------------------------------------
          100.00   * Total *                [* = reference structure]

According to NBO, the terminal nitrogen has an $\mathrm{sp}^{0.5}$ σ-type lone pair orbital, two $\mathrm{p}$ π-type bonding orbitals, and a $\mathrm{sp}^2$ σ-type bonding orbital.

Hybridisation is a mathematical tool to analyse a bonding situation. There is no correct way of applying it. It is also very important to understand an atom is never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised.


Is the activity coefficient always a constant?

In my copy of Atkins Physical Chemistry, 6th edition (it's old!) there is a chart that shows a calculation of the activity and the activity coefficient for an example system (chloroform and acetone). It shows the results using a calculation based on Henry's law (approximating ideal-dilution) and again with one based on Raoult's law (approximating a pure substance).

In both cases, the activity coefficient is not constant over the range of concentrations.

For some evidence from a less-reputable but more easily accessible source, check out the wikipedia article on activity coefficients:

  $\gamma_B$ is the activity coefficient, which may itself depend on $x_B$

($x_B$ is the mole fraction of the solute).

Now this means that the activity coefficient is not constant for a given substance in a given mixture at all concentrations, but that isn't the same as saying that the definition of activities is not exact - it is exact, because it is defined that way.

In other words,

$$
\mu_i = \mu_i^\ominus + RT\log \gamma_i x_i
$$

is an exact relationship, even though $\gamma_i$ is not constant.


Electronegativity trends: Is it more important to prioritize families or periods?

There's no hard and fast rule.  In fact, there's actually no rule per se that states that electronegativity does increase along the periods and decreases down the groups.  It's just a general trend, which can be and is broken on occasion.  For instance, Zinc, to the right of Copper, has an electronegativity of 1.66, while copper is 1.75, which "breaks" the rule that electronegativity increases along the periods.

As far as which one takes precedence, it's a case by case scenario.  In some cases, down one and right one would be very close, like you guessed, as with Scandium (1.20) and Zirconium (1.22).  However, if you go over to the non-metals, some of them increase radically in a down-one-right-one scenario (particularly from 3rd period to 4th period, such as Aluminum [1.47] to Germanium [2.02], Phosphorus [2.06] to Selenium [2.48], etc.), while some of them decrease radically (particularly from 2nd to 3rd period and 4th to 5th, such as Boron to Silicon, Oxygen to Chlorine, Arsenic to Tellurium, etc.)

Bottom line:  There's no rule.  You just have to either learn them by heart or check a periodic table such as this one:


How to control current and voltage in electrolysis

You can't control both voltage and current electrically:  Your power supply will hold one or the other constant, but the properties of the electrolytic cell will then determine the other one.

So you need to hold all of the chemical parts of the experiment constant:  Have a large ion bath so that the concentration doesn't change much as the plating takes place.  Stir well to keep things mixed. Have large electrodes (i.e. small current density) so that small surface changes don't dominate.  Etc.


What effects would an electron in the nucleus have?

An electron being at/in the nucleus results in the Fermi Contact Interaction.

This interaction can be observed experimentally through NMR, EPR, and electron capture.

In NMR of paramagnetic compounds, chemical shifts can be far outside the usual diamagentic range. The contribution to chemical shift of the electron density at the nucleus is referred to as the "contact shift".

For more information on Fermi contact, see "Origin and meaning of the Fermi contact interaction" Theor Chim Acta (1988) 73:173-200.


An E1 mechanism is highly unlikely. I don't know what level you're at, but one could invoke the Hammond Postulate in this case, and look at the high energy cation intermediate. There are three chloride leaving groups that could generate one of two intermediates. Both of these intermediates are primary cations. A chloride substituent isn't likely to be able to stabilize this much via resonance but more likely destabilize due to inductive effects.

The other potential mechanisms to consider (that do not involve a cation intermediate) are E2 and E1cb. E1cb might work for the first step depending on where the anion is formed, especially if the inductive effect of the chlorines is enough to help stabilize the carbanion. E2 just requires an anti-periplanar arrangement, which isn't hard here with free rotation.

The reason I'd favor E2 and the reason the reaction likely requires heat is the fact that hydroxide is just an OK base. Add to that the fact that it's in a polar protic solvent that might further dampen basicity. If you tried this reaction with an amide base (e.g., LDA) in toluene, I would guess no heating is necessary.

Heating is generally helpful for E1 because the dissociation of the leaving group is slow. However, the need to heat a reaction is by no means indicative that it is E1 (it might not even be an elimination!). Any number of factors could conspire to make the reaction slow, in which case, heating might help make the reaction proceed at a faster, more practical rate.


Does the initial rate in the reaction of magnesium powder with nitric acid depend on the volume or the concentration of the acid?

The magnesium is present in form of a solid powder. Therefore, the reaction of magnesium with acid can only occur at the solid–liquid interface between magnesium and the aqueous nitric acid solution. Accordingly, the initial reaction rate depends on the surface area of the solid–liquid interface. For the experiments described in the question, the same mass of magnesium powder is used; therefore, the available surface area of the magnesium powder is approximately the same, too. Hence, as long as the added volume $V$ of the nitric acid solution is sufficient to cover the entire available surface of the magnesium powder, the initial reaction rate does not depend on the added volume of the solution. (Assuming a reasonable geometry of the reaction vessel, a volume of $V=100\ \mathrm{ml}$ should be more than enough to cover a mass of $m=5.0\ \mathrm g$ of magnesium powder.)

Furthermore, the reaction rate depends on the concentration $c$ of the acid at the solid–liquid interface. Therefore, the initial reaction rate depends on the concentration of the added nitric acid solution.

And finally, the reaction rate also depends on temperature $T$. Therefore, the initial reaction rate depends on the temperature of the reactants, i.e. the magnesium powder and the added nitric acid solution.

Hence, in order to obtain the same initial reaction rate, the concentration of the nitric acid solution $(c=1.0\ \mathrm{mol\ l^{-1}})$ and the temperature $(T=25\ \mathrm{^\circ C})$ should be the same during both experiments.


How can I deduce the structure of my compound from this NMR spectra?

The hydrogens in the structure you drew definitely should not be in the same environment — so while overlapping or close signals would be okay, the way your signals look isn’t. To get two signals to be identical, there needs to be some symmetrical way of corresponding the hydrogens to each other.

I can confirm a trans-configured double bond (I assume you can, too?) and two disubstituted aromatic rings. The question remains which type of substitution they have (ortho, meta or para). If you expect two rings with the substituents you drew, remember they must be pretty different in their chemical shifts!

For ortho, as you drew, one would indeed expect a set of doublet, triplet, triplet, doublet for each ring. meta would expect one set of doublet, triplet, doublet and a singlet somewhere. para, finally, would expect two doublets.

Remember that triplets must have a $1\,:\,2\,:\,1$ ratio, or they aren’t triplets. With this information in hand, go back into the aromatic region and think what you could have gotten. I have dropped enough clues, I hope.


Predicting electronic geometry, observable geometry, and hybridization for any atom in a molecule

Do we just always look for the "lowest" possible VSEPR in any resonance structure?

Yes. Consider the two resonance contributors below for a N,N-dimethylacetamide, which has a functional group in common with your molecule. The rule of thumb I teach is: **if an atom appears to be $sp^2$-hybridized in one resonance contributor and $sp^3$ in another, then that atom is $sp^2$-hybridized always.

In the figure above, I have used the curved arrow notation to represent how the lone pair on the nitrogen atom is delocalized into the $\pi$-system. In order for this delocalization to occur, that lone pair needs to be in a $p$ orbital. In order for that nitrogen atom to have a $p$ orbital, it must be $sp^2$-hybridized (or possibly $sp$-hybridized if applicable). All three of the atoms (N, C, and O) in the delocalized system need a p orbital, and thus must be $sp^2$-hybridized.

Once you have hybridization, then you have limited your choices of geometry:

$sp^3 \implies$ tetrahedral geometry group: tetrahedral, trigonal pyramidal, bent, "none"
$sp^2 \implies$ trigonal planar group: trigonal planar, bent, "none"
$sp^{\ \ \  }  \implies$ linear group: linear, "none"

  Why isn't the observable geometry for "X" linear instead of none?

You are confusing observable geometry at an atom with molecular geometry.

Observable geometries at an atom are a relationship describing the relative geometric positions of other atoms bonded to that central atom. The central atom is not counted.A linear arrangement means that the atom in question is bonded to two other atoms or groups and those two atoms or groups are $180^\circ$ apart. If there is only one atom or group bonded to the atom in question, we cannot define the geometric relationship between that atom or group and the other atoms or groups bonded to that atom because there are not other atoms or groups. Consider the following examples from the tetrahedral group. The last case is "none" or "undefined, because you need at least two things to have a relationship.

This is different than molecular geometry which deals with the shape of the entire molecule. Let's look at HF. The shape of the molecule is linear, because the two points define a line (segment). The geometry around F is "none" or undefined. We cannot describe the arrangement in space of the hydrogen atom relative to other atoms bonded to fluorine because there are none.


Measure ethanol volume fraction in a beverage

Your question: Given a sample of an alcoholic beverage (be it beer, wine, etc..), what are possible methods to find out the ABV (alcohol by volume) of it with moderate accuracy?

There are a few method used in winery industry as  Karl mentioned a comment in elsewhere. I assume you want to find out better methods to find the alcohol content:

The best technique is the gas chromatography method, which is the most accurate method for determining alcohol content. Gas chromatography has the added benefit of providing additional information about the wine, beer, or any other liquor product (e.g., Ref.1).
Seemingly most popular technique is the determination of ethanol in alcoholic beverages by titrimetric and/or spectrophotometric dichromate methods (e.g., Ref.2-Ref.4).
Another method is enzymatic assay using spectrophotometry: Ethanol, in an alkaline environment, is oxidized through an enzymatic reaction and converted in acetaldehyde in presence of $\mathrm{NAD^+}$, which in turns produces $\mathrm{NADH}$. The increase of the absorbance, measured at $\pu{366 nm}$ (end-point), is proportional to the concentration of ethanol in the sample (e.g., Ref.5).
In addition, you can also use NMR technology to determine the ethanol content in alcoholic beverages (e.g., Ref.6).

References:

M.-L. Wang, Y.-M. Choong, N.-W. Su, and M.-H. Lee, “A rapid method for determination of ethanol in alcoholic beverages using capillary gas chromatography,” Journal of Food and Drug Analysis 2003, 11(2), 133–140.
G. J. Pilone, “Determination of ethanol in wine by titrimetric and spectrophotometric dichromate methods: collaborative study,” J. Assoc. Off. Anal. Chem. 1985, 68(2), 188-190 (https://www.ncbi.nlm.nih.gov/pubmed/3988696).
Arthur Caputi, Masao Ueda, Thomas Brown, "Spectrophotometric Determination of Ethanol in Wine," Am. J. Enol. Vitic. 1968, 19(3), 160-165 (https://www.ajevonline.org/content/19/3/160).
Malinee Sriariyanun, Parita Mutrakulcharoen, Surapun Tepaamorndech, Kraipat Cheenkachorn, Kittipong Rattanaporn, "A Rapid Spectrophotometric Method for Quantitative Determination of Ethanol in Fermentation Products," Oriental Journal of Chemistry  2019, 35(2), 744-750 (http://dx.doi.org/10.13005/ojc/350234).
Amy A. Caudy, "Spectrophotometric Analysis of Ethanol and Glucose Concentrations in Yeast Culture Media," Cold Spring Harbor Protocol  2017, (9), 735-741 (doi: 10.1101/pdb.prot089102).
Meden F. Isaac-Lam, Thomas Brown, "Determination of Alcohol Content in Alcoholic Beverages Using 45 MHz Benchtop NMR Spectrometer," International Journal of Spectroscopy 2016,  Article ID 2526946, 8 pages (https://doi.org/10.1155/2016/2526946).


From the limited excerpt of your book (title? author?) it is likely it just deals with the interaction of linearly polarized light with condensed matter.  If there is near- and far range order (as regular consecution of building elements), the solid is crystalline.  Among the crystaline states, there are (Bravais) lattice systems, that differ in symmetry, i.e. because they possess different symmetry elements, some are of higher symmetry (like cubic) than others (like monoclinic).  And vectorial physical properties of the material are coherent with the crystalline symmetry met.

(image source).

By means of an index ellipsoid, you may map the orientational dependance of physical properties, in other words, anisotropy.  In the case of mineral- and more general in polarized light microscopy, such an ellipsoid may be used as representation to map refractive indices along three directions orthogonal to each other. Now

either the three axes of the indicatrix differ from each other: optical biaxial system, characteristic Bravais lattice of lower symmetry: orthorhombic, monoclinic and triclinic.
two of the three axes of the indicatrix are identic, the third is different: optical mono- or uniaxial system, characteristic for hexagonal, trigonal and tetragonal Bravais lattice.
all three axes of the optical indicatrix are identic, in other words, the material is optical isotropic: this is characteristic for the cubic Bravais lattice.  (And: for the state of randomness, the glasses).

So in short: it is the number, the degree, and the relative orientation of the symmetry elements of the cubic Bravais latice that renders the more general case of an ellipse into the special case of a sphere.  And this means, the question should be "cubic crystal's isotropy".

Further information may be found for example here.


Decomposition through thermolysis, electrolysis, photolysis

You should analyse the chemical bindings in your compound, e.g. ion bindings will easily break-up in water, stronger bindings may require You to use heating or electric current to dissociate the molecule. Also if you know the structure of the compound, look for some similar structures on the internet, the more similar the structures, the more similar properties they have, so similar factors will dissociate them.


How much Carbon Dioxide would be required to displace enough oxygen to prevent ignition?

The parameter you are asking about is called the Minimum Oxygen Concentration (MOC), and is a measurement of the minimum amount of oxygen present in air to propagate a flame. The amount depends on the fuel being burned, and varies from about 4% for hydrogen to about 18% for some solid combustibles.

Let us consider your example of butane, and round some numbers to make our calculations simple. Butane has a MOC of 12%. Normal oxygen content in air is 20%, with nitrogen and sundries making up the remaining 80%. Assume uniform mixing and displacement of gases is equal. In order to reduce the oxygen content to 12%, you need to introduce enough CO2 to make up 40% of the air content:

$O2 + N2 = 100$  =>20 + 80 = 100%

$O2 + N2 + CO2 = 100$    => 12 + (12*80/20) + CO2 =100%

However, let us consider a couple of other very significant points.

An oxygen level of 12% is unsafe. Anything below 19.5% is unsafe. At levels of 12%, a person will have impaired judgement and coordination, and experience increased pulse rates. Prolonged exposure can have fatal consequences.
A CO2 level of 40% is fatal. Full stop. Levels above 7% can cause loss of consciousness and just 10% can be fatal.
CO2 is actually denser than air, so mixing of the main components is unlikely to be uniform. This means that higher CO2 concentrations will be found lower down in the room.

I hope you stay well clear of your basement until you have resolved this problem and have appropriate active ventilation installed, as well as possibly an oxygen monitor.

In the very least, get yourself a canary.


Why aren't the copper(II) ions in CuSO4 precipitated by hydroxide ions from water?

In aqueous solution, the copper(II) ion forms the aqua complex $\ce{[Cu(H2O)6]^2+}$:

$\hspace{55 mm}$

Because the central $\ce{Cu^2+}$ ion is positively charged, it polarises the $\ce{O-H}$ bonds towards oxygen, which makes the hydrogens more acidic. Therefore, the ion acts as a weak Brønsted acid, with a $\mathrm{p}K_\mathrm{a}$ of approximately $8$ (J. Phys. Chem. A, 2015, 119 (12), pp 2926–2939):

$$\ce{[Cu(H2O)6]^2+ <=> H+ + [Cu(H2O)5(OH)]^+}$$

Suppose we have a $0.1~\mathrm{mol~dm^{-3}}$ solution of $\ce{CuSO4}$. To a first approximation,

$$\begin{align}
[\ce{Cu(H2O)6^2+}] &= 0.1~\mathrm{M} \\
[\ce{H+}] &= \sqrt{(0.1~\mathrm{M})(10^{-8}~\mathrm{M})} \\
&= 3.162 \times 10^{-5}~\mathrm{M} \\
[\ce{OH-}] &= \frac{10^{-14}~\mathrm{M^2}}{[\ce{H+}]} \\
&= 3.162 \times 10^{-10}~\mathrm{M}
\end{align}$$

Therefore,

$$\begin{align}
[\ce{Cu^2+}][\ce{OH-}]^2 &= (0.1~\mathrm{M})(3.162\times 10^{-10}~\mathrm{M})^2 \\
&= 10^{-20}~\mathrm{M^3} < K_{\mathrm{sp}}
\end{align}$$

and no precipitation of $\ce{Cu(OH)2}$ occurs.

If the pH of a $\ce{Cu^2+}$ solution is neutralised by careful addition of base, then precipitation will be observed as Ivan noted in his comment. The exact pH at which precipitation will occur is of course dependent on the concentration of $\ce{Cu^2+}$. The same argument here actually applies to all of the transition metals: all of them form rather insoluble hydroxides, but they also form acidic aqua complexes, which prevent the precipitation of the hydroxide.

The calculations could have been done more accurately by introducing fewer assumptions; however, they should not affect the final conclusion. (If this was a textbook, it would probably say: this exercise is left to the reader!)


What is the rationale of the IUPAC name for DDT?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure of the DDT molecule fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 1,1'

(b) the name of the linking multiplicative substituent atom or group;

Here: 2,2,2-trichloroethane-1,1-diyl

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: bis

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: 4-chlorobenzene

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Here: 1,1'-(…)bis(4-chlorobenzene) (not 4,4'-(…)bis(1-chlorobenzene))
Therefore, the preferred IUPAC name is 1,1'-(2,2,2-trichloroethane-1,1-diyl)bis(4-chlorobenzene)


Are electrostatic effects temperature dependent?

The strength of the electric charge does not change with temperature (unless approaching that of the Big Bang). However, effects of the charges do change with temperature. For example, at low temperatures (near 0 K), ions can form arrays in space, though their charges repel. The history of Bose-Einstein condensates begins with trapped ions, extending to neutral gases.


Why doesn't fluorine harm the ozone layer the way chlorine and bromine do?

TL;DR: The mechanism is different for fluorine and chlorine atoms. Fluorine atoms react and forms stable HF molecules while chlorine atoms turns into a radical by the action of UV which helps in the destruction of ozone.
Long answer:
Ozone depleting substance(ODS) are gases that take part in ozone depletion process. Most of the ODS are primarily chlorofluorocarbons(CFC) and halons that contain chlorine and bromine atoms respectively which reach the stratosphere and reacts with ozone and lead to ozone depletion. In this context, the chemical component responsible are chlorine radical (Cl·) and bromine radical (Br·). The ODS travel to the stratosphere without being destroyed in the troposphere due to their low reactivity and once in the stratosphere, the Cl and Br atoms are released from the parent compounds by the action of ultraviolet light, e.g.
$$\ce{CFCl3 ->[UV] Cl· + ·CFCl2}$$
Ozone is a highly reactive molecule that easily reduces to the more stable oxygen form by the catalytic action from the halogen radicals. Cl and Br atoms helps in destruction of ozone molecules through a variety of complex cycles. A simple cycle is as follows:
$$\ce{Cl· + O3 → ClO + O2}$$
$$\ce{ClO + O3 → Cl· + 2 O2}$$
Chlorine radical is released and the process is continued and the overall effect result in ozone depletion. Halon gases contains bromine and the process is similar to that of what CFCs does but the action is more potent and is more efficient in this ozone depletion process.

What about fluorine and iodine?

CFCs also contains fluorine atoms which gets released along with chlorine and bromine on photodissociation in the stratosphere. But this fluorine by itself does not contribute to ozone depletion. Fluorine atoms released are quickly sequestered into carbonyl compounds and subsequently into hydrogen fluoride by reacting with water and methane molecules which is very stable in the stratosphere. As such, CFCs are now being replaced with Hydrofluorocarbons (HFCs) according to Montreal protocol (but they do contribute to global warming).
Iodine containing organic molecules react so rapidly in the lower atmosphere that they do not reach the stratosphere in significant quantities and does not play role in ozone depletion.
References

https://en.wikipedia.org/wiki/Ozone_depletion
https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q7.pdf
https://hal.archives-ouvertes.fr/hal-00256296/document


Is there any reason to fear personal exposure to rain with a pH of 3.1?

According to a United States Environmental Protection Agency report on the effects of acid rain on human health:

Walking in acid rain, or even swimming in a lake affected by acid rain, is no more dangerous to humans than walking in normal rain or swimming in non-acidic lakes. However, when the pollutants that cause acid rain $\ce{—SO2}$ and $\ce{NOx}$, as well assulfate and nitrate particles— are in the air, they can be harmful to humans.
$\ce{SO2}$ and $\ce{NOx}$ react in the atmosphere to form fine sulfate and nitrate particles that people can inhale into their lungs.  Many scientific studies have shown a relationship between these particles and effects on heart and lung function, such as asthma and bronchitis.


Is chromium (III) green or violet in solution?

Even in water this depends on the hydration process, $\ce{[Cr(H2O)6]Cl3}$ is purple. Also the anhydrous $\ce{CrCl3}$ when you buy it is a purple compound. The chloride-ligands exchange for water ligands like $\ce{[Cr(H2O)5Cl]Cl2 * H2O}$ which is blue-green. And in the end the  $\ce{[Cr(H2O)3Cl3]*(H2O)3}$ is green. Some ions like fluoride are said to influence this process but I have not been able to confirm that yet. For Cr(III) there's this interesting phenomenon, and I think I already asked for that, that this exchange seems to be quite slow. I once prepared a Cr(III) citrate solution and it was blue-green, but after about a week it was deep purple. So this exchange may sometimes take some days.


Effect of magnetism on dissolved Na+ and Cl- ions in water

There are a couple of issues: first, a uniform magnetic field doesn't exert a force on stationary charges and secondly, when the charges are moving the force is at right angles to the magnetic field. There are a few ways to get the charges moving. You could have an electric current flow from the bottom of your figure to the top. The $\text{Na}^+$ ions would move with the current, up the page and since the magnetic field points to the left in your figure, the $\text{Na}^+$ ions would be pushed towards the reader by the right hand rule resulting in higher electric potential near the reader. This is called a positive Hall effect. The $\text{Cl}^-$ ions would move against the current, down the page and paradoxically would be pushed towards the reader again by the right hand rule, this time resulting in a lower electric potential near the reader: a negative Hall effect.
Another way is to have the electric current directed out of the page. Positive $\text{Na}^+$ ions move towards the reader and then are pushed down the page by the magnetic field according to the right hand rule. Negative $\text{Cl}^-$ ions move away from the reader and are also pushed down the page by the magnetic field via the right hand rule. This is the principle of magnetohydrodynamic propulsion.
Another way is to have the brine flowing up the page which forces $\text{Na}^+$ charges towards the reader and $\text{Cl}^-$ charges away from the reader by the right hand rule, both resulting in higher electrostatic potential on the near side of the stream. This would be magnetohydrodynamic propulsion in reverse operating as a generator.


Hydrogen bonds to chlorine atoms are known. In the gas phase HCl dimers are hydrogen bonded but the bond has only about half the dissociation energy of $\ce{(HF)2}$ which has a larger electrostatic attraction energy. The adduct $\ce{HCl\cdots OH2}$ is also known and has a similar hydrogen bond energy to that of  $\ce{HOH\cdots OH2}$.

Using neutron diffraction hydrogen bonds have been studied in many crystals. From a study of many types of amino acid crystals (and derivatives thereof) the normal $\ce{HOH\cdots O}$ hydrogen bond is found to have  a range of lengths, from about 0.175 to 0.2 nm with most lengths at around 0.185 nm. The $\ce{OH\cdots Cl}$ hydrogen bond peaks about 0.2 nm with an range of $\approx \pm$ 0.05 nm. The $\ce{NH^+\cdots Cl}$ bond is longer peaking at 0.21 nm.

Calculations (Singh & Kollman, J. Chem. Phys. v83, p4033, 1985) show that the linear hydrogen bond, X-H-X, has an energy profile that is not that dissimilar to any chemical bond when energy is plotted against X to X separation. There are electrostatic attraction terms and exchange repulsion terms. These form a potential well with repulsion dominating (positive energy) at close separation of the two X atoms, a minimum energy (negative) at about 0.18 nm for water dimers, and a gradual increase in energy as the distance X - X increases reaching zero at large separation. However, there are also polarisation terms and those due to dispersion and charge transfer to take into account. These terms are all attractive and become important only at small separation of the X atoms. However, they are not insignificant and  contribute  a few tens of percentage to the total energy. The result of these terms is to move the minimum of the potential energy to smaller values that that due to exchange and electrostatic terms alone. Thus one can imagine the H atom residing in the potential energy between the two X atoms and being close to the minimum energy. Having described this ideal case,  IR spectroscopy does  indicate that in some cases the potential may have two minima rather than just one as described so the nature of these bonds is still a subject of debate.


My answer would be to refuse to answer this case.
The 4n+2 rule applies only if the electrons are conjugated in a cycle -- nothing more, nothing less.  If you have pendant pi bonds in the conjugated system, like the carbonyl group in cytosine and many similarly structured compounds, you don't have the right conditions to apply the 4n+2 rule at all.
Interior conjugated atoms, as in pyrene, also mean you don't have specifically conjugated cycle.


What is the proper name for the but-1-ene's isomer methylpropene?

Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).
Ignoring the rules for omission of locants, the complete systematic name of the parent compound would be ‘prop-1-ene’. According to Rule P-14.3.4.2, however, the locant ‘1’ is omitted. (Note that no other isomer can be generated by moving the double bond from its position to another.)

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(d) in unsubstituted dinuclear and trinuclear alkenes and alkynes and monounsaturated cycloalkenes and cycloalkynes; (…)

Thus, the preferred IUPAC name for the unsubstituted parent structure is propene.
Ignoring cis/trans isomerism, the substitutable hydrogen atoms of propene have three different locants (‘1’, ‘2’, and ‘3’). Therefore, the locants for substituent prefixes (e.g. ‘1-chloro’, ‘2-chloro’, or ‘3-chloro’) cannot be omitted in substituted propene.
Since, the locants for the substituent prefixes cannot be omitted, also the locant ‘1’ of ‘prop-1-ene’ can no longer be omitted for preferred IUPAC names.

P-14.3.3 Citation of locants
In preferred IUPAC names, if any locants are essential for defining the structure of the
parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. (…)

For example, (1E)-1-chloroprop-1-ene, (1Z)-1-chloroprop-1-ene, 2-chloroprop-1-ene, and 3-chloroprop-1-ene are the preferred IUPAC names for some substituteded propenes.
This principle applies to any substituent. The special situation that replacing the chloro substituent in (1E)-1-chloroprop-1-ene, (1Z)-1-chloroprop-1-ene, and 3-chloroprop-1-ene with a methyl substituent would actually lead to the names (2E)-but-2-ene, (2Z)-but-2-ene, and but-1-ene instead of ‘(1E)-1-methylprop-1-ene’, ‘(1Z)-1-methylprop-1-ene’, and ‘3-methylprop-1-ene’, respectively, does not change the preferred name for 2-methylprop-1-ene. (Note that the substitutable hydrogen atoms of propene have three different locants irrespective of the kind of the substituent.)
Therefore, the preferred IUPAC name is 2-methylprop-1-ene.


Combining Lead pipe with Zinc or Copper pipes in a domestic water installation

The focus of the analysis should be first on the anodic index (see, for example, table here), which relates to the electrochemical corrosion potential of the metal serving as the anode. Second, materially important is the current state of the old municipality Lead pipes.
Per the source, the anodic index of Pb is high (see chart) above copper, so historically it has already electrochemically corroded likely inducing a protective Lead oxide coating in the presence of dissolved oxygen.
However, Zn metal is more anodic than the Pb anode, and as such, would readily corrode and may introduce high levels of Zn ions (even levels considered problematic/toxic).
As such, copper metal, which is lower in the anodic index relative to the Pb electrode, is likely a better choice to preserve the current state of affairs, which is in apparent agreement with the reputed cited textbook 'correct' answer.


When is a reaction said to be under kinetic control, is it not easily reversible?

A product formed under kinetic conditions may or not be reversible under the conditions of the reaction. Here is an example of one that is reversible under the reaction conditions. Kinetic control is determined by the difference in activation energy (ΔEa) of the two pathways. The thermodynamic equilibrium is determined by ΔΔGo. Acetoacetic ester (AAE 1) is a mixture of keto and enol forms. Bromination of AAE in an inert solvent rapidly gives α-bromoAAE 2 in addition to HBr. This is the product that is formed faster having a lower activation energy. This reaction is also  reversible, initiated by HBr (microscopic reversibility). Eventually, the more stable γ-bromoAAE 3 is formed from AAE 1 by enolization toward C4 (γ-position) of AAE under the influence of HBr.


Semipinacol-type rearrangement leading to ring expansion

Three-membered rings are not exceptionally stable, on the contrary, they suffer
a lot of ring strain and readily take part in ring-opening reactions.
Exceptions include the Hückel-aromatic cyclopropenyl cation, and the
resonance-stabilized cyclopropylmethyl cation.

Your assumption that the alkene is protonated is correct, as this generates a
substituted cyclopropylethyl cation. The regioselectivity of the protonation occurs to give the more stable secondary carbocation instead of a a primary carbocation. A 1,2-alkyl shift then occurs, which is in this case assisted by the lone pair on oxygen. Deprotonation of the carbonyl then yields the four-membered ring ketone C (2-methylcyclobutanone). [The carbons are numbered in blue to make it easier to see what's going on: the C1-C3 bond is broken and C3-C4 bond is formed.]

This is similar to the semipinacol rearrangement; in this case, the carbocation is generated via protonation of an alkene instead of by loss of a leaving group.

The formation of products A or B would require protonation of the alcohol group, followed by loss of water in an E1 reaction. In either case, the final deprotonation yields a three-membered ring with an endocyclic or exocyclic double bond, which is more strained than a saturated three-membered ring. This reaction pathway is therefore less favorable than ring expansion. Protonation of the alcohol can certainly happen, but since it does not lead to the formation of any stable products, it is said to be unproductive.


Take the example of methyl bromide reacting with chloride ions:

In protic solvents the chloride ion is strongly solvated by hydrogen bonding and so the energy of the reactants is lowered, compared to in aprotic solvents. For the SN2 mechanism, the transition state is less polar than the reactants because the negative charge is spread out across the chlorine and the bromine and so the transition state is less stabilised by protic solvents than the reactants, increasing the activation energy of the reaction.

In aprotic solvents, the reactants and transition state are less stabilised because of the lack of hydrogen bonding but crucially the difference between the stabilisation of the reactants and the transition state is less and so the activation energy is lower, meaning that the SN2 reaction proceeds more quickly in aprotic solvents than in protic solvents.

The reverse argument applies to SN1 reactions where the first transition state is more polar than the reactants and so is more stabilised by protic solvents.


Not knowing what concentrations of the other chemicals that are present, it is impossible to do a theoretical calculation. The only way to do this would be experimentally.


With 6 components and 4 reactions, you must have two conservation rules (that is, if the reactions are independent, which they are).

Mathematically, we find ourselves in a 6D space (as there are 6 components), we move in a 4D subspace defined by the 4 reactions, and we need a basis for the 2D subspace normal to it, which is precisely where we can't go. The basis vectors will be our conservation laws.

You may use any CAS or your own hands to solve the problem. In Wolfram Mathematica, this is done as follows:

NullSpace[{
{1, -1, 0, 0, 0, 0},
{0, 2, -1, 0, 0, 0},
{0, 0, 1, -1, -1, 0},
{2, 0, 1, 0, 0, -1}
}]

As expected, the answer includes two vectors. One is $\{1, 1, 2, 2, 0, 4\}$. Another is $\{0, 0, 0, -1, 1, 0\}$, which translates to $\rm [D] - [E] = const$ ($0$ if there were no initial amounts), which is obviously true, since both are produced in the same reaction and never spent. That's your second conservation law.

BTW, the sum of the two vectors corresponds to the conservation law you already know. A basis can be chosen in a multitude of ways, after all.

That's what you do in any problem of this sort. That's linear algebra; it is about as simple and robust as a cleaving axe. It will never run out of gas, or freeze because its usage license has expired, or refuse to chop the incompatible firewood. It just works.

So you prepare the stoichiometry matrix (which you already did; I just transposed it), find the normal subspace, translate the basis back to your compounds. That would be it. The notion of time is irrelevant. The notion of differential equations is irrelevant. The notion of limiting reactants is irrelevant. It is just vectors and spaces. They do not even know you apply them to chemistry, nor do they need to.

So it goes.


How does adding acids (H+ ions) to limited soluble salts help it further dissolve?

In general, the solubility of a sparingly soluble salt having a basic anion will increase with the addition of a strong acid.  Given:

$$\ce{X+A- <--> X+ + A-}$$

where $K_{sp}$ is small and A- is a weak base, the addition of a strong acid will form HA, driving the reaction to the right and increasing $K_{sp}$.

In the specific case of

$$\ce{CaF2 <--> Ca^2+ + 2F−}$$

the addition of $\ce{HCl}$ gives

$$\ce{CaF2 + 2HCl <--> Ca^2+ + 2Cl- +2HF}$$

driving the reaction to the right, thus increasing the solubility of $\ce{CaF2}$.  Note that $\ce{HF}$ itself is a weak acid, having a pKa of $3.17$, and will be in equilibrium with $\ce{H+}$ and $\ce{F-}$.


Consider that the carbon in carbon dioxide is bonded to two electronegative oxygen atoms. This will cause shift the electrons towards the oxygens and away from the carbon, making it electron deficient:

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This makes the central carbon electrophilic, and explains the reactivity of the carbonyl moiety towards nucleophiles. Similarly, $\ce{CO2}$ undergoes addition reactions with Grignard reagents to form carboxylate anions.

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