H2Te dipole and electronegativities of H and Te (high school level chemistry)

Information in this question is all over the place and often out of context. Let me get a few things straight:

Van der Waals forces, also known as London interactions, occur between every pair of molecules regardless of polarity. They are the only intermolecular force observed for non-polar molecules such as $\ce{I2}$. They are explained by induced and spontaneous dipoles at a molecular level and femtosecond timescale.
Dipole interactions are present between macroscopic dipoles and account for the boiling point difference between e.g. tetrahydrofurane and cyclopentane ($\Delta \vartheta_\mathrm{b} \approx 15~\mathrm{^\circ C}$). For those, you need a notable difference in electronegativity resulting in an overall dipole (so $\ce{CO2}$ is not a dipole because the dipole moments of both $\ce{C=O}$ bonds cancel each other out). For strong dipoles, these interactions are orders of magnitude larger than the underlying van der Waals forces.
Electronegativities are, of course, atomic properties (and also depend on things like oxidation state if one is truly accurate). For no two different elements is the electronegativity difference ever zero. Every $\ce{X-Z}$ bond is always polarised ever so slightly in one of the two directions. Sometimes, as in the case of $\ce{NCl3}$, this is best determined experimentally. ($\ce{N-Cl}$ bonds in $\ce{NCl3}$ are polarised towards nitrogen.)
However, only bonds which are polarised to a certain extent (usually a cutoff value of $\approx 0.4$) are called polar on a high school level as per convention.
$\ce{H2Te}$ has non-polar bonds as per convention of the fourth bullet point. Therefore, you should consider it a non-dipole and hence only consider van der Waals forces/London dispersion forces. (DavePhD’s source is likely correct, but the dipole moment is neglegible at high school level.)
I would love to see a link to the pages that say $\ce{H2Te}$ be notably polar/display dipole dipole interactions. (Compare the $0.2~\mathrm{D}$ to the $1.78~\mathrm{D}$ dipole moment of ethyl acetate.) Calling van der Waals forces dipole interactions is usually considered wrong (although they build up upon the same principles: electrostatic interaction).

I tried to stick to a high-school level as well as possible. Hence a number of simplifications in this answer. At research levels, every molecule without improper rotation (of which inversion and symmetry planes are a special case) should be considered to have a non-zero dipole moment as DavePhD’s answer states. Thus, every molecule of that type will display dipole-dipole interactions however weak they may be.


My understanding of catalyst poisons and inhibitors is that:

Inhibitors curb the catalyzing powers of a catalyst but do not react with any of the reactants and the reaction is reversible.
Poisons react with the catalyst and/or the reactants and the reaction is irreversible.


Why is it aqueous and ethanolic conditions for nucleophillc substitution of cyanide ion with benzyl chloride?

Since $\mathrm{p}K_\mathrm{a}$ of $\ce{HCN}$ is 9.21 (Wikipedia), the cyanide ion in water is pretty alkaline and contains significant amounts of hydroxide ions, according to the following equilibrium:
$$\ce{NC- + H2O <=> OH- + HCN}$$
That's why you smell strong $\ce{HCN}$ in aqueous  $\ce{KCN}$ or $\ce{NaCN}$ solution.  Thus, when you perform the reaction of $\ce{KCN}$ or $\ce{NaCN}$ with the haloalkane (including allyl and benzyl) in water you tend to get the substitution by $\ce{^-OH}$  instead of $\ce{^-CN}$. To get major product as cyanoalkane, you may need to do the reaction in absolute alcohol.

Not relevant, but note worthy that $\ce{^-CN}$ is an ambident nucleophile, meaning a nucleophile which can attack acitve site of a molecule through more than one sites (here either by $\ce{N}$ or $\ce{C}$). Thus, a reaction of alkyl halide ($\ce{R-X}$) with $\ce{KCN}$ gives alkyl nitrile ($\ce{R-CN}$) while a reaction of alkyl halide with $\ce{AgCN}$ gives an alkyl isonitrile ($\ce{R-NC}$).


Your redox equations look awkward (and wrong, too) because you are trying apply the half-reactions method as if it  were happening in aquatic media to the combustion reaction, which effectively occurs in the gas phase.
Of course, there should be no requirements for water to be present.
As it's been noted in the comments, ionic half-reactions method is not the most elegant way of find coefficients in this case.

If you really want to use it, I'd suggest to write the elements undergoing redox reactions separately from the corresponding compounds (and yes, we do consider oxygen being reduced into carbon dioxide too):

$$
\begin{align}
\ce{2 \overset{-3}{C} &→ 2 \overset{+4}{C} + 14 e-} &|\cdot 2 \tag{ox} \\
\ce{\overset{0}{O}_2 + 4 e- &→ 2 \overset{-2}{O}} &|\cdot 7 \tag{red} \\
\hline
\ce{4 \overset{-3}{C} + 7 \overset{0}{O}_2 &→ 4 \overset{+4}{C} + 14 \overset{-2}{O}}
\end{align}
$$

resulting in a balanced equation:

$$\ce{2 C2H6 + 7 O2 → 4 CO2 + 6 H2O}$$

Note that in general there is an average oxidation number of carbon involved, and it might as well be fractional.


Chemical Reagent Test sensitive at low concentrations

Do you have the pure compounds on hand?  If you do, the two target compounds can be crudely isolated by extracting the broth with diethyl ether, concentrating it down, and spotting on SiO2 TLC plate. Compare RF with that of the pure compounds.
Alternatively, the same general extraction methodology could prepare a sample for GC-MS or crude NMR, if such instrumentation is available


Why are the "extents of reaction" the same for all reactants and products?


Finding the energy given out when 2 moles of reactants are mixed and ignited from a balanced equation

The N2O4 is going to define the maximum amount that will be used up in the reaction, since it has the higher stoichiometric number of the reagents. That is, the second term will always be less and the amount consumed will be defined by the N2O4 because that two moles will limit the N2O4 and in doing so reduce the amount of the second term consumed- there will be excess second compound after the N2O4 has been consumed.

For example, the ratio of first to second term is 5:4, so if we consume all of the second term (2 mols) this requires 2.5 mols of N2O4, which we don't have. So we are limited with the N2O4.

So, knowing that, with two mols of N2O4 the amount is reduced to 2/5 of the enthalpic calculation. That becomes 5116*2/5 = 2046, or answer B.

Apologies for awkwardness of this answer, on a mobile device and have to keep swapping from question to answer.


Does adding mass to a liquid (not water) increase the volume

"The rule" is that an ideal mixture will have the volume of the sum of its components and the density will be the molar averaged density of the components. Real mixtures deviate from this rule to varying degrees.

The name of the property you're looking for that describes this deviation is "excess volume". Excess volume can be either positive (1L of A and 1L of B creates >2L of AB mixture) or negative (1L of C and 1L of D creates <2L of CD mixture). The deviation from the rule depends mostly on how the molecules of the components interact with themselves compared to with the each other. Without knowing much about this specific case, I would guess that the big agar polymer molecules interrupt the local organization of the water molecules. This causes the density to decrease.

For many fluids, such as ethanol and water, the densities have been well studied and you can look up tables of excess volume. On a quick google search, it doesn't appear that anyone has ever done this for coconut water and agar. Unfortunately, that means more work for you. Fortunately, it means you get to break new scientific ground!


Why do we need to add FeBr3 for bromination of anisole, but not for bromination of phenol?

I don't think polarization is the issue. Adding $\ce{FeBr3}$ is just to reduce the $\ce{Br-Br}$ bond strength, speeding up the reaction (attack of the aromatic ring on the bromine is part of the rate determining step, which is slow in the bromination of benzene). It functions as a catalyst being recycled when the $\ce{Br-}$ grabs a $\ce{H+}$ to restore aromaticity in the ring. I'm pretty sure — at least I see no reason to doubt — that the catalyst requirements for the bromination of anisole and the bromination of phenol do not differ much, if at all. That is, no catalyst is required for the bromination of anisole.


Why do liquid crystal displays (LCDs) visually distort under pressure?

Liquid crystal displays work by using liquid crystals where the direction of alignment of the molecules in a plane rotates as you go from layer to layer. Usually this is done with twisted nematic crystals, however cholesteric LCs can also be used to achieve similar effects (I don't think smectic LCs are useful here as they don't align).

To create a liquid crystal cell, twisted nematic liquid crystal is filled between two crossed polarizers1, and a backlight is sent through. The important thing about liquid crystals is that the specific rotation is easy to control by varying the potential difference (more potential difference leads to more anisotropic disorder).

Now, if you make a grid of these and put color filters on them, we get an RGB LCD display.

What happens when you press on them? You change the concentration/length of the liquid crystal, as well as making it bulge out slightly2.. This changes the optical rotation it induces, causing the amount of light that comes out to change (as the amount of light is dependant on how aligned the rotated light is with the second polarizer)

I'm not too sure of the exact reasons behind the phenomena you list, but I can guess:

Firstly, the color changes are caused because each type of crystal cell is affected differently by the pressure. I think there's a difference in concentration between the different color cells to better suit human eyes. When I do it to LCD displays accessible to me, I only get a darkening, not a color change, so it seems to not be a universal feature of LCD displays. If the rotation of each color cell is affected differently by pressure, then pressure can easily cause color distortions as each color will not darken by the same amount.

I think the ripples are caused by high pressure where you change the optical rotation of the inner cells (the ones closer to your finger) so much that they go through a at least one full 360o rotation, leading to regions of both bright and dark light.

The persistence is because liquid crystals are viscous — they have to be, as they consist of long molecules. They seem to align quickly to potential differences (potential differences align the molecules without moving the liquid), but when we add pressure (where the liquid itself distorts) I think it takes some time to bounce back. The ripples move due to this same persistence.

1. If you're ever at a 3D movie, have a look at your phone screen with the 3D polarised glasses on. It looks pretty cool!

2. If this did not occur, then the rotation would stay the same as $\theta\propto lc$ and $lc$ does not change if there is no bulge.


What is the relationship between equilibrium concentrations in two similar experiments?

Note that you are not given the equilibrium constant so you do not know the absolute equilibrium concentration of any species at equilibrium.  This eliminates 2 and 4.

Since all conditions and total concentration of reactants + products are the same between the two experiments, you do know that the concentration of any species in experiment A is equal to that species in experiment B.  Again, only at equilibrium, and you don't know to what degree that equilibrium lies in terms of absolute concentrations.


Can we classify all the d-block elements as transition metals?

The idea that the properties of d-block elements are transitional between those of s-block and p-block elements, and that is the reason for calling them transition element is absolutely correct, but the thing is transition metals are defined is slightly different manner.

Definition of a transition metal/Criteria for an element to be a transition metal

  A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals

Appreciate that the general electronic configuration for d-block elements are slightly different for those of transition metals.

EC of d-block elements: $\ce{(n-1)d^{1-10}ns^{1-2}}$
EC of transition metal/ion : $\ce{(n-1)d^{1-9}ns^{1-2}}$

Some examples of elements which are in d-block but not a transition metal

Scandium has the electronic structure $\ce{[Ar] 3d^{1} 4s^2}$. When it forms ions, it always loses the 3 outer electrons and ends up with an argon structure. The $\ce{Sc^{+3}}$ ion has no d electrons and so does not meet the definition.
Zinc has the electronic structure $\ce{[Ar] 3d^{10} 4s^2}$. When it forms ions, it always loses the two $\ce{4s}$ electrons to give a 2+ ion with the electronic structure $\ce{[Ar] 3d^{10}}$. The zinc ion has full d levels and does not meet the definition either.

There's a small thing that one must take into account that some elements are capable of forming multiple metal ions in those cases we must consider more common ion (the most stable one)

The pefect example for this case is $\ce{Cu}$ ($\ce{[Ar] 3d^{10} 4s^1}$), $\ce{Cu}$ is capable of forming two ions :

$\ce{Cu^{+1}}$ ($\ce{[Ar] 3d^{10}}$) (not the most common ion of $\ce{Cu}$)

$\ce{Cu^{+2}}$ ($\ce{[Ar] 3d^{9}}$)  (most common ion formed by $\ce{Cu}$)

Hence, copper is definitely a transition metal because the $\ce{Cu^{+2}}$ ion has an incomplete d subshell.

Using this idea we can also say  $\ce{Zn}$, $\ce{Cd}$ and $\ce{Hg}$ are not considered as transition elements.

How are these metals different from rest of the d block elements?

$\ce{Zn}$, $\ce{Cd}$ and $\ce{Hg}$ are not hard metals (as compared to other d block elements), which can be attributed to the fact that they have no unpaired electrons which makes their metallic bonding weak.


What is the identity of the unknown compound using GCMS and NMR?

Hints:

If you counted carbon signals in $\ce{^{13}C}\mathrm{~NMR}$, you'd find the compound has at least 10 $\ce{C}$ atoms.  Thus, it's indicated boiling point is incorrect, unless it's determined at reduced pressure. If so, you need to give that pressure.
You are correct about saying your compound should be aromatic. Did you count how many aromatic hydrogens are there? At least, by doing that, you may able to speculate some framework for your compound, as a beginning point.
Now, if you are careful, you see no aliphatic carbon signals in your $\ce{^{13}C}\mathrm{~NMR}$. Then, what is the signal at $\mathrm{\delta~1.6}$ in your $\ce{^{1}H}\mathrm{~NMR}$ doing there?
Now, go to $\mathrm{GC/MS}$. In there, your parent peak is almost looks like a doublet. Do you know why? Can you recall what element have the two stable isotopes about 50:50 natural abundance with $2~m/z$ units? If you don't remember, then subtract each of your parent peaks from your base peak in your $\mathrm{MS}$. Now, you almost have the empirical formula of your compound. Now it's matter of analyzing peak patterns in $\ce{^{1}H}\mathrm{~NMR}$ and fragmentation pattern in $\mathrm{MS}$.


The given dihydroborinine is indeed anti-aromatic if is it fully conjugated and planar, but counting pi electrons in a conjugated ring is not as simple as it seems when you have boron in the ring.  If the boron is not joined to a ligand, as in (mono)boronine or unhydrogenated diborinine, it could pass its electron deficiency into the sigma bonding framework to preserve pi electron aromaticity.
The basic (or Lewis acidic) case of borinine
Borinine or borabenzene, $\ce{C5H5B}$, is two electrons short of a well-known isostructural aromatic compound, pyridine.  We might suppose that as in pyridine, the sigma bonding and non-bonding orbitals including the outward-directed orbital from the boron must be filled before pi electrons can be introduced, which would leave borinine with only four pi electrons.  And we might be wrong.
Raabe et al. [1] perform an extensive quantum mechanical study of borinine molecules.  They find that in fact there are six pi electrons, the ring is aromatic, and it is the sigma electron framework that is electron-deficient because the outward-pointing boron-based orbital is empty.
To understand this, we look at the structure of borinine given in the above reference, from which the figure below is taken:

The above figure is one of several calculation results, but the all show similar results.  Not only is the ring highly distorted, it is distorted oppositely from what one might expect from the boron atom being inherently larger than the carbon atoms.  The boron is displaced inwards from where it "should be".
This inward displacement means the lobes from the adjacent carbon atoms are overlapping the "empty" outward boron-based orbital.  This overlap creates delocalized bonding around the boron and makes the empty orbital sigma antibonding, whereupon it becomes the LUMO.  The low-energy sigma orbitals are filled after all, but they do not include that outward-pointing boron orbital.  Thus borinine retains six pi electrons.
We should not conclude that borinine is therefore stable.  The pi electron component is stabilized by aromaticity, but with the overall electron deficiency in borinine this comes with an electron-deficient and highly distorted sigma-electron framework.  The antibonding orbital referenced above is too high in energy to be occupied in borinine as such, but it is low enough to combine with an electron pair from a Lewis base, such an an amine.  The resulting adduct (called a boratabenzene when the base is an anion) is the typically isolated product.  With the outward-pointing boron lobe now occupied, boratabenzenes and neutral Lewis-base adducts resume a more typical benzene-ring structure.  They still have six pi electrons, and thus remain aromatic, because the base contributes both electrons to its bond with the boron.
Going for two
I do not have a similar reference for diborinines or their hydrogenared derivatives.  However, Wahab et al. [2] provide nucleus-independent chemical shift (NICS) values for several monocyclic rings which they use in the simplified computational scheme they discuss; these are shown in the figure below, taken from this reference.

As many readers know, NICS models the inherent magnetic shielding (negative value) or deshielding (positive value) within a conjugated ring, these opposing signs then being interpreted as showing aromaticity or anti-aromaticity respectively.  Thus the unhydrogenated 1,4-diborinine is aromatic and even beats several more familiar aromatic heterocycles.  The delocalized bonding around the boron noted above apparently applies at each boron atom here, enabling the aromatic ring structure to be retained.
However, the dihydro compound shown at the right of the figure has become antiaromatic.  As was also noted earlier, when the boron is bound to a ligand the delocalized bonding around the boron is lost and we can go back to counting electrons as we would in an ordinary benzene ring.  However, whereas in Lewis base adducts both electrons needed to bond the boron to its ligand come from the ligand, in dihydrodiborinenes the hydrogen ligands provide only one electron apiece and thus the remaining two electrons to form these bonds have to come from "robbing" the pi system.  Thereby aromaticity in the pi system of dihydrodiborinines is lost, as seen in the NICS becoming positive.
Nobody else like you
Ultimately, we have to be careful rendering molecular-orbital structures in boron compounds because boron, uniquely among nonmetallic elements, lacks sufficient valence electrons to fill its orbitals through pairwise electron sharing alone. The unique bonding characteristics boron often adopts to compensate for this deficiency must be taken into account in order to achieve a proper rendering.
References
1.
Gerhard Raabe, Wolfgang Schleker, Eberhard Heyne, and Jörg Fleischhauer , "Borinine (Borabenzene): Its Structure and Vibrational Spectrum. A Quantum chemical Study", Z. Naturforsch. 42a, 352-360 (1987).  Link.
2.
Alexandra Wahab, Felix Fleckenstein, Stefan Feusi and Renana Gershoni-Poranne, "Predi-XY: a python program for automated generation of NICS-XY-scans based on an additivity scheme" Electronic Structure 2, 47002 (2020).  Link


Why does MgCl2 lower the freezing temperature, and how does it do so?

The $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$ ions disrupt hydrogen bonding, preventing ice from forming a lattice structure. Therefor, the water must be colder so those weak hydrogen bonds will be acting on water molecules with less kinetic energy (i.e., colder), finally allowing them to create a crystal around the dissolved "impurities" $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$.

This is a basic colligative property. Its more effective for electrolytes because they dissociate, creating more lattice disruptions. As a general resource, see: Ebbing, Darrell D., General Chemistry, 3rd ed, Houghton Mifflin, 1990.


Calculating the concentration of a diluted solution

The answer is basically correct (see note at the end of my answer about significant figures) but there is a simpler method, using a dilution factor.

The original volume of your solution was 2 mL, and the final volume was 12 mL, so the dilution factor is simply $\frac{2}{12}$.

The original concentration times the dilution factor gives the final concentration:

$$\pu{10\frac{mg}{L}}\cdot\frac{2}{12} = \pu{1.7\frac{mg}{L}}$$

Note that only 2 significant figures were given in the problem and your answer reported 3.


Comparison of two spectra in order to find whether the irradiated sample has significantly different resulting spectra

Why don't you start with the simplest possible approach. I am sure there are very fancy spectral matching statistical algorithms, but simplicity has its own beauty. Comparing an inteferogram (not many features there) might be difficult but what about absorbance rather?
Look at the absorbance difference spectrum: (Spectrum after irradiation)-(Original sample spectrum). How does the plot look like? What is the signal which you are collecting? Sometimes the human eye can save a lot of time rather than doing significance testing on data points and forming null hypotheses.
The features which remain the same will be zero now and the features which change will be visible in the the difference spectrum.
Alternatively, you can take the first derivative of both spectra and smooth it with Savitsky-Golay or study the difference of "first derivatives". Noise will be an issue but I guess you can take care of that.


"Number of moles of electrons involved in the reaction" for hydrogen cell

Nernst equation is used for finding the cell potential when the various reactants, i.e. anodic and cathodic electrolytes are not in their standard states ($298\text{K}, 1\text{mol l}^{-1}$). Thus, in such cases, you first write down any balanced form of equation, which for our hydrogen electrode, I choose to write down $\ce{3H2 -> 6H+ +6e-}$ and these electrons are used in a standard $\ce{H2}$ cell on the other electrode. Hence the Nernst equation would give the cell potential for this concentration cell as ($E^\circ=0$ since the the standard potentials of both sides is the same since the materials are same, and the non-standard hydrogen cell is the anode)
$$E_{cell}=-\frac{RT}{6F}\ln\frac{[H^+]^6}{1}$$
By using the properties of logarithm, you can show this is equivalent to $-\frac{RT}{1F}\ln[H^+]^1$, and this is the equation if you took only 1 electron transfer in the reaction.
It does not matter what equation you write, since in the final analysis, the number of electrons taken are compensated by the power of the concentration term in the logarithm.


Aren't isothiocyanate and thiocyanate ligands monodentate?

The answer to your title question is: Yes, they are monodentate.
But I think you have a misconception about when to use which type of counting: It does not depend on whether the ligand is polydentate or not.
The IUPAC guide states that you use the bis-tris-tetrakis-etc. prefix when you want to avoid confusion:

The  prefixes are ‘di’, ‘tri’, ‘tetra’,etc., for  use  withnames for  simple  entities,  or ‘bis( )’, ‘tris( )’, ‘tetrakis( )’,etc., for names for most entities  which themselves contain multiplicative  prefixes or locants.  Care must  also  be  taken  in  situations when  use  of  a simple multiplicative prefix may be misinterpreted, e.g., tris(iodide) must  be  used  for  $\ce{3I−}$ rather  than  triiodide  (which  is  used  for  $\ce{I3^−}$), and  bis(phosphate)  rather  than  diphosphate  (which  is  used  for $\ce{P2O7^{4−}}$).
Source: https://iupac.org/cms/wp-content/uploads/2018/05/Inorganic-Brief-Guide-V1-3.pdf

So while in the case of thiocyanates and isothiocyanates I don't see that there is any confusion avoided by using bis( ), it is certainly not wrong.


What is the mathematical relation for the change in water density at different temperatures?

To do this one must measure density accurately at many temperatures and then fit a curve to the measurements. This has been done by several investigators in several industries. For example in the sugar industry we have

$$\rho =(((((-281.03006e-12*t +105.84601e-9)*t-46.241757e-6)*t-7.9905127e-3)*t+16.952577)*t +999.83952)/(1+16.887236e-3*t)$$

where $t$ is the temperature in °C and the density $g·cc^-1$. The coefficients were taken from the ICUMSA formula with the sugar concentrations set to 0.

And

$$\rho = 0.99984+t*(6.7715e-05-t*(+9.0735e-06-t*(1.015e-07-t*(+1.3356e-09-t*(1.4421e-11-t*(+1.0896e-13-t*(4.9038e-16-9.7531e-19*t)))))))$$

returns the density of water, also in $g·cc^-1$,  as a function of centigrade temperature (ITS 1990). It is based on a fit to data from
Bettin, H.; Spieweck,F.: "Die Dichte des Wassers als Funktion der Temperatur nach Einführung der Internationalen
Temperaturskala von 1990. PTB-Mitt. 100 (1990) pg 195-196
The referenced data set lists densities for each 0.1°C over the range (0,100). The data used in determining this
polynomial are the subset on integer degree values. Sporadic checking shows that the fit appears to be
accurate to 1 count in the 6th decimal place. The rms residual wrt the fit points is 3.1E-7 and the peak
residual 5.5E-7. The residuals appear noiselike.

You ought to be able to paste those formulas into a visualization program and plot values. I don't see any logarithmic behaviour. I've left the * in the formulas so you can do that. Note, when reading them, that 16.887236e−3∗t means (16.887236e−3)∗t.


Physical interpretation of the Gibbs-Duhem equation

If you write down the equation in terms of the very definition of $\mu_i=\frac{\rm dG}{{\rm d}n_i}$ and impose ${\rm d}n_\text{totoal}=0$, it's just the conservation of the energy potential ${\rm d}G=0$.

Say you own a theme park where, every hour, each adult has to pay ${\rm d}\mu_a=\$7$ (money=energy) and each kid has to pay ${\rm d}\mu_k=\$3$.
We assign positive signs to when they put money out of their pockets.

If there are $n_a=10$ adults and $n_k=5$ kids in the park, and if this number of people stays fixed (${\rm d}n_\text{totoal}=0$ over some time), then they together have to spend

$$n_a\cdot {\rm d}\mu_a + n_k\cdot {\rm d}\mu_k = \$70 + \$15 = \$85$$

Now say you, the park owner, are not allowed to actually make any money (${\rm d}G=0$) and you relax the condition that the kids have to pay any money at all. That is, now just the adult have to pay ... and since the money has to go somewhere, the kids are on the receiving end. Then the ten parents still spend $\$70$ per hour and now the five kids will split that money among each other. Each kids receives
$$-{\rm d}\mu_k = \frac{1}{n_k}\cdot n_a\cdot {\rm d}\mu_a = \frac{1}{5}\cdot \$70 = \$14$$

The comparison with "money per time" lacks in that here people don't bring in money just from coming to the park (as it's the case with particles coming into the system and energy). But I hope this clear up the meaning of factor $\frac{1}{n_k}$ clear.


Why should a complexometric titration be kept neutral?

Your last guess is exact. This is a classical method, the equivalent point is seen by our eyes, so in our interest is for the complex to be formed so we can recognize the colour, and we should add $\ce{NH3}$ so the colour won't change immediately after the equivalent point.


Partial reactions in redox formation of rust

Bringing up rust seems to indicate confusion between the concept of corrosion and the concept of oxidation. Practically, iron corrosion is the breakdown of iron as a result of oxidation reactions.

To explain corrosion, understand water can self-ionize producing hydronium, $H_3O^+$, and hydroxide, $OH^-$ as well as ionize small amounts of iron causing it to dissolve. The dissolved iron ions collide with the hydroxide ions to form iron hydroxide.  Iron hydroxide may fall out of the solution as it is not very soluble, or if the concentration is low enough the surplus of hydronium donates a proton to the iron hydroxide converting it to iron oxide.

Oxidation, which is a more abstract concept, involves any reaction which results in the loss of an electron.  Instead of considering corrosion, consider other reactions.  In photosynthesis water is oxidized because it loses an electron.  In combustion, fuel is oxidized because it looses an electron.

Chemical reactions can get extremely complex, so it may be prudent to master the language of chemistry before delving into the complexities of specific reactions.  Many reactions are described with the language of redox (Reduction and Oxidation) chemistry.


Why are tertiary carbocations the most reactive, if they're already stable?

This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at:

First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.

Second, transition states involving tertiary carbocations as opposed to primary carbocations are more favorable precisely because tertiary carbocations are more stable than primary carbocations.

Note that this does not mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the transition state ... of course it's going to be reactive - of course it'll change very rapidly!

We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation.

Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons).

Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question!


Production of hydrogen using vinegar and metal for experiments

Acetic acid will probably react to slow. If you want to produce hydrogen in the litre scale I would suggest to use diluted hydrochloric acid and zinc flakes or iron. But be aware that it can be difficult to control the reaction. In a laboratory you could use a Kipp's apparatus for this.


A picture is worth a thousand words so I've redrawn your compound and have numbered the areas to be considered.

Double bond "1" is not capable of displaying cis-trans
stereoisomers since there are two identical substituents (hydrogens)
on one end of the double bond.
Double bond "2" is capable of displaying cis and trans stereoisomers since there are two different substituents on each end of the double bond.
Double bond "3" can only exist as a cis isomer due to the geometric constraints imposed by the 6-membered ring.
Double bond "4" is a 1,2,3-triene (allene would be a 1,2-diene); such trienes  display the same stereochemical cis-trans possibilities as a double bond.
The substituents on the two cyclohexene carbons contained in circle "5" also have cis-trans arrangements; these are further illustrated in the cyclohexene drawings pictured below.
Finally, each of the carbons contained in circle "5" are chiral (they each have 4 different substituents) and can exist as R and S stereoisomers.

Considering only the circles that can have stereoisomers (2, 4 and 5) we can have the following spatial arrangements:

2-cis, 4-cis, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-cis, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-trans, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-trans, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-cis, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-cis, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-trans, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-trans, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.

Therefore, there are a total of 16 stereoisomers for your molecule, 8 diastereomers, each of which can exist as an enantiomeric pair.

Thanks to Loong for pointing out that the cis and trans geometry in circle "5" limits the R and S possibilities at the two chiral carbon atoms in circle "5".  That is, these carbons in (for example) the cis isomer cannot be R,S R,R S,R and S,S, - in the cis isomer they can only be either R,S and S,R or R,R and S,S.


Yes.  But it is a bit of a joke of a sort.  In any system with energy levels, the lower energy levels never have a lower population than the higher ones.  In fact there is an equilibrium arrangement of the populations of such levels characterized by the temperature.  The higher the temperature, the more evenly populated the levels are.  There is even a formula for calculating the populations in this case.

The energy levels of some systems depends on a magnetic field.  You can let the populations of the energy levels come to equilibrium at some temperature $T$ and then quickly invert the magnetic field.  What was a low energy level now becomes a high one and vice versa.  You have an upside down distribution of populations.

Soon enough that upside down population decays to a right side up population.

But while the distribution is "upside down" one finds that it is given by the same formula used for an ordinary distribution, but with a negative temperature.


Why does the size of hybrid orbitals vary as sp³>sp²>sp?

Looking at the shapes of atomic orbitals shown in many textbooks can be deceptive. The probability density of the radial part of the wavefunction ($r^2|R_{n,l}(r)|^2$)  shows that the maximum probability for 2s orbitals is noticeably greater than for 2p, and for d orbitals the same applies with p greater than d.  (As the principle quantum number increases so does the position of the maximum probability density of an orbital, i.e. increasing $n$ increases the size of the orbital.)

The hybrid orbitals form from  the atomic orbitals and are added together in different proportions. So the position of maximum probability is a type of average value of the s and p. The sp hybrid has $2s \pm 2p$, the $sp^2$ in the z direction $2s+\sqrt{2}.2p$ and the $sp^3$ also in the z direction just as for the sp orbital. It seems to be that the maximum position of probability taken along, say, the z direction is the same in each case.


Can a very high concentration of hydrogen peroxide, inhibit or denature the enzyme catalase?

20% hydrogen peroxide sounds HUGE. But something can happen that is called substrate inhibition. It can be detected if the substrate acts as a non-competitive reversible inhibitor (i.e. binds only to the active ES Michaelis complex). In the case I studied during my PhD, the inhibition constant ($K_\mathrm{i}$) was roughly 1000 times higher that the Michaelis constant ($K_\mathrm{M}$).
The steady-state rate was well fit with :
$$r = \frac{r_\infty}{1+\frac{K_\mathrm{M}}{s} + \frac{s}{K_\mathrm{i}}}$$
So the activity would go through a maximum ($s$ is substrate concentration).

Tip : if it is possible, do you assays with substrate concentrations growing exponentially, like $1\,\mathrm{\mu M}$, then 2, 5, 10, 20, etc.

A good way to decide between enzyme denaturation and substrate inhibition would be to know if it is reversible : denaturation must not be reversible. You could first make a baseline assay (at low conc.), expose the enzyme to a high concentration, then change the composition of the solution (eg dialysis, dilution) and run another assay at low concentration.

A "volume of bubbles" does not sound very accurate. Could you use a Clark electrode to measure the activity instead ?


Can carboxylic acids be oxidised further?

In carboxylic acid, the carbon atom of the COOH group is already in high oxidation state, namely +3. In order to reach a higher oxidation state (+4), it requires breaking the C—C bond, typically to form molecular CO2. Thus, oxidation of carboxylic acid in strongly oxidizing conditions is known as a decarboxylation reaction.


Aspartame poisoning: Can it modify aminoacids?

Aspartame is one of the most well studied food additives there is. It seems to be controversial in the same way that evolution is—scientific research is overwhelmingly on the side of the substance being safe to consume, even at very high levels, but a Google search on it brings up things like "Aspartame is, by Far, the Most Dangerous Substance on the Market that is Added To Foods" (from a site trying to sell natural health products). A lot of this comes perhaps from the naturalistic fallacy: aspartame is artificial and thus unsafe. A huge number of regulatory authorities have deemed it safe to consume: FDA, Health Canada, EFSA, FSA, etc.

There is also a good deal of bad biochemistry thrown about: aspartame is metabolized into the two amino acids aspartic acid and phenylalanine, as well as methanol. There have been some hypotheses that aspartic acid intake could induce some amount of toxicity, but experimental evidence is lacking. Methanol, of course, is well known to be toxic. The key is, that aspartame is very sweet and only tiny amounts are added to food. So even drinking enough diet cola to hit the FDA's acceptable daily intake of aspartame (which works out to ~7.5 L), the quantity of methanol consumed is only around 300 mg, and things like fruits are larger sources of methanol in people who drink sane amounts of diet soft drinks.

The one condition in which aspartame is known to do harm at reasonable intake is in people who have phenylketonuria (PKU). This might be what you're thinking of by "faults in the vital proteins of their body". PKU is a genetic disease caused by a defective gene for the enzyme phenylalanine hydroxylase which results in an inability to metabolize phenylalanine correctly; thus, consuming phenylalanine from any source is hazardous to sufferers of the disease.


How can diamond have a fixed refractive index?

Your textbook is wrong. Some crystalline solids are optically anisotropic. Many crystalline solids are indeed optically isotropic in nature. In solids like this the light beam experiences the same electron density regardless of the direction the crystal is oriented. Some examples of that are (obviously) diamond, halite and sylvite (sodium and potassium chlorides), garnets, spinels and more. Note that this is a feature derived from the electronic structure and it has nothing to do with the actual shape of the crystal.

Even in the case of anisotropic crystals, some are so weakly anisotropic they may considered isotropic in practice. Zeolites are a perfect example of those.


Why decarbonate soda for spectroscopy experiment?

The primary reason is that when soda warms up, the $\ce{CO2}$ would escape slowly as bubbles. These bubbles would significantly scatter light affecting the absorption spectra. Usually, the effect of such scattering is quite pronounced and the absorption bands of individual molecules will be (might be) buried below the broadened feature in the lower wavelength region.

In colloidal and suspension systems light is scattered by individual particles. These particles are larger in size than the wavelength of light and such scattering phenomena is termed as Tyndal scattering or Tyndall effect.  For such a process, the scattering coefficient has $wavenumber^{4}$ dependence which means that for lower wavelengths the scattering increases significantly. Thus when the sample has particles then absorption spectra is skewed and we see an increasing absorption peak when going to a lower wavelength. (You can yourself try this with milk which is a suspension).

About the absorption spectrum of $\ce{CO2}$:

$\ce{CO2}$ being a simple molecule is expected to have electronic absorption bands below 300nm. It does have electronic absorption between 200 to 270 nm. It is more difficult to find on the internet since usual absorption of $\ce{CO2}$ is referred to the vibrational absorption observed in the in the infrared region. Hence, 'UV absorption of $\ce{CO2}$' is a better search term.

For your reference: T. Joutsenoja, A. D'Anna, A. D'Alessio, and M. I. Nazzaro, "Ultraviolet Absorption Spectra of Carbon Dioxide and Oxygen at Elevated Temperatures," Appl. Spectrosc. 55, 130-135 (2001)


Anodisation of aluminium in sulfuric acid - Electrolysis reactions

As the experiment is described, in the solution we have :
$$H^{+}, H_{2}O, Al_{(s)}, SO_{4}^{2-}, OH^{-}$$
Taking that into account and seeing how the environment is acidic ( $pH<7$ ), we can see that the reaction of reduction reaction that can take place is $H^{+}/H_{2}$
Following the same kind of reasoning, reaction $(1)$ in the anode is caused by the odydation of the aluminium plate when placed in an acidic environment, $O^{-2}$ is present because of the oxygen in the atmosphere and $OH^{-}$ exists in the solution anyway(water autoionization).

As far as the second question is concerned, notice that the aluminium foil in the anode does not participate in any reaction. It is the $H^{+}$ that is reduced, while the aluminium foil serves as a conductor for the current. Knowing that, we could replace the aluminium foil with either $Cu_{(s)}$ or $Pb_{(s)}$, that may also be used as conductors ($Cu^{2+}, Pb^{2+}$ will not be present in the solution and we will have the same reduction reaction in the anode), no matter their standard electrode potentials.

Finally, it is interesting to point out that in this experiment the anode and cathode are in the same solution which is used at the electrolytic reactions (acidic environment ) and at the same time plays the role of the electrolytic bridge constantly providing $H^{+}$ and keeping the solution electrically neutral.


Why m-hydroxybenzoic acid has higher acidity then p-hydroxybenzoic acid?

The -$\ce{OH}$ group has two effects, namely +R and -I effect i.e. Resonance and Inductive effect. When -$\ce{OH}$ group is attached to the para position, it has a tendency to delocalize it's electrons towards the aromatic ring (i.e., benzene). Thus the electron density on oxygen atoms of carboxylic group increases and the polarity between $\ce{O}$ and $\ce{H}$ decreases leading to less acidity. Whereas when the -$\ce{OH}$ group is attached to the meta position, +R effect is not applicable and only -I effect is effective. Thus it withdraws the electrons from the ring, and increasing the polarity between $\ce{O}$ and $\ce{H}$ in carboxylic group, making it more acidic.


What kind of metal alloys can resist corrosion from molten salt?

George Lai's High-Temperature Corrosion and Materials Applications (excerpts available on Google Books) is a good source. Very often for molten salts, Hastelloys are used; they're special-purpose alloys developed by Haynes International. I can't find a reference, but I believe Oak Ridge National Laboratories used Hastelloys for their molten salt reactors (which had fluoride salts, very reactive).

If you're working with sodium chloride (little potassium), according to Lai, Hastelloy-188 seems to be the best bet, although many others would work as well. One reality of chemical engineering is that there will always be corrosion, it's just a matter of keeping it at an acceptable rate for the application.


Are bulkier molecules more likely to react?

The relative rates are going to depend not on molecular weight so much but the structure of the molecule.

The specific reaction is also very important.

Nucleophilic substitution reactions (some of the first reactions most students learn about in an organic chemistry course).

Let us compare 1-chlorobutane and 2-chloro-2-methylpropane, both $\ce{C4H9Cl}$. However, the 1-chlorobutane is a primary alkyl halide (not very bulky) and 2-chloro-2-methylpropane is a tertiary alkyl halide (bulky).

The bulkier compound reacts faster in a solvolysis reaction:

$$\ce{(CH3)3CCl ->[\ce{H2O}] (CH3)3COH}\ \ \mathrm{fast}$$
$$\ce{CH3CH2CH2CH2Cl ->[\ce{H2O}] CH3CH2CH2CH2OH}\ \ \mathrm{slow\approx no\ reaction}$$

The reason that the tertiary halide reacts faster in this reaction is that its mechanism involves the formation of a carbocation intermediate and the extra carbon groups at the tertiary center provide more inductive stabilization to that carbocaion than the single alkyl group attached to a primary position. In general, in reactions that generate carbocation intermediates, tertiary substrates will react faster.

However, in many reactions bulkier molecules have steric interactions that negatively effect the rate. In a nucleophilic displacement with $\ce{NaI}$ in acetone, 1-chlorobutane will react faster.

$$\ce{(CH3)2CCl ->[\ce{NaI}][\mathrm{acetone}](CH3)3CI}\ \ \mathrm{slow\approx no\ reaction}$$
$$\ce{CH3CH2CH2CH2Cl ->[\ce{NaI}][\mathrm{acetone}]CH3CH2CH2CH2I}\ \ \mathrm{fast}$$


Alcohols have labile protons which are exchangeable through acid-base equilibria with other similar protons, such as other alcohols, amines, amides, thiols and especially water. This reaction is fast enough, on the NMR timescale, that it can be observed as a reversible exchange process. The most common scenario is that there will be some amount of water in the sample, and so the alcohol and water 1H atoms exchange. The reaction is typically so fast that you don't see the water and alcohol as separate peaks, but as a population-weighted average. Therefore the actual position of the peak depends on a number of variables; temperature, solvent, starting chemical shifts of the exchanging species, populations of the exchanging species, rate of exchange, solvent and so on.

When you say 'thicker', I assume you mean a broader linewidth. Again, this is a consequence of the exchange process, and that you are seeing an average peak for the -OH and (say) H2O. Sometimes this broadening is so large that you can only observe it as a broad hump in the baseline of the spectrum. Again, the linewidth will be a function of the starting chemical shifts of the two components, the rates of exchange, and their relative populations.

It is actually entirely possible to see the alcohol peak with a narrow linewidth, similar to the CH2 and CH3 linewidths. To do this, you need to be able to eliminate the exchange process, either by excluding water from the sample( run ethanol in very dry dmso for instance), or cooling down the sample sufficiently to slow down the exchange (probably not feasible for most common solvents with ethanol and water though). In this case, you will observe your alcohol as a narrow triplet (approx 1.7Hz coupling by memory), with the CH2 appearing as a doublet of quartets. A great example of observing alcohol protons is to run sucrose dissolved in pyridine-d5. All OH protons, and their couplings are observable.

It is really important to understand that the integrals of these peaks are still valid. In the absence of any exchange (where your alcohol peaks are sharp), the integration of an  ethanol 1H spectrum should still give you 3:2:1. Where there is an exchange process occurring with (say) water, the integrals will be 3:2:X, where X=(sum of -OH plus water). This method of analysis can be used to verify alcohol content of spirits and wine, for instance.


Catalytic poison in Pt catalysed manufacture of H2SO4

$\ce{Fe}$ and $\ce{V}$ are used in the contact process.
$\ce{Cu}$ is used in the preparation of chlorine gas.
Thus $\ce{Cr}$ can be used as a catalyst poison.


Ionize Hydrogen and Oxygen Gas in Electrolytic Cell?

Since the monatomic hydrogen and oxygen have combined, respectively, in H2 and O2 gases, those molecular bonds must be broken in order to get them to recombine as water and produce electricity. That is precisely what a fuel cell does. You can lease a hydrogen fuel cell powered Hyundai Tucson, if you're close to H2 filling stations, and Iceland is looking into setting up a hydrothermal-powered H2 fuel station system.


Difference between any arene and an aromatic compound?

Arene — Compound which contains a benzene ring

Aromatic Compounds — Compounds having aroma (wait, the modern definition is different, I know)

Every arene is an aromatic compound but every aromatic compound need not be an arene.

Mathematically,

$$\text{arene} \subset \text{aromatic compounds}$$

therefore anything which contains a benzene ring has to be aromatic

Some examples of arenes:

Getting into aromatic compounds, the modern criteria for aromaticity are:

The molecule is cyclic (a ring of atoms)
The molecule is planar (all atoms in the molecule lie in the same plane)
The molecule is fully conjugated (p orbitals at every atom in the ring)
The molecule has $4n+2$ π electrons ($n=0$ or any positive integer)*

Examples (non-benzenoids):

Now, benzenoids (compounds which contain benzene) do match with this. Therefore, benzene derivatives are aromatic compounds.


The partial pressure of a component in a mixture is always less than that of the pure component at the same temperature. (Calculating the chemical potential shows that if it were not the case the mixture would be unstable compared to the pure liquid).  The question then becomes what is the partial pressure of the component A above a mixture $p_A$ related to that above the pure liquid $p_A^*$.  Raoult's law states that the ratio is equal to the mole fraction $x_A$ of A, $p_A/p_A^*=x_A$.

Physically, this is justified by considering that the chance of finding a molecule of A in the vapour, relative to the chance for the pure liquid, is equal to its mole fraction. This is reasonable if dealing with ideal mixtures and assuming that the vapour behaves as an ideal gas. Thus there is no derivation as such, we proceed directly from the physical argument to the law.


Why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane?

The difference in boiling point, between methanol and methane on one hand and between 1-decanol and decane on the other hand, is primary due to the presence of hydrogen bonds (HB) in methanol and 1-decanol and not in methane and decane. The density of these intermolecular forces (HB) is higher in methanol than in 1-decanol (In 1-decanol, for every ten atoms of carbon there's one possibility to have HB. While in methanol for every atom of carbon there's one  possibility to have HB.


How does hyperconjugation in alkenes affect bond lengths?

The MO interpretation is not really all that complicated: you are allowing the C–H σ orbital to overlap with the C=C π* orbital. This leads to a very slight population of the π* orbital, which is antibonding with respect to the doubly bonded carbons. Therefore, that C=C bond length is slightly lengthened. Likewise, the C–H σ orbital is slightly depopulated, which makes the bond length longer.

At the same time, you are forming a slightly stronger C(sp2)–C(H) bond because these new MOs have bonding interactions between the C–H σ orbital and the p orbital on C(sp2). This is not entirely made clear in the diagram above from Fleming's text, which I have addressed separately here. From the diagram, it seems that there are equal amounts of bonding character (from ψ1) and antibonding character (from ψ2) However, the main point which I wrote in that post is that the π* orbital mixes (very slightly) into both ψ1 and ψ2. This increases the C–C bonding character in ψ1 and decreases the antibonding character in ψ2, such that there is a (very slightly) shorter bond.

The actual text in the book makes this clear, but the diagram does not. I suggest reading the text carefully.


How can chlorine have 10 electrons around it in chlorine trifluoride?

Interhalogens consist of one $\ce{X-F}$ pair whith a regular two electron bond, and pairs of flourine atoms taking over one free electron pair of X (chlorine, bromine, iodine).

What forms are three-centre four-electron bonds (one pair of electrons from the chlorine, two single electrons from two fluorine atoms). That's why interhalogens always have an uneven number of fluorine atoms (except for the dimer $\ce{Cl_2F_6}$).

https://en.wikipedia.org/wiki/Three-center_four-electron_bond

If this is still octet rule is debatable, or at least not directly obvious (see the wp article). Let's just say that fluorine is a bad, bad guy who has things his way.


The role of temperature in Nernst equation

The standard cell potential is temperature-dependent itself, just like the Gibbs free energy. The easiest way to see this is to write the Gibbs free energy in terms of enthalpy, entropy and temperature:

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

This shows you that $\Delta G^\circ$ is strongly temperature dependent (even if $ \Delta H^\circ$ and $\Delta S^\circ$ were independent of temperature, which they are not).

To relate this to the cell potential, remember that $\Delta G^\circ$ is a measure of the maximal work a system can do, and that the electrical work is electrical potential times charge moved across that potential.

$$\Delta G^\circ = w_\text{electrical} = -E^\circ n F$$

So $E^\circ$ is temperature dependent as well.


Coordination compounds with alkene ligands

Ethylene is a monodentate ligand and there are 4 ligands total: ethylene,  two chlorides, and water.

Regarding hapticity, you can find a great answer here: What does the eta notation mean in the naming of a transition metal complex?

While a chapter can be dedicated to the nature of organometallic bonding, I think it is practical to say that it is the two electrons in the pi-bond that are involved in bonding to the metal center.


The group itself unprotonated is called a carboximidate as noted by the Wiki page.  The acid that the group is derived from is carboximidic acid.  As is common practice with imines, the nomenclature for the corresponding protonated halide is done by removing -es, adding -ium, and the halide name.  If this is done using the acid name, the group will be called a carboximinium chloride.


Why does a 1:1 mixture of Polyvinyl Alcohol (PVA):H2O decompose/carbonize during heating?

I experienced similar phenomena in my experiments.
In my case, I put PVA pallets (M.W. is 84000~12500, 87~89% hydrolyze degree) in the oven at 120∘C. After over 12 hours, the color became yellowish (partially brown). But after dissolving the pallets in the water, there was no difference between control sample(without heat treatment).

There are several papers about formation of polyene in PVA chain under acid catalyst. And, almost commericial PVA product has always small amount of acetic acid which hydrolyzed from PVAc spontaneously. Therefore partial polyene-nization (semicarbonization) was possible with heating under melting temperature.

Also, I made acidic PVA film by solution casting (10% PVA, pH 2 by adding HCl). Then, the color of PVA film was changed to black at 120∘C in the oven.

So color change of PVA is common while experiments. Maybe it is related with carbonization.

maybe polyenenization or carbonization.
I remembered that PVA becomes viscous near 150-180∘C. Measuring Tm of polymer exactly is not easy. Especially hot plate is not suitable tool to analysis physical properties of polymers. Maybe due to cooling by air, melting of PVA is indistinguishable to glass-transition. I recommand to use DMA instead of your experiment.
To measure Tm, you don't have to mix PVA with water. DSC or DMA is better.


Why do coefficients become exponents in the speed equation when dealing with elementary reactions

A reaction occurs when its reactives are in the same region of the space. If you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^b$.
The probability of encounter is proportional to the number of encounters in some time interval. And the later to the reaction velocity.


http://www.sigmaaldrich.com/catalog/product/aldrich/z516333?lang=en&region=US

This condenses, and depending on what you have the knob turned too will either send it back to it's origin or to a collector. (I think, I have something similar in my lab but the link doesnt have a great drawing)


How to determine the number of water ligands attached to Pd in palladium chloride aquacomplex PdClₓ · yH₂O?

I hope the following two references would help you find the way. I honestly think Ref.1 shows much more promise, abstract of which states that:

Monomeric palladium(II) chloro aqua complexes of the form $\ce{PdCl_r(H2O)_{4-r}^{2-r}}$ ($r = [0,4]$) were studied both experimentally and theoretically to gain insight on both the stabilities and the nature of palladium-chloride interactions.
The thermodynamic stabilities of these complexes were studied in aqueous solutions from $5$ to $\pu{125 ^{\circ}C}$ with UV-vis spectrophotometry using a quartz flow-through cell. Tentative measurements up to $\pu{200 ^{\circ}C}$  were also carried out in pressurized titanium and gold-lined optical cells but revealed important losses in soluble palladium. The strong ligand-to-metal charge transfer bands of the palladium complexes below $\pu{350 nm}$ were used to constrain the step-wise thermodynamic formation constants at each temperature, using results of singular value decomposition of the spectra over a broad range of palladium:chloride ratios and wavelengths. The temperature-dependent constants were used to obtain changes in enthalpy and in entropy for each reaction. The thermodynamic stabilities of $\ce{PdCl(H2O)3^+}$, $\ce{PdCl2(H2O)2^0}$, and $\ce{PdCl3(H2O)-}$ are larger at higher temperatures, whilst the one of $\ce{PdCl4^2-}$ is smaller. All changes in entropies are positive for the former three species, but negative for the latter, presumably due to a larger solvent reorganization around the doubly charged $\ce{PdCl4^2-}$ species. The room temperature thermodynamic values derived from this study are also in agreement with previously published calorimetric data.
Theoretical calculations on the intramolecular distributions of electrons in the different palladium(II) chloro aqua complexes, using the methods of atoms in molecules and of the electron localisation function, showed $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ interactions to be of largely closed-shell/ionic nature. These interactions induce an important distortion of the outer core shell electrons of $\ce{Pd}$, as well as stable accumulations of electrons between adjacent $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ bonds known as ligand opposed core charge concentrations.

References:

J.-F. Boily, T. M. Seward, “Palladium(II) chloride complexation: Spectrophotometric investigation in aqueous solutions from $5$ to $\pu{125 ^{\circ}C}$ and theoretical insight into $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ interactions,” Geochimica et Cosmochimica Acta 2005, 69(15), 3773-3789 (https://doi.org/10.1016/j.gca.2005.03.015).
L. Espinosa-Alonso, K. P. de Jong, B. M. Weckhuysen, “A UV-Vis micro-spectroscopic study to rationalize the influence of $\ce{Cl-(aq)}$ on the formation of different $\ce{Pd}$ macro-distributions on $\gamma$-$\ce{Al2O3}$ catalyst bodies,” Phys. Chem. Chem. Phys. 2010, 12(15), 97-107 (DOI: 10.1039/b915753k).


What type(s) of gloves are effective against DCM and acetone?

I just did a bit of an experiment. My lab uses Kimberly Clark nitrile gloves, and we have a huge jug of acetone. First, I put a glove on my hand, and poured acetone all over it. My hand felt cold, but not nearly as cold as it feels when I spill acetone on my bare hand. I allowed the acetone to evaporate from the glove (approximately 30 s), and removed the glove. There was absolutely no acetone on my hand or the inside of the glove. I then took a fresh glove, and poured enough acetone into it to fill the fingers with acetone. I held the opening shut for about 5 minutes, and poured out the acetone. During that time, the glove slowly felt thinner and easier to stretch. At the end, it was so thin that it tore with almost no effort. So for the particular pair of nitrile gloves and acetone, it looks like somewhere between 1 and 5 minutes' exposure will render the glove useless.


This is a classic science experiment. Where saltwater (using $\ce{NaCl}$) is heated, as it does, the water evaporates and most, if not all, of the salt remains - increasing the concentration of salt.

Evaporating half the water would approximately double the concentration of $\ce{NaCl}$.

Eventually, the salt concentration becomes supersaturated and crystallises. The reason is due to:

  As the water evaporates less and less water molecules are present to keep the salt particles apart. The salt therefore recrystalises and can be collected.

Source

This process is also used in salt evaporation ponds, where

  seawater or brine is fed into large ponds and water is drawn out through natural evaporation which allows the salt to be subsequently harvested.

As Curt F mentioned in the comments, different salts and mixtures of salts can have different outcomes in terms of vaporisation.


Z* effective charge and Ionization Energy

Using the common idiom: "full subshells are stable"

It is a little more compact to use the incorrect explanation and correct it than to explain in terms of correct. Also, when you hear this incorrect explanation, you will understand what is meant.

Yes, size is a factor (thus the "up" part- IE is higher in lower periods), but the observed anomaly in second-ionization energies in $\ce{Na}$ and $\ce{Al}$ is better related to removal of an electron from a full subshell.

$\ce{Na}$ has an electron configuration of $\ce{[Ne] 3s^1}.$ The first ionization gives $\ce{Na}$ makes it $\ce{[Ne]}$, the same as a noble gas- full 2p subshell. Because full subshells are stable, it takes a lot of energy to remove the first electron from it. $\ce{Al}$ is analogous; the second ionization means removing an electron from a full subshell.

The same is observed in the first ionization energies (in $kJ/mol$):
$$\ce{Na}:495$$$$\ce{Mg}:737$$$$\ce{Al}:577$$
Ionization of $\ce{Mg}$ removes an electron from a full $\ce{3s}$ subshell, so it is a little higher than the trend.

There are many factors that contribute to exact values, but these are difficult to predict, which is why they taught a simple version in your class.

Correcting the myth

While it is common to refer to full subshells as being more stable, it isn't really what is going on. In reality, the next electron added in destabilized. This occurs because of shielding. Outer electrons feel repulsion from inner electrons, thus outer electrons are easier to remove than inner ones. The saying that "half-filled subshells are stable" is also used. Again, it is that the next added electron is destabilized, but in this case it is because the energy needed to spin pair the two electrons.


Is sublimation a property of all substances?

Look up the term “phase diagram”. At different temperatures and pressures different phases are favored. Think about water. At 0 °C and 1 atm water and ice are equally favorable. There can be freezing or melting. At other temperatures and pressures there can be sublimation and deposition.

The sublimation of water is involved in freeze drying and freezer burn of ice cream.

It turns out that iodine sublimes at room temperature and atmospheric pressure but there’s nothing special about iodine compared to other substances on a fundamental level.


What are the non-aqueous solvents for sodium polyacrylate?

Sodium polyacrylate can locally organize solvent molecules to become soluble. Since the backbone is huge compared to water, the water aligns with its more positively charged end towards the carboxylate site while the sodium ions end up on the negative end of the water. I can't find the good papers right now, but if you search google scholar for using polyacrylic acid as a draw solute for forward osmosis you can dig way deeper into the mechanisms.

What this means practically is that you need really polar solvents that can line up with the ions. Wikipedia has a good page explaining solvent classification in more detail with a table of solvents including their dipole moment (quantifying the polarity). As suggested by Mithoron, DMSO seems to be a good option. Unfortunately, after that you start getting into much nastier solvents like acetonitrile and DMF.


Mechanism and type of polymerization of chloroethene

You got it wrong - monomer is chloroethene (vinyl chloride) and nothing is eliminated in addition polymerization in contrary to polycondensation. This monomer is made via addition of chlorine molecule to ethene and elimination of hydrogen chloride. Alternative methods are addition of hydrogen chloride to acetylene or oxychlorination - in this reaction $\ce{HCl}$ and $\ce{O2}$ react with ethylene to form the monomer and water.


The curves are rather obtained clinically, statistically evaluating diagnose, symptoms and results of clinical lab tests.
Metabolic curve follows malfunctioning metabolism keeping too low ( m. acidosis ) or too high ( m. alkalosis ) level of bicarbonate.
Respiratury curves follows states of too intense breathing ( r. alkalosis ) or too shallow breathing ( r. acidosis ), because of too low respectively high level of carbon dioxide partial pressure, caused by abnormal breathing.
Note that the $K_{\ce{H, CO2}}$ Henry constant  or CO2 is $\pu{29.41 atm/(mol/L)}$. the value will be higher because of the body temperature, can be corrected by temperature dependent $\ce{CO2}$ solubility.
$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)=\mathrm{p} K_{\mathrm{a}}+\log\left(\frac{[\ce{HCO3-}] \cdot K_{\ce{H, CO2}}}{p_{\ce{CO2}} }\right)$$
Where pKa of carbonic acid is 6.35, if counting the total $\ce{CO2}$ concentration.
There are 3 variables ( $\mathrm{pH}$, $p$, $\ce{[HCO3-]}$ ) and 2 degrees of freedom. $\mathrm{p}$ is controlled by breathing, $\ce{[HCO3-]}$ is controlled by metabolism. $\mathrm{pH}$ is result.


How are the atomic orbitals for multi electron atoms obtained?

Already in year 1928 Hartree proposed an approximate method of solving the Schrödinger equation for a many-electron atom that became known as the Hartree method (1, 2, 3). The Hartree method is based on representing the many-electron wave function as a product of one-electron ones with the subsequent application of the variational principle. Two years later, Slater (4) and Fock (5) independently corrected the Hartree method for that did not respect the antisymmetry of a many-electron wave function by representing the wave function by a Slater determinant, rather that a simple product of orbitals. The resulting physically more accurate method became know as the Hartree-Fock method.

In the Hartree–Fock method one solve a set of one-electron equations, called the Hartree–Fock equations, of the form
$$
  \newcommand{\op}{\hat}
  \newcommand{\core}{^{\mathrm{core}}}
  \op{F} \psi_i(\vec{q}_{1}) = \varepsilon_i \psi_i(\vec{q}_{1}) \, ,
$$
where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator and $\psi_i$ are the spin-orbitals of an $n$-electron system. It is customary to suppose that spin orbitals come in pairs: for each pair of electrons two spin orbitals corresponding to two different pure spin states are constructed out of the same spatial orbital,
$$
    \psi_{2i-1}(\vec{q}_{1}) = \phi_{i}(\vec{r}_{1}) \alpha(m_{s1}) \, ,
    \quad
    \psi_{2i}(\vec{q}_{1})   = \phi_{i}(\vec{r}_{1}) \beta(m_{s1}) \, .
$$
Substitution of such spin orbitals into the Hartree-Fock equations results in a similarly looking system of equations for the corresponding spatial orbitals,
$$
    \op{F} \phi_{i}(\vec{r}_{1}) = \epsilon_{i} \phi_{i}(\vec{r}_{1}) \, ,
    \quad
    i = 1, 2, \dotsc, n/2 \, ,
$$
although, the expression for the Fock operator is different $\op{F} = \op{H}\core + \sum_{j=1}^{n/2} (2 \op{J}_{j} - \op{K}_{j})$. Note here, that for the case of the original Hartree method there was no exchange terms $\op{K}_{j}$.

So, essentially, atomic orbitals of many-electron atoms were obtained by solving the Hartree and Hartree-Fock equations "by hands". Of course, the spherical symmetry of atomic systems greatly simplified the problem, as it can be seen already in the Hartree's original work (1), plus some additional simplification, such as the central field approximation, were often used. But still, in general, a numerical integration was required and it was usually done on some calculating machines (6). Below is the photo (courtesy of AIP) of Douglas Hartree (left) and Arthur Porter (right) viewing one such machine, the meccano differential analyzer.

And the results of numerical integrations were simply tabulated as the values of the radial part of an orbital for different values of distance from the nucleus (7).

1) Hartree, D. R. The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part I. Theory and Methods. Math. Proc. Cambridge Philos. Soc. 1928, 24 (1), 89–110. DOI: 10.1017/S0305004100011919.

2) Hartree, D. R. The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part II. Some Results and Discussion. Math. Proc. Cambridge Philos. Soc. 1928, 24 (1), 111–132. DOI: 10.1017/S0305004100011920.

3) Hartree, D. R. The Wave Mechanics of an Atom with a non-Coulomb Central Field. Part III. Term Values and Intensities in Series in Optical Spectra. Mathematical Proceedings of the Cambridge Philosophical Society, 1928, 24 (3), 426–437. DOI: 10.1017/S0305004100015954.

4) Slater, J. C. The Theory of Complex Spectra. Phys. Rev. 1929, 34, 1293. DOI: 10.1103/PhysRev.34.1293.

5) Fock V. Näherungsmethode zur Lösung des quantenmechanischen Mehrkörperproblems. Z. Physik. 1930, 61 (1), 126–148. DOI: 10.1007/BF01340294.

6) Hartree, D. R. The Differential Analyser. Nature. 1935, 135, 940-943. DOI: 10.1038/135940a0.

7) Hartree, D. R. and Hartree, W. Results of Calculations of Atomic Wave Functions. III. Results for Be, Ca and Hg. Proc. R. Soc. London, Ser. A. 1935, 149 (867), 210-231. DOI: 10.1098/rspa.1935.0058.


Reaction of nitrobenzene with sodium metal in ethanol

Benzene Rings or Substituted Phenyl rings undergoes reduction in presence of $\ce{Na}$ in liq. $\ce{NH3}$ in presence of $\ce{EtOH}$ solvent, and generally forms substituted Cyclohexa-$1,4$-dienes.

This reduction generally follows Single Electron Transfer (SET) Mechanism.The single unpaired electron of Sodium is transferred to the Benzene ring forming a radical which subsequently forms a Carbanion. This carbanion eventually takes the Proton from $\ce{EtOH}$ and becomes Cyclohexa-$1,4$-diene.

But in the first step, the transfer of electron by $\ce{Na}$ occurs in that place, where the intermidiate formed after this step gets maximum stabilisation. In case of Electron-withdrawing group, such as $\ce{-NO2, - COOH, -SO3H, -CHO, -COOR, -COR }$ etc., that transfer occurs para to that group, as the negative charge generated is highly stabilised by that electron-withdrawing group. The mechanism thus in case of Nitrobenzene is as follows,

But if the group is electron donating such as $\ce{-CH3, -OH, -OMe, -NMe2 }$ etc. The transfer will occur in such a way to stabilise the generated negative charge by keeping it in meta to the group (where the effect of electron donation by the group is least). For those compounds, the mechanism and corresponding product will be as follows,

Thus, in your case, the product will be $\text{3-nitrocyclohexa-1,4-diene}$ as described in the first mechanism.


What do GGN, AAP, TCN, CAP, CCN, TGQ, etc. mean in DNA analysis?

It is not always possible to obtain an absolutely precise result from automated DNA sequencers, mostly due to too much background noise or two very close peaks which the computer has difficulty reading.

An example of a very noisy four-color chromatogram from a sequencer (sanger):

As you would notice, there are several ambiguous spots and the computer algorithm can be confused or mis-call the wrong nucleotide. Often humans are called in to resolve such ambiguities. It's a fairly common practice.

Sometimes the ambiguity goes beyond a solid conclusion, like when two or more peaks have almost the same intensity.

In that case it is better to explicity label those ambiguous nucleotides as either, say A or B, rather than report a possibly incorrect sequence. Thus was born the incompletely specified nucleic acids.

Here's where IUPAC comes in: the single letter code recommendations.

The "N" in GGN, TCN and CCN would mean the third codon was completely ambiguous and could be any one of tbe four bases. Y would mean pyrimidine (C, U or T).

This would satisfy your query with the exception of the letters P and Q, which I assume were wrongly picked up, or the code was for proteins (amino acid) sequence single letter code.


What old chemistry textbooks are good until present day?

Charles E. Mortimer's Chemistry book was very good for our General Chemistry I & II courses. It has detailed Molecular Orbital diagrams, coordination chemistry information, sufficient examples of electrochemical cells and also a good text of nuclear chemistry. It does not have too many colorful images and pages so it is easier to focus on the knowledge.

In general, old books are only for general chemistry and history of chemistry research etc. When one starts to delve into things, it is dangerous not to use the most up-to-date textbook.


Why is the vanadium(3+) ion paramagnetic?

In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $\mathrm{d}$-block (a.k.a. the transition metals) obey one special rule:

In general, electrons are removed from the valence-shell $\mathrm{s}$-orbitals before they are removed from valence $\mathrm{d}$-orbitals when transition metals are ionized.

(I took this formulation from these online lecture notes, but you will find equivalent statements in your textbooks.)
So, what that does mean is that if you remove electrons from vanadium(0), you will remove the $\mathrm{4s}$ electrons before you remove the $\mathrm{3d}$-electrons. So, you have the following electronic configurations:

$\ce{V}$ is $\ce{[Ar]} \mathrm{4s^2 3d^3}$
$\ce{V^2+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^3}$
$\ce{V^3+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^2}$
$\ce{V^4+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^1}$
$\ce{V^5+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^0}$

And thus, $\ce{V^3+}$ is paramagnetic, because it has two unpaired $\mathrm{3d}$-electrons. In fact, all the ions above are paramagnetic, except $\ce{V^5+}$.


Does the position of a functional group in an organic compound have any effect on the compound's heat of combustion?

The exhaustive combustion of an organic compound consisting of carbon and hydrogen only will yield carbon dioxide and water.  Thus, if the heat of formation for isomers differ from each other, then the heat of combustion will differ, too.
This entry in the English Wikipedia states, for example
$\pu{−146.9 kJ mol^{-1}}$ for n-pentane, $\pu{−154.4 kJ mol^{-1}}$ about isopentane (methylbutane), and $\pu{−167.8 kJ mol^{-1}}$ for neopentane / dimethylpropane; all of same number of atoms $\ce{C5H12}$ (pentane), but with different connection of methyl groups to a backbone in common.  The very table continues for other branched isomers of alkanes, too.

Possibly there is a misunderstanding between the computation of heat of formations from paper (or, in a computer program), on one hand, and the experimental determination of these properties, too.
For the former, the approach typically taught in schools to determine the enthalpy of formation is

to imagine the reaction of $\ce{C_xH_y}$ with $\ce{O2}$ to yield $\ce{CO2}$ and $\ce{H2O}$
to look up the standard heat of formation of $\ce{CO2}$ and $\ce{H2O}$, hopefully with a discern between water as gas and water as liquid
to apply Hess's law to compute $\Delta_fH^\circ$ for $\ce{C_xH_y}$ with the convention that $\Delta_fH^\circ$ for elements in their most stable form (here, oxygen) is set zero

This approach however is an approximation which does not account how the atoms are connected with each other and their neighbours, or bond order; as if e.g., a carbon in acetylene does not differ from the carbon in methane.
If one leaves the imaginary reaction to $\ce{CO2}$ and $\ce{H2O}$, there are empirical increment systems based on experimentation, set to improve this approximation when computing this on paper.$^*$  With the advent of quantum chemistry, it is up to you to select a theory suitable enough to compute this in silico, too (example by MOPAC).
Among the classic approaches to determine experimentally the heat of formation via the heat of reaction (again, Hess' law is applied) is bomb calorimetry.  Here, exhaustive combustion a known quantity of your compound takes place in an atmosphere of pure high-pressure oxygen.  It is one of the experiments students may encounter early when studying physical chemistry.
$^*$  Publications like 1993ChemRev2419, 2011JPhysChemA10576, or 2012JPhysChemA7196 review / present extensions, or are basis for more recent work, too (cited by below the landing page of the publication) in addition to the increment system above mentioned Wikipedia entry currently presents.  Of course, one equally may  adjust existing / establish a new increment systems, e.g., as combination of own work (bomb calorimetry) and using data already recorded by others and presented in primary references (articles) and tables (e.g., CRC Handbook of Chemistry and Physics, NIST Webbook Chemistry)


Why does ozone have a dipole moment different than 0

The electrons are not shared equally among the oxygen atoms.  The central oxygen atom alone donates three electrons to the covalent bonding whereas the other two oxygen atoms combined donate the other three (there is a 3-center, 4-electron pi bond in which two of the rlectrons are shared between just the end atoms).  Since the central oxygen atom is providing an excess of bonding electrons, it becomes positively charged.  The other oxygen atoms, contributing fewer electrons to the bonds and retaining more "lone pairs", take the compensating negative charge.
Trisulfur is a similarly polar elemental species, with a bent structure and unequal electron sharing similar to ozone.  Sulfur dioxide and disulfur monoxide, the latter found along with trisulfur on the Jovian moon Io, have a similar unequally-shared-electron structure with sulfur in the middle.  But we take little notice in the latter two cases because the sulfur, being less electronegative than oxygen, would be positively charged anyway.


Trying to understand the causes and implications of kinks in fatty acid chains

No, trans fats rarely occur in nature, and are generally thought to be very unhealthy.
No, like all neutral atoms, hydrogen atoms attract each other, albeit weakly (by London Dispersion Forces). That's why neatly stacked saturated fats are solid. The kink is to do with the structure of a hydrocarbon chain; their structural formula is usually drawn as a straight line, but it's actually more of a zigzag, each carbon at an angle to the last. Loosely speaking, the double bond interrupts that, so you get one more zig than zag, resulting in a kink.
See 2.
They don't need to straighten out to become solid; they just need to attract each other enough to overcome the tendency of warm molecules to keep moving around.


Solvents adequate for dissolving overheated polyethylene terephthalate glycol-modified (PETG) residue?

It is difficult to tell when you consider PETG overheated (how long? what temperature?). Thermal degradation likely produces several reaction products each of which having different solubility in different solvents. So I don't know whether I ever encountered "overheated" PETG during my own 3D printing experiments.

But I have had very good experience with Tetrahydrofuran (THF), which I happened to have in my basement for a long time (beware of peroxides!) and I have finally found an application for. It dissolves freshly printed PETG (and glue parts together) as well as the residues on the print nozzle.

Side notes: Acetone is wellknown for being absolutely useless for dissolving PETG and I can confirm this from my own experience. All you can expect is to roughen the surface a little bit.

You can read several reports that say dichloromethane is not working as a solvent for PETG. I have found a report that claims Cyclohexanone is working. But I haven't got personal experience with these both. Toluene neither.

But in general, the best method for cleaning nozzles is brute force, i.e. pyrolysis by bringing the (brass) nozzle to red heat with a gas torch. Fast (takes only a few minutes) and reliable (absolutely no residues, expecially at the nozzle hole). One thing I have noticed is that it is a good idea to polish the nozzle with finest steel wool after pyrolysis, especially the inside. On the inside this polish reduces friction for the filament (the brass gets "hazy" after heating it with the gas torch). On the outside, having a shiny surface may reduce emissivity/absorbtivity a little bit, which helps keeping set temperature during printing.


What is the form of the Hamiltonian for solids?

There are multiple possibilities. First of all, you can construct them like the ones for molecules, since a solid is also made from nuclei and electrons. So the Hamiltonian looks identical, but with much more terms.
On the practical side, you would do something different. If, for example, your system is periodic (think of a crystal), then you would choose a single elementary cell from the crystal and build the Hamiltonian for that (ignoring all the other cells), so you arrive at a small Hamiltonian.
Then to get the wavefunction for the whole system you do not change the definition of the Hamiltonian but the one of the wavefunction. While for molecules the wavefunction has to vanish at infinity, you now want the wavefunction to have special properties, called periodic boundary conditions:
For example, consider a one-dimensional crystal with unit cell from 0 to L. You now want the wavefunction $\psi$ at 0 and L to be equal, as well as the derivatives of $\psi$ to be equal at these points:
$\psi\left(0\right)=\psi\left(L\right), \frac{\partial\psi}{\partial x}|_{x=0}=\frac{\partial\psi}{\partial x}|_{x=L}$
With this information you can determine the wavefunction in this cell with any method you want. Now you could put this exact wavefunction into the cell from L to 2L and and the given wavefunction would be a solution for this second cell, since the cells are identical. This way you can get a wavefunction $\Psi$ for the whole system, by periodically putting the wavefunctions for the unit cells $\psi$ together:
$\Psi\left(x\right)=\psi\left(x-n\cdot L\right), x\in\left]n\cdot L,(n+1)\cdot L\right], n\in \mathbb{Z}$
The interaction of the wavefunction with the other nuclei in the crystal is taken care of by the periodic boundary conditions, so the wavefunction will be the real solution to the Problem.


Why is the CASSCF method multi-configurational, while the CI method is not?

When you say multiconfigurational I think you mean that the ground state wavefunction does not have one single dominant MO configuration.

CI can be multiconfigurational depending on the type of excitation included.  For example, Full CI is of course multiconfigurational since it is the exact solution for the basis set used.  However, CIS (single excitations) or CISD (single + double excitations) calculations are the only two CI methods that are computationally efficient enough to be used in practice.  So that’s probably what you mean when you say CI. You also probably mean CIS or CISD based on an RHF wavefunction, to distinguish it from multi-reference CI.

The lowest energy solution of CIS based on RHF is the RHF energy, so you can think of the SCF procedure as a form of CIS.

So a CASSCF(2,2) calculation (e.g. where the active space is HOMO, LUMO) is a bit like a CISDT calculation.  For example, there will be contributions to the wavefunction where the LUMO is doubly occupied and another MO outside the active space is “singly excited”.  Thus, the LUMO is allowed to change when electrons are put into it, which becomes more and more important as the HOMO-LUMO gap decreases.  CISD cannot account for this, so in that sense it is not multiconfigurational.


What is the product of a Dieckmann condensation of dimethyl adipate?

I've given a strong hint by changing the title of your question!

Methoxide as a base can abstract a proton from an acidic $\alpha$-position of an ester, that is from $\ce{-C\color\red{H}_2COOMe}$.
The resulting carbanion/enolate can add to the other ester group of your starting material.
The cyclic intermediate eventually loses methoxide to form ...

This Dieckmann condensation is the intramolecular variant of a Claisen condensation.

I suggest to do it on paper!


What is the definition of a "Simple enzyme" exactly? Is it an enzyme that does not require a co-factor?

What is the definition of a “Simple enzyme” exactly?

This is not a term that is commonly used. There are commonly used terms for an enzyme lacking its cofactor (apoenzyme) and having its cofactor (holoenzyme). Also, the term "enzyme complex" is used for the enzyme bound (non-covalently) to other molecules.

  Does this mean that any protein that utilizes a co-factor, even if that co-factor is simply a co-enzyme or an activating metal, i.e., not an actual part of the enzyme's structure, is considered a complex enzyme? Or are the two terms restricted to the structure of the enzyme (i.e., whether they have prosthetic groups, be they organic or inorganic, or not)?

Yes, just apply the definition you were given. If the metal/co-enzyme/prosthetic factor is absent, the enzyme does not work.


How many isotopologues of water are there?

In an example as small as this, you could hypothetically construct all the cases, or at least enough of them to see that $9$ is incorrect.

The only thing you need to change to get the right answer is instead of using $3\choose{2}$ which is a combination without replacement, you need to do combinations with replacement so ${2+3-1}\choose{2}$$=6$ ways to choose the hydrogens. We can very quickly verify this result by writing out the pairs of masses and seeing how many unique combinations we can make: 11 22 33 12 13 23. And then of course each of those combinations can have one of the three different types of oxygen, so 18 possibilities total. If you haven't seen combinations with replacement before, go here to learn more about them.


Can good quality crystal meth really be made and sold as in Breaking Bad?

Yes, most procedures are accurate, although the whole methylamine is so hard to get idea in Breaking Bad is nonsense. It is DEA List I chemical, but can be easily manufactured and no sane kingpin would go steal it from a cargo train (although it makes a good scene). The real problem is getting the P2P (phenyl-2-propanone, or phenylacetone) which the authors of breaking bad seem to ignore (probably on purpose, so that stupid kids won't try and make it at home).

  Can a pharmaceutical drug really be reverse engineered?

Reversed engineered? Assuming you are talking about the transition of pseudoephedrine into methamphetamine, this is just a chemical reduction, following this scheme (if this kid wants to make meth, he would have found this procedure anyway, I'm not giving out any special information that would help anyone manufacture methamphetamine).


Home foundry — melting aluminum, mitigating undesirables

I never encountered S or C as problems in Al. The problem with hydrogen is that it can cause porosity as it is dissolved in the molten Al and comes out as a gas during freezing. Engine blocks and heads are usually Si alloys ( 5 to 12 %) like the wheels. They may have some Cu to make it hardenable. I expect contamination problems would be more likely Fe and Cu ( getting too high ). The cans usually have very little alloy.


Since you understand why it works for monovalent surrounding atoms, I will assume   that we have a molecule $\ce{MA_{$m$} B_{$n$}}$ where $\ce{M}$ is the central atom, $\ce{A}$ represents the monovalent atoms which may or may not be identical, and $\ce{B}$ represents not necessarily identical atoms which are polyvalent. First up, there are some assumptions to be made:

All "surrounding atoms" are directly bonded to the central atom $\ce{M}$, and with no other atom;
All atoms form as many covalent bonds as is indicated by valency;

Now say the $n\ \ce{B}$ atoms are absent. Let the steric number be $x$,
Then, $x=0.5\times (V+m-C+A)$
Now add the $\ce{B}$ atoms. Now the central atom forms $n$ more sigma bonds, but if the sigma bonds were not present, these would exist as lone pairs, and hence already counted in $V$. Take for example, $\ce{PCl_3}$ and $\ce{POCl_3}$. For $\ce{PCl_3}$ you would get $0.5\times (5+3)=4$ and for $\ce{POCl_3}$ you would get $0.5\times (5+3)=4$. The reason the $\ce{O}$ made no difference is because phosphorus uses two of its electrons to bond with both of which are nothing but valence electrons of phosphorus, only present as a lone pair in one case and a sigma bond in the other. Also, this formula does not work when you have triple bonds, say for $\ce{HCN}$.
As to how it works for mono valent atoms being the only surrounding atoms, it is simple.The number of sigma bonds is the number of mono valent atoms, and thus by adding that to the number of lone pairs we get the hybridization. Say you have $ n $ lone pairs and $s$ sigma bonds.You can see that $V=2n+s$, and that $s=H$. So $0.5*(H+V)=n+s$, as required.


What all the comments boil down to is we are comparing different things:

Allyl vs benzyl cations:  Both contain onlybhydrogen and carbon, with the positive charge formally distributed over two carbon atoms in allyl versus four carbon atoms.  Putting the positive charge on more carbon atoms makes the benzyl cation more stable.

Acetate vs phenolate anions:  Here, unlike the hydrocarbon actions mentioned above, oxygen enters the picture.  The negative charge in the acetate ion is distributed over two oxygen atoms, versus phenolate having its negative charge on one oxygen plus multiple carbons.  Having the second oxygen atom wins out because of the superior electronegaytivity of oxygen, which is a nonfactor for the cation case mentioned above.

Note that the "rules" above are not absolute laws.  You have to consider other factors such as aromatic or antiaromatic coupling that involves the charge; e.g. cyclopropenyl cation is much more stable than benzyl cation because of the positive charge being inherent to an aromatic ring, even though cyclopropenyl has only three instead of four  carbons to take up the charge.


Difference between thermal reactions and photochemical reactions

Yes, you are right thermal reactions take place exclusively on the ground state surface, while a photochemical reaction takes place, at least in part  on the excited state surface.

But there is another difference between them as
photochemical processes offer an advantage over thermal methods by forming thermodynamically disfavored products. Therefore, they overcome large activation barriers in a short period of time, and allow reactions that were inaccessible by thermal processes.


Are there any guidelines for what constitutes an “organic acid” regarding storage requirements?

@Jan is correct.

Organic vs. inorganic is not the main concern for acid storage.  The strength and oxidizing ability of the acids is a much more serious issue.

Your institution/company may already have internal safety guidelines that you should follow.  If not, your country/province/state will have occupational health and safety regulations that you will need to respect.

You can find the American Chemical Society's safe storage guidelines here.


Classical reason for good heat conductance in diamond?

I think that there is no really good classical reason.  Heat conductivity in a solid depends on bonding in that solid and bonding is inherently a quantum mechanical phenomenon.

For instance, recall that graphite is a good conductor of heat (and electricity) in two dimensions and a fairly good insulator (heatwise and electrically) in the third.  This is due to the layered structure in graphite and that is due to the bonding in graphite.


How to proceed when percentage "exchange" is given in a reaction?

We have this equation of reaction:

$$\ce{Cu + Cl2 -> CuCl2}$$

Which means that $1~\mathrm{mol}$ of $\ce{Cu}$ reacts with $1~\mathrm{mol}$ of $\ce{Cl2}$ to form $1~\mathrm{mol}$ of $\ce{CuCl2}$.

Therefore, we would expect $3.7~\mathrm{mol}$ of $\ce{Cu}$ to produce $3.7~\mathrm{mol}$ of $\ce{CuCl2}$ if there is excess $\ce{Cl2}$, which is stated in the question.

However, the exchange is only $45\%$, which means that only $45\%$ of $\ce{Cu}$ is converted, making we can only create $3.7\times45\%=1.665~\mathrm{mol}$ of $\ce{CuCl2}$.


Has a carbon analogue of poly(dimethylsiloxane) been synthesized?

The comments to the original question do well to demonstrate that acetone polyacetal (or polyacetone; names in the literature vary, but these are the most common) is extremely difficult to produce by typical polymerization methods; this is mainly due to thermodynamic reasons, though acetal dimers and trimers of acetone have been produced at very low temperatures.1

Polyoxymethylene is the polymeric form of paraformaldehyde (formaldehyde oligomers), which forms spontaneously in aqueous solutions of formaldehyde. The methyl-capped paraformaldehyde and polyoxymethylene would form if methanol is added to the solution; indeed, this is actually typically done to limit the length of paraformaldehyde oligomers. Polyoxymethylene, as a homopolymer, is quite stable and machinable, provided end-capping is performed; this is typically done with acetic anhydride.2

Burnop, V. C. E. Polymer 1965, 6 (8), 411-417. DOI: 10.1016/0032-3861(65)90025-X
Boyer, Lyndon D.; Sorenson, Wayne R.. Polyoxymethylene process. U.S. Patent 3355428, issued November 28, 1967.


Synthesis of 1-methylcyclopentene from cyclopentene

In Friedel–Crafts alkylation you form a strong electrophile to break the aromaticity of a benzene ring, resulting in a species that is also electrophilic, but not on the carbon atom. It is much more energetically favorable to reform the aromatic ring by losing a proton to $\ce{AlCl4-}$, evolving $\ce{HCl}$ gas.

The reaction of the electrophile with an alkene forms another carbon centered electrophile, however, which can rearrange and will go on to react with other alkenes or with $\ce{AlCl4-}$. There is also nothing to stop the reformed $\ce{AlCl3}$ from continuing to react with the newly formed chloroalkanes, so overall it would probably be a big mess. I found a resource here that discusses the reaction of $\ce{AlCl3}$ with alkenes:

For your reaction, though, I would propose something a little simpler and probably higher yielding like this:


How to accurately calculate the electromotive force for various conditions in a salt water battery?

I suspect that the reason you see a decay in the cell potential of $\ce{Zn/1M~ NaCl/Cu}$ with increasing temperature is due to increased corrosion of zinc. In $\ce{NaCl}$ solutions the anodic corrosion reaction is dissolution of $\ce{Zn}$ as $\ce{Zn^{2+}}$ and the cathodic reaction is reduction of dissolved oxygen to form an insoluble zinc hydroxide (Ichiro Suzuki, Corrosion Science 1985,25 (11), 1029-1034). The insoluble hydroxide passivates the metal surface. So, you have two effects which would contribute to EMF decay: 1) increased $\ce{Zn^{2+}}$ concentration and 2) passivation of the electrode.

One last thing which may or may not be helpful to you - if you are planning on utilizing the EMF from this cell to drive any kind of current you must account for the hydrogen overpotential on the copper electrode. It takes a lot of extra voltage to make hydrogen on pretty much any metal except platinum. You may find that you have a cell that looks good at open circuit (i.e. high potential with no current flow) only to see it crumble to nothing when you try connecting it to a load.