the chemical mechanism for deviations from Raoult's law

First some general comments about Raoults law before discussing the particular solutions in question.

Raoult’s law suggests that the partial pressure of each substance above a solution is proportional to its mole fraction x, thus $p=p^ox$ where $p^o$ is the vapour pressure of the pure substance.

Experimentally there are deviations from Raoult’s law and they can be in both the ‘positive’ and ‘negative’ direction. A positive deviation means that p is greater than expected from Raoult’s law and thus a negative deviation has pressure that is less than expected. Alcohol/water mixtures show positive deviations and for example acetone/chloroform negative deviations.

The reason for these deviations is due to the different interaction between the molecules. The change in interaction on adding one liquid to another is due to the difference in interaction energies between them, thus if the energies  between the same type of molecules of type 1 and 2 is $E_{11}$ and $E_{22}$ (which can be positive or negative) we can write the difference in energy as
$$ \Delta E = c(E_{12} - E_{11}/2 - E_{22}/2)$$
where c is a constant to account or the fact that molecule of type 1 and 2 are different.  A positive value of $\Delta E$ indicates that the interaction between types 1 and 2 is less attractive than either molecule to itself and a negative value that types 1 and types 2 will prefer to interact with the other type of molecules than themselves.

All atoms have dispersion interaction (induced dipole-induced-dipole) due to their polarisability, and in addition may have permanent dipoles, (causing dipole-dipole  and induced-dipole energies) and hydrogen bond interactions. Clearly which contribute depends on the particular case being studied.

Starting with the chemical potential and the interaction energy it is possible to calculate the partial pressure as a function of mole fraction and energy $\Delta E$ and this is

$$ \frac{p}{p^o}=x\exp(+(1-x)^2\Delta E^o)$$

where $\Delta E^o = \Delta E/kT$ where k is Boltzmann’s constant and T temperature. This function is shown in the figure for different values of $\Delta E^o$. When this is positive there is a positive deviation from Raoult’s law, a $\Delta E^o = 1$ approximately describes the ethanol/ water data and $\Delta E^o = -0.5$ the acetone/chloroform data. (The maximum $\Delta E^o = 2$ since above this phase separation occurs)

figure: calculated pressure vs molec fraction x. Positive deviation $\Delta E^o > 0$ and negative deviation $\Delta E^o < 0$. Raoult's law $\Delta E^o = 0$

Now for the particular solutions mentioned, ethanol-acetone has a positive deviation and thus $\Delta E^o > 0$. It is not really possible without some (spectroscopic) experimental data to say what interaction causes this; most likely it is slight changes of each type of interaction, but probably mainly changes in dipole-dipole and hydrogen bonding. The data indicates that acetone-ethanol interactions are less favourable than either molecule with itself.

In the acetone - water data the interaction between these two types of molecules is more favourable in such a way as to overcome the sum of water-water and acetone-acetone interaction energies. This is somewhat unusual for water as it is a highly associated (i.e. hydrogen bonded) liquid and a second component tends to break up this association.

We must conclude, however, that the acetone is able to become solvated in such a way as to disrupt the water structure but still be of lower energy than the sum of water-water and acetone-acetone interactions. Again only spectroscopic evidence will indicate what the interactions are; without this we can only speculate but it could be that the main gain comes from removing acetone-acetone interactions and replacing them by water-acetone even though the water-water interaction is favourable, losing some of this interaction must be compensated for by gaining acetone-water ones.


You are neglecting the autoionization of water.

The autoionization of water has a $K_\text{eq}$ of $10^{-14}$ at room temperature and hence contributes $10^{-7}$M $\ce{H+}$ to solution, which is small (and thus negligible) when we deal with "regular" quantities of acid and base. Since we are not dealing with regular quantities of acid and base here, we must take into account the protons generated from autoionization; the addition of $10^{-7}$M $\ce{H+}$ will resolve any anomalies in pH calculations.


Compounds containing hydrogen are not only acidic but we have other concepts also. Like in your case it can be explained through Lewis acid-base concept.

Lewis concept: A base is defined as a substance which can furnish a pair of electrons to form a coordinate bond whereas an acid is a substance which can accept a pair of electron.

For example, $\ce{NH3}$ is a Lewis base since on nitrogen we have one electron pair which easily be donated to form a coordinate bond. As for acid, Take any two oxides of nitrogen for example $\ce{NO2}$ and $\ce{N2O5}$ . First check the oxidation state of nitrogen in two cases, here it is +4 and +5 . Since a substance which can accept a pair of electrons is an acid, so the nitrogen with more positive oxidation state will relatively accept pair of electrons with more easily, therefore $\ce{N2O5}$ is a better acid then $\ce{NO2}$.


Why is temperature regulation an important factor for producing explosives?

My advice on TATP is simple:

If you're an inexperienced amateur researcher, do the synthesis on paper - and only there!

Actually, the synthesis on TATP can be performed at temperatures above 0 °C. When Richard Wolffenstein accidently found in 1895 during his works on coniine that TATP had been formed in a mixture of 10% hydrogen peroxide and acetone, he did not use any acidic catalyst and had kept the mixture at room temperature for weeks:

  Ueber die Einwirkung von Wasserstoffsuperoxyd auf Aceton und Mesityloxyd, Ber. dtsch. chem. Ges., 1895, 28, 2265-2269 (DOI)

When he had collected the precipitate and recrystallized it from diethyl ether, it exploded, which Wolffenstein attributed to to the formation of peroxides in ether.

One of the first reports on the synthesis of TATP under acidic conditions was given by Milas and Golubovic:

  Studies in Organic Peroxides. XXVI. Organic Peroxides Derived from Acetone and Hydrogen Peroxide,
  J. Am. Chem. Soc., 1959, 81, 6461-6462 (DOI)

Btw, the experimental part of this article starts with

  These peroxides are dangerously explosive and should be handled with extreme care.

Under controlled lab conditions, the decomposition of TATP in solution seemingly has been examined:

  J.C. Oxley, J.L. Smith and H. Chen,
  Decomposition of a Multi-Peroxidic Compound: Triacetone Triperoxide (TATP), Propellants, Explosives, Pyrotechnics, 2002, 27, 209-216

and the activation energy for this first-order reaction was calculated to be around $\mathrm{150\,kJ\cdot mol^{-1}}$.

What makes TATP different from most of the other known explosives is

the molecule (obviously) does not contain nitro groups. Typical detectors for explosives thus fail here. Together with the simplicity of the synthesis, this is a real problem in law enforcement!
the explosion does not generate a lot of heat, but is an entropic explosion.

The latter was througly examined by Ehud Keinan and coworkers:

  Decomposition of Triacetone Triperoxide Is an Entropic Explosion, J. Am. Chem. Soc., 2005, 127, 1146-1159. (DOI)

If you don't want to go though all their calculations of a possible decomposition mechanism, the following hopefully is impressive enough:

  The approximate pressure increase, assuming an
  infinitely rapid conversion that occurs in the volume of 1 mol
  of solid 1 (=TATP), is of the order of 100 bar. The associated energy
  content is approximately 2500 kcal/mol. This is the origin of
  the explosive power of solid 1.

I'm aware that I probably haven't fully anwered your question, but as a take-home message I can only suggest again to not attempt any experiment with this peroxide!


I'm not a gun owner or military expert, but from what I understand the requirements for the armor material are more or less uniform; it should:

Be able to withstand an impact and quickly dissipate the kinetic energy from a projectile. Diamonds are are hard, but very fragile. They cannot redistribute incoming impact from a projectile in the opposite or perpendicular direction. Once they crack, there is no more protection, and they cannot self-heal.
Have low weight. The density of diamond is $\pu{3.51 g cm-3}$, which is nearly 2.5 times heavier than kevlar (ca. $\pu{1.44 g cm-3}$).
Be easily molded in desired form, and/or be flexible (relevant for body armor). Process of crystal growing is rarely precisely controlled and often requires mechanical post-processing, which is ineffective due to hardness of the diamond and can potentially weaken the structure.
Have adequate thermal interface. Due to exceptionally high phonon thermal conductivity, the person wearing this armor will suffer from cold in winter and from heat in summer time, not to mention what damage an opponent with a flamethrower can cause.
Be modular and easily repairable. Diamond plates can only be replaced, so one needs to carry a stockpile of the shards.
Be suitable for camouflage (absorb light and other EM radiation, optionally). Diamond surface is hard to modify, it has low adhesion, so there is no easy way to tarnish the object if the diamond layer is the outer one.
Be optically transparent and don't cast any distortions (relevant for the bullet-proof windows). Due to high refraction index it's a questionable choice for a transparent shield.
Be inexpensive in terms of production and maintenance. No comments here.

To wrap it all up, for now classical macro-scale diamonds seem to be a very expensive and ineffective material for use as an armor material due to their physical properties. However, at a nanoscale the situation can be different, as nanodiamonds show somewhat different properties, yet not nearly fully investigated for military applications.

Also, you might want to check this topic on Worldbuilding.SE: Is diamond armor better than traditional armor?


Reaction of hydrogen peroxide and catalase

Manganese (IV) oxide, iron (III) chloride, and lead (IV) oxide are common examples for a catalyst that isn't found inside living organisms and which can break down $\ce{H2O2}$. Potassium iodide can also be used as a catalyst, along with other transition metal compounds, of metals such as silver or platinum.

This didn't take much effort to find so I'm sure you can find more examples easily.

As for catalase, it seems to just decompose $\ce{H2O2}$, but this is useful in many different areas. For example, decomposing $\ce{H2O2}$ in contact lens cleaning solutions, preventing foods from being oxidised in wrappers (in conjunction with glucose oxidase, another enzyme), and removing $\ce{H2O2}$ from milk before cheese is made. The decomposition reaction is also used as a test for catalase in bacteria.

Sources: General use - National Center of Biotechnology Education - and more uses (including the cheese example) - Worthington Enzyme Manual


How do I name a molecule with a compound substituent?

The given compound includes the characteristic group $\ce{-OH}$, which is expressed as suffix ‘ol’. Since there is only one suffix, the order of seniority of suffixes is not relevant in this case. Functional suffixes are always attached to the name of the parent structure. Note that alcohols are named by attaching the suffix ‘ol’ to the name of the parent structure, with elision of the final letter ‘e’ in the parent structure, if present.

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. The longest chain in the given structure has seven skeletal atoms; i.e., it is a ‘heptane’.

In addition to the characteristic group $\ce{-OH}$, the parent structure has another substituent ($\ce{-CH2-Cl}$), which is expressed as the substitutive prefix ‘chloromethyl’. Note that parentheses are used around such compound prefixes.

Finally, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound. In particular, the order of seniority provides that low locants are given first to the principal characteristic group (here: ol) before low locants are given to prefixes (here: chloromethyl).

Therefore, the resulting name is 5-(chloromethyl)heptan-3-ol.


Does the van der Waals equation remain valid when repulsive intermolecular forces dominate?

You need to consider the $b$ term, as well as $a$.

Rearranging the van der Waals equation to solve for $p$ gives

$$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$

You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result.

But a positive $b$ will act to increase the pressure compared to the ideal gas result.

At high pressures (small molar volumes), the effect from finite molecular volume ($b$) outweighs the effect of intermolecular attraction ($a$).


Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

The sentence means that only one, either the  bond or the lone pair will participate in resonance if the atom has both.

For example, in a pyridine, the nitrogen has a lone pair and it is also attached to a  bond.

Here, however, only the  bond participates in resonance because it is in the plane perpendicular to the molecule, while the lone pair is not.

Methyl isocyanate is best represented by a group of three resonance structures:

Here is the 3D structure of the most significant resonance structure of methyl isocyanate:


Potential energy curve for dissociation of cation of carbon dioxide

Few of the literatures are here:
http://iopscience.iop.org/article/10.1088/0953-4075/47/12/124025/meta

https://pubs.acs.org/doi/pdf/10.1021/jp907351s

https://aip.scitation.org/doi/10.1063/1.444413


Assignment of the IR spectrum of 2,4-pentanedione (acetylacetone)

OH str for beta diketones is tabulated from 3200 to 2400 1/cm. The shift is explained by resonance of the hydrogen bonded structure.

Its absorption features is small or unseen as for

commonly the ketonic form is highly dominant; i.e. we won't expect to see OH str in the acetaldehyde spectrum, even
owing to the above mentioned resonance,  the dipole change associated to symmetric OH str is null, and it is minimal for the AS one.

The symmetry of 2,4-pentanedione makes the last point even more important.

Concerning the second part. The fact that the O of the carbonyl of the enoform is H bonded does further shift down the related CO str. Moreover, in addition to resonance with the C=C, note that again there will be resonance with the CO in beta as above, and this further shift the frequency to lower wn.


What does the "3,4" mean in 3,4-methylenedioxymethamphetamine (MDMA)?

Numbers in organic molecules tell you where so-called functional groups (important configurations of elements) are attached to a central structure.  The central structure in MDMA is a benzene ring $\ce{(C6H6)}$, with the main functional group attached at a carbon which makes methamphetamine.  This carbon is labeled 1 as the "most important" functional group.  To get to MDMA from methamphetamine, you add a methelynedioxyl group, and this attaches at carbons 3 and 4.  So that's where those numbers come from!

EDIT: Below is a molecule of MDMA:

The hexagon shape near the center is the benzene ring.  The group to the right, attached to the benzene ring at about 2 o'clock is the main functional group that turns benzene into methamphetamine.  That position on the ring is labeled 1.  Now all other functional groups are numbered sequentially in the direction that makes the next functional group nearest to 1.  So that means we go counterclockwise, and the group on the far left with the 2 O's (oxygens) is attached at positions 3 and 4.


What is the difference between low-density and high-density Polymethyl methacrylate (PMMA) on a molecular level?

Low density and high density PMMA differ by density, which is given as the ratio of mass and volume.

Given that they are prepared from the same monomer, two polymers with the same number of monomers will have the same mass.

If the polymers differ in density, this must be due to the fact that they occupy different volumes.

Note that we are talking about polymers here and not the ordered packing of molecules in a crystal. Nevertheless, inter- and intramolecular interactions (and thus "packing") is important here as well.

As a rule of thumb, polymers with higher branching will "pack" less tightly and thus occupy a higher volume.

This results in a lower density.


Assume that the initial temperature is room temperature and neglect the initial amount of acetonitrile in the gas phase.  Calculate the equilibrium vapor pressure of acetonitrile at 140 C and compare it with the partial pressure you would calculate from the ideal gas law if all the acetonitrile had evaporated (so that its volume is 100 cc).  If the latter is greater than the former, then there will still be some acetonitrile remaining as liquid.  The partial pressure of the acetonitrile in the gas phase will then be equal to the equilibrium vapor pressure.  Assuming that x gm of acetonitrile have evaporated, you can then calculate, in terms of x, the new mass and volume of the liquid and the new volume of the head space.  You can then determine x from the requirement that the total amount of acetonitrile has not changed.


What is the physical state of vitamin E acetate at room temperature and normal pressure

Vitamin E acetate, or α-tocopheryl acetate, exists in two forms: D-form (solid/highly viscous at 20 °C) and a racemate (viscous liquid at 20 °C).
Interestingly enough, the difference in physical properties is well denoted in German Wikipedia article, whereas English Wikipedia omits this.
Also, I noticed many authoritative sources arbitrarily listing a melting point for one or another isomer without denoting the exact form, e.g. even CRC Handbook of Chemistry and Physics provides a single entry for vitamin E acetate with melting point −27.5 °C [1, p. 3-550], so no wonder online sources and sellers mix the things up a little.

More relevant details can be found in Handbook of Pharmaceutical Excipients [2, p. 32]:

  d-Alpha tocopheryl acetate

  Empirical formula $\ce{C31H52O3}$
  Molecular weight 472.73
  CAS number [58-95-7]
  Synonyms (+)-(2R,4'R,8'R)-2,5,7,8-Tetramethyl-2-(4',8',12'-trimethyltridecyl)-6-chromanyl acetate; d-α-tocopheryl acetate; vitamin E.
  Appearance A practically odorless, clear, yellow, or greenish-yellow colored viscous oil that may solidify in the cold.
  Melting point 28 °C
  Solubility Practically insoluble in water; soluble in ethanol (95%). Miscible with acetone, chloroform, ether, and vegetable oils.
  Specific rotation $[α]_\mathrm{D}^{25} = +0.25^\circ$ (10% w/v solution in chloroform)
  Comments Unstable to alkalis.

  dl-Alpha tocopheryl acetate

  Empirical formula $\ce{C31H52O3}$
  Molecular weight 472.73
  CAS number [7695-91-2]
  Synonyms (±)-3,4-Dihydro-2,5,7,8-tetramethyl-2-(4,8,12-trimethyltridecyl)-2H-1-benzopyran-6-ol acetate; (±)-
  (2RS,4'RS,8'RS)-2,5,7,8-tetramethyl-2-(4',8',12'-trimethyltridecyl)-6-chromanyl acetate; (±)-a-tocopherol acetate; α-tocopheroli acetas; all-rac-α-tocopheryl acetate; dl-α-tocopheryl acetate; vitamin E.
  Appearance A practically odorless, clear, yellow, or greenish-yellow viscous oil.
  Density $\pu{0.953 g/cm3}$
  Melting point –27.5 °C
  Refractive index $n_\mathrm{D}^{20} = \pu{1.4950 – 1.4972}$
  Solubility Practically insoluble in water; freely soluble in acetone, chloroform, ethanol, ether, and vegetable oils; soluble in ethanol (95%).
  Comments Unstable to alkali. However, unlike alpha tocopherol, the acetate is much less susceptible to the effects of air, light, or ultraviolet light. Alpha tocopherol acetate concentrate, a powdered form of alpha tocopherol acetate, is described in the
  PhEur 6.0. The concentrate may be prepared by either dispersing alpha tocopherol acetate in a suitable carrier such as acacia or gelatin, or by adsorbing alpha tocopherol acetate on silicic acid.

D-form may appear as a highly viscous oil, however it will become solid over time [3, p. 232] (emphasis mine):

  D-α-Tocopheryl acetate. CAS: 58-95-7. Obtained by vacuum steam distillation
  and acetylation of edible vegetable oil products. Occurs colorless to yellow, nearly
  odorless, clear, viscous oil. May solidify on standing, and melts at about 25 °C.
  Unstable in presence of alkalies.

  DL-α-Tocopheryl acetate. CAS:7695-91-2. Colorless to yellow or greenish yellow, nearly odorless, clear, viscous oil. Unstable in presence of alkalies.

  D-α-Tocopheryl acetate concentrate. Obtained by vacuum steam distillation and
  acetylation of edible vegetable oil products. May be adjusted by suitable physical
  or chemical means. Occurs as light brown to light yellow, nearly odorless, clear,
  viscous oil.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; CRC Press, 2016. ISBN 978-1-4987-5429-3.
Handbook of Pharmaceutical Excipients, 6th Ed.; Rowe, R. C., Ed.; APhA, (PhP) Pharmaceutical Press: London, 2009. ISBN 978-0-85369-792-3.
Food Additives, 2nd Ed., rev. expanded.; Branen, A. L., Ed.; Food science and technology; Marcel Dekker: New York, 2002. ISBN 978-0-8247-9343-2.


$\ce{CaCO3}$ is sedimenting when mixed with water, and therefore is not easy to remove with hot water. Nevertheless several methods exist that actually dissolve it that are better than vinegar:

You could add hydrochloric acid again to convert it into soluble Calcium chloride, according to equilibrium:
$$\ce{CaCO3 (s) + 2 HCl (aq) <=> CaCl2 (aq) + H2CO3(aq)}$$
You could use carbonic acid instead
$$\ce{CaCO3 (s) + H2CO3 <=> Ca(HCO3)2}$$
this is the reason shells are getting dissolved by an increase of the amount of carbon dioxyde in solution:
$$\ce{CaCO3 (s) + CO2 (aq) + H2O <=> Ca(HCO3)2}$$
But I recommend Sulfamic acid ($\ce{H3NSO3}$), is frequently used for removing rust and limescale, replacing the more volatile and irritating hydrochloric acid, it may be a little more expensive but is certainly not toxic.


I have the impression that you a very close to find the answer yourself!

I will not give a full answer here, but some hints.

  [...] I know I added Sulfuric acid and sodium carbonate [...]

Let's assume that you did not add them simultaneously. Then, it seems likely that sulfuric acid does the magic and sodium carbonate is added to neutralize the reaction mixture after your product has been formed.

  I am certain there is some type of internal reaction to create the five membered ring.

Good point! Let's draw the starting material and the product.

3-nonenoic acid is a carboxylic acid with a non-conjugated $\ce{C=C}$ double bond. With other words, there are two functional groups.

Now think in the sulfuric acid! It's a strong acid while its anion is a rather weak nucleophile.

"Coconut aldehyde" is  not an aldehyde at all, but a furanone, a five-membered lactone. Think intramolecular ester.

EDIT

Is it conceivable that the $\ce{C=C}$ double bond is protonated?
If so, what could possibly add to the resulting cation?


What is the difference between D and L configuration, and + and −?

The D-L system corresponds to the configuration of the molecule: spatial arrangement of its atoms around the chirality center.

While (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane of polarized light clockwise (+) or counterclockwise (-).

D-L system tells us about the relative configuration of the molecule, compared to the enantiomers of glyceraldehyde as the standard compound. Compounds with the same relative configuration as (+)-glyceraldehyde are assigned the D prefix, and those with the relative configuration of (-)-glyceraldehyde are given the L prefix.

It's kind of another way to tell the configuration of molecules beside the Cahn–Ingold–Prelog convention (R/S system), with little difference. (D-L system labels the whole molecule, while R/S system labels the absolute configuration of each chirality center.)

In short, the D-L system doesn't have direct connection to (+)/(-) notation. It only relates the stereochemistry of the compound with that of glyceraldehyde, but says nothing about its optical activity. We may have compound with same relative configuration as (+)-glyceraldehyde (thus, it's given the D prefix), yet it rotates the polarized light counterclockwise (-), such as D-(-)-ribose.

And also, don't confuse the D-L system with d- and l- naming. d- and l- is the exact same with (+) and (-) notation.

Additional explanation

D-L system (also called Fischer–Rosanoff convention) is mainly used for naming α-amino acids and sugars. It compares the relative configurations of molecules to the enantiomers of glyceraldehyde. This convention is still in common use today.

Rosanoff in 1906 selected the enantiomeric glyceraldehydes as the point of reference[1]; any sugar derivable by chain lengthening from what is now known as (+)-glyceraldehyde (or named D-glyceraldehyde) belongs to the D series. In other words, we used a D to designate the sugars that degrade to (+)-glyceraldehyde and an L for those that degrade to (-)-glyceraldehyde.

In assigning the D and L configurations of sugars, we could direcly look for the OH group of the bottom asymmetric carbon in the Fischer projection. If it's located on the right, we designate it with D, and vice versa, since they would have the same relative configurations with glyceraldehyde for the bottom asymmetric carbon.

Reference

[1]: IUPAC and IUBMB. Joint Commission On Biochemical Nomenclature. Nomenclature of Carbohydrates. 1996, 7.


Chondroitin Sulfate (greater volume in solution than in crystal form)

$\hspace{4.5cm}$

In solution, the negative charges on the chondroitin sulfate repel each other and force the molecule into an extended conformation. The polar molecule attracts many water molecules, which increases its molecular volume. In the dehydrated solid, each negative charge is counterbalanced by a positive ion, and the molecule is condensed.


How does one experimentally determine virial coefficients?

These coefficients can be calculated based on 2 ways:

1) Based on experimental data

In this method, from pressure-volume-temperature (PVT) data these coefficients will be calculated. Consider the following equation:
$$\vartheta\left(\frac{P\vartheta}{RT} - 1\right) = B(T) + \frac{C(T)}{\vartheta} +\ldots$$
$$\vartheta = \frac{1}{\rho} =\mathrm{Molar\ volume}$$

This equation is a re arange form of Virial equation. Now accroding to this equation, if you plot the left hand side of aforementioned equation versus molar density, the intercept of that line represents second virial coefficient and the slope of that line is the third virial coefficient. If the curve was not a straight line, we should consider more virial coefficients for them.

2) From potential functions.

In this method, the virial coefficients are calculated by potential functions such as L-J model, Kihara model, Exp-6 model and etc. The basis of This method is statistical thermodynamics.


I think you answer comes from the part of the sentence that you left of the quote from Wikipedia:

  ... i.e. it's value can't be inferred by looking just at the initial and final states of the given system.

Inexact differential are the result of integrating nonstate functions (functions which aren't path independent). The fundamental theorem of calculus, which allows the integration of a function over a range by determining the antiderivative at the end points of that range, requires path independence in order to be true. So it won't work for inexact differentials that are path dependent.

As for integrating over $dU$, $U$ is a state function and so you can apply the fundamental theorem and solve the integral using the antiderivatives at the end points.


According to This article, vinylsilanes are stable in TBAF solutions, allowing other silyl groups to be removed selectively. They can be activated in the presence of TBAF using palladium similarly to a halide, though.

This link shows C2-trialkylsilylalkenes being cleaved with potassium t-butoxide, 16-crown-8 and TBAF. I'll look more thoroughly when I get to a computer, though.

I also found this example, but it had no reference, so I'll try to find one:


Benzenesulfonyl chloride dehydration of an amide to give nitrile

One key detail is that this reaction is performed in the presence of base. Typically pyridine is the solvent. Amides are nucleophilic at oxygen, explained by the resonance structure shown. Displacement of chloride by the amide oxygen gives the O-sulfonylated intermediate. After neutralization, the O-sulfonylimidate is set up for an E2 elimination across the C-N bond to give the nitrile.

In addition to sulfonyl chlorides, $\ce{P4O10}$ and $\ce{POCl3}$ are common reagents for this reaction. In general, the reagent should be electrophilic and be able to act as a leaving group.


Determining the pH of an organic solvent

I'm not sure why you're wanting to exert so much effort to find the charge of compounds.

If you really want to know the charge, determine it more directly. Electrophoresis can allow you to do so. A charged species is put in a capillary tube that is in between a cathode an an anode. When a charged species is subjected to an electric field, it will experience a force proportional to its charge and the field strength.

Drag forces present in the solvent are also a consideration in this setup and they depend on the "size" of your ion as well as viscosity of your solvent. One also has to worry about Joule heating, as an increase in temperature will cause poor reproducibility. Picking a suitable detector is fun, but hopefully you can get by with the common UV-vis detector.

So if you can find a suitable setup, know the radius of your ion and know the time between injection of your sample and its arrival at a detector, you can at least determine its effective charge.


How to calculate an enthalpy change for this reaction

To preface, the reason that this is not a combustion reaction is because this is a redox reaction that occurs at room temperature. In fact, chloroform is noncombustible.

Though at first glance, this may simply appear to be an unbalanced equation that could be rewritten:
$$\ce{CHCl3 + 1/2O2 -> COCl2 + HCl}$$

It is better to take it for what the question presents it as, an oxygen atom, so I would therefore recommend you to leave the equation as:

$$\ce{CHCl3 + O -> COCl2 + HCl}$$

Though $\ce{O2}$ is a pure element, and its $\Delta H^0_\mathrm f=0\:\mathrm{kJ\:mol^{-1}}$, for the atom $\ce{O}$, $\Delta H^0_\mathrm f=58.99\:\mathrm{kJ\:mol^{-1}}^{[1]}$ (provided thanks to user1420303).

To calculate $\Delta H$, the equation is:  $$\Delta H=\Delta H^0_\mathrm f(\ce{COCl2}) + \Delta H^0_\mathrm f(\ce{HCl})-\Delta H^0_\mathrm f(\ce{CHCl3})-\Delta H^0_\mathrm f(\ce{O})$$

Though enthalpy is a state function, this question is asking you to calculate $\Delta H$ from a specific set of reactants, and their individual energies must be considered. There is more than just one way to calculate $\Delta H$, and an analysis of the $\Delta H^0_\mathrm f$ (the enthalpy of formation of a compound from its constituent elements) for each compound accounts for which bonds are being broken and formed, and will yield (about) the same results as a bond by bond analysis. I find this method to be preferable because it bond by bond analyses are based on average bond enthalpies, while $\Delta H^0_\mathrm f$ is often painstakingly measured for each compound, and will yield the most accurate $\Delta H$ calculation.

[1] Curtiss, L. A.; Raghavachari, K.; Redfern, P. C.; Pople, J. A. Assessment Of Gaussian-2 and Density Functional Theories for the Computation of Enthalpies of Formation. The Journal of Chemical Physics J. Chem. Phys. 1997, 106, 1063.


Representation of bromelain molecule in two dimensional diagram

I think I got it. The diagram in PubChem is not bromelain molecule, but a tiny fragment of it. This thing is indeed too small to be an enzyme. Also, it has nothing to do with proteins. Look at the formula, there is not a single amino acid residue here. Indeed, scroll down to 3.4.2 and read: "carbohydrate moiety of bromelain". That's what it is, and not the entire enzyme.

If you want to know what the rest of it is made of, then google "bromelain amino acid sequence". It seems to be known for quite a while. No nice pics, though.

On a side note, there is no such thing as "the smallest protein" or even "the smallest known protein", for the border between proteins and polypeptides is rather arbitrary. You are right, though, in that most enzymes are huge proteins, too big to fit on a 2D diagram. So is bromelain.

So it goes.


Equilibrium, Kp constant, Partial Pressures

The method referred to in the comment is not correct and is a common mistake. One may think,

$$p_1'+p_2'=p_t\frac{V_t}{V_t'}=2p_t=0.88 \rm atm,$$

but this is invalid since $n_1+n_2\neq n_1'+n_2'$, thus $p_tV_t\neq p_t'V_t'$

The correct way is to use the fact that the number of iodine atoms is constant, from which we get,
$$2n_1+n_2=2n_1'+n_2',$$
$$(2p_1+p_2)V_t=(2p_1'+p_2')V_t',$$
$$1.3 {\rm atm}=4p_1+2p_2=2p_1'+p_2',$$

Then you can go on by combining this with the equation you have already obtained from $K_p$ (listed in your question) and solve for $p_1'$ and $p_2'$.


What really happens atomically in an explosion?

The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF).

A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some way to get the reaction going. Typically the reaction is started by the generation of free radicals. These will react with stable molecules to split them up and generate more free radicals so in effect we have a chain reaction.

So the ignition process is basically the generation of free radicals, and that's what the lit cigarette does. The heat of the burning tobacco generates free radicals that then start the butane oxygen reaction.  You can also generate free radicals in lots of other ways e.g. using shock, from an electric spark, or by using a suitable catalyst such as the catalytic convertor on your car exhaust.


For myself, ideal means it is composed of ideal gas particles. This means that the ideal gas law applies. This means that there is no accounting for particles interacting which is a grave mistake for liquids and solids, and often for gases too.
$dU = dH - d(pV)$ is always true. In an real system however, the variables involved may take on different quantitative values for a given state point, thus the final change in eternal energy $\Delta U$ may be different.

In general $\Delta U = \Delta  H -V \Delta p - p \Delta V - \Delta p \Delta V$

If pressure is constant during the change then $\Delta p \Delta V = V \Delta p = 0$ which then leads to the term

$$\Delta U = \Delta H - p \Delta V $$

If volume is constant during the change then $\Delta p \Delta V = p \Delta V = 0$ which would then lead to

$$\Delta U = \Delta H - V \Delta p $$

If neither volume or pressure is constant the simplest form is

$$\Delta U = \Delta  H -V \Delta p - p \Delta V - \Delta p \Delta V$$

a.k.a.

$$\Delta U = \Delta H - (p_f V_f - p_i V_i) $$

see Andrew's answer for more details on the last step.

And I do correct you, non-ideal systems maintain equilibrium just as ideal systems do. If they didn't, we wouldn't be able to use thermodynamics on them since thermo only applies to equilibrium.


Is an SN1 reaction an elementary reaction/one step reaction?

From http://goldbook.iupac.org/E02035.html:

  A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state.

So you're right that elementary reactions occur in a single step/have a single transition state. I'm not sure what you mean by "obey the rate law", but necessarily, a reaction that has only a single transition state is first order in each reactant. Because there is only a single step, the concentration of every molecule involved in that step effects the rate.

In practice, because the probability of three molecules colliding together in a way that can cause a reaction is very low, all elementary reactions involve one or two molecules. A single molecule reaction is just a dissociation, e.g. $A_2 \rightarrow A + A$, with a first order rate in A and an overall order of one. A two molecule reaction could be something like $A+B\rightarrow AB$, with still a first order rate in each of the reactants for an overall order of two.

In your question, you've identified that SN1 reactions have multiple transition states, so SN1 cannot be an elementary reaction. In fact, the 1 in SN1 means that the overall order of the reaction is one, but since it involves two reactants, the order would be two for a single step reaction. Naturally, an SN2 reaction has an overall order of two and is an elementary reaction, proceeding through a single transition state where a bond between the incoming nucleophile and the carbon centre is formed at the same time as the leaving group's bond is broken.


Separate toluene from acetone and isopropyl alcohol

Multiple washes with smaller quantities of water will be more efficient than "just adding a lot of water" in a single wash. The toluene will be "purified" further with each successive wash. Some toluene will be lost with each wash also to the degree of it's solubility in water at the temperature of the extraction (about 0.03% soluble in water at room temperature). As stated in a comment however, distillation would be a superior method for recovering the pure toluene.


$$E_n = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

You can easily verify that if you replace $Z$ with $2Z$, then the energy of the corresponding transition will be four times the original energy. Now if you replace both $n_1$ and $n_2$ with $2n_1$ and $2n_2$, then you can take out a factor of $1/4$; so the transition $2n_1 \leftarrow 2n_2$ in an ion with nuclear charge $2Z$ occurs at the same energy/wavelength/wavenumber as the transition $n_1 \leftarrow n_2$ in an ion with nuclear charge $Z$.

Same applies for any other factor $k$: the $1 \leftarrow 2$ transition in hydrogen occurs at the same wavelength as the $4 \leftarrow 8$ transition in $\ce{Be^3+}$. Obviously this discussion ignores the variation in the Rydberg constants for different nuclei.


What does nutella add to the Coke+Mentos reaction?

Nutella was used just to protect cola from incidental reaction with Mentos. And it makes more effective insertion - all mentos tabs are inside at the same time.

The reaction is completely the same with simple Mentos. (this video also is quite noisy)

This is not chemical reaction, is carbon dioxide evaporation.


Why is the water column empty in ICE table

Presumably water is the solvent for this reaction, which means that it will be in extreme excess and the amount will not change to a measurable extent.


Why are the total energies of transition states so commonly corrected for zero-point vibrational energy?

At the transition state, you have exactly one vibration mode with imaginary frequency (which corresponds locally to the reaction coordinate), and $3N-1$ other “normal” vibration modes. In the transition state theory, you separate out the reaction coordinate (consider it infinitely slow compared to the other movements). Thus, when you consider the energy of the transition state, you have to include the contributions of ground state energy of all the other vibration modes.

The more rigorous way of looking at it is to use statistical physics, and to consider the partition function of the transition state. See here or there for such treatments.


Why is the Gibbs free energy for phase changes zero at constant temperature and pressure?

The initial and final thermodynamic equilibrium states of your system are as follows:

State 1:  $\pu{1kg}$ liquid water at $\pu{0 ^\circ C}$ and $\pu{1 atm}$

State 2:  $\pu{1 kg}$ water ice at $\pu{0 ^\circ C}$ and $\pu{1 atm}$

You want to find the change in enthalpy, entropy, and Gibbs free energy between these two states.

Now, we know that, to get the change in entropy between two thermodynamic equilibrium states of a closed system, we need to identify a reversible path between these states and then calculate the integral of $\mathrm dq/T$ for that path.  For the present situation, this can be accomplished by putting the system into contact with a constant temperature reservoir at a temperature only slightly lower than $\pu{0 ^\circ C}$.  As the liquid water freezes, its molecules lock into place (losing potential energy) and this results in a release of heat to the reservoir.  So the temperature of the system never deviates significantly from $\pu{0 ^\circ C}$. Since the process is at constant pressure, the change in enthalpy is equal to the heat transferred from the surroundings to the system:$$\Delta H = q$$where both $q$ and $\Delta H$ are negative for this process.  Since the process is also at constant temperature, the change in entropy, which is given by the integral of $\mathrm dq/T$, is:$$\Delta S=\frac{q}{T}=\frac{\Delta H}{T}$$Thus, for this change from State 1 to State 2, $\Delta S$ is also negative.  If we now use the change in enthalpy and the change in entropy to calculate the change in free energy between the two thermodynamic equilibrium states, we obtain:
$$\Delta G = \Delta H-T\Delta S=\Delta H-T\frac{\Delta H}{T}=0$$


Cis and Trans Isomerism given three different groups around the reference plane

Am I on the right track?

Generally, yes. For this question, yes but you have over thought this question.

For starters, any answer with Y can be totally eliminated as Y is neither cis nor trans. Now you are down to answers 'A.' and 'D.'

Here you correctly determined the E and Z configurations of molecules with identical functional groups, but could have saved yourself some effort if you had recognized W and Z are identical molecules that are inversions/180° rotations of each other. Thus it can't be D. and the answer is in fact:

  A.


The short answer is no.

The long answer is:

Henry's law works for small concentrations of ideal mixtures at equilibrium. Henry's Law constant varies with temperature according to the Van't Hoff equation:

$$
k(T)=k(T^{\circ})exp\left[-C\left(\frac{1}{T}-\frac{1}{T\circ}\right)\right]
$$

Here $C$ is a constant related to the enthalpy of solvation for each gas, and $^{\circ}$ represents the standard state, T = 298 K and 1 atm.

Using standard state data, you could get an estimate of the partial pressure at a given temperature. However, two key assumptions that are implicit to Henry's law and the Van't Hoff equation do not hold for global atmospheric gas composition:

The atmosphere and the oceans are not at equilibrium (not even close when you consider atmospheric layers and convection)
$\ce{CO2}$ in seawater is not an ideal mixture (very non-ideal since $\ce{CO2}$ reacts with water).

This is one of the many reasons that climate modeling is difficult - the oceans act as huge carbon sinks with a very complex and not-very-stable relationship between dissolved $\ce{CO2}$ content, temperature, composition of rock on the sea floor, convection currents in the atmosphere and the ocean, amount of shellfish (no kidding), and probably lots of other factors (I'm not a climate expert).

The net result is that while Henry's law can get you a quick idea of how much $\ce{CO2}$ could be dissolved at a given temperature under pseudo-equilibrium conditions (or what the corresponding partial pressure would be for a given concentration in water), it isn't enough to get you anywhere close to a good prediction of what the global atmospheric composition will be, particularly for $\ce{CO2}$ (due to its reactivity with water and incorporation into biomass).

Now, if you really just want to convert $\ce{CO2}$ concentration in ppm to a partial pressure, you can do that.

$\ce{CO2}$ in air is close enough to ideal that we can use the ideal gas law. For an ideal gas, the partial pressure is the same as whatever the pressure would be if it were in a container by itself:

$$
P_1 = \frac{n_1RT}{V}
$$

You can find V for a given mass of air, find the mass of $\ce{CO2}$ from the concentration, then convert to moles and plug it into the above.


Why is lithium bromide used in absorption refrigerators?

Your summary is correct; $\ce{NaCl}$ is more soluble than $\ce{LiCl}$ or $\ce{CaCl2}$, for that matter, but both $\ce{LiCl}$ or $\ce{CaCl2}$ are far  more hygroscopic. The explanation of hygroscopicity involves various mechanisms, including adsorption (surface forces) and absorption (requiring transport to the interior of the substance) due to hydrogen bonding, coordinate covalent bonding etc.

So these are distinct, though related, properties: solubility (limit on how much solute the solvent holds) and hygroscopicity (how avidly the solute attaches to the initial molecules of solvent).

For that reason, $\ce{CaCl2}$ is used to absorb water in damp closets, though it will eventually deliquesce into a sodden mess.

BTW, Isaac Asimov wrote of a compound so hygroscopic that it dissolved before it hit water: The Endochronic Properties of Resublimated Thiotimoline, often cited.


Why is a hydroxyl group more activating than a methoxy group in electrophilic aromatic substitution?

First of all we must have to note that the difference in reactivity between anisole (methoxybenzene) and phenol is small. Small enough for the undergraduate texts I own to not mention anisole separately from phenol, just referring to it as having "phenol-like reactivity" (i.e. strong ortho-para directing). We can more or less measure the difference in ortho-para directing strength by looking at the proton NMR spectra of anisole and phenol.

Source.

Source.

As you can see they are almost identical to the point where you can ask yourself if there is any practical difference in reactivity at all. The A protons nicely reflect the ortho-para direction since they haven't shifted from benzene's 7.26 ppm signal and the B and C protons are all upfield as expected.

Also I don't think steric hindrance is a big factor either since aromatic electrophilic substitution goes first trough a $\pi$-intermediate and a single methyl-group is not that bulky.


Does a molecule need to be placed symmetrically in the unit cell?

Don't think of it so much as putting the center mass of an atom on a vertex but as matching the unit cell symmetry to the crystal symmetry. Take the definition from google below, I've bolded key parts:

  The smallest group of atoms of a substance that has the overall symmetry of a crystal of that substance, and from which the entire lattice can be built up by repetition in three dimensions.

While the red boxes satisfy the first and last bolded criteria, they fail the second. The red boxes have 2-fold rotational symmetry and inversion symmetry, but the green boxes have 2 mirror planes that the red boxes do not. This lack of mirror symmetry with the red boxes does not have the same overall symmetry of the lattice and therefore the green box is the unit cell.

Think of Sodium Chloride; its unit cell face is represented as:

$\ce{\color{red}{Na}\  Cl\ \  \color{red}{Na}}$
$\ce{Cl\ \  \color{red}{Na}\ \ Cl}$
$\ce{\color{red}{Na}\  Cl\ \  \color{red}{Na}}$

instead of

$\ce{\color{red}{Na}\  Cl}$
$\ce{Cl\ \  \color{red}{Na}}$

Because the first example has 4-fold rotational symmetry where as the second has 2-fold and the first one has 2 mirror planes that the second example does not. Both have inversion symmetry. That said the atoms at edges/corners are centered because that is what allows for the symmetry otherwise the unit cell would be unsymmetrical which does not reflect the symmetry of the lattice.

As an example of a sort of off centered lattice there is perovskite. You can't see it in the image but the black atom is $\ce{Ti^4+}$ and is too large to actually center in the lattice so it will reside in an off-centered configuration (the titanium ion is shifted off-center towards a face). This property of off-centering makes perovskites excellent as dielectric materials for capacitors as the ion can change is off-centering by an electric field.


It does.  You would find the average percentage of the atmosphere that is argon is very slightly higher at the floor of valleys.  However, bear in mind first of all it wouldn't be anywhere near a complete stratification -- a layer of pure argon, then another of pure N2, and so on.  A mixture of nearly ideal gases doesn't do that, at least at equilibrium, because it would eliminate the considerable entropy of mixing.  (It can happen in liquids because liquids have strong intermolecular forces that normally favor separation and oppose the entropy of mixing.)  Another way to think about it is that since the atoms and molecules in gases don't (much) interact, there's nothing stopping an individual argon atom going slightly faster than nearby nitrogen and oxygen molecules from bouncing up higher than they do.

What you would get in a theoretical ideal (uniform gravitational field, complete stillness -- no wind -- and uniform temperature) would be an exponential fall of pressure with altitude, and the exponential for heavier gases would be steeper than for lighter gases.  That would result in enrichment of the heavier gases at lower altitudes.  A little work starting from the Boltzmann distribution of gravitational potential energies of each type of atom and molecule would get you an ideal estimate of the argon excess as a function of altitude.

In practice the lower atmosphere has so much mixing due to wind and big thermal gradients that I doubt you could even measure the mild excess of argon and other heavy gases.

There is one fascinating short-term exception, which bears directly on your question.  Sometimes volcanoes will belch out a considerable quantity of CO2, which is significantly denser than air, and this CO2 can accumulate briefly in a thick layer at the bottom of a valley or over a lake, if there isn't much wind.  It can persist for some hours, perhaps days, before it diffuses away and is mixed with the rest of the atmosphere.

Then indeed the valley bottom becomes an invisible death trap for humans and animals: walk into the valley, or be unable to exit fast enough when it happens, and you will suffocate for no reason you can see.  The most famous example of this is the Lake Nyos disaster in 1986 which killed thousands of humans and animals.  I think the government now has mixing devices installed in that lake to prevent any future sudden release of CO2.


The further to the left a peak in an NMR spectrum is, the more deshielded it is. Deshielding or shielding are terms to describe how much a certain nucleus is shielded from the outer magnetic field that is applied during NMR measurements. Shielded nuclei will experience less of a magnetic field and thus resonate at a lower frequency meaning lower ppm-values; and vice-versa. In general, shielding can be explained by more electrons being in the vicinity of an atom because the electrons — in layman’s terms — can be thought to create an electric current induced by the outer magnetic field, which in turn generates an opposing magnetic field and lowers the effective magnetic field at a nucleus.

If we have an electronegative atom close by, that atom will draw electrons towards it, reducing the electron density around the nucleus we are observing and thus deshielding it. That not only works for electronegative atoms alone, but also for electron-withdrawing groups like the carbonyl group.

In your specific example we have two electron-withdrawing groups that will deshield neighbouring protons. However, the deshielding that the acetyl group is responsible for is stronger than that of the chlorine atom. So both pairs of protons are moved downfield (to higher ppm values) when compared to toluene or para-xylol, but the protons closer to the acetyl group are deshielded more.

It ultimately boils down to oxygen’s electronegativity in combination with an $\ce{sp^2}$ carbon that is not attached to any hydrogen, but it may be easier to think of carbonyl groups being electron-withdrawing per se.


In case of p-nitrophenol the lone pair on oxygen of OH takes part in an extended resonance upto NO2 , more delocalisation of lone pair results into more delta minus on H atom ,so stronger H bonding, whereas  in case of m-nitrophenol it cant have resonace structures upto NO2. Since less delocalisation results in comparitively less delta minus, H bonding is comparitively weak.


Why is the boiling point of a solution higher than that of a pure solvent?

The change in vapour pressure with added solute is called Raoults law. The vapour pressure of a species i above a solution is proportional to mole fraction of this species (solute) $x_i$ and its pure vapour pressure $p_i^*$ or $p_i=x_i.p_i^*$.

As the solution now contains a fraction of a low vapour pressure solute then this component will not contribute much to the overall vapour pressure which will be less (in this case) than pure solvent.

It is not correct to say that solvent molecules are replaced at the surface as both solute and solvent are distributed uniformly in the solution. Thermal diffusion moves the solute and solvent around in a random manner. It is that there are now fewer solvent molecules as some have been replaced by low vapour pressure solute (in this example) and because they are of low volatility they need more energy to evaporate than the solvent does. As the temperature is constant, by the Boltzmann (energy) distribution fewer solute molecules populate higher energy levels compared to solvent and so have a lower vapour pressure.


Calculating percentages of microstructures in equilibrium phase diagrams (engineering)

Assuming you mean this phase diagram, it looks like your calculations are correct.

Based on my understanding of the diagram, at $727\ ^\circ \text{C}$, you have 100% pearlite at $C_b = 0.76\ \%C$ and 0% pearlite at $C_a=0.022\ \%C$. $W_a$ then appears to be how far $C_o$ is between $C_a$ and $C_b$, so $W_a$ should be the amount of pearlite.

The units of $C_a$, $C_b$, and $C_o$ are $\% C\  by\  mass$. That means when you calculate $W_a$ initially, you are calculating the fraction of pearlite and not the percent of pearlite. All of the $\% C\  by\  mass$ units cancel:

$$W_a=\frac{0.76\ \%C-0.4\ \%C}{0.76\ \%C - 0.022\ \%C}=\frac{{\%C}(0.76-0.022)}{{\%C}(0.76-0.022)}=0.487805\implies48.7805\%$$


It is correct that white phosphorus is usually used as the standard state for tabulating thermodynamic data, in particular by the Committee on Data for Science and Technology (CODATA).

A comparison of thermodynamic data for phosphorus and some of its compounds can be found in Rard, J. A.; Wolery; T. J. The Standard Chemical-Thermodynamic Properties of Phosphorus and Some of its Key Compounds and Aqueous Species: An Evaluation of Differences between the Previous Recommendations of NBS/NIST and CODATA. J. Solution Chem. 2007, 36 (11), 1585–1599.. This publication also includes an explanation for the use of white phosphorus as the standard state:

  Although white phosphorus is not the thermodynamically stable allotrope, the red and black forms are difficult to prepare in pure form, which makes them less suitable for quantitative thermodynamic measurements.

Similarly, from the other side of the Atlantic, A.T. Dinsdales' SGTE Data for Pure Elements, originally published as CALPHAD 15(4) 317-425 (1991), says:

  [The] reference phase is normally the phase stable at 10$^{5}$ Pa and 298.15 K. The exception to this rule is phosphorus for which, by convention, the white form is chosen as the reference phase because the more stable red form is difficult to characterise.


Any example of colloidal solution where liquid Helium is solvent?

A solution implies that there are solvent-solute interactions. Helium is about the most chemically inert atom there is. In the case of water ice, the hydrogen bonding will be orders of magnitude stronger than the van der Waals forces, so colloids will not be stable at 'normal' pressures. I guess if I were looking for a colloid in He(liq), I'd look at N2 or perhaps H2. We've made a lot of progress since I had colloidal (surface) chemistry class, but I doubt that we could predict size of particles in any such system. But I don't know.


Why is a metabisulfite / starch / HCl solution unstable?

I can partially answer the question. I know what is happening to solution B. I don't know of a more stable version.

One potential problem is the acid-catalyzed hydrolysis of the starch into glucose, which does
not form the blue complex.
$$\ce{nH2O + (C6H10O5)}_n \ce{->nC6H12O6}$$

For the timing to be off, the metabisulfite must also be decomposing, either by reducing glucose or by reacting with dissolved oxygen. I'm not sure what the specific process is, nor do I know how to stop it.


Py and Cl both ligands are monodentate . Therefore, the coordination number of Pt in both the cases are 4 and Pt in general shows +2 oxidation state for 4 coordination number. Again py is neutral ligand whereas Cl is anionic (-1). So, $\ce{[Pt(py)4]^2+}$ is positively charged and $\ce{[PtCl4]^2-}$ is negatively charged. In IUPAC nomenclature, for a coordination complex, the name of cation is written first and of anion at end (cation or anion may be simple or any coordination entity). Therefore the second name, that you told can't be accepted.

Now , the prefixes bis, tris, tetrakis are used-

When the name of the ligand is lengthy or complicated (polysyllabic or multiplicative prefixes are already present in the name of the ligand)
In cases of organic ligand.
In all other cases where using bi, tri, tetra, ... creates ambiguity because in that case a single name may indicate more than one compound .

In case of py first it is an organic ligand and despite that if you use bi/tri/tetra that may indicate a different ligand (for example bipyridine and pyridine are two different ligands) . So, the prefix tetrakis will be used here.


Why does the cyclization of open-chain glucose occur via the C5 hydroxyl group?

Due to the internal ring strain the best possible conformation is a six-membered heterocycle in this case called pyranose . A five-membered heterocycle may in rare cases also be formed, called furanose. (From left to right: α-D-Glucopyranose, β-D-Glucopyranose, α-D-Glucofuranose, β-D-Glucofuranose; courtesy of wikipedia)

Including the linear chain form they are all isomers (wikipedia), as they have the same chemical formula. More specifically the relationship between $\ce{linear <-> pyranose <-> furanose}$ is called constitutional or structural isomer.
The $\ce{\alpha-D <-> \beta-D}$ forms are diastereomers, or more specific diastereomeric conformers.


Storage of methane in mineral oil or other liquid hydrocarbons

"Practical" can mean different things to different people, so I am not going to comment on that.

Instead I am going to address the question, of solubility of methane in hydrocarbons, without any thought about "practicality" :)

This graph is taken from the book: Fuel and Lubricants Handbook (a simple google search lead me to it, and the relevant section appears in the preview :))

The data is reported at $50\ \mathrm{^\circ C}$ and $ 5000 \ \mathrm{psig}$ (pounds per square inch gauge, ungodly non-SI units. ca. $3.4 \times 10^7\ \mathrm{Pa}$ according to an online conversion)

Also, the following table from this publication:

I did not read it in detail, but the data was collected in the vicinity of 1 atm and between $25$ and $200\ \mathrm{^\circ C}$

FYI: Of possible interest could be this paper.


A simplistic answer is as follows (though, please do refer to the links given by Greg in a comment to the question):

According to the University of Wisconsin page Coupling Reactions, coupling reactions are

  a reactant-favored reaction is linked to a product-favored reaction so that both reactions yield products.

A main point in the related University of Wisconsin website is that if the reaction is reactant-favoured ($\Delta{G} > 0$), then it is not going to happen, so it is coupled with a product-favoured reaction ($\Delta{G} < 0$), leading to coupled reactions are

  When chemical reactions are coupled, the free energy released by the exergonic reaction is used to drive the endergonic one toward the formation of products.

where the overall reation is product-favoured.

(The website uses an analogy of a coupled train).

A great explanation is in this YouTube clip from Dr. John Pollard, of the University of Arizona.


How to calculate the relationship between the pressures of zinc, carbon monoxide, and carbon dioxide based on mole balance?

First of all, let me explain why your approach is a bit wrong...
Addition of steps will only give you the net reaction, which is what you got. But in the process, the mechanism becomes blurred. For example, the net reaction which you have expressed, doesn't tell us the full story. One completely overlooks the fact the CO is also recombining to give CO2, and that is exactly where the fallacy arises.

My solution to the problem. Hope it helps.


Why the formation of a fog is observed when triethylamine is added?

You are adding Et3N to a solution that contains HCl as a reaction product. That means there will be HCl vapour above the solution. The Et3N is fairly volatile so there will be Et3N vapour around the addition stream/droplets and this will react with the HCl vapour as it is added to give triethylammonium chloride. This gives you the fog.


how is the esterification of amino acid carried on in this reaction

I think you need a protection strategy and there are several choices. This one uses O'Donnell's benzophenone imine strategy details here.
First you prepare glycine ethyl ester hydrochloride from glycine with EtOH/HCl, a method is here patent.
From the ethyl ester hydrochloride prepare the benzophenone imine by O'Donnell's method as described in ref 1. This gives a protected species that can be easily alkylated under phase-transfer conditions with allyl bromide as described in ref 1
The allyl product is then epoxidised (MCPBA) and the epoxide opened with iodide procedure here
Deprotection with acid hydrolysis gives your target molecule.


What is the number of nodes in the molecular orbitals of cyclic H3?

In linear $\ce{H_3}$ you have one MO with zero nodes, one with one node and one with two nodes.  When you bend the molecule around to make a cycle, the two nodes in that last orbital merge to make a second orbital with one node.  The two orbitals of cyclic $\ce{H_3}$ with one node are, of course, degenerate when the ring has threefold rotational symmetry.


Finding pKa from equivalence point on titration curve

You've got a weak acid, since you're contemplating a positive pKa, which means when you're halfway to the end point you're in the buffer region and you can use the Henderson-Hasselbalch equation:

pH = pKa + log [A-]/[HA]

You've titrated half your initial HA, so half of it is still around and half got turned into A-, which means [A-] = [HA].  For example, if you started with 0.1 mol/L of HA, you now have 0.05 mol/L HA and 0.05 mol/L of A-.

That means [A-]/[HA] = 1, and the log of 1 is zero.  The HH equation then tells you pH = pKa.


Does the mechanism of the Schmidt reaction involve a nitrene as an intermediate?

Nowhere in that link is there a nitrene formed. All steps are concerted, that is, you see that alkyl shift happens at the same time as dinitrogen fragmentation.

So indeed, nitrenes are not intermediates. I suspect that mechanistic studies have been done. You might consider looking at this paper.


What is the IUPAC preferred name of this tetracyclic natural product?

The pubchem website has given Platensimycin IUPAC name, $3$-$[3$-$[(1S,5S,6R,7S,9S,10S)$-$5,9$-dimethyl-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-en-$5$-yl]propanoylamino]-$2,4$-dihydroxybenzoic acid (see attached picture):

Accordingly, since https://pubchem.ncbi.nlm.nih.gov is usually updating data according to the changes, I believe the name of the cyclic system (without 5,9-dimethyl groups) OP is asking is same as one in the comment by @KarstenTheis (including the correct stereochemistry): $(1S,5S,6R,7S,9S,10S)$-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-ene (see the blue numbering in the picture).

Late correction: After exchanging few comments with Loong, I realized that the given system should be treated as a chemical compound. Accordingly, the name should have been: $(1S,5S,6R,7S,9S,10S)$-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-en-$4$-one (as suggested by Loong).


Synthesis of the tetrachloroaluminate(III) anion

Note that the first reaction would require substantial amount of heat to assure the designated product $\ce{NaAlO2}$ is formed (nowhere near STP):

$$\ce{2 Al(s) + 2 NaOH(l) + 2 H2O(g) ->[\pu{\approx 450 °C}] 2 NaAlO2(s) + 3 H2(g)}$$

In aqueous solution the product would be tetrahydroxoaluminate $\ce{[Al(OH)4]-}:$

$$\ce{2 Al(s) + 2 NaOH(aq, conc) + 6 H2O (l) ->[Δ] 2 Na[Al(OH)4](aq) + 3 H2(g)}$$

Synthetic procedure from "common laboratory reagents" is described in Brauer's Handbook of Preparative Inorganic Chemistry [1, p. 816]:

  Sodium Tetrachloroaluminate

  $\ce{NaAlCl4}$

  $$\ce{\underset{58.4}{NaCl} + \underset{133.4}{AlCl3} = \underset{191.8}{NaAlCl4}}$$

  The reaction is carried out in a Pyrex vessel (Fig. 245), which should be as compact as possible.
  The stoichiometric amounts of C.P. $\ce{NaCl}$ and freshly sublimed $\ce{AlCl3}$ are introduced into the pear-shaped part of the reaction vessel.

  Fig. 245. Preparation of sodium tetrachloroaluminate.

  The filling is done under nitrogen and the vessel should be very clean and dry.
  A melting point capillary is then affixed underneath the ground joint; the apparatus is evacuated to a high vacuum and torch-sealed.
  It is then immersed as deeply as possible in an oil bath at 200–240 °C.
  The $\ce{AlCl3},$ which tends to sublime onto the cool parts of the wall, is driven back by passing a luminous flame over the condensate spots.
  When the flask contents have become a nearly clear melt, they are poured hot into the side flask of the vessel.
  The product solidifies on cooling.

References

Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1963; Vol. 1.


What are the physico-chemical processes at play in ironing of clothes?

Ironing of clothes is a physical change (if you believe that physical changes and chemical changes are not part of a single continuum of change).

Most polymers (including the cellulose fibers that comprise cotton) have a glass transition. The glass transition is a reversible phase change from a rigid fixed solid amorphous "glass" to a solid phase where there is some mobility of the polymer chains, but not enough to be a liquid (this phase is sometimes called a "rubber," "plastic," or "elastomer" phase). The glass transition is sometimes used to refer to the phase change between an order crystalline (or semicrystalline) solid to a disordered amorphous solid.

Above the glass transition temperature, the solid material has some mobility - it can be shaped, molded, flattened, whatever. When it is cooled down again, it becomes rigid and holds its shape.

Ironing raises the temperature of the fabric to above the glass transition. The action of ironing presses out the wrinkles in the now more flexible fabric. When the fabric cools, it retains its new wrinkle-free shape.

Water is used during ironing for one important purpose. I tried to find the glass transition temperature of cellulose, but am unable to do so. Presumably it is close to the temperature at which cellulose begins to burn. Water prevents the burning, but not just in the way you would expect. Water does absorb heat from the iron in order to keep the fabric from getting too hot. However, if this were the only thing water was doing, the fabric would not reach the glass transition and ironing would not work. Water also acts as a plasticizer, lowering the glass transition temperature. Most fabrics are hydrophilic (and cellulose as a carbohydrate is especially so). The water binds to the chains, swells the fibers, and changes the physical and chemical properties of the material.


Change in Enthalpy in Free Expansion of Ideal and Real gas against Vacuum

In the Joule experiment, $\Delta H$ is not zero for a real gas.  $\Delta H=\Delta U+\Delta (PV)$, and even if $\Delta U$ is zero, there is no physical reason to expect that $\Delta (PV)$ would be zero in this process.


Difference between atomic mass, mass number, atomic weight, and relative atomic mass

Please correct me if I am wrong but in my understanding, the following are the definitions for the aforementioned terms:

Mass number: The total number of nucleons present in the nucleus of an atom(Dimensionless).

Atomic mass: The combined mass of all the nucleons present in the atom of any given element(Unit: amu/u).

Relative isotopic mass: Ratio of atomic mass to one amu(Dimensionless)

Average atomic mass: Weighted average of all the isotopes for the given element(amu/u).

Relative atomic mass/atomic weight: Ratio of atomic weight to one amu(Dimensionless).

Here is the Wikipedia page, explaining that "atomic weight" is simply the older term used for "relative atomic mass":

Relative atomic mass


Calculate Work Done for Reversible and Irreversible Adiabatic process

For an irreversible process, the pressure of the gas within the system is typically non-uniform spatially, and so there is no unique value that can be used in conjunction with the ideal gas law to calculate the work.  But, at the interface with the surroundings, the force per unit area exerted by the gas on the interface always exactly matches the pressure of the surroundings, which is usually referred to as $P_{ext}$.  So, if $P_{ext}$ can be controlled externally as a function of the volume, one can calculate the amount of work done by the gas on the surroundings.  In many homework problems encountered in thermodynamics, the value of $P_{ext}$ is often specified as being constant during an irreversible process.

In addition to the effects of pressure and temperature nonuniformities in a gas experiencing a rapid irreversible expansion or compression, there is also another physical mechanism acting.  This is related to the viscous behavior of a gas that is deforming.  In the case of a rapid irreversible deformation, the gas pressure depends not only on the gas volume, but also on the rate of change of volume.  The behavior is analogous to that of an ideal spring and damper in parallel (as in the front suspension of a car).  The force exerted on a spring/damper combination during a rapid stretching or compression depends both on (a) how much the combination is stretched, but also on (b) how rapidly it is stretched (because of the effect of the damper).  A very crude approximation to the behavior of an ideal gas that is experiencing a rapid (irreversible) rate of deformation is provided by the equation:  $$P=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}=P_{ext}$$ where k is a positive quantity related to the viscosity of the gas.  So the gas pressure on the piston face depends both on the amount of volume change and on the rate of volume change.  The equation indicates that, during rapid expansion (i.e., dV/dt > 0), the pressure exerted by the gas on the surroundings is less than that given by the ideal gas law, while, during rapid compression (i.e., dV/dt < 0), the pressure exerted by the gas on the surroundings is greater than that given by the ideal gas law.


Are all stable Xenes (graphene-like 2D honeycomb sheets) buckled?

According to the authors of Ref. 1 this is a general property. They provide an explanation for the stability of such sheets based on formation of particular buckled geometries:

  The discovery of a flat two-dimensional crystal known as graphene has contradicted Landau−Peierls−Mermin−Wagner arguments that there is no stable flat form of such crystals. Here, we show that the “flat” shape of graphene arises due to a microscopic buckling at the smallest possible interatomic scale. We show that the graphene, silicene, and other two-dimensional crystals are stable due to transverse short-range displacements of appropriate atoms.

This explanation is corroborated by experimental observations:

  On the other hand, the generality of our prediction of the buckled universal shape has been revealed in recent experiments on silicene [...]

The authors note in the introduction that their explanation is only one among three that at that time were being seriously considered. The two other arguments are that (1) the sheets are trapped in a quenched metastable state that is not subject to large thermal perturbations compared to the lattice strength and (2) the sheets are slightly warped in the third dimension, the deformation suppressing thermal excursions.

The reference cited above is from 2012. Low-buckled 2D lattices have been reported more recently, for instance for germanene in Ref. 2.

Reference

A. O’Hare, F. V. Kusmartsev, and K. I. Kugel .  A Stable “Flat″ Form of Two-Dimensional Crystals: Could Graphene, Silicene, Germanene Be Minigap Semiconductors?
Nano Letters 2012 12 (2), 1045-1052. DOI: 10.1021/nl204283q
Jincheng Zhuang, Nan Gao, Zhi Li, Xun Xu, Jiaou Wang, Jijun Zhao, Shi Xue Dou, and Yi Du . Cooperative Electron–Phonon Coupling and Buckled Structure in Germanene on Au(111). ACS Nano 2017, 11 (4) , 3553-3559. DOI: 10.1021/acsnano.7b00687.


The distinction between dynamic & static equilibrium

My answer starts with the premise that a static equilibrium exists (although that is debatable).

  How could we be sure that one certain reaction in equilibrium is dynamic?

If your reaction is at dynamic equilibrium, changing the temperature or the concentration of one of the reactant or product species (by adding some solute, for example) should disturb the equilibrium. You would see a net reaction, i.e. change in the concentrations of species in solution or in the gas phase. The reason you did not see any changes before is that the forward and reverse reaction were occurring simultaneously at the same rate.

  Since there would be no visible change in vessel, which property of the reaction would provide a proof besides color? Which laboratory technique could be used?

Color would not help either, unless you disturb the equilibrium first. Once you disturb it, any method sensitive to changes in concentration will give you evidence, as will calorimetric methods.

  [Devils advocate question]: What about static reaction equilibrium?

If your reaction is at static equilibrium, changing the temperature or the concentration of one of the reactant or product species, in contrast, would not result in any macroscopic change. For example, if you lower the temperature such that the forward and reverse rate are practically zero, and then add some reactant or product, nothing would happen. Not sure why you would call it a reaction equilibrium though when the reaction never occurs under these conditions. In other words, I know of no reactions where forward and reverse reactions suddenly stop with reactants and products still present under conditions where the reaction proceeded to reach equilibrium.


It rapidly oxidized. If the ball mill is sealed for long periods of time the oxygen inside the ball mill is consumed coverting aluminum to aluminum oxide. When the lid is opened the aluminum is rapidly oxidized with a exothermic reaction. If the temperature gets hot enough it can spontaneously ignite.
The reason for the carbon is it coats the aluminum particles to prevent them from oxidizing.


Derivation of relationship between Gibbs free energy and electrochemical cell potential

There are two ways to understand this equation.

One is to realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$.

The fact that $\Delta G = W_\text{non-exp}$ can be derived as under:
\begin{align}
\mathrm{d}S
  &= \frac{\delta q}{T}  && \text{(reversible case)}\\
\mathrm{d}U
  &= \delta q + W_\text{non-exp} + W_\text{exp}\\
W_\text{non-exp}
  &= \mathrm{d}U - W_\text{exp} - \delta q \\
  &= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S
  && \text{(const. $p$ and $T$)} \\
  &= \mathrm{d}H - T\,\mathrm{d}S \\
  &= \mathrm{d}G
\end{align}

Another approach is to use
$$\mu_\mathrm{electrochemical} = \mu_\mathrm{chemical} + zF\phi,$$
where $\phi$ is the electric potential at the point, $z$ is the charge on the species. Using this and
$$\mathrm{d}G_{T,P} = \sum_{i=1}^n \mu_i\,\mathrm{d}n_i,$$
we get the required equation for $z = 1$ for an electron, and considering the electrochemical potential of both the electrodes. I am unsure of a rigorous derivation for the electrochemical potential but Wikipedia has some basics on it.


Why is it obligatory to cool down the container of a sample to measure its mass in a lab?

A container that is not at ambient temperature will generate air currents around it.  If you place such a non-ambient container in a balance, air currents will develop around the container as it heats or cools to ambient.  These air currents will cause the balance to read incorrectly.


How to get (2R,3R)-2,3-dibromobutane from but-2-yne?

I think I figured it out. Lindlar is a cis-addition so one would not obtain a (2R,3S), but rather (2S,3S) and (2R,3R) which happen to be enantiomers.


Ambimodal transition states in some organic reaction pathways

Why do these ambimodal transition states occur?

I don't know that there is a general reason for that. Potential energy surfaces are complicated. Very often, there is an abundance of local minima, separated by potential energy barriers. It is similar to what happens with mountain chains and valleys, but with many dimensions.
One could argue that the opposite could be asked: why would there be a single minimum associated with each and every barrier? Why would there be exactly a single downhill valley behind each mountain chain, instead of, sometimes, two different valleys being similarly accessible, once you climb over the top of a hill?

How does the situation differ from normal transition state theory?

Transition state theory is a useful simplification, assuming a simple potential energy surface. It is useful because, at room temperature, the energy differences in the different possible paths are very often sufficient for a single reaction pathway to predominate, so that others effectively do not happen. When the temperature is comparable to the difference between the energies involved in different possible reaction pathways, the situation is more complicated.


Unfortunately Pubchem is right, the two structures have the same InChI string and key, since the protonation state is the same in the zwitterion and the neutral form. So the reason for the discrepancy is by design.
I also always thought, InChI was designed for distinguishing between these conformations, but it turns out to just be one of the limitations of the system. The issue is addressed in section 13.2 of the technical FAQ of the InChI trust:

  The different protonation states of the same compound will have InChIKeys differing only by the protonation indicator (unless both states have a number of inserted/removed protons greater than 12; in this case the protonation flag will also be the same, ‘A’).
  This is exemplified below by standard InChIKeys as well as standard InChI strings for neutral, zwitterionic, anionic and cationic states of glycine (note that neutral and zwitterionic states do not differ in the total number of protons so they have the same standard InChI/InChIKey):


You find #2 difficult because #1 and #3 are ridiculously easy!

In these two questions the mass of every element present just happens to match exactly with the atomic weight of that element.  So when you divide each mass by the corresponding atomic weight, the resulting ratios are very close to integers, or integers divided by 100.  So the (almost) final step of dividing each ratio by the smallest ratio goes by almost unnoticed.

But consider for the moment what would happen if you used three or four decimal place atomic weights.  All those nice neat exact answers would disappear in a haze of "sort of" whole numbers, and you would have to use some judgement as to whether a number is close enough to a whole number.

In #2, the questioner made the "drastic" change of using 4.0 and 40.1 as the two numbers for an element.  If you had carried on mechanically, dividing all ratios by the smallest ratio, you would have gotten:

Ca ... $1.0$
O .....$2.005$
H......$2.005$

Is $2.005$ close enough to a whole number?  Or should you multiply everything by $1000$ to give $1000, 2005, 2005$ as the empirical formula?


Jan has already covered this topic in a related question: Why is the cyanide ion toxic?

$\ce{CN-}$ and $\ce{CO}$ are isoelectronic, and one might expect them to behave exactly similarly. However, there is a key difference: the orbital energies in $\ce{CN-}$ are higher than in $\ce{CO}$, due to the lower electronegativity of $\ce{N}$ compared to $\ce{O}$.

This means that $\ce{CN-}$ is a much better σ donor, and a much poorer π acceptor than $\ce{CO}$. Consequently $\ce{CN-}$ is much happier binding to a metal in a higher oxidation state, whereas $\ce{CO}$ is much happier binding to a metal in a lower oxidation state.


What products did this aldol condensation create?

Straightforward aldol between benzaldehyde and acetone. Enolate of acetone adds to benzaldehyde (which cannot enolise) then eliminates water to give the enone. Product recrystallised from EtOAc.


Which gas makes Orange juice bottles swell?

I can't imagine the gas was anything except $\ce{CO2}$ formed by the action of microorganisms contaminating the orange juice.  They degrade the organic components of the juice to power their metabolisms, and produce carbon dioxide as a byproduct.

A common contaminant of pasteurized fruit juice is a genus of bacteria called Alicyclobacillus, but this paper says that Alicyclobacillus bacteria do not cause bottle swelling and do cause a medicinal smell.

By searching Google for ["bottle swelling" "orange juice"], I found an English-language abstract of a thesis from China which implies that the culprit could be a yeast.


Can Cellulose Molecules be Synthesized from Glucose?

Cellulose may be broken into smaller pieces (though not necessarily into separate glucose molecules) and regenerated into cellophane with the viscose process.

For the home lab, rather than using the poisonous and extremely flammable $\ce{CS2}$ in the viscose process, the somewhat safer cuprammonium process is a better choice. Do be careful -- don't wind up with blue stains on skin or clothing, it shows poor technique. See this video for a demonstration.


Glassy carbon may be a good choice. Looking at this, electrooxidation of methanol doesn't seem to be possible at reasonable potentials on glassy carbon, and it's a common working electrode material, relatively inexpensive, durable, and available in many shapes and sizes. Graphite could also work, though it may need to be replaced periodically as graphite electrodes tend to be less robust.


Your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, does not have line symmetry (plane of symmetry), and hence $\ce{C}$3 and $\ce{C}$6 are chiral centers. Thus, maximum possible stereoisomers are $2^2 = 4$. However, there is a symmetry element in the structure, which is order 2 ($180^\circ$) rotational symmetry (also called point symmetry). Therefore, one stereoisomer, $(3S,6R)$ would be superimposable with it's mirror image $(3R,6S)$, thus it is a meso-isomer:

To understand this, I put "$\bullet$" symbol to indicate relevant hydrogen is above the plane of paper. And, "$\bf{.}$" symbol to indicate relevant hydrogen is below the plane of paper. Thus, first two structures have both 3,6-dimethyl groups below the plane of the paper while the third structure has 3-methyl group below and 6-methyl group above the plane of the paper. Accordingly, all chiral centers have CIP nomenclature as indicated under each structure:

The first structure is $(3S,6S)$ and the second structure is $(3R,6R)$, thus they are non-superimposable mirror images as indicated by the mirror. Therefore, they are enantiomers. To make all the groups to coincide, you may need to rotate the second structure $180^\circ$ horizontally along the axis going through $\ce{C}$3 and $\ce{C}$6. However, after the rotation, the two methyl groups at $\ce{C}$3 and $\ce{C}$6 will be now above the plane. Therefore, these two structures cannot be superimposable.
On the other hand, the third structure is $(3S,6R)$ ($\ce{C}$3 methyl is below the plane and $\ce{C}$6 methyl is above the plane). To the mirror image of this molecule, if you follow the same operations as previous steep and do another $180^\circ$ rotation on the plane of the paper (or vise versa), you'd get the same compound.  That means, the third compound is not optically active and called meso-isomer. This can be found easily by looking at the two chiral centers: They are mirror images of each other, $(3S)$ and $(6R)$ or $(3R)$ and $(6S)$.

Therefore, your compound, 1,3,4,6-tetramethylcyclohex-1,4-diene, has three stereoisomers, but only two of them are optically active.


What are the degrees of freedom that define the temperature of an ionic solid (such as sodium azide)?

Degrees of freedom describe the different ways atoms move in a sample.

The classical equations of motion for a system of N atoms describe the time evolution of 3N positions and 3N momentum coordinates for the atomic nuclei (ignoring electronic degrees of freedom). The total energy is a function of those 6N degrees of freedom. Sometimes some of those coordinates do not provide a significant way of storing energy so they can be ignored. For instance, in a dilute gas, translational momentum provides a means of storing energy, but the potential is essentially flat, so position does not.

  How many degrees of freedom does solid sodium azide have (per formula unit) and what are these degrees of freedom classified as?

Sodium azide ($\ce{NaN3}$) has $4\times6\times \text{N}_\text{A}$ degrees of freedom per mole of substance. Half are associated with momentum coordinates, the other half with position coordinates (ignoring once again electronic degrees of freedom). Positions can be expressed in terms of 3N-6 internal and 6 global coordinates, and internal coordinates expressed approximately as harmonic oscillator modes. Each oscillator mode is a linear combination of the 3N-6 spatial coordinates, the entire set being orthonormal. If the solid does not move we can ignore global coordinates.

  I am more [concerned] with how these degrees of freedom change with temperature, giving rise to the same black body radiation at the same temperature, and being at thermal equilibrium with other samples of the same temperature through collisions at the interface between them.

Ignoring electronic degrees of freedom (that is, focusing only on nuclear motions), each atom contributes 6N degrees of freedom. Unlike in a dilute gas, here both momentum and position provide a means of storing energy. The amount of energy that can be stored in  each degree of freedom is, following classical statistical mechanics (high T limit), equal to kT/2, for a total of 3RT for one mole. However, quantum mechanics limits occupation to certain states rather than a continuum, and this can reduce the amount of energy each degree of freedom can store (the heat capacity).

There are some concepts in your question that need to be disentangled:

1) Equipartition, a principle that states that energy is dispersed equally among all degrees of freedom. At low T quantization may determine to what extent different degrees of freedom are thermally available. Too cold and only a reduced number of states may be occupied, and there is no equipartition. This is well explained in the wikipedia. The implication for the example you present is that, when compared to a metallic solid with softer vibrational modes, a molecular solid, in which atoms form covalent bonds having large force constants, might be expected to have a lower heat capacity: degrees of freedom corresponding to high frequency vibrational modes that distort covalent bonds can accommodate less energy (on average) at lower temperatures. Such arguments justify some trends in the table below, showing the heat capacity of various solids on a per atom basis (computed from heat capacities from Ref. 1).  Sodium azide is a combination of our definitions of ionic solid and molecular solid. For solid sodium azide the heat capacity (on a per atom basis) is $\pu{19.2 J/molK}$, a value intermediate to that of metallic and molecular solids. Note also that $3\text{R} \approx \pu{24.9 J/mol K}$, the heat capacity of a solid (on a per atom basis) predicted by the partition theorem. This value excludes an estimate of an electronic contribution to the heat capacity.

2) Temperature, an equilibrium thermodynamic property: bodies that exchange energy as heat are not in thermal equilibrium, their temperatures differ. The concept of thermal equilibrium also applies to degrees of freedom. For instance, if you (somehow) freeze particular motions (translational motion in a gas, say) you lower the temperature associated with those particular modes. If the system is then allowed to return to equilibrium, heat flows from warmer into the (accessible) colder degrees of freedom.

3) Blackbody radiation, which is intuitively thought of as a "photon gas" in a cavity. The concept of thermal equilibrium can be applied to this "gas". You can consult the wikipedia for more information on these last aspects:

  Just as the Maxwell–Boltzmann distribution is the unique maximum entropy energy distribution for a gas of material particles at thermal equilibrium, so is Planck's distribution for a gas of photons.[8][9] By contrast to a material gas where the masses and number of particles play a role, the spectral radiance, pressure and energy density of a photon gas at thermal equilibrium are entirely determined by the temperature.

References

Engineering ToolBox, (2003). Specific Heat of Solids . [online] Available at: https://www.engineeringtoolbox.com/specific-heat-solids-d_154.html [Accessed 23/9/2019]. The source does not state the measurement temperature but it is presumably ~room T.


Does reduction/ oxidation take place separately at a half cell electrode?

In a (standard) Zn/Zn++ half cell, no oxidation (and no reduction) is taking place. A piece of zinc is sitting in a 1M solution of Zn++ ions (with their accompanying anions). It is not imaginary, and on a micro scale, an equilibrium is occurring (could be occurring!) where Zn + Zn++  --> Zn++ + Zn. No overall change occurs; this is a reference point. If it were not stable, some process would be taking place that degrades the half cell and it would be reacting itself away. In reality, half cells are only useful to the extent that they don't react themselves away until they are connected to the other half of the overall cell.

Same for Cu/Cu++. However, when the half cells are connected (thru a salt bridge so ions can flow, and a wire so electrons can flow), two processes occur: electrons leave the zinc (the zinc oxidizes) and electrons flow to the copper (reducing it). No "process" or reaction occurs in the half cells until they are connected.

The reality of oxidation and reduction require both to occur simultaneously - WHEN THEY OCCUR - but we can speak of each process independently of the other. When electrons leave an atom (oxidation), they have to go somewhere (reduction); if electrons are not flowing (no actual reaction) there is no oxidation or reduction. The half cell potentials then define places from which electrons will flow, or locations to which they will flow.


How exactly are molecules detected in space?

The research article corresponding to the observation to which the question is referring is Detection of the simplest sugar, glycolaldehyde, in a solar-type protostar with ALMA. Note that the molecule is glycolaldehyde rather than glycoaldehyde.

ALMA is the Atacama Large Millimeter/submillimeter Array, a radio telescope array in the Atacama desert.  The extremely low humidity of the Atacama desert minimizes water vapor interference.

This observation of glycolaldehyde in space is not the first.  It was earlier observed by Greenbank radio telescope, in West Virginia, USA, for example.  The Greenbank observation is reported in INTERSTELLAR GLYCOLALDEHYDE: THE FIRST SUGAR.  Glycolaldehyde was detected through radio signals in the 71-103 GHz range that are photons emitted when the molecule transitions from an excited rotation state to a lower energy rotational state.

In the ALMA observation, the molecule was detected based upon radio signals near 220 GHz and 690 GHz.  These correspond to photons emitted by the molecule when transitioning from an excited rotational state to a lower energy rotational state, but are a different set of transitions from those observed at Greenbank, including some for excited vibrational states.

I'll try to add more information later, there are a wide variety of spectroscopic techniques: absorbance vs. emission; rotation, vibration and electronics transitions, and many different frequency ranges.


How will citric acid be deprotonated by pyridine?

The table on the right looks like a standard group of acids shown to illustrate the effect of inductive withdrawing on carboxylic acid pKa. The first two structures are relevant to the question. The table may have been copied from somewhere else because it had both structures on it. Two of the pKa values in the table are every similar to two of the pKa values given in the question. The fluoroacetic acid is a fine distraction; you are correct in thinking that it is not involved.

  As for figuring out which of citric acid's three carboxylic acids is deprotonated first...

Citric acid has three carboxylic acid groups of two different types. Can you see that? One of the acids is closer to the OH group than the other two. Which acid or acids look more like the first entry in the table (except jammed onto the citric acid structure)? Which acid or acids look more like the second entry on the table? Once you have made that distinction, you should be able to use the pKa data on the table to determine which of the three carboxylic acid groups in citric acid is the most acidic (goes with the 3.13 pKa).


When you balance the equation for a redox reaction, the process of balancing it or the balanced equation itself say nothing about kinetics.  That said, redox reactions, as the name implies, deal with electron transfer.  Unless you are dealing with unusual molecules (bulky ligands for example) or unusual conditions (solids, glasses, very low temperature, etc.) electron transfer reactions are close to diffusion controlled, so kinetics is usually not an issue and is often glossed over in such reactions.


What nomenclature do names like fumarate and malate derive from?

Unfortunately, you will have to memorize the structures and the names because there is no chemical structure hidden in those names.
From Elsevier's Dictionary of Chemoetymology

fumaric acid $\ce{C4H4O4}$, derived from the genus name Fumaria (fumitory), from fumus (Latin:
smoke)

Fumaria plant

malic acid $\ce{C4H6O5}$, derived from malum (Latin: apple)

The same goes for all other common names like oxalic acid, acetic acid etc.


Acidity of hydrochloric acid in acetic acid

So basically I am asking, is this value totally wrong or is it actually quite reasonable?

It's quite reasonable. Compare to page 32 of this reference.

  This was much higher than I expected. Also doesn't this imply that
  Cl− is stronger as a base than HCl is as an acid in acetic acid, which doesn't sound right.

To answer whether Cl- is a stronger base than HCl is an acid, you need to know the self-dissociation constant of acetic acid, the equivalent of Kw for water. It is about $10^{-10}$. So whereas the middle of the pH scale is thought of as 7, the middle for acetic acid would be 5. So, yes, Cl- is a stronger base than HCl is an acid in acetic acid.


Do lithium batteries 'explode'/'burn' because of the electric energy stored within?

"Electrochemical" shows the correspondence between chemical and electrical energy. The energy in a cell can be released slowly over time, or rapidly when short-circuited internally or externally, or the ingredients meet under the wrong conditions. No additional chemicals, such as oxygen from the air, are needed.

You state, "there wouldn't be enough electrochemical energy in the battery to cause such an intense reaction." Calculate how much energy is stored, instead of guessing:

The common Li-ion 18650 cell stores about 2.5 ampere-hours at 4 volts (rounded). That's 10 watt-hours, and multiply by 3,600 sec/hr to get 36,000 watt-seconds or joules (J). That is a lot of energy compared to some projectile weapons. For example, the largest cartridge shown in this reference, for the .50 caliber Browning Machine Gun, packs 18,000 J, half that of the inexpensive 18650 cell.

So the question is, does the energy release come from thermal effects due to ohmic heating, or from the direct combination of chemicals. Most likely, once the cell is physically damaged, whether by overheating or mechanical stress, direct chemical combination takes place. Though it can be exacerbated by water or atmospheric oxygen, there is more than enough energy in even small Li-ion cells to cause spectacular fires or explosions.