Why is methyl parathion less toxic than ethyl parathion?

Both methyl parathion (O,O-Dimethyl O-(4-nitrophenyl) phosphorothioate) and ethyl parathion (O,O-Diethyl O-(4-nitrophenyl) phosphorothioate) are organothiophosphates with identical structure except for dimethoxy group in methyl analog has been replaced by diethoxy group in ethyl analog. Both analogs are acetylcholinesterase (AChE) inhibitors, although phosphorothioates ($\ce{P=S}$) are less potent inhibitors compared to corresponding phosphate ($\ce{P=O}$) analogs. The active site of AChE is depicted in following Figure (Ref.1) where acetylcholine (ACh) get hydrolyzed:

The hydrolysis mechanism invoves the $\ce{OH}$ of serine residue and imidazole moiety of histidine residue in the active site as shown. The major inhibition mechanism of all organophosphate esters is involved phosphorylation of the $\ce{OH}$ group of serine residue in the active side. The aging (breakage of the $\ce{O-R}$ ester linkages leaving $\ce{P-O-}$ groups) is followed this phosphorylation making the inhibition irreversible.
Although aging can be visualized by the process of hydrolyzing ester group by water molecule attacking phosphorus atom first, replacing $\ce{R-O-}$ group followed by proton exchange, the literature has shown preferably $\ce{O-R}$ breakage. If that's the case, ethyl breakage is facilitated over methyl breakage just by stability of the resultant products ($\ce{CH3CH2^+}$ vs $\ce{CH3^+}$). Hence, we can conclude that ethyl parathion is more potent than methyl parathion.
An early study of inhibitory activity of various batches of ethyl parathion has revealed that a dithio analog,  O,S-Diethyl O-(4-nitrophenyl) phosphorothioate is much potent inhibitor compared to ethyl parathion (Ref.2). It is possible that this dithio analog is a common contaminate during ethyl parathion sysnthesis. Thus, this factor may contributed to the higher activity of ethyl parathion compared to that of methyl parathion.
It is also worth noting that recent studies on versatility of imidazoles or in reactions with organophosphates (OPs, Ref.3). According to this study, OPs from the $\ce{P=O}$ family, imidazole attacks the phosphorus atom exclusively, making $\ce{P-N}$ bond. With the less reactive $\ce{P=S}$ family, an unusual N-alkylation predominates. Surprisingly, imidazole reacts with methyl parathion exclusively at the aliphatic carbon, whilst for ethyl parathion, imidazole reacts by two pathways: at both the phosphorus and aliphatic carbon, with predominance for the latter:

Since immidazole moiety is present in active site of the AChE, one can suggest the different reactivity of ethyl parathion over methyl parathion with imidazole may play a role in their different activity in inhibition.
References:

Hay Dvir, Israel Silman, Michal Harel, Terrone L. Rosenberry, Joel L. Sussman, “Acetylcholinesterase: from 3D structure to function,” Chemico-biological Interactions 2010, 187(1-3), 10-22 (DOI: 10.1016/j.cbi.2010.01.042).
W. M. Diggle, J. C. Gage, “Cholinesterase inhibition in vitro by OO-diethyl O-p-nitrophenyl thiophosphate (Parathion, E 605),” Biochemical Journal 1951, 49(4), 491-494 (https://doi.org/10.1042/bj0490491).
Valmir B. Silva, Elisa S. Orth, “Are Imidazoles Versatile or Promiscuous in Reactions with Organophosphates? Insights from the Case of Parathion,” J. Braz. Chem. Soc. 2019, 30(10), 2114-2124 (https://doi.org/10.21577/0103-5053.20190084 ).


Paint changes color when applied to Stucco. Chemical rxn assumed, but what?

You haven't told us the color of the stucco you are painting, nor the type of paint.  Chemical reaction is very unlikely, paint that changes color would make for bad paint.  Pigments are chosen to be stable.  Old stucco would be only a little alkaline, but it is porous.  This sounds more like color leaching out of the stucco and into the new paint.  You need paint with better coverage (covers contrasting colors) or better sealing ability.  Seriously, don't ask us, talk to your paint supplier at a paint store.  Maybe get them to refer you to their in-house experts.  You probably will have to spend more for the paint than you are planning on.


light Intensity and its effects on energy and emitted electron on metal surface

what if i double the intensity of light , what would happen to the
  number of electrons emitted and the average energy of the electrons?

This is known as the "Photoelectric Effect" and is described by the following equation

$$\ce{E_{max}=h\nu- WF_{M}}$$

where $\ce{E_{max}}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\ce{WF_{M}}$ is the workfunction for the particular metal. Your guess is part right, doubling the intensity of the incoming beam doubles the number of electrons ejected from the metal surface, but their kinetic energy distribution remains unchanged.  The maximum kinetic energy of an emitted electron is still determined by the above equation and therefor still capped by $\ce{E_{max}}$ and the kinetic energies of all electrons emitted are still distributed from 0 to $\ce{E_{max}}$.  If you'd like more information, here's a good, concise reference.


Regioselectivity in opening of cationic three-membered rings

There are three factors that you should consider in nucleophilic attack.

Steric effects
Electronic effects
Stereoelectronic effects

Two of these are related to lowering the overall energy of the transition state. The third relates to how productive a collision can be.

The one that matters less here (or is already accounted for) is stereoelectronic effects. Notice that you proposed a backside attack for a $\mathrm{S_N2}$ reaction. Therefore, you have considered the proper alignment of orbitals, which is what is meant by stereoelectronic effects.

Steric effects government repulsion of large groups. In your example, attacking the more substituted side would result in greater steric repulsion since the $\ce{R}$ group is larger than $\ce{H}$. Steric arguments therefore argue for attack at the other carbon.

Frequently, the three factors do not say the same thing, and you need to use some intuition and/or experience to figure out which one dominates (in reality, the relative energies will determine the "winner").

In this case, it turns out with a carbocation and a negative nucleophile, perhaps not surprisingly, electronic effects dominate. There are actually two related electronic effects. One is that the partial positive is greater on the more substituted carbon (cf. Markovnikov rule). This carbon is hyperconjugatively stabilized (relative to the other carbon) by greater substitution. In addition, the greater positive charge also implies a weaker $\ce{C-E}$ bond, which further implies a lower energy $\ce{C-E}$ $\sigma^{*}$ orbital. We can then argue that the transition state energy will be lower given that the two interacting orbitals are closer in energy. And indeed, the electronic effects dominate this reaction giving the reactivity shown in your diagram.


Galvanic cell using NaCl, Cu and Zn, why does it work?

I want you to try and work out the details yourself, but there's one aspect of this kind of cell that's not straightforward. At one electrode, you have a reduction reaction and at the other, an oxidation, just like any other galvanic cell. In the oxidation reaction, one of the metal electrodes is oxidized to put cations in solution. (think about which metal is more easily oxidized) In the type of galvanic cell that you may be more familiar with, a metal cation in solution is reduced onto the other electrode, but in this case, the only metal cation is $\ce{Na+}$, which is extremely difficult to reduce. Therefore, some other reduction reaction must be happening. The only other reducible substance present is water.

The two reactions occurring can be found on this table, and using the potentials in the table, you should be able to work out which reactions are occurring and at which electrodes.


What is the priority of a phenyl group in a compound such as this?

For preferred IUPAC names, the ring is preferred to the chain.
Therefore, the preferred IUPAC name for the given structure is but-3-en-1-ylbenzene.

However, the context may favour the chain, so that, for example, an unsaturated acyclic structure may be recognized.

When one double bond is present, an unbranched alkene is named by changing the ending to ‘ene’. The double bond is assigned the lower locant possible, which is placed immediately in front of the ending ‘ene’. When several structural features appear, low locants are assigned to ‘ene’ endings before detachable alphabetized prefixes.
Therefore, a correct name for the given structure is 4-phenylbut-1-ene.


tap water and distilled water for electrolysis

This article tells you to use distilled water because the salt in tap water will damage the booster. The author also tell you to add NaOH or KOH to the distilled water. Did you add these salts?


Why is Acetic acid (pKa = 4.76) stronger than carbonic acid (pKa = 6.36)?

Well actually the $\mathrm{pK}_\mathrm{a1}$ for carbonic acid at 25 °C is 3.6. The pka = 6.36 is also including the equilibrium with dissolved $\ce{CO2}$ in aqueous solution.


Why does the conductivity of a LiOH solution decrease when boric acid is added?

TL;DR: The addition of boric acid effectively switches the conducting anionic solute from hydroxide to borate, which has a lesser contribution to the conductivity due to its larger size and corresponding lower diffusivity. Most solute systems will not behave this way.

In sufficiently dilute solution, the contribution $\kappa_i$ of each dissolved ion to the overall conductivity is proportional to the concentration, diffusivity, and squared charge of that ion:

$$
\kappa_i \propto z_i^2D_iC_i
$$

And, the total conductivity is just the sum of these contributions:

$$
\kappa = \sum_i{\kappa_i}
$$

In this case, you're starting with $\ce{LiOH}$, which dissociates  more or less completely into aqueous $\ce{Li^+}$ and $\ce{OH^-}$, and the conductivity rises as you add more of it. Once you start adding boric acid, though, you're no longer just dissociating ions into the solution.

Boric acid is a funny creature.  Even though it looks like it might be a triprotic Arrhenius acid if we were to write it as $\ce{H3BO3}$, in reality it's a monoprotic Brønsted acid that either exchanges a hydroxide ion with the water solvent:

$$
\ce{B(OH)3 + 2H2O <=> B(OH)4^- + H3O^+}
$$

or, at sufficiently high $\mathrm{pH}$ $(\mathrm pK_\mathrm a = 9.2)$, it just takes up a hydroxide ion from solution:

$$
\ce{B(OH)3 + OH^- <=> B(OH)4^-}
$$

So, when added to a solution of $\ce{LiOH}$, boric acid just substitutes dissolved $\ce{OH^-}$ for dissolved $\ce{B(OH)4^-}$, with no net addition of anions to the solution!  Thus, the $C_i$ contribution to $\kappa$ doesn't really change much. As well, both $\ce{OH-}$ and $\ce{B(OH)4^-}$ are monovalent anions, so there's no effect on any of the $z_i^2$ terms either.

The overall decrease in $\kappa$ comes from the change in diffusivity that results from the swap of hydroxide for borate.  The diffusivity of an anion is roughly proportional to its ionic volume—bigger ions experience more "drag" from the surrounding solvent molecules, so they diffuse (move around) more slowly. I couldn't find an ionic radius for borate on a quick search, but just from their compositions, it seems sensible to assume that the borate anion is considerably larger than the hydroxide anion, and thus its diffusivity should be correspondingly less.


Is the Hydrogen spectrum only an emission spectrum?

Emission and absorption are inverse processes.
So, naturally, if an atom exhibits some emission spectrum, it will also give an absorption spectrum. So you were right in assuming that.
However, the processes by which these two spectra are generated are completely the opposite of each other.
An atom emits a photon when an electron goes from a higher energy state to a lower energy state, thereby releasing energy in the form of light. When a photon is absorbed, the electron uses this energy to excite to a higher energy state.
What kind of spectrum you have at hand depends on which instrument and detector you are using. When atoms releases photons (light), the detector catches the production of the photon and gives out emission spectra. For absorption spectroscopy, varying wavelengths of photons are incident on the atom and the corresponding photon is absorbed by the atom resulting in no photon of that particular wavelength being detected.
I am attaching the images of the emission and absorption spectra of hydrogen atom. Note how they are complete inverses of each other.

Image source: https://chem.libretexts.org/Courses/Solano_Community_College/Chem_160/Chapter_07%3A_Atomic_Structure_and_Periodicity/7.03_The_Atomic_Spectrum_of_Hydrogen


Is the middle ring of this organic compound aromatic?

You don't have to force a positive charge on the carbon between the central and left rings.  The methoxy oxygen on the left ring can donate an electron pair to take up the positive charge.  The left ring then becomes quinoid and the central ring becomes aromatic.  Such switching between quinoid and aromatic rings is common in conjugated polycyclic structures.


Why are the two bonds in sulphur dioxide identical?

There are two ways to look at the $\ce{SO2}$ molecule. The valence bond way is this:

$\ce{SO2}$ is a bent molecule with C2v symmetry point
group. A valence bond theory approach considering just s and p
orbitals describes the bonding in terms of resonance between two
resonance structures:

The sulfur–oxygen bond has a bond order of 1.5 and does not invoke d
orbital participation. In terms of electron-counting formalism, the
sulfur atom has an oxidation state of +4 and a formal charge of +1.
(Wikipedia)

However, there is a molecular orbital approach which seems to give a deeper appreciation of the bonding. The picture is this:

(Picture source with detail explanation)
This is my simple explanation:

The sulfur is hybridized sp$^2$-p. The bond angle is 119$^o$ - you
can hardly get closer to 120$^o$!

We will assume that the oxygens are also hybridized sp$^2$-p.

Then the sigma bonding is designated in the diagram by the numbers 1 and 2 for the sigma bonds between sulfur and oxygen. Lone pairs are designated by numbers 3 (for S), and 4, 5, 6 and 7 (for O).
The pi system has three molecular orbitals, one from each atom and four electrons: one from each oxygen, two from sulfur. The lobes in these orbitals are drawn at the lower right side of the picture. Orbital 8 has the lowest energy, with all lobes bonding. The non-bonding pi orbital (#8) has reversed signs for the lobes of oxygens and therefore a node at sulfur. The antibonding pi orbital (#10) has two nodes and is unoccupied.
So the final count is two filled sigma bonds and one filled pi bond; five lone pairs and one filled non-bonding pi orbital. The formal bond order is 1.5, same as in the valence bond explanation with resonance.
The only thing I might change in the diagram is the relative energies of the sulfur orbitals and the oxygen orbitals: I would think they might be closed to the same energy. The diagram might be drawn as it is in order to make it clearer and give more room for all the numbers.
We just don't need d orbitals to explain anything, but their involvement cannot be entirely eliminated.


Are these molecular structures of cyclodecapentaene identical?

If you make spring-and-ball models of both compounds, you can see that the two compounds are entirely different, and both are contorted.

The all-cis model on the left has all its hydrogens pointing outward, so the contortions of the carbon chain involve just carbon-carbon twisting to come close to 120 degree angles. The model of the cis-trans molecule on the right has two inward-pointing hydrogens (all the rest point out) which are sterically interacting so much that they introduce strain that is not evident from the carbon-only diagram in the original post. (If I put in all the hydrogens, the image is too cluttered.)
The fact that one is all cis and the other cis-trans certifies that the molecules are different, but the 2-D drawing can seem vague. The NMR spectra of these two compounds would be different. I can imagine the molecule on the left flipping all around (the model sure did!), but the molecule on the right seems pretty stuck.
Thinking in 3-D is easy after you've done it many times. But when you are starting out, it is helpful to make molecular models. And even if you've done it a million times, there are situations where your mental image is so fluid that you need to solidify it with a model, no matter how inadequate the model is. And then you mentally add onto the solid model.


Why does potassium react more violently with water than lithium?

For the reaction,
$$\ce{M -> M+ + e-}$$

the heat liberated is highest for lithium owing to its high negative $E^\circ$ value so one would think that the reaction must be most vigorous.

The reason behind the more violent reactivity of potassium rather than lithium lies in kinetics and not in thermodynamics.

No doubt, maximum energy is evolved with lithium but the vaporization and ionization will also consume maximum energy (the melting point and ionization energy of lithium are the highest) and so the reaction proceeds gently. On the other hand, potassium has a lower melting point and ionization enthalpy. The heat of reaction is sufficient to melt it. The molten metal spreads over the water and exposes a larger surface to water. Also, the hydrated radius of lithium is the greatest out of all alkali metals. This reduces the ionic mobility which in turn reduces the speed of the molten metal.

That's why potassium gives a more violent reaction with water.

Reference:

Kumar, Prabhat Conceptual Inorganic Chemistry; Shri Balaji Publications: Muzaffarnagar, U.P., 2014.


Neopentyl halides are not good electrophiles in the way that other primary alkyl halides are, primarily due to steric effects but also because of rearrangents and eliminations.

This paper 1 notes in its abstract that "As a rule, the SN2 reactions of neopentyl bromide and tosylate with sodium alkoxides are extremely retarded by steric hindrance, and the yields of alkyl neopentyl ethers are poor because of side reactions such as neopentyl rearrangement, elimination, and decomposition." This is supportive of the view that an SN1/E1 process may be occurring.

This paper 2 confirms rearrangements occur.

I don't think this leads to us being able to say neopentyl bromide undergoes SN1 reactions directly or easily. It is unreactive to nucleophiles; when it does react it does so through mulitple pathways leading to low yields.


What are the isomers of a complex having a structure of M(AB)2C2?

Referring to the replies of Mr. Ivan Neretin, complexes that has a form of M(AB)2C2 posses 8 stereoisomers, 2 of which are optically inactive. The other 6 isomers can be interchanged to their respective enantiomers.


Rules to follow while writing a chemical formula for polyatomic ions

What they are trying to say is that hydroxide $\ce{OH^−}$ and nitrate $\ce{NO3^−}$ ions, which are polyatomic anions react bond with cations as a group.
When $\ce{NO3^−}$ bonds to $\ce{Na^+}$ only a single nitrate anion bonds to a single sodium cation, to form $\ce{NaNO_3}$ so brackets are not needed. You can still use them, in which case it would be $\ce{NaNO_3}$ but it is redundant and current convention doesn't use it.
However, when $\ce{OH^−}$ bonds to $\ce{Ca^{2+}}$ two hydroxide anions bond to a single calcium cation, to form $\ce{Ca(OH)_2}$ so brackets are needed to avoid confusion and show the structure of the compound.
Old ways of writing it (before my time) before the conventions were standardised, such as $\ce{CaO2H2}$ or $\ce{CaH2O2}$ don't immediately show that it is a hydroxide, so would have been confusing to non-experts.
To reiterate: the brackets are used when there is more than one polyatomic ion in a compound.  So if you write down sodium hydroxide it is $\ce{NaOH}$ but calcium hydroxide is $\ce{Ca(OH)2}$


How to calculate probability of finding an electron at a point?

Consider the 1D case for simplicity.

The modulus square of the wavefunction is interpreted as a probability density in the sense that
$$
|\psi(x)|^2\mathrm{d}x
$$
gives you the probability of finding the electron in within $x$ and $x+\mathrm{d}x$. Now if you want to consider the probability to find the electron exactly in $x$ you have to take the limit $\mathrm{d}x\to 0$ and this gives you zero probability. Formally, this corresponds to compute
$$
P(x=x_0) = \int_{x_0}^{x_0} |\psi(x)|^2\mathrm{d}x = 0,
$$
which is zero since the limits of the integral are the same. Therefore there is $0$ probability to find an electron at exactly one point.


From the values on the Wikipedia page for sodium carbonate, the Gibbs free energy of formation is negative and thus, formation is thermodynamically preferred over decomposition at $\pu{298K}$. Your given values seem to be wrong after all.

By the way, $\Delta H$ and $\Delta S$ should be the negative of those given on that page for your case, making $\Delta G$ a positive number.


Electrode potentials vs interfacial potential differences

IUPAC defines the electrode potential of an electrode to be the EMF of a cell in which the electrode in question is on the right hand side of the line representation of the cell and a SHE is on the left hand side.

As such, the electrode potential of the $M^+/M$ couple is

$E^o_{M^+/M} = \phi_{M} - \phi_{R}$

where $R$ refers to the reference electrode, in this case the SHE. This is equivalent to

$E^o_{M^+/M} = (\phi_{M}-\phi_{S}) - (\phi_{R}-\phi_{S})$

Where $\phi_{S}$ is the potential of the solution. Essentially, this means that a relative electrode potential really means

$E^o_{M^+/M} = \Delta \phi_{M^+/M} - \Delta \phi_{H^+/H_2}$

That is, the potential difference of the electrode in question and the solution relative to the potential difference between the SHE electrode and the solution.

It doesn't help the cause that electrochemists refer to potential differences when they use the word potential, however this is just a case of getting to grips with the terminology used in the field!


What solvents can get rid of oil stains?

Depending on where you live different dry-cleaning solvents are indeed the best choice. Perchloroethylene $\ce{Cl2C=CCl2}$ is probably still the easiest one to get.


IUPAC nomenclature of alicyclic compound with an equal number of cyclic and acyclic carbons

According to current IUPAC recommendations (2013), for the preferred IUPAC name, a ring has seniority over a chain. Therefore, the given structure is a propylcyclopropane. (However, the IUPAC recommendations allow that the context may favour the chain.)

The two prefixes of cyclopropane (chloro- and 2-bromopropyl-) are arranged alphabetically. Thus, the name is 1-(2-bromopropyl)-2-chlorocyclopropane.


I agree with Poutnik's point that this is a poor question.

Oxalic acid has two acidic protons so its equivalent weight is $\frac{1}{2}$ its molecular weight. Thus:

Answer A is false.
Answer B is true.
Answer C is false since there are 2*100*0.2 = 40 milliequivalents of acid, but only 100*0.2 = 20 milliequivalents of base.

Answer D requires some extra work. First it is not clear if Answer D applies to both reaction (i) and reaction (ii)

For reaction (i), in 100 ml of 0.2 M oxalic acid there are 40 milliequivalents of acid
For reaction (ii), the reaction is a redox reaction, so the number of electrons exchanged by oxalic acid needs to be determined.

Here is another sloppy point about the question. Reaction (ii) is not balanced.

However to figure out the number of electrons exchanged by oxalic acid, we only need the half cell reaction which is:

$\quad\quad\ce{2CO2(g) + 2 H+ + 2e− <=> HOOCCOOH(aq)}$

$\quad\quad$ Since oxalic acid exchanges two electrons in the redox reaction,
its "redox" equivalent$\quad\quad$ weight is $\frac{1}{2}$ its molecular weight.Thus there are also there are 40 milliequivalents of$\quad\quad$ oxalic acid redox wise too.

Answer D is false, because there are 40 milliequivalents of oxalic acid, not 40 equivalents


Beer's Law: Are the graphs of concentration/time and absorbance/time supposed to look very similar?

Well, since Beer's Law says

$A = \epsilon⋅l⋅c$

And $A$ hence is directly proportional to $c$, as Beer's Law actually represents a line equation.


Rate of substitution and elimination reactions at high temperatures

With temperature, the rates of substitution reactions and elimination reactions both increase.

For the purpose of answering this question, let us consider $S_N2$ and $E_2$ type reactions.

For $S_N2$ type reactions, the HOMO of the nucleophile has to be oriented with the LUMO of the electrophile. Here, a linear orientation is required. So, the entropic requirement is that the nucleophile should approach the electrophile in a certain way.

For $E2$ type reactions, the base attacks the $\alpha$-proton, which finally leads to the elimination. So, the HOMO of the base interacts with the $\sigma^*$ orbital of the $C-H$ bond, and then the $\sigma$ orbital interacts with the $\sigma^*$ orbital of the leaving group. So here the entropic requirement is the linear orientation of the base, the proton and the carbon it is attached to, and the leaving group should be antiperiplanar with respect to the proton.

With an increase in temperature, the "rate of bond rotation" increases and hence, the probability that a collision with a proper orientation will occur in a given time frame increases. This increase is more prominent in the case of the elimination reaction because it is quite dependent on the orientation of bonds (2 constraints vs. 1 constraint) compared to the substitution reaction.

So, a rise in the temperature will not only affect the thermodynamics of the equation, but it will also directly affect the rate of reaction.


Why must a voltaic cell take place in a conductive solution?

The electrolyte must be conductive to ions, not electrons. In fact, if there's an easy path for electrons to move between the two electrodes, the cell will immediately self-discharge. The exact reason for needing ionic-conductivity kind of depends on the cell. For example, in lithium-ion batteries, lithium-ions are generated at the anode and consumed at the cathode (or vice-versa). If ions cannot move freely between the two (through the electrolyte), then you will quickly build up an ion-rich region at one electrode and an ion-deficient region at the other at which point the reaction will stop.

It's not even a requirement that the electrolyte be liquid, as long as it conducts ions. Eg. the sodium-sulfur battery uses a form of solid alumina that conducts sodium ions.


Relative rate of hydrolysis of the carbon chlorine bond in thioethers and ethers

This is a classic case of neighboring group participation. The hydrolysis of the thioether is significantly faster because the sulfur can displace the chlorine to give a three member ring with a sulfur in it. The ring is quite reactive and will easily be opened by water.

For reference, you can also look at a compound like mustard gas. It has the same functional groups and is so reactive that is a war gas.


Significant Figures: To expand or not to expand?

In the first example, the 4(2+2), if you were ABSOLUTELY going to adhere to significant figure rules, then 20 would be the correct answer, though I don't believe many people would fault you for saying 16. In your second case, 2.0(0.5+0.5), the appropriate answer here would indeed be 2.0.

In any case, significant figures are not really valuable when referring to purely mathematical equations like you did. We use them in order to avoid making errors when it comes to the accuracy of a result, in other words, significant figures are there so that your answer is not more specific than your measurement was. In mathematics, this is not an issue, as you are not making any measurements. So, while you COULD use significant figures in a situation like this, it really is not quite the appropriate situation in which to do so.


Biological Consequences of Asteroid Mining—Death by Isotope?

This is an interesting question and you raise a number of points, let's step through them.

  A consequence of this is that relative atomic masses of elements
  mined—those with two or more stable isotopes—will no longer be
  faithful to our current periodic table.

But this is already happening.  $\ce{^235U}$ constitutes 0.72% of uranium found on earth and decays to the stable isotope $\ce{^207Pb}$, which is found in a natural abundance of 22.1%.  Before we get to the asteroid the abundance of various isotopes and their weighted average mass is already shifting.

  98 Ru , then this would be expected to diffuse much faster than its
  heavier counterparts

Diffuse much faster?  First off, we're talking about diffusion, not chemical reaction.  Elements or compounds usually enter the body by ingestion of one sort or another (eating, breathing) rather than diffusion, and these ingestion pathways wouldn't involve isotopic discrimination.  That is, if we breath air that contains isotopes in a certain ratio, then that's the ratio that will initially appear in our lungs.  Nonetheless if diffusion were found to be an issue, my guess is that isotopic discrimination by diffusion would be a very small effect.  For a gas the maximum separation of two isotopes is given by
$$\mathrm{\sqrt{\frac{[MW~of~compound ~with ~isotope~1]}{[MW~of~compound ~with ~isotope~2]}}}$$
In the case of separating $\ce{^235UF6/^238UF6}$ this amounts to a fractionation ratio of 1.0043 after 1 pass.  Ruthenium is lighter, so the effect would be larger, about 1.02 ($\ce{^102Ru/^98Ru}$) after 1 pass, even less if it is in a molecule.

Chemical reaction within our body of ingested isotopic compounds would also show  discrimination due to primary kinetic isotope effects.  The maximum primary kinetic isotope effect is proportional to the reduced mass as follows:
$$k~ \thicksim ~\sqrt{\frac{m_1 + m_2}{m_1 m_2}}$$
Applying this to $\ce{^102Ru}$ and $\ce{^98Ru}$ yields a primary kinetic isotope effect of 1.020, again, even less if the element is incorporated into a compound.

To me, the effects look small and since shifts in natural abundance have been occurring for a long time here on earth, with no one raising a flag, my guess (and that's all it is, a guess) is that it's not something to worry about.  Still, I'd feel better if a NASA toxicologist looked at the question and said "no problem"


How long can iodine bottle left open? Can it go bad sooner than expiration date?

A solution of iodine antiseptic is a medical item that is used for improving health. There are two ways to treat your cracked cap.
#1. If you inspected the bottle in the drug store before you bought it, you would not buy this bottle. Almost all over-the-counter medicines have warnings: Do not use if the safety seal is broken. So, out of utmost caution, do not use this iodine, because it might be contaminated (even tho you are the only one who could have contaminated it). Look at it this way: you bought a bottle of iodine solution, you had its potential usefulness for a year to keep your skin abrasions antiseptic (even tho you did not have the need to use it). It can be considered to have served its purpose usefully; get a new bottle and be more careful.
#2. Being the cheapskate that I am, I would examine the bottle: has the liquid level dropped and dried out? Is it gooey so as to be difficult to apply? If not, it might be OK to keep and use. On the other hand, a cracked lid is an invitation to disaster - if the bottle gets knocked, it will leak - probably onto the item you just laid down, which will, of course, be very important and/or delicate. To avoid catastrophes of this sort, dispose of the iodine solution. You will feel more secure. Get a new bottle.
Or, get a tube of antiseptic ointment, which will have less tendency to spill, unless you step on it.


Are any commercial plastics made from biomass/plant matter?

The original answer only gave a link to a commercial bioplastic producer.  The answer has been edited to provide more information.

Yes there are.  Here is an example: Biome Bioplastics

European Bioplastics (an industry group) estimates that the global production of bioplastics was about 2 million metric tonnes in 2017.  They forecast that this will increase to almost 2.5 million metric tonnes by 2022.Source

One estimate puts the market value of bioplastics at around 6 billion USD in 2018 and growing to almost 15 billion USD by 2023.  Source

Common sources of bioplastics are starches (like corn), cellulose and aliphatic esters.

Commonly, bioplastics are used to make compostable/biodegradable packaging and films, as well as disposable cups and utensils.


How can CO2 exist in two different phases?

The $\ce{CO2}$ exists in two phases when you buy a partially-filled $\ce{CO2}$ cylinder because there is extra volume (which is needed to allow expansion as the container warms). Therefore, the pressure in the container is not 100 bar, but closer to 80 bar at 293K.
If the container had only enough room for the $\ce{CO2}$(l), then, indeed, there could be no gas phase. But if space is allowed above the liquid, then $\ce{CO2}$ evaporates, filling that space, and the $\ce{CO2}$(g) and $\ce{CO2}$(l) reaches equilibrium pressure.
On the molecular level, consider it an ongoing exchange between the molecules of liquid and gas. For the same reason, there is humidity in the air and oceans full of water, a 1 bar at sea level.


Why is the rate of reaction calculated as d[A]/dt?

Although the appearance of C is due to the combined disappearance of A and B, the relation is still 1:1 between A and C. Therefore, the rate that A disappears = the rate that B disappears = the rate that C appears.

Let's look at an analogy in stoichiometry. Say I have 1 mole of A and 1 mole of B. Will I make 1 or 2 moles of C? By your logic, I should make 2 moles of C. But I don't.


Growing crystals that are in a magnetic field or subjected to an electric field

The main influence on resonant frequency of a crystal is mechanical: the mass of the crystal and the restoring force of elasticity cause the crystal to vibrate at a specific frequency when disturbed. To fine-tune the crystal, bits may be chemically or abrasively removed or the mass may be increased slightly by rubbing with graphite.

That said, crystal orientation does influence resonant frequency because the modulus of elasticity of a crystal such as quartz is anisotropic, i.e. it varies significantly in which plane it is measured. Perhaps applying a strong magnetic or electrostatic field would change the resonant frequency -- but you would need to isolate the effect on the crystal from that of its mounting, particularly if there are metallic contacts.


What happens to the pressure of gas when only one molecule of it is placed in a very large container?

I think you have forgotten what pressure represents in terms of the kinetic theory of gases. When studying gas kinetics, you start by understanding the behavior of a single molecule/atom moving around a container and colliding with the surface and bouncing in random directions. Then you add a few more molecules/atoms so that now there are collisions and as long as the mean free path of the gas is less than the dimensions of your vessel you still have motion that acts in random diffusion-like behavior. Once sufficiently many atoms have been added that a given amount of area experiences a regular number of collisions per unit of time, then you have something that can be given as an average rather than a count of collisions. This is what pressure represents, an average force on a surface per unit area of molecular collisions on that surface. The instantaneous force can change for a given spot nanosecond to nanosecond, but averaging all spots over a span of microseconds will give an average force per area can be resolved which we call pressure.

The problem here that gives you a misunderstanding is that you have taken a behavior of an aggregation of many molecules and tried to apply it to a single molecule. This approach will disappoint. But to summarize and actually answer the question, pressure cannot describe a system of only one gas molecule, and is therefore undefined.


Why can one double bond exist in some resonance structures but two can't?

In your drawn example, the red-boxed structure requires the carbon atom to have five bonds, which, to channel my secondary school teacher, says "yuck, I don't like that".

This is a case of breaching the octet rule which has to with how the orbitals hybridise. Five full bonds on a carbon is not "allowed". However three full bonds and two partial bonds is common in both resonance and transition structures, and is in fact a much more accurate representation of the molecule you're drawing.

Don't forget too, that the location of the negative charge matters, and I can't see it in any of your drawings.


Propagation of the uncertainty in measurements

The volume $V$ can be calculated from the length $x$, width $y$, and height $z$.

$$\begin{align}
V&=x\cdot y\cdot z\tag1\\[6pt]
&=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times 12.7\ \mathrm{cm}\tag2\\[6pt]
&=2048.383\ \mathrm{cm^3}\tag3
\end{align}$$

If we assume that the uncertainties of length $x$, width $y$, and height $z$ are not correlated (which is not necessarily true when the same ruler was used to measure length, width, and height), i.e. their correlation coefficient is $\rho=0$, the relative uncertainty of the volume may be estimated as

$$\begin{align}
\left(\frac{u(V)}{V}\right)^2&=\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2\tag4\\[6pt]
\frac{u(V)}{V}&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\tag5
\end{align}$$

Thus, the absolute uncertainty of the volume is

$$\begin{align}
u(V)&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot V\tag6\\[6pt]
&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot x\cdot y\cdot z\tag7\\[6pt]
&=\sqrt{\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag8\\[6pt]
&=83.809\ \mathrm{cm^3}\tag9
\end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align}
u(V)&=\sqrt{\left(\frac{\partial V}{\partial x}\right)^2\cdot u^2(x)+\left(\frac{\partial V}{\partial y}\right)^2\cdot u^2(y)+\left(\frac{\partial V}{\partial z}\right)^2\cdot u^2(z)}\tag{10}\\[6pt]
&=\sqrt{\left(y\cdot z\right)^2\cdot u^2(x)+\left(x\cdot z\right)^2\cdot u^2(y)+\left(x\cdot y\right)^2\cdot u^2(z)}\tag{11}\\[6pt]
&=\sqrt{\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2\\+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2}\tag{12}\\[6pt]
&= 83.809\ \mathrm{cm^3}\tag{13}
\end{align}$$

However, if we have to assume that the uncertainties of length $x$, width $y$, and height $z$ are correlated, i.e. their correlation coefficient is $\rho=1$, then the equation for the estimate of the relative uncertainty of the volume simplifies to

$$\frac{u(V)}{V}=\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\tag{14}$$

and the absolute uncertainty of the volume is

$$\begin{align}
u(V)&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot V\tag{15}\\[6pt]
&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot x\cdot y\cdot z\tag{16}\\[6pt]
&=\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag{17}\\[6pt]
&=145.161\ \mathrm{cm^3}\tag{18}
\end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align}
u(V)&=\frac{\partial V}{\partial x}\cdot u(x)+\frac{\partial V}{\partial y}\cdot u(y)+\frac{\partial V}{\partial z}\cdot u(z)\tag{19}\\[6pt]
&=y\cdot z\cdot u(x)+x\cdot z\cdot u(y)+ x\cdot y\cdot u(z)\tag{20}\\[6pt]
&=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\\
&\hphantom{={}}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\tag{21}\\[6pt]
&=145.161\ \mathrm{cm^3}\tag{22}
\end{align}$$

The numerical values of an estimate $y$ and its standard uncertainty $u(y)$ should not be given with an excessive number of digits. It usually suffices to quote $u(y)$ to at most two significant digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.
Thus, the result for the volume $V$ according to $\text{(3)}$ and its standard uncertainty $u(V)$ according to $\text{(9)}$ or $\text{(13)}$ may be expressed as

$$V=\left(2048\pm84\right)\ \mathrm{cm^3}\tag{23}$$

or

$$V=2048(84)\ \mathrm{cm^3}\tag{24}$$

Using the uncertainty according to $\text{(18)}$ or $\text{(22)}$, the result may be expressed as

$$V=\left(2.05\pm0.15\right)\times10^3\ \mathrm{cm^3}\tag{25}$$

or

$$V=2.05(15)\times10^3\ \mathrm{cm^3}\tag{26}$$

(According to various international standards, the ± format should be avoided.)


Yes, in your example, it is possible to form acetylene. E1 mechanism does not depend on whether the leaving group is anti to $\ce{H}$ or not.

All vicinal halides will not give alkynes, as E1 mechanism involves a carbocation intermediate which can rearrange to the most stable position. Example: In 4-methyl-2,3-dibromopentane, the product will be a conjugated alkene not alkyne.

Rate of E1 elimination depends on the stability of the carbocation whereas rate of E2 mechanism does not. Since in your example, the carbocation formed is a 1 degree, it is not very stable so the rate will be higher for E2 mechanism.


How to assign the signals in the H-NMR of Aspirin?

Here is a link to the actual assigned proton nmr spectrum for acetylsalicylic acid.  Your peak assignments are correct.

  Would E and B be doublet of doublets because they couple with D and C
  so they would have ortho and meta coupling?

Yes, that is correct. Here is another link with more detail on the coupling constants (see p. 13 for example)

  Would D and C be coupling with each other and B and E so would they be
  doublet of doublets of doublets, with two ortho and one meta coupling?

Conceptually you are correct, however in the case where two coupling constants are very similar, the expected quartet collapses to a triplet. Since $\ce{J_{DC}}$ is very close in magnitude to $\ce{J_{BC}}$ the expected quartet appears as a triplet.  Therefore, proton C appears as a triplet from these two protons and then each of these lines is further split into a tight doublet by the smaller coupling to proton E.  The overall pattern appears as a triplet of doublets.  The same reasoning applies to proton D, and again a triplet of doublets results.


To find the value of x in a metal chloride MClx

$$\ce{MCl_x + xAgNO3 -> xAgCl + M(NO3)_x}$$

$500$ mL of $0.8$ M is $0.4$ mol, so the mole ratio between $\ce{MCl_x}$ and $\ce{AgNO3}$ used is $1:4$.

From the equation above, the mole ratio is $1:x$.

Therefore, $1:x=1:4 \implies x=4$.


How does the chemical potential allow for open systems to be considered?

Actually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)

The correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):

$$
\mu_i=(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}
$$

So the total exact differential is expressible as you had originally expected:

$$
dH=TdS+VdP+\sum_i\mu_idn_i=(\frac {\partial H}{\partial S})_{p,n_i}dS+(\frac {\partial H}{\partial p})_{S,n_i}dS+\sum_i(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}dn_i
$$
  The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.

Sorry for any confusion my initial errors may have caused you.  Hope this clears it up.


Solubility when multiple salts are involved

I'm not sure what your train of thought was on this problem. Maybe annotating the problem would help see where you've gone wrong.

We're given that $K_{\mathrm{sp}}$ = $[\ce{Ag+}][\ce{Cl-}] = 1.6\times10^{-10}~\ce{mol^2/L^2}$. Since the solution already contains a known concentration of the chloride ion, we will solve for the amount of $[\ce{Ag+}]$ that will dissolve from the added $\ce{AgCl}$ to figure out the molar solubility of $\ce{AgCl}$.

$$[\ce{Ag+}] = \frac{K_{\mathrm{sp}}}{[\ce{Cl-}]} = \frac{1.6\times10^{-10}~\mathrm{mol^2/L^2}}{8\times10^{-3}~\mathrm{mol/L}} = 2\times10^{-8}\mathrm{mol/L}$$

Note that $[\ce{Cl-}]$ is double that of $\ce{[CaCl2]}$ since $$\ce{CaCl2<=>Ca^2+ + 2Cl-}$$

Since the dissolution of $\ce{AgCl}$ is in a one-to-one ratio with respect to the silver ion, $\ce{Ag+}$, the above answer is sufficient to show the molar solubility of $\ce{AgCl}$ in the given solution.

Any further chloride ions that dissolve will be negligible to the amount that is already in solution (according to the final solubility, only about $2\times10^{-8}~\mathrm{M}$ would be added, which is negligible to $8\times10^{-3}~\mathrm{M}$ by a factor of $400000:1$).


Calculating the dissociation constant of one acid in a mixture

I just wanted to add an answer based on "no" assumptions, except (and that is occasionally a big assumption) that concentration equals activity. At first it does not seem very practical, as of its size, but I couldn't stop me from adding it here as I like to see the whole picture. In the end, it is actually not that hard.

The exact equation for the proton concentration is:

$$[\ce{H+}] = [\ce{OH-}] + [\ce{Cl-}] + [\ce{CHCl2COO-}] + [\ce{CH3COO-}]$$

which is the sum of the hydroxide ions from the autodissociation of water and the sum of the resulting acid residue ions from the three acids. As is shown here [1], this system expands to the following (and can be reduced again, if appropriate simplifications are taken into account for each acid, as is shown here [2]):

$$
[\ce{H+}] = [\ce{OH-}]
+ \frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}
$$

where $K_{\mathrm a1}$, $K_{\mathrm a2}$, and $K_{\mathrm a3}$ are the $K_{\mathrm a}$'s of $\ce{HCl}$, $\ce{CHCl2COOH}$, and $\ce{CH3COOH}$ respectively.

Now what you need to find is $K_{\mathrm a2}$. Solving it at this stage for $K_{\mathrm a2}$ is probably not a good idea aka nothing nice to handle, as the system is quite large. But as you know all values except $K_{\mathrm a2}$, it's actually not that bad.

$$
\underbrace{[\ce{H+}]}_\pu{0.1 mol/l} = \underbrace{[\ce{OH-}]}_\pu{10^{-13} mol/l}
+ \underbrace{\frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}}_\pu{0.09 mol/l}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \underbrace{\frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}}_\pu{10^{-5} mol/l}
$$

$\ldots$ I'll leave units from here on (everything is in $\pu{mol/l}$). This simplifies quite much into:
$$\begin{align}
0.1 &= 10^{-13} + 0.09 + \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}} + 10^{-5}\\
0.1 - 10^{-13} - 0.09 - 10^{-5} &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01^* &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01\,(0.1 + K_{\mathrm a2}) &= 0.09\,K_{\mathrm a2}\\
0.001 + 0.01 K_{\mathrm a2} &= 0.09 K_{\mathrm a2}\\
0.001 &= 0.09 K_{\mathrm a2} - 0.01 K_{\mathrm a2}\\
0.001 &= (0.09 - 0.01)\,K_{\mathrm a2}\\
0.001 &= 0.08\,K_{\mathrm a2}\\
K_{\mathrm a2} &= \frac{0.001}{0.08}\\
&= 0.0125
\end{align}$$

* I rounded to $0.01$; it was actually $0.00999\ldots$ which would lead to $\pu{0.0124861... mol/l}$

[1] Equations 2.1 and 2.2
[2] Assumption 1 for HCl, Assumption 3 for acetic acid


How to differentiate hydrous and anhydrous compound?

There is no general answer to this question.
Heating is probably the best approach. It is not the only one. For some compounds, heating the hydrated variety produces an emission of water vapor, and nothing else, like $\ce{CuSO4·5H2O}$ or $\ce{Na2CO3·10H2O}$. For others compounds, it may produce a total decomposition like $\ce{AlCl3·6H2O}$, which is transformed into $\ce{Al(OH)3 + HCl}$ at high temperature. Considering the corresponding anhydrous compounds, $\ce{CuSO4}$ is decomposed into $\ce{CuO + SO2 + O2}$ at $650$°C. $\ce{Na2CO3}$ resists decomposition up to $1200$°C. $\ce{AlCl3}$ sublimes at $177$°C. You see that there is no general rule.
You may also check the behaviour of your hydrated compound with substances that react violently with water like thionyl chloride $\ce{SOCl2}$. if there is a reaction, it is probably a hydrate. But it may also be an anhydrous compound that react with $\ce{SOCl2}$, a metal for example, or an organic acid.
If you want a better answer, please give us the names of the particular compounds you are interested in.


Is the activation barrier of a reaction related to the free energy or heat of formation of its reactants?

They are certainly related, but not necessarily the same. If all bond-breaking and forming would be sequential than the activation energy would equal the bond-breaking energy.

However, it is possible in some reaction mechanisms that the forming bonds 'donate' energy to the break-up and therefore the overall activation energy will be lower. An example is the reaction of hydrogeniodide to hydrogen and iodine show in the figure below. Instead of a full dissocation, two molecules of $\ce{HI}$ line up and the breakup occurs simultaneously with the formation of the new bonds thus allowing a lower energy level.

The only thing you can safely say is that the sum of breaking energies is an upper bound for the activation energy, because single atom level is 'as bad as it gets' (excluding nuclear reactions).


Why does an acid salt possess acidity as well as basicity?

This is because Barium Hydrogen Phosphatev ($\ce{BaHPO4}$) is an amphoteric species:

  In chemistry, an amphoteric species is a molecule or ion that can
  react as an acid as well as a base

(Wikipedia)

Whether it acts as a base or as an acid depends on the medium the salt is dissolved. In a pure water solution, it's an acid, since the following occurs:

$$\ce{BaHPO4 + H2O <=> Ba^2+ + PO4^3- + H3O+}$$

The phosphate ion also has its reactions with water forming other species:

$$\ce{PO4^3- + H3O+ <=> HPO4^2- + H2O}$$

$$\ce{HPO4^2- + H3O+ <=> H2PO4^- + H2O}$$

Since these are all weak bases, you won't "consume" all $\ce{H3O+}$ and the solution will be acid.

If, however, you add a strong acid, the scenario changes:

$$\ce{BaHPO4 + H3O+ <=> Ba^2+ + H2PO4^- + H2O}$$

You'll still have all phosphate species as above, but now Barium Hydrogen Phosphate has acted as a base, since it received some hydrogen atoms, which is a Brønsted-Lowry base definition.

Remember the same kind of behaviour happens with water, even if pure, because of self ionization:

$$\ce{2H2O <=> H3O+ + OH-}$$

Here, water acted as both the base and the acid, which is why it's an amphoteric species.


How does Al-Hg react with carbonyl in presence of concentrated HCl?

In comparision to Clemmensen reduction, both $\ce{Na-Hg}$ and $\ce{Al-Hg}$ cannot be used as direct substitutes for $\ce{Zn-Hg}$.
Reaction with $\ce{Na-Hg}$ amalgam:
$\ce{Na-Hg}$ amalgam is also a good reducing agent and is capable of reducing ketones/aldehydes to alcohols1. However, Do not use conc. $\ce{HCl}$ with $\ce{Na-Hg}$ amalgam, as there would be a vigorous reaction between the acid and the sodium in the amalgam. Instead, use an organic solvent such as absolute alcohol.
The mechanism for this reduction is a bit different than standard Clemmensen reduction:

\begin{align}
\ce{2(C6H5)2CO +2Na &->2(C6H5)2C-ONa} \\
\ce{2(C6H5)2C-ONa&<=>(C6H5)2C(ONa)-C(C6H5)2-ONa\tag{1}} \\
\ce{2(C6H5)2C-ONa + 2C2H5OH &-> 2(C6H5)2C-OH + 2C2H5ONa} \tag{2a} \\
\ce{2(C6H5)2CO &-> (C6H5)2CO +(C6H5)2CHOH \tag{2b}}
\end{align}

Reaction with $\ce{Al-Hg}$ amalgam:
Aluminium amalgam, like Sodium amalgam, vigorously reacts with water, and you cannot use concentrated $\ce{HCl}$ with it. With ethanol as a solvent, it is a mild reducing agent, can reduce cycloalkanones and aldehydes to alcohols. Acyclic ketones remain almost inert2

In some cases, it behaves similarily to $\ce{Mg-Hg}$ amalgam and promotes dimerization of ketones to form pinacols. Thus, your reaction would not take place, and instead pinacol would be the major product.

This question and this one cover some other reduction uses of $\ce{Al-Hg}$, such as reduction of imines to amines.
References:

Bachmann, W. E. “The Mechanism of Reduction by Sodium Amalgam and Alcohol. I. The Reduction of Aromatic Ketones to Hydrols.” Journal of the American Chemical Society, vol. 55, no. 2, Feb. 1933, pp. 770–74. doi:10.1021/ja01329a051.
Troyansky, Emmanuil I., and Meghan Baker. “Aluminum Amalgam.” Encyclopedia of Reagents for Organic Synthesis, John Wiley & Sons, Ltd, 2016, pp. 1–6. doi:10.1002/047084289X.ra076.pub2.


How to now if the OH group on the C5 carbon of a cyclic sugar is up or down in Fischer Projection

The OH group you've circled is always on the right hand side for D-sugars and on the left for L-sugars as its position is in the very definition of D/L sugars.


The issue is: what parameters can we introduce? As for changing the form of the trial wavefunction to something else to achieve a lower energy, this is prohibited by the variational principle, which HF theory satisfies.
In short (and within the context of Hartree-Fock), the variational principle guarantees that any given trial function will produce an energy equal to or greater than the solution provided by Hartree-Fock.


How can residual entropies be reconciled with the Third Law?

First of all, entropy is an absolute value; you don't use different reference scales and reference states for it. Internal energy is another story: it never appears in the equations as an absolute value, but only as a difference of starting and ending values, hence you may define a scale for it and place zero on that scale more or less arbitrarily. Ditto for enthalpy and Gibbs energy. But entropy is not like that.

Now to the point. The only thing that has zero entropy at $0\,\mathrm{K}$ is an ideal crystal, which is also the equilibrium state (whatever that might mean) of crystalline compounds at that temperature. Getting to equilibrium in fluids might be anything from slow to very fast; in solids it is usually rather slow, and in solids near absolute zero it is so slow that the Sun might well become a red giant before the last defects leave your real crystal. See, thermodynamics is all about eternity. Wait until then, and you'll see the Third law at work.

That's what we mean when we are talking of residual entropy.

   I know about the possible degeneracy of the ground state, but this post is already getting too long.

Does the said law have no bearing on earthly things, then? Why, it does. Many crystals just get good enough while growing at reasonable temperature. They have very few defects, hence their entropy is already pretty low and gets lower as we cool them down. On the other hand, many other crystals are not that good.

So why would the book list entropy of all compounds at $0\,\mathrm{K}$ as zero? Well, getting to absolute zero is infinitely hard, and getting close to it is pretty hard too. Maybe the authors just put all those zeros there in a well-intentioned attempt to remind the readers of the Third law? Too bad for those who by that point have heard of residual entropy.

Then again, I don't see why would $\ce{CO2}$ have much of residual entropy in the first place. It is a good rigid molecule with nice polar bonds and without nasty things like nuclear spin. Perhaps its entropy near the absolute zero is really $0.0$ within the given precision. Water ice is another story, with all its hydrogen disorder. But wait, isn't it transforming into a hydrogen-ordered form at low temperatures? So maybe the zeros are true, after all?


Why is the polar-nonpolar electronegativity difference defined to be 0.4?

Both the $0.4$ threshold for nonpolar/polar bonds and the $1.7$ threshold for polar/ionic bonds are arbitrary choices. When defining nonpolar, polar and ionic bonds, it is important to allow $\ce{C-H}$ bonds ($\Delta \chi = 0.4$) to be nonpolar — it is the best description of alkanes’ low polarities and it facilitates discussions so much. Likewise, a few compounds commonly known as salts should fall just within the ionic boundary for ease of discussion. Teaching introductory levels is just so much easier if you can draw things black and white.

At higher levels, it should always be remembered that not only the thresholds are arbitrary but even electronegativity as a concept follows quite arbitrary definitions.

Except for one electronegativity scale that links electronegativity to ionisation enthalpy and electron affinity, all definitions and all defined reference points (typically fluorine having $\chi = 4.0$) are arbitrary.
Choosing a definite value for a specific element is also arbitrary since the physical implication of electronegativity depends on:

bonding partners
structure (i.e. hybridisation)


Butane blowtorch condensing liquid on heating surface

This is a guess - but I think it might be water formed as a byproduct of butane combustion precipitating on the plate:

$$
\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)}
$$

Aluminum has very high thermal conductivity, so it is possible that the plate is rapidly absorbing heat from the water vapor, causing it to condense. If this stops happening right around the point when the aluminum temperature reaches 100 C, that would be evidence to support this hypothesis.


What's the relation between the octet rule and quantum exclusion and how is it related to oxidation state?

What in the quantum nature of electrons gives rise to the octet rule?

Ultimately, the octet rule is justified by the fact that in many cases an $n s^2 n p^6$ electron configuration is the most energetically favorable one. Recall that electrons are arranged in different electron shells around the nucleus. Each shell corresponds to a principal quantum number $n$ and can hold only a certain number of electrons, namely up to $2n^2$. Each shell is composed of subshells each of which corresponds to an azimuthal quantum number $l$.

Now recall also that the binding energy of the electron decreases with both $n$ and $l$, and thus, once $n s$ and $n p$ orbitals of an $n$-th shell are filled, in many cases it is indeed energetically unfavorable to fill the subsequent orbitals: $n d$ if (there are any) or $(n+1)s$. The energy gain from binding an electron to an atom might not be enough to compensate a cost of separating it from a different atom.

  How is it related to the fact that an element cannot have an oxidation
  state higher/lower than the complementary to 8?

Suppose an atom has $x$ electrons in the outermost shell, where with the exception of transition metals $x \leq 8$. In the limiting cases the atom will either gain $8-x$ electrons to completely fill $ns$ and $np$ subshells or lose all $x$ electrons. In the first case we will end up with the atom in $-(8-x)$ oxidation state while in the second case with the atom in $+x$ oxidation state. But these are only the limiting cases, while in principle atom can gain few electrons and loose few others at the same time. Besides, atoms of the same type can evenly share electrons with each without satisfying the octet rule without gaining or loosing electrons. Thus, usually elements have few possible oxidation states, though, in the above mentioned limiting $[-(8-x), +x]$ range.

Excerpt from Wikipedia's list of oxidation states of the elements.


How does black CuO impart a green color to glazes and glass?

This is due to the oxidation of cupric oxide.

The main componds believed to cause this colour comprise a mixture of 3 compounds: $\ce{Cu4SO4(OH)6}$ (green); $\ce{Cu2CO3(OH)2}$ (green); and
$\ce{Cu3(CO3)2(OH)2}$(blue).

The following reactions are believed to take place:

Copper (I) oxidised to the black copper (II) sulfide ($\ce{CuS}$)  in the presence of sulfur impurities. Under accelerated conditions, the described process occur at a faster rate.

$\ce{CuO}$ and $\ce{CuS}$  reacts with carbon dioxide ($\ce{CO2}$) and hydroxide ions ($\ce{OH-}$) in water (in presence of air) to form $\ce{Cu2CO3(OH)2}$.

The extent of humidity and the level of sulfur have a significant impact on how fast the compounds develop, (under controlled conditions these reactants are varied to produce a favourable hue) as well as the relative
ratio of the three components.

$$\ce{2CuO + CO2 + H2O → Cu2CO3(OH)2~~~~~~~(1)}$$

$$\ce{3CuO + 2CO2 + H2O → Cu3(CO3)2(OH)2~~~~~~~~(2)}$$

$$\ce{4CuO + SO3 +3H2O → Cu4SO4(OH)6~~~~~~~~~(3)}$$

References

http://www.wskc.org/documents/281621/282063/ENGAGE_E3S_Chemistry_Statue+of+Liberty.pdf/e4f24c7e-3666-425e-9c41-7dbdd0065eb4


Oxidation State in Tetraaquadichlorochromium(III) chloride

I think the easiest way to see this is to count the formal charges on the ligands.

The four waters do not have net charge.  The remaining ligands--the three Cl's--are likely $\ce{Cl-}$, and so their total charge contribution is $-3$.

The only way the complex can be neutral is if the chromium has a $+3$ charge:  hence Chromium (III).


What is the origin of the colour of azo dyes?

Yes, it is all about the absorption of light at specific wavelength.

Azobenzene, the parent compound has an absorption maximum around $\lambda$= 430 nm in the visible spectrum.

The interesting part is: The absorption can be tuned by substitution of the arenes. This is done before the azo coupling.

Some examples are Allura Red (1), Chrysoine Resorcinol (2), Janus Green B (3) and Direct Blue 1 (4).

The colours of the molecules are chosen to resemble the colour in solution.

In order to make the dyes soluble in water, either sulfonate groups are attached or the dye molecule is just one fat cation.

As already mentioned by GM in another answer, +M substituents elongate the $\pi$ system (auxochromic effect).


How to add hydrogens to a crystal structure?

I've used Reduce before, which is excellent. I'm not sure what you mean by ' does not always seem to be giving the correct result', though? Perhaps you could file a bug report with the PyMOL team if you think there is a bug in h_add.


Lemon juice contains sugars (carbohydrates) and acid (lemon acid). Sugars turn into carbon+water when heated in presence of acid. This is a standard thin layer chromatography analysis for sugars (spray plate with acid and heat up - black spots are sugars). The reaction is

C12H22O11 (sucrose) + H$^+$ (catalytic) -> 12C (charcoal) + 11 H2O (steam).

Paper also has polysugars (cellulose), but cellulose is more robust. Eventually even paper will turn black by the same reaction.

You can use sugar + vinegar as the ink. Result will be the same. It proves that decomposition of sugars in presence of acid is the reaction responsible for revealing this ink.


If the polarity of the solvent increases, will the Rf value increase or decrease?

I wish we would stop teaching chromatography in terms of "polar" and "nonpolar."

The aspirin will interact fairly strongly with the silica due to hydrogen bonding/electrostatic interactions of the carboxylic acid and the ester with the silica. If you increased the "polar" component of the mobile phase, it would travel further due to the mobile phase competing more for binding spots of the silica gel.

So increasing the polarity of the mobile phase does raise $\mathrm{R_f}$ of aspirin (or anything) on regular silica gel.


The reaction is exothermic left to right and endothermic right to left. Yes, as you noted, it's wise to note which way you are looking at the reaction, but generally, we look at reactions from left to right (because the arrow generally points left to right). For example, the following reaction is endothermic as written, i.e. very hot water (steam) is very corrosive and adding heat to water encourages further ionization of water.

$$\ce{2H2O ⟶ H3O+ + HO-}$$


Visualizing PDB file in VMD, how to upload molecules?

I found a shortcut, and that is if I go to (click on):

file
New Molecule
Enter in the "browser" window the direct PDB accession code "3WD5" (for my protein) then VMD retrieved the protein structure form the PDB database on its own, and is showing it as it should in the visualization window.

However, this does still not resolve why it will not load a downloaded PDB file, but I suspect its most likely due to some problem with downloading of the files or installing the program its self that has gone wrong.


Yes, adding additional solute to a super saturated solution can be done without crystallization.

Adding additional solute to a super saturated solution and preventing crystallization is dependent on temperature, degree of agitation, and rate of solute addition. Changing the temperature is usually the easiest way to control the degree of super saturation, but you can experiment with the other variables as well. The degree to which these three variables affect any given system is dependent on the identity of the system and the amount of impurities (generally you do a lot of work to minimize impurities before you create a supersaturated solution).


Find out of the friend is getting a good deal

The jewelry gold usually comes historically rated in grades 14 or 18 karats, where 24 means pure gold. The rest is mostly silver.

Find out specific heat capacity for the complementary metals like silver or copper, so you could roughly estimate via interpolation.

Another method would be pyknometry, determining the ring density by differential weighting the mass of displaced water - old good method of Archimedes. Gold has almost double density, compared to other expected metals.

Edit:  list of densities:

Gold 19.3.
Silver 10.5.

If linearly interpolated by mass fraction:.

14 karat gold  15.63.
18 karat gold  17.1


The use of concentrated H2SO4 is not amenable. Per a source:

Concentrated sulphuric acid produces messy results. Not only is it an acid, but it is also a strong oxidising agent. It oxidises some of the alcohol to carbon dioxide and at the same time is reduced itself to sulphur dioxide. ... It also reacts with the alcohol to produce a mass of carbon.

On the other hand, per Wikipedia, to quote:

  The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides via an SN1 reaction:[3]...The differing reactivity reflects the differing ease of formation of the corresponding carbocations. Tertiary carbocations are far more stable than secondary carbocations, and primary carbocations are the least stable.

And further:

  An equimolar mixture of ZnCl2 and HCl is the reagent. The alcohol is protonated by this mixture, and H2O group attached to carbon is replaced by the nucleophile Cl−, which is present in excess. Tertiary alcohols react immediately with Lucas reagent as evidenced by turbidity owing to the low solubility of the organic chloride in the aqueous mixture. Secondary alcohols react within five or so minutes (depending on their solubility). Primary alcohols do not react appreciably with Lucas reagent at room temperature.[3] Hence, the time taken for turbidity to appear is a measure of the reactivity of the class of alcohol, and this time difference is used to differentiate among the three classes of alcohols:

Now, with respect to HClO4, there is at least one catalytic study of HClO4-SiO2 during the reaction of benzyl alcohol (1 mmol) and thiophenol (1.1mmol) at room temperature under solvent-free conditions (see).


Determining the chemicals in potash and crystallizing them

I think it would be quite difficult to separate all of the different compounds dissolved in the sample into their solids because they are all in the same phase. I do not think any method exists for this kind of separation.

However, determination is absolutely possible.

Firstly, regarding your deductions...

Although carbonates give carbon dioxide gas, which is colourless and odourless, it is nevertheless important to test that the gas is indeed carbon dioxide by bubbling it into limewater and looking out for white precipitate. If you have not conducted that test, then you cannot claim that the gas produced is carbon dioxide. Otherwise, you can conclude that there is carbonate in the sample.

The high pH is indeed indicative of high hydroxide ion concentration in the sample. That is for certain.

However, you have not proved that the cation is indeed potassium. Perhaps, you want to conduct a flame test or do analysis using atomic absorption spectroscopy (AAS). Procedure for flame test can be found here: https://www.soinc.org/sites/default/files/uploaded_files/flametest.pdf

If you would like to find out what other ions are present in the sample, carry out a simple qualitative analysis using some common lab reagents. Procedure can be found here: http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis

By boiling off the water, you would likely get some ionic solids but I believe that it should contain a lot of impurities.


What is the precise definition of standard molar entropy?

Not quite an answer, but too long for a comment:

Here's a link to the official CODATA site showing the values:
codata.info (via the Internet Archive).
It references the following source:
Cox, J. D., Wagman, D. D., and Medvedev, V. A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1989.
Unfortunately, that source isn't readily viewable online (all I found was what appears to be a German version, at Berichte der Bunsengesellschaft für physikalische Chemie 1990, 94 (1), 93–93 (It's behind a paywall.)  But if you could get your hands on it (interlibrary loan, say), it should contain a detailed description of how the values were actually calculated.

FWIW, here's a link showing how standard molar entropies might be calculated (but I don't know if the procedure described here is what is used to determine the official CODATA values): Standard Molar Entropy of Solid Aluminum Oxide, Comments to Tandy Grubbs, Stetson University (via the Internet Archive).
Essentially, it mentions using the Debye extrapolation I mentioned in an earlier comment up to $\approx\pu{15 K}$, and then using a differential scanning calorimeter to determine the reversible heat flow between $\pu{15 K}$ and $\pu{298.15 K}$

And here's a purely theoretical approach (unfortunately, everything except the abstract is behind a paywall):
Pascal, T. A.; Lin, S.; Goddard Iii, W. A. Thermodynamics of liquids: standard molar entropies and heat capacities of common solvents from 2PT molecular dynamics. Phys. Chem. Chem. Phys. 2011, 13 (1), 169–181. DOI: 10.1039/C0CP01549K.

Also, some general comments on the issue of the entropy of a substance at absolute zero:

It is possible the question of whether substances need to be perfect crystals to have zero entropy at $\pu{0 K}$ does not enter into the calculation of standard molar entropies.  Consider the following:

Any calculation of standard molar entropy requires determining the initial reversible heat flow starting at $\pu{0 K}$, at standard pressure:
$$
S^\circ(\pu{298.15 K}) =
\int_{0}^{\pu{298.15 K}}\frac{\mathrm{đ}q_\mathrm{rev}}{T}
$$
The initial part of this integral (starting at $\pu{0 K}$) can't be done experimentally, and thus has to be based upon a model.
We may not know how to model whether a substance being a perfect crystal at $\pu{0 K}$ will affect the amount of reversible heat flow needed to increase its temperature.  In addition, we may not know how to predict the actual state of substances at $\pu{0 K}$ (i.e., the extent to which they are not perfect crystals–and to the extent that they are not, what the distribution of their forms actually is).
Further, even if there were an effect on heat flow from not being a perfect crystal, and we did know how to model it, the effect may be small compared to the experimental errors that are present in determining standard molar entropies.
It's still an open question whether a substance needs to be a perfect crystal to have zero entropy at $\pu{0 K}$.  Some in the field argue it does, some argue it doesn't.  Those in the latter camp argue that, given that no other states are accessible at $\pu{0 K}$, the fact that the system is locked into only one microstate (even if it's not a perfect crystal) gives it zero entropy.


It isn't.  We're just used to working in the lab where solutions are generally 1 M or less in concentration.  So we get no more than 1M (solvated) hydrogen ions where pH is 0, or at the other end no more than 1 M hydroxide ions corresponding to pH 14.

These 1 M limits are, however, just for our convenience and not any law of nature.  Right in our own labs we likely have concentrated acid stock solutions, well above 1M, where if we attempted to measure pH we would get less than zero.  Acid pickling solutions in the steel industry are concentrated enough to do the same thing, as manufacturers are more interested in getting hot mill scale cleaned off than in a convenient pH scale.  Bases like sodium hydroxide also have concentrated stock solutions with pH above 14.


Giving too much energy to an exothermic reaction

There are two separate effects we need to consider here: thermodynamic and kinetic.

Let's assume you are only providing the energy thermally. So providing more energy means increasing the temperature (T).  Thermodynamically, if a reaction is exothermic, and you increase T, the reaction becomes less favorable (assuming it stays exothermic over that temperature range) according to, as you mentioned, Le Chatelier's principle.  I.e., the equilibrium shifts to the left.

However, we also need to consider kinetic effects.  Here, as you increase T, the rate of reaction increases, and thus the rate of thermal energy production increases, thus further increasing T, thus further increasing the rate, and so on, leading to the thermal runaway effect that Buck Thorn mentioned in his comment.  I.e., we get a positive feedback loop.


Strange Sticky Substance on Digital Camera

After the lengthy back and forth perhaps an official answer is due: the blue salt deposit on the camera battery holder door is most likely copper(II) hydroxide by reaction of leaked alkaline battery electrolyte (likely $\ce{KOH}$) with copper metal in the battery door contact. It could possibly also consist of copper(II) carbonate hydroxide formed by reaction of the copper(II) hydroxide with carbon dioxide, but the deep blue rather than green color in the picture suggests it is the previous. Copper hydroxide is regarded as moderately hazardous, but you are unlikely to be at risk unless you ingest it. Feel free to consult a MSDS for this compound (such as here) for more information.

The sticky substance coating the camera is possibly a degradation product of plastic components in the camera housing. It is not likely to be toxic unless ingested or if you are in contact with it persistently.

Without more information about how the camera was maintained it is difficult to say much more.


What happens with ammonia ions in tap water after they are separated from chlorine?

Chloramines (mono-, di-, and tri-) are used as a disinfectant for drinking water treatment (the source water, not the tap water) - this process is called combined chlorine. Upon contact with pathogens, organics, etc., chloramine is reduced to chloride (not chlorine), ammonia, and hydroxide. However, this half reaction is for alkaline conditions. The pKa of ammonia is around 9.4, so around circumneutral pH, some ammonia will exist as ammonium, the protonated state.

So, will activated carbon (AC) remove the products? First, we need to clarify something about AC - it is not catalytic, it is a physical adsorption removal process. AC will remove chloride ions, but it will not remove ammonium. Ammonia is quite volatile and will move from the water to the gas phase if e.g., a pitcher of water is left at room temperature. If you wanted to remove ammonium as well you could use something called biologically activated carbon (BAC). One last note, chloride will not "bind" will AC, it will be adsorbed through electrostatic forces.


If an electron moves closer to the nucleus in a given value of n, does it lose energy?

You are trying to apply macroscopic understanding to the world of quantum mechanics. This thinking is doomed to fail. The energy of an electron occupying a specific orbital is that orbital’s energy. Full stop. Unless it changes its orbital it does not change energy.


Pyrex glass labware can be dropped, scratched, even used if cracked (I've started filling a beaker, then noticing that there was a crack - but it still seemed OK).
Borosilicate glass for commercial (non-lab) use is often tempered, so that a scratch is a potential hazard - the item might fail unexpectedly. The issue is not so much the borosilicate but the tempering.


Why is dU an exact differential and dq an inexact differential?

As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$.
And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$.

Consequently, for a process in which a system goes from state $1$ to state $2$ a change in a state function $a$ can be evaluated simply as
$$\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} \, ,$$
while a change in a path function $b$ can not be evaluated in such a simple way,
$$\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \, .$$
And for a state function $a$ in a thermodynamic cycle
$$\oint \mathrm{d} a = 0 \, ,$$
while for a path function $b$
$$\oint \text{đ} b \neq 0 \, .$$
The last mathematical relations are important, for instance, for the first law of thermodynamics, because while $\oint \text{đ} q \neq 0$ and $\oint \text{đ} w \neq 0$ it was experimentally found that $\oint (\text{đ} q + \text{đ} w) = 0$ for a closed system, which implies that there exist a state function $U$ such that $\mathrm{d} U = \text{đ} q + \text{đ} w$.


Why is hydrogen bonding more significant than any other interaction between dipoles?

Hydrogen bonds often have a significant covalent component. This can be seen in many aspects of the hydrogen bond: it is directional, unlike Coulombic interactions, and the bond length is generally significantly shorter than predicted by Coulombic models. There is great variation in the covalent character of hydrogen bonds depending on their environment, with e.g. bond lengths between 2.2 and 4.0 Angstroms known.

Other intermolecular interactions have significant covalent components; there is nothing special about hydrogen per se but it is the most important example.


Why does aluminum foil "electrocute" dental fillings?

Ouch...  actually a known phenomenon.  Between the foil of aluminum/Al, and the metal filling of amalgam in your tooth, you created a little electrochemical cell.  Your saliva served as an electrolyte and your nerves around sensibly detected the current. This already "works" with little splinters of Al foil wrapping chocolate (obviously, it was not a small Hershey bar), yet is less powerful than the said 9 V block.

(one reference)


What are good books about MO theory that don't require a lot of math background?

I'd suggest the seminal Orbital Interactions in Chemistry by Albright, Burdett and Myung-Hwan Whangbo.

The second edition came out in 2013 and I'm sure there are many copies of the first edition around as well.

They focus a lot on pictoral MO theory - indicating how orbitals of different shapes and symmetry combine.

To truly understand MO theory at a fundamental level, it really helps to study some quantum mechanics (and thus mathematics) but you can grasp a lot from the approach used in this book.


Unfortunately, many aqueous endothermic reactions involve some chemicals you may wish to avoid.  Ammonium nitrate is the classic example, but in the US it is becoming more and more unavailable because of its potential use as an explosive.  Another frequently mentioned possibility is reacting hydrated barium hydroxide with ammonium chloride, but $\ce{Ba(OH)2}$ is very caustic and poisonous.

So of what I've read, your best choices might be mixing water with a chloride salt like ammonium or potassium.  I do not know how cold this will get your container. I suspect it will not be as good as ammonium nitrate, which would probably be ideal given the former use of this compound in medicinal cold packs.


Classification/Nomenclature of aroma and flavour compounds - terpenes, aldehydes, and alcohols, etc

Perhaps by classification/nomenclature you are referring to odour descriptors. If I remember correctly I have already covered a similar question about possibility of defining smell, so I will extract relevant parts here:

Odour descriptors and odour profiles

Odour descriptors are the words that come to mind when smelling a substance.
The more generally understood the words are, the more useful they are as descriptors. In practice, it is easy for any observer, after a little training, to use the standard descriptors of fragrance chemistry. Example of descriptors include musky, camphoraceous  etc

Smelling chemical groups

A fact that has, in our opinion, received too little attention from olfaction researchers is the ability of humans to detect the presence of functional groups with great reliability.

The case of thiols ($\ce{-SH}$) is familiar, but groups ($\ce{NO2}$), aldehydes ($\ce{C=O(H)}$), can be reliably identified once the odour character the functional group character confers is known. When nitriles are used as chemically stable replacement for aldehydes, they impart a metallic character to any smell: cumin nitrile smells like metallic cumin (cuminaldehyde), citronellyl nitrile smells like metallic lemongrass (citronellal), and nonadienylnitrile smells like metallic cucumber (nonadienal). Oximes give a green-camphoraceous character, isonitriles a flat metallic character of great power and unpleasantness, nitro groups a sweet-ethereal character,etc.

Here are some odour categories and their representative molecules, chosen to illustrate
structural diversity:

Musk

  Musk odour descriptors might be “smooth clean, sweet and powdery”. The
  molecules that possess this odour character are exceptionally diverse
  in structure. Macrocyclic musks contain a 15-18 carbon cycle closed
  either by a carbonyl or by a lactone and smell similar but fresher and
  more natural, often with fruity overtones (cyclopentadecanolide,
  ambrettolide). Nitro musks, discovered originally as a byproduct of
  explosives chemistry, smell sweeter and are reminiscent of
  old-fashioned barbershop smells.

Representatives from five chemical classes which yield musk odors. 1 androst-16-en-3a-
ol, a steroid musk. 2: ambrettolide, a macrocyclic musk. 3: Musk Bauer, a nitro musk. 4: Tonalid, a
tetralin musk. 4: Traseolide, a indane musk.

Ambergris

  Originally derived from concretions spat out by whales and aged in the
  sun, ambergris odorants smell nothing like natural ambergris tincture,
  which has a weak animalic marine smell. The smell of ambergris
  odorants was once aptly described to us by a chemist-perfumer as
  “glorified isopropanol”. Ambergris odourants provide an interesting
  combination of very closely related smells with widely different
  structures: amberketal, timberol, karanal and cedramber

Two ambergris odorants, timberol (left) and cedramber (right)

Bitter almonds

This easily-recognized category is interesting  because it includes a small molecule (HCN) which, however, is perceived by a large fraction of observers  to smell as metallic not almond-like to. Benzaldehyde, nitrobenzene,trans-2-hexenal (but see above) are good examples.

Green

  Cut grass, fresh green bean notes with a sharp, almost aggressive
  feel. Diverse compounds possess this descriptor, ranging from classic
  grassy notes of cis-3-hexenol and ligustral, to the cucumber peel of
  nonadienal and the bell-pepper green note of some pyrazines.

Some examples of green odorants. Clockwise from top left cis-3-hexenol, ligustral, nonadienal and ethylmethoxypyrazine.

Camphoraceous

Camphor, cyclooctane, cineole are good
examples of camphoraceous smells, and smell rather similar to each other.

Three camphoraceous odorants: 1,8 cineole, camphor and cyclooctane

None of the above

  Many other categories such as musty, spicy, aldehydic, lactonic,
  indolic, marine, etc exist, each with subdivisions. It must be
  emphasized that the odor categories above are merely convenient
  descriptors and only cover a very small fraction of odor “space”. In
  fact, particularly when one steps out of perfumery materials proper
  into smells noticed by chemists in the course of organic and inorganic
  syntheses the most frequent descriptor appears to be sui generis, i.e.
  a smell associated with nothing in particular

The complexity of structure-odour relations, and the fact that the three dimensional structure of the receptor site is unknown, make it very difficult to apply conventional quantitative structure activity relationships. For detailed SAR and theories please refer to above linked post.

References

Structure-odour relations: a modern perspective: Luca Turin et al


Question about oxygen and beta carbons in chemical shifts

I think perhaps you are misunderstanding some basic principles of determining chemical shifts. I have no idea where you are getting your +0.5/+0.6 factors

Chemical shifts are a product of all contributing functional groups; those that contribute to the local electron density either through direct electronic effects or through steric effects. Looking at the 3 proton environments, we can roughly predict their chemical shift using additive tables (I use Pretsch, Buhlmann and Badertscher).

First, let's look at the types of protons. Methyl protons and methylene protons do not have the same basis chemical shift to work from when doing additive calculations; the shielding from a C-C bond is not the same as for a C-H bond, and so the local shielding for the proton of interest will be different. Generally, when doing additive calaculations for chemical shifts, we can use the starting shifts of 0.86 (-CH3), 1.37 (-CH2-) and 1.50 (>CH-).

For this molecule, there are 3 proton environments. I have shown their predicted chemical shifts below.

For methyl protons at site 1

Each -CH3 proton is alpha to a quaternary C, beta to an oxygen-bound ester and beta to a carbon-bound ester.
So that gives us 0.86+0.05+0.44+0.07= 1.42ppm

For methylene protons at site 2

alpha to a ether oxygen, alpha to a quaternary carbon, beta to an oxygen-bound ester and beta to 2 methyl groups.
So that gives: 1.37+2.04+0+0.24-0.06-0.06= 3.53ppm

For methylene protons at site 3

alpha to ether oxygen and alpha to carbonyl-bound ester. The ester oxygen is taken into consideration as part of the ester group as a whole. Also, the atom on the other side of the ether (in this case the carbon) is taken into consideration - The only important thing here is whether the ester is oxygen-bound or carbonyl-bound.
So, that gives us 1.37+2.04+0.92= 4.33ppm


Yes, carbon-2 is a stereocenter, and by (arbitrarily) giving the R conformation higher priority than the S, we can assign its stereochemistry using s/r notation:

$\hspace{2cm}$

Pseudochirality simply means that a molecule has two substituents of a stereocenter are mirror images of each other, and the other substituents are either achiral or also mirror images of each other. Despite having multiple stereocenters, pseudochiral molecules are ahciral (optically inactive) because the mirror images of both the s and r conformations are identical.


Where is the center of symmetry in the PbO crystal?

The best thing to do is load up the CIF at the Crystallography Open Database (COD), e.g.
http://www.crystallography.net/1010977.html?cif=1010977

JSmol used for the visualization lets you visualize the symmetry and adjust the packing. In this case, look for the inversion center, e.g.:

The inversion center itself is the yellow ball. But it's probably easier to see it in the inverted Pb centers I've circled in yellow. You can see that there's an exact inversion center there.


If polyethylene glycol (PEG) is hygroscopic, how does it protect wood?

The reason why PEG is suitable for this purpose has many reasons including the wood chemistry and degradation of wood in water.

Degraded wood usually very characteristic chemical losses. The cellulose of the wood is usually lost due to bacterial action and/or due to the action of the water itself. Over long period of time wood looses water-soluble components like starch and complex sugar like molecules to water. What remains in there is the water-insoluble component in the cell walls and wood tissues primarily consisting of lignin.

If the wood is in water, lignin provides structural stability to the wood. But wood becomes spongy and easy to break. If this wooden piece is exposed to air, water evaporated and wood becomes dry and shrinks. Most of the spaces between lignin layers is nor air, which makes it easy to break.

Here, comes in PEG. PEG is remarkable because it can dissolve in water.
PEG solution has very high osmotic pressure and this is one of the reason why PEG solution treatment works. PEG seeps in the tissues and provides the required osmotic pressure to keep lignin (and wood tissue) stable. This prevents the wood from expanding and shrinkage. While treatment, it is ensured that PEG penetrates the wood tissue which takes considerable times, and water is removed from the wood. In the end a large part of the original water in wood is replaced by PEG in the tissues providing structural stability.

Coming to your point on the hygroscopic property (also known as deliquescence). Higher molecular weight PEG has lower hygroscopic property because of more alkyl chained part. So it depends on the type of PEG used. But a balance has to be made since, lower MW chains are easier to penetrate tissue.

In case the relative humidity changes, for example increases, the effect due to absorption due to PEG inside tissue is negligible. Only the PEG which is outside tissues (excess, which was not removed during PEG treatment) might cause additional increase in weight. The removal of excess PEG is usually taken care of by experts.

Still however, it is the norm that >80 % relative humidity is not good for PEG treated wood.

Reference

D.W. Grattan, "A Practical Comparative Study of Treatments for
Waterlogged Wood, Part II, The Effect of Humidity on Treated Wood",
(Ottawa: ICOM Waterlogged Wood Working Group, 1981), pp. 243-252.
Taduteru Nishiura and Setsuo Imazu, "Experimental study of the
dimensional change of highly degraded waterloged wood according to
ambient humidity after preservation treatement"Bremerhaven: ICOM
Waterlogged Wood Working Group, 1990) pp. 107-118.


Cheap but efficient disinfectant / biocide for home lab

The Amazon website gives the composition of Peressigsaure/Alkohol: Inhaltsstoffe pro 100g:

● 19g 2-Propanol

● 5g Ethanol

● <1% Wasserstoffperoxid

● <1% Peroxyessigsäure

● <1% Essigsäure

Working time 30 seconds, shelf life 12 months.

Don't expect that H2O2 plus acetic acid to give peroxyacetic acid. The full Wikipedia reference is:

Peracetic acid is produced industrially by the autoxidation of acetaldehyde:[2]

O2 + CH3CHO → CH3CO3H

It forms upon treatment of acetic acid with hydrogen peroxide with a strong acid catalyst:[3]

H2O2 + CH3CO2H ⇌ CH3CO3H + H2O

Making the peroxy acid can be done from acetic acid, but requires a strong acid catalyst (to generate CH3CO+).

So, you will have to get peroxyacetic acid from somewhere in some form, in addition to your other chemicals. Duplicating this formula requires more isopropanol than ethanol, but either alcohol should perform equally.

An overview document titled Chemical Disinfectants at https://www.cdc.gov/infectioncontrol/guidelines/disinfection/disinfection-methods/chemical.html mentions two chemical sterilants are available that contain peracetic acid plus hydrogen peroxide (i.e., 0.08% peracetic acid plus 1.0% hydrogen peroxide [no longer marketed]; and 0.23% peracetic acid plus 7.35% hydrogen peroxide. This document also lists the benefits of H2O2, peracetic acid, alcohol(s) and chlorine bleaches. And formaldehyde! About peracetic acid, it says: It is considered unstable, particularly when diluted; for example, a 1% solution loses half its strength through hydrolysis in 6 days, whereas 40% peracetic acid loses 1%–2% of its active ingredients per month.

As regards explosivity, a document of ECETOC at http://www.ecetoc.org/wp-content/uploads/2014/08/JACC-040.pdf gives the formulas of two cleaners as:

1: PAA 15%,H2O2 23%,HOAc 16%,H2O 45% and

2: PAA 15%,H2O2 14%,HOAc 25%,H2O 42%.

I gather from these two formulations that they and anything less in H2O2 or PAA would not even remotely be explosive. No alcohol was listed in these formulations.

Now if you could obtain either of these concentrated sterilants, you would have a reasonable ratio of peracetic acid, H2O2 and acetic acid. The only item missing would be the alcohol. Let's use ethanol, at 70% (or 69%, if you must). The peracetic acid cleaners would have a good shelf life because of their high concentrations. When you need to make up some sterilant, you could take ~800 grams of ethanol and add ~350 grams of the cleaner you choose to get 1 L of final sterilant (density ~1.150 g/cc), which will be 69-70% ethanol, 4.5% PAA, 4.3-7% H2O2, 7.6-4.9% HOAc, remainder H2O.

These levels are 5 or more times higher than the concentrations listed for Peressigsaure?Alkohol Pro, so you could consider adding much less of the peracid cleaner to get in the range of 1% PAA and H2O2.

Overall, it seems that H2O2 by itself is good at killing bacteria, and so is alcohol, by itself, and so is peracetic acid, by itself, but PAA is not stable for long at low concentrations (12 months at 1%??).

And perhaps it would be easier to buy the Peressigsaure/Alkohol Pro, or one of the other sterilants offered by that company, because it already comes in a spray bottle.


Which alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol?

In order to arrive at an unambiguous description of stereoisomers, the priority for the groups attached to the chirality center is established by the application of ‘Sequence Rules’. In the hierarchical Sequence Rules given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), we find

P-92.1.3.3 Sequence Rule 3
When considering double bonds and planar tetraligand atoms ‘seqcis’ = ‘Z’ precedes ‘seqtrans’ = ‘E’ and this precedes nonstereogenic double bonds.

In the given example, however, the stereochemical configuration at the central carbon atom is not defined. Note that solid wedges should be used to indicate bonds that project above the plane of the paper and hashed wedges to indicate bonds that project below the plane of the paper.


What happens to potassium iodide in the "elephant Toothpaste" reaction?

The potassium iodide serves as a catalyst for the decomposition of the hydrogen peroxide by facilitating oxygen removal from the solution so the reaction is carried to completion. The rapid decomposition allows you generate a lot of gas quickly and the long cylindrical shape of the graduated cylinder directs the momentum of the gas in a vertical direction to achieve the gushing effect. Since potassium iodide is a catalyst in the reaction, it is not consumed in the process of making the foam, but exists as the ions that come from dissolving the salt in the hydrogen peroxide solution.

Reaction mechanism on wikipedia

P.S. This reaction is exothermic, so if you go really large scale, expect your vessels and the foam to be hot to the touch.


Is the surface of glass a fine layer of silanol?

Yes, it is correct that the surface of glass has silanol ($\ce{Si-O-H}$) at the surface. This is because water will react at the surface to relieve bond strain of a $\ce{Si-O-Si}$ bond to form $\ce{Si-O-H H-O-Si}$ pairs. additionally there is sodium siloxide at the surface ($\ce{Si-O-Na}$) which will release sodium hydroxide on contact with water $(\ce{Si-O-Na + H2O -> Si-O-H + NaOH (aq)})$ and is major reason why normal glass cannot be used for biomedical implants.

Additionally these surface silanols can react with chlorosilanes to produce hydrophobic surfaces $$\ce{Si-O-H + Cl-SiR3 -> Si-O-SiR3 + HCl}$$
where the R is a non-polar organic pendant group that is hydrophobic. This coating causes water to bead off of treated glass that would priory be wetting.