Choice of base for malonic ester synthesis

Understanding how this reaction is actually performed in the lab is key to answering your question.

Typically in the malonic ester synthesis, a full equivalent (or 2 equivalents if you are carrying out a dialkylation) of base is used, so little, if any, ester remains with the $\ce{\alpha}$-proton still in place.

(image source)

Also, keep in mind that both the alkylation and Claisen reaction have an activation energy. This means that it is usually necessary to heat the solution to bring about reaction (see here for a lab example).

The reaction is carried out by first forming the deprotonated malonate ester at room temperature and then dripping it into a refluxing solution of the alkylating agent ($\ce{RX}$ in the above diagram). This procedure achieves  the following:

concentration of non-deprotonated ester is very low
deprotonated ester is formed at room temperature, so even if any non-deprotonated ester remains there is not enough energy (at room temperature) to make it over the energy barrier and react
the deprotonated ester is dripped into the solution of refluxing alkylating agent; therefore, now that there is enough thermal energy to bring about reaction, the relative concentration of alkylating agent (the entire solution) is much higher than the relative concentration of deprotonated ester (added drop by drop) making the desired alkylation reaction much more probable than reaction of deprotonated ester with protonated ester (which, for reasons explained above, is in very low concentration in any case).


Why don't electrons simply stick to the nucleus on account of being *oppositely* charged, instead of moving around it?

Richard Feynman alludes to this exact problem in the first volume of the Feynman Lectures on Physics. I'll present his argument here (based on the Uncertainty Principle), albeit, in my own words ;)

The German physicist [since he dealt with atomic chemistry/physics...by all means, go ahead and call him a "chemist" ;) ], Werner Karl Heisenberg came up with what we now know as "The Heisenberg Uncertainty Principle"... a revolutionary idea that points to an "inherent fuzziness", that exists within quantum systems (which is a fancy word that refers to any region of space that is sufficiently "small" enough to introduce wave-particle duality), and becomes apparent when we try to "measure" the various parameters that constitute them.

Basically, what (one version of) the Uncertainty Principle states is that:

  The product of the uncertainty in measurement of velocity and the uncertainty in measurement of the position of a particle can never be less than a certain constant; i.e- ħ/2

Rephrasing this

  If we were to know a particle's position to a very high precision (i.e- small uncertainty in measurement of position) then the corresponding uncertainty in the measurement of the particle's velocity (or momentum, if you know its mass) increases greatly. The same holds true vice versa.

In other words,

  You cannot know both a particle's position and momentum in a quantum system to a great precision simultaneously.

Here's a good analogy for this (Part 3: Atomic Structure - Analogy with a ballon)

From this point on, I guess the question can best be dealt with by "proof by contradiction".

  First part of the question: Why don't electrons stick to the nucleus?

We know for a fact, that Heisenberg's Uncertainty Principle holds true (yeah, I didn't prove it...but for now, just take my word for it).

So, let's assume that electrons do stick to the nuclei of their respective atoms. In that event, we would know the electrons' positions and momenta (with respect to the nucleus) with almost absolute certainty...something that Heisenberg's Uncertainty Principle does not allow.

Seeing that this situation is impossible, it is obvious that our little "assumption" must be wrong (Knowing the the Uncertainty principle is indeed, correct).

So yeah, electrons don't just go and simply "stick" to atoms.

  Second part of the question: Why do electrons follow a whole bunch of wacky paths?

The Dane, Niels Bohr postulated that electrons move in certain, fixed orbits around a nucleus...a fairly convincing (but not very accurate) model that is still often brought up in high-school and under-grad physics and chemistry courses. The Bohr Model was developed before the Uncertainty Principle.

But electrons do not move in fixed orbits ...the Uncertainty Principle triumphs again!

So if electrons don't stick to the nucleus, and if they don't move in fixed orbits around the nucleus, then where exactly are they?

We can never exactly point out an electron's position in an atom. However, we can determine the probability of finding electrons in a given region of space around the nucleus. These regions of space with a very high electron probability density are called orbitals.

This answer has been greatly over-simplified. Thought it does deal with the gist of the topic, I'm personally looking forward to seeing more detailed answers myself.


Usage of TFA in Pfitzner-Moffat oxidation and Collins vs Jones oxidation

Moffat & Pfitzner have accidentally found their famous oxidation while working on mono-nucleotides (acually 5'-phosphates). Breifly, when pyromidium thymidine-5'-O-phosphate was reacted with dicyclohexylcaebodiimide  in anhydrous DMSO at room temperature resulted a violent reaction to give 90% of 5'-aldehyde product (Ref.1). They further pursued this finding and published their finding as specific reaction (Ref.2). The acid they have used in first finding was anhydrous orthophosphoric acid, which have given 90% yield. Hence, it is not necessary to have TFA as the acid. Following is the suggested mechanism for Pfitzner-Moffat oxidation:

About Collins vs Jones oxidation, both of them used chromium-based oxidants. Oxidations with chromium oxidants proceed through a similar mechanism. Different between Collins and Jones are conditions: Anhydrous (in pyridine/$\ce{CH2Cl2}$) vs aqueous (use of $\ce{CH2Cl2}$). Following is the suggested mechanism for chromium-based oxidants:

After first oxidation to aldehyde, chromium oxidant needs $\ce{CH-OH}$ part available to further oxidize to carboxylic acid (see in both mechanism, how oxidation initiated and finished). In anhydrous conditions, this feature is lacked. So, oxidation stops at aldehyde moiety. However, trace of water make aldehyde in equlibrium with its gem-diol moiety, which gives the necessary requirement of $\ce{CH-OH}$ feature. So, in Jones oxidation condition, oxidation will further continue to give carboxylic acids.

References:

K. E. Pfitzner, J. G. Moffatt, “The Synthesis of Nucleoside-5' Aldehydes,” J. Am. Chem. Soc. 1963, 85(19), 3027-3027 (https://doi.org/10.1021/ja00902a035).
K. E. Pfitzner, J. G. Moffatt, “A New and Selective Oxidation of Alcohols,” J. Am. Chem. Soc. 1963, 85(19), 3027-3028 (https://doi.org/10.1021/ja00902a036).


How do Precipitation Reactions behave in the Absence of Gravity?

In gravityless environment, all directions are equivalent save the very vicinity of the surface(*), where just few water molecules are toward the surface.

Therefore, in the bulk volume, the precipitate would stay where it is, unless some currents exist for whatever  reason.

At the very surface, it depends on nature of precipitate.

If it is rather hydrophilic, water molecules would have tendency to surround it, so it would be rather pulled from surface inwards.

If it is rather hydrophobic, water molecules would have tendency to expelled it, so it would be rather pushed outwards to the surface.

If it is far enough ( tens molecules ) from surface, all effects will get spherically symmetric, but one....

(*)  .....  as in a very very long term, precipitate denser than water would be more concentrated near the centre by the own gravity of the "bubble" and vice versa.

I am not aware of such an ISS experiment, but it makes little sense to do it for me. In the microgravity context, it is very very long term experiment.


after some searching and a little help from google I have found that Potassium (K) Bromide (Br). In a low concentration of an aqueous solution it will taste sweet. In higher concentrations it will bitter and even salty. wiki
The toxicity of Bromide isn't very high and it can be tolerated by a healthy human body under some doses. Yet it's long half-life in the human body make it hard to avoid intoxication (as written in the wiki page).

A question arises, why does the change in concentration affect the taste? I would guess that in higher conc. the KBr distorts the water structure and thus affect the interaction of the ions and cations with the taste buds.
I wouldn't use it as a sweetner, and wouldn't suggest tasting it to see the effects of concentration change on the taste.

If we don't restrict ourselves to heteroatomic salts then Sucrose Na is a sweet salt as can be seen here: http://dx.doi.org/10.1016/0031-9384(80)90077-3


How can pKas differ across similar protons of the same atom?

Here is a picture of citric acid.

As you noted, there are 3 carboxylic acid groups in citric acid, so there are 3 acidic protons. A chemist looking at this unionized molecule would say that the carboxylic acid group drawn on the right side of the molecule is equivalent to (the same as) the carboxylic acid group drawn on the left.  Therefore, these 2 equivalent carboxylic acid groups should have the same pKa.  The carboxylic acid group drawn in the middle of the diagram is different from the other two (it is not the same as the other two) and should therefore have a different $\ce{pKa}$.  In chemistry, groups are either the same or different.  If they are the same they must be indistinguishable in all aspects, including physical properties like $\ce{pKa}$

That said, we might expect when we treat citric acid with more and more base to see only 2 different $\ce{pKa}$'s, but that is not the case.  In actuality when we treat citric acid with base one of the two different (right and left or middle) carboxylic acid groups will ionize.  Once that happens the ionized citric acid molecule contains a negative charge.  It will be harder to ionize the molecule again and remove the second proton and place a second negative charge on the molecule (bringing like charges together is destabilizing and costs energy).  The same argument tells us that removing the third and final proton will be even more difficult because we wind up with 3 negative charges on the molecule.  As a consequence of this increasing difficulty to remove protons, we observe 3 different $\ce{pKa}$'s.


Some substances do exist that slow down reactions and they're different from catalytic inhibitors/poisons. Such substances are called negative catalysts. Here is one example I can think of right now:
$$\ce{2H2O2 \xrightarrow{\large\mathrm{glycerol}} 2H2O + O2}$$

Glycerol, in this case behaves as a negative catalyst and slows down the reaction.

Another notable example you might have heard is tetraethyllead (TEL) which was used in petrol to prevent knocking.

It may be tempting to think that a negative catalyst will raise the energy of activation for a given reaction, however this isn't true. It takes away the internal energy of reactants and deactivates them to reduce the rate of a reaction.


What is the differences between partial pressure and vapour pressure?

In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature.

What does this mean? For example, if we have a mixture of gases $A$, $B$ and $C$ in an isolated room, then, according to Dalton's law, the pressure exerted by the gases will be the sum of their partial pressures :
$$P = p_A + p_B + p_C$$
where $p_A$, $p_B$ and $p_C$ are the partial pressures of each gas. Also, if we have a moles of $A$, b moles of $B$ and c moles of $C$, we can express the partial pressure of each gas as below:
$$p_A = \frac{a}{a+b+c} P$$
$$p_B = \frac{b}{a+b+c} P$$
$$p_B = \frac{c}{a+b+c} P$$

  Vapor pressure or equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

As the definition states, if we took an equilibrium as $\ce{A(l) <=> A(g)}$, where $A(l)$ and $A(g)$ are the liquid and the vapor state of the compound $A$ . The vapor pressure of $A$ is the pressure exerted by the gaseous state ($A(g)$) over the liquid state ($A(l)$).
Now, if we have a mixture of $A$, $B$ and $C$, we will have the equilibrium between the liquid and gaseous state : $\ce{(A, B, C)(l) <=> (A, B, C)(g)}$ and the vapor pressure $P_v$ (I noted $P_v$ so there won't be any confusion between it and the total pressure $P$ I mentioned before).
According to Raoult's law, the vapor pressure $P_v$ is the sum of the products between the vapor pressures of each compound taken separately and the molar fraction of the compound in the mixture (i.e. in the solution) :
$$P_v = x_A p_{v^A} + x_B p_{v^B} + x_C p_{v^C}$$

This may look just like Dalton's law, but there are different pressures involved.
The partial pressure of a gas is the pressure exerted by a gas in the volume occupied by a mixture of gases, while the vapor pressure of a gas is the pressure exerted by a gas over its condensed phase.
Although :

  The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial pressure.

The vapor pressure still refers to the pressure of a gas over its liquid (or solid) phase in a phase transition (condensed phase$\ce{<=>}$non-condensed phase), while the notion of partial pressure can be extended to a mixture of gases only.

All the definitions are linked to Wikipedia, and they can really cause confusion throughout chemistry learners..


InChI is the acronym for the IUPAC International Chemical Identifier. It is basically as system, to catalogue molecular information. Its prime advantage is, that it is just a text string. That basically means, that it can easily be understood by machines. The other advantage is, that it is also human readable. It consists of several layers: the molecular formula, connectivity of the heavy atoms, placement of the hydrogen atoms/ stereochemical information. This shall not be confused with the IChIKey, which is a string of 27 characters derived from the hash code of the InChI string.

A quite thorough presentation/ documentation of the software and its algorithms (including the generation of the InChI string and the InChIKey) is available as an open access paper (currently asap): Stephen R Heller, Alan McNaught, Igor Pletnev, Stephen Stein and Dmitrii Tchekhovskoi, Journal of Cheminformatics, 2015, 7:23. Since the abstract is not very comprehensive, I will quote a short passage from the introduction, which outlines the contents of the article:

  We then describe the workflow of InChI generation (normalization, canonicalization, and serialization stages), as well as generation of the compact hashed code derived from InChI (InChIKey); the related algorithms and implementation details are briefly discussed.

In response to R.M.'s comment I am also including the introductory sentence of the linked paper, which describes what an IChIKey is:

  InChIKey is a compact chemical identifier derived from InChI. The InChIKey is always only 27-characters long.

For a more technical description of the algorithms and the scope of the InChI string and the InChIKey, the reader is pointed towards the technical manual: Stein S, Heller S, Tchekhovskoi D, Pletnev I. IUPAC International Chemical Identifier (InChI) InChI version 1, software version 1.04 (2011) Technical Manual. The description of the InChIKey can be found in section V, pp 53-62:

  The InChIKey is a character signature based on a hash code of the InChI string. A hash code is a fixed length condensed digital representation of a variable length character string. Providing a signature derived from an InChI string is helpful for search applications, including Web searching and chemical structure database indexing; also, it may serve as a checksum for verifying InChI, for example, after transmission over a network.

Please note that the InChIKey by itself does not contain enough information to restore the molecular structure. This is also mentioned in the open access paper:

  A disadvantage of the InChIKey is that one loses the ability to algorithmically restore a structure from a textual label: InChIKey is a structure-based registry-lookup identifier [...].

But these are not the only sources. As an IUPAC member, most information relating to the developement and implementation can be accessed freely at iupac.org. Since itself is an non-profit organisation, it also has its own website, www.inchi-trust.org. This includes a very extensive technical FAQ (html and pdf) and since it is open source software also the source code.

The software is still actively developed by the chemical community and the trust itself. As of the current version (1.04, Sept. 2011) there are still several shortcomings, which have bean addressed in Structure that breaks InChI.


Is there intramolecular hydrogen bonding in salicylaldehyde?

It can indeed hydrogen bond, this can (to a certain extent) be observed by NMR. Due to the internal hydrogen bonding, the phenolic proton is surprisingly consistent, and shows no dependence on concentration.

  Image reproduced from Basic one- and two-dimensional NMR spectroscopy, Wiley, 2010.


You  might want to try BioVia's (formerly Accelrys) Discovery Studio Visualizer. This is free software and the "gateway drug" to their sophisticated suite of molecular modeling and chem/bio informatics software. It is a step above the various drawing programs out there and lets the user generate realistic 3-D models of an infinite variety of chemical structures. http://accelrys.com/products/discovery-studio/visualization-download.php


Are there any nonflammable thermal insulators which can be made at home?

Consider using a ceramic container.  You could adjust the density, thickness, and size to match your insulating requirements.

"home made' is loose, here, since you would need to acquire oven bake-able ceramics, or a similar product.  (unless you have a kiln lying around).


Yes, sodium metal is also going to react exothermically with salt water or any other aqueous solution as long as it comes in contact with water:

$$\ce{Na (s) + H2O -> Na+ (aq) + OH- (aq) + 0.5 H2 (g)}$$

eventually leading to explosion of hydrogen-oxygen mix forming near the water surface. Presence of sodium chloride in salt water isn't going to influence the reaction as it neither capable of shifting equilibrium to the left (gas formation), nor passivate sodium metal surface as it would do, for instance, some viscous oil.


I don't like to speak in generalities if I can help it.

But we can talk, for instance, about hydrogen peroxide. One of the most familiar schemes is a disproportionation where,
$$2 \ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2}$$
and as you study this particular species, you will see this reaction more and more often. Actually in solutions of hydrogen peroxide you can take a 30% concentrated solution and add some Mn and subsequently measure a significant increase in oxygen concentration over time. Although you should know the solution itself decomposes slowly in practice when exposed to light or due to trace impurities, but this pales in comparison to the case involving a catalyst. This reaction is, in part, why researchers believe a peroxide-type species is important for the Oxygen Evolving Center of Photosystem II. Another fun demonstration of this process is often where foam is generated from a large graduation cylinder, see here, upon addition of KI.

Now 30% is weak, if instead you look at 80% solutions you will notice that the decomposition will evolve steam because it is so exothermic. In fact, the first liquid-fueled rocket in 1939 used an 80% concentrated solution and decomposed it with calcium permanganate. Later engines utilized another important property of hydrogen peroxide for its decomposition until around 1980: the fact that it is a stronger oxidizer than chlorine/chloride.

Thermodynamically we say the disproportionation is favorable since it has a fairly negative (~ -100 kJ/mol) enthalpy change and a considerable change in entropy ( ~ + 70 J/(mol K)). The basis of this is that it is highly favorable to form dioxygen and this is a large driving force of this reaction in terms of enthalpy and entropy. You'll also see this danger in species that are really close to forming dinitrogen too, see here and here. It only takes a catalyst for this down-hill process to take off. Which could be quite deadly if unintentionally concentrated and then introduced to a catalyst in a closed system. Perhaps the story is similar for other peroxides.


Is there a pH at which all proteins are negatively charged? How can it be determined?

That depends highly on what you mean by "all." Since acid-base chemistry is equilibrium based, there is in principle always some of the less preferred species around even though it might be quite small. At the very least, there is a nonzero probability of finding the disfavored species, in your case the neutral species.


What are the components of a labware setup for short-path distillation?

Of course the answer depends a bit on the quantity to distill and the conditions (temperature and vacuum applied, difference in the boiling points of the components to separate).

There are about palm size short-path distillation heads, which may be connected easily directly to your round bottom flask and receivers like the following example:

(source)

At chemglass.com, this device (CG-1239) is listed for USS 178.95 and allows you to mount as usual thermometer, water-in and -out for the condenser, and the vacuum by an oil pump.  The efficiency of separation may be increased by mounting a vigreux column below the head, of course, which are available in different (shortened) lengths, too.  There are "jacketed versions" (e.g., CG-1240) which are more suitable if the head temperature to reach is quite elevated (say, 150 $^\circ\mathrm{C}$) already while applying vacuum, at a bit more elevated price tag, too.

Of course, the advantage of the Kugelrohr shortpath's design is that you may moderate the heat gradient better, because the bulb outside the oven may be cooled with a bath of acetone / dry ice, and the rotation reducing sudden evaporation.  The one by Buchi B-585 is more suitable for smaller quantities, while the ones by Aldrich obviously allow distillations out of a 250 mL flask, which truly are both considerably more expensive.  (The larger Aldrich ovens require more bench place, too.)

Do not forget that there are e-bay-like companies specialized in selling used but still working lab equipment, too, e.g. Richmond, LabX to mention two.


Is it possible to perform ROMP2 numerical optimizations in Gaussian09?

I've seen this error when I want to combine an optimization with an orbital deletion (NBODel) in Gaussian. It cannot handle an optimization with more than 50 degrees of freedom, so you need to freeze as much as you can, or pre-optimize your molecule with a more traditional method and then freeze whatever you can and do the ROMP2 optimization on the parameters of interest.


How would someone in 1917 describe elements 43, 61, 72, 75, 84, 85, 87, 89, and 91?

Unknown elements were discussed according to Mendeleiev's nomenclature. The prefix eka- was used for the element just under a known element, and it was dvi- for the following one, two lines under. For example, the element 31 under aluminum was celled eka-aluminum, just before it was discovered as gallium. With this same technique, the element 43 was named eka-manganese before it was discovered as technetium, and element 75 was dvi-manganese before its discovery as rhenium.
In your list, the element 61 was not expected. The 72 was eka-zirconium. The radioactive elements (Z > 83) were not expected.
Ref. : M. Fontani, M. Costa, M. Orna, The Lost Elements, Oxford University Press 2015, 530 p., ISBN 978-0-19-938334-4.


Why is it a good idea to put some salt in the bread dough?

Common table salt is not necessarily an antiseptic in the conventional "sterilizing" sense. What gives it weakly antiseptic properties in biological context is its ability to absorb water from cells through osmosis, should the concentration of the salt outside the cell be too high, the cell would lose too much water through its membrane and die.

However this is strongly dependent on concentration. For example, according to this source, a concentration upwards of 20% is required to kill bacteria; at lower concentrations the salt will only hinder growth, which is most likely the reason it is added (in moderation) to bread.


How does dehydration of an alcohol occur in alkaline medium?

The commonly accepted mechanism for the base-catalysed aldol example here is that the alpha hydrogen (shown in blue on your scheme) is removed by OH- present in the reaction medium, and the negative charge on the enolate then forms the alkene by loss of OH-. This mechanism falls under the general heading of E1cB, further information on this mechanism may be found here and here 3.

Thanks to @orthocresol for the note about this


Comparing the solubility of glucaric acid and mucic acid

Wikipedia lists the melting points of the sugar acids as:

Glucaric (or saccharic) acid: $\pu{125-126 ^\circ C}$

mucic (galactaric) acid: $\pu{230 ^\circ C}$.

The very great solubility of all the sugary materials (glucaric acid: $\pu{63 mg/mL}$1, or perhaps $\pu{912 mg/mL}$2) is due to their hydroxyl groups, which are very compatible with $\ce{H2O}$. Mucic acid, even though, it has as many hydroxyls as glucaric, is soluble only to the extent of $\pu{3.3 mg/mL}$, although it was predicted from its structure to dissolve reasonably well ($\pu{63.1 g/L}$3). (Note the differences in units; I used the same numbers and units as in the original reference.)
The likely reason for the great difference in solubility and melting point is the way the mucic acid crystal packs. This has shown up in density and x-ray crystallography4.
Looking at the structures, or playing with molecular models (typically, one molecule at a time - you never have enough atoms to make a half a dozen molecules, nor the time to rotate them all and pack them, nor the straight-line vision to see how well they pack) doesn't give you the same sense of difference that the solubility and melting point give.
Maybe this is not terribly unusual: tartaric acid(s), a simpler diacid, shows similar differences in solubility and melting point just because of the chirality of the halves of the molecule. Melting points: D or L: $\pu{171-174 ^\circ C}$; DL $\pu{206 ^\circ C}$; meso: $\pu{147/ 165 ^{\circ} C}$. Solubilities (T=?): D or L: $\pu{1.33 kg/L}$; DL: $\pu{0.21 kg/L}$; meso: $\pu{1.25 kg/L}$ (Wikipedia).
References

https://www.drugbank.ca/drugs/DB03603
https://pubchem.ncbi.nlm.nih.gov/compound/Glucaric-acid#section=Experimental-Properties
https://hmdb.ca/metabolites/HMDB0000639
The crystal structure of galactaric acid (mucic acid) at −147 °C: An unusually dense, hydrogen-bonded structure by George A.Jeffrey, Richard A.Wood, Carbohydrate Research, Volume 108, Issue 2, 16 October 1982, Pages 205-211, DOI: 10.1016/S0008-6215(00)81790-5


What is the mechanism by which reactants in endothermic reactions absorb energy from their environments?

In your first paragraph you state at the end that photons are released. In all reactions, except the rather few chemiluminescent ones, this is not the case.

The energy difference is made up of vibrational and rotational energy of the products, which by collisions with solvent (or inert buffer gas) removes this energy.  This energy transfer occurs as the product is initially 'hot' with respect to the surroundings and thus energy flows out of this into the surroundings, heating it a little, until equilibrium is reached. In femtosecond laser experiments it is possible to observe the flow of energy when in solution and it is very rapid, not much longer than a few picoseconds. Ir photons could be emitted but the radiative lifetime of levels is much, much longer than collision lifetime so that i.r. emission yield will be tiny.

In the low pressure gas phase (where collisions with buffer gas are absent) infra-red can be observed (called infra-red chemiluminescence as it comes from excited vibrational levels) in reactions such as $\ce{H + H2 \rightarrow  HF(v'J') + H} $ where $\ce{v'J'}$ are excited vibrational and rotational levels. Many experiments have been reported by Polanyi and co workers for which Polanyi received the Chemistry Nobel prize. This reference https://www.nobelprize.org/nobel_prizes/chemistry/laureates/1986/polanyi-lecture.pdf gives a clear account of this important work

In the reverse case it is the few molecules that by the Boltzmann distribution have enough energy to surmount the activation barrier that are important not promotion by ir photons.


What are the biochemical ramifications of ingesting of distilled water?

As @DavePhD has noted, drinking small amounts of distilled water does not pose any significant adverse health consequences.  However, the World Health Organization (WHO) has studied the effects of long-term drinking of distilled water and issued a report.  They investigated the following five areas of potential adverse human health impact:

Direct effects on the intestinal mucous membrane, metabolism and mineral homeostasis or other body functions.
Little or no intake of calcium and magnesium from low-mineral water.
Low intake of other essential elements and microelements.
Loss of calcium, magnesium and other essential elements in prepared food.
Possible increased dietary intake of toxic metals.

In their conclusion, they state,

  "Drinking water should contain minimum levels of certain essential
  minerals (and other components such as carbonates). Unfortunately,
  over the two past decades, little research attention has been given to
  the beneficial or protective effects of drinking water substances."

and

  "Demineralised water that has not been remineralized, or low-mineral
  content water – in the light of the absence or substantial lack of
  essential minerals in it – is not considered ideal drinking water."

(emphasis mine)


There is a crystal structure of human gastric lipase (PDB ID 1hlg). The accompanying paper does not mention any quaternary structure and shows the protein as single chain. Automated analysis of the coordinates support that. The enzyme is glycosylated, so the molar mass does not match the one expected from the sequence. The apparent molar mass determined experimentally is about 50,000 g/mol, corresponding to a monomer.

However, the annotation in the structure (probably erroneously) say that the biological assembly consists of the two chains in the asymmetric unit. Also, there is a report that the size of the molecule changes with pH (90,000 g/mol at pH 6 vs 45000 g/mol at pH 3).

The same group that solved the human lipase structure later solved the dog lipase structure (in an open conformation with a bound ligand) where the annotation in the PDB deposition says it is a monomer. I doubt that there are interspecies differences between dog and human in terms of quaternary structure, so that would be another piece of evidence supporting that the human enzyme is a monomer in its native form.


I don't believe you have any ferric in the colorless picture since you have a metallic iron which would reduce the ferric to a ferrous. I believe you have a ferrous (Iron(II)) acetate characterized as an off-white precipitant.


How can water below a pH of 7 still be alkaline?

Classically, alkalinity is determined by $\ce{HCl}$ titration with methyl orange as indicator ( $\mathrm{pH}$ transition range $3.1 - 4.4$ ). IF some $\ce{HCl}$ is spent by lowering $\mathrm{pH}$ from $5.5-6.6$ to $\mathrm{pH}$ of noticing red colour, there is some nonzero alkalinity.
Note that at $\mathrm{pH} \approx 6.3$, the ratio $\dfrac{[\ce{CO2}]}{[\ce{HCO3-}]}$ is 1 : 1. (*)
Generaly, there is no direct relation between $\ce{pH}$ and maximal alkalinity, similarly as there is no direct relation between $\ce{pH}$ and $\mathrm{pH}$ buffer capacity.
But for typical water composition, containing $\ce{CO2/HCO3-}$ $\mathrm{pH}$ buffer, the alkalinity can be estimated from $\ce{pH}$ and expected $\ce{CO2(aq)}$ concentration, which is supposed to be in equilibrium with gaseous $\ce{CO2}$.
$$[\ce{HCO3-(aq)}] = K_{a1,\ce{CO2}}^{*} \cdot \frac {[\ce{CO2(aq)}]}{[\ce{H+(aq)}]}$$

(*) The  Wikipedia page for $\ce{H2CO3}$ has been recently misedited,  confusing $K_\mathrm{a2}$ and $K_\mathrm{a1}^*$(assuming 100% conversion of $\ce{CO2(aq)}$ to $\ce{H2CO3}$.


How does fat affect the precipitation of proteins?

Casein is the main protein in milk. When you perform the polymerization experiment, the result is an "aggregation" of said protein, which gives the product a solid consistence.

Products such as butter and cream are made with the fats from milk (not the proteins!), hence they are almost casein-free. Of course, being them almost without proteins, the polymerization doesn't occurr.

If you look up the nutrition facts of milk and heavy cream, you see that proteins constitute the 20% of calories of milk, and only the 2% of heavy cream.


Zinc Picolinate - Positive, negative or neutral charge?

Zinc picolinate with two ligands per metal ion ($\ce{Zn(C6H4NO2)2}$) is a neutral chelate complex. Deprotonated picolinic acid has one negative charge at the carboxylate group ($\ce{NC5H4-COO^-}$). Two picolinate ligands thus compensate the two positive charges of $\ce{Zn^2+}$, forming a neutral complex in which they act as bidentate ligands. The nitrogen of the pyridine ring and the carboxylate coordinate the Zn center ion.

Image source: http://www.lookchem.com/Zinc-picolinate/


In the $\ce{sp^2}$ transition state for an $\mathrm{S_{N}}2$ reaction

the attacking nucleophile, the central carbon and the leaving group prefer a linear arrangement
the 3 remaining groups around carbon are arranged with 120° bond angles

(image source)

Now, let's look at ethylene (as a model for a vinyl halide) and let's use the bent bond description of ethylene because it makes it easy to visualize what's happening (the bent bond model has been shown to be equivalent to the pi bond model in all aspects, see below for the same explanation phrased in terms of the pi bond model).

(image source)

Replace one of the ethylene hydrogens with halide and imagine what happens as we approach the transition state.  The bond angle in the two-membered ring wants to be small, but as we approach the transition state it needs to open up wider to 120°!  This would create a lot of strain.  It is this large strain energy that causes the $\mathrm{S_{N}}2$ reaction at $\ce{sp^2}$ hybdridized vinylic carbons to be a very high energy pathway and therefore less likely to occur.

On the other hand, the elimination reaction at a vinylic carbon is quite facile.  Here as the base removes a proton we build up negative charge on an $\ce{sp^2}$ hybridized carbon.  An $\ce{sp^2}$ orbital has more s-character than an $\ce{sp^3}$ orbital; the more s-character in an orbital the lower the energy of an electron in that orbital.  Therefore, negative charge is more stable in an $\ce{sp^2}$ orbital than a $\ce{sp^3}$ orbital.  Consequently it is easier to remove a proton when the negative charge will reside on an $\ce{sp^2}$ hybridized carbon compared to an $\ce{sp^3}$ hybridized carbon (and that's why it is even easier [lower energy] to remove a proton from an $\ce{sp}$ hybridized carbon).

Explanation Based on Pi Bond Model

As expected for equivalent models, the bent bond explanation can be rephrased in terms of traditional $\ce{sp^2}$ sigma-pi hybridization, by noting that we start with an $\ce{sp^2}$ hybridized carbon which becomes $\ce{sp}$ hybridized in the transition state. It is known that $\ce{sp}$ hybridized carbons are roughly 5 kcal/m less stable than $\ce{sp^2}$ hybridized carbons (more electrons are in higher energy unhybridized p orbitals and this is not offset by the energy lowering experienced by the electrons as some bonds rehybridize  from $\ce{sp^2}$ to $\ce{sp}$) thereby increasing the energy required to reach the transition state.


Is osmosis really a special case of diffusion across a semipermeable membrane?

I've listed a series of points about diffusion and Osmosis.

The cause of osmosis is simply diffusion: the solvent can diffuse through the membrane and the solute cannot.

Thermodynamically, we say that the reason that the diffusion occurs is that the chemical potential of the solution is at a lower chemical potential than the pure solvent and so there is a tendency for the solvent to pass into its solution and to dilute it. The osmotic pressure is the excess pressure just needed to stop that flow. If more pressure were to be applied the flow of solvent would be reversed and the solution would become more concentrated.

Osmosis occurs because there is an increase in entropy overall when the solvent dilutes the solution. There is a decrease in entropy in the pure solvent side because its volume decreases but this is balanced by an increase in entropy on the solution side as the volume increases so this is not the reason osmosis occurs. The reason is the increase in entropy that is experienced by the solute now in a larger volume or equivalently more solvent molecules in which to dissolve. This means that there are more ways of arranging solute and solvent molecules compared to that before osmosis stared, and this represents an increase in entropy. (Left unchecked this will continue until the solution is so dilute it is indistinguishable from the solvent).

When we say that a solution has an osmotic pressure of, say, 5 atm, it does not mean that the solution exerts this pressure but that if it is in contact with its solvent through a semi-permeable membrane then an applied pressure of 5 atm is needed to maintain equilibrium.

It is possible for a solute in a solution to diffuse spontaneously from regions of higher to lower concentration so that uniform concentration is reached and the solution is at equilibrium.  Naturally, diffusion does not stop at this point but is always present both for solute and solvent.

In some cases, however, diffusion can occur form lower to higher concentrations. Such a situation is that when a solute is dissolved in two immiscible solvents in contact with one another, and in which the solute has different solubilities. Solute diffusion will occur until the chemical potential in each solvent is equal which can mean diffusion into the more concentrated solution. Thus at equilibrium there will be a large difference in solute concentration but no overall flow of solute due to diffusion.


Derivation of the Orgel diagram for octahedral d2 complexes

1. Weak-field and strong-field limits

I will adopt the description used in Figgis and Hitchman's Ligand Field Theory and Its Applications (p 5), because I cannot really phrase it better:

  It is useful to recognise the relative importance of the terms in the Hamiltonian for different systems:

  First- and second-transition series: $\mathbf{H}_\mathrm{LF} \approx \mathbf{H}_\mathrm{ER} > \mathbf{H}_\mathrm{LS}$

  $\mathbf{H}_\mathrm{LF}$ (ligand field) takes into account the presence of the surrounding ligand molecules [...]. $\mathbf{H}_\mathrm{ER}$ (electron repulsion) takes into account the repulsions within the d or f electron set [...]. $\mathbf{H}_\mathrm{LS}$ (L-S coupling, also known as Russell-Saunders coupling) takes into account the formal coupling between the spin and the [orbital] angular momenta of the d or f electrons.

Spin-orbit coupling (which scales as $Z^4$, $Z$ being the atomic number) plays a nonzero, but relatively small, role in the properties of the 3d or 4d transition metals (compared to the 5d, 4f, or 5f metals). Therefore, it is not reflected in the Orgel diagrams above.

The weak-field limit simply refers to the case where the ligand field is weak, i.e. $\mathbf{H}_\mathrm{LF} < \mathbf{H}_\mathrm{ER}$. In this regime, the spherical symmetry of the free ion (e.g. $\ce{V^3+}$) is almost unperturbed by the six ligands (assuming octahedral symmetry). Therefore, the electronic states are similar to those of the free ion.

The strong-field limit simply refers to the opposite case, i.e. $\mathbf{H}_\mathrm{LF} > \mathbf{H}_\mathrm{ER}$. In this regime, the electron-electron repulsions are relatively small compared to the ligand-field splitting $\Delta$. As such, the electronic states are entirely determined by the electronic configuration $(\mathrm{t_{2g}})^m(\mathrm{e_g})^n$ (for a $\mathrm{d^2}$ complex ion, clearly $m + n = 2$).

As $\Delta$ increases (i.e. going from left to right), we move from the free ion limit where there is no ligand field and $\Delta = 0$, through the weak-field regime where $\Delta$ is small, into the strong-field regime where $\Delta$ is large.

2. The electronic states: their symmetries and their energies

2.1. Free-ion terms

For a $\mathrm{d^2}$ configuration, one can either enumerate microstates (a nasty, tedious method that I do not recommend), or one can use group theory (wait for the MathJax to load, then scroll to direct product tables: $R_3$). The wavefunction of a d-orbital, in the spherical symmetry of the free ion, transforms as $D \equiv \Gamma^{(2)}$. The electronic wavefunction of two d-electrons transforms as the direct product of $D$ with itself:

$$\begin{align}
\Gamma^{(2)} \otimes \Gamma^{(2)} &= \Gamma^{(4)} \oplus \Gamma^{(2)} \oplus \Gamma^{(0)} + [\Gamma^{(3)} \oplus \Gamma^{(1)}] \\
\mathrm{D \otimes D} &= \mathrm{G \oplus D \oplus S \oplus [F \oplus P]}
\end{align}$$

Square brackets indicate spatial wavefunctions that are antisymmetric with respect to interchange of the two electrons; to satisfy the Pauli exclusion principle, these have to be paired with symmetric spin wavefunctions. Adding up the spins of two electrons ($s_1 = s_2 = 1/2$), we can either get $S = 1$ ("triplet") or $S = 0$ ("singlet"). The triplet spin wavefunctions are symmetric, so these go with the $\mathrm{F}$ and $\mathrm{P}$ terms. Likewise, the singlet spin wavefunction is antisymmetric, and goes with the $\mathrm{G}$, $\mathrm{D}$, and $\mathrm{S}$ terms. This yields the term symbols $\mathrm{^1G}$, $\mathrm{^1D}$, $\mathrm{^1S}$, $\mathrm{^3F}$, and $\mathrm{^3P}$ for a $\mathrm{d^2}$ configuration, where the superscript indicates the spin multiplicity $M = 2S + 1$.

The singlet states are higher in energy due to exchange energy (Hund's first rule). As such, they get ignored in our treatment here. This is not always wise, and the more sophisticated Tanabe-Sugano diagrams do plot electronic states of different multiplicity, but as far as we are concerned, we can neglect them here.

Between the two triplet states, the state with the larger value of $L$ (total angular momentum quantum number) is the ground state (Hund's second rule). Therefore, the $\mathrm{^3F}$ state with $L = 3$ is the ground state, and the $\mathrm{^3P}$ state with $L = 1$ is the excited state.

The difference in energy between these two states is solely attributable to electron-electron repulsions.[1] The two states are separated by an energy difference $15B$, where $B$ is a Racah parameter that acts as a measure of electron-electron repulsions. I believe there is a quantum mechanical expression for it, but in practice the value of $B$ is empirically obtained (via Tanabe-Sugano diagrams). It is easier to treat it as an empirical parameter that is experimentally obtained, similar to how one would treat the ligand-field splitting $\Delta$.

The free-ion terms and the difference in energy $15B$ are shown in the middle of the diagram, where $\Delta = 0$.

2.2. Strong-field limit terms: no electron repulsions

In an octahedral $\mathrm{d^2}$ complex, there are three ways of arranging the two d electrons: $\mathrm{(t_{2g})^2}$; $\mathrm{(t_{2g})^1(e_g)^1}$; and $\mathrm{(e_g)^2}$. These are the three electronic states under consideration and are what you see at the right-hand side of the diagram. Hopefully the energy ordering is straightforward; the energy gap between each state is equal to $\Delta$ since it requires promotion of one electron from $\mathrm{t_{2g}}$ to $
\mathrm{e_g}$.

2.3. Weak-field limit terms

As can be seen from the Orgel diagram, there are four states: two $\mathrm{T_{1g}}$ states, one $\mathrm{T_{2g}}$ state, and one $\mathrm{A_{2g}}$ state. (The subscript g, as well as the spin multiplicity, are omitted in the diagram. This is to allow it to be generalised to tetrahedral complexes, where the g/u labels are not applicable, as well as complexes with other d-electron counts, where the spin multiplicity will be different but the symmetries the same.) The question we will have to look at is how these states are obtained, and how they correlate with the two extreme limits described earlier.

Going from the free ion limit, we need to find out what happens to $\mathrm{F}$ and $\mathrm{P}$ terms when the symmetry of the compound is reduced from spherical to octahedral. Luckily, there are group theory tables that do just that, called tables of descent in symmetry. These tell us that:

$$\begin{align}
\mathrm{F} &\to \mathrm{A_2 \oplus T_1 \oplus T_2} \\
\mathrm{P} &\to \mathrm{T_1}
\end{align}$$

and you can see this correlation in the lines spreading out from the centre. Conventionally, the $\mathrm{T_1}$ state arising from the $\mathrm{P}$ state is labelled as $\mathrm{T_1(P)}$ and likewise the other $\mathrm{T_1}$ state is labelled $\mathrm{T_1(F)}$. This isn't shown on this particular diagram, but may or may not appear in other texts (and it is very helpful because it eliminates any ambiguity about which $\mathrm{T_1}$ state you are talking about!).

Now, we need to figure out which final configurations to join them to. This is done by taking more direct products (yes, group theory tables again). For the $\mathrm{(t_{2g})^2}$ configuration we have

$$\mathrm{T_2 \otimes T_2 = A_1 \oplus E \oplus [T_1] \oplus T_2}$$

however since we are only interested in the triplet states, we are only going to take the $\mathrm{T_1}$ term, which has an antisymmetric spatial component to go with the symmetric triplet spin component. Likewise

$$\begin{align}
\mathrm{T_2 \otimes E} &= \mathrm{T_1 \oplus T_2} \\
\mathrm{E \otimes E} &= \mathrm{A_1 \oplus [A_2] \oplus E}
\end{align}$$

So, the $\mathrm{(t_{2g})^1(e_g)^1}$ configuration gives rise to $\mathrm{T_1}$ and $\mathrm{T_2}$ terms,[2] whereas the $\mathrm{(e_g)^2}$ configuration gives rise to an $\mathrm{A_2}$ term. The only remaining question is, which of the $\mathrm{T_1}$ states correlates with which of the strong-field configurations? The answer is given by the non-crossing rule: two states of the same symmetry cannot cross each other.[3] Therefore, we can confidently join the $\mathrm{T_1(F)}$ state to the $\mathrm{(t_{2g})}$ configuration, knowing that the alternative would lead to a forbidden crossing.

Incidentally, this is also why the two $\mathrm{T_1}$ electronic states have curved lines: quantum mechanical mixing introduces an extra perturbation to the energy. The upper energy level is always pushed upwards, and the lower energy level pushed downwards, which leads to the curvature shown in the diagram. The other two states do not have such effects and are therefore depicted as straight lines.

3. Other geometries and configurations

Since a $\mathrm{(t_{2g})^3(e_g)^2}$ configuration is spherically symmetric, the electronic states arising from an octahedral (high-spin) $\mathrm{d^7}$ configuration are exactly the same as those arising from an octahedral $\mathrm{d^2}$ configuration. (The spin multiplicity is different: $S = 3/2$ gives $2S + 1 = 4$.)

Furthermore, the octahedral $\mathrm{d^3}$ configuration can be thought of as having two holes in a spherically symmetric $\mathrm{(t_{2g})^3(e_g)^2}$ arrangement. These holes give rise to exactly the same symmetry states as the $\mathrm{d^2}$ configuration; however, since holes are essentially "negative electrons", the relative energies are reversed. For example, the $\mathrm{A_2}$ state, which arises from placing the two holes in the $\mathrm{e_g}$ orbitals (i.e. a $\mathrm{(t_{2g})^3}$ electronic configuration), is actually the ground state instead of the highest excited state. This is reflected in the left half of the diagram.

(If the above two paragraphs were confusing, they were covered in more detail in another answer of mine: How do I find the ground state term symbol for transition metal complexes?)

As for tetrahedral complexes, note that the orbital splitting in a tetrahedral complex is exactly the same as in an octahedral complex, except that the relative energies of the $\mathrm{e}$ and $\mathrm{t_2}$ sets are reversed (and the g subscripts are gone).

Therefore, if we consider (for example) the $\mathrm{(e)^2(t_2)^1}$ ground state of a tetrahedral $\mathrm{d^3}$ configuration, this can be considered to have two holes in the $\mathrm{t_2}$ orbitals. These two holes give rise to exactly the same states as the octahedral $\mathrm{d^2}$ case, where in the ground state there are two electrons in the $\mathrm{t_{2g}}$ orbitals. The tetrahedral $\mathrm{d^3}$ ground state therefore has a $\mathrm{^4T_1}$ label, exactly analogous to the octahedral $\mathrm{d^2}$ ground state which is a $\mathrm{^3T_{1g}}$ term. Of course the spin multiplicity is different and the g label is gone, but we've already established that the diagram specifically says nothing on these two points for a reason.

Consequently, tetrahedral $\mathrm{d^3}$ and $\mathrm{d^8}$ complexes fit into the right half of the diagram. Using similar arguments, tetrahedral $\mathrm{d^2}$ and $\mathrm{d^7}$ complexes fit into the left half of the diagram.

Notes

[1] Quoting myself from another answer

  The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.

[2] As an aside: why is the $\mathrm{(t_{2g})^1(e_g)^1}$ configuration split into two terms? The answer lies in, unsurprisingly, electron-electron repulsions (which were neglected in the strong-field limit). Consider the electronic configuration $(\mathrm{d}_{xy})^1(\mathrm{d}_{x^2-y^2})^1$, which has electron density concentrated in the $xy$-plane. We might expect this configuration to experience more electron-electron repulsion than, for example, the $(\mathrm{d}_{xy})^1(\mathrm{d}_{z^2})^1$ configuration. While molecular terms do not necessarily have a one-to-one correspondence with orbital microstates, one can see from this discussion how it is plausible for the degeneracy to be lifted when electron repulsions are taken into account.

[3] The Hamiltonian $\hat{H}$ transforms as the totally symmetric irreducible representation of the relevant point group (which makes sense, since the energy of a system has to be constant under reflection, inversion, rotation, etc.). Consequently, the matrix element $\mathbf{H}_{mn} = \langle m | \hat{H} | n \rangle \neq 0$ if $m$ and $n$ have the same symmetry. This directly leads to the non-crossing rule, that energy levels corresponding to wavefunctions with the same symmetry do not cross under any strength of a perturbation (in this case, the ligand field).


Container to minimise radiation exposure from U-238

Due to its very long half-life, U-238 is not particularly radioactive; the specific activity is only about 12 kBq/g. The emission probability of gamma radiation is less than 0.1 % per disintegration of U-238. Furthermore, the gamma energy is relatively low. Therefore, the direct gamma radiation from pure U-238 (i.e. disregarding Th-234 and further decay products) is usually insignificant.

For example, the gamma dose rate in 1 m distance from 1 kg of pure U-238 (about 12 MBq) is only approximately 0.1 nSv/h – probably less due to self-absorption. By way of comparison, the dose rate from natural background radiation amounts to about 100 nSv/h.

Nearly all of the alpha radiation from U-238 is absorbed within the sample. Only a small fraction originating from the surface of the sample can escape but is stopped by air within a distance of about 3 cm.

Therefore, you do not need any shielding. Just use a suitable closed container to prevent contamination or incorporation.


Given an unbalanced chemical equation, is it possible to determine the largest possible coefficient that could be used in the balanced equation?

In short

Just use matrices.

In long

Consider this equation to be balanced:

$$\ce{CuSCN + KIO3 + HCl -> CuSO4 + KCl + HCN + ICl + H2O}$$

(source with spoiler)

We are trying to solve this equation:

$$\ce{aCuSCN + bKIO3 + cHCl -> dCuSO4 + eKCl + fHCN + gICl + hH2O}$$

Now, we move every term to the left hand side (warning: handwavey chemistry):

$$\ce{aCuSCN + bKIO3 + cHCl - dCuSO4 - eKCl - fHCN - gICl - hH2O = 0}$$

Transform into matrix form (the element names are just for reference):

$$
\begin{matrix}
\ce{Cu}\\
\ce{S}\\
\ce{C}\\
\ce{N}\\
\ce{K}\\
\ce{I}\\
\ce{O}\\
\ce{H}\\
\ce{Cl}
\end{matrix}
\left(
\begin{matrix}
1 &0 &0 &-1&0 &0 &0 &0 \\
1 &0 &0 &-1&0 &0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 \\
0 &1 &0 &0 &-1&0 &0 &0 \\
0 &1 &0 &0 &0 &0 &-1&0 \\
0 &3 &0 &-4&0 &0 &0 &-1\\
0 &0 &1 &0 &0 &-1&0 &-2\\
0 &0 &1 &0 &-1&0 &-1&0
\end{matrix}
\right)
\left(
\begin{matrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h
\end{matrix}
\right)
=
\left(
\begin{matrix}
0\\0\\0\\0\\0\\0\\0\\0\\0\\
\end{matrix}
\right)
$$

Notice that the biggest matrix is not a square. You must transform it to a square. If the elements are in excess (just like this case), add variables; otherwise, add elements. Set the extra variables/elements to zero:

$$
\begin{matrix}
\ce{Cu}\\
\ce{S}\\
\ce{C}\\
\ce{N}\\
\ce{K}\\
\ce{I}\\
\ce{O}\\
\ce{H}\\
\ce{Cl}
\end{matrix}
\left(
\begin{matrix}
1 &0 &0 &-1&0 &0 &0 &0 &0 \\
1 &0 &0 &-1&0 &0 &0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 &0 \\
0 &1 &0 &0 &-1&0 &0 &0 &0 \\
0 &1 &0 &0 &0 &0 &-1&0 &0 \\
0 &3 &0 &-4&0 &0 &0 &-1&0 \\
0 &0 &1 &0 &0 &-1&0 &-2&0 \\
0 &0 &1 &0 &-1&0 &-1&0 &0
\end{matrix}
\right)
\left(
\begin{matrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h\\
i
\end{matrix}
\right)
=
\left(
\begin{matrix}
0\\0\\0\\0\\0\\0\\0\\0\\0\\
\end{matrix}
\right)
$$

Note: $i$ must be zero.

Now you can use any algorithm for solving homogeneous square matrix equations to continue.


Molarity of hydrogen gas in standard hydrogen electrode

The standard hydrogen electrode is a "reference electrode." It is supposed to be used with ultra-low currents.

In operation, hydrogen gas is bubbled through a 1 N acid solution, but there is little hydrogen gas in the solution. In order to get the hydrogen gas to bubble through the solution, a pressure of a bit more than 1 bar must be used.


The blue flame from butane (or many other hydrocarbons) is due to the complete combustion of the gas.  From the Wikipedia article Flame color it mentions,

  In the most common type of flame, hydrocarbon flames, the most important factor determining color is oxygen supply and the extent of fuel-oxygen pre-mixing, which determines the rate of combustion and thus the temperature and reaction paths, thereby producing different color hues.

Specifically, making the link to the common tool in the laboratory, a Bunsen Burner, which typically has two 'types of flame (as seen below):

Image Source - which states that:

  The Bunsen Burners we use at school use a mixture of alkane gases like propane and butane

The yellow flame is from an incomplete combustion, producing $\ce{CO}$ and the colour is due to (according to the Wikipedia article):

  incandescence of very fine soot particles that are produced in the flame.

With more oxygen, a complete combustion of butane is possible (without the soot):

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

Which provides

  enough energy to excite and ionize gas molecules in the flame.

and the blue flame colour is due to the

  emission of excited molecular radicals in the flame, which emit most of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.


How is the cavity of a cyclodextrin chiral?

It is because the constituent glucose molecules are chiral.

In the image you can see that all the $\ce{CH2OH}$ groups in the center are out of the plane (towards you). These groups are on the upper side of the cavity. If they were below the plane, the cyclodextrin would have other interactions with the host molecule.


I believe it is about perspective. Note that "covalent molecule with ionic character" and "ionic compound with covalent character" are different.

If we are looking at a covalent molecule and assessing its ionic character, we cannot talk about polarising power, because in a covalent molecule, there would be no cation and anion to polarise in the first place. Thus, we would go by the rule: the bigger the electronegativity difference, the higher the ionic character and therefore, the more polarised the bond.

If we are looking at a ionic compound and assessing its covalent character, then we will go by the rule: the more polarising the central atom, the higher the covalent character.

Now, from the way you phrased your question, I believe that you are conflating the concepts "big electronegativity difference" and "polarising power". They are distinct concepts and should be thought of separately.

Although bond polarity is associated with electronegativity difference, "polarising power" is about the polarisation of the electron cloud of the anion by the cation and not the polarisation of the bond. This is also why "polarising power" is a concept associated with ionic compounds with covalent character.

Hope this clarifies your doubt! If you are still unclear and require some examples for clearer understanding, please leave a comment on my post.


Calculating the mass of acceptable carbon monoxide in a room

You have to be careful with your conversions - you are missing a step in your conversions.

You have 19000 mg

As 1 mg = 10-3 g  and
1 g = 10-3 kg

So:

19000 mg = 19 g = 1.9 x 10-2 kg


Can carboxylic acids like act as nucleophiles in substitution reactions?

Carboxylic acids can act as nucleophiles of course. Consider reactions with isobutene or thionyl chloride (although base certainly helps in this latter case). However, whether it's in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reaction mechanisms is another matter. If you beast something with enough heat it will usually react. I would not be surprised if a carboxylic acid reacted with triflouromethanesulfonic anhydride without base. It's still $\mathrm{S_N2}$ as mechanistic studies have shown the S=O bonds do not open up like acyl electrophiles.
I suspect you would need to resort to a continuous as opposed to a batch approach to push a reaction between the acid and an alkyl halide without using any base. Expose the reaction to high temperature (I'm thinking 250C and higher)  for 5 seconds or less under continuous liquid flow.


'Trace crossing' is in the literal sense, where 'trace' is another word for 'line'. 'Trace crossing' basically means that the line in the voltammogram crosses itself.


Are there any examples of activating meta directors?

Note: This requires an extremely high energy scenario to occur.

This is another example that looks promising, it is essentially an ylide that has enough EWing power to meta direct while giving some density to the ring in another resonance structure. []

The phosphonium ylide derivative (phenylphosphonic acid) may provide the same effect if the hypervalent nature of nitrogen is challenged, or if this substance is unobtainable under any circumstances.


What's the relationship between dew point and temperature?

There are two separate effects:

The average temperature above urban land is greater than above rural land. This could be because of the cooling effect of plants evaporating water, or because of the higher albedo of vegetation than pavement and roofing.
The absolute humidity decreases; that is, the total amount of water vapor in a given volume of air decreases, perhaps due to the capture and run-off of rainwater into storm drains, rather than evaporating.


Will alkoxide ion in protic medium attack on double bond or carbonyl group?

Your title should read "protic medium" as opposed to "acidic medium". Acidic medium would be EtOH/H+. Your reaction conditions are basic and protic. Nothing much will happen to the unsaturated aldehyde 1. 1.2-Addition leads to hemiacetal 2. Hemiacetals are unstable to base and reverse to the carbonyl compound. Adduct 3 would eliminate to form 1 in the presence of base. If you employed EtOH-d1 as the solvent, you would incorporate deuterium at the γ-position of 1-d2 via enolate 4. If heptanedial 5 were exposed to EtOK/EtOH, an aldol condensation would ensue to form unsaturated aldehyde 1. Under truly acidic conditions, the diethyl acetal (in the red box) of aldehyde 1 would form. Formation of the β, γ-isomer of 1 is unlikely because the α, β-form 1 is much more stable.


A blueberry contains ~85% water, 5% glucose, and 5% fructose (and it is generally considered a low glycemic index food, see e.g. this site). The solubility of fructose in water is extremely high (wikipedia says ~4000 g/L (25 °C), pubchem says "freely soluble in water"), and that of glucose is pretty high too, so there is essentially zero chance of removing these sugars through crystallization or other method without significantly altering the fruit or its juice.

If you are still interested in generating an extract of the fruit, this article might be useful. It suggests two different approaches, juicing and drying. The first will generate juice and a press cake, and since the sugars are highly water-soluble the cake should be the reduced-sugar component of interest to you. You could also freeze dry the juice, assuming this contains the important components, then partially rehydrate the dry product and attempt to crystalize out sugars. Or you could dry the whole fruit and then attempt a separation with ethanol, with sugar extracted into the alcohol to obtain a reduced-sugar extract which can then be dried. Further separations with more apolar solvents might extract hydrophobic "phytonutrients" from the juice. But you are likely to remain somewhat in the dark with respect to what you've removed from the berry without further analytical tools to help you characterize the products.

There are alternatives to extraction of the sugar that involve biochemical conversion of the monosaccharides. One alternative is to treat crushed blueberry with xylose isomerase, which would convert the glucose to fructose. Fructose has a low GI ranking, which might be a benefit. Another alternative is to treat the blueberry with a mixture of enzymes (perhaps including sucrose-sucrose 1-fructosyltransferase (1-SST), see e.g. this example) that converts the glucose/fructose mixture into largely indigestible polysaccharides (soluble dietary fiber).

Otherwise the best solution might remain to eat the fruit raw or to seek other foods with similar composition and without sugar (perhaps you should consult a dietitian).


What is the solublity of water in liquid carbon dioxide at low temperature?

The lower the temperature and higher the pressure, the more soluble CO2 is in water.  Pepsi and Coke have staudiesd this for years.


For two compounds that are symmetrical, why does the more compact one have a higher melting point?

The terms used by your teacher are quite in line with at least a qualitative explanation.

I really suspect that there are various ways to express compactness and sort of "sphericity" of molecules. I am not a specialist so forgive me for being imprecise on this.

However, if we forget a rigorous classification in symmetry group, a potato is more spherical symmetrical than a banana.

In such a sense the ramified isomer is definitely more  symmetric and "round" than its linear counterpart.

This is even more evident if all conformers are taken into account: in the linear molecule a chain end can even turn toward the center while the opposite being extended.

Now this does not necessarily explain the observed behaviour as for the more spherical molecule can pack easily but at expense of VdW forces that depend on molecular surface.

However, it relates to the fact that the linear isomer has a number of conformers much higher than the ramified one (rotation about 4 bonds versus 1 for each side of the ether)*** and results in the first having a bigger conformational entropic term.

Although the crystals will generally contain one isomer, the crystal formation take place at expense of the intramolecular forces. As such, when keeping entalpy terms comparable, bigger is the entropy term then less stronger is the binding.

$\Delta G = \Delta H - T\Delta S$,

For comparable enthalpies and very different conformational entropy terms, than the melting occurs at lower T for the entropy rich molecule, i.e. that one for wich more conformations become accessible.

This obviously is not the case for the liquid to gas transition, as the conformational S term is similar in both phases and negligible faced to the gas entropic content.

***Conformers due to the rotation around the bonds in the terbutyl groups do not really alter the shape of the molecules (connection with the compactness mentioned above). More rigorously at each side we should have (calling x the number of conformers around a bond) that the number of conformations is circa $x^4$ and about $3x^2$ for the linear and the branched molecules, respectively.

Additional note: conformational entropic terms are crucial to many phenomena in polymers and proteins (solubility, crystallization, protein folding/unfolding, etc).
A treatment of this requires statistics that I have forgotten. So take my numerical estimated about conformers with a grain of salt.  I am satisfied if I did convey the message.

Add. Note 2. Another useful example is, for instance, a stiff and insoluble chain such as polythiophene.

It is insoluble as for electron delocalisation makes it flat and rigid. Once side substituted with alkyl chains, the entropic term brought in by the flexible chain is the main responsible for the acquired solubility.


Does quantum yield of reaction factor in number of electrons needed?

I'm not aware of a definition of quantum yield that factors the electrons in. The normative reference was/is Silvia Braslavsky's Glossary of terms used in photochemistry (SRC).

Unless I misunderstood, you do have an overall four electron reduction, but this happens in a sequence of four excitation events.

From a formal point of view, the quantum yield(s) should be given for each of these processes. I reckon that this isn't possible, so an overall quantum yield of 0.25 is all you can get. In the end, I'd simply explain that, rather that looking for some "tricks" to come up with a more impressive number.


I would recommend Zumdahl Chemistry. It discusses most of the basic topics, and contains lot of exercises.


How does the relationship between carbonate, pH, and dissolved carbon dioxide work in water?

$^{\circ}KH$ or $dKH$ is called (degree of )carbonate hardness, but it is usually about bicarbonates $\ce{HCO3-}$, not carbonates $\ce{CO3^2-}$ . Carbonates are present in water just in traces, unless it is e.g. highly alkaline water from soda lakes of East Africa rift (like where cichlidae from lake Malawi live).
There is the equilibrium for carbon dioxide solution acidity, keeping the expression below constant:
$$K_\mathrm{a1,H2CO3^{*}} = \  \frac{\ce{[H+][HCO3-]}}{[\ce{CO2(aq,total}]}=10^{-\mathrm{pH}}\cdot \frac{\ce{[HCO3-]}}{[\ce{CO2(aq,total}]}=const \cdot 10^{(7-\mathrm{pH})}\cdot \frac{KH}{\mathrm{ppm}  \ce{CO2}},$$
where [] denotes the molar concentration.
Therefore $$\mathrm{ppm}  \ce{CO2} = \frac{\mathrm{const}}{K_\mathrm{a1,H2CO3^{*}}} \cdot 10^{(7-\mathrm{pH})}\cdot KH \approx 3 \cdot 10^{(7-\mathrm{pH})}\cdot KH$$

Pure water in equilibrium with air has theoretical $\mathrm{pH=5.6}$, given by dissolved $\ce{CO2}$ from air. There is established equilibrium
$\ce{CO2(g) <=>[H2O] CO2(aq) <=>[H2O] H2CO3(aq) <=> H+(aq) + HCO3-(aq)}$, but all components are present just in traces.

Addressing feedback:
Carbonate hardness implies there are dissolved bicarbonates $\ce{Ca(HCO3)2}$ resp. $\ce{Mg(HCO3)2}$. These are determined indirectly by titration, that determines bicarbonate content.
This of course fails, if there is sodium bicarbonate or carbonate presence, or if all bicarbonate comes from $\ce{CO2}$. Both cases have carbonate hardness lower or zero, compared to what would say its measurement.
In fact, measuring carbonate hardness ( dKH ) is measuring alkality:
$$\ce{HCO3-(aq) + H3O+(aq) -> 2 H2O + CO2(aq,g)},$$
assuming it is equivalent to content of $\ce{Ca(HCO3)2}$ or $\ce{Mg(HCO3)2}$. But it can be aslo sodium carbonate/bicarbonate, if added to water to increase alkality.

pH of water is determined by this equation:
$$\mathrm{pH} = 6.35 + \log {\frac{[\ce{HCO3-}]}{[\ce{CO2}]}}$$
both in molar concentration.
Calcium carbonate has solubility about 15 mg/L, what makes roughly 1 dKH.
$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-}$$
$$\ce{CO3^2-(aq) + H2O <=>HCO3-(aq) + OH-(aq)}$$
This solubility increases in presence of dissolved carbon dioxide,which decreases concentration of carbonate:
$$\ce{CO3^2-(aq) + CO2(aq)  + H2O <=> 2 HCO3-(aq)}$$
There is direct relation between pH and relative ratio carbon dioxide : bicarbonate : carbonate.
pH ( in usual aquarium range ) is directly determined carbon dioxide / bicarbonate ratio. If pH change is forced  by other chemicals, the new ratio is established.
BTW, there are just 2 degrees of freedom. You choose 2 of 3 parameters. The third one is determined by the former 2.


Is cis-1,2-dimethylcyclohexane a meso compound?

cis-1,2-Dimethylcyclohexane is achiral, not because there is a plane of symmetry, but because it consists of two enantiomeric conformations which interconvert rapidly via ring flipping at normal temperatures.

This is exactly the same case as amine inversion. "Chiral nitrogens" such as that in $\ce{NHMeEt}$ do not lead to chirality or optical activity because of rapid inversion of configuration at the nitrogen atom, leading to interconversion of the two enantiomeric forms 1a and 1b.

Likewise, the ring flip in cis-1,2-dimethylcyclohexane leads to two different conformers. I have deliberately chosen to depict the ring flip in the following fashion, to make the mirror image relationship more obvious. The green methyl group, equatorial in conformer 2a, is changed into an axial methyl group in conformer 2b. Likewise, the blue methyl group goes from axial to equatorial. The 1,2-cis relationship between the two methyl groups is retained in both conformers.

Each individual conformer can be said to be chiral, but just like how the amine is considered achiral, cis-1,2-dimethylcyclohexane as a whole is considered achiral.

Is the compound meso? According to the IUPAC Gold Book, a meso-compound is defined as:

  A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

1,2-Dimethylcyclohexane possesses two diastereomers, one cis and one trans form. The trans form is chiral, but the cis form is achiral, as explained above. Therefore, the cis form satisfies the above definition and is considered a meso-compound.


Your second idea, that they have the same volume, is correct.  Each gas in a mixture of gases stored in the same container has the volume of that container.  Remember that gases are mostly empty space, so it is not a problem that their volumes overlap.

As is implied by your comment about subtracting the vapor pressure of water from the pressure of wet hydrogen, the gases in a mixture can have different partial pressures. Perhaps that is your point of confusion.

The only case where they would have different volumes would be in a more accurate gas model than ideal gases, something like the Van der Waals model, where gas volume is corrected by a small amount to account for the portion of space that is taken up by the gas particles.

In short, each component of a mixture of gases stored in a given container shares the same volume and temperature and may (or may not) have different partial pressures and number of moles.


Why hydrogen bonding in some acids make them a stronger acid when it is present even before deprotonation?

Some of the conformers of salicylic acid are given below, all of which are conformers planar in configuration.$\mathrm{^{[1]}}$

Although the compound can adopt several conformations the first conformer(in the first image, the structure on the left) is of the lowest energy and has a strong intramolecular hydrogen bond. The second conformer(first image, right structure) is the conformer with the second lowest energy as described in the published paper as given below:

The molecule can adopt several conformations, although only one of them is highly populated and was assigned in IR experiments. The spectrum was recorded by Fiedler et al., $\mathrm{^{[2]}}$ in tetrachloride solution and indicated a strong intramolecular H-bond. The deviation of the OH
−1
stretching frequency from that in phenol was 395 cm . The theoretically calculated deviation is
359 cm−1 at the B3LYP/6-311+G(d,p) level. The lowest-energy conformer is planar, the phenolic OH is a H-bond donor to the carbonyl oxygen of the syn carboxylic group. The =O...H distance and the =O...H–O bond angle were calculated at 176 pm and 145°, respectively. The second-most-stable conformer is higher in energy by 14.3 kJ/mol, where the phenolic OH is the H-bond donor to the syn carboxylic OH. Similar conclusions were drawn by Yahagi et al.,$\mathrm{^{[3]}}$ by interpreting the gas-phase IR frequencies of the phenolic OH.

Much more convincing is the fact that the lowest energy conformer has a fractional population distribution$\mathrm{^*}$ of nearly 99.7% which further supports the argument regarding the higher acidity of salicylic acid in comparison to its para isomer. The higher acidity of the compound thus is contributed by the first conformer. The lesser interaction of the carboxyl group’s acidic proton and the stability of the conjugate base due to hydrogen bonding thus increases the acidity of the compound. Regarding detailed mechanism of acidity of salicylic acid you may refer this and this post in chemistry SE. You may further note that the other structures might have decreased the acidity of the compound if their population distributions would have been higher. The same has been mentioned in the published paper as given below:

In a former calculation by Nagy et al.$\mathrm{^{[4]}}$, the two conformers above were found also to be the most stable with MP2/6-31G*//HF/6-31G* energy separation of 13.7 kJ/mol and free energy difference of 12.1 kJ/mol at T = 298 K. All other conformers are much higher in free energy, supporting the estimate of Fiedler that the population of the lowest-energy form is 99.7%.

Notes and References

Nagy, Peter. Competing Intramolecular vs. Intermolecular Hydrogen Bonds in Solution. Int. J. Mol. Sci. 2014, DOI: 10.3390/ijms151119562
Fiedler, P.; Böhm, S.; Kulhánek, J.; Exner, O. Acidity of ortho-substituted benzoic acids: An infrared and theoretical study of the intramolecular hydrogen bonds. Org. Biomol. Chem. 2006, 4, 2003–2011. DOI: https://doi.org/10.1039/B601875K
Yahagi, T.; Fujii, A.; Ebata, T.; Mikam, N. Infrared spectroscopy of the OH stretching vibrations of jet-cooled salicylic acid and its dimer in S0 and S1. J. Phys. Chem. A 2001, 105, 10673–10680. DOI: https://doi.org/10.1021/jp0126199
Nagy, P.I.; Dunn, W.J, III.; Alagona, G.; Ghio, C. Theoretical studies of the 2- and 4-hydroxybenzoic acids with competing hydrogen bonds in the gas phase and aqueous solution. J. Phys. Chem. 1993, 97, 4628–4642. DOI: https://doi.org/10.1039/B601875K
$\mathrm{*}$ Note that here fractional population distribution refers to the population of each conformer in the compound. You may find more information about it on the Wikipedia article.


Why isn't chlorine reduced or nitrogen oxidised in the reaction between ammonium perchlorate and aluminium?

Here,

The oxidation number of chlorine in the perchlorate radical is +7 whereas, in the product side, chlorine has an oxidation number of  minus 1 in AlCl3. As the oxidation number is reduced by 8 here (gaining 8 electrons), chlorine is, indeed, reduced.

Nitrogen has an oxidation state of minus 3 in NH4ClO4 whereas, it has an oxidation state of 0 in N2 molecule in the product side. Here, the oxidation number of nitrogen has increased by 3 via losing 3 electrons. So, Nitrogen here is oxidised.

Hence, all the options are correct.


Raman Scattered (inelastic) photons rescattering outside of capture area

Yes, of course the Raman scattering can suffer from additional scattering on its way through the sample, just like it will generally be attenuated on its way to the detector by any element in its path (mirrors, lenses, glass surfaces, etc.).

If scattering is a problem or not very much depends on your sample. The ideal sample will not scatter your excitation laser or Raman scattering much, like for instance a clear solution, and then I would say the influence on the measurement is negligable. If you have something opaque, such as an emulsion, additional scattering will be relevant and you would want to measure near the sample surface.

If you have problems with a glass-affected background you could switch to quartz glass or use a confocal Raman setup, if available.

And also depending on your sample, absorption might be something you should consider as well.

By the way: I would probably rather go for a planar glass surface in order to avoid refraction as it occurs on a curved glass surface.

Edit due to comments:
Pure water does neither absorb nor scatter light with a wavelength around 532 nm much. If it would, you would be able to see this with your eyes. Wikipedia claims that a fraction of 10-7 of the incoming light is scattered inelastically. Although this should be only used as a rule of thumb, this tells you that a loss of Raman intensity due to "secondary" Raman scattering is not a practical problem you would usually be able to observe. Also, I would not be worried about elastic scattering although it is much more relevant than inelastic scattering (Wikipedia claims 0.1% to 0.01%).

I do not know your Raman setup in detail. But if you have anything like a confocal aperture (pinhole, slit...) on the path to the detector, refraction on the container/water surface will definitely reduce your signal intensity with increasing penetration depth, especially if you do not measure with an immersion objective lens. This is because your focus is severely distorted especially in length when focussing deeper into the sample, it will become much longer. Your confocal aperture then blocks part of your signal because it originates from a part of the sample which is far from the focal plane. This is descibed in detail for example in this paper, and also in this paper. However, the papers are behind a paywall.


The mass of the sulfate is $m_{\ce{SO4}} = 0.610~\mathrm{g}$, the mass of the metal is $m_\text{M} = 1.250~\mathrm{g}$.

With the molar mass of the sulfate anion $M_{\ce{SO4}} = 96.061~\mathrm{g\, mol^{-1}}$ we can calculate the amount of sulfate present in the final compound:
$$ n_{\ce{SO4}} = \frac{m_{\ce{SO4}}}{M_{\ce{SO4}}} = 6.35~\mathrm{mmol} $$

Since the sulfate has the formula $\ce{M2(SO4)2}$ we know that we have the same amount of metal as sulfate: $n_\text{M} = n_{\ce{SO4}}$.

We can now calculate the molar mass of the metal using the information above:
$$ M_\text{M} = \frac{m_\text{M}}{n_\text{M}} = 196.85~\mathrm{g\, mol^{-1}} $$

This number resembles most closely gold with a molar mass of $M_\text{Au} = 196.97~\mathrm{g\, mol^{-1}}$.

Note: It is important that you don't round your numbers while doing the calculations, but round them at the end. So you should keep as much precision as possible in between to minimize rounding errors along the way.

If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.35\ mmol}$ you will get $\mathrm{196.85\ g/mol}$ (gold).
If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.4\ mmol}$ you will get $\mathrm{195.31 g/mol}$ (platinum).


So far I feel like it is endothermic because heat had to be added to
  the water in order for its temperature to go up, so the water
  essentially absorbed heat, but I'm really not sure.

This is mostly semantics. Such exam questions are annoying when they make you argue about words. The term endothermic and exothermic are reserved for reactions by IUPAC. An endothermic reaction is one where the overall enthalpy change is positive. However one can extend it to endothermic process, which is any process that acquires energy (from external surroundings). Atkin's physical chemistry clarifies it nicely "it must never be forgotten that heat is a process (transfer of energy as a result of temperature difference), not an entity"


Why Heating of certain salts like Lead salts produce crackling sound?

Frankly speaking, you're reading too much into this phenomenon. Have you ever heard the satisfying crackling noise when you burn logs of wood in a bonfire? I'm sure you must've got the answer by now:
Decrepitation is more of a physical process rather than a chemical process.
When we write chemical equations, we treat compounds as single molecules, but in reality this is not so.
You specifically mentioned lead nitrate. You must be aware that it is a  white/colorless crystalline salt. To expound further,  solid lead(II) nitrate exists in a face centered cubic structure. Coming back to your question, when we heat Pb(NO3)2, we don't heat a single molecule, we heat a sample of the salt.
When we do so, the crystal structure breaks as the entropy increases. The deformation of the crystalline structure causes sound energy to be evolved.
TL;DR: Breaking down of large crystals into smaller crystals and/or intermediates causes the crackling sound known as decrepitation.


Do 1 mole of both dioxygen gas and its atoms occupy 22.4 litres at STP?

Does this mean that both

1 mole of $\ce O$ would occupy $22.4~\mathrm L$ (or if this doesn't usually occur in nature, say 1 mole of $\ce{He}$ or another monoatomic gas)
1 mole of $\ce{O2}$ would occupy $22.4~\mathrm L$

Yes, it means exactly that.  And you're right, a stable gas of $\ce O$ atoms is a pretty exotic thing, so $\ce{He}$ is a much better example.

Wouldn't $\ce{O2}$ contain twice the amount of $\ce O$ matter

Yes and no. The $\ce{O2}$ gas would indeed contain twice the mass of the $\ce O$ gas, but it would contain the same number of particles ("molecular entities") as the $\ce O$ gas.

and hence take up twice the volume?

Nope.  There is a lot of empty space in gases. A lot. So much, in fact, that molecules have to get pretty darn large in the gas phase$^\dagger$ before they start to get "cramped" and "insist" on taking up more space (viz., occupying a larger volume).
To give you an idea of just how much room there is, consider what's called the van der Waals volume of $\ce{O2}$.  This is an estimate of the total volume actually occupied by one molecule of $\ce{O2}$, and has a value of about $0.053~\mathrm{nm}^3$ (that's 'cubic nanometers').  For one mole of $\ce{O2}$ molecules, that works out to just under $32~\mathrm{cm^3\over mol}$ of volume actually occupied by the oxygen molecules themselves.
However, as you note, that mole of $\ce{O2}$ gas actually expands to fill a full $22.4~\mathrm L$ of space at STP. So, that space is actually only about ${32~\mathrm{cm^3}\over 22.4~\mathrm{L}} = 0.14\%$ full of $\ce{O2}$.
So, even though one $\ce{O2}$ molecule does take up more space than one $\ce O$ atom, compared to the amount of space available, they're both really, really tiny.  And, thus, the fact that the two $\ce O$'s in $\ce{O2}$ are bound so closely to one another means that they both behave essentially identically in the pressure they exert when they fill up a container of a given size.
$^\dagger$ At ambient temperature and pressure, anyways.


Question:

  Which one is larger: an atomic mass unit based on the current standard or one based on the mass of a Be-9 atom?

Answer

It is easy enough to look up the weights of $\ce{^{12}C}$ (12.00000000...) and $\ce{^9Be}$ (9.0121822) based on $\ce{^{12}C}$ scale and make a decision from that data. Thus a mass scale based on $\ce{^9Be} =9.00...$ would be heavier than an mass scale based on $\ce{^{12}C} = 12.00...$.

"Justification"

The OP has asserted that the problem is not supposed to be just a "lookup problem" but that some reasoning should be offered.

I content that the question simply asks which is larger. It doesn't ask why. Furthermore I'd assert that there isn't any "simple" reasoning by which the problem can be analyzed other than lookup.

Neutron to proton ratio reasoning

In his answer QuarkyLittleThing's reasoned that a neutron is heavier than a proton and since $\ce{^9Be}$ has a higher neutron to proton ratio (5:4) than $\ce{^{12}C}$ (6:6), it ought to be heavier.

Using the neutron to proton ratio reasoning, $\ce{^9B}$ only has 4 neutrons but 5 protons, so a $\ce{^9B}$ scale should be lighter than the $\ce{^{12}C}$ scale. But the mass of $\ce{^9B}$ is 9.0133288 u, so the neuron to proton ratio isn't entirely sound, and a mass scale based on $\ce{^9B} =9.00...$ would in fact be heavier than a mass scale based on $\ce{^{12}C} = 12.00...$.

Lower nuclear binding energy reasoning

Mithoron's made a comment on QuarkyLittleThing's answer asserting that the reason was "lower nuclear binding energy."

To properly analyze Mithoron's comment one must note that the nuclear binding energy is based on the energy differences of the actual fission/fusion of nuclei. Thus as NicolauSakerNeto pointed out on an earlier version of this answer, this is really not a "reason", but a tautology due to $\text{E}=mc^2$. So a decrease in the mass of the nucleus is proportional to an increase in the nuclear binding energy for that nucleus.


Why large basis sets give better approximations to the exact solution of the Schrödinger equation?

The statistics analogy may not be applicable, but it illustrates what
  goes on in my head.

The overfitting analogy is indeed not applicable here. It is a totally different and unrelated phenomenon. Basically, when we are doing regression analysis we do not know the relationship between a dependent variable and independent variables. Then we can indeed have an overfitting problem when instead of fitting the actual relationship we start to describe some random noise. It generally happens when we do have too many parameters in a model, and we never know how many is too many.

But when we are doing LCAO-MO we know the relationship between a dependent variable (say, energy) and independent variables (LCAO-MO fficients). ANd we are certain that the bigger the number of AOs (and correspondingly of the coefficients), the better result could be expected with the the exact solution corresponding to using infinitely many AOs. So we should not worry about anything like overfitting. Yes, by increasing the number of AOs in a basis set we increase the number of variational parameters, but we know it can't do any harm: in the worst case1 we will get the same energy as before the just another expansion of the basis.

This is similar to a Taylor series expansion of a function. The exact result corresponds to an infinite series, and by using a finite number of terms a function can only be approximated. But the more terms we use, the better approximation is. There is even a formal proof, but I think it should be intuitively clear that if an infinite series is the exact solution, than the later you truncate it the better approximation you get.

1) The worst case would be, for instance, when you expand a basis by including new functions that can be expressed as linear combinations of basis functions from the previously used basis set. You won't improve energy by doing so, but only increase the computational time. Note, that most of the programs have a simple defense mechanism against such situations: they are smart enough to detect the linear dependence in the basis and get rid of it prior to doing the calculations.


Why does sodium sulfate have an unusual solubility-temperature curve?

"Sodium sulfate" is a mixture of hydrated phases in shifting equilibrium given temperature and concentration.


How can FeS2 be the formula of iron sulphide and CaC2 be the formula of calcium carbide?

What you call iron sulphide, in my opinion, is more appropriately referred to as or iron disulphide.

If one were to assign oxidation states to each atom, an appropriate description would be  $\ce{Fe^2+}$ and $\ce{S_2^2−}$. This formalism recognizes that the sulfur atoms in pyrite occur in pairs with clear $\ce{S–S}$ bonds. These disulphide units can be viewed as derived from hydrogen disulfide, i.e $\ce{H2S2}$ (similar to hydrogen peroxide)

Take a look at the crystal structure (of pyrite):

In the center of the cell a $\ce{S_2^2−}$ pair is seen in yellow. The $\ce{S}$ atoms have bonds with three $\ce{Fe}$ and one other $\ce{S}$ atom.

Contrast this with, $\ce{FeS}$, i.e Iron (II) Sulphide where the counter anion is what you would typically expect $\ce{S^2-}$. This is reflected in its crystal structure, which is the nickel arsenide structure:

The bonding situation is completely different in these two compounds and this is reflected in their chemical formulas.

Similarly, for calcium carbide, the crystal structure is a tetragonally distorted $\ce{NaCl}$ lattice, comprised of discrete $\ce{Ca^2+}$
and discrete $\ce{C_2^-}$ ions.

Also, bear in mind what @Jon Custer said in his comment:

  Because the 'rules' on valences are not firm rules. On the $\ce{Fe-S}$ binary phase diagram, $\ce{FeS2}$ exists as a line compound

Also, I'll briefly mention what he probably meant by a line compound:

Vertical lines such as the one seen here for $\ce{FeS2}$ indicate inter metallic compounds, which have precise chemical composition.

References

This article for the crystal structure of calcium carbide: http://pubs.acs.org.ejgw.nul.nagoya-u.ac.jp/doi/abs/10.1021/j100010a022
The crystal structures of pyrite and Iron (II) sulphide both come from their respective wikipedia pages here and here
The phase diagram comes from: http://www.minsocam.org/msa/collectors_corner/arc/scn1.htm (which further credits it to Ehlers, 1972, after Kullerud, 1967 The Interpretation of Geological Phase Diagrams, Fig. 217, p. 232)


Can you write (with and ink pen) on hydrophobic paper?

This depends on what your ink is based on. If it can be washed off ordinary paper using water, it will be repelled on hydrophobic paper. However, if it can't, it has a chance of staying on.

Note that the paper in particular might repel all or most liquids, not just water. In this case, the ink has a very low chance of staying on the paper.


Rotovibrational selection rules for symmetric tops

Some of the statements you make above are not quite correct, rather that correct individual points I think that it is simpler to start from scratch, as it were, and try to explain what happens.

Two quantum numbers are needed to define the rotational energy of a symmetric top molecule, such as CH$_3$I or benzene. The total angular momentum is characterised by $J$ and this generally does not lie along the 'figure' axis of the molecule which is the C-I bond axis in CH$_3$I for example. This quantum number has integer values starting at zero, i.e. $0,\, 1,\,2\cdots$  The total angular momentum is $ \hbar\sqrt{J(J+1)}$. The second quantum number $K$ is the projection of $J$ along the figure axis and so has $2J+1$ values ranging from $-J$ to $+J$ and therefore has values $0,\,\pm 1,\, \pm 2, \cdots \pm J$. The size of the projection quantum number is $K\hbar$ and clearly as it is the projection along the figure axis $K$ cannot exceed $J$.

The energy levels of any vibrational level form stacks of $J$ levels rather like those of a rigid rotor but where each $K$ value has its own stack beginning with $J=K$. Each type of vibration has sets of $K$ and associated $J$ levels for each vibrational quantum number.

We are considering rotation-vibrational transitions. The selection rule for parallel transitions, i.e. those with a dipole moment along the figure axis always have $\Delta K=0$ and then $\Delta J = 0, \pm 1$ unless $K=0$ then transitions with $\Delta J=0 $ are forbidden. The reason that $\Delta K = 0$ is that radiation cannot induce transitions between different $K$ values because rotation about the figure axis induces no rotating dipole moment. (The dipole lies along the axis in this type of transition). When $K=0$ there is no angular momentum component along the figure axis and so if the total angular momentum is also zero there is no rotating dipole to couple to the radiation.

In the case of perpendicular transitions the dipole has a component perpendicular to the figure axis. Bending vibrations can induce perpendicular transitions. The selection rules are $\Delta K=\pm 1$ and $\Delta J = 0,\pm1$. In this case changing $K$ does induce a change in the dipole and so transitions can occur.


Why do some diagrams of some molecules show the bonds of only some of their submolecules?

Your example is one of a variety of shorthands that exist because every chemist after the third year of studies knows what is meant so they are too lazy to draw.

In a nutshell, the $\ce{NH2}$ group will always be a nitrogen with a lone pair, two single bonds (one to each hydrogen) and one bond to the molecular skeleton. Much like $\ce{-OH}$ will always be $\ce{-O-H}$. It is also common practice to omit any insignificant hydrogen atoms in organic molecules; in the image above, a total of eight hydrogens has been omitted because every chemist will know where they are (and that they are there).

In principle, the phosphate chain could also be shortened to $\ce{HO3P-PO2H-PO2H-{}}$ rather than writing out all of it. Any chemist will mentally ‘expand’ the molecule in the same (correct) way.

As you continue to study, you will notice when which groups are generally shortened, which rules to observe while doing so and which hydrogens are so superfluous that they are not drawn — and, most importantly maybe, how atoms are labelled that actually are missing hydrogens. it is beyond the scope of a single answer to note them all.


Why is formic acid a stronger acid than acetic acid?

We are discussing the following equilibrium

We can make the acid a stronger acid by pushing the equilibrium to the right.  To push the equilibrium to the right we can

destabilize the starting acid pictured on the left side of the
equation, and \ or
stabilize the carboxylate anion pictured on the right side of the
equation.

Comparing acetic acid ($\ce{R~ =~ CH3}$) to formic acid ($\ce{R~ =~ H}$), the methyl group is electron releasing compared to hydrogen.  Therefore the methyl group will stabilize the dipolar resonance form of the starting acid where there is a partial positive charge on the carbonyl carbon.  This should stabilize the starting acid. Further, this electron releasing ability of the methyl group will tend to destabilize the resultant carboxylate anion which already has a full unit of negative charge.

Therefore, because the methyl group 1) stabilizes the starting acid and 2) destabilizes the carboxylate anion product, the methyl group will push the equilibrium to the left, compared to the case where the methyl group is replaced by a hydrogen.  Consequently, acetic acid is a weaker acid than formic acid.


Why don't ions have one equilibrium potential? (Nernst equation)

Good question.  The answer (as I understand it) is that if there were only a single type of permeable ion, then there would be only one equilibrium ratio, and it would depend on the transmembrane potential.

However, in real life, there (a) is usually more than one membrane permeable ion, and more importantly (b) the transmembrane potential is not a fully independent variable but is affected by the transmembrane permeation of all those ions.

Imagine sovling a single instance of the Nernst equation when you know the electrical potential $\Delta \psi$:

$\frac{\Delta G}{F}=\text{ion motive force}=0=-m \Delta \psi+ RT\ln{\frac{[X_{out}]}{[X_{in}]}}$

If you know $\Delta \psi$ (and $R$ and $T$) and you are at equilibrium (so $\Delta G$ is 0), then the ion concentration ratio will always go to the same value, as you suggested in your question.

But now suppose there are many equilibrating ions, say sodium, potassium, protons, and chloride.  And suppose you don't know $\Delta \psi$.  Now, there is a Nernst equation for each species, as well as Gauss's law that $\nabla \centerdot \bf{E} = \frac{\rho}{\epsilon}$.  And probably also total material balances like $V_{out}[X_{out}] + V_{in}[X_{in}] = N_{tot}$ for each ion.   To fully solve the system to determine equilibrium, all equations have to be solved at once!  I hope you can see that (a) doing this is complicated and (b) if parameters like the volume of the cell $V_{in}$ or even more likely the total amount $N_{tot}$ of ions in the system changes for one ion, it could completely change the ratio $\frac{[X_{out}]}{[X_{in}]}$ for other ions.  That's why in general, the ratio is not the same every time.