According to the United States Environmental Protection Agency, there are 12 principles of green chemistry. The second method of synthesis is more green than the first method of synthesis as it adheres to the first, second and fourth principles while the first method does not. We can easily observe why this is so.

Firstly, the second synthesis method prevents wastage of any reactants, all the reactants go into the making of the product, copper (I) iodide while for the first method, there are other products generated along with the desired product. In other words, there is a wastage of reactants if the first synthesis method is used.

Secondly, the second synthesis method maximises atom economy (i.e. waste few or no atoms). This is similar to the first principle. Atoms are wasted in the first synthesis method in making the other products.

Thirdly, the second synthesis method produces what is desired and does not produce any other chemicals that may be hazardous. According to this and this, sodium tetrathionate and sodium thiosulfate are irritants to the human body.

Hope I have answered your question.


Mechanisms for Reimer-Tiemann and Kolbe-Schmidt reaction on deuterium labelled phenol

This is the mechanism of the Reimer-Tiemann reaction (source Wikipedia here)

The polyhaloform reacts with strong base to create a carbene, a highly reactive species which reacts with the phenoxide anion through the alpha carbon to give an anionic intermediate with the -ve charge on the halogenated carbon. It is this species that picks up the deuterium.
In Kolbe-Schmidt reaction, the process does not proceed through a carbene. It is nucleophilic attack (through carbon) of the phenoxide on the polyhaloform (or carbon dioxide) and the deuterium shifts to the phenol oxygen as the product re-establishes aromaticity. Mechanism from Wikipedia here


Yes, x << $\mathrm{0.118 M}$ that it is not gonna hurt even if it is ignored. But for a rigid perfectionist,
$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}][\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{x\times (0.118+x)}{0.118-x}$$
Hence we get a quadratic equation,
$$x^2 + 0.118000062 - 7.316\times 10^{-9}=0$$
Solving for $x$, we get $x = 6.199993484584665\times 10^{-8}$ which is approximately equal to the value of $K_\mathrm{a2}$.


Most people seems to think that hydrogen bonds are limited to $\ce{X-H\bond{~}Y}$ interactions, where $\ce{X}$ and $\ce{Y}$ are heteroatoms.

In the crystal, a $\ce{C-H}$ of an aromatic system or a $\ce{C=C}$ double bond can serve as a donor as well and interact with heteroatoms, such as oxygen atoms in $\ce{C=O}$ or $\ce{C_{arene}-O-C_{alkyl}}$.

These interactions can form complex networks of rings and strands.

We've examined and published that a while ago for a couple of different phthalimides, such as N-(4-methoxybenzyl)phthalimide or N-allylphthalimide, and some more.

While the articles are behind a paywall (they were published before Acta Cryst. E went open source), the CIF files can be downloaded and played with/examined.

A crucial parameter for hydrogen bonds is distance, if the acceptor is too far away, there is (obviously) no hydrogen bond.

As a rule of thumb, $2.3\,\mathrm{\mathring{A}}$ is a distance where hydrogen bonds can be expected.


Which coloring agents are responsible for the color in red tea and yellow tea?

I believe that it is the different polyphenols (Flavonoids are a type of polyphenol) in tea that give it its colors. Black tea has thearubigins that give it a red-brown color. The yellow color in tea comes from theaflavins. The color of the tea also depends on the acidity of the water used, because thearubigins are very weak acids, and the anions they produce have strong colors. If the water you use is alkaline, the color will be stronger. If it is acidic, the color will be lighter.


The method I've used is a simple propane torch, which you can get at Home Depot for $14 and magnesium powder. The mixture of powdered aluminium, iron oxide, and magnesium is thoroughly mixed and then lit using the propane torch. It lights all the time with the magnesium embers burning brightly and then the main reaction takes hold. Getting powdered magnesium is cheap as well. Using this method, the reaction takes all the time and I haven't had a single problem with it the numerous times I've done it for an audience.


Comparison of chemical shifts among alkanes in NMR spectroscopy

A carbon atom is slightly more electronegative than a hydrogen atom. For tertiary hydrogens, the extra bonded carbons will inductively withdraw more electron density from the hydrogen. This means the hydrogen will be deshielded slightly more, and resonate at a higher ppm.

I would classify cyclopropane in its own class, as its molecular orbitals make it somewhat unique. The C-C bonds do not exhibit exclusive sigma bond characteristics, and due to that phenomenon the hydrogens will hold slightly more electron density.


Why can the lone pair not align with the phenyl moiety in 2,6-xylidine?

There is no prove for this statement, as it is most likely incorrect for 2,6-xylidene. This is based on an electronic structure theory approach. However, there are non-negligible steric interactions, that affect the basicity as demonstrated in Ortho-effect in substituted aromatic acids and bases.

Whenever a distance between two atoms becomes (significantly) shorter than the sum of their respective van der Waals radii, this has to be considered. Because of the strain in the xylidene molecules, this is the case.

As a very rough estimate you can use the pen and pencil approach with tabulated values for van der Waals radii (german - much better overview) and covalent radii. Another approach is building the molecule with a molecular modeling set (Molekülbaukasten -  german wikipedia), rotate and wiggle bonds, to see if there are any close contacts.
There is also some free software that lets you build molecules and check them, they usually come equipped with a set of standard bond lengths and angles. It is always a good approach to visualise a molecule.
If you have access to quantum chemical tools you can perform calculations, but this is usually too much of an effort. For this educational purpose I ran some quick calculations, though.

The following table is composed from values from the german wikipedia pages (values in Å):
\begin{array}{rrr}\hline
             & \text{van der Waals} & \text{covalent}\\\hline
\ce{H}       & 1.10  & 0.32 \\
\ce{C}       & 1.70  & 0.77 \\
\ce{N}       & 1.55  & 0.71 \\\hline
\sum\ce{H-H} & 2.20  & 0.64 \\
\sum\ce{C-H} & 2.80  & 1.09 \\
\sum\ce{N-H} & 2.65  & 1.03 \\
\sum\ce{C-N} & 3.20  & 1.48 \\
\sum\ce{C-C} & 3.40  & 1.54 \\\hline
\end{array}

Now we can have a look at the optimised (DF-BP86/def2-TZVPP) geometry of the molecule. (Of course the pen and paper approach will result in a different geometry.)

Here you can see, that the $\ce{H\cdots{}H}$ distance is very close to the sum of the van der Waals radii. The $\ce{C\cdots{}H}$ distance with about $2.5~\mathrm{Å}$ and the $\ce{C\cdots{}N}$ distance with about $2.9~\mathrm{Å}$ are already shorter than the sum f their van der Waals radii. While these interactions are not strong, the will effect the behaviour of this molecule in solution (and gas phase). It is very important to understand, that these are only snapshot values. In reality the molecule is very flexible and bonds will rotate and vibrate. Steric interactions are basically electronic and dispersive interactions in disguise - they are most likely to influence the kinetic behaviour of a molecule.

The best evidence for this is the transition state of the rotation of the amine moiety. Here we can clearly see a bonding interaction between the nitrogen and the neighbouring hydrogen (see linked post), which is evidently influencing the basicity of this molecule.

This state is only about $7~\mathrm{kcal/mol}$ higher in energy, which means it is readily accessible at room temperature. This will mean that there are many different conformations present, there is no true one. This affects the reactivity of a molecule (in this case lowered basicity).

In conclusion, your gut feeling was correct. However, there are more effects at play and the whole picture is often not easy to see. It is usually a case by case decision/ interpretation how strong which effects affect reactions and molecules (properties and barriers). When substituents get larger the more dominant steric effects will obviously be.

I hope this answers your question sufficiently.


Is it safe to hold silicon nitride ceramic without protection?

I have a few silicon nitride balls in my drawer. They look pretty cool, that's why I keep them and touch them all of the time.

It is a very hard, glasslike material. It won't be easy to crack. There is nothing coming off the surface, ball bearings have extremely smooth surfaces (nm roughness), if anything would rub off, they would not be usable as ball bearings. Besides that, I don't think it is toxic. It is an extremely inert material, much more inert than steel, does not dissolve in much. It is even used in prosthetic parts that go into the body. I assume it needs a lot of certification for that.

If you break it (which won't be easy), you'll have some sharp pieces that might cut you, that's all.

The only risk would be from a ball bearing that was ground to a very fine dust.


Can a flame start on water (video included)?

Watch this video from 1:50 onwards to see the reactions of sodium and potassium with water. https://www.youtube.com/watch?v=H6ZDiRIvc2E

This video, 0:56, shows  gasoline (petrol) burning on water. (Note that the batteries have nothing to do with it). https://www.youtube.com/watch?v=Bs13jBWqFjA


Does cathodic protection work for air oxidation?

Not according to Wikipedia (current version of that page as of writing this answer) https://en.m.wikipedia.org/wiki/Cathodic_protection

  Galvanizing, while using the electrochemical principle of cathodic protection, is not actually cathodic protection. Cathodic protection requires the anode to be separate from the metal surface to be protected, with an ionic connection through the electrolyte and an electron connection through a connecting cable, bolt or similar.

So, the portions of your material exposed to air won't actually be receiving any "cathodic protection" formally per se, though galvanization may still help out quite a bit.


How can a chemical structure be stored in a matrix?

Are the use of MATLAB (or OCTAVE) and a self-designed molecular representation mandatory for you project?

Established 2D- and 3D representations for molecules beyond a connectivity matrix, i.e. with coordinates, (partial) charges, annotation of isotopes, etc. exist, such as in the form of the Chemical Markup Language (CML).

In addition, well-documented toolkits, frameworks and modules are available to operate on these data are available for C++, Java, Perl, Python, and Ruby.


Hybridization and bonding in ethenone H2C=C=O?

You are right.the hibrydization is the same with allenes but in allenes there are 2 CH2 groups but in ethenone there is a CH2 and there's a O with two none bonding pairs. The hydrogens will be in the same plane as C=C=O but  the electron pairs on oxygen are not in the same plane


What is the SNi mechanism of the reaction of moist silver oxide with a halide?

If you suspect that there is something different going on here, and I believe you are right.
My take, in the presence of lab light, $\ce{Ag2O}$ and $\ce{H2O}$ actually results, in the formation of the hydroxyl radical ($\ce{.OH}$) and the hydrogen atom radical ($\ce{.H}$) via the creation of a powerful photocatalyst couple $\ce{Ag2O-Ag}$ as outlined in this article: Ag2O as a New Visible‐Light Photocatalyst: Self‐Stability and High Photocatalytic Activity.
The photoelectric introduced electron hole $\ce{h+}$ and electron $\ce{e-}$ can act as follows in the present of water:
$\ce{H2O <=> H+ + OH-}$
$\ce{H+ + e- -> .H}$
$\ce{OH- + h+ -> .OH}$
The hydrogen atom radical is likely responsible for the extraction of the chlorine atom from the organic forming a product of $\ce{HCl}$.


How much more salt can I dissolve in hot water compared to cold water?

There seems to be two parts to your question. First, not all substances have increased solubility as water temperature rises. Gasses for example have decreased solubility in this case.

As for the second part of the question, increased heat adds energy to the water molecules which makes it easier to overcome the bonds between $\ce{Na}$ and $\ce{Cl}$. According to Wikipedia, the solubility of $\ce{NaCl}$ in nearly freezing water is $35.65 \, \mathrm{g} / 100 \, \mathrm{mL} \, \ce{H2O}$ versus boiling water which is $38.99 \,\mathrm{g} / 100 \, \mathrm{mL}\, \ce{H2O}$. Hope this helped.


What is the difference between a desiccant and a deliquescent?

A deliquescent solid will eventually dissolve in water it absorbed, like, for example, calcium chloride does. Molecular sieves are very good dessicants but they do not dissolve in water, so they are not deliquescent. As you can probably guess, operating with dessicants that are deliquescent may be more troublesome than with non-deliquscent ones, but it doesn't mean that they are inferior.

The choice of a proper drying agent depends on the material you want to dry. For example, non-deliquescent dessicants, like aluminium oxide, silicagel or molecular sieves are best to fill columns for solvent drying, because they won't clog when their drying capacity is exceeded. Deliquescent drying agents are perfect for the use in dessicators, because it is easy to remove them when they are exhausted.


Why do you use catalysts if they increase the rate of forwards and backwards reactions equally?

In a simplified representation, a catalyst lowers the overall activation energy while going from initial state A to the of the products B along the reaction coordinate. The interference of the catalyst on the reaction alters the reaction mechanism, renders alternative pathways accessible, which then may be  more favourable than for the reaction in absence of the catalyst. Catalysts are deployed regardless if the overall reaction is exothermic, or endothermic. And, because of the relationship described in the Eyring equation, changes in the free activation enthalpy influence rate constants.

For the instance of a heterogeneous catalysis, the picture is simplified as assuming energy chances for adsorption of the starting material(s) to and subsequent desorption of the product(s) from the catalyst were non-significant in the overall picture.

You are right that it may ease both the forward, as well as the backward reaction.  However, in absence of catalyst, overcoming the activation energy at all were just too expensive, even if the reaction $\ce{A -> B}$ drawn here happens to be an exothermic one.

As examples, think about the hydrogenation of benzene; in absence of Nickel there were practically no reaction at all, even at elevated temperatures.  Numerous biochemical reactions were unable to proceed in absence of a catalyst at ambient temperature, too.


An ionic equilibrium equation is simply one that involves ions.

Example : $\ce {HA <=> H^+ + A^-}$


How do you get contents of large flask into a smaller separatory funnel?

You should proceed as follows:

Charge the funnel with the mixture until it is almost full.
Let the content settle.
Separate the lower phase.
Separate the upper phase.
Repeat until you have processed all the mixture.

The ground joint is to close the funnel hermetically.
This is necessary when you need to perform an extraction.
A liquid with a mixture of components is charged in the funnel and an immiscible liquid is added.
The funnel is closed with a stopper and it is vigorously shaken.

BE VERY CAREFUL BECAUSE PRESSURE CAN DEVELOP INSIDE THE FUNNEL.

It is better to shake a bit first, take off the pressure, shake again, etc.
To release the pressure, put the funnel upside down and open the valve.
Put the funnel upright and let settle.
When there are two clean phases, release the stopper and take them out separately.
The components will be distributed between the phases in different quantities, so it is a way to purify those substances.


Equilibrium for very small amounts of reactants

Thermodynamics is only applicable to the systems of size in between astronomic and microscopic scales. Systems with a small number of particles can spontaneously abandon
equilibrium state due to fluctuations. Statistical physics tells us [1] that the relative magnitude of the fluctuations of thermodynamic quantities $x$ (e.g. pressure, temperature, volume, internal energy, entropy etc.) is inversely proportional to the root of the number of particles in the system:

$$
\frac{
\sqrt{\left\langle\left(\Delta x \right)^2\right\rangle}
}{
\left\langle x \right\rangle
}
\sim
\frac{1}{\sqrt{N}}
$$

Lets arbitrarily set the detection limit for experimental determination of the relative change of $x$ to optimistic value of $10^{-8}$. In this case we would need at least $N = (10^{-8})^{-2} = 10^{16}$ particles belonging to the thermodynamic system before we can speak of equilibrium at all.

References

Landau, L. D.; Lifshitz, E. M. Statistical Physics, 3rd Edition, Part 1: Vol. 5; Butterworth-Heinemann: Amsterdam u.a, 1980.


Cause of uniform glow in cathode ray tubes

The uniform glow is due to ionization and recombination of the residual gas; it's called a glow discharge. At higher pressure, a spark or arc discharge occurs at much higher current density. Fluorescent lamps and neon lamps operate in the glow-discharge region, and high-pressure xenon lamps use an arc discharge.

Elements of the residual gas can be identified by the color of the glow discharge. Nitrogen and argon in air, for example, glow purplish blue.

As pressure decreases further, density is too low (and path too short) to produce a visible glow discharge, but electrons, AKA "cathode rays", hurled off the electrodes impact the walls of the container, which may glow green in borosilicate glass, and in very low-pressure vacuum tubes, such as the 1G3GT high-voltage rectifier, the electrons impacting the anode produce X-rays (high energy photons), which may also create fluorescence in the glass shell.

The difference is easy to see in this video, which demonstrates the change in appearance from an arc discharge to glow discharge, and finally to the fluorescence of the glass envelope due to X-rays or electron bombardment. The arc starts at ~20 seconds, the glow discharge at ~26 seconds, striations form ~36 seconds, and by ~46 seconds, the glass envelope glows green from electron bombardment (though it might rather be the green of residual oxygen). This was one of my favorite demonstrations to show students the basis for spectroscopy!


Interpretation of H as T goes to absolute zero

Short answer: No. There's no reason to expect that this should be the case.

My standard reference for classical thermodynamics is Callen's Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Ch. 11 is relevant here.

Planck's statement of the third law is, according to Callen, "The $T=0$ isotherm coincides with the $S=0$ adiabat." In addition to the theoretical ramifications of having an isotherm and an adiabat coincide, the Planck statement also assigns to this adiabat $S=0$. This is physically grounded in statistical mechanics.

Let $\Delta$ represent the change in a quantity of a reversible isothermal process. Then, at zero temperature:

$S = 0$, by the third law.
$\Delta S = 0,$ again by the third law.
$\Delta G-\Delta H = T\Delta S = 0,$ because $T \to 0$.
$$\frac{\text{d}\Delta H}{\text{d}T}-\frac{\text{d}\Delta G}{\text{d}T} = \lim_{T\to0}\frac{\Delta H-\Delta G}{T} = \lim_{T\to0}\Delta S = 0,$$ once again by the third law. (This shows why the Gibbs free energy and the enthalpy are essentially equivalent thermodynamic potentials at low temperatures!)

But there is no reason to expect $\Delta H = \Delta G = 0$. Regardless of temperature, some energy is still required to effect a change in the system.


Is readily available ammonia a substitute for ammonia water?

Ammonia at standard temperature and pressure is a colourless gas. So when your recipe says "powdered ammonia" there must be some kind of mistake. It might refer to a salt, that can be produced with ammonia and hence serve as a source of it. One of such salts is ammonium chloride.

In household stores available ammonia is a solution of the gas in water. Depending on the product there might be other ingredients as well, for example surfactants if used as a cleaner. In any case there should be some information about the concentration of the solution. Often given in percent and it strength may vary, but I doubt it contains more than 10% in any case.

In conclusion, the available ammonia is ammonia water.

Lastly a word of warning. As an artist you are probably handling paint thinners and other chemicals all the time. Treat ammonia just as carefully as these. And keep your place well ventilated. You will probably do this anyway, because the smell is just horrible.


Why is the 5d orbital instead of the 4f orbital filled in lanthanum?

According to Ground-state configurations of ionic species I through XVI for   Z = 57-74 and the interpretation of 4d-4f emission resonances in laser-produced plasmas Phys. Rev. A 25, 275 (where "I" means the neutral atom) :

  We begin therefore by showing in the left-hand
  columns of Table I the configurations given by
  Martin et al. for lanthanum through lutetium.
  Here it is seen that at the fourth spectrum the 4f
  level has in all cases fallen below 5d and 6s. This
  is due to the well-known phenomenon of 4f wavefunction
  collapse [reference 5]. The effective potential in
  which the 4f electron moves is made up of an attractive
  Coulomb term and a repulsive centrifugal
  term, which, in general, combine to give a double
  well potential, the height of the intervening barrier
  depending on the nuclear charge Z. For neutral
  atoms the field in the region of the outer potential
  is essentially hydrogenic. The occurrence of the
  neutral lanthanides coincides with the appearance
  of bound 4f levels in the inner well, which, in turn,
  is due to the deepening of the well and the lowering
  of the potential barrier as a result of the increasing nuclear charge.

Here, "reference 5" is Spectral Distribution of Atomic Oscillator Strengths Rev. Mod. Phys. 40, 441 which goes into quantitative equations for the attractive Coulomb term and a repulsive centrifugal term.


Davies and Rideal1 published the following graph. This graph nicely describes the action of various solutes on a solution.

It's clearly visible that:

Addition of $\ce{MeOH}$ reduces the surface tension nonlinearly
Addition of $\ce{NaCl}$ increases the surface tension linearly, but by a small amount
Addition of a surfactant such as $\ce{RSO3Na}$ decreases the surface tension drastically, up until Critical Micelle Concentration is reached.

Smith and Gillham2 present a more detailed graph of the effect 1-butanol has on the surface tension of water.

References:

Hansen, Robert S. “Interfacial Phenomena (Davies, J. T.; Rideal, E. K.).” Journal of Chemical Education, vol. 39, no. 7, 1962, p. A552. doi:10.1021/ed039pA552.
Smith, James E., and Robert W. Gillham. “Effects of Solute Concentration-Dependent Surface Tension on Unsaturated Flow: Laboratory Sand Column Experiments.” Water Resources Research, vol. 35, no. 4, 1999, pp. 973–82. doi:10.1029/1998WR900106.


Why are tertiary carbon atoms of polymers more prone to photo-oxidation and -degradation?

This is admittedly a very brief and somewhat tempting answer and it might not be complete.

Photo-oxidation and -degradation occurs mainly via a radical mechanism. This is outlined on Wikipedia in more detail. In principle the initial step produces bond cleavage via UV radiation and with it radicals. Let $\ce{[P]_{$n$}}$ be a polymer, then $\ce{.[P]_{$n'$}}$ and $\ce{.[P]_{$n''$}}$ are decomposition products of this polymer, both of radical nature.
$$\ce{P ->[h\nu] .[P]_{$n'$} + .[P]_{$n''$}}$$

In a saturated non chained polymer a homolytical bond cleavage would lead two primary radicals, i.e. $\ce{R-CH2.}$, which are very unstable. In a chained polymer, like PP, if the bond break occurs at the tertiary carbon, there would be a secondary radical $\ce{R-\overset{\bullet}{C}H-CH3}$ and another primary radical. Homolytical bond cleavage of a hydrogen carbon bond can even lead to a tertiary radical, $\ce{R-\overset{\bullet}{C}(CH3)2}$.  All in all this can lead to a higher concentration of radicals, and therefore more access points for oxygen to attack, hence propagating the chain. One of the driving forces is the gain in entropy of this process.

$$\begin{align}
\ce{.[P]_{$n''$} + O2 &-> [P]_{$n''$}OO.}\\
\ce{[P]_{$n$} + [P]_{$n''$}OO. &-> [P]_{$n''$}OOH + .[P]_{$n$}}\\
\ce{[P]_{$n''$}OOH &->[h\nu] [P]_{$n''$}O. + .OH}\\
\ce{[P]_{$n$} + [P]_{$n''$}O. &-> [P]_{$n''$}OH + .[P]_{$n$}}\\
\ce{[P]_{$n$} + .OH &-> H2O + .[P]_{$n$}}\\
\text{etc.}&\phantom{\ce{->}~~}\text{pp.}
\end{align}$$


What does a comma signify in inorganic chemistry?

Quoting [1]:

  Comma. In accord with general usage, the comma "," should be used to
  indicate continuous substitution, partial as well as complete, if
  there is no change of CN. Vacancies (symbol $\square$) are treated here just
  like atoms. This usage is restricted to chemical formulas for atoms on
  a specific site. The comma is not appropriate for structure site
  formulas because different symbols indicate a change in the
  coordination number. Examples for proper use of the comma include
  forsterite-fayalite $\ce{[(Mg{,}Fe)2(SiO4)]}$ and hornblende $\ce{\{(Na{,}\square)Ca2(Mg{,}Fe{,}Al)_5[(Si{,}Al)4O11]2(OH)2\}}$. In structure type formulas, the
  comma is used to mean "or" where substitution may occur. An example is
  in the zeolites where H2O and M units may both occur.

Comma representation are often used in writing chemical compositions of minerals as the elements and its concentration vary to a certain extent. See minerals.net:

  A number of minerals contain a varying amount of two or more elements like Aurichalcite, which has a chemical formula of
  $\ce{(Zn{,}Cu)5(CO3)2(OH)6}$ contains an unspecific varying amount of
  zinc (Zn) and copper (Cu). This is indicated by comma separating the
  Zn from the Cu. If a chemical formula with two elements in
  parenthesis is separated by a comma, the number of those elements
  vary. Aurichalcite has a variable amount of zinc of copper where the
  combination of both these elements totals five. The more dominant
  element is usually listed first. (Emphasis added.)

For more information, see this previous Chemistry.SE question and see here to find steps regarding how to write such type of formulae using chemical markups.

References:

Smith, D. K.; Roberts, A. C.; Bayliss, P.; Liebau, F. A systematic approach to general and structure-type formulas for minerals and other inorganic phases. Am. Mineral. 1998, 83 (1-2), 126–132. DOI: 10.2138/am-1998-1-212. Online version: http://database.iem.ac.ru/mincryst/fst.html


Why is barium used instead of bismuth as a contrast in gastrointestinal xrays?

Is it less expensive?

Yes, but not so much that it would be favored, i.e. you don't worry about saving \$5 when your doing a \$5000 procedure.

  Is there a particular property of barium that makes it a better contrast?

No both absorb x-rays as a result of their high atomic numbers, in fact bismuth has more electrons per atom and thus is a better contrast agent..

  Bismuth is a heavy metal that has a relatively low toxicity in comparison to barium.

Yes but this is for the soluble metal ion. Barium sulfate is inert under most conditions, importantly the acidic ones in the digestive tract. Because the sulfate binds the barium ions to the solid and thus barium cannot enter the blood stream to cause a physiological response. Contrast this to Bismuth (no pun intended) which is non-toxic but typically soluble in acid and if not soluble, the bismuth is at least leachable. Now typically bismuth ions are not an issue as they will harmlessly be excreted, but there is a potential it could interact with a patient or the drugs of a patient in a negative or uncontrolled way. Since doctors don't like to introduce variables when making diagnosis's nor do the like surprises, especially when it's caused by them, the fully inert barium sulfate is the favorite for radio contrast.


Your reasoning is correct that $K$ is independent of concentration. But I think you are misreading the question: it wants you to tell how you will miscalculate K with respect to the correct value if you put in the wrong concentrations in the formula, that is, if there was an overshoot in your titration.


When a carboxylic acid acts as a nucleoplile, why does it attack through the carbonyl bond?

For both the examples that you mentioned, the carbonyl oxygen has its lone pair localized while the other oxygen's lone pair is delocalized with the double bond. The better nucleophile is the one which has greater electron density.


Can water be ignited in an ozone atmosphere?

We are basically talking about a derivation of the $\ce{H2/O2}$ system, which involves a radical mechanism. You maybe want to have a look at this answer of mine going a bit more into detail.

The reaction you are interested about is only a total sum of reactions and is most likely not to occur as other reactions will be favourable. Ozone itself is not a very stable compound and many other molecules may actively catalyse decomposition. There are various publications[1-3] on the mechanism of this reaction:
\begin{align}\ce{
  M + O3 &<=> M + O2 + O\\
  O + O3 &<=> 2O2\tag1
}\end{align}

The range of $\ce{M}$ is in $\ce{O3, O2, CO2, N2, He, etc}$.[1] Although not stated explicitly, there should basically no reason why water is not be able to accomplish the same. Hence the driving force of the total reaction would be entropy, giving the overall equation:
$$\ce{2 O3 ->[\ce{H2O}] 3 O2}\tag2$$

It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}\tag3$$

In presence of Hydrogen donors ($\ce{RH}$ apparently any kind will do) a chain reaction will be initiated.[1] $$\ce{O + RH -> OH + R}\tag4$$

The main chain carriers will be $\ce{HO, HO2}$.
\begin{align}\ce{
  HO + O3 &-> HO2 + O2\\
  HO2 + O3 &-> HO + 2O2\tag5
}\end{align}

The process will be terminated by this (quite unspecific) reaction. This may then also involve $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[1]
$$\ce{2HO2 ->}\text{stable products}\tag6$$

In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}\tag7$$

Steahelin and Hoigné propose several hypothesis in which way ozone will be decomposed, but I will not go into absurd detail.[4] The core assumptions involve $\ce{{}^{-}OH}$ as initiator of a radical chain.
\begin{align}\ce{
  O3 + {}^{-}OH &-> .O2- + HO2. &  HO2. &<=> H+ + .O2-\\
  O3 + {}^{-}OH &-> HO2- + O2   &  H2O2 &<=> HO2- + H+\\
                               &&  O3 + HO2- &-> .OH + .O2- + O2\\
  O3 + {}^{-}OH &-> .O3- + .OH\tag8
}\end{align}

Which is basically followed by another set of radical chain reactions.
\begin{align}\ce{
  O3 + .O2- &-> .O3- + O2\\
  .O3- + H2O &-> .OH + {}^{-}OH + O2\tag9
}\end{align}

Which is all again resulting in $(2)$ again.

In summary the proposed reaction might occur, but will immediately be followed by another set of reactions resulting in the formation of biatomic oxygen.

References

Sidney W. Benson, and Arthur E. Axworthy Jr., J. Chem. Phys., 1957, 26, 1718.
J. V. Michael, J. Chem. Phys., 1971, 54, 4455.
Oliver R. Wulf, and Richard C. Tolman, *Proc. Natl. Acad. Sci. USA, 1927, 13 (5), 272–275.
Johannes. Staehelin, and Juerg. Hoigne, Environ. Sci. Technol., 1982, 16 (10), 676–681.


What you are seeing is most likely PVC plasticiser.

Pure PVC is a hard, inflexible solid. It doesn't make for good flexible objects. To make it flexible, plasticisers are usually added. These plasticisers are often compounds like diisononyl phthalate (DINP), diisodecyl phthalate (DIDP) and di-2-ethylhexyl phthalate. These are, when pure, oily liquids. They often slowly migrate from the PVC object and can do so faster in certain circumstances (eg they will be pulled out when the PVC is in contact with hydrocarbon solvents).

So, most likely, you are seeing small amounts of plasticiser migrate from the PVC and not any degradation of the polymer itself. Eventually, of course, if enough plasticiser leaves the PVC object it will become brittle and inflexible. But it may be quite some time before that happens.


describing in words a primitive heated water vapor lift

To carry out the B transformation, you have to heat a lot. If you don't heat, the volume will of course increase, but the pressure will decrease. B is a process difficult to carry out, because it is not easy to  heat enough a gas who is inflated to as to maintain the inner pressure. Then, at the end of B, the temperature is very high. You block the position of the piston, and you cool the system : the pressure will decrease from P2 to P1, during the process C. The last process D is also difficult to carry out. Because in D you must compress the gaz so as to decrease its volume, but the pressure must not increase. It means that the system should be cooled during D. At the end of D, the temperature will be extremely low.


Entropy change when indistinguishable particles suddenly become distinguishable

Very interesting question. The issue is that your formula for $\ln Q_\mathrm{indis}$ does not hold for $N = 1$.

Since the rotational, vibrational and electronic degrees of freedom do not come into play I will just ignore them. The way you derived the term $\ln (q_\mathrm{tr}e/N)$ comes from the use of Stirling's approximation

$$\begin{align}
Q_\mathrm{tr,indis} &= \frac{q_\mathrm{tr}^N}{N!} \\
\ln Q_\mathrm{tr,indis} &= N \ln q_\mathrm{tr} - \ln(N!) \\
&\approx N \ln q_\mathrm{tr} - N \ln N + N \\
&= N\ln\left(\frac{q_\mathrm{tr}e}{N}\right)
\end{align}$$

Plugging $N = 1$ into Stirling's approximation however tells you

$$\ln (1!) \approx 1\ln 1 - 1$$

or $0 \approx -1$, which is obviously a problem. Stirling's approximation is really meant to be used for large $N$ only!

Now, you could try working it out without using the approximation: plug $N = 1$ into the line before the approximation $\ln Q_\mathrm{tr,indis} = N \ln q_\mathrm{tr} - \ln(N!)$ and you will get the result

$$\ln Q_\mathrm{tr,indis} = \ln q_\mathrm{tr}$$

which matches the expression for the distinguishable case, so all your terms will cancel out and you will get the desired answer of $\Delta S = 0$.


Difference in work done by a gas in reversible and irreversible isothermal expansion

For a massless piston, the force per unit area exerted by the gas on the inner face of the piston is exactly equal to the external pressure exerted by the surroundings on the outer face of the piston, irrespective of whether the expansion is reversible or irreversible (by Newton's 2nd law).  The work done on the surroundings is thus always equal to the integral of the external pressure integrated over the volume change.

However, for an irreversible expansion, the force per unit area exerted by the gas is comprised of two components: (1) the local gas pressure at the piston face (as determined by applying the ideal gas law locally) and (2) viscous stresses that are proportional to the rate at which the gas is expanding.  Furthermore, in an irreversible expansion, the gas pressure and temperature are typically not even uniform within the cylinder.

In a reversible expansion, which is carried out very slowly, the viscous stresses are negligible (since the rate of expansion is negligible), and the gas pressure and temperature are uniform throughout the cylinder. Thus the external pressure is equal to the uniform pressure of the gas calculated from the ideal gas law.  Consequently, for a reversible expansion, one can use the pressure calculated from the ideal gas law in place of the external pressure to calculate the work done.

To get a better feel for how this all plays out, particularly with regard to viscous stresses and viscous dissipation, please see the following link:  https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/.  This article points out the close analogy between the transient behavior of an ideal gas experiencing an irreversible expansion or compression, and that of a spring and damper connected in parallel. It demonstrates that the physical mechanism responsible for the difference between the irreversible work and the irreversible work is viscous dissipation of mechanical energy to thermal energy (yes, even in the ideal gas region, gases exhibit non-zero viscosity).


Inorganic Chemistry: the order of MO diagram

The numbers that come in front of each molecular orbital irrep are simply referring to the energy order in which the MO's are arranged, with the maximum number (for example, $5a_1'$ for $BF_3$) corresponding to the highest energy MO that has an $a_1'$ symmetry.

For understanding the molecular orbital energy order, you need to explicitly consider the combination of the fluorine ligand group orbitals and the boron AO of the same symmetry. If you do that you'll find $4a_1'$ has perfect bonding interactions between all the ligand p-orbitals involved in $\sigma$-bonding and the boron s orbital (both transform as $A_1'$ in $D_{3h}$). The identity of the p-orbitals would depend on the co-ordinate axis you choose to define as the molecular axis. $3e'$ would arise because of $\sigma$ interactions between the same ligand p-orbitals, this time slightly differently oriented and the boron p-orbitals. The $a_2''$ MO is pi-bonding in nature, so on and so forth.

But without further ado, I'll point you to this excerpt from Purcell and Kotz's Inorganic Chemistry, which is self contained and provides a detailed pictorial description of the various ligand group orbitals very nicely.

Best of luck!


Periodically oscillating chemical reactions?

A couple of points:

The demo is probably an oscillating chemical reaction, in which the concentration of a given species increases and decreases repeatedly over time. However, the energy of the overall system is dissipating as that is happening - not oscillating.

Resonance, on the other hand, is a (widely misunderstood) concept that applies to a single chemical species and not a bulk reaction mixture. Resonance does not consist of energy flowing back and forth between two chemical states, but rather is meant to represent a single energy state that is a hybrid between two (or more) classical Lewis structures. A double headed arrow between the two classical structures indicates resonance stabilization — mostly resulting from delocalization of electrons relative to the classical structures. The double headed arrow does not indicate oscillation between those two structures.


How did Williard Gibbs come up with the Gibbs equation?

He derived this formula by fully appreciating the fact that entropy tends to a maximum.

As the first paragraph of the relevant paper says,

  It is an inference naturally suggested by the general increase
  of entropy which accompanies the changes occurring in any
  isolated material system that when the entropy of the system
  has reached a maximum, the system will be in a state of equilibrium.
  Although this principle has by no means escaped
  the attention of physicists, its importance does not appear to
  have been duly appreciated. Little has been done to develop
  the principle as a foundation for the general theory of thermodynamic
  equilibrium.

That is to say, he saw it as obvious that entropy increases until it cannot increase anymore and this makes a for a good definition of equilibrium. Gibbs offers a second definition of equilibrium, however, by saying,

  For the equilibrium of any isolated system it is necessary and
  sufficient that in all possible variations of the state of the system
  which do not alter its entropy, the variation of its energy shall
  either vanish or be positive.

The new move, however, is to construct some kind of function which makes these two definitions simply the same definition stated in a different way. That is, Gibbs has introduced two conditions:

$$(\delta S)_{\epsilon} \le 0$$
$$(\delta\epsilon)_{S} \ge 0$$

In words, at equilibrium a system's entropy (for a constant amount of energy) can only change by decreasing, and at equilibrium a system's energy (for a constant entropy) can only increase. Or simply, entropy is a maximum and energy is a minimum.

Then, because these two properties are opposites in the sense that one tends to a maximum and the other a minimum, it is a reasonable thing to subtract them and say that when this subtraction is equal to zero, we are at equilibrium. This is exactly what Gibbs did. He literally just multiplied the entropy by temperature without justifying it ($TS$ is common on thermodynamics though so he may have thought it obvious), and wrote down:

$$\psi=\epsilon-TS$$

where in this case $\psi$ is what we now call the Gibb's free energy. With this definition, the two conditions for equilibrium above are turned into one condition which is,

$$(\delta\psi)_T \ge 0$$

That is to say, at a constant temperature physical systems tend to the minimum of a free energy surface.

The reasoning Gibbs' has walked through thus far is sufficient to derive the equation as you have written it if we say that $\epsilon=\epsilon(U,P,V)$. That is, the energy of the system is determined fully by the internal energy (imagine microscopic motions) and the pressure on the substance and volume of the substance. Then we can simply write (again he just guesses a proportionality of one and uses units to figure out the rest),

$$\zeta=U+PV-TS=H-TS$$

where $\zeta$ is again the Gibb's free energy in a more explicit form. He takes the next couple of pages to take the total differential and then simple integration would give the familiar form but this is just a mathematical exercise not worth going through here.

A more formal, but not much more complicated, derivation of Gibbs' energy from the same condition that entropy increases can be found here under free energy of reactions. In this case two conditions for the entropy are used. Namely, the internal and external entropies. That's worth looking at as it contains the same principle which Gibbs used.


There are a few widespread misconceptions about this topic. The original question, edits, and comments give the impression that such misconceptions may also contribue to the confusion in this case. Therefore, we should like to clear up some of the misconceptions first.

High-octane fuel does not contain a high concentration of octane.

The composition of gasoline varies widely; anyway, typical gasoline contains less than 1 % octane.

The octane number of gasoline does not describe the composition of gasoline. It refers to the measurement of the knock resistance of gasoline, which is usually expressed by their research octane number (RON) or motor octane number (MON). Both parameters are determined in single-cylinder four-stroke test engines with an adjustable compression ratio according to standardized test procedures, e.g. EN 25164 or ASTM D 2699 for RON, and EN 25163 or ASTM D 2700 for MON. During the measurement, the compression ratio is raised until a certain standardized level of knocking is reached. The corresponding octane number is determined by finding a reference mixture of heptane and “octane” that gives the same knock level. The octane number is the fraction of “octane” (in %) in the “octane”-heptane mixture. For example, gasoline with an octane number of 95 has the same knocking performance as a mixture of 95 % “octane” and 5 % heptane; however, this does not mean that the gasoline contains 95 % “octane”.

Nevertheless, the wrong idea that gasoline would contain a high concentration of octane is widespread. For example, in the first relevant Google hit, Ritter, S. Chem. Eng. News 2005, 83 (8), 37:

  Octanes are an important component of gasoline because they help provide smoother combustion in the car’s cylinders and prevent knocking.

Also remarkable is the fact that many textbooks and lectures select the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion:

  $$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$

The reference substance for the octane number is not octane.

The octane number refers to a mixture of heptane and “octane”. However, the “octane” used for this mixture is not the unbranched alkane octane. It is actually a branched isomer of octane with the traditional name “isooctane”. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % “isooctane” and 5 % heptane; however, this also does not mean that the gasoline contains 95 % “isooctane”.

The octane number of pure octane is actually very low. Its knock resistance is worse than the performance of pure heptane, which has a octane number of zero by definition.

The “isooctane” that is used as a reference substance for the octane number is not isooctane.

The name isooctane was in accordance with older nomenclature rules. The corresponding systematic name according to current IUPAC recommendations is 2-methylheptane. However, the “isooctane” that is used for the determination of the octane number is not isooctane, i.e. it is not 2-methylheptane. This “isooctane” actually is 2,2,4-trimethylpentane, which is another isomer of octane. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % 2,2,4-trimethylpentane and 5 % heptane; and once again, this does not mean that the gasoline contains 95 % 2,2,4-trimethylpentane.

By way of comparison, the octane number of pure isooctane (2-methylheptane) is only about 23.

Irrespective of which compound is really meant in the original question (octane, 2-methylheptane, or 2,2,4-trimethylpentane; the question and comments also mention the similar alkenes 6-methylhept-1-ene and 2,4,4-trimethylpent-1-ene, which might be a mistake), such components of gasoline are usually not selectively synthesized in a pure form.

Gasoline consists of hundreds of different compounds, and the exact composition can vary widely. It is produced by blending of different product streams of oil refining. Today, only a small part (so-called straight-run gasoline) is directly obtained by distillation of crude oil. Other important product streams are thermally cracked gasoline, catalytically cracked gasoline, catalytic reformate (so-called platformate), isomerate, alkylate, polymer gasoline, and oxygenates.


Qualitative explanation of how COSY works

In the 1D-proton experiment a hard 90° pulse is applied to the sample before the FID is acquired over some time, t. This set of events is known as the ‘pulse sequence’, and the resulting data is in the time domain, and needs to subjected to a Fourier transform in order to get a 1D spectrum in the frequency domain.

2D experiments are measured much in the same way, except instead of running the pulse sequence once, we run it multiple times, changing some variable and collecting all of the data into one set which will later become the 2D spectrum. [*]

The  standard COSY experiment is one of the simplest 2D experiments to understand. It’s pulse sequence differs from the 1D-proton experiment only by an additional hard 90° pulse. We can explain most (though not all) properties of the COSY experiment using a vector model with no mathematics.

—

Consider a single magnetisation vector, M, which isn't coupled to anything. $\ce{CHCl3}$ would be a good example of this, and leads to a single cross-peak along the diagonal of the  resulting COSY spectrum (a trivial result, but easier to explain).

The experiment consists of 4 steps:

An initial 90° pulse is applied, which transfers M from the z-axis down to the y-axis. This is no different to the pulse used when acquiring a 1D-proton spectrum.
M is allowed to precess for some time, t1. During this time, M ends up somewhere in the xy-plane (having both an x and a y component). The exact position will depend on the length of t1, which is the variable we will differ during the acquisition of our COSY.
A second 90° pulse is applied. This 90° pulse transfers the y-component back onto the z-axis (-z to be precise). The x-component is unaffected (this requires more than a qualitative explanation).
Immediately following the second 90° pulse, we detect. The FID is acquired during t2, which remains constant (just as multiple scans of a 1D proton are acquired for the same length of time). What we are actually acquiring is the residual magnetisation in the x-axis (which wasn't affected by the previous 90° pulse), similar to the standard 1D-proton experiment. If we applied a Fourier transform at this stage, we would indeed just end up with a 1D spectrum. FID in memory to be combined with subsequenc
The above procedure (steps 1-4) is repeated, changing t1 and keeping all other variables constant. For each value of t1, the FID is recorded during t2 - this is the basis of the COSY experiment. The data is combined into a single set, and a Fourier transform applied twice which gives us the 2D-COSY spectrum in the frequency domain. In this simple case, we'd end up with a single cross-peak along the diagonal corresponding to the chemical shift of the proton (in $\ce{CHCl3}$) as expected:

In more complex systems, where multiple protons are present, we observe additional cross peaks in the COSY spectrum off the diagonal corresponding to the coupling between pairs of spins.

The appearance of these additional cross peaks cannot be explained qualitatively using a vector diagram (as per above), but is a consequence of the second 90° pulse that we apply prior to acquisition of the FID.

Where coupling exists, the second 90° pulse doesn't just change the position of the magnetisation vector M, but also causes a population change for the transitions of the spin system through a phenomenon known as coherence transfer.

[*]: There is some potential confusion here, as even in a 1D experiment, we typically run the pulse sequence multiple times in order to improve signal:noise ratio. In this case, all of the variables are kept constant. In a 2D experiment, we run the pulse sequence multiple times, changing some variable in order to build up a 2D data set. We may also run the set of pulse sequences multiple times, to improve signal:noise ratio as per the 1D experiment.


What is the chemical composition of the human body?

oxygen 61.3533
carbon  22.8291
hydrogen 9.9877
nitrogen 2.5683
calcium 1.4268
phosphorus 1.1129
potassium 0.20
sulfur 0.20
sodium 0.14
chlorine 0.14
magnesium 0.03
iron 0.01
fluorine 0.004
zinc 0.003
silicon 0.001
rubidium 0.001
strontium 0.0005
bromine 0.0004
lead 0.0002
copper 0.0001
aluminum 0.0001
cadmium 0.0001
cerium 0.0001
barium 0.00003
iodine 0.00003
tin 0.00003
titanium 0.00003
boron 0.00003
nickel 0.00002
selenium 0.00002
chromium 0.00002
manganese 0.00002
arsenic 0.000010
lithium 0.000010
cesium 0.000009
mercury 0.000009
germanium 0.000007
molybdenum 0.000007
cobalt 0.000004
antimony 0.000003
silver 0.000003
niobium 0.000002
zirconium 0.000001
lanthanium 0.000001
gallium 0.000001
tellurium 0.000001
yttrium 0.0000009
bismuth 0.0000007
thallium 0.0000007
indium 0.0000006
gold 0.0000003
scandium 0.0000003
tantalum 0.0000003
vanadium 0.0000002
thorium 0.0000001
uranium 0.0000001
samarium 0.00000007
beryllium 0.00000005
tungsten 0.00000003

Reference: Emsley, John, The Elements , 3rd ed., Clarendon Press, Oxford, 1998


Is it possible to calculate specific heat capacity of a compound based on its elements' heat capacities?

There is not much connection between the heat-capacity of a compound and the heat-capacity of the elements it is made of. Heat capacity is primarily coming from the low frequency vibrations of the lattice (if it is solid), and vibration+rotation+translation if it is liquid or gas. These have no connection with the properties of the lattice etc properties of the given elements.

So short answer: no


Is is possible to do mass spectrometry without coupled liquid chromatography?

The practical answer really depends on two things: do you know exactly what you're looking for and can your instrument do tandem MS?

If you know exactly which metabolites you're looking for and they're present in great enough abundance, it may be possible to resolve them from the the background, depending on the resolving power of the instrument and what else is in there (if you have an ICR instrument, it may not be a problem at all, but if it's just a single quadrupole, you may be out of luck).

If you know exactly which metabolites you're looking for (or know that they share some common structural feature that can be knocked off by CID or the like) and your instrument can do tandem MS, you're in a much better position. Tandem MS is often used to avoid needing to do a slow LC step as it's possible to identify two compounds that aren't resolved on a precursor ion scan by looking at their fragment ions: A messy primary ion scan can be dealt with if you know that your analyte has a [M-18] transition or whatever, by CID. Also, a given class of compounds often shares certain CID reactions, which can be picked up with a neutral loss scan.

If you have tandem, but aren't sure exactly what you want to find, it's a bit more work, but it just means you need to look at product ion scans of peaks that are ambiguous and try to work out what they are.


How much heat is needed to break CO2 down to CO?

Heating CO2 at atmospheric pressure:

At 1000K it is still essentially all CO2.

At 2000K: 98% CO2, 1.4% CO, 0.7% O2

At 3000K: 44% CO2, 36% CO, 16% O2, 4% O

At 5000K: 50% CO, 50% O

only at even higher temperature does significant atomic C appear (9% at 6000K).

See Thermal decomposition of carbon dioxide in an argon plasma jet for more information.


Based on what I gathered from this Wikipedia article, Yes.

Drinking copious amounts of water can prove fatal. The proper term is "Water intoxication".

When you start taking in a lot of water (by "a lot" I mean more water than your body can excrete via sweat or urine), the interstitial fluid that bathe the cells that form your (living) tissue end up getting "diluted", i.e- the concentration of ions like $\mathrm{Na^{+}}$ is greatly reduced. The result? The concentration of ions in the interstitial fluid is far lower than the concentration of ions inside the cells it surrounds. Thinking of this in another way; the concentration of water in the interstitial fluid is far higher than the concentration of water inside the cells.

Ever soaked dry raisins in water? Over time they begin to absorb water and swell up as a consequence of the concentration gradient set up between the tissue in the raisin (Low concentration of water) and the water in the bowl. This phenomenon, where water flows from a region of higher concentration (of water) to a region of lower concentration (of water) across a semi-permeable membrane is called osmosis.

It's pretty much the same thing in the case of water intoxication. A concentration gradient has been set up between cells and the interstitial fluid that surrounds it. So water begins to flow into the cell, and the cell swells up quite a bit, resulting in a build-up of turgor pressure

This increase in turgor pressure will be seen to varying degrees in all sorts of (living) tissue, especially tissue that is highly vascular. An important example would be neural tissue, particularly that in the brain. This swelling up of tissue results in an increase in intracranial pressure which can lead to loss of functionality over time. This is what makes water intoxication so lethal.

There isn't really much about the chemistry of water that's responsible for this, apart from osmosis that is.

Addendum:

Drinking a ton of water isn't the only way for water to build up in potentially lethal amounts in the body. Hyponatremia, which refers to a condition of low (below-normal) sodium levels in body fluids, can also lead to water accumulating in the body in excess.


Sodium azide to Nitrogen ratio in airbags

Assuming you have an excess of $\ce{KNO3}$, the ratio of $\ce{NaN3}$ consumed to $\ce{N2}$ produced would be 5:8.

Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10.  By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $\ce{N_2}$ are produced for every molecule of  $\ce{NaN3}$ consumed.  If you prefer an integer ratio, it's 1:(8/5) = 5:8.

  $\ce{NaN3 -> Na + 3/2 N_2}$
  $\ce{Na + 1/5 KNO3 -> 1/10K2O + 1/2 Na2O + 1/10 N2}$


Are there any molecular modeling programs that can build materials made up of three or more elements into a cylindrical shape?

Various custom software exists to create nanotubes.

The following appears suitable:
http://www.jcrystal.com/products/wincnt/

A Wolfram Demonstrations application may also be of some use:
https://demonstrations.wolfram.com/NanotubeBuilder/

Furthermore, you may have further considerations like how to generate molecular mechanics parameters. For example, you may be interested in how to generate the nanotubes to be compatible with AMBER, or GROMACS. The following may be of use if that is the case:
https://github.com/bio-phys/cnt-gaff
http://www.gromacs.org/Documentation/How-tos/Carbon_Nanotube

I should also mention that VMD and Avogadro, which are more easy to use, can create carbon nanotubes. VMD can create boron-nitride nanotubes
http://www.ks.uiuc.edu/Research/vmd/plugins/nanotube/


On the molybdenum(III) oxide-catalysed oxidation of alkanes to carbonyls

The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow.  Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$.
On industrial scale, formaldehyde is prepared from methanol by an  heterogeneous redox-reaction.  The oxygen needed may come from oxygen gas, or simply air.  Thus, the last step known as Formox process is
$$ \ce{2 CH3OH + O2 -> 2 CH2O + 2 H2O} $$
(from wikipedia)
$\ce{Mo2O3}$ is deployed ease the advancement of the reaction, i.e., to perform the wanted reaction at lower temperature (heating costs money) than without this help, and to further increase rate or / and yield of said reaction compared performing this conversion at room temperature.


Polar and non-polar liquids do not mix, but that doesn't tell you anything about their ability to react, which depends on the thermodynamics and kinetics of the reaction. The reaction shown is in the gaseous state.

The reaction of methane gas with gaseous water, which you show, is a common reaction called Methane reformation. Methane gas is heated with steam (gaseous water) to produce hydrogen (and $\ce{CO2}$).

As shown in that link, the reaction takes place in two steps:
$$\begin{align}\ce{CH4_{(g)} + H2O_{(g)} &->  CO_{(g)}  + H2_{(g)}} &&\text{(Endothermic)}\\
\ce{CO_{(g)} + H2O_{(g)} &-> CO2_{(g)} + H2_{(g)}} && \text{(Exothermic)}\end{align}$$

It appears it probably requires some heating to get this reaction to proceed (the first step is endothermic), I'll edit when I have a chance to look up the thermodynamic properties and see if we can glean what drives the reaction, or if it requires catalysis to avoid the energy barrier.

Side-note: by "bipolar movement" I think you mean "dipole moment"

Edit: According to this source, the first step of the reaction happens at ~3–20 atmospheres of pressure. The second step uses a catalyst to make the reaction more energetically favorable.

The thermodynamic variables for the individual species (at $25~\mathrm{^\circ C}$) in the net reaction are:

$$
\begin{array}{c|cc}
 & \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) & \Delta H_\mathrm{f}^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right) \\
\hline
\ce{CH4_{(g)}} & 0.1862 & -74.81 \\
\ce{H2O_{(g)}} & 0.1887 & -241.8 \\
\ce{CO2_{(g)}} & 0.2136 & -393.5 \\
\ce{H2_{(g)}} & 0.1306 & 0 \\
\end{array}
$$

The balanced net reaction is:

$$ \ce{CH4_{(g)} + 2 H2O_{(g)} -> CO2_{(g)} + 4 H2_{(g)}}$$

Therefore:

$$
\begin{array}{l|ccc}
 & \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) & \Delta H^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right)  \\
\hline
\text{Reactants} & 0.5636 & -558.41 \\
\text{Products} & 0.7576 & -393.5 \\
\hline
\text{Net Reaction} & 0.2136 & 164.9 \\
\end{array}
$$

Right away, it's apparent that the reaction is entropically favorable ($S_{\text{rxn}} > 0$), but enthalpically unfavorable ($H_{\text{rxn}} > 0$) at room temperature. To find out if the reaction is spontaneous, consider:

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

The reaction is spontaneous when $\Delta G < 0$. Because $\Delta S$ is so close to $0$ ($0.194~\mathrm{kJ/(K\,mol)}$), and $\Delta H$ is so high ($164.91~\mathrm{kJ/mol}$), we should expect already that this reaction is non-spontaneous at room temperature. This is consistent with the fact that methane doesn't readily decompose in moist air, as any cow pasture will tell you.  But we can check:

$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-298.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=107.07\ \frac{\mathrm{kJ}}{\mathrm{mol}}$$

Quite non-spontaneous. However, since the overall reaction is entropically favorable, we can use $\Delta S$ to overpower the $\Delta H$, by jacking up $T$. Since the industrial reaction takes place between $700$ and $1000~\mathrm{^\circ C}$, let's plug in $700~\mathrm{^\circ C}\ (973.15~\mathrm{K})$ and check that the reaction is indeed spontaneous:

$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-973.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=-23.85 \frac{\mathrm{kJ}}{\mathrm{mol}}$$

Voilà! A spontaneous reaction!

Do note, however, that this calculation:

ignores any temperature-dependence of the thermodynamic variables;
ignores the role of pressure.

Tldr: This reaction is not energetically favorable at room temperature. However, the reaction is entropically favorable. If you heat it up enough (i.e. $\gtrapprox 700~\mathrm{^\circ C}$), the reaction will proceed spontaneously.

Source for thermodynamic values: Oxtoby, Principles of Modern Chemistry Seventh Edition


Calculate the rate of decomposition of H2O2 at 25 °C when [H2O2] is exactly 10.0 mmol/L

It is a matter of definition:

The rate of the reaction is defined as: $$r=-\frac{1}{2}\frac{d[\ce{H2O2}]}{dt}$$
The rate of the decomposition of $\ce{H2O2}$ is defined as:$$r_{\mathrm{decomp}}=-\frac{d[\ce{H2O2}]}{dt}= 2r$$


Why is the dipole moment of acetone higher than ethanol?

Dipole moment is just one of aspects of solvent polarity, as is well known in HPLC separation  analytical technique where the following effects take part:

Dipole moment interaction, permanent and induced
The above for the whole molecule and for particular bonds
Electron pair donor
Electron pair acceptor (including $\ce{H+}$ donation)

Depending on solute molecules specifics, the solvent polarity order is variable, as various molecules have different responses to above interactions.

See also Using solvent triangle in HPLC


What's the difference between isoionic point and isoelectric point?

The difference, according the Journal of Research of the [US] National Bureau of Standards (1940) is:

  The isoionic point is defined as the point at which dissociable groups of the substance combine equally and only, with hydrogen and hydroxyl ions [6]. This is identical with the isoelectric point only when the substance does not combine with ions other than hydrogen or hydroxyl. It follows then that although it may be possible to determine the isoelectric point by several methods, as by titration, provided that the substance combines with only hydrogen or hydroxyl ions, only an only an electrokinetic method can determine the isoelectric point in every case. The isoionic and isoelectric points of some soluble proteins have been found to be approximately the same. However, in the case of the insoluble materials, especially those which are more or less crystalline in nature, the isoelectric and isoionic points may be far apart. This is true of crystals of even relatively simple substances, such as amino acids [8]. It has been suggested that in addition to the amphoteric properties at the surface, negative ions are also adsorbed at the surface, which results in a shift of the isoelectric point to lower pH values.

In other words, "isoionic" only applies to a zwitterionic substance in pure water, whereas "isoelectric" can be used when there are ions such as Na+ or Cl- present in the solution along with the zwitterion.


Can isotopes of a given element be represented by different symbols?

IR-3.3.1 Isotopes of an element
The isotopes of an element all bear the same name (but see Section IR-3.3.2) and are
designated by mass numbers (see Section IR-3.2). For example, the atom of atomic number
8 and mass number 18 is named oxygen-18 and has the symbol $\ce{^{18}_{}O}$.
IR-3.3.2 Isotopes of hydrogen
Hydrogen is an exception to the rule in Section IR-3.3.1 in that the three isotopes $\ce{^{1}_{}H}$, $\ce{^{2}_{}H}$ and $\ce{^{3}_{}H}$ can have the alternative names protium, deuterium and tritium, respectively. The symbols D and T may be used for deuterium and tritium but $\ce{^{2}_{}H}$ and $\ce{^{3}_{}H}$ are preferred because
D and T can disturb the alphabetical ordering in formulae (see Section IR-4.5). The
combination of a muon and an electron behaves like a light isotope of hydrogen and is
named muonium, symbol $\ce{Mu}$.⁵
These names give rise to the names proton, deuteron, triton and muon for the cations $\ce{^{1}_{}H+}$, $\ce{^{2}_{}H+}$, $\ce{^{3}_{}H+}$ and $\ce{Mu+}$, respectively. Because the name proton is often used in contradictory senses, i.e. for isotopically pure $\ce{^{1}_{}H+}$ ions on the one hand, and for the naturally occurring undifferentiated isotope mixture on the other, it is recommended that the undifferentiated mixture be designated generally by the name hydron, derived from hydrogen.

Source:
N.G. Connelly, T. Damhus, R.M. Hartshorn, A.T. Hutton (eds) (2005). Nomenclature of Inorganic Chemistry (PDF). RSC–IUPAC. ISBN 0-85404-438-8.

Addendum:
The small subscript ⁵ present in the source is a reference to Names for Muonium and Hydrogen Atoms and Their Ions, W.H. Koppenol, Pure Appl. Chem., 73, 377–379 (2001) which can be viewed over at:
https://www.iupac.org/publications/pac/pdf/2001/pdf/7302x0377.pdf

$\cdots$ A particle consisting of a positive muon and an
electron ($\pu{\mu^+ e^–}$) is named “muonium” and has the symbol $\ce{Mu}$. Examples: “muonium
chloride,” $\ce{MuCl}$, is the equivalent of deuterium chloride $\cdots$


You'v answered the question yourself.

In general, any enthalpy of reaction is the measure of the energy remainder after breaking the bonds (positive) and what energy we got back from the formation of the bonds (negative).

The net sum can be positive or negative, and it depends on what the reactants and products were. As you correctly said yourself, $\ce{NO}$ happens to have a positive standard enthalpy of formation because, as experimentally determined, more energy went into breaking the bonds of reactants than what we got back from forming the products.

So,

  Is the standard enthalpy of formation always non-positive?

The answer is no.


Calculate the atomic radius of iridium atom

Iridium has a face-center cubic (FCC) structure.

You assume that each iridium sphere has a radius $r$. But the edge of a unit cell has the dimension $a$. The volume of the unit cell, which is a cube, is $a^3$. The relationship between $a%$ and $r$ is that $a = 2.83r$.

FOLLOWING THE TEACHER"S(?) WORK...(or how you copied it?)

The following step is ok...
$$V = (2.83r)^3\tag{1}$$

But the second step here is wrong. The 2.83 should be cubed too.
$$5.7 \times 10^{-23} = 2.83r^3\tag{2}$$

(3) is totally wrong if we accept (2).
$$r = 1.34 \times 10^{-8}\tag{3}$$

From (2)
$$r = \sqrt[3]{\dfrac{5.7\times 10^{-23}}{2.83}} = 2.72 \times 10^{-8}\tag{4}$$

Starting over...

$$V = (2.83r)^3\tag{5}$$
$$5.7 \times 10^{-23} = (2.83r)^3\tag{6}$$
$$r = \dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = 1.36 \times 10^{-8}\tag{7}$$

After calculating the the volume of the unit cell, I'd solved for $a$ by taking the cube root, then divided by 2.83 to get $r$.

You have to be careful with significant figures too. It is not $5.7\times10^{-23}$ but rather $5.70\times10^{-23}$. Using a calculator I'd carried 5 significant figures in all the calculations then rounded to 3 at the end. In my day I used a slide rule so everything was to 3 significant figures.


Explicit differential equation based model of protein folding

What you are looking for is a force field: http://en.wikipedia.org/wiki/Force_field_(chemistry)

Standard force fields for proteins include CHARRM and AMBER. These have relatively simple, well-defined expressions for defining the various types of interactions between bonded and non-bonded atoms. Parameters are available for all standard atom types, and are fit to experimental and/or electronic structure calculations. These force fields are considered "classical" in that they do not explicitly treat electronic interactions, but rather try to capture these interactions in a coarse-grained way. Typically, there are expressions for describing the following interactions:

bonded interactions (bond stretching)
angle interactions (bond bending)
torsion interactions (dihedral bending)
12-6 Lennard-Jones interactions (non-bonded, dispersion forces)
Short-range charge-charge interactions (Coulombic interactions between charged or partially charged atoms)
Long-range charge-charge interactions (see Ewald summation method for more details)

Note that in these classical models, partial charges are generally statically assigned to each atom. The functional form of these different terms varies depending on the force field and sometimes implementation. These interactions are typically pairwise additive, so to get the total potential energy of the system you simply add together all the various interactions between the appropriate pairs of atoms.

There are certainly more rigorous models that attempt to capture other electronic structure affects, however for large molecules such as proteins, relatively simple force fields are generally applied.

Please note that finding the global minimum in potential energy of a protein is not a trivial problem. People have been working on this for years, it's extremely difficult to develop algorithms that do not get trapped in local minima.


Hardness and brittleness are not contradictory properties. There are plenty of substances that are both hard and brittle (cast iron is a good example, glass is another).

And "strength" is different as well. It depends on context. Tensile strength is not necessarily related to ability to survive a blow: some compounds with high tensile strength (which basically means the compounds resist being stretched) are incredibly fragile (or brittle if you prefer that term). Some compounds with low tensile strength are very hard to break (nylon rope, for example).

Part of the problem is that many of the properties of materials are based on the structural properties of the bulk material not the idealised properties of the chemical bonds making up the material. Some metals, for example, are tough (not brittle) and are hard to break but this property arises because defects in the crystal structure can move around absorbing external stress. Brittleness is often caused because no such mechanism exists and, once a surface bond breaks, the stress on the next bond gets worse, releasing energy and propagating a crack, tearing the material apart. Glass often has many small surface cracks which, under tension, concentrate stress making them brittle and subject to falling apart. But very pure, carefully grown glass fibres can be nearly crack-free making them much stronger under the same stress. In short to explain brittleness, toughness or strength, you need to know the macroscopic makeup of the bulk material not just the chemical composition.

So it is with graphite. There is no easy mechanism in a single graphite crystal to dissipate stress if a bond breaks, so it will tend to be brittle. But most samples of graphite are not single crystals: in most cases they will consist of a bunch of small units with the layers oriented in different directions. So bulk graphite isn't just brittle but also soft as some orientations will slide easily around. And it is hard to even talk about bulk graphene because by definition it is a single layer or a very thin bundle of layers. When people say graphene is strong they usually mean that it has high tensile strength along the plane of the layers. This doesn't mean it isn't either brittle or hard and even if it were in some orientations, it probably would not be in others.


Can nanotechnology reduce electrode losses in electrochemical cells?

Nanotechnology offers enormous opportunities for improving a huge range of electrochemical processes.

The use of platinum as a catalyst is been widespread throughout the electrochemical industry, helping reduce the activation energy of reactions such as in hydrogen and methanol fuel cells. However, platinum is a rare and expensive metallic element. Given that the reactions only take place on the surface of the catalyst, the use of nanoparticles of platinum reduces the amount of platinum needed to achieve the same surface area, thereby achieving the same performance at a fraction of the cost.

Another interesting application is in lowering the operating temperature in solid oxide fuel cells.

A good reference on Nanotechnology in Electrocatalysis for Energy, which covers background fundamentals and some applications, can be read on Google Books for free.


Hydrolysing large molecules to reveal amino acids

You are on the right track.  $\ce{NH_2-C(H)R-COOH}$ is the formula for an alpha amino acid, one where the amino group of the molecule is attached to a carbon atom one carbon atom away from the carboxylate carbon.  This is the most common type of amino acid found in biochemistry.  (A beta amino acid would be $\ce{NH_2-C(H)R^2-C(H)R^1-COOH}$).

When amino acids condense into a polymer or oligomer they do so by forming peptide bonds, also called amide bonds.

$\ce{R^1COOH + NH2R^2 -> R^1CONH2R^2 + H2O}$

So to find the places in oxytocin that would be hyrolyzed into amino acids, look for individual carbon atoms that are attached to both (i) an oxygen atom via a double bond and (ii) a nitrogen atom while (iii) not being attached to any other heteroatoms.  (Item (iii) is generally necessary to exclude other functional groups such as carbamides or carbamates, but these groups are not found in oxytocin.)  For oxytocin, your impulse to look just at carbon double-bonded to oxygen (item (i) only) is OK, but be warned that this procedure is not precise enough in general, because molecules with e.g. ketone moieties will have C=O bonds but will not be hydrolyzed.

The red-circled amino group is a free amino group:  the carbon attached to it is not bound to a carbonyl oxygen, so it is not a place where hydrolysis will happen.  (However, look one carbon atom to the left of C atom attached to the circled N....)

@DavePhD is right; there are more than five amide bonds in oxytocin.  I think there are a total of ten.  Eight are the result of amide bond formation between alpha amino acids and two are the result of amide bond formation between an acid side chain and unsubstituted ammonia.


How much CO2 is emitted by burning coal to produce 1 kWh of energy

The value from EIA is pounds of $\ce{CO2}$ produced per kWh of thermal energy produced, while those from the latter two sources are pounds of $\ce{CO2}$ produced per kWh of electrical energy produced.  Since the thermal efficiency of coal-fired plants is ~37% (~37% of the coal's thermal energy is converted to electrical energy), converting the EIA figure to pounds of $\ce{CO2}$ produced per kWh of electrical energy produced gives us:

$$ 0.735 \frac{\text{lbs }\ce{CO2}}{\text{kWh thermal energy}} x \frac{\text{1 kWh thermal energy}}{\text{0.37 kWh electrical energy }} = 1.99 \frac{\text{lbs }\ce{CO2}}{\text{kWh electrical energy}}, $$

i.e., about the same as the NY Times and shrinkthatfootprint figures.


A few things first...
Arrhenius Theory:

An "acid" is a proton donor; a "base" is a hydroxide donor.

So for example, $\ce{HCl}$ and $\ce{NaOH}$ (both in water) ionize to produce $\ce{H+}$ and $\ce{OH-}$ respectively.
$$\ce{HCl -> H^+ + Cl^-}$$
$$\ce{ NaOH -> Na^+ + OH^-}$$
Which gives us the all-too-familiar neutralization reaction:
$$\ce{HCl + NaOH -> NaCl + H2O}$$
Brønsted–Lowry Theory:

An "acid" is a proton donor; a "base" is a proton acceptor.

At first, the base being a "proton acceptor", may appear to be in contradiction with the Arrhenius Theory... but it isn't.

 The following's from Chemguide. It's been put so well, that I didn't even bother with paraphrasing it
The Brønsted–Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.
Hydroxide ions are still bases because they accept hydrogen ions from acids and form water.
An acid produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them.

When an acid in solution reacts with a base, what is actually functioning as the acid is the hydronium ion. For example, a proton is transferred from a hydronium ion to a hydroxide ion to make water.
$$\ce{H3O^+ + OH^- -> 2H2O}$$

 I'm assuming you already know what the hydronium ion is...so I won't get into that

Or, meh, I'll re-phrase that last bit. Remember that example/equation we first  brought up?
$$\ce{HCl + NaOH -> NaCl + H2O}$$
Looking at it from the Brønsted–Lowry POV, your "acid" is really just the hydronium ion, because that's the real proton-donor here; the "base" is really the hydroxide ion, because that's the real proton-acceptor here.
The hydronium ion (acid) passes on a proton to the hydroxide ion (base).
$$\ce{H3O^+ + OH^- -> 2H2O}$$

Back to your question:

Which definition of Acids and Bases conforms to the pH scale?

Both the Arrhenius and Brønsted–Lowry definitions are "compatible" with $\mathrm{pH}$ . They both point in the same direction ;)

Our teacher told us that pH stands for power of Hydrogen, but some sources say it is a number. Which is it?

$\mathrm{pH}$ is, by definition, (approximately) the negative logarithm (base 10) of the molar concentration of $\ce{H+}$ ions in a solution.
Therefore, $\mathrm{pH}$, is indicative of the $\ce{H+}$ strength, in a solution.
I've been led to believe that the $\mathrm{p}$ in $\mathrm{pH}$, comes from the German "Potenz", or the French "Puissance", both which mean the English "Power". I'm under the impression it was originally "Potenz" since German was the language of science well up to the end of the 19th century (and then the whole World War thing ruined that...); besides, the man who came up with the $\mathrm{pH}$ notation was a Dane, and Danish is a Germanic language with little semblance to French. However, Wikipedia suggests it's more likely "Puissance", since the laboratory the man worked for was French-speaking ...that's still just a guess though.
Your teacher isn't wrong, the $\mathrm{p}$ does mean "power" (or even "potential").
However, $\mathrm{pH}$ is a number (it is dimensionless/unit-less...but not without meaning either), so your "sources" are correct too ;)


How would using different salts in a salt bridge effect the voltage generated by a voltaic cell?

The salt bridges used by mine (so far) were filled with aqueous electrolytes with solutions of either potassium chloride, potassium nitrate, or sodium chloride.  For an electrochemical cell of $\ce{Zn | Zn^2+||Cu^2+|Cu}$ described by you, i.e. based on

$\ce{Cu^2+ + Zn \rightleftharpoons Cu + Zn^2+}$

these appear to me a resonable choice, because

Their ions (of KCl, KNO3, NaCl) would not interfer with the redox reactions, so the potential between the two electrodes (the "voltage" you mention) were not affected.  Referring to the standard potentials, $\ce{Fe^3+}$ of iron nitrate in contrast, is more likely to interfer with your system than the $\ce{Ca^2+}$ of calcium nitrate, esp. in aqueous solution; but the design of the salt brigde (diaphragm) may reduce this electrochemical interference.
Beside electrochemical inertia in the window of investigation, an additional criterion to select electrolytes for salt bridges is their cations and their anions should possess a similar mobility in the electrolyte of tehe salt bridge (the flow you mention).  The rate of the ions to move not only depends on the viscosity of the medium and the charge of the specis, but on the efficient radius of the ion (the ion plus the hydration sphere), too; yet this should not influence the potential between the electrodes, rather than the current between the electrodes if the set up is intended to transform chemical into electrical energy.

In short: within reason, keep the KCl in the salt bridge.


Yes, and no.

Acid and base chemistry is a neat balance between equilibria. As you well know, water dissociates readily by itself. A tendency that can be observed in any compound with $\ce{-OH}$ groups, yet not to a similar extend.
In other words, a hydroxyl group tends to be slightly acidic. Given a base, i.e. sodium hydroxide in this particular case, the following equilibrium will develop:
$$\ce{EtOH + NaOH <<=> EtO- + Na+ + H2O}$$

While the equilibrium will mostly be on the left side, depending on what other substrates are available, a tiny amount of alkoxide might already be enough to drive the reaction forward. In this case it is important to look at what actually drives the reaction.

In the example of 4-bromo-1-butanol that your book uses I am assuming there is a cyclisation happening.
$$\ce{BrH2C(CH2)2CH2OH + NaOH <=>> THF + H2O + NaBr}$$
Unfortunately I cannot quickly find a reference, but sodium bromide is not very soluble in diethylether, so I would assume it's not very soluble in THF either. The driving force here would be having a well enough leaving group and the continuous removal of product, basically making it a one-way reaction.

In some cases you need a higher concentration of alkoxide, and that's where hydrides come into play.
$$\ce{EtOH + NaH <=>> EtO^-\bond{~}Na+ + H2 ^}$$

In this case dihydrogen, as it is a gas, will be removed from the reaction mixture, basically ensuring full conversion to the alkoxide.


Quotable Literature for the Enthalpy of a Proton

A quite complete procedure in the calculation of proton affinities from ab initio calculations is referenced in an article by H. Audier et al., Int. J. Mass Spectrom. 1996. It goes much further in the procedure than only on the calculation of the enthalpy of a proton (including corrections due to the basis set superposition error and so on), but confirms that you are right in assuming that the enthalpy of a proton is $3/2 RT$.


Mass Spectroscopy peak assignments in but-3-en-1-ol

As you have pointed out, the peak at M = 73 is M+H, protonation is however more likely to occur at the oxygen atom.

In my opinion, peaks at M = 66 and 67, i.e. with a rather small mass difference to $\mathrm{M^+}$, are negligible.

Common sense and the link in permeakra's comment suggest that the major fragmentation of but-3-en-1-ol occurs along the red dotted line, yielding an allylic and a hydroxymethyl fragment.


What's the reason for the instability of ethylene dione?

Expanding what Tan Yong Boon said, there is a full scientific article detailing why ethylene dione is unstable Chem. Eur. J., 1998, 4, 2550-2557.

Long story short: Two CO molecules are far more stable than OCCO. The ground state of ethylene dione is a "diradical". This is, OCCO has two unpaired electrons instead of forming the intuitive chain of double bonds (O=C=C=O). However, if you add one or two additional electrons, the resulting anion becomes stable and adopts the shape of glyoxal (but without hydrogen atoms).


Which metals hydroxides dissolve in excess NH4OH?

The Wikipedia article for "Metal amine complex" gives many examples.  According to the article:

  Almost all metal ions bind ammonia as a ligand, but the most prevalent
  examples of ammine complexes are for Cr(III), Co(III), Ni(II), Cu(II)
  as well as several platinum group metals.

Specific examples include:

$\ce{[Pt(NH3)4][PtCl4]}$
$\ce{[RhCl(NH3)5]Cl2}$
$\ce{[Cr(NH3)4Cl2]Cl}$
$\ce{[Co(NH3)6]Cl3}$
$\ce{NH4[Cr(NCS)4(NH3)2]}$
$\ce{[Zn(NH3)4]^2+}$
$\ce{[Ag(NH3)2]+}$
$\ce{[Ni(NH3)6]^2+}$
$\ce{[Cu(NH3)4(H2O)2]^2+}$

Several others are listed just in that one Wikipedia article, along with a fair list of references.


Why is chloridric acid used in salt bridges?

Any ionic solution can function as a salt bridge. Having it acid based helps with the fast conduction paths of H+ in water.

The critical question is the leak rate. Any salt bridge will leak some amount of the ions into the solution. Good ones leak less, bad ones leak more.

So if you cannot tolerate any acidity in your solution, don't use an acid. Similarly, if you cannot tolerate any chloride in your solution, this would be a bad choice.


Most of the $\ce{CaCO3}$ in commercial use today is mined not synthesized. The one exception I know a bit about is the ultra-small particle sized materials. However, few of those are pure, they have surface treatments which constitute a significant portion of their total weight. Pharmaceutical and perhaps food grades are also synthesized.

The major synthetic pathway is circular: limestone is mined, then it is heated which produces lime and carbon dioxide. The lime is then mixed with water and $\ce{CO2}$ is added (bubbled in/through) and $\ce{CaCO3}$ precipitates. It is then dried.

Purification depends on the source; how impure it is to start with. It is generally difficult to separate magnesium carbonate from calcium carbonate, and given that limestone and aragonite as well as dolomite ($\ce{CaMg(CO3)2}$) are taken out of the ground (i.e. are 'dirt' cheap) there's usually no need.

As far as the risks of making it, well you could burn yourself, adding lime too fast to water could splash and it's infamous for dissolving bodies in murder mysteries, as well as being used in real life to help in outhouses and gardens where organic material is desired to rapidly decompose. Meaning you don't want to get it in your eyes, nor do you want it left on your skin for very long. It can, theoretically, fume, while being decomposed and that wouldn't be good to breathe. And if a centrifuge or mechanical press is used to squeeze the water out of it (first step in drying), then you have the assorted mechanical dangers. Its synthesis is probably one of the safest chemical syntheses you could undertake.

$\ce{CaCO3}$ decomposes with fairly mild heating (I think around 840 °C), so any material which is predominantly that (like seashells, bones) could be used to make the lime.


Can there be more than one combination of reactant/product concentration for a specific equilibrium constant?

A single "elementary reaction" does not involve an equilibrium.

There must be a forward reaction and a reverse reaction to have an equilibrium.  This is at least two elementary reactions.

Considering an equilbrium reaction: Just as a square encloses the most area for a rectangle of given perimeter, if the sum [A] + [B] is kept constant,  [A][B] is maximal when [A] = [B].  Deviation from this maximum will shift the equilibrium away from the product and toward the reactants.  In other words, if the system is at equilibrium with [A] = [B], and [A] is decreased and [B] increased, keeping [A] + [B] constant, there will be net reverse reaction until equilibrium is again reached, and [AB] at equilibrium will be less than orginally.


Why does DMF (dimethylformamide) have 3 peaks in its 1-H spectrum?

If you understand the partial double bond character induced by the N-CO moeity, then you can draw out the molecule as shown below.

The three signals you get are from the CH and the two CH3 groups. When the energy barrier for rotation is high to prevent averaging (fast rotation), you can see that the two methyl groups are non-equivalent: one is trans to the C-H, and the other is cis to it. This is a very common characteristic of many amide-containing molecules. The energy barrier to rotation is highly dependent on steric interactions. The number of possible conformers arising from even small peptide molecules increases very rapidly, when you consider the combination of cis/trans type orientations of the amide bonds.

Now, as you start to heat this system (or observe at a lower magnetic field - that may be for another question), you will start to see signal averaging, as the two methyl groups begin to exchange on the NMR timescale. As this rate of exchange becomes faster, you will get signal coalescing and finally complete signal averaging to the point where the two CH3 groups are equivalent. This is seen below, with the bottom spectrum being at low temperature (and/or high magnetic field) and the top spectrum being at high temperature (and/or low magnetic field).

Signal averaging via rotation about bonds happens even in single bonds. If you cool a CH3 group down far enough, you will be able to slow down the rotation of the C-C bond enough to be able to distinguish all protons as non-equivalent.


Is the redox method applicable to all chemical reactions?

Sometimes you aren't able to use the redox method to balance a chemical equation. This is because the redox method is only used to balanced an equation where a element gets oxidised or reduced. However if there is a chemical equation where no element is getting oxidised or reduced, then you won't be able to use the redox method to balance the equation. However, fortunately most of the time these type of equations can be easily balanced. Consider the following unbalanced equation: $$\ce{Ba(OH)2.8H2O (s) + NH4SCN (s) -> Ba(SCN)2 (s) +  H2O (l) + NH3 (g)}$$

Here none of the elements are reduced or oxidised, hence you would not be able to balance this equation using the redox method.


Which dicarboxylic acid has the most acidic hydrogen?

The $\mathrm pK_\mathrm a$ of the $\ce{CH2}$ group in a dicarbonyl compound is roughly $10$, whereas the $\mathrm pK_\mathrm a$ of a carboxylic acid is roughly $5$. For example, dimethyl malonate has a $\mathrm pK_\mathrm a$ of $13$ whereas acetic acid has a $\mathrm pK_\mathrm a$ of $4.7$. So, the C–H bond acidity is not likely to be under consideration in any of your compounds.

Let's draw up some $\mathrm pK_\mathrm a$ data (only first ionisations, all data from Wikipedia):

$$\begin{array}{cc} \hline
\text{Carboxylic acid} & \mathrm pK_\mathrm a \\ \hline
\text{Maleic} & 1.9  \\
\text{Fumaric} & 3.03 \\
\text{Succinic} & 4.2 \\
\text{Malonic} & 2.83 \\ \hline
\end{array}$$

Clearly, all of these relate to ionisation of the $\ce{CO2H}$ group; if the acidity of dimethyl malonate is anything to go by, the $\ce{CH2}$ group in malonic acid is less acidic than the $\ce{CO2H}$ group by a factor of $10^{10}$.

What remains is to analyse this trend. Why is maleic acid the most acidic? The answer is – at least partly – intramolecular hydrogen bonding. When the first carboxylic acid group is deprotonated, the resultant monoanion is stabilised as such:

You are probably familiar with the mantra that a more stable conjugate base implies a more acidic compound. This is precisely what happens here.

Obviously, for fumaric acid, this is not possible because the double bond holds the two carboxyl groups apart. One could, however, argue that similar stabilisation could be derived for malonic and succinic acids. Why is it not a factor, then? It has to do with conformation: for the malonate or succinate anions to enjoy the benefits of intramolecular hydrogen bonding, they have to twist themselves into one specific conformation where the carboxyl groups are close to each other. This results in a loss of configurational entropy (the molecule is not as "free" as it would like to be to explore different conformations). For maleic acid, though, this is not a problem because the (Z)-double bond plonks those two carboxyl groups next to each other, and they don't have a choice about it. This bears some similarity to the concept of preorganisation, where two or more groups are "organised" in a fashion which alters their properties.

That's not the only reason why maleic acid is especially acidic, however. Part of the reason is also because an $\mathrm{sp^2}$ carbon is more electronegative than an $\mathrm{sp^3}$ carbon. An $\mathrm{sp^2}$ orbital contains 33% s-character, compared to an $\mathrm{sp^3}$ orbital which has 25% s-character. s-Orbitals are closer to the nuclei than p-orbitals, and therefore, an electron in an $\mathrm{sp^2}$ orbital experiences a greater effective nuclear charge from carbon; this translates into a greater electronegativity. (For similar reasons, alkenes are more acidic than alkanes.)

So, relative to the succinate anion, both the maleate and fumarate anions are stabilised by electron withdrawal via the inductive effect. This explains why fumaric acid is more acidic than succinic acid; the intramolecular hydrogen bonding explains why maleic acid is more acidic than fumaric acid.

Lastly, of course, malonic acid is more acidic than succinic acid because the electron-withdrawing $\ce{CO2H}$ substituent is fewer bonds away. In fact, succinic acid is barely more acidic than acetic acid, which doesn't have an electron-withdrawing substituent on it.


How to determine the mass of ammonium nitrate needed and enthalpy of solution?

I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.

The result will depend on the amount of water where you're dissolving the salt. For instance, the data you present refers to different solvent mass. And by the way, for your temperature difference ($\pu{20^\circ C}$) the problem is already solved: a $\pu{20^\circ C}$ temperature decrease, will require $\pu{10 g}$ of the salt.

It is important to distinguish the mass of the salt from the mass of the solvent.
\begin{align}
Q &= m_\mathrm{salt} \times \Delta_\mathrm{solution} H^\circ_\mathrm{salt}\\
Q &= m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T\\
\therefore m_\mathrm{salt} &= \frac{m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T}{\Delta_\mathrm{solution} H^\circ_\mathrm{salt}}
\end{align}

Hope this helps.


Strong exothermic mixing of two 'safe' liquids

If you need a quick source of moderate heat, why not use a single liquid? Sodium acetate trihydrate ($\ce{CH3COONa·3H2O}$), for example, releases heat energy on crystallization... and it takes very little to trigger that crystallization! Commercial heat packs contain a "clicker", which makes a sharp vibration. You can make your own heat pack fairly safely and inexpensively: sodium acetate can be found for ~US$8/450 g, and a Safety Data Sheet states, "it's  not classified for physical or health hazards under GHS."

Not only can you get a noticeable temperature rise quickly, it's reusable.

It would be more difficult to find liquids that quickly release energy that do not have physical hazards (e.g. ignition) or health hazards. Hypergolic fuels, for example, tend to be rather toxic, as well as explosive. You might get a bang out of reading John Drury Clark's Ignition!, though.


Modeling of photocatalytic and photoelectric devices on a macroscopic device level

Look,
I try to give you a broad answer.
If you want band gaps, you need electronic structure that means DFT!

If you do MD that means classical mechanics means no electronic structure no band gaps.

You want to have solvent effects for your, I presume large system, then DFT is costly, you would be left with QM/MM or sometimes called double resolution.

Reactive forcefields such as Reaxff use some valence bond theory to treat electronic structure which is low level of theory and can not be extended to inorganic compound unless it's parametrized.

Finally all these methods are highly approximate, but for engineering purposes everything is approximate.


CIF fields that are available for some atoms, but not others

Your loop defines 8 records. That works fine for heavy elements. However, the lines with hydrogen atoms have only 6 items. That's the reason why this error is reported (all you records are not multiplier of 8).

Usually in refinement the hydrogen atom's $U_{iso}$ is 1.2–1.5 bigger than this of atom to which the H is bound (smaller 1.2 for more constrained movements like on aromatic carbons, bigger 1.5 for aliphatic)

Try something like this:

N1   1    -0.8852(9) 0.120(1)   -1.0300(8) Biso 3.1(2) N
H1a  2    -0.9650000 0.08500000 -1.0010000 Biso 4.     H
H1b  3    -0.8800000 -0.0800000 -1.0330000 Biso 4.     H
C2   4    -0.739(1)  0.049(2)   -0.940(1)  Biso 3.9(2) C
H2   5    -0.7460000 -0.0800000 -0.9360000 Biso 4.     H
C3   6    -0.614(1)  0.092(1)   -0.988(1)  Biso 5.1(3) C


Why only F, O and N form Hydrogen Bonds?

No, that's not necessary. Amongst the strongest hydrogen bonds are formed by $\ce{ N, O, F}$ because of their high electronegativity.

There's evidence that the ions $\ce{Cl-, I-, Br- }$ form hydrogen bonds that are much stronger than those of the covalently bonded atoms

$\ce{Cl}$ can form weak hydrogen bonds, but $\ce{Br}$ and $\ce{I}$ form very weak bonds if at all.

A sulfur atom can also function as weak acceptor., but the $\ce{SH-}$ ion forms much stronger bonds.

Hydrogen bonding has been directly observed between a negatively charged carbon and an $\ce{-OH}$ group in the same molecule

Isocyanides' ($\ce{R- N+ #C- }$) carbon atom can also act as an acceptor (forming a rather strong Hydrogen bond).

Reference: March's Advanced Organic Chemistry, Chapter 3: Bonding weaker than Covalent