Mechanism of formaldehyde / phenol condensation

@Harry Holmes: Here is a simplified example of how the formation of Bakelite may occur under base-catalyzed conditions to incorporate formaldehyde as -CH2- at the ortho and para positions of phenol. There are a myriad of permutations as to the order of condensations. I have selected one that illustrates the Michael-like addition that seems to have eluded you.
Phenoxide 1 adds to formaldehyde at the para position in what amounts to the first step in an aldol addition. The addition is likely adding to the ortho position of phenoxide as well. Species 3 may condense with phenoxide at this point but I have chosen to incorporate two more equivalents of formaldehyde at the ortho positions of 3 to form 4. Now elimination of hydroxide from any of the three positions of 4 leads to a Michael acceptor, such as 5, which reacts with phenoxide or any substituted phenoxide having an open ortho or para position. Structure 6 simply illustrates the mechanism for incorporating formaldehyde. The o-methylenedienone 5 is the product of an aldol condensation wherein phenoxide is the enolate.


Why do the atoms in molecular models have the colors they have?

There's an interesting article here - Illustrating Atoms and Molecules - that discusses atom colors:

  In 1865, the chemist August Hoffman gave a Friday Evening Discourse at London’s Royal Institution on the “Combining Power of Atoms.” In order to demonstrate chemical bonding of atoms, he drilled holes in croquet balls and connected them with metal pipes (Figure 2). His choices were limited by the available colors of croquet balls and he relied on many of the same color conventions that had been adopted centuries earlier

These colors are listed as :

  Carbon is colored black because it’s the color of charcoal.
  Oxygen is red because it’s necessary for combustion.
  Nitrogen is blue because it’s the most abundant element in the
  Earth’s atmosphere and the sky appears blue.
  Hydrogen is white because it forms a colorless gas.
  Chlorine is green because it forms a greenish gas.
  Sulfur is yellow because that’s its color in mineral form.
  Phosphorus is orange because it glows orange in a flame.
  Iron is reddish brown because it rusts.

Which is interesting, but I've no idea where this list comes from...


Citric acid is not a defoaming agent in itself. It is routinely used for pH adjustment in many commercial hand-dishwashing agents.

But it depends on the surfactant in the product. As Ivan tells, anionics will at some point be protonated by acids, and thereby loose their surface activity –and foaming capability.

Commercial anionic surfactants like ethersulfates will only do so at rather low pH, outside what is relevant for dish-washing products. But if your product is based on saponified oils it’s another story. Many fatty acids are weak acids (pK in the rage of 4-5) and will be significantly protonated at pH even as high as 5-6.

By the way: making good foam is as much an art as a science :-). If you want to dig deeper in the subject, I can recommend “Prof Steven Abbot –Surfactant Science, Principles and practice”. It is freely available on the web.


After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)?

It's not clear to me that $\ce{N2O}$ is produced from $\ce{N2}$: Rather, it appears that commercial quantities of $\ce{N2O}$ are produced by thermally decomposing ammonium nitrate:
$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$
as described (at least) by this manufacturer and supplier of nitrous oxide gas plants. See the comprehensive Wikipedia entry for more information.

  I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air?

I'm not sure what you mean here.

  When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other?

Nitrogen is liquid in the range 63 to 77 K; for oxygen, it's 54 to 90 K. Thus, the temperature range for which both are liquids is bounded by the range on nitrogen, or roughly between 63 and 77 K.

The temperature at which nitrous oxide solidifies (182 K per Wikipedia) is much higher than the temperature range at which oxygen and nitrogen are in the liquid phase, and thus it would have come out of solution as a solid long before the requisite temperature was reached. Thus, we would not expect to see liquid $\ce{N2O}$ co-existing with liquid oxygen and nitrogen.

Note bene: I am using the pressure range of 5 to 10 bars as cited here in assuming that liquid and solid phases of the above moieties do not coexist during industrial gas production at these relatively low pressures.


Why is volume inversely proportional to pressure?

Given you know and understand Charles' and Gay-Lussac's laws, it's not about chemistry, rather, simple ratios:

$$
\begin{cases}
T \propto V\\
T \propto p
\end{cases}
\implies
p \propto \frac 1 V
$$

which, as Zenix commented, is a math form of Boyle's law.


Why aren't the Anton computers used more widely?

Protein dynamics occur in multiple different time scales. I myself is a person using MD simulation for my research. One reason I am not so much interested in ANTON is not it's cost, limited resource etc. It's rather because one extremely long simulation wouldn't help much at least for my research.
Protein's thermodynamics as well as kinetics all depend on free energy landscape which is a priori unknown. To directly compute free energy from simulation, you generally need multiple order of magnitude faster machine even than ANTON. However, there has been many techniques such as Umbrella Sampling that helps resolve this issue on computers in our lab.
I would appreciate it if I can run my Umbrella Sampling in ANTON but it would be so much waste of time to wait on the line if I can run the same simulation for a week in my workstation.
Another reason I can guess why ANTON has not been widely used is because of its specificity. ANTON does not run typical MD simulation packages such as GROMACS, NAMD, AMBER etc. All packages use same algorithm to run the simulation but the commands, file structures etc, are all different. Typical MD user would prefer to use package that he/she is familiar with. I am not sure how much I would have to study to get used to ANTON MD package but why bother if I am not desperate?
So in summary, only those users who are interested in actual trajectories of protein and water in millisecond time scale would want to use ANTON but not many others. Other than such purpose, there are many techniques and algorithms that would work better than running a single extremely long simulation.


Are compounds containing active methylene groups positive for the haloform test?

As orthocresol already commented, apart from some nitpicks, the general direction is correct.

The only issue really is the addition of bromine, which does not occur via a [2+2]-cyclo-addition (2nd from below in scheme).

Ordinarily you would expect addition the addition of bromine to a π-bond to occur first via a polarisation complex and then via a bromonium ion (bottom of scheme).[1] In the case of the enolate, the π-bond is already readily polarised, and sufficiently asymmetric that a direct addition will occur.[2] The following depicts the HOMO -1 calculated at the DF-BP86/def2-SVP level of theory, the HOMO is an (delocalised) in-plane lone pair of the carbonyl-oxygen atoms in the C2v conformation (1st row). In the Cs conformation the π-bond is the HOMO, it is also the low-energy conformer (2nd row). (On mobile the rows might differ.)

Please note that the following scheme is just a proposed mechanism, and as such probably a reasonable simplification; the actual one might depend on more factors like solvent, temperature, pressure, etc., and might have more, or even less steps, other intermediates (ion pairs), hydrogen abstraction might be via water/base bridges, etc. For educational purposes I think this is sufficient though.

See for example Aromaticity-Bromonium Ion, Mechanism for bromoetherification of olefinic alcohol, Why does cyclopropane react with bromine?, What is trans-addition? and more...
Thanks to orthocresol pointing it out in the comments.

  Enolate reaction with bromine (as opposed to ordinary alkene) is traditionally taught to not go via a bromonium ion (img), presumably because the electronic character of the two doubly bonded carbons is very different. But I have not consulted any literature, this is just based off undergrad teaching + textbooks, so...

Since I was unlucky to find any specific literature on the topic (I did not search very thoroughly), I have tried optimising a bromonium analogue at the DF-BP86/def2-SVP level of theory. I have been unable to obtain a stationary point (no local minimum and no transition state). Which is not unexpected given the HOMO/ HOMO -1 is strongly polarised.


To find the mass of chromium, one needs to know the amount $n$ ($m = nM$). Each formula unit of ammonium dichromate $\ce{(NH4)2Cr2O7}$ contains $2$ $\ce{Cr}$, therefore

$$n(\ce{Cr}) = 2 \cdot n(\ce{(NH4)2Cr2O7})$$

The amount of ammonium dichromate can be found from its concentration (as you correctly did):

$$n(\ce{(NH4)2Cr2O7}) = C(\ce{(NH4)2Cr2O7}) \cdot V$$

Putting it all together, the final expression for the mass of chromium is

\begin{align}
m(\ce{Cr}) &= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{Cr}) \\
&= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{51.996 g mol-1} \\
&= \pu{0.416 g}
\end{align}

The answer deviates with the one provided by the textbook; however, if you were asked to find the mass of chromium(VI) oxide $\ce{CrO3}$ instead, it fits perfectly:

\begin{align}
m(\ce{CrO3}) &= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{CrO3}) \\
&= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{99.99 g mol-1} \\
&= \pu{0.80 g}
\end{align}

I deduced $\ce{CrO3}$ by backtracking to the molecular weight of the compound, which should be around $\pu{100 g mol-1}$, and among common chromium substances this is the only one that comes to mind.


How to reduce the turbidity of a water sample?

Apart from the article linked in a comment above, possible options are

wait and let settle, then decant
use an established flocculant, such as potassium alum $\ce{KAl(SO4)2*12H2O}$


De-/protonation of R-OH group by acids and bases

Your first statement is correct. The basicity increases with $pK_a$, so a base with $pK_a > 16$ will be able to deprotonate your alcohol.

Your second statement is also in principle correct; however, the value of the sum $pK_a+pK_b$ is solvent-dependent, and 14 is the value for water at standard conditions. $pK_a$ and $pK_b$ values of acids and bases depend on the solvent as well. For example, sodium amide is a strong base in liquid ammonia, but not so in water due to the leveling effect of the solvent. Dissolving sodium amide in water will give a solution of sodium hydroxide, which is not basic enough ($pK_a=13$) to deprotonate the alcohol.

Furthermore, the amount of deprotonated alcohol will depend on the difference between $pK_a$ values of alcohol and base, since the acid-base reaction is an equilibrium. For example, a base with $pK_a=33$ will deprotonate your alcohol to a greater extent than one with a $pK_a$ of 16.5 .


What is the net ionic equation of the following?

You are making a mistake splitting up your reaction equation from \eqref{full} to \eqref{wrong} in order to filter out the spectator ions.

\begin{align}
\ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)}\tag1\label{full}\\
\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) &-> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}\tag{wrong}\label{wrong}
\end{align}

What you are essentially doing is not conserving the charges. Dichlorine $\ce{Cl2}$ is a neutral molecule, and the following is what your transition implies, but is also wrong:
$$\ce{Cl2 -> 2Cl-}\tag{also wrong}$$

The correct way to transform \eqref{full} is to split (dissociate) only the ionic compounds $\ce{AlBr3}$ and \ce{AlCl3}, and then cancel the spectator ions:

\begin{align}
\ce{3 Cl2 (aq) + 2 AlBr3 (aq) &-> 2 AlCl3 (aq) + 3 Br2 (l)}\tag1\\
\ce{3 Cl2 (aq) + 2 Al^3+ (aq) + 6 Br- (aq) &->
    2 Al^3+ (aq) + 6 Cl- (aq) + 3 Br2 (l)}\tag2\label{split}\\
\end{align}

Next step is to cancel the aluminium ions on each side, and reduce the equation to \eqref{net-ionic}:

\begin{align}
\ce{3 Cl2 (aq) + 6 Br- (aq) &-> 6 Cl- (aq) + 3 Br2 (l)}\tag3\\
\ce{Cl2 (aq) + 2 Br- (aq) &-> 2 Cl- (aq) + Br2 (l)}\tag4\label{net-ionic}
\end{align}

I don't think the reaction will actually produce liquid bromine, but aqueous bromine instead; that doesn't matter to the solution though.


Digest a sample of the metal in aqua regia, I have found that the gauze in a kitchen sieve can be digested in a very short time with a mixture of 16 M nitric acid and concentrated hydrochloric acid.

Next you should dilute it with water and filter it to remove any particles.

I would use either flame AAS or ICPOES to measure the metals after dilution to a suitable concentration in plastic ware. I would want to measure the Fe, Ni, Cr and Mo in the sample. The Fe to Ni and Fe to Cr ratio may help you distingish between a lot of grades of stainless steel.

One alternative for iron is to use 2,2'-bipy and hydroxylamine to form the deep red iron(II) tris bipy complex.

This can be measured with a UV / vis machine in a 1 cm cuvettes, you will have to make a series of solutions in volumetric flasks. You make 0, 1, 2, 5, 10 and 20 ppm iron by adding a 1000 ppm standard to the flasks, then add a solution of 2,2'-bipy in water, the hydroxylamine and then dilute to the line. Mix them well and then measure the red colour with the UV / vis machine.

I am sure that there are colourmetric methods for Mo and Cr but sadly I do not know them.


Why must the forward reaction and the backward reaction follow the same pathway?

A reaction could proceed by many different pathways, for example an extreme way would be splitting the reactants into atoms and combining them to form products, but this is essentially impossible, the energy required would be huge and how would the atoms reassemble just into the product required?

In any reaction the molecules follow the lowest energy pathway, which is to say that the lowest activation barrier between reactants and products will be crossed in preference to any other pathway. The same argument applies to a catalysed as to a 'normal' reaction even though a catalysed reaction proceeds by a different mechanism. The lowest energy pathway also ensures in some general way that close to minimum number of bonds are broken in the reaction.

The reason for this is that the activation energy (barrier height) is usually far higher than average thermal energy and many collisions happen between reactants before by chance there is enough energy to cross the barrier. The probability $p$ of having sufficient energy $E$ is given by the Boltzmann exponential distribution, $p \approx exp(-E/(RT))$. This is a rapidly falling distribution as $E$ increases thus the minimum energy pathway is generally hugely favoured over others.

In biochemical reactions it is possible for the same molecules to be synthesised by different pathways, involving several steps, but the same arguments apply to each stage in the reaction within the enzymes used.Synthetic chemists will have many examples of the same products being formed from the same starting molecules and progressing by different sequences of pathways in a series of $separate$ reactions.  However, the yields will be generally be different; one of the aims of synthetic chemistry is often to optimise yields, but I take the question to ask that the reaction is not that of a separate sequence of steps but in one process.


What does "spin degeneracy" mean in this context?

I'm not a solid-state chemist, but I think the meaning seems reasonably clear from context to me. "Spin degeneracy" here means that each energy level is capable of holding a spin-up electron as well as  a spin-down electron, i.e. in each orbital there are two different spin states which are degenerate (some would call these spin orbitals). Thus, $N$ atoms form $N$ different energy levels which $N$ electrons need to be distributed between. Because of this degeneracy, only the lowest-energy $N/2$ energy levels are actually populated: each energy level has $2$ electrons.


Is the conjugate base of pyrrole resonance stabilised?

There is like you assumed no reasonable amount of resonance stabilisation. However, the aromatic π-system contributes to the stability of σ-skeleton.

Below you can find the MO of pyrrole and its N-deprotonated anion (click it to enlarge). You can see that in MO 17 the electron density is shifted further into the centre of the ring, accounting for the larger negative charge on the nitrogen. It is a bit more difficult to see, but that is also true for MO 18 and 14 (pyrrole), which becomes 15 (anion). The most significant change is that the upon deprotonation resulting lone pair becomes much higher in energy (12 to 16), and it is as you note perpendicular to the π-system in a σ-type orbital. Note that the image only shows the trend and has no information of the actual orbital energies, apart from that they increase from bottom to top.


What determines the density of an element?

There are a couple of factors here that you need to consider.

First of all, increasing Z does mean that there are more particles in the nucleus; however, you're also adding electrons to the systems as well. Electrons dictate the size (i.e. volumen) of the atom, so the size of the atom could increase with increasing Z, which would mean that fewer atoms could fit into the same volume. To use your example, the covalent radius  of Os is 144 pm; the covalent radius of U is 196 pm. So just from this property alone, you would expect more Os than U atoms to pack into the same volume.

There is another factor to consider as well: the crystal system of the solid. At standard temperature and pressure, atoms for different elements in the solid state pack into different arrangements, which are called crystal systems. Some of these are much more efficient than others. For example, there are three variants of the cubic crystal system: primative, body-centered, and face-centered. From Wikipedia:

  Assuming one atom per lattice point, in a primitive cubic lattice with
  cube side length a, the sphere radius would be a⁄2 and the atomic
  packing factor turns out to be about 0.524 (which is quite low).
  Similarly, in a bcc lattice, the atomic packing factor is 0.680, and
  in fcc it is 0.740. The fcc value is the highest theoretically
  possible value for any lattice, although there are other lattices
  which also achieve the same value, such as hexagonal close packed and
  one version of tetrahedral bcc.

Now, depending on which crystal system the atoms in the solid take, you could get more or less atoms per unit volume. Unfortunately, determining which crystal system is favored for different elements is not straightforward, and requires some knowledge of the bonding between the atoms and other factors.


Effect of temperature change in an equilibrium reaction

As per Le Chatelier's principle, when the temperature is increased, for an exothermic reaction , the reaction shifts in the backward direction. This is clear from the van't Hoff's equation:

$$ \log\left( \frac{K'_\mathrm{p}}{K_\mathrm{p}}\right) = \frac{\Delta H^0}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

where $T_1$ corresponds to $K_p$ and $T_2$ corresponds to $K'_p$ . Also $T_2 > T_1$.

This is because $K'_\mathrm{p}$ becomes less than $K_\mathrm{p}$.  Hence the reaction goes in the reverse direction.


Is there any software that can do geometry optimization using machine learning?

There is one Python package that I know: AMP (Atomistic Machine-learning Package). It is based on ASE (Atomic Simulation Environment), a homogeneous interface to a lot of computational chemistry packages.

Basically, after you obtain geometry trajectories through ASE (using any computational chemistry package you like), an approximation to the potential energy surface (PES) is machine-learned. It can be used as an ASE interface.

You can theoretically do anything ASE is capable of with this, from saddle point optimization to quantum dynamics.
I am not an expert in this field, but they seem to apply ideas from convolutional neural networks to molecular structures for describing the chemical environment surrounding atoms.
Their method is described in Khorshidi, A.; Peterson, A. A. Amp: A Modular Approach to Machine Learning in Atomistic Simulations. Computer Physics Communications 2016, 207, 310–324.


Aluminium foil on windows: what is the residue, and how to remove it?

Finally I used a glass scraper. Bleach didn't help, and I didn't dare to use even stronger caustic soda.

Fuller details at the answer I wrote at the DIY SE site: https://diy.stackexchange.com/questions/151160/how-can-i-remove-residue-left-after-removing-aluminum-foil-from-my-windows#answer-152162


With alkanols, sodium will form alkoxides (and hydrogen gas).
$$\ce{2CH3OH + 2Na -> 2 CH3ONa +H2 ^}$$
With alkanoic acids, sodium will form alkanoates (and hydrogen gas).
$$\ce{2CH3COOH + 2Na -> 2 CH3COONa +H2 ^}$$

If your starting material contains both an $\ce{OH}$ group and a bromo substituent, the alkoxide formed upon the reaction with sodium may attack the carbon bearing the $\ce{Br}$ atom in a substitution reaction to yield a cyclic ether.


Not only I've never heard of a name for such a class, but the very choice of compounds seems quite artificial to me. What feature do they have in common?

Consisting of H and O? That would exclude oxygen and ozone.
Consisting of H and/or O? That would include $\mathrm H_2$ as
well.
Containing "active oxygen", whatever that might mean? That
would exclude water, but include a bunch of inorganic and organic
peroxides.
Oxide is indeed too generic; also, it does not include oxygen and ozone. Oxygenate is even farther off.

Why would you want to have such a class, really?


Reaction between silver nitrate and aluminum chloride

Le Châtelier's principle only states that a system previously at equilibrium will want to stay at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is initially not at equilibrium, and therefore we don't need his principle.

I think the reaction proceeds because the formation of AgCl(s) is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO3, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO3 since it is insoluble in water.

This reaction is a type of metathesis reaction. It can be called double displacement reaction or salt metathesis reaction, depending on who you're talking with.

EDIT

The other answers make an error, I think. Both suggest that the solubility product of AgCl is of importance in determining why the reaction takes place. My opinion is that the argument, as presented, is circular, in the following manner:

The solubility product of AgCl in water is very low, therefore we observe formation of a precipitate.

vs

AgCl is insoluble in water, therefore its solubility product is low.

It's easy to see that the fact AgCl is insoluble in water implies that it has a low Ksp in water. If someone can predict and quantify the solubility product of AgCl in water from first principles, then that's another story, but the explanations offered only rely on observation (experiment) to quantify the Ksp.

My argument goes down the ladder. If you can answer why AgCl is insoluble in water, then you can definitely predict that Ag+ and Cl- ions from different sources (i.e. in different solutions) put into contact will spontaneously form a precipitate.

Now you maybe thinking this argument is also circular, but it's not.

We have access to experiment, which clearly informs us that AgCl is insoluble in water. We can confidently assume (but we might be wrong!) that the AgCl bond is less ionic than the AgNO3 bond, since it practically does not dissociate/has a low Ksp in water. A bond with a greater covalent character is more stable, since the electrons are shared and not transfered.

Perhaps the crystal structure of AgCl makes it especially stable to solubilization from H2O. I just think it's obvious that the enthalpy of formation of AgCl is favorable, since it has to counter an unfavorable entropy loss in the formation of an ordered crystal lattice.

As you can see, I cannot provide you with a thorough answer, it would be extremely time consuming (for me) and I'd have to conduct many many calculations. I really hope someone can clarify the why. Maybe you can do your own research from the answers provided here and come back with a more "elaborate" answer.


How to correct the pH and composition of bottled water?

Long time ago, I supervised an undergraduate on a summer tea project (for fun). In South Asia there is a natural strawberry colored tea made from slightly matured green tea. You boil the leaves in water with a small amount of sodium bicarbonate. Ice cool it, and boil again and aerate it with a ladle. The tea becomes reddish on aeration and after adding milk, it becomes pinkish. So I can recall your experiences because I tried this recipe in another country. No matter what I did, the right color would not develop. It remained murky brown. It turned out that the chlorinated water was ruining the natural colors, it was particularly destroying the reddish pigment. Passing the tap through a carbon filter+ion-exchanger did not work as calcium also seemed important for the right color.
Finally a mineral water bottle worked and the undergrad saved a project for his poster presentation.
Anyway, it seems your tap water is of bad quality or it is hard. This tea "scum" are Ca/Mg salts. You guessed it right. My suggestion is do not add any acid such as citric acid. Tea is not meant to be made in the presence of citrate salts especially if you add milk to it. I do not know what the taste be like. Most likely horrible!
I do not know how easy it is find potable distilled water in stores. Ask the manufacturer if it is fit for human consumption. Distilled water is not ultrapure water as it may have organic impurities or volatile impurities. It is only free from water soluble salts. All these tea scum problems should vanish.


How to evenly ionize in mass spectrometry

I know that in mass spectrometry they charge different molecules and then run them through a magnetic field and then see where they land.

That is indeed how some mass spectrometers work, but not most of them.  Most mass spectrometers in the world do not use magnets of any type.  Instead they rely on electric fields to move the ions (i.e. charged molecules) and to see where they land.

  Depending on this, they calculate their masses relative to each other. To do this however, they have to ionize the different particles evenly. If one molecule is charged +2 and the other +1, then the results would be thrown off.

Multiple charge states definitely happen in mass spectrometry.  Mass spectrometrists like to say that name "mass spectrometry" is a misnomer, since nearly every "mass" spectrometer in the world is really a mass-to-charge ratio spectrometer.  That is, MS doesn't measure the mass of molecules, it measure the mass-to-charge ratio (or $m/z$) of an ion.

The problem of multiple charge states is especially common in the mass spectrometry of proteins, because proteins are very large molecules with many sites that can become charged.  For example, the image below shows the "mass" spectrum (i.e. $m/z$ spectrum) of horse myoglobin, which is a compound with the formula of $\ce{C769 H1212 N210 O218 S2}$.  The spectrum was obtained from this web page which further describes the multiple charge state phenomenon.

The different peaks correspond to different charge states.  The uncharged molecule has a mass of 16951 Daltons.  The peak at an $m/z$ 737.99 corresponds to a +23 charge state, and the peak at 1542.02 corresponds to a charge state of +11.  The peaks in between correspond to charge states from +22 to +12.

  How do they do they make sure that the charges of all the ions are the same?

There is no foolproof way to do this chemically or physically.  Some modes of ionization, such as electron ionization or APCI tend to favor the formation of singly-charged ions relative to electrospray ionization, but these modes of ionization are sadly ineffective on large molecules such as proteins.  And there are many exceptions to this tendency.

The mass spectral data can be manipulated computationally through a process called deconvolution in an attempt to computationally estimate the true mass spectrum at a given point.  A 1992 paper describes one computational method for spectral deconvolution.  This and/or very similar algorithms are often provided by mass spectrometer vendors in their software for browsing and analyzing MS data.


How much can the pH change through dilution?

Water undergoes autoionization, i.e., it reacts as follows:

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

The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$.

Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as:

$$
[\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V}
$$

We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$:

$$
\lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7}
$$

Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$.


What is the correct formula for diamminechloridonitrito-n-platinum(II)?

So I did some digging, and here's what I found:
According to the current version of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005,

IR-9.2.3.1 Sequence of symbols within the coordination formula
(i) The central atom symbol(s) is (are) listed first.
(ii) The ligand symbols (line formulae, abbreviations or acronyms) are then listed in
alphabetical order (see Section IR-4.4.2.2).Thus, CH3CN, MeCN and NCMe would
be ordered under C, M and N respectively, and CO precedes Cl because single letter
symbols precede two letter symbols. The placement of the ligand in the list does not
depend on the charge of the ligand.
(iii) More information is conveyed by formulae that show ligands with the donor atom
nearest the central atom; this procedure is recommended wherever possible, even for
coordinated water.

So in this case, $\ce{Cl}$ should come before $\ce{NH3}$ and the correct formula would be $\ce{[PtCl(NH3)2(NO2)]}$.


The culprit in the undesired production of hydrogen gas from electric-arc metalizing is water vapor, as it is the only significant source of hydrogen atoms in the atmosphere. The concentration of water vapor is then of course a major factor in how much hydrogen gas can be produced. In very humid conditions the air can contain as much as a few percent water vapor. The chemical reaction for the conversion of water vapor to hydrogen gas under the conditions of this process is as follows:

$$\ce{H2O <=> H2 + 1/2O2}$$

A few percent water vapor may not seem like enough to produce explosive levels of hydrogen gas, but hydrogen gas becomes explosive in air at levels as low as $\pu{4\%}$.

A recirculating dust filtration system, dry or wet, is not an appropriate system for removing hydrogen gas. The only reliable way to prevent the possibility of hydrogen buildup in a closed area is with an appropriate fume hood (or similar approved device) that vents the gases out of the building.

I can't give an exhaustive list of the metals/alloys that could result in hydrogen production via this process, although base metals like copper, lead, nickel and zinc have the potential to do so.


Conversion of molecular structure into Fischer projection for compounds having more than two chiral carbons

To do this exercise, you must need to know how to assign stereo configuration according to Cahn–Ingold–Prelog priority rules, and what Fischer projection represent. First, get correct assignment of three stereo centers in the given molecule. That should be $(1R,2R,3S)$ as depicted in the image:

Now, sketch a Fischer projection with three stereo centers without assigning any groups but number the stereo centers as $1,2,3$. Put the least priory group of stereo center 1 (that would be $\ce{CH3}$ group) on top of the vertical line. By definition, top and bottom vertical lines are wedges-down (dashes) while all horizontal lines are wedges-up (solid wedges). If you put highest priority group ($\ce{Cl}$ group) on left horizontal line and second priority group ($\ce{OH}$ group) on right horizontal line, you get the correct clock-wise rotation for $(1R)$ configuration (See the image). You can follow same procedure or the one you are familiar to assign other stereo centers correctly. For example, for stereo center 3, the least priority group is $\ce{H}$ while the highest one is $\ce{F}$ group. The second priority group is substituted carbon chain, which is already in the projection (the last priority to consider is $\ce{CH3}$ group). Thus, I put $\ce{H}$ group on bottom vertical line (should be a dash) as did for stereo center 1. Then, to get correct counter-clock-wise rotation to give $(3S)$ configuration, you must put $\ce{F}$ group on right horizontal line so last priory group $\ce{CH3}$ would occupy the remaining left horizontal point.
Note: You can exchange any group at a stereo center with another group at same center to reverse the configuration, e.g., $(R) \rightarrow (S)$. If you repeat same with two different group, you get back the original configuration. Once I was done with assigning the groups, I have performed these operation twice to change the positions of three groups at stereo center 3 to my likeness. Still, the configuration remains same.
Finally, you may able to assign remaining two groups ($\ce{Br}$ and $\ce{F}$ groups) for stereo center 2 to get $(2R)$ configuration.
To get the enantiomer, you can sketch the mirror image of the compound, which is $(1S,2S,3R)$ as depicted in the image and also tally with the given answer for the corresponding enanthiomer.


Why do different metal ions release different colors of light?

You are right, it depends on the energy level of valance electrons, but more precisely it depends on the difference between the highest occupied and lowest unoccupied levels' energies. If the energy difference is $\Delta E$ then the light emitted by the excited atom has a frequency $\nu = \frac{\Delta E}{h}$ so a wavelength $\lambda = \frac{c}{\nu}=\frac{h c}{\Delta E}$


Which is the most stable out of the following linked cyclohexadienes?

You must be aware about the concept of resonance. Similar ly over here as you can see in option number 3 there is longer conjugation between double bonds then compared with any other option because Three double bonds are in conjugation and this conjugation supplies stability to the structure


If you're just curious for practical purposes (i.e., "Can I put this KOH solution in this PVC container or will it ruin it?"), there are many material compatibility tables for this type of thing. This one suggests that your solution will be OK at low temperatures for a few days at least. It may react after more time or at higher temperatures.

If you want to know specifically about the chemistry, KOH is a strong base. Strong bases are known to crack polymer chains through solvolysis. PVC in particular loses some chlorine atoms to the solvent which weakens and ultimately shortens the polymer chains, as explained in more detail in the link.


Can all water soluble ionic compounds conduct electricity?

DavePhD is right! Your material is an organic compound.

In order to be soluble in water, it should have some polar substituents. I'd however exclude carboxylic acids or phenolic $\ce{OH}$ (except maybe in the proximity of a carbonyl group) since these will partly dissociate and yield to minor conductivity.

If the melting point wouldn't be that high, inositol would be a candidate.

The nucleobases thymine and uracil do show melting points in the 300 °C range and are both soluble in water.


How to calculate concentration from reaction quotient and initial concentration?

You have everything you need.

$$\ce{A -> B}$$

From the initial condition, stoichiometry and the conservation of mass, always:

$$\ce{[A] + [B]=1}$$

At time $t$:

$$\ce{\frac{[B]}{[A]}=1\to [B]=[A]}$$

Substitute into the first equation:

$$\ce{\to [A] + [A]=1\to 2\times [A]=1\to [A]=\frac12=[B]}$$

For a more general case:

$$\ce{[A] + [B]=n}$$

At time $t$:

$$\ce{\frac{[B]}{[A]}=Q\to [B]=Q\times [A]}$$

Substituting as above, we get:

$$\ce{[A]=\frac{n}{1+Q}}$$

$$\ce{[B]=\frac{n\times Q}{1+Q}}$$


What do different dissociation constants mean?

The first, second, third, etc... acid dissociation constants refer to the equilibrium constant for loss of the first, second, third, and so on proton. For example, $H _2SO_4$ can lose one proton to make HSO4–,  which can then lose another proton to generate SO42–. Ka1 and Ka2 would be the equilibrium constants for these reactions. pKa1 and pKa2 are, of course, -ln(Ka1) and -ln(Ka2).


Why n-alkanes can be found in organic solvent extract of rubber?

Although natural rubber and synthetic rubber display a poor chemical resistance to alkanes among others, they do not  dissolve in alkanes. Your GC/MS traces (I believe that so because of alkane/alkene identification) have confirmed that since they have shown n-heptane as the main peak (I believe everything else are trace amounts exhibiting below your set-up threshold).

There is a reason for your finding of straight chain hydrocarbons in synthetic rubber such as SBR in your solvent extracts. For example, SBR is the shorten form for synthetic material, styrene butadiene rubber, which is made by anionic copolymerization of styrene & butadiene. For this anionic copolymerization, frequently used initiator is a solution of lithium alkyl compound in small paraffins, for instance, n-butyllithium in hexanes. The rubber may still have few molecules of these small paraffins, which may diffuse into the n-alkane you use to extract, and hence, would show up in your GC/MS chromatogram.


How to mix small amount of alcohol into coconut oil

When you consider that rum and most other hard liquors contain at least 50% water, it is not surprising that the result is a two-phase system. If you began with absolute or 95% ethanol, or perhaps even grain neutral spirits, which vary but often come as 180 proof (90% ethanol) you might get the homogeneous solution you want. Another approach would be to add an edible emulsifier such as lecithin, which will help the ingredients to form a stable emulsion.

The earlier suggestion to hydrogenate the oil is not likely to work, as hydrogenation does not make oil able to mix with water. Coconut oil already has a high saturated fat content, as can be seen by the fact that it is often a solid at room temperature.


Ethanol is an organic solvent. I would guess that it makes wood shine by:

Cleaning the surface, dissolving away organic compounds
Maybe slightly dissolving (?) a thin layer of cellulose at the surface of the wood, removing irregularities and making it more malleable, hence easier to polish

There may be more specific interactions between wood and ethanol, but I don't know about them.


Why is Pyridinium Chlorochromate soluble in organic solvents?

Apart from the fact noted by Mithoron that it's soluble in polar organic solvents, we may consider what lattice binding energy the solvation forces must overcome.  In a compound like sodium chloride with compact, monatomic ions the lattice forces are too strong for all but the most aggressive solvents (but lithium salts have a way around this problem).  With pyridinium chlorochromate (PCC), however, the ions are multiatomic, bulky, and differently shaped as well as having only one charge, so the crystal lattice is only weakly bound compared with sodium chloride.  PCC is far from unique; to cite another example, the triangular cyclopropenyl cation was identified through its nmr specta when salts were dissolved in polar solvents.


Acidities of amines and carboxylic acids

Compound 1's conjugate base IS compound 2. The positive NH3 loses a proton to become the neutral NH2, but deprotonation of the NH2 group is extremely unlikely. Compound 2 will lose a proton at the carboxylic acid group first, so your friend is wrong to say deprotonation will occur at the nitrogen in both.

This will likely form a zwitterion, where the nitrogen is NH3+ and the carboxylic acid is COO-. This compound will have 2 pKas, one for the carboxylic acid, 1 for the amine. The carboxylic acid will be the lower of the 2 pKas. So if we start at very low pH and add base, the compound will start like compound 1. As we approach the pKa for the carboxylic acid, it will deprotonate to form the zwitterion. Then, as we approach the pKa of the NH3+, it will also deprotonate, forming a compound with an NH2 and COO-.


One of the effects could be that vibrations increase mixing on phase boundary between wine itself and air above it.

Other thing is that vibration are generally not good for crystal growing process (instead big and nice monocrystals one gets polycrystalline solid).  Usually you can find nice crystals of tartaric acid in the bottom of bottles and barrels. I do not know chemistry behind it but certainly it will change taste of the wine if the tartaric acid stays longer in wine (slow crystallisation) instead of falling rapidly out of solution when you shake it.

Of course if there is still a sediment earlier during fermentation, certainly that will be not good if this mixes with clear wine.


You could proceed from either end member of the olivine solid series and yield $\ce{SiO2}$ as you suggest.

However, I'd consider mechanisms that have been researched in the course of studying so-called mineral sequestration in addition to what you've written, especially considering the energy requirements you propose: the mineral sequestration reactions are (slightly) exothermic and would thus also yield heat as a by-product which I suspect would be desirable in a cold place like the Moon. That said, the reactions are slow at standard conditions and can be accelerated at the cost of supplying energy. How that balance works in your scenario is ultimately up to you, of course.

Additionally, carbon dioxide is a reactant (but needs to be in a supercritical state, which costs energy), which you'd get from human respiration and other organic sources "for free," which relieves you of having to have a lot of hydrogen gas on-hand.

The idea with mineral sequestration is to allow supercritical carbon dioxide to react with certain minerals to yield carbonates that are stable over some long time period:

  Mineral carbonation reactions are known to geologists and occur spontaneously on geological time scales. For example, the reaction of $\ce{CO2}$ with common mineral silicates to form carbonates like magnesite or calcite is exothermic and thermodynamically favored.

An example is:

$$\ce{Mg2SiO4 + 2CO2 -> 2MgCO3 + SiO2}$$

  (which) illustrates the transformation of forsterite, which is the end member of the common silicate mineral olivine. One ton of olivine can dispose of approximately two-thirds of a ton of $\ce{CO2}$. Again, the reaction is
  exothermic and releases 90 kJ/mole of $\ce{CO2}$.

In summary: There's nothing wrong with your chemistry or geology and there is more than one way to yield silica from mafic/ultramafic minerals.

The reference I have quoted above and cited below also details the process in general terms for minerals such as olivine and serpentine, and illustrates schematics for implementing such schemes on industrial scales, which might also be of interest to you.

Reference from the National Energy Technology Laboratory within the DOE:

  Goldberg, P. ,Chen, Z. Y. ,O’Connor, W. ,Walters, R., and Ziock, H. $\ce{CO2}$ Mineral Sequestration Studies in US. Technology. 1 (1): 1–10 (2000).


Why are signals of carbon and hydrogen atoms (in NMR spectra) split by coupling to 19F?

You always have a coupling if you have to NMR-active nuclei connected, there is nothing unusual about fluor here. But in the most common and simplest NMR experiments on small organic molecules you usually don't encounter such heteronuclear couplings.

The first important aspect is that many common nuclei in organic molecules are present mostly in an NMR-inactive isotope. Carbon and nitrogen for example are only present in the NMR-active isotopes 13C and 15N in tiny amounts. So you'll be able to see small satellite peaks sometimes, but no complete splitting. If only 1% of the nuclei is NMR-active, only 1% of the signal will be affected by the coupling.

The second aspect is that some experiments like 13C 1D experiments are typically recorded with decoupling. There is a coupling between 1H and 13C, but it is intentionally suppressed as it would complicate the spectra.


How to calculate the number of atoms of each element in a urea?

You were correct in converting the grams of urea to mol of urea, but the next step would be to convert the mol of urea to molecules of urea using avogadro's number (which converts mol to molecules or atoms).  Then you want to multiply the molecules of urea by the number of atoms of the element in question that are present in Urea.  So for example there are 2 atoms of N per molecule of Urea, therefore multiplying the molecules of urea you calculated earlier by 2 will yield the total number on nitrogen atoms present.  As was mentioned in the comments, the value for N atoms present given by your book was incorrect and should have been 3.37 × 10^26 atoms of Nitrogen.  See if you can calculate the correct number of each of the remaining elements, as the answers given were correct.


Stability comparison of benzylic carbocation and benzene

First refer to https://en.m.wikipedia.org/wiki/Chemical_stability to clean up the situation.

There is not much in comparing the stability of individual different molecules which are not among the reactants or the product of a specific reaction or in contact to a specific and same environment.

The question arises from a common confusion for we are used to compare the "stability" of different limiting forms (resonance structures ) of a hybrid of resonance.

Or, differently, but still analyzing entities with the same constitutional formula, the stability of various conformers.

An analysis in the terms meant by the question could be done to compare the stability of different carbocations, for instance, and see if there could be mesomeric effects able to delocalise the charge. In such a sense the presence of the phenyl ring has a stabilising effect as compared to, e.g., a linear alkyl chain.

This state, but anyway answer the question, the number of resonance structure does not even count as for all those of the benzyl cation are to be supposed much less stable  than those of benzene, the carbonation is obviously charged so this will be the main difference anyway.

You won't ask what is more stable between methane and ethyl carbocation. The fact that in benzyl cation there is a phenyl ring can only tell you that its formation under a specific mechanism will be favoured as compared to the case of ethane.


Paper mostly consists of cellulose, form which wood is a good source.
Another interesting source is the hemp plant and it seems that hemp was indeed the major source in paper manufacturing up to the 19th century.

Nowadays, recycling of paper probably is the most reasonable way to reduce the wood use.

  Do we even need paper anymore with all electronic devices and stylus'

Yes, for different reasons:

Think in how long old books and handwritings have lasted. Compare that with the durability of magnetic (tapes, hard disks) or optical recordings (CD, DVD, Bluray, etc.)
Supposed that the ability to read doesn't get lost and a document is written in a well-defined alphabet, you do not need any additional technology to access the information. Who can read data from a 9 track tape nowadays (the technology is only fifty years old) or an "old" MFM hard drive? 8" or 5.25" floppy disks, anyone? ZipDrives, supposed they survived the "click-od-death"?
Most people do not use electronic signatures for documents. Papers are signed, sealed, etc. Consequently, written contracts, bills, etc. are considered more forge-proof than electronic documents. Storing them is therefore mandatory in business.


How do chemists decide on which detergent to use?

If it's polar, then it won't pick up grease because lipids (fats and oils) are generally non-polar. I don't know specifically how a detergent is chosen, but soaps are chemically the salts of a fatty acid. When choosing a detergent, you want a molecule with a polar head with a long, nonpolar tail.

Recall that water is polar and polar substances dissolve polar substances, and nonpolar dissolve nonpolar. The detergent in a sense forms a "bridge" between the water and the nonpolar substances (dirt, grease, oil, etc.) which would not dissolve in pure water. The detergent molecules form little "cells" (my choice of this word is not a coincidence - this is also how your cell membranes work) with the polar heads facing out and the nonpolar tails mingling with nonpolar dirt molecules inside. This allows the dirt be washed away by water.

As to why certain detergents are chosen, it is probably a balance struck between safety, effectiveness, cost to mass-produce, possibly environmental considerations associated with production, etc.


I believe that in the examples where a 1,2 hydride shift was mentioned, the hydrogen atom would have 'shifted' between two adjacent carbon atoms. I think the 1,2 refers to the fact that the carbon atoms involved in the shift are adjacent.
A 1,3 shift would mean a the hydrogen atom shifts to the second carbon from the original carbon atom it was bonded to, if you get what I mean.
The same applies to methyl shifts.


Is it rigorous enough to change a DFT functional after geometry optimization to calculate excited state properties?

What is sometimes done is use the geometry and thermochemical corrections (i.e. the results of the frequency calculation) of method X, but use the electronic energy (SCF energy) of another method Y. This is generally noted Y//X. Since the geometries and thermochemical corrections typically are not much influenced by the quality of the method (beyond a certain point), this scheme gives good result. The electronic energy can however benefit from a larger basis set or a better computational method (e.g. MP2 with DFT geometries).
I am assuming this is what you have seen in the literature, as it is relatively common, especially for large molecules like organometallic complexes. I am not too familiar with excited states, but the same thing likely applies.
However, you must always use the exact same method for the geometry optimisation and the frequency calculation. The frequency calculation uses the curvature of the potential energy surface (PES) to calculate approximate corrections. If you are not at an extremum (minimum for equilibrium structures or saddle point for transition state), the results are meaningless. Changing the method will move you off the extremum sufficiently and your frequency calculation will likely have several imaginary frequencies.


reaction coordinate, kinetics, equilibrium in example

You have chosen an unusual scheme in that A appears to be an intermediate as well as a reactant. If you use the simpler scheme things are easier and so rather than try to follow yours I have examined the scheme below which is very nonetheless very similar

$$RX\underset{k_{-1}} {\stackrel{k_1}{\leftrightharpoons}}R+X$$
$$R+Y\underset{k_{-2}} {\stackrel{k_2}{\leftrightharpoons}}RY$$

where species RX looses X and is replaced by Y. The intermediate species is R. The overall reaction is $RX+Y=RY+X$ and the equilibrium constant for the two steps are

$$K_1=\frac{k_1}{k_{-1}}=\frac{\mathrm{[R]_e[X]_e}}{\mathrm{[RX]_e}}$$
$$K_2=\frac{k_2}{k_{-2}}=\frac{\mathrm{[RY]_e}}{\mathrm{[R]_e[Y]_e}}$$

and the overall equilibrium constant

$$K=K_1K_2=\frac{k_{1}k_{2}}{k_{-1}k_{-2}}=\frac{\mathrm{[RY]_e[X]_e} }{\mathrm{[RX]_e[Y]_e }}$$

and all the concentrations in square brackets with subscript $e$ are equilibrium values. [This type of equation is true if there are many equilibria one after the other, $\displaystyle K=K_1K_2K_3K_4\cdots=\frac{k_1k_2k_3k_4}{k_{-1}k_{-2}k_{-3}k_{-4}}\cdots$].

In a rate equation approach we can apply a steady state approach to the intermediate species R. The steady state assumes that the rate of change of R is zero;

$$\frac{d[R]}{dt}= k_1[RX]-k_{-1}[R][X]-k_2[R][Y]+k_{-2}[RY]=0$$

from which

$$ [R]_{ss}= \frac{k_{-2}[RY]+k_1[RX]}{k_{-1}[X]+k_2[Y]}$$

The rate $r$ can be given by

$$\begin{align}
r=-\frac{d[Y]}{dt}=k_2[R][Y]-k_{-2}[RY] &= \frac{k_2(k_{-2}[RY]+k_1[RX])[Y]-k_{-2}[RY](k_{-1}[X]+k_2[Y])}{k_{-1}[X]+k_2[Y]}\\&=\frac{k_1k_2[RX][Y]-k_{-1}k_{-2}[RY][X]}{k_{-1}[X]+k_2[Y]}\\
\end{align}$$

and the numerator is zero when equilibrium concentrations are used making the rate equal to zero also. So this connects the equilibrium to the rate equations.

Initially just after the reactants are mixed the amount of product is very small and $k_1k_2[RX][Y]>>k_{-1}k_{-2}[RY][X] $ and the the rate is

$$r=\frac{k_1k_2[RX][Y]}{k_{-1}[X]+k_2[Y]}$$

which can be confirmed by experiment. Hope this helps.


How to determine the empirical formula given mass percentages of the elements and the molecular weight?

Rather than starting out with 100 g of unknown, I would start out with 1 mol of unknown, i.e. 283 g. That way, you can directly calculate how many moles of each element are in one mole of compound:

$$ n = 67.3 \%  \cdot \frac{\pu{283 g}}{\pu{12 g mol-1}} = \pu{15.86 mol}$$

Or to get the stoichiometric coefficient $\nu_C$ directly, take the percentage and multiply it by the ratio of compound molar mass to atomic mass:

$$ \nu_C =  67.3\% \cdot \frac{\pu{283 g mol-1}}{\pu{12 g mol-1}} = \pu{15.86}$$

If you do that for C, N, O and H, you get the formula:

$$\ce{C_{15.86}H_{19.4}N_{0.93}O_{3.80}}$$

You can try rounding this to the nearest integers:

$$\ce{C16H19NO4}$$

This compound has a molar mass of 293 g/mol. Playing around with the formula, you can find coefficients that match the molar mass exactly, e.g. $\ce{C16H13NO4}$ or $\ce{C17H17NO3}$. In any case, neither one of these nor the formula suggested by the OP matches both the molar mass and the mass percentages.

  Is this correct?

$\ce{C17H21NO4}$ is a good answer, but the data are not consistent, so it is not the complete answer. I would say depending on whether the errors on the molar mass determination or the elemental composition are larger, you would have a different top candidate, and should report the numerically exact formula with non-integer coefficients. That way, the reader can critically evaluate the meaning of the data themselves.


What does reaction exothermicity say about the stability of molecules

Good question. There exists two types of stability - thermodynamic stability (determined by $\Delta G_{reaction}$, or $\Delta H_{reaction}$ approximately) and kinetic stability (determined by $\Delta G ^{\ddagger}$, or $E_{activation}$ approximately). The reactant compound in your first reaction is thermodynamically more stable than the second, since it sits at a lower potential energy, but it is kinetically less stable, since there is a smaller barrier for decomposition.

Usually context is sufficient to determine what kind of stability is being talked about, though in general I think stability is more often mentioned in the thermodynamic sense than the kinetic, since thermodynamic stability is related to the reaction free energy (a sum of its enthalpy change and a factor containing its entropy change) and is more commonly tabulated and more readily measured than activation free energies.

For example, all explosives are quite thermodynamically unstable (otherwise they wouldn't be very good at exploding!), but we still speak of "stable explosives", such as TNT or RDX and "unstable explosives", such as $\ce{NI_3}$ or $\ce{Hg(CNO)_2}$. Clearly kinetic stability is what is taken into account.

In constrast, when studying acidity and basicity of substances, it's common to justify reactions based on the "stability" of the acids/bases/conjugate species, but in this case, we're talking about thermodynamic stability, as it defines the acidity/basicity constants.


Reduced-pressure (vacuum) evaporation/drying of 15 liters of aqueous solution

The aspirator pump would be unlikely to be damaged by the distillate (and can be cheaply repaired or replaced), but the electric vacuum pump likely will require a cold-trap to prevent the vapors from destroying it.

The aspirator pump would seem the way to go, but you must factor in the cost of water in your location. This Nalgene aspirator uses ~6.5 liter/minute. A recirculating aspirator pump might be the best compromise for a production environment, where efficiency is important, and it might be feasible to save and recycle the distillate.


To know what's happening here you should know what "emulsifying" is.

First we need to know that fat is "non-polar" substance. what does that mean?

It means that all fats can't be dissolved in water which is "polar" substance.

So how can we make fat -which is non-polar- dissolve in water or any other polar substances?

We add "emulsifiers" like: egg yolk and many other chemical substances like Lecithin.

These emulsifiers are (polar and non-polar) so they can link fat with water together.

We add butter slowly so that the molecules of emulsifier can link the molecules of butter with molecules of sauce -which is polar-.

If there's no free molecules of emulsifier, every molecule of added butter will not be dissolved in the sauce.


In a reaction between Calcium Ammonium Nitrate or Urea with water, does Nitrogen Content and Purity affect how cold the solution gets?

The endothermic process is just the dissolution of the calcium ammonium nitrate (CAN) or urea in water.
The most likely explanation for the lack of cooling when you used your pack is that over time, by getting partially squished at some point, or from defective manufacturing, some water leaked out so that the CAN or urea had already been partially/fully dissolved by the time you tried to use it.
These reagents are pretty stable and it is very unlikely that the nitrogen had changed form or that the system had been contaminated.


How much sodium hydroxide would it take to dissolve one tonne of bauxite?

Bauxite contains usually 45 - 60 % Al2O3. To get an order of magnitude, one ton bauxite contains about 500 kg Al2O3, or about 5000 mole Al2O3. If 2 NaOH are needed to dissolve 1 Al2O3, it means that a minimum of 10'000 mol NaOH is needed to dissolve Al2O3, which weighs 400 kg. But more NaOH is necessary, because the bauxite contains also some silica, which is also dissolved in NaOH. So the amount of NaOH must be higher than 400 kg.


Which temperature is meant in the Gibbs free energy equation?

If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment.

You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this must occur equally in both directions, otherwise a temperature gradient would settle between system and environment. We could describe thermal equilibrium as the result of "a zero balance of rate of heat transfer between system and environment". That's what goes on during a phase transition (e.g. solid to liquid, where entropy of the system increases due to heat given by the environment, whereas temperature stays constant).

Now, applying the second law of thermodynamics to an ensemble system + environment, undergoing an infinitesimal transformation, we write:

$$\mathrm dS_\text{syst} + \mathrm dS_\text{env} \geq 0 \tag1$$

As said above, the entropy of the environment changes if there's a  reversible heat exchange with the system:

$$\mathrm dS_\text{env} = \frac{\delta Q_{env}}{T}=-\frac{\delta Q_{syst}}{T} \tag2$$

Merging eq. $\text{(1)}$ and $\text{(2)}$, you write:

$$\mathrm dS_\text{syst} - \frac{\delta Q_{syst}}{T} \geq 0 \tag3$$

Now the first law allows you to write:

$$\delta Q - P\mathrm dV_\text{syst}=\mathrm dE_\text{syst} \tag4$$

and then

$$T\mathrm dS_\text{syst} - \left(\mathrm dE_\text{syst}+P\mathrm dV_\text{syst}\right) \geq 0  \tag5$$

Now (and this is crucial), assuming both pressure and temperature constant, you end up with

$$\mathrm d{\left(E_\text{syst}+PV_\text{syst}-TS_\text{syst}\right)_{T,P}} \leq 0 \tag6$$

The expression within the brackets is a state function of the only system and corresponds to the free energy. Quoting Wikipedia:

  A state function describes the equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state.

Considering that $E+PV=H$, we write:

$$\left(\mathrm dG_\text{syst}\right)_{T,P} = \left(\mathrm dH_\text{syst}-T\mathrm dS_\text{syst}\right)_{T,P} \leq 0 \tag7$$

Now, while equation $\text{(1)}$ is much more general, equation $\text{(7)}$ applies only to the situation where the work is expansion or contraction and for constant temperature and pression. Nonetheless, since for most situations of chemical interest, those conditions are satisfied, eq $\text{(7)}$ maintains a well deserved relevance and is very convenient to establish whether a process is spontaneous or not, within the limitation of constant temperature and pressure.

A chemical reaction occurs down to the point where $\left(\Delta G\right)_{T, P}=0$.

For a general reaction:

$$\ce{$a$A + $b$B <=> $l$L + $m$M}$$

For constant temperature and pressure:

$$\left(\Delta G\right)_{T, P}=l\mu_{\ce L} + m\mu_{\ce M} - a\mu_{\ce A} - b\mu_{\ce B}  \tag8$$

being $\mu$ the chemical potential for each of the species involved.
The chemical potential relates to activity, according to:

$$\mu=\mu^\circ+RT\ln a \tag9$$

Equation $\text{(8)}$ and $\text{(9)}$ shows that free energy varies with the composition (expressed as activity) of our system: that's exactly what happens during a chemical reaction.

Proceeding further, you will end up with Van't Hoff isotherm:

$$\left(\Delta G\right)_{T, P}=\left(\Delta G^\circ\right)_{T}+RT\ln\frac{a_{\ce L}^{l}a_{\ce M}^{m}}{a_{\ce A}^{a}a_{\ce B}^{b}}$$

Activities vary until a chemical equilibrium is established, so that $\left(\Delta G\right)_{T, P}=0$.


Why do chemistry and physics have different sign convention in thermodynamics?

This is not a simple physics versus chemistry distinction. I taught Physics for 25 years and saw many examples of either usage in multiple textbooks. In fact, at some point in my tenure, the AP Physics committee swapped conventions on the equation sheet for the AP Exam.

Just my take here: I've always attributed the work-done-by-the-system camp as being more prone to be used by engineering types who want to know "what the system can do for us" in practical applications. On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from without.


What is the extent of a reaction between an acid and its conjugate base?

The reaction you've shown is termed an "identity reaction", the products are the same as the reactants so $\ce{K_{eq}}$ must equal one.  In the particular case you've drawn the reaction involves a simple proton transfer and will likely be diffusion controlled.  Other examples of identity reactions are shown below.  That they are actually occurring can be demonstrated using isotopically labeled substrates.  The last one is one of my favorites, the degenerate rearrangement of bullvalene into itself via a Cope rearrangement.  My drawing of bullvalene doesn't do it justice, just google bullvalene and prepare to be fascinated.  If you number each of the ten carbon atoms in bullvalene you'll find that there are over 1.2 million possible arrangements.  Sequential Cope rearrangements will convert each of these arragments into every other possible rearrangement.  Connectivity in bullvalene is lost, each carbon atom is rapidly bound to every other carbon atom.  At a bit over 100 °C, both the proton and carbon-13 NMR of bullvalene show only 1 sharp peak.


Bleaching of chlorophyll stains using hydrogen peroxide

Horseradish peroxidase is not hydrogen peroxide, $\ce{H2O2}$, but rather an enzyme that breaks $\ce{H2O2}$ into $\ce{H2O}$ and $\ce{O2}$. In general, enzymes ending in -ase are lytic enzymes, catalyzing the breakdown of a similar-sounding substance. So, if you add ground horseradish to hydrogen peroxide, bubbles of oxygen are released. BTW, horseradish roots differ in peroxidase potency, so test beforehand with a particular batch before doing this as a demonstration... sometimes it fizzles out, or not.

Another example of lytic enzyme, amylase in the mouth, breaks apart amylose in starch, so a starchy cracker tastes sweeter as it melts in your mouth.

As @Buttonwood suggests, alcohols, ether and oils are more effective solvents than water.


Gibbs free energy in standard state vs. equilibrium

As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\Delta G=\Delta G^\circ+ RT\ln{1}=\Delta G^\circ$$ On the other hand at equilibrium $Q=K$ and so $$\Delta G=\Delta G^\circ + RT\ln{K}$$ This of course leads to $\Delta G^\circ = -RT\ln{K}$ since at equilibrium $\Delta G=0$.

So you might want to think of it as multiple statements:

For the conversion of reactants to products in their standard states $Q=1$
At equilibrium  $\Delta G=0$
At equilibrium $Q=K$
$\Delta G^\circ$ is the free energy change for conversion of reactants to products in their standard states.

The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

$K$ and $\Delta G^\circ$ are very much connected, but $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $\Delta G^\circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.


How to derive the conductivity titration curve which accounts for salt formation

I believe have found a answer.
  It's valid mathematically, make sense physically but I don't know if chemically is true.
  I posted to community appreciation.
  There we go!

The reactions

ionization of weak acid:
$$\ce{HA + H2O <=> H3O+ + A-}\qquad K_\ce{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}
\label{eq:KWeakAcid}$$
ionization of weak base:
$$\ce{B + H2O <=> BH+ + OH-}\qquad K_\ce{b}=\frac{\ce{[HB+][OH-]}}{\ce{[B]}}
\tag{2}
\label{eq:KWeakBase}$$
self-ionization of water:
$$\ce{2 H2O <=> H3O + OH-}\qquad K_\ce{w}=\ce{[H3O+][OH-]}
\tag{3}
\label{eq:KWater}$$
equilibrium of poorly soluble salt:
$$\ce{HBA <=> HB+ + A-}\qquad K_\ce{sp}=\ce{[HB+][A-]}
\tag{4}
\label{eq:KSalt}$$

Charge balance

$$\ce{[H3O+] + [HB+] = [OH-] + [A-]}\tag{5}\label{eq:ChangeBalance}$$

Mass balance

To mass balance we can use an expression based on amount of substance:

Weak acid:

\begin{align}
{n_\ce{HA}}_\text{total}&={n_\ce{HA}}_\text{solution}+{n_\ce{A-}}_\text{solution}+{n_\ce{A-}}_\text{ppt}\\
C_\ce{HA}V_\ce{HA}&=\left(\ce{[HA] + [A-]}\right)V_\text{solution}+{n_\ce{A-}}_\text{ppt}\\
\\
\ce{[HA] + [A-]}&=\frac{C_\ce{HA}V_\ce{HA}-{n_\ce{A-}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{6}
\label{eq:AcidMassBalance}\\
\end{align}

Weak base:

\begin{align}
{n_\ce{B}}_\text{total}&={n_\ce{B}}_\text{solution}+{n_\ce{HB+}}_\text{solution}+{n_\ce{HB+}}_\text{ppt}\\
C_\ce{B}V_\ce{B}&=\left(\ce{[B] + [HB+]}\right)V_\text{solution}+{n_\ce{HB+}}_\text{ppt}\\
\\
\ce{[B] + [HB+]}&=\frac{C_\ce{B}V_\ce{B}-{n_\ce{HB+}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{7}
\label{eq:BaseMassBalance}\\
\end{align}

The general equation

Combining ($\ref{eq:KWeakAcid}$) and ($\ref{eq:AcidMassBalance}$) equations we have:

$$\ce{[A-]}=\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}\frac{C_\ce{HA}V_\ce{HA}-{n_\ce{A-}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{8}
\label{eq:Anion}$$

Combining ($\ref{eq:KWeakBase}$), ($\ref{eq:KWater}$) and ($\ref{eq:BaseMassBalance}$) equations we have:

$$\ce{[HB+]}=\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}\frac{C_\ce{B}V_\ce{B}-
{n_\ce{HB+}}_\text{ppt}}{V_\ce{HA}+V_\ce{B}}
\tag{9}
\label{eq:Cation}$$

From ($\ref{eq:KSalt}$) equation we have ${n_\ce{HB+}}_\text{ppt}={n_\ce{A-}}_\text{ppt}=
n_\ce{HBA}=n_\text{ppt}=n$ and replacing ($\ref{eq:KWater}$), ($\ref{eq:Anion}$) and
($\ref{eq:Cation}$) equations in ($\ref{eq:ChangeBalance}$) equation we have:

$$\ce{[H3O+]} + \frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}
\frac{C_\ce{B}V_\ce{B}-n}{V_\ce{HA}+V_\ce{B}}
= \frac{K_\ce{w}}{\ce{[H3O+]}} + \frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}
\frac{C_\ce{HA}V_\ce{HA}-n}{V_\ce{HA}+V_\ce{B}}
\tag{10}
\label{eq:TitrationCurve}$$

Besides that combining the ($\ref{eq:KSalt}$), ($\ref{eq:Anion}$) and ($\ref{eq:Cation}$) equations we have:

$$K_\ce{sp}=\overbrace{
\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}\frac{C_\ce{B}V_\ce{B}-n}{V_\ce{HA}+V_\ce{B}}
}^{\ce{[HB+]}}
\overbrace{
\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}\frac{C_\ce{HA}V_\ce{HA}-n}{V_\ce{HA}+V_\ce{B}}
}^{\ce{[A-]}}
\tag{11}
\label{eq:Kppt}$$

Reproducing your data

If we ignore the poorly soluble salt formation, the ($\ref{eq:TitrationCurve}$)
equation is exactly the same which you deduced.

From your image we have
$V_\ce{HA} = \pu{1.00 L}$,
$C_{A} = \pu{0.01 mol L^{-1}}$,
$K_\ce{a} = \pu{2.0E-4}$,
$C_\ce{B} = \pu{0.20 mol L^{-1}}$,
$K_\ce{b} = \pu{2.0E-4}$, and
$K_\ce{w} = \pu{1.0E-14}$
for which the titration curve would be:

The plot of the concentrations of $\ce{[HB+]}$ and $\ce{[A-]}$ through titration would be:

Using the conductivity equation of the solution

$$\kappa_\ce{solution} = \lambda_\ce{A-}\ce{[A-]} + \lambda_\ce{HB+}\ce{[HB+]}
+ \lambda_\ce{H3O+}\ce{[H3O+]} + \lambda_\ce{OH-}\ce{[OH-]}
\tag{12}
\label{eq:ConductivitySolution}$$

and
$\lambda_\ce{A-} = \pu{ 90.0E-4 m^2 S mol^{-1}}$,
$\lambda_\ce{HB+} = \pu{172.0E-4 m^2 S mol^{-1}}$,
$\lambda_\ce{H3O+} = \pu{849.6E-4 m^2 S mol^{-1}}$, and
$\lambda_\ce{OH-} = \pu{398.0E-4 m^2 S mol^{-1}}$
from your image is possible to plot the conductivity curve:

If we look to contribution of each species to conductivity curve
we can understand its final shape.

Some considerations

To take account the formation of $\ce{HBA}$, suppose that $K_\ce{sp} = \pu{7.5E-5}$.
The $Q_\ce{sp}$ plot through titration would be:

While $Q_\ce{sp}<K_\ce{sp}$ we have $n=0$, because $\ce{[HB+][A-]}$ is less than
critical concentrations to start precipitation and the titration curve follows as usual.
When $Q_\ce{sp}=K_\ce{sp}$ we have the precipitation eminence and $n=0$.
When $Q_\ce{sp}>K_\ce{sp}$ I'm not sure what will happen.

The $V_\text{critical}$ in which $Q_\ce{sp}=K_\ce{sp}$ can be found graphical or analytical way.
The last way in not so easy, but I believe then can be done.
Personally I will use the graphical way.

Solving ($\ref{eq:TitrationCurve}$) equation to $V_\ce{B}$ (with $n=0$) we can write:

$$V_\ce{B}=\frac{V_\ce{HA}\left(\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}-\ce{[H3O+] +
C_\ce{HA}\alpha_\ce{A-}}\right)}{C_\ce{B}\alpha_\ce{HB+} -
\left(\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}-\ce{[H3O+]}\right)}
\tag{13}
\label{eq:Vtit}$$

Again, the ($\ref{eq:Vtit}$) equation is exactly the same which you deduced.

Solving ($\ref{eq:Kppt}$) to $V_\ce{B}$ (with $n=0$) we have:

$$V_\ce{B}=\displaystyle\frac{-\left[V_\ce{A}\left(2 -
\displaystyle\frac{C_\ce{HA}\alpha_\ce{A-}C_\ce{B}\alpha_\ce{HB+}}{K_\ce{sp}}\right)\right]-
\sqrt{\left[V_\ce{A}\left(2 - \displaystyle\frac{C_\ce{HA}\alpha_\ce{A-}C_\ce{B}\alpha_\ce{HB+}}
{K_\ce{sp}}\right)\right]^2 - 4V_\ce{HA}^2}}{2}
\tag{14}
\label{eq:Vcritical}$$
where
$${\alpha}_\ce{HB+}=\frac{K_\ce{b}}{\displaystyle\frac{K_\ce{w}}{\ce{[H3O+]}}+K_\ce{b}}$$
and
$${\alpha}_\ce{A-}=\frac{K_\ce{a}}{\ce{[H3O+]}+K_\ce{a}}$$

The $V_\ce{B}\ vs.\ \ce{pH}$ plot to ($\ref{eq:Vtit}$) and ($\ref{eq:Vcritical}$) equations
allows to find $V$ and $\ce{pH}$ for which the curves cut across each other.

Another (graphical) way could be the $Q_\ce{sp}$ plot.
Any way the values found are $\pu{45.20 mL}$ and $4.685$ respectively.
What happens with $\ce{pH}$ when $V_\ce{B}$ is greater than $V_\text{critical}$?
Some hypothesis can been proposed.

Hypothesis #1

  $\ce{HBA}$ will precipitate and the titration curve don't be affected.

In this case the ($\ref{eq:Kppt}$) equation can be solved to $n$:

$$n=\displaystyle\frac{\left(n_\ce{HA}+n_\ce{B}\right)-
\sqrt{\left(-\left(n_\ce{HA}+n_\ce{B}\right)\right)^2 -
4\left(n_\ce{HA}n_\ce{B}-
\displaystyle\frac{K_\ce{sp}\left(V_\ce{HA}+V_\ce{B}\right)^2}
{{\alpha}_\ce{HB+}{\alpha}_\ce{A-}}\right)}}{2}
\tag{15}
\label{eq:nppt}$$

where $n_\ce{HA} = C_\ce{HA}V_\ce{HA}$ and $n_\ce{B} = C_\ce{B}V_\ce{B}$.

The ($\ref{eq:nppt}$) equation allows calculate the amount of precipitate is formed
given a $V_\ce{B}$ and $\ce{pH}$ value. But given a $V_\ce{B}$, what is $\ce{pH}$ value?
I believe that, if this hypothesis is true, the $\ce{pH}$ value can be calculated
from ($\ref{eq:TitrationCurve}$) equation and the titration curve don't be affected.

The the $n\ vs.\ V_\ce{B}$ plot of ($\ref{eq:nppt}$) equation is

and the $n\ vs.\ \ce{pH}$ plot of ($\ref{eq:nppt}$) equation is

The effect of precipitation can be seen in plots below.
The $\ce{[HB+]}$ and $\ce{[A-]}$ plot

The $Q_\ce{sp}$ plot

The plot of contribution of each species to conductivity

The total conductivity plot

Hypothesis #2

  $\ce{HBA}$ don't precipitates and the $\ce{pH}$ rises.

In this case the $V_\ce{B}\ vs.\ \ce{pH}$ plot gives some clue.

The ($\ref{eq:Kppt}$) equation can be rearranged to give:

$$\ce{[H3O+]}^2 - \ce{[H3O+]}\left(
\frac{K_\ce{a}C_\ce{HA}V_\ce{HA}C_\ce{B}V_\ce{B}}
{K_\ce{sp}\left(V_\ce{HA}+V_\ce{B}\right)^2} - K_\ce{a} -
\frac{K_\ce{w}}{K_\ce{b}}\right) +
\frac{K_\ce{a}K_\ce{w}}{K_\ce{b}}=0
\tag{16}
\label{eq:TitrationCurvePpt}$$

Note that, using one of the roots from ($\ref{eq:TitrationCurvePpt}$) equation,
the $\ce{pH}$ jumps from $4.685$ to $9.315$ (see titration curve below) when
$V_\ce{B}=V_\text{critical}$ and, while $V_\ce{B}$ increases, the titration curve
is look like as usual.

Mathematically in this hypothesis no precipitate is formed, but the
solution cross the region where there should be precipitate.
The consequence this is the concentrations of $\ce{[A-]}$ and $\ce{[HB+]}$,
which depends on $\ce{pH}$ values, jumps too.

Hypothesis #3

  $\ce{HBA}$ don't precipitates and the $\ce{pH}$ falls.

Again the $V_\ce{B}\ vs.\ \ce{pH}$ plot gives some clue.

Using the other root from ($\ref{eq:TitrationCurvePpt}$) equation, the titration curve
have a discontinuity at $V_\text{critical}$ and then the $\ce{pH}$ values falls
while $V_\ce{B}$ increases. No jumps, neither on concentrations or pH, are observed.

If hypothesis #3 is true, the plot of the concentrations of $\ce{[HB+]}$ and $\ce{[A-]}$
through titration is

the $Q_\ce{sp}$ plot is

the conductivity curve to each species

and finally the conductivity curve is

Addendum

To those interested follows the gnuplot code to produce these graphs. All equations needed to reproduce this answer are available.

The present code creates several graphs used in this answer but no all. Some of them are combination of two or more data sets or a graph from an equation available.

#
#  Implements a lot of equations and create a couple of graphs showing what
#  happens with pH, concentrations, conductivity, and Qsp when a poorly soluble
#  salt is formed during a titration.
#
#  To more details see the original question/answer
#
#  Question: https://chemistry.stackexchange.com/q/117984/80491
#  Answer:   https://chemistry.stackexchange.com/a/118760/80491
#
#  Author: GRSousaJr (https://chemistry.stackexchange.com/users/80491/grsousajr)
#  gnuplot 5.2 patchlevel 7
#

# ========== INITIAL CONFIGURATION ===============================================
reset               # return all graph-related options their default values
set encoding utf8   # selects a character encoding

# Save each graph as pdf file
set terminal pdfcairo size 8 in, 6 in font "Times New Roman,18" enhanced

# Colors of lines
set style line 1 lc "#e41a1c"   # red
set style line 2 lc "#377eb8"   # blue
set style line 3 lc "#4daf4a"   # green
set style line 4 lc "#984ea3"   # purple

# Margins
set lmargin screen 0.110    # left margin
set tmargin screen 0.940    # top margin
set bmargin screen 0.130    # bottom margin

set grid ls -1 lc "gray"    # style for grid
set tics out nomirror       # tics marks on axis
set samples 1000            # number of points to each curve
set dummy pH                # "pH" as "x" for all plot-commands
unset key                   # turn-off the legend (key) exhibition

# ========== INPUT PARAMETERS ====================================================
# ---------- Weak acid (HA) ------------------------------------------------------
VA    = 1.0     # L, HA volume (mandatory)
CA0   = 0.01    # mol/L, HA concentration (mandatory)
pKa   = 3.69897 # Ka = 0.00020 (mandatory)

# ----------- Weak Base (B) ------------------------------------------------------
CB0   = 0.2     # mol/L, B concentration (mandatory)
pKb   = 3.69897 # Kb = 0.00020 (mandatory)
VBmax = 120.0   # mL, B volume (optional)

# ----------- Poor soluble salt (HBA) --------------------------------------------
pptn  = "True"  # "True/False" (mandatory) indicates if occurs precipitation
Ksp   = 7.5E-5  # HBA poor soluble salt (mandatory if pptn = "True")

# ----------- Solvent ------------------------------------------------------------
pKw   = 14.0    # H2O self-ionization

# ----------- Conductivity constants ---------------------------------------------
array lambda[4]
lambda[1] =  90.00 # (E-4) A-
lambda[2] = 172.00 # (E-4) HB+
lambda[3] = 849.60 # (E-4) H+
lambda[4] = 398.00 # (E-4) OH-

# ========== DERIVATE PARAMETERS =================================================
# ---------- Equilibrium constants -----------------------------------------------
Ka     = 10**(-pKa)     # Ka from pKa
Kb     = 10**(-pKb)     # Kb from pKa
Kw     = 10**(-pKw)     # Kw from pKw
# ---------- Logical test to VBmax------------------------------------------------
if (!exists("VBmax")){
    VBmax = 2.0*VA*CA0/CB0  # Double that required for neutralization
} else{
    VBmax = VBmax/1000.0    # Converting mL to L
}

# ========== FUNCTIONS ===========================================================
# ---------- H+ and OH- concentrations (as a function of pH) ---------------------
H(pH)  = 10**(-pH)      # H+ from pH
OH(pH) = 10**(pH-14.0)  # OH- from pH

# ---------- Ionization fractions (as a function of pH) --------------------------
aA(pH)  = Ka/(H(pH) + Ka)       # A-
aB(pH)  = Kb/(Kw/H(pH) + Kb)    # HB+

# ---------- B volume (as a function of pH) --------------------------------------
VB(pH) = (VA*(Kw/H(pH) - H(pH) + aA(pH)*CA0))/(aB(pH)*CB0 - (Kw/H(pH) - H(pH)))

# ---------- Amount of substance (as a function of pH) ---------------------------
nA       = CA0*VA         # HA
nB(pH)   = CB0*VB(pH)  # B

# Amount of precipitate formed
nppt(pH) = ((nA + nB(pH)) - sqrt((-(nA + nB(pH)))**2 - \
    4*(nA*nB(pH) - (Ksp*(VA + VB(pH))**2/(aB(pH)*aA(pH))))))/2.0

# Logical test to know if precipitation must be taken account
if (pptn eq "True" ){
    n(pH) = nppt(pH) <= 0 ? 0 : nppt(pH)  # Disregards negative values of 'n'.
    sufix = "_with_ppt"
} else {
    n(pH) = 0   # No occurs precipitation
    sufix = "_without_ppt"
}

# ---------- A- and HB+ concentrations (as a function of pH) ---------------------
CA(pH) = aA(pH)*(nA - n(pH))/(VA + VB(pH))       # [A-]
CB(pH) = aB(pH)*(nB(pH) - n(pH))/(VA + VB(pH))   # [HB+]

# ---------- Qsp (as a function of pH) -------------------------------------------
Qsp(pH) = CB(pH)*CA(pH)

# ---------- Individual and Total conductivities (as a function of pH) -----------
kappaA(pH)   = lambda[1]*CA(pH)
kappaB(pH)   = lambda[2]*CB(pH)
kappaH(pH)   = lambda[3]*H(pH)
kappaOH(pH)  = lambda[4]*OH(pH)
kappaSol(pH) = kappaA(pH) + kappaB(pH) + kappaH(pH) + kappaOH(pH)

# ---------- Initial and final pH ------------------------------------------------
pH0   = -log10((-Ka + sqrt(Ka**2 + 4*CA0*Ka))/2.0)    # Initial pH
pHmax = pKw + log10((CA0*2*VA - CA0*VA)/(VA + VBmax)) # Estimate for pHmax at VBmax

#---------- Finding pHmax by iteration -------------------------------------------
while (VB(pHmax) > VBmax || VB(pHmax) < 0){
    pHmax = pHmax - 0.0001
}

# ========== CREATING THE GRAPHS =================================================
# ---------- Titration curve -----------------------------------------------------
filename = "Titration_Curve" . sufix # Define a filename to pdf file

# Plotting using a data block (like a temp file)
set table $Titration_Data
    plot [pH0:pHmax] 1000*VB(pH)
unset table

set xlabel "Volume of titrant / mL" # name for x-axis (the same to all plot-commands)
set ylabel "pH"                     # name for y-axis
set xrange [0:VBmax*1000]           # VBmax in mL
set yrange [0:14]                   # pH 0 to 14
set xtics 10                        # tics at each 10 mL
set ytics 1                         # tics at each 1 pH unit

set output filename . ".pdf"        # Create the pdf file

plot $Titration_Data using 2:1 with lines linestyle 1 linewidth 2

# ---------- Concentrations curves ------------------------------------------------
filename = "Concentrations_Curve" . sufix

set table $ConcB_Data
    plot [pH0:pHmax] CB(pH)
unset table

set table $ConcA_Data
    plot [pH0:pHmax] CA(pH)
unset table

set ylabel "Concentration / mol L^{–1}"
set yrange [0:CA0]
set ytics .001
set format y "%.3f" # number with 3 decimal places

# Labels
set label 1 "{/:Bold [HB^{+}]}" at 20.0, 0.003 tc ls 1 font ',20' offset 0,0.75
set label 2 "{/:Bold [A^{–}]}"  at 05.0, 0.003 tc ls 2 font ',20' offset 0,0.75

set output filename . ".pdf"

plot \
    $ConcB_Data u (1000*VB($1)):2 w l ls 1 lw 2 ,\
    $ConcA_Data u (1000*VB($1)):2 w l ls 2 lw 2

unset for [i=1:2] label i   # Clean the labels
unset format                # Clean the format

# ---------- Individual conductivity curves --------------------------------------
filename = "Individual_Conductivity_Curve" . sufix

set table $ConducA_Data
    plot [pH0:pHmax] kappaA(pH)
unset table

set table $ConducB_Data
    plot [pH0:pHmax] kappaB(pH)
unset table

set table $ConducH_Data
    plot [pH0:pHmax] kappaH(pH)
unset table

set table $ConducOH_Data
    plot [pH0:pHmax] kappaOH(pH)
unset table

set ylabel "{/:Italic κ} / S m^{–1}"
set yrange [0:2]
set ytics .2
set format y "%.2f"

set label 1 "{/=24 {/:Italic κ}_{/=12 A^{–}}}"      at 95, 0.70 center tc ls 1
set label 2 "{/=24 {/:Italic κ}_{/=12 HB^{+}}}"     at 95, 1.50 center tc ls 2
set label 3 "{/=24 {/:Italic κ}_{/=12 H_{3}O^{+}}}" at 07, 0.90 center tc ls 3
set label 4 "{/=24 {/:Italic κ}_{/=12 OH^{–}}}"     at 95, 0.15 center tc ls 4

set output filename . ".pdf"

plot \
    $ConducA_Data  u (1000*VB($1)):2 w l ls 1 lw 2 ,\
    $ConducB_Data  u (1000*VB($1)):2 w l ls 2 lw 2 ,\
    $ConducH_Data  u (1000*VB($1)):2 w l ls 3 lw 2 ,\
    $ConducOH_Data u (1000*VB($1)):2 w l ls 4 lw 2

unset for [i=1:4] label i
unset format

# ---------- Total conductivity curves -------------------------------------------
filename = "Total_Conductivity_Curve" . sufix

set table $ConducSolution_Data
    plot [pH0:pHmax] kappaSol(pH)
unset table

set yrange [0.75:3]
set ytics .25
set format y "%.2f"

set output filename . ".pdf"

plot $ConducSolution_Data u (1000*VB($1)):2 w l ls 1 lw 2
unset format

# ---------- Qsp curves ----------------------------------------------------------
filename = "Qsp_Curve" . sufix

set table $Qsp_Data
    plot [pH0:pHmax] Qsp(pH)
unset table

set yrange [0:0.000100]
set ytics 0.00001
set ylabel "{/:Italic Q}_{sp}"
set format y "%h" # scientific notation format

set label 1 "{/:Bold {/:Italic Q}_{sp}}"\
    at 32.0, 3.0E-5 tc ls 1 font ",20" offset 0,0.75
set label 2 "{/:Italic Q}_{sp} = {/:Italic K}_{sp}" \
    at 70.0, Ksp    tc ls 1 font ",20" offset 0,0.75

set output filename . ".pdf"

plot $Qsp_Data u (1000*VB($1)):2 w l ls 1 lw 2


Is there a compound that binds with salt (NaCl) in solution?

$\ce{NaCl}$ itself does not exist as a compound in water. The dissolution of salt in water means that both the sodium cation and the chloride anion are separated from each other and are each surrounded by water molecules. Therefore, no compound can bind to NaCl in water because $\ce{NaCl}$ does not exist in water.

It is, however, possible that a compound "binds" with either sodium or chlorine. For example, by adding copper to the solution and performing a bit of more electrochemical magic (out of the scope of this answer), you can get the solid compound copper chloride ($\ce{CuCl}$).


What is the chemistry behind smell of human perspiration(sweat)?

Why does sweat smell?!

Sweat released from sweat glands odourless. Bad smell of sweat is due to certain bacteria on your skin mix with sweat(That is why sweat smell bad after sometime) and also due to certain hormones released during puberty.

I have tried my best to keep this part chemistry point of view(Actually this belongs to biology).

  Chemical composition of sweat

The main reason people perspire is so evaporation of water can cool the body. Thus water is main component.

The chemical composition of perspiration varies between individuals and also depends on what they have been eating and drinking, why they are sweating, how long they have been perspiring and several other factors.

On average, the mineral composition is:

sodium (0.9 gram/liter)
potassium (0.2 g/l)
calcium (0.015 g/l)
magnesium (0.0013 g/l)

Trace metals that the body excretes in sweat include:

zinc (0.4 milligrams/liter)
copper (0.3–0.8 mg/l)
iron (1 mg/l)
chromium (0.1 mg/l)
nickel (0.05 mg/l)
lead (0.05 mg/l)

Source: ncbi and chemistry.about


Can acid-base reactions occur in a non-aqueous medium?

If you dissolve $\ce{NH4Cl}$ in water, your solution contains $\ce{NH4+}$ and $\ce{Cl-}$ ions. If you dissolve $\ce{NaOH}$ in water, your solution contains $\ce{Na+}$ and $\ce{OH-}$ ions.
Now if you mix those two solutions, nearly all $\ce{NH4+}$ and $\ce{OH-}$ will react and produce $\ce{NH3}$ and $\ce{H2O}.$ This shows that $\ce{NH4OH}$ cannot exist. It is decomposed into a solution of $\ce{NH3}$ in water, with only a few percent of $\ce{NH4+}$ and $\ce{OH-}$ ions.

You will certainly object and say that $\ce{NH4OH}$ is a commercial substance, that you can buy anywhere in the market. You are right. But you may not know that the reason of this mistake is commercial and goes back to the 19th century. In the 19th century, the ions were unknown. All substances had a chemical formula, whatever pure or in solution. $\ce{NaOH}$ in solution was made of $\ce{NaOH}$ and that's all. Charged ions like $\ce{H3O+}$ and $\ce{OH-}$ did not exist.

At this time, chemists were often studying the neutralisation reaction

$$\ce{Acid + Base -> Salt + Water}$$

They succeeded in carrying out this reaction with mineral or organic acids, and with many hydroxides (hydroxide was synonymous to base). When a new organic acid was discovered, they tried to make salts with all available hydroxides. Each new salt deserved a publication. And this was fine!

Unfortunately, ammonia was a problem: $\ce{NH3}$ reacts with any acid, but it does not produce any water. $\ce{NH3 + HCl}$ makes $\ce{NH4Cl}$ and that's all: no water! This was difficult to admit. So the chemists decided that, when dissolved in water, $\ce{NH3}$ is transformed into $\ce{NH4OH}.$ This makes sense, because $\ce{NH3}$ was very soluble in water, as if a chemical reaction happens. It was fine. When neutralizing a solution of ammonia by reaction with any acid, the equation of the reaction was:

$$\ce{NH4OH + HCl -> NH4Cl + H2O}$$

$\ce{NH3}$ was only existing in a gaseous state. It was considered as a sort of "dehydrated ammonium hydroxyde". A little bit like other hydroxides, which can easily lose water. At high temperature, $\ce{Cu(OH)2}$ becomes $\ce{CuO}$ and $\ce{Al(OH)3}$ becomes $\ce{Al2O3}.$

So the merchants started to sell ammonia solution with the label $\ce{NH4OH}.$ It was soon discovered that this solution is helpful unblocking blocked up pipes. Everything was OK!

Unfortunately at the end of the century, Arrhenius discovered the ions, and showed that $\ce{NH4+}$ and $\ce{OH-}$ ions cannot exist simultaneously in solution. Disaster! The so called "ammonium hydroxyde" was mainly containing $\ce{NH3}$ and $\ce{H2O},$ with only a few percent of remaining ions $\ce{NH4+}$ and $\ce{OH-}.$ So the merchants changed their labels and started to sell "concentrated ammonia solution". It was OK for the chemists.

But this change was not accepted by the caretakers, managers and contractors. They protested to the manufacturers against this new "ammonia solution" which was useless to unblock blocked pipes. They said they have tried the new and the old solutions. They demanded to get the old ammonium hydroxide, which was "much better" than this stupid ammonia solution.

Well! The merchants are not stupid. They want to sell and get money. So they renounced to modernize their labels, and continued to sell $\ce{NH4OH}$ solutions. This was accepted by chemists and non-chemists.

That is why $\ce{NH4OH}$ is still on sale today. I am an old man, 82. This story was reported to me by my father, who was a chemist who remembered this old time.


What would you call the polymerised sheet-like polymer arising from Fittig reaction of 1,3,5-trichlorobenzene?

Ignoring the complex challenges it would take to make such a structure, the name is rather unclever with the literature referring to it as "holey graphene". You could probably take this a step further with your hypothetical material and call it :

well-ordered holey graphene
or better yet, well-ordered monodisperse-C6-holey graphene.

with C6 alluding to the missing $\ce{C6}$ ring that would be present in perfect graphene. Materials science has very unrigid naming conventions. If you make it first, you can greatly influence the literature terms.

As far as conductivity goes, I agreed that it would likely be conductive or semi-conductive at a minimum similar to poly(9,10-anthracene diylidene).


You're right, they don't really have concentrations or pressure to work with, which would lead to 0/0 craziness.

I think we can go back a step and look at how the equilibrium constant is defined. When you derive the equilibrium constant you don't use pressure or concentrations. You actually use the substances' activities which we can relate to change in Standard Gibbs Free Energy for the reaction (which is easy to find with tabulated data):

For a full derivation refer to Wikipedia's article on the equilibrium constant.


What does the signals in a XPS spectrum mean?

you are quite right, XPS does measure the 'binding energy' of electrons. More specifically, XPS bombards a surface with X-rays and when those X-rays have sufficient energy they are able to sufficiently excite electrons to be emitted from the sample. Each peak in an XPS spectrum corresponds to the energy necessary to emit a given electron.

So what would you expect to see for Mo?

Outer lying electrons (for Mo, we are talking about electrons in the
4p and 4s orbitals) will emit at a lower energy. You should observe
low-energy peaks that correspond to these emissions (approx. 30 - 60
eV).
you should see a cluster of peaks at approx. 230 eV that correspond to the d-orbital electrons.
Finally, you might see discrete peaks at very high energy, approx. 2600 - 2800 eV that    correspond to emission from the 2s and 2p orbitals

In general, outer orbitals emit at lower energy, and inner orbitals emit at higher energy.

Hope this helps!


Gaussian: Relaxed scan with modredundant optimization and dummy atoms

Disclaimer: This answer concerns the use of Gaussian 09. While the ModRedundant interface is retained and therefore the same restrictions also apply in newer versions, there is now a newer way of using internal coordinates. See my other answer for this option.

Like I posted earlier in the comments, you can indeed not use dummy atoms with a relaxed scan via the ModRedundant interface, because they are not existent in that coordinate system. Internal redundant coordinates are generated only from actual atoms. I tried to find an explanation of that, but failed. I believe the reason for that behaviour is, that dummy atoms do not have basis functions and are therefore not included in the wave function.

In this case you need alternative means to what you actually want to achieve. That of course depends on what you would like to model. From the input you provided, I can see, that you are trying to move the middle points of the carbon-oxygen and nitrogen-hydrogen bonds closer together, possibly to simulate the carbon nitrogen bond formation and proton transfer. While I seriously doubt this will happen in one step, this is at least one way to try to go about it.

If this method had worked, I believe you would have gotten a different than expected result. There are some more issues, that you might have not considered with your approach.
You might think that the following two lines keep the nitrogen hydrogen bond distance fixed. They do not.

B 22 7 F
B 2 7 F

Because the bond angle A 22 7 2 is still flexible, and the bond length B 22 2 is still flexible, this can possibly do whatever it wants. The same does hold for the carbon oxygen bond.

B 6 19 F
B 21 19 F

If I am correct with my assumption, then what you want look at is the carbon nitrogen bond formation. Why not take this internal coordinate and scan it? It is easy and it should pretty much do what you like. The following input should take you there. Note that I changed the method to pm6 to massively cut down computer time - and even with that, it took to long for my patience. (I did also not include any trailing lines about resources - see my notes below.)

#P pm6
opt=(modredundant)
scrf=(pcm)
nosymm

4MeAminoButanal scan

0 1
C        0.000000     0.000000     0.000000
N        0.000000     0.000000     1.451901
C        1.325038     0.000000     2.055476
C        1.991320     1.377897     2.032970
C        1.229182     2.449098     2.809397
C        1.309917     2.335179     4.306919
H        1.218960    -0.323776     3.083767
H        1.990264    -0.714465     1.564774
H        2.993872     1.281489     2.436811
H        2.106328     1.711600     1.006802
H        1.602469     3.443240     2.566292
H        0.178120     2.455789     2.536445
H        0.708887     3.065770     4.854632
H       -1.016528    -0.113770    -0.358506
H        0.375985     0.943036    -0.379032
H        0.606031    -0.798850    -0.431982
O        1.974063     1.543637     4.907309
H       -0.496745    -0.808017     1.771266

B 6 17 F
B 2 18 F
B 2  6 S 60 -0.05

Also note, that I did not check that the N-C bond length will become shorter than an Ångström, so the last steps take forever to converge and it will die in the last step running out of cycles. Run it with something more reasonable.
The calculation itself produced a somewhat reasonable result. The following animation cycles through all optimised structures. It was made with ChemCraft.

Some more notes on your input:

%nprocshared=16 is quite a number you are requesting for a Hartree-Fock calculation. Consider that parallelisation is not linear. While it became better throughout the years, I guess 4 threads is the most reasonable for this system size. You can go higher, I'd not recommend to go above ten though.
%mem=5gb is probably too much. I doubt that any of these processes will exceed 200MB of memory, probably a lot less.
%chk=4MeAminoButanal You should consider adding the .chk ending in all cases. Gaussian is known to mess things up when there are non-ordinary characters like _-. in a file name.
It is maybe a good idea to not write the whole route section in one line. Consider this as improving the readability of your input.  You can cluster similar commands (like pop statements) in one line, etc..
In previous versions, Gaussian might have had trouble reading in anything beyond the  80th char. I cannot reproduce this bug anymore and it's been a long time since I was told about it, so this should not be an issue. However, a well structured input file can help you finding possible syntax errors that result from typos.


What is the keto form of a hexahydroxybenzene?

The question is badly phrased because it uses the term ‘the keto form’, implying that there be only one. In the case of hexahydroxybenzene, you are correct in that there is a monoketo form, at least two different diketo forms and potentially even more triketo forms.

All of these polyketo forms are created in the same way as the monoketo form except you do the isomerisation multiple times. So one way of arriving at the answer would be to check whether you can get to any of those molecules by tautomerising the hexahydroxy form often enough.

It may, however, be easier to think in the other direction, i.e. which ketones can I tautomerise to give a fully aromatic ring. Obviously, as your monoketo example shows, a ketone needs an α-hydrogen atom, so a) and b) cannot tautomerise at all and d) can tautomerise only twice, not three times. Upon careful examination of c) you should be able to realise that this can tautomerise three times to give hexahydroxybenzene.

Entirely out of scope of the textbook question is whether this tautomerisation is probable. None of the keto forms provides any advantage over the hexa-ol isomer in no solvent. We cannot argue with better hydrogen bonding and we cannot argue with polarity (both would favour the hexa-ol in all environments I can think of). Thus, the question in itself is somewhat academic. Yet it remains helpful to see how to get from polyketo to poly-ol forms and back.