How does a water molecule bond to an anhydrous salt to form a hydrate?

"Loose bond" is really a loose term which engenders loose understanding. The bonds that hold water molecules in the structure are usually dative bonds, which is a variety of polar covalent bonds. They are formed between water's O and the metal ion, using spare electron pairs of the former. There is nothing unusual about having covalent bonds in an ionic crystal. Think, for example, of $\ce{S-O}$ bonds in the sulphate anions within $\ce{MgSO4}$ or any other ionic sulphate.

Alternatively, in some relatively rare cases water molecules may be held by hydrogen bonds to the anions, like you suggested. This is the case with that fifth water molecule in $\ce{CuSO4\cdot5H2O}$, as explained by @ringo.


The sodium hydroxide will not decompose appreciably, and leaves an area of concrete dangerously caustic. A person or pet walking across it could suffer serious burns.

You can neutralize the alkali with a mild acid such as vinegar or citric acid. Vinegar is only 5% acetic acid, so you'd need about 20 times as much as the amount of drain cleaner. Too much acid will attack concrete, so use pH indicator strips (e.g. for fish tanks or swimming pools) to adjust pH to neutral (about 7).

During the neutralization, a bit of chlorine gas will be driven off from the hypochlorite, but that smell should no last long. Any other smell would be from waste in the drain... that might last quite a while.


How to choose a proper chromatography way to separate substrates?

As you already mentioned, there are numerous types of chromatography, and they vary in several different ways. I can't tell you what would be best ways to separate each of these pair of substances, but I can give you some insights. Using chromatography always requires optimization of experimental procedures, and the first step is usually searching on papers and other references for information on how the specific separation is usually done.

N-phenylethanamide/N-phenylpropionamide and methoxymethane/ethoxyethane

Despite being very similar, the presence of one or two extra $\ce{CH2}$ group in each of the second molecules is enough to give the molecules different properties, specially considering they aren't big. For simple detection purposes, I'm sure it would be possible to find the appropriate solvent mixture to separate them using TLC or column chromatography.

A much more advanced technique, gas chromatrography is very powerful and could also be used to separate them, as long as the appropriate setup was used (injection and oven temperatures w or w/o gradients, column type and length, etc).

Aβ36/Aβ42

These would probably decompose in elevated temperatures, so gas chromatography is not a good idea. Peptides are, nowadays, mostly separated using liquid chromatography, usually in HPLC or UPLC systems using reversed phase columns. This type of chromatography will separate the peptides mostly by their polarity and size, and once again, reference search and experimental trials are essential to optimize conditions. HPLC and UPLC are used to separate mixtures of thousands of peptides (with varying efficiency, of course), but that should be a sign that two wouldn't be a problem, even if very similar.

Depending on the objective of the separation (identification and/or recovery), using CNBr or any other digestion step could be useful or not, but I don't think this would be the case for only two peptides.

PAGE and the variations of electrophoresis were for a long time the goto separation method for peptides and proteins, but recent developments in chromatography (and mass spectrometry) have made them fall off preference.

I'm not familiar with affinity chromatography (or the whole antibody thing, to be honest), so I can't say much about that, but, what I do know is that you won't always find or be able to manufacture the chromatographic column with the desired specificity.


How to prevent dripping when using a Pasteur pipette with solvents like diethyl ether?

Usually I avoid pipetting diethyl ether because it can be a bit annoying, but recently I've realized there is a very simple technique which markedly increases ease of transfer (though admittedly I've been using it mostly for DCM). All you have to do is saturate the interior of the pipette with the vapour of the low-boiling solvent before performing the transfer.

The problem arises because the low-boiling solvent will quickly (but not instantly) evaporate into the partial vacuum inside the pipette after pulling the solvent. If you try to transfer the liquid while it's still evaporating significantly, the large volume change from the phase transition will push the liquid beneath through the pipette exit.

So how do you get around this problem in practice? Using the pipette, simply pull and dispense the solvent a few times in succession over 5-10 seconds inside the outgoing vial. Then pull what you need and transfer. In my experience this greatly mitigates dripping. If you're not too slow between transfers, you can get away with doing this only once, at the start.

Other things which can be done to help:

As mentioned in the comments, avoid dragging the filled pipette over a  large distance. Keep the outgoing and ingoing vials right next to each other, at about the same level.
Pull a little bit of air into the pipette, keeping it between the exit and the solvent. This gives you a small time buffer before it drips out.
If you have a steady hand, keep very gently pulling air while you move the pipette between vials, matching the speed at which the liquid tries to drip out. This can be somewhat fiddly, so train to get a feel for it.
Don't fill the pipette to its maximum capacity on each transfer; do more transfers, each with less volume.
There's always a drip left on the very tip of the pipette, and that can easily fall. To avoid it, contact the remaining drop onto one of the outgoing vial's walls above the surface of the liquid, then pull a little bit of air into the pipette, sucking the last droplet along and leaving the tip dry.
Practice.


All proteins have complex structures. There are three broad classes: globular, membrane and scleroproteins. Globular proteins are generally water soluble while membrane and scleroproteins are not.

Albumins and haemoglobin are globular and water soluble.


Hydrogenolysis typically involves a metal surface ($\ce{Pd, Pt}$, etc.) and hydrogen gas. In the reaction a weak bond, that is, a bond that is either strained or a bond that can generate a stabilized radical is broken. In your examples there is no strain, but a resonance stabilized benzylic radical can be generated in one case.

In the benzylic amine pictured on the left, the benzylic carbon can easily adopt a geometry such that when the p-orbital on the benzylic carbon is generated it is already aligned with the p-orbitals in the aromatic ring and is therefore stabilized while it develops.

The molecule on the right is different. I built a model and as the C-N bond starts to break and the p-orbital starts to form, the developing p-orbital is twisted 45-60° away from alignment with the p-orbitals in the aromatic pi system - it is not effectively stabilized due to conformational constraints imposed by the ring system.

Therefore this latter molecule will not undergo hydrogenolysis of the C-N bond as readily as the molecule on the left.

Here are some photos of my model:

the molecule - note the blue $\ce{N-CH3}$ nitrogen as a point of reference

conformation A - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is axial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted about (90-30) 60° away from being aligned with the p-orbitals in the aromatic ring - poor overlap potential

conformation B - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is equatorial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30-45° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted at least (90-45) 45° away from being aligned with the p-orbitals in the aromatic ring - poor overlap potential


(Energetics) Potential Energy and Kinetic Energy

In classical physics, kinetic energy and gravitational potential energy are almost interchangeable.  You do work on a ball and throw it vertically into the air.  Neglecting friction with the air all the kinetic energy becomes potential energy.  However, it then begins to fall, converting the potential energy back into kinetic energy.  Again, with no friction, the kinetic energy is exactly the amount that you began with.

However, we have to be careful to distinguish between ordered kinetic energy and random kinetic energy.  The ball, thrown into the air has both ordered and random kinetic energy.  The energy that becomes potential energy was the ordered form where-as the random thermal motions of all the individual atoms, which add up to no NET motion of the ball, remained the same throughout the entire flight of the ball.

But did they?  Again and again we have to say in these situations: Ignoring friction.  Friction does in fact exist when there is air present, and friction is very relevant to this discussion.  Friction actually converts ordered kinetic energy of motion into random kinetic energy of various particles (air molecules and ball molecules).

The creation of random motion from ANY other form of energy increases entropy and reduces the amount of energy that is available to do useful work.

I'm not sure that totally answers the question.  I think that the statement needs to clarify that random motion and its associated kinetic energy is, as your note claims, a preferred state over potential energy.


How to determine the specific heat of aluminum?

If the water and aluminum are regarded as incompressible, then the internal energy is a function only of temperature.  In addition, the heat capacity at constant pressure is equal to the heat capacity at constant volume.  If the pressure is constant, then the change in enthalpy will be equal to the change in internal energy.  So the change in enthalpy of the aluminum is equal to its heat capacity times its temperature change.  The same goes for the water.  So the total change in internal energy and/or enthalpy for the combination of water and aluminum is zero.


Clarifications on thermodynamic cell efficiency

${\Delta G}$ and ${\Delta H}$ are normally expressed on a per mole basis.  Your assertion that ${\Delta G}$ is related to concentration is not precise.  It is related to the ratio of concentrations, and as such the incongruity with ${\Delta H}$ is not troubling.
If you found a fuel cell with a gain in entropy, according to your equation you would have more than 100% efficiency.  I seriously doubt such a situation exists.


I recommend you to put a large amount of activated charcoal into the fridge and left it several days. If you've still sensed the smell, replace the old activated charcoal with the new one. I hope it help you get rid of the smell.


No offence to you (certainly to the original author of this question), I call such home-work questions such as garbage in-garbage out type questions. No wonder a student will be confused if they see such open ended questions. The reason is that the author is not telling you or us of 2.0 grams of what?
Even if we assume it was 2 grams of salt, several things can affect the measured current
(1) What are the electrode areas?
(2) What are the electrodes made of?
(3) What is the distance between the electrodes?
Last but not the least, the identity of the salt matters a lot. Did both groups use the same salt? Monovalent, divalent, etc


Do really the electrons could only orbit in specific orbits with a fixed radius?

Bohr said electrons are quantized. It must be in a fixed Orbital with a fixed radius. Isn't that contradictory?

No it is not contradictory.

"Quantized" means that a continuous range of a variable, such as energy, is not allowed.  Only discrete values of the variable are allowed.

In Bohr's model, the radii and energies were quantized.

  Which one is true?

It is not true that electrons orbit in fixed circular orbits.  Subsequent to the Bohr theory, there was Sommerfeld theory, which involved elliptical orbits and was a relativistic theory that correctly predicted the fine structure of the hydrogen atom (which Bohr's theory did not). Then there was Schrodinger's theory, which did away with the fixed orbits, but was non-relativistic and gave the same energy levels as Bohr's theory.  This was followed by Dirac's theory which was a relativistic theory and gave the same energy levels as Sommerfeld's theory.  However, the hyperfine structure of the hydrogen atom, such as the Lamb shift, was still not explained, until quantum electrodynamics was developed. Further developments included the recognition that the proton is not a point particle, but is composite.  So there has been no finding the truth, just more accurate, but more complex, approximations of the truth.

  Sal said that orbitals means that if we took a number of snapshots of atoms, we would see that the electrons are denser in those orbitals but that does not necessarily mean that we wouldn't find it outside of the orbitals. The probability of finding electrons is higher in the orbitals but not 100%

Orbitals have an infinite spatial extent.  There are nodes where the wavefunction is zero for the Schrodinger equation solutions of the hydrogen atom, but nodes are specific planes, conical surfaces or spherical surfaces, not 3-dimensional regions.  People depict the orbitals as having finite volumes, but these depictions are the high probability regions not the entire orbital.  So if Sal said that, he should have said that the electrons can be outside of the high-probability regions of orbitals that people draw, rather than saying outside the orbitals.


Are the values for enthalpy, internal energy and Gibbs free energy the same for a particular process?

While enthalpy, internal energy, and Gibbs's energy are all types of energy, they are not all measuring the same quantities. Internal energy is the energy of the molecules in your system. Enthalpy takes into account the internal energy plus whatever work is being done by or on your system. And Gibbs's energy is the maximum amount of non-PV work that you can get from your system. While these three quantities are all in units of energy, they are not just different terms for the total energy. Since enthalpy and Gibbs's energy involve work, and hence exchange of energy with the surroundings, these quantities will be different than just the internal energy of the system alone.


How can the size of atom be the cause of octet expansion?

Your revised version of Atkins’ statement is correct, no hypervalent main group compounds require the involvement of d-orbitals. Rather, all of those molecules formerly explained by d-orbital participation can just as well be explained with 4-electron-3-centre bonds (including resonance structures) and charge separation (e.g. in sulphate; no need for resonance), giving rise to all-octet Lewis structures. In fact, calculations have shown the d-orbital participation to be minimal and neglegible at best.

A much more relevant question is whether more than $x$ substituents actually fit around the central atom. This is especially notable for the difference between the second and third periods: The reason why nitric acid is $\ce{HNO3}$ while phosphoric acid is $\ce{H3PO4}$ is likely due to the much smaller size of nitrogen, not allowing for tetracoordination by oxygens in the $\mathrm{+V}$ oxidation state while the larger phosphorus atom can support four oxygen atoms. If I remember correctly, that was also discussed as a reason why $\ce{NF5}$ hadn’t been found at the time of Klapötke’s article while $\ce{PF5}$ has been known for quite a long time.

It gets more pronounced when you go even lower in the periodic table, i.e. to higher periods. Antimony is known to be coordinated by six oxygens in the $\mathrm{+V}$ oxidation state ($\ce{H7SbO6}$, if you wish, although $\ce{H[Sb(OH)6]}$ is a more accurate description) — something phosphorus could technically do ($\ce{PF6-}$ with small fluorine atoms is known) but usually doesn’t. It is mainly the greater atom size of antimony that allows the coordination of two additional oxygen atoms.


When to permute two times to find optical configuration

If you switch any two substituents, that will change the stereochemistry. If you do a second switch (any two), you are back to the original stereochemistry. You can try it out with substituent 4 in the back, and seeing what switching around 1, 2, and 3 do to the clockwise or counterclockwise orientation (i.e. whether it is R or S).

Two switches (for example switching 2 and 4 and then switching 3 and 2, see animated GIF) amounts to a rotation around the bond of the "unswitched substituent (here substituent 1). This does not change the stereochemistry.

If you are spatially gifted, you can just do the rotation in your head to get substituent 4 in the back. If not, you can do one switch (realizing you just interchanged R and S) or two switches (R stays R, and S stays S). It is personal preference (oops, I think I just said the question should be closed because the answer is opinion-based).

If you want to try it with 3D models yourself, check out http://proteopedia.org/wiki/index.php/R/S_nomenclature.


An equal amount of hydrogen gas is dissolved in a 1L vessel and a 2L vessel, both filled with water. Which vessel has higher entropy?

Denote by Vessel I the vessel with volume $V$, moles of water $N$, and moles of hydrogen $N'$. Denote by Vessel II the vessel with volume $2V$, moles of water $2N$, and moles of hydrogen $N'$.

Increasing the volume will increase the number of microstates available to the system, and hence also the entropy. Increasing the moles of water will increase the number of microstates available to the system, and hence also the entropy.

Both statements imply that Vessel II has the greater entropy.


Why is −78 °C a magic temperature for organic reactions?

Dry ice (solid carbon dioxide) sublimes at −78 °C. Dry ice and acetone are a common cold bath for chemical reactions.  The melting point of acetone is -95 °C so the bath never gets cold enough to freeze the acetone. The bubbling of the carbon dioxide gas as the dry ice sublimes keeps the cold bath well stirred.

Typically, though, the temperature in the flask with an ongoing reaction is at least about 5 °C higher than the one in the cooling bath.  (Except if using a thermocouple, working at a scale of 10...25 mL and less often implies that the thermometer is in cooling bath, and not in the reaction mixture. Thus, this temperature difference isn't recorded.)


Quasi-steady state hypothesis application

Thanks for the interesting question.  Before we get into the details, let me note a few things:

Your system is a model of substrate inhibition, a phenomenon where higher-than-optimal concentrations of a substrate for an enzyme decrease its own rate of conversion to product.
It sounds like you have a good handle on the enzyme mass balance, i.e. that $e_0 = e + c_1 + c_2$ is a conserved quantity whose time derivative is everywhere zero (no approximations or hypotheses required) in this kinetic scheme.  This makes sense because enzymes are catalysts that accelerate reactions without any net participation in them.

OK, onward to your questions.

The assumption of fast equilibrium for $c_2$ is a different assumption than the PSSH hypothesis.  Assuming a fast equilibrium for $c_2$ means that reactions 3 and 4 happen much more quickly than reactions 1 or 2.  Thus the concentration of $c_2$ is determined solely by the thermodynamics of reactions 3 and 4: $$k_4 c_2 = k_3 s c_1 \;\;\; \Rightarrow  \;\;\; c_2 = \frac{k_3}{k_4} s c_1$$
The pseudo-steady-state hypothesis could be applied to $c_1$.  $$\frac{
dc_1}{
dt} = k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)c_2 = 0$$  Substituting in the result of the fast equilibrium assumption (point #3) gives $$k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)\frac{k_3}{k_4}s c_1 = 0$$  You can solve this equation for $c_1$ and plug the result into your $\frac{
ds}{
dt}$ equation.
If you are interested in understanding more about the mathematical validity of the PSSH, you should know that it is related to singular perturbation.  For a treatment of a "simple" Michaelis-Menten system by singular perturbation, see this book excerpt.  The PSSH solution is an "outer" singular perturbation solution.


Why does the addition of PCl5 increase the rate of dissociation of PCl5?

Imagine a overpopulated kingdom suddenly gaining a large area of land (Start of Reaction). Immediately, groups of people start migrating to the area of land (Forward Reaction). However, as people migrate to this new land, some decide that, for whatever reason, they don't like this area (Reverse Reaction). As more and more people move there, more start coming back until there is an equal rate of people moving to the territory and people moving back (Dynamic Equilibrium).

  To your question: To model this system using my example of immigration, we suddenly add an enourmous influx of refugees into the old kingdom (Increase in $\ce{PCl5}$).

So the land in the outer territory seems more valuable now (less crowding, more open space, better air, etc.). This causes a temporary increase in the forward direction until the reverse reaction increases to compensate. This is called Le Chatelier's principle in chemistry.

References: Le Chatelier's Principle, My Handy Old 2011 Introductory Chemistry Textbook.


What is the molecular mechanism of ammonia's toxicity?

Ammonia is reportedly not toxic in bacteria even at molar conentrations (PMID 16604417).

However, yeast, plants, and animals (presumably, all eukaryotes) are unable to tolerate high intracellular ammonia concentrations.  The best explanation for this phenomena I've seen is a hand-waving assertion that intracellular ammonia taxes overtaxes the cell's ability to maintain membrane gradients (PMID 7764110).  However, that isn't a particularly detailed or satisfying response.

I agree with your assertion that there's a paucity of literature on the subject - we probably don't know the answer yet!


How to use Kohlrausch's law with ionic equivalent conductances?

You textbook writer has explained it in a very complicated way. I think you are confused by the concept of equivalent. Gram equivalent is an obsolete unit now but it is still used in conductance measurements.

First understand the definition of gram equivalent weight for a salt (and it applies to salts only). Acids/ Bases/Redox agents have different definitions.

Gram Equivalent weight of a salt or an ion = gram molecular weight/charge on the ion

This is the definition of gram equivalent.

From the context of your book, ionic equivalent conductance is same as "molar conductance" of aluminum ion. Here a and b refer to molar conductance. This is the mistake you are making. Thus, a 1 Molar Al3+ solution has 3 gram equivalents of Al/ 1 L or it is a 3 N solution (N = Normal).

So if you are given ionic equivalent conductance of Al2(SO4)3 (at infinite dilution) then your textbook is right = (ionic equiv. conductance Al)/ charge on Al + (ionic equiv. conductance sulfate)/ charge on sulfate = a/3 + b/2

By the way, equivalent conductance can only be added at infinite dilution. Whatever you see in tables are usually given at "infinite dilution".


How to calculate n factor for salts in this reaction?

$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$
Now this is clearly evident that dividing whole equation by 2 we get 1 mole of $\ce{Na3PO4}$ and at the right hand side 3 moles of $\ce{NaCl}.$ Therefore, 1 mole of phosphate ion is replaced by 3 moles of $\ce{Cl-}$ ion.
Hence $n\text{-factor} = 3.$
In reactions like this where there's no change in oxidation number, the $n\text{-factor}$ will always be the net positive charge in 1 molecule of that compound, in this case since there are 3 $\ce{Na}$ ions in sodium phosphate, the $n\text{-factor} = 3 × 1 = 3.$


Are buckminsterfullerene-corannulene complexes actually stabilised by π stacking?

In short, I think the answer is that although the overall symmetry of the buckminsterfullerene is almost spherical, the quadrupole–quadrupole interaction with other molecules with aromatic cycles are local (in addition to the ever-present van der Waals interactions).

I tried to think of a way to exemplify that, and drew the 2D picture below:

In this 2D model, the “double bonds” of the circular structure represent the individual aromatic rings of your buckminsterfullerene (which isn't superaromatic, i.e. electrons are not delocalized over its entire structure; think may be a key point), bearing each a local quadrupole moment (symbolized in the red/blue thingy). The other particle with a quadrupole is close to one of the ball's quadrupoles, and this local interaction dominates. In fact, other interactions maybe be stabilizing or not, depending on the orientation of the individual quadrupoles, but they are weaker that the local quadrupole–quadrupole interaction.


Why do most carboxylic acids have high pKa (~5) in spite of having a conjugate base ion that is stabilized by resonance?

Everything is relative in acid-base reactions; the equilibrium will favor the side with the weaker acid. A carboxylic acid with $\ce{pKa = 5}$ will not protonate $\ce{H2O}$ significantly because $\ce{pKa(H3O+) = -0.7}$, but it will protonate something more basic like $\ce{NH3}$ because $\ce{pKa(NH4+) = 9.24}$.

Note: the pKa of $\ce{H3O+}$ is debated, see What is the pKa of the hydronium, or oxonium, ion (H3O+)? for more discussion.


Is there a crystal structure of ammonia in ionic form?

According to the work of Ninet et al. published in 2014 [1], ammonia has a β-phase at 150 GPa, where it transitions into ionic crystalline form, forming $\ce{[NH4+][NH2-]}.$

References

Ninet, S.; Datchi, F.; Dumas, P.; Mezouar, M.; Garbarino, G.; Mafety, A.; Pickard, C. J.; Needs, R. J.; Saitta, A. M. Experimental and Theoretical Evidence for an Ionic Crystal of Ammonia at High Pressure. Phys. Rev. B 2014, 89 (17), 174103. DOI: 10.1103/PhysRevB.89.174103.


Dissenter's answer is not quite right, since, strictly speaking, as defined there, the scientific notation will work for positive numbers only. Obviously, it can be easily fixed to work for negative as well, but the case of 0 will continue to be problematic. Just try to follow this rule to understand why.

All in all, it is the matter of definitions, but usually, scientific notation is defined as just a way of writing numbers in the form $a \cdot 10^b$ without any restrictions on $a$ which can be any real number. So, both $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are considered to be numbers written in scientific notation.

However, in the normalized scientific notation, indeed,

  the exponent $b$ is chosen so that the absolute value of $a$ remains at
  least one but less than ten ($1 ≤ |a| < 10$),

and thus, only $6.023 \cdot 10^{23}$ is written in this notation. Normalized scientific notation  is just one special case of scientific notation. And yeah, you could not write down zero in this notation, which is kind of awkward.

Another special case of scientific notation is the so-called engineering notation in which the exponent $b$ is restricted to multiples of 3. Neither $6.023 \cdot 10^{23}$ and $6023 \cdot 10^{20}$ are written in engineering notation, while $602.3 \cdot 10^{21}$ is. Although engineering notation is a special case of scientific one as defined above, is rarely called scientific notation to avoid an ambiguity.


If you need a good answer, you should not put an information embargo on the context of your question.

This is to be taken rather as an expanded comment than a full answer:
$\ce{HBrO2}$ oxidizes $\ce{Br-}$ to $\ce{Br2}$.
$$\ce{HBrO2 + 3 Br- + 3 H+ -> 2 Br2 + 2 H2O}$$
To tell if it may compete with $\ce{Br-}$ to react with the "chemical", we would have to know that chemical.
Regarding standard redox potentials, an exhaustive list is on Wikipedia, but $\ce{Br^{III}}$ species are not there.
But be aware $\ce{HBrO2}$ is unstable, strong oxidizer, while $\ce{Br-}$ is a weak reducer. I personally think the info may have been misinterpreted, having meaning $\ce{HBrO2}$ may compete with the chemical to react with $\ce{Br-}$.


Why do different H+ in polyprotic acids have different K?

You are correct that all of the $\ce{H}$'s in $\ce{H3PO4}$ are equivalent.

However, once the first H is removed, the remaining two $\ce{H}$'s are no longer equivalent to the $\ce{H}$'s in the original $\ce{H3PO4}$. After the initial deprotination, the molecule gains an overall negative charge making it more difficult to remove the second $\ce{H}$.  Then when the second $\ce{H}$ is removed the molecule gains a second negative charge making it even more difficult to remove the final $\ce{H}$.  Thus, the $K_a$ values for the originally identical $\ce{H}$'s change each time the molecule is deprotinated.


Can we predict the IMF strength of Acetone vs Glycerol?

So the intermolecular forces would be stronger in glycerol then in acetone. Glycerol can form hydrogen bonds while acetone is not very polar and can’t significantly form h bonds with itself.
The surface tension for glycerol is 63.4 $\frac{mN}{m}$ while acetone is 25.2 $ \frac{mN}{m}$ (both @ 20 C)
I supposed the error in your experiment arises from using pipette to determine the amount of material on the coin. Unless your using very precise pipettes, counting drops isn’t necessarily the best way to quantify the amount of material added. Additionally the density of glycerol is higher then acetone which needs to be accounted for in the test.


In the molecule 2,3,7,8-tetrachlorodibenzo-p-dioxin, the "p" means para because each chlorine is para to an oxygen?

No. The p (for “para”) indicates the dioxine isomer.

  In IUPAC nomenclature, shouldn’t “para” be replaced by a number?

Yes. The preferred IUPAC name for this isomer is 1,4-dioxine.

Therefore, the corresponding name of the parent structure of the compound that is given in the question is 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine. The complete name is 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine.

The preferred IUPAC name, however, is 2,3,7,8-tetrachlorooxanthrene.


Why do alcohols and ethers have approximately the same solubility in water but different boiling points?

Ivan's comment is spot on. It take two to tango (or hydrogen bond) - a donor $\ce{Z-H}$, where $\ce{Z}$ is some electronegative atom (usually $\ce{O},\ \ce{N},$ or $\ce{F}$) and an acceptor (something with just $\ce{Z}$). 1-Butanol can hydrogen bond with itself, since it has an $\ce{O-H}$ group capable of being both donor and acceptor. Deithyl ether cannot hydrogen bond with itself since it lacks an $\ce{O-H}$ group. However, both can form hydrogen bonds with water. Thus, 1-butanol has the higher boiling point, but they both are similarly soluble in water.

Actually, Morrison and Boyd oversimplify the solubility difference (and there is one). 1-Butanol is slightly more soluble in water than diethyl ether, perhaps because it can interact with water in more than one way, although the dipole moment of the alcohol group is certainky larger than that of the ether group as well.

The NIH PubChem entries for diethyl ether and 1-butanol list several solubilities in water, but I want to focus on two entries that share the same source for both compounds:

  Solubilities in water at $25\ ^\circ \mathrm{C}$ from the NIH TOXNET Hazardous Substances Database

$$\begin{array}{c|c|c|c}
\mathrm{substance} & \mathrm{Solubility\ in\ water\ (g/100\ mL)} & \mathrm{density\ (g/mL)} &\mathrm{volume\ percent}\\ \hline
\text{1-butanol} & 6.8 & 0.8098 & 8.4\\
\text{diethyl ether} & 6.0 & 0.7134 & 8.4
\end{array}$$

  Solubilities in water at $20\ ^\circ \mathrm{C}$ from the International Labor Organization's International Chemical Safety Cards

$$\begin{array}{c|c|c|c}
\mathrm{substance} & \mathrm{Solubility\ in\ water\ (g/100\ mL)} & \mathrm{density\ (g/mL)} &\mathrm{volume\ percent}\\ \hline
\text{1-butanol} & 7.7 & 0.8098 & 9.5\\
\text{diethyl ether} & 6.9 & 0.7134 & 9.7
\end{array}$$

Note that I have used the solubility data to calculate percent compositions by volume in both cases. It was and still is common to report low solubilities as percents composition. It also is and was common to see percents composition listed without specifying whether they were mass/mass (which should be the default), mass/volume (convenient for dilute solutions), or volume/volume (convenient for solutions, but unhelpful due to colligative properties).

At both temperatures, the data supports that 1-butanol and diethyl ether have nearly the same percent solubility by volume. Morrison and Boyd is an older generation of textbook, first published before the world wide web. If they found a source of solubility data for these compounds that listed percents (but did not specify volume/volume), then the authors made a reasonable assumption that those percents were mass/mass or mass/volume and reported the same solubility.

The slight solubility edge for the n-alkanols persists over a larger range of compounds. The table below lists the solubilities in grams per 100 mL of water of the n-alkanol and the symmetric ether, with data drawn from the NIH Hazardous Substance Database.

$$\begin{array}{c|c|c|c|c}
\text{sol. in water in g/100 mL} & \ce{C2} & \ce{C4} & \ce{C6}  & \ce{C8}\\ \hline
\text{n-alkanol} & \text{miscible}  & 6.8 & 5.9  & 0.54 \\
\text{symmetric ether} & 4.6  & 6.0 & 4.9 & 0.30
\end{array}$$


Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature?

You can convert RFNA to WFNA by removing dissolved nitrogen oxides ($\ce{N2O4}$ and $\ce{NO2}$) from the RFNA.

Given the volatility of $\ce{N2O4}$ and $\ce{NO2}$ it might be possible to just let the gases evaporate from the solution at low temperature. However, since the boiling point of $\ce{N2O4}$ if about $21.69~\pu{°C}$, I do not think that this is a very practical method.

Another path that is theoretically possible is the reaction of residual water with excess $\ce{N2O4}$ and $\ce{O2}$ from air to $\ce{HNO3}$.

$$\ce{2 H2O + 2 N2O4 + O2 -> 4 HNO3}$$

Again, to me this seems not to be a very plausible way. Since the Wikipedia article does not give a source for its claim, we cannot decide which of the two paths, if at all, is the correct one.


Is 4-nitrobenzaldehyde or 4-aminobenzaldehyde more likely to be hydrated?

I think you're asking about the following equilibrium

This equilibrium will shift to the left as the aldehyde becomes more stabilized.  So the question becomes, which R group, $\ce{NO2}$ or $\ce{NH2}$, will tend to stabilize the aldehyde the most?  A substituent that further delocalizes the positive charge shown in the dipolar resonance form would tend to be stabilizing.  Just like in electrophilic aromatic substitution, additional stabilizing resonance structures can be drawn involving the amino substituent, but similar structures involving the nitro group are not as significant due to electrostatics (2 adjacent positive charges). Therefore, since the amino will stabilize the aldehyde better than the nitro substituent, the equilibrium involving the amino substituent should lie further to the left, and the aldehyde with the nitro substituent should be more hydrated.


Help with finding the right catalyst and reaction conditions for hydroamination reaction

You may follow the procedure in this reference. I included the abstract for your convenience:

  Abstract: The hydroamination of styrene with aniline catalyzed by phosphine-ligated palladium triflates exhibits a substantial $\ce{^{13}C}$ isotope effect at the benzylic carbon. This supports rate-determining nucleophilic attack of amine on a $\eta^3$-phenethyl palladium complex. Deuterium exchange observations and predicted isotope effects based on DFT calculations support this mechanism. Selectivity in these reactions is determined by the facility of palladium displacement after reversible hydropalladation of the alkene.

The Graphical Abstract:

Reference:

L. K. Vo, D. A. Singleton, “Isotope Effects and the Nature of Stereo- and Regioselectivity in Hydroaminations of Vinylarenes Catalyzed by Palladium(II)−Diphosphine Complexes,” Org. Lett. 2004, 6(14), 2469–2472 (DOI: 10.1021/ol049137a).


How does a Brunauer–Emmett–Teller device function?

BET analysis on zeolithes represents a typical example.  Dependent on the size of the particles you have, you select a glass cell of about a centimetre; for a powder a smaller cell, for more coarse pellets you don't want to grind a bigger one.  You need to outgas the samples first, i.e. by applying vacuum and elevated temperatures already adsorbed gases and moisture is removed from the surface of the material to be analyzed as well as the inner walls of the sample cell. This already may take time to be accomplish -- because you do not want to destroy or alter the material in question by too harsh conditions.

The actual measurement requires the sample cell being kept at a constant temperature.  An initial determination takes place while the sample cell is immersed in a Dewar of liquid nitrogen.  This low temperature provides better conditions for the adsorption experiment.

A first blank is measured with Helium gas, entered via a graduated piston. Helium typically does not adsorb on surfaces (at this temperature).  This allows to determine the dead volume of the measurement cell.  In the second step, Nitrogen gas enters the measurement cell, again in a controlled manner (graduated piston, pressure control).  Again, this takes time as you want to establish the equilibrium between the adsorbent and the amount of gas adsorbed in function of gas pressure and sample temperature.
A controlled entry of Helium gas is necessary as a second determination of the dead volume to complete the calibration eventually leading to the read-out for our substance at given temperature.

Of course, dependent on what kind of isotherme (BET/Langmuir) is to be determined, Nitrogen gas as a load may be replaced by other gases, like methane.  Beside determination of the surface area of a solid, the adsorption properties may be determined in a series of more elevated temperatures, too.


Reaction direction change based on phase of reactants and products

Quite correct.  And manipulating which direction a reaction tends to go by changing the phase of reactants or products is a common trick in chemical engineering.


Can ML algorithms be applied to predict the products of a chemical reaction given a set of reactants?

Adding on top of Emre's answer, product formation depends on certain reactive groups say for e.g., $-OH$ (Alcohol) having more chance of releasing a Hydrogen than $-CHO$ (Aldehyde) or a Methyl $-CH_{3}$. Instead of using reactants as categorical variables you should use reactive groups on that, as variables.

For example if you take a vector of say (Chlorine, Alcoholic_group, Amine) to be coded for Para-Amino Phenol, it will be $(0,1,1)$ as chlorine is absent and for $CCl_{4}$, it will be $(1,0,0)$. But remember the first vector will be same for O-Amino Phenol as they are positional isomers. But both have different properties and reactivities. So categorical is kind of wasted here.

In my more deeper and frank opinion, these values should be continuous. You should be kind of looking for a value to quantify these each dimension of the vector coded. Then you can use classifiers from this vectors to find the probability of reaction happening or for product formation as @Emre mentioned.

So, take a simple, small set of molecules and make a vector having fewer dimensions and test it with categorical and continuous values(Give a value to each dimension based on your chemistry gut opinion). That is the better way to find whether your hypothesis/assumptions are reasonable.

EDIT 1:
My answer basically answers whether reaction is feasible or not but does not answer which products are returned. I am sorry for that. But I hope it kind of points to what you can achieve with ML if you want to take your first step.

Hope this helps.


If the programmers did not make a mistake it seems that you are right - ChemDraw is on your side.

$\hskip.5in$


I used Zn + S as a teenager. It is a lot safer than Al, and you can mix it cold.

Stick to solid propelants because it's simpler, safer and smaller. You don't want an explosion but something very quick burning.

Mix the two fine powders well.

You can mix them in a thick plastic bag or wide mouth bottle and shake until you see an even color in all the mix with no lumps in sight.

(You can make a paste with alcohol 100% but I never tried this.)

Then tap it in with wood or plastic. You don't want a spark! Nozzle reduction is 1/3 and not smaller. The mix is 3:1. Light with a filament, it is safer than a spark.

In those days all I had was a book but you can find a lot of info out there.

We switched to this after a (not too serious) accident with rocket candy.

Safety first allways, the rocket can go anywhere!


What is the point group of hexachloridotitanate(III)?

Getting straight to the point, I believe whichever reference you have that claims $D_\mathrm{3h}$ is incorrect.

Titanium(III) complexes were the first example presented in my coordination chemistry class to examine the energy difference between the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbitals, explain why it exists and where the colour of coordination compounds derives from including carefully examining the UV spectrum of titanium(III). Our lecturer, Professor Klüfers of the LMU Munich, explained that the hexaaquacomplex is practically octahedral in shape. However, there is not just a clean absorption band but rather a band with a shoulder. This is because the excited state of $\ce{[Ti(H2O)6]^3+}$ is Jahn–Teller distorted, having an uneven $\mathrm{e_g}$ population, meaning that two slightly different excitation energies exist.

One might go a step further and argue that even in the ground state a certain distortion should exist. If one were to pull (not push) the ligands in $z$-direction slightly closer to the central metal, this would destabilise any orbitals with $z$ contribution meaning that $\mathrm d_{xy}$ suddenly becomes the single most stable d orbital. This slight distortion which one might call anti-Jahn–Teller (since it is opposite to the classic Jahn–Teller distortion) could explain a reduction of symmetry from $O_\mathrm{h}$ to $D_{\mathrm{4h}}$.

$D_\mathrm{3h}$ symmetry is not consistent with octahedraloid coordination spheres at all. It is very common for pentacoordinated centres such as $\ce{PF5}$. The only way to have six ligands — as in hexachloridotitanate(III) — around a central metal in this point group would be a trigonal prism. That would be a very uncommon coordination sphere and I am sure I would have heard of it in the context of titanium(III) in said course above should it exist for this species. Indeed, the source orthocresol found and linked in the comments makes no mention of any distorted geometry at all:

  The low-temperature structure results from rotations of the $\ce{[TiCl6]^3-}$ octahedra around a 4-fold axis. (Emphasis and square brackets mine)

Source: Amels, R.; Kremer, S.; Reinen, D. Jahn-Teller effect of titanium(3+) in octahedral coordination: a spectroscopic study of hexachlorotitanate (TiCl63-) complexes. Inorg. Chem. 1985, 24, 2751–2754. DOI: 10.1021/ic00212a009.


Proton NMR splitting in 2-substituted furan

I agree with Buck Thorn's comment. I suspect the coupling in the quartet is just not fully resolved. I'm not 100% sure why, but one possible reason is because there is further unresolved coupling with the OH peak: instead of being resolved as a splitting in the spectrum this just broadens the individual lines. Or it could just be that that proton relaxes faster.
You may be able to play around with window functions (or apodisation as it's referred to in Mestrenova) to try and tease out this small coupling. My suggestion would be: try enabling/selecting both "Gaussian" and "Exponential", set the exponential value to -2 Hz and the gaussian value to 0.2 GF, or something similar. See if the quartet is still a quartet. [From the comments, it seems that this has worked and that the quartet is actually a quartet of doublets under the hood.]
The window functions are basically mathematical functions that you multiply the FID by prior to Fourier transformation. Peaks in the FID generally have a damped sinusoidal form. The faster a signal decays to zero, the larger the corresponding linewidth will be, i.e. the broader the peak will be.
The two "big" window functions for 1D processing are exponential and Gaussian window functions. I won't go into too much detail because these are very well-described elsewhere. See e.g. Chapter 5 of Keeler's Understanding NMR Spectroscopy (2nd ed.), or Glenn Facey's blog posts on Exponential Line Broadening and Resolution Enhancement (which also include some helpful pictures). Googling for either "window function nmr" or "apodisation nmr"  will also throw up a huge number of documents.
The exponential one basically multiplies the FID by another decaying exponential. This is largest at the start and decreases as time passes, so effectively causes peaks to look as if they're decaying even faster. This leads to broader lines, but because the decaying exponential is largest at small $t$ where the signal is large, it "emphasises" the parts of the FID that have more signal. Effectively, you lose spectral resolution but gain some signal-to-noise.
A Gaussian window function (or a combination of Gaussian + exponential, which is more common) starts small, reaches a maximum in the middle, and tails off again towards the end. In a way it has an opposite effect to the exponential: by trying to "compensate" for the natural signal decay, it makes the peaks look as if they are decaying more slowly, i.e. you get narrower lineshapes. On the other hand, you lose signal-to-noise because you de-emphasise the start of the FID which has the greatest signal.


Lithium can react with nitrogen giving a dark product.  https://en.wikipedia.org/wiki/Lithium


Spatial Wavefunction Symmetries and Identical Particles

When we talk about exchanging electron $i$ with electron $j$, we are actually changing the wavefunction according to

$$\Psi(..., x_i, ..., x_j, ...) \to \Psi(..., x_j, ..., x_i, ...).$$

The operation is taken by the parity operator $P$. Applying it twice would
return the wavefunction to its original form. So the following eigenvalue equation is satisfied

$$ P^2 \Psi = 1 \Psi. $$

Then the $\lambda$ eigenvalue of the $P$ operator ($P\Psi = \lambda \Psi$)
also satisfies an analogous equation

$$ \lambda^2 = 1. $$

$\lambda$ can only be either $\pm 1$. If it is positive the wavefunction is symmetric under swapping and if the eigenvalue is negative the wavefunction is antisymmetric under swapping. It can also be said that the symmetric (antisymmetric) wavefunction has even (odd) parity but do not confuse it with spatial symmetry.

It is assumed that fermions must be described by antisymmetric wavefunctions
and bosons by symmetric wavefunctions. Many have argumented this fact alluding
to the spin-statistics theorem but others have, as well, stated that such justification is not satisfactory. Besides, it is not completely clear why.


Inorganic IUPAC for Compounds with bridged ligands?

Here is an example of a similar ligand and it can be named in three ways.
All you have to do is to put hydroxido/hydroxo in place of superoxido.

So yes, bridged ligands can be named in a few ways, combining any amount of recurring atoms separately, but most common used would be (if the LHS and RHS are the same) μ-amido-μ-hydroxidobis(tetraammine)dicobalt.
Reference

Crystal structure of .mu.-amido-.mu.-superoxo-bis[tetraamminecobalt(III)] tetranitrate, $\ce{[(NH3)4Co(NH2)(O2)Co(NH3)4](NO3)4}$ by
Gary G. Christoph, Richard E. Marsh, William P. Schaefer, Inorg. Chem. 1969, 8, 2, 291-297, DOI: https://doi.org/10.1021/ic50072a022


How do I find out the number of bonds an atom must link to?

It's named a covalent bond. The amount of hydrogen atoms that can be bond (or any other atom) can be  calculated most of the time using the octet rule, that states that any atom will try to get the electron configuration of a noble gas (i.e. 8 electrons in the outer shell). The electrons from the covalent bond count for both atoms that take part in the bonding.

As an example consider $\ce{N}$ that has five electrons in its outer shell, and $\ce{H}$ that has just one. $\ce{N}$ wants three more electrons and can get them from covalent bond with hydrogen. $\ce{H}$ want one more electron (this is an exception) in its outer shell to fill it up, so will kindly receive them from nitrogen.


Does storing a hygroscopic reagent in a bottle makes it prone to absorb humidity from the air trapped inside?

Yes, there is that tendency. I'll make a couple of points which I'll number for possible further discussion.

(1) 99.5% acetone is probably 0.5% water. So the "extra" water from the air above the acetone in the bottle (from one opening and closing of the cap) would be negligible compared to the amount of water already in the acetone. Obviously when using the acetone, uncap it and recap it ASAP to keep it as dry as possible. The "problem" is opening and closing the bottle many times.

(2) I looked quickly at Wikipedia, and do a quick a google search and didn't find any water-acetone azeotrope. So the distillate would be "pure".

(3) I'm not sure how pure a simple distillation in air of the acetone would be in removing water. I'd guess that the resulting purity would be like your 99.5% starting purity.

(4) Vapor Pressure would play a role in that the acetone would have a partial pressure in the bottle. So not all of the void in the bottle would be air. Acetone doesn't "boil" at room temperature though so the partial pressure won't be enough to keep some air from seeping in and out of the bottle. In other words the void in the bottle is mostly air with just some acetone.

(5) The !@#$%^&*( bottles "breath" as the air pressure changes if they are not absolutely sealed.


Any free tool for Calculating Gasteiger-Hückel charges for ligands?

The method is not fully published according to Gasteiger.

There is an online server for PETRA run by the Gasteiger group, and in principal, it's available from Molecular Networks. I note that PETRA is no longer listed on the products page, but it's probably part of some other product now.

The open source CDK package has an implementation of the PEPE method, although I don't know if it has been validated against Gasteiger's code.


When Compounds are Considered Compounds and Their Types

A molecule is formed when two or more atoms join together chemically e.g. Na and Cl.

A compound is a molecule that contains at least two different elements e.g. Na and Cl.

All compounds are molecules but not all molecules are compounds e.g. H2 and O2 are molecules but are not compounds because bond is between two of the same elements.

Im not sure what you are asking in question 3. Gay-Lussacs Law is generally associated with the combining volumes principle. e.g. 1 volume of H2(g) + 1 volume of O2(g) = 2 volumes of H2O(g). However, you may want to consider the combined gas law, not just Gay-Lussac law. This combines Gat-Lussacs law with Boyles' and Charles' Law.


There is still the old WINDNMR programme developed by Hans Reich: https://www.chem.wisc.edu/areas/reich/plt/windnmr.htm which apparently should still work on modern versions of Windows.

It is not quite as sophisticated as Spinach, but if all you want is a 1D spectrum with various multiplets, then it is probably good enough.

If you have the computational resources, then there is software developed by Stefan Grimme's group https://xtb-docs.readthedocs.io/en/latest/enso_doc/enso.html, but you can't specify shifts manually, you need to generate them via a quantum chemical calculation using ORCA. Technically, all of these are free to use, but obviously it is not always practical. But in any case, if you are interested, the relevant publication is Angew. Chem. Int. Ed. 2017, 56 (46), 14763–14769. It should be possible now to run all the steps using just one ORCA input file; I suggest looking in the ORCA manual for more information.

For what it's worth, if you have the programming knowledge required, it's easy enough to write a (e.g.) Python script that is capable of simulating two- or three-spin (maybe four-spin) systems. At the end of the day, an NMR experiment is "just" a bunch of unitary operators acting on density matrices, i.e. lots and lots of matrix multiplication. For example, you could take a look at some of my MATLAB code here which simulates the first FID of a sensitivity-enhanced HSQC experiment. It is fairly trivial to port this to Python, since numpy provides you with pretty much every function you might need.

[Disclaimer: the code will not necessarily stay there forever.]

The only problem is that this scales exponentially with the number of spins, so unless you do some serious optimisation of the code (like that done in Spinach), it becomes intractable very quickly. But if your molecule can be "decomposed" into several spin systems which are small, then the overall spectrum is just the sum of the spectrum of each spin system, so you might be able to get away with simpler code. I've never thought about how you might do that, although my guess is that you would want to construct a Hamiltonian matrix, then pass it to a function which would call itself recursively if the matrix can be written in block-diagonal form.


Viability of reduction of oil paint binder and solvent flammability upon disposal via addition of water

It really depends exactly on what you're mixing the water with. Soaking a rag or paper towel will reduce the flammability and is recommended by many safety organizations online (here is one). However, if you are washing off a large amount of solvents or flammables with water and the chemicals are immiscible in water, it can sit on the top of the water and still burn if exposed to some ignition source.
The best thing to do is to look at SDS or MSDS sheets if you are unsure what the best disposal procedure is for some chemical you are using. Additionally, a google search should offer additional information on disposal and hazards of the substances painters used.


The Wikipedia article has a nice picture of the stages of deprotonation of phenolphthalein:

https://en.wikipedia.org/wiki/Phenolphthalein

However, there is a big jump here between H2In and In2-. In fact, smaller steps, showing just one deprotonation, are very reasonable. Whoever heard of two protons coming off at the very same time?

http://www.ch.ic.ac.uk/vchemlib/course/indi/indicator.html

Of course, a fuller explanation is needed.

Thymol Blue has a structure siilar to phenolphthalein in that it has two phenolic groups. They deprotonate at different pHs and thymol blue has two color changes.(https://en.wikipedia.org/wiki/Thymol_blue):

The uk reference also lists the pH for a color change of various indicators and also the range over which the color change occurs.

And litmus. Litmus is a water-soluble mixture of different dyes extracted from lichens. The litmus mixture has the CAS number 1393-92-6 and contains 10 to 15 different dyes. All of the chemical components of litmus are likely to be the same as those of the related mixture known as orcein, but in different proportions. In contrast with orcein, the principal constituent of litmus has an average molecular mass of 3300.
https://en.wikipedia.org/wiki/Litmus

In the list of indicators, all have a range of color change of 1.6 pH units, even thymol blue, which has two phenolic groups. Except phenolphthalein, which changes with a broader range of 1.8 pH units. Then there's litmus, with a color change range of 3.8!

The range for litmus can be ascribed to its being an extract of many similar dyes, a natural product, with slight differences between individual compounds. Thymol blue has a sulfonate moiety which will ionize more easily than the carboxyl in phenolphthalein. It seems that the colorless form is singly ionized, being a single quinone moiety and two aromatic rings, without extended conjugation.

In thymol blue, the second deprotonation gives a phenolate which more certainly conjugates with the quinone, establishing a larger conjugated system, thus highly colored. (Yeah, the unconjugated neutral compound is red, how do I explain that - I guess it's just the starting color of a pretty aromatic compound.)

Now in phenolphthalein, the first deprotonation yields an ion, shown in the uk reference. (Although they say it is pink, I am going to disregard that. In addition, I'm going to postulate that the mono-ion would be better described as an ester with a phenolate group, not as a quinone-carboxylate.) This first-deprotonated ion resembles the colorless form of thymol blue: a quinonoid structure with two phenyl rings attached to a central carbon. I might guess/conclude that this form could be colorless and that the second deprotonated form is the pink/fuchsia compound. The (slightly) larger range for the color transformation could well come from the lower tendency of the carboxylate (compared to the sulfonate in thymol blue)  to dissociate after the first proton is drawn off; as long as the central carbon is tetrahedral, the molecule has one phenol and one phenolate. The central carbon goes planar in phenolphthalein AFTER the second proton comes off, whereas the second proton comes off in thymol blue from an already planar structure.

To restate: in phenolphthalein, the central carbon is tetrahedral and the two phenolic groups have very similar pKa; after both phenolic protons are removed, the carboxyl detaches and the extended resonant system is formed (pink/fuchsia). In thymol blue, the first proton removal is easy because the sulfonate is a more stable ionic group than carboxyl, so this proton is more acidic than phenol, but the second proton is just part of a substituted phenol.

Just for reference, the pKa of phenol is 9.95 (https://en.wikipedia.org/wiki/Phenol).


Which ligands would be exchanged in this reaction?

Complexes like $\ce{M(CO)5(solv)}$ are typically generated by photolysis ($\lambda$ = 350 nm) of $\ce{M(CO)6}$ in the presence of a non-inert solvent. This proceeds through a highly reactive "naked" $\ce{M(CO)5}$ intermediate.

$$\ce{W(CO)6 + solv ->[h\nu] W(CO)5(solv) + CO}$$

When offered another ligand $\mathrm{L}$, the weakest bond  ligand in the $\ce{W(CO)5(solv)}$ complex is replaced. This is THF.
Taking the lone pair on the phosphorous atom into account, is is conceivable that $\ce{(CF3)2PH}$ will do the job:

$$\ce{W(CO)5(thf) + (CF3)2PH -> W(CO)5((CF3)2PH) + THF}$$


How to calculate the amount of substance of nitrogen that dissolves in water?

Unfortunately, various definitions of Henry’s law and the corresponding proportionality constant $H$ or $K_\mathrm H$ exist. (For several definitions and corresponding parameter values, see Sander, R. Compilation of Henry’s law constants (version 4.0) for water as solvent. Atmos. Chem. Phys. 2015, 15, 4399–4981.) Therefore, it is important to identify the dimensions of the given data. In the question, the proportionality constant is given as Henry volatility $K_\mathrm H$ with

$$K_\mathrm H = 10^5\ \mathrm{atm}$$

Apparently,* since the given Henry volatility $K_\mathrm H$ is expressed in terms of pressure (the unit symbol “atm” stands for “standard atmosphere”, which is an obsolete unit of pressure; the use of this unit is actually deprecated), the used definition is

$$K_\mathrm H=\frac{p_{\ce{N2}}}{x_{\ce{N2}}}$$

where $p_{\ce{N2}}$ is partial pressure of nitrogen and $x_{\ce{N2}}$ is amount-of-substance fraction (the use of the unsystematic name “mole fraction” is deprecated) of nitrogen in the aqueous phase.

The amount-of-substance fraction $x_{\ce{N2}}$ is defined as

$$x_{\ce{N2}}=\frac{n_{\ce{N2}}}{n}$$

where $n_{\ce{N2}}$ is the amount of substance of nitrogen and $n$ is the total amount of substance. For dilute aqueous solutions, the total amount of substance is approximately equal to the amount of water

$$n\approx n_{\ce{H2O}}$$

which is given as $n_{\ce{H2O}}=10\ \mathrm{mol}$.

Thus, the amount of nitrogen $n_{\ce{N2}}$ can be calculated from the partial pressure of nitrogen $p_{\ce{N2}}$ as

$$\begin{aligned}
n_{\ce{N2}}&=n\cdot x_{\ce{N2}}\\
&=n\cdot \frac{p_{\ce{N2}}}{K_\mathrm H}\\
&\approx n_{\ce{H2O}}\cdot \frac{p_{\ce{N2}}}{K_\mathrm H}
\end{aligned}$$

For a mixture of gases, the partial pressure of nitrogen $p_{\ce{N2}}$ is defined as

$$p_{\ce{N2}}=x_{\ce{N2}}\cdot p$$

where $x_{\ce{N2}}$ is the amount-of-substance fraction of nitrogen in the gaseous phase (as apposed to the above-mentioned parameter in Henry’s law) and p is the total pressure.

Since the total pressure is given as $p=5\ \mathrm{atm}$ and the amount-of-substance fraction of nitrogen in air is given as $x_{\ce{N2}}=0.8$, the partial pressure of nitrogen in air is

$$\begin{aligned}
p_{\ce{N2}}&=x_{\ce{N2}}\cdot p\\
&=0.8\times5\ \mathrm{atm}\\
&=4\ \mathrm{atm}
\end{aligned}$$

* This use of the unit is in the question is actually not permissible. The unit symbol should not be used to provide specific information about the quantity and should never be the sole source of information on the quantity.


Would a quarternary ammonium electride conduct electricity?

You can't conclude that by thinking alone; some experiments are necessary.

True, an electron in an electride is kinda "free", in that it isn't connected to any particular atom. But that doesn't mean it can roam the entire structure, free as the wind. In fact, it sits in a potential well formed by the neighboring ions and molecules, only the well is so wide and shallow that it looks more like a puddle. To decide the conductivity question, you want to know how one puddle relates to the others, and the answer may be any of the following:

connected to form a continuous sea, much like in a metal;
connected to form a network of channels;
isolated but still close enough for electrons to tunnel between them;
isolated and spaced wide apart.

All in all, theoretically an electride can be anything from a metallic conductor to an insulator, depending on its composition and structure.


Adding hydrogen ion concentration while calculating pH

pH is calculated by a logarithmic formula given by:

$$\text{pH}=-\log{\ce{[H+]}}$$

Lets say, the initial hydrogen ion concentration is $\ce{[H+]_0}$, and change in concentration is $d\ce{[H+]}$. The change in pH would be:

$$d\text{pH}=-\log{(\ce{[H+]_0}+d\ce{[H+]})}$$

This can otherwise be written as:

$$d\text{pH}=-\log{\ce{[H+]_0}}×\frac{d\ce{[H+]}}{\ce{[H+]_0}}$$

You can see that the change in pH depends on the initial hydrogen ion concentration. It's in fact inversely proportional to the initial concentration. Adding a drop of concentrated acid to a base would give a much bigger change in pH compared to adding a drop to pure water, which in turn is bigger compared to adding a drop to an acid.


What will happen to the entropy and free energy of the gasses when the partition is removed?

A Spontaneous process is characterized by an increase in the total entropy (for both system and surroundings).

Spontaneous processes are characterized by a decrease in free energy (analogous to the decrease in gravitational potential energy occurring for a ball rolling downhill).


Does all processes that consume energy from outside result in negative entropy change?

Entropy increases in a spontaneous evolution of an isolated system… other than that, there is no general law on the evolution of entropy.

  does that mean all processes that consume energy from outside result in negative entropy change?

No. When you heat a solid, it increases its entropy.

  endothermic reactions causing the entropy to decrease?

Endothermic reactions which cause entropy to decrease have both ∆H positive and ∆S negative, so they will have a positive ∆G, i.e. their are not thermodynamically favorable.


Why are acids usually depicted in fiction as green?

The green is the color of bile which usually accompanies gastric acid.  Acid  in its original context referred to a bitter taste.  Bitter tasting compounds, such as vinegar, is what led to the modern chemical definition.  The chemical definition is less likely to have influenced the populace than the experience associated the somewhat bitter, acidic, taste of gastric acid accompanied by bile.


How to calculate the pH of an acetic acid/ calcium acetate buffer?

In general: $\ce{acid + H2O <=> base + H_3O^+}$
It is called the Henderson-Hasselbalch equation
$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$
I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$


Do external electronic shells influence internal ones?

Not only do "outer" electrons impart some shielding to "inner" ones, but an analytical technique is based on that fact.

In X-ray photoelectron spectroscopy, binding energies of electrons just below the outer shell are measured.  These binding energies are typically hundreds of electron volts, roughly 280-300 eV for carbon as a typical example.  An important feature is that this binding energy is measurably not fixed for a given element, rather is varies according to how the atom is chemically bound.  As an example, the 1s electrons in carbon lie at about 285 eV binding energy in hydrocarbons, but shifted by as much as 10 eV by bonding to fluorine (see here), which is more electronegative and draws electrons away from the carbon 2s and 2p  shells.  Similarly, the 1s electrons in oxygen range from about 529 to 534 eV in various oxides, with the higher values tending to occur in nonmetal oxides (see here).


Does the temperature of liquid ammonia stored at high pressure, room temperature suddenly drop to −33.35 °C when poured into a bowl?

The key to understanding what happens is that evaporation costs energy.

Changing the state of a liquid to a gas requires the input of energy: this is basic thermodynamics. This is in addition to any effects related to a change in temperature which also costs energy or releases energy.

When liquid ammonia evaporates to gaseous ammonia, energy is required from somewhere. If liquid ammonia at room temperature is poured into a bowl at room temperature it will want to evaporate as it is above its boiling point. But that evaporation requires an input of energy. That energy will come from the environment which consists of the bowl and the rest of the liquid. The only way to provide that energy is to cool the rest of the liquid and the bowl. This happens quickly and the temperature will often fall locally to below the boiling point of ammonia as there isn't enough time to equilibrate the temperature of the liquid with the temperature of the environment surrounding the bowl.

The net effect is that you will end up with some liquid ammonia at some temperature below its boiling point and a lot of gaseous ammonia (don't do this outside a good fume hood).

Forced evaporation of volatile liquids used to be used to freeze water via a similar process. For example, place a small bowl of water in a larger bowl of ether. Blow air onto the ether, which will evaporate, drawing energy from its environment (including the remaining ether and the bowl of water). The water will freeze if you used enough ether. I don't recommend doing this as ether is very flammable, but it was once a common demonstration of this process in chemistry classes.


Will distillation do the job of purifying the water?

The old school method of generating ultra-pure water by distillation was to add a small amount of potassium permanganate in a slightly alkaline solution. All the organics were oxidized by the permanganate ion in boiling water. A small of $\ce{MnO2}$ so formed during boiling would take care of the acid(s). This water was good enough for electrochemical experiments. HCl only forms an azeotrope at 20.2% composition and it boils around 110 Celsius.

Today ultrapure water is produced by using quite expensive deionizers with UV lamps. UV destroys all the organics, and then an activated carbon filter absorbs them. Water produced from such deionizer has a specific resistance of $18.2  \rm{M\Omega.cm}$.


Strongest negative inductive effect group between trimethylammonium, ammonium and dimethylsulfonium groups

The inductive effect can be quantitatively measured by the Hammett equation

$$\sigma(\ce{X}) = \mathrm{p}K_\mathrm{a}(\ce{H}) - \mathrm{p}K_\mathrm{a}(\ce{X})$$

where $\mathrm{p}K_\mathrm{a}(\ce{X})$ refers to the $\mathrm{p}K_\mathrm{a}$ of $\ce{X}$-substituted benzoic acid and $\mathrm{p}K_\mathrm{a}(\ce{H})$ to the $\mathrm{p}K_\mathrm{a}$ of unsubstituted benzoic acid. A more electron-withdrawing group will lead to a more acidic carboxylic acid group, a lower $\mathrm{p}K_\mathrm{a}(\ce{X})$, and hence a more positive $\sigma$ coefficient.

Values for a wide range of substituents in the meta and para positions have been tabulated in Table I of Ref. 1. Also included are Swain and Lupton's $F$ ($F$ for field) parameters which are designed to remove the influence of resonance effects on the original Hammett coefficients $\sigma$ (given in the same table; see Ref. 2 for original discussion).

$$\begin{array}{ccccc} \hline
\text{Entry number} & \text{Substituent} & \sigma_\mathrm{m} & \sigma_\mathrm{p} & F \\ \hline
\text{main text, p 167} & \ce{NMe3+} & 0.88 & 0.82 & 0.89 \\
58 & \ce{NH3+} & 0.86 & 0.60 & 0.92 \\
241 & \ce{SMe2+} & 1.00 & 0.90 & 0.98 \\
15 & \ce{F} & 0.34 & 0.06 & 0.45 \\ \hline
\end{array}$$

So, perhaps the answer should be $\ce{SMe2+}$. I don't have a good explanation for why, given that sulfur is less electronegative than nitrogen. I also do not think that the difference between $\ce{NMe3+}$ and $\ce{NH3+}$ can solely be attributed to the inductive effect of the methyl groups, as it is quite possible that there are solvation or hydrogen bonding effects that affect the $\mathrm{p}K_\mathrm{a}$ values. The bottom line is that I don't think this can be rationalised as simply as the question seems to want.

References

Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. Chem. Rev. 1991, 91 (2), 165–195. DOI: 10.1021/cr00002a004. (there is a non-paywall copy here)
Gardner Swain, C.; Lupton, E. C. Field and resonance components of substituent effects. J. Am. Chem. Soc. 1968, 90 (16), 4328–4337. DOI: 10.1021/ja01018a024.


Consider a strong acid such as $\ce{HCl}$ which — within experimental error — can be considered fully dissociated in aquaeous solution. This can be expressed as:

$$c(\ce{H3O+}) = c_0(\ce{HCl})$$

Assume you have a concentration $c = 1~\mathrm{\frac{mol}{l}}$. pH is defined as:

$$\mathrm{pH} = - \lg [\ce{H3O+}]$$

Thus our solution has a pH of:

$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 1 = - 0 = 0$$

(Remember that to calculate pH we need dimensionless concentration. In this post, $c$ shall always contain a dimension while square brackets shall not.)

Take $1~\mathrm{l}$ of this solution and add $1~\mathrm{l}$ of water. Our new solution has a concentration of $c = \frac{1~\mathrm{mol}}{2~\mathrm{l}} = 0.5~\mathrm{\frac{mol}{l}}$. Our pH is:

$$\mathrm{pH} = - \lg [\ce{H3O+}] = - \lg 0.5 = - (-0.301) = 0.301$$

The solution’s pH value obviously changed with concentration change.

Now, let’s do the same thing with a weak acid such as acetic acid ($\mathrm{p}K_\mathrm{a} = 4.76$). But first, I need to do a bit of maths. Remember that we cannot use the Henderson–Hasselbalch equation, because it assumes a buffered system. A simple acid is unbuffered. Instead, I will start at the definition of the acid constant:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{OAc-}]}{[\ce{HOAc}]}$$

And since chemistry dictates that $[\ce{H3O+}] = [\ce{A-}]$.

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]}$$

Because $c_0(\ce{HOAc}) = c(\ce{HOAc}) + c(\ce{H3O+})$, we can transform to:

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}$$

Which finally gives us an equation with only one unknown. Solving this:

$$K_\mathrm{a} = \frac{[\ce{H3O+}]^2}{[\ce{HOAc}]_0 - [\ce{H3O+}]}\\
K_\mathrm{a} \left ( [\ce{HOAc}]_0 - [\ce{H3O+}] \right ) = [\ce{H3O+}]^2\\
[\ce{H3O+}]^2 + K_\mathrm{a} [\ce{H3O+}] - K_\mathrm{a} [\ce{HOAc}]_0 = 0$$

Since the determinant is positive, we can immediately write:

$$[\ce{H3O+}]_{1/2} = \frac{- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0}}{2}$$

Now plug the values $c_{0,1}(\ce{HOAc}) = 1~\mathrm{\frac{mol}{l}}$ and $c_{0,2}(\ce{HOAc}) = 0.5~\mathrm{\frac{mol}{l}}$ as above. You will need to add the value of the square root or you will get a negative concentration. This is what you should be getting (assuming OpenOffice isn’t buggy):

$$\begin{array}{ccc}
\hline
\text{expression} & c_0 = 1~\mathrm{\frac{mol}{l}} & c_0 = 0.5~\mathrm{\frac{mol}{l}}\\
\hline
K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0 & 6.95 \cdot 10^{-5} & 3.48 \cdot 10^{-5}\\
\sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.34 \cdot 10^{-3} & 5.90 \cdot 10^{-3}\\
- K_\mathrm{a} \pm \sqrt{K_\mathrm{a}^2 + 4 K_\mathrm{a} [\ce{HOAc}]_0} & 8.32 \cdot 10^{-3} & 5.88 \cdot 10^{-3}\\
[\ce{H3O+}] & 4.16 \cdot 10^{-3} & 2.94 \cdot 10^{-3}\\
-\lg [\ce{H3O+}] & 2.38 & 2.53\\
\hline
\end{array}$$

Obviously, once again, the pH went up when diluting the solution.

The Henderson–Hasselbalch equation, as I skimmed past earlier, is meant for buffer systems. The concentrations $[\ce{A-}]$ and $[\ce{HA}]$ are those concentrations you added to the buffer system. Since we did not add any acetate ions (other than acetic acid), the logarithmic argument would become zero causing an undefined error. We cannot use Henderson–Hasselbalch here.

However, if you have a solution containing $1~\mathrm{\frac{mol}{l}}$ acetic acid and $1~\mathrm{\frac{mol}{l}}$ sodium acetate, the logarithm’s argument will be $1$, therefore the logarithm will disappear and the pH value will be $4.76$ — exactly the $\mathrm{p}K_\mathrm{a}$ value of acetic acid. This buffer system is concentration independent. Diluted acids to whom the conjugate base has not been added as a salt are not!


Removing sulfur in Barton-Kellogg reaction without triphenylphosphine

After contacting the authors of the paper, one author, Prof. Wezenberg said that a conclusive mechanism is not yet available for this case but he believes that the unstable episulfide undergoes desulfurization by itself under reflux.


Why are the absorption and emission spectra mirror images of each other?

The sketch below explains how the mirror image spectrum comes about. (the figure is not at all to scale, its just illustrative, but does show the salient features.)

There are some conditions attached.

(a) mirror image spectrum is only observed in the condensed phase. This is because solvent collisions cause vibrational relaxation from the vibrational level excited and removes$^*$ energy to the solvent much faster than fluorescence can occur, thus all emission occurs from v=0. This vibrational relaxation occurs with a rate constant of $\approx 10^{13}$ s$^{-1}$.(This is not true in the vapour phase where if the pressure is low enough no collisions may occur within the excited state lifetime and fluorescence occurs from the level excited) The intensity of each transition (not wavelength) in absorption or emission is governed by the Frack-Condon Principle. The strength of a transition is proportional to the squared modulus of the two wavefunctions. i.e. $|\psi _{s0} \psi _{s1}  |^2$. In practice this means that overlap between vibrational levels in the two states is only large at the turning points of the wavefunctions, so transitions appear to be vertical. Classically this makes sense as momentum has to be conserved and this can only be achieved from vertical transitions starting and ending at the potential curve where kinetic energy is zero. In the quantum world the uncertainty principle limits change in position and momentum and so Franck -Condon Factors.

$^*$(The collisions preferentially remove energy as generally the excited state is at a higher 'temperature' than the surrounding solvent, i.e. it has more vibrational energy that the average $3k_BT/2$ of the solvent)

(b)The geometry of the excited state has to be similar to that of the ground state, thus mirror image spectra are observed for 'rigid'$^{**}$ molecules such as anthracene (3 benzene rings attached side by side) and not for flexible molecules such as cis or trans stilbene (phenyl=phenyl, where = is a double bond and this molecule isomerizes in the excited singlet state).

Relative displacement of potential energy surfaces (for the normal modes mode involved in the transitions) is not the only consideration, for the shape of the potential has to be similar in both states, thus the vibrational frequencies are similar and so gaps between vibrational peaks in absorption and fluorescence are similar. (Gaps between peaks in absorption give excited state vibrational frequencies and vice versa)

If the lowest triplet state has a similar potential to the singlet, which is usual if it derives from that singlet state (and not another higher singlet state) the the phosphorescence spectrum is very similar to the fluorescence, but shifter to longer wavelengths

$^{**}$ By 'rigid' is meant that the molecule has only its normal vibrational modes and no large torsional motion as is the case in stilbenes.


Ab initio molecular properties which can be experimentally measured and don't require derivatives/gradients

Many thermochemistry benchmarks, like the GMTKN55 (homepage listing most results, DOI), use fixed geometries and require only ground state total energies. Some of the subsets probably have back-corrected experimental references, but you would have to check each paper, I'm afraid. Note that for some properties, a dispersion correction is vital.


Finding the enthalpy of formation of magnesium oxide using Hess' Law

You are trying to find out the enthalpy of formation.
The enthalpy of formation is the energy change when one mole of a substance is formed from its constituent elements in their standard state.

The last equation matches this, so that should be your 'main' or 'target' equation. From there you should be able to use Hess' law to calculate the enthalpy change of the formation of magnesium oxide.


Temperature in the Gibbs Free Energy equation

In fact, Gibbs free energy is not defined at exact temperature. It is defined for standard states of all compounds in equation. So you really can use this solution of problem. However, it is a very simplified solution, because you assume that standard entropy and enthalpy do not change with temperature. This is not entirely correct assumption but if you took change of enthapy and entropy with temperature into account, it would be much more complicated.
You can find information about Gibbs free energy here
http://www.chem1.com/acad/webtext/thermeq/TE4.html#3A


Molarity Numerical on Water Gas Shift Reaction


For combustion to occur you need 3 things at the same place: fuel, oxygen and heat.

If the air intake is dirty or not big enough, you get a flame that has an orange color, and on a cooler surface deposits C before it has a chance to burn.

The reaction with the pot is only a physical one of heat transfer and maybe C deposition.


Why does ozone, being heavier than air, not settle down?

While gravity does cause some separation of atmospheric constituents at high altitudes, the reason why ozone doesn't migrate to the ground has more to do with its chemistry.

Ozone is pretty unstable and is formed from some chemical reactions and from the interaction of oxygen with UV light. But it is unstable and decays fairly quickly. This means that is only appears anywhere when the reaction forming it keeps happening.  This is true in the upper atmosphere which is why we have an ozone layer. Some hydrocarbons (not just from pollution from cars but also from volatile stuff released by trees) can react to generate ozone at ground-level. But we only see a significant concentration when the reaction forming it is happening.

The effect of gravity on the separation of the components of the atmosphere is small, especially in the lower atmosphere where diffusion and turbulent circulation mix things up thoroughly. In the upper atmosphere the dominant process is diffusion and gravity can have more of an effect. This paper describes the process and has some details of the calculations about the size of the effect. It summarises the situation like this:

  Turbulent mixing keeps the relative concentrations of gases nearly constant in the lowest 100 km. At higher altitudes, molecular diffusion controls the concentrations, with the lighter gases becoming relatively more abundant with increasing altitude.

It is a worthwhile read if you want the detail.


How to rationalise the major product formed in a nucleophile promoted epoxide cleavage reaction?

Let's do this step by step and figure your questions out.
First, what a nucleophile needs is positive terminal in the compound.

The other carbon attached to oxygen is not as positive as the one where the attack has been done due to inductive effect of $\ce{-CH2}$ adjacent to it. The product (1) is not formed as there is no source providing $\ce{H+}$ to form alcohol. If possible a very small amount  may be formed by $\ce{H+}$ extraction from $\ce{EtO-}$. Next the intramolecular nucleophilic substitution takes place.

I do agree that product (2) is formed, but as a minor product. Here what we observe is that there is neighboring group participation in the mechanism. Generally these neighboring groups increase the rate of the reaction by switching on abilities like dipole in the bond. Whenever we observe compounds with such epoxides or sometimes even alcohols, we better think of such consequences which can help us predict the product.


How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?

There is something called an average oxidation state, which is typically not very meaningful chemically, but can (usually) easily be calculated by just knowing the structural formula. In your case, since chlorine is the most electronegative element (save nitrogen), you can assume it to be in $\mathrm{-I}$ state while nitrogen of course would have $\mathrm{-III}$ and hydrogen $\mathrm{+I}$. Thus to get platinum’s average oxidation state just sum up to the compound’s charge:

$$\begin{align}2 x + 4 \cdot (-3) + 3\cdot 4 \cdot (+1) + 6 \cdot (-1) = 0\\
2x - 12 + 12 - 6 = 0\\
2x -6 = 0\\
2x = 6\\
x = 3\end{align}$$

Which gives you an average oxidation state of $\mathrm{+III}$.

But as I said before, average oxidation states do not carry any physical or chemical meaning and can only serve as exercises to students and pupils. Take for example tetrathionate’s ($\ce{S4O6^2-}$) average sulphur oxidation state of $+\frac{5}{2}$.

Instead, one should always attempt to allocate per-atom oxidation states you cannot do that without knowing the charge of each of those complexes, and you need to do it per complex; ideally by drawing it, heterolytically cleaving all bonds (or homolytically for homoatom bonds) and then counting electrons. In the case of your compound — tetraamminplatinum(II) hexachloridoplatinate(IV) — you should arrive at the numbers in brackets in the IUPAC name I gave you, knowing that the cation has $2+$ charge and the anion $2-$.


You start of by calculating solubility with one approximation: that all $\ce{AsO4^3-}$ ions are obtained by dissolving the potassium salt:
Let $S$ be the molar solubility:
$$K_\mathrm{sp} = 0.02 \times 27 S^3$$
$$S = \pu{0.0223 M}$$
Then, you calculate the solubility by taking in account the concentration of arsenate ion obtained by dissolution of silver arsenate:
$$K_\mathrm{sp} = 27 S^3 \times (S+0.02) = 27 S^4 + 0.54S^3$$
Solving this equation by applying the iteration method gives the value of $S = \pu{0.018M}$.
So the relative error is $23.9~\%$.


What would happen if you heat volatile/flammable material in a vacuum?

If there's any trace of oxygen, and a spark, the vapors would burn like normal.

If there isn't any oxidising agent, it'll just vaporize and stay that way (maybe the vapors will attain equilibrium with the liquid if the vapor pressure is high enough, basically if there's enough liquid to start with).

That's mostly it... Nothing spectacular :)


Stability of carbocations with -M groups in beta position

These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question:
The order (B > A > C > D)  was said to be correct for the conc. $\ce{H2SO4}$/ $\Delta$ case since there would be no effect of $\ce{CH-}$ acidity since the protonation of the $\ce{-OH}$ group takes place and so the carbocation stability takes precedence meaning that the stronger the -I group, the slower the rate since the transition barrier increases.
However, for the scenario where only heat is used in dehydration (I assume basic conditions here), the reaction proceeds via E1cB  since the two main requirements for E1cB is that there is a relatively acidic hydrogen and a poor leaving group. Here, the RDS of the reaction would be the $\ce{C-H}$ bond cleavage. This would be stabilised by the presence of a strong -I group since a carbanion is being generated here which is stabilised by the delocalisation of the electrons via the inductive effect.
So, for such a scenario the rate order would be D > C > A > B.
The answers depend upon the medium as you can see.
This entire argument seems hand wavy with the assumption taken but otherwise as Martin pointed out, this would not be trivial and would require quantum modelling:

Protonation and deprotonation reactions in a catalytic cycle are notoriously hard to calculate. The RDS approximation will break down.


Molar conductivity of coordination compound

You are right. But for some reason, your compound has never been synthesized, although theoretically ti should exist. It may be a problem of entropy.


How to speed up flash powder flaming reaction

First, we postulate safety is not an issue. If you try any of this you will probably die.  If you need a short duration event-triggered flash, explode a wire.  A Pockels cell plus a polarizer is a nanoseconds shutter.

Ball mill potassium perchlorate all by itself.  Mix well with micronized aluminum.  Compact.  (Impact, friction, or a static discharge may kill you.)

Aluminum has a high boiling point, 2519 C.  Magnesium is 1091 C and burns under nitrogen.  As above, with micronized Mg, adjusting stoichiometry for the new equivalent weight.  Same warning.

Use a more potent oxophile, like titanium, adjusting stoichiometry for the new equivalent weight.  Micronized titanium can flash in air, nitrogen, or carbon dioxide all by itself.  Same warning.

DO NOT TRY CHLORATE, as opposed to perchlorate.  There is almost no activation energy barrier to reaction.


Why is a tin can that appears like galvanised steel rusting?

First of all, any 'tin' can you are likely to find is actually made of steel. Tin is too expensive. Also, tin does not 'rust', although it oxidizes. Your rust is iron oxide. Galvanized steel is steel with a thin zinc coating, likely hot-dip galvanization. What you describe (crystalline surface or spangles sort of appearance) does sound like galvanization, although I am not aware of any steel cans that are galvanized.

I thought galvanised steel only rust when the layer of galvanisation wears out. This is true. If your can is indeed galvanized, then it has likely chipped off, possibly on the rims of the can where the lid it crimped or pressed together, or by cutting the can open. Strong acids will dissolve zinc  Also interesting to note: sodium chloride (salt) will dissolve zinc over time by converting the galvanization to zinc chloride, but that's not applicable here.

So, is the tin made with galvanised steel?
  Well, that does sound like what you were describing, but galvanized steel is not fit for foodstuffs. Zinc is mildly toxic and ingesting it can make you ill. I like feetwet's suggestion; that universally the inside of metal food cans are lined, by actually an epoxy that is sprayed on. This is almost certainly to be the case. Perhaps this makes sense: if your can was indeed galvanized, then the rational by the manufacturer was that the lining will keep the zinc separated and the food safe. Sounds legit. It might be good to be aware that the type of epoxy being used here is the one that uses as its plasticizer Bisphenol A and Bisphenol S (BPA/BPS). Ill leave it up to you to decide if that should bother you or not.

If yes, why are there concentrated spots of rust on the wall and other rusting?
Well, if there is rust, then we can deduce that water must have be coming in contact with the iron in the steel, so whatever was protecting the steel has some failure.

And BTW, why does the emptied and washed can smell "metallish"?
While zinc does have a metallic smell to it, iron oxide's is much stronger. Since you visually identified rust, this is what you are smelling.


Why can higher voltages lead to less efficient electroplating?

When you apply the required voltage for the desired cell, then your desired product is being produced, but if you provide an overpotential then you forced the other reaction to also occur(the undesired one).

For eg. If in a reaction two species can be oxidized then, of the required potential is applied then you will get the required product but if an over potential is applied then the other species will also get oxidized and you will have two oxidation products.


It depends on the concentration of Pt in your alloy.

If the concentration of Pt is to high the HCl will not or not sufficently attack the alloy.

In this case you could try to melt your alloy with some additional Fe, to lower the Pt concentration and then hit it with HCl again.

You could also try to boil it in HCl to increase the reactivity of the acid (do it in a fumehood or outdoors).


Why do some crystal structure files not include bonds?

If you solve an x-ray crystal structure, you simply get a set of atomic positions. No more, no less.

Often, the atomic positions don't include hydrogen atoms, only the "heavy" atoms.

Thus, most CIF (crystallographic interchange format) files, there are no defined bonds. Somewhere along the way, software perceives bonding information because most chemists prefer to have a valence-bond graph of atoms and bonds (and bond orders).

There are many algorithms for determining a bond graph from a set of atomic positions with varying "accuracy." I put the word in quotes because it's sometimes unclear what the goal should be when determining bond orders. For example, is a molecule in a keto or enol tautomer form? In some cases, the $\ce{C-O}$ bond is not clearly a single bond distance or a $\ce{C=O}$ distance. Maybe there's an ensemble average of both forms.

The standard seems to be if the CIF file looks more "ionic" (e.g., a salt like $\ce{NaCl}$) bonds probably will be ignored, but otherwise, bonds are interpreted from interatomic distances.


Can you have addition polymerisation with monomers other than alkenes?

$\ce{N2H2}$ is a very unstable molecule, prone to disproportionation into hydrazine and nitrogen. $\ce{N-N}$ bonds except one in $\ce{N2}$ molecule are unstable and generally unfavorable. So, no there are no polymers with nitrogen as a sole component of the main chain.

Several classes of compounds of 3-row elements demonstrate tendency for oligomerization, namely $\ce{(HP)_{n}}$ derivatives usually exist as a ring. It is possible to obtain metastable at room temperature plastic sulfur, containing linear sulfur chains terminated by free radicals.

However, generally when in search of easily polymerizing molecule that produce main chain containing elements other than carbon, one usually have to look into compounds producing chains with two elements. The most common and known one is formaldehyde $\ce{CH2=O}$, producing $\ce{HO-(CH2-O)_{n}-H}$ chains. A less known, but interesting example is hexachlorotriphosphazene, the source material for production of Polyphosphazenes.

In  general, metastable double bonds are rare for elements other than carbon (sometimes with nitrogen or oxygen partners), so usually reactions intended to produce such bonds, say, $\ce{Si=Si}$ usually end with olygomer instead, typically of ring structure. Moreover, single homoelemental bonds are usually much weaker for elements other than CHNOB, so typically when a chains containing other elements is occurred, it is usually hetero-atomic chain, produced by means other than olygomerization of double-bond containing monomer. For example, an extremely wide-used family of polymers (polysiloxanes, or silicones) is produce by hydrolisys of dialkyldicholorsilanes $\ce{R2SiCl2}$.


Enthalpy and Internal Energy in Constant Volume Calorimeter

The definition of enthalpy is
$$H = U + pV$$

which gives

$$\begin{align}H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
\Delta H &= \Delta U + \Delta(pV)\end{align}$$

The equation you cited is only true at constant pressure: if $p_1 = p_2 = p$,

$$\begin{align}
H_2 - H_1 &= (U_2 + pV_2) - (U_1 + pV_1) \\
\Delta H &= \Delta U + p(V_2 - V_1) \\
&= \Delta U + p\Delta V
\end{align}$$

The condition for that equation is that the pressure is constant. That equation is not applicable for constant-volume (isochoric) processes unless the process is also constant-pressure (isobaric).

However, if you do have a process that is both isobaric and isochoric, you are right to say that $\Delta H = \Delta U$.


Adjustment of calcium, magnesium and alkalinity in drinking water?

The parallel thoughts:

The source water I am working with is:

12.6 mg/L or 0.126 mmol/L GH as CaCO3 equivalent
11.3 mg/L or 0.113 mmol/L KH as CaCO3 equivalent
  4.8 mg/L or 0.12 mmol/L Ca
  0.2 mg/L or 0.0082 mmol/L Mg
  1.5 mg/L or 0.065 mmol/L Na

And the desired target water is:

70    mg/L or 0.7 mmol/L GH
40    mg/L or 0.4 mmol/L KH
20    mg/L or 0.5 mmol/L Ca
  4.9 mg/L or 0.2 mmol/L Mg
  8    mg/L or  0.35 mmol/L Na

What is needed to add:

0.574 mmol/L GH, as Ca+Mg
0.287 mmol/L KH  as Ca+Mg as bicarbonates, I.e. 2x0.287=0.574 mmol/L of bicarbonate
0.38 mmol/L Ca
0.192 mmol/L Mg
0.285 mmol/L Na

Reaction of 0.285 mmol/L sodium bicarbonate
with 0.285 mmol/L lactic acid
releases 0.285 mmol/L carbon dioxide,
dissolving 0.285 mmol/L calcium carbonate
forming 0.285 mmol/L calcium bicarbonate.

( Not exact due equilibrium reactions but good enough )

What is needed to add:

0.287 mmol/L GH, i.e Ca+Mg

0.093 mmol/L Ca
0.192 mmol/L Mg

Adding 0.192 mmol/L MgSO4.

What is needed to add:

0.093 mmol/L GH, i.e Ca only

0.093 mmol/L Ca

This can to be added as 0.093 mmol/L calcium carbonate(optionally supported by pressured carbon dioxide) + 0.186 mmol/L lactic acid.


Do Faraday's gold colloidal solutions still exist?

Please note that I haven't seen the sample myself!

However, the Royal Institution claims that a bottle with the collidal gold solution is on display at Michael Faraday's Magnetic Laboratory (see bottle at the bottom of the image) in the museum of the Royal Institution.

Update

This might be nitpicking and English isn't my native language, but David Thompson does not explicitly state in his article in Gold Bulletin that all samples are lost.

On the other hand, Ryan D. Tweney in his article Discovering Discovery: How Faraday Found the First Metallic Colloid, published in Perspectives on Science, 2006, 14, 97-121 (DOI) not only describes the finding of 700 surviving microscopic slides in Faraday's reconstructed magnetic with gold film samples and numbered in Faraday's hand and that the numbers correspond to the numbering in Faraday's 1856 Diary (lab journal) but continues:

  It had long been known that some of Faraday's gold colloids survive from the 1856 research: Four bottles of these are on display at the Royal Institution in London […]. These few colloids are all that survive of many dozens referred to in the Diary.


Curious about the chemistry of Carbon-14 after it's produced in the atmosphere

wikipedia article cites: "After production in the upper atmosphere, the carbon-14 atoms react rapidly to form mostly (about 93%) 14CO (carbon monoxide), which subsequently oxidizes at a slower rate to form 14CO2, radioactive carbon dioxide. The gas mixes rapidly and becomes evenly distributed throughout the atmosphere (the mixing timescale in the order of weeks)."


Why does oxygen have less negative electron gain enthalpy than sulphur?

It is because Oxygen is first element of Oxygen family in its table and Electron gain enthaply decreasing from top to bottom in the group.

This exemption happens because of the reduced 2p orbital of Oxygen, there is enormous interelectronic r between the electrons in 2p orbitals thus Oxygen has less proclivity for an additional electron and the EGE of Oxygen is more positive than Sulphur.


Free radical bromonation of 1-ethyl-4-nitrobenzene

This is would be a free radical bromination reaction. Bromine is very selective. It is going to replace the hydrogen from the highest substituted carbon. So according to me second one should be the answer i.e. Br attached to secondary carbon.

Hope it helps