How could the ideal gas law be discovered from experiments on real gases?

The ideal gas law is a very good approximation of how gases behave most of the time

There is no logical flaw in the laws. Most gases most of the time behave in a way that is close to the ideal gas equation. And, as long as you recognise the times they don't, the equation is good description of the way they behave.

The ideal gas equations assume that the molecules making up the gas occupy no volume; they have no attractive forces between them and their interactions consists entirely of elastic collisions.

These rules can't explain, for example, why gases ever liquefy (this requires attractive forces). But most of the common gases that were used to develop the laws in the first place (normal atmospheric gases like oxygen or nitrogen) are usually observed far from the point where they do liquefy.

As for the volume taken up by the molecules of the gas, consider this. A mole of liquid nitrogen occupies about 35mL and this is a fair approximation of the volume occupied by the molecules. at STP that same mole of gas occupies a volume of about 22,701mL or about 650 times as much. So, at least to a couple of decimal places, the volume occupied by the nitrogen molecules is negligible.

The point is that for gases not close to the point where they liquefy (and few components of the atmosphere are), the ideal gas laws are a very good approximation for how gases behave and that is what we observe in experiments on them. The fancy and more sophisticated equations describing them are only really required when the gas gets close to liquefaction.


interaction of grape juice or wine with marble

CaCl2 and magnesium are not toxic at these levels. In fact NIH suggests around 1 g calcium per day: https://medlineplus.gov/magazine/issues/winter11/articles/winter11pg12.html

Marble itself is predominantly calcite or dolomite, that is carbonates of calcium and magnesium. Marble also contains other elements. We can't say if these other chemicals are safe just by a photo because composition of marble varies.

When you bought this cup - did they say it is safe to drink from it? Supplier should know if the material is safe or not. If you didn't get assurance that it is safe then you should assume it is not.


Which is the parent chain in Neopentyl chloride? and why?

A chemist would say that all 3 methyl groups are equivalent (identical by symmetry).  Therefore they all produce the same chain, and that chain is the longest chain in the molecule.  Consequently, since the methyl groups are all the same, it doesn't matter which one you pick.


I think a cross Cannizzaro with the Ketone and Formaldehyde takes place. The ketone give the alcohol and Formaldehyde formiate. In a second step a full acetal will be created with formaldehyd and now the Diole.
Sorry I cannot add comments here. So use answer function.


What is the internal energy at constant pressure?

If you are working with an ideal gas mixture, for which the internal energy of the reactants and products is not a function of pressure, the change in internal energy in going from reactants to products at constant temperature and volume is the same as the change in internal energy in going from reactants to products at constant temperature and pressure.  With regard to the changes in enthalpy, the same is true.  If the number of moles of reactants and products is the same, then the change in enthalpy for each of these processes will be the same as the change in internal energy.  However, if the number of moles changes, the change in enthalpy will differ from the change in internal energy in proportion to the change in the number of moles.


It depends on the temperature, specifically the 1700 F +/- boiling point of zinc , not the flux. At 1700 F the zinc will vaporize regardless of any flux . I never saw a single component flux, likely the the flux also contains soda ash, silica, etc. and it does protect the molten metal from atmosphere by forming a physical layer. Footnote: many years age I worked with copper alloy fluxes and found that for some alloys, the boron promoted the formation of spinels in the metal ; These are exceptionally hard materials that essentially prevent machining of the bronze/ brass.


You are quoting from my answer here.

Looking at the source reference Inexpensive Equipment for the Preparation
and Concentration of Pure D2O or of
Deuterium-Rich Water. Description of
Construction and Operation of the Former
Ohio State University Heavy Water Plant The Ohio Journal of Science. v41 n5 (September, 1941), 357-365:

  With NaOH electrolyte we obtained, with Armco iron, an electrolytic
  separation factor[footnote 8] of a little more than 8, whose magnitude was
  practically independent of current density and of the deuterium concentration
  in the electrolyte. This is the true separation factor applicable
  to the dry gas. The practical separation factor is somewhat less than
  this (about 6 or 7) due to water vapor carried out with the electrolytic
  gases.

So, like Ivan Neretin is commenting, it is not optimal to electrolyze most of the first stage solution if the goal is obtaining deuterium depleted hydrogen.  The initial hydrogen coming from natural abundance water will be deuterium depleted by a factor of "about 6 or 7".  Since natural abundance water is essentially free, it would be best to only electrolyze a small fraction of a large amount of water.  Then you will deplete deuterium down to about 25 ppm in one stage.

After the first stage, if you attempt a second stage, burning the water to make deuterium depleted water, you will then have to add electrolyte, which, if it is NaOH, would introduce more deuterium.  So if you used a hydrogen-free electrolyte it would be better.  By the second stage, the starting material is valuable, and you would want to electrolyze a significant fraction of it, but not 31/32.


It is simply because the wave nature of the functions.

Imagine a tight rope with fixed ends (like a string in a guitar). If you shake it it will vibrate. Both ends remains fixed. If you take a high frame rate video and see it in slow motion, at a fixed horizontal length (let say $x$ positions) the rope will oscillate up and down with some amplitude. It can be seen one or more intermediate $x$ positions that remains always with null amplitude (if it is in state close to a stationary one). There you have nodes in one dimensional string.

Visual example

You would have more complex patterns going to a two dimensional system, and even more in a three dimensional case. They all have nodes, like the 1D case. The 3D case can be represented mathematically by using the spherical harmonics. The angular part of the solutions for the hydrogen atom are those spherical harmonics. The only intuitive explaining that I found is their wave nature. Of course it is not as visual as in the 1D case which is easily found in the daily life. And clearly, what is more weird is the wave nature of matter.

I hope that this answer address your question.


Why does Ammonium sulfate cause proteins to precipitate?

Well, I wouldn't know how to sum it up better than chemwiki.ucdavis:

  To achieve this conformation the molecule folds in such a way that all of the hydrophobic parts of a molecule are aggregated together and the hydrophillic groups are left to interact with the water. In the case of proteins it is the charged amino acids that allow selective salting out to occur. Charged and polar amino acids such as glutamate, lysine, and tyrosine require water molecules to surround them to remain dissolved. In an aqueous environment with a high ionic strength, the water molecules surround the charges of the ions and proteins. At a certain ionic strength, the water molecules are no longer able to support the charges of both the ions and the proteins. The result is the precipitation of the least soluble solute, such as proteins and large organic molecules.

If you ask what the aminoacid content of bovine serum albumin is, I'll cite:

  Glycine,
  illanine,
  Valine,
  Leucine,
  Isoleucineh,
  Proline,
  Phenylalanine,
  Tyrosine,
  Tryptophan,
  Serine,
  Threonine,
  Half-cystinee,
  Methionine,
  Arginine,
  Histidine,
  Lysine,
  Aspartic acid and
  Glutamic acid. Source


GOC allylic,vinylic, benzylic positions, carbocation stability

allylic position explanation
benzylic position explanation
vinylic position explanation

For carbocation stability, think about which case has opportunity for electron resonance, i.e., which case(s) allows the positive charge to be delocalized to adjacent pi bonds. This resource might help.


How to identify if a chemical equation has a 'fractional order'?

It is not possible to determine the rate law of a reaction based on solely the balanced reaction. We must determine the rate law by experiment or by an established mechanism for the reaction.

What is meant by a mechanism is a series of elementary steps, a set of reactions whose linear combination yields the balanced reaction. Each elementary step represents a microscopic collision that happens during the process of the reaction, and these elementary steps obey the law of mass action, which allows us to write the rate law for each step in the reaction.

We then make a few approximations, perhaps guided by experimental observation, to obtain the rate law for the reaction. For example, we may conclude that one elementary step is rate-determining, so that the rate law for the reaction is governed by the rate law for that step; we may make a quasi-equilibrium approximation, which means that the species involved in an elementary step are equilibrated; or we may make a pseudo-steady-state assumption, which states that the net rate of production of an intermediate, presumed to be short-lived, is zero.

How do we know that a proposed mechanism is correct? Well, we don't. All we can do is prove that a mechanism is incorrect, by establishing that it posits the wrong rate law as found from experiment; we cannot distinguish between different mechanisms that result in the correct rate law. Nevertheless, we can often obtain much indirect evidence to support a given rate law, and we can frequently make good guesses based on patterns from experimental observations.

For example, while I wouldn't be able to accurately guess the rate law for the reaction you've provided, I can at least say that it is likely to be of half-integer order in chlorine, because such reactions often follow a radical mechanism, which often lead to non-integral powers in rate laws.


What is the density of carbohydrates/protein/fat?

I assume the "energy density" you are referring to are the widely known statistics that:

fat contains 9 "calories" per gram (actually kilocalories)
protein and carbohydrates both contain 4 "calories" per gram

How can all fats contain "9" kcal per gram but have different mass densities?  Here are a couple of reasons:

The 9 kcal/g figure is an approximation that assumes a typical dietary blend of fats.  The figure would if considering chemically pure fat molecules.
The mass densities of most common fats doesn't vary that much.  The densest "fat" molecules are probably the extremely short chain triglycerides like triformin and triacetin.  They have densities of approx. $1.3~ \frac{\mathrm g}{\mathrm{cm}^3}$.  These triglycerides have chain lengths of 1 or 2 carbons per fatty acid chain, a value far lower than common in natural fats.  Triolein is more representative of dietary fats and has a density of 0.95$~\frac{\mathrm g}{\mathrm{cm}^3}$.  So the mass density variation among different fat molecules is pretty small: a 30% increase is all that is really possible.


Is the scientific explanation behind Smartkleen legit?

I think the idea of "smaller water clusters" penetrating the cloth more efficiently is actually quite smart as the same idea is utilised in steam cleaners. But the rest of the article is a bit too psychotic. How can they say that colours are preserved while the Ball constantly generates H$_2$O$_2$ (in every second)? And why do they recommend the Stain Remover for stains if the ball was such a miracle...


The short answer is that I need to come up with better wording to describe the interaction and that Hückel is surprisingly bad for these systems.

I'll give a slightly longer answer for now, and add a reference as we submit the paper.

First, an example system:

So "A" is a strongly electron-deficient dinitrothiophene ring (an electron acceptor) and "B" is a strongly electron-rich diaminothiophene ring. The figure in the question is a subset of the entire MO diagram, focusing on the HOMOs of the monomers and the HOMO and HOMO-1 of the dimer.

Why do I think this is a strange system? Well, the particle-in-a-box and Hückel models are usually pretty good at explaining conjugated organic molecules. If you make a longer conjugated system, the HOMO should go up in energy, and the HOMO-LUMO gap goes down (delocalization).

The alternative, localization, would have the orbitals remain at the same energy level:

Both of those "limits" can be explained in the Hückel model by changing the $H_{ab}$ or $\beta$ parameter (depending on how you write the equation, the symbol is different). In the localized case, $H_{ab} = 0$.

What's strange about these strong donor-acceptor cases is that as I said, it's not really delocalized, or localized.

Instead, as one comment correctly deduced, the DFT calculation (B3LYP/6-31G*) shows a large degree of charge transfer, but not necessarily localization. The dipole moment computed to be >7 Debye.

If one attempts to use Hückel to analyze these results, you must infer that $H_{ab}$ is imaginary. This indicates an "open system" because electrons are indeed transferred - from monomer "B" to monomer "A" creating the large dipole moment.

My overly-simplified question was probably a bit confusing, but:

It's interesting that no simple model captures charge transfer unless you allow Hückel to have imaginary (non-physical) $H_{ab}$ parameters. (NB, my colleagues aren't too happy with this concept, suggesting it simply indicates Hückel is a poor choice.)
It's surprising that in prototypical conjugated systems, I can't (yet) find a discussion about this alternative to delocalization and localization. We find it commonly as the difference in energy level increases.


How do meth lab test kits detect the presence of a meth lab?

Simple self-made or commercially available tests use the Marquis reagent, which is a mixture of sulfuric acid and formaldehyde. For example, Marquis reagent can be prepared by adding 100 ml of concentrated sulfuric acid to 5 ml of a saturated aqueous solution of formaldehyde (40 % formaldehyde by volume or 37 % by mass).[reference]

The Marquis reagent reacts with many alkaloid drugs to form a variety of colours. Methamphetamine creates an orange dye (deep reddish orange to dark reddish brown).


Rigorous reason behind internal energy change being zero while mixing

The process in question is presumably adiabatic, since we are told that the temperature changes and are not given any information about the temperature of the surroundings (adiabaticity can also sometimes be assumed if gases are mixed rapidly). Secondly, the final volume of the system (the entire container) is the same as the initial volume (and in addition we can assume that each gas expands without performing any work, i.e. the expansions can be assumed to be free). That means that neither heat nor work exchanged with the system during the process, which means that in accordance with the first law $\Delta U = 0$.


How to determine the Bravais lattice and atom basis vectors from a CIF file?


Could PNP (Silicon) transistor be toxic when exposed to air\light?

The semiconductor itself is harmless, but, since you heard some sizzling, it is possible that a getter, such as an alkali metal, was in the metal-cased component. There is a patent for getters to be used with semiconductors, http://patentimages.storage.googleapis.com/pdfs/US3259490.pdf, though I've never seen one in use.

More likely, you heard the glass header breaking.


Why do internal combustion engines only burn to carbon monoxide, not dioxide?

I think this is sort of covered by What really happens atomically in an explosion? because the burning of the fuel air mixture is basically an explosion. The point is that there isn't a nice simple answer because the reaction is fearsomely complex.

Actually most of the fuel does burn to carbon dioxide. The proportion of carbon monoxide is quite small. In a normally tuned engine there is enough oxygen to burn all the fuel to carbon dioxide but the fuel/air mixture isn't in the cylinder long enough for the reaction to go to completion. You probably know that all modern cars are fitted with catalytic convertors, and what they do is to complete the reaction as the exhaust gases pass through them.


Equilibrium constant calculation from concentrations

Well, I haven't done that in a while, but we seem to agree that the equilibrium constant is expressed as

$$ K_c = \frac{[AB]^2}{[A_2][B]^2}$$

What else do we know?

Initially, $[A_2]$ = 8.0M, $[B]$ = 10.0M
In equilibrium, $[AB]$ = 8.0M
For every mole of $B$, one mole of $AB$ is produced.
For every mole of $A_2$, two moles of $AB$ are produced.

With just one cup of coffee so far, I get

$$ K_c = \frac{8^2}{(8 - \frac{8}{2})(10-8)^2} = 4$$


The main parts of the mechanism that will affect the rate of the reaction are the following steps (full mechanism on Wikipedia): abstraction of a hydrogen to form a benzylic radical and the rearrangement of the of the protonated peroxide.
These steps will be slower as they involve the formation of the higher energy species when ethylbenzene is used.

Secondary benzylic hydrogen have a higher bond bond dissociation energies ($357.5\: \mathrm{kJ\: mol^{-1}}$) than tertiary benzylic hydrogen ($353.3\: \mathrm{kJ\: mol^{-1}}$)$^{[1]}$. This will diminish the rate at which the initiation step proceeds, though a difference of $4.2\: \mathrm{kJ\: mol^{-1}}$ shouldn't make or break the reaction.

The formation of a primary carbocation would otherwise be highly unfavorable, except for that it is bonded to an oxygen atom. A lone pair on the oxygen can therefore be shared with the carbocation, thereby stabilizing it. This step will have a higher activation energy, but likely all that will be necessary for this reaction to occur is a little more heat and time.

[1] Baciocchi, E.; D'acunzo, F.; Galli, C.; Lanzalunga, O. Tertiary : Secondary : Primary C–H Bond Relative Reactivity in the One-Electron Oxidation of Alkylbenzenes. A Tool to Distinguish Electron Transfer from Hydrogen Atom Transfer Mechanisms. J. Chem. Soc., Perkin Trans. 2. 1996, 133–140.


Are there methods for iron content determination in spinach that do not require thiocyanate?

There are iron testing kits in water which are available at an affordable price. I have not used them but read more about them here Testing iron in well water. You can test your spinach extracts for a yes no type answer if iron is present in significant quantities or not. Just keep in mind that PopEye's spinach is really not that rich in iron. Too much iron is a liver toxin.
I am afraid to share the bad news for you that with the resources you have, you cannot reliably determine iron content in spinach. Micronutrients or minerals cannot be determined by classical methods because they are present in a complex matrix and in very small quantities (parts per million range). They need instrumental analysis (absorption by molecular complexes, atomic emission or atomic absorption). I saw a couple of papers from ResearchGate which talk about the titration of spinach extract with permanganate. Nothing can be more wrong than this procedure in analytical chemistry. They are basically titrating all the organic stuff with it. Those methods are horribly wrong. Someone on SE Chemistry posted a query on how to determine iron in spinach by titration Iron in spinach...that is all incorrect. Yes, you will generate some numbers with those burette readings but those numbers have no significance.
Regarding your ashing method, that will not work either, because you need at several kgs of spinach to get a small amount (milligram) of iron. What should be the ashing temperature? How would you isolate iron from the ashes? Selective precipitation to get rid of other minerals? Nope, that is a bigger headache. For that you would need more reagents.
If this is a school project, make sure you have the resources after searching the literature. If this is an home experiment- classical methods will not work.


What makes for a good 'swimmer' in order see the filling level of Xenon gas in a vessel?

You can easily blow large, incredibly thin bubbles of soft glass with a propane torch or alcohol lamp -- so thin that interference patterns color them. Though fragile, it might be fun to make one as part of the demonstration. You'll need to experiment with surface-to-volume ratio to get it to float in Xe (you could also use halocarbon computer duster or $\ce{CO2}$, rather than expensive Xe). N.B. Use goggles, not just safety glasses, because the thin glass, when it breaks, lofts into the air... and possibly, your eye.

Rehearse, though, because removing the bubble from the tubing is likely to break it. After you get a large, thin bubble, aim the flame at the point where it connects to the tubing, and heat it until the tubing droops to a hair-thin tube. Do not melt this thin tube shut, because outside air-pressure will crush the bubble as it cools! Break that thin tube, leaving a short stem on the bubble. Yes, there will be some slow diffusion through the neck, but the bubble should last for weeks. If the demo is in a temperature-stable location, you might be able to seal the neck without the bubble busting.

There are videos showing lampwork here and here, but I suggest using soft glass (not borosilicate "Kimax" or "Pyrex") for your first bubbles, as it takes much less heat to melt.

BTW, a similar demonstration of density is the Cartesian diver, which also makes a good first student project in flamework:

With ~6 - 8 mm diameter soft glass, blow a bubble which can fit through the mouth of a plastic soda bottle.
Cut the bubble off with a triangular file, leaving a ~10 - 20 mm stem.
Optional: flame-polish the cut stem.
Dribble water into the (cooled) bubble until it just floats in a beaker of water.
Put the bubble in a filled plastic soda bottle, cap the bottle, and squeeze.

When I've had student 11-16 years old do this, it went exceedingly well (I actually had to remind them when lunch-break came that they needed to stop).


Significance of single point energy when calculating interaction energies

Single point energy arises in the framework of the Born–Oppenheimer approximation and corresponds to just one point on the potential energy surface. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-called clamped nuclei approximation.

Mathematically, if you develop the Born–Oppenheimer approximation step-by-step you can easily see that single point energy it is the sum of the electronic energy and nuclear repulsion potential energy,
$$
U = E_{\mathrm{e}} + V_{\mathrm{nn}} \, ,
$$
where the electronic energy $E_{\mathrm{e}}$ is the solution of the electronic Schrödinger equation,
$$
    \hat{H}_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) = E_{\mathrm{e}} \psi_{\mathrm{e}}(\vec{r}_{\mathrm{e}}) \, .
$$
The fact that at this point we use the symbol $U$ which (alongside with $V$) is usually used for potential energy to mean the single point energy is justified a little later. Namely, when we introduce the Born-Oppenheimer approximation which give rise to the nuclear Schrödinger equation,
$$
    \Big( \hat{T}_{\mathrm{n}} + U(\vec{r}_{\mathrm{n}}) \Big) \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) = E \psi_{\mathrm{n}}(\vec{r}_{\mathrm{n}}) \, ,
$$
it is easy to recognize that the values of the single point energy $U$ for all possible nuclear configurations define the potential energy for nuclear motion. So, it is in this sense that the single point energy is related to the potential energy.

Update: it became clear that OP misunderstood the notion of the single point energy $U$. Indeed, once we do few single point calculations for different nuclear configurations, the resulting $U(\vec{r}_{\mathrm{n}})$ is the potential energy for nuclear motion. However, it is not the interaction energy between some fragments, though, the interaction energy contributes to it. So if one wants to obtain the interaction energy one has to decompose the potential energy $U(\vec{r}_{\mathrm{n}})$ into its parts.

There different ways to perform the energy decomposition, just to name a few without any particular order:

SAPT (Symmetry-Adapted Perturbation Theory) a separate program (few of them, to be more precise) which can be interfaced with different quantum chemistry codes.
NEDA (Natural Energy Decomposition Analysis) which is available as a part of NBO package.
Morokuma decomposition already available in some quantum chemistry codes (GAMESS-US, for instance).
LMO-EDA (Localised Molecular Orbital Energy Decomposition Analysis) also available in some quantum chemistry codes (GAMESS-US, for instance).


The short answer: no, there is no such a general distance.

The long answer: atoms and molecules are not just hanging is the vacuum and waiting someone to grab them. They actually have rather high energy in translation, rotation and vibration modes and particles actually collide during a reaction some of these energies can be involved. The science that study these collisions is called reaction dynamics, and you can read e.g. Polanyi's Nobel lecture.You should note that:

It is a quantum event, so technically speaking it is not deterministic.
The orientation of the collision and the geometry of the potential energy surface along the reaction can be favourable, and rotation, vibration etc can help the reaction, or can prevent it.
The details are also influenced by the medium (is it a reaction in gas or condensed phase, on surface?) and what kind of orbitals are involved in the reaction. About frontier orbitals and their roles you also can find lot of materials in textbooks, eg. any text that discusses the work of Fukui and Hoffmann.


How is the configuration of an ester enolate double bond assigned according to CIP?

Whilst the (E)/(Z) nomenclature works fine for enolates of ketones and aldehydes, there is, as you've discovered an issue with the assignment of enolate geometry for esters (an amides and thioesters) since depending on the nature of the ester and the metal counter-ion used, we end up with a different answer.

Whilst the CIP rules are trivial to apply to any enolate, their use is not always conducive to describing the chemistry going on, especially when we end up with (E) and (Z) enolates both giving syn-adducts (for example) merely because we've changed a counter-ion (the actual spacial configuration of the enolate is identical in both cases).

To avoid this switching of configurations, we usually choose some other way of definition such that at the beginning of any thesis or SI where enolates are used, there is usually a statement that explains the way in which the author is planning to define their enolates (i.e. define them however you wish, but make sure you define them).

This is often not in agreement with formal CIP rules for the reasons outlined above, or at least not consistently in agreement with CIP rules (things may happen to end up named correctly regardless of how we define them).

This statement defining the enolate geometry is generally a re-hash of Prof Evans who stated:

  ...irrespective of the formal CIP criteria, the O-atom bearing the metal (the OM residue) is given a higher priority, and the ipso-substituent X (in enolates) the lower one...

Whilst this is widely accepted, there are some issues, mainly because there are inconsistencies between groups: Evans' calls it (E) but Myers' calls it (Z) with Furstner calling it (E) but for a totally different reason than Evans'.  For this reason alone, its always worth looking at the actual schemes and defining the enolate yourself rather than trusting that enolate X gives Y.

This issue is the source of the major objection to Evans' idea about the nomenclature. The whole reason we use E/Z/R/S is because they are well defined, I can tell you that I have (R)-..... and you'll know what I'm talking about, because we both follow the same rules, and if you don't, you're wrong.

The proposed solution to Evans' solution (Ernest Eliel) is to stop messing with these well defined descriptors, and instead go back to using cis and trans for enolate geometry, as these terms are less well defined and as such abusing them in this way doesn't lead to any significant confusion.


When surface area of electrodes is increased, why does the current in the galvanic cell increase?

Yeah, you're on the right track. In any battery reaction there is some rate-limiting step that determines the maximum power the battery can supply. Most of the possible rate-limiting steps involve the surface of the electrode.

As an example, imagine that the rate-limiting step is the diffusion of new ions from the electrolyte bulk to the cathode (or diffusion of ions away from the surface at the anode. The physics of diffusion in this case is governed by Nernst-Planck equation, where the quantity of interest is the ion flux (number of ions of unit area). However, the total current is determined by the total number of ions. To get from flux to total number of ions you have to integrate over the area of the electrode: if you double the area of the electrode you'll double the total number of ions (ie. current) if all else stays equal.

That's just one example that depends on the specifics of your cell, but many of the surface-related limited steps in a battery follow this type of behavior.


I remember learning that these two reactions proceed via different mechanisms. I'm going to use an alkenylborane to illustrate them.

Organoboranes are electron-deficient and electrophilic; when hydroxide is added together with iodide, the first step is the formation of an anionic ate complex. This is now nucleophilic via the $\ce{B-C}$ $\sigma$ bond, and can therefore react with the electrophilic iodine:

This mechanism is somewhat analogous to the hydroboration-oxidation $\ce{H2O2}$ step, where the carbon "nucleophile" is intramolecularly transferred from boron to an electrophilic oxygen atom. This step is an intermolecular transfer of carbon from boron to iodine instead.

Incidentally, I remember that with alkylboranes the migration occurs with inversion of stereochemistry, but I cannot seem to find a reference. Given the premise of the question, I would hypothesise that alkenylboranes react with retention of stereochemistry (which would not be surprising since stereochemistry at $\mathrm{sp^2}$ and $\mathrm{sp^3}$ centres, and the stereoelectronic requirements for inversion/retention, are entirely different).

As for the other reaction, hydroxide isn't added first, so iodine (electrophilic) has to react with the double bond (nucleophilic). The subsequent steps are best described by a diagram instead.

Again I am a little frustrated at not being able to find a reference, but I am fairly sure this was how I learnt it. I literally looked up all reactions of $\ce{BR3}$ with iodine on Reaxys but couldn't find anything on it, so...


Zinc is higher in the reactivity series than copper, so simply by placing a piece of zinc metal into a copper sulphate solution, you will get a displacement reaction where the zinc metal loses electrons and becomes $\ce{Zn^2+}$ ions, and $\ce{Cu^2+}$ ions gain those two electrons to become copper metal.  This will happen until the zinc electrode is coated with copper and no more zinc ions can get into the solution.

The standard half-cell potentials are measured using standard 1M solutions, and deviations from this concentration cause the equilibria to shift according to Le Chatelier's principle:

$\ce{ Cu^{2+}_{(aq)} + 2e^{-}<=> Cu_{(s)}} \quad  E^{o}=+0.337V$

Since the $\ce{[Cu^{2+}_{(aq)}]}$ is zero for the copper electrode, this system is far from equilibrium and the half-cell potential will be closer to zero.  In the other half of the system, the $\ce{[Cu^{2+}_{(aq)}]}$ is less than 1M due to the reaction with the zinc electrode, but it is still significantly greater than zero.  The result is a slight potential for copper metal to become ions and for the electrons to travel through the circuit to reduce some of of the copper(II) ions in the other half cell.  Using the Nernst equation for this isn't valid since concentrations close to 1M are too high for accurate results on one hand (concentrations greater than $10^{-3}$M deviate from experimental values), and the use of the equation is on the assumption that the ion is in equilibrium.  If we use it naively we do get something:

$E_{cell} = E^{o}_{cell} - (\frac{0.059}{2})log(Q) \approx 0-0.0295log(\frac{\ce{[0.0000000001]}}{\ce{[0.5]}}) = 0.37V$

As the approximation of the copper(II) ion concentration approaches zero the potential difference for the cell approaches infinity, which goes some way to demonstrate this is only valid within limits since you don't get a 100,000 Volts electric shock when you put a piece of metal into pure water.  Nevertheless it describes a trend and supports the Le Chatelier interpretation.

However, if the circuit is connected using a voltmeter this has a sufficiently high resistance to prevent current from flowing, so although there is potential for the copper to reduce itself through the cell, this can only happen if electrons can flow.  Even without the high resistance the reaction of the copper metal forming ions may be so slow, that very little current flows and few ions are formed which may explain the inconclusive test result.


Is it possible to create weaker version of guncotton by replacing cotton wool with wood chips or linen cloth?

This article mentions hemicellulose from wood chips being hydrolyzed, not nitrated by a mixture of ethanol and nitric acid, while cellulose does get nitrated: Isolation of cellulose from poplar wood by nitric acid-ethanol treatment and its effect on the quality of films cast from ionic liquid

I also found a patent for Preparation of wood pulp cellulose for nitration which from the name alone suggests that cellulose from wood can be nitrated (the preparation described is purely physical, no chemicals applied).

Regarding linen and cotton fabric, they will work same as cotton wool. Cotton fabric is how nitrocellulose was first accidentally discovered (wikipedia article) and linen is made of cellulose fiber just like cotton (these fabrics have 94% and 95% cellulose respectively: flax, cotton).

Tldr:

Linen fabric, cotton fabric and cotton wool work equally well. It is possible to nitrate the cellulose in wood chips. Hemicellulose will (most likely) not get nitrated. No idea about smoke production and effectiveness so far though.


Le Chatelier's Principle and Rate constants

It is true that the forward reaction rate increases as you increase pressure. However, the reverse reaction rate will increase as well, and it will increase more than the forward reaction rate.

The key here is that the system is in equilibrium; the forward and backward reaction rates are equal. Altering the pressure affects both reactions.


What would be the reaction if melting iron is put in normal water?

Unlike the other metals we usually see in the reactivity series, such as calcium, magnesium, zinc, etc., hot iron does not form iron hydroxide, instead it forms iron oxide. The reaction is:

$\ce{3Fe + 4H2O <=> FeO.Fe2O3 + 4H2}$

Iron is similar to magnesium and zinc as they react only with hot water (not cold water). But the reaction of iron with hot water is less vigorous than that of magnesium and zinc.


Oxidation states of iron in Roussin's salts

I suppose it's asked about formal oxidation numbers of iron as there is not enough information to make assumptions about the ones that reflect electronic structure.
For instance, analysis of interatomic distances could point to a delocalized electron pair, but you are not given any linear or angular geometrical parameters to begin with.

It's only the symmetry that's evident from the crystal structure: the thioferrate cluster $\ce{[Fe4S3(NO)7]-}$ belongs to $C_\mathrm{3v}$ point group, hence the only possible arrangement of four iron atoms is a tetrahedral one.
Another conclusion that can be made is that three equatorial iron atoms are equivalent ($\ce{Fe(B)}$, $\ce{Fe(C)}$, $\ce{Fe(D)}$), and the last apical one ($\ce{Fe(A)}$) most likely deviates in terms of oxidation state.

Supposing formal oxidation number ($\text{O.N.}$) of three equivalent irons and the last one are $y$ and $x$, respectively, and the fact that the average $\text{O.N.}$ is $-0.5$, the following holds true:

$$x + 3y = -0.5\cdot 4 \implies y = -\frac{x + 2}{3}$$

As formal oxidation numbers are integers, there are several pairs giving the solution:

$$
\begin{array}{l|rrrrr}
\hline
x & \ldots & -5 & -2 & \color{green}{1} & 4 & 7 & \ldots \\
\hline
y & \ldots & 1 & 0 & \color{green}{-1} & -2 & -3 & \ldots\\
\hline
\end{array}
$$

among which the least "extreme" (e.g. the one with the absolute lowest negative $\text{O.N.}$) for a metal would be the $-1;+1$ pair; so that for $\overset{0}{\ce{Fe}}:~[\ce{Ar}]\,(\mathrm{3d})^6\,(\mathrm{4s})^2$ your answer gives two configurations, $(\mathrm{3d})^7$ for $\overset{+1}{\ce{Fe}}\ce{(A)}$ and $(\mathrm{3d})^9$ for $\overset{-1}{\ce{Fe}}\ce{(B,C,D)}$.


Carbon forms large molecules such as the ring structure of graphite, or tetrahedrons of diamond. Boron, like carbon, forms relatively large covalently-bonded molecules.

Nitrogen, oxygen and fluorine are diatomic, so molecules are less massive. The inert gases He and Ne are monatomic and therefore have lower b.p. yet.

In a similar fashion, metallic bonds hold together Li, Be and other metals.


What is the preferred IUPAC name of o-toluidine?

In principle, your rationale is correct.

Generally, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes. Since o-toluidine (note that this name is no longer recommended) contains only one characteristic group $(\ce{{}-NH2})$, it is named as amine.

According to Subsection P-62.2.1.1.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained name “aniline” is the preferred IUPAC name (PIN) for the functional parent compound. The systematic name “benzenamine” may be used in general nomenclature.

For the retained name “aniline”, substitution is permitted at any position. Substitution of such functional parent compounds is limited to substituent groups having a lower seniority than that denoted by the (explicitly expressed or implied) suffix of the functional parent compound. Anyway, in this case, this limitation is not relevant since o-toluidine contains only one characteristic group and carbon compounds (rings or chains) have a lower seniority than amines.

Therefore, the PIN for o-toluidine is “2-methylaniline”. The systematic name “2-methylbenzenamine” may be used in general nomenclature.


Last step of nicotine extraction using DCM

If you are worried only with DCM, then you should be fine since it is quite volatile. To be sure you have no DCM left in your nicotine leave it for a few days in a closed recipient filled with some dried silica gel or calcium chloride (here in my country those are easily found, even for domestic use), and if you are in a proper laboratory just leave it in the desiccator with the vacuum on for a few hours.

However, nicotine is hard to isolate since other alkaloids from tobacco are also extracted in the most commonly used methods. I don't know which extraction methodology you used but just for comparison, here are two papers with some information for isolating nicotine:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3263647/
https://www.ncbi.nlm.nih.gov/pubmed/19135499


How to find the moles of carbonic acid I need to add to make a solution of KOH acidic

You have not stated the volume of your $\ce{KOH}$ solution. I therefore have to make some assumptions.

I will use $\mathrm{p}K_\mathrm{a1} (\ce{H2CO3}) = 6.37$. I further assume you have $\pu{1 L}$ of the $\ce{KOH}$ solution and that this volume will not change by the addition of the carbon dioxide.

For the following, we have also to assume that all the solid carbon dioxide will react, i.e. no evaporation of any gas -- tricky, but perhaps not impossible.

You start with a solution of KOH that has the $\mathrm{pH} = 13.2$. From this we calculate $\mathrm{pOH}$:

$$\mathrm{pOH} = 0.8 \quad \to \quad [\ce{OH-}] = \pu{0.16 M}~(\text{or}~\pu{0.1584 M})$$

First, we suppose that the reaction between the solid carbon dioxide
and the solution of KOH will result in only potassium hydrocarbonate:

$$\ce{CO2 + OH- -> HCO3-}$$

OK, how much carbon dioxide is needed to transform all hydroxide ions into potassium hydrocarbonate? You will need (per liter $\ce{KOH}$): $\pu{0.16 mol}$.

After addition of $\pu{0.16 mol}$ carbon dioxide we will face the same situation as if you had dissolved $\pu{0.16 mol}$ of potassium hydrogencarbonate in $\pu{1 L}$ of water. The $\mathrm{pH}$ of such a solution will still be alkaline $(\mathrm{pH} > 8)$.

You wanted the solution to be acidic, but you did not specify the pH. I suppose you will be happy with $\mathrm{pH} = 6.37$.

Accordingly, we have to add more carbon dioxide. The charge balance is:

$$\ce{[H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO3^2-]}$$

At $\mathrm{pH} = 6.37$, both $[\ce{OH-}]$ and $[\ce{CO3^2-}] << [\ce{HCO3-}]$.
Therefore, we re-write the charge balance as:

$$\ce{[H3O+] + [K+] = [HCO3-]}$$

We know the values of $[\ce{H3O+}]$ and $[\ce{K+}]$, so we calculate:

$$[\ce{HCO3-}] = 10^{-6.37} + 0.16 \approx \pu{0.16 M}$$

We can now calculate $[\ce{H2CO3}]$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} \quad \to \quad [\ce{HCO3-}] = [\ce{H2CO3}] = \pu{0.16 M} \quad \to \quad [\ce{HCO3-}] + [\ce{H2CO3}] = \pu{0.32 M}$$

Accordingly, in total we need $\pu{0.32 mol} \approx \pu{0.3 mol}$ of carbon dioxide to reach $\mathrm{pH} = 6.37$. Per liter KOH solution this will correspond to $\pu{0.3 g} \times \pu{44 g mol-1} = \pu{13 g}$ solid carbon dioxide.


Buffer Solution Problem (Acid/Conjugate Base)

There is no such rule that you need equal concentrations of weak acid and salt of conjugate base to make a buffer. (In this case acidic buffer)

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

In this case, if the solution contained equal molar concentrations of both the acid and the salt, it would have a $\mathrm{pH}$ of $\mathrm{4.76}$ because $\mathrm{pK_a}$ of acetic acid is $\mathrm{4.76}$. It doesn’t matter what the concentrations were, as long as they were the same, the $\mathrm{pH}$ remains at $\mathrm{4.76}$.

You can change the pH of the buffer solution by changing the ratio of salt to acid, or by taking a different acid and one of its salts.

Note: The $\mathrm{pH}$ of a buffer is calculated using Henderson-Haselbalch equation, which is,
$$\mathrm{pH = pK_a + log\frac{[Conjugate Base]}{[Acid]} }$$

Also, as indicated by @VicNeal in the comments, the buffer solution has highest buffer capacity at equal concentrations of the constituents, meaning its more effective at resisting the change in $\mathrm{pH}$.


Wavelength and Colour in Coordination complexes

Low spin (formed from strong field ligands) complexes result in a larger difference in energy between the $t_{2}g$ and $e_{g}$ orbitals. High spin (caused by weak field ligands) complexes are, evidently, the opposite.
This means thats high spin complexes require lower energy photons than low spin. Since wavelength and frequency are inversely proportional, high spin complexes require lower frequency (longer wavelength) photons to promote transitions.


Equilibrium Constant in terms of Molar Concentration of Gases versus Equilibrium Constant

I believe that the real $K_c$ would be higher. This is because in ideal gases london dispersion forces, hydrogen bonding, etc. is ignored. These would increase the number of collisions between particles since they would be attracted to each other more.
$$\ce k=zp*e^{-E_a/RT}$$
where k= rate constant, z= frequency of collisions, p= fractions of collisions with proper orientation to cause a reaction, $E_a$= activation energy, T= temperature. So Since z would increase, the K would increase. Hence the $K_c$ would increase.


What is the carbon dioxide content of a soda can or bottle?

General estimates have placed a can of Coca-Cola to have 2.2 grams of $\ce{CO2} $ in a single can. As a can is around 12 fluid ounces, or 355 ml, the amount of $\ce{CO2}$ in a can is:
$$\text{2.2 g} \ \ce{CO2}* \frac{\text{1 mol} \ \ce{CO2}}{\text{44 g} \ \ce{CO2} } = 0.05 \ \text{mol}$$
$$ \text{355 mL} * \frac{\text{1 L}}{\text{1000 mL}} = 0.355 \ \text{L} $$
So here we can see we have about 0.05 mol/0.355 L or about 0.14 mol of carbon dioxide per liter of soda. Of course this value varies by manufacturer, type of drink, container, etc.
Looking at Wikipedia, inside Coca-Cola is:

Carbonated water, Sugar (sucrose or high-fructose corn syrup depending on country of origin), Caffeine, Phosphoric acid, Caramel color (E150d), Natural flavorings
A can of Coke (12 fl ounces/355 ml) has 39 grams of carbohydrates (all from sugar, approximately 10 teaspoons), 50 mg of sodium, 0 grams fat, 0 grams potassium, and 140 calories.

Thus, we can calculate the pressure of $CO_2$ gas using the Ideal Gas equation if we store our coke at, say, 20 Celsius:
$$ P = \frac{nRT}{V} $$
$$ P = \frac{\text{0.05 mol} * \text{0.08206} \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K} } * \text{293.15 K}}{\text{0.355 L}}$$
$$ P = 3.39 \ \text{atm} $$
According to this website, http://hypertextbook.com/facts/2000/SeemaMeraj.shtml:

On average, the 12 ounce soda cans sold in the US tend to have a pressure of roughly 120 kPa when canned at 4 °C, and 250 kPa when stored at 20 °C.

$$\text{3.39 atm} * \frac{\text{760 torr}}{\text{1 atm}} * \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 342.66 \ \text{kPa}$$
Water vapor exerts it's own partial pressure. Looking at standard tabulated values for water vapor pressure, water exerts a pressure of 17.5 torr at 20 Celsius.
$$\text{17.5 torr}* \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 2.3275 \ \text{kPa} $$
Knowing that our total pressure is the sum of all our pressures:
$$P_{total} = \text{342.66 kPa} + \text{2.3275 kPa} = 344.99 \ \text{kPa}$$
Here, we are roughly about 100 kPa off from the data provided by the website. This is just an approximation. A more accurate way would be to calculate the moles of each product inside the soda, and knowing the total pressure or partial pressure of one of the parts, we can calculate the pressures more accurately. However, that information is proprietary. It's their secret recipe!
Headspace:
http://extension.psu.edu/food/preservation/news/2012/why-allow-headspace
We have to determine the volume of the headspace - again, I am not sure of exact data - which is between 1/2 inch to 1 and 1/2 inches depending on the container and what it holds. I will assume that the headpsace occupies 6% of the total volume of the can which is 21.3 mL.
At manufacturing and at storage, the can is at different temperatures. Taking the above data, we'll say it is manufactured at $4^\circ C$ and say, stored at $20^\circ C$
Furthermore, when carbon dioxide is solubilized in water, it forms carbonic acid. I will neglect that as the ionization constant is small.
Assuming our 2.2 grams of carbon dioxide is the maximum amount of carbon dioxide that can be placed inside, some of the carbon dioxide is soluble in water while the rest exerts pressure inside the headspace to force the carbon dioxide inside the liquid. The pressure is necessary inside this closed volume as, once you open the cap, the carbon dioxide tries to achieve equilibrium.
$$ \ce{CO_2 (solution) <=> CO2 (g)} $$
In general, the solubility of gases increases at lower temperatures and decreases at higher temperatures. A notable exception is the noble gases. In regards to pressure, Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.
The solubility of carbon dioxide in water shifts according to Le Chatlier's principle.
For the solution to this problem, we would need to know several things about the manufacturing of soda's, or rather, more quantitative data. At manufacturing, the pressure is high and temperature low, so as much carbon dioxide as possible is solubilized in the water.
Once shipped out and stored, the pressure inside the headspace increases as the pressure is decreased from manufacturing, the carbon dioxide then leaves solution and enters the headspace. Increasing temperature also decreases solubility.
If you can provide me with more information, I will be more than happy to help. :)


Does intramolecular hydrogen bonding occur in dicarboxylic acids?

No, this would not happen the way your second drawing depicts because the hydrogens do not orient themselves that way. Intramolecular hydrogen bonding would be no more feasible in ethandioic acid than any other monocarboxylic acid. See picture below:

Source: http://commons.wikimedia.org/wiki/File:Oxalic-acid-3D-balls-B.png


Why can an s orbital only hold two electrons?

Electrons are magnets, they have magnetic fields. Those fields have only two possible orientations, and a single orbital can only be occupied by two electrons if those orientations are mutually opposed.

OK, but why? Why are electrons magnets? Why are there only two possible orientations for the fields? Why can the two electrons in the orbital only exist with opposite directions for the magnetic moments?

Well, there are answers to those questions, but they don't make sense a lot of sense. The magnetic moment of electrons is due to a property called "spin"; the spin value of an electron is 1/2, so it can adopt one of two spin quantum states, +1/2 and -1/2, which correspond to the "up" and "down" orientations of the magnetic moment. Since it has a half-integer spin, an electron is a member of a class of sub-atomic particles called "fermions" which obey rules called the Pauli Exclusion Principle and Fermi-Dirac statistics - one key result of these rules is that no two identical fermions can simultaneously occupy the same quantum state: you cannot have two electrons in the same orbital if they have the same spin orientation, so if one is +1/2, the other must be -1/2, and no more can be added because a third would have to adopt the same quantum state as the one of the first two.

OK, but why? Why does an electron have a spin value of 1/2? Why can't two fermions adopt the same quantum state (when integer spin particles, called bosons, can do so)? Why does the property of "spin" even exist? You need quantum mechanics to even begin to answer those questions -- again, that's more math than anybody is prepared to show you for at least five years -- and even then, the "why" of the Pauli Exclusion Principle is an empirical observation (it has no theoretical basis, it's just an after-the-fact thing that happens) without relativity, but there is a firm theoretical rationale and a reason "why" if one revisits the postulates of quantum mechanics and takes into general relativity into account. But relativistic quantum field theory is more math than any mortal should be forced to endure.

OK, but why? Why is quantum mechanics how the universe behaves? Why is general relativity the way the universe behaves?

You can play the "why" game for a very long time. None of your high school teachers will be able to give you more than one or two repetitions. University professors will be able to go a lot farther. Eventually, you get to questions to which the only possible answer is "nobody knows.... yet". And those are the questions that scientists -- including those university professors -- spend their day trying to answer.

Which is terrific fun. So keep asking "why", even to the answers your teachers give you. And if they can't answer, go ask someone else, or read about it, or go to university. And if they don't know, be a scientist, and figure it out yourself, and tell the rest of us.


Can physical absorption happen despite the formation of products?

When you consider the dissolution of a gas like $\ce{SO3}$ , $\ce{CO2}$, $\ce{SO2}$ etc. there are three equilibria existing at the same time:
$$\ce{SO2(g)\overset{H2O}{<=>}SO2(aq)\overset{H2O}{<=>}H2SO3(aq)<=>H+(aq) +HSO3-(aq)}$$
The first equilibrium is the dissolution equilibrium of the gas, the second is the hydration equilibrium of the dissolved gas, and the third is the dissociation equilibrium of the acid.
The pKa associated with the first deprotonation of $\ce{H2SO3}$ is 1.81 from Wikipedia, so you can pretty much consider the last equilibrium as completely dissociated in dilute solutions.
You can consider $\ce{SO2(aq)}$ as the state where the $\ce{SO2}$ molecules are physically adsorbed in the solution, the rest exist as $\ce{H+(aq)}$ and $\ce{HSO3-(aq)}$.
How much is physcially adsorbed and how much is chemically dissociated would depend on the particular compound. Unfortunately, I cannot find any data on $\ce{SO2}$. However, in case of dissolving $\ce{CO2}$ in water, 99% exist as $\ce{CO2(aq)}$ i.e. the physically absorbed gas, and 1% as $\ce{H2CO3(aq)}$, $\ce{H+(aq)}$ and $\ce{CO3^2-(aq)}$. (reference)
So, the absorption of $\ce{SO2}$ is partly chemical and partly physical, but note that all of these reactions are reversible. Also keep in mind that the distinction between physical and chemical change is not clear-cut, and gets less and less useful the more you delve into chemistry.
The website you mention in the question gets a lot of other things wrong, for example, dissolution of $\ce{SO3}$ in water gets you sulfuric acid, which is quite stable and hygroscopic, so you would be hard pressed to reverse it, but the table lists it as physical absorption. Whereas chlorine does not dissolve that well in water, and you can drive it off by heating, but it's listed as a reversible chemical change. It doesn't make much sense to me, so I would advise you to get a more reliable source of information.


Since you want to remove the tarnished silver ($\ce {Ag2O}$), you would convert it into silver metal (or reduce it). Therefore, you would oxidize the aluminum ($\ce {Al -> Al^3+}$). The silver and aluminum need to be in physical contact and the sodium chloride or the carbonated water is just providing the medium to transfer the ions; the aluminum does not dissolve first.


What is meant by the "probability density" of finding an electron?

In quantum mechanics we often need to find the probability at some position $x$. However, $x$ is a continuous variable with an infinite number of values so it does not make sense to ask what is the probability of being at some exact position , say $x=1/4$, as this will be vanishingly small. Instead we imagine that the probability is that of being at position x to $x + \mathrm{d}x$ where $\mathrm{d}x$ has some very small value. If $f(x)\,\mathrm{d}x$ is that probability then $f(x)$ is the probability density which must be a real non-negative number. It turns out that quantum mechanics postulates that the probability density is given in terms of the wavefunction is $|\psi|^2$.


Disposing of hazardous waste indicated by numbers on chemicals

The numbering clearly is not a part of the international GHS (Globally Harmonized System of Classification and Labelling of Chemicals) labeling scheme.

On a side note, GHS requires

hazard pictograms,
coded hazard statements (Hnnn and EUHnnn)
coded precautionary statements (Pnnn)

but typically does not give any information on the how/where to dispose chemicals.

For classroom settings, the questions are:

How many different waste containers are needed?
How to label them?
Which chemicals go where?

To my knowledge, a global or EC-wide (European Community) scheme does not exist for this and I'd be astonished to find that the African Union has agreed upon a common scheme.

Instead, this is typically regulated on a nation-wise basis only.

To give an example: In Germany, the classification scheme is provided by the Deutsche Gesetzliche Unfallversicherung (mandatory accident insurance) and you may find their suggestions for waste disposal in schools here (PDF in German, Stoffliste zur Regel
„Unterricht in Schulen
mit gefährlichen Stoffen“, 2017).

Consequently, without knowing the (national) origin of the source you are citing in your question, it is very difficult to say which number refers to which compound class.


Why does benzaldehyde not respond to Fehling's test?

The rate-limiting step of the Fehling’s test reaction with aldehydes is the formation of the corresponding enolate:

The subsequent reaction of the enolate with copper(II) proceeds through a single electron transfer mechanism.

Aldehydes that lack alpha hydrogens, such as benzaldehyde or pivalaldehyde (2,2-dimethylpropanal) cannot form an enolate and thus do not give a positive Fehling’s test result under usual conditions.


In general, to break one bond vs another on demand in the presence of many others  is a very difficult thing to achieve. Not possible at all by heating molecules thus there has been much effort to do so using lasers.

But this becomes very difficult also. This is  because on exciting a molecule into an excited electronic state, such that it has enough energy to break a bond, and into the type of vibration involving the bond you wish to break,  the energy then leaks out to other different vibrational normal modes and can then end up anywhere. So it does not just stay where the energy was initially placed.

This 'leaking' occurs due to vibrational anharmonicity which provides a 'coupling' with which one normal mode can swap energy with another. The process can be very fast and even if femtosecond laser pulses are used energy still moves all over the place and competes with accumulating energy into the 'victim' bond effectively defeating the process. The area is, however, one of very active study so progress in this area is to be anticipated.


Sodium cyanoborohydride and iminium reductions

I believe the reaction effected in the third step is a reductive amination. In this case, the ketone and amine moieties are both present within the same molecule, so the reaction proceeds intramolecularly. The $\ce{TsOH}$ is an acid, presumably serving to protonate the carbonyl oxygen and activate the carbon toward nucleophilic attack by the amine. As the iminium group forms (with concomitant loss of water), it is rapidly reduced to an amine by the $\ce{NaBH3CN}$. After workup, the final product is:


This all falls back to the hydrogen absorption spectra. There are multiple "series" discovered by different scientists. The series in the visible region is the Balmer series.

Since the electronic energy levels of the hydrogen atom are quantized, there can only be certain discrete transitions.

As it turns out, if you solve the Balmer series (n=2), you get:

$$\frac{1}{\lambda} = \frac{4}{B}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) = R_\mathrm{H}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \quad \mathrm{for~} n=3,4,5,...$$

You will then get lines at $\lambda$ = 656.3 nm, 486.1 nm, 434.1 nm, 410.2 nm...

Only a few of those fall in the visible. The Lyman series (n=1) occurs in the UV spectra, and the Paschen series (n=3) occurs in the IR.


This turns out to be very difficult to do for two reasons. One problem is that the voltage of a silver cell is different from that of Ni-Cd cell, so it would require a specialized, multi-cell configuration to emulate the voltage characteristics of the Ni-Cd rechargeables.

The other problem is that silver cells generally have a sophisticated frame inside of them that is produced by an intricate high-temperature welding process. To duplicate this process and produce a suitable frame would require a significant amount of experimentation and work.


First note that the process is an irreversible adiabatic expansion. If it were a reversible one, the gas has to expand against a varying pressure because pressure of gas decreases during expansion and this varying gas pressure must be matched with external pressure in all infinitesimal steps of expansion. "The gas expanded against constant pressure" ⇒ the gas must have a pressure which is always higher than the constant external pressure during the expansion ⇒ irreversible.
Will entropy definitely increase?

ΔSsur = 0 as q = 0. Irreversible ⇒ ΔStot > 0. ΔStot = ΔSsys + ΔSsur = 0 + ΔSsys = ΔSsys. So ΔSsys > 0, so entropy of system will definitely increase.

Will compressibility factor definitely increase?

In adiabatic irreversible expansion all p, V, T changes. If Z is to increase, the increment in pV must be greater than the increment in T, or the magnitude of decrement in T must be greater than that in pV, which cannot be sure will happen as all p, V, T change.

What did you do wrong?

First of all, TV equation is for perfect gas doing reversible adiabatic expansion only. Second of all, a gas expands against constant external pressure does not mean the pressure of gas does not change. All p, V, T will change. You said as volume increases, temperature decreases, but pressure of gas will decrease too.


How exactly can an increase temperature be used to break the chemical bonds of a gas [SO2, for example]?

As we provide heat to any molecule, the energy of the molecule increases. Specifically, the energy of atoms increases. Using the fact that stability is inversely proportional to energy, we can conclude that as energy increases, the bonds between the atoms become more unstable and eventually break and constituent atoms disassociate.

This heat provided is known as bond disassociation enthalpy, or bond disassociation energy [BDE].


Why is ammonium a weak acid if ammonia is a weak base?

First, let’s get the definition of weak and strong acids or bases out of the way. The way I learnt it (and the way everybody seems to be using it) is:

$\displaystyle \mathrm{p}K_\mathrm{a} < 0$ for a strong acid
$\displaystyle \mathrm{p}K_\mathrm{b} < 0$ for a strong base
$\displaystyle \mathrm{p}K_\mathrm{a} > 0$ for a weak acid
$\displaystyle \mathrm{p}K_\mathrm{b} > 0$ for a weak base

Thus strong acid and weak base are not arbitrary labels but clear definitions based on an arbitrary measurable physical value — which becomes a lot less arbitrary if you remember that this conincides with acids stronger than $\ce{H3O+}$ or acids weaker than $\ce{H3O+}$.

Your point of confusion seems to be a statement that is commonly taught and unquestionably physically correct, which, however, students have a knack of misusing:

  The conjugate base of a strong acid is a weak base.

Maybe we should write that in a more mathematical way:

  If an acid is strong, its conjugate base is a weak base.

Or in mathematical symbolism:

  $$\mathrm{p}K_\mathrm{a} (\ce{HA}) < 0 \Longrightarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1}$$

Note that I used a one-sided arrow. These two expressions are not equivalent. One is the consequence of another. This is in line with another statement that we can write pseudomathematically:

  If it is raining heavily the street will be wet.

  $$n(\text{raindrops}) \gg 0 \Longrightarrow \text{state}(\text{street}) = \text{wet}\tag{2}$$

I think we immediately all agree that this is true. And we should also all agree that the reverse is not necessarily true: if I empty a bucket of water on the street, then te street will be wet but it is not raining. Thus:

  $$\text{state}(\text{street}) = \text{wet} \rlap{\hspace{0.7em}/}\Longrightarrow n(\text{raindrops}) \gg 0\tag{2'}$$

This should serve to show that sometimes, consequences are only true in one direction. Spoiler: this is also the case for the strength of conjugate acids and bases.

Why is the clause above on strength and weakness only true in one direction? Well, remember the way how $\mathrm{p}K_\mathrm{a}$ values are defined:

$$\begin{align}\ce{HA + H2O &<=> H3O+ + A-} && K_\mathrm{a} (\ce{HA}) = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{HA}]}\tag{3}\\[0.6em]
\ce{A- + H2O &<=> HA + OH-} && K_\mathrm{b} (\ce{A-}) = \frac{[\ce{HA}][\ce{OH-}]}{[\ce{A-}]}\tag{4}\end{align}$$

Mathematically and physically, we can add equations $(3)$ and $(4)$ together giving us $(5)$:

$$\begin{align}\ce{HA + H2O + A- + H2O &<=> A- + H3O+ + HA + OH-}&& K = K_\mathrm{a}\times K_\mathrm{b}\tag{5.1}\\[0.6em]
\ce{2 H2O &<=> H3O+ + OH-}&&K = K_\mathrm{w}\tag{5.2}\end{align}$$

We see that everything connected to the acid $\ce{HA}$ cancels out in equation $(5)$ (see $(\text{5.2})$) and thus that the equilibrium constant of that reaction is the autodissociation constant of water $K_\mathrm{w}$. From that, equations $(6)$ and $(7)$ show us how to arrive at a well-known and important formula:

$$\begin{align}K_\mathrm{w} &= K_\mathrm{a} \times K_\mathrm{b}\tag{6}\\[0.6em]
10^{-14} &= K_\mathrm{a} \times K_\mathrm{b}\\[0.6em]
14 &= \mathrm{p}K_\mathrm{a} (\ce{HA}) + \mathrm{p}K_\mathrm{b} (\ce{A-})\tag{7}\end{align}$$

Now let us assume the acid in question is strong, e.g. $\mathrm{p}K_\mathrm{a} (\ce{HA}) = -1$. Then, by definition the conjugate base must be (very) weak: $$\mathrm{p}K_\mathrm{b}(\ce{A-}) = 14- \mathrm{p}K_\mathrm{a}(\ce{HA}) = 14-(-1) = 15\tag{8}$$

Hence, our forward direction of statement $(1)$ holds true. However, the same is not true if we add an arbitrary weak acid to the equation; say $\mathrm{p}K_\mathrm{a} (\ce{HB}) = 5$. Then we get:

$$\mathrm{p}K_\mathrm{b} (\ce{B-}) = 14-\mathrm{p}K_\mathrm{a}(\ce{HB}) = 14-5 = 9\tag{9}$$

A base with a $\mathrm{p}K_\mathrm{b} = 9$ is a weak base. Thus, the conjugate base of the weak acid $\ce{HB}$ is a weak base.

We realise that we can generate a weak base in two ways: by plugging a strong acid into equation $(7)$ or by plugging a certain weak base. Since the sum of $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$ must equal $14$, it is easy to see that both cannot be strong. However, it is very possible that both the base and the acid are weak.

Thus, the reverse statement of $(1)$ is not true.

$$\mathrm{p}K_\mathrm{a}(\ce{HA}) < 0 \rlap{\hspace{1em}/}\Longleftarrow \mathrm{p}K_\mathrm{b} (\ce{A-}) > 0\tag{1'}$$


What is the definition of of 'compound', 'mixture', 'element' and 'molecule'?

I can't really provide a systematic approach, but I can attempt to clarify (as a student myself).

Elements are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of Isotopes which are physical variations among atoms of the same element).
A definite (I suppose, systematic) way to distinguish elements is that their corresponding atoms have different numbers of electrons orbiting the nucleus.
The term is also used to describe a collective of the same atom (element). A bar of gold (Gold being an 'element'; a chemical class) is said to be an element itself.
Molecules are a group of atoms covalently bonded to each other (which can be considered a 'direct connection', if you will). The molecule can consist of atoms of the same element, or atoms of different elements.
If you joined to molecules via covalent bonding, you'd have created a new molecule (it's nothing special)
Compounds are (in the chemical sense of the term) made of two or more elements. This means a lot of molecules (like H2O) are considered compounds (though H2 is not). 'Compound' is often used interchangeably with the word substance, which describes a collective of molecules (water is a substance. It is a large quantity of neighboring H2O molecules).
A substance (and a compound, when a compound is used to describe a quantity of molecules) only consists of the same molecule. In water, all the molecules are H2O. If there were also some H2 molecules among them, it would no longer be considered a compound/substance (but instead; a mixture).
These molecules are not directly bonded to each other (or they'd be one big molecule), they are just in close* proximity to one another and are attracted to one another, keeping them in proximity. Even when the substance is a solid, like ice, they are not directly connected, the molecules are just slower (less energetic) and closer together (and in some special cases, like ice, stronger, such as Hydrogen Bonding. Don't be confused, though it is referred to as 'bonding', there is no direct covalent bond, just an attraction).
Mixtures are collectives of different molecules or atoms. The molecules and atoms are like that in a substance; they're not bonded to each other, but are attracted, and in a mixture specifically, the molecules/atoms are not all the same. Juice, for instance, is a mixture. It contains some Vitamin C molecules, some water molecules, some sugar molecules, etc.

Here is a horribly drawn diagram to help clarify the distinctions.

Reflecting on this, here is what in your understanding needs specific attention.

Compound: two or more different atoms bonded together A molecule or a group of identical molecules.
Mixture: two or more different atoms or molecules together but not joined (they're not covalently bonded).
Molecule: two or more atoms(same or different) bonded together.


Home chemistry can be difficult sometimes due to the expense of the essential working materials and instruments required. As a mathematician all I need is a sharp pen (which is sometimes still quite difficult to obtain!) and blank paper. I can give you the following general (personal) advice regarding purchasing chemistry lab equipment:

Determine your budget and the needs for the experiments you want to perform. Do you really need a scale that has to weight with an accuracy of 0.001g or is 0.01g sufficient (or even 0.1g)? Ask yourself what you REALLY need for an experiment BEFORE searching for instruments. If you do it in the converse order you often buy something way more fancier (and pricier) than what you actually needed in the first place.
Look at specialized chemistry equipment supply websites to orientate on what is available and in what price range certain products are (think of scales, chemicals, glassware, etc.)
A lot of stuff can be bought incredibly cheap on websites where you not expect (at least I did not) people to sell lab supplies/instruments. Think of ebay.com and amazon.com as examples. They sometimes list items at a fifth of the usual price that would have to be paid at a specialized store.
Take a look at local non-specialized stores like the construction market (you know, the place where they sell all kinds of construction material). They sometimes hide hidden (cheap) gems! As an example I was able to obtain a vacuum source (membrane pump) for < 100\$, which would have cost me easily > 400\$ at a chemical supply website.

About your particular question: I really recommend a 0.01g precision 200g (or 600g) scale. It brings the benefit of having relatively accurate weighting for even small samples, while still being able to weight bulk mass. Furthermore, they are most common in the lab/home environment and hence relatively cheap in comparison with professional analytical scales. If you plan to do a lot of microscale experiments (yielding a product < 200 mg) the analytical scale (precision of 0.001g) may suit your needs better.

I wish you much success, think and buy wisely!


Where can I find example MS, IR, NMR spectra for teaching molecular spectroscopy?

An elder way would be the via book examples, like «Organic Structures from Spectra» by Field, Sternhell, Kalman, with dozens and dozens of combined exercises in a format like the following:

and currently available in the 5th edition by Wiley (e.g., here). There is a  related project of «Interactive Spectroscopy Problems» by the University of Calgary (here), too.

German VCH published years ago Peter Bigler's training course «NMR-Spectroscopy: Processing Strategies» (e.g., here) including a CD full of 1D and 2D NMR data, which comes close to your aim of interactive sessions to process fids till the plot.  Bruker's WIN-NMR included is old (think Windows 95 era, could be still 16bit), but the experiment-based NMR data should still be processed by current software.

Almost forgotton:  NIST webbook chemistry has for many compounds which may qualify as «easy» IR, UV (condensed phase) and mass spectra, too.  While not aiming to be a teaching set, there are options to download the spectra not only as picture, but equally in the JCAMP-DX format to process, e.g., benzaldehyde.


Relation between oxoanion acidity and number of oxygen double bonds

The strength of an acid is determined by how far the reaction

$$
\ce{HA} \rightarrow \ce{H^+} + \ce{A^-}
$$

proceeds. One of the products will always be the same for all acids ($\ce{H+}$), so no influence there. The other product is the anion ($\ce{A-}$) and the basic question is: "How stable is that anion?", or equivalently, "how well can that anion stabilize its negative charge?"

Many factors go into answering that question, the one under consideration in the rule alluded to in the question is: Are there atoms surrounding the central atom that can stabilize the charge effectively? Oxygen atoms bound only to the central atom (in the protonated form) are a prime example, because oxygen is a very electronegative element. (Another example not involving oxyanions/oxyacids is trifluoroacetic acid, which is more acidic than acetic acid.) The more such oxygens are in your acid, the stronger it will be - $\ce{H2SO3}$ is weaker than $\ce{H2SO4}$.

Jan commented on the question that double bonds between the central atom and oxygens only exist in the first row of main group elements (i.e. $\ce{C},\ce{N}$). This is true, however, the rather basic rule cited in the question does not go into this level of detail and simply assumes that the outdated picture of double bonds in say, $\ce{ClO4-}$, is correct.


According to the Iowa State University page Carbon Monoxide Poisoning: Checking for Complete Combustion (AEN-175) suggest the following mechanisms that cause incomplete combustion:

  Insufficient mixing of air and fuel.
  Insufficient air supply to the flame.
  Insufficient time to burn.
  Cooling of the flame temperature before combustion is complete.

The main notion is the lack of air to cause complete combustion (either through lack of supply, mixing or time for combustion).

However, it should be noted that, practically speaking, according to the ChemWiki page Burning Alkanes, the size of the hydrocarbon has an affect in 'normal' conditions:

  Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.

The reason being (according to ChemWiki) is:

  If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.

So, the larger the hydrocarbon, the less likely it would vaporise sufficiently, thus less chance of the optimal mixing of air and fuel, resulting in incomplete combustion.

However, provide the right conditions with sufficient supply of air and optimal hydrocarbon-air mixing ratio, it is still possible to cause complete combustion, in the general formula (for many hydrocarbons):

$$\ce{C_{x}H_{y} +O2 ->H2O + CO2}$$

A quite comprehensive resource explaining each of the factors is the University of Tulsa chapter Fuels and Combustion.


How to read mutual solubility curves of salts with a common ion?

My interpretation is:
I assume both cross solubilities are related to the volume of water used for the initial solution, not of the initial solution volume, i.e. both in $\pu{g}/\pu{100 mL}$ of water.
If you start at $x$-axis with the $\ce{KCl}$ solution, you can dissolve $\ce{NaCl}$, climbing upwards until you reach the curve for the given temperature. If you cross it, some $\ce{NaCl}$ will not dissolve, and some $\ce{KCl}$ may precipitate, because of the common ion effect.
The same of the initial $\ce{NaCl}$ solution, starting on $y$-axis, and dissolving $\ce{KCl}$, going to the right, again until you reach or cross the curve.
$\ce{KCl}$ solubility is lower, if some extra $\ce{Cl-}$ ions from $\ce{NaCl}$ are present, because of the solubility product $K_{\mathrm{sp}, \ce{KCl}} = a_\ce{K+} \cdot a_\ce{Cl-}$.
Similarly for $\ce{NaCl}$,  $K_{\mathrm{sp},\ce{NaCl}} = a_\ce{Na+} \cdot a_\ce{Cl-}$.
Problem is difficulty with both measuring and predicting ion activities or activity coefficients for concentrated ionic solution. Replacing activities with concentrations could be done only with lot of grains of salt and both eyes closed tightly.
I do not have clear interpretation of the about horizontal edge in the upper chart part. It looks like a kind of "precipitation eutecticum", an analogue to a binary mixture phase chart.

Feedback to comments:
If the system is at the cross-saturation curve, and there is excess of one solid salt, then due dynamic nature of processes, this salt dissolves and the other precipitates. The system is moving slightly along the curve, until for each of salts the rates of dissolving and precipitating are equal. This process is slow.
If a cross-saturated solution cools down, both salts precipitate, according to the ratio of kinetic rates of both processes, until the curve for new temperature is reached.
The ratio can be only very roughly estimated from

the chart position
the direction of the normal vector of the curve
the different ion mobility.


how do i synthesize (E) 2-methyl-1-phenylpent-1-en-3-one?

Looks like a crossed aldol condensation between benzaldehyde and pentan-3-one.

In a classical approach, you would (under stirring)

Add your base to a dry solvent of your choice
Cool down (ice bath or dry ice/isopropanol)
Add pentan-3-one dropwise to make sure that it's all immediately converted to the corresponding enolate. You do not want self-condensation of the ketone!
Let it come to room temperature (and stir for a while)
Cool down again
Slowly add benzaldehyde. Freshly distilled benzaldehyde is recommended.
Let it come to room temperature again (and stir for a another while)

Check the progress of the reaction by TLC. You should have at least the aldol addition product by now. In order to facilitate elimination of water and thus complete the aldol condensation, it might be necessary to acidify the mixture and heat it.

There are other more modern (e.g. Mukayama aldol reaction, using the silyl enol ether of the ketone) and/or "greener" methods, but they are mostly a variation of the basic approach described above.


Half cells: difference between E(cell) = E(cathode}-E(anode) & E(cell) = E(oxidation) + E(reduction)

Ah, I see what you did wrong.

The first equation is correct.

$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} $$

The second equation should be:

$$ E^\circ_\text{cell} = E^\circ_\text{reduction} - E^\circ_\text{oxidation} $$

Alternatively, you can write it as:

$$ E^\circ_\text{cell} = - E^\circ_\text{oxidation} + E^\circ_\text{reduction} $$

The cathode is actually the reduction step, and the anode is the oxidation step.

A useful mnemonic I learned from my chemistry professor is a red cat and an ox. Interestingly, cathodes are red and are positive. So I have found many uses for this along the way.


You can also use the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{\ce{[A-]}}{\ce{[HA]}}$$

rearranges to:

$$\mathrm{pH} - \mathrm{p}K_{\mathrm{a}} = \log \frac{\ce{[A-]}}{\ce{[HA]}}$$

$\log \frac{\ce{[A-]}}{\ce{[HA]}} > 0$ for conjugate base dominating.

$\log \frac{\ce{[A-]}}{\ce{[HA]}} < 0$ for acid dominating.

$\log \frac{\ce{[A-]}}{\ce{[HA]}} = 0$ if they're present in the same concentrations.

The equilibrium of interest is the third dissociation.

Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($\ce{HA^{2-}}$) dominates over the conjugate base ($\ce{A^{3-}}$). You should be able to extend this for each equilibrium.


There are multiple packages capable of designing QSAR models.

At a basic level, QSAR is about building a statistical model fitting some property/activity based on a set of multiple descriptors. You can use any multivariate statistical package to do that, including R (as indicated in the other answer) or Python.

If you're new, I would highly suggest KNIME an open "platform" for connecting multiple software packages for analyzing data. It's heavily used for QSAR.

There are several examples for KNIME QSAR, e.g. using CheS-Mapper - a quick search turns up several.

A critical concern, though is validation, either through separate training and test sets, cross-validation / bootstrapping methods, or other techniques. It's easy to design models that fit noise.


The one solid answer is that everything you've heard is right, given the appropriate conditions.

If the cathode and anode spaces are properly separated with a porous membrane or a salt bridge, then it is just like that:
$$\ce{2NaCl + 2H2O -> \underbrace{2NaOH + H2}_{cathode} + \underbrace{Cl2}_{anode}}$$
If they are not, then the products inevitably mix with each other and react further:
$$\ce{2NaOH + Cl2 -> NaCl + NaClO + H2O}$$
or, at elevated temperatures,
$$\ce{6NaOH + 3Cl2 -> 5NaCl + NaClO3 + 3H2O}$$
See, chemistry is diverse. Oh wait, and there are more options. Think of the mercury cell.


Does the change in Gibbs free energy depend on initial and final concentrations?

$$\mathrm dG = -S\mathrm dT+ V\mathrm dP + \sum \mu_i \mathrm dN_i$$
If (products) and (reactants) are at the same temperature and pressure then $\mathrm dT$ and $\mathrm dP$ are both $0$, and $\mathrm dG = \sum \mu_i\mathrm dN_i$. Integration gives $\Delta G = \sum \mu_i\Delta N_i$. $\mu$ is the chemical potential which is just the specific molar Gibbs free energy. For the decomposition of $\ce{AB -> B + A}$ this is $$(\mu_\ce{A}\Delta N_\ce{A} + \mu_\ce{B}\Delta N_\ce{B}) - \mu_\ce{AB}\Delta N_\ce{AB}$$


The key to your question lies in the word "bound". Without mentioning the state of your species we can say in general, species which are joined by covalent bonds are single molecules, there are countless examples of such reactions in organic chemistry where two small molecules condense into a single molecule.

Species with intramolecular forces (hydrogen bonds, van der waals etc) can be called bound or perhaps "clusters", which is quite a broad term. Real bonding however is not so easily defined as the molecular orbitals may  generate a wide spectrum of charge distributions etc. Quantum mechanics gives a much clearer description of the bonding of a particular molecule or extended system.

The terminology you write your $\ce{H2OHCl}$ is similar to how we write water of crystallisation in a solid. If this is where your idea came from, it should be remarked that this is a distinct concept.

To conclude; species that are covalently bound (that could have strong polarisation even to bordering ionic bonding) are molecules. Intermolecular forces are precisely that, between distinct molecules. We should remark that we can also have extended solids with covalent bonding (e.g. group 14). In addition some regard metals as molecules too, molecules whose molecular orbitals are distributed over the entire structure etc (research tight binding model for more information). But I feel that wasn't the focus of your question?


Absorption and emission at same wavelength?

Yes it is very common particularly in the more rigid type of molecule. The best example is chlorophyll and this overlap of absorption and emission leads to energy transfer in photosynthesis.  The reason for the overlap is that the excited state potential energy profile is very similar in shape but slightly shifted (as displacement) to that of the ground state and this results in a 'mirror image' spectrum. The spectra below are from anthracene.

You can see even in this rather crude figure that there can be absorption and emission at the same wavelength.


Polymers with no molecular mass distribution?

Living polymerizations can afford nearly zero polymer polydispersivity.  You can go anion (Kraton thermoplastic elastomers - and tell us about the one silicon atom in each polymer molecule), free radical (TEMPO or PROXYL capping of the free radical end, thermally activated), redox (copper halide), olefin metathesis polymerization (ROMP and ADMET), and other clevernesses.

Proteins are made from a DNA template via RNA, one AA residue at a time. There is no dispersivity.  Post-synthesis modification, e.g., glycosylation in erythropoietin, can be polydisperse.


Why does Naphthalene turn into needle-shaped crystals by sublimation?

Every solid material has a crystal structure.  Sublimation tends to produce good crystals because the solid can grow slowly.  In bulk form naphthalene crystals are broken and compressed and so bulk solid naphthalene looks quite different.

The exact same situation applies to snowflakes.  They are flat hexagonal structures when seen individually.  Crunch them into a snowball and they lose that pretty structure.


Why is the molozonide adduct thermally unstable?

The molozonide adduct (1,2,3-trioxalane; the initial adduct by 1,3-dipolar cycloaddition of alkene and ozone) is thermally unstable because it contains two weak $\ce{O-O}$ bonds (Ref.1). Mechanistic studies reveal that the ozonide (1,2,4-trioxalane) is formed by a sequence of three pericyclic steps involving 1,3-dipolar cycloaddition, retro-1,3-dipolar cycloaddition, and again 1,3-dipolar cycloaddition as shown in following scheme:

Also, I'd like to point out that the initial molozonide adduct is not always symmetric. It is symmetric only if the alkene is symmetric. And, the rearrangement of 1,2,3-trioxalane to 1,2,4-trioxalane is not about the stability of $\ce{C-C}$ in 1,2,3-trioxalane as OP argued. It's about thermal stability of consecutive $\ce{O-O}$ bonds in 1,2,3-trioxalane (recall that one $\ce{O-O}$ bond in peroxy compounds is not thermally stable, e.g., $\ce{HO-OH}$).

Knowing that there are two symmetrically placed $\ce{O-O}$ bond in the adduct, the cleavage of the only one bond is weird.

The accepted mechanism (Criegee Mechanism) says one of either $\ce{O-O}$ bond in the initial adduct can be broken to give a corresponding carbonyl oxide (a dipole) and carbonyl compound (dipolarophile), which undergo second 1,3-dipolar cycloaddition to give stable ozonide (see above scheme). None of the mechanism says one $\ce{O-O}$ bond cleaves selectively. it could be either one:

However, the second 1,3-dipolar cycloaddition of either set of carbonyl oxide (a dipole) and carbonyl compound (dipolarophile) gives identical ozonide. This second dipolsar cycloaddition $([_\pi2_\mathrm{s} + _\pi\!4_\mathrm{s}])$ has also been confirmed by $\ce{^{17}O}$-NMR studies (Ref.2):

The mechanism of ozonolysis was revisited with the use of $\ce{^{17}O}$-NMR spectroscopy. In a crossover experiment with $\ce{^{17}O}$‐labelled benzaldehyde and the ozonides of styrene and ethylidenecyclohexane it was shown that only the ether bridge of the secondary ozonides is carrying the $\ce{^{17}O}$ label. This is contrary to results reported earlier and confirms the Criegee mechanism.

References:

Dipak K. Mandal, “Chapter 4: Cycloadditions 1: Perturbation Theory of Reactivity, Regioselectivity and Periselectivity,” In Pericyclic Chemistry: Orbital Mechanisms and Stereochemistry; First Edition, Elsevier Inc.: Amsterdam, Netherlands, 2018, pp. 107-190 (ISBN: 978-0-12-814958-4).
Christian Geletneky, Stefan Berger, “The Mechanism of Ozonolysis Revisited by $\ce{^{17}O}$-NMR Spectroscopy,” Eur. J. Chem. 1998, (8), 1625–1627 (DOI: https://doi.org/10.1002/(SICI)1099-0690(199808)1998:8<1625::AID-EJOC1625>3.0.CO;2-L)).


Why do the halogenations of benzene and toluene lead to different products?

Both halogenations can be, according to procedures on SciFinder, performed at ambient temperature, simple sunlight and in reasonable reaction times.

The first step in both transformations is the formation of chlorine radicals by photoinduced homolytic cleavage of the $\ce{Cl-Cl}$ bond (equation $(1)$).

$$\ce{Cl-Cl ->[$h \nu$] 2 Cl^.}\tag{1}$$

These chlorine radicals want to react with something to lose their radical configuration. In the case of toluene, the kinetically favoured attack is that generating the stabilised benzyl radical as shown in equation $(2)$.

$$\ce{Ph-CH3 + Cl^. -> Ph-CH2^. + HCl}\tag{2}$$

This benzyl radical can then react with another chlorine molecule forming benzyl chloride (equation$(3)$), and so on in two further steps to form trichloromethylbenzene (equations not shown but analogous to $(2)$ and $(3)$).

$$\ce{Ph-CH2^. + Cl2 -> Ph-CH2-Cl + Cl^.}\tag{3}$$

Once we have reached the fully chlorinated methyl group, we can again imagine a chlorine radical looking for a reaction partner. Again, we should ask us which the favoured attack is — the one leading to the most stable products. And rather than attacking the aromatic ring, the benzylic carbon can again be attacked as shown in equation $(4)$:

$$\ce{Ph-CCl3 + Cl^. -> Ph-CCl2^. + Cl2}\tag{4}$$

Thus, we are ‘destroying’ our product because this attack is still better (leads to more stable products) than attacking the aromatic system.

Why does benzene react in the way it does? Well, equation $(1)$ is identical. However, there is no benzylic group; there are only two possibilities how a radical can react with benzene:

abstract a hydrogen radical forming a phenyl ($\mathrm{sp^2}$-hybridised) radical
attack the aromatic ring generating a 6-chlorocyclohexa-1,3-diene-5-yl radical

The second reaction is clearly favoured over the first, since while it does break the aromatic system, it still generates a p-orbital centred pentadienyl-type radical which is much more stabilised than a radical in an $\mathrm{sp^2}$ orbital.

However, toluene or its reaction products will hardly ever go this route since a benzyl radical is still much more stabilised than a pentadienyl-type one. The minute quantities of toluene products that do chlorinate this way will be fully chlorinated, but their yields will be very insignificant.


What is the molecular interpretation of Raoult's law?

I believe what the quote is saying is that once equilibrium is reached between the vapor-liquid mixture, then there will be no tendency to increase entropy. This is because entropy is not the only factor to consider. In any system, one must always consider the interplay between both entropy and energy. For instance, at equilibrium, $$\Delta G=\Delta H-T\Delta S=0$$ which means that at equilibrium, $$\Delta H=T\Delta S$$So, while you raise a good point that in most cases, a system will tend to maximize entropy, in a closed system, one must consider the interplay between entropy and energy which in this case are exactly equal (technically they differ by a scaling factor which is the absolute temperature).

Additionally, what is meant by "vapour pressure represents the tendency of the system and its surroundings to reach a higher entropy" is that one can make an inference about the energetic and entropic properties of a solution based on the vapor pressure of that solution. For instance, if a system has a very low vapor pressure (consider a solid as an extreme example) then the solution is very energetically stable and thus it has a very low tendency to reach higher entropy. But, if you think about acetone, which has a vapor pressure of $30.6 ~\text{kPa}$ at $25^{\circ}~\text{C}$ (ten times that of water at the same temperature), one could say that this then has a strong tendency to increase its own entropy and the entropy of the surroundings.


A lot of rubbers possess C=C double bonds. Alkenes are very easily oxidized by KMnO4 or KClO3, therefore in the case where you're not aware of the chemical nature of the rubber, it's best not to expose it to strong oxidizers.


Claisen Condensation vs. Trans-Esterification

In the Claisen condensation, notice that the alkoxide base usually matches the ester given. For example, if an ethyl ester is used, the base will be ethoxide. The reason for this is that trans-esterification is occuring, except that the product is the same as the reactant.

In general, the Claisen condensation is slow compared to trans-esterification. The rate limiting step of the Claisen is the C-C bond formation. Keep in mind that there will be very little of the active ester enolate nucleophile present, because the equilibrium of alkoxide deprotonating an ester's alpha-position will lie dramatically to the ester (by ~ 10 billion fold).

If one tried to run a Claisen condensation where the base does not match the ester (e.g., an ethyl ester with methoxide as the base), I would expect a mixture of methyl and ethyl esters as the products.


Determining the amount of water to create a given pressure in a sealed container

As I understand your question, you are limited to a total volume of $\pu{1L}$ and you can only add water (which I assume should remain liquid) and heat to the system, and are able to monitor it's pressure and temperature.

Your limitation is the vapor pressure of water at the desired temperature.

At $\pu{453K}$, the vapor pressure of water is $\pu{9.88 atm}$. At $\pu{9.95 atm}$, water boils at $\pu{456K}$. So, if you heat the water to $\pu{453K}$, you will have the desired temperature and be very close to your desired pressure. Or, you could heat the water to $\pu{456K}$, and you would have the desired pressure.  Or you could heat the water to a temperature between $\pu{453K}$ and $\pu{456K}$ to get some trade off of being even closer to the desired temperature and pressure, but with neither of them being exactly what you want.

Again,this trade-off is a limitation arising from the constraint of using water to control the pressure.  If you selected a liquid with a higher boiling point at $\pu{9.95 atm}$, then you could select a temperature at which the vapor pressure of the liquid would be equal to your desired pressure.


Can ionic and covalent bond be explained in terms of orbitals?

The "transfered" electron will be in an orbital that is mostly on the side of only one atom, i.e. the anion. That orbital will be slightly deformed by the presence of the cations around it, but no significant overlap with any of their outer orbitals (those which are as good as empty, because the electron has been lost to the other atom).

(Im writing "mostly" and "as good as" because there can always be a covalent contribution to the bond.)


You have to think about the whole process.  When a metal loses electrons to make a metal ion the following happens:

The metallic bonds holding the metal atoms together are broken.
The metal atom loses the electrons.
The resulting metal ion is hydrated.

In your analysis you are only focusing on step 2.  The enthalpy and entropy of the entire process factor into the reduction potential.