What would the chemical name be for C13H8Cl3NO

Well, let's reconstruct that starting from the very right side, where it says a $\ce{C_6H_5}$. The ring and the $\ce{CO}$ group would be a benzaldehyde if it had an $\ce{H}$ instead of an $\ce{N}$, right? Or a benzoic acid if it was $\ce{OH}$ instead of $\ce{N}$. So what would it be if it had an $\ce{NH_2}$-group? It would be a benzamide. If the $\ce{N}$ is substituted with for example a chloride we call that N-chlorobenzamide. And now we have another $\ce{N}$-centered ligand, the second phenyl ring. The ring has three positions, the one where it's connected to the rest of the molecule would be 1, so that makes a 2,4-dichlorophenyl.

Summarizing we get N-chloro-N-(2,4-dichlorophenyl)benzamide


Cyanide from seeds breaking down in a pond

Loquat seeds contain Amygdaline, which is also found in apricot kernels. Amygdaline belongs to the group of cyanogenic glycosides.

In these compounds, a potentially cyanide releasing aglycon, such as mandelic nitrile, is bound to a carbohydrate (= a "sugar"). These compounds are quite abundant in nature; other known examples are Prunasin from the japanese bush cherry or Sambunigrin from elderberry seeds.

In some cases, cyanide is released by enzymatic cleavage of the glycoside, that is when the plant material is digested by animals. Consequently, some of the plants (or at least some parts of them) are known to be toxic. In other cases, the glycosides do degrade in water.

If hydrogen cyanide is released to the pond, I'd assume it stays there - $\ce{HCN}$ is highly soluble in water.

However, I wouldn't be very concerned. Seeds aren't like paper bags full of toxic material.

It will likely take a while until the seeds break up to release Amygdaline.
To my knowledge, Amygdaline is rather stable in water, hydrogen cyanide isn't released until a sugar-splitting enzyme or acid acts on the molecule.


Regarding 6-311G**:

The * notation (one for polarization functions on the heavy atoms and ** for polarization functions on the heavy and hydrogen atoms) is an older notation belonging to the split-valence basis sets of John Pople. Many people still use this notation today but if you want to be more precise and more descriptive, the (d,p) notation is what you would want to use.  In fact, we only ever use the explicit (d,p) type notation and never use the *'s in my lab.

X-YZg

Split-valence basis sets have the aforementioned structure for us to quickly and easily determine how that basis set was constructed. The X simply is the number of primitive Gaussians that describe the core electron shells of an atom. Most of the time you will be freezing core in your computations so you split the core and valence regions of your atoms and assign various basis functions to each one.  The Y and Z components mean you have two sets of primitive Gaussian functions which describe your valence region. Because you have two sets here, you can denote this basis set as a double-zeta. In your case, a 6-311G** is in fact a triple-zeta quality basis set, which is better than a double-zeta.

For the sake of interest, a 6-31+G* includes a + symbol. This simply means that diffuse functions would be added to the heavy atoms.  A ++ symbol would extend these diffuse functions to the hydrogen atoms. Of course we would never want to use the '*' notation so I should have written this as 6-31+G(d,p).  Of course polarization functions can come in sets as well such as 6-31+G(2df,2pd), and other combinations exist.

Lastly, when using split-valence basis sets such as 6-311G*, you must be very careful to check the manual and see what function space the program is going to use by default.  I will use Gaussian 09 as an example.  By default, Gaussian 09 uses spherical harmonics as its default... EXCEPT FOR A FEW DIFFERENT TYPES OF BASIS SETS including the family of 6-31G* type basis sets which uses the CARTESIAN SPACE BY DEFAULT.  This means that rather than the 5D, 7F stuff that you may be used to using on a regular basis, invoking the 6-31G* type basis set will automatically use the 6D, 10F function set which inevitably changes the size of your basis set which means that you may have data that is not readily comparable to other things.  I always opt out for the 5D, 7F type functions.


When our hard water is boiled, it leaves a blue-green residue in our kettle.

If your building's water pipes are made of copper, that could explain the blue-green coloration. Hopefully the inside of your pipe doesn't look like this, though:


Why is hydrogen evolved at the copper side of a lemon battery?

So you have a zinc electrode and a copper electrode and a lemon. The lemon is not a reagent grade chemical, so let's say it's the citric acid in the lemon that provides the corrosive action.

The copper essentially just sits there; as you indicated, the low reactivity of copper in acid means it doesn't dissolve readily. On the other hand, the zinc is much more reactive, and as you suggested, should just react all by itself. If it did, it would make a terrible battery; it would have a very short shelf life because the zinc would dissolve without providing an exterior current. But zinc has a high hydrogen overvoltage; that is, hydrogen does not readily evolve from zinc. The reasons are complex, but you may imagine that hydrogen ions or atoms are adhered to the zinc surface in such a way that they do not easily combine to form molecules which can form bubbles and go away. So the hydrogen evolution reaction is stifled.

However, the copper surface is different, and does permit hydrogen ions to react with an electron to form an atom which can then find another atom to form a molecule, and then more, to form bubbles. Alas! Copper has no extra electrons to spare; after you take off a few, the copper becomes positively charged and repels hydrogen atoms.

But the zinc electrode has gained a few electrons from the few zinc atoms that have dissolved in the citric acid. Let's connect the zinc electrode to the copper electrode: electrons flow from the zinc to the copper where they attach to hydrogen ions. The copper provides a surface for hydrogen ions to grab an electron and eventually become a bubble. The zinc provides electrons by dissolving as ions and leaving the electrons behind. And the battery could be useful for a long time if you disconnect the wire, because the zinc stops dissolving.

If this is still not clear, imagine a "battery" with a copper electrode and a sodium electrode. When you put the sodium into the lemon, it will react immediately without needing the copper to distribute its electrons. This "battery" might last about 2 seconds and would not be very useful.

And, as it turns out, impure zinc may have inclusions of different materials that provide a cathode surface right in the midst of the anode,  so the zinc could dissolve rather rapidly and make a poor anode. In the distant past, a tiny bit of mercury would be amalgamated onto the zinc, covering up the impurities. Mercury has an even higher hydrogen evolution overvoltage, so the zinc was protected. The electrochemical reaction then was the zinc dissolves in the mercury and travels to the mercury-solution interface before it ionizes and dissolves, but since the mercury layer was so thin, this was no problem. Eventually mercury was eliminated for potential human health reasons.


How does the presence of chlorine atoms in PVC make it flame resistant?

PVC is not truly flame resistant, it does decompose on relatively gentle heating (around 150 °C) However, the process produces significant volumes of hydrogen chloride, that is not flammable and isolates the plastic from the flame. This makes makes PVC self-extinguishing or fire-retardant.


Chemical compatibility of polycarbonate and carbonate esters

Yes, you need to verify that the housing of the battery is made of a material that is unaffected by its contents. Polycarbonate is a good material for rigidity and transparency, but may be dissolved or weakened over time by organic solvents (carbonate esters or otherwise) and also by pH extremes and high temperatures that could come into play when the battery is charged and discharged. Thermosetting resins might be a better bet for this type of application.


Can an improvised torch be made to last for a prolonged period of time?

The key is whether or not he added some kind of fuel to the cloth. Without that, a flammable cloth would burn itself out very quickly. If he had kerosene or some other fuel to soak the cloth, then the cloth would act somewhat like a wick in an oil lamp (except that it would not be "wicking" fuel up from any fuel source) in that mostly the fuel would burn rather than the cloth.

If he had a liquid fuel source, even a non-flammable cloth like his clothing would have been made of would have sufficed. In fact, he probably wouldn't have had to keep adding cloth in that case, but he would have had to add more fuel periodically.

So, the bottom line answer is: Without a liquid fuel source to soak the cloth, no chance. With a liquid fuel source to soak the cloth, no problem.


Bibenzimidazole nomenclature: Indicated H and locants

Your understanding of the numbering of the individual cyclic systems is correct. If we consider the following arbitrarily simplified separate compounds, the numbering is in accordance with your expectations; low locants are given first to the indicated hydrogen and then to the substituent according to the usual rule P-14.4.

In ring assemblies of two identical cyclic systems that are joined with a single bond, however, low locants are given first to the junction according to Rule P-28.2.1.

  P-28.2.1 Ring assemblies with a single bond junction

  (…)

  Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
  (…)

After that, indicated hydrogen is named and numbered according to P-28.2.3.

Therefore, the systematic name is 1H,3'H-2,4'-bibenzimidazole and not 1H,1'H-2,7'-bibenzimidazole since the locants 2,4' for the junction are lower than 2,7'.


Why do not metals gain electrons and non-metals donate electrons?

First, it is a high-school rubbish that was kept only because of lack of desire to revise basic courses.

Can metal gain electrons? Certainly, YES. Sodium has several compounds with negative oxidation state, many transition metals have extensive chemistry where the metal is in formal negative oxidation state and so forth.

Can nonmetal lose electron? Certainly, as in nonmetal-nonmetal compounds it is unavoidable.

So, what is it really about? What is the difference between metals and non-metals?

Before we continue, we have to revisit definition of what metal is.

Metallic solid (sometimes shorthanded as metal) is a solid with metallic conductivity. It arises from having continuous half-filled orbitals allowing electrons to move freely. Metallic solid perfectly can be a compound (say, $\ce{Ag2F}$). Furthermore, many non-metals under heavy pressure undergo transition into metallic phase.

However, when we are talking from chemical PoV, metal is an element. But what kind of element? The problem is, that the definition of metal in chemistry has a history, and the term was introduced long before many of metals known today were discovered. As such, only common metals were considered and no exotic compounds was known, and certainly before extremely high pressures became available.

Consequently, a metal was characterized by its ability to lose electron (but many elements are capable of doing so) and having metallic conductivity when in form of simple compound. The border cases were either discarded or counted as 'metalloids' - a special kind of non-metal.

For example, tin has metallic and non-metallic allotropes at ambient pressure. Oups, is it a metal or non-metal ? Well, legally it is considered as a metal. However, antimony, with pretty much the same case, is typically considered as metalloid.

There is however a strict difference between tin and antimony in than tin is capable of forming normal salts and having a basic oxide, while antimony has only slightly acidic oxides. The reason for discrimination DOES exist. On the other hand, some heavy transition metals, such as rhenium and tungsten do not have true basic oxides.

So, it would be best to define metals by enumerating them. It is, however, easier to enumerate non-metals: commonly recognized non-metals are boron, silicon, arsenic, tellurium, iodine, everything to right and up of that elements and hydrogen. Optionally, germanium and antimony may be included. (Polonium, astatine and some others are questionable cases, since their chemistry is virtually unexplored thanks to their high radioactivity. But for that reason they can be safely ignored)

Said elements have some things in common: they have relatively high electronegativity, form covalently bound or molecular solids, form acidic oxides (if form at all) and do not form simple cations stable in water. All this comes from them having relatively high number of electrons in valence shell and tight binding of said electrons.

By no mean you should assume that sodium, for example, is eager to lose an electron. Nope, this process results in energy consumption. Only subsequent stabilization of by electron affinity of its partners and ionic packing results in net energy gain. On the other hand, addition of electron to an atom often results in slight energy release.

TL; DR. Non-metals typically has compact electron shells tightly tied to their nucleus and as such are uneager to loose them.


How do I use the priority list of substituent groups to decide the parent chain for this compound?

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

So, yes, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound according to Rule P-44.1.1. However, you have already done that in both of your suggested examples: The principal characteristic group that is expressed as a suffix is the acid group ($\ce{-COOH}$). Therefore, the parent structure is a butanoic acid.
Looking at your two examples, your actual question is about which of the two possible butanoic acids is the principal chain: 3-chloro-2-(2-hydroxyethyl)butanoic acid or 2-(1-chloroethyl)-4-hydroxybutanoic acid. Thus, your actual question is about how to identify the principal chain after you have already considered the priority order of functional groups.
Both possible chains have the same length; therefore, the next rule (P-44.3.2) cannot be used to make a decision.
Both possible chains don’t have any multiple bonds; therefore, also Rule P-44.4.1 (a and b) doesn’t help.
In both possible chains, the principal characteristic group ($\ce{-COOH}$) is located at the start of the chain; thus, also Rule P-44.4.1 (h) doesn’t make any difference.
The number of substituents expressed as prefixes is the same (two) in both cases: chloro and 2-hydroxyethyl compared to chloroethyl and hydroxy; therefore, Rule P-45.2.1 cannot decide which chain is the principal one.
The decision is finally made using Rule P-45.2.2 since the locants et ‘2,3’ in 3-chloro-2-(2-hydroxyethyl)butanoic acid is lower than ‘2,4’ in 2-(1-chloroethyl)-4-hydroxybutanoic acid.
Therefore, the preferred IUPAC name for the compound that is given in the question is 3-chloro-2-(2-hydroxyethyl)butanoic acid.


As other answers have noted, the only gas lighter than helium is hydrogen, which has some flammability issues that make it more difficult to handle safely than helium.

Also, in practice, hydrogen is not significantly "lighter" than helium.  While the molecular mass (and thus, per the ideal gas law, the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the difference between the density of the gas inside the balloon and the air outside.

The density of air at STP is about $\rho_{\ce{air}}=\pu{1.2754 kg m-3}$ , while the densities of hydrogen and helium gas are $\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$ and $\rho_{\ce{He}}=\pu{0.1786 kg m-3}$  respectively.  The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences  $\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$  and $\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$.  Thus, helium is only about $7.5\%$ less buoyant in air than hydrogen.

Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant.  For example, if you wished to go ballooning on Jupiter, which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply sink, and even a pure hydrogen balloon (at ambient temperature) would not lift much weight.  Of course, you could always just fill the balloon with ambient Jovian air and heat it up to produce a hot hydrogen balloon (not to be confused with a Rozière balloon, which are used on Earth and have separate chambers for hot air and hydrogen / helium).

Ps. A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about $\pu{2 Da}$, while a helium atom has two protons and two neutrons, for a total mass of about $\pu{4 Da}$.

Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about  $\pu{32 Da}$ (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to  $\pu{28 Da}$ (one proton and one neutron per atom less than oxygen).  Thus, the average molecular mass of air should be between $28$ and  $\pu{32 Da}$; in fact, since air is about three quarters nitrogen, it's about  $\pu{29 Da}$, and so the buoyancies of hydrogen and helium in air are proportional to $29 - 2 = 27$ and $29 - 4 = 25$ respectively.  Thus, hydrogen should be about $\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$ more buoyant than helium, or, in other words, helium should be about $\frac{2}{27} \approx 7.5\%$ less buoyant than hydrogen.

Pps. To summarize some of the comments below, there are other possible lifting gases as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices.

For example, methane (molecular mass  $\approx \pu{16 Da}$) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas.  However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's more dangerous by others (total energy content per volume).  In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium.

A much less flammable choice would be water vapor which, with a molecular mass of  $\approx \pu{18 Da}$, is only slightly less buoyant than methane at the same temperature and pressure.  The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all.  This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster — whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses.

Despite these drawbacks, hot steam balloons have certainly been suggested, studied and tried in the past — alas, not always particularly successfully (although, apparently, there have been much more successful attempts as well).  There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope.  So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas.

Other potential lifting gases, with molecular mass similar to methane and water, include ammonia and neon.  Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium.

Ammonia, on the other hand, while much less flammable than methane, is rather toxic and corrosive (not to mention really stinky, which, given its other properties, is probably a good thing).  I don't think I'd like to fly in an ammonia balloon, but apparently, some people do!  It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form.

Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second.  If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical.  Then again, so would hydrogen — its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.


Reaction between manganese(IV) oxide and sulfuric acid

I am afraid I cannot provide the exact mechanism. Nevertheless, here is what I know.

The reaction is most certainly more complex than the equation

$$\ce{2 MnO2_{(s)} + 2 H2SO4_{(aq)} -> 2 MnSO4_{(aq?)} + 3 O2_{(g)} + H2O_{(l,g)}}$$

suggests. One might say it is even misleading, but it is definitely not the whole story.

A purple smoke

After adding concentrated sulfuric acid to manganese(IV) oxide, a purplish smoke manifests. This suggests to the species $\ce{MnO4-}$. Yet, what is the balancing particle? Is it $\ce{Mn^2+}$, or rather $\ce{H}$? I have no clue.
$$\ce{2 Mn_{(aq)}^2+ + 5 MnO2_{(s)} + 6 H3O+_{(aq)} -> 2 HMnO4_{(aq)} + 5 Mn^2+_{(aq)} + 8 H2O_{(l,g)}}$$

A green to brown mixture

The initial mixture is green. This points to either $\ce{MnO4^2-}$ or $\ce{Mn2O7}$, even $\ce{MnO4^3-}$ could be it. Again, which one is it? Maybe it is all of the above? Upon standing, we observe a brown precipitate. The manganese(IV) oxide appears to reform … Speculation (more in the next chapter):

$$\ce{MnO4_{(aq)}^2- + H2SO4_{(aq)} -> H2MnO4_{(aq)} + SO4_{(aq)}^2-}\\\ce{3 H2MnO4_{(aq)} ->  2 HMnO4_{(aq)} + MnO2_{(s)} + 2 H2O_{(l,g)}}$$

$$\ce{3 MnO4^2- + 2 H2O_{(l)} -> 2 MnO4- + MnO2_{(s)} + OH_{(aq)}-}$$

Oxygen, where's that from?

I did the classic match experiment; it did indeed burn more vigorously near the reaction mixture.

$$\ce{HMnO4_{(aq)} -> MnO2_{(s)} + 3 O2_{(g)} + 2 H2O_{(l,g)}}$$
$$\ce{2 MnO4^2- -> MnO4^3- + MnO2_{(s)} + O2_{(g)}}$$
$$\ce{2 Mn2O7_{(l)} -> 4 MnO2_{(s)} + 3 O2_{(g)}}$$

I think you get the idea

Sorry I could not be of more assistance. The mixture is indeed complex with multiple processes happening in parallel. (This is actually the case with every reaction.) It should at least point you in some direction.

Extra

I was a bit loosey–goosey with the ionic equations; you should not do that ;). In my opinion, it made the process more clear.
One might also wish to consider $\ce{Mn2MnO4}$ or $\ce{Mn3O4}$ which, I believe, are also brown.


Is visible spectroscopy the only non-electronic method of all the spectroscopy method?

No, Ultraviolet spectroscopy and Infra-red spectroscopy were practiced without any use of electricity.

Infra-red and ultraviolet light were both known by 1801.

Ultraviolet absorption lines had been observed by 1843.


The precatalyst system [RhCp*Cl2]2 was developed by Peter Maitlis and his group at the University of Sheffield in the 1970's and 80's. It has been used as a hydrogenation catalyst under a number of conditions, and is readily commercially available. Under the right conditions, ethanol can act as a suitable transfer hydrogenation source (not quite as good as 2-propanol, but better than methanol), but in this case, in the presence of a suitable base (sodium carbonate), the rhodium dimer will react to give a stable product.

$$\begin{array}{|c|c|c|}
\hline
\textbf{label} & \delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline
\textbf{A} & 1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline
\textbf{B} & 1.73 & 15 & \text{s} \\ \hline
\textbf{C} & 2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline
\textbf{D} & 3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline
\textbf{E} & 3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline
\textbf{F} & 3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline
\end{array}$$

From the NMR, it is possible to deduce what structure is likely, if we construct a valid spin system from the available data.

signal B: δ1.73 integral of 15 : we still have a Cp*
signal A: δ1.65 integral of 3 : looks a lot like we have a methyl group. It has a doublet coupling of 6.2Hz, which corresponds to signal E at δ3.58.
signal E: δ3.58 is single 1H which appears as a dq. We can account for the q coupling; it is next to a methyl group, which leaves the doublet coupling of 7.0Hz, which corresponds to signal F at δ3.96
signal F: δ3.96 is a single 1H with additional coupling to 2 other single 1H nuclei; one of 11.0Hz (signal C), which is pseudo-trans, and the other at 6.6Hz (pseudo-cis, signal D). This two signals have a mutual coupling of 1.5Hz, typical of a pseudo-gem couping on a double bond.

So that gives a spin system skeleton that looks like:

which, is able to form a stable π complex.
Homework spoiler alert:

and a useful reference from many years ago....

   K. Moseley,   J. W. Kang and    P. M. Maitlis J. Chem. Soc. A, 1970, 2875-2883. DOI: 10.1039/J19700002875


Your problems are caused by using the Arrhenius theory, which is based on electrolytic dissociation.

According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:
$$\ce{HCl -> H+ + Cl-}$$
Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:
$$\ce{NaOH -> Na+ + OH-}$$
The products of a neutralization reaction of an acid with a base are a salt and water:
$$\ce{HCl + NaOH -> NaCl + H2O}$$
This definition works for metal hydroxides such as $\ce{NaOH}$, which can exist as pure compounds. However, it doesn’t work for bases such as $\ce{NH3}$.

In order to solve this problem for $\ce{NH3}$, the fictitious compound “ammonium hydroxide” $\ce{NH4OH}$ was proposed (see also this related answer). Assuming this formula, the Arrhenius definitions of bases and neutralization reactions work again:
$$\ce{NH4OH -> NH4+ + OH-}$$
$$\ce{HCl + NH4OH -> NH4Cl + H2O}$$
However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.

In order to really solve your problems with $\ce{NH3}$, you need to switch from the definition of bases according to the Arrhenius theory to the definition of bases according to the Brønsted–Lowry theory.


You obviously already have a general idea of what is going on in a distillation, so I will spare you the picture and everything.

Your question can be broken into two parts:

Do I need heat for vapourisation?
Do I need cooling for condensation?

I am going to address them one after the other.

Vapourisation

As you probably also know, reducing pressure is one means of evaporating liquids. This is used in rotavaps which use a $40~\mathrm{^\circ C}$ bath and reduce the pressure to make sure the solvent evaporates. You could think that just reducing the pressure should be enough to vapourise any liquid.

But if you vapourise a liquid only by reducing pressure, you will reduce the temperature of the liquid. This is the evaporation cooling that your body uses when sweating: If the water in sweat evaporates it cools the surface it evaporates from. Thus, after a certain time and especially if you attempt to distill large quantities of water, it will freeze and further evaporation will become exponentially harder. Therefore, a practical setup will always require heating.

Condensation

Effectively, this is the same discussion but in reverse. Note that if you boil water in a closed pot, the lid of the pot will get successively hotter due to condensing water releasing heat. After a certain time, the temperature of the surface you want to condense on will be so hot that the liquid will no longer condense.

This is especially a problem if you decided to reduce pressure for evaporation: If you have to reduce the pressure so far that the liquid would evaporate at $10~\mathrm{^\circ C}$, then there is no way that it will condense at a cooling trap of $25~\mathrm{^\circ C}$. Therefore, the condensation area must be cooled which is typically achieved with Liebig coolers or similar.

What about vapourisation followed by condensation in the same vessel with changing pressure?

Well, that is possible. It is not practical, though, since the inside of a single vessel can only have a certain pressure (within slight margins). And vapour spreads out uniformly. Your vessel would have to have an initial ‘bowl’ and a collecting one. You reduce pressure to evaporate out of the initial bowl and then re-raise pressure. The liquid will condense uniformly and you will lose a significant amount of it because it will drip back into the initial bowl. Your yield with a typical common setup is much better.


How is it possible to obtain a value of 8/3 as an oxidation state?

The formula of $\ce{Fe3O4}$ is a shorthand for $\ce{Fe^{II}Fe^{III}2O4}$, so there are two sorts of atoms of iron here in +2 and +3 oxidation state.

However, it is still possible to gain truly fractional oxidation states. The simplest example probably would be potassium superoxide, $\ce{KO2}$. Here, the anion is symmetrical $\ce{O2^{-}}$ anion, so obviously both oxygen atoms are in $\ce{-1/2}$ oxidation state. There are more complicated examples as well.

Often fractional oxidation state comes from symmetrical structures with charge, that would have unpaired electrons otherwise. However, systems with truly fractional oxidation states usually have bonding, hard to describe in terms of common Lewis structures (and with little relevance of idea of oxidation state), and mostly are considered in semi-advanced organic chemistry and advanced inorganic chemistry.


This is mostly the case for sulfuric acid. Commercially available sulfuric acid is dense (~1.8 g/ml) and when water is added, it may not mix. In this case a layer of hot weak acid solution is formed, which boils and sprays around. When acid is poured into water, it flows down the flask and mixes much better, so no boiling occurs.

The reason this occurs is due to the large amount of energy released in the hydration reaction of sulfuric acid ions. Do not believe that heat comes from dissociation, as the dissociation of acids, bases, and salts always consumes energy. The energy is released from subsequent hydration, and the release may be high, especially if $\ce{H+}$ or $\ce{OH-}$ ions are hydrated.


Regioselectivity in coupling reactions of α-naphthol

The product at the para-position is more stable as it allows the compound to exist in the quinoid form, rather than only in the benzoid form:

This is important as in your case, if the substituent $\ce{X}$ is $\ce{H}$, then depending on the solvent, as much as 89% of it may exist in the quinoid form ($\bf{2}$).
References:

Satoshi Kishimoto, Shinya Kitahara, Osamu Manabe, Hachiro Hiyama, “Studies on coupling reactions of 1-naphthol. 6. Tautomerism and dissociation of 4-arylazo-1-naphthols in various solvents,” J. Org. Chem. 1978, 43(20), 3882–3886 (https://doi.org/10.1021/jo00414a020).


We can find the average molar mass with the density formula involving pressure and temperature $\overline M=\frac{\rho TR}{p}$.

We also know that, in this case, $\overline M=\frac{n_1M_1+n_2M_2}{n_1+n_2}$. Since we only "care" about $M_1$ we can write the average molar mass as $\overline M=\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2$ and then find $\frac{n_1}{n_1+n_2}$.
$$\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2}^{(*)}=\frac{\frac{n_1}{n_1+n_2}M_1}{\frac{n_1}{n_1+n_2}M_1+\frac{n_2}{n_1+n_2}M_2}=\frac{n_1M_1}{n_1M_1+n_2M_2}=\frac{m_1}{m_{total}}$$
Compute $(*)\cdot100\%$. The development of the fraction was to show why it works.

Another possible solution is this :
$$\%w_{m_1}=\frac{m_1}{\rho V}*100\%\Leftrightarrow m_1=\%w_{m_1}\rho V/100\%$$
$$\%w_{m_2}=\frac{m_2}{\rho V}*100\%=100\%-\%w_{m_1}\Leftrightarrow m_2=(100\%-\%w_{m_1})\rho V/100\%$$
$$n=\frac{m_1}{M_1}+\frac{m_2}{M_2}=\frac{\rho V}{100\%}\left(\frac{\%w_{m_1}}{M_1}+\frac{100\%-\%w_{m_1}}{M_2}\right)=\frac{PV}{RT}$$


When I burn isopropyl alcohol (IPA), it burns orange. But when I burn ethyl alcohol, it burns totally blue. Why is this?

IPA has a different carbon:hydrogen ratio than ethanol. There is more incomplete combustion occurring with IPA, hence the smoky orange flame and smell of soot. Ethanol combusts more completely, leading to a blue (soot-free) flame and no smell.

In response to your second question, ethanol likely has a lower latent heat of vaporisation than IPA, resulting in it evaporating rapidly. It takes a lot of heat energy away from the tray when doing this, leading to the tray cooling. A similar effect can be observed if you accidentally get some types of solvent on your hand and feel a sudden coldness as they vaporise, taking heat from your skin.


What does it mean to define entropy as a measure of a systems internal energy per unit temperature which is unavailable to do work?

The answer to this question goes back to the Gibbs equations according to which $$\mathrm{d}U=T\,\mathrm{d}S-p\,\mathrm{d}V$$ and for a reversible process $$đq = T\,\mathrm{d}S$$ So, if you have a change in internal energy $\mathrm{d}U$ and it does work $p\,\mathrm{d}V$ then it has also absorbed $T\,\mathrm{d}S$. The verbiage you are quoting is very awkward and is not very useful, nevertheless for your future studies you will be better off if you will never think of entropy as "disorder" in any sense. Entropy is not disorder and such concept will never help you solve any real problem. Also, never think of a body containing "heat"; a body has energy depending on its state but independent of external constraints; and it has entropy that depends both on its state and its external constraints. For example, an ideal gas after it has expanded into a larger volume has the same energy as before it has done no work and absorbed no heat but its entropy is larger than it was in a smaller available volume. The entropy is the same in all possible molecular configuration consistent with the volume, so it is the same whether the molecules occupy uniformly the available volume or because of a large fluctuation spontaneously collapse into a smaller volume form the larger one. So it is the constraint (the available) volume that determines the entropy not how the molecules are distributed in it in any given instant of time. No disorder.


What is the difference between dissolution and absorption?

Casual use of "absorption" is different from the normal, technical, use of the term
The key difference between "absorption" and "solution" is confused in your examples because you use the casual use of the term not the technical one. Even so, the two processes are very distinct.
Solutions are homogeneous. But when things are absorbed, either in casual use or technical use, there are distinct phases. Absorption is usually a technical term for a surface phenomenon where, for example, gas molecules stick to the surface of a catalyst. This is a widely studied process as it is a vital part of many major industrial processes. Water absorbed onto cotton wool is a more casual use of the term absorption but the result is still not a homogeneous mixture. Liquid water sticks to the surface of the fine fibres of cotton wool by surface tension. There are still two phases: liquid water and solid fibre.
Only in the most careless and casual use should you ever describe the process of dissolving sugar in water (or sugar taking up water) as one "absorbing" the other.


What is the reaction between piperine and concentrated sulfuric acid?

I think OP is confused with the colorimetric method of detection of piperine, which uses the concentrated sulfuric acid as one of reactant with chromotropic acid, 1,8-dihydroxynaphthalene-3,6-disulfonic acid (Chromotropic Acid Test; Ref.1 & 2). Accordingly, this colorimetric method, developed in 1965, relied on hydrolyzing the methylenedioxy group of piperine by chromotropic acid in the presence of concentrated sulfuric acid to formaldehyde, upon which a definite color was developed with chromotropic acid (Ref.3). Color intensity is proportional to the amount of piperine present. Water inhibits color development, but a small amount of methyl alcohol is necessary.

Note: It is good to know that chromotropic acid test is very sensitive to presence of formaldehyde and detect them in very low concentrations (Ref.4). It is also fact that chromotropic acid is specific to formaldehyde (Ref.2).

References:

Leila Gorgani, Maedeh Mohammadi, Ghasem D. Najafpour, Maryam Nikzad, “Piperine—The Bioactive Compound of Black Pepper: From Isolation to Medicinal Formulations,” Comprehensive Reviews in Food Science & Food Safety 2017, 16(1), 124-140 (https://doi.org/10.1111/1541-4337.12246).
Clark E. Bricker, Hilding R. Johnson, "Spectrophotometric Method for Determining Formaldehyde," Ind. Eng. Chem. Anal. Ed. 1945, 17(6), 400–402 (https://doi.org/10.1021/i560142a021).
Horace D. Graham, “Quantitative Determination of Piperine. I. The Komarowsky Reaction,” J. Food Sci. 1965, 30(4), 644–650 (https://doi.org/10.1111/j.1365-2621.1965.tb01818.x).
Jingping Zhang, David Thickett, Lorna Green, "Two Tests for the Detection of Volatile Organic Acids and Formaldehyde," Journal of the American Institute for Conservation 1994, 33(1), 47–53 (doi:10.2307/3179669).


Multiphasic liquid "flotation" experiment

https://www.youtube.com/watch?v=GRPufqPgTeI

I played around a bit with this in the lab and this is what I came up with.
Basically, I made a three phase system with hexanes on top, water with fluorescein and green food colouring in the middle, and a perfluoroalkylether HT-110 on the bottom.
Just finding three mutually immiscible liquids was pretty challenging and unfortunately, like most highly fluorinated liquids, HT-110 is almost twice as dense as the other two liquids, but if you look at the video carefully, you can see that some bubbles of HT-110 are getting trapped in the hexanes as its surface energy isn't quite as high in hexanes as it is in water.
When I tilt the vial on its side, you can see these bubbles snapping back across in the interface.

It's not perfect, uses a goofy solvent, and I probably could have played with the liquid ratios/container shape to trap more HT-110 at the top, but it seems that something at least similar to what you're after can be made fairly simply.


Can't there ever be hydrogen bonds between 2 purines or between 2 pyrimidines?

Yes, it is possible to have hydrogen-bonded base pairs (and triplets and quadruplexes) beyond the canonical Watson-Crick base pairs A:T and G:C. You can explore these in 3D on the DSSR-Jmol site for example. If you look at the 1ehz structure (a tRNA from yeast), which loads automatically on DSSR-Jmol, and click on non-canonical base pairs, you will see some examples. Below is an A:A pair with two symmetric hydrogen bonds from that structure:


Different names for Fidaxomicin (OPT-80)?

Fidaxomicin is the International nonproprietary name (INN) of a pharmaceutical, which was/is distributed by different companies under different trade names.

To my knowledge, OPT-80 was a provisional name/signature used by Optimer Pharmaceuticals Inc.. The drug was later sold as Dificid.


Your initial statement is wrong. There are organic chlorates. When you want to search compounds, try chemical search engine like PubChem or ChemSpider instead of google. Even google lead you to something if you read carefully. It is no reason that chlorate ester can not exist and they do exist. Below is the link to the PubChem entry of Choric Acid, Methyl Ester, which is Methyl Chlorate.

https://pubchem.ncbi.nlm.nih.gov/compound/18695386#section=Top


What does all chromatography have in common?

Here are some similarities:

There is always a mobile (e. g. Gas - GC, Liquid - HPLC; GPC) and a stationary phase (liquid; gel - GC, solid - LCs, GPC).
Compounds in the sample interact different with both phases and are therefore held back stronger or lesser. This results in the accumulation of compounds that interact similar at some point in the system $-$ with some chromatographys, especially affinity chromatography, one might even get pure substances.
The mobile phase carrys the sample and is commonly less polar than the stationary phase, except for RPCs.
The structure of a chromatograph is basically like this:

transport system: reproductive transport of the mobile phase, adjustable (e. g. piston pump - LCs, compressed gas cylinder - GC)
sample application: import of the sample into the flowing mobile phase (e. g. injector, valve)
column: stationary phase and its mounting (e. g. capillary column in the column oven)
detector: transformation of the chemical or physical properties, which change with substance concentration, into electrical signals (e. g. UV/VIS-detector, thermal conductivity detector)
plotting unit: visualisation of the electrical signals (e. g. integrator, computer)


Why are there multiple lines in the hydrogen line spectrum?

As the electrons fall from higher levels to lower levels, they release photons.  Different "falls" create different colors of light.  A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength).

The visible wavelengths are caused a by single electron making the different transitions shown below.  There are even more transitions that release invisible wavelengths.

 Wavelength   Transition  Color
 (nm)
---------------------------------------------
 383.5384      9 -> 2    Ultra Violet
 388.9049      8 -> 2    Ultra Violet
 397.0072      7 -> 2    Ultra Violet
 410.174       6 -> 2    Violet
 434.047       5 -> 2    Violet
 486.133       4 -> 2    Bluegreen (cyan)
 656.272       3 -> 2    Red
 656.2852      3 -> 2    Red

*Values taken from Hyperphysics: Hydrogen Energies and Spectrum

(Why are there two different 3-> 2 transitions?  See here: Hydrogen Fine Structure )


Is Gibbs Free Energy of formation equal to Gibbs Free Energy of transition state?

The Arrhenius equation

$$k=A\exp\left(-\frac{E_a}{RT}\right)$$
places all of the $T$-dependence in the exponential factor. The pre-exponential factor is not assumed to be temperature-dependent. By contrast, the Eyring equation

$$k=\frac{\kappa k_\mathrm{B}T}{h}\exp\left(-\frac{\Delta ^\ddagger G^⦵}{RT}\right)$$

which is justified on the basis of statistical-mechanical principles, displays a a temperature-dependence in the pre-exponential factor. So the exponential terms are not expected to be equal, at least theoretically, unless the pre-exponential factor in the Eyring equation becomes T-independent by some coincidental cancellation of terms.

As regards the second equation, I would define the free energy of the transition state more accurately as

$$\Delta ^\ddagger G :=\Delta _\mathrm{f} G(\mathrm{ts})-\Delta _\mathrm{f} G(\mathrm{ini}) $$


Relationship between dipole moment of a molecule and electronegativities of constituent atoms

Yes you can. By looking at the different electronegativities of each element in a molecule, you are able to determine the dipole moment for each bond. Then by adding the dipole moments for each bond, you are able to determine whether a molecule is non-polar or polar and if it is polar, how polar it is.

For example, in $\ce{CH4}$, the carbon atom is bonded to four hydrogen atoms. The electronegativity of carbon is 2.55, while the electronegativity of hydrogen is 2.2 This means that for each $\ce{C-H}$ bond, there will be a dipole moment from the hydrogen to carbon atom. However, since methane has a perfect tetrahedral structure, the four dipole moments cancel each other out so that there is no net dipole moment. Hence, methane despite having polar bonds, is actually not a polar molecule, shown below.

Another example that we can consider is ammonia, $$\ce{NH3}$$. Ammonia consists of 3 $\ce{N-H}$ bonds. The electronegativity of nitrogen is 3.04. That means that ammonia consists of 3 polar $\ce{N-H}$ bonds. The molecular structure of ammonia is trigonal pyrimadal and hence unlike methane, when you add up the 3 dipole moments, they don't cancel each other out. Instead, they result in a net upward dipole moment. Hence, the molecule itself is also polar, shown below


Is planarity really necessary for conjugation?

this is a rather good question several students have actually asked me about it. it is not any exeption or contradiction .i have written the explanation below


Safety concerns - Heating PMMA (acrylic glass) and fluorinated polymers

One problem is bonding of the bundle.  The contact surface is the fluororesin cladding that allows more nearly perpendicular rays to be totally internally reflected.  If you nick the cladding you get a light leak.  Will the cladding fuse when heated?

If you ligthly bond a bunch of cylinders you get air gaps where they touch.  Those will be a triangular grid off dark spots, 9.3% of the area of the image.

http://en.wikipedia.org/wiki/Circle_packing


Why is the bond order in the SO₃ molecule 1.33 and not 2?

The bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for more guidance: Hypervalency and the octet rule.

I stated in Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

  The bonding picture is usually trivialised as each $\ce{S−O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=−\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three π bonding orbitals, resulting in electron density deficiency.

There are a couple of molecules that belong roughly to the same class. I call them Y aromatic systems: $\ce{CO3^2-}$, $\ce{NO3^-}$, and $\ce{SO3}$. The description is best carried out in the framework of resonance. (Please also see Why is there a need for resonance?)

I have explained the delocalisation in the nitrate ion, but I am happy to repeat it here for $\ce{SO3}$. The molecule has an overall symmetry of $D_\mathrm{3h}$, hence the central sulfur can be regarded $sp^2$ hybridised. Since oxygen has only one bonding partner, it can sufficiently be described as $sp$ hybridised (compare here). The combination of these orbitals form a $\sigma$ bond each. The remaining $p$ orbitals perpendicular to the molecular plane form the delocalised $\pi$ system. Since delocalisation is not part of the original Lewis description it is impossible to address this with such structures, which ultimately lead to the concept of resonance. It is important, that the bonding situation can only be described by all resonance structures, as there is no predominant. The blue structure tries to emulate delocalisation, but here formal charges and charges are quite messed up.

However, we can take home an approximate bonding order for the $\ce{S-O}$ bond. Since all of the resonance structures are equally dominant, we can see that all of these bonds consist of a $\sigma$ bond. We also see, that there is a $\pi$ bond in the first three structures. Each of them contributes equally, therefore it makes a third of a full contribution per bond, hence the bond order is expected to be about $1\frac13$.

Now the molecular orbital scheme is quite similar, but it suffices with one structure. The explanation of the $\sigma$ system remains unchanged. From the four $p$ orbitals we can form four symmetry adapted linear combinations. Two of the are non-bonding degenerate, the completely symmetric one accounts for the huge stabilisation. The antibonding orbital is not occupied. See below for the schematics of $\ce{NO3^-}$, which is analogous to $\ce{SO3}$. You can find the complete MO scheme here.

Since each of the $\ce{S-O}$ bond contributes equally to the lowest lying $\pi$ orbital, the bond order is found to be about $1.33$. The calculated bond order is a little bit higher, most likely due to attractive electrostatic interactions.


Is there an experiment that visibly shows the effects of metastability?

The standard demo for this is crystallization of sodium acetate trihydrate from its melt.  The trihydrate melts at 58 °C.  The molten liquid is easily supercooled, after which point dumping in a few crystals of unmelted stuff results in a rapid solidification of the mixture.

Several videos on YouTube demonstrate the reaction; Nurdrage's is my personal favorite.


$\ce{CaCO3.MgCO3}$ reacts with $\ce{4 HCl}$, therefore $m_\mathrm{eq}=m/4$.
$\ce{CaCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.
$\ce{MgCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.
$\ce{Na2CO3.10 H2O}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/4$.
$\ce{1 CO2}$ of carbonate origin is released by $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.


Cadmium Selenide, Cadmium Sulfide, Cinnabar and Lead Oxide toxicity

I found an answer to my question. Typically oil paint manufacturers will write what pigments are used in their paints. The pigments are classified according to the 'color index' or C.I. These pigments may actually have different chemical composition, but nevertheless the color index can be used to search the pigment database here: http://www.artiscreation.com/red.html

For example, PR106 stands for genuine Vermillion, which is true Cinnabar and has a toxicity rating of C according to the table linked above, where A is non-toxic and D is extremely toxic.

Most colors that are marketed as 'cinnabar' are actually composed of various different pigments, most often not actually containing mercury.


Tertiary carbanion in more alkylated alkene

There are two types of description of hyperconjugation: (1) Valence bond approach & (2) Molecular orbital approach. Your question is pertaining to valence bond approach as you are interested in the stability of the canonical forms.

In Valence bond approach, hyperconjugation is said to happen due to conjugation of C—H bond electron pair with adjacent double bond. We call this phenomenon 'hyper'-conjugation because there is no real negative charge that is getting delocalized. Rather, the partial negative charge which is accumulated on the carbon due to electronegativity difference with hydrogen is getting delocalized.

In a real carbanion, the carbon has ~100% share of the negative charge. However, in this case, the carbon has only partial share of the negative charge. To be precise, about 7%, because 7% is the ionic character of a C—H bond (Calculated using Hannay-Smith's Relationship for Calculating Percentage Ionic Character [1]).

In case of conjugation of real carbanion, the ~100% negative charge is getting delocalized, and you can give importance to the stability of canonical forms to detrmine whether that conjugation is relevant. However, when ~7% negative charge is getting delocalized, it is more important to give importance to the number of such conjugations, than the relevance of each conjugation based on stability of canonical forms, because anyway, ~7% negative charge won't cause much destabilization even in a tertiary carbanion!

In Molecular orbital approach, hyperconjugation is said to happen due to overlap of bonding σ oribitals of C—H bond with adjacent antibonding π* orbital (in case of double bond). Owing to energy & orientation mismatch, the overlap is minimal and thus stabilization arising from hyperconjugation becomes significant only when the number of hyperconjugations is greater. So, this explains why 2,3-dimethyl-but-2-ene assumes much greater stabilization from hyperconjugation compared to 2-methyl-propene.


Because the O atoms are sp3 hybrid and as a result they adopt a tetrahedral geometry, now the lone lairs also repel each other and also the bond pairs so according to VSEPR theory to minimize the repulsion and attain a confirmation of minimum possible energy, the molecular structure is non planar, although it deceptively looks planar in a flash! The actual structure is like that of an open book


Formation of polymers - addition polymerisation

During radical polymerization, we use molecules called initiators $I$ for forming free radicals. The molecules of initiators contain  at least one fragile bond that can easily dissociate upon heating, and yield free radicals.
$$I\rightarrow2A^.$$
The as-formed radicals  have enough energy to attack the monomer molecule $M$ and transfer the radical to the monomer molecule:$$A^.+ M\rightarrow  AM^.$$
More precisely, one electron of the double bond in the monomer forms a covalent bond with the free electron of the radical $A^.$. The other electron of the double bond is now free. We have  a radical  on the monomer  $AM^.$.

The radical monomer  attacks another monomer molecule, and the polymerization reaction propagates:
$$  AM^. + M \rightarrow AMM^.  $$
In the above reaction, one electron of the double bond in the monomer molecule forms a covalent bond with the free electron of the radical $AM^.$.The other electron of the double bond is now free $AMM^.  $.
and so on..


What is the cause for Crystal Field Splitting Parameter?

According to this theory ligands are considered point charges. These points acquire positions as those you've presented above. This leads to some d orbitals being closer to the ligands. That means that a higher amount of energy is needed for an eletron to be placed there, because it would be closer to the negative charged ligand.

However, which orbital is going to be more affected, is a matter of an "agreement" you may say. In order to be easier to explain situations, structures etc it is inclined to use a conformation such as the one you've shown. It would be much more difficult to discuss matters if everyone considered different naming and configurations.


$\ce{Mg2SnO4}$ Magnesium Orthostannate

$\ce{MgSnO3}$ Magnesium Metastannate


Entropy change of surroundings and system

In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process.  The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings.  To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings).  No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.

On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $dq/T_{boundary}$ than the integral of $dq/T_{boundary}$ for the original irreversible process.


Does borax do anything more for "boosting" detergent than adding active oxygen would?

Borax was used as a "builder" in washing formulations long before perborate was used as a bleach. "Builder" is the terminology for additives which reduce the influence of water hardness. Other examples of builders include trisodium phosphate, soda, polyphosphates (like Graham's salt) or today most often aluminosilicate ion exchangers (Phosphates are more or less forbidden nowadays because of eutrophication of rivers and seas).


What is the difference between reversible and irreversible work done in adiabatic process?

Your problem lies in your analysis of the irreversible expansion.

In the case of both a reversible expansion and an irreversible expansion, the gas pressure at the interface with the piston matches the externally applied pressure during the deformation. However, in the irreversible expansion, the gas within the cylinder is not uniform with respect to pressure or with respect to temperature.

So the interface pressure (which determines the amount of work done) does not match the average pressure of the gas within the cylinder. In addition, the pressure at the interface depends not only on the gas volume but also on the rate at which the gas volume is changing. So for an irreversible process, we have very little control over what is happening inside the cylinder.

The only thing we can control is the pressure at the interface $p_{ext}$ where the work is being done. We do this by whatever means necessary, including feedback control systems. In the typical scenario, we drop the pressure at the interface to a constant value, less than the initial pressure in the cylinder, and then hold it at that value until the gas volume has increased to the point where the gas in the cylinder is again at equilibrium. At this point, the gas pressure throughout the cylinder is described by the ideal gas law again.

If we follow this game plan for the irreversible process, even if the final pressure matches the value be obtain in the reversible process, the final temperatures and the final volumes will not match.

If you would like me to provide a detailed analysis of the irreversible case, illustrating in detail what transpires, I will be glad to do so. But right now, I'll give you a chance to ask some questions.

Irreversible and Reversible Expansions

Initial conditions:  $T_i, P_i, V_i$

Irreversible expansion:

$$\Delta U=-\int_{V_i}^{V_f}{P_{ext}dV}$$
In the irreversible expansion we are considering, $P_{ext}$ is controlled to be constant at the final pressure $P_f$ with $P_f<P_i$. The expansion is allowed to continue until the gas re-equilibrates at the final pressure and volume. Therefore, we have:
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)\tag{1}$$

As noted in my discussion above, the ideal gas law cannot be applied to the intermediate states during the irreversible expansion, but it can be applied to the two equilibrium end states. Therefore, $$V_f=\frac{nRT_f}{P_f} \tag{2}$$
and
$$V_i=\frac{nRT_i}{P_i}\tag{3}$$If we combine equation $(1)$, $(2)$ and $(3)$, we obtain:$$C_v(T_f-T_i)=-R\left(T_f-\frac{P_f}{P_i}T_i\right)\tag{4}$$

We can solve this equation for $T_f/T_i$ and a function of $P_f/P_i$ to obtain:
$$\frac{T_f}{T_i}=\frac{1+(\gamma-1)(P_f/P_i)}{\gamma}\tag{5}$$
Note that this equation differs from the relationship one obtains if the expansion is carried out reversibly.

Reversible expansion:
For a reversible expansion, we have $$dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$$ or $$C_vd\ln T=-Rd\ln V=-R(d\ln T-d\ln P)$$This integrates to:
$$\frac{T_f}{T_i}=\left(\frac{P_f}{P_i}\right)^{\frac{(\gamma-1)}{\gamma}}\tag{6}$$
Note from equations 5 and 6, that if the temperature ratios are the same in both cases (equal work), the pressure ratios are not, and if the pressure ratios are the same in both cases, then the temperature ratios are not (unequal work).


NGP mechanism vs the simple carbocation mechanism

From Clayden (Chapter 37 Rearrangements page 976):

  Intramolecular reactions (including participation of a neighbouring group) that give three-,
  five-, or six-membered rings are usually faster than intermolecular reactions.

It also gives the exact same example as your question:

Notice that both products are given to be exactly 50% each.

NGP is very often faster than bimolecular reactions as for the latter to take place two molecules have to collide first.

1:1 ratio can be attributed to the fact that in NGP the transition state involves loss of aromaticity BUT NGP supports the ionization of the starting material.

There are a lot of examples  this one reaction isn't a black swan. Interested readers may see chapter 36: Participation, rearrangement and fragmentation in Clayden to find more details and examples.


If you want to analyse a compound, it is best to start with the $d$-electron count of the metal. For $\ce{Cu^{2+}}$ we get $d^9$, since $\ce{Cu^0}$ would be $d^{11}$.

Now, consider the following (simplified) molecular orbital diagram for a $d^1$ compound:

As you can clearly see, the ligand electron pairs mix with some of the $d$, the $s$ and $p$ orbitals of the metal.

Now, instead of $d^1$ we have $d^9$, so we just start filling in 8 more electrons into the MOs. This gives us a full $t_\text{2g}$ MO and 3 electrons in the $e_\text{g}$ MO. ($t_\text{2g}$ and $e_\text{g}$ are ways of describing the symmetry of the orbitals. This comes from applied group theory.)

So you see, the $5s$ orbital never comes into play here, only the $3d$, $4s$ and $4p$ orbitals of the metal take part in bonding.

You might want to look at Ligand Field Theory and Molecular Orbital Theory to further your knowledge in inorganic chemistry. Not everything is as simple as the "fill in the boxes" method.


Equation for overall heat transfer coefficient

This is the heat equation analogue to Ohm's law, in which resistances are additive (in series), and hence the inverse conductances are additive.

The way to reason is that the full equation (1) for the heat conductivity (1) contains three terms: one for the interior of the tube, one for the tube wall, and a third for the exterior. When you look at the profile, it becomes evident that, since temperatures immediately adjacent inside and outside the tube are equal, the tube has infinite heat conductance ($k=\infty$).

The right answer is therefore 3.


Steady-state approach - how to find rate equation for intermediate?

Split the scheme into three parts and write down each step. Start by definition with
$$\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=+\frac{d[\ce{N2O2}]}{dt}$$
then the first step $\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=k_1[\ce{NO}]^2\quad$ or $\quad\displaystyle \frac{d[\ce{NO}]}{dt}=-2k_1[\ce{NO}]^2$
and so $\ce{N_2O_2}$ is produced as
$$\displaystyle \frac{d[\ce{N2O2}]}{dt} = k_1[\ce{NO}]^2$$
In the reverse equilibrium step $\ce{N2O2}$ is lost (by dissociatiing) so $\displaystyle \frac{d[\ce{N2O2}]}{dt} = -k_1^{'}[\ce{N2O2}]$
and in the last reaction it is also lost by reaction with oxygen therefore $\displaystyle \frac{d[\ce{N2O2}]}{dt}=-k_2[\ce{N2O2}][\ce{O2}]$.
Adding up all three rates for $\ce{N_2O_2}$ gives
$$\displaystyle \frac{d[\ce{N2O2}]}{dt}=k_1[\ce{NO}]^2-k_1^{'}[\ce{N2O2}]-k_2[\ce{N2O2}][\ce{O2}]$$
You can now find the steady-state conditions.
(In your last equation there is $2k_1$ and it is quite common to see it written this way, all it does is to change $k_1$.)


Nomenclature of complex benzene-based substituents

Intro

Your statement, that complex substituent names are placed in square brackets, is incorrect. These are to be placed in parentheses (round brackets). Correct preferred IUPAC name (PIN)[1] example:

  2-(3-methylbutyl)phenol (PIN)

Square brackets are used when the substituent name already contains parentheses, e.g.:

  2-[(2R)-2-methylbutyl]phenol (PIN)

Retained aromatic compounds allowed for substitution

You can see that the names above are based on phenol, which is one of retained aromatic functional parent compounds, with allowed substitution, in preferred names.

The other ones are aniline and benzoic acid (see IUPAC rule P-34.1.2.[1]). E.g.:

  2-hydroxybenzoic acid (PIN)
  (not 2-carboxyphenol, as $\ce{-COOH}$ group is preferred over $\ce{-OH}$)

Retained alkyl-substituted aromatic hydrocarbons (not allowed for substitution in PINs)

There are also retained alkyl substituted aromatic parent hydrocarbons (P-22.1.3):

  toluene (PIN)
  methylbenzene

and

  1,2-xylene (PIN)
  o-xylene
  1,2-dimethylbenzene

(and the two other xylenes).
But none of them is allowed for substitution in preferred names, e.g.:

  1-chloro-4-(chloromethyl)benzene (PIN)
  α,4-dichlorotoluene

Xylenes are not allowed for substitution even in the general non-preferred names:

  1,2-bis(bromomethyl)benzene (PIN)
  α-bromo-2-(bromomethyl)toluene
  (not α,α′-dibromo-o-xylene)

(Yet another retained name mesitylene for 1,3,5-trimethylbenzene (PIN) is allowed only for general names, not allowed for substitution.)

Complex aromatic substituents

Now if a complex substituent is made from substituted phenol, aniline or benzoic acid, the substituent name is no longer based on these names. They can be thought as suffixes, but now we have another higher-priority “suffix”, -…yl. E.g.:

  3-hydroxy-4,5-dimethylphenyl acetate (PIN)
  (not 2,3-dimethylphenol-5-yl acetate)

(In case you didn't hear ‘phenyl’ yet – it's not to be confused with ‘phenol’, it's the only, notorious, name for “benzenyl”, $\ce{-C6H5}$)

References:

Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)


Is there a relationship between pH indicator colors and the visible light spectrum?

It has to do with the way pH indicators and particularly large-range pH indicators work.

In simplest terms, a pH indicator is simply a weak acid/base which has one color in its acidic form, and one color in its basic form. When the solution is highly acidic, all of the indicator molecules are protonated, and the color of the indicator is the color of the protonated form. When the solution is highly basic, all of the indicator molecules are deprotonated, and the color of the indicator is the color of the deprotonated form.

Where it gets interesting is where the pH of the solution is close to the pKa of the indicator. In those situations, you have some of the indicator molecules in the protonated form, and some of them in the deprotonated form. You don't ever get "fractionally protonated" forms - each individual molecule is either one or the other. Since light absorption happens on a molecular level, you get

You can see this with a titration of something like Bromothymol Blue [image]. The protonated form is yellow, so at low pH, everything is yellow. The deprotonated form is blue, so at high pH, everything is blue. At around the pKa (~7), there's near equal amounts of the two forms, so it acts like you mixed equal amounts of a yellow pigment/dye and a blue pigment/dye. That is, you get a green solution due to the subtractive color mixing model. As you titrate the pH, you get a range of colors as you add and remove the amount of protonated and deprotonated forms, as would be indicated by the Henderson–Hasselbalch equation.

Okay, so that's a single pH indicator, which is normally limited to a small range. (Because more than ±1 pH unit to either side of the pKa you typically have only a single species present.) To get broad range pH indicators, you have to mix multiple different pH indicators, each with their own different colors and changes. But you have to do this intelligently, as the indicators are always present, and will always contribute to the color.

Here's a chart of some of the major pH indicators, their pH ranges and their colors. You can see what might happen with various indicator combinations. For example, with an ethyl violet/ethyl orange mixture, at low pH you'd have an orange solution (yellow+red), which would transition to be purple (blue+red) at around pH 3 and then to be green at high pH (blue+yellow). But those intermediate colors are going to be pretty muddy. At around 1.5 you'll have an orangy-purple, and at around 4.5 you'll have a greeny-purple.  Imagine what things would look like if you mixed orange paint with purple paint, or purple paint with orange paint.

What we need is a layout of color that can gradually transition one to the other and look good all the way through the approximately 7-ish transitons you'd need to segment up a 14 unit pH scale ... right, the rainbow spectrum has that property.

So there's no real chemical reason for the connection. It's more a practical human consideration, where the people combining the indicator molecules to make a large-range pH indicator need to pick particular ones which make a clear, unambiguous color sequence, and the visual color spectrum has the same useful properties they need.


Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium

From Proceedings of the Symposium on Oxygen Electrochemistry and Water Research 2002, 36 (1), 85-94:

$$
\begin{array}{lcc}
\text{Formation}&:&~\ce{2H+ + O2 + 2e- &->& H2O2} &\quad E^0=\pu{0.440V} \text{ vs. SCE}\\
\text{Reduction}&:& ~\ce{H2O2 + 2H+ + 2e- &->& 2H2O} &\quad E^0=\pu{1.534V} \text{ vs. SCE}\\
\text{Direct water}&:& ~\ce{2H+ + 1/2O2 + 2e- &->& H2O}\\
\text{Decomposition}&:&  ~\ce{H2O2 &->& H2O + 1/2 O2}
\end{array}$$

($\text{SCE}$ stands for Saturated Calomel Electrode)

Another strange suggested reaction involves the anion $\ce{HO2-}$ (Abel E.: Über die Selbstzersetzung von Wasserstoffsuperoxyd. Monatshefte für Chemie 1952 83(2) 422–39.):

$$\text{Decomposition 2}:\ce{H2O2 + H2O- -> H2O + O2 + OH-}$$

At the cathode occur at the same time the hydrogen gas evolution:

$$\ce{2H+ + 2e- -> H2~} \qquad\qquad\phantom{00} E^0=\pu{0.242 V} \text{ vs. SCE}$$

Meanwhile, at the anode, oxygen gas is evolved:

$$~\ce{2H2O -> 4H+ + O2 + 4e-} \quad E^0=\pu{0.987V} \text{ vs. SCE}$$


Reaction mechanism for reduction of nitrobenzene to aniline in catalytic and acidic media

The first reaction is called a catalytic reduction.  Here is a picture of how it works with an olefin like ethylene.

You can see that the first step involves adsorption of hydrogen onto the catalyst surface followed by dissociation of the hydrogen molecule into its catalytically active form.  The same things happen with nitrobenzene.  As the following diagram shows, the reduction passes through the nitroso and hydroxylamine intermediates.

The mechanism for the Sn\HCl reduction involves reductive electron transfer of the nitro group while the tin is being oxidized. This diagram shows the electron transfers involved

and here is the mechanistic scheme.


The position of equilibrium depends on temperature and pressure, as Nick Burns describes in his answer. The equilibrium constant depends on temperature only. The equilibrium constant is not a measure of the position of equilibrium, but rather which positions (representing different compositions) along the the reaction coordinate surface are equilibrium positions. Changing the temperature changes the shape of the reaction coordinate surface. Changing the pressure just moves you to a different point on the surface.

Example: consider a hypothetical reaction in which two molecules of gas react to form a third.

$$\ce{A(g) + B(g)<=>C(g)}$$

The reaction quotient $Q$ is simply the ratio of the product of the partial pressure(s) of the product(s) to the product of the partial pressure(s) of the reactant(s). $Q$ can have any value representing different points on the reaction coordinate surface.

$$Q=\frac{P_C}{P_A P_B}$$

The equilibrium constant represents special values of Q in which the reaction coordinate surface has a free energy minimum. For the sake of our example, let's make the equilibrium constant an easy value to work with, say $K=2$.

$$K=Q_{eq}=\frac{P_{C,eq}}{P_{A,eq} P_{B,eq}}=2$$

Since we are talking about gases, we also have the constraint the sum of the partial pressures of the reagents must equal the total pressure of the system. Again, for ease, let us set this value to an easy value to work with, say $P_T = 7 \text{ kPa}$.

$$\frac{P_{C}}{P_{A} P_{B}}=2$$
$$P_A + P_B + P_C = 7$$

Now we have two equations of three variables, which means there are an infinite number of possible values of the partial pressures of the reactions components that fit these two conditions. One is $P_A = 1$, $P_B =2$, and $P_C =4$. Another equally valid combo is $P_A = 2$, $P_B=1$, and $P_C=4$. The position of the equilibrium in this case is better given by a ratio of the sum of the partial pressure(s) of the product(s) to the sum of the partial pressure(s) of the reactant(s):

$$\frac{P_C}{P_A + P_B}=\frac{4}{3}$$

If we change the total pressure to, say $P_T = 4$, we change the allowable combinations of $P_A$, $P_B$, and $P_C$ that can get us to the equilibrium condition of $Q=2$. One allowable combination (of an infinite number) is $P_A=1$, $P_B=1$, and $P_C=2$. The position of the equilibrium is now:

$$\frac{P_C}{P_A + P_B}=\frac{2}{2}=1$$

Le Chatelier's principle suggests that When pressure decreases, the equilibrium shifts toward the side of the reaction with more equivalents of gas.

If we consider changing the situation from $P_A = 1$, $P_B =2$, $P_C =4$, and $P_T=7$ to a situation with $P_T=2$, we can do some math to figure out new values of $P_A$, $P_B$, and $P_C$ that fit this new situation. First, we set the partial pressures of the compoents to $\frac{2}{7}$ of their original value, and calculate Q.
$$P_A =1\times \frac{2}{7}=\frac{2}{7}$$
$$P_B =2\times \frac{2}{7}=\frac{4}{7}$$
$$P_A =4\times \frac{2}{7}=\frac{8}{7}$$
$$Q=\frac{P_C}{P_A P_B}=\frac{\frac{8}{7}}{\frac{2}{7} \times \frac{4}{7}}=\frac{\frac{8}{7}}{\frac{8}{49}}=49$$

Since $C>K$ the equilibrium will shift to decrease $Q$. This can happen by decreasing $P_C$ and increasing $P_A$ and $P_B$. Since there is an reaction equation relating these three components in a 1:1:1 stoichiometric ratio, the change must have the same value vut in different directions:

$$K=2= \frac{\frac{8}{7}-x}{\left(\frac{2}{7}+x \right)\left( \frac{4}{7}+x\right)}$$

Rather than goof up the algebra, I let WolframAlpha do it. The solution for $x$ is:

$$x=\frac{1}{128}\left(-19 \pm \sqrt{681} \right)\approx \{+0.2534..., -1.6106...\}$$

The first answer for $x$ is nonsense, since it would put one partial pressure less than $0$. If we change the partial pressures appropriately, we get new values of

$$P_A = \frac{2}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.5391...$$
$$P_B = \frac{4}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8248...$$
$$P_C = \frac{8}{7}-\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8894...$$

The position of equilibrium is

$$\frac{0.8894...}{0.5391...+0.8248...}\approx 0.65$$

which is less than the value of $1.33$ from when $P_T=7$ and different from the value of $1$, when the partial pressures were different (but still met the criteria than $P_T=2$ and $Q=2$).

Changing the temperature changes the values of $Q$ which define the equilibrium state:

$$K=e^{-\Delta G^\circ/RT}$$


Chlorine has 17 protons and Sulfur has 16. In the 1s orbital there is no ineer electron to provide shielding effect of repulsion. So greater the positive charge, greater is the attraction and better is the binding.


Why is the benzyne triple bond distorted?

First, Wikipedia doesn't really say what is the extent of this change. Here are superimposed a benzene and a benzyne molecule (own HF calculation, don't ask for the details):

The carbon atoms involved in the triple bond are displaced by less than $\pu{0.2 Å}$. In benzene, we have with this level of calculation:

$\ce{C–C}$ distance = $\pu{1.39 Å}$
$\ce{C–H}$ distance = $\pu{1.08 Å}$
$\ce{C–C–C}$ angle = $120^\circ$

while in benzyne, we have:

$\ce{C–C}$ triple bond = $\pu{1.22 Å}$
other $\ce{C–C}$ bonds = $\pu{1.38 Å}$ (almost all equal)

So, merely because the triple bond is stronger, the distance will be shorted and the benzyne structure is distorted.

Regarding orbitals, well, the p orbitals of the carbon atoms involved in the triple bond are “naturally” directed towards where the hydrogens would be. Slightly distortion of the system is possible, and indeed evidence above on the structure, but they will not be parallel, as that would disrupt the structure too much and diminish the $\sigma$ bond. So it's a competition between: making the new p orbitals “more parallel” increases overlap, but it weakens the $\sigma$ bonding pattern, which is much stronger. As such, while some distortion happens, you stay mostly close to the benzene-type orbitals.


That structure is incorrect. Several online chemical distributors have similar structures posted in their entries for aluminum carbide; so does PubChem! Some sources show the structure as linear; some show it as bent.

That structure basically assumes that aluminum forms primarily covalent bonds with an "incomplete octet" in VSEPR structures, rather like boron. But aluminum atoms are a lot larger and more metallic than boron atoms, and its compounds will have comparatively more ionic character. $\rm Al_4C_3$ isn't a molecular compound.

Cotton & Wilkinson's Advanced Inorganic Chemistry says that aluminum carbide "reacts instantly with water to produce methane, and X-ray studies have shown it to contain discrete carbon atoms (C-C = 3.16 Å); for these reasons it is sometimes considered to be a "methanide", that is, a salt containing $\rm C^{4-}$, but this is probably an oversimplification."

This is the structure that's posted on Wikipedia:

It says that "Aluminium carbide has an unusual crystal structure that consists of two types of layers. It is based on AlC$_4$ tetrahedra of two types and thus two types of carbon atoms. One is surrounded by a deformed octahedron of 6 Al atoms at a distance of 217 pm. The other is surrounded by 4 Al atoms at 190–194 pm and a fifth Al atom at 221 pm." This is attributed to Greenwood's Chemistry of the Elements, but I don't see it in the edition I have.


Is dissolving salt a chemical reaction or a physical change?

I would think that dissolving a non-reactive salt, like $\ce{NaCl}$ in water can be considered as a physical change. This is because, you can recover the salt by evaporating the water away.

From the Wikipedia page of physical change

  In general a physical change is reversible using physical means. For example, salt dissolved in water can be recovered by allowing the water to evaporate.

But the distinction between physical and chemical change can be misleading and confusing at times. In general, anything that does not change the chemical constitution of the substance, is a physical change.

When you dissolve $\ce{NH3}$ in water, the following equilibrium is set up:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

But if you heat the water up, you will get back the original ammonia without any chemical change in its composition. However, the ammonia does change its chemical composition in water, as $\ce{NH4+}$ ions are formed. By that argument, you can say that the dissolution of an ionizable salt is a chemical change.

In some cases, dissolution of a compound can be called a true chemical reaction. For example, when you dissolve Calcium oxide ($\ce{CaO}$) in water, a vigorous chemical reaction takes place, forming $\ce{Ca(OH)2}$. You cannot get back the original $\ce{CaO}$ by just evaporating the water. In case of aqueous solution of $\ce{AlCl3}$, if you try to evaporate the water by heating the solution, you will get back only $\ce{Al(OH)3}$ as anhydrous $\ce{AlCl3}$ reacts with water as soon as it is in contact with water, to form $\ce{Al[H2O]6Cl3}$.

What I am trying to say is that some changes can be classified into one of these types, while some other changes can not be given a clear classification. Melting of water is clearly a physical change, while reaction of $\ce{H2}$ and $\ce{O2}$ is clearly a chemical change. One another fact, putting paint on a canvas is not always a physical change. In fact some paints are manufactured in such a way that during drying, some chemical reactions occur, due to which the paint fixes firmly on the surface. This is the reason many water thinnable paints can not be dissolved in water once they become dry.

To conclude, you should not hold hard classifications like that in your mind. However, my final verdict: Dissolving a salt is physical change (mostly).


Custom-defined functionals in Gaussian 09

Take a look at the IOp manual. Under IOp(3/74),

  IOp(3/74) Type of exchange and correlation potentials.

  -5 Becke3 using VWN/LYP for correlation.
  02 Lee-Yang-Parr correlation.
  05 VWN 80 (LSD) correlation.
  200 Hartree-Fock-Slater exchange (Alpha = 2/3).
  400 Becke 1988 exchange.

  So 100 is Hartree-Fock, 200 is Hartree-Fock-Slater, 205 is Local Spin Density, and 402 is BLYP.

They're unclear about it, but this means you can take the "sum" of a 1- or 2-digit number (which specifies the correlation) and a 3- or 4-digit number (which specifies the exchange) to build an exchange-correlation functional. So, IOp(3/74=-5) == IOp(3/74=402), and an identical way to define B3LYP is

IOp(3/74=402) IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

  Does that mean, that the $E_{c}^{\text{local}}$-part

  cannot be defined,
  is always the VWN functional as in B3LYP, or
  is also the chosen correlation functional?

Looking at the available correlation functionals (incomplete, these are the lowest few),

01 VWN5
02 LYP
03 P81 (PL in keyword list)
04 P81 + P86
05 VWN3
06 VWN3 + P86
07 OS1
08 PW91
09 PBE
10 VSXC
11 Bc96
18 VWN5 + P86
19 LYP + VWN5 for scaling

your choices for local correlation functionals are rather limited. Either you choose a correlation functional that's purely local (VWN3, VWN5, or P81), or a non-local/gradient-corrected functional which in Gaussian all seem to have the local component as part of their definition (as in LYP), or they're forcibly combined (as in P86).


How would an increase in the concentration of a sucrose solution affect the rate of its acidic hydrolysis?

Looking at the reaction mechanism (of which I understand you are aware)
is good to keep in mind that the reaction rate really depends on the amount of intermediate here, provided the second step is rate limiting.

In the first step, $H^+$ is consumed. Since it is not rate limiting, I can assume the intermediate $[sucrose.H^+]$ to be in pseudo-equilibrium with $[H^+]$&$[sucrose]$, on the timescale of the second reaction step.

$$r_2 = k_2[sucrose.H^+]$$

This means adding more sucrose (provided you have enough acid to protonate it) will indead result in the formation of more [sucrose.H^+] intermediate, and thus, higher kinetics.

Btw, I'm simpifying stuff here by not taking into account the reaction backwards for the second step.


Do electronic balance weighings still need to be corrected for buoyancy error?

A balance cannot determine the density of the weighted material and air automatically. For this reason a buoyancy error correction must be applied for high precision measurements.

Suppose you weigh $\pu{100.0000 g}$ water and have $\pu{100.0000 g}$ reference weights with a density of $\pu{8.0000 g cm^{-3}}$. Then the static buoyancy in air is equivalent to about $\pu{120 mg}$ and $\pu{15 mg}$ respectively, accounting for a difference of about $\pu{105 mg}$ in the reading of the balance.

Besides the static buoyancy there may also be dynamic buoyancy due to air flow caused by temperature differences between the weighted material and the surrounding air.

Dynamic buoyancy can be avoided be weighing in thermal equilibrium. Static buoyancy can be corrected by formulae involving the density of air, the density of the weighted material and the density of the reference weights.


Preferred IUPAC name of Protoporphyrin IX

I think both EPA and ChemSpider names are acceptable. However, I don't agree with dihydroporphyrin part in PubChem name. This is because the Gold Book gives following framework for porphyrin nucleus:

According to the numbering the IUPAC name of protoporphyrin IX is 3,3'-(7,12-Diethenyl-3,8,13,17-tetramethylporphyrin-2,18-diyl)dipropanoic acid. This name based on least locant numbers for ethynyl group, which have higher priority before methyl groups (3,3'- is based on positions of propanoic acid). The same name is given to protoporphyrin IX in Ref.2, without 3,3'- locants.

References:

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
IUPAC-IUB Joint Commission on Biochemical Nomenclature (JCBN): Nomenclature of tetrapyrroles (Recommendations 1986); Prepared for publication by Gerard P. Moss, Queen Mary College, London, Eur. J. Biochem. 1988, 178(2), 277-328 (https://doi.org/10.1111/j.1432-1033.1988.tb14453.x) (PDF).


The tl;dr answer is that, as the question is worded (and without an actual copy of the question or the data provided to you), its not possible to definitively assign a BCS class.

Nifedipine is a class II drug:

  class II is low solubility and high permeability. Formulation typically is used to enhance solubility of compounds in this class.

The data in your question is a little confusing, but you state that the compound is 'practically insoluble' which means that it must be either class II or class IV (practically insoluble definitely suggests low solubility).

With the information provided (i.e. nothing about permeability), you cannot distinguish between II and IV. Permeability, whilst often correlated to solubility, is not directly linked, such that you cannot attempt (as I think you have tried) to calculate one based on the other (how permeable something is depends upon structure, and whether it can get through the membrane).


How can I find out how much caffeine is in my decaf tea and coffee?

You may try to extract the caffeine with DCM and determine its concentration by spectrophotometry.


Is there a rule of thumb to predict when coupled cluster might dip below the variational limit?

Let us begin with the following principle: the CC ansatz always returns an appropriate electronic wavefunction (meaning a combination of Slater determinants). Still, at least in a non-relativistic approach, the Hamiltonian does commute with the Spin operator
\begin{equation}
S^{2}=S_{+}S_{-}-S_{z}+S_{z}^{2}
\end{equation}
(I'm using atomic units here)
and we want the wavefunction to be an eigenfunction of total spin. The difficulty in doing so is due to the fact that the excitation operator to ensure spin purity is not easy to parametrize unless you are lucky enough to have a $S^{2}=0$ state and your system wavefunction is represented by a $S^{2}=0$.
So, when you're starting with a wavefunction which is not a spin eigenfunction (for example the ground state of the nitrogen atom with the unpaired electrons), CC might not be able to recover the spin purity and, by allowing superposition of functions with different spin, you might get an energy lower than the correct one because you're looking for the minimum in a space bigger than the one you are physically allowed to .
Hope I was clear


Logical reasoning behind the formula for molar conductance of an electrolyte

For most physical properties of substances, it's useful to know how those properties depend on other parameters.  For example, the various coefficients of thermal expansion (CTEs) indicate how rapidly the dimension(s) of a substance change when the temperature is changed by a given amount. A common CTE used when working with materials is the linear CTE:

$$
\alpha_\mathrm L = {1\over L} {dL \over dT}
$$

When thermal expansion occurs, if the temperature change is relatively small the amount of thermal expansion is roughly proportional to the original length $L$. So, the definition of $\alpha_\mathrm L$ divides the length dependence $dL/dT$ by $L$ to yield a parameter that is approximately constant within a particular temperature range.

A similar rationale underlies the definition of the molar conductance (also called the 'molar conductivity') as:

$$
\Lambda_\mathrm m = {\kappa \over c}
$$

The conductivity $\kappa$ of an electrolyte is a strong function of the concentration $c$ of the electrolyte of interest. However, for electrolytes containing only strong acids and/or bases (or salts of these), $\kappa$ is nearly a linear function of $c$ over a very wide concentration range, and thus $\Lambda_\mathrm m$ for these electrolytes is nearly constant.

Thus, for species that show this strong linearity, defining $\mathbf{\Lambda_m}\equiv\kappa / c$ is useful because a single $\mathbf{\Lambda_m}$ value can be employed to calculate the absolute conductivity $\mathbf\kappa$ across a wide range of electrolyte concentrations $c$, as:

$$\kappa = \Lambda_\mathrm m c$$


There is also a disproportionation reaction that converts toluene to a mixture of benzene and xylene (Bawa et al), also, see wikipedia.  It has commercial applications (because the benzene/xylene mixture can often be sold for more than toluene) and can be licensed.  However, it is more of an industrial process than one you would use in the lab, as it is really driven by an exothermic dealkylation in vapor phase over zeolite catalyst.

Interestingly, as zeolites act as a Lewis acid, this is strangely similar to Friedel-Crafts alkylation, which uses Lewis acid catalysis in liquid phase.  The paper by Bawa et al discusses experiments in which the trans alkylation was done with Frieden-Crafts catalysts in liquid phase.


How to determine the theoretical carbonic acid concentration in soda water from a Soda Stream

The equilibrium you describe is quite well documented (especially with entire industries depending on their ability to carbonate drinks), so the Wikipedia page for carbonic acid is often a good place to start. The equilibrium constant, ${K_\mathrm{h}}$, is given by:

$$\ce{\frac{[H_2CO_3]}{[CO_2]}}=\pu{1.7e-3}$$

Now, for your first question:

  How can one determine the gaseous $\ce{CO_2}$ concentration knowing only the temperature (assume room temperature) and the pressure (above).

Well, for starters you should probably check the volume of the container and from there you could work out the amount of $\ce{CO2}$ that is above the solution. But if you are interested in the amount inside solution, you could check this paper published in 1940 which has basically compiled the data for you.

And the second part:

  How can one determine the carbonic acid concentration knowing only the dissolved $\ce{CO2(g)}$ concentration?

Now that you know the equilibrium constant, I'm sure you can work out the amount of carbonic acid.


Effects of Groups on the Nucleophilcity of Nitrogen

Your analysis of "pulls more electron density away" is a bit too simplistic. Both of them are approximately equally electron-withdrawing via the inductive effect, but in terms of resonance, which is the main factor here, they are quite different.

Compound I is an amide. Amides are non-nucleophilic on nitrogen because the N lone pair is delocalised into the C=O π* orbital. In fact, amides act as nucleophiles via the carbonyl oxygen and not the nitrogen.

In compound III, there is also the same delocalisation, but a way of looking at it is that the other oxygen (on the -OMe group) also wants to donate its lone pair into the C=O π*. In a sense this means that there is some degree of competition between the nitrogen and the oxygen, and the nitrogen lone pair is no longer delocalised to such a great extent.

Formally speaking, I think the best way of putting it is that the HOMO of compound III has a greater coefficient on nitrogen than the HOMO of compound I. You can also draw resonance forms to show that there is a smaller contribution from the resonance form with positively charged nitrogen.


How does Le Chatelier's principle apply to a gas phase reaction at constant volume?

Let's see your formula (with the difference of enthalpy being negative):

$$\ce{2CO (g) + O2 (g) <=> 2CO2 (g)}\qquad(\Delta H^\circ<0)$$

At constant temperature, for Le Chatêlier purposes,

Adding $\ce{CO}$ produces that it react with $\ce{O2}$ to produce more $\ce{CO2}$ to counteract the effect of the applied change and a new equilibrium is established.
Removing $\ce{CO2}$ produce that $\ce{CO}$ react with $\ce{O2}$ to produce more $\ce{CO2}$.

When you are adding or removing a reagent or product, and all components are in a gaseous state, you are changing the pressure and you can study the reaction by $K_p=\dfrac{P^2_\ce{CO2}}{P^2_\ce{CO}P_\ce{O2}}$.

We also have $K_p=K_c$ $RT^{Δn}$:

$K_p$ is the equilibrium constant written with partial pressure of gases.
$K_c$ is the constant but being written by concentration of substances.
$R$ is universal gas constant.
$T$ is the absolute temperature (Kelvin degrees).
$\Delta n$ are moles $(\Delta n=n_{\text{products}} - n_{\text{reagents}})$.

If you change concentration of $\ce{CO}$ you are changing the pressure it, too.

But you can change also the pressure, by changing the volume. (For Boyle's Law: the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system).

For example, if you increase the pressure, volume will decrease and the system will shift towards it will be less molecules to counteract the change. In this case, to products. If you in decrease pressure, volume will increase and the system will shift towards reagents.

Other option is to change the pressure by adding or removing an inert gas, but since it don't intervenes in the reaction, the equilibrium is not affected.


A good textbook to balance chemical equations?

This isn't a textbook, but there is a way to balance pretty much any chemical equation algebraically.

Consider the reaction $$\ce{C4H10 + O2-> CO2 + H2O}$$

To start, you would put a variable in front of each molecule here.
That would become:  $$\ce{aC4H10 + bO2 -> cCO2 + dH2O}$$

Now, count the amount of the same elements as they appear in the different molecules. In this example, there are 4 carbons in the 'a' molecule, and 1 carbon in the 'c' molecule. I would thus write 4a=1c
Repeating this process for the other elements, I get:
10a=2d and 2b=2c+d (because oxygen appears in both molecules on the right side).

After this step, you have your three equations.
4a=c
5a=d (This is simplified from 10a=2d)
2b=2c+d

Now, choose one of those variables, and make it equal 1. It can be any of your choosing, the end outcome will be the same. In this case, making 'a' 1 will make the whole process much easier, so let's do that. a=1.
From this, it follows that c=4, d=5, and b= (8+5)/2 or 13/2.
Now if there are any fractions, multiply every single number by the denominator to make everything a whole number. Multiplying by 2 in this case yields:
a=2 b=13 c=8 and d=10
Now put those values exactly where you originally put the letters.
It becomes $$\ce{ (2)C4H10 + (13)O2  ->  (8)CO2 + (10)H2O}$$

So the steps are:
Assign variables to each molecule in the reaction.
Write out those variables as equations corresponding to the amount of certain elements in each molecule.
Make one variable equal 1.
Solve for the rest of the variables.
Multiply by the denominators of any fractions to leave whole numbers.
Those numbers are your answers. You just plug them in :)

If you want any clarification, or another example done, just comment and I'll be happy to.

Note: If someone could edit my post, or comment how to use the molecular writing thing, that would be awesome.