Is there a way to fine tune a laser to excite the electrons of a particular molecular bond into dissociating?

Obviously if you have a diatomic then yes, you can put energy in and the bond will break. There are three cases; (a) as long as the photon has enough energy it can excite a dissociative excited state, (grey in sketch) that is a state that is only repulsive. The absorption is continuous here so its energy has to be above a certain value. (b) The second case has a curve crossing in the excited state. This means that two excited states cross , a bound state and a dissociative one. Then if you hit the energy at the crossing point, dissociation will occur, but not at higher or lower energy. This is called pre-dissociation. (The read & grey line crossing show pre-dissociation). It occurs in many molecules not just diatomics, e.g. benzene. (c) The third case is when a bound state is excited above its dissociation limit, e.g. above approx 10000 in the blue curve, 5500 on the black curve . However, reaching these energies may be hard because the Franck-Condon factors may be vanishingly small. (roughly speaking the FC factors ensure vertical transitions)

Now the case with polyatomic molecules is tricky. Many experiments have been tried with varying success, the area is sometimes called 'mode selective chemistry' and/or 'coherent control'.

In a molecules there are many vibrations, called normal modes. Although we think generally of one bond vibrating, in actuality many bonds vibrate in unison (i.e. in a fixed phase relationship) in a normal mode. These modes are governed by the symmetry species of the molecule. Each normal mode is supposed to be orthogonal, so that vibration of one does not effect vibration of another. But in fact vibrational anharmonicities allow vibrational quanta to flow from one mode to another. Thus exciting one normal mode soon involves energy spreading out into all other modes and thus even if there is enough energy to break a bond it soon becomes diluted.

In coherent control experiments one uses one or more very short light pulse ( < 100 femtoseconds) to excite the molecules but control the phase of the light (hence pulse shape) to control how long the molecule takes absorb the photon. This is done in the hope that once the energy starts to spread out (as a vibrational wavepacket) it can be controlled to end up in just the bond you want to break, i.e. to try and use the vibrational spreading out tendency to advantage. The pulse shaping is done using an electro-optic or acousto-optic device in a feedback loop using say, a genetic algorithm linked to a detector that observed product formation.


Why does copper form bubbles in vinegar in this situation?

When you put a zinc rod in a acid solution containing $\ce{H^+}$ ions, the zinc piece loses $\ce{Zn^{2+}}$ ions to the solution leaving electrons behind. These electrons attract hydrogen atoms to the zinc rod, and an electron transfer occurs, resulting in bubbles of hydrogen forming on the surface of the zinc. However, the evolution of gaseous hydrogen from zinc surface is rather slow, and that is why pure zinc reacts slowly with dilute acids.

The situation changes when a copper,or any other metal less electropositive than hydrogen is attached to zinc, by a wire. The copper can now be considered as an extension of the zinc rod: a place form where hydrogen can take electrons quickly and become gas. As the rate of evolution of hydrogen from the copper piece is faster, hydrogen bubbles out from the copper piece. This does not mean, however, that the zinc stops reacting with acid; it does, but at a very slow rate.

A common mistake is to think that the copper rod participates in the electrochemical reaction taking place. The copper is merely a pathway to release electrons to H+ ions in the solution. The only reaction taking place is

$\ce{Zn + 2H^+ -> Zn^{2+} +H2\uparrow}$

Copper has no place in the reaction.

So, to answer your questions,

  Why does this happen?

This happens because copper can do away with the electrons quicker than zinc, and when electrons get a pathway, they will take the pathway.

  Why do some electrons travel through the conductor instead of being attracted immediately by the hydrogen ions in the vinegar?

Well, some electrons do attach to H+ ions from vinegar directly, and hydrogen bubbles should form on the zinc plates, although they are less in amount. And I believe my earlier explanation answers this question.

  What if the conductor instead was very long and the pieces of copper and zinc far apart?

It does not have anything to do with distance. If you take a large solution, the reaction will certainly occur. On the other hand, if you separate the poles into two solutions the copper rod will not give off electrons as the circuit is not complete. Think about it this way, if the reaction was to occur, the solution with zinc will become progressively positive and the solution with copper will become negative. This can never happen.

Footnotes:

(1) As per your comment below, I now attempt to answer the question "What makes copper better than zinc in releasing electrons?"

The answer lies in a factor called overpotential, sometimes referred to as overvoltage. There is one type of overpotential which is important in this case— activation overpotential. This is caused due to the fact that when hydrogen atoms are produced on the surfce of a metal ($\ce{H+ +e-> [H]}$), they take some time before they can combine to form diatomic hydrogen and some more time before the hydrogen molecules can be released as gas. This results in the formation of a layer of hydrogen on the surface of metal effectively reducing the rate of reaction and halting it; this effect is known as polarization.

As you can see from the data chart in the Wikipedia page, Zn electrode has -0.77 V overpotential for evolution of hydrogen while there is no overpotential for copper electrode. This means that copper is not polarized easily.

(2) I mentioned in the beginning of my answer that pure zinc reacts with dilute acid slowly. Commercial zinc, on the other hand, reacts quickly with acids, owing of the same effect of polarization. In commercial zinc, impurities of other metals form cathode and the pure zinc forms anode in the acid medium, essentially producing a galvanic cell, where the anode and cathode are connected. Thus the reaction occurs quickly, without the hindrance of polarization.

(3) You should read this paper : http://pubs.acs.org/doi/abs/10.1021/ed078p516. It has some facts which you may find interesting.


Is it safe to mix potassium alum powder with magnesium oil (made from water and magnesium flakes)?

If you are not sure and you have to do the reaction, try it with small sample of $\ce{Mg}$ such as one flake. $\ce{Mg}$ flakes should react in water and will react with potassium alum.


Extracting elemental ratios from X-ray photoelectron spectroscopy (XPS)

I found some information in Skoog, Principles of Instrumental Analysis, 6th ed.  On page 597:

  Quantitative Applications.  Once, XPS was not considered to be a very useful quantitative technique. However, there has been increasing
  use of XPS for determining the chemical composition of the surface
  region of solids.  If the solid is homogenous to a depth of several
  electron mean free paths, we can express the number of photoelectrons
  detected each second $I$ as $$I = n \phi \sigma \epsilon \eta A T \ell$$ where $n$ is the number density of atoms ($\text{atoms cm}^{-3}$) of the sample, $\phi$ is the flux of the incident X-ray
  beam ($\text{photons cm}^{-2} \text{ s}^{-1}$), $\sigma$ is the
  photoelectric cross section for the transition
  ($\text{cm}^{2}/\text{atom}$), $\epsilon$ is the angular efficiency
  factor for the instrument, $\eta$ is the efficiency of producing photo
  electrons ($\text{photoelectrons/photon}$), $A$ is the area of the
  sample from which photoelectrons are detected ($\text{cm}^{2}$), $T$
  is the efficiency of detection of the photoelectrons, and $\ell$ is
  the mean free path of the photoelectrons in the sample ($\text{cm}$).

  For a given transition, the last six terms are constant, and we can
  write the atomic sensitivity factor $S$ as $$S = \sigma \epsilon \eta A T \ell$$  For a given spectrometer, a set of relative values of $S$
  can be developed for the elements of interest.  Note that the $I/S$ is
  directly porportional to the concentration $n$ on the surface.  The
  quantity $I$ is usually taken as the peak area, although peak heights
  are also used.  Often, for quantitative work, internal standards are
  used.  Relative precisions of about 5% are typical.  For the analysis
  of solids and liquids, it is necessary to assume that the surface
  composition of the sample is the same as its bulk composition.  For
  many applications this assumption can lead to significant errors.
  Detection of an element by XPS requires that it be present at a level
  of at least 0.1%.  Quantitative analysis can usually be performed if
  5% of the element is present.


How does Dichlorofluoromethane get into the stratosphere?

This is one of the more frequently asked questions about ozone depletion. I will try to give a short summary.

First of all, it is important to know that Chlorofluorocarbons (CFCs) are extremely stable at room temperature and pressure. It is therefore only natural, that in the lower atmosphere these gases are enriched, therefore they have plenty of time getting in higher altitudes via diffusion.

The earths atmosphere is in constant motion, you can experience that on a macroscopic level via different winds. The sun heats up the earth's surface, causing all sorts of liquids to be evaporated. While evaporating water, these small molecules may form little conglomerates, taking other molecules up to higher heights. This will happen to dust, so it will also happen to CFCs. There will be local areas (or volumes, layers) of the air in different heights of the atmosphere that have different densities, concentrations and composition. As a result they will also have different air pressures. You can experience this on a plane at higher altitudes. Assuming that all reactions tend to react towards equilibrium, there will be forces acting between these areas. In a consequence there will be also upward forces acting on the CFC containing gas mixture. Slowly these molecules will move upwards to the stratosphere.

As you can see, this process is heavily dependent on many different factors and may well require years. I think it is actually very hard to find any accurate numbers for this process. My research on the subject turned up nothing about that. In most of the cases it was just stated, that these gasses are there.

Some further reading:

Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?
A collection of nature articles. Some are free access.
Ozonewatch @ NASA
http://www.ozzyozone.org/ (for children, in english, french, spanish)
A hole in the sky @ Science in school (different languages available)


Alternating axes are not axes, much like a Guinea pig is not a pig.

Having axes of symmetry is irrelevant to chirality.

Having alternating axes is a necessary and sufficient condition of non-chirality, that's why we look for them.

So it goes.


Why is p-methoxyphenol more acidic than p-methylphenol?

First off, it is a very small difference in $\mathrm{p}K_\mathrm{a}$ that you are asking about.  In fact, I would say that the two compounds have similar acidity.  It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium.  To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion).  If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group.  Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will.  This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side  of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect.  The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring.  Certainly such an effect would be greater with methoxy than methyl.  If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_\mathrm{a}$'s similar to phenol ($\mathrm{p}K_\mathrm{a}=10$), which they do.

The approach used here is sound.  I think the explanation of the substituent effects on the phenoxide are also sound.  The explanation of the substituent effects on the phenol side is (admittedly) weak.  Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.


Naming compounds with multiple -COX functional groups

Systematic names for acyl halides from suffix acids are formed according to Subsection P-65.5.1.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

  P-65.5.1.1 Acid halides in which hydroxy groups of all acid groups expressed as the suffix denoting the principal characteristic group (carboxylic, sulfonic, sulfinic, selenonic, etc. acids) have been replaced by halogen atoms (F, Cl, Br, and I) are named by citing the name of the acyl group (…) followed by the name(s) of the specific class(es) as a separate word(s), in alphabetical order, each preceded by a multiplicative prefix, as needed.

Therefore, the preferred IUPAC name (PIN) for the examples given the question is ‘propanoyl chloride’, ‘propanedioyl dichloride’, and ‘propanedioyl chloride fluoride’, respectively.

However, note that various carboxylic acids were often given trivial names (e.g. reminiscent of their animal or vegetable origin), which are retained for general nomenclature (including functionalization).


How can we justify setting the affinity $\sum_i \mu_i \nu_i$ equal to zero at chemical equilibrium?

Chemical equilibrium is a stable equilibrium, i.e. the system returns to equilibrium to whatever perturbations. This also leads to the condition of equilibrium, as it should  contain this information.

The short answer to your question:

Due to the stability, the system stays/returns to equilibrium at any kind of small perturbation. It also means, $\left(\sum_i \mu_i \nu_i\right)\; d\zeta = 0$ not only in the equilibrium point, but also for any small $d\zeta$. Therefore the $\left(\sum_i \mu_i \nu_i\right)$ itself must be zero.

Somewhat longer answer:

The reaction goes from one direction to the other ($\sum_A \nu_A A \rightarrow \sum_B \nu_B B$)  spontaneously only till it reaches the equilibrium condition. In this direction  $\sum_i \mu_i dN_i \le 0$.

We followed the reaction from a given concentration toward equilibrium. If we follow the same $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$  transformation BEYOND the equilibrium point, $ \sum_i \mu_i dN_i $ must change sign at the equilibrium point, and it must be positive should be valid (since it is not a spontaneous reaction).

Therefore the condition of equilibrium is not $d\zeta = 0 $ , but sign change: $ \sum_i \mu_i dN_i =0$.


How can semi-permeable membrane be so smart?

Most of the time what you say about semi-permeable membranes is correct.

  a semi-permeable membrane is a membrane that allows only solvent molecules to pass between the membrane; solute molecules are blocked. Solvent molecules flow from the membrane side that has from low solute concentration (i.e. high solvent concentration) to high solute concentration (i.e. low solvent concentration).

However, you have to keep in mind several assumptions that are inherent in the definition as you have given it.  The most important of these is that thermodynamic activity differences in the membrane-permeable solvents determine the membrane behavior.  When phases are immiscible, such as the water/oil system in your example, it is essential to think in terms of activities, not in terms of concentration.

  I take oil as a solvent in one side and water as a solvent in other side. I take the same solute in both sides.

So we are imagining a solute that can be dissolved by both water and by oil.  What do we know about the solubility of the solute in these two solvents?  What is the partition coefficient?  And what can we say about the thermodynamic activity difference across the membrane for the three species?  Let's go step by step:

Water.  Water is not soluble in oil to an appreciable degree.  In equilibrated phases, the thermodynamic activity of each chemical species is the same in each phase.  Therefore in an oil-water mixture, the thermodynamic activity of water in the aqueous phase is the same as the thermodynamic activity of water in the oil phase. The extremely low solubility of water in oil means that if there were to be a difference in thermodynamic activities across the membrane, only a few molecules would need to cross over for the condition of equal water activities to be satisfied.
Oil.  Oil is not soluble in water to any appreciable agree.  Therefore the same argument applies.
Solute.  The solute could have a different concentration on each side of the membrane, but because we don't know the partition coefficient, we don't know what the equilibrium state is.  Suppose the solute were membrane permeable.  Even if it was twice as concentrated in water than in oil, if it's partition coefficient favored an equilibrium where the oil-to-water concentration ratio was 10, then it would flow from the water to the oil.  But, in your example the solute is not permeable to the membrane.

  Initially, the water side is filled to height $h$ and the oil side to height $2h$. The amount of solute is same in both the sides. So the solute concentration is more in water than in the oil.

Now hopefully the discussion above makes you realize that there aren't many chemiosmotic forces in your example.  The only driving force left is hydrostatic pressure.  So the oil will cross the membrane, not the water, just because of hydrostatic pressure differences.  This is regardless of the solute concentration.

  Now, I add more water on the water side to reduce the solute concentration on the oil side. Accordingly, water should now flow towards the oil side.

Well adding more water raises the height $h$ to some higher height and thus increases the hydrostatic pressure.  So the increased pressure forces water back through the membrane.


Remember that temperature is a macroscopic quantity, telling the heat content of system. So if the crystalline solid is at its fusion temperature, and every atom in the crystals have equal interaction energy with each other (ignoring all defects of lattice), the whole solid should fuse simultaneously, thus having sharp melting point.
On the other hand, amorphous solids don't have uniform lattice and interaction energies of atoms/molecules may differ. So even at the same temperature, the whole solid might not melt, so we say amorphous compounds melt at range of temperatures.


Separating lanthanides using ion-exchange method

You must be reading this from really old ion exchange references. It is always good to mention the resources and references for other readers.

Ion-exchange retention can be very roughly approximated with Coulomb's law, which says that the force of interaction between two "ions" is directly proportional. The force of interaction is inversely proportional to the square of the hydrated radius of the ions. Here the word interaction means attraction or repulsion.

So you have to look up the hydrated radii of the lanthanides and try to see a trend on a simple sulfonated cation exchanger.

You second point is a misconception. The "trick" to add complexing agents works when you wish to alter the sign of the species you wish to separate. For example Cu(II) has a +2 charge, if I complex it with EDTA, the [Cu(EDTA)] has a charge of -2 Think about the Coulomb's law now. Can we use a cation exchanger now? This time we must try an anion exchanger.

One can still use a cation exchangers, in that case the complexed metallic anions will be repelled out and those ions which did not complex well will stay on the resin.


Miscibility of Hexane and Ethanol (Anhydrous vs 96%)

Theory and Application of solubility Parameters are beyond our scope. However, an interesting viewer should have at least read through collection of research by Charles M. Hansen, published during 60s (Ref.1). One of the parameters discussed in this book is Hildebrand solubility parameter ($\delta$), which is a good indication of solubility, according to Wikipedia. The principal utility of Hildebrand solubility parameter is that it provides simple predictions of phase equilibrium, based on a single parameter. These predictions are often useful, particularly for nonpolar and slightly polar systems without hydrogen bonding. According to the tabulated $\delta$ values, ethanol is polar, and has relatively higher value ($\pu {12.92 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$) than that of hexane ($\pu {7.24 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$) (Ref.1). Yet, it is interesting to note that hexane is soluble in absolute ethanol, but is insoluble in methanol ($\delta_{\ce{MeOH}}=\pu {14.28 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$), albeit the difference of relevant Hildebrand solubility parameters (${\Delta\delta}$) is just $\pu {1.36 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$ (from Ref.1 and 2).

It is an interesting aspect of the Hildebrand solvent parameters that the $\delta$ value of a solvent mixture can be determined by averaging the $\delta$ values of the individual solvents by volume. For example, if you mix $V_\mathrm{A}$ of solvent A with $V_\mathrm{B}$ of solvent B, the mixture will have Hildebrand value of $\delta_\mathrm{Mix}$, which would be given by:

$$\delta_\mathrm{Mix} = \delta_\mathrm{A}\left( \frac{V_\mathrm{A}}{ V_\mathrm{A}+ V_\mathrm{B}}\right) + \delta_\mathrm{B}\left( \frac{V_\mathrm{B}}{ V_\mathrm{A}+ V_\mathrm{B}}\right)$$

Let's see what is the $\delta$ of 96% ethanol ($\delta_{\ce{H2O}}=\pu {23.5 cal^{\frac{1}{2}}cm^{\frac{3}{2}}}$):
$$ 23.5\left(\frac{4}{100}\right) + 12.92\left(\frac{96}{100}\right) = 13.34 = \delta_{96\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$$

Let's see now, what is the value of $\delta$ of 90% ethanol:
$$ 23.5\left(\frac{10}{100}\right) + 12.92\left(\frac{90}{100}\right) = 14.0 = \delta_{90\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$$

Even though still $\delta_{90\%\ce{EtOH}} \lt \delta_{\ce{MeOH}}$, the concept is clear. The value is more than $\delta_{\ce{EtOH}}$ and getting closer by each addition of water droplet and to been not miscible with hexane. Therefore, it is safe to say that these data would gives you some insight to what you are looking for.

Note: For a review of solution systems: See ref.3

References:

C. M. Hansen, In The Three Dimensional Solubility Parameter and Solvent Diffusion Coefficient: Their Importance in Surface Coating Formulation; Danish Technical Press: Copenhagen, Denmark, 1967.
C. M. Hansen, "The Three Dimensional Solubility Parameter Key to Paint Component Affinities: 1. Solvents Plasticizers, Polymers, and Resins," Journal of Paint Technology 1967, 39(505), 104-117.
R. Wiśniewski, E. Śmieszek, E. Kamińska, " Three-dimensional solubility parameters: simple and effective determination of compatibility regions," Progress in Organic Coatings 1995, 26(2-4), 265-274 (https://doi.org/10.1016/0300-9440(96)81583-9).


Does this equation - involving calcium carbonate and a lead (II) cation - balance?

Unfortunately your equation does not add up, let me briefly explain why.

Calcium carbonate is a not very soluble salt. There is ionic binding between the calcium cation and the molecular carbonate anion. If you dissolve it in water you, the following equilibrium will establish: $$\ce{CaCO3~(s) <=>[\ce{H2O}] Ca^{2+}~(aq) + CO3^{2-}~(aq)}$$

Lead carbonate is also a salt, it is even less soluble. Only a little bit will dissolve according to the following equilibrium:$$\ce{PbCO3~(s) <=>[\ce{H2O}] Pb^{2+}~(aq) + CO3^{2-}~(aq)}$$

Calcium hydroxide is also a salt. It is a little bit more soluble than calcium carbonate. The following equilibrium is established when put in water:$$\ce{Ca(OH)2~(s) <=>[\ce{H2O}] Ca^{2+}~(aq) + 2{}^{-}OH~(aq)}$$

You write the following:

  $$\ce{CaCO3 + 2H2O + Pb^{2+} -> Pb(CO3) + Ca(OH)2 + H2}$$

So in principle you have $\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$ and $\ce{CO3^{2-}}$ on both sides of the equation. What is left is not a balanced equation, since there are two electrons missing:
$$\ce{2H2O ~\nrightarrow 2{}^{-}OH + H2}$$

You need a source for the electrons, one way of balancing the equation is
$$\ce{CaCO3 + 2H2O + Pb -> Pb(CO3) + Ca(OH)2 + H2}$$

Given the insolubilities of all the species involved, it might be hard to see any reaction at all. And I have not checked if the redox reaction is favourable, with potentials.


Is "stainless" a bulk or a surface property of stainless steel?

Let's divide the steel world into two classes: 1) rusting steel and 2) stainless steel.

Rusting steel, in the presence of oxygen and moisture, will oxidize, forming hydrated iron oxides/hydroxides which have a greater volume than the original iron, and which have relatively little adhesion to the metal. They curl up and continue to expose bare metal, and so rusting iron/steel will continue to rust.

It was discovered in the early 1900s that addition of 12% or more chromium to iron would produce an alloy that oxidized very slowly, forming a very thin oxide layer that was adherent and did not continue to thicken. It resembled chromium in that respect, and the oxide was probably rich in chromium. Nickel was also found to increase the corrosion resistance of the alloy.

Now the bulk composition affects the surface oxide, but it is the surface that stains, or rusts - or doesn't. If you broke a block of stainless steel in half, in a vacuum, the fresh surfaces would be bare metal, active to many reagents. Exposure of the fresh surfaces to oxygen will oxidize them fairly rapidly (minutes, hours) to a passive state which resists many chemicals, like acids, especially oxidizing acids like nitric. (Interestingly, plain iron dissolves in dilute HNO3, but in concentrated HNO3 (>30%), the oxidizing power of the acid is so great that the surface of the iron is rendered passive: an oxide coat is produced which does not flake off and does not dissolve in the acid.)

Stainless steels are less resistant to chloride ion, which has a way to infiltrate the oxide layer and corrode the base metal. Stainless steels which have been brushed with a wire wheel made of ordinary steel will have a surface contaminated by tiny fragments of regular steel, which will rust. The oxide layer on the stainless steel body will be imperfect, and oxidation will progress thru and under the passive oxide until the whole stainless steel is corroded.

Since it is the surface of the steel which contacts the active reagent, it would seem that the surface (oxide) is the determining factor, but, of course, the bulk composition strongly affects the surface oxide layer. And the surface needs oxygen to resist further corrosion!


Explain the difference between reaction intermediate and reactive intermediate

A reaction intermediate is nothing more than a plain old intermediate, which you probably understand. There is no stipulation as to how stable the intermediate must be. The IUPAC Gold Book defines intermediate as:

  A molecular entity with a lifetime appreciably longer than a molecular vibration (corresponding to a local potential energy minimum of depth greater than $RT$) that is formed (directly or indirectly) from the reactants and reacts further to give (either directly or indirectly) the products of a chemical reaction; also the corresponding chemical species.

I am not aware of a corresponding "official" definition for a reactive intermediate, but in my experience it simply refers to, well, intermediates that display exceptionally high reactivity. Typically these correspond to high-energy species such as carbocations, radicals, carbanions, or carbenes, which exist only transiently, and are difficult or impossible to isolate. A simple example is the carbocation intermediate in an SN1 reaction.

Obviously all intermediates are reactive to one degree or another – which is why they are intermediates and not simply thermodynamic products – but the term reactive intermediate is usually reserved for those which are particularly reactive. So, reactive intermediates are a subset of reaction intermediates.

As for an example of an intermediate which is not particularly reactive, consider the Staudinger reduction of an azide. The initial steps leading to loss of nitrogen gas are extremely fast and (very vigorous) effervescence is observed immediately upon addition of PPh3:

but the iminophosphorane, i.e. RN=PPh3, reacts with water only very slowly to produce triphenylphosphine oxide and the primary amine. Typically you need to heat the reaction mixture for a number of hours, and if you didn't add water it would probably be quite happy on its own. I would therefore consider the iminophosphorane to be an intermediate (or reaction intermediate if you prefer), but I would not refer to it as a reactive intermediate.


Bond length comparison between two carbon atoms

In the case of a carbon-carbon single bond, 2 electrons are shared in the bond connecting the two carbon atoms. With a carbon-carbon double bond, 4 electrons are shared between the two carbon atoms and 6 electrons are shared in a triple bond.  Having additional electrons between the two atoms 1) improves the bonding overlap (makes the bond stronger) between the two carbon atoms and 2) better screens the two carbon nuclei form each other.

Both of these factors, better bonding overlap and better nuclear screening, will allow the two carbon atoms to approach closer together.  Consequently, the more electrons (or the more bonds) between two carbon atoms, the shorter the distance between them.


Significant figures addition vs multiplication

Your calculations are in accordance with the rules of the so-called significance arithmetic even though they yield different results.

When multiplying numerical values, according to the rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. When you multiply the value of the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ by 2, the value of the molar mass $M_{\ce{Br}}$ has four significant digits, whereas the factor 2 is assumed to be exact. Accordingly, the result is rounded to four significant digits:

$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.8\ \mathrm{g/mol}\tag1$$

When adding numerical values, according to the rules of significance arithmetic, the result is rounded to the position of the least significant digit in the most uncertain of the numbers being summed. When you add the molar mass $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ to itself, the value of the molar mass $M_{\ce{Br}}$ is significant to the hundredth place. Accordingly, the result is rounded to the hundredth place:

$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag2$$

However, this concept of significant digits is just an oversimplified form of the concept of uncertainty as a quantifiable attribute. In your question, you are testing the limits of the applicability of this simplification, which is certainly a good idea.

When a number is given without any further information, it is generally interpreted so that the last digit is rounded. Thus, the molar mass of $M_{\ce{Br}}=79.90\ \mathrm{g/mol}$ is generally assumed to represent a value between $79.895\ \mathrm{g/mol}$ and $79.905\ \mathrm{g/mol}$, or $M_{\ce{Br}}=\left(79.900\pm0.005\right)\ \mathrm{g/mol}$.

Therefore, the assumed uncertainty of the molar mass $M_{\ce{Br}}$ is $u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}$. (Note that, normally, the uncertainty corresponds to the estimated standard deviation; in this case, however, the uncertainty just indicates an interval corresponding to a high level of confidence.)

When you multiply the value of the molar mass $M_{\ce{Br}}$ by 2, you also have to multiply its uncertainty by 2 (the factor 2 is assumed to be exact):

$$M_{\ce{Br2}}=2M_{\ce{Br}}=2\times79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{3a}$$

$$u(M_{\ce{Br2}})=u(2M_{\ce{Br}})=2u(M_{\ce{Br}})=2\times0.005\ \mathrm{g/mol}=0.010\ \mathrm{g/mol}\tag{3b}$$

When you add the value of the molar mass $M_{\ce{Br}}$ to itself, you also have to add the uncertainty to itself:

$$M_{\ce{Br2}}=M_{\ce{Br}}+M_{\ce{Br}}=79.90\ \mathrm{g/mol}+79.90\ \mathrm{g/mol}=159.80\ \mathrm{g/mol}\tag{4a}$$

$$\begin{align}u(M_{\ce{Br2}})&=u(M_{\ce{Br}}+M_{\ce{Br}})=u(M_{\ce{Br}})+u(M_{\ce{Br}})=0.005\ \mathrm{g/mol}+0.005\ \mathrm{g/mol}\\&=0.010\ \mathrm{g/mol}\tag{4b}\end{align}$$

(Note that such a simple addition of uncertainties is not correct when the errors of the added values are uncorrelated, for example when calculating $M_{\ce{HBr}}=M_{\ce{H}}+M_{\ce{Br}}$.)

Both calculations yield the same results for the molar mass $M_{\ce{Br2}}$ ($\text{(3a)}$ and $\text{(4a)}$) as well for its uncertainty ($\text{(3b)}$ and $\text{(4b)}$). Thus, the molar mass of $M_{\ce{Br2}}$ is estimated to represent a value between $159.79\ \mathrm{g/mol}$ and $159.81\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.01\right)\ \mathrm{g/mol}$$

By way of comparison, the result of the multiplication according to the rules of significance arithmetic $\text{(1)}$, $M_{\ce{Br2}}=159.8\ \mathrm{g/mol}$, represents a value between $159.75\ \mathrm{g/mol}$ and $159.85\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.80\pm0.05\right)\ \mathrm{g/mol}$$ Thus, this approach overestimates the uncertainty of the result.

The result of the addition according to the rules of significance arithmetic $\text{(2)}$, $M_{\ce{Br2}}=159.80\ \mathrm{g/mol}$, represents a value between $159.795\ \mathrm{g/mol}$ and $159.805\ \mathrm{g/mol}$, or $$M_{\ce{Br2}}=\left(159.800\pm0.005\right)\ \mathrm{g/mol}$$ Thus, this approach underestimates the uncertainty of the result.


No reaction in net ionic equation for lithium nitrate and sodium chloride?

Assuming all these ions are in an aqueous solution, according to solubility rules nitrates are soluble, chlorides (with the exception of AgCl, PbCl₂, and Hg₂Cl₂) are soluble, and salts containing group 1 elements are soluble so they all remain as individual ions as nothing can come together to form an insoluble compound and precipitate out of the solution.


Homemade spectrophotometer to exemplify Beer-Lamberts law

The molar absorptivity is a proportionality constant. You do not know what is the compound responsible by the color. So, this constant will not be in L/mol.cm but will be in terms of the concentration units. For example, if you use the mass to express the concentration (g/L) the constant will be express in this unit L/g.cm. With this system, you can estimate concentrations and so on. To known the molar absorptivity you must know the concentration in mol/L and it is not possible with this experiment.


Predict photochemical decomposition of organic compound?

Just as an experienced synthetic chemist can make an educated guess about how compound A and compound B will react, an experienced photochemist can usually do the same.  There are many, many well-known photochemical reactions that are mechanistically understood and sometimes even used in synthetic procedures, some commercially.  A few such reactions are the

Norrish Type I reaction

Norrish Type II reaction

Paterno-Buchi reaction

di-pi-methane reaction

photochemical induced halogenation, olefin dimerization and so many more.  So yes, an experienced photochemist can make reasonable predictions about which chemicals will react and what the possible products might be.


Why should I not use different battery types together?

Serial usage

For the serial usage, one must avoid combination of cells of different kind and health. They should be as identical as possible.

The best is using

the same chemistry
the same format
the same vendor
the same brand
the same lot
the same aging

Otherwise the cells will have the different capacity and different discharge profiles under the load.

If the serial cell chain is deeply discharged, the "weakest" cell gets discharged completely.

And what is worse, other cells start to electrolyze the unlucky cell in the opposite direction that would be its charging.

This would lead almost inevitably to heavy damage or more likely destruction of the cell, what is unwanted for rechargebles.

Physical destruction with electrolyte leakage is also possible.

For primaries, it would be rather annoyance of not fully used better cells and no easy possibility to find a matching cell.

Parallel usage

It is generally adviced not to use it, unless you know exactly, what you are doing. See also notes for non rechargeable cells below.

Dangers

If cells of different chemistry and therefore voltage range are used in parallel, they overcharge or overdischarge each other when their common voltage does not fit the overlapping of their save voltage ranges.

But even within the overlapping voltage range, cells can be overheated and thermally functionally damaged/destroyed or can even explode, depending on the cell voltage difference and cell internal resistence.

The last paragraph is also valid for cells of the same chemistry. They cannot overcharge each other, but the thermal damage from above applies to them as well.

How to do it well

Only cells of the same chemistry and performance class with about the same actual voltage (check it ) should be connected in parallel.

The voltage difference must respect expected internal resistance and acceptable resulting balancing current. Note that typical internal resistance of Modern AA NiMH cells is $\pu{30-100 m\Omega}$, so $\pu{60 mV}$ difference can cause the current $\pu{1 A}$.

Once connected and balanced, they are safe, as they autobalance themselves during charging and discharging, but consider notes below.

Cell autobalancing of already connected parallel cells

This analysis is aimed for rechargeable cells., supported by Excel simulation.

Cells in an open circurt

Balanced parallel batteries have always the same external (behind their internal resistance ) and internal voltage.

Cells under load

The initial current is in reciprocal ratio of their resistances, so high performance cells would get drained initially faster despite their lower capacity.

The cells with lower $resistance * capacity$ parameter gradually get the internal voltage lover than the others and provide therefore gradually lower current than the other cells.

This leads to the cell balancing, with the current ratio possibly even getting reversed to the initial one. The current ratio slowly converges to the capacity ratio.

Important parameters are the ratio of cell capacities and the ratio of cell internal resistances.

The best results are achieved if these ratios are reciprocal. In such an ideal case, cells are discharging in ratio of their capacities and no internal autobalancing ever occurs.

The good thing is, that the cell resistance increases with the degrading the cell capacity, as cells are aging.

Best autobalancing is achieved on low currents.  High currents lead to higher internal voltage disbalance and are more demanding for good cell matching.

If the cell resistance ratio is not optimal for the capacity ratio, autobalancing kicks on, but does not fully compensate. The internal voltages get therefore little disbalanced.

After disconnection, there is ongoing small temporary compensating current.

It is not good to combine lower capacity high performance cells with low resistance with high capacity low performance cells, especially for high loads. As it significantly raises the internal voltage disbalance and after load compensation

Cells under charging

It is generally adviced against parallel charging, as it affects charge stop detection and can cause overcharging of some cells.

That affects serial charging as well.

Non rechargable cells

Parallel connection is not generally good for different . As there can occur small balancing load after disconnection of the powered device.

OTOH, if the identical cells ( like in the serial section) are used in parallel connection all their lifetime, it is not a big deal, as these currents are minimal and short time.


What emits the least radiation when hot?

Technically, a smooth or molten, unoxidized metal surface (regardless of chemistry) makes a good candidate.  This thesis from Auburn University reports the emissivities of both oxidized and unoxidized metals at temperatures typically encountered in industrial processes; note the low values for all the unoxidized metals studied in the IR range where most radiation is given off.


Interpreting notation format 1.64E-02 from particulate emission data

The format refers to a quite common scientific notation, where the $\mathrm{E}$ refers to exponent of $10$. It is often used when superscripted values become tedious to read and or write. It is widely known in computational sciences and was introduced for an easier display of huge or tiny numbers.

You can interpret the $\mathrm{E}$ as "times ten raised to the power of", hence the format can be interpreted like this: $$n\mathrm{E}m = n\times10^{m}$$

Therefore $1.64\mathrm{E}-02 = 1.64\times10^{-02}= 0.0164$ or $2.25\mathrm{E}+00=2.25\times10^0=2.25$.


What makes alcohol a good conservation product?

Alcohol kills bacteria and fungus, put simply. It does this as its metabolism products cause damage directly to the DNA, and directly to the cell respiration process. It also happens to tend to dry it out.

We can drink alcohol safely because of our alcohol dehydrogenase enzyme, which breaks the toxic alcohol into useful byproducts, that we can use for energy.

Also, it's a matter of scale. Consider submerging yourself in alcohol for a period of time. Don't you think you'd probably die, too?

It's very important to note that alcohol strength matters in this case. It also depends strongly on type. For example, ethyl alcohol: we drink this kind. It's not as toxic as say, methyl alcohol, or isopropyl alcohol. Those last ones are used as preserving and disinfecting agents, respectively.


How to calculate the pH of a buffer after adding HCl?

I'm not sure I follow your argumentation, but I guess you don't really need $K_\mathrm{b}$ at all since addition of strong acid ($\ce{HCl}$) influences the dissociation of the one buffer component - the weak acid ($\ce{NH4+}$):

$$\ce{NH4+ <<=>[K_\mathrm{a}] NH3 + H+} \qquad K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}$$

Henderson–Hasselbalch equation applied to this buffer system before the addition of acid allows to find initial $\mathrm{pH}$ (not required by the problem, I do this solely for demonstration):

\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}} \\
&= -\log{(\pu{5.56e-10})} + \log{\frac{\pu{0.25 M}}{\pu{0.40 M}}} \\
&= 9.05
\end{align}

Once the strong acid ($\ce{HCl}$, assuming complete dissociation) is added, the equilibrium shifts accordingly:

\begin{align}
\mathrm{pH_1} &= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}] - [\ce{HCl}]}{[\ce{NH4+}] + [\ce{HCl}]}} \\
&= -\log{(\pu{5.56e-10})} + \log{\frac{\pu{0.25 M} - \pu{0.10 M}}{\pu{0.40 M} + \pu{0.10 M}}} \\
&= 8.73
\end{align}

You would've needed $\mathrm{p}K_\mathrm{b}$ though when a strong base (e.g. $\ce{NaOH}$) were added.


Why do we use J(J+1) instead of J² in the rigid body rotor energy equation?

This arises from the quantum mechanical properties of the angular momentum operator (vide infra). Since the operators for angular momentum and translational energy are different, there is no reason to expect a similarity between the form of the eigenvalues.

Spectra of the angular momentum operators

First we note the classical relationship between the angular momentum $L$ and the rotational kinetic energy

$$E_\mathrm{rot} = \frac{1}{2}I\omega^2 = \frac{L^2}{2I} \tag{1}$$

since $L = I\omega$, with $I$ being the moment of inertia and $\omega$ being the angular velocity. By analogy, the quantum mechanical Hamiltonian for rotational motion is

$$\hat{H} = \frac{\hat{L}^2}{2I} \tag{2}$$

where $\hat{L}^2$ is the operator for (the square modulus of) orbital angular momentum. Because $L$ is a vector, we have

$$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 \tag{3}$$

where $L_x$ is the operator for the projection of orbital angular momentum onto the $x$-axis, and so on. The classical angular momentum is given by $\vec{L} = \vec{r} \times \vec{p}$; by expanding the cross product we can obtain the equations $L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, and $L_z = xp_y - yp_x$. Using these relations as well as the classical commutation relationship $[x,p_x] = [y,p_y] = [z,p_z] = \mathrm{i}\hbar$, we can derive the commutators of all four operators in equation $(3)$:

$$\begin{align}
[L_x,L_y] &= \mathrm{i}\hbar L_z; & [L_y,L_z] &= \mathrm{i}\hbar L_x; & [L_z,L_x] &= \mathrm{i}\hbar L_y; \tag{4} \\
[L^2,L_x] &= 0; & [L^2,L_y] &= 0; & [L^2,L_z] &= 0 \tag{5}
\end{align}$$

Conventionally, angular momentum eigenstates are chosen to be simultaneous eigenstates of $L^2$ and $L_z$: since $[L^2,L_z] = 0$, it is guaranteed that there is a complete set of states which are eigenstates of both operators. We denote these states by $|\lambda,\mu\rangle$ where $\lambda$ and $\mu$ are related to the eigenvalues of these states:

$$\begin{align}
L^2|\lambda,\mu\rangle &= \lambda\hbar^2 \tag{6} \\
L_z|\lambda,\mu\rangle &= \mu\hbar \tag{7}
\end{align}$$

In this paradigm, we then define the raising and lowering operators $L_+$ and $L_-$ (the meaning will become clear in due time)

$$L_\pm \equiv L_x \pm \mathrm{i}L_y \tag{8} $$

Now consider the effect of $L_\pm$ on any arbitrary state $|l,m\rangle$ which is simultaneously an eigenstate of $L^2$ and $L_z$. Since $L^2$ commutes with both $L_x$ and $L_y$, we have

$$\begin{align}
L^2L_\pm|\lambda,\mu\rangle &= L_\pm L^2|\lambda,\mu\rangle \\
&= L_\pm \lambda\hbar^2|\lambda,\mu\rangle \\
&= \lambda\hbar^2 L_\pm|\lambda,\mu\rangle \tag{9}
\end{align}$$

or in other words, $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L^2$ with the same eigenvalue $\lambda\hbar^2$ as $|\lambda,\mu\rangle$ itself. Also, using the commutators in equation $(4)$,

$$\begin{align}
L_z L_\pm |\lambda,\mu\rangle &= (L_zL_x \pm \mathrm{i}L_zL_y)|\lambda,\mu\rangle \\
&= [(\mathrm{i}\hbar L_y + L_xL_z) \pm \mathrm{i}(L_yL_z - \mathrm{i}\hbar L_x)]|\lambda,\mu\rangle \\
&= [\hbar(\mathrm{i}L_y \pm L_x) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\
&= [\pm\hbar(L_x \pm \mathrm{i}L_y) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\
&= L_\pm(L_z \pm \hbar)|\lambda,\mu\rangle \\
&= L_\pm(\mu \pm 1)\hbar|\lambda,\mu\rangle \\
&= (\mu \pm 1)\hbar L_\pm |\lambda,\mu\rangle \tag{11}
\end{align}$$

which tells us that $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L_z$, but with a different eigenvalue: the new eigenvalue is $(\mu \pm 1)\hbar$. The raising operator (for example) therefore has the effect of conserving the total angular momentum $L^2$, but increasing the $z$-projection of this angular momentum $L_z$.

However, for this to make physical sense, there has to be a limit to how far this can go: or else we will have no maximum value of $L_z$, even for a given value of $L^2$. Note that since $L_x^2$ and $L_y^2$ are both real and non-negative (which follows from the Hermiticity of $L_x$ and $L_y$), equation $(3)$ sets a maximum limit on $L_z$:

$$L_z^2 \leq L^2 \tag{12}$$

Mathematically, the only way this can be fulfilled is if there is a "highest" state $|\lambda,\mu_\mathrm{max}\rangle$ which the raising operator turns into zero:

$$L_+|\lambda,\mu_\mathrm{max}\rangle = 0 \tag{13}$$

Now consider

$$\begin{align}
L_-L_+|\lambda,\mu_\mathrm{max}\rangle &= (L_x - \mathrm{i}L_y)(L_x + \mathrm{i}L_y)|\lambda,\mu_\mathrm{max}\rangle \\
&= [L_x^2 + L_y^2 + \mathrm{i}(L_xL_y - L_yL_x)]|\lambda,\mu_\mathrm{max}\rangle \\
&= [L^2 - L_z^2 + \mathrm{i}(\mathrm{i}\hbar L_z)]|\lambda,\mu_\mathrm{max}\rangle \\
&= (L^2 - L_z^2 - \hbar L_z)|\lambda,\mu_\mathrm{max}\rangle \\
&= (\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2)|\lambda,\mu_\mathrm{max}\rangle \tag{14}
\end{align}$$

However, we already know that $L_+|\lambda,\mu_\mathrm{max}\rangle = 0$. When $L_-$ acts on zero, it simply returns zero, and so the "eigenvalue" in equation $(14)$ must be equal to zero:

$$\begin{align}
\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2 &= 0 \\
\lambda &= \mu_\mathrm{max}^2 + \mu_\mathrm{max} \\
&= \mu_\mathrm{max}(\mu_\mathrm{max}+1) \tag{15}
\end{align}$$

Fundamentally, that is where the term comes from. There are several steps now to close some loopholes. I will not go through them in detail because it is far too long, but I will outline what is needed.

You need to show that for orbital angular momentum the value of $\mu_\mathrm{max}$ is restricted to non-negative integer values; we define this quantity to be $l$ and we can replace the quantum number $\lambda$ with $l$ because $\lambda$ is related to $\mu_\mathrm{max}$, which is in turn equivalent to $l$. The eigenvalues of the total angular momentum operator $L^2$ are $\lambda \hbar^2 = \mu_\mathrm{max}(\mu_\mathrm{max}+1)\hbar^2 \equiv l(l+1)\hbar^2$.

Note that for orbital angular momentum only integer values of $l$ are allowed; half-integer values are not allowed. This condition can only be derived by solving the differential equations corresponding to the eigenvalue equations of the angular momentum operators (see further reading). For example, you need to solve this:

$$\begin{align}
L_z f(x,y,z) &= (xp_y + yp_x)f(x,y,z) \\
&= \left(-x\mathrm{i}\hbar\frac{\partial}{\partial y} - y\mathrm{i}\hbar\frac{\partial}{\partial x}\right)f(x,y,z) = \mu\hbar f(x,y,z) \tag{16}
\end{align}$$

Half-integer values for spin are allowed because the exact form of the spin angular momentum operators are not known. However, they obey the same commutation relations as the orbital angular momentum operators, so the rules governing them are the same except for the constraint on integer values.
You need to show that the allowed values of $\mu$ are integers between $-l$ and $l$ respectively. We have already shown that $\mu$ increases in integer steps, and that $\mu \leq l$, so what remains is to show that $\mu \geq -l$: the proof is exactly analogous to that in equations $(13)$ through $(15)$, except that you need to consider $L_+L_-|\lambda, \mu_\mathrm{min}\rangle$.

Conventionally the quantum number $\mu$ is renamed to $m_l$ and the eigenstates are labelled with $|l,m_l\rangle$. So now we have shown that $l$ is the maximal value of $m_l$, and that $m_l$ ranges from $-l$ to $l$ in integer steps, as we expected
The total angular momentum $J$ is the sum of the orbital and spin angular momenta, i.e. $J = L + S$. For the rigid rotor system where you ignore spin, $J = L$ and hence we can simply replace $L$, $l$, and $m_l$ with $J$, $j$, and $m_j$: we therefore have $J^2|j,m_j\rangle = j(j+1)\hbar^2|j,m_j\rangle$.
Finally, the energies are then given by

$$\begin{align}
E &= \langle j, m_j|H|j,m_j \rangle \\
&= \left< j,m_j \middle| \frac{J^2}{2I} \middle| j,m_j \right> \\
&= j(j+1)\frac{\hbar^2}{2I} \tag{17}
\end{align}$$

where $j$ is a non-negative integer (i.e. $j = 0,1,2, \ldots$).

Further reading

Chapter 4 of Molecular Quantum Mechanics, 5th ed., Atkins.

Chapter 7 of The Physics of Quantum Mechanics, Binney & Skinner for an alternative derivation.


The simplified version of why radiation occurs is a balance between attraction of the protons and neutrons in the nucleus and the positive charge of the protons repulsing each other. Get the balance right, and you end up with a stable isotope, get it wrong, and the isotope is unstable and will decay.

As for the exponential decay: a nucleus falling apart is stochastic process, meaning it is only dictated by chance! When you take an unstable nucleus, after it's half-life there is a 50% chance that it will have decayed. Then, you can add more nuclei the chance of half of them decaying after it's half life is far, far larger than all of them being stable. The larger amount you start of with, the better this theory describes the decay. If you then measure a vast amount of atoms (a mole for example) the curve you find will describe an exponential decay.


The questioner asks that exactly enough energy is added to a molecule of hydrogen just to dissociate it and will energy be shared. The answer is both yes and no. It is better to take a step back and consider what happens when a photon is absorbed. In the case of hydrogen, dissociation must occur from an electronically excited state as it is a homo-nuclear diatomic and does not have a permanent or varying dipole. This means that the ground state cannot absorb a photon's energy from v=0,1,2 etc to any other vibrational level. (I'm assuming absorption into continuum just above v=$\infty$ is zero.) So excitation has to be with a UV photon. If the photon is just at dissociation, the H atoms will have effectively zero kinetic energy and so not depart from one another. In fact at this point life becomes complicated and interesting. The potential energy between two atoms increases very slowly with increasing separation as one reaches dissociation. In fact it is possible to form Rydberg atoms which may be a micron in diameter when excited to within fractions of wavenumbers of dissociation. This size is huge, about ten thousand times larger than the minimum H$_2$ atom separation, bigger than a protein and approaching the size of a bacterium! Clearly the two atoms hardly influence one another and consequently a small perturbation, a minute magnetic field for example, and certainly a nearby molecule may cause the Rydberg molecule either to break apart or recombine depending on whether some energy is added or subtracted in the interaction. If there is not enough energy then the H$_2$ will remain in a highly excited vibrational level until it suffers a collision with another molecule and some vibrational energy is transferred to this molecule. Alternatively the electronically excited H$_2$ could fluoresce and any vibrational energy left in the ground state shared on collision with another H$_2$.  If there is far too much energy than needed to dissociate, then the H atoms ballistically fly apart in opposite directions and transfer some of their energy at each collision with other H$_2$ molecules. Possibly reaction also occurs, but H$_3$ is almost certainly unstable.


Why is benzene called "aromatic"? Why not "circular" instead?

Aromaticity is a electronic property of a molecule, not a description of its shape. An aromatic molecule contains a circular system of conjugated π bonds and p orbitals that contains 4n+2 electrons, where n is an integer.

Benzene is aromatic: it has 6 electrons in a circular conjugated system (4n+2=6, n=1).

Cyclooctatetraene is not aromatic: it appears to have a circular conjugated system, but it contains 8 electrons (4n+2=8 would require n=3/2). This molecule actually adopts a non-planar geometry to avoid antiaromaticity.

Cyclohexane is not aromatic, even though it is circular: it does not have a conjugated electron system.


An early synthesis of borazon from hexagonal boron nitride would be one possible example.  When other approaches did not work, Robert H. Wentorf Jr. added magnesium and voila!  The magnesium had reacted with some of the hexagonal nitride to form $\ce{Mg3N2}$, and the latter then catalyzed the hexagonal to cubic conversion.  (Today other methods have taken over.)


Why do simple molecular substances have weak intermolecular forces and why do giant covalent substances have strong intermolecular forces?

In simple molecules the intermolecular forces are, in order of increasing strength, dispersion forces, permanent dipole interactions and hydrogen bonding (which has significant covalent character but is generally considered to be an intermolecular force).

By contrast giant covalent repeating structures such as diamond and $\ce{SiO2}$ are not molecular in the same sense as they can theoretically be infinitely large. Therefore they do not really have intermolecular forces but they are simply held together by covalent bonds between the atoms in the structure.


Expected accuracy of experimental IR spectra compared to literature reference

In addition to @porphyrin's comment, the interaction of your sample with the surrounding may overall cause some changes in shape and place of the absorption bands; namely if your analysis

follows the classical sample preparation, grinding your compound altogether with excess of KCl to press a pellet (ionic environment) / mixed in nujol (apolar environment) that is then characterised in transmission

.or.

is an analysis by ATR with single or multiple reflection where your sample basically is a neat film on top of your optical window; either probe head or between crystal and anvil. (The ATR correction offered by the spectrometer's software, not every time active, basically will account for the penetration depth varying by the wavelength used, hence influencing rather the displayed %transmittance, then the position of the barycentre of the absorption band in question.)

If you aim for a qualitative IR analysis of a small organic molecule, I would not worry too much if the reported IR band was reported  at 3106 cm$^{-1}$ and is now spot at 3103 cm$^{-1}$.  The iS10 ATR-FT-IR spectrometer by Nicolet/ThermoScientific, for example, provides a data spacing of around 0.7 cm$^{-1}$.  Performing a library search in an electronic database to identify / narrow which compound the sample in question is hopefully would not be affected, either.

(A quantification, like determining the composition of gasoline by PCA, would of course need some calibration.)

The original question however may be read as if you aim to monitor the (partial) digestion of larger proteins and polysaccharides, to say something about their secondary structures.  A shift of 10...20 cm$^{-1}$ of an amide absorption band may be relevant; yet then more often the ones around 1600...1700 cm$^{-1}$ are taken into consideration.


Is there a nomenclature for linear alkanes with n carbons?

This does not answer you question, just mentions an alternative name that does have that count arabic numeral in the name, but is not eliminating the greek/latin prefixes. One could use IR-7.4 INORGANIC CHAINS AND RINGS nomenclature that would be impractical, e.g. the mentioned n-pentacontane would become dohectahydridopentacontacarby[50]catena or maybe even 1,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,14,15,15,16,16,17,17,18,18,19,19,20,20,21,21,22,22,23,23,24,24,25,25,26,26,27,27,28,28,29,29,30,30,31,31,32,32,33,33,34,34,35,35,36,36,37,37,38,38,39,39,40,40,41,41,42,42,43,43,44,44,45,45,46,46,47,47,48,48,49,49,50,50,50-dohectahydridopentacontacarby[50]catena (not to confuse with catenanes).


Let's derive the appropriate realtionship...

$$PV = nRT\tag{1}$$

so

$$\dfrac{PV}{T} = nR\tag{2}$$

but $nR$ is constant for the given sample of gas, thus

$$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\tag{3}$$

For the sample the volume is 137 ml, the dry gas has a pressure of 753-22=731 mm Hg, and the temperature is 22 °C = 295 °K.

Now here is a bit of a wrinkle. What is STP? Until 1982 IUPAC defined STP as 760 mm of mercury at 0 °C = 273 °K. After 1982 IUPAC has defined STP as  750.06 mm of mercury at 0 °C = 273 °K.

For old IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{760} = 121.95 = 122\text{ ml}$$

For current IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{750} = 123.57 = 124 \text{ ml}$$

In looking at some possible variations, it seems the prof is using the wrong initial pressure, and the old IUPAC definition.

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{753\times137}{295}\times\dfrac{273}{760} = 125.62 = 126\text{ ml}$$

NOTE

When collecting a gas over water you must subtract the partial pressure of water at the collection temperature from the atmospheric pressure, but you don't adjust the volume too.

Also this assumes that the atmospheric pressure given, 753 mm Hg, is for dry air.
If you were measuring the pressure with a laboratory barometer then you'd have to adjust the measured laboratory pressure for the relative humidity (partial pressure of water).


Can Fe(CO)5 adopt both square pyramidal and trigonal bipyramidal geometries?

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...).

Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, though MO theory and crystal field theory can help understand which might be more favourable.

The structure of iron pentacarbonyl is widely agreed to be trigonal bipyramidal, with three CO ligands occupying equatorial positions, and two CO ligands occupying axial positions.

  Trigonal bipyramidal structure of iron pentacarbonyl

Whilst both geometries are possible (and indeed observed, IF5 is square pyramidal), I suspect that your confusion arises from the fact that iron pentacarbonyl is able to undergo a phenomenon known as Berry psuedorotation.

  Mechanism of Berry psuedorotation

Berry psuedoroation is the process in which two of the equatorial ligands are switched out for the two axial ligands via a square planar intermediate. This process is sufficiently rapid that if observed by NMR, only one environment for the carbon of the CO ligand is observed (i.e. the psuedorotation is faster than the NMR timescale).


Why is zinc deposited on copper when boiling an aqueous zinc sulfate solution?

Zinc, being in contact with copper, is easily oxidized, because copper drives electrons out of it.

Metallic copper becomes electron-rich and is therefore capable to yield electrons to $\ce{Zn^2+}$ (stemming from $\ce{ZnSO4}$), which reduces to $\ce{Zn^0}$. The latter is finally observed to adsorb at the copper strips.

In other words, copper metal works as an auxiliary electrode whose function is to transport electrons, without being $-$ by no means $-$ transformed during the experiment.

In conclusion, the overall reaction uniquely involves zinc and its ions.

This is similar to what is observed when copper, in contact with zinc, is dipped into an acidic solutions. Hydrogen bubbles are seen even at the copper surface, without the latter being oxidized.


Difficulty in cis, trans isomers in Alkene

You may think of the two lines representing the double bond like the bird's eye view on a door of a saloon:  If the two wings, the one on the left of the double bond, and the one of right flap in the same direction, then the double bond is in cis-configuration.  On the contrary, if the two flap in opposing directions, than the double bond is in trans-configuration.

For future reference -- taking into account the third picture shown by you -- cis and trans may be ambiguous; and triple and quadruple substituted double bonds' configuration are described only with E- or Z-configuration (cf. here).  Eventually, as correctly pointed by @Zhe, in your last example, the double bond is neither E-, nor Z-configurated.


Does a grain of salt have exactly equal amount of sodium and chlorine?

Crystal solids, however small, have defects in them. These defects are basically irregularities in the arrangement of constituent particles. They are of two types:

Point defects
Line defects.

Point defects themselves are divided into other kinds of defects. Here, I'll explain only the relevant one.

One type of point defect is a vacancy defect in which some of the lattice sites in the crystal are vacant. This means that some atoms of Na or Cl or both are missing from their lattice sites.

So the answer to your question is, yes. That scenario is very much possible.


Potassium Atom can act as reducing agent, yet ion cannot?

The ability of K to act as a reducing agent is mainly due to its electronic structure.
$K = [Ar] 4s^1$ . In this case, such atom tends to donate his $4s^1$ electron (acting as a reducing agent). This is due to the gain in energy that produces to have its octet complete and stay in the cationic form $K^{+}=[Ar]$ which is quite more stable that the pure element one.

This is common to all elements in the first column of the periodic table, just because all of them have a quite similar electronic structure, just with more electrons on their inner orbitals.


Classifying acid-base reaction between aluminium hydride and lithium hydride

This is indeed a Lewis acid-base reaction.

The hydride ion has a lone pair which is, in general, not bound to anything. The aluminium atom has at least one unpopulated orbital (typically taken to be a $\mathrm p_z$ orbital in a planar $\ce{AlX3}$ environment). This makes hydride a Lewis base and aluminium a Lewis acid. The association of hydride to aluminium is thus a Lewis acid-base reaction.


What is the link between hydrogen production and potassium oxide?

When you react potassium and potassium hydroxide, you will produce potassium oxide and hydrogen as a byproduct.

$$\ce{2K + 2KOH ->[\Delta] 2K2O + H2}$$

  Potassium react with potassium hydroxide to produce potassium oxide
  and hydrogen. This reaction takes place at a temperature near 450°C.(Chemiday)

The Wikipedia page of potassium oxide states:

  Potassium hydroxide cannot be further dehydrated to the oxide but it
  can react with molten potassium to produce it, releasing hydrogen as a
  byproduct.


Why is phloroglucinol (1,3,5-benzenetriol) not very soluble in water?

Why does it have such a low solubility? I would have thought the OH group would make it more soluble.

Interesting question but there is a logical fallacy. You are connecting the concept of solute's polarity with its water solubility. Note the word polarity is a catch-all term for many interactions. Solubility is the hardest thing to predict or model! It is very hard to predict melting point as well as boiling point. Not sure why general chemistry textbooks only misguide students and they have to unlearn so many things at a later stage.

You made very good observation. Let us look at your list from another angle i.e., chromatography- a special type of chromatography. Lets us say we have a polar stationary phase like silica and use a mobile phase which contains acetonitrile + water at a neutral pH. In this chromatography mode, the most polar compound elutes in the last and the least polar first. If you inject a mixture of phenol, resorcinol and phloroglucinol into a column, what is the elution order?

Phenol, then resorcinol and then phloroglucinol. Lo and behold, this is exactly like your prediction! So there is indeed a trend. However we cannot correlate this single observation with the water solubility.

Let us think of a cyclic sugar, cyclofructan: 21 hydroxyl groups, yet it is insoluble in water! It is insoluble in many organic solvents as well. So it is very hard to correlate the number of individual functional groups and predict water solubility. Exceptions like these will pop up again and again.

However there are softwares, such as Advanced Chemistry Development (ACD) which can predict solubility. No idea what models they use. You can look up their documentation or even contact them.


Is it correct to term boron trifluoride a "catalyst" in electrophilic aromatic substitution?

On the formal definition, I think $\ce{BF3}$ may fail the remains unconsumed part. It will form $\ce{[BF3X]-}$, which would need to revert to $\ce{BF3 + X^-}$ in order to complete the catalytic cycle. I don't think the $K_\mathrm{d}$ of this complex is particularly high, so this is not likely to happen, hence it won't be a proper catalyst. This is also why you need a stoichiometric ammount of $\ce{BF3}$ for the reaction to proceed to completion.


Which computational method/basis set should I use for small molecular ions?

$\ce{H2+}$ or $\ce{H3+}$ have unpaired electrons because they are ions. Due to this, you should be using an unrestricted shell theoretical model which doesn't restrict electrons to paired (closed shell) orbitals. You should start with unrestricted Hartree-Fock (UHF) and can increase the level of theory to include electron correlation with something like 2nd order Møller Plesset (MP2) or 4th order (MP4).

The electron correlation models acknowledge the fact that electron locations are not totally random, repulsion between electrons is a real effect and can give more realistic results.

With regard to basis set, STO-3G would be considered a starting point. If time and resources allow, you could build up to something like a 6-31G(d,p) which adds p-orbitals to the hydrogen atoms allowing more optimization of molecular orbitals. Increasing the basis set quality approximates the radial components of the atomic orbitals with more Gaussian curves, thereby increasing the possible accuracy of the calculation.

Exploring Chemistry with Electronic Structure Methods - Foresman and Frisch


How can I calculate molality from molarity?

The key here is to find the mass of the solvent. If you take one mL of the solution, it will contain $10^{-3} \mathrm{mol}$ of KBr.

The mass of $10^{-3} \mathrm{mol}$ of KBr is  $10^{-3}\times(39+79,9)=10^{-3}\times 118.9=0.1189 \mathrm{g}$.

Or, the mass of one mL of the solution is $1.06 \mathrm{g}$. This means that the mass of the solvent in one mL of the solution is  $1.06 -0.1189=0.9411 \mathrm{g}=0.9411\times 10^{-3}\mathrm{kg} $.

The molality is:
$$m=\frac{10^{-3} \mathrm{mol}}{0.9411\times 10^{-3}\mathrm{kg}}=1.063\,\, \mathrm{mol/kg} $$


How to predict one equilibrium constant given two others?

Remember that when you reverse the reaction the equilibrium constant changes. For the general gas-phase reaction $\ce{A(g) + B(g) <=> C(g) + D(g)}$ the equilibrium constant expression is $$K_\text{f} = \frac{p_\ce{C} p_\ce{D} }{p_\ce{A} p_\ce{B}}$$ (Strictly, it's activities, not partial pressures, but the principle is the same.)
The reversed reaction $\ce{C(g) + D(g) <=> A(g) + B(g)}$ has the equilibrium constant expression $$K_\text{r} = \frac{p_\ce{A} p_\ce{B} }{p_\ce{C} p_\ce{D}} = \frac{1}{K_\text{f}}$$

You are correct in multiplying the equilibrium constants to get the equilibrium constant for the combined reaction.


Why are 2 ways to calculate the energy of the ground state?

Both formulas have exactly the same meaning. They are just two different ways of not double counting the Coulomb and the exchange interactions. In principle, the goal can be achieved in two ways:

double counting in the first place and then dividing by two (first formula);
counting wisely so that there is no double counting in the first place (second formula).

Obviously, for any practical purposes when you one does not want to double count one would almost always choose the second procedure, but the first representation in some cases provides more apparent picture when dealing with the formulas analytically.

Where does all this come from?

The whole derivation of the expression for the ground state Hartree-Fock energy basically starts from writing down the expectation value of the ground state electronic energy for the ground state Slater determinant $\Phi_{0}$
$$
\langle \Phi_{0} \mid \hat{H}_{\mathrm{e}} \mid \Phi_{0} \rangle \, ,
$$
in terms of the individual spin-orbitals of which $\Phi_{0}$ is constructed.
And in doing so we use the so-called Slater rules and it happens so that the electronic Hamiltonian $\hat{H}_{\mathrm{e}}$ contains only two different kinds of operators:

one-electron operators of the form
$$
    \hat{F} = \sum\limits_{i=1}^{N}\ \hat{f}(i)\,,
$$
and two-electron operators of the form
$$
    \hat{G} = \frac 12 \sum\limits_{i=1}^{N} \sum\limits_{j=1\atop{j\neq i}}^{N}\ \hat{g}(i,j)\,.
$$

Already at this stage an expression for any two-electron operator $\hat{G}$ could as well be written as
$$
    \hat{G} = \sum\limits_{i=1}^{N} \sum\limits_{j>i}^{N}\ \hat{g}(i,j)\,.
$$
The mathematical meaning is exactly the same: we do not want to double count pairwise interactions, so that if, say $\hat{g}(1,2)$ is already counted, $\hat{g}(2,1)$ has to be excluded.

Note carefully that we also have to exclude physically meaningless self-interactions, such as, say, $\hat{g}(1,1)$, which is done in the first case by an additional restriction on summation $j\neq i$ and in the second case by the strict inequality in the lower limit of summation $j>i$.

With respect to the exclusion of this non-physical self-interactions note now that they it is missing from the HF energy expression being written in the first form. But that is perfectly fine since each and every Coulomb self-interaction is perfectly canceled by the corresponding exchange self-interaction. Thus, you do not necessarily need the above mentioned additional restriction of the form $j\neq i$ ($b \neq a$ this time), although you could include it if you would like to,
$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b=1\atop{b \neq a}}^N \langle ab||ab\rangle \, ,$$
or adopting a rather usual convention for such summations, as follows,
$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b \neq a}^N \langle ab||ab\rangle \, .$$


Role of NaOH in synthesis of Dithiocarbamate?

$\ce{NaOH}$ is used as a base to remove the amine proton, which improves the nucleophilic power of the molecule. This way the addition to the double bond in $\ce{CS2}$ is favoured.


Are there any good examples of commonly ingested molecules that contain particular toxic individual elements?

Table salt! What would be worse than putting sodium (it can spontaneously combust if you get it wet) and chlorine (used as a war gas in WWI) all over your food?

Then there's water which always contains some $\ce{OH-}$ (an active ingredient in Drano) and $\ce{H+}$ (that stuff eats through metal, man). Although those aren't elements.

And hiding inside every $\ce{O2}$ molecule are two deadly oxygen atoms; once freed, they'll wreak havoc on just about anything they come into contact with.


What question could be answered by "because the metabolites in the cell are in ionized form" in the topic of anaerobic metabolism?

Interesting reverse question. I could figure out three questions:

Why milk changes from a liquid to a gel when it turns into yoghurt?
Why venous blood has lower pH than arterial blood?
Why do we get tired on intense physical exertion?

The answer to the first one would be because the metabolite of lactose fermentation in the lactobacilli cells is lactic acid, that in ionized form releases $\ce{H+}$, lowering the medium pH until the isoelectric point of the milk proteines is reached, causing their agglutination. The answer to the second one is that because the venous blood carries away carbonic acid produced as a metabolite in cellular respiration, that has two ionized forms, $\ce{HCO3-}$(bicarbonate ion) and $\ce{CO3^2-}$(carbonate ion). The answer to the third one would be that, as in lactobacilli case, our body resorts to anaerobic lactic fermentation in activities that require a large power output, beyond the oxygen delivery capacity of the circulatory system. So lactic acid builds up in the muscles, lowering the pH and causing the fatigue sensation. This last explanation was broadly accepted, but seems to be controversial in light of recent studies, as we can see in the abstract for this article:

  In addition, many textbooks report that muscle fatigue is mainly the
  result of a decrease in pH within the muscle cell due to a rise in
  hydrogen ion concentration ([H+]) resulting from anaerobic metabolism
  and the accumulation of lactic acid.6–8 Recent literature, however,
  contradicts this assertion.


Can reaction of gastric acid with swallowed things be dangerously exothermic?

For an average healthy adult person maximum concentration of hydrochloric acid
in gastric juice is $C = \pu{0.16 M}$, and on average there is $V = \pu{2 L}$ of gastric acid in the stomach, assuming it's daily secretion of roughly $\pu{1.5 L}$ and some leftovers. Considering that the acid hasn't been used for the purpose yet, and the following reaction

$$\ce{Al + 3HCl -> AlCl3 +1.5H2},\tag{1}$$

one can make an assumption that the mass of aluminium metal that can possibly be digested is

$$m_\ce{Al} = \frac{1}{3} CVM_\ce{Al} = \frac{1}{3} \times \pu{0.16 mol L^{-1}} \times \pu{2 L} \times \pu{26.98 g mol^{-1}} = \pu{2.85 g}.$$

In reality the outcome depends on the health conditions and biometrics of the given individual. People suffering from acid deficiency won't be able to digest even this amount. Larger or overweight specimen can have larger/stretched stomach, so the amount of acid and its pH may vary quite a lot.

On the other hand, having certain decease such as gastric ulce may amplify the effect of digestive system reacting to the produced $\ce{AlCl3}$ and heat. Also, aluminium(III) chloride is considered to be a neurotoxin. Acute oral
toxicity ($\ce{LD_{50}}$, mouse) of $\ce{AlCl3}$ is $\pu{3805 mg/kg}$. Reaction (1) produces approx. $\pu{14 g}$ of $\ce{AlCl3}$, which is not all that bad.

As @MaxW noticed in the comments, swallowing any metal is potentially dangerous and may or may not cause unwanted effects.


This NIST publication lists the dipole moment as zero, while drugfuture.com (I've never heard of this site) reports a value of 0.50 as measured in benzene at $\pu{25^oC}$.


How do I run MP2 and/or CCSD-level hyperpolarizability calculations with GAMESS?

As you guessed, to calculate $\beta$ within GAMESS for MP2 and CCSD, you would need to perform first- and second-order finite difference, respectively, as there is no coupled cluster gradient needed, so no true CCSD dipole moment.

However, there is a more fundamental problem: it is impossible to calculate time-dependent response using a finite field approach. Take the general form of 1-dimensional 2nd-order central difference,
$$
f''(x) \approx \lim_{h \to 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2},
$$

which can directly translate to a diagonal element of the polarizability tensor formed from energies at finite electric fields:

$$
\alpha_{zz}(x) \approx \lim_{h \to 0} \frac{E(x+h)-2E(x)+E(x-h)}{h^2}
$$

where $h$ is a finite electric dipole field of some strength applied along the $z$-direction. What is $x$? Is it as simple as converting the desired frequency $\omega$ to units of electric field strength? No; that still corresponds to a static field rather than a time-varying field. How are the different perturbation directions handled? $x$ must be set to 0.

From a molecular properties review by Jurgen Gauss, page 5:

  (...) disadvantages of the numerical differentiation scheme are

  a) that there is no straightforward extension to the computation of frequency-dependent properties (...)

Page 37:

  While analytic derivative theory is sufficient for the theoretical treatment of time-independent (static) properties, the underlying theory needs to be extended for the calculation of time-dependent (dynamical) properties. In particular, the fact that there is -- unlike for the static case -- in the time-dependent case no well-defined energy explains why the simple derivative theory discussed so far is not applicable.

Page 39 (with some manipulation by me):

  It can be shown that the linear, quadratic, etc. response functions $\left<\left<A;B\right>\right>_{w_{b}}$, $\left<\left<A;B,C\right>\right>_{w_{b},w_{c}}$, etc. and thus the frequency-dependent properties of interest can be determined as derivatives of the so-called time-averaged quasi energy

  $$
Q(t) = \left<\tilde{\psi}\left|\left(H - i\frac{\partial}{\partial t}\right)\right|\tilde{\psi}\right>
$$

  with the phase-isolated wavefunction

  $$
\tilde{\psi}(t) = e^{iF(t)} \psi(t)
$$

where $F$ is a time-varying external field, and $\psi(t)$ is the usual wavefunction where MO coefficients and thus the density are explicitly time-dependent.

The only numerical (rather than analytic) method for calculating dynamic properties I am aware of is direct time-propagation of the wavefunction. If you want to calculate frequency-dependent CCSD hyperpolarizabilities, your best (free) bet is DALTON. Note that their coupled cluster module is closed-shell only.


how would people give estimates of the actual bonding angle of water
What physics would be involved in the calculation

Background
That's a very good question. In many cases Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.
$$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

For example, in the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
How can we use this approach to answer your question?
Unlike methane, water is not a perfectly tetrahedral molecule, so the oxygen bonding orbitals and the oxygen lone pair orbitals will not be exactly $\ce{sp^3}$ hybridized.  Since addition of s-character to an orbital stabilizes electrons in that orbital (because the s orbital is lower in energy than the p orbital) and since the electron density is higher in the lone pair orbital than the $\ce{O-H}$ orbital (because the electrons in the lone pair orbital are not being shared with another atom), we might expect that oxygen lone pair orbital will have more s-character and the oxygen $\ce{O-H}$ orbital will have less s-character.
If we examine the case where the $\ce{O-H}$ bond is $\ce{sp^4}$ hybridized, we find from Coulson's Theorem that the $\ce{H-O-H}$ angle is predicted to be around 104.5°. So indeed, in agreement with our prediction, removing s-character from the oxygen $\ce{O-H}$ orbital gives rise to the observed bond angle.
Note on reality
For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H sigma bonds and two equivalent lone pairs orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the hybridization changes as described above and the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.
With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^4}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital


Where does the 9th electron go in a N=O bond?

Nitric oxide is a tough molecule to represent as a Lewis structure. However, you have made one common mistake in your structures. I also want to clear up a misconception about resonance that's present in your post. Finally, I will need to introduce a different bonding picture of $\ce{NO}$ to answer your question about which orbital the unpaired electron is in.

  The Lewis Structure of $\ce{NO}$

The mistake you made in drawing those Lewis Structures is that you used all of the electrons in the nitrogen and oxygen atoms and not just the valence or outer-shell electrons.

A Nitrogen atom has 7 total electrons in the following configuration:
$$1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$$

Of these seven electrons, only the $5$ in the second shell (energy level) are available for bonding. These are the valence electrons. The $1s$ electrons are core electrons. They are not available for bonding because their probability density function maxes at less than the atomic radius.

Similarly, an oxygen atom can only use the six electrons in its second shell for bonding: $$2s^2 2p_x^2 2p_y^1 2p_z^1$$

The bonding in $\ce{NO}$ thus only has 11 electrons and not 13 (like you have). With two fewer electrons, it's easy to place the unpaired electron in a way that does not violate the octet rule.

$$\begin{aligned}
&.. \ \ \  .. &&\  &&&\ ..\ \ \ ..\ \\
\ce{.} &\ce{N=O:} &&\ce{<->} &&&\ce{:N=O.}
\end{aligned}$$

  Resonance

The most common misconception related to resonance is unfortunately suggested by the name of the phenomenon. The two resonance structures drawn do not exist as separate structures inter-converting. They are attempts to represent a compound using Lewis structures that has bonding that is not localized. The true structure of nitric oxide is some hybrid of these resonance contributors.

  Where is the unpaired electron?

To answer this question, we need molecular orbital theory. A molecular orbital diagram describes the bonding in terms of constructive and destructive overlap of atomic orbitals. A MO diagram dispenses with the need for resonance - all electrons are clearly shown in their orbitals: $\sigma$ or $\pi$, bonding or anti-bonding. Since this answer is running long already, I won't go into how to construct such a diagram.

Here is the MO diagram for nitric oxide:

The unpaired electron (highlighted in blue)is in a $\pi$ anti-bonding orbital that is formed from the destructive overlap of out of phase $p$ orbitals, one from nitrogen and one from oxygen.


How can absolute binding free energies be calculated?

Free energies are computed by transforming one state into another, e.g. by transforming ligand A into a ligand B (inside a protein in our case). This transformation yields the difference in binding free energies between ligands A and B: $\Delta \Delta G^{bind}_{A \rightarrow B}$. This is a relative binding free energy.

On the other hand, if ligand A is transformed into nothing, it's absolute binding free energy is calculated: $ \Delta G^{bind}_A$.

In both cases ($\Delta G$ and $\Delta \Delta G$), we are calculating differences between two states, and never calculating the absolute free energy $G$ of a single state. This is probably what you meant by: "only relative free energies can be calculated."

PS: since binding occurs in a solvent, it would be necessary to take ligand solvation and protein solvation into account, in order to properly describe the unbound state. For example, in the calculation of $\Delta G^{bind}_A$ described above, ligand A should appear in bulk water instead of disappearing completely.


Reference value for two-electron repulsion integral over GTO's

You don't state which method you implemented in order to compute two-electron integrals, therefore I will list all the main references at first.

Cook's book [1] contains analytical formulas for overlap, kinetic, electron-nucleus attraction and electron-electron repulsion integrals. The analytical formula for electron-electron repulsion integrals is wrong in the book, but you can look at this discussion for the errata.

However, a Full CI calculation is computationally costly. For this reason I believe a better approach would be to use a more efficient scheme to compute integrals. You can look at the following:

Obara-Saika scheme
McMurchie-Davidson scheme
Rys quadrature

You can find a good explanation of these methods in Ref. [2].

Now, if you want to check what you already implemented, you can find a list of the two-electron integrals in the STO-3G basis set for $\ce{HeH+}$ in Szabo's book appendix [3]. Here I can give you a set of two-electron integrals I obtain for $\ce{H2}$ (always in the STO-3G basis set) for a bond length of $1.4$ Bohr:

 (           1           1           1           1 )   0.77460834925515787
 (           1           1           1           2 )   0.44410904384277344
 (           1           1           2           1 )   0.44410904384277350
 (           1           1           2           2 )   0.56967771733030592
 (           1           2           1           1 )   0.44410904384277361
 (           1           2           1           2 )   0.29702946944511982
 (           1           2           2           1 )   0.29702946944511982
 (           1           2           2           2 )   0.44410904384277333
 (           2           1           1           1 )   0.44410904384277333
 (           2           1           1           2 )   0.29702946944511982
 (           2           1           2           1 )   0.29702946944511982
 (           2           1           2           2 )   0.44410904384277361
 (           2           2           1           1 )   0.56967771733030592
 (           2           2           1           2 )   0.44410904384277350
 (           2           2           2           1 )   0.44410904384277344
 (           2           2           2           2 )   0.77460834925515787

Note that these results came from a program I wrote myself, but it usually match very well Szabo's [3] and Gaussian09 values for the total energy. If these results match your calculations, then the problem might be in integrals with higher angular momentum.

[1] Cook, Handbook of Computational Chemistry, Oxford University Press, 1998.

[2] T. Helgaker, P. Jørgensen and J. Olsen, Molecular Electronic-Structure Theory, Wiley, 2000.

[3]  A. Szabo and N. Ostlund, Modern Quantum Chemistry, Dover, 1996.


Calculating entropy: why consider a reversible path?

There are two parts to this answer.  The first is that the entropy change is defined only for reversible processes.  The second is that the entropy is a state function.  Many, but not all, the functions in thermodynamics (including the internal energy) are state functions.

A state function is one that is independent of path.  That is, if one goes from state $A$ to state $B$, the change in internal energy is independent of how the change takes place.  The same is true for the entropy.

So the entropy change in going from state $A$ to state $B$ is independent of how the change takes place.  But we can calculate the change ONLY for reversible changes.

So we mentally find a reversible path from state $A$ to state $B$ and calculate the entropy change.  Since entropy is a state function, that must also be the change for a non-reversible path.


Does the hybridization of sp3d2 type exist?

First, hybridization is a model used in Bond Valence Theory to explain the bond nature between some atoms in some molecules. Actually, the current binding theory is the Molecular Orbital Theory, in which we no longer use hybridisation. But, we study hybridisation because it is a simple and useful tool to explain a lot of compounds.

But, what the professor said in your question, it is not that the sp3d2 hybridization does not exist and cannot be used to explain any molecule. What he is saying is: In hypervalent compounds, there is evidence that the d functions do not participate in the bond.
After reading your question, I have searched a lot about the binding in hypervalent compounds. And what I found was that the nature of bond in hypervalent compounds like XeOF4 has been the subject of a lot of controversy for a long time, because in the represetative atoms, the energy gap between the nsp and nd orbitals is too large, so the d orbitals cannot hold extra electrons. And, nowadays we have evidence through ab initio calculations that the d-functions do not participate in bonding in hypervalent molecules.

However, this is a recent evidence, so, if you have to awnser in a test about the hypervalent molecules, use the d-orbitals to explain them!

I am leaving some links with some articles and books that I read about this subject:

Chemistry of Hypervalent Compounds

The three-center-four-electron (3c-4e) bond nature revisited. An atoms-in-molecules theory (AIM) and ELF study

New bonding concept for Hypervalent molecules, including electron poor and electron odd compounds


Can't we reduce climate change by converting CO2 & CH4?

There are a few problems with what you're proposing:

The reaction requires energy.
Where do get the methane from?
What will you do with the carbon you're generating?

The energy problem can be solved by using surplus energy from renewable source like solar and wind, but, as airhuff also mentiond in a comment, there is much more CO2 in the atmosphere than methane. Also capturing CO2 from the air isn't simple,  and finally why would you use a valuable fuel like methane to create carbon or propane?

In any case I think a much more useful reaction is
$$\ce{CO2 + 2H2O -> CH4 + 2O2}$$

Water is split into oxygen and hydrogen using electrolysis. The hydrogen reacts with carbon-dioxide to form methane.
There are already small plants that do this, but AFAIK the idea hasn't really caught on yet. This may have something to do with the cost and/or the availability of carbon dioxide or surplus energy.

More recently scientists did something even cooler; they captured CO2 directly from the atmosphere and turned it into methanol (CH3OH) using Ruthenium as a catalyst. A drawback is that the reaction works at high temperatures (about 150 degrees C).


What is the major product of the reaction given?

Your reasoning is quite correct for an uncatalysed reaction.
I think you have forgotten about the hydrochloric acid you also add (in catalytic amounts). You will therefore protonate the double bond first, with the carbo cation being more stable in the ortho position. The alcohol can now act as a nucleophile and attack at that carbon. Therefore (c) is the correct answer.


A bidentate ligand is one that forms bonds to the metal center through two different atoms, not through two lone pairs. While chloride does have more than one pair of valence electrons, it only has one atom. Thus it cannot be bidentate.

If the chloride shares a second pair of electrons with the metal center, there would still only be one bond site between the ligand and the metal:

$$\ce{[M+]-Cl <-> [M]=Cl+}$$

A bidentate ligand needs two lone pairs on two separate atoms. Consider ethylenediamine (en), one of the classic bidentate ligands: $\ce{H2NCH2CH2NH2}$. Each nitrogen atom has a lone pair that can be used to form a bond to the metal center, creating two bonds from two different sites on the ligand. See the structure of the tris(ethylenediamine)cobalt(III) ion from tris(ethylenediamine)cobalt(III) chloride $(\ce{Co(en)3Cl3})$.

Update:
As Nicolau points out in a comment, chloride can bridge two metal centers. This is not the denticity phenomenon, which is one ligand with two bonding sites to the same metal. This is one ligand with bonds to two metal centers. See the structure of niobium pentachloride (really $\ce{Nb2Cl10}$ or better $\ce{(NbCl5)2}$.


iso-Borneol is a fairly common reagent, and methods for its preparation and purification are readily accessible (even without journal access). For information, the entry in Purification of Laboratory Chemicals suggests recrystallising from ethanol or petroleum ether, melting point data is also provided as a way of assaying the resulting material  (though NMR would be better).

  dl-Isoborneol [124-76-5] M 154.3, m 212° (sealed tube). Crystallise isoborneol from EtOH or petroleum ether (b 60-80°). It sublimes in a vacuum. The 4-nitrobenzoyl derivative has m 153°. [Yager and Morgan J Am Chem Soc 57 2081 1935, Beilstein 6 II 80, 6 III 299, 6 IV 281.]

  Taken from Purification of Laboratory Chemicals, 7 ed. Armagero and Perrin

Its worth noting that a powder and flakes are essentially just two amorphous forms, and there isn't really any benefit of having one over the other (if anything, the powder is more convenient to handle).


Could metal hydride storage tanks for hydrogen store helium as well, given their repective atomic radii?

Hydrogen does not just insert itself physically into the metal.  Although only a few metals form stoichiometric hydrides with predominantly ionic character, most of them (and all those involved in hydrogen/hydride storage) form some type of bond, generally involving saturation of the hydrogen 1s orbital with ionic or polar covalent bonding.  Helium simply does not have such capability for the same reasons it does not bond with much else, and it is generally not significantly soluble in metals.

An introduction to various methods to store hydrogen is given by Wikipedia.  The
US Department of Energy reviews some proposed materials here.