diH2O, dH2O, and DI H2O. What do they mean?

I don't think there is a commonly accepted standard. If I had to guess, I'd say more often than not $\mathrm{di}\ce{H2O}$ is used to mean deionized water, but I've nothing but my own experience to base that on. Sorry for the lack of references. Also, $\mathrm{dd}\ce{H2O}$ is a common abbreviation that used to mean "double-distilled water" but now is used to refer to highly pure water produced by laboratory purification systems such as those from Milli-Q or similar.

$\mathrm{DI}\ce{H2O}$ = deionized water
$\mathrm{di}\ce{H2O}$ = deionized water

Such water is often available from centralized water purification systems in buildings that host many chemical laboratories.  It isn't as pure as "$\mathrm{dd}\ce{H2O}$" (see below) but is subtanstially more pure than tap water.

$\mathrm{d}\ce{H2O}$ = distilled water
$\mathrm{dd}\ce{H2O}$ = doubly distilled water
$\mathrm{mq}\ce{H2O}$ = Milli-Q water

The brand name of a popular type of water purifier that gives purity similar to $\mathrm{dd}\ce{H2O}$


Except for the missing equilibrium arrow (these reaction are acid-base equilibria), they are fine. Or do you have any objections :) (what makes you doubt) ?

Also take @jerepierre's comment in account. This would probably be considered wrong if at an exam.


How to calculate the equilibration time of two connected chambers of salt solution, initially at different concentrations?

In the future please specify where your equation came from, justify its use and specify all of your parameters.

When only a concentration gradient is present, the diffusion coefficient, $D$, is typically described by Fick's first law of diffusion. To determine $D$ in these terms we must know the flux, $j$, and the gradient, $d c / dx $.  $c=$ concentration of our species divided by the volume and $x$ is the distance this occurs over.

Flux, $j$, is the (flow quantity/time)/area or (flow rate)/area. Fluxes depending on potential gradients are molecular, but bulk flow flux is called convective. Here we have molecular flux present and the units of this are kmol/(m$^2 \cdot$ s). (It's true that diffusion creates its own convection but there is no bulk flow so we ignore this diffusion-induced convection.) Properly our flux is the product of a transport property by the gradient of the driving force, $ j = (D) ($Gradient of driving force$)$. Note that $D$ has units of m$^2$/s. For example, $D$ is $10^{-5}$, $10^{-9}$, $10^{-10}$ m$^2$/s for some gases, liquids and solids respectively at STP.

Now let's think back to basic physics. Say we have a feather at $h$ height here on Earth and we drop it. It will experience an acceleration due to gravity, $g$. Remember that gravity has units m/s$^2$. We can estimate the time it takes to travel distance $h$ crudely by dimensional analysis: knowing that $g$ has units of m/s$^2$ and $h$ has units of m then time is simply $\sqrt{\frac{h}{g}}$. In fact the equation that describes this is $h = \frac{1}{2}gt^2$ and so we actually needed to multiply our original answer by $\sqrt{2}$. In the same way we can crudely estimate the time by dimensional analysis on the diffusion coefficient, $D$ with units m$^2$/s and going a distance with units of m. So we need to take the distance$^2$ divided by $D$ to get seconds. Say ( $0.01$m )$^2$ divided by $10^{-9}$ m$^2$/s gives $10^5$ seconds which is about 1.1 days. This could easily be 4 days if  we needed to multiply by 3.7 which is certainly possible if your orifice is small. It is calculating these other things which turns out to be tricky. Although it's obvious the single easiest thing you can do is switch to a gas when considering just diffusion because a gas would take on the order of 10 seconds because $0.01^2/(10^{-5}) = 10$. You can also change the time by introducing another flux, such as by stirring the solution.

A better answer requires differential equations. Are you sure you want to venture down that road?


Why is Maleic Acid more soluble in water than its trans isomer Fumaric Acid?

Determining the crystal lattice energy would be helpful in comparing the starting points for a solubility investigation. Perhaps an approximation to the crystal stability is provided by the melting points: for fumaric acid, the mp is 287$^o$C; for maleic acid, 135$^o$C. 135$^o$C is a sort of normal mp for organic compounds. 287$^o$ seems quite high, suggesting a high lattice energy, and therefore, lower solubility for this isomer.

Even if molecules have the possibility of intramolecular bonding, as indicated by molecular models or drawings on paper, the existence of intermolecular bonding is not precluded.
Sometimes explanations of an observation are used as predictions which then turn out to be incorrect. The explanations may still be useful, but as explanations of actual data, rather than as reliable predictions.


When a bond is formed between an electophile and a nucleophile, they both lose energy and become more stable right?

In the context of the picture provided, it seems just a matter of language choice. When energy is lost, it is gained somewhere else because of conservation of energy. So a change in energy corresponding to a gain is at the same time a loss elsewhere. The author of that diagram could have used interchangeably energy loss without loss of meaning.


Calculation of an equilibrium reaction quotient

The reaction quotient, Q, is calculated the same way you would calculate the equilibrium constant, K. That is, Q is the product of the concentrations of the products of the reaction divided by the product of the concentration of the reactants,

$$
Q = \frac{[SO_3]^2}{[SO_2]^2 [O_2]}
$$

Plugging the concentrations in $([SO_2] = 2 \times 0.102, [O_2] = 0.0132, [SO_3] = 0.148)$ should give you the answer for $Q$. I think you may have the concentrations written down incorrectly $([SO_3] = 0.184$ gives $Q = 61.6$, for instance$)$.


Oil-water separation is a big money industrial process, and there are a number of methods for successfully separating the two. So the answer to your question is best approached in two parts: a theoretical answer, and a practical answer.

Theoretically, yes. A number of systems exist and the best are described below. There are off-the-shelf applications, as well as major capital infrastructure for major water treatment plants. Whether these solutions are for you will depend on your application, as well as budget. Hence, the practical answer may be different if you are looking for a cheap, backyard-employable device to solve the problem created by Granny filling the radiator through the sump oil inlet again.

Wikipedia has a very nice summary of water treatment, and includes a lot of information about the API (American Petroleum Institute) oil-water separator, which has been in effective use for decades. It is a very effective device, albeit costly, and can treat upwards of 50000 litres of water per hour. You can read about it in detail here, but I don't doubt it is what you are after.

Something more in the lines of what you describe is the Vertical Gravity separator. It is a pressurized system which promotes separation through convection mixing as the liquid is pushed though a complex path. The differences in liquid densities encourages separation of the two components. VGS systems can treat up to 3000L per hour. A document describing the operation of this system can be found on a link to a pdf document on this page. There are some problems associated with this type of device, in that the 'filter' mesh can get blocked if the material is very dirty.

Of course, any treated water is not 100% pure, and may need additional treatment depending on your requirements.

There are a number of interesting web resources where you can read more about oil-water separation methods. Pielkenrood have an excellent page of various methods, past and present. This blog also has a number of useful pages, although it is a commercial site promoting their own preferred devices.


Can terminal enols be reduced by using LiAlH4?

Not only can that enol easily tautomerize to the aldehyde, the aldehyde is the major component at equilibrium at room temperature (by a lot!). The equilibrium constant for the similar system of ethanal/vinyl alcohol (one carbon shorter than yours) is $K \approx 3\times 10^{-7}$ (source: Wikipedia page on keto-enol tautomerism):

$$\ce{ CH3CHO <=> CH2=CHOH}\\
K = \dfrac{[\ce{CH2=CHOH}]}{[\ce{CH3CHO}]}\approx 3\times 10^{-7}$$

Thus, for every molecule of enol in this system, there are $\sim 3 \times 10^6$ molecules of aldehyde. $\ce{LiAlH4}$ will reduce the aldehyde and then LeChâtelier's principle will drag the enol along for the ride.


What is the structural formula of alkali hypohalite: MOX or MXO?

I felt tempted to close this as a duplicate of the sodium acetate notation question, because of my answer there, but the question in itself does not ask the same thing, so I’ll just rewrite the relevant parts of my answer.

In short, there is no correct formula. Depending on which point you want to bring across, you can use either, and depending on the context, $\ce{ClNaO}$ may also be acceptable.

Structually, you are correct with your assumption; the oxygen bears the negative charge. But that doesn’t include any implication that the formula should be written that way; compare $\ce{SO4^2-}$ where the charge is also on the oxygen atoms and yet they come last.

In fact, that leads us to our first clue. By convention, oxygen atoms are typically written last; compare $\ce{ClO3- , ClO4- , BrO3-}$ and others. You could probably explain that with electronegativity, the convention being that electronegative elements come last. This assumption argues for $\ce{NaClO}$.

On the other hand, the corresponding acid $\ce{HOCl}$ is often understood as derived from water $\ce{HOH}$, replacing a hydrogen with a chlorine atom. From this starting point, it makes sense to write $\ce{NaOCl}$, because then you replaced the other hydrogen; this time with sodium. It also includes a resemblance to $\ce{NaOH}$.

Both ways can thus be supported by the reasons given above. In practice, I think I have always seen the substance being referred to as $\ce{NaOCl}$ but I could simply have ignored the occurances of $\ce{NaClO}$.


What's the composition of gun smoke and is it dangerous?

Toxic smoke from propellant combustion has been known about for over 500 years..
The propellants used today are Nitrogen based, thus nitro glycerin, nitrocellulose and nitroguanidine . These produce gases containing oxides of nitrogen, carbon oxides, carbon monoxide, hydrogen cyanide, ammonia and several other gases..
The oxides of nitrogen can be inhaled deeply into the lung thus causing damage as deep as the Alveoli, thus causing bronchiolitis obliterins resulting in a "Pulmonary Edema". This can happen 5 years after exposure..
If you show signs within 24 hours of exposure then oxygen and Arterial Blood gas tests are the solution to treating the condition and knowning the effect. The symptons show just like influenza but no sign of virus.
This is known as an Acute Toxic Smoke Inhalation Injury.
WW1 and WW2 this was studied to reduce the effect on gun crews on both land and sea. It is still a problem today with Armoured vehicles and Artillery and one of the reason the Navies use unmanned turrets.
Look up this file and you will find quite a bit of information about this condition.

Occupational Health - The Soldier and the Industrial base

I had first hand experience of this injury in 1988 and am still alive to tell my story, I had a "Pulmonary Edema" in 1991 and survived that, so I am in the 2% of survivors from that period. In the mid 1990's they discovered how to treat a pulmonary edema thus it has increased the survival rate.

There is thousands of pages of information on this subject, just look for "Propellant exhaust gas exposure" or "Toxic gas inhalation" that will bring up more information.

The military threshold limits are higher than civilian but a tank will expose the crew to 1,300 the TLV for less than 5 minutes if the fume extractor is working. In my case it was not!

http://www.public.navy.mil/surfor/documents/p_5041.pdf  Chapter 10

http://www.cs.amedd.army.mil/FileDownloadpublic.aspx?docid=a1d49fb8-3a16-4489-9408-1089f7044b4c

https://permanent.access.gpo.gov/lps101453/www.bordeninstitute.army.mil/published_volumes/occ_health/OHch10.pdf

There are 3 links for you to go to...
Enjoy reading it...
Oh just as an idea, if you had 1,100 .308(7.62mm) rifles fired in a 100 m3 room, its would be the equivalent of 5x L7 105mm tank rounds...
Enjoy..


You might be confusing $\ce{\Delta G}$ and $\ce{\Delta H}$.

$$\ce{\Delta G = \Delta H - T \Delta S}$$

$\ce{\Delta G}$, free energy, measures the "usefulness" of a system. If $\ce{\Delta G \gt 0}$, then the reaction is not thermodynamically favored - it's NOT SPONTANEOUS.

$\ce{\Delta H}$ being positive means that the reaction is endothermic, or makes the surroundings cooler. While this would CONTRIBUTE to a positive value of $\ce{\Delta G}$, it DOES NOT GUARANTEE it.

In the case of your example above, I would assume the $\ce{T \Delta S}$ factor would make $\ce{\Delta G \lt 0}$ if that reaction really is SPONTANEOUS (I don't have an entropy table with me).


What is the result when the polarity of the solvent is increased in nucleophilic substitution?

The standard textbook explanation is based on a comparison of charge separation in the ground state and the transition state (TS).  If there is more charge separation in the TS than the ground state, then polar solvents stabilize the TS more than the ground state and the reaction will be accelerated. On the other hand, if there is more charge in the ground state than in the TS, then polar solvents will stabilize the ground state more than the TS and consequently the reaction will slow down.

Let's look further at the effect of charge separation and solvent on these two specific reactions and see how they are affected.  Let's start with the $\ce{S_{N}1}$ reaction.

  I found in another book that the SN1 reaction rate increases in polar
  protic solvent. Why does this occur?

In the $\ce{S_{N}1}$ reaction, the rate determining step (rds) is formation of the carbocation intermediate. Let's say we start with a charged nucleophile and a neutral compound.  Moving along the reaction coordinate, by the time we reach the carbocation intermediate, the starting materials (charged nucleophile plus neutral compound) have transformed into a charged carbocation plus a charged leaving group plus the charged nucleophile.  Moving from starting materials to the intermediate has created additional charged species (carbocation plus leaving group).

$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [R^+ +X^{-} + Nu^- ]~~ (the~ TS)}$$

Since the transition state leading to these intermediate species will strongly resemble the intermediate, it will also have more charge separation then the starting materials and therefor, will be more stabilized by polar solvents then the starting materials.  This stabilization of the transition state will lower its energy and speed up the rds.  In other words a polar solvent has sped up the rate of the $\ce{S_{N}1}$ because more charge exists in the transition state than in the ground state.

  I read in a (book) that increasing the polarity of the solvent
  slightly reduces the rate of SN2 reaction. Why does this happen?

Now, lets perform the same analysis on the $\ce{S_{N}2}$ reaction.  Again we start out with a charged nucleophile and a neutral compound.  The rds in this case is the formation of a trigonal bipyramid species.

$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [\delta^{-} ~~ X...R...Nu ~~\delta^{-} ]~~ (the~ TS)}$$

In the ground state there was one unit of negative charge residing fully on the nucleophile. In the TS that unit of charge is now spread out over both the nuceophile and the leaving group; the charge has become much more diffuse, the charge density per unit volume has decreased in the TS compared to the ground state.  In this case, we would expect a polar solvent to stabilize the ground state (starting materials) slightly more than the still charged - but more diffusely charged - TS.  Stabilizing the starting materials slightly more than the TS would slightly increase the energy required to pass over the TS.  Hence we would expect polar solvents to slightly slow down the rate of an $\ce{S_{N}2}$ reaction.


Why does Fe(II) have electrons in the 3d subshell?

You might want to compare iron(II) not with neutral iron, but with neutral chromium.
Both iron(II) and neutral chromium have 24 electrons, but there the similarity ends:
$\ce{Fe^{2+}}:  [\text{Ar}]3d^6$
$\ce{Cr^{0}}:  [\text{Ar}]3d^54s^1$
Think of the Bohr model.  According to this model the energy level of electrons depends only on the shell number $n$, so we would expect 24 electrons to follow the $[\text{Ar}]3d^6$ configuration.  In real life that happens exactly only for single-electron atoms where there are only electron-nucleus interactions.  When there are electron-electron interactions they could fill the shells not in order, like the chromium atom described above.
But in a multielectron atom, if you add more nuclear charge you make the electron-nucleus interaction stronger, and the Bohr-model configuration becomes more favorable.  One might suppose that dozens of added protons might be needed to get chromium's $[\text{Ar}]3d^54s^1$ to the Bohr-predicted $[\text{Ar}]3d^6$.  Instead, the energies of the two configurations are so closely spaced that two extra protons, converting neutral chromium to ferrous iron, is enough.


"...chemical equilibrium is governed immensely by activation energy of the reaction."

This sentence is incorrect. The relative amounts of species in a chemical equilibrium are not at all influenced by the kinetics; if anything, kinetics are influenced by the chemical equilibrium (such as in the Marcus theory of electron transfer reactions). A rigorous derivation of the expression for an equilibrium constant can be made with no direct reference to reaction kinetics.

Chemical equilibria are what happens to systems after they've been left without interference for an infinite time; any system must reach a steady state in finite time, otherwise it would essentially be a perpetual motion device. In other words, for a chemical equilibrium to be achieved, the system is given more than enough time for any finite activation energy barrier to be overcome.


Common ingredient between two disinfectants?

Dettol's active ingredient is chloroxylenol.
Jeye's fluid's is 4-chloro-m-cresol.

It's still possible that both disinfectants simply have an added scent to make them smell "clean". However the active ingredients above are both phenols and will likely have a similar phenolic smell (think hospitals), though I haven't sampled them both myself.


Why is ethanoic acid more acidic than phthalimide?

The deprotonated phthalimide anion is indeed highly delocalized, but the neutral molecule with the N-H proton reattached retains most of that delocalization.  Protonating acetate ion back to acetic acid causes a much greater loss of delocalization.


I read your question as asking why the C-C pi bond is reduced while the C-O pi bond is not. It appears to be possible to change the selectivity of hydrogenation based on the catalyst used (reference), but for simple palladium on carbon, reduction of alkenes (even conjugated to carbonyls) is much faster than an aldehyde or ketone. Coupled with complementary hydride reagents, which are selective for carbonyls, synthetic chemists can achieve the desired reduction without worrying about reaction of other reducible groups.


How do quaternary ammonium salts catalyse SN2 reactions?

Trimethylbenzylammonium chloride is in this reaction a good example of a phase transfer catalyst. It first forms the trimethylbenzylammonium cyanide in aqueous phase (or on the surface of solid phase) which is soluble in the organic phase. Because of the polar nature of the cyanide ion it is not going to be heavily solvated by organic solvent (in this case benzene) molecules so the result is 'naked' cyanide which is highly reactive. This reacts with the substrate forming, in this case, the cyanodecane and trimethylbenzylammonium bromide. The trimethylbenzylammonium bromide is then free to migrate back to the aqueous phase and exchange anion to reform trimethylbenzylammonium cyanide. Most phase transfer reactions benefit from vigorous stirring to accelerate the exchange between the phases.

Further reading in this PDF.


How do we turn PTFE powder into a paste we can dilute in a solvent

You might try making a solution, rather than suspension. PTFE is very resistant to most solvents, but can form a 3% solution in perfluorinated hydrocarbon solvents and, if I remember correctly (I could not find it on the web), in $\ce{Na}$ dissolved in liquid $\ce{NH3}$.


I'd say no, because there is also a reactivity dimension to desiccants. Some drying agents react with some solvents and therefore can't be used as a desiccant. This happens for many organic solvent and desiccant pairs. Some desiccants also react as desiccants and this changes their drying power. This is the case for phosphorous pentoxide, for example, which gradually forms a viscous layer of phosphoric acid which stops any drying. So any linear ordering you are attempting is likely to be overwhelmed by the reactivity side effects of the many desiccants.

On the other hand, if you are only interested in some theoretical point in time estimate of desiccant efficiency, this might be possible but would have virtually no practical import. A desiccant has to react over time and the side effects cannot be ignored.


Combustion is a gas phase reaction. The heat of the flame vapourises the substrate and it's the vapour that reacts with the air. That's why heat is needed to get combustion started.

Anyhow, wood contains lots of relatively volatile compounds so it's not too hard to get combustion started. Once combustion has started the heat of the flame keeps the reaction going.

However sugar dehydrates and emits water when you heat it. Water isn't flammable (obviously) so there's no way to get combustion started. Dehydration leaves behind pure carbon and that is non-volatile so again there's no way to get this to burn. Carbon will burn of course, but you need a high temperature to get it going.


Ag-Sn-Cu Phase diagram to formulate homogenization temperature / duration

Much of this answer is based on a document available from NIST at Properties of Ternary Copper-Silver Systems [originally J. Phys. Chem. Ref. Data, 6(3) 621-673 (1977)] - this has an equivalent to the diagram in the question (which is just the surface of the liquidus). Now, since you are talking about balances of solid phases, it is also worth looking at the binary diagrams to discuss those.

For Ag-Sn, one gets:

For Cu-Sn there is:

Now I'll throw out a few observations:
Ag$_{3}$Sn is a line compound, which in the binary system is not stable above 480C. In the NIST report this phase is referenced as $\theta$, but called Ag$_{3}$Sn($\epsilon$), causing great confusion since the $\zeta$ Ag-Sn phase is designated $\epsilon_{2}$, and Cu$_{3}$Sn($\epsilon$) is designated $\epsilon_{1}$. So, watch it...

The Cu$_{3}$Sn line compound converts to the fcc $\gamma$ phase at around 676C. The $\gamma$ phase is sometimes also referred to as Cu$_{3}$Sn (yet more confusion possible). The Cu$_{3}$Sn and Ag$_{3}$Sn line compounds are different crystal structures, and do not form a solid solution across the ternary.

For both Ag-Sn and Cu-Sn, another thing to notice is the various other phases that are in close proximity to the Ag$_{3}$Sn and Cu$_{3}$Sn line compounds. Going into the ternary system, you get questions like how adding Cu into $\zeta$Ag-Sn impacts the free energy.

So, what now? In the NIST report, on page 664 is a pseudo-binary diagram where the Sn wt% is held at 20%, and Cu wt% is varied from 0 to 80 (balance Ag). Personally I hate weight percent, but...  The point that one can eventually pick out is that the Ag$_{3}$Sn phase, designated $\theta$ appears almost nowhere - there is just a little sliver way over near 0 wt% Cu and under 250C. It appears that with the addition of Cu, the $\zeta$Ag-Sn phase rapidly replaces the Ag$_{3}$Sn phase.

So, as far as I can tell, I'm not sure that the kinetics even matter here. You basically can't get much Ag$_{3}$Sn to form within the ternary parameters you have chosen - you will get $\zeta$Ag-Sn instead.


How can we tell the particle size of solutes in solutions?

I assure you there are many reasons to believe that the ions really are solvated individually. The first thing to look at is the relative strength of interactions between the ions in a lattice and the ions when in a solution. For water, these interactions must be very strong because a number of hydrogen bonds must be broken, the interaction between ion and water must overcome this initial deficit and the loss of some entropy. So, the interaction is surely quite strong.

It is of course possible for two ions which dissolve to get back together and interact with each other. They are oppositely charged ions after all. By looking at the relative size of equilibrium constants for these two processes, however, I think you'll be convinced it is far preferable for the ions to be solvated. For instance, a quick google search will tell you that the $K_{sp}$ of $\ce{NaCl}$ is around $36$. That's an astronomically large solubility product. This means we dump sodium chloride into water until the concentration of both ions is around $6\ M$.

Thus, the easiest experiment to do is to just saturate a solution of water with salt and write the mass of salt added in molarity units. This doesn't really answer your question though because we assumed that the salt existed as two ions and then made a correspondence between that concentration of ions and the concentration of water which each ion is occupying. Plus, we would generally be interested in more dilute solutions than this.

This is why people have used methods which probe the structure of ions in a solution directly. Namely, using x-ray diffraction and neutron scattering experiments, it is possible to determine the local environment around water (or something else solvating an ion) and calculate the hydration number. This is the number of ions, of a specific ion, solvated per molecule of solvent.

Here is a review which discusses x-ray diffraction and neutron scattering experiments and one which discusses coordination numbers of ions solvated in water.

Using these structural methods, which show it is extremely unlikely there are chunks floating around in solution, the following data can be found for hydration numbers of ions in water:
$$n_{\ce{Li+}}=4$$
$$n_{\ce{Na+}}=5$$
$$n_{\ce{K+}}=6$$

Here is also an old review which describes many simple but outdated methods for determining the solvation/hydration numbers of ions in a solvent.

Hopefully some of this is helpful and convincing. The experiments are quite definitive, but if you believe the equilibrium constants, then it is very hard to believe these lattices would remain intact in solution.


Can benzaldehyde undergo chlorination (aka halogenation)?

The experimental process of chlorination of benzaldehyde to form benzoyl chloride can be found here:

  A round bottom flask is fitted with a reflux condenser and a gas inlet
  tube. 100 grams of cold and dry benzaldehyde are added to the flask
  and saturated with the current of dry chlorine. The chlorine gas is
  readily absorbed and the evolution hydrogen chloride takes place. When
  the reaction has moderated, the external heat is applied by
  keeping gentle boiling. The stream of chlorine is continued until no
  more hydrogen chloride is evolved. A stream of dry air or carbon
  dioxide is then passed through the apparatus in order to remove
  an excess of chlorine. Benzoyl chloride is obtained by distillation as
  a colorless fuming liquid with a very irritating smell. Yield almost
  quantitative.

  Preparation of organic compounds, E. de. Barry Barnett, 52, 1912

There is also a German paper written by Freidrich Wohler and Justus von Leibig which mentions an early preparation of benzoyl chloride by the action of chlorine over benzaldehyde.


What red and blue area represent in a orbital software?

To answer this question, we need to start from the beginning.

Movement of an electron is described by wave function. The best layman description for it is that wave function is a quantum equivalent of trajectory. It isn't exactly trajectory, but it serves as equivalent for quantum mechanics.
A trajectory repeating itself and wrapped around an object is called an orbit. A quantum mechanical equivalent is a wave function, remaining constant in time. Such a function for hydrogen atom is called orbital.
Wave function is a function. It means, that it can be positive and negative. (Well, in case of a stationary function, see below). It usually doesn't matter in simple chemistry, but once we start talking about orbital interactions it becomes important. In short, orbitals interact with formation of orbitals with reduced energy as long as lobes with same sign overlap. Thus, it is important to know sign of the wave function in lobes of the orbitals.
So, whenever you see an orbital depicted using two colors, it means two different signs of the wave function. Note, since math, it is not important which color depicts what sign. If one inverses sign of all lobes simultaneously, the result will depict the same system, actually.
Side note 1: the surfaces depicted in the picture likely are isosurfaces (i.e. the function in consideration has same value in all surface points) of square of wave function that enclose volume containing some fixed amount of electron density (usually somewhere around 90-95%). There is no good way to depict wave function/orbital in 3 dimensions really.
Side note 2: wave function has well defined sign only in case of stationary system (i.e. unchanged with time). Whenever we talk about system evolving in time, we need to consider arcane math of complex numbers.


Why can solid lubricants operate only at lower temperature in oxidizing environment but higher temperature in reducing environment?

Graphite is elemental carbon, and carbon is unstable in the presence of oxygen because they react to form carbon dioxide. The rate of this reaction: $\ce{C + O2 = CO2}$ is very slow and has a high activation energy, so a piece of graphite can sit in oxygen at room temperature for millions of years without anything happening to it. Once temperature increases, the reaction becomes much more rapid. From my own experience, a fist sized chunk of graphite can survive for several hours when heated to above 1000 °C in atmospheric oxygen.

Molybdenum sulfide is composed of two things that are likewise not stable in Earth's surface atmosphere: molybdenum and sulfur. It reacts by: $\ce{2MoS2 + 7O2 → 2MoO3 + 4SO2}$. As with graphite, this reaction is strongly temperature dependent.

There isn't a single temperature at which the lubricant is "not stable" any more. It is a balance of the required lifetime of the lubricant, the temperature, and how oxidising or reducing the environment is.


Can nuclear fission or fusion occur together with bond formation in a reaction?

It has been done, not with bond formation but rather with bond breaking, and is called the Szilard–Chalmers reaction (credit to Loong for pointing out the name).

When a biphasic mixture of ethyl iodide (with stable $\ce{^{127}I}$) and water is irradiated with neutrons, all the radioactivity from $\ce{^{128}I}$ is found as iodide in the aqueous phase; there is none in the organic phase.  This is because $\ce{^{128}I}$ is initially formed in an excited state that emits a γ-photon, and the recoil from that is more than enough to break the C–I bond.

The energy of a γ-photon is > 10 keV; in comparison, the strength of a typical chemical bond is ~ 1 to 10 eV. The C–I bond strength is roughly 2.5 eV.


How is the 'rate law' and 'order' for the ozone decomposition reaction can be defined?

Chemical kinetic describes both reversible and irreversible reactions.

The reaction scheme and rate expression do not refer to the same thing. The reaction scheme is an overall one whereas the rate equation refers to a particular multi-step reaction. Two examples are given below:

Ozone can be destroyed by UV light in the $240$ to $\pu{320 nm}$ region with the scheme $(\ce{M}$ is an inert gas needed to take away excess energy):

$$
\begin{align}
\ce{O3 &->[$hν$] O2 + O} \\
\ce{O3 + O + M &-> O2 + O2 + M} \\
\hline
\ce{2 O3 &-> 3 O2}
\end{align}
$$

Similarly, oxygen molecules can react with short wavelength UV as

$$
\begin{align}
\ce{O2 &->[$hν$][$(<\pu{242 nm})$] O + O} & \\
\ce{O + O2 + M &-> O3 + M} &|\cdot 2 \\
\hline
\ce{3 O2 &-> 2 O3}
\end{align}
$$

which also gives a similar net reaction. This is the Chapman mechanism and is very important in atmospheric chemistry.


Calculating mass of krypton from mole fraction

Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:

$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$

If we can get a second equation in both variables, you can solved the system for each variable.

You were able to determine the mole fractions $\chi$:

$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$

$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$

Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.

  From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).

$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$

$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$
$$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$
$$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$

  From mole fractions

$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$
$$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$

  Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$.


Electronic strucure of cyclopropene radical cation

You do not really have a singly bonded carbon atom. When you remove the hydrogens from the originally saturated carbon the pi electrons delocalize to make an aromatic ring.  In effect you have a cyclopropenyl cation with one of the hydrogen atoms homolytically removed.  The positive charge is thus delocalized in the ring but the odd electron is localized to the carbon with no hydrogen, in the outward pointing $sp^2$ hybridized orbital.


Neutralization reaction between ammonia and nitric acid

Your answer is very close to the answer given, except for the following two tidbits (the first being more significant).

An acid-base reaction is not the exchange of a hydrogen atom $\ce{H}$. It is the exchange of a hydrogen ion (or proton) $\ce{H+}$. Thus your answer should be: $$\ce{NH3(aq) +HNO3(aq) -> NH4+(aq) + NO3-(aq)}$$
The given answer combines the two ions produced into a single compound. $$\ce{NH4+(aq) + NO3-(aq) ->NH4NO3(aq)}$$

The result of #2 would be reasonable if you had not been told that the reaction was occurring in aqueous solution. In aqueous solution, the products should have been those of #1, since ammonium nitrate is freely soluble in water.


Heat of combustion among substituted cyclohexanes

Ashish: Welcome to SE! You are quite correct that heats of combustion and formation are inversely related. Since all of your compounds are trimethylcyclohexanes (C9H18), their sum will be equal to the heat of combustion of 9 moles each of graphite and hydrogen. To approach the solution of this problem, we can start with methylcyclohexane and evaluate the effect of adding ultimately two more methyl groups. In the somewhat busy diagram below, all numbers are in kcals/mol. Known heats of formation (NIST) are located inside the ring. Calculated heats of formation (ΔHfo) are in red and values over the arrows are the incremental changes in heats of formation (ΔΔHfo). The more stable conformation of all compounds is assumed. Equatorial (e) and axial (a) assignments are made.
There is no methyl/methyl interaction in the formation of 1,4-isomers A (cis) and B (trans) or 1,3-isomers E (cis) and F (trans). The addition of an equatorial methyl (B and F) is worth -7.0 and -7.1 kcal/mol, respectively, while addition of an axial methyl (A and E) has a value of -5.2 kcal/mol. Note that the difference between an axial and equatorial methyl is worth 1.9 kcal mol. The axial methyl necessarily makes the ΔHfo value less negative than addition of an equatorial methyl. However, there are gauche butane interactions in 1,2-isomers. Because C is less stable than D, the addition G --> C (cis) is -4.1 kcal/mol and G --> D equals -6.0 kcal/mol.
Using a value of -7.0 kcal/mol for the addition of an equatorial methyl, the conversion G --> R gives a value of ΔHfo = -51.0 kcal/mol [NIST = -51.48 kcal/mol]. Note that 1,3,5-isomer U's value is what is expected given the difference between a non-interacting equatorial vs. axial methyl group.
The reader should be able to corroborate the values assigned to P, Q, S and T. The stability (heats of formation) of P through T are R>S>T>Q>P. Therefore, the heats of combustion are in the order R<S<T<Q<P.


Why is para-aminobenzoic acid more acidic than ortho-aminobenzoic acid?

The problem in ortho-aminobenzoic acid is that the acidic hydrogen of carboxylic group is H-bonded with the lone pair of nitrogen in amino group. As a result it is more difficult to extract it compare to that in para-aminobenzoic acid since the H-bond must also be broken during acid-base reaction. Para-aminobenzoic acid does not have a H-bond due to the distance between the two groups.

I would like to add that since the hydrogen of the $\ce{-COOH}$ group is more acidic and $\ce{N}$ has a lone pair. This is why H-bond occurs with $\ce{-NH2}$ group as electron donor and not the other way round.
Also, I presume that by ortho effect you intended to tell that due to repulsion of $\ce{-NH2}$ group $\ce{-COOH}$ group would go out of the plane. However, as we have seen, it is difficult as it is already H-bonded with the nitrogen atom and moving out of the plane would put strain into the H-bond. Moreover, the repulsion by $\ce{-NH2}$ group is not too intense so it would (in most part) avoid the strain.


What determines a substance's energy of activation with air?

Energy of activation is pretty much an experimental thing. You can try to explain obtained results, but predicting exact values without numerical analysis is not really possible.

Basically, given a compound to react with air, you first devise a reaction mechanism -- how the reaction proceeds. Now, using some rules, you determine the highest energy transition state. Finally, you try to approximate the energy difference between this and the reactants by comparing with known quantities.

About the rules that are used to compare energies and energy differences: There are many of them, and out of the scope of this answer (indeed, one would be hard pressed to fit them all on a single bookshelf). Basically, one uses general rules that compare stability. For example, a positively-charged carbon is more stable when it has more alkyl groups attached to it (the reason usually given for this is hyperconjugation). A carbon conjugated with a double bond is considered even more stable, due to resonance.

Regarding paper, paper is made of molecular fibres which are effectively infinitely long. To melt paper, one would have to disentangle these (as molecules need to be much farther apart and with much less attraction in liquid form), and even then one would be left with a very, very viscous liquid (which is not too different from a solid). So paper would be extremely hard to melt, if not impossible.


Using electronegativity and atomic size to compare acidity of cysteine with serine

You are correct that Oxygen is more electronegative than Sulfur.  However, the relative size of the atoms affects more than their ability to carry charge.  In this case, the larger radius of the Sulfur atom actually causes the H-S bond in Cysteine to be longer than the H-O bond in Serine.  The longer bond is weaker and, thus, more easy to break, making Cysteine easier to deprotenate.


$\ce{-CH3}$ group has +I effect; it has electron donating effect. It increases basicity and reduces acidity.
Moreover due to resonance in phenyl group, the availability of electrons over N decreases.


In general, salts are more soluble in water than organic acids or organic bases, because they are made of ions. It is a general trend. There are plenty of exceptions. So usually salts are quickly dissolved in the stomach and absorbed in the intestine later on.


Producing sulphur dioxide for using in a reaction

Note: Sulfur dioxide causes dyspnea. Prolonged inhalation of it may causes respiratory problems. So, reaction should be done under laboratory supervision.

The most common laboratory method of producing sulfur dioxide is the reaction of copper and sulfuric acid, both available in laboratories.

$$\ce{Cu(s) + 2H2SO4(l) → CuSO4(aq) + 2H2O(l) + SO2(g)}$$

Process: To a round bottom flask, 100 g of copper (small pieces, turnings, etc.) are placed. From a dropping funnel 100 ml of concentrated sulfuric acid is slowly added dropwise. The reaction flask is gently heated and obtained sulfur dioxide is dried by passing through a wash bottle with concentrated sulfuric acid. (Process taken from here. Read more to know how the reaction can be controlled and the precaution needed)

Alternatively, sodium sulfite can also be used.

$$\ce{Na2SO3 + H2SO4 -> SO2 + Na2SO4 + H2O}$$

Process: Sodium sulfite is placed in the bottom of the 300 ml Erlenmeyer flask of the apparatus. The dropping-funnel is filled with dilute sulfuric acid, made by pouring one volume of concentrated sulfuric acid into an equal volume of water and cooling the mixture. The glass elbow is connected with a gas washing-bottle containing concentrated sulfuric acid, and the dry gas is collected by displacement in two or three empty cylinders. On allowing the dilute sulfuric acid to drop slowly upon the sodium acid sulphite, a very regular evolution of sulfur dioxide is obtained.

For full setup, check here.


Clarification of Kinetic Energy during Phase Change

The temperature is not the average kinetic energy.

This is a bad habit which most chemists and physicists pick up because for an ideal gas, and for most systems at normal parts of the phase diagram, this is true. In thermodynamics, however, temperature is defined in terms of the thermodyanic beta as,
$$
\frac{1}{k_b}\frac{\partial S}{\partial E}=\beta
$$
where $\beta=1/(k_bT)$.

That is to say that during a phase change, because the temperature is constant, the ratio of the change in entropy to the change in total energy (kinetic plus potential) is constant. This means that when the phase change just begins and just ends, the ratio of entropy and energy must be identical, and unless this is quite an unusual phase change, this means both need to have changed from their initial values. It wouldn't be a phase change if neither entropy nor energy changed. Because temperature is defined in this way and not as $T\propto\langle \text{KE}\rangle$, then it is perfectly reasonable for the kinetic energy to either increase or decrease during a phase change.

For instance, in normal systems one would expect that the entropy of the liquid would almost certainly be larger than the entropy of the solid. This means that the total energy must decrease by an amount $T_m\Delta S$, which makes sense as one would expect the potential energy (and perhaps the kinetic energy) to decrease when going to the liquid phase. In other words, the entropic gain allows the system to energetically relax. With this picture, it is not too surprising that the average kinetic energy can decrease during a phase change.


There are various electronegativity scales.

Electronegativity vales from Wikipedia

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Element} & \text{Pauling} & \text{Allen}& \\
\hline
\text{H}  & 2.20  & 2.30 \\
\text{C}  & 2.55  & 2.544 \\
\text{N}  & 3.04  & 3.066 \\
\text{O}  & 3.44  & 3.610 \\
\text{F}  & 3.98  & 4.193 \\
\text{P}  & 2.19 & 2.253 \\
\text{S}  & 2.58 & 2.589\\
\text{Cl} & 3.16 & 2.896 \\
\hline
\end{array}
$$

The compounds are all binaries, so the polarity, $\mu$, is equal to the difference in charge, $\delta$, times the distance, $d$, between ions ($\mu = \delta \cdot d$). Only given a table of electronegativities, the first level of analysis would be to assume that all the bond lengths were equal and that that the polarity would simply be a function of the difference in electronegativity. The differences are thus:

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Element} & \text{Pauling} & \text{Allen}& \\
\hline
\Delta\text{Cl-F}  & 0.82  & 1.297 \\
\Delta\text{S-O}  & 0.86  & 1.021 \\
\Delta\text{P-N}  & 0.85  & 0.813 \\
\Delta\text{C-Cl}  & 0.61  & 0.352 \\
\hline
\end{array}
$$

So on the Pauling scale S-O should be the most polar, but on the Allen scale Cl-F should be the most polar.


Determining whether something is an acid, base, or amphoteric

An amphoteric substance can act both as an acid and a base. That is, an amphoteric substance can donate protons and also accept protons.

$\ce{CO3^{-2}}$ does not have any protons so it cannot act as an acid. It is the conjugate base of $\ce{HCO3^{-}}$.

$\ce{HPO4^-{2}}$ is the conjugate base of $\ce{H2PO4^{-}}$. It can accept a proton to go back to $\ce{H2PO4^{-}}$ or it can donate a proton to become $\ce{PO4^{-3}}$. Thus, it is amphoteric.

$\ce{H3O^{+}}$ is the conjugate acid of $\ce{H2O}$. It can donate a proton to go back to $\ce{H2O}$ but it cannot accept any other protons. Thus, it not amphoteric.

$\ce{NH4^{+}}$ is similar to $\ce{H2O}$. In this case it is the conjugate acid of $\ce{NH3}$. As such it can donate a proton to go back to $\ce{NH3}$ but it cannot accept any protons. It is not amphoteric.

So, out of your list only $\ce{HPO4^-{2}}$ is apmhoteric.


Intermolecular forces in secondary amines compared to primary amines

A primary amine has two $\ce{N-H}$ bonds and one $\ce{N-C}$ bond. A secondary amine has two $\ce{N-C}$ bonds and one $\ce{N-H}$ bond. Since $\ce{N-H}$ is more polarised than $\ce{N-C}$ we can expect ethyl amine to be more polar than dimethyl amine.


What materials for a boiler, if any, do not require a deaerator?

There are many more common alloys in the Inconel , Hastelloy , and Incoloy families. And some odd balls like MP35N, 5o:5o Cr- Ni, Vitallium , and others. Titanum won't take the heat. For all of these materials; if you need to ask the price ,you can't afford them.


Why is the valency of the Calcium (Ca) Bohr model, set to 2 electrons, instead of 1 electron for stability?

After following Mithoron's link, I familiarized my self with the writing on the following ChemGuide site (link).

I went into Excel spreadsheet, and had each 'sub orbital' it's own column, and came up with the following.

+---------+---------+----+---------+----+----+---------+-----+
| Shell 1 | Shell 2      | Shell 3           | Shell 4       |
+---------+---------+----+---------+----+----+---------+-----+
| 1s      | 2s      | 2p | 3s      | 3p | 3d | 4s      | SUM |
| 2       | 2       | 6  | 2       | 6  | 0  | 1       | 19  |
| 2       | 2       | 6  | 2       | 6  | 0  | 2       | 20  |
+---------+---------+----+---------+----+----+---------+-----+

This is what the overall table (link) looks like in excel; I can't attach the image directly to my post.

While googling, to get a clear understanding of S,P,D,F I came across ThoughtCo's article (link). This let me know the stability precedence, i.e. electron distribution, more clearer, as follows:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f

4s is filled before, 3d. 3s and 3p, can hold a combined maximum of 8 electrons.

The following (link) has a diagram with downward/diagonal arrows, showing how one can better remember the stability precedence.


Self Answer

Additionally, I plotted $\frac{1}{V}$ and $\frac{1}{p}$, as suggested by @Chester Miller in the comments below, and indeed I get a straight line.

A linear model fit was performed and the equation $4.39274 x+0.214246$ fits the data with an $R^2 = 0.999929 $. From my limited knowledge on this topic (I haven't formally studied this subject), the intercept should give me $\frac{1}{\text{adsorption capacity}}$, and the slope should give some indication of the affinity of the adsorption sites?

So going back to the initial equation, $$\theta = \frac{V}{V_\mathrm m} = \frac{Kp_\mathrm a}{1+Kp_\mathrm a} $$

And now considering $$\frac{V_\mathrm m}{V} = \frac{1+Kp_\mathrm a}{Kp_\mathrm a} = \frac{1}{Kp_\mathrm a} + 1$$

and dividing throughout by $V_\mathrm m$ we get,

$$\frac{1}{V} = \frac{1}{V_\mathrm m} + \frac{1}{V_\mathrm mKp_\mathrm a}$$

And we get the appropriate values for $V_\mathrm m$ and $K$ from the slope and intercept of the $1/V$ vs $1/p$ plot, and then obtaining $\theta$ and plotting vs. $p$ is a trivial exercise.

Anyway, using the calculated parameters, I once again plotted $V$ vs $p$ and fit a curve of the form $\frac{V_\mathrm mKp_\mathrm a}{1+Kp_\mathrm a}$


If you add up the enthalpies of the following reactions with the given factor and divide the result by two you will get the desired result.

$$\ce{CH4 + 2 O2 -> 2 H2O + CO2} \tag{· 2}$$
$$\ce{2 C + H2 + 3 Cl2 -> 2 CHCl3} \tag{· 1}$$
$$\ce{C + O2 -> CO2} \tag{· -2}$$
$$\ce{2H2 + O2 -> 2 H2O} \tag{· -2}$$
$$\ce{2 HCl -> H2 + Cl2} \tag{· -3}$$

$$\ce{2 CH4 + 6 Cl2 -> 6 HCl + 2 CHCl3}$$


Do amines or alcohols have stronger intermolecular hydrogen bonds?

The boiling point of non-ionic compounds are highly depend on their H-bonding abilities. For example, boiling point of water (molar mass: $\pu{18.02 g/mol}$) is $\pu{100 ^\circ C}$ at $\pu{1 atm}$ while that of ethanol (molar mass: $\pu{46.07 g/mol}$) is $\pu{78.4 ^\circ C}$ at $\pu{1 atm}$, even though ethanol is heavier and have more other intermolecular attractions, excluding H-bonding, such as van der Waals interactions than water. This difference in boiling points displays clear dominamce of H-bonding (HB) on boiling point over other forces on these non-ionic compounds.

Therefore, since molar masses of methanol and methylamine is very close ($\pu{32.04 g/mol}$ for $\ce{CH3OH}$ versus $\pu{31.05 g/mol}$ for $\ce{CH3NH2}$), comparition of their boiling point would be excellent indicator of the comparison of their H-bonding strength: Boiling point of $\ce{CH3OH}$ is $\pu{64.7 ^\circ C}$ at $\pu{1 atm}$ while that of $\ce{CH3NH2}$  is $\pu{-6 ^\circ C}$ at $\pu{1 atm}$. Therefore, it is safe one to say that $\ce{CH3OH}$  has stronger H-bonding capabilities than that of $\ce{CH3NH2}$ even though $\ce{CH3NH2}$ has extra $\ce{H}$ compared to only one in $\ce{CH3OH}$. The best comparison measures of these two types of hydrogen bonding (HB) are the $\Delta H_\mathrm{HB}$ values in Wikipedia (Hydrogen Bonding):

  $\Delta H_\mathrm{HB} = \pu{21 kJ/mol}$, listed for $\ce{O−H···:O}$ in water-water.
  $\Delta H_\mathrm{HB} = \pu{13 kJ/mol}$, listed for $\ce{N−H···:N}$ in ammonia-ammonia.


Is ground-state structure of a molecule entirely determined by the positions of the nuclei?

Yes, these methods (both wavefunction-based and electron density-based) typically solve a purely electronic system defined by an electrostatic potential which is produced by the nuclei; the Born-Oppenheimer approximation can be understood as treating the electronic system as a function of nuclear geometry, which implies that (1) nuclear dynamics are not strongly affected by electrons and (2) electrons will almost immediately adapt to any new position of the nuclei. Born-Oppenheimer can also be understood in terms of electronic and nuclear motion (and the corresponding energies) being separable; that is, electronic structure takes shape after (and as a consequence of) nuclear dynamics, but does not directly affect nuclear dynamics themselves. Since chemistry is usually interested in stationary solutions as snapshots, these approximations are fitting for pretty much all chemical questions.

In practice, nuclei are not even particles in electronic structure calculations - in both Schrödinger equation methods (HF and post-HF) and Hohenberg-Kohn methods (DFT), the only role of nuclei is to generate an external electrostatic potential, $V_{ext}$, that will be part of the Hamiltonian/energy functional of the system. The specification of nuclear positions that you mention is just a more convenient, and chemically intuitive, way of describing that external electrostatic potential, but there is no mathematical or conceptual difference if we consider that this external potential is arbitrarily imposed from outside; it's just that it's easier to define a number of point charges (with their charge defined by their atomic number) than to provide a complex 3D potential function.

Once we assume that Born-Oppenheimer is a valid approximation and that nuclei merely exert an external electrostatic potential on electrons, the fact that electronic structure for the ground state is entirely determined by the positions of the nuclei is axiomatic: it is a consequence of the postulates of quantum mechanics (and the basis of all variational methods) for wavefunction-based calculations, and one of the Hohenberg-Kohn theorems for density functional calculations.

Geometry optimisation is also performed under the same assumptions - electronic structure is calculated under a given geometry, and then resulting forces over nuclei are estimated; a small motion is performed along these tensions, and the process is repeated iteratively until forces experienced by nuclei are small enough to fall below a certain tolerance level. Note that there is no explicit treatment of the interaction between electrons and nuclei; only two process that interact iteratively through electrostatics: electronic structure is minimised for each geometry, and then geometry is relaxed, but not minimised, according to that electronic structure. We obtain a quantum mechanics depiction of the electronic system for each geometry, but we do not perform a quantum mechanics treatment of the nuclei. Ab initio molecular dynamics is a similar approach, in which nuclei move through classical dynamics but electronic structure is calculated through QM at each step.

In all these cases, chemical bonds are only emergent interpretations of the energetics of electronic structure, but their identification is largely arbitrary, as energetic effects are not localised. While we can calculate the energy of two molecular fragments as they approach from infinity, and we'll observe interactions that we identify with bonding, the limits of what is a bond are not clear, and at a quantum chemical level, it makes more sense to talk about energies associated with different geometries than to think in terms of bond "objects".

It is in principle possible to perform QM computations which treat both nuclei and electrons as quantum particles, thus obtaining a single wavefunction describing both, but the complexity of this treatment (especially given the large difference in scale for electronic and nuclear motion) restricts calculations to a very small number of particles - the highest number I remember reading about is 8, although I might have missed recent studies. In addition, the problem of describing interactions accurately (which is behind the variety of quantum chemistry methods used just to describe electrons) grows considerably worse, to the point that quantum Monte Carlo may be a better option than quantum chemistry methods, as happens with bosons.


Rate-determining step and steady state approximation failure

When can't the rate-determining step be applied, and why? How to recognize those mechanisms?

The rate determining step (often abbreviated as "rds") of a reaction is the slowest step in a reaction.  Every reaction has a "slowest step", therefore, every reaction has an rds.  The slowest step will be the step where we pass over a transition state with the highest energy, e.g. the highest energy point along the reaction coordinate. Look at the figure below, the reaction coordinate shown in the top left diagram

illustrates a reaction with a transition state, but no intermediate.  There is only one step in this reaction, passing over that transition state (marked by the double-headed arrow) is the rds for that reaction.  The drawing in middle and right side of the top row show a reaction with an intermediate, and consequently, two transition states.  In each case the double-headed arrow identifies the rds - the highest energy point along the reaction coordinate.  The 3 drawings in the bottom row illustrate the case of a reaction with 2 intermediates (and therefore 3 transition states) along the reaction coordinate - in each case the arrow identifies the transition state with the highest energy, the rds.

  When can't the steady state approximation be applied, and why?

Consider the following reaction
$$\ce{A <=>C[{k_1}]\   [B] <=>C[{k_2}]\ C}$$
$\ce{A}$ is the starting material, $\ce{[B]}$ is an intermediate and $\ce{C}$ represents the product.  Either the drawing in the middle or right in the top row represent the reaction coordinate for this reaction.  The steady state assumption applies to reactions that have one or more intermediates along the reaction coordinate.  The steady state assumption states that the rate of change of the concentration of the intermediate is approximately zero.  In other words, the intermediate is being created at about the same rate at which it is
$$\ce{\frac{d[B]}{dt} = 0}$$
being destroyed.  The steady state assumption is best applied to situations where the intermediate is present in low concentration (e.g. $\ce{k_2 >> k_1}$).  A general rule of thumb for the assumption to be valid is for
$$\ce{\frac{k_2}{k_1} > 10 }$$
Looking back at the figure, we now see that this condition is met in the top row, center drawing.  Here is a Wikipedia article on the steady state assumption, the section on "Validity" is nice and clear if you'd like to read further on the subject.


How are unstable transition states determined?

Experimentally, and surprisingly,  there is only circumstantial evidence for most short lived species in any type of reaction, i.e. no one has spectral or structural evidence for a transition state, such as often portrayed in textbooks, other than for example NaI dissociation or atom-diatomic molecule reactions e.g. H+OH  etc. done in ultra-low vacuum.

For longer lived intermediates (picoseconds to longer) short pulse laser pump-probe (flash photolysis) methods have to be used to obtain spectral signatures but give no structure, and this method only works for light triggered reactions which are rare. An alternative is stopped flow, but the time scale is slow, many microseconds so usually too slow for most intermediates, and again does not give structure. NMR is immensely slow also. Time resolved x-ray only works on crystals. So far light driven reactions in proteins have been studied down to a few tens of femtoseconds.  There is a huge and exciting field here to be explored.


How to dock aggregated structures comprised of "elementary" protein units like LEGO pieces?

Try DNA origami instead

While protein assemblies are well known, rationally using proteins to assemble larger supramolecular structures is still a difficult problem:

Predicting protein folding is an unsolved problem. Sometimes even small modifications to a sequence can change the folding propensity.
Predicting protein-protein interactions is often difficult, due to challenges in understanding protein dynamics, intermolecular interactions, and accurate predictions of protein electrostatics.

There has been some progress in terms of forcing particular protein-protein assembly through intentional side-chain modification and cross-linking, and reactive end-groups (e.g protein-based metal-organic frameworks (MOFs).

That said, use of proteins for supramolecular assemblies is far harder and less prevalent than using DNA assemblies. At the moment, DNA origami is far superior for these purposes.

Edit: If you're really looking to just copy and displace atoms, I don't think you need any particular software. I'd write a Python (or other script) that reads in the PDB files, and writes out duplicate atoms displaced in the XYZ directions as needed. Group theory can help predict the angles, but you'll still need to know the protein size to work out the geometry.


How to find mole fraction of solvent and solute in a vapor pressure lowering problem?

$\displaystyle \ce{x_{solute}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}}$

Thus, we have that $\displaystyle\ce{x_{solute}} + \ce{x_{solvent}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}} + \ce{\frac{n_{solvent}}{n_{solute} + n_{solvent}}} = \ce{\frac{n_{solute} + n_{solvent}}{n_{solute} + n_{solvent}}} = 1,$ so $\ce{x_{solute} + 0.974 = 1\implies x_{solute} = 0.026}$


Calculus is the basis of physical chemistry. Have a look to Maxwell relations,Clapeyron equation. How stated Uncle Al you should take a book of physical chemistry and look into, furthermore some books (see Atkins) have an appendix showing you the mathematical skills required to understand the content of the book. Here what you should know only for thermodynamics in my opinion:

algebra
logarithms
vectors
calculus
Partial derivative
differential equations

If you go deep into the atoms more mathematical skills are required.


Why do research papers mention experimental details in such detail?

Experimental details are very, very important: they are used to ensure that results are reproducible, or at least that is the aim.

Let's talk about this specific example of suppliers and sources. One might naively think that a chemical purchased from supplier X is the same as the chemical from supplier Y. In an ideal world, that would be how it is. Unfortunately, this isn't the case, due to various considerations e.g. method of preparation, packaging, and so on. Therefore, even though the majority of the compound may be the same, there can be different impurities. And unfortunately, these impurities can lead to reactions working or failing. Many experimental chemists will personally know of cases where a reaction doesn't work until a reagent from a different company is used, or a reaction that stops working after buying a different bottle of a reagent.

There are many examples of this in the literature, but I'll choose the first one that came to mind, which is a reasonably recent paper from my old group (open access):

Mekareeya, A.; Walker, P. R.; Couce-Rios, A.; Campbell, C. D.; Steven, A.; Paton, R. S.; Anderson, E. A. Mechanistic Insight into Palladium-Catalyzed Cycloisomerization: A Combined Experimental and Theoretical Study. J. Am. Chem. Soc. 2017, 139 (29), 10104–10114. DOI: 10.1021/jacs.7b05436.

  During these investigations, we also uncovered a critical dependence of the stereochemical outcome of the reaction on the batch of Pd(OAc)2 employed as catalyst. This discovery was made through the chance purchase of Pd(OAc)2 from a different supplier, which led to an unexpected
  ratio of enamide alkene geometries ((Z):(E) = 80:20), rather than the typical ratio of ∼97:3.  Screening of further samples of Pd(OAc)2 gave variable results, the most extreme being a reversal of stereoselectivity to 40:60 in favor of the (E)-isomer.

The exact supplier isn't named, but 1H NMR spectra of various batches are given and the performance in the reaction can be correlated with the exact form of Pd(II) in the catalyst, which ensures that future readers are aware of what to look out for. This is very important for good science, because what's the point of publishing your fancy new method if nobody can replicate it?

Of course, this doesn't always happen. However, it's far better to include the information, just in case it is needed.

As mentioned by Waylander in the comments, commercial samples of samarium diiodide (SmI2) also have a reputation for being inconsistent: see e.g. this report by Szostak, M.; Spain, M.; Procter, D. J. J. Org. Chem. 2012, 77 (7), 3049–3059. DOI: 10.1021/jo300135v.


Name of a proton bonded to a hydroxyl carbon

I would call it the $\alpha$-proton to the $\ce{-OH}$. The alpha proton is the proton bonded to the alpha carbon (i.e., the hydroxyl carbon).


NMR spectrum of 3-hexylthiophene: why is the methyl group not a triplet?

The reason why the terminal methyl group is not a simple triplet is because of strong coupling, which is the scenario whereby the coupling constant between two nuclei $J$ is comparable to the difference in resonance frequency $\Delta \nu$.

As labelled in your question, protons B and C have almost the same resonance frequency (0.01 ppm difference corresponds to a 3 Hz difference on a 300 MHz spectrometer). The coupling constant between these protons, though, is probably on the order of 7 Hz or so. Hence, protons B and C are considered to be strongly coupled.

What does this have to do with protons A, which your question is asking about? Well, by virtue of the fact that protons A are coupled to protons B, which are in turn strongly coupled to protons C, there will be some "extra" coupling observed in the peak corresponding to proton A. The term for this is virtual coupling.

It turns out that this is the case no matter what nuclei they are or exactly how they are connected in the molecule - as long as X and Y are strongly coupled, any nucleus A that is coupled to either X or Y will have some degree of virtual coupling. The reason for this is fundamentally quantum mechanical in nature. Suffice it to say, though, that the n+1 rule will not predict multiplet shapes accurately under such conditions.

Further reading on Hans Reich's excellent website. In particular you may want to take a look at the "Methyl Region" section which is pretty similar to your example (although in your case it is actually even more complicated).


How are subshells and shells arranged in an atom?

I will start off by offering an apology for anything I might have misunderstood - if this is the case, just tell me in the comments below.

  If 2s and 2p are in the second energy level, and 3s, 3p, and 3d are in the next (3rd) energy level

This is, in general, not the best choice of words. The 2s and 2p orbitals are only degenerate, meaning that they have the same energy, in a hydrogenic atom: a species which contains one nucleus and one electron. The hydrogen atom itself satisfies this requirement. Other hydrogenic species include $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, ...

For any species that has more than one electron, the 2s and 2p orbitals have different energies (see this question, although it's a bit of a read) and therefore it would not be correct to refer to them belonging to the same "energy level".

What you can say is that 2s and 2p orbitals belong to the same principal quantum shell. All this means is that the number in front of them - called the principal quantum number - is the same, in this case, 2.

  How are these subshells arranged in space in relation to one another?

Orbitals are not actually containers, or tangible objects that occupy space. They are mathematical functions obtained by the solution of the Schrodinger equation. The pictures of orbitals that you are probably familiar with are just pictorial representations of these functions. Without going into too much detail, one quantity that is used in the functions is the radial distance between the electron and the nucleus, denoted $r$. $r = 0$ corresponds to the nucleus's point in space. What this means is that all orbitals - regardless of their shape or energy - are centred at the nucleus. The $\text{s}$ orbitals are spherical in nature, and the centre of the sphere corresponds to the nucleus. And the same goes for $\text{p}$ orbitals, $\text{d}$ orbitals, and so on.

Therefore, to put it very succinctly, everything is superimposed on everything. You could think of it as shining different colours and shapes of light onto a screen - the projected images can all simultaneously exist.

  3s, 3p, and 3d orbitals all of a relatively larger size

You're right about this - in general, orbitals with a larger principal quantum number tend to be larger, as long as we restrict the discussion to the same atom. For example, the $\text{3s}$ orbital is larger than the $\text{2s}$ orbital, which is in turn larger than the $\text{1s}$ orbital.

Now since I did say that orbitals are mathematical functions, this does raise the issue of how you define an orbital to be larger than another. If you are interested, you can read up about the radial distribution function, although I shall not expand on it as that would really be beyond the scope of this question.


Effect of direct exposure to sunlight on various drugs

Conversion of D to L isomer or vice versa is about the last thing to worry about in this situation, even if it may be facilitated by sunlight to a significant extent (which I doubt). Also, storage of chemicals in dark coloured bottles is by no means limited to compounds that have optical isomers, or even to drugs in general. Other examples are numerous; permanganate solutions and nitric acid come to mind.

Photons of visible light have pretty decent energy of a few eV. That's way higher than the energy of thermal motion at room temperature. When a molecule is hit by a photon, it is hit hard. If a molecule is fragile or strained in any way, it will probably break. Photons come in droves, and after a while your precious organic sample turns into a brown goo. (It would probably do so even in the dark, but that would take a bit longer.)

That's what's wrong about sunlight in the first place.


Mechanism of substitution reaction with no change in stereochemistry

I can explain the reason why the configuration does not change by using neighboring group participation of the hydroxyl group. This is the mechanism that I propose:

The hydroxyl group attacks C-3 and $\ce{I-}$ leaves.
Then $\ce{N3-}$, the nucleophile, attacks C-3.
Since two successive inversions occur at C-3, the net result is retention.

As you can see, the configuration has not changed.

@bon brought up a point: The formation of a mixture of products when an epoxide opening takes place. If the nucleophile were to attack C-2 instead,

C-3 will still have the S configuration.

As noted by @Jan in the comments, the epoxide intermediate is symmetrical and both epoxide carbons are equivalent. This can be shown by a $C_2$ rotation:


As you can see on this data page on Wikipedia helium-4 has critical pressure of 2.24 atm, but helium-3 has 1.13 atm, only slightly above atmospheric pressure. Hyperbaric chambers work with 1,3-1,5 atm, so you could try there.

There is, unfortunately, question if it has sense, as you wouldn't notice any serious difference in properties between subcritical helium-3 and slightly above critical point in pressure, but so far above in temperature. It would still be lighter than air, for example.

I don't think that much higher pressure would be a huge problem - the highest pressure survived by human is supposedly 54 atm - so says Wikipedia, but I didn't find proof. If there is a compound with critical parameters sufficiently near standard, you could touch liquid-like, not gas-like supercritical fluid.

Ionic liquids have relatively low critical pressures (even only 14 bar), but too high crit. temps (source). On the other hand ethane's crit. temp. is below 30°C, but pressure over 48 atm. Maybe there's sweet spot somewhere, but it would be still high pressure and may be quite hot or cold.


Why does dry ice "dissolve" so quickly in water?

It's just about heat capacity + heat conductance, nothing more. $\ce{CO2}$ sublimates much quicker in the water because water is much better at heat exchange than air. Anybody who ever experienced swimming in water at 15°C and walking in air at 15°C can attest to that.

Catalysts for heat exchange are usually made of copper and other metals. (Also, usually they are not called catalysts.) You have one in your computer, attached to the CPU.

As to formation of carbonic acid, it hardly has any influence on the process (though it does happen to a certain extent, of course). Most $\ce{CO2}$ escapes in gaseous form anyway.


Efforts made to reduce the amount of CFC

After learning more about environmental chemistry I got this.

On 16 september 1987, 95 countries accepted not to use ODS(Ozone Depleting Substances)  after signing Mortal Protocol. Later other countries also accepted it.

After that United Nation have decided to celebrate 16 september as "Ozone protection day" at international level to spread awareness.


You are right.  The × in 8× refers to the fold-change in concentration between the concentrated source and diluted solution.

$$\mathrm{DF = \frac{C_{stock}}{C_{working}}}$$ where $\mathrm{C_{stock}}$ and $\mathrm{C_{working}}$ are the concentrations of a stock solution and the desired final concentration, respectively.

The concentration of the working solution can be expanded:

$$\mathrm{C_{working} = \frac{C_{stock} V_{stock}}{V_{working}} = \frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}}$$ where $\mathrm{V_{stock}}$ is the volume of the stock solution and $\mathrm{V_{diluent}}$ is the volume of diluent (e.g. pure water) added.

Plugging back into the DF equation:

$$\mathrm{DF = \frac{C_{stock}}{\frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}} = \frac{V_{stock}+V_{diluent}}{V_{stock}}}$$


If you have a fuel cell, only hydrogen needs to be carried with the car; the air provides all the oxygen you need. Hydrogen is an extremely light fuel, but there are some issues with current means to contain it (e.g. heavy high pressure container, or massive hydride tanks).

According to Iceland is Bullish on Hydrogen, they've started to use their plentiful geothermal and hydroelectric power to electrolyze water, using the hydrogen to power cars and the oxygen byproduct for other uses.


What does "oxidized form" and "reduced form" mean?

In this case, the adjective "reduced" does not imply something is actually being reduced. It simply refers to the form of the element in question which has the lower oxidation state. Let's say something is getting oxidised in a half-reaction, for example

$$\ce{Fe^2+ -> Fe^3+ + e-}$$

Iron takes two forms here, $\ce{Fe^2+}$ and $\ce{Fe^3+}$. $\ce{Fe^3+}$ is the oxidised form, that much is obvious, right? Then, $\ce{Fe^2+}$ is the reduced form.

No $\ce{Fe^3+}$ is actually being reduced to $\ce{Fe^2+}$. However, it is still correct to refer to $\ce{Fe^2+}$ as the "reduced form", since it is the form of iron with the lower oxidation state.


3-ethyl-2-methylpentane-2,3-diol with sulfuric acid

Edit: This answer is based on a typo in the original question referring to a different starting compound than the OP intended.

A quick analysis of this reaction gives me the following:

3-ethyl-2-methylpentan-1,2-diol can be protonated by acid to produce cation A or B. Either of these cations can lose water to form a carbocation

The carbocation formed from A is a tertiary carbocation C that might not be likely to rearrange, but we'll consider the possibility in a bit.
The carbocation formed from B is a primary(!) carbocation D that seems unlikely to form.

  Pinacol rearrangement from cation A

A pinacol rearrangement from cation A through carbocation C forms an aldehyde: 3-ethyl-2-methylpentanal.

  Pinacol rearrangement from cation B

Having established that carbocation D is unlikely to form, we might suspect no rearrangement, but the rearrangement can occur in a concerted fashion with the loss of water. The end result is a ketone: 4-ethyl-3-hexanone. These kinds of concerted rearrangements to avoid primary cations are also known to occur in Friedel-Crafts reactions.

Neither product seems to match what you provided as the book's answer. Is it possible you miss-named the product?

  Which product is more likely the major product?

Rather than tell you, I will give you a hint:

One rearrangement produces a thermodynamically more stable cation than the other.


What you are referring to as perturbation theory sounds like a very crude approximation to the 'charge-transfer' or CT intermolecular interaction.  In quantum chemistry lingo, the CT interaction is the mixing of occupied orbitals on one molecule/fragment with the virtual orbitals on another fragment (this is the 2-body CT interaction).  This permits the transfer of electron density from one fragment to another (charge-transfer).

Instead of redoing the HF (variational) calculations with the new orbitals included, perturbation theory is instead used to calculate the nth order correction as a result of occ(a)+vir(b) mixing.  Differencing the HOMO on fragment A with the LUMO on fragment is just a way to quickly approximate the more complicated (and expensive) perturbation theory expressions for the CT iteraction.

So this seems to make sense in the context of HSAB theory and Lewis acids where acid/base exchange electron pairs and thus participate in a strong charge-transfer interaction.

*EDIT:  I'd like to add that the orbital energy of the LUMO orbital (in fact, all virtual orbitals) is sensitive to the basis-set size. When a large enough basis-set is used, the LUMO can always be forced to near 0.00.  Orbital energies, at the end of the day, are not nature.


The adenosine part is needed to make DNA and RNA.  It is logical for the triphospate to be attached to ADP, GDP, UDP or CDP since they are needed to make DNA and RNA.

In some instances, GTP is used instead of ATP.  In fact the Krebs cycle does involve phosphorylation of GDP.   GTP is used in the formation of microtubes by tubulin.

I've seen some give the rate of hydrolysis being greater for ATP than GTP as a reason for ATP being more utilized.

There is a related question here: https://biology.stackexchange.com/questions/11286/why-is-atp-the-preferred-choice-for-energy-carriers


Derivation of Ideal Gas Law with Boltzmann's Constant

$R$ is defined as $k_{B}N_{A}$ where $N_{A}$ is Avogradro's number. Notice that these forms are equivalent once you realize that $N = N_{A}n$.

I think $k_{B}$ is measured experimentally. I learned the derivation via statistical mechanics.


The first part of your question has been answered in the comment, as for the part about what does $\ce{Me2S}$ do:
The mechanism used here is reductive ozonolysis and $\ce{Me2S}$ acts a reducing reducing agent itself getting oxidised to dimethysulphoxide $\ce{(Me)2SO}$ (DMSO).
As for the mechanism of reductive ozonolysis with $\ce{Me2S}$ here's a helpful link.


It seems the rust reaction was not well explained.  The following is the formula for iron rusting with water.
$$\ce{4Fe + 3O2 -> 4Fe^3+_(aq) + 6O^2- _(aq) -> 2Fe2O3}$$
Water will dissolve the iron because often carbon dioxide is available in the atmosphere to make the water slightly acidic. With the help of acidity electrons can be transported much easier from from dissolved iron to dissolved oxygen. Where oxygen, iron, and water meet is most often where rust pits start to form.

Without water the formula is the following:
$$\ce{4Fe + 3O2 -> 2Fe2O3}$$
This reaction is very slow.  More energy is needed to initiate the reaction as in the case with burning steel wool, or another means is needed to transport electrons.  Dirt, grime, residues, or any number of things can transport electrons resulting in rust.


According to this Wikipedia article:

  Carbonated water (also known as club soda, soda water, sparkling water, seltzer water, or fizzy water) is water into which carbon dioxide gas under pressure has been dissolved. Some of these have additives, such as sodium chloride, sodium bicarbonate or similar, but seltzer water is almost always composed of water and carbon dioxide with no other additives.

So the only possible candidates for a chemical reaction from the sparkling water are $\ce{CO2}$, and water itself. The other salts present at low levels would not have sufficient reactivity to produce the relatively high-energy results observed. And water can be ruled out as a reactant because ginger itself of course is a plant that contains water (or contained water previously if dried ginger was used). For the sake of thoroughness, I went to my kitchen and added tap water to dried, ground ginger and observed no effervescence.

This leaves $\ce{CO2}$ as the only candidate for causing the observed reaction upon the addition of ginger to sparkling water. A chemical reaction can be ruled out from a couple perspectives. First, as $\ce{CO2}$ is a very stable, completely oxidized compound, ginger would have to contain some strong reducing agent for it to react. We know this is not the case as the ginger is produced and stored in an oxidizing environment (our atmosphere of $\pu{21\%}$ oxygen). The atmosphere also contains on the order of $\pu{0.04\%}$ $\ce{CO2}$, which is less than that found in sparkling water but would still have quenched any $\ce{CO2}$ reactive component, albeit in slower less dramatic fashion.

At this point we've determined the effervescence must be a result of the $\ce{CO2}$ in the sparkling water, but we've eliminated the possibility of a chemical reaction. Because the carbonation process leaves the sparkling water supersaturated in $\ce{CO2}$, the only thing preventing the $\ce{CO2}$ from bursting out all at once is the lack of sites for $\ce{CO2}$ bubbles to nucleate. And this is what the ginger provides, particularly if it were in a ground, high-surface area form, but even in the case of raw ginger slices there are many more surface sites available than a glass container.

Bottom line, TL/DR: There is no chemical reaction, rather the ginger just provided a relatively high surface area for the rapid formation of $\ce{CO2}$ bubbles from the $\ce{CO2}$ supersaturated sparkling water.


Why do we use pV term independent of U in the equation H = U + pV?

The energy U of a system consisting of a single component and phase is

$ U = -PV + TS + \mu N$

As you rightly point out, U includes the energy required to create the volume by expanding against the surrounding (constant) pressure, namely the -PV term. Removing this contribution gives H.

Therefore

$ H = U + PV = TS + \mu N$

Equating TS with the "heat content" of the system leads to the statement in the textbook (this is my interpretation of this textbook's author's meaning). This kind of algebraic manipulation can lead to all sorts of confusion.

EDIT

Some people recoil (and downvote) when they see an expression such as

$ U = -PV + TS + \mu N$

as if this were somehow unphysical. Well, it does suggest that there is such a thing as an absolute value of U rather than only relative values, and it is hammered into our heads in school that there are only relative, not absolute, energies. But the point here is that there is an absolute value, relative to an empty system, and you can use N (the number of particles) as your simple scaling constant. The expression above captures the extensive property of U. You could equally write

$ \Delta U = -P \Delta V + T \Delta S + \mu \Delta N$

and no one will blink. Well, if you define

$\Delta U = \Delta N \times U_m$

$\Delta S = \Delta N \times S_m$

$\Delta V = \Delta N \times V_m$

with

$\Delta N = N-0$

you obtain an expression equal in spirit to the sacriligeous one above.

Usage of terminology such as "heat content" is potentially confusing, but we can use similar arguments to those used in explaining the PV term's meaning as work to create the system from scratch by increasing the particle number, while keeping intensive properties such as T and P constant, to suggest that the term TS is the heat that must be exchanged with the surroundings during creation of the system (from 0 to N particles). This is more of theoretical than experimental interest, in the same way we can hypothetically charge an object from 0 to a finite charge to compute its energy (relative to no charge).

Note that saying that H represents the heat content is confusing, since the application of the equation is not usually in the spirit I have just outlined, and if you look at

$ H = U + PV = TS + \mu N$

there is an additional term $\mu N$ term which cannot be easily associated with the concept of "heat content".

So, being generous, I would think the author of the book meant by the statement was, very loosely, that changes in H, that is $\Delta H$, are usually associated with the heat exchanged by a system at constant P (subject also to other restraints but more on that in other questions and answers referring specifically to the meaning of H).


How may the water from rain be tested to determine it is natural, or from cloud seeding?

The AgI would act as a cloud condensation nuclei (CCN) on which water vapour would condense allowing cloud droplet formation. Presumably, you could use scanning electron microscopy to determine the composition of the CCN and determine whether cloud seeding has taken place.

Apparently, his has already been done with cloud chamber (so, artificial scenario) measurements (Zimmerman et al., 2007 - "Environmental scanning electron microscopy (ESEM) as a new technique to determine the ice nucleation capability of individual atmospheric aerosol particles").

However, considering that there has been ESEM measurements in field situations before (Pöschl et al., 2010 - "Rainforest Aerosols as Biogenic Nuclei of Clouds and Precipitation in the Amazon") to determine CCN composition, I suppose a similar set up could be performed.


How to write half equations based on oxidizing/reducing agents pairs?

Firstly, you need to understand that there is a maximum and minimum oxidation number.

For metals, the minimum is usually $0$ while the maximum is usually the number of their outermost electrons. For example, $\ce{Na}$ which has $1$ outermost shell electron would have a minimum of $0$ and a maximum of $1$.

For non-metals, the minimum is usually the number of their outermost shell electrons minus $8$, while the maximum is usually the number of their outermost shell electrons. For example, $\ce{S}$ which has 6 outermost shell electrons would have a minimum of $-2$ and a maximum of $6$.

For the pair $\ce{CO2}$ and $\ce{H2C2O4}$:

Firstly, note that the most electronegative element in each compound is $\ce{O}$. You can learn more about electronegativity here.

Therefore, it takes the minimum, which is $-2$.

Secondly, usually hydrogen is $1$ (in most covalent compounds) or $-1$ (in most ionic compounds). In this case, hydrogen is $1$.

It follows from calculation (that the sum of the electronegativity of each element in a compound must be zero) that the $\ce{C}$ in the first compound has oxidation number $4$ while that in the second compound has oxidation number $3$.

Since the oxidation number has dropped, this is a reduction reaction.

We write the unbalanced equation as a starter:

$$\ce{2CO2 -> H2C2O4}$$

Balancing the number of $\ce{C}$.

Now, we are in an acidic environment, so we add $\ce{H2O}$ to balance the oxygen and $\ce{H+} to balance the hydrogen. The oxygen is already balanced, so:

$$\ce{2H+ + 2CO2 -> H2C2O4}$$

Note that the charges are still not balanced while the elements are balanced, so we add electrons:

$$\ce{2H+ + 2CO2 + 2e- -> H2C2O4}$$

Thus our half equation is complete.

The method to find the oxidation number of each element in $\ce{C6H12O6}$ is detailed above, and I leave this for you as an exercise.

PS: in an alkaline environment, we add $\ce{OH-}$ to both sides of the equation to eliminate the $\ce{H+}$, i.e. (the following is wrong because we are in an acidic environment thanks to the acid $\ce{H2C2O4}$ (acetic acid):

$$\ce{2H+ + 2OH- + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$
$$\ce{2H2O + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$


On the angle subtended between the angular momentum vector and the applied field for a single spin

You are correct that a given value of $m_I$ and $I$ does correspond to one single value of the angle $\theta$. For the most commonly used nuclei in NMR spectroscopy, the "allowed" values are $I = 1/2$ and $m_I = \pm 1/2$. The two states are commonly labelled $\alpha$ and $\beta$; these are the eigenstates of the angular momentum projection operator $\hat{I}_z$, such that

$$\begin{align}
\hat{I}_z|\alpha\rangle &= \frac{1}{2}|\alpha\rangle \\
\hat{I}_z|\beta\rangle &= -\frac{1}{2}|\beta\rangle
\end{align}$$

(treating $\hbar \equiv 1$). The problem is that you are assuming that the wavefunction (of a spin) can only be in an eigenstate. This is not true. It is perfectly valid, and indeed necessary (see below), to have states that are superpositions of eigenstates. In general, a spin can have any state

$$|\psi\rangle = c_\alpha |\alpha\rangle + c_\beta |\beta\rangle$$

where $|c_\alpha|^2 + |c_\beta|^2 = 1$ (normalisation), and in general the coefficients are time-dependent (see below). This is no longer an eigenstate of $\hat{I}_z$, and consequently is no longer well-described by one value of $m_I$, or for that matter, $\theta$. But that is not intrinsically a problem. The only circumstance in which a wavefunction must be described by an eigenstate is immediately following an observation, where the wavefunction $|\psi\rangle$ collapses to the eigenstate $|i\rangle$ with probability $|c_i|^2$. Otherwise, superposition states are entirely consistent with quantum mechanics.

A more generalised example is to think of the time-independent Schrödinger equation:

$$\hat{H}|\psi\rangle = E|\psi\rangle$$

which is an eigenvalue equation for the Hamiltonian, $\hat{H}$. It is true that only eigenstates of the Hamiltonian are admissible solutions to this equation. But that does not mean that the only physically permissible states for the system are eigenstates of the Hamiltonian. The time-independent SE only governs systems that are not evolving in time, or stationary states. We need something else to describe systems that do evolve with time, which is the time-dependent Schrödinger equation:

$$\hat{H}|\Psi\rangle = i\hbar\frac{\partial |\Psi\rangle}{\partial t}$$

The most general solution to the TDSE is, indeed, a superposition of stationary states (the Internet has lots of more thorough explanations, or read Griffiths' Introduction to Quantum Mechanics):

$$|\Psi\rangle = \sum_i c_i \exp\left(-\frac{iE_it}{\hbar}\right) |\psi_i\rangle $$

where the $c_i$'s are time-independent. Indeed, it is necessary for us to have superposition states, because this allows the coefficients of each eigenstate to evolve in a time-dependent fashion (because of the exponential term, which evolves at different frequencies for different eigenstates). If a system is prepared in an eigenstate of the Hamiltonian, then it can never evolve into another eigenstate (the exponential evolves, but if there is only one eigenstate, it is no more than a simple phase factor).

This might not seem immediately relevant for NMR, but recall that the (free) Hamiltonian in NMR is simply $-\gamma B_0 \hat{I}_z$. So, the eigenstates of $\hat{I}_z$, $|\alpha\rangle$ and $|\beta\rangle$, are also the eigenstates of the Hamiltonian. If we never had superposition eigenstates, then the spins would simply sit there forever and do nothing.*

* As long as the Hamiltonian doesn't change. But if you change the Hamiltonian, say from $\hat{H}_0$ to $\hat{H}_1$, then the eigenstates change, and so your system (which was originally an eigenstate of $\hat{H}_0$ is no longer an eigenstate of $\hat{H}_1$). Indeed, it would be in a superposition of eigenstates of $\hat{H}_1$, and would evolve in time according to the TDSE, so it doesn't invalidate any of the preceding discussion.


Pericyclic ring opening, cyclobutene to butadiene (Woodward-Hoffmann rules).
Claisen rearrangement, vinyl ether to ketone (re bullvalene).
Diels-Alder addition.
Sn2 displacement; AgF will move that along (not a soft Lewis base solvent).
Alkene to expoxide is not done with water.  Direct catalytic oxidation, base elimination from the cyanohydrin (first adding HOX), peracids.
Primary alcohol to aldehyde is Jones reagent (in water), Collins reagent (in anhydrous pyrdine, and much better. There is no product in the sludge), Swern oxidation (makes you unpopular downwind), fancier stuff.
Markovnikov addition of HCl to an alkene.

The Woodward-Hoffmann rules tell you what conditions are required (thermal vs. photochemical) and the associated stereochemistry (suprafacial vs. antarafacial) re conservation of orbital symmetry.


Why are lab coats required in chemical laboratories?

It is protection. It should be worn whenever you work with potentially dangerous chemicals. In German, it is even called (part of the) persönliche Schutzausrüstung, i.e. personal protective gear. Lab coats:

protect you against contamination with chemicals — acids, bases, toxic molecules and salts. They will ‘hit’ your coat and ‘stick’ to it rather than reaching your clothes. Don’t say you are careful. Your labmate may not be or an accident may happen. This is also the reason why labcoats are never allowed where food or drink are consumed.
protect you against spills of liquids. You can quickly take off a lab coat and not be contaminated by the liquid in question. This is basically an extension of the first bullet point.
protect you against fire. This might not be obvious, but lab coats are generally cotton which does not burn well. If you are wearing synthetic fibre below the lab coat, ripping that off could give you valuable seconds for not burning to death.
show other people that you are working with something potentially dangerous. Assuming you are dealing with adults that means they will know better than startle you because they can.

Everything else a lab coat supplies is basically just extension of the protection or meant to help people actually use it. Pockets: If it didn’t have any, people would keep using their trousers’ ones potentially contaminating them. Long sleeves: You arms need protection, too. Being long: so do your legs (at least partially).

And of course, it’s pretty cool if you walk in a lab coat, isn’t it? (That was my feeling at least.)


Can you make R-Candy without heating the propellant mix?

First of all, when it comes to something that may impact your future well-being in a bad way, do a thorough research on the topic. Wikipedia is a good starting point as it highlights the role of homogenization in the fuel–oxidizer mix preparation:

There are many different methods for preparation of a sugar-based rocket propellant. Dry compression does not require heating, only the grinding of the components and then packing into the motor. However, this method is not recommended for serious experimenting.[2] Dry heating does not actually melt the $\ce{KNO3},$ but it melts the sugar and then the $\ce{KNO3}$ grains become suspended in the sugar.

Heating assures homogenization and packing of the fuel–oxidizer mix to create constant burn rate. If you decide to heat the mix, it is imperative that you look up the properties of all components, especially melting points. For example, in the most common nitrate–sorbitol mix sorbitol (m.p. 95 °C) would act as a flux for potassium nitrate (m.p. 334 °C). Make sure you are using a well-functioning hot plate with a temperature controller and not open flames, otherwise the mix will ignite.
Make sure the powders are dry before you mix them. Otherwise trapped water inside molten sorbitol (remember, m.p. 95 °C and both sorbitol and nitrates are hygroscopic) will create steam pockets which may cause local explosion with the stuff dispersed around and potentially catching fire from the hot plate. Needless to say, the mix must be carefully stirred and slowly heated until fully homogenized; only then it can be poured into the firm template or the rocket body. The mix is going to be at about 150 °C and sticky, so wear heat-resisting protective gloves.
If a hot plate is not an option, you may opt out for the worse alternative of grinding the powders and pressing into shape. First, you have to produce fine powders of the fuel and oxidizer separately. Normally, separated sorbitol and potassium nitrates should not ignite because of friction, but work with small batches at a time regardless. Second, they must be thoroughly mixed and pressed to assure good contact between particles. Work with dry components in small quantities without producing too much dust flying around to prevent dust explosion.
General notes:

Work with small portions of materials at a time in a well-ventilated area using PPE.
Check your equipment thoroughly and make sure there are no sources of sparks and open flames nearby. Molten mix will ignite when overheated, and the dust is explosive.
Remaining water will quench burning due to its tremendous heat capacity and may cause unstable engine operation with a non-zero chance of explosion, so I'd avoid adding water at all regardless what preparation method you pick.


Why doesn't a halide anion react with primary or secondary alcohols using SN2?

The issue is that when carrying out the Lucas test you have a polar protic solution (due to the alcohol which you are testing), in which case Cl$^-$ is well solvated and thus not a very good nucleophile. As you are not using a strong nucleophile, you disfavor the S$_\text{N}$2 mechanism, but since you have a primary alcohol, you are not really favoring S$_\text{N}$1 either. Hence, why no reaction happens and the Lucas test works to identify primary alcohols. I think your question of Why is it biased towards an S$_\text{N}$1 reaction? might be a bit misleading because the reaction is neither biased towards S$_\text{N}$1 nor S$_\text{N}$2 for primary alcohols - it simply does not happen at all (at room temperature). In the case of secondary and tertiary alcohols, however, I would definitely say the reaction is biased towards S$_\text{N}$1 since you have polar protic conditions, a weak nucleophile, and a more stable carbocation intermediate.

Upon raising the temperature, I suspect that the reaction would then proceed via an S$_\text{N}$2 mechanism for a primary alcohol (as, despite not having a strong nucleophile, the barrier would be lower than via S$_\text{N}$1), but remain S$_\text{N}$1 for secondary and tertiary alcohols. Hope this helps!