Radical formation via tritium to helium decay

Here are two reasons why this approach would be impractical for organic synthesis:

(1) The half life of tritium is 12.3 years, so the decay is too slow to be practical as part of a synthetic method

(2) While on a scale of nuclear reactions, tritium decay is relatively low in energy emitted, it still dwarfs a typical chemical reaction (18.6 keV or 429,000 kCal/mole) and may cause degradation of the molecule (or the bulk material) in unpredictable ways.

But it could be useful for other types of chemical studies. I will defer to others in the forum to elaborate.


What does the prime symbol mean in organic nomenclature (e.g 2')?

Primes, double primes, triple primes, etc., are added to differentiate between the same locant in different parts of the structure. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), they are used

to differentiate the nitrogen atoms of hydrazides, imidamides (amidines), amidrazones, and hydrazidines;

  N'-acetyl-N'-ethyl-N-methylpropanehydrazide

with superscripted numerical locants indicating the position of the suffix on the parent structure, to differentiate the nitrogen atoms in other cases, especially when symmetry conditions are not fulfilled;*

  N3-ethyl-N'3-methylhexane-3,3-diamine

in multiplicative nomenclature to denote multiplied units and modify locants accordingy;

  2,2',2''-nitrilotri(ethan-1-ol)

in spiro-fused compounds, to denote positions in polycyclic systems

  7,7'-spirobi[bicyclo[4.1.0]heptane]

in ring assemblies, to number identical ring components;

  1,1'-biphenyl

in fusion nomenclature, to identify first and higher attached components, identical attached components, and multiparent names;

  difuro[3,2-b:2',3'-e]pyridine

in fullerenes ortho-fused to organic ring or ring systems, to identify positions in the nonfullerene component;

  3'H-cyclopropa[1,9](C60-Ih)[5,6]fullerene

in natural product nomenclature, to identify positions in ring(s) fused to a fundamental parent hydride;

  naphtho[2',1':2,3]-5α-androstane

The example that is given in the question is a ring assembly. The numbering of locants corresponds to the numbering of 1,1'-biphenyl, where primes are used to differentiate between the same locant in different phenyl groups. Therefore, the correct systematic name is 4,4',5,5',6,6'-hexahydroxy-[1,1'-biphenyl]-2,2'-dicarboxylic acid.

* Note that, according to current IUPAC recommendations, primes are no longer used to differentiate the nitrogen atoms of symmetrical geminal diamines, diimines, diamides, di(imidamides), (diamidines), di(hydrazonamides), di(amidrazones), di(hydrazonohydrazides), imidohydrazides, and poylimidopolycarbonic and inorganic oxoacids.


Group 13 elements reactivity with halogens

This is due to the Inert Pair Effect. As we move down the 13th group, addition of d orbitals and f orbitals are taking place. Since the d orbitals and f orbitals are away from the nucleus, they have poor sheilding power or capability and thus the s orbitals of the outermost shell are attracted towards the nucleai, making it difficult for s orbital electrons to leave the orbitals and only pairing of the 1 electron in p orbital is possible. Thus +1 valency of Tl is more stable than +3 as it is not energetically favored. This,ofcourse, does not mean that formation TlI3 is not possible. It's just that it requires more energy.


Why don't ligands affect the oxidation state of the central metal ion?

If you take a glimpse at the molecular orbital scheme, you will notice that the electrons donated by the water molecules still reside in orbitals that are largely ligand-centred (and have a relatively low energy). On the other hand, iron(III) also has five d electron itself; these reside in the $\mathrm{t_{2g}}$ and $\mathrm{e_g}^*$ orbitals, which are largely metal-centred (and have a relatively high energy).

This is true for most complexes. The metal tends to have rather high-lying orbitals while the ligands tend to donate rather low-lying ones to the overall complex. As long as this order is retained, there is no chemical redox behaviour going on.

For some systems, however, the orbitals are either so similar or even crossed-over that suddenly a metal orbital will become lower-lying than a ligand orbital. In this case, an electron that used to reside in a ligand orbital will relax into a metal orbital — a redox process. You can for example think of the reaction of iodide with copper(II) in this manner: copper(II) will immediately get reduced to copper(I) by iodide ions. While this is a unidirectional process, some systems are also known to bind ligands in a redox fashion reversively, meaning that the electrons will be exchanged both ways. The binding mode of oxygen to haemoglobin is understood in this manner.

Finally, sometimes there is a change in oxidation state that is merely formal. Tetracarbonylhydridoiron(0) $\ce{[Fe(CO)4H]}$ is such a case. It can be protonated to formally give tetracarbonyldihydridoiron(II) $\ce{[Fe(CO)4H2]+}$ or deprotonated to give tetracarbonylferrate(–II) $\ce{[Fe(CO)4]-}$. Both changes in oxidation state are merely formal and chemically all that is happening is a protonation/deprotonation.


Why are molecular reactions endo- or exo-thermic?

You are thinking it in a pseudo classical way. In those terms, imagine bonds like some kind of springs with some potential energy. During the reaction there can be rupture or formation of springs, changing the potential energy. In consequence, due to energy conservation kinetic energy must change.

A pictorial argument, that may help comparing your view point about collisions and the chemical reaction case, that is not always true, can be go like this: If we think of two balls bonded by an ordinary spring, in the "relaxed" position (that occurs somewhere between max and min elongation) we have the minimum potential energy. This is also true for a diatomic molecule (look some graph of potential energy curve for a diatomic molecule). Imagine that two of this balls bonded by the spring collide in the region of the (ideal, non mass, non perturbed by the collision etc.) spring and cut them out:

If you make this collision the speed of each ball will remain the same due to the spring is mass less. If molecules were like the balls no change in temperature would occurs, because kinetic energy does not change.

The key point is that molecules does not behave like that, spring never cut so: potential energy always is present. So if they where molecules, as products the separation between balls is larger and so the potential energy is greater and the kinetic energy turns to be smaller.

Now, think in the backward reaction, this reaction would be an exothermic reaction.

I am not sure if that is the kind of answer you need. If not or if something is unclear leave a comment.

Edit
I forgot answer that last questions. Yes, the temperature change can be calculated. Normally what is calculated is the enthalpy change, but they can be related trough heat capacity. Of course there are limitations in precision and size of the molecule, but in essence the answer is yes, it can be calculated following the quantum mechanics principles.


Different shapes of fluorescence spectra in Raman

There is no reason why fluorescence emission background should look similar at all with green and red excitation.

Different substances

You say the green excitation spectrum is of Coke.
According to the video, the red excitation spectrum is of the peel of a tangerine.

Those samples consist of totally different substances so they will have different fluorescence spectra.

Different emission wavelength ranges

Assume this black line is the fluorescence emission spectrum of one sample:

the 532 nm excitation (dark green line) Raman spectrum is in the wavelength range 532 to 781 nm (green shaded area)
the 785 nm excitation (red line) Raman spectrum is in the wavelength range 801 to 1014 nm (red shaded area)

There isn't even any overlap between the spectral ranges you look at.

Different Excitation wavelengths

In first approximation (slight) changes in the excitation wavelength won't change the shape of the fluorescence emission, only the intensity.
However, if the excitiation wavelength changes sufficiently to excite a different fluorophore, also the shape of the fluorescence emission spectrum can (will) change.

About buying a Raman system

Before going deeper into the buying decision I suspect you'll need to learn more about the basics of Raman spectroscopy (and maybe about fluorescence). I recommend starting maybe with
some textbook chapters about spectroscopy for chemical analysis in general e.g. Skoog et al.: Principles of Instrumental Analysis if you directly like to have a focus on instrumentation (possibly also get some introductory text about spectroscopy for structure determination), and then to get deeper into Raman McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5.

Any serious advise about which system is better for you will need a whole lot of further background information: e.g.
what samples will you measure (phase, analytes, matrices), what applications (qualitative, quantitative, spatially resolved, restrictions on measurement time, lab use vs. field use, need for polarized measurements/depolarization ratio, ...),
will these specification change in future (other research projects).

As you comment that there is no Raman instrument in your country,

maybe you could visit a reputable lab that does Raman on similar systems as the one you want to study?
Raman system manufacturers are usually quite willing to demonstrate the ability of their instrument to measure what you need with a few of your samples, either by you sending them a couple of samples or even by you visiting them with your samples so you can try out the handling of the instrument yourself.
Do this at least with the few instuments are in your shortlist so you do have a basis for comparing actual performance for your samples.

A few comments about the Raman experiments in the linked youtube-video

Vodka-in-bottle spectrum: measuring through or on glass usually doesn't disturbn Raman spectra for green (e.g. 532 or 514 nm excitation) and also 633 nm red excitation is OK. With 785 nm excitation you'll get very bad background.
The reason why the handheld measurement works is its long working distance: the focus is deep inside the bottle, thus hardly any scattered light from the bottle gets to the detector.
Sugar Raman measurement background: not all background in Raman spectroscopy is fluorescence. White powders (that are actually clear substances with lots of refraction and reflexion due to the small particle size) will give a lot of elastically scattered light (i.e. still excitation wavelength). Depending on how good the spectrograph is, such scattered light may end up in the detector at the wrong pixel and thus be interpreted to have another wavelength.
Such scattering background will usually decrease the further you get from the Rayleigh line. A conclusive experiment to show that it is scattering and not fluorescence is: dissolve (or melt) the powder and deposit the substance as a clear film (after drying). Scattering will then (mostly) be gone.
Sugar vs. Sugar in fruit and the all-same-food spectra: solid sugar vs. sugar dissolved in water won't have the same vibrational spectrum (the spectra of some substances do not change much when dissolving in certain solvents, but sugar has tons of interaction with the surrounding water molecules due to its -OH groups - and that does affect the spectrum.)
As for the "almost all food has the same NIR spectrum": right in that wavelength range of 940 - 99 nm are overtones of -OH (water, carbohydrates, alcohols, H-bridges to oxygen). As all those fruits and veggies (including the peel) are in first approximation a mix of water and carbohydrate, there shouldn't be any surprise in those spectra being very similar.
With a good spectrometer and good data analysis (which includes good design of the calibration experiments) you can approach questions like "is this apple ripe for harvesting yet", but this requires the spectra to be available with a noise level (reproducibility) that allows measuring small shoulders of the absorption band.
Besides what the video discusses, there are a some other important things Tellspec claimed but cannot measure for physico-chemical reasons: mono-atomic ions such as Na⁺. They don't have molecular vibrations (again with super-good spectroscopic equipement you may be able to measure the effect of the ion on the water, but that's not going to work in practice). Which means, if you scan a soup and then add salt, it won't see the difference. Yet they claim to tell you sodium intake...


As far as I remember from my lectures nucleophilicity has the same definition problems as electronegativity.
So unlike e.g. ionisation energy you cannot define an absolute scale of ordering which you can verify experimentally
and just as in the case of electronegativity there are several possible definitions (compare https://en.wikipedia.org/wiki/Nucleophile).
Similar to electronegativity the definitions arrive on quite similar results. ($\ce{Br^-}$ is a good nucleophile in polar, protic solvents...)
Other formulations should be "qualitatively consistent" with previous ones.

Your alternative formulation has two big problems:

There is no solvent dependency.
You completely omit steric effects and nucleophilicity is also a kinetic phenomenon.


Which process would separate the mixture of toluene, cyclohexane and water?

Distillation
The most effective and easiest way to separate the mixture of toluene and cyclohexane is simple distillation. Cyclohexane has boiling point is about 81°C, and toluene’s boiling point is 111°C.
References:

http://origin-www.thermoscientific.jp/content/dam/tfs/ATG/CAD/CAD%20Documents/Product%20Support%20Bulletins/NMR%20Support%20Documents/pS45-pS80-Simple-Distillation-of-Cyclohexane-and-Toluene.pdf
https://www.sas.upenn.edu/~dbalmer/chem502/Labs/lab2_distillations/balmerexp2.pdf


can i use an isobutane stove (like jetboil) or butane blowtorch with borosilicate glass?

While this would have been considered normal in the past (I used such a burner when I was a kid), today bunsen burners are considered outdated as heat sources.
Of course it will still work fine fine, but our view on risk has changed.
Normally one wants to avoid having any source of ignition in the lab, and burners are also difficult to control.

That being said, a burner like this one  will make for a very nice burner with a focused flame. This type of burner works really well when you need an open flame, like for melting salts onto magnesia rods, sterilizing inoculation spatulas, and showing how metals change flame colors.

You can even buy a portable bunsen burner that is pretty much the same:

One big downside is that the burner will transform into a flame thrower if is falls over.
Also, you will use a lot of gas. If you like working with an open flame, I would recommend to buy a small propane tank with a pressure regulator and a regular bunsen burner, a setup like this:

Just get the smalles propane tank feasible, like a 2 or 5kg tank, just to keep the total amount of flammable material as low as possible.
Any store that sells gas supplies will be able to set you up for little money, just buy a bunsen burner for propane.

Doing a distillation of salt water will work, but it will be difficult to control and you can easily shatter the flask by uneven heating. Maybe that is a good lerning experience!

Distilling anything flammable, such as alcohol, on an open flame would be reckless.

Some general advice:

Never leave a burner unattended, not for the shortest moment
Remove anything flammable nearby. Work outdoors if possible.
In any distillation setup, always be able to remove the source of heat very quickly. Don't use the burner as a support, so that you can take out the burner without your setup falling apart. That is especially important for a burner with an attached gas tank.
Make sure the gas tank does not get hot.
Always be prepared for any flask to break at any time. What would happen if the boiling liquid would be released. Would you get injured?
Wear safety glasses
Wear non flammable clothing that is closed at the neck. A traditional lab coat is not a good protection for your chest. Buy something like this
Learn about superheating and use boiling stones


Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?

Disclaimer: I now believe this answer to be fully incorrect.

  Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22.

  However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver & Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles:

  Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. J. Chem. Educ. 1978, 55 (1), 2 DOI: 10.1021/ed055p2.
  Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. J. Chem. Educ. 1996, 73 (6), 498 DOI: 10.1021/ed073p498.
  Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. J. Chem. Educ. 1994, 71 (6), 469 DOI: 10.1021/ed071p469.
  Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. J. Chem. Educ. 1989, 66 (6), 481 DOI: 10.1021/ed066p481.
  Some criticism of Atkins' books by Eric Scerri.

While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver & Atkins 5th ed:

The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d > 4s but for Sc onwards, 4s < 3d.

As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by
$$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$
where $hcR$ is a collection of constants, $Z_\text{eff}$ is the effective nuclear charge experienced by the electron, and $n$ is the principal quantum number. Since $n = 4$ for the 4s orbital and $n = 3$ for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more penetrating than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as $R(r)^2 r^2$ where $R(r)$ is the radial wavefunction obtained from the Schrodinger equation:

The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes $Z_\text{eff}$ larger. Going from the 3d to the 4s orbital, the increase in $Z_\text{eff}$ wins ever so slightly over the increase in $n$, which makes the energy of the 4s orbital lower.

Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively radially diffuse (the maximum in the radial distribution function occurs at a larger value of $r$). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy $U = -\frac{kq_1q_2}{r}$. The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by $n$, and if this were to be the case, you would expect 3d < 4s in terms of energies, as we said at the very start.

However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more "personal space" and experience less repulsion. One way of putting it is that an empty 4s orbital in Sc has a higher energy than an empty 3d orbital, but a filled 4s orbital has a lower energy than a filled 3d orbital. The fact that 4s > 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation ($\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$.)

I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an "anomalous" configuration that maximises exchange energy, whereas we don't get carbon adopting a $(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$ configuration in order to have "stable half-filled shells".


Suitability of liquid plastic resin for plumbing application

Epoxy should work OK and you have different options that are easy to get.


How does heptafluoropropane suppress fire?

Curt F. is probably correct in his estimate that heptafluoropropane extinguishes fire primarily by physical means. According to an overview by Choy and Fong, 'An Introduction to Clean Agents: Heptafluoropropane' (Int. J. on Eng. Performance-Based Fire Codes, vol. 5, nr 4, p. 181$-$184, $2003$),

For heptafluoropropane, the contribution of physical mechanisms to the extinguishment of fires predominates over the chemical mechanism. It suppresses  fires  primarily  by

extracting  heat  from  the flame reaction zone,
reducing the flame temperature below that which is necessary to maintain sufficiently high reaction rates by a combination of heat of vaporization [and] heat capacity.

They add that,

Oxygen depletion plays an important role in reducing flame temperature. The energy absorbed in decomposing the agent by breaking fluorine bond[s] is quite important, particularly with respect to decomposition production formation.

The chemical effect does indeed involve tying up active radical species. As an example, consider the mixture $\ce{H2/O2}$. The primary active species are $\ce{OH^\bullet}$ radicals.
$$\ce{H2 + O2 -> 2OH^\bullet}$$
The chain starts to grow.
$$\ce{OH^{\bullet} + H2 -> H2O + H^\bullet}$$
$$\tag{two reactive sp. instead of one}\ce{H^{\bullet} + O2 -> OH^{\bullet} + O^\bullet}$$
$$\tag{two reactive sp. instead of one}\ce{O^{\bullet} + H2 -> OH^{\bullet} + H^\bullet}$$
The bottom two steps show especially clearly why fires are violent, quick reactions (after incubation period is over). Heptafluoropropane's chemical extinguishing effect arises from

[----] the thermal decomposition of small amounts of heptafluoropropane in the flame which form fluorinated fragments such as $\ce{CF3}$ and $\ce{CF2}$. These will then consume the key combustion chain-propagating species $\ce{H}$ and $\ce{O}$, but to a lesser extent on $\ce{OH}$ radicals. The rates of chain-branching combustion reaction will decrease, the chemical flame is inhibited and the flame propagation is halted. (Choy, Fong)

To increase the chemical inhibition, heptafluoropropane (or HFC-227ea) is sometimes mixed with $\ce{NaHCO3}$.

Flame inhibition by sodium species is believed to be due to chemical scavenging of major radical species (e.g., $\ce{OH}$, $\ce{H}$) in the flame.
source: Skaags, 'Assessment of the Fire Suppression Mechanics for HFC-227ea Combined with $\ce{NaHCO3}$', U.S. Army Research Laboratory, link

One commercial source, unlike Wikipedia, groups pentafluoroethane together with heptafluoropropane.

source: Group of Companies "RealSnabService"

Further reading, more in-depth discussion and related topics:

Luo et al.. 'Effect of Hydrofluorocarbons and Perfluorocabons on Suppression Efficiency of $\ce{CH3I}$'. Process Safety and Environment Protection Research Group. link
Williams, et al.. 'Intermediate Species Profiles in Low Pressure Methane/Oxygen Flames Inhibited by 2-H Heptafluoropropane: Comparison of Experimental Data with Kinetic Modeling'. Annual Conference on Fire Research, $1998$. link
Sheinson. 'Heptafluoropropane With Water Spray Cooling System As A Total Flooding Halon 1301 Replacement: System Implementation Parameters'. Navy Technology Center for Safety and Survivability. link
Grosshandler, et al.. 'Extinction of Hydrofluorocarbon Flames with F/H Ratios of Unity and Greater'. Annual Conference on Fire Research, $1998$. link


Difference between an absorbance, emission, and excitation spectrum for uv-visible spectroscopy

When you have an excitation spectrum and an emission spectrum, it is for fluorescence spectroscopy. To obtain these spectra, you hold either the emission wavelength or the excitation wavelength and scan the other to gather the information. The excitation spectrum shows at what wavelengths the solution uses to produce its fluorescence. The emission spectrum shows what wavelengths are given off from the solution.

For absorbency spectrum, the solution has all the wavelengths shining on the solution, and you monitor the wavelengths of interest. The absorbency shows which wavelengths are absorbed by the solution.

The excitation spectrum has the same peaks as the absorbency spectrum; although the absorbency spectrum may have more peaks.


Cis Orientation of dienophiles orientation of groups maintained in Diels Alder

To answer you question, imagine (or: build a molecular model) that the newly formed six-membered ring may be simplified as a flat disk.  Then

either you start with a (trans)-configurated diene, as in the first example drawn by you.  Since the Diels-Alder simultaneously forms the new $\sigma$-bonds on the expense of $\pi$-bonds and moves the location of the remaining double bond, the substitutents R are either both below, or both above this reference plane.  Relatively to each other, these substitutents R in the product conserved their (cis)-relationship from the dienophile.
or, if your dienophile is (trans)-configurated, yielding a product were one of the substitutents is above, and the other is below this reference plane; corresponding to the second example drawn by you where are R conserved their (trans)-relationship to each other.

On paper, the observed conservation of sterochemical information is both the synchronicity of the events (pushing the pairs of electrons, absence of [charged] intermediates as e.g. in nucleophilic substitutions), as well as the principle of conserving the symmetry of molecular orbitals envolved here.


What causes straight fatty molecules to form a lattice (e.g. saturated fats)?

Nature abhors a vacuum, and your sticks aren't sticky.

When you have a jumble of sticks, there's lots of voids and spaces between them - the pile of sticks takes up much more room than the bundle. For wooden sticks, this is no problem, as air can easily fill in the voids between them. With saturated fats, this isn't possible. There really isn't anything that can fill the voids. The only way to fill them is to line up the chains in an ordered lattice.

Also, when the chains are in an ordered lattice they're making a large number of van der Waals interactions between them. Because the chains are straight, the best way to maximize these favorable attractive interactions is to pack the chains together in an ordered array. The van der Waals forces are comparatively small, but over a long chain it adds up. (I haven't ever seen this done, but I'm guessing that if you coated your wooden sticks with something sticky like honey, and shook them around a bit, your wooden sticks would also want to line up with each other and stick together in an ordered array.)

Temperature, of course, can overcome both effects. The additional kinetic energy in the molecules allows them to overcome the attractive interactions between the molecules, and higher temperatures mean that the molecules are moving faster, pushing away other molecules and effectively filling in the voids to some extent. When you get to a high enough temperature, the fat melts.

In unsaturated fats, the chains are kinked. This means that it's much harder to line them up to form all those van der Waals interactions, and there are intrinsically voids in even the most ideal packing. Thus it's much easier to break those interactions and add the additional voids needed for a liquid state, so the solid to liquid transition happens at a lower temperature.

One thing to keep in mind is that things like solid/liquid transitions are not absolute things. Where the solid to liquid phase transition happens is based on the relative stability of the solid versus the liquid. It isn't that the saturated fat can form a solid and the unsaturated fat can't, it's just that the solid saturated fat is more stabilize relative to the liquid saturated fat, than the solid unsaturated fat is to the liquid unsaturated fat. As the change in stability is much greater for saturated fat, it takes a higher temperature to melt the saturated fat than it does the unsaturated one.


Systematic Approach to Finding Number of Isomers

To my knowledge, there is no direct method or formula to find out how many isomers for given framework of an organic compound. Interestingly, this could be a good question for a mathematician. The only formula of this kind is the one to find theoretical optically active stereoisomers of a compound if the number of stereocenters (chiral centers) are given.

For instance, the given framework has 4 stereo centers, hence theoretically, you have $2^4 = 16$ stereoisomers for that compound alone. If your book says it has 64 stereoisomers, it probably has counted geometric isomers of conjugated double bonds (E/Z isomers) as well. All these 64 isomers are illustrated as compound 1 in the scheme below where each red asterisk represent a chiral center. Note that, since the ring has three different substitution groups, none would lead to symmetry in the molecule. Thus, all theoretical amount is equal to the actual number of isomers. Now, question is whether you need to count the stereoisomers for this framework only or not. If it is only for this framework, then the number is 64. But, if you want to change the framework by moving the substitution around, that would be a different story.

Now, for example, if you just move the carbonyl group around three other available carbons (say, move to the carbon between two hydroxy groups as shown in the compound 2 in the diagram), then you have additional 64 stereoisomers by that single move (compound 2). There are other 4 movable groups in the ring and all of them create multiple combinations such as 2,3,4,5-substituted cycloocta-1-one. Also note that chloro group and one of the hydroxyl groups can switch their positions to give another option. Or, the chloro group can just occupy the ring as well, creating another chiral center (e.g., 6-alkenyl-3-chloro-1,4-dihydroxy-8-methylcycloocta-1-one) or move to side chain. Upon all, there are changeable positions for the two double bond on heptyl chain as well (see compound 3). Double bond(s) in the side chain and an alkyl radical(s) can also move to ring (as methy, ethyl, etc.) are also possibilities. Thus, you are going to have tons of stereo isomer as any moving to the ring creating more stereo centers. Good luck on that, if you are not a mathematician.


Clausius–Clapeyron relation for states of water and vapor

The answer may be somewhat counterintuitive. Yes, an inert gas does very much exert pressure on a solid or liquid. The problem is that the chemical potential of the substance in the gas phase is not affected by the inert gas (if the gas is assumed ideal).
Assume you have a substance (liquid or solid) in a rigid isothermal but otherwise empty box. The result is as expected evaporation or sublimation of the substance until the pressure of the vapour has increased to an equilibrium value (the saturation vapor pressure). We assume there is plenty of substance so that some remains in the condensed phase while the rest fills the box as vapour. What drives evaporation/sublimation? The fact that the chemical potential of the substance is (initially) lower in the vapour. Once equilibrium is achieved the chemical potential is the same in both phases.
Now assume an alternate scenario: in addition to the substance (solid or liquid) there is now some inert gas in the box. As suspected this gas exerts a pressure on the liquid and thereby affects its chemical potential (free energy). But the free energy of the substance in the gas phase is unaffected by the inert gas (assuming the gases behave ideally). And here the somewhat counterintuitive result: the substance again evaporates/sublimes until reaching the saturation vapour pressure, but precisely because the inert gas exerts a pressure on the condensed phase and increases its chemical potential in this phase, but does not alter the chemical potential of the vapor, it increases the saturation vapor pressure. The increased vapour pressure raises the chemical potential of the vapour until it matches the heightened potential of the condensed phase.
And the short answer to the question is that the inability of vapor A to 'feel' the total pressure including the inert gas is part of the definition of an ideal or perfect gas, which fails if the pressure is raised sufficiently.


Will propane liquefy under these conditions?

First of all, you might want to consider using proper SI units.

If you know the saturation temperature (boiling point) of propane at a given pressure:

At a pressure of p = 75 psi (0.52 MPa), the saturation temperature of pure propane is 37.13 °F (2.85 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.

At a pressure of p = 200 psi (1.38 MPa), the saturation temperature of pure propane is 104.54 °F (40.30 °C); thus, at a temperature of 60 °F (16 °C), propane would be liquid.

However, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.

For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p

Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa).

At a pressure of p = 78 psi (1.38 MPa), the saturation temperature of pure propane is 39 °F (4 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.

If you know the vapour pressure of propane at a given temperature:

At a temperature of 60 °F (16 °C), the vapour pressure of propane is 107.71 psi (0.74 MPa).
Thus, at a pressure of p = 75 psi (0.52 MPa), pure propane would be gaseous. At a pressure of p = 200 psi (1.38 MPa), it would be liquid.

However, once again, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.

For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p

Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa). Since this partial pressure is below the vapour pressure of 107.71 psi (0.74 MPa) at a temperature of 60 °F (16 °C), propane would be gaseous.


For a hydrogen atom, according to Hertel and Schulz [1]:
$$\langle r\rangle_{n,\,l}=\frac{a_0 n^2 \left(\frac{1}{2} \left(1-\frac{l (l+1)}{n^2}\right)+1\right)}{Z}$$
and
$$\langle r^2\rangle_{n,\,l}=\frac{a_0^2 n^4 \left(\frac{3}{2} \left(1-\frac{l
   (l+1)-\frac{1}{3}}{n^2}\right)+1\right)}{Z^2},$$
where $\langle r\rangle_{n,\,l}$ is the average distance of the electron from the nucleus as a function of n and l, and  $\langle r^2\rangle_{n,\,l}$ is the average distance squared of the electron from the nucleus as a function of n and l, $a_0$ is the Bohr radius, and Z is the nuclear charge (Z = 1 for hydrogen).  These are known as expectation values, because these are the expected values based on the probabilistic distribution of electron density.
Using these, we can compare $\langle r\rangle_{n,\,l}$ and $\langle r^2\rangle_{n,\,l}$ for the $3s \,(n = 3, l = 0), 3p \,(n = 3, l = 1) \text{ and } 3d\, (n = 3, l = 2)$ subshells:
$$\langle r\rangle_{3,\,0}= \frac{27 a_0}{2 Z}$$
$$\langle r\rangle_{3,\,1}= \frac{25 a_0}{2 Z}$$
$$\langle r\rangle_{3,\,2}= \frac{21 a_0}{2 Z}$$
$$\langle r^2\rangle_{3,\,0}= \frac{207 a_0^2}{Z^2}$$
$$\langle r^2\rangle_{3,\,1}= \frac{180 a_0^2}{Z^2}$$
$$\langle r^2\rangle_{3,\,2}= \frac{126 a_0^2}{Z^2}$$
You can see that, as the azimuthal quantum number (l) increases, the electron's average distance from the nucleus decreases.  Note, however, that in hydrogen (or any single-electron atom, such as $\ce{He^+}$ or $\ce{Li^{2+}}$, etc.), all orbitals in the same shell have the same energy. Thus the energies of the 3s, 3p, and 3d orbitals are identical.  As a consequence, a hydrogen electron excited to the third energy level is actually in a weighted linear combination of all of these orbitals, and its actual average distance would likewise be a weighted average of these.
But still, why does the average distance decrease with increasing azimuthal quantum number?  Well, quantum is not my field, but I believe this is what's going on (perhaps a quantum expert can let us know if this is correct):
As I noted, in any single-electron atom, the 3s, 3p, and 3d orbitals have the same energy. The electron's energy is a combination of potential energy due to its distance from the nucleus, and the energy due to its angular momentum.  As the azimuthal quantum number increases, the electron's angular momentum increases (it has no angular momentum in 3s, some in 3p, and more in 3d), and thus the energy associated with this angular momentum increases.  But since the total energy is constant, as the energy due to angular momentum increases, the potential energy associated with distance from the nucleus must decrease, and hence the average distance from the nucleus gets lower.
Finally, you can see from the formula that $\langle r\rangle$ is much more sensitive to changes in n than l.  For instance, if we compare the sizes of the s, p, and d subshells in the 3rd shell (n = 1, l = 0 to 2), $\langle r\rangle$ ranges from $\frac{27 a_0}{2 Z}$ to  $\frac{25 a_0}{2 Z}$ to $\frac{21 a_0}{2 Z}$.  By contrast, if we compare the sizes of the s-subshells in the first three shells (l = 0, n = 1 to 3), $\langle r\rangle$ ranges from $\frac{3 a_0}{2 Z}$ to $\frac{12 a_0}{2 Z}$ to $\frac{27 a_0}{2 Z}$.

Hertel, Ingolf V., and Schulz, Claus-Peter. Atoms, molecules and optical physics. Berlin: Springer, 2015


Nomenclature and chemical formulae for bridged dinuclear chromium complexes

$\require{unicode}$The rules for formulating and naming coordination compounds are presented in Chapter IR-9 of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.

The line formula for a simple coordination entity is constructed within square brackets. The symbol for the central atom is placed first and is then followed by the symbols or abbreviations for the ligands (in alphabetical order according to the way they are presented in the formula).

  IR-9.2.3.1 Sequence of symbols within the coordination formula

  (i) The central atom symbol(s) is (are) listed first.

  (ii) The ligand symbols (line formulae, abbreviations or acronyms) are then listed in alphabetical order (…). Thus, $\ce{CH3CN}$, $\ce{MeCN}$ and $\ce{NCMe}$ would be ordered under C, M and N respectively, and $\ce{CO}$ precedes $\ce{Cl}$ because single letter symbols precede two letter symbols. The placement of the ligand in the list does not depend on the charge of the ligand.

  (iii) More information is conveyed by formulae that show ligands with the donor atom nearest the central atom; this procedure is recommended wherever possible, even for coordinated water.

Note that according to the recommendations in Item (iii), where possible, the coordinating (ligating) atom should be placed nearer the central atom in order to provide more information about the structure of the complex. Therefore, the line formula of the hexaaquachromium(3+) complex ion should be written as $\ce{[Cr(OH2)6]^3+}$ rather than $\ce{[Cr(H2O)6]^3+}$.

For the same reason, if possible, bridging ligands should be placed between the central atom symbols. Bridging ligands are those bound to more than one central atom. They are differentiated in names by the addition of the prefix ‘μ’.

  IR-9.2.5.2 Bridging ligands

  Bridging ligands, as far as they can be specified, are indicated by the Greek letter μ appearing before the ligand symbol or name and separated from it by a hyphen; (…)

  (…) In formulae, bridging ligands are placed after terminal ligands of the same kind. (…)

  The bridging index $n$, the number of coordination centres connected by a bridging ligand, is placed as a right subscript. The bridging index $2$ is not normally indicated. (…)

Polynuclear complexes may be named differently according to whether only stoichiometry is to be specified or partial or complete structural information is to be included.

  IR-9.2.3 Formulae of coordination compounds

  (…) Different applications may require flexibility in the writing of formulae. Thus, on occasion it may be desirable to violate the (…) guidelines in order to provide more information about the structure of the compound that the formula represents. In particular, this is the case for dinuclear compounds where a great deal of structural information can be provided by relaxing the ordering principles outlined in Section IR-9.2.3.1. (…)

  IR-9.2.5.1 General

  (…)

  As a general principle, as much structural information as possible should be presented when writing the formula or name of a polynuclear complex. However, polynuclear complexes may have structures so large and extended as to make a rational structure-based nomenclature impractical. Furthermore, their structures may be undefined or not suitably elucidated. In such cases, the principal function of the name or formula is to convey the stoichiometric proportions of the various moieties present.

  (…)

  Note, however, that the rules for formula writing may be relaxed in various ways in order better to display particular features of the structures in question. (…)

As a first approximation, the systematic line formula for the complex ion shown in Picture 1 is $\ce{[(H2O)3Cr($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)Cr(OH2)3]^2+}$. Note: pay attention to the Greek letter; it is to be written upright.

However, since this complex ion is symmetric, its formula may be further simplified.

  IR-9.2.5.4 Symmetrical dinuclear entities

  For symmetrical dinuclear entities, the name may be simplified by employing multiplicative
  prefixes.

  (…)

The resulting formula is: $\ce{[\{Cr(OH2)3\}2($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)]^2+}$. The corresponding name is di-μ-hydroxido-μ-sulfato-bis(triaquachromium)(2+).

The corresponding formula for the complex ion shown in Picture 2 is $\ce{[\{Cr(OH2)2(SO4)\}2($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)]^2-}$. The corresponding name is
di-μ-hydroxido-μ-sulfato-bis(diaquasulfatochromate)(2−).

Furthermore, if necessary, you can indicate the connectivity of the sulfato ligand using the kappa convention (sulfato-2κO).


Ketone prefix organic chemistry nomenclature

The prefix ‘keto’ doesn’t exist in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). Already in the 1979 Recommendations, the prefix ‘oxo’ is used to indicate $\ce{=O}$ when the group cannot be cited as suffix.

Nevertheless, the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995) still remembers the use of the prefix ‘keto’:

  To indicate a double-bonded oxygen that is part of a ketonic structure, the term keto is sometimes used as a prefix, but such use has been abandoned by IUPAC for naming specific compounds. A traditional use of keto is for indicating oxidation of $\ce{CHOH}$ to $\ce{C=O}$ in a parent compound that contains $\ce{OH}$ groups, such as carbohydrates. E.g. 3-ketoglucose.


How do I find the hybridisation of the orbitals of carbon in a molecule?

My method for finding the hybridization of the orbitals of an atom in a molecule:

Look at the molecular structure (geometry), apply the following:

Tetrahedral: sp3
Trigonal planar: sp2
Linear: sp


How to make Poinsettia pigment into ink?

I think that would be hard, you could try to use a mordant to convert a dye to a pigment, you can use Alumina $\ce{FeSO4}$  or $\ce{CuSO4}$. There is an article that explains it in detail how to use Poinsettia for colouring silk fabric.

  Rawat, Bhawana, et al. "Color fastness properties of silk fabric dyed
  with poinsettia leaves." MAN MADE TEXTILES IN INDIA 48.11 (2005): 418.

You can try to soak a thick paper in a solution of 1-5% of $\ce{FeSO4}$  with your dye heating gently.

Another solution could involve a different medium, like arabic gum mixed with citric acid.


The energy levels are in increasing energy  3S, 3P, 3D and the final state is 2P. The P-P transition is not allowed by $\Delta L =\pm1$ selection rule. The two transitions are 3S-2P at approx 812 nm and 3D-2P at approx 610 nm. As you are asked about the lowest energy transition this must be the 3S-2P. (Li also has non-zero nuclear spin depending  on its isotope which lead to hyperfine transitions and are ignored )

With some spin-orbit coupling the (Russell-Saunders) terms symbols are for the upper level $^2S_{1/2}, $ and for the lower level $^2P_{3/2}, \, ^2P_{1/2}$. The $M_J$ levels are assumed to be split so the degeneracy is 2 for the S and 2 or 4 for the P states as given by $2J+1$.  The selection rules on $M_J$ levels are $M_J = 0,\, \pm 1$.

Thus there are four transitions possible from $^2S_{1/2}$ to $^2P_{1/2}$ starting in  $  M_J = 1/2 $ in $^2S$ to $(1/2, -1/2)$ in $^2P$ and similarly $  M_J = -1/2 \rightarrow (1/2,\, -1/2)$, making 4 transitions in all as the splitting of the S and P levels will be different. (This is the same case as the splitting of sodium $D_1$ line but 'upside down' in energy to that case).

The other transitions are from $^2S_{1/2}$ to $^2P_{3/2}$. In this case there are 6 transitions as the P splits into 4 levels. The transition between the $\pm 1/2$ make four of these (as just outlined above) and in addition there are transitions between $M_J = 3/2 \rightarrow 1/2$ and $M_J = -3/2 \rightarrow -1/2$.

This makes ten transitions in all, thus it seems I have calculated the wrong transitions (or misunderstood completely),  however, you should be able to follow the method. Note that the observed spectral lines may not correspond to transition as distinct transitions may have the same energy gaps.

I've added a sketch below, it is hand drawn but should be clear enough. The $S_{1/2} - P_{1/2}$  transitions are a subset of these as you can see.


Does sulfuric acid formed in a previous step dehydrate a secondary alcohol?

It could, but the addition of water would push equilibrium toward the hydrated product, and the $\ce{H2SO4}$ would likely transfer into the water layer that you made by adding $\ce{H2O}$.

Plus, you would only make stoichiometric sulfuric acid. Depending on the amount of water, the pH of the solution formed could be fairly neutral. Also, usually, the acid (despite being a catalyst) is used in large excess for dehydration reactions.

Therefore, it is possible (given the right conditions) but unlikely.


Who figured out that different isotopes have different numbers of neutrons?

From Pauling's "General Chemistry" (ref. 1):

In 1921 the American chemist W.D. Harkins described nuclei as being built of protons and neutrons; he used the word "neutron" for a hypothetical particle with mass equal to that of the proton and with no electric charge. Ernest Rutherford made a similar suggestion during the same year. In 1932 the discovery of tne neutron was reported by the English physicist James Chadwick (born 1891).

The original article in which Harkins (ref. 2) reported this hypothesis is linked by the Wikipedia:
References

Linus Pauling, General Chemistry, second edition, 1970, p. 102
Harkins, William (1921). "The constitution and stability of atomic nuclei. (A contribution to the subject of inorganic evolution.)". Philos. Mag. 42 (249): 305.


Piece of equipment similar to syringe for gas collection

This piece of laboratory glassware is a gas syringe. Apparently, gas syringes are not standardized; however, typical models have a capacity of 100 ml. Some models include a GL 14 screwthread tube, a stopcock, or a three-way stopcock.


Express the maximum work from a voltaic cell

There is a simpler approach to this problem.

First, since both initial concentrations are not 1 M, you need to determine $E$ for the initial conditions, where $Q$ is the reaction quotient $Q=\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^{+}}]^2}$:

$$E=E^\circ-\dfrac{RT}{nF}\ln{Q}$$

The value of $n$ for this reaction is $n=2$, and assuming $T=298\ \mathrm{K}$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ K})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.1 \text{ M}}{(1.0 \text{ M})^2}}=3.22 \text{ V}$$

You can then go and convert this potential into free energy or maximum work or whatever else you want.

At equilbrium $E=0$ and $Q=K_c$, so we can calculate the equilibrium constant if we want to compare it to $Q$ and determine which direction is favored. We don't need to. Since $E_i>0$, this reaction procedures toward the products.

Now, as the reaction proceeds the concentrations of $\ce{Ag+}$ and $\ce{Mg^{2+}}$ will change. From the law of mass action:

$$-\Delta \ce{[Ag+]}= 2\Delta \ce{[Mg^{2+}]}$$

After some time $\Delta \ce{[Ag+]} = 0.25 \text{ M}$, so $\Delta \ce{[Mg^{2+}]}=0.125\text{ M}$ and $[\ce{Ag+}]=0.75 \text{ M}$ and $[\ce{Mg{2+}}]=0.225 \text{ M}$. The value of $Q$ has changed and so has the value of $E$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ V})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.225 \text{ M}}{(0.75 \text{ M})^2}}=3.18 \text{ V}$$

As the reaction proceeds, the value of $E$ decreases, which means the amount of work that can be done by the reaction also decreases. The maximum value of $E$ occurs at the initial state, and so the maximum work that this cell can do happens at the initial state. Since $E$ decreases as the reaction proceeds, so too does the work the cell can do at any moment.

If you want the total work, then you need to integrate with respect to $Q$, not worrying about the actual concentrations:

$$\int_{Q_i}^{K_c}E\cdot\mathrm{d}Q$$


Why is temperature constant when an ideal gas expands into a vacuum?

This process is called an expansion into a vacuum. As you described, the volume of the gas increases and the pressure drops, such that $pV$ does not change at all. Since $n$ is constant, $T = \frac{pV}{nR}$ is also constant.

I think the fact that $pV$ is constant is not very intuitive if one only uses the ideal gas law. A more formal way of explaining that $T$ is constant would use thermodynamics. If you assume that the large container does not allow any transfer of heat (i.e. it is thermally insulating), then the large container is an isolated system: $\mathrm{d}q = 0$. Since the gas is expanding into a vacuum, $p_{\text{ext}} = 0$ and $\mathrm{d}w = -p_{\mathrm{ext}}\,\mathrm{d}V = 0$. By the First Law, $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w = 0$ and temperature is therefore constant (since the temperature of an ideal gas depends fully on $U$).


What happens when chemical equilibrium is restored after being disturbed?

Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

Let's say it is a homogeneous equilibrium. You have to consider different cases:

The equilibrium was disturbed by changing one of the concentrations (removing or adding a substance). You can't get back to the same set of concentrations because there are too few or too many atoms.
The equilibrium was disturbed by adding solvent. Again, you can't get back to the same set of concentrations (all are lower than at equilibrium, and a net reaction in one or the other direction can't increase all of them at once).
You disturbed the equilibrium by changing the temperature. The equilibrium constant is at a different temperature will be different, so the old set of concentrations will not satisfy the new equilibrium constant.
You disturbed the equilibrium by intermittently changing the temperature (or - more elaborate - by attaching a power source to an electrochemical cell, and then replacing it by a wire shortcutting the cell). In this case, the concentrations change exclusively because of the chemical reaction (no "outside" disturbance), and they can return back to the original concentrations once the disturbance is removed.

In cases 1. and 2., however, the reaction quotient Q will go back to the original one (=K). In case 3., Q will be different.


Why sp3 hybridisation in NH3 but bonding with pure p orbitals in case of other hydrides of group 15

The answer to this question is actually much more complicated than most textbooks make it seem. There's a great explanation in Abright, Burdett and Whango "Orbital Interactions in Chemistry" Ch. 9, which I'll try to summarize here.

As explained in this answer: Why is the inversion barrier larger in PH3 than it is in NH3?, the lone pair orbital in a planar XH3 molecule would be pure $p$. Thus the bonding orbitals average to $sp^2$. As the bond angle decreases from 180, the loss of symmetry makes possible mixing between that lone pair orbital (originally a pure central atom $p$ orbital) and an anti-bonding orbital involving the central atom $s$ orbital. This "s-p mixing" stabilizes the lone pair orbital at the expense of the (empty) antibonding orbital. This mixing necessarily also involves the bonding orbital that involves the central atom $s$ atomic orbital.

In nitrogen, the low energy of the atomic $s$ orbital and the strong overlap interaction of that $s$ orbital with the hydrogen $s$ orbitals means that the bonding orbital made from the atomic $s$ orbital is quite low energy and the antibonding orbital is quite high energy. As a result, mixing with the $p$ lone pair orbital is less favorable, so less mixing occurs and the angle does not deviate as far from 180.

With the larger central atoms, the interaction of the $s$ orbital with the hydrogen $s$ orbitals is less, so the antibonding orbital is not as high energy. Likewise, the $s$ bonding orbital is not as low energy. So mixing with the lone pair $p$ orbital is more favorable, and the bond angle deviates farther from 180. The figure below (from Albright, figure 9.9) plots the energy of the occupied orbitals. The lowest line represents the $s$ bonding orbital and the top line the lone pair.

The key point here is that more $s$ character is mixed into the lone pair orbital as you move down from NH3.


AlMg3 is not the chemical formula of the alloy, it's the product name for a wrought alloy. Its technical sheet specifies the following composition:

Product name
AlMg3
Class of product
Al-Mg alloy for MIG/TIG welding.
Corresponding standards
DIN 1732, SG-AlMg3, AWS A5.10, ER 5754
Nominal composition (weight %)
Al: Bal. Si: 0.4 Mg: 3 Mn: 0.5

Thus, the 3 in the product name refers to 3 % (in weight) of magnesium.


What is the difference between a time of flight mass spectrometer and a spectrometer that uses a magnetic field?

Both types of mass spectrometers separate different chemical species based on the mass-to-charge ratio.

The time-of-flight mass spectrometer (TOF MS) is based on the ions being released at a given instant, such as when struck by a laser, accelerating the ions electrostatically and measuring how long it take for them to strike the cathode target. This is ideal for a lightweight interplanetary or cometary mission such as Rosetta.

Magnetic and quadrupole MS work by bending the beam magnetically  or electrostatically so as to spread out the target beam along the width of the target. This is like sending a light-beam through a prism to spread out the spectrum. An advantage of this technique is that it can resolve multiple species simultaneously, but at the expense of fine resolution.

An excellent explanation of TOF and other MS is presented by Young, Waite and Miller.


Difference between Phase, heterogenous and state of matter and solution

A deep and real understanding?!?

I don't think you'll find any definitions that you can't pick at.

So if in a large bottle you mix some water and benzene that don't fill the bottle then you'll end up with three phases. A gas phase that has air/water/benzene, a liquid phase that is mostly benzene with a tiny amount of water, and another liquid phase that is mostly water with a tiny amount of benzene.

So if I just take some random rocks (different kinds of rocks) and dump then into a bowl of water, then yes there is a solid phase. But as a chemist I wouldn't really think of the random rocks as a phase because there really isn't much of an equilibrium. Picking at this I'd have to admit that a minuscule amount of rock would dissolve in the water. Hence there would be an equilibrium of sorts between the various rocks too.

What we'd normally consider a glass is a mixture of silica and other minerals. However amorphous solids are also referred to as glasses.


Writing a chemical equation with a reactant of unspecified chain length

Tips and hints would be greatly appreciated.

My tips and hints for you (I apologize for if my tips and hints direct to me treating OP as a novice):

Your dibasic acid is $\ce{HO2C-(CH2)n-CO2H}$. Thus, $\ce{A^2-}$ in the equation $(2)$ of Karsten Theis's answer is $\ce{^-OOC-(CH2)n-COO^-}$.
Molar mass of this dibasic acid is: $2 \times (12+32+1) + 14n = 90 + 14n$
Find the concentration of FA2 solution in $\pu{mol dm-3}$ (in other words, in $\pu{M}$. Hence, you can find the the amount of $\ce{NaOH}$ (in $\pu{mol}$) in FA2 solution ($\pu{17.0 cm3}$).
Determine the amount of dibasic acid (in $m_\ce{H2A} \ \pu{ mol}$) in FA1 solution ($\pu{25.0 cm3}$) using the equation $(2)$ of Karsten Theis's answer.
Now, you can find $n$ using following equation:
$$m_\ce{H2A} \ \pu{mol} \times \pu{(90 + 14n) g mol-1} = \pu{5.00 g dm-3} \times \pu{0.025 dm-3}$$

Since you have already calculated $m_\ce{H2A}$ in step 4, you can calculate the $n$ $(\pu{1.0 dm-3} = \pu{1000 cm-3})$.
Note that $n$ is an integer so final answer should be rounded to closest integer.


How to identify the anode and cathode in a concentration cell?

Let's restate the problem: we have two hydrogen half-cells with other stuff in them. Hydrogen electrodes will have voltages related to their pH, because that is the definition of a hydrogen electrode. Then it goes haywire: does one half-cell contain BaCl2 and the other contain Be(OH)2 at 0.5 M - certainly not BaCl and BeOH. In fact, Be(OH)2 is amphoteric and soluble at only 2 ppm, so even estimating the pH of that cell would be a blind guess; you could guess the BaCl2 solution to be about 7.0, but might be a tiny bit different.

I suspect the question is not one that has actually been created and measured, but is rather a made-up question with no dependence on reality, to ask how to visualize  two hydrogen electrodes in pH 7 (the BaCl2 solution) and pH 13.69 (pH of 0.5 M OH), which could not occur because of the low solubility and amphotericity of Be(OH)2. So the whole cell could be written:

Pt(H2,1 atm)|H+(aq,10e-7 M)||(H+(aq,10e-13.69 M)|Pt(H2,1 atm)

or, just consider both half cells to be at pH 7 and there's no overall cell voltage (my preference).


Causing a metal or glass marble to float in an aqueous solution

The best current non-toxic alternative to Clerici Solution appears to be Sodium Polytungstate (SPT).  It also has a neutral (~7) pH, which is important when working with aluminum. By dilution/evaporation, density covers the range $1.1\!\!-\!\!3.1$g/cm$^2$, but above $2.5$g/cm$^2$ it can develop a harmless reversible blue tint when exposed to reduced metals. Aluminum has an oxide patina, so I am not sure if it qualifies as a reduced metal in this setting. SPT is rather expensive though.
Aluminum is $2.7$ g/cm$^2$ and Common glass is $2.52$ g/cm$^2$. So both of those materials, many gemstones and some minerals, will float in SPT.


Why concentrated sulfuric acid for dehydration and dilute for hydration?

Think of the reaction as an equilibrium equation:

$$\ce{CH3CH2OH <=>[H2SO4] CH2=CH2 + H2O}$$

Now, using Le Châtelier's Principle, we can deduce Dissenter's comments.

In concentrated sulfuric acid there is little to no water (note that we also start off with little to no ethylene). The equilibrium shifts to produce products.

In dilute sulfuric acid there is lots of water (and little to no ethanol). The equilibrium shifts to produce reactants.

As for the role of sulfuric acid, it acts as a proton source to enable the loss of leaving group in the forward direction and the electrophilic addition in the reverse direction.


In the International System of Quantities, i.e. the system of quantities that corresponds to the International System of Units (SI), as described in ISO 80000, the symbols for quantities are generally single letters from the Latin or Greek alphabet, sometimes with subscripts or other modifying signs. However, symbols for so-called characteristic numbers are written with two letters from the Latin alphabet, the initial of which is always capital. For example:

Reynolds number $Re$
$$Re=\frac{\rho vl}\eta=\frac{vl}\nu$$
Nusselt number $Nu$
$$Nu=\frac{Kl}\lambda$$
Prandtl number $Pr$
$$Pr=\frac{\eta c_p}\lambda=\frac\nu a$$

(Note that such two-letter symbols should be separated from other symbols if they occur as factors in a product.)

Quantity symbols are always written in italic type.

Various other widespread multi-letter symbols are not quantity symbols in accordance with the above-mentioned systems; they are actually just abbreviations. For example in toxicology, LD50 is used as abbreviation for “median lethal dose” (or “lethal dose, 50 %“). Such abbreviations are typeset normally, i.e. generally upright.

Also note that abbreviations for quantities shall not be used in equations.


I will try to explain the difficulty of this problem by using a simpler reaction, the reduction of nitrobenzene, which is already difficult enough. The reduction of dinitrobenzene should produce more different possibilities.
Nitrobenzene can be reduced into at least five different compounds. It depends on the choice of the reducing agent. I am afraid nobody can justify the particularly mechanisms used by a given reducing agent to get to each final result.
With ($\ce{Sn + HCl}$) or with ($\ce{Fe + HCl}$), it gives aniline, $\ce{C6H5NH2}$.
With sodium methoxide ($\ce{NaOCH3}$), it gives azoxybenzene, $\ce{C6H5N(O)NC6H5}$
With zinc, it can produce three different products:

With zinc + $\ce{NH4Cl}$ at elevated temperature, it gives
phenylhydroxylamine, $\ce{C6H5NHOH}$.

With zinc and an alcoholic solution of $\ce{NaOH}$, it gives azobenzene
$\ce{C6H5N=NC6H5}$.

With zinc and an aqueous solution of $\ce{NaOH}$ at $\pu{60 ^\circ C}$, it give hydrazobenzene $\ce{C6H5NH-NHC6H5}$.


chemical equilibrium reaction for burned gases in engine

It's simpler than it might look. As a random example:

$$\ce{3A + 2B->2C + 4D}$$

If I were to ask you how much of $\ce{D}$ would be produced after $5\  mol$ of $\ce{B}$ were consumed, then you would immediately answer $10\  mol$ "due to the stoichiometry". In more elaborate terms, for a $\Delta n_B=n_B^{after}-n_B^{before}=-5\ mol$, one has $\Delta n_D=n_D^{after}-n_D^{before}=+10\ mol$. Any infinitesimal change in quantity of any species $i$ ($\delta n_i$) is accompanied by a shift in the quantities of all other species precisely determined by the stoichiometry of the reaction (the stoichiometric coefficients, $v_i$), such that $\frac{\delta n_i}{v_i}$ is constant for all species $i$. For the above reaction, $-\frac{\delta n_A}{3}=-\frac{\delta n_B}{2}=\frac{\delta n_C}{2}=\frac{\delta n_D}{4}$. The ratio $\frac{\delta n_i}{v_i}$ is defined as $\delta n$, and you can think of it as a counter for "how many times the reaction happened" ($\delta n>0$ if left to right, $\delta n<0$ if right to left).

$\delta n$ is a general change in amount of substances, but the particular case where changes are calculated relative to the equilibrium condition ($\Delta n=n^{equilibrium}-n^{before}$), then it represents the extent of reaction.


Wood burning and carbon dioxide or monoxide?

A good reference to read is Laboratory and field investigations of particulate and carbon monoxide emissions from traditional and improved cookstoves Atmospheric Environment February 2009, Pages 1170–1181.

Unforntunately, there is a significant amount of carbon monoxide, from 29 to 118 grams of CO per kg of wood.

They find that the drier the wood, the less CO, but don't expect to completely eliminate CO.


The relative molecular mass ($M_r$) of the gas determines the rate of diffusion of the gases from a region of higher concentration to a region of lower concentration.

The smaller the $M_r$ of the gas, the greater the rate of diffusion (movement of gas molecules).

  $M_r$ of $\ce{CO2}$ = 40

$M_r$ of $\ce{H2}$ = 2 (Balloon A)
Since hydrogen gas has a smaller $M_r$ than carbon dioxide gas, more hydrogen gas diffuse out of the balloon than carbon dioxide gas into the balloon per unit time. Hence, this balloon will shrink.

$M_r$ of $\ce{N2O}$ = 44  (Balloon B)
Since nitrous oxide gas has a greater $M_r$ than carbon dioxide gas, more carbon dioxide gas diffuse into the balloon than nitrous oxide gas out of the balloon per unit time. Hence, this balloon will inflate

$M_r$ of $\ce{CO}$ = 28  (Balloon C)
This balloon will shrink (same logic as $\ce{H2}$).

$M_r$ of $\ce{CO2}$ = 40  (Balloon D)
Since the balloon and the tank both contains $\ce{CO2}$ (same $M_r$ of gases), there is no net diffusion of gases in/out of the balloon, thus the balloon will not inflate/shrink (theoretically).


Can lead(II) oxidize tin metal in aqueous solution?

Are you certain that your statement from the textbook is correct? The cation of X will oxidize element Y if element Y is more active than X, which is usually labeled (may be towards the top or the bottom of the list). For reference, see the activity series at Wikipedia, where the more active metals are above the less active metals in the list.

  How to use the activity series:

For example, what happens when iron metal is placed into an aqueous solution of copper (II) nitrate?

Let's write the potential equation:

$$\ce{2Fe(s) + 3Cu^{2+} -> 2Fe^{3+} + 3Cu(s)}$$

Since iron is more active than copper, iron will reduce copper cations (and copper cations will oxidize iron metal). The iron will dissolve.

What happens if copper metal is placed into an aqueous solution of iron (III) nitrate?

Since copper is less active than ion, copper metal will not reduce iron cations. No reaction occurs and the copper will not dissolve.

  Your question

Lead is less active than tin in all activity series that I have seen. Lead metal should reduce tin cations. This reaction should work. It is possible that the textbook/solutions manual has a typo. I find them in my books all the time.


If you need to start with ortho-xylene, then you should use fazekazs's approach. I searched through SciFinder and found that the most commonly reported synthesis of 2-indanone involves oxidation of indene using a variety of conditions.

The following Organic Synthesis prep highlights this approach, although it certainly uses some outdated methodology (distillation by aspirator vs. rotary evaporation, for example). Newer reports use a variety of oxidations, including oxone or the the Wacker process.

In this example, indene 1 is added to a mixture of formic acid and hydrogen peroxide to afford the monoformate ester of indene 1,2-diol 2. The monoformate ester was hydrolyze with sulfuric acid to give 2-indanone 3.

However, when I first saw your question, I wanted to propose a synthesis based on a Dieckmann condensation on diethyl phenylene-1,2-diacetate 4 followed by decarboxylation of the $\beta$-ketoester 5. However, I did not find this approach in the literature, and both the diacetate ester and its parent diacid are of comparaple price to 2-indanone.

Another interesting idea that I did not find in the literature is a pericyclic approach based on photolysis or thermolysis of beznocyclobutene 5 in the presence of carbon monoxide. Following electrocyclic ring opening, intermediate 6 can react with CO in 4+1 cycloaddition. An isocyanide could be used instead of CO, and hydrolysis of the imine would give 2-indanone. Alternatively, there may be a transition metal species that could mediate this transformation.


What is the hybridisation of trifluoromethyl free radical?

Hyperconjugation between the orbital that contains the lone pair and the $\ce{C-F}$ antibonding orbital contributes to the pyramidalization of $\ce{CF3^·}$ radical:

In the planar geometry the orbitals are perpendicular and no overlap occurs.

References

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th edition.; Springer: New York, 2008. ISBN 978-0-387-68346-1.


Can the colour change in litmus paper be explained by conjugated systems?

You have shown the structural change in bromothymol blue upon deprotonation.

For the anion of 7-hydroxyphenoxazone, the following resonance structures involving a p-quinone imine are conceivable:

The assumption that the $\pi$-system of the neutral species and the corresponding anion are identical is not justified.

A smaller system, in which a similar effect can be observed, is 4-nitrophenol. A solution of the phenol is colourless, whereas the solution of the phenolate is yellow ($\lambda_{\mathrm{max}}$ = 405 nm).

Again, loss of a proton leads to tautomerism with contribution of a quinoid structure.


Plasma is made up of ionized gas, so molecules have a positive electric charge and valency electrons are totally or partially separated by their nuclei.
Plasma is different from a gas since it has a high temperature and is radiation emitting; think of the Sun and other stars, which are made up of plasma and show both these properties.
Plasma hasn't got a proper volume, like gases; e.g. stars can expand or contract under the opposite effects of gravity and nuclear fusion. For example, this property is important to comprehend the formation of white dwarfs and neutron stars, process caused by high pression due to gravity.
Little trivia: there is also a fifth state of matter, whose name is Bose-Einstein Condensate (BEC).


From the image above and from what you're asking, I presume you are referring to a particle confined in a certain space region. You can find everything you need on the internet but I wanna clarify a few things about QM:

There is no concept in QM for "path of a particle": position and momentum are incompatible observables, which means that they cannot be measured simultaneously each with zero standard deviation: in fact the product of standard deviations for position and momentum must be greater than $\dfrac{\hbar}{2}$.
$$\sigma_{x}\sigma_{p}\geq\dfrac{\hbar}{2}$$
So it's meaningless in QM to talk about trajectories.
Meaning of the wave function: Let's assume that your system consists of a single particle in a one dimensional space. The wave function $\Psi\left(x,t\right)$
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared $\left|\Psi\left(x,t\right)\right|^{2}$
, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx
, gives you the probability for finding the particle between $x$
and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$
Of course you can integrate the previous eq. to find the probability over a certain region of space $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$
Stationary state: If the potential is time independent (in your case it's zero), the time dependence of the wave function can be factored out like
$$\Psi\left(x,t\right)=\mathrm{e}^{-i\omega t}\psi\left(x\right)$$
In this case it's easy to prove that the probability density is time independent.
The quantization: For this simple problem, you must solve the eigenvalue equation for the energy
$$-\dfrac{\hbar^{2}}{2m}\dfrac{d\psi\left(x\right)}{dx}=E\psi\left(x\right)$$
By solving it and by imposing that the particle cannot exists outside a certain region (outside the region $\left[0,3\right]$ in your picture), you find a natural quantization condition for the energy: the energy can assume only certain discrete values, which are called eigenvalues, $E_{n}$
. Of course, this means that for every eigenvalue there is an associated wave function, which is called eigenfunction $\psi_{n}\left(x\right)$
. The general solution to the eigenvalue equation is in fact a linear combination of eigenfunctions $$\psi\left(x\right)=\sum_{n}c_{n}\psi_{n}\left(x\right)$$
In the above picture, you can see the first three $\left|\psi_{n}\left(x\right)\right|^{2}$
. They of course represent probability density functions: note that where $\left|\psi_{n}\left(x\right)\right|^{2}$
is zero the probability of finding the particle is of course zero (for example in 0
and 3
, like we requested before).
So where is the electron? Like said before, in QM locating a particle it's a non-sense. All you can do is to compute the average position and momentum, which for a system in the state $\psi\left(x\right)$ are given by
$$\left\langle x\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)dx$$
$$\left\langle p\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]dx$$


Relationship between the first and the second quantum number

Does the secondary quantum number tell how many subshells a specific
  principal quantum number shell has?

No, it does not. It is the principal quantum number $n$ which does it (as you actually mentioned yourself later in the question). The second quantum number $l$ describes the shape of the orbitals in a particular subshell. Say, $l=0$ for $\mathrm{s}$-subshell of any shell and all $\mathrm{s}$-orbitals have a spherical shape.

  If the principal quantum number is $n$, there are $(n−1)$ subshells.

Why $(n−1)$? Think about the most trivial example: how many subshells the first shell ($n=1$) have? One, not zero, so $(n−1)$ is obviously wrong formula. The situation is in fact simpler: $n$-th shell has $n$ subshells.


If the human body consists of 60% water why can't we put out fire with our body?

We need to consider something. Are the people being truly burnt at the stake or are they being killed by the effects of fire? If the fire is very prolonged then their bodies will dry out and then finally be calcined to ashes.

It is an interesting (but a rather morbid) fact that cremation furnaces typically run at about 800 degrees Celsius for many hours, yet they tend to be unable to burn bodies to dust. Typically after cremation, some large fragments remain. These have to be crushed.

There are tales of spontaneous human combustion, these have been circulating for many years. I recall that the Victorians thought it was associated with extreme alcohol abuse. They thought a build-up of alcohol in the body would lead to the body being able to catch fire. A modern-day explanation is that humans wear clothing, the clothing can act as a wick.

A person dies as a result of something, then their clothing comes into contact with an ignition source. The clothing starts to burn, this heats up fat in their body. The fat melts and runs into the clothing. The clothing then acts like a candle wick and then the burning fat makes more heat which melts more fat thus turning the person into a candle.

I think that this type of human candle effect is very rare, it is likely to require a combination of several things at once. Many people who are killed in serious fires (who suffer burns as well as being exposed to hot and toxic smoke) leave behind bodies which have not been burnt to ashes.


Since the room temperature butane didn't instantly cool when it hit the container, it was still above the boiling point.

If you had a large freezer that was roughly at equilibrium at a temperature below $-1^\circ C$, and you put both the pressurized container and the empty container in there and let them reach equilibrium, then when you opened the pressurized container it would not boil and you could conceivably pour it into the other container.

However, butane has a large vapor pressure (the enthalpy of vaporization is relatively small). This means that even at low temperatures, it will evaporate very quickly. This is especially true because the atmosphere does not normally have any butane in it - so even a small partial pressure combined with convective transport in the air will cause it to evaporate even faster. If you imagine pouring some rubbing alcohol out on a table and then putting a fan next to it, it is the same idea. Butane molecules randomly have enough energy to escape the liquid, and air currents carry them away. Since there is no butane vapor nearby to establish an equilibrium with the liquid, more butane evaporates to take its place.

The net result of this is that even if you did the experiment in a freezer and made sure everything was below $-1^\circ C$ (which includes you wearing well-insulated gloves and not breathing on it), it might still evaporate so fast that it would look as if it were boiling.

From wikipedia, here is a plot of vapor pressure vs T for butane:

The boiling point is the point where the vapor pressure line exceeds atmospheric pressure. Atmospheric pressure is usually around 760 mmHg. This is a log plot for the P axis, which means one large grid step is ten times larger than the previous one. From this graph, you can see that at $-30^\circ C$, the vapor pressure is still ~200 mmHg. This means that roughly 26% of the air at equilibrium would be butane. Unless you have a very tiny freezer, that is a lot. This means that you might have to wear a gas mask and get a 55 gallon drum of butane before you could really do any "pouring" in the traditional sense.

At $-100^\circ C$ you are down to 1 mmHg vapor pressure. At this point I would say that even a small amount (like what is in a lighter) would stay liquid long enough that you could pour it.


Oxygen or carbon with a positive charge? (comparison of canonical structures)

Which of the contributing structures of the resonance below is more stable?

Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities.

What the question is asking is "Which contributor is more important in describing the structure and behavior of the hybrid?" Contributor B probably approximates the structure better (i.e. the positive charge is more on the oxygen than on the carbon), but structure A represents the reactivity better (i.e. nucleophiles want to attack the carbon and not the oxygen).


What is the logic behind the division into groups of anions and cations in qualitative inorganic analysis?

There is a somewhat established classic procedure of qualitative and quantitative chemical analysis (also known as "systematic analysis") based on treatment of the solution by specific set of reagents in specific order. Cations are divided into groups based on the stage they are detected in the procedure. There is NO other logic behind this division, nor it has any meaningful chemical significance.

  Also, how does one get at the various tests, like, did people use to test every possible reagent or combination of reagents and see if something happens or is it deducible?

Somewhat deducible. Solubility product constant can be used to check if the solution is prone to precipitation of particular salt barring cases of redox reaction. For redox reactions standard electrode potentials and Nernst equation might be used for a quick check. Still, knowledge of these particular constants is mostly (if not completely) empirical.


How can nitrogen be both inert and bond with everything in periodic table?

Nitrogen as an atom can be bound to any other atom of the periodic table. But nitrogen as a molecule is something else. It is $\ce{N2}$ and the two atoms are strongly bound in the molecule. It is even extremely difficult to separate the two atoms from the molecule $\ce{N2}$, because $\ce{N2}$ contains a triple bond : there are three covalences between the two atoms. That is why $\ce{N2}$ is nearly inert. A huge amount of energy is needed to break these three bonds. And it is usually impossible to do it in a lab at ordinary temperatures. Some plants (beans, peas) are able to do it and to metabolize $\ce{N2}$ by a somewhat mysterious process.


The most obvious approach for me would be to find a partition of the set of points into two subsets and two straight lines, one for each subset, such that the sum of the squared residuals over both partitions is minimized.


Electron Ionization and the Franck-Condon Diagram: vibrationally excited and vibrationally ground states

Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?

It's not so much about whether the bond length in $\ce{M^.+}$ is larger or smaller than that in $\ce{M}$; it's more about whether it's different. If you think pictorially about the vibration of a diatomic molecule or a spring, it makes sense that as you increase the energy (i.e. go to higher vibrational states), the amplitude of the vibration increases. The only way for a molecule to access bond lengths that are substantially different from its equilibrium bond length is therefore to be in a vibrational excited state.

Ionisation of $\ce{M}$ effectively always takes place from the vibrational ground state, so the initial bond length will always be very close to the equilibrium bond length of $\ce{M}$. If ionisation is almost instantaneous, such that the bond length cannot change during the ionisation, then $\ce{M^.+}$ is going to be generated with a bond length that is far away from its own equilibrium. This corresponds to an excited vibrational state. Whether it's larger or smaller than equilibrium doesn't matter; it just matters that it's different.

Of course, to say that this is the case presupposes that the equilibrium bond lengths of $\ce{M}$ and $\ce{M^.+}$ are substantially different. But this appears to be the assumption that the author makes. For more insight into this, @atbm's comments are spot on: the change in the equilibrium bond length will depend on the bonding character of the electron being removed. If a bonding electron is removed, then the bonds holding atoms together are weakened, and consequently $\ce{M^.+}$ will have a larger equilibrium bond length than $\ce{M}$; similar considerations hold for nonbonding and antibonding electrons.

  How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)

I suspect that the FC principle is no longer relevant to the adiabatic ionisation process. Adiabatic processes are associated with "slow" transitions: see e.g. https://en.wikipedia.org/wiki/Adiabatic_theorem, whereas the FC principle is to do with "instantaneous" processes where there is no possibility of the bond length changing during the ionisation.

This is not my area of expertise, but I will go out on a limb and say that the adiabatic ionisation process is largely hypothetical (at least in the present context). Wikipedia writes that

  However, due to experimental limitations, the adiabatic ionization energy is often difficult to determine, whereas the vertical detachment energy is easily identifiable and measurable.

I am not sure what "error" the author is talking about, but my interpretation is that the adiabatic ionisation energy is the "true" or "actual" ionisation energy of the molecule; however, because physical ionisation does not take place in an adiabatic fashion, the measured ionisation energy differs from the "true" ionisation energy by ca. 0.1–0.7 eV.

So I would not interpret this as being two different physical processes that actually occur in the mass spectrometer; rather, it seems to be more of a thought experiment to me. Of course, I am open to being corrected on this.


I am not a physics major, this post will be rather free of equations and spherical chickens. This is a chemist's view of the situation ;)

You need more than an unpaired electron to create color in a substance. You need two places for it to be, with a given energy difference in the visible color range. Looking at the molecular orbital diagram of NO we see that no such organisation is possible (within reasons). The only choice is to go from one $\pi^*$ to the other, but those are at the same energy level. There is no way for this diatomic molecule to deform in any meaningful way so as to create a difference in energy levels between them. Metal complexes, especially iron- and cobalt-based ones, can do this, but not with the $\pi^*$ in any case. There is only $\sigma^*_2$ left, but the distance from $\pi^*$ to $\sigma^*_2$ is in huge, larger than visible spectrum (an event that knocks an electron up there is likely to unravel the entire molecule).

All in all, there is no possibility for the electron to move anywhere that requires the emission of a photon in visible range. It cannot produce the color you seek.


Why does the transverse magnetization in Fourier Transform NMR consist of Ix+iIy ($\hat{I}_{\!x}+\mathrm{i}\hat{I}_{\!y}$)?

As Mad Scientist said in the comments, measurement of the transverse signal in NMR occurs in both the $x$- and $y$-directions. This is known as quadrature detection. In practice this is not achieved with two detector coils, but rather with a "trick" which is described more fully in Keeler's Understanding NMR Spectroscopy, 1st ed., section 5.6.

Instead of having two separate quantities $M_x$ and $M_y$, it's easier to lump them into one complex quantity $M$:

$$M = M_x + \mathrm{i}M_y$$

For example, if we only have one spin with equilibrium magnetisation $M_0$ pointing along the $z$-axis, then following a $90^\circ (y)$ pulse (which rotates the magnetisation vector to $x$),

$$\begin{align}
M_x &= M_0 & M_y = 0
\end{align}$$

and after a period of free precession $t$

$$\begin{align}
M_x &= M_0\cos(\Omega t)\mathrm{e}^{-t/T_2} & M_y = M_0\sin(\Omega t)\mathrm{e}^{-t/T_2}
\end{align}$$

and $M = M_x + \mathrm{i}M_y = M_0\mathrm{e}^{\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2}$, where $T_2$ is the spin-spin relaxation time. Fourier transformation of this complex signal gives a spectrum $S(\omega) = A(\omega) - \mathrm{i}D(\omega)$. There is a constant of proportionality which doesn't matter and has been dropped. The real and complex components are known as the absorption and dispersion Lorentzians respectively:

$$\begin{align}
A(\omega) &= \frac{1/T_2}{(1/T_2)^2 + (\omega - \Omega)^2} & D(\omega) = \frac{\omega - \Omega}{(1/T_2)^2 + (\omega - \Omega)^2}
\end{align}$$

(Diagram from Keeler's Understanding NMR Spectroscopy, 2nd ed. (p 84).) The absorption component is usually what is shown in the actual NMR spectrum. There are some complexities because the spectrum has to be phase shifted, but the idea holds.

For a single spin it is very easy to use product operators instead of the vector model. After that, to obtain the expressions in your question, you just generalise to multiple spins. I might come back to this some time later, but for now, this will suffice.

The benefit of quadrature recording (which seems to have been cut from Keeler's 2nd ed., but remains in the 1st ed.) is that it allows us to find out the sign of the offset $\Omega$. Let's say that instead of having quadrature recording we only measure along the $x$-axis. Our signal is

$$\begin{align}
M &= M_x \\
&= M_0\cos(\Omega t)\mathrm{e}^{-t/T_2} \\
&= \frac{1}{2}(M_0\mathrm{e}^{\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2} + M_0\mathrm{e}^{-\mathrm{i}\Omega t}\mathrm{e}^{-t/T_2})
\end{align}$$

It's relatively straightforward to show that for a function $M = M(t)$, if $\mathcal{F}[M(t)] = f(\omega)$:

$$\mathcal{F}[M^*(t)] = [f(-\omega)]^*$$

and using this, we can show that if we take the Fourier transform of the one-dimensional magnetisation, we get

$$\begin{align}
S(\omega) = \frac{1}{2}[A(\omega) - \mathrm{i}D(\omega) + A(-\omega) + \mathrm{i}D(-\omega)]
\end{align}$$

and if you take the full expression for this and replace $\Omega$ by $-\Omega$, you will find that $S(\omega)$ is the same regardless of the sign of the offset $\Omega$. This means that our spin of interest could have either a positive or a negative offset, relative to the spectrometer reference frequency $\omega_\mathrm{rf}$, but we would not be able to determine it.

In theory, if you could control the phase of the signal you are detecting, single detection would allow you to extract the sign of the offset. For example, in our example above, if you measure $M_y$ instead of $M_x$, you'll find that there is a difference when you replace $\Omega$ by $-\Omega$. Here's a schematic from Gunther's NMR Spectroscopy, 3rd ed. (p 265):

Here, the author Fourier transformed $M_x$ and $M_y$ separately before adding them up. Since the Fourier transform is linear, this has the same effect as adding $M_x$ and $M_y$ and Fourier transforming the sum. In any case, it shows that if the magnetisation vector starts off along the $y$-axis, you could, in theory, measure only the $x$-magnetisation to determine whether the offset is negative (case a) or positive (case b).

Such a spectrum would be very difficult to interpret, and would probably be impossible to analyse if positive and negative peaks overlapped. On top of that, it would also require you to know the exact phase difference between the initial signal and your single detection coil. I believe this is (due to several reasons) not experimentally possible, so unfortunately, there's no way to extract this information using single detection. (This is also part of the reason why NMR spectra have to be phase corrected – for the average NMR user the computer does this automatically.)

Now, we've established that you need quadrature detection to determine the sign of the offset. However, in the final NMR spectrum, we don't really look at offsets; we look at resonance frequencies. (To be precise, chemical shifts, but those are linearly related to resonance frequencies.) So, if you really wanted to, you could use single detection while placing the reference frequency at the far end of the spectrum. For example, if you want to measure peaks from $400.000$ to $400.001~\mathrm{MHz}$, you can place the reference frequency at $400.002~\mathrm{MHz}$. Then, by virtue of choosing an overly high value for $\omega_\mathrm{rf}$, you know that your offset $\Omega$ has to be negative.

However, a rf pulse centred at $400.002~\mathrm{MHz}$ would be less effective at exciting the spins (as it is further away from being on-resonance), and essentially you are throwing away half of the rf pulse (any components of the rf pulse with frequency greater than $400.002~\mathrm{MHz}$ are wasted). So, it's preferable to put $\omega_\mathrm{rf}$ right in the middle of the spectrum; but as said before, that would necessitate quadrature detection so that you can obtain information about the sign of the offset.


Ring expansion and carbocation formation

To understand which rearrangement pathway is preferred, we must consider the relative stabilities of the intermediates. The rearrangement that puts the carbocation one carbon away from the the oxygen does not have any stabilizing resonance structures:

$\hspace{5.4cm}$

The rearrangement that places the carbocation adjacent to the oxygen delocalizes the positive charge to two atoms, and is much more stable:

$\hspace{3.7cm}$


Vapor pressure and boiling point for solutions in open containers

Boiling is not about the upper surface at all. What happens there is but of little interest to us. Boiling means that bubbles are forming at the bottom and probably within the bulk of the liquid. Guess what is inside these bubbles? It is pure water vapor. That's where the equilibrium between vapor and water actually comes into play.

As for the second question, the bubbles must withstand the external pressure, otherwise they would not be able to form in the first place (or if they will, they will be instantly squeezed into nothing). That's why they actually appear only when vapor pressure becomes equal to atmospheric pressure.


Why does the reaction between potassium permanganate and oxalic acid result in a yellow/brown solution?

The reaction between $\ce{MnO4-}$ and $\ce{C2O4^2-}$ is performed under acidic conditions.

In acidic conditions only, $\ce{MnO4-}$ gets reduced to colourless $\ce{Mn ^{2+}}$ as, $$\ce{MnO4- + 8H+  +5e- -> Mn^2+ + 4H2O}$$

But if you don't maintain proper acidic condition of the medium, it continuously absorbs $\ce{H+}$. Eventually, due to lack of protons, it gets reduced to brown coloured $\ce{MnO2}$ under neutral conditions, where it needs less no of protons to get reduced than acidic conditions, as $$\ce{MnO4- + 4H+ + 3e--> MnO2 + 2H2O}$$ That's why you see the brown colour which is due to $\ce{MnO_2}$.


What is the relationship between entropy and heat capacity?

The entropy depends on the mass. The translational part is given by the Sakur-Tetrode eqn., the rotational part depends on the moment of inertia, and the vibrational part on the vibrational frequency. In each case the partition function $z$ has to be calculated then the entropy via $\displaystyle S=R\ln(z)+RT\left(\frac{\partial z}{\partial T} \right)_V -R\ln(N)+R$ (for a perfect gas). The physical reason is that for the heavier molecule there are more energy levels up to a given energy that can be occupied and so more ways of populating these levels and so more entropy.

The heat capacity is the slope of internal energy $U$ with temperature (at constant volume) $\displaystyle C_V=\left({\partial U}{\partial T}\right)_V$ and the internal energy is related to the partition function as $\displaystyle \frac{U}{T}=RT\left(\frac{\partial z}{\partial T}\right)_V$ and so also to the entropy. The classical thermodynamic relationship is $\displaystyle S=S_0+\int_0^T \frac{C_p}{T}dT$ where $C_p$ is a function of temperature.


What does the IUPAC definition of the emf of a galvanic cell mean?

We think of electrochemical processes happening immediately in one step, but that's not really true. In a galvanic cell, you have to have the dissolved ions move from the bulk to the surface, adsorb to the surface, and possibly undergo an activation step before they can reduce or oxidize. The last step is for the electron to actually move from the reduced to the oxidized species. This specific step is the "local charge transfer" reaction.

You are right that when the cell reaches equilibrium, the $E_{cell}$ would be zero. However, here we're interested in  the equilibrium of each local process separately: eg. diffusion of ions from the bulk electrolyte to the surface will reach an equilibrium.

When you first submerge the electrode into the electrolyte, all of these steps are able to reach an equilibrium, and things like concentration gradients between the electrolyte bulk and electrolyte surface can effect the potential experienced by the electrode. In fact, some charge transfer even occurs initially, but because the half-cells are not connected, electrostatic forces build up and the concentrations are such that the chemical potential is quickly opposed and a steady-state is achieved.