Why are group 1 elements so low in density?

Because it is energetically easier for an atom (other than hydrogen) with only one outer electron to lose that electron, than to gain seven more, in order to have an inert gas configuration, that electron has a high probability of wandering off. In bulk samples, that causes metallic conduction. Having that lone, loosely-attached electron increases the effective atomic radius.

Hydrogen is an interesting special case. At the ionization potential is more than twice that of lithium, for example, and $\ce{H2}$ has a covalent, rather than metallic, bond. The gas is diatomic, with very little attraction between molecules, so its density is low. However, at sufficient pressure, it is believed that hydrogen would behave as a metal.


CO2 conversion to a harmless gas like oxygen

Submarines currently use NaOH/Ca(OH)2 to absorb CO2. They generate O2 separately either from gas tanks or by electrolysis of water. Peroxides seem attractive at first, but they are a fire hazard and you don't want fire on a submarine. British subs are using Zeolite. The role of zeolite isn't to deal with CO2, but to remove the smell of volitile organic compounds (of dirty socks, so to say). Air is first dried and purified.

What exactly are you trying to achieve?


Calculation of dispersion energy in Gaussian 09

The value of the empirical dispersion correction is printed when the print level is turned up using #p in the route line. For any method that contains an empirical dispersion correction, search for R6Disp or Dispersion energy, where other parts of the line will vary depending on the type of correction (-D2, -D3, or -D3(BJ)):

R6Disp:  Grimme-D2 Dispersion energy=       -0.0174233223 Hartrees.

This line is not printed with # or #n at the start of the route section.

For wavefunction methods that include dispersion via electron correlation, the dispersion contribution is non-separable. For a method like B2PLYP-D3, the printed "Dispersion energy" is the empirical correction, not the MP2 contribution!


Equilibrium constant and thermodynamic relationship

Well first think of thermodynamics as determining what a system wants to do, and kinetics as what the system is allowed to do.  The rate  constant gives how fast something will react (kinetic rate).  The thermodynamic equilibrium constant gives how much of something will react.  Kinetics are dictated by the movements of atoms, Thermodynamics is dictated by the changes in energy and entropy.


How can a neutral atom attract electrons when it's supposed to have zero charge?

An atom is only neutral when viewed as a single object from large enough distance. But as an electron comes closer to the atom, it "notices" the electron cloud first. This cloud also "notices" the electron and deforms—the atom polarizes—so as to keep the atomic electrons farther on average from the extra electron, since like charges repel. But this deformation leads to less shielding of the nucleus than in isolated atom, so the extra electron begins to feel a non-negligible attracting electric field.


What is the purpose of halogen in oxidation of thiols?

Try drawing a mechanism without halogens present, starting with your base DE-protonating one thiol. See how far you get. The problem is that you'd have to somehow kick out H- (hydride), which is a terrible leaving group (enzymes can accomplish this, but standard lab reactions cannot do this easily). The fact is, combing two thiols to give the disulfide is an oxidation reaction, and there is an electron pair being lost from the reactants. It has to go somewhere, i.e. to reduce a diatomic halogen (playing the role of an oxidizing agent).

The thiol-disulfide redox equilibrium you are asking about has great biological significance; see disulfide bridges (for protein structure) and glutathione (an antioxidant).


Short answer: use crushed watermelon seeds (urease)
Combine urea in solution with urease, get some ammonia with carbamate intermediary, which rapidly decomposes to ammonia and carbon dioxide.
$$\ce{(NH2)2CO + H2O ->[urease] NH3 + H2NCOOH -> 2NH3(gas) + CO2(gas)}$$
Of course the ammonia gas dissolves in the water and increases the pH.
Sources:

https://www.hemmaodlat.se/research/citrullus%20lanatus%20seeds%20as%20a%20urine%20catalyst%20for%20anthroponic%20use.pdf
https://en.wikipedia.org/wiki/Urease
https://en.wikipedia.org/wiki/Ammonia_volatilization_from_urea


How is entropy change defined for a chemical reaction?

The simplest way is to calculate  $\Delta$H and $\Delta$G of the reaction, then calculate the difference $\Delta$H - $\Delta$G. And you divide this difference by the temperature T. The result is $\Delta$S of the reaction.


Equations for UV degradation of condensation polymers

Having spoken to my Chemistry teacher, the idea is that the UV causes the molecules to split; after the fission, the molecules may rejoin or react with 'other radicals' to break the polymer down, hence degrading it.

E.g. for a polyester:

$R_1COOR_2 \xrightarrow{hv} R_1CO\boldsymbol\cdot + R_2O\boldsymbol\cdot$


Is "propanol" a correct name in IUPAC nomenclature?

The last IUPAC nomenclature recommendations were published in 1993. The current draft documents (release date uncertain) are far more lenient in their acceptance of semi-systematic naming of molecules. Please make sure you understand the differences between the Preferred IUPAC Name (PIN), the retained name, and trivial name.

Propanol is not be the preferred IUPAC name, but significantly, neither is it listed in the tables of Trivial and semisystematic names retained for naming organic compounds. It is, however, a widely accepted name for this chemical, and you are far more likely to encounter propanol in industry than propan-1-ol. Nobody uses propan-1-ol in the real world, even if it is the preferred IUPAC systematic name.


Difference between water solubilities of phenol and cyclohexanol?

In addition to the acidity difference described in the comments, there is a more subtle effect involving conjugation in the undissociated phenol.  When the phenyl group withdraws some electron density from the π-donating hydroxyl group, it gains some negative charge and so is better able to attract the hydrogen ends of the water dipoles.  Ultimately this is a manifestation of the polarizability of the delocalized π electrons.


Generally speaking, the typical auxochromic groups possess (at least) one pair of non-bonded $n$-electrons and -- if taken alone -- do not absorb in the UV.  If attached, for example to a benzene ring, their pair(s) of $n$-electron interact(s) with the $\pi$-electrons, and the delocalization of the $\pi$-system is extended.

Now the catch:

Bathchromic auxochromic groups (like the mentioned $\ce{-OH}$ and $\ce{-NH2}$) increase the electron density in the $\pi$-system. This is typically caused by a strong $+M$-effect. As they lower the excited $\pi^*$-level, a longer wavelength is needed for the excitation along $\pi{} \rightarrow {} \pi^*$, hence the redshift.
In contrast to the former, substitution equally may alter the $n \rightarrow{} \pi^*$ transition, inducing the opposite effect, moving the absorption towards higher absorption (blue) and a hypsochromic shift.  One example is the to compare the absorption of indigo yielding "jeans blue" with 6,6'-dibromoindigo, main constitutent of tyrian purple. A second, that heteroatoms next to a $\ce{C=O}$ carbonyl group induce a hypsochromic shift like in the case of acetaldehyde example here in comparison to acetic acid here

Note the use of the terms of "bathochromic shift" and "hypsochromic shift" along solvatochromism.


The IUPAC Blue Book defines rules where locants can be ommitted in IUPAC preferred names.

Unlike terminal locants in names for aldehydes that are omitted, there is no rule for homogeneous chains with three or more identical atoms that would allow for the ommission of the locant $2$ in propan-2-one.

This leads to the result that propanal is a IUPAC preferred name while propanone isn't.


$\ce{CO2}$, being non-polar (due to the 180 degree bond angles),  will probably be more soluble in solvents that are non-polar as well, although the quadrupole moment might cause higher solubility in polar solvents as well (thanks to @georg for this addition). Also temperature will play a significant role, the higher the temperature the lower the solubility.

These qualitative remarks however only take you so far. If you really want to know the solubility in a specific liquid you will have to look it up or, if that is not possible, test it yourself. In this paper they test the solubility of $\ce{CO2}$ in different types of crude oil and make predictions based on the structure of the oil.

I cannot find anything specific for corn oil, although the reverse process of dissolving palm oil in supercritical $\ce{CO2}$ might set you on the right track to find what you want to know.

Just a fun addition: don't ever try to carbonate milk as these guys do - it will get messy! (it is in Dutch, but you can still watch what happens). Click here an English video with similar footage


Benefits of a Buchner funnel w/ a fritted disc instead of Quantitative Filter Paper Circles

Buchner + filter paper cons:

The filtering surface is not uniform. It filters only where the holes are, so most of the surface is not working.
Some solid can slip between the paper and the filter.  Then the filtrate must be filtered again.
You have to cut the paper.

Fritted glass filter pros:

Uniform filtering surface.
Pore sizes from number 1 (coarse) to number 4 (fine). In organic chemistry, number 3 is the standard.
Can be cleaned easily when the solubility of the product is known.
Can be sterilized at high temperature.

I use them almost for everything except for filtering activated carbon and very dirty unknown samples. If the solid does not dissolve in solvents, we try with concentrated caustic, concentrated sulfuric acid or strong oxidants; in that order.

It is better to buy the ones without the vacuum connection.
They are cheaper and have less points where they can break.


Nitrogen inversion barriers of trimethylamine and N,N-diisopropylmethylamine

So I thought about this for a bit and I think I have an answer. Be warned, I might be wrong.

The inversion process involves a conversion from $\mathrm{sp^2}$ to $\mathrm{sp^3}$. So $\mathrm{sp^2}$ hybridization increases angle between these bulky groups, which would be in fact favorable. Hence the low activation energy.


What happens when Hinsberg derivative is added to acidic solution?

" compound produces a solid" indicates that a reaction has occured.(hence no 3° amine group(according to this)). Further this is insoluble in base so there is a 2° amine group.

"This solid dissolves in acid" directly implies that there is a 3° amine group.

Hence there is both 2° amine group as well as 3° amine group in the compound.

Hence option b is the correct answer.

References:
1. Solomon and Fryle Hinsberg Test


Confusion regarding a stoichiometric formula

You misread the formula. For compound $\ce{M_{x}N_{y}}$

$$y\cdot \ce{mol M} = x\cdot \ce{mol N}$$

This is because the formula essentially tells you that the ratio between moles of $\ce{M}$ and $\ce{N}$ is:

$$\frac{\ce{mol M}}{\ce{mol N}} = \frac{x}{y}$$

This rearranges to the equation above.


Retention of surface texture after combustion

The curling up has a simple reason: the burning was not uniform. Paper contains water (really!), and heating it makes it lose the water.

Now, paper is a bad conductor of heat. When you burn it, inevitably the flame will be mainly on one side(the side that's pointing up). The flame will transfer heat to nearby regions of paper, and evaporate some (or possibly all) of the water. The loss of water leads to a contraction of the cellulose fibers in the paper (I'm not too sure of this, comments appreciated). But, this only happens on one side of the paper. Even though paper is razor-thin, the fact that it still has two "faces" is important here. Now, due to the contraction, the paper curls in that direction (similar to how a bimetallic strip works). That's it.

The reason for retaining texture, is something I'm not too sure about. The burning reaction throws up a lot of soot, but a lot of the products stay right where they are. This results in the charred/burnt paper occupying the previous position of the cellulose. While charred paper isn't a polymer and thus isn't too strong, it can "mimic" the initial structure just by occupying the position of the cellulose. But, unlike cellulose, if we try bending it, it will crumble since there are no long polymer "chains"/intermolecular forces holding it together. It's like the difference between a house of cards and a similar "house", made of a single piece of rubber. The structure is the same, but the former case is only staying together because it had the "luck" to exist in a stable configuration. Disturbing it will destroy it.


7-Methylbicyclo[2.2.1]hex-2-ene exists as two diastereomers, anti-1 and syn-1, which differ by the relationship of the methyl group and the double bond. Both compounds are meso with the bridgehead carbons stereogenic and chirotopic (R/S) while C7 is stereogenic but achirotopic (r/s). The plane of symmetry  in each one does not produce an enantiomer of itself. Stereoisomers have the same molecular formula and the same atom connectivity. You probably know that the stereoisomers (R,R)- and (S,S)-tartaric acid are enantiomers and bear a diastereomeric relationship with (R,S)-tartaric acid, which is a meso-compound. A pair of meso-compounds with the same molecular formula and the same atom connectivity may also be diastereomers. Just like structures 1, the central carbons (C3) of ribitol and xylitol are stereogenic and achirotopic. The properties of these two pentaols are different as are the properties of anti-1 and syn-1.


Which affects acidity more, in relation to induction? (Distance vs. quantity)

Inductive effects decrease strongly with increasing distance. Thus, the further away an inductively acting group is, the lower its influence will be. For practical intents and purposes, the otherwise very comprehensive Brodwell $\mathrm{p}K_\mathrm{a}$ database does not even list any β-halo compounds; only the difference between α-hydrogen, carbon and halogen is typically given.

Thus, we can expect compound IV to have practically the same acidity as compound II, and that compound V’s acidity is comparable to the unhalogenated pentan-4-thiol. Considering that all chlorines are very far away from the acidic proton, the effects would be smaller still.


How much water evaporates in a vacuum before freezing?

You will need to know:

the heat capacity of water $C_p$
the latent heat of fusion of ice $L_f$
the latent heat of vaporization of water $L_v$

First determine how much heat must be lost to change the temperature from 20 to 0 using the heat capacity.

Next determine how much heat must be lost to change from 0 degree water to 0 degree ice using the latent heat of fusion.

Finally, add the two heats from above together to get a total heat loss, and use the heat of vaporization of water to calculate the amount of water evaporated that corresponds to this total heat loss.

Note: The above involves an approximation that the amount of water evaporated is relatively small.  In reality, once an infinitesimal portion of water evaporates, the amount remaining that needs to cool will be infinitesimally less than one kg.  If you are expected to know and use intergral calculus for your class, you may need to take into consideration how the mass is varying thoughout the process.

To calculate more accurately, break the problem is into two parts.

First, a mass m1 evaporates, to cool water from 20 degres to 0 degrees.

Let $m$ be the varible liquild mass during the process and temperature be $t$

$mC_pdt=L_vdm$

$C_p\int_{20}^{0}dt = L_v \int_{1}^{1-m1}(dm/m)$

The remaining mass is 1kg-m1

Second, a mass m2 evaporates, to freeze the remaining water.

$L_v m2 = L_f (1kg - m1 - m2)$


In addition to the equation for the preservation of the amount of substance you've written, there is one more equation you can use for the system, namely the one reflecting preservation of volumes: $x + y = 3$:

$$
\left\{
\begin{align}
x + y &= 3 \\
2.5x + 0.4y &= 3
\end{align}
\right.
$$

$$x + y = 2.5x + 0.4y$$

$$1.5x = 0.6y$$

$$\frac{x}{y} = \frac{6\cdot 2}{10\cdot 3} = \frac{2}{5}$$

E.g. the answer is $x : y = 1 : 2.5$.


Why is the maximal percent concentration of acids similar to their molar mass

Because of its solubility.
More concentrated solutions of, for example, nitric acid or hydrochloric acid, will result in fuming solutions where acid concentration evaporate out until the solution reaches an equilibrium.
Do not get confused. You can actually dissolve more than 370g of hydrochloirc acid in 630 mL of water.
In effect:

Solubility in water (Wikipedia)

823 g/L (0 °C)

720 g/L (20 °C)

561 g/L (60 °C)

But if we talk about a commercial solution, which should stay at the same concentration in time, you need to find a compromise between most quantity of acid possible in solution and less lose of acid due to vaporization.

Remember that pure hydrogen chloride is a gas at room temperature. If you continue increasing its concentration, you end up with man equilibrium shifted towards gas phase.

Also, as you can see, if you increase the temperature, solubility decreases, so if you heat a 50% aqueous solution of hydrochloric acid, you would end up not knowing what quantity of the reactant are you actually using.

Nitric acid is less volatile, and sulphuric acid is much less volatile, so you can have the latter almost in a pure form (98% is not an aqueous solution anymore! is just sulphuric acid with some water and other trace compounds)


Diamond (carbon) does not melt at 1 atm. It sublimes to vapor.

Using carbon's theoretical phase diagram below (from Wikimedia), "liquid diamond" could be achieved at about 10 GPa (99 thousand atmospheres) and 5000 K (4700 °C).

Edit:

In fact, heating up diamonds at 1 atm turns them into graphite first. Only further heating would cause sublimation, as expected by observing the phase diagram. Here's a video:

http://www.youtube.com/watch?v=7L7BV3IBfFA


The sign of the wave function does not have any physical meaning since for every solution $\Psi$ of the Schrödinger equation $-\Psi$ is also a solution. In fact there are infinitely many equivalent solutions that only differ in their overall phase.

So the (meaningless) answer to your question is yes.

It should be noted that $\Psi \Psi^*$, that can be physically interpreted as a probability distribution for finding a particle at a certain point, is always a non-negative real number.


It's not certain properties, but a very careful observation if and with what products glucose reacts to other chemicals. Took many years. And even then Fischer did not know the absolute configuration, because at his time nobody knew the absolute configuration of anything.

Only when the first absolute configurations were known from X-ray diffraction experiments, one could start comparing.

Today we can calculate the absolute optical activity via quantum mechanics, but Fischer solved the Glucose structure in 1891, that's a quarter century before quantum mechanics.


Why is Borax used as a primary standard?

Yes borax is a hydrate, but the reason it is favored for titration is that it gives consistent results, which to most scientists, is far more important than accuracy.

The $\ce{ .\!10H2O}$ hydration number will remain constant and will not fluctuate to $\mathrm{11}$ or $\mathrm{9}$ within normal temperature and humidity ranges. This means that per mass the same amount of sodium ions are dispensed every time and gives the same results time after time and experiment after experiment even with different batches of powder. Also the tetraborate conjugate base is very weak, which is ideal for a titration agent as well.


Usually, if we're doing these for basic application or experimentation, standardization doesn't really matter. But when it comes to anything analytical where you start to involve calculations, standardization is a must. This is done with NaOH because it's hygroscopic and readily sucks up the moisture in the air. So what is being weighed isn't totally NaOH, but also the moisture that it has absorbed. So almost always the concentration will be lesser than what is sought to be prepared because of this. Tedious as it may be (because you also have to heat up the KHP for around an hour before titrating), standardization brings you closer to the true concentration, but not exactly on the mark. At least in the end, calculations using the standardized concentrations will be analytical. This is why they never introduce standardization in general chemistry courses, I only got to grips with it when I took up Analytical Chemistry, and back then it was really tedious to do.

For example, I actually experienced this before where we had to prepare a 1M stock of NaOH, but after standardization it turned out to be around 0.8 (maybe due to the humidity in the lab at the time).


Is a galvanic cell only called that if copper and zinc are used?

Voltaic cell and galvanic cell are both synonyms for the same thing: An electrochemical cell, which produces an external electric current. So the two terms can be used interchangeably.

Check out also a site from Georgia state University, where it says:

  Electrochemical cells which generate an electric current are called voltaic cells or galvanic cells, and common batteries consist of one or more such cells.

Addendum: Nevertheless there might be some galvanic cells, like the Daniell Cell, which have a defined electrode material and defined electrolyte concentrations to supply a specific voltage.

For the Daniell cell these would be:

$$\begin{array}{lc}
\text{Electrode material:} & \ce{Zn|Zn^{2+}||Cu^{2+}|Cu}\\
\text{Electrolyte concentrations:} & c(\ce{Zn^{2+}}) = c(\ce{Cu^{2+}}) = 1\ \mathrm{mol\ l^{-1}}\\
\text{Voltage:} & U(25\ \mathrm{°C}) = 1.10\ \mathrm{V}
\end{array}$$


I (might) have accidentally combined these mold-killers at home: Any adverse reactions?

I wouldn't be too worried about any chemicals that came from either of the products, or from reactions between the products. Any of those would have been volatile enough that they're long gone by now.

From your description, it sounds like you may be having a reaction to either residual mold or another allergen present in the trailer.

Mold is very difficult to fully eradicate. Even if you cleaned it off of all visible surfaces, it could very easily be growing behind something and you'd never know it without ripping out pretty much everything that's not glass or metal. Also, even if you've managed to kill all the most that was originally present, there could still be residual mold spores or proteins that get kicked up into the air very easily. If you're sensitive to mold, it doesn't take much for your immune system to react, giving you eye and upper respiratory tract irritation.

If this is the case, it may be easier to get rid of the trailer. To fully decontaminate, you'd want to go in and do a full tear out of everything that could have mold on it. After you're down to bear metal, you'd need to wet-wipe everything with a good biocide, making sure that you don't spread possible contamination with your wiping cloth. The point isn't only to kill the mold - it's to get rid off every possible protein fragment that may have been generated by the mold or mold spores.

While it's true that some people are sensitive to mold generated mycotoxins, more common is that people have immune reactions to the mold proteins, and it is not a small job to fully get rid of them.

Now, before you go full bore, keep in mind that it could also be something other than mold. Dust mites and roaches are other common allergens that come to mind. It may be worth going to an allergist to get tested for these and other common environmental allergens. You may find out that you're actually allergic to some kind of pollen (like poppies!) and it just happens that the trailer was driven through a poppy field at some point...


Coloured compounds on reacting with oxygen

The feature that makes colored compounds colored is the one that makes the compounds easily oxidized. When the compound is oxidized, something does not exist anymore. I'm not giving you a straight answer since you will experience a nice enlightenment when you figure that out. Take a look at the structure of beta-carotene, for example. Where would the oxygen go and what would the structure look like after that?


What is the correct way to clean spectrophotometry cuvettes?

I always managed to wash out the remaining solution with the corresponding solvent a couple of times, subsequently rinsing with acetone/isopropyl alcohol and leave it dry on air for a few minutes. The method of spectrophotometry requires that you are using homogeneous solution anyway, so there shouldn't be any precipitation/deposition, otherwise you cannot interpret the data with Beer-Lambert law.

Also, don't soak cells in KOH-bath or similar alkaline media. And I'm always strongly against abrasives, especially when it comes to cleaning optics. Even quartz cuvettes can be occasionally damaged with the brush pick. Disposable PS-cells can be easily scratched. Plus, there is a little risk of hydraulic pressure knocking out the cuvette's bottom (like it happens with test tubes when cleaned too vigorously). Leave mechanical cleaning for the professors of mechanics *.

*From a story published in The Oxford Book of Scientific Anecdotes:

  Victor Moritz Goldschmidt, a distinguished geochemist, was reputed to have carried a capsule of potassium cyanide when he was planning his escape from Nazi Germany. When a friend in the engineering department of the university expressed interest, Goldschmidt was supposed to have replied that cyanide was for professors of chemistry; his friend, a professor of mechanics, would have to carry a rope.


Why does pyridine make such a big change from SNi to SN2?

Both cases share, that the reaction of the alcohol and the thionyl chloride yields an alkyl chlorosulfite under liberation of HCl. Then

Either the C-O bond of the alkyl chlorosulfite brakes, a (tight) ion pair of carbo-cation and chlorosulfit anion is formed.  The later decomposes into $\ce{SO2}$ and $\ce{Cl-}$, the latter reacting with the carbo-cation.  This yields the rentention of configuration observed.

.OR.

In presence of pyridine, HCl generated does not leave the reaction mixture (as above), but reacts with pyridine to yield pyridinium chloride.  Pyridinium chloride is able to react along SN2 on the alkyl chlorosulfite.  Seeing this variant from the beginning till the SN2 of the alkyl chlorosulfite yields the inversion of configuration observed.


Wavenumber range reversed in IR spectroscopy

Sorry, I can't comment yet. I thought I'd add a historical perspective, which I don't know if it is true, but it occurred to me as a good reason for why this might be the case.

UV should historically preceed IR. In UV, the energies axis has been historically reported in wavelength (nm), which is inversely proportional to energy, meaning you have high energies on the left and low energies on the right.
It might be that for consistence with UV, the wavenumber axis has been reversed so to have higher energies on the left.

This is highly speculative, though.


Ozone resonance structure: differing opinions

Those are all valid resonance structures for ozone (except the second from the left in the "four-form structure" row, that one has two negative charges, maybe it's a typo).  In fact you can draw even more structures like this one, where this is no single bond between the middle and terminal-right oxygens.

All of these structures contribute to the "real" structure of ozone - however, some resonance structures contribute more and some resonance structures contribute less.  Here is a link to a nice set of rules (see page 2, the "Resonance Contributor Preference Rule" section) that describes how to weight the contribution of various resonance structures.  The most important rule is that each atom in the resonance structure should have an octet of electrons.  Only the structures labeled I and II in your "three-form structure" row meet this requirement, so they are the two most important resonance structures for ozone - but all of the others do contribute, just to a lesser extent.


In reality, when you're saying that the molality is infinite, you are indeed basically asking for solute with infinitesimal amounts of solvent, and it will be true that $T_\text{solution} \approx T_\text{solute}$.

This equation describes the boiling point of the solvent under the effects of the non-volatile solute. Ie. you're making the assumption that the solute does not enter the gas phase at the temperatures you're considering.

A substance boils when the chemical potential of the liquid phase equals the chemical potential of the gas phase: $\mu_\text{liquid} = \mu_\text{gas}$, or when the vapor pressure equals the surrounding pressure.

So to elevate the boiling point, we're making the claim that for a given temperature, the chemical potential of the liquid phase is lowered (it's made more stable), or equivalently, that the vapor pressure is lowered. The effect is entropy based.

When particles are interspersed between the solvent molecules, the solvent molecules become less ordered, and are thus at a state of higher entropy than the pure solvent. This translates into a lower chemical potential. The gas phase will have only the solvent molecules, because we made the assumption that the solute is non-volatile. Therefore, the gas-phase is unaffected by this effect, and the chemical potential of the gas phase remains the same.

It's easy to visualise from the vapour pressure perspective. Imagine the liquid-gas interface. At a given temperature, a certain portion of the liquid solvent molecules at the liquid-gas-interface will enter the gas phase. The gaseous solvent molecules at the liquid-gas-interface will likewise enter the liquid phase when they come in contact with it.

The amount of solvent molecules going from gas->liquid depends on the atmospheric pressure. The amount of solvent molecules going from liquid->gas depends on the kinetic energy of the molecules (the temperature). A molecule may have enough energy to break free of the solvent-solvent interactions, or it may not. The higher the temperature, the more likely.

Now imagine that we add solute molecules. There will be less liquid solvent-molecules present at the liquid-gas interface, and thus a lower rate of liquid -> gas (meaning a lower vapour pressure), simply because the probability of a liquid molecule hitting the phase interface is lower...  So the temperature has to be higher to compensate, if we want to counter the unchanged rate solvent molecules going from gas -> liquid.

Why boiling point goes infinite: At the limit where molality goes towards infinity, there will be an infinitesimal amount of liquid solvent molecules at the phase interface... And if we want that infinitesimal amount of molecules to sustain a vapour pressure that equals the atmospheric pressure, we need a temperature that approaches infinity as well.


Does 1-phenyl-1-keto-but-2-ene show positive iodoform test?

Enones react with with iodine and catalytic base under aqueous conditions to give $\alpha$-iodoenones (Ref.1). Pyridines are most often used for this, the reference quoted also uses $\ce{K2CO3/DMAP}$. They further show that reaction works for a range of enones including those with the $\ce{RCOCH=CH-CH3}$ grouping and no mention is made of $\gamma$-iodination. A "background" reaction was run with $\ce{K2CO3/I2}$ in aq. THF and no iodination was observed.

Proposed mechanism from the referenced paper:

References:

Marie E. Krafft, John W. Cran, "A Convenient Protocol for the α-Iodination of α,β-Unsaturated Carbonyl Compounds with $\ce{I2}$ in an Aqueous Medium," Synlett 2005, (8), 1263-1266 (DOI: 10.1055/s-2005-868495).


There are some good examples in Multiwavelength Spectrophotometric Determination of Acid Dissociation Constants Pharaceutical Research Volume 16, Issue 3, pp 374-381.

Benzonic acid (pka = 3.98), phenol (pka 9.73), phthalic acid (pkas 2.70 and 4.86), nicotinic acid (pKas 2.1 and 4.63) and p-aminosalicylic acid (pKas 1.79 and 3.58) would be good in the ~220nm-300nm range.


So the reason for carbonates and phosphates being insoluble in water because in an acidic environment, they are both protonated and lose their negative charges. The symmetry of their binding groups produces identical resonance structures at high pH values (since rotating one double bond will shape it into a superimposable structure of another). This allows any given oxygen at a time to posses the double bond (in reality the pi bond to the phosphorus or carbon is shared throughout all the bonded oxygens and each oxygen maintains an equal distance away from the central atom).

This partial negative charge (created by lack of firm placement of the pi electrons) can be neutralized by bonded hydrogens (this will create a small positive charge, however hydrogen's ionization energy is much less than oxygen's electron affinity). The lack of polar-ness of the phosphates and carbonates makes them insoluble in water.

In a basic environment, (created by hydroxides) the carbonates and phosphates will become deprotonated and charged. The charge on these molecules will allow them to be more miscible with water.

https://en.wikipedia.org/wiki/Carbonate


One possibility:  Starch/iodine solution + sodium thiosulfate.  The starch/iodine solution goes from dark blue/black to clear when sodium thiosulfate reduces the iodine.


Unusual Friedel–Crafts alkylation with pivaloyl chloride

You were on right path; tert-butyl carbocation is quite stable$\ldots$ so stable that the acylium cation, which normally reacts as an electrophile itself, instead decomposes via decarbonylation (loss of stable carbon monoxide molecule).

The t-butyl carbocation isn't as stable as the acylium cation (which is stabilised by resonance), but the difference is small enough that with the help of the non-reversibility of CO loss (which is removed from the mixture as a gas) and reduction of the steric strain (which is lower in t-butyl than in acylium), it is enough to make it the main reaction route instead of a side route.


How do ligands and coordination complexes work in this case with copper (II) chloride?

When solid $\ce{CuCl2}$ is dissolved in water, aquo complexes are formed.

$\ce{CuCl2_{(s)} + H2O -> [Cu(H2O)6]^2+ + 2 Cl-}$

Addition of ammonia results in a ligand exchange reaction to give tetrammine complexes.

$\ce{[Cu(H2O)6]^2+ + 4 NH3 -> [Cu(NH3)4(H2O)2]^2+ + 2 H2O}$


Can we define a new unit instead of specific heat that is constant with temperature rise?

There isn't a simple redefinition that would work in all cases, and this is probably why nobody does this.  Of course, you COULD define such a unit for a given system by defining it as a function of temperature, but it would be applicable only to the specific system.

As you noted, the physics behind the temperature dependence of a material's heat capacity lies in the quantum states which become available for excitation at different energies.  Consider an insulator which stores heat primarily in the lattice vibrations.  As you increase the temperature, more vibrational modes are accessible.  This is often described using the Debye model which gives the heat capacity as such:

$C_V = k(\frac{T}{\theta})^3$

where $C_V$ is the heat capacity, $k$ is a constant, $T$ is the temperature in Kelvin, and $\theta$ is a constant pertinent to a specific solid.[1]

However, a metal can also store a significant amount of heat in the motions of electrons since conduction electrons are not bound to specific sites.  Heat capacity for the electron 'gas' is often described like this:

$C_V' = k' \frac{T}{\theta'}$

where $C_V' \neq C_V$, $k' \neq k$ and $\theta' \neq \theta$.[2]  Depending on $\theta$ and $\theta'$ for a given material, at low temperatures (often cryogenic) the electron contribution can be the dominant mode of storing heat in the solid.  Thus, if you chose a unit which was sensible for an insulator, it wouldn't apply to a conductor (and vice versa).

Even within a specific material, the heat capacity on either side of a phase transition is often different since the phase transition has changed the quantum states available in the material.  Therefore, your new unit would become discontinuous at phase transitions and would need a different mathematical expression for each segment of the heat capacity function, not to mention a different parameterization for each material.  Hence, no real gain.

[1] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 106, eq. 47b

[2] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 193, eq. 37.  Note that folks don't actually use the symbol $\theta'$ in this case, but choosing so makes the explanation above clearer.


What non-toxic non-water substances have a freezing point very close to water's?

I tried to use the curated dataset from Mathematica to answer this question.  You can ignore the following if you have no interest in Mathematica and just jump down to the end for my ideas.
The following code gets all of chemicals currently in the curated dataset and grabs the melting points and NFPA health rating.  The second line finds the chemicals with melting points between -7 and 0 degrees and the last line looks for chemicals with a health rating of 0, which is low risk.
out = {#,  ChemicalData[#,"Name"], ChemicalData[#,"MeltingPoint"],
   ChemicalData[#,"NFPAHealthRating"])& ChemicalData[];
cull1 = Cases[out, x_ /; -7 < x[[3]] < 0];
cull2 = Cases[cull1, x_ /; x[[4]] == 0];

We go from about 43000 chemicals in the first list to 220 in the second and finally 6 in the last.  I'm not sure I agree with the health ratings of these compounds but here you go (melting points in parentheses):

2-methyl-2-propanethiol (-0.5)
prehnitene (-6)
trans-1,2-dichlorocyclohexane (-6)
hexadecene (-6)
trans-1,2-dibromocyclohexane (-6)
olive oil (-6)

The only chemical I would deem "food safe" would be the last one.  Perhaps some type of food-grade oil is the mystery substance.


Why is the photon energy of lithium flame coloring lower than that of cesium?

The colours you get in flame tests / burning metals, comes from the electrons in one shell being 'knocked' (by the heat), into a higher shell; then shortly after the electron 'drops' back (in one or more steps). The electron which dropped down a shell has a lower energy state, so to conserve energy, a photon is emitted. The further it 'drops' the higher energy/frequency/ colour it has. (Lower energy Infrared -> ROYGBV -> UltraViolet -> X-rays Higher Energy)

An element with more protons has a 'deeper' bottom shell. More protons => more attraction to electrons => more energy required to 'knock' up to a higher shell => higher energy/frequency photons.

So in general, a larger element would give higher energy photons off; in a flame test.
However it is a little more complex than that.

Because elements have many shells, an electron may jump back in more than one hop; giving off several different colours. For example Calcium has an orange flame colour, but heaver Strontium has a Scarlet red colour. If you look at a Strontium flame through a prism, you get Red, Green, and Violet.

Note: You may get infrared and ultraviolet in the mix too.

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron. Normally an electron in its outermost (filled) electron shell.
Lithium only has electrons in its bottom 2 shells. Cesium has electrons in its bottom 6 shells. Its outer most one has a single electron and is relatively "further away" from the protons in the nucleus. So it is easier to liberate Cesium's outer most electron than Lithium's outer most. In a flame test, if the flame is hot enough, electrons from deeper shells may be 'knocked' out of place and so you get higher energy photons.


compression set of tin catalyzed silicone better than platinum?

Tin catalyzed silicone lasts for a couple of years and platinum catalyzed lasts for 10 years, after which they lose their elasticity, becoming brittle and splitting. So while they may have equal properties for the first year, After a while the Platinum silicone will perform better. Using less catalyst for longer cure times also increases durability, and vacuum treating the silicone prior to casting and heat treatng it afterwards increases it's compression set performance.


Why does the mixing of sigma 2s and 2p orbitals lower the energy of the sigma 2s orbital and raise the energy of the sigma 2p orbital?

This is a frequent source of confusion in my experience - you'll see there are at least two other very similar questions here already.

Let's take the first point: why do the $\sigma_{2s}$ orbitals get lower in energy when "mixing" with the $\sigma_{2p}$?

I'm going to ignore "mixing" and use the word "interact" instead. If I consider the 1s orbitals in $\ce{H2}$, I take two 1s atomic orbitals and they (of course) interact. So one orbital goes down in energy, and one orbital goes up in energy.

This is a common theme - if there's any degree of overlap between two orbitals, one will become more stable and one will become less stable. In the diagram, we claim "bonding" and "anti bonding" but in more complex molecules, these terms become hard to identify (e.g. benzene).

The theme remains: if orbitals interact (i.e., have the same symmetry), some will go down in energy, some will go up in energy.

That's the answer to your first question. Since the $\sigma_{2s}$ orbitals do interact with $\sigma_{2p}$ the 2s will go down in energy.

Your second question is about the $\sigma_2p$ going up in energy.

I could answer with a formula from quantum mechanics, but let's argue qualitatively. Here's a diagram for two interacting orbitals:

This is much like our case with the 2s and 2p orbitals. We see that the result is a system that's most likely more stable - the orbital lower in energy is even lower because of the interaction.

If we consider the reverse case - that the 2s goes up in energy, and the 2p goes down in energy, we would have a system that's less stable. In theory, this could happen - if the interaction is negative. In other words, the system would be more stable without the interaction than with.

In short, the "mixing" or interaction between 2s and 2p orbitals occurs because the species becomes more stable.


Why are catabolic reactions exergonic and anabolic reactions endergonic?

In most chemical reactions, including those of catabolism and anabolism, some bonds break and others are formed. For example, if you hydrolize a biopolymer such as a polysaccharide or protein into its building blocks, you break bonds in the polymer and in water, but form new bonds between parts of the water and parts of the polymer.

Making a polymer from building blocks sometimes is endergonic (like some anabolic reactions) and sometimes is exergonic (like making polyethylene from ethylene). There is no inherent rule that says more complex molecules have higher or lower Gibbs free energy than the small molecules they are made from.

You could say there are three reasons why we eat food: To directly use the substances we ingest (e.g. vitamins), to use the substances we ingest to make new substances we need (e.g. amino acids in proteins), and to oxidize substances (in humans with $\ce{O_2}$ as the terminal electron acceptor) and harvest the Gibbs free energy. We choose food accordingly (e.g. we don't eat sand). Also, we don't eat Grignard reagents or other highly reactive substances because they are usually not available and we are not equipped to handle very exothermic reactions in our bodies. So the food we eat is fairly stable, and often it requires Gibbs free energy to drive biosynthetic reactions (explaining why many anabolic reactions are endergonic)

Molecules that can be hydrolyzed in the digestive tract using exergonic reactions are good as food. It is difficult to run endergonic reactions in the digestive tract (the reactants often used to drive endergonic reactions, most notably ATP, are found inside cells, not in the stomach or intestine). For those reasons, any large molecules we digest can be broken down to smaller molecules in exergonic reactions (these smaller molecules then enter cells, where further reactions could be exergonic or endergonic).

Specifically for hydrolysis (breaking biopolymers into building blocks) and condensation reactions (making biopolymers from building blocks), the hydrolysis reactions tend to be exergonic and the condensation reactions tend to be endergonic. This might be related to the high concentration of water throughout the body.


Show the acid nature of hydrogen sulfide

As orthocresol suggested, to illustrate "classic" acidity as it's understood on a school level you unquestionably want to start with Arrhenius definition, which attributes acidic properties to  any substance which increase concentration of hydronium $\ce{H3O+}$ ions in water.

That's said, the very first reaction you probably want to write is dissociation of $\ce{H2S}$ in water and demonstrating that it is a weak acid by showing the corresponding acid dissociation constants $K_\mathrm{a}$ (the larger the $\mathrm{p}K_\mathrm{a},$ the weaker the acid; data from [1, p. 5-87]):

$$
\begin{align}
\ce{H2S + H2O &<=> HS- + H3O+} &\quad \mathrm{p}K_\mathrm{a1} &= 7.05\\
\ce{HS- + H2O &<=> S^2- + H3O+} &\quad \mathrm{p}K_\mathrm{a2} &= 19
\end{align}
$$

Some reactions illustrating acidic properties of $\ce{H2S}$:

Reaction with sodium hydroxide

Note that the reaction

$$\ce{H2S(aq) + 2 NaOH(aq) -> Na2S(aq) + 2 H2O(l)}$$

is for complete neutralization (concentrated $\ce{NaOH}$ solution).
If the reactants are taken in 1:1 ratio (diluted $\ce{NaOH}$ solution), sodium hydrosulfide is formed:

$$\ce{H2S(aq) + NaOH(aq) -> NaHS(aq) + H2O(l)}$$

Reaction with sodium carbonate

Sodium hydrosulfide is also formed when saturated $\ce{H2S}$ solution reacts with $\ce{Na2CO3}$:

$$\ce{H2S(aq) + Na2CO3(aq) -> NaHS(aq) + NaHCO3(aq)}$$

Reaction with ammonia

Reaction with aqueous ammonia solution yields exclusively in ammonium hydrogen sulfide $\ce{NH4HS},$ even when concentrated ammonia solution is used:

$$\ce{H2S(aq) + NH4OH(aq) -> NH4HS(aq) + H2O(l)}$$

Ammonium sulfide $\ce{(NH4)2S}$ doesn't form as ammonium $\ce{NH4+}$ $(\mathrm{p}K_\mathrm{a} = 9.25)$ is a much stronger acid than hydrosulfide $\ce{HS-}$ by a factor of about $10^{19 - 9.25}\approx\pu{6E9}.$

So far the reactions between well-soluble compounds that were carried in aqueous solutions where Arrhenius definition works fine.
However, it cannot be used to explain acidity in non-aqueous media, where a Brønsted–Lowry definition of an acid comes in handy: the acid is a proton donor.
This can be illustrated by the reaction with liquid ammonia:

$$\ce{H2S(g) + 2 NH3(l) ->[\pu{-40 °C}] (NH4)2S(s)}$$

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J., Eds, CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data; 97th Edn.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016-2017. ISBN 978-1-4987-5429-3.


Why is latent heat of vaporization not exactly proportional to boiling point?

Hm, the proportionality is not so very good. (data from the wikipedia link above)

The line is a linear regression through all ten data points.

The answer to your question is given on the same wp article. The quotient of enthalpy of vaporisation and temperature is the difference in entropy between the liquid and gaseous phase.

According to Trouton´s rule, it is about 85 to 88 J/molK (ca. $10.5 R$) for most liquids. https://en.wikipedia.org/wiki/Trouton%27s_rule explains the reasons for deviations, like hydrogen bonding, small rotational excitation, dimerisation in the gas phase etc.


Can in any case the faster step of the reaction be rate determining?

I can provide you an example.
The oxidation of formate ion by peroxydisulfate in water solution:
$\ce{HCOO^{-} +  S_2O_8^{2-} -> CO_2 + 2SO_4^{2-} + H^+}$
has the following mechanism:

As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: $r=k[\ce{HCOO^-}]^{1/2}[\ce{S_2O_8^{2-}}]$. The  first  reaction  is  very  slow,  so  most  of  peroxydisulfate  is consumed  in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.
(This isn't the true mechanism but it is good for showing the point. A  more complex  mechanism  includes  the  formation  of  OH  radicals  and  several  chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.)


What is the correct name for this lab technique in crystallization?

I think the most specific you can get on this phenomenon is nucleation , specifically heterogenous nucleation / surface nucleation.

The rough surface formed creates scope for nucleation to occur, and  this creates the crystals.

I don't think there is a specific name for this technique as a lab tenchique .


Are there softeners in synthetic materials?

   Yes. All plastics have plasticizers in them. The particular plasticizers chosen depends on what properties are desired of the plastic. Some can make the plastic much stronger but also more brittle at the same time, so that it would be much more difficult to break but when/if it did snap, it would tend to shatter or snap violently. Others make the plastic softer and less brittle, so that the plastic form has less structural strength, but will bend, stretch and disform under heavy load instead of snapping.

   Plasticizers can also affect the  temperature resistance to deformation, volatility (temperature resistance to deflagragation, and properties of said polymer if ignited, such as, will it go out on its own or will it continue to burn the whole thing). Plasticizers can be added to affect the polymer's UV resistance, resistance to chemicals or resistance to abrasion.

   Once again, all plastics have plasticizers in them. If one were to magically remove all of the plasticizers from a plastic bottle (PETE or PE), the plastic would become brittle, dry, and probably reduce to tiny pieces (think shattered tempered glass, but smaller) or a powder upon touching the plastic form which no longer contained any plasticizers. In fact, you can observe this with some old plastics from the 1970s. Even though the plastic may have never been exposed to UV light, a plastic that is losing its plasticizer can take on the same cracked, opaque appearance. Most plasticizers are volatile at room temperature and will slowly evaporate away just by simply being exposed to air.

   Most plastics will leach their plasticizers more rapidly after being washed with soap and water, after being microwaved, while being heated, or while at a low temperature (such as in the freezer). Even still, some plasticizers will take 20 years or more to evaporate away enough to change the properties. Some plastics will never lose their plasticizer (that we know of).

According to Wikipedia:
Some of the most common plasticizers and their uses are:

Phthalate-based plasticizers - used in situations where good resistance to water and oils is required. Examples include Bis(2-ethylhexyl) phthalate (DEHP) and Butyl benzyl phthalate (BBzP).
Adipate-based plasticizers - used for low-temperature performance or resistance to ultraviolet light. Some examples are Bis(2-ethylhexyl)adipate (DEHA) and Dimethyl adipate (DMAD)
Trimellitates - used in automobile interiors and other applications where low volatility and resistance to high temperature is desired.
Other common plasticizers are Polybutene, Benzoates, Sulfonamides, Glycols/polyethers and poxidized vegetable oils


Nucleophilic Substitution with ethyl bromide and NaOH

With the help from Waylander (s.o.) we get the following mechanism.


How to determine which salt will precipitate from a solution containing multiple ions?

You need to know that $K_\mathrm{sp}$ is used to determine the solubilities of sparingly soluble salts usually. You can get the values of $\ce{AgSO4,AgBr}$ and as $K_\mathrm{sp}=[\ce{Ag+}]^2[\ce{SO4^2-}]$ or $[\ce{Ag+}][\ce{Br-}]$. Now concentration of silver ion is constant so try calculating the other two anions and try comparing them.


Primarily it is used as an anti-microbial agent. There is lots of data out there showing that the efficacy of ethanol at killing bacteria peaks at the 70% mixture. As they are making tons of it (literally) for the microbiologists, it's easier to just scale production then set up a new production facility for chemists to use at an arbitrary concentration.


Can mechanical agitation increase the reaction rate?

If the reaction is heterogeneous (multiple phases involved), mechanical agitation helps. If the reaction is homogeneous (all reactants and products in the same phase), mechanical agitation usually does not help. An exception is when the process is auto-catalytic (i.e. starts in one corner and then propagates).

In the example mentioned, agitation will help to transport concrete away from the brick surface and to bring fresh acid to the surface. Scrubbing also helps if there is no chemical reaction and the removal is mechanical in nature.


Why and how is this stability maintained?

Beryllium carbonate decomposes into Beryllium oxide and Carbon dioxide.
$$\ce{BeCO3 <=> BeO + CO2}$$

According to Le Chatelier's principle one simply has to store it under a $\ce{CO2}$ atmosphere to push the equilibrium to the left.
$$\ce{BeCO3 <<=> BeO + CO2}~\text{in excess}$$

  Why this compound decomposes when kept in atmosphere of oxygen?

When there is no $\ce{CO2}$ atmosphere the equilibrium is pushed to the right, which would be the "normal" case under standard conditions.
$$\ce{BeCO3 <=>> BeO + CO2}$$


nickel-copper fabric: shielding viability after being subjected to weathering

Your alloy should be sufficient. For one, cupronickel alloys are very corrosion resistant and is used to make many different coins across the world. occasionally yes, a nickel or dime will tarnish but this is usually from exposure to soda, salt or some other corrosive liquid, other wise they can sit in a fountain for weeks without tarnishing. AS long as the acid rain isn't too severe, you should be okay.

As for the product you are considering, the description on amazon describes it as:

  Copper+Nickel+Polyester Plated Fabric

Now plated metals do not necessarily have the same properties as the pure casted alloy since to plate on polymer requires using electroless plating and does not guarantee the same microstructure as a casted alloy, but corrosion should be unaffected.

See also: Monel


What is the difference between steric strain and torsional strain?

TL;DR Torsional strain can be thought as the repulsion due to electrostatice forces between electrons in adjacent MOs. Meanwhile steric strain (also known as van der Waals strain) can be thought as the repulsion when two bulky groups which are not directly bonded to each other become too close to each other and hence there isn't enough space for them.

Here is the more detailed version.

Torsional Strain

Lets consider an ethane molecule. The C-C sigma bond is free to rotate and in principle there are an infinite number of possible conformations. However only 2 are significant, these are staggered and eclipse conformations. Different conformers are usually drawn as Newman projections as they can easily be compared with each other. Below are the Newman projections for the eclipsed and staggered conformer:

The staggered conformer is the most stable conformer while the eclipsed conformer is the least stable conformer. The staggered conformer is approximately $\mathrm{12~kJ~mol^{-1}}$ more stable than the eclipsed conformer. This energy difference between this maxima and minima is known as the torsional barrier.

So what is torsional strain? The reason for the eclipse conformer being higher energy than any other conformers is due to the destabilising electrostatic repulsion between the electron pairs of the C-H sigma bonds on the two carbons. Furthermore, there also exists a stabilising feature which is greater in the staggered conformer. In the staggered conformer there is a constructive orbital interaction involving the bonding and anti-bonding MOs of the adjacent H atoms. This results in hyper conjugation which stabilises the compound.

The two effects that I have mentioned above is what torsional strain refer to. So you can think of torsional strain as the strain that is the result of electrostatic forces.

Steric Strain

Now lets consider butane. Rotation of the $\mathrm{C_2-C_3}$ sigma bond also lead to infinite possible conformers. However there are 4 main conformers shown below:

Here there are 2 types of eclipsed conformers which are the least stable of the conformers. The most unstable one is known as the eclipsed syn form. The syn form is approximately $\mathrm{20~kJ~mol^{-1}}$ higher in energy than the staggered conformer. The reason for this can partly be attributed to torsional strain as there is repulsion between the electrons in the sigma bonding orbitals. However also it can largely be attributed to the repulsion of the two relatively bulky methyl groups as they become too close to each other and there isn't enough space for them.

This repulsion is known as steric strain. Therefore steric strain can be define as the repulsion which occurs when non-bonded groups which are not directly bonded approach each other too closely. This repulsion only exists for bulky substituents, such as methyl or ethyl groups. So in ethane there is no steric hindrance as the hydrogen atoms are not that bulky.


Does hydrochloric acid change in concentration over time?

First of all, check the expiry date. In general HCl solutions are secondary standards. 1 N is just a approximation. You need to standardize it every time with a primary standard such as sodium carbonate. On the other hand, constant boiling HCl is indeed a primary standard. It is not very well known, perhaps because modern analytical chemists have become lazier.

Constant-boiling hydrochloric acid
John L. Sheldon
J. Chem. Educ., 1935, 12 (8), p 377
https://pubs.acs.org/doi/abs/10.1021/ed012p377


Why do metals tend to lose electrons, as opposed to maintaining electric neutrality?

Firstly, atoms "want" to achieve the noble gas configuration of 8 valence electrons because it is the most stable form. All that means is that it doesn't tend to react under normal conditions that we experience on Earth, therefore it will stay in that configuration for quite a while and are less likely to react. There is a more complex quantum physical answer for that but you'll have to go elsewhere for than.

The main force that keeps electrons in atoms is the electrical attraction between the electrons and the protons in the nucleus and so, if it is more energetically favourable to lose that electron in order to form a bond, then that is what will happen.

Focusing on the Alkali metals as an example, as you move down the group, they get more and more reactive. This is because of two main reasons that are a result of the electrons being further away from the nucleus:

Because they're further away, the attraction between the protons and the outer most electron is less
Secondly, taking Rubidium as an example, it has 37 electrons and 37 protons. From the perspective of the outer-most electron, there are 36 electrons repelling it, and 37 protons attracting it, therefore acting as a net charge of 1. However, if you take into account the first point, the repulsion of the closer electrons is stronger than the attraction of the protons so it could even be less than one


Why do electrons in sigma bonds have zero orbital angular momentum about the internuclear axis?

My quantum is slightly rusty, but I think this is how it goes (I would appreciate any corrections or clarifications). We're talking about diatomic molecules here. As you no doubt know, σ-orbitals are invariant about rotation around the internuclear axis: that is to say, you can rotate the molecule as much as you like, but the sigma orbital will still look the same. This is true whether the sigma orbital is made from s, p, or d orbitals.

If we define the $z$-axis to be the internuclear axis and $\phi$ to be the angle around this axis (the azimuthal angle), then we can also say that σ-orbitals have no dependence on the angle $\phi$.

In a diatomic molecule, because the orbital angular momentum (OAM) operator $\hat{L}$ does not commute with the Hamiltonian, the quantum number $l$ is no longer a 'good quantum number': that is to say, the eigenstates of the Hamiltonian (which are the MOs) are no longer simultaneously eigenstates of $\hat{L}$, so the MOs do not have a specific value of $l$ that can be assigned to them. However, for a one-electron molecule (e.g. $\ce{H2+}$) it turns out that $\hat{L}_z$ still commutes with the Hamiltonian, where the $z$-axis is the internuclear axis (or molecular axis). This operator represents the projection of the OAM onto the internuclear axis, and the corresponding quantum number is denoted $m_L$ or $\lambda$ (depending on where you read).

Now, let's denote the wavefunction of the σ-orbital by $\psi$. We know that the eigenvalues of $\hat{L}_z$ are unit multiples of $\hbar$, or in other words, $\hat{L}_z\psi = m_L\hbar\psi$. [Note that this is no different to $\hat{L_z}\psi = m_l\hbar\psi$ in an atom, except that we use a capital L because the context is different, i.e. we are talking about a diatomic molecule.]

It turns out that in spherical coordinates, $\hat{L}_z$ is equal to $-\mathrm i \hbar (\partial/\partial \phi)$. Since we've established that $\psi$ has no dependence on $\phi$, the partial derivative $(\partial \psi/\partial \phi)$ is zero and hence $\hat{L}_z\psi = 0$, meaning that the quantum number $m_L = 0$, as desired.

It doesn't matter what the constituent atomic orbitals are. However, I thought a bit of clarification would be useful. For the atomic p-orbitals, you are correct that these possess $l = 1$. However, note that $l$ is a quantum number which characterises the magnitude of the angular momentum, whereas $\lambda$, or $m_L$, is a projection quantum number. That is to say, it is possible to "add" two atomic orbitals with $l = 1$ and have the projection cancel out.

If you think about the projection of the orbital angular momenta in the original p-orbitals, they are of course $m_l = (+1,0,-1)$. However, these are the complex p-orbitals which are obtained as eigenstates of $\hat{L}_z$. The real p-orbitals, i.e. the dumbbell-shaped p-orbitals that you and I think about when constructing sigma bonds, are either $m_l = 0$ (i.e. the $\mathrm p_z$ orbital) or are linear combinations of $m_l = \pm 1$ (i.e. the $\mathrm p_x$ and $\mathrm p_y$ orbitals). So, it's not at all counterintuitive that when you add these together to form sigma bonds, the projection $m_L$ is simply zero.


Forging steel inside a zero oxygen environment

It is possible to forge iron in a low atmosphere environment, but it isn't worth it because there are few if any benefits to doing so.

This isn't true for some other metals, though, or even for some specialist steels. In many other metals and alloys the presence of oxygen or water vapour has a very bad effect on the quality of welds. But the common way to exclude air from these systems is to use an apparatus that flows some inert gas around the spot being welded rather than to fill the room with an inert gas. Some examples are given here.

This is mainly worth doing for many specialist aluminium, copper and titanium alloys where the metals are oxidised more easily than typical irons or steels and effect of impurities in the welds is much larger. It isn't as simple as just having an inert gas blanket though, as this brochure for commonly used gas mixtures shows.


Thermodynamics: Combining the 1st and 2nd Law

Starting with $$dH=TdS+VdP\tag{1}$$H is a function of T and P, so$$dH=\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP\tag{2}$$Substituting Eqn. 2 into Eqn. 1,
$$\left(\frac{\partial H}{\partial T}\right)_PdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{3}$$But, from the definition of $C_p$, $$C_p\equiv \left(\frac{\partial H}{\partial T}\right)_P \tag{4}$$Substituting Eqn. 4 into Eqn. 3 gives:$$C_pdT+\left(\frac{\partial H}{\partial P}\right)_TdP=TdS+VdP\tag{5}$$Solving for dS then gives the desired relationship.


How to label atoms or groups "backwards" in ChemDraw

ChemDraw is a piece of software that attempts to be as intelligent as possible. In many cases, that is truly helpful, in others it can be annoying.

When labelling an atom in text mode, ChemDraw will automagically assume a label from left to right (i.e. in standard North American/European writing direction) and interpret the label as such. If it finds that the atom has bonds to the right but none to the left, as tyrosine’s hydroxy group has in your example, it will group the label into ‘sensible’ fragments[1] and reverse them. This makes sense if you think of rotating a molecule: If you had drawn tyrosine rotated by $180^\circ$, you would have wanted to add OH on the rightmost atom, not HO. If you then manually rotate the entire molecule within ChemDraw, you will notice the OH flipping into an HO as soon as it graphically makes sense to do so.

Note that if you just hover over the atom in lasso mode or bond adding mode and then hit o (lowercase!) it will automatically add OH (or HO, depending on the orientation) at that position. Hitting o again will remove the additional hydrogen (this holds true for all chemical elements that are attached to single letters regardless whether the atom is attached to a lowercase letter (i.e. o) or an uppercase one (B for boron)).

[1]: Consider OTBS: ChemDraw automatically recognises TBS as an abbreviation for the $\ce{TBS}$ group so when reversing, it becomes TBSO. However ChemDraw does not standardly recognise the triethylsilyl group $\ce{TES}$, so when reversing OTES it arrives at SETO. The preferred way would be to add an abbreviation to ChemDraw.


How does a molecule search work by drawing in visual editor?

Usually, structure search is implemented using a graph model. For example, let's take cyclohexane. On screen this is a hexagon - but this is converted on the server to a simple atom/bond graph model of six carbons in a ring.

This graph can then be compared to all the molecules in the database by various means. For example, subgraph isomorphism or fingerprints or even SMILES string comparison.

As far as I know, ChemSpider allows a variety of searches - with/without hydrogens, substructure search, element fingerprints, and so on. Each of these is a whole question by themselves, so I'm not sure how useful an algorithm that covers them all would be.


Ad 1) I am not sure what do you mean by "equivalent". But you add Zinc in a copper sulphate solution, the "energy" will be consumed through cementation.

to a recation Cu$^2$$^+$(aq) + Zn(s) → Cu(s) + Zn$^2$$^+$(aq)

Ad 2) Please see Equilibrium Constant of an Electrochemical Cell Reaction Example Problem where you can find ansewer to your second question.


Why do we see complementary colors in UV/Vis Spectroscopy?

There are several points

  From what I read, a compound that absorbs visible light will produce a
  complementary color (using the color wheel), that can be seen. Why is
  this the case?

This has to do with the physiology of the eye and how it has been designed to precept the colors in the visible range. As you stated, if we remove green from the visible spectrum, the transmitted light appears reddish or purplish to our eye. May be a cat or dog sees it differently. Color is a human perception. This complementary chart is just an approximation. If we take potassium permanganate solution, and if you look at what colors does it allow to pass through, you will see that it absorbs mostly in the green. You can easily do such experiments with a cardboard box and a diffraction grating made from CDs and a white LED.

  I thought if we excite an electron to its excited state, with it
  absorbing a certain wavelength, when it drops back down to its ground
  state, it will give off a photon of that same wavelength (and same
  color).

This does not happen in majority of the cases i.e, in solutions and solids. The phenomenon you are referring to is called resonance fluorescence and it only happens to atoms in the gas phase. You shine 589 nm to Na atoms in gas phase, and you will see emission of 589 nm from Na atoms. This is accomplished in a flame. See Atomic Fluorescence Spectroscopy.

Normally, once light is absorbed, the absorbing material does not emit light. Instead, the absorbed energy is dissipated as heat.

  Also, when a molecule absorbs UV radiation and reflects all of the
  visible wavelengths, I read that it will appear white or colorless
  (isn't black the absence of all colors?), is this just two ways of
  saying the substance appears white or does that mean that it can be
  white or clear?

Clear and white are two different things. Something clear like water does not absorb much in the visible range, perhaps a very very small portion from the red region. Therefore when you have huge amount of water you start to see the effects of absorbing red portion of the light. Do you recall that ocean water looks bluish? Even liquid oxygen is light blue!

You can say that water transmits most of the visible light. Majority of the substances will also absorb UV. How deep UV do you want? Most of the substance will start absorbing deep UV. Even air does, otherwise we would have been fried by the Sun.

A white piece of plastic or a cotton cloth, is reflecting most of the visible light, hence  it appears white to us. If you look the same piece in red light, it will look red, and in blue light it will appear blue and so on.


Regarding its ionic form, I don' think it is as simple as $\ce{I^3+(IO3^-)3}$. It is much more complex than that. From Ref.1:

This compound is likewise to be considered as an $\ce{I(III,V)}$ oxide and reacts with alkali hydroxide to give $\ce{I-}$ and $\ce{IO3-}$.
$$\ce{3I4O9 + 12OH- -> I- + 11IO3- + 6H2O}$$
Structurally, $\ce{I4O9}$ is possibly an iodate $\ce{I3O6+IO3-}$ (more
precisely $\ce{(I3O6+)_n.nIO3-}$) in which the isopolycation
$\ce{I3O6+}$ has a polymeric structure and is formulated as
$\ce{I_^{III}(I^{V}O3)2^+}$ consisting of twice as many pyramidal
$\ce{I^{V}O3}$ groups as square-planar $\ce{I^{III}O4}$ groups thus
$\ce{I4O9}$ would correspond to $\ce{I(IO3)2^+IO3-}$ which would
become $\ce{I(IO3)3}$.

Reference

Inorganic Chemistry By Egon Wiberg, A. F. Holleman, Nils Wiberg, Academic Press, 2001


Why Atomic size decreases left to right in a period in MPT?

The Modern Periodic table is made according to increasing number of their atomic numbers . So if we move from left to right in a period the no of electrons as well as the no of protons goes on increasing . The electrons are added to the same orbit or energylevel but the nuclear charge increases by one in each step due to increase in the number of proton by one in each step. So the inward pull increases and the outer orbital shrinks. As a result the atomic size decreases moving towards right in the modern periodic table.


Group theoretical condition for an integral to be zero

By Maschke's Theorem, every direct product of representations is decomposable into a direct sum of representations, that is, the function you are integrating can be rewritten as a sum of functions with symmetry properties equal to specific irreducible representations.

As you note, integrating an odd function (which has a certain symmetry, namely antisymmetry with respect to $\sigma_y$, the mirror "plane" of the y-axis) over all space is zero, while integrating an even function (which is symmetric with respect to $\sigma_y$) is generally not. Since integration is linear (that is, the integral of a sum is the sum of an integral), we can consider the integral of each irrep separately.

If the function being integrated is antisymmetric with respect to some symmetry, it will be zero when integrated over all space. The only way for the integral not to be zero is if it is symmetric with respect to all possible symmetries, i.e. it is the totally symmetric irrep.

Expressed in equation form:
$$\int{\prod_{i}\Gamma_i\,\mathrm d\tau}=\int{\sum_j\Gamma_j\,\mathrm d\tau}=\sum_j\int{\Gamma_j\,\mathrm d\tau}$$
and
$$\int{\Gamma}\,\mathrm d\tau=\left\{
  \begin{array}{@{}ll@{}}
    0, & \text{if}\ \Gamma \text{ contains any odd (antisymmetry)} \\
    \text{not necessarily } 0, & \text{otherwise}
  \end{array}\right.$$

Then, it is clear that the only time the integral is nonzero is when, amongst the irreducible representations of the direct product (regardless of its dimensionality), the totally symmetric one is present, all other ones integrate out to zero.

I should clarify that when I write the integral of $\Gamma$, I really mean the integral of some awful, complicated looking function that has the symmetry properties of $\Gamma$, which is why I used $\prod$ and $\sum$ instead of $\otimes$ and $\oplus$.

If your question was instead how do you know that the functions with a given antisymmetry really do integrate to zero, let's consider what it means to have a certain symmetry. It means that there is a particular line or plane or point about which the value of the function is redundant. That is, if you specify it on one side, then the value is exactly determined on the other. We are integrating over all space, which means we can choose bounds of integration that utilize this symmetry. This then separates your integral into a non-symmetric piece and a piece using the symmetry. You first integrate over the non-symmetric piece and get something, and then you integrate that over the symmetry-including part. If that symmetry is antisymmetry, as will be the case with at least one symmetry of any irrep other than the totally symmetric one, this will cancel out the non-symmetric part with an equal and opposite non-symmetric part and the whole integral will be zero. Regardless of the dimensionality of the irrep being integrated it has at least one symmetry element over which it can be integrated that is antisymmetric. Hence, only integrals of functions that are totally symmetric are non-zero.

As a basic example, consider your $\int{x\,\mathrm dx}$ integral that you started with. Since this function has a symmetry (namely antisymmetry with respect to $\sigma_y$), I can make the following choice of bounds:
$$\int_{-\infty}^\infty{x \,\mathrm dx} = \int_{-\infty}^0{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}  =-\int_{0}^\infty{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}=0$$

I can choose to integrate only over the nonsymmetric part and then use the symmetry for the rest of space. The presence of symmetry guarantees that I can do that. Remember, the irrep label is just telling you the symmetry properties of some complicated function. If $x$ in the above equation were some function $f$ that I knew nothing about except that is has this symmetry, I could still make this choice of bounds and use this property to get a 0 integral.

EDIT: To address orthocresol's question below:

How does this work for the $E$ set in $C_{3v}$? Let us first think of a function with $E$ symmetry. The $p_x+p_y$ orbitals are such a candidate. Let's consider a top-down view of a $C_{3v}$ molecule. Now, I will consider one of the mirror planes to lie in the $xz$-plane (see Figure 1). I can do this because I can rotate the $p_x+p_y$ ensemble anywhere in the $xy$-plane. That is, there is some linear combination of $p_x+p_y$ such that there is a new set $p_{x'}+p_{y'}$ where the mirror plane lies on the $x'y'$-plane.

Now we set about integrating the upper half plane. Clearly, the half of the negative lobe of the $p_x$ orbital will cancel the half of the positive lobe, leaving only the integral of the positive lobe of $p_y$. Now we consider the lower half plane, the two halves of the $p_x$ lobes cancel again and we have only the integral of the negative lobe of $p_y$. Since the integral of the upper-half plane is the negative of the lower-half plane, the integral over all space is 0.

Now, why was it important to consider $p_x+p_y$ together? This is what allowed us to orient the orbitals with respect to the mirror plane. Consider a $p_x$ orbital alone rotated away from the axis given. It will have no defined symmetries with respect to the plane (reflecting across it will not map the orbital to itself or its negative) and there is no way to rotate it without the $p_y$. This is why they need to transform together in $C_{3v}$. Indeed, the reason why $E$ functions have 0 character under this symmetry is because the $p_x$ got sent to $p_x$ and the $p_y$ to $-p_y$ and so the whole ensemble was sent to $1-1=0$.

Now, hold on you say, can't I use similar arguments to imply that if I integrated a $p_z$ orbital (which is $A_1$ in $C_{3v}$) in all space, I will also get zero. The answer is yes. The $p_z$ orbital is not zero by symmetry in $C_{3v}$. I alluded to this above, the integral of a totally symmetric function in a given point group can still be zero. However, by describing the point group as $C_{3v}$, we are indicating that there is something special about the $z$-axis that breaks the symmetry and therefore precludes coming to this conclusion based on symmetry alone. Consider the deformation of planar $\ce{NH_3}$ to pyramidal $\ce{NH_3}$. In the former ($D_{3h}$), nothing distinguishes the positive $z$ from negative $z$ axis and a $p_z$ orbital is not totally symmetric ($A_2''$) and so we could conclude that it is zero by symmetry. As we pyramidalize to $C_{3v}$ though, the negative $z$ axis becomes different (it's got hydrogens there while the postive $z$ does not) and we can no longer conclude by symmetry that the integral of the $p_z$ orbital will be 0 (indeed, if it mixes even slightly with those hydrogens, it won't be). If it didn't mix though, it's integral would still be 0 (it's still just a $p_z$ orbital), we just couldn't tell that assuming $C_{3v}$ symmetry.


Can entropy increase during bond formation?

I do not think that an entropy decrease is a requirement to form a bond.

Think about a "normal" case: $\ce{HCl}$ (a gas) and $\ce{NH3}$ (another gas) react together to form $\ce{NH4Cl}$, which is a solid. This is an example of bond formation where entropy clearly decreases.

But now think what would happen if two liquid or solid compounds react together to form a gaseous compound. An increase of entropy has taken place. Can't think of a concrete example, but I think this might be the case of, for example, an esterification between a liquid alcohol and a liquid carboxylic acid that form an ester where, due to the lose of $\ce{-OH}$ groups, intermolecular interactions are lower and it can result with the product molecule being more volatile than the original compounds.

It's all about the number of micro states in which the compounds can exist, and gas formation from solids or liquids importantly increases entropy.

(PS: These are just some thoughts, really, I'm no physical-chemist.)


What can be a harmless electrolytic additive for water for creating an oxy-hydrogen flame?

Electrolysis of water has become an increasingly common means of creating torches for soldering and welding, and there are commercial devices available as well as plenty of internet do-it-yourself sites. According to this site regarding the SafeFlame consortium:

  "[...]the European Union-funded SafeFlame consortium has developed a
  torch system that generates a flame using nothing but H2O and
  electricity.   SafeFlame utilizes an electrical current to electrolyze
  ordinary water, separating it into hydrogen and oxygen gas. Those
  gases are then mixed and ignited as they exit the torch's nozzle. By
  fine-tuning the proportions of the two gases, different types of
  flames can be produced for different applications."

A commonly used electrolyte / electrode combination is $\ce{KOH}$ or $\ce{Na2CO3}$ with stainless steel electrodes. I believe this would be pretty corrosive toward aluminum, but stainless steel for such a setup should be cheap and easy to come by.


A chamber to control concentration of a vapor: Is there a name for this equipment?

There is a book The Design of Controlled-atmosphere Chambers for the Study of Oxygen Toxicity, so I suggest "contolled-atmosphere chamber".


Isotope pairs in different states at room temperature?

It’s not exactly room temperature but it’s probably the closest you can get to two isotopes having different states so that they can be separated by physical means:

At very low temperatures (below $1~\mathrm{K}$) liquefied helium will separate into a superfluidic helium-4 phase and a normal liquid phase containing mostly helium-3. The superfluid and the normal liquid can be considered different phases and are immiscible.


Why is ammonia less acidic than terminal alkynes?

It is the alkyne that is not in line.
In the alkynes, the carbon is sp hybridized and this gives it much more s character.
Because of this, the electrons are closer to the carbon nucleus and the $\ce{C-H}$ bond is slightly polarized.
The carbon has some δ - and the hydrogen δ +.

Therefore the terminal alkynes are more acidic and have a lower $\mathrm{p}K_\mathrm{a}$ than ammonia.


The first answer is wrong, and the second answer is somewhat true (alcohols are more polar than carbonyls, but not for this reason), but unhelpful.

  You use protonated ethanol because it is the most likely acid present.

Let me provide an analogy.

Sulfuric acid is a strong acid. When sulfuric acid is dissolved in water, it ionizes (reacts with water)to an extent that only the tiniest fraction of $\ce{H2SO4}$ is left:

$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

In an aqueous solution, its fair to say that the concentration of hydronium cation outweighs the concentration of free sulfuric acid by quite a lot. If you were drawing an acid-catalyzed mechanism in water, would you use $\ce{H2SO4}$ or would you use $\ce{H3O+}$?

This reaction is done in ethanol as the solvent. The basicity of ethanol is pretty similar to water, and we have a lot of ethanol, so we can expect the sulfuric acid to ionize to a great extent. These acid-base equilibrium shenanigans reaction would occur regardless of whatever else we added to the mix.

$$\ce{CH3CH2OH + H2SO4 -> CH3CH2OH2+ + HSO4-}$$

Now, if we are doing an acid-catalyzed reaction in ethanol using a strong acid, which species should we use as the proton source when drawing a mechanism: $\ce{H2SO4}$ or $\ce{CH3CH2OH2+}$?

Note that this is an example of the "leveling effect". The strongest acid that can exist in your solution is the conjugate acid of your solvent, and the strongest base that can exist in your solvent is the conjugate base of your solvent. Anything stronger reacts with your solvent.


Relationship Atmospheric Gases and Ocean Gases

Q1: The equilibrium concentration c of a gas in a solution (typically water) which is present in the atmosphere with partial pressure p dictated by Henry's Law:

$H_{cp} = c/p$

where $H_{cp}$ is Henry's coefficient for that specific gas.

This law states that the concentration of the gas in the solution is directly proportional to its pressure in the atmosphere above the liquid phase and it is a simplification generally applicable to inert, ideal gases.

Corrections must be taken into account when considering the following factors:

Temperature: The enthalpy of dissolution of the gas changes according to van't Hoff's equation. The general result is that the solubility of most gases in water decline with temperature (i.e. solubility is inversely proportional to temperature).
Chemical equilibria: Any transformations (ionization, deprotonation, etc.) that happens to the gas in solution must be taken into account by expressing its concentration by the according equilibrium equation.
Ionic strength: Generally the higher the ionic strength of the solution, the smaller the solubility of the gases in it (Sechenov equation).
Non-ideal solutions: Instead of concentrations, activities of the substances must be considered.

Q2: How much of each gas will dissolve into the water?

The calculation of the concentration for carbon dioxide, for example, is:
$$c = [(0.45\%)\times(\pu{0.986923 atm})]\times[\pu{3.4 \times 10^{-2} M/atm}] = \pu{0.0001509957 M}$$

This calculation is a direct application of Henry's Law, where the first term is the partial pressure and the second is Henry's constant. The same type of calculation goes for the others.

Observation:

It is important to notice that those volume percentages given should already be considering measures taken for those gases in our atmosphere, so they are already taking in account the solubilization. The concentration of each gas in the atmosphere can be calculated by the ideal gas equation, where the number of moles divided by the volume will be the concentration.


Let me list some equations first: $$ w = -p\Delta V,\qquad\Delta U = q + w,\qquad \Delta H = \Delta U + \Delta (pV)$$
where $w$ is work, $U$ is internal energy, $q$ is heat transfer and $H$ is enthalpy

Now, when you are trying to calculate the internal energy, enthalpy or anything else, you should first note what values are constant and what values change through out the reaction. For example, if I burnt a sample of glucose in an open beaker, the volume would obviously change, but the pressure would remain constant. This is because the beaker is open, which means that pressure exerted on the products and reactants is the external atmospheric pressure which is always remains a constant value. Noting the fact that pressure is constant is important as it allows us to simply the equations that I have listed above. $$q = \Delta U - w = \Delta U + p\Delta V$$ $$\Delta H = \Delta U + p\Delta V = q$$
Therefore, by realising that pressure is constant, we are able to state the $\Delta H = q$

Now lets consider the case of burning glucose in a bomb calorimeter. Here is a diagram of one below:

So lets consider what values remain constant and what values change throughout the combustion. As you should be able to see, the bomb calorimeter has a fixed volume (it can't expand or contract), hence $\Delta V = 0$. Since the bomb calorimeter is a closed object, the pressure on the system is the internal pressure, not the atmospheric pressure which is a constant. This internal pressure will change as the reaction proceeds (in this case it will increase as gases are produced from the combustion). Therefore $\Delta p \neq 0$.

Now, using the data from above, we can obtain the following expressions: $$w = 0,\qquad\Delta U = q,\qquad\Delta H = q + V\Delta p$$

So when you do the calculation: $641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K$ you are finding the total heat transfer ($q$) from the system to the surroundings (the bomb calorimeter). Now form the above information we know that the total heat transfer is also equal to change in internal energy, therefore: $$641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K = \Delta U$$

I think you were trying to find the molar internal energy for the reaction, which the calculation will be then: $$\Delta U_\mathrm m = \frac{641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K\times180\ \mathrm{g\ mol^{-1}}}{0.3212\ \mathrm g}$$


The most variant groups of periodic table

Sure Group 11 is a group. And if you look in the right areas you see similarities between copper, silver and gold:

1)  Among the best electrical conductors even compared with other metals (http://eddy-current.com/conductivity-of-metals-sorted-by-resistivity/).

2)  Tend to prefer combining with "soft base" nonmetals:  Cu and Ag found as sulfides, gold is oxidized by $\ce{HNO_3}$ when a chloride source is added.

3)  Have some chemistry in the +1 oxidation state, especially wuth said soft-base nonmetals.


Buffer made of salts of polyprotic acid: reactions

For use in a bacterial culture, you no doubt want a buffer solution with a $\mathrm{pH = 7.20}$, which corresponds to the $pK_{a2}$ for phosphoric acid. At that pH $\ce{[H2PO4-] = [HPO4^2-]}$. The $\ce{K+}$ ion is basically a spectator ion and doesn't influence the pH directly. So typically you'd start with a solution which has equal molarities of $\ce{KH2PO4}$ and $\ce{K2HPO4}$ and then adjust the pH with $\ce{H3PO4}$ and $\ce{KOH}$ using a pH meter to get $7.20$ "exactly."


Source for thermodynamic and/or transport properties

I have a CHEMKIN files with HCL:

HCL                1   344.7       3.339     0.000     0.000     1.000  !(singh)

344.7 should be $\varepsilon/k$ in K and 3.339 should be $\sigma$ in Angstroms.

! Species: HCL              CAS Number: 7647-01-0
! Name:    Hydrogen Chloride
! Source:  SNLL fit to 1986 JANAF tables, 4/21/89
! H0(298K) =      -22.0600 (Kcal/mole),  S0(298K) =       44.6500 (cal/mole-K)
HCL                42189CL  1H   1          G   300.000  5000.000 1000.00      1
 2.75533500e+00 1.47358100e-03-4.97125400e-07 8.10865800e-11-5.07206300e-15    2
-1.19180600e+04 6.51511600e+00 3.33853400e+00 1.26820700e-03-3.66691700e-06    3
 4.70399200e-09-1.83601100e-12-1.21315100e+04 3.19355500e+00                   4

Also, Poling [1] has some data on HCl, Cl$_2$, ClNO. You could estimate the other properties using group contributions to define critical properties and then corresponding state principle methods (like Chung's [2]) to get your temperature/pressure dependencies. You can then make your own curvefit for your conditions of interest. Not as good as curvefits from experimental data but it should get you in the ballpark...

[1] B. E. Poling, J. M. Prausnitz, and J. P. O’Connell. The Properties of Gases and Liquids, Fifth Edition. McGraw-Hill, 2001.

[2] T. H. Chung, M. Ajlan, L. L. Lee, and K. E. Starling. Generalized multiparameter corresponding state correlation for polyatomic, polar fluid transport properties. 27:671–679, 1988.


Does this activation of carbonyls usually involve a SN1 or a SN2?

It is not an $\ce{S_{N}1}$ or an $\ce{S_{N}2}$ reaction.  It is commonly referred to as an "addition-elimination" reaction.  You've drawn the key piece of evidence in your diagram - the tetrahedral intermediate.  Not an $\ce{S_{N}1}$ reaction because there is no carbonium ion formed in this reaction.  This is evidenced by many facts such as the lack of carbonium ion type rearrangements when substituents are present alpha to the carbonyl.  Nor is it an $\ce{S_{N}2}$ reaction, the attacking group does not start to bond and the leaving group does not start to leave and break its bond simultaneously.  Instead there is a real tetrahedral intermediate that exists in equilibrium with the reactant as well as the product.  You "add" a nucleophile to the protonated carbonyl compound to form the tetrahedral intermediate, then you "eliminate" one the substituents in the tetrahedral intermediate and either go back to reactant or forward to product.

Here's another, more specific drawing for the case at hand.


Interpreting a Formula Tattoo (Bipolar?)

This is lamotrigine, a sodium channel blocker prescribed against epilepsy.  Apparently it's also prescribed for bipolar disorder.


For a benzene shown in a skeletal structure, what does a substituent to the center of the ring mean?

Here the bond connecting the substituent with the center of the ring suggests that the substituent may be located on any of the three positions (ortho-, meta-, para-). From Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), section GR-9.1 Small substituents [1, p. 393] (emphasis mine):

  Parent structures with a substituent that is attached at one of several ring atoms may be depicted by drawing that substituent attached to a bond that extends into the center of a ring. In printed works, it is assumed that the substituent may then be attached to any of the ring atoms as is allowed by normal bonding rules. When working in electronic formats, it may be possible to specify that the substituent
  can be attached to certain of the ring atoms and not to others, depending on the capabilities of the software being used. If more than one substituent may be independently connected to various ring atoms, the bonds attached to those substituents should both extend toward the center of the ring, but should not
  themselves be connected to each other.

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277. (Free Access)


Does Potassium Superoxide also absorb Carbon Monoxide?

As the other answer and comments imply, potassium superoxide is not a good reagent for absorbing carbon monoxide.  Cuprous chloride solution works better, as in this study[1] where the salt is dissolved in an ammonia-bearing solution.
Reference:

R. V. Gholap and R. V. Chaudhari, "Absorption of carbon monoxide with reversible reaction in cuprous chloride solutions", Ind. Eng. Chem. Res. 27, 11, 2105–2110 (1988).


Inverse electron demand reactions are inefficient and slow. Even the prototypical Diels-Alder reaction of 1,3-butadiene and ethylene is quite sluggish. From here you can see the reaction only proceeds to 93.5% yield when ran with ethylene in excess by 50:1 at 200ºC and 350 atm for 4 hours. Yikes!
Now consider that some normal electron demand D-A reactions can occur at room temperature. In fact, 1,3-butadiene spontaneously dimerizes to 4-vinylcyclohexene while in storage, meaning the dimer must be cracked by heating immediately before use. This should give you an idea of the sensitivity of D-A reactions to substituent groups. By lowering the HOMO-LUMO gap, orbital overlap is improved, lowering the activation energy and resulting in a faster reaction.
To supplement and support my answer, here is a paraphrased excerpt from Modern Physical Organic Chemistry by Anslyn and Dougherty (2006):

Electron-withdrawing groups on the dienophile lower its LUMO. This occurs because electronegative substituents have correspondingly lower energy MOs, and lower the energies of all MOs in which they participate. Inversely, electron-donating groups raise the diene MOs.

In support of this analysis, the log($k$) for Diels-Alder reactions correlate quite well with the inverse of the difference in energy between the ionization potential of the diene (related to the energy of the HOMO) and the electron affinity of the dienophile (related to the energy of the LUMO). To further enhance the electrophilicity of the dienophile, it is common to add Lewis acids that can complex the electron withdrawing groups on the dienophile, further lowering the LUMO.


Determine the position of an equilibrium process

You are right, you can use Le Chatelier's principle to predict the shift of the equilibrium when temperature or pressure is in- or decreased. It is stated that the reaction is exothermic, and this tells you on which side of the reaction equation the heat goes. And as both products and educts include gases, it would also be easy to predict how the equilibrium shifts when pressure is varied. I presume that the question, in its current form, is asking for the shifts that happen in all possible combinations of in- or decreased pressure or temperature.


why HI isn't fully miscible with cold water when HF is?

Hydrogen iodide is not listed as miscible with water because it is a gas. Miscibility is a property generally ascribed to pairs of liquids. Hydrogen fluoride is a liquid, and so it can be described as miscible with water. HF is a weak acid, but its high solubility can be explained by the high polarity of the H-F bond and the ability of HF to be both a hydrogen bond donor and accepter.

Hydrogen iodide is a gas at room temperature. It's boiling point is $-35\ ^\circ\text{C}$. Just because it is not miscible does not mean that it is not highly soluble. The solubility of HI in water is 245 g per 100 mL. In other words, a saturated solution has more HI by mass than water. HI is a strong acid, and the solubility of the gas is due to the nearly complete ionization of HI. Since HI is a gas, I presume that the solubility limit occurs when the vapor pressure of HI above the solution is high enough that vaporization begins to compete with ionization.

Physical data on HI from Wikipedia