Most stable resonating structure of 1-nitro-4-nitrosobenzene

The answer is not that difficult. You should have a certain number of rules to follow in determining the most stable resonance (not "resonating") structure.

It probably starts with check all for octets (which you correctly did). 1 has ten electrons on the nitroso nitrogen, so that rules it out.

It probably then goes on to ask you to check for the least number of formal charges (which, judging by your description, you probably did).

The next step should probably say something about the negative charges being on the most electronegative elements. Electronegative atoms like O, F... (and N to a slightly lesser extent) are quite happy to take on a negative charge whereas carbon will not quite be so pleased.

Here, resonance forms 2 and 4 have the negative charge on carbon, whereas structure 3 has it on oxygen. 3 is therefore the most stable resonance structure.

Incidentally, all of the resonance forms 2 through 4 have complete conjugation throughout the molecule so that's not a factor. The $\mathrm{sp^2}$-hybridised anionic carbon can take part in conjugation equally well. Conjugation does not require alternating single and double bonds - it only requires an extended, continuous, $\pi$ system. Generally this means that all atoms along the chain must have unhybridised p orbitals parallel to each other.

Side note: "Most stable" resonance structure is a slightly sloppy description. None of these resonance structures actually exist and the true structure is a linear combination of all of these resonance structures. It should more properly be called the "greatest resonance contributor" or "greatest contributor to the resonance hybrid".


Why cobalt reacts favorably with nitrogen?

According to the recent first principles ab initio DFT calculations done with VASP of structural, electronic and mechanical properties of 3d transition metals mononitrides $\ce{MN}$ carried out by Liu et al. [1], there is a local maximum of cohesion energy $E_\mathrm{coh}$ at cobalt for all possible/discovered structure types:

Ultra-soft Vanderbilt pseudo-potentials (US-PP) … as supplied by Kresse and Hafner …, with local density approximation (LDA) … and PW91 general gradient approximation (GGA) … were both used. It is known that LDA underestimates the lattice constant and overestimates the elastic constants and cohesive energy, while PW91 GGA does the opposite for the lattice constant and elastic constants by a smaller amount and yields more accurate values of cohesive energy.

Figure 1.
Cohesive energy per atom of the nitrides $(\ce{MN})$ versus their corresponding 3d transition metals $(\ce{M}).$
The three structures are zincblende (zb, black square), rocksalt (rs, red circle) and
cesium chloride (cc, blue triangle).
Experimental values for the rocksalt structure (green ×) … are also presented.

Since we are discussing synthesis from the elements, it's especially handy to use cohesion energy since for solids it signifies the energy required to disassemble the structure into an array of neutral free atoms.
As a rule of thumb, the higher the atomic cohesion energy is, the more stable the solid is.
Numerical values of $E_\mathrm{coh}^\mathrm{GGA}$ for cobalt nitrides are always exceeding the ones for iron, nickel, copper and zinc, and also $\ce{CoN}$ demonstrates mechanical stability across all structure types:

There is still a lot of controversy between the studies of preferred phase and magnetism [2]; I'm not sure though it is currently possible to interpret/rationalize or let alone predict this behavior by other, simplified means.
References

Liu, Z. T. Y.; Zhou, X.; Khare, S. V.; Gall, D. Structural, Mechanical and Electronic Properties of 3d Transition Metal Nitrides in Cubic Zincblende, Rocksalt and Cesium Chloride Structures: A First-Principles Investigation. J. Phys.: Condens. Matter 2013, 26 (2), 025404. DOI: 10.1088/0953-8984/26/2/025404.
Soni, H. R.; Mankad, V.; Gupta, S. K.; Jha, P. K. A First Principles Calculations of Structural, Electronic, Magnetic and Dynamical Properties of Mononitrides $\ce{FeN}$ and $\ce{CoN}.$ Journal of Alloys and Compounds 2012, 522, 106–113. DOI: 10.1016/j.jallcom.2012.01.100.


How many carbons in this skeletal structure?

Your intuition is not wrong.

There are indeed 6 carbon atoms in the molecule (3-methyl-pentane).

With bondline/skeletal notation, there's a carbon atom at all the vertices; at the ends, and every time the line changes direction.

Counting the carbon atoms then becomes as simple as counting the vertices.

This becomes more apparent if you explicitly write all the atoms like illustrated below:


Understanding what is happening when something is dissovled

Your question seems to be exclusively about salts dissolving to form ions in water. The answer is fairly trivial: a saturated solution of a salt in water is not water.

At (thermodynamic) equilibrium, the solubility depends on the net free energy of the system. When the interactions in solution no longer energetically justify charge separation of the ionic solid into aqueous ions, then you've got equilibrium, right?

So, consider a solution as being (in a very crude and simplified way) Solvent-M1 interactions, Solvent-X1 interactions, solvent-solvent interactions, and M1(aq)-X1(aq) interactions as well as M1(aq)-M1(aq) and X1(aq)-X1(aq) interactions. Which of those will change when you add M2X2 ions to it? Well, at low [M2X2] none of those will change significantly, but you'll add the interactions between M2 and X2 and ALL of the species present. (as already explained, silver nitrate being added to NaCl is a very poor choice, AgCl2 is quite insoluble)

At high enough concentrations you'll begin to significantly change the energetics. In other words, adding M2X2 to a solution of M1X1 changes the interactions of all species present, so you would expect that the mixed salt solution will have different equilibrium concentrations for both M1X1 and M2X2.

If you were to carefully choose highly soluble salts, I think what would be surprising is how well their solubility was predicted by just using each independent Ksp, rather than the (expected) deviations from those predictions. Note that nothing I've said here should be taken to mean that adding a different salt to a saturated solution will decrease the solute concentration of the first salt. (Although it is the most likely scenario, just as the common ion effect would suggest. In some cases solubility will increase!) The reality is that highly concentrated solutions are far more idiosyncratic and generalizations (especially "laws") are much harder to come by.

While there probably is justification to claim that water is dissolved in sodium nitrate (at equilibrium), I've never found such foolishness useful. And I've been around for a long while. (ultra) Pure water does not freeze at 0°C, negative absolute temperatures (negative Kelvin) are attainable, and a bunch of other stuff which while true, doesn't really aid understanding (for the novice) probably should be ignored until one wants to deep dive into a subject.


Confusion on when to use mass of entire solution when performing q=mc∆T

In mcΔT, you must use the total mass of the system that is changing temperature. In this case, that is the solution.

The same thing is occurring in the second problem. The temperature of the solution changes, and therefore the solution mass is the mass you should use.

It wouldn't make sense if the solute changes temperature but the solution it is dissolved in doesn't change temperature, does it? I find it always helps me to think about the physical process that is happening for questions like this. Always identify what your system of interest is, and the initial and final states of the system. Doing so can clear up a lot of these problems, and will also help you get through more difficult problems if you ever take a thermodynamics course.


What happens when a super-critical fluid is compressed?

When you heat up a liquid at constant volume (leaving sufficient space for the gas phase), the density of the liquid will decrease and the intermolecular interactions will weaken. Some of the liquid will transition to vapor, so the vapor above the liquid will get denser and the frequency of intermolecular interactions will increase (it behaves less and less as an ideal gas as the number of collisions increases with increasing density). At the point where the density of the liquid and the gas are the same, there is no more phase boundary (with gravity, we picture liquid on the bottom and gas on the top, but if the density is identical, that is no longer the case).

This is all pretty outlandish for most of us because we are used to constant pressure, not constant volume. Here are two videos showing the process:

Liquid chlorine: Here, it looks like the liquid level stays constant - the expansion of the liquid and the evaporation balance each other out. If you look carefully, though, you can see how the vapor phase color gets more intense, indicating more molecules in the vapor phase. At the same time, the color of the liquid gets less intense, indicating a decrease in density.
https://chem.libretexts.org/Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/10%3A_Solids%2C_Liquids_and_Solutions/10.13%3A_Critical_Temperature_and_Pressure

Carbon dioxide: When approaching the critical point, the system gets turbid because the difference in density between liquid and gas is no longer sufficient to keep the liquid down and the vapor up. The system is a heterogeneous mixture of liquid and vapor for a while, and then a single pure phase.

https://www.youtube.com/watch?v=GEr3NxsPTOA

Making of video: https://www.youtube.com/watch?v=-gCTKteN5Y4


https://en.wikipedia.org/wiki/Potassium_nitrate gives a melting point of $\pu{334°C}%edit$ without decomposition, but then decomposition sets in at $\pu{400°C}$ so you have a narrow liquid range.

If you avoid decomposition the oxidizing property still works.


How did Michael Faraday melt Sodium Chloride?

The answer is given in Faraday’s own work, Experimental Researches in Electricity, Volume 1, freely available thanks to The Project Gutenberg:

  399. Other substances, which could not be melted on glass, were fused by the lamp and blowpipe on platina connected with one pole of the battery, and then a wire, connected with the other, dipped into them. In this way chloride of sodium, sulphate of soda, protoxide of lead, mixed carbonates of potash and soda, &c. &c., exhibited exactly the same phenomena as those already described: whilst liquid, they conducted and were decomposed; whilst solid, though very hot, they insulated the battery current even when four troughs were used.

The text above shows that the lamp in question is the spirit lamp, and the whole aparatus is described here:

The crux of it is that an open flame might not have enough heat, so a conventional flame is simply contained into a more restricted space.


Is the SN1 reaction faster with an axial or equatorial leaving group?

In compund 2, both substituents can be placed in equatorial positions, whereas in 1 the $\ce{Cl}$ group is forced into an axial position since the bulky t-butyl group has to be placed equatorial. This makes compound 1 more unstable (higher energy), leading to a faster rate of formation of the carbocation.


Which reaction occurs first when 5-hexen-1-ol reacts with sulfuric acid?

In general, intramolecular reactions occur faster than their intermolecular counterparts. Most commonly this is rationalized by an entropic argument: in the case of the intermolecular reaction, two molecules (substrate + water) combine to form one (diol). This is entropically disfavored relative to the intramolecular case, in which the molecule cyclizes.

However, the biggest factor at play in determining which will occur in this case is the size of the resulting ring. If the product contains a 5, 6, or 7-membered ring, these rings are considered be unstrained (bond angles ~ 109.5 degrees). If the rings are smaller than 5 atoms or larger than 8, the rule of thumb is that the thermodynamic (enthalpic) cost of the ring strain associated with cyclization overcomes the entropic driving force to cyclize, and the intermolecular reaction is preferred.


Finding partial pressure of a gas in a mixture

Your intial step is correct, finding the number of moles of argon in the 1.1 vessel. The next step however is incorrect. Because the pressure and temperature are both in different conditions for argon in the end and argon and carbon dioxide together in the beginning, you cnnot simply subtract the volume 1.72 - 1.1 L and then find the number of moles of carbon dioxide there. Remember, subtracting the volume and finding the moles of co2 would only work if the conditions are the same, but in this case, the pressures and temepratures are different. Thus, you can just simply go about finding the total n moles of Argon and carbon dioxide through PV=nrt.

So, 1.72 (1.72L) / 0.08206 x (273 + 17) = 0.1243 total moles. Subtract 0.06869 moles from earlier from 0.1243. You will get 0.055356 moles of CO2. Now, you can use your partial pressure method.

Hope this helps.


Would I use PBr3 or HBr in this synthesis?

As @permeakra has said neither reagent will get you the stereochemistry you want in the product. The HBr route goes through a carbocation which scrambles the stereochemistry completely.

PBr3 will invert the stereochemistry, then SN2 -CN addition will invert it again, giving you the wrong product.

You need something that turns the -OH group into a leaving group while preserving the stereochemistry, so that the CN can displace it by a SN2 process. This will invert the stereochemistry giving you the product, after hydrolysis, you want.

I suggest forming a sulfonate, either tosylate, triflate or (my choice) mesylate. This should be displaced smoothly by cyanide under the DMF conditions you refer to.


How does physical adsorption of mixed gas occur on solid? (eg I mean how the gas molecules of compete each other for a slot on solid surface)

Your supposition that gases do not have equal access to teh surface does not seem to me to be correct.

At the surface  physisorption and chemisorption can occur. The former is due to inter-molecular interaction with the gas molecule and the surface molecules so clearly will depend on the nature of both of them. In chemisorption a chemical bond is formed. The physisorption occurs at longer range and has a smaller potential well (as measured by potential energy vs distance from surface) than chemisorption. (See Atkins & DePaula 'Physical Chemistry' for a figure.)

The gases approach the surface at random and the number of collisions of each will be in proportion to their partial pressure. What happens on collision depends, as mentioned above, on the strength of the intermolecular interaction of gas and surface and so depends on what they are. At higher temperature the collision energy is greater than at lower ones and so one would expect less physisorption as the collision energy will be greater than that of most available physisorption sites.

It is also possible for multiple layers to be formed (as in BET isotherm) and for a physisorbed molecule to diffuse around on the surface until it leaves again (by random thermal activation) of finds a more stable site or becomes chemisorbed.

As to your particular case we can only surmise but $\ce{CO2}$ is far more polarisable than methane so should have the larger dispersion forces so larger induced dipole - induced dipole intermolecular interaction and so preferentially occupy the surface. Both gases have transient dipoles due to molecular vibrations of which the $\ce{CO2}$ should be the larger. This may also be important.


How do I convert salinity measured in mg/L Cl into parts per thousand as NaCl?


Solubility of methanol and ethanol in water

For primary alcohols, the trend is the longer the chain, the less soluble.

However, at room temperature, methanol, ethanol and propanol are all completely miscible with water.

At lower temperature, below the freezing point of water, there is finite solubility.

See the phase diagram that is Fig. 1 in CRYSTAL GROWTH OF WATER ETHANOL MIXTURE and that in Fig. 1 of Solid-Liquid Phase Diagram of the System Methanol-Water

Which alcohol would you choose for windshield deicing solution?


Finding the charge passed in electrolysis of water using volume of hydrogen released

$PV = nRT$ and $\ce{2H+ + 2e- -> H2}$ are your best friends.
Step 1: Find the number of moles of hydrogen gas released
Use the formula: $PV = nRT$, where $P =$ Pressure exerted by the
gas, $V =$ Volume of gas, $n = $ number of moles of gas, $R = $ universal gas constant, $T = $ temperature of gas.
Using this formula with $P$ in atm and $V$ in liters we get:
$$n =\frac{PV}{RT} = \pu{0.0044 mol}$$
Step 2: Find the number of farads of charge used
We know that $$\ce{2H+ + 2e- -> H2}$$
This means that for every mole of hydrogen released, $\pu{2 moles}$ of electrons are used. A farad is the charge on $\pu{1 mol}$ of electrons and so number of farads of would be two times the number of moles of electrons. So charge would be $\pu{0.0088 F}$.
Step 3: One farad = 96500 coulombs
Therefore, $\pu{0.0088 F = 850.32 C}$ and that should be the answer.


Your description of hyperconjugation is a very minimalistic one. More generally, hyperconjugation should be seen as the interaction of occupied bonding MOs with unoccupied ones — these unoccupied ones can be both antibonding MOs or empty (cationic) p atomic orbitals.

In principle, $\ce{Si-H}$ or $\ce{Si-Si}$ bonds would also be able to donate electron density in a hyperconjugative matter. In pracitce, due to silicon’s low electronegativity and greater size these orbitals simply do not overlap as well and their energy levels are such that less stabilisation may be obtainable.

Carbon-silicon bonds are of course great at providing hyperconjugation since silicon — much like hydrogen — has a lower electronegativity than carbon, meaning the occupied orbitals have a greater contribution on carbon.


Is the proper formula for phosphorus oxide(V) P4O10 or P2O5?

The simplest types of chemical formulas are called empirical formulas, which use only letters and numbers indicating atomic proportional ratios (the numerical proportions of atoms of one type to those of other types). Molecular formulas indicate the simple numbers of each type of atom in a molecule of a molecular substance, and are thus sometimes the same as empirical formulas (for molecules that only have one atom of a particular type), and at other times require larger numbers than do empirical formulas. (Wikipedia)

For phosphorus pentoxide the molecule seems to be an adamantane-like compound with 4 phosphorus atoms.

So the correct molecular formula of the compound would be $\ce{P4O10}$.


Using combined gas law to calculate the volume of dry hydrogen gas

Consider the composition of the gas in your eudiometer.  Most of it is hydrogen, but some of it is water vapour.  In other words, not all the pressure is exerted by hydrogen, part of it is due to water.  You could ask yourself: what would happen to the volume of gas produced, had you passed it through a drying tube.

You should have been given vapour pressure of water in a table.  You should also have taken down value of air pressure and ambient temperature.


Do minerals and/or metals found in water tend to settle to the bottom?

I have experimented with this phenomenon and found that allowing water to sit for a day may have a non-trivial effect on the TDS gradient in the water. This effect could be due to various reasons, however.

Materials: Glass pitcher, plastic & porcelain & glass containers; tap water from near Chicago, IL, USA; TDS meter

Methods: I poured tap water into various containers and allowed the water to stand for about 30 hours. I then measured the TDS at the top of the water. Then I stirred the water vigorously and let the water sit for about a minute. I took a TDS reading again at the top of the water. I repeated the experiment a total of 4 times.

Results (Updated 12/23/2016):
$$\begin{array}{ccc}
 & \text{TDS Before Stirring} & \text{TDS After Stirring} \\ \hline
\text{Glass} & 138 & 141 \\
\text{Plastic} & 150 & 146 \\
\text{Porcelain} & 153 & 152 \\
\text{Glass} & 147 & 149 \\
\end{array}$$

Conclusion: The results are interesting. I will discuss a few possible reasons behind the results, but I'm not certain.

TDS (after) was higher both times for glass. Maybe this is because when the water sits, a small number of ions adsorb to the glass surface.
For plastic and porcelain, TDA (after) was lower for both plastic and porcelain. This may be due to dust that settled and dissolved into the water, as I did not cover the containers (looking back, covering the containers would probably have made the experiment more controlled). In addition, maybe ion adsorption happens less with porcelain and plastic.


How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂?

As noted in the comments, to attain mercuric iodide, you need to maintain a balance between the concentration of $\ce{Hg^2+}$ and iodide ion. If there is slight excess of iodide ion, it will form a complex (much like $\ce{KI3}$ from $\ce{KI}$). The complex is popularly known as Nessler's reagent and is prepared pretty much in this way.

Another solution is to decompose the complex to revert back into original products. But I won't recommend this solution. Though, it is expected to give the original reactants, occasionally, decomposition reactions give rise to undesired side products and may release toxic fumes of mercury, iodine, iodide, potassium oxide($\ce{K2O)}$ etc. Do not try to heat the complex. Instead opt for the 1st solution.


Electron configurations for d-block and p-block

I did not completely understand you question here, but I am gonna give it a try, so maybe I might answer your question.

I do not think that it matters how you write them, but its just depends on how you fill them. The order for filling orbitals is as follows:
$$\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s}$$

In high school, I just used to write the $(n+1)s$ before the $nd$. Because with some exceptions, the rule says that you will always fill the $(n+1)s$ orbital before you fill the $nd$ orbital. Meaning that you fill $\ce{4s}$ before you fill $\ce{3d}$, $\ce{5s}$ before you fill $\ce{4d}$, and so on.

The reason is that the $\ce{s}$ actually has a lower energy than the $\ce{d}$ orbital. You wrote that wrong in your question. That is the only you fill the s before the d, and when you make ions, you will have to remove the electrons from s before you do from $\ce{d}$.

But like I said, you can write them like you want as long as you have filled them correctly in the right order. Likewise once you have completely filled your $ns$, only then you can move to fill the $(n-1)d$ orbital and then again proceed to fill the $\ce{np}$ orbital. Its just how we fill them.

In some books, you will find that they write the s as the outermost but some write d as the outermost. Frankly, in my book and myself have learned to write $\ce{d}$ at the end. But I do not think that writing it matters.

For example, take a look at this site. They write $s$ orbital at the end of the configuration and also note the energy diagram. But in the following Wikipedia website, they have written the $\ce{d}$ at the end instead of $\ce{s}$. So basically, the writing of the orbitals doesnt matter, but the order of filling matters.


Can precipitation occur in states of matter other than liquids?

As stated in my previous answer on precipitation, formation of precipitates in supercritical is a common issue. Ask those who do sub/supercritical fluid chromatography. Sometimes you inject a methanol soluble compound into a chromatography column and the mobile phase is supercritical carbon dioxide (often mixed with a small amount of methanol) and the compound precipitates on the head of the column. So there is no restriction on using the term precipitate in fluids other than liquids.
Wikipedia also talks about precipitation in solids as

In solids, precipitation occurs if the concentration of one solid is
above the solubility limit in the host solid, due to e.g. rapid
quenching or ion implantation, and the temperature is high enough that
diffusion can lead to segregation into precipitates. Precipitation in
solids is routinely used to synthesize nanoclusters

The issue of calling a mixture of gases as a solution or not, it purely a matter of semantics. There is no rule or law which restricts that the solvent and the solute must have (strong or a certain type of) interaction in order to be labelled as a solution. Even gases interact with other gas phase molecules. Microwave spectroscopists know that very well. Many authors use the term gaseous solutions. Even one IUPAC's terminology related document uses this term. Heavens don't asunder apart if a trivial gas mixture is termed as a solution. Keep in mind that if you look up IUPAC definition of solvent, it does not include the usage of solution for gases. Only solids and liquid solutions are defined. So it is better to follow IUPAC's nomenclature.
In short, keep the meteorological precipitation away from chemistry.


Why does higher temperature affect the rate of electrolysis?

Rate of electrolysis is only dependent on the current (Faraday's law) passing through the solution. Remember that in an electrochemical experiment you cannot independently vary voltage or current. One factor has to be fixed. These two modes are called constant current electrolysis or constant potential electrolysis.
Once you have fixed voltage, you can ask you yourself, what factors affect current rather than electrolysis. The key factor is solution's resistance or better conductance (reciprocal of resistance). Conductance increases with temperature.


How to calculate the enthalpy change from the reaction of silver nitrate and sodium iodide?

When the reaction takes place and releases its heat, everything in the calorimeter  heats up: the water, the $\ce{AgI}$ and the $\ce{NaNO3}$. But of these the water is by far the most massive and for that reason, as a reasonable approximation, we consider that mass that absorbs the heat to be the water, i.e. $40.00\ \mathrm{g}$.

Use this to calculate $Q$.

Then determine how many moles of $\ce{AgI}$  was formed, by stoichiometry. The quotient of these two numbers gives you the reaction enthalpy per mole of $\ce{AgI}$


Solution to Raoult's Law for a volatile binary solution (no pun intended)

As you correctly pointed out, you can connect the moles of each compound in the gas phase, $a$ and $b$, to the composition of the liquid phase, $A-a$ and $B-b$, through their partial pressures, $p_A$ and $p_B$:

$p_A=a\cfrac{RT}{V}=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}$

$p_B=b\cfrac{RT}{V}=p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$

And, as you did, we can use the total number of moles in the gas phase to rewrite these equations in terms of $p_e$. Since

$p_e=p_A+p_B=\left(a+b\right)\cfrac{RT}{V}$

and therefore

$\cfrac{RT}{V}=\cfrac{p_e}{a+b}$

we can write

$p_e\cfrac{a}{a+b}=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}$

$p_e\cfrac{b}{a+b}=p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$

Note that this is a two-equation system with two unknowns, so, barring incompatible values of the constants $A$, $B$, $p_e$, $p_A^o$ and $p_B^o$, it's determined (as you essentially pointed out).

Let's find a relationship between $a$ and $b$. A straightforward way to do that is by dividing the equations:

$\cfrac{p_e\cfrac{a}{a+b}}{p_e\cfrac{b}{a+b}}=\cfrac{p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}}{p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}}$

$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}$

This expression is key; we will return to it later.

Clearing $a$:

$a=\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}\left(A-a\right)$

$a\left(1+\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}\right)=\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}A$

$a=\cfrac{\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}}{1+\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}}A=\cfrac{b}{\cfrac{p_B^o}{p_A^o}\left(B-b\right)+b}A$

We haven't made any additional assumptions in the derivation (beyond Raoult's law itself), so this is valid for all the situations where Raoult holds. Let's interpret the numbers to see what they mean.

With this expression you can check out various situations to get a feeling of how does Raoult behave. Some limit cases:

· If $p_A^o \approx p_B^o$, $\cfrac{a}{A} \approx \cfrac{b}{B}$ and both components are distributed to the same extent between the gas and liquid phase.

· If $p_A^o \gg p_B^o$, the vapour pressure quotient $p_B^o/p_A^o \ll 1$ and therefore $a \approx A$ (the more volatile component will be almost entirely in the gas phase).

· And, contrarily, if $p_A^o \ll p_B^o$, $a \approx 0$ (the less volatile component will be almost entirely in the liquid phase).

Let's return to the key expression above:

$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}$

If you ponder carefully what this means, it's telling you that for ideal mixtures (Raoult mixtures), the relationship between the components in the gas and liquid phases is always fixed as long as there is equilibrium, and the proportionality factor of that relationship is the vapour pressure ratio of the two components. So, if you have $p_A^o=2p_B^o$, the ratio of moles of A and B in the gas will always be twice the ratio of moles of A and B in the liquid. Or, equivalently, Raoult predicts that the gas phase is enriched in the most volatile component, precisely in proportion to the relative vapour pressures. The more volatile a component is, the more "additional fraction" of the gas phase it represents.

So, what values for $p_e$ are compatible with Raoult? Let's find out:

$p_e=p_A+p_B=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}+p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$

$p_e\left(\left(A-a\right)+\left(B-b\right)\right)=p_A^o\left(A-a\right)+p_B^o\left(B-b\right)$

$\left(A-a\right)\left(p_e-p_A^o\right)=\left(B-b\right)\left(p_B^o-p_e\right)$

$\cfrac{A-a}{B-b}=\cfrac{p_B^o-p_e}{p_e-p_A^o}$

Note that this is a particular case of the lever rule and, if all the quantities are positive, it demands that $p_A^o \leq p_e \leq p_B^o$ or $p_B^o \leq p_e \leq p_A^o$. A value of $p_e$ outside that range would correspond to a value of $a$ or $b$ outside of the range $0 \leq a \leq A$ or $0 \leq b \leq B$ - which doesn't make physical sense.

But is any value of $p_e$ between $p_A^o$ and $p_B^o$ valid? No, it depends on $A$ and $B$. Every value of $p_e$ corresponds to a certain composition - so you can't set both to arbitrary values.

Take for example $p_A^o=100$, $p_B^o=200$, $p_e=125$, $A=1$ and $B=4$. Applying the lever rule expression we just derived,

$\cfrac{A-a}{B-b}=\cfrac{p_B^o-p_e}{p_e-p_Ao}=\cfrac{200-125}{125-100}=3$

So, the ratio of moles of A to moles of B in the liquid in equilibrium is 3. In other words, 3/4 of the liquid molecules are of A.

Let's introduce this value in the key expression we derived earlier:

$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}=\cfrac{100}{200}3=1.5$

So, the ratio of moles of A to moles of B in the gas in equilibrium is 1.5. In other words, 3/5 of the gas molecules are of A.

All of this is compatible with Raoult's law... but it is incompatible with a total system composition of 1 mole of A and 4 moles of B!

Hope this helps.


The update of your question changes everything!

In Germany, around 25% of the base load is generated in lignite-fired power plants.
Lignite (brown coal) is awfully wet (50% water) and it is definitely a good idea to apply drying.

In the power plants, powdered lignite is used. Traditionally, the milled lignite is dried using the hot exhaust gas at temperatures up to 1000 °C, but modern plant like the one in Niederaußem operated the fluid bed drying at temperatures as low as 110 °C.

In summary, pre-drying of lignite is a good idea, if not essential, but it open to question how efficient the drying process of stacked lignite is.


Total number of stereoisomers of cyclobutane-1,2,3,4-tetracarboxylic acid

You can view 1,2,3,4-tetrasomethingcyclobutane as a square with black and/or white corners (black representing e.g. up, then white representing down configuration of the substituent). Two squares represent the same object when after applying some of the following operations on the first you get the second one:

rotation in the paper plane (of course)
mirroring AND color inversion

By drawing all 24 = 16 possible squares and using these two rules, you quickly find that they represent only 4 different objects, as seen on the following sketch.

Note that the second "controversial" operation mirroring AND color inversion is in our scenario, because of D2 symmetry of all squares, equivalent with more proper, "mirror-free" operation - rotation around some in-plane axis, considering that obverse of the square has inverse colors. (Because of the symmetry, the operation can even be reduced to color inversion only.)


Is is possible to add NO2X, AlX3 to achieve nitration of benzene?

Nitration using nitryl chloride($\ce{NO2Cl}$) was studied extensively by Price and Sears where they found $\ce{AlCl3}$ to be a suitable reagent for the reaction. They also noted that nitration was difficult for deactivating aromatics such as halobenzene and benzotrifluoride and hence this method was considered of limited value.
On further experimentation, it was found that $\ce{TiCl4}$ was the best suitable catalyst. Other reagents used were $\ce{FeCl3,ZrCl4,AlCl3}$ and $\ce{AlBr3}$ but the reaction conditions were difficult to control. $\ce{BCl3}$ gave smaller yield with possible ring chlorination and $\ce{SbCl5}$ were effective for only aromatic compounds. The following compounds were nitrated using nitryl chloride and $\ce{TiCl4}$ as catalyst and data was noted:
\begin{array}{c|c}
\mathbf{Compound} & \mathbf{Yield (percent)} \\\hline
\text{benzene} & \text{88%}\\
\text{toluene} & \text{81.5%} \\
\text{ethylbenzene} & \text{79%}\\
\text{fluorobenzene} & \text{91%}\\
\text{chlorobenzene} & \text{89%}\\
\text{bromobenzene} & \text{86.5%}\\
\text{o-dichlorobenzene} & \text{41.5%}\\
\text{benzotrifluoride} & \text{32%}\end{array}
An excess of solvent was used for the reaction. Tetramethyl sulfone was suitable for the reaction. Other solvent experimented with was carbon tetrachloride but many chlorinated byproducts were observed. The possible mechanism for the reaction (with $\ce{AlCl3}$ as catalyst) was noted:

or a possible salt formation mechanism was indicated:
$$\ce{NOCl2 + AlCl3 <=> NO2+AlCl4-}$$
Nitration using Nitryl chloride was once important in making chloronitrosoalkane where carbon-carbon double/triple bond were broken by nitryl chloride and an free radical addition reaction occured but the yield was not that high. Dinitromethane was used as solvent as nitryl chloride breaked into chlorine and nitrous oxide:
$$\ce{NO2Cl -> NO2 + Cl.}$$
$$\ce{CH2=CH2 ->[NO2Cl] Cl-CH2-CH2-NO2(45 {%}) + Cl-CH2-CH2-ONO(35 {%}) + CH2Cl2 + N2O4}$$
$$\ce{C2H5-C#C-C2H5 + NO2Cl -> C2H5-ClC=CNO2-C2H5}$$
As for nitryl bromide, it was found to be unstable. Zuskin et. al. claimed to have synthesized the product$\ce{^{[4]}}$ but were invalidated due to their instability. Many attempts$\ce{^{[5]}}$ of making nitryl bromide were made but failed as they decomposed to nitric oxide and bromine. So, nitryl bromide was observed as a combined entity of (nitryl chloride + $\ce{KBr}$ in sulfur dioxide) at $\pu{-20 ^{\circ}C}$ and using $\ce{TiBr4}$ as catalyst, nitration was done but yields were low(lower than that observed with nitryl chloride) with possible ring-brominated side products. This is due to interference of side products from partly decomposed nitryl bromide.
Notes and References

Across Conventional Lines: Selected Papers by George A Olah, G K Surya Prakash

Chemistry of Energetic Materials by George A. Olah, David R. Squire

Nitrate Esters Chemistry and Technology by Jiping Liu

Bromine vapor saturated with nitrogen peroxide passed through a tube filled with bone-ash at $\pu{200-250^{\circ}C}$.

Halogen exchange of nitryl chloride with $\ce{KBr}$ in liquid $\ce{SO2}$, reaction of $\ce{BBr3}$ with anhydrous niyric acid or dinitrgen pentoxide, reaction of bromosulfonic acid with nitric acid and ozonolysis of nitrosyl bromide.


Why does NO2 react with HCl ? What is the mechanism of the reaction?

Yes, I believe it is a typical redox where nitrogen dioxide is an oxidising agent and hydrochloric is a mild reducing agent as Ivan noted.

The reaction proceed in two phases:

$$\ce{NO2 + 2HCl ->[\Delta] NO + Cl2 + H2O}$$

  Nitrogen dioxide react with hydrogen chloride to produce nitric oxide,
  chlorine and water. This reaction is proceeds by heating.(source)

Now some nitrogen monoxide react with water to form nitrosyl chloride.

$$\ce{Cl2 + 2NO → 2NOCl}$$

Finally, we get our main reaction

$$\ce{2NO2 + 4HCl -> 2NO + 2Cl2 + 2H2O ~~~~~~ (1)*2}$$

$$\ce{2NO2 + 4HCl -> (2NO + Cl2) + Cl2 + 2H2O}$$

$$\ce{2NO2 + 4HCl → 2NOCl + 2H2O + Cl2}$$

  Nitrogen dioxide react with hydrogen chloride to produce nitrosyl
  chloride, water and chlorine.(source)


Determining the order of basicity of acetate, phenoxide, hydroperoxide, and hydroxide ions

You are partly wrong and partly right. The correct order is $4 > 3 > 2 > 1$ ($\mathrm pK_\mathrm a$ data taken from the Evans table):

$$\begin{array}{cc}
\hline
\text{Molecule} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{CH3COOH} & 4.76 \\
\ce{C6H5OH} & 9.95 \\
\ce{H2O2} & 11.6 \\
\ce{H2O} & 15.7 \\
\hline
\end{array}$$

Between acetate and phenoxide, phenoxide does have more (non-trivial) resonance forms, but all of them involve putting a negative charge on carbon which is hardly favourable. So, the stabilisation derived from resonance is not much at all.

On the other hand, in acetate, the other major resonance form has the negative charge on an electronegative oxygen. The moral of the story is that the actual number of resonance contributors is not the only deciding factor:

Between hydroperoxide and hydroxide, the $\mathrm{p}K_\mathrm{a}$ values show that you are right in saying that the second oxygen in the hydroperoxide ion stabilises the negative charge. This operates entirely via an inductive effect, though. There is no resonance involved, which partly explains why both ions are more basic than phenoxide and acetate.


What easy to conduct (at home) reactions could yield solid carbon?

A classic reaction to yield carbon is the dehydration of sucrose (table sugar) with concentrated sulfuric acid: Put some sugar in a beaker, carefully add sulfuric acid and watch a voluminous "snake" of porous carbon (blow up due to the steam released during the reaction) from the beaker.

You'll find lots of videos on this on youtube or other platforms.

Note that this is performed with concentrated sulfuric acid! If you decide that you must do this at home, protect yourself (googles, gloves, lab coat) and keep the sulfuric acid locked away.

But there's a more simple way to yield carbon:

Hold a spoon (with a long handle) into the flame of a candle. The black stuff that deposits on the spoon is carbon. Play with the distance for optimal results.

As far as the reactivity of carbon is concerned: There's no danger in that! Millions of people stockpile carbon in the form of coal at home in their cellars or sheds for heating.

And actually, this points to the easiest and most safe option: Don't perform any reaction at home and grind some charcoal or anthracite coal instead.


The most basic kinetic scheme for enzymes is represented as
$$\ce{E + S <=>[K_m] ES ->[k_{cat}] E + P}$$
As should be clear, the $k_{cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The resulting rate (kcat[E]tot) is only achieved when every molecule of enzyme essentially always is in the act of converting substrate to product. That is, every time a product molecule is released, a substrate molecule immediately binds. Even if the substrate molecule dissociates before reacting, another immediately takes its place. We describe this situation as the enzyme being "saturated" with substrate.
The Km is a measure of how tightly the substrate binds to the enzyme, approximately equal to the equilibrium constant for the dissociation of the substrate from the enzyme. If the substrate is at a low concentration relative to this Km value, then many of the enzyme molecules will not have substrate molecules bound to them and will be unproductive as a result. The overall rate will be substantially below the maximum  kcat[E]tot.
All of this is captured in the basic Michaelis-Menten kinetics equation:
$$Rate = \frac{k_{cat}[E]_{tot}[S]}{K_m + [S]}$$
You can see that the impact of $K_m$ on the rate increases substantially as [S] decreases relative to $K_m$.
Since most substrates exist physiologically at concentrations below what is required for maximum rate, it is the combination of $k_{cat}$ and $K_m$ relative to [S] that determines the in vivo rate of reaction (along with the amount of enzyme of course).


How can one synthesise a conjugated alkyne chain?

A carbenoid Fritsch-Buttenberg-Wiechell (FBW) rearrangement of a substituted dibromoolefinic precursor is used to generate a lithium acetylide, and subsequent trapping with carbon-based electrophiles provides a wide range of di- and triynes. The lithium acetylide formed from the FBW reaction can also undergo transmetalation to provide zinc, copper, tin, or platinum acetylides [1]. The paper described the preparation of poly-ynes of up to 6 triple bond units.

Image source: organic-chemistry.org

References

Luu, T.; Morisaki, Y.; Cunningham, N.; Tykwinski, R. R. One-Pot Formation and Derivatization of Di- and Triynes Based on the Fritsch−Buttenberg−Wiechell Rearrangement. J. Org. Chem. 2007, 72 (25), 9622–9629. https://doi.org/10/ffbmwt.


The question implies a misunderstanding about orbitals and their representation as dumbbell-shaped lobes. These plots show the relative orientation of orbitals to each other by showing regions that represent some threshold probability (say, 90%) of finding an electron contained therein. The underlying probability distribution has the same symmetry as the orbital itself because it defines the orbital. Any (spatial) orbital can hold up to two electrons$^1$, regardless of its shape. It thus makes little sense to ask in which lobe (or, further, which half of the dumbbell, since there is a nodal plane in between the halves) an electron is.

Moreover, any single atom is spherically symmetric in the absence of external fields, meaning that the shapes of the orbitals are not actually important to understand electron distribution in a free atom. The shapes are useful tools to understand certain bonding concepts such as crystal/ligand field theory nevertheless.

$^1$ a spin orbital holds one.


change in state (sublimation and deposition)

Firstly, in regards to the diagrams, as mentioned in the comments, both diagrams are correct. One thing that both diagrams imply is that sublimation/deposition is equivalent to the combination of melting/freezing and evaporation/condensation - another way to look at this is the following diagram from the UC Davis ChemWiki page Heat of Sublimation:

An explanation from the webpage is that:

  Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize.

In regards to your related questions:

  Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?

Not quite, liquids can occur in the right conditions, the explanation is below.

  Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?

To understand why sublimation occurs requires an understanding of phase diagrams. A generalised phase diagram from UC Davis Chemwiki page Phase Diagrams is shown below:

The sublimation/deposition curve is the line from the origin to the triple point, represented as the red line in the diagram above. A generalisation that can be made is that a substance undergoes sublimation/deposition when either (or both) the temperature and pressure are lower than at the triple point. Consider the following examples comparing the phase diagrams of carbon dioxide and water (from the UC Davis Chemwiki page linked before):

First, carbon dioxide - at 'normal' atmospheric pressure (such as in the typical laboratory - 1 atm), it is lower than the 5.2 atm at the triple point (labelled 'D'). Assuming minimal pressure changes from standard laboratory conditions, as the $\ce{CO2}$ is cooled, it will eventually undergo deposition to the solid phase.

Now, for comparison, consider the phase diagram for water, the triple state (also labelled 'D') occurs at a very low pressure 0.006 atm, pressures lower than that would result in water subliming. Note for the standard laboratory conditions, water maintains the familiar solid, liquid and gas characteristics (depending on temperature).

Note, both do have a liquid phase.


Why are Erlenmeyer flasks shaped the way they are?

Imagine a time when magnetic stirrers were unknown. You, or your lab assistant, would mix solutions in suitable glassware, shake manually, put it aside and then observe a change.

An Erlenmeyer flask is much more suitable than a simple beaker

the flat bottom gives it a stable stand on the bench
the narrrow neck allows to hold and shake a large flask with one hand
the conical shape prevents spills when the flask is shaken circularly


Derived units are defined as products of powers of the base units. Some of the coherent derived units in the SI are given special names and symbols.

The joule (unit symbol: $\mathrm J$) and the pascal (unit symbol: $\mathrm{Pa}$) are such coherent derived units.

Expressed in terms of SI base units, one joule is

$$\mathrm{1\ J=1\ \frac{kg\ m^2}{s^2}}$$

and one pascal is

$$\mathrm{1\ Pa=1\ \frac{kg}{m\ s^2}}$$

(see: BIPM: The International System of Units (SI))

Hence,

$$\mathrm{1\ \frac J{m^3}=1\ \frac{\frac{kg\ m^2}{s^2}}{m^3}=1\ \frac{kg\ m^2}{m^3\ s^2}=1\ \frac{kg}{m\ s^2}=1\ Pa}$$

SI prefixes may be used in forming the decimal multiples and submultiples of units. Note that, when SI prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor.

$$\begin{align}
\mathrm{1\ m}&=\mathrm{10\ dm}\\[6pt]
\mathrm{1\ m^3}&=\mathrm{1000\ dm^3}\\[6pt]
\mathrm{\frac1{m^3}}&=\mathrm{\frac1{1000\ dm^3}}=\mathrm{0.001\ \frac1{dm^3}}\\[6pt]
\mathrm{1\ \frac J{m^3}}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt]
\mathrm{1\ Pa}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt]
\mathrm{1000\ Pa}&=\mathrm{1\ \frac J{dm^3}}
\end{align}$$


As we in the trade say "rust never sleeps". Rust develops on the surface , rust is somewhat protective in some "weathering" steels such as Corten. Small additions of Cu ,Cr, and P , mostly provide limited slowing of rust . Otherwise there are many tables , based on conditions and location of exposure . The tables will give rust rates like 5 mpy ( mils per year ; mil is a thousandth of an inch , pardon my US units). The loss rates will be much higher on seacoasts or where rain often occurs. So with no other information ,you can guesstimate your steel will disappear in (THICKNESS divided by 0.005 ) years. Half that time for two sides exposed.


Oscillation of particle in harmonic oscillator

Classically speaking, the (idealised) harmonic oscillator is defined by Hooke's law. Quoting from Wikipedia:

In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force $F$ proportional to the displacement $x$:
$$\vec{F} = -k\vec{x}$$
where $k$ is a positive constant. If $F$ is the only force acting on the system, the system is called a simple harmonic oscillator.

In one dimension we can drop the vectors and write this as $F = -kx$. But the force is also the (negative) derivative of the potential, i.e. $F = -\mathrm dV/\mathrm dx$, and integrating this to get $V = kx^2/2 + C$ is trivial. The constant of integration $C$ is arbitrary, but it doesn't fundamentally change the physics (it just moves the energy scale), so we can ditch it by setting it to zero.
If you had a potential energy with higher-order terms in $x$ (i.e. $x^3$ and above), then the derivative with respect to $x$ would contain terms that are $x^2$ and above. This directly contradicts the definition of the harmonic oscillator. It could be some other kind of oscillator, but its motion would not be called simple harmonic.
Consequently, for this particular oscillator, trying to perform a Taylor expansion is a little pointless: there are no terms in $x^3$ or above, so all the higher-order derivatives $\mathrm d^3V/\mathrm dx^3, \cdots$ are all zero, and all those terms in the Taylor expansion are zero. Of course, in the general case, Taylor analysis of a different oscillator could be very useful. Just not here.
The potential energy of the quantum harmonic oscillator is directly inherited from the classical simple harmonic oscillator, by replacing the classical quantities $x$ and $p~ (= mv)$ with their quantum analogues $\hat{x}$ and $\hat{p} ~(= -\mathrm i\hbar(\partial/\partial x))$. So given that the classical potential energy is $V = kx^2/2$, it's not too much of a stretch to get the quantum potential energy $V = kx^2/2$.


Calculation of the pH of a mixture of a strong acid and weak acid

Chemical thermodynamics is easy in that it always has enough equations to derive all the variables involved, and guarantees to produce a unique solution. The derivation itself may be not that easy, but that's another story.

In short, we have some weak acid $\ce{HA}$ and the products of its dissociation. Mind you, $\ce{H+}$ comes also from the strong acid, which dissociates completely. Now, the equations:

$$
\ce{[HA] + [A-] = }C_1 \\
\ce{[H+]=[A-]}+C_2 \\
{\ce{[H+][A-]}\over\ce{[HA]}} = K_\mathrm{a}
$$

Is the rest clear?

   Use the second equation to exclude $\ce{[A-]}$, then use the first to exclude $\ce{[HA]}$, and end up with a quadratic equation on $\ce{H+}$.


Percentages of radical constitutional isomers

Your question is a bit hard to understand, you should change the wording or add more explanation. But I believe you want to know how to calculate the percentage of a particular isomer from the relative rates of chlorination. Well the calculation is as follows:

First, derive the ratio of different isomers. To find the terms of ratio, multiply the number of hydrogens which can be substitued to form the isomer, by the rate of halogenation of that type of hydrogen.

Next, calculate percentage from ratio. This is very easy arithmetic; you can do this by creating a fraction with the value of that particular isomer on the numerator and sum of all values from the ratio on the denominator. Then multiply the fraction by 100.

The ratio of reactivity of tertiary($\ce{3^\circ}$), secondary($\ce{2^\circ}$) and primary($\ce{1^\circ}$) H-atoms for chlorination is $\ce{5 : 3.8 : 1}$  at 298K. For bromination of $\ce{3^\circ}$, $\ce{2^\circ}$, $\ce{1^\circ}$ H, the ratio of reactivity is $\ce{1600:82:1}$ at 400K

Example: Monochlorination of n-butane. Here, two isomers are formed, 1-chlorobutane, and 2-chlorobutane.

Now, substitution of six $\ce{1^\circ}$ H  (shown by blue box) results in 1-chlorobutane

and substitution of four $\ce{2^\circ}$ H (shown by red box) results in 2-chlorobutane

So, the ratio is 1-chlorobutane : 2-chlorobutane  $\ce{=6\times1 : 4\times3.8= 6:15.2}$

Calculate percentage: 1-chlorobutane is $\ce{ \frac{6}{6+15.2}\times 100\%= 28.3\%}$

and similarly, for 2-chlorobutane, percentage is $\ce{71.7\%}$


Why the radius of noble gases is more than the halogens or the previous groups?

Unlike a ball, an atom doesn't have a fixed radius. The radius of atom
  can only be found by measuring the distance between the nuclei of two
  touching atoms and then halving that distance.

According to my findings, I think the reasoning lies in the way in which the measurements are done:

i. Metallic/Covalent radius (left)

ii. Van der Waal radius (right)

It is believed that Noble gases don't form (or rather don't easily form bonds). This way only Van der Waals radius is measured.

The rationale here is based on the fact that noble gases have very low chemical reactivity and their atomic radii are non bonded.
Another explanation lies on electronic configuration of noble gases. They have completely filled outer orbitals with increased repulsion.

In addition, following Hund's rule, in a set of degenerate orbitals, spin-paired electrons exhibit more repulsion compared to singly occupying orbitals which have parallel spins (and consenquently less repulsion).

This also implies that increased electron repulsion between the completely-filled orbitals (as seen in noble gases which all have spin paired electronic configuration) has an effect of increasing "Van der Waals radius" as electrons tend to move as further away as possible to minimise repulsion.

However there is some disparity when comparing Noble gases radii with Group 1 elements (metallic radii) or Halogens (covalent radii) because you are comparing two different unrelated properties,
nevertheless it makes sense to compare radii between different noble gases.

References

Atomic and Ionic radius
Conceptual Chemistry


What are the factors affecting osmosis (bio)?

Osmotic flow is the movement of a substance (typically water), across a semi-permeable membrane, from the side on which the chemical potential of the substance (water) is higher to the side on which it is lower.  I.e., just as temperature provides a gradient for the flow of thermal energy, chemical potential provides a gradient for the flow of matter.

Thus anything that affects the chemical potential of water will affect osmotic flow.  This includes dissolved solutes, pH (which of course comes from a particular dissolved solute—H$^+$), temperature, and pressure. For instance, increasing the solute concentration decreases water's chemical potential, while increasing the pressure increases it.

Indeed, that's the basis of osmotic pressure—water will flow toward the side with the greater solute concentration (activity) until the pressure on that side increases enough such that the reduction in water's chemical potential due to the presence of the dissolved solutes is exactly negated by the increase in water's chemical potential due to the increased pressure.


Why is the +3 oxidation state of Cerium more stable than +4?

Are you sure $\ce {Ce}$ (III) is really considered more stable than $\ce {Ce}$ (IV)? Since the most common cerium based compound is its dioxide so $\ce {Ce}$ (IV) is actually more stable.

In addition to that, $\ce{Ce}$ has 4 electrons in its outer orbital, thus leaving only one unpaired electron on the $\ce{6s}$ orbital doesn't sound really stable to me.


Why do acids usually completely react with bases?

Acids and bases don't necessarily completely react with each other. Consider your hypothetical reaction between hydrochloric acid and ammonia.

\begin{align}
\ce{HCl + H3N &-> H4N+ + Cl-}&
 \text{Extent:}& \approx100\%
\end{align}

The products are ammonium ion and chloride ion - the conjugate acid and conjugate base, respectively. So if the forward (above) reaction is complete, how can the reverse reaction between the conjugate acid and conjugate base be anything but insignificant? I.e.

\begin{align}
\ce{H4N+ + Cl- &<<=> HCl +H3N}&
 \text{Extent:}& \approx0\%
\end{align}

An equilibrium is established whenever the extent of reaction is not complete. Another good example would be an acetic acid solution. Acetic acid and water exist in equilibrium with acetate anion and hydronium ion. So no not all acid-base reactions go to an extent of 100%. Things to consider: stoichiometry and acid and base strengths.


Why is lanthanum a lanthanide and actinium an actinide?

According to www.webelements.com, "There is no one single or best structure for the periodic table..."

I believe that the table was split at lanthanum since it kept the elements of the d orbitals together. You could argue that the table should be split with Sc, Y, La and Ac together, with the elements Ce through Lu and Th through Lr as the f orbital elements. The next column would have Ti, Zr, Hf and Rf.

Actually, all elements have f orbitals, below lanthanum they are not usually populated. Lanthanum is the first element that could have an electron in the f orbital in the ground state. Since the transition energy of the electron between in the d orbital and the f orbital is quite low, there should be a significant number of electrons occupying the higher energy orbital, even at room temperature.


Reason for shorter bond between sulphur and oxygen?

If you would consider it more carefully, you would see that the bond order in SO2 and SO3 are, respectively, 1.5 and 4/3. This means that they are shorter than a single bond of sulphur and oxygen.


Your textbook description is correct. Recall the principle of potentiometry, the potential difference is measured when there is no current flowing in the system. This is true under ideal conditions. Do you recall the pH meter? What is the electrode made of? It is made of glass, do you feel that current can flow across glass when you are measuring pH of your solution?
BTW, the working of the glass electrode puzzled electrochemists for ages and the still do not know how it works exactly. So if you are confused, welcome to the club.
You should rather ask how do high impedance voltmeters work? Such voltmeters work when there is no current flow and you are solely interested in potential difference between the electrodes.

Since there is no transfer of electrons, how does this apparatus
actually work?

This is perhaps a misconception. Potential difference only develops when electrons are transferred (only once) but the is no continuous flow of electrons (which constitutes current). If you are combing your dry hair in winter, a huge potential difference develops between your hair and the comb. You comb is still "charged" when you move it away from your hair. The electrons were transferred but only at one instant. There is no current flow between your comb and the hairs.


Modern open-source tools for simulation of NMR spectra

In Org Biomol Chem 2016, 14, 3943, Goodman reports the replacement of some "tradition" (expensive) programs with free/open-source ones, as applied to their dp4 method.

Specifically:

Molecular mechanics with tinker (I believe as a replacement for MacroModel), you may or may not want this step depending on what you're doing. The paper is looking at molecules with stereocentres and as such, they appear to have wanted to ensure the NMR values calculated matched the most likely conformation of the molecule experimentally studied.
NMR prediction using NWChem (replacing Jaguar as the gold standard) to do the DFT calculations. There is some mention that this was slower, but again, it would depend on what you're interested in studying.

I think that fulfils your requirement for being able to do the prediction using open-source programs.

As for scripting, there is a mention in the paper of wrapping the whole thing up in a python script (and to my knowledge, there isn't actually a GUI for either piece of the aforementioned programs), so it should be amenable to this.

One thing that it doesn't appear to do, is to graphically plot the resulting data generated (I believe they just compare a list of values to a list of experimental values), but you should be able to plot this in something like MatLab, or, if you play enough, it should be importable into TopSpin via something to turn it into a set of suitably written values.


Calculating self-diffusion coefficient from MSD programmatically

In 1D diffusion along the $x$ axis by Brownian motion, let the mean jump lengths be $a$ each taking a time $\tau$ and this should be large compared to the time between successive collisions, which are $\approx 10$ fs, i.e. $a$ is the sum of all displacements during time $\tau$. Assuming this means that there is no correlation between collisions at time $t$ and $t+\tau$. From this the diffusion equation (Ficks Law)  can be derived and the diffusion coefficient defined by $D_x=\langle \Delta x^2\rangle/2\tau$ where $\langle \Delta x^2\rangle$ is the variance of the jump lengths $a$ or $\langle\Delta x^2\rangle = ( x-\langle x\rangle)^2$.  In 3D as there is no correlation between the motions on each axis,  $D=\langle \Delta x^2\rangle/6\tau$. This equation is subtly different from your equation and so you now need to connect these terms with what you have calculated. You should find that your diffusion coefficient settles down to a constant value (plus or minus statistical fluctuations) as the calculation proceeds.


Will adding apple cider vinegar to freshly pressed fruit juice preserve it

Apple cider vinegar is commonly used as a food preservative so I'm sure you would see some extended shelf life.

The acetic acid (the "active ingredient" of vinegar, usually present at around $\pu{5 - 20\%}$) in the vinegar would not "initiate" the production of more vinegar from the juice.  The production of acetic acid comes from a two-step fermentation process. First, alcohol is produced from the sugar in the juice by adding yeast and the appropriate bacteria. Then, acetic acid forming bacteria are added to oxidize the alcohol to acetic acid.

So, simply adding the apple cider vinegar to your juice will likely increase it's shelf life, but will not result in the production of more vinegar.


Where can I find CAD blocks for lab glassware?

ChemDraw has a couple of templates with all kinds of laboratory glasswere.
Here is a screenshot of some of the possibilities:


Differences between formulae for dipole–dipole interaction energy

They are both dipole-dipole energies as from the link but their contexts are different. Eq. 3 (as numbered in the LibreTexts link),
$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}} \tag{3}$$
is specifically for the case when the dipoles are aligned. The more general Eq. 4
$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}_{12}} (\cos\theta_{12} - 3\cos\theta_{1}\cos\theta_{2}) \tag{4}$$
is preferable because it includes the dependence on the orientation of the dipoles in space. And I believe Eq. 7
$$V = - \frac{2\mu_{\mathrm{A}}^{2}\mu_{\mathrm{B}}^{2}}{3(4\pi\epsilon_{0})^{2}r^{6}} \frac{1}{k_{\mathrm{B}}T} \tag{7}$$
is the energy derived by thermally averaging the dipole–dipole energies when the dipoles are fixed in space but are free to rotate.
So Eq. 3 is the dipole–dipole energy when the dipoles are both in a specific orientation, Eq. 4 reveals the orientational dependence of the dipole-dipole energy, and Eq. 7 is the thermally averaged energy for two dipoles that have a fixed separation $r$ but are free to rotate.


Is there an energy cost associated with flipping the spin of an electron?

I will try to describe what happens when two hydrogen atoms approach each other from infinity. At infinite separation the hydrogen atoms don't feel their mutual presence and each atom has one electron localized in its atomic 1s orbital. In the absence of magnetic fields it will not matter whether the spins of the electrons are parallel or antiparallel and you will basically have an ensemble of systems with an equal amount of triplet (parallel spins, $S=1$) and singlet (antiparallel spins, $S=0$) states, since $E_{\uparrow \uparrow} = E_{\uparrow \downarrow}$. But when hydrogen atoms are brought closer together the situation changes: For one thing each atom has a magnetic moment associated with its electronic spin and there will be a dipolar interaction between those magnetic moments. However this dipolar interaction will be very weak (in the order of $10^{-4} \, \mathrm{eV}$ at close interatomic distance) and won't have much influence. Another factor that comes into the game is spin-orbit coupling. Yet again, since $\ce{H}$ is such a light element spin-orbit coupling will be very, very weak and so it can be ignored too. But the predominant factor to be taken into acount is the exchange energy (its effect is in the order of $0.1$ to $1 \, \mathrm{eV}$). Here you can find a very nice treatment of the exchange energy for the hydrogen molecule which comes to the conclusion that

  For the hydrogen molecule the exchange energy is negative [...] so the state when the resultant spin is $S=0$ has the lower energy than the state when $S=1$.

This source also provides the potential energy curves for the hydrogen singlet and triplet states.

These curves amount to the following electron configuration diagrams:

${}$

So, the situation for the approaching hydrogen atoms is like this: If both atoms start out with electrons of opposite spin (i.e. the system is in a singlet state) you have the common case that the hydrogen atoms will bond together thus lowering the total energy of the system. But if both atoms start out with parallel spins (i.e. the system is in a triplet state) the total energy will rise when they get closer to one another and the atoms will not bond together but keep their distance. This can be understood from the electron configuration diagram of the triplet hydrogen molecule: The antibonding $\sigma^{*}$ MO is more destabilized than the bonding $\sigma$ orbital is stabilized by the interaction (this is generally true for 2-orbital interactions) and the energy splitting becomes larger the closer the hydrogen atoms get. But because of the Pauli principle the two electrons of like spin in the system can't both occupy the $\sigma$ MO but one is forced to occupy the $\sigma^{*}$ MO and thus the system's total energy rises when the hydrogen atoms approach each other. And because there is virtually no spin-orbit coupling for hydrogen the total angular momentum (represented by $L$) and the total spin (represented by $S$) of the electrons are essentially conserved quantities and there is no mechanism to facilitate intersystem crossing, i.e. the process of changing from the triplet to the singlet state has such a low probability that it virtually doesn't happen. The situation changes of course if you add an external bias to the system, e.g. irradiation with high-energetic light. Then a triplet-singlet transition might be possible but as I read it that was not your question.

So, the way MO-diagrams are often pictured with electrons of like spin pairing up in a bonding MO does not reflect the "real" situation of two atoms approaching each other from an infinite distance. But it is not meant that way although it is often pictured as such.

Update

So, in summary I established that if you have two hydrogen atoms in free space without spin-orbit coupling and relativistic effects they are stuck with the spin state they initially start out with and can't change it during their time evolution, i.e. they have to stay either on the triplet or the singlet Born-Oppenheimer surface and no switching between those is allowed. But I want to delve a little deeper into the reasons for it. If the hydrogen atoms wanted to change from triplet to singlet state, the atoms would have to transfer $1 \hbar$ of spin angular momentum somewhere else in order to comply with the law of conservation of momentum. In the absence of any particles to collide with the only possible mechanism by which this can be achieved is to emit a photon whose energy is equal to the energy difference between the triplet and the singlet state, i.e. $\Delta E = \hbar \omega = E_{\uparrow \uparrow} - E_{\uparrow \downarrow}$, and which would "carry away" the $1 \hbar$ of spin. But in order to emit a photon there must be a coupling between the electrons and the electromagnetic field. The magnitude of this coupling or in other words the probability for a transition from an initial to a final state under the emission of a photon is described by the transition dipole moment

\begin{equation}
  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \Psi_{\mathrm{i}} \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \Psi_{\mathrm{f}} \big\rangle
\end{equation}

where $q_{n}$ and $\mathbf{\hat{r}}_{n}$ are the charge and the position operator of the $n^{\mathrm{th}}$ particle in the system respectively, and the subscripts $\mathrm{i}$ and $\mathrm{f}$ signify the initial and final states of the transition. Under the conditions of the Born-Oppenheimer approximation the contribution of the nuclei to $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be zero for the situation at hand, so that we only need to be concerned with the electrons in the system. If there is a sizable spin-orbit coupling then the spin degrees of freedom and the spatial degrees of freedom are coupled via the total angular momentum operator $\mathbf{\hat{J}} = \mathbf{\hat{L}} + \mathbf{\hat{S}}$ because

\begin{align}
  \mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2} + \mathbf{\hat{L}} \cdot \mathbf{\hat{S}}
\end{align}

and $\mathbf{\hat{L}} = \mathbf{\hat{r}} \times \mathbf{\hat{p}}$ where $\mathbf{\hat{p}}$ is the momentum operator. However, if there is no spin-orbit coupling then $\mathbf{\hat{L}} \cdot \mathbf{\hat{S}} = 0$ and $\mathbf{\hat{J}}{}^{2} = \mathbf{\hat{L}}{}^{2} + \mathbf{\hat{S}}{}^{2}$, i.e.
the spin degrees of freedom and the spatial degrees of freedom are decoupled. That means you can make a product ansatz for the wave function where you seperate the spin-dependent part $| \psi(S) \rangle$ and the spatial part $| \psi(\mathbf{r}) \rangle$ such that $| \Psi \rangle = | \psi(\mathbf{r}) \rangle | \psi(S) \rangle$. So, for the transition dipole moment you get

\begin{equation}
  \Theta_{\mathrm{i} \to \mathrm{f}} = \big\langle \psi_{\mathrm{i}} (\mathbf{r}) \big| \sum_{n} q_{n} \mathbf{\hat{r}}_{n} \, \big| \psi_{\mathrm{f}}(\mathbf{r}) \big\rangle \underbrace{\big\langle \psi_{\mathrm{i}} (S) \big| \psi_{\mathrm{f}}(S) \big\rangle}_{= \, 0} = 0
\end{equation}

where $\langle \psi_{\mathrm{i}} (S) | \psi_{\mathrm{f}}(S) \rangle = 0$ because the initial state would be the triplet state ($S =1$) and the final state would be the singlet state ($S=0$) and spin wave functions for different $S$ are orthogonal.
So, no spin-orbit coupling means that the probability for a transition from a triplet to a singlet state under the emission of a photon is zero.

In real atoms and molecules there is of course some spin-orbit coupling, so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be larger than zero. But the spin-orbit coupling in hydrogen is tiny and so $\Theta_{\mathrm{i} \to \mathrm{f}}$ will be tiny too and thus the probability of a triplet-singlet transition will be tiny as well and can be neglected for all practical purposes.


How can temperature be constant during a reaction?

But how can the reaction occur at constant temperature

Despite the dynamic name, thermodynamics is not concerned about time. Or more precisely, the theoretical model/world-view of thermodynamics allows the reaction to take place over an infinite amount of time, so you have all the time you need to dissipate the heat and thus keep the same temperature constant (within arbitrarily small limits).

To get to practical reality, we have a whole arsenal of possibilities that cover a range from very close to the definition here to experimental conditions that are quite different from those stated here:

The closest one is: The reaction takes place in a thermostat, i.e. in a device that keeps the temperature constant. You may measure the amount of heat that needs to be transported into or out of the reaction vessel to keep the temperature there constant (within practical limits).

As @MaxW says in their comment, you can also approximate this by using a reaction vessel that keeps the temperature approximately constant.

@Maurices comment goes yet a step further: as long as the outcome doesn't change (e.g., no irreversible reations at the intermediate temperature), you may deliberately allow intermediate temperature changes as long as you can keep track of the heat exchange until the reaction products reach the starting temperature again.

And lastly, you could allow temperature change and do a series of experiments that then allows you separate in the data analysis temperature-dependent from temperature-independent parts.


Is there another name for compound "delta-7 sterine", does it exist alone?

"delta-7 sterine" seems to be an adaptation of German term "Sterine" (plural for "Sterin") used by German scientists who elucidated the structures of those steroids using NMR [1].
In English, it seems, the name should be written as Δ⁷–sterol and is not a single compound but a group of compounds (for more info I'd suggest searching Google Scholar).

References

Sucrow, W.; Reimerdes, A. Δ⁷-Sterine Aus Cucurbitaceen. Zeitschrift für Naturforschung B 1967, 23 (1), 42–45. https://doi.org/10.1515/znb-1968-0109. (Open Access)


Is there any chemical that can catalyze "reverse combustion?"

By principle, all catalysts accelerate both forward and backward reactions
in the same extent, so the equilibrium with and without the catalyst is the same.

So the catalyst catalysing the "reversed combustion" would catalyze the normal combustion as well. The equilibrium would be  far at the combustion products side, as if the catalyst was not used.

If it had not been so, one could have constructed a chemical perpetuum mobile. It could generate energy by shifting equilibrium forward and backward just by adding and removing the catalyst.


Why don't use the definition just above the image?

  meso-compound

  A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

Achiral, i.e. the molecule is superimposable with its mirror image. So, if finding a plane of symmetry, or other symmetry elements in the molecule is difficult for you, a simpler method (and according to the definition) might be drawing a mirror image, i.e. all chiral elements inverted

or unchanged but drawn "right-to-left", i.e. flipped

or flipped vertically

.. (just use any of the mirroring methods) and deciding whether it's the same molecule (stereoisomer), by rotating it, in 3D, in your head. (Some internal rotations around single bonds might be needed as well.)

In the first case, the mirrored structure can be superimposed with the original one by in-plane 180° rotation around an axis perpendicular to the paper plane.

In the second case it can be done by 180° rotation around the in-plane horizontal axis.

In the third case it can be done by 180° rotation around the in-plane vertical axis.


How does one compute the number of unique 2-electron integrals for a given basis set?

To find the number of unique $2e^-$ integrals $\left<AB|CD\right>$, it is useful to first find the number of unique $1e^-$ integrals $\left<A|B\right>$.

If $n$ is the number of real basis functions, we can find the number of $1e^-$ integrals as ${n+1}\choose{2}$. This stems from the general formula for combinations with replacement: $${n+k-1}\choose{k}$$ where $n$ is the selection pool and $k$ is how many are being chosen. Combinations with replacement are needed to ensure we count the cases where the two Gaussians are the same.

As an aside, this is only true for real basis functions, since we can only say the order doesn't matter (what we do when we compute the number of combinations) because all the basis functions are explicitly real and thus the complex conjugate equals the original function. This explains one of my own questions on the site regarding why we don't commonly use complex basis functions.

For this particular case ($n=100$), we find that the number of unique $1e^-$ integrals is $${{101}\choose{2}}=\frac{(101)*(100)}{2}=5050$$

So how does this allow us to find the number of unique $2e^-$ integrals? Well, for the $1e^-$ integrals, all we did was find the number of unique pairs of real basis functions. So, for a $2e^-$ integral, we might expect that all we need to do is find the unique pairs of basis function pairs. This winds up being exactly the case (again, only for real basis functions).

To find the pairs of pairs, we need to again find the number of combinations with replacement, but instead of sampling from the $100$ basis functions, we are now sampling from the $5050$ basis function pairs. We again need replacement to account for the fact that we can pick the same pair twice. So, we obtain: $${{5051}\choose {2}} =\frac{(5051)*(5050)}{2}=12,753,775$$ as the number of unique $2e^-$ integrals. This method avoids the checking of specific cases (i.e none of the basis functions are the same, two are the same, etc.) and gives a general way of finding the number unique integrals for any amount of real basis functions.

To show explicitly why complex basis functions aren't used, we can look at the same problem, but assume all the functions are complex instead. In bra-ket notation, the bra is the complex conjugate of the function inside, so for $\left<A|B\right>$, the order would matter if $A$ and $B$ are complex. So finding the number of pairs requires finding the number of permutations rather than combinations. This only requires a subtle change of the original calculation $$2!{101\choose2}= (101)*(100)=10100$$ So, we now have twice as many unique pairs, so there are twice as many $1e^-$ integrals.
The equation for the $2e^-$ integrals remains the same, but with the new amount of unique pairs filled in. $${{10101}\choose {2}} =\frac{(10101)*(10100)}{2}=51,010,050$$ We still need to do a combination for the pairs because regardless of whether the functions are complex or real $\left<AB|CD\right>=\left<CD|AB\right>$ via exchange of electron labels. So using complex basis functions would double the number of $1e^-$ integrals and nearly quadruple the number of $2e^-$ integrals. So even if some feature of complex basis functions made them easier to integrate, it would have to make them easier enough to warrant doing and storing 2-4 times as many integrals.


α,β-unsaturated carbonyl compounds and alkyl addition

If I understand the question correctly (I do not have the book), it goes like this:

(Ref. J. Am. Chem. Soc.  85, 2, 207-222)

In short I would say that it is not possible to perform the reaction in question in the same way with the Stork enamine alkylation simply because—due to the α,β unsaturation of the ketone—the enamine can not be produced with pyrrolidine. Also, I performed a Reaxys search and found no relevant results. (Of course, I might have missed something.) However, I did find a couple of interesting reactions between cyclohexenone and pyrrolidine which I submit below for completion:

And the following L-Proline organocatalytic reaction taken from: Tetrahedron Letters, 2010, vol. 51, #38, p. 5086 - 5090. Note that compounds 1 and 2 in the scheme were isolated.

However it must be noted that generally, using a different α, β unsaturated ketone and a different amine to the ones used above, the reaction in question is possible as demonstrated by user55119 in his answer here: Why enamine formation does not occur with α,β unsaturated ketones?


Why should the salt being used in salting out process be neutral?

"Its [sic] been said" is inadequate contextualization. Many organic solutes are potentially capable of being ionized, especially in aqueous solution. Maintaining aqueous media at a neutral pH will for most solutes minimize this (undesired) effect. For instance, in the following reference (note date and distinguish between what is observational (and thus probably "true") and theoretical (based on what is likely to be incorrect chemistry)), read the last sentence on page 531 over to 532. http://www.jbc.org/content/33/3/531.full.pdf


Is it correct to say that “some degree of pollution is an inescapable consequence of the Second Law of Thermodynamics”?

Yes, this is true. However, the question is where that increase in entropy occurs.

Since the industrial revolution, life on earth is a bit one-sided, with humans controlling the chemistry going on, and a lot of it. When we burn fossil fuels, there is an increase in entropy on earth. Before that, one organism's waste was some other organisms food, and the entropy increase occurred mostly in the sun (giant nuclear reaction slowly going towards equilibrium).

Humans (and all other animals) benefit from the $\ce{O2}$ "pollution" from cyanobacteria and plants. Plants benefit - to a degree - from the $\ce{CO2}$ "pollution" from animals. However, when global concentrations change too fast, some species might benefit a lot, but many other species will go extinct.

If the "plastic ocean" had built up over millions of years instead of 50, some microorganism that uses the polymers as its primary source of food might have evolved, and there would be a cycle instead of a build-up.


Inter-classification of organic chemistry functional groups (carboxyl vs alcohol & ether vs ester)

Would it make sense to classify a pistol as a type of pipe? The barrel of a pistol is a cylindrical metal tube, after all.

Clearly, no. The other key defining components of a pistol (stock, trigger, etc.) combine with the cylindrical metal tube to radically change its function and behavior.

Similarly, even though a carboxyl group does contain an $\ce{-OH}$ group bonded to a carbon atom, its behavior is sufficiently different from that of an alcohol that it makes no sense to group them together. A parallel argument holds for ethers and esters as well, as ably noted by MaxW.


Under what circumstances would azeotrope exist in non ideal solutions? (when would the dew point curve and the bubble point curve meet?)

Azeotropes do not exist for all non-ideal solvent solutions. All solvent solutions are more or less non-ideal, but not all solvent combinations form azeotropes.

E.g. ethanol and isopropanol form azeotropes with water, but methanol does not.

Azeotropes exist if there is large enough deviation from the Raoult law. It means if the azeotrope point of the diagram is pushed enough away from the endpoint connections.

For near ideal solutions, like n-pentane + n-hexane. there is no bottleneck and onle one convex/concave chart segment.

Back to the meeting of boiling and condensation curves, they meet at the pure solvent points and eventually at the azeotrope point, if it exists.

Existance of such minimums ( or maximums ) ob the vapour diagram means there is an azeotrope. An azeotrope behaves like if it were the 3rd substance, with  the same composition of liquid and vapour above it, so the curves meet there.

Minimums meet if and only if liquid and vapour compositions are the same. That happens if and only if either the liquid is pure solvent either it has the composition of the azeotrope ( if it exists ).

You can create simulations yourself by defining nonideal vapour pressure equations as $p_i(T)<>p_{i,0}(T) \cdot x_i$. It cannot be exactly calculated, as neither the mixture of pure solvent and azeotropic mixture has the ideas behaviour.


A rather quick googling shows up that there is some Longuet-Higgins theorem, which states almost what you are looking for. For instance, the abstract of this paper says:

  It is proved that if the wave function of a given electronic state
  changes sign when transported adiabatically round a loop in nuclear
  configuration space, then the state must become degenerate with
  another one at some point within the loop.

In the paper itself it is further stated that this in turn implies that

  [the corresponding] potential energy surface intersects one of another electronic state

and that the theorem

  makes it possible to diagnose the presence of an intersection between potential energy surfaces from the behaviour of the electronic wave function as it is transported adiabatically round a closed loop remote from the intersection

which answers both your questions, if I understand them correctly.


As it stands, your question is very general. It is unlikely that a single answer here will describe spin coupling and multiplicity in a simple and detailed manner any better than any publicly available reference source. There are hundreds available. I suggest you read one or more of the following extremely useful introductory NMR references, and ask more specific questions that you do not understand from them.

Reich (http://www.chem.wisc.edu/areas/reich/chem605/)
Hornak (http://www.cis.rit.edu/htbooks/nmr/)
Libretexts (https://chem.libretexts.org/Core/Organic_Chemistry/Spectroscopy/Nuclear_Magnetic_Resonance_Spectroscopy
and
https://chem.libretexts.org/LibreTexts/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Chapter_12._Nuclear_Magnetic_Resonance)


Question on deflection of cathode rays in a magnetic field

1)  for individual charges Lorentz force F =q[v*B], so curl your right hand from velocity vector of positive charge towards magnetic field so that your thumb points at the force acting on the charge. Here we have considered positive charge

2)It is better use this mnemonic to find out the direction of force to avoid confusion

Stretch the thumb, forefinger and middle finger of your right hand such that they are mutually perpendicular

fore finger denote direction of electron moving
middle finger denotes the direction of magnetic field (north to south on outside the magnet)
finally force direction can be found out using your the direction where your thumb is pointing

The book is correct

Tip- If current is mentioned then Use your left hand, to avoid twisting your hand in weird ways! .


What is the concentration of H2 for the equation H2+I2=2HI if the equilibrium constant is 64 and the initial concentrations are both 1.9

let the amount of $\ce{H2}$ reacted = $\pu{x~mol}$
Use the molar ratio to detemine the amount of $\ce{I2}$ reacted and the amount of $\ce{HI}$ formed:
$$\begin{align}
&\text{The amount of }\ce{H2}~\text{reacts to reach equilibrium }&=\pu{x~mol}\\
&\text{The amount of }\ce{I2}~\text{reacts to reach equilibrium}&=\pu{x~mol}\\
&\text{The amount of }\ce{HI}~\text{formed at equilibrium }&=\pu{2x~mol}\\
\end{align}$$
Use the following table to determine the amount and the concentration
of each species at equilibrium :
$$\begin{align}
\ce{&H2 &&+ &&O2 &<=> &&2HI\\
I~~~~  &1.9   &&&&1.9      &&&0 \\
C ~~~  &-x    &&&&-x       &&&2x \\
E ~~~  &(1.9-x) &&&&(1.9-x) &&&2x}
\end{align}$$
Th concentrations at equilibrium :
$$[\ce{H2}]_\mathrm{e}=(1.9-x)  , [\ce{I2}]_\mathrm{e}=(1.9-x) ,  [\ce{HI}]_\mathrm{e}= \pu{2x~M} $$
Sustitute the concentrations in the following formula:
$$\begin{align}
K_\mathrm{C} = \frac{ [\ce{HI}]_\mathrm{e}   }  {[\ce{H2}]_\mathrm{e}[\ce{I2}]_\mathrm{e} }\\
64= \frac{(2x)^2}{(1.9-x)(1.9-x)}
\end{align}$$
Take the square root of both sides :
$$\sqrt{64}=\sqrt{\frac{(2x)^2}{(1.9-x)^2}}$$

$$8=\frac{2x}{(1.9-x)}$$
- Solve for $x : x=1.52$

Calculate the concentrations at equilibrium :
$$\begin{align}
[\ce{H2}]_\mathrm{e}&=(1.9-x)&=(1.9-1.52)&=\pu{0.38~M}  \\
[\ce{I2}]_\mathrm{e}&=(1.9-x)&= (1.9-1.52)&=\pu{0.38~M} \\
 [\ce{HI}]_\mathrm{e}&= \pu{2\times{x}}&= 2\times{1.52}&=\pu{3.04~M}
\end{align} $$


Electrophilic Aromatic Substitution of phenols

Way overthinking this.  Kohn reports that 2,4,6-tribromophenol can be further brominated to pentabromophenol by adding more bromine and an iron catalyst.  Subsequently the pentabromophenol is oxidized by fuming nitric acid to give a fully brominated quinone.


It doesn’t matter. Both units are basically identical, much like there is no everyday difference between the units mbar and hPa. There may be situations where one is preferred — e.g. hPa are, in my opinion, rarely used outside weather reports — but they should be understood nonetheless.

If there is anything to be said about the two, it seems using dalton is more popular in biologic contexts. But that wouldn’t be a tendency to base anything on.

Being consistent within one’s lab, within the journal one is publishing in or throughout one’s thesis is far more important.


Controlling Current Density in Electrowinning Cell

The system is complicated by many nonlinear processes.

As the topic is electrode focused, the primary controlled parameter should be the electrode surface current density $\sigma$, especially on the depositing electrode.

the total coltage applied on the electrolytic cell is devided on several potential differences:

$$U_\mathrm{applied} = U_{td}(\sigma) + U_\mathrm{kin}(\sigma) + U_\mathrm{diff}(\sigma) + U_\mathrm{cond}(\sigma)$$

where the particular terms are

electrode potential difference given by reaction thermodynamics
electrode potential difference given by reaction kinetics
electrode potential difference given by near electrode diffusion gradient.
free solution potential difference controlling ion migration.

For practical reasons, I advice using empiric approach :

either constant current power source, if applicable
either manual voltage control, using visual ampermeter feedback.

For further reading, I recommend:

For starters, Electrochemistry on hyperphysics
Wikipedia articles,  following internal and external links, also searching for mentioned topics.
for more advanced matters, textbooks for Physical chemistry.


Can enamines be formed from α,β-unsaturated ketones?

Stork and Birnbaum have reported (Tetrahedron Lett. 1961, 10, 313) the formation of dienamine 2 from octalone 1. Alkylation with methyl iodide leads to α-alkylation of the dienamine. Buffer conditions are required for hydrolysis of the dienamine to octalone 4.


Which kind of plastic material would be suitable for a clutch cover on a motorcycle transition?

Most plastics consist of linear molecules with little inter-molecular interactions, so most plastics melts too easy. The only ones you mentioned having a chance are polyuretanes and polyamides. They have (CO)(NH) fragments, that forms strong intermolecular hydrogen bonds. So, they are ones of the strongest polymeric materials

And I will recommend aluminum anyway as it is much harder to screw things with aluminum.


I found 3 properties:

The molecular weight (MolWt)
Wildman-Crippen partition coefficient (LogP)
Topological Polar Surface Area (TPSA)


Solubility in different pressure conditions

As noted by Wikipedia,

  "Henry's Law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid."

So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved.

The answer key for the problem must be wrong.

  User Delta_G made an interesting observation/correction. The problem is asking about the volume of gas that dissolves, not the concentration of oxygen in the water. So 28 ml at 4 atm of oxygen gas contains four times as much oxygen as 28 ml at 1 atm of the gas. So in either case 28 ml of the gas phase dissolves into the water.


What is the difference between moisture regain and water holding capacity?

According to the FabricLink Textile Dictionary:

  Moisture Regain - The amount of water a completely dry fiber will absorb from the air at a standard condition of 70 degrees F and a relative humidity of 65%. Expressed as a % of the dry fiber weight.

Water Holding Capacity is the amount of water a completely dry fiber will retain after soaking in liquid water and allowed to "drip dry".


Why do electrons follow from Zn to Cu but not Cu to Zn in the lemon battery experiment?

In fact $\ce{Cu}$ doesn't release electrons, only $\ce{Zn}$ does. For the usual explanation see this Wikipedia article, I will try to explain it in other words to make it much easier.

One important thing that you should know is that energy comes from metal not from the acid.
Some species have the tendency to donate electrons other species the tendency to acquire them. Chemically speaking some species have more internal energy when their are in the reduced form (when they have the maximum number of electrons) others species reach their minimum internal energy when they are in the oxidized form (when they have the least number of electron). In this world every system tend to go to a state of lower energy spontaneously so if two species are not in their lowest energy form if they can exchange their electrons freely they acquire (are reduced) or give (are oxidized) electrons until they reach the lowest energy.  There are also species that do not have any particular tendency or are only little more prone to acquire electrons or to donate them. In all these cases the exchange can occur only if the electrons can move easily from one specie to another, for this reason a good electrolyte is many times needed.
An easy comparison
We could imagine chemical species as men and electrons as money.
Imagine that there are different kind of men at the antipodes there are avaricious men that needs more money and charitable men that feel the urge to give money. I think that you have no doubt which of them do what when one meet the other. The problem is that many times the transition occurs only if there is the possibility that the flow of money goes from who want to give them to who want them so you need a third agents to create the right environment and conditions for the transition. Furthermore the loss of money must be counterbalanced by something similar.
Then there are others type of men that simply don't want to give or take money we can call them noble men and then there are other men that can take money without actually feeling the need, they simply take them, I will call them hydron men.
Your case
Lemon batteries are in fact one of the more complex and debated battery you could examine. When you plug the zinc and the copper in your lemon we can say that we have three main active species: copper, zinc and acid (citric acid). Zinc is one of that metals we compared to a charitable man it tends to donate electrons:
$$\ce{Zn <=> Zn^{2+} + 2e^{-} }$$
Copper however is a noble metal, physics explain this because it has the d-bands of the electronic structure filled, so it doesn't accept zinc's electrons. Instead hydrogen from acid, can accept the electrons in this way:
$$\ce{2H+ + 2e^{-} -> H2 ^}$$
And this is in fact what actual experimental results show, see J. Chem. Educ. 2001, 78 (4), 516 for reference.


The partition coefficient for a chemical species across a membrane

Partitioning across a membrane is better known as Osmotic pressure. Think of a tube with a "bag" made out of the membrane attached to it. The more concentrated solution goes inside the bag, the less concentrated solution suspends the bag. The liquid from the suspending liquid then diffuses into the bag causes the fluid level to rise against gravity. The height of the rise yields the Osmotic pressure. So it is the solvent that diffuses not the solute.

If the solute diffuses through the membrane then it is just a diffusion problem such that the rate of diffusion is related to pore size. Bigger pores give faster diffusion.

To a chemist a "partition coefficient" would also really seem to apply best to a system at equilibrium. If you're considering drugs in the body then an equilibrium probably isn't obtained. The blood sweeps the drug away to react somewhere else in the body. So the drug is continually transported into the blood.

The other side of this is something like urea in urine and blood. The kidneys actually concentrate urea from the blood plasma into the urine. However this biological process consumes energy. So this sort of process is different than a simple equilibrium.


Why do the amino groups in EDTA deprotonate first, in contrast to amino acids?

The carboxylates deprotonate first followed by the amines.  The pKa values for the carboxylates ar 0.0, 1.5, 2.0, and 2.66.  The amines have pKa values of 6.16 and 10.24.  Ref. Harris.  Quantitative Chemical Analysis.


Identifying an extra NMR peak in silyl protection reaction

I believe some of your triisopropylsilyl chloride may be hydrolyzed triisopropylsilylol, $\ce{((CH3)2CH)3SiOH}$. That may gives you $\delta$ 1.04 and 12.4 (for $\ce{C}$ attached to $\ce{Si}$) in $\mathrm{^1H}$- and $\mathrm{^{13}C}$-NMR spectra, respectively. Compare the spectral values in Ref.1 and 2 (see below; corresponding $\ce{^{13}C}$ values are in red):

References:

A. Goti, M. Cacciarini, F. Cardona, A. Brandi, “A convenient access to (3S)-3-(triisopropylsilyl)oxy-1-pyrroline N-oxide, a useful intermediate for polyfunctionalized enantiopure indolizidine and pyrrolizidine synthesis,” Tetrahedron Letters 1999, 40(14), 2853–2856 (https://doi.org/10.1016/S0040-4039(99)00310-X).
R. F. Cunico, L. Bedell, “The triisopropylsilyl group as a hydroxyl-protecting function,” J. Org. Chem. 1980, 45(23), 4797–4798 (https://doi.org/10.1021/jo01311a058).


How to properly denote imaginary magnitudes of vibrational frequencies

The suffix i to the positive wavenumber to denote an imaginary frequency is seemingly accepted by reviewers, see Photochemical transformations of 5-methyltetrazole. Matrix isolation FTIR and DFT studies as one example I could find in a very quick search.