Why water vapors exerts pressure rather than diffusing in air before reaching the boiling point of water?

I be mistaken, but are you referring to vapor pressure?

Vapor pressure  or equilibrium vapour pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A substance with a high vapor pressure at normal temperatures is often referred to as volatile. (http://en.wikipedia.org/wiki/Vapor_pressure)

What is a closed system?

A system in which no mass is transferred.

Thus, vapor pressure would not be exerted if the container is open and the liquid is constantly diffusing. It would only occur in a closed system.

Therefore, liquids DO diffuse (evaporate) in air, rather than exert vapor pressure.

You can leave a bottle of water, one with a lid and one without. The one without the lid will soon be without water, but the one with a lid would remain with water.

*Vapor pressure does relate to evaporation rate, which occurs in the open systems we normally observe in life


What happens in the NH4Cr2O7 with HgSCN reaction?

I can't comment on the metaphysical aspects ;)

This experiment is known as Pharaoh's snakes, and it is indeed impressive.

Upon heating, mercury thiocyanate decomposes to yield mercury vapour, carbon sulfide, and dicyan (ethanedinitrile):

$$\ce{2Hg(SCN)2 ->[\Delta] 2 Hg + 2CS2 + (CN)2}$$

As a result, the solid starting turns to a foam-like mass.

The effect is amplified due to the presence of ammonium dichromate, which oxidizes the carbon disulfide to sulfur dioxide and carbon dioxide.

The experimenter is apperently aware that quite a lot of rather harmful components are released and therefore has decided to run the show in the fumehood and closes the window immediately after igniting the mixture.


Why do selenium , tellurium and polonium have more negative electron gain enthalpy than oxygen?

Electron Gain Enthalpy becomes less negative down the group.

This is true, as a general trend. But the 2nd period is an exception to this trend.
In p block, it is observed that the EGE of some 2nd period elements is actually less negative then their 3rd period counterparts.
In the case of oxygen, it’s EGE is even less than selenium, tellurium and polonium. It’s because oxygen has very small radius and thus have high electron density. Due to this, the incoming electron faces a lot of repulsion from the already present electrons.
So, it takes more energy to add an electron to an oxygen atom i.e. oxygen has less negative EGE. On the other hand, selenium and tellurium have much larger radius and thus the electron density isn’t very high. Hence, their EGE is more negative as its easier to add an electron to them.


Number of Sigma Bonds in this Structure Containing a Ring

There're are 13 sigma bonds there (count the red lines),

You've clearly tried to attempt this yourself, so good on you (and by virtue of that, it's equally clear that you counted the bonds too hastily).

I counted (quickly, because it looked simple enough) and got 12 too at first. Counted a little slower the second time (I was actually going to attempt to prove to you that the number of bonds is 12), turns out the ring is not entirely conjugated, so one carbon atom in it actually has two hydrogen atoms bonded to it. ;-)


Biological activity of the isotopes of the same element

I would not expect significant differences in the chemical properties (reactivity) of these lead isotopes, the differences in atomic weight being as you point out very small. There are reports  that claim an effect. Ref 1 for instance makes such a claim, but there is no plausible mechanism presented to explain the reported isotopic discrimination. Double check what peer reviewers think (see e.g. the response in Ref 2 to the results presented in Ref 1).
References

Wu J, Liu D, Xie Q, Wang J (2012) Biological Fractionation of Lead Isotopes in Sprague-Dawley Rats Lead Poisoned via the Respiratory Tract. PLOS ONE 7(12): e52462. https://doi.org/10.1371/journal.pone.0052462

Gulson, Brian and Kamenov, George D. and Manton, William and Rabinowitz, Michael (2018) Concerns about Quadrupole ICP-MS Lead Isotopic Data and Interpretations in the Environment and Health Fields. International Journal of Environmental Research and Public Health 15(4): 723. https://doi.org/10.3390/ijerph15040723


Ethanol and acetone are not non-polar organic solvents. Each one has  a slight dipole moment; due to the difference of electronegativity between $\ce{H}$ and  $\ce{O}$ in ethanol and between $\ce{C}$ and  $\ce{O}$ in acetone.  Wax is composed of heavy, long-chain alkanes. And as "Like dissolves like" try to dissolve your wax in toluene or in xylene.


Firstly, the reactivity being referred to is not reactivity in the sense of the reactivity series of metals, but rather in the sense of standard electrode potentials (standard reduction potentials).

This comes into play when setting up an electrolytic cell in determining the reaction that will take place at the cathode/anode. In this case, the reaction that requires the least amount of work (or voltage in this case, since voltage is essentially a measure of work) will take place. So, assuming that this is done aqueously, if the reduction/oxidation reaction to take place requires more voltage than the reduction/oxidation of the water, then the water will simply be reduced/oxidized instead. (I think this is where the reactivity of hydrogen comes into play in how you were taught).

So to answer your question more directly, it is actually lead that forms at the cathode.

Edit: To add some detail, the reduction potential of H2O is -0.8277 V, while the lead(II) ion is -0.13. In this case, less energy means closer to zero, the negative being essentially irrelevant. Hope this helped.


Does naphthalene in moth balls really sublimate or does it evaporate?

Naphthalene, as well as any other solid, may sublimate at any temperature and pressure. The triple point parameters are irrelevant. The pressure of other components (that is, air) is irrelevant as well. What is relevant is the vapor pressure of naphthalene itself in the surrounding air, and that is pretty close to $0$. Until it reaches the saturated vapor pressure (or until your mothball is all gone, whatever happens sooner), naphthalene will sublimate.

The solids which we perceive as incapable of sublimation (like steel, glass, or rocks) have very low vapor pressure, so their sublimation would take forever. Other than that, everything above applies to them as well.

Then again, some solids melt upon heating and some do not, but that's another story. We aren't talking about any heating, are we?


You have correctly identified the given compound as amine. Amines can be named in various ways. According to Subsection P-62.2.1.2 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC names (PINs) are formed by using systematic substitutive nomenclature. Primary amines are systematically named by adding the suffix ‘amine’ to the name of the parent hydride; secondary and tertiary amines are named by the same method with further N-substitution (see Subsection P-62.2.2.1).

When there is a choice among acyclic parent hydrides, a principal chain must be chosen. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain (see Subsections P-44.3 and P-44.3.2).

In this case, the longest chain has five carbon atoms; i.e. the unmodified parent hydride is ‘pentane’.

Addition of the suffix ‘amine’ and the corresponding locant to the name of the parent hydride yields the name ‘pentan-3-amine’.

Finally, the substituent prefix ‘methyl’, the multiplicative prefix ‘tri’, and the corresponding locants are added to express substitution by three methyl groups, which yields the name ‘N,N,3-trimethylpentan-3-amine’ (note that ‘N’ is lower than ‘3’).


Let's say the final temperature is $T$. Also we assume the system is isolated and the process is adiabatic, as you said. Also all energy that is released by toluene or gained by benzene is heat.

The absolute amount of heat then released by toluene equals to the absolute amount of heat gained by benzene:

$|Q_\mathrm t| = |Q_\mathrm b|$

Considering signs, it will look like this:

$Q_\mathrm t = -Q_\mathrm b$

By definition, $Q = nc\Delta T$. Thus,

$Q_\mathrm t = 2\cdot96\cdot(40-T)$, and

$Q_\mathrm b = 1\cdot84\cdot(20-T)$

Plug in the latter two equations into the second one, and here's what you get:

$2\cdot96\cdot(40-T) = -1\cdot84\cdot(20-T)$

Now just solve it for $T$. And plug that in your equation as $T_\mathrm f$.

Note that it's not really important to do the temperature in kelvin, since we are dealing with changes in temperature. And yeap, your $T$ should be between $20\ \mathrm{^\circ C}$ and $40\ \mathrm{^\circ C}$, because there are such things as thermodynamics laws. So if it's not, you know for sure you did something not really correct.


Cinnamaldehyde listed as (2E)-3-phenylprop-2-enal , but is the first 2 redundant?

The preferred IUPAC name (PIN) for cinnamaldehyde is (2E)-3-phenylprop-2-enal.

With regard to numbering of locants that describe the position of stereogenic units, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-91.3 NAMING OF STEREOISOMERS

  (…)
  In preferred IUPAC names, stereodescriptors are placed immediately at the front of the part of the name to which they relate. They are placed at the front of the complete name when related to the parent structure; they are cited in parentheses followed by a hyphen. When they relate to substituent groups, they are cited at the front of the corresponding prefix. They are preceded by a numerical or letter locant to describe the position of the stereogenic unit when such locants are present; general rules of numbering are applied (see P-14.4).
  (…)

Accordingly, all stereodiscriptors have to be preceded by a locant.
In case of the PIN for cinnamaldehyde, the stereodiscriptor ‘E’ is preceded by the numerical locant ‘2’.

In general nomenclature, however, locants are usually omitted when there is no ambiguity.

Nevertheless, the current IUPAC recommendations are prescriptive about when omission of locants is permissible in PINs. Locants are omitted in PINs only in the exceptional cases that are explicitly mentioned in the IUPAC recommendations. None of these cases applies to the locant of the stereodiscriptor in (2E)-3-phenylprop-2-enal.

As a rule of thumb, if any locants are essential for defining the structure, then all locants must be cited in the PIN.

Note that the current recommendations given in the 2013 Blue Book incorporate a change from earlier recommendations given in the 1993 Guide. By way of comparison, the following examples are explicitly mentioned in the 1993 Recommendations as well as in the 2013 Recommendations:

(2E)-but-2-ene (PIN, 2013 Recommendations)
(E)-but-2-ene (1993 Recommendations)

(1Z)-1,2-dibromo-1-chloro-2-iodoethene (PIN, 2013 Recommendations)
(Z)-1,2-dibromo-1-chloro-2-iodoethene (1993 Recommendations)

(2Z)-2-methylbut-2-enoic acid (PIN, 2013 Recommendations)
(Z)-2-methylbut-2-enoic acid (1993 Recommendations)

(2E)-but-2-enedioic acid (PIN, 2013 Recommendations)
(E)-but-2-enedioic acid (1993 Recommendations)


How does a Frost diagram reproduce the solutions to the wave equation?

Algebraically, the levels of cyclic polyenes may be derived using simple
Hückel theory (see also: Pi molecular orbitals of polyenes). The general result for the energy of the $j$-th level for a cyclic system containing $N$ atoms is:

$$e_{j} = \alpha + 2 \beta \cos\left(\frac{2j\pi}{N}\right)$$

where $\alpha$ is the energy of each carbon $\mathrm p_{\pi}$ orbital before interaction (Coulomb integral), $\beta$ is the interaction energy between two adjacent $\mathrm p_{\pi}$ orbitals (the resonance integral) and $j= 0, \pm 1, \pm 2,\ldots, \pm \frac{N - 1}{2}, +\frac{N}{2}$ for even $N$, and $j= 0, \pm 1, \pm 2, \ldots, \pm \frac{N - 1}{2}$ for odd $N$. The very simple form of this equation leads to a useful mnemonic for remembering the energy levels of these molecules. Draw a circle of radius $2\beta$ and inscribe an $N$-vertex polygon such that one vertex lies at the bottom position. The points at which the two figures touch define the Hückel energy levels. And that is what is called a Frost diagram.


If a gas absorbs heat, then how can its temperature remain constant?

From the first law, $\Delta U=Q-W$.  If the process is isothermal, then $\Delta U=0$ (assuming an ideal gas), so Q = W.  The heat energy is consumed by your gas doing work, so there is no change in internal energy (temperature).


Experiments in chemistry involving use of stereo microscope?

For starters, it might be fun to grow different kinds of crystals in shallow dishes and watch them grow in real time under a microscope. Either a low melting solid, melted and resolidified, or growing from a saturated solution. Also, looking at how a metal surface is etched by acid using a microscope would be fascinating.


Does titration usually always involve neutralization?

While neutralization of an acid (or a base) is a frequent example of titration, titration is more general. IUPAC's gold book mentions  "Also contains definitions of: acidimetric titration, acid–base titration, alkalimetric titration, chelatometric titration, complexometric titration, coulometric titration, equivalence point, precipitation titration" right in the beginning.  Even the emblematic glass buret is not a must, as publications like My Dear Buret, Your Time Has Indeed Come! (Journal of Chemical Education) showcase.


Effect on mass when decreasing pressure in equilibrium reaction

This is an approach to Question 2.
The reaction equilibrium condition for the system is:  $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$.  Then we have
$$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$.  For a small change in volume dV, we can write the new number of moles of A, B, and C at equilibrium as
$$n_A=n_{A0}-dn\tag{3a}$$
$$n_B=n_{B0}-dn\tag{3b}$$
$$n_C=n_{C0}+dn\tag{3c}$$Combining Eqns. 1-3 yields:$$\left(1+\frac{dV}{V_0}\right)=\frac{(1-dn/n_{A0})(1-dn/n_{B0})}{(1+dn/n_{C0})}$$And, linearizing this equation with respect to dn yields:$$\frac{dV}{V_0}=-\left(\frac{1}{n_{A0}}+\frac{1}{n_{B0}}+\frac{1}{n_{C0}}\right)dn$$This equation indicates that if dV is negative, then dn is positive, and the reactants decrease while the product increases.
If we apply this same methodology to the concentration of A, we obtain
$$dC_A=-\frac{n_{A)}}{V_0}\left(\frac{dn}{n_{A0}}+\frac{dV}{V_0}\right)$$Combining this with the previous equation then yields:
$$\frac{dC_A}{dV}=-\frac{n_{A0}}{V_0^2}\left[\frac{1/n_{B0}+1/n_{C0}}{1/n_{A0}+1/n_{B0}+1/n_{C0}}\right]$$The term in brackets is always positive, so, when the volume decreases, the concentration of A increases at all points along the equilibrium contour.  This conclusion is independent of the specific value of the equilibrium constant.


Are there IP limitations on (M)SDS documents?

A really nice question. First, let's point out that specifics about these issues will vary from country to country, as they all relate to the laws of that country. Secondly, let's clarify that copyright and IP aren't exactly the same. Copyright covers the exclusive right of that owner to make copies, license and otherwise exploit their works (printed, artistic etc). Intellectual property is property that results from creative thought, and can include patents, copyrighted material and trademarks. So, IP is a bit like an umbrella that encompasses copyright material.

The specifics relating to copyright and IP differ slightly from country to country, but it is safe to say that the (M)SDS document is a copyrighted document. Most SDS documents will have this specified, usually at the very end of the document - I just checked some of the ones I have on file from commercial suppliers, and they all have a copyright statement. This means that you cannot make copies etc under the usual copyright limitations.

There are requirements about what information goes into a SDS. Again, there will be differences from country to country, but for the US, for example, I think all hazardous components above 1% must be specified, and known carcinogens above 0.1% must also be listed. The exact numbers might be a little different, but the general gist is that hazardous components must be listed.

HOWEVER, every country has a clause in their code governing SDS requirements that specifies, essentially, an exemption from listing ingredients that may be regarded as a 'trade secret', or puts at risk an identified Intellectual Property claim. How this is expressed varies; in some countries, a trade name may be used to 'hide' the true identity of a IP-protected ingredient. In other countries, a summary of functional groups can be used to represent that key ingredient. Usually, though, the nature of the hazard must be represented somehow on the SDS document. I know in some countries, even if an ingredient is protected by an IP clause, its true identity must be revealed to health practitioners if requested. I'd have to dig around for the exact laws for each country/zone if you needed more specific answers.


No.

In the equation $C_x$ simply means the concentration at depth x. It does not mean the change of concentration. There is no need to add its initial concentration.

Besides, concentration cannot be linearly added. (Does adding a 1M solution to a 2M solution give 3M?)


In general, a equilibrium will shift in a direction that opposes the stress that was placed on the system. For example, if you add heat to an endothermic reaction, you will receive more products, as this will lower the temperature of the solution. When you have the dissociation of an acid, the acid will dissociate more (act like an acid) when you make the solution more basic. An equilibrium will shift to minimize any change in a system, so when evaluating what will happen after a system is stressed, look to see which side of the equilibrium relives that stress, and that side will be favored.


Do antimicrobial abilities of copper boilers decrease over time?

It is possible that the ability to release soluble copper ions could be reduced with age.
I am assuming, for example, that a very small amount of the cuprous oxide coating could react with, say, carbonic or hypochlorous acid (from chlorinated water) to create a poorly soluble basic salt.
However, with time, to quote a source, namely Handbook of Industrial Water Treatment, Boiler Water Systems, Chapter 12 Boiler Deposits: Occurence And Control Contact:

DEPOSITS:
Common feedwater contaminants that can form boiler deposits include calcium, magnesium, iron, copper, aluminum, silica, and (to a lesser extent) silt and oil. Most deposits can be classified as one of two types (Figure 12-1)...scale that crystallized directly onto tube surfaces
sludge deposits that precipitated elsewhere and were transported to the metal surface by the flowing water
Scale is formed by salts that have limited solubility but are not totally insoluble in boiler water. These salts reach the deposit site in a soluble form and precipitate when concentrated by evaporation. The precipitates formed usually have a fairly homogeneous composition and crystal structure.

So, this growing deposit could likely reduce the concentration of copper ions that in even micro quantities has been reported as having an oligodynamic effect.
For example, per a reference a 2011 work published in Applied Environmental Microbiology, Metallic Copper as an Antimicrobial Surface, to quote:

The antimicrobial activity of copper and copper alloys is now well established, and copper has recently been registered at the U.S. Environmental Protection Agency as the first solid antimicrobial material...In connection to these new applications of copper, it is important to understand the mechanism of contact killing since it may bear on central issues...

As to possible mechanics, the article further notes:

In these enzymes, copper serves as an electron donor/acceptor by alternating between the redox states Cu(I) and Cu(II) (15)... the redox properties of copper can also cause cellular damage. A number or mechanisms have been suggested. Reactive hydroxyl radicals can be generated in a Fenton-type reaction:

$\ce{ Cu+ + H2O2 -> Cu++ + OH- + .OH}$ (1)

The extremely reactive hydroxyl radical can participate in a number of reactions detrimental to cellular molecules, such as the oxidation of proteins and lipids (45).

So at least partially, suspected Fenton-like properties of copper could contribute to its reputed antimicrobial properties. However, this requires a surface presence of either metal copper or cuprous salt, but a growing scale coating may be inhibiting to this action.


Why is bromobenzene more stable than benzene?

The corresponding section in Wade, L. G. Organic Chemistry, 8th Edition; Pearson Education: Glenview, IL, 2013; p 758. reads

  Bromination of Benzene Bromination follows the general mechanism for electrophilic aromatic substitution. (…)

  Formation of the sigma complex is rate-limiting, and the transition state leading to it occupies the highest-energy point on the energy diagram (Figure 17-1). This step is strongly endothermic because it forms a nonaromatic carbocation. The second step is exothermic because aromaticity is regained and a molecule of HBr is evolved. The overall reaction is exothermic by 45 kJ/mol (10.8 kcal mol).

Thus, the given value of 45 kJ/mol refers to the reaction enthalpy of the total reaction
$$\ce{Br2 + C6H6 -> C6H5Br + HBr}$$

The reaction is mainly so exothermic because of the formation of HBr. This can be shown using values for the standard molar enthalpy of formation taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align}
&\Delta_\mathrm f H^\circ(\ce{Br2(l)})&&=0\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{Br2(g)})&&=30.9\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{C6H6(l)})&&=49.1\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{C6H5Br(l)})&&=60.9\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{HBr(g)})&&=-36.3\ \mathrm{kJ/mol}\\
\end{align}$$

although the exact value of 45 kJ/mol cannot be reproduced using these values.


What is the most complex molecule we can manufacture without the help of proteins? (soft queston)

I guess in the field of inorganic chemistry polyoxometalates (POMs) are comparable with proteins in terms of structural complexity. Macroanionic POMs such as Keplerates $\ce{\{Mo132\}}$

are also capable of further self-assembly in diluted solutions, producing so-called "blackberries" [1] of a size of around $\pu{0.05 .. 0.1 \mu m}$:

Also, see the post Measure the complexity of a molecule and the corresponding paper [2] of Prof. Lee Cronin, one of the most productive polyoxometalate chemist of current era.

Kistler, M. L.; Bhatt, A.; Liu, G.; Casa, D.; Liu, T. J. Am. Chem. Soc. 2007, 129 (20), 6453–6460. DOI 10.1021/ja0685809.
Marshall, S. M.; Murray, A. R. G.; Cronin, L. arXiv:1705.03460 [q-bio.OT] 2017.


The picture in Wikipedia doesn't give you the primitive cell. In Yttria, $\ce{Y2O3}$, 8 metal ions are in the special positions $\frac{1}{4}$, $\frac{1}{4}$, $\frac{1}{4}$; the remaining 24 metals occupy the sites $u$, $0$, $\frac{1}{4}$. The 48 oxygen ions are in general positions $x$, $y$, $z$, and are arranged in distorted octahedra around the metal ions, the metal-oxygen bonding distances being unequal. So, the stoichiometry is $\ce{Y32O48}$ which gets simplified to $\ce{Y2O3}$. This structure is generally referred to as the C-type metal oxide structure. All the sesquioxides of the rare earths belong to this system, as do $\ce{Ga2O3}$, $\ce{In2O3}$, $\ce{Tl2O3}$ and the mineral bixbyite, $\ce{(Fe, Mn)2O3}$. The primitive cell looks like this (oxygen ions are red, metal ions are gray)

Here another view which might show more easily that indeed the positions of all the 48 oxygen atoms are different in such a way that no smaller primitive cell can be found.

Reference

M. G. Paton, E. N. Maslen, Acta Cryst. 1965, 19, 307-310


What is difference between intramolecular redox and disproportionation redox

Disproportionation is a redox reaction where one element works both ways, with part of it being oxidized and another part reduced. It does not have to be intramolecular in the strict sense, since the oxidizing and reducing agents do not necessarily come from the same molecule (though they come from similar molecules, of course). Think of the well-known reaction:
$$\ce{3KClO -> KClO3 + 2KCl}$$
It is certainly a disproportionation, but one may hardly call it intramolecular; there is not much going on within one molecule (or rather within one ion, for all of these are ionic compounds). Chlorine oxidizes chlorine; now, one ion contains one atom of chlorine, so to oxidize anything it needs to find another ion in the first place.

Intramolecular redox reaction occurs when we have the oxidizing and reducing agents within one molecule. They may or may not be of the same element. In the former case it is going to be a disproportionation, but not in the latter. Think of TNT:
$$\ce{2 C6H2(NO2)3CH3 → 3 N2 + 5 H2O + 7 CO + 7 C}$$
See what's going on? Basically, N oxidizes C. They are indeed from the same molecule, so this may be intramolecular, but they are not the same element, hence this is not a disproportionation.


There are several ways that enantiomers can be separated, but none of them are particularly simple.

The first way to separate them is chiral chromatography.  In chiral chromatography, silica gel is bonded to chiral molecules to form what is called a chiral stationary phase.  The enantiomers will then separate as they run down the column because one of the enantiomers will interact more strongly with the column and "stick" in place.  Chiral sugars (ex. cellulose) are frequently used in chiral chromatography.

The second common method is to react the enantiomers with another chemical to form diastereomers.  While enantiomers are identical in terms of chemical properties, diastereomers are not.  Diastereomers can be created by reacting a mixture of both the enantiomers with another chiral molecule, such as s-brucine, which is commonly used because it is cheap.  Diastereomers have different chemical properties (for example melting points), so it is much easier to separate them.  Then, after separation, the enantiomers can be recovered from the single diastereomer.


Why do vinylic carbocations generally not undergo hydride rearrangement from neighbouring sp3 carbon to get more stability?

If you think about protonating 1-propyne on the terminal acetylenic carbon, you get a vinylic cation that could apparently form an allylic cation by a hydride shift. The bond to the hydride that would shift has to be lined up with the empty orbital to which it is migrating.

So far, no problem. But after the shift, the empty $\mathrm{2p}$ orbital of the new carbocation is orthogonal (at right angles) to the $\mathrm{2p}$ orbitals of the adjacent double bond. Therefore, what looks like an allylic carbocation is actually a twisted allylic carbocation.

The $\mathrm{2p}$ orbitals of the double bond can't stabilize the new cation by resonance. Worse yet, the $\mathrm{2p^2}$ carbons of the double bond are relatively electronegative, so they actually destabilize the new carbocation. In other words, the migration is forming another unstable carbocation.

This is why rearrangements in vinylic carbocations don't occur. (There is a Russian Chemical Reviews article [1] that explains all this, but it's hard to get a copy of this, and their PDFs seem to be corrupted.)

References

Shchegolev, A. A.; Kanishchev, M. I. Rearrangements in Vinyl Cations. Russ. Chem. Rev. 1981, 50 (6), 553–564. https://doi.org/10/b7rrk8. (mathnet.ru mirror; PDF (in Russian))


The classic catch-all term is molecular entity, often abbreviated to just entity. There's a Wikipedia page for it, which references the IUPAC Gold Book entry. Quoting from the latter:

  Any constitutionally or isotopically distinct atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity.

  Molecular entity is used in this Compendium as a general term for singular entities, irrespective of their nature, while chemical species stands for sets or ensembles of molecular entities. Note that the name of a compound may refer to the respective molecular entity or to the chemical species, e.g., methane, may mean a single molecule of $\ce{CH4}$ (molecular entity) or a molar amount, specified or not (chemical species), participating in a reaction. The degree of precision
  necessary to describe a molecular entity depends on the context. For example, 'hydrogen molecule' is an adequate definition of a certain molecular entity for some purposes, whereas for others it is necessary to distinguish the electronic state and/or vibrational state and/or nuclear spin, etc. of the hydrogen molecule.

  Source: PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1142.

As you can see, the definition is exceptionally general. It also specifically includes atoms. Usage of the term is widespread, even if it seems a bit exotic when first encountered.

Another option which can work is chemical species (often abbreviated to just species), which is even mentioned in the definition above. Again, there is a Wikipedia page. IUPAC actually has two definitions for chemical species, but the relevant definition is the following:

  An ensemble of chemically identical molecular entities that can explore the same set of molecular energy levels on the time scale of the experiment. The term is applied equally to a set of chemically identical atomic or molecular structural units in a solid array.

  For example, two conformational isomers may be interconverted sufficiently slowly to be detectable by separate NMR spectra and hence to be considered to be separate chemical species on a time scale governed by the radiofrequency of the spectrometer
  used. On the other hand, in a slow chemical reaction the same mixture of conformers may behave as a single chemical species, i.e. there is virtually complete equilibrium population of the total set of molecular energy levels belonging to the two conformers. Except where the context requires otherwise, the term is taken to refer to a set of molecular entities containing isotopes in their natural abundance. The wording of the definition given in the first paragraph is intended to embrace both cases such as graphite, sodium chloride or a surface oxide, where the basic structural units may not be capable of isolated existence, as well as those cases where they are. In common chemical usage generic and specific chemical names (such as radical or hydroxide ion) or chemical formulae refer either to a chemical species or to a molecular entity.

  See also: chemical species (of an element)

  Sources:
  PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1096.
  PAC, 1996, 68, 2193. (Basic terminology of stereochemistry (IUPAC Recommendations 1996)) on page 2202.

In this case, there is a notion of an ensemble (a collection/multiple) of whatever is being referred to. This is why "chemical specimen" (singular) is never used.

In short, a (chemical) species is composed of a collection of (molecular) entities. Both are umbrella terms covering virtually everything, and therefore can be used in almost all contexts. Perhaps in your case, (molecular) entity is the preferred term, as it emphasizes the differences in identity of what interacts with the cells.


The $\ce{Br-}$ would make less $\ce{HBr}$ dissociate, But $\ce{HBr}$ is such a strong acid ($\mathrm{pKa}$ is $\mathrm{-9}$) that still only a trivial amount would be molecular.


Find the initial volume of the weak acid given a titration graph

Outline

There are three main points of interest in the titration graphic, and two you have found yourself. The three points are

The start at $V_\text{titr}=0$
Halfway to the equivalence point at $V_\text{titr}=10~\mathrm{mL}$
The equivalence point at $V_\text{titr} = 20~\mathrm{mL}$

Why is number 2 important? Because that is where we can read out the $\text{p}K_\mathrm{a}$ value of the acid, because at that point $[\ce{HA}] = [\ce{A-}]$.

Now, armed with this information, here is how to proceed:

Find out total amount of acid
Find out total concentration of acid
Calculate the initial volume from the data gathered in steps 1 and 2

1  Total amount of acid

The total amount of acid can be gleaned from information point 3, the equivalence point.

$$ n_\text{eq} = V_\text{titr,eq} \cdot c_\text{titr} = 20~\mathrm{mL} \cdot 0.150~\mathrm{mol\,L^{-1}} = 0.003~\mathrm{mol} $$

2  Total concentration of acid

The total concentration can be calculated from information points 1 and 2. We know the initial concentrations (denoted by subscript $_0$) because of the pH value at the beginning:
$$ [\ce{A-}]_0 = [\ce{H+}]_0 = 10^{-3}$$

Using the equilibrium constant that we got at point two, we can calculate the initial concentration of the associated (un-dissociated) acid species:
$$ [\ce{HA}]_0 = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}} = \frac{10^{-6}}{10^{-5}} = 10^{-1} $$

The total initial concentration of all acid is then trivial:
$$ [\ce{A}]_\text{tot,0} = [\ce{HA}]_0 + [\ce{A-}]_0 = 0.101 $$

Note that we here actually calculated the activities and not concentrations. Since we assume an activity coefficient of $\gamma = 1$ because we only consider dissolved species, the concentrations are simply the activity multiplied with the standard concentration:
$$ c(\ce{A})_\text{tot,0} = \gamma \cdot [\ce{A}]_\text{tot,0} \cdot 1~\mathrm{mol\,L^{-1}} = 0.101~\mathrm{mol\,L^{-1}} $$

3  Calculation of the initial volume

The initial volume is now calculated following the simple concentration formula:
$$ c_0 = \frac{n_0}{V_0} \Leftrightarrow V_0 = \frac{n_0}{c_0}$$

Since the amount of acid is constant ($n = n_0\; \mathrm{const}$), we can easily fill in the values:
$$ V_0 = \frac{0.003~\mathrm{mol}}{0.101~\mathrm{mol\, L^{-1}}} \approx 0.0297~\mathrm{L} = 29.7~\mathrm{mL} $$


The mistake is in the charge $q$, that you have put in:

$
D_A=\frac{\mu_{q}k_BT}{q}
$

$q$ is simply the particle charge (and not the whole charge, as you've unfortunately considered) and should then be expressed as:

$q=\frac{zF}{N_{A}}$

with $F$ the Faraday constant (charge / mole ), ${N_{A}}$ the Avogadro's number (number of particle / mole) and $z$ the charge valence.

${N_{A}}$ will finally cancel out when you divide Boltzmann's constant $k_B$ by the the charge $q$, in the definition of diffusion coefficient. That will allow you to recover the Goldman equation found on Wikipedia.

By checking all that through an always useful dimension analysis for $D_A$ and using the defintions you have given in your question for ($\mu_q$ = $\frac{v_d}{E} = \frac{v_d L}{V_m}$ and $k_B = \frac{R}{N_A}$):

$
\frac{\mu_{q}k_BT}{q}=\frac{m/s\,\times m}{V}\times \frac{J/(mol\times K)\,\times K}{{N_A}\times C/mol}\times{N_A}=\frac{m^2}{s}
$

you find the correct units for a diffusion coefficient.


To be accurate you will need standard gas cylinder and pneumatic connections (pressure regulator, needle valves or better mass flow controllers, pipes). You will need a measuring chamber with an overflow. As a rule of thumb, if the chamber geometry is not exotic (lets say a cube with no dramatic dead volume), you need to flush it, at least  10 times, with each concentration you want to measure.

You can do a 2 points calibration for O2 by measuring atmosphere where O2 concentration can be assumed constant (span) and then flush with pure nitrogen (zero). For CO2 and CO you will need standard cylinders.

Anyway a 2 points calibration is a little bit risky because you will not be able to assess linearity and then you will have no idea in which extent your sensor is linear (dynamic range). Properly calibrating sensor requires multiple points calibration and therefore a dilution device and standards.

Caution: When dealing with gas you must take care of exhaust and prevents risk such as intoxication (especially CO and in a fewer extent CO2) or asphyxiation (due to a lack of oxygen) and explosion (CO is explosive).

You should find those materials, resources and skills in a University (Chemistry or Physics Department) or an Air Quality Laboratory.

Note: If your sensors are electrochemical ones, you need also to assess their correlations/deviations with temperature and humidity.


Electrolysis of water with aluminum at the negative pole

If you are using salt to make the solution more conductive the process will cause the creation of Sodium Hydroxide, which will react with aluminium.


What happens if I have an acid/base equillibrium and remove hydronium ions until there is 50/50 of acid and conjugate base?

You can't just remove hydronium ions because the solution needs to remain electrically neutral.

You can't just add hydroxide ions, but you could add sodium hydroxide until the HA/A- ratio was 50/50.

  Is this a very good buffer?

It is the optimum ratio of HA and A- for buffering.


Separation of sulfur dioxide and sulfur trioxide in the contact process

You are true that $\ce{SO2}$ is very soluble in water but it depends on the temperature. Also, $\ce{SO3}$ reacts with water then dissolves in it in a very exothermic way. My understanding is that, if the quantity of water is well adjusted, you will actually get hot concentrated sulfuric acid and sulfur dioxide is not quite soluble in it (ca. 8 g/kg at 120 °C). So $\ce{SO3}$ reacts with water and sulfur dioxide gas can be recirculated.

HTH


Optional determination techniques in Atomic Absorption Spectroscopy

Remember this is "atomic" spectroscopy so it is expected that the "sample" is being totally ripped apart. So one "best technique" per element.

$\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ are analyzed by the graphite tube technique rather than in a flame because the technique provides more sensitivity.

$\ce{Hg}$ is an odd one since you use a cold vapor technique.

Of course depending on the particular sample matrix what you might need to change.


Why do exothermic dissolution reactions occur?

Obviously the intramolecular bonds are not stronger than the intermolecular bonds if the majority of the solute dissociates when placed in water. There are some salts ($\ce{AgCl}$, $\ce{BaSO4}$, $\ce{ZnS}$) that have extremely low $\ce{K_{sp}}$ values, meaning the majority of the salt placed in the solution does not dissociate, and are therefore considered insoluble salts. In these salts, the intramolecular bonds are stronger than the intermolecular bonds formed with the water, which explains why only extremely small amounts of these salts dissociate in water.

There are, on the other hand, salts that do dissociate in water. Dissolution requires energy, and this energy is supplied by the kinetic motion of the water. This energy is used in breaking apart lattice structure of the salts, thereby lowering the temperature of the water.

The difference between salts with a negative enthalpy of solution and a positive enthalpy of solution, as you stated, is in formation of bonds between the water and the ions of the salt. The formation of a hydrogen bond is an extremely exothermic process, and accounts for the extremely negative enthalpies of solution of $\ce{HF}$, $\ce{NaOH}$, and $\ce{CsOH}$. Weaker intermolecular bonds also form in the dissolution of these salts and many others as a result of the interaction of the water with the electron clouds of the ions. Interactions between the water and the ions always occur, though to a greater extent in some salts than others.


I'm not sure I have answer to your question maybe, if you consider natural colorant too, these are the articles that go more close to what you want:

Prediction of phenolic compounds in red wine fermentations by visible and near infrared spectroscopy

Spectroscopic, stability and radical-scavenging properties of a novel pigment from gardenia

Try to have a look a these too:

Mimani, T., and Samrat Ghosh. "Combustion synthesis of cobalt pigments: Blue and pink." Current Science 78.7 (2000): 892-896.

Bernardi, M. I. B., et al. "Comparison of blue pigments prepared by two different methods." Journal of the European Ceramic Society 22.16 (2002): 2911-2919.

Visinescu, Diana, et al. "The environmentally benign synthesis of nanosized Co< sub> x Zn< sub> 1− x Al< sub> 2 O< sub> 4 blue pigments." Dyes and Pigments 87.2 (2010): 125-131.


Dependence of change in entropy on temperature

An analogy that I like is one paraphrased from Peter Atkins' great book "The Laws of Thermodynamics - a VSI":

  Imagine two rather different places in terms of noise - a silent
  library and a busy train station. Imagine that the temperature is the amount of noise at the given place - a measure of the "state of noise". Let's say that you laugh out loud,
  corresponding to supplying energy as heat.

  In the library, where there is little background noise (low $T$) this would cause a large amount of disturbance and
  disorder - corresponding to a large increase in entropy. On the busy
  train station, on the other hand, you laughing would go largely
  unnoticed - there is lots of background noise already (high $T$). The (relative) amount of additional disturbance and
  disorder would be much smaller, corresponding to a smaller increase of
  entropy.

In a very cold system, even a tiny amount of supplied energy in terms of heat would cause a large change in entropy. On the other hand, supplying the same amount of energy to a really hot system - with lots of molecular motion going on - would cause a minuscule change in entropy.

Disclaimer: Note that the analogy and explanation above are two ways of convincing yourself that the entropy change due to added heat is inversely proportional to the system temperature. They do not claim to be correct or physical in any other sense, so read them with a (large) pinch of salt.


Do pipettes always outclass glass syringes?

Glass syringes and glass pipettes are high precision and high accuracy tools of an analytical chemist. Like any tool at home, such as a cloth cutting scissor and a plant pruning clipper do the same job of cutting but their applications are different.
As an analytical chemist, I have not seen anyone using microsyringes for standard preparations in volumetric flasks. One uses a class A pipette for volumetric transfers and to make different standards.
Now you wish inject known microliter or nanoliter volumes in a GC injection port or an HPLC rheodyne, can we use glass pipettes? You would choose a glass syringe to pierce the GC septum.
The tip is that you would use the highest quality glass pipette (class A) and highest quality syringes in each case. class A 1 mL pipettes are commercially available.
If you really wish to compare, you would have to do a calibration yourself on a analytical balance under tightly controlled conditions.


There are many ways in which bond lengths can be changed, and these will be accompanied by changes in bond energy.  I will give examples for four different classes of substances, though they may not all count as "forcing" the bonds to be shorter in the way in which you mean.  You can look these over, and perhaps clarify your question:
Glasses: Toughened glass is made by rapidly cooling glass, which creates internal stresses in which the interior is under tension (increasing bond lengths), while the surface is under compression (reducing bond lengths).  So while the changes in bond length are still a result of pressure, the pressure is internally, rather than externally, applied.
Molecules: You can get the molecule to undergo a chemical change.  For instance, changing ethane to ethene will shorten the carbon-carbon bond by converting it from a single bond to a double bond.
Metals: Different metallic crystal structures will have different densities, and thus likely also different average bond lengths (the change in density is not due purely to a change in bond lengths, but also to changes in atomic arrangements, including the number of nearest neighbors, next-nearest-neighbors, etc.). Thus you can alter bond length by inducing the substance to change from one crystal structure to another.  As an example, I'll quote from https://www.conservation-wiki.com/wiki/Tin_pest :

Solid tin exists in two allotropic forms, white and grey, each of which contains different chemical properties. White tin has a body centered tetragonal crystalline structure and a density of 7.28, while grey tin has a cubic structure and a density of 5.84 (Faulkner 1965). This transformation is generally accepted to occur around 13C (Gilberg 1991:3). Below this temperature, grey tin is more stable, while white tin is more stable at temperatures above 13C.

Covalent network solids:  Essentially the same effect as what was described for metals, above, except that the transformation involves making and breaking covalent bonds.  A good example is the transformation between diamond and graphite; though note that, in this case, the structure changes from 3D (diamond) to 2D (graphite)
Here's a graphic by Wikipedia contributor Nath321 illustrating this: https://en.wikipedia.org/wiki/Atomic_spacing -- though here the comparison is between a 2D structure:


Difference between catenation and allotropy

Catenation refers to the ability of an element to form long chains. The most notable example is carbon (catenation of carbon is what gives rise to the entire field of organic chemistry).

Allotropy is an entirely different concept; it refers to the natural existence of an element in different forms, i.e. bonded differently. For example, solid carbon exists in the forms of diamond and graphite (and more).

In a sense, the two are related because the ability of an element to catenate often leads to the formation of many different allotropes, since the chains can be linked in different ways. Apart from carbon, another example of an element that can catenate well and forms multiple allotropes would be phosphorus. On the other hand, nitrogen and oxygen prefer to form triple and double bonds respectively, instead of single bonds; this means that catenation is unlikely and that is why these elements have few allotropes (nitrogen exists as diatomic $\ce{N2}$, oxygen as diatomic $\ce{O2}$ and ozone).


How can a nonlinear field gradient be generated in NMR?

Regular off-the-shelve pulsed field gradient (PFG) systems only give linear gradients, obviously. However you can produce higher-order gradients with your shim coils, normally to compensate similar opposing unwanted gradients in the field of the main magnet.
So, by manipulating the shim set, you can create any odd gradient field that you can sufficiently approximate with the terms that your shim system supplies (usually up to 5th order in $z$). To find out what exactly the terms are (coordinates and absolute amplitudes), you can image them using the regular PFG system.
There are two hitches: You cannot switch the shim system quite so fast as the dedicated PFG set, and the "room temperature" shim system isn't nearly as powerful as a PFG, so you are limited in terms of gradient strength. (The "cryoshims" are stronger, but you cannot switch them, as they are superconducting.)
So, if you're OK with a static gradient of low amplitude, you can go straight ahead. If you need it to be pulsed, give it a try. It's been done before, switching shim currents during a pulse sequence. Very good chance to really get to know the insides of your spectrometer system.

If you're secretly an engineer also, check my other answer. ;)


Is dissolution of H2S in water exothermic?

Koschel et al. (2013, p. 14486) report that

  dissolution of gaseous hydrogen sulfide in aqueous phase is exothermic at conditions where $\ce{H_2S}$ is gaseous, i.e., at a temperature of 323 K and pressure 2 MPa, the saturation pressure $p_\text{sat}(\mathrm{H_2S})$ being 3.5 MPa. The dissolution process is endothermic at the medium and highest temperature where hydrogen sulfide is liquid and supercritical, respectively, the critical parameters of $\ce{H2S}$ being $T_c = 373.4$ and $p_c = 8.94$ MPa.


You cannot obtain $\ce{CuCl2}$ mixing $\ce{CuSO4}$ and $\ce{NaCl}$ because both are very soluble, the second strategy seems good because $\ce{CuCO3}$ is insoluble in water, and you can react the new salt with $\ce{HCl}$.

Good luck.


If there are no other ingredients like citric acid- 1800mg. It is probably magnesium BISglycinate. For each Mg ion, there are two glycine molecules.

After a calculation, if it's MagBisglycinate, for 200mg Mg you should have around 1235.5mg glycine. It probably means there are other ingredients or simply glycine that left unreacted. If it's simple Magnesium glycinate than you should have 617.7mg glycine for 200mg Mg.

Correct me if I'm wrong.


Why is cis-1,2-dichloroethene more stable than trans-1,2-dichloroethene?

Here's a pre-MOT rationalisation of the cis form being more stable than the trans form for 1,2-dihaloethylenes:

The lone pair of chlorine atoms is involved in resonance with the double bond, as it does so positive charge appears on one chlorine and negative on another. A cis geometry allows for a stabilizing interaction (attraction) between these positive and negative charges, which makes this resonance form a greater contributor to the resonance hybrid than it is for the trans isomer.

MOT-based explanation:

Eyring's work suggests that there is maximum delocalisation possible in trans form.

Quoting, from the same paper:

  In a recent paper Binghams proposes that the main effect responsible for the “cis effect” of ethylenes is an effect which tends to destabilize the trans form rather than provide extra stabilization for the cis form. In short, this is an unfavorable conjugation effect which stems from the fact that the conjugation in these cases is between two filled orbitals (halogen lone pair and $\pi$ bond). The interaction gives rise to two new orbitals, one bonding and one antibonding, the latter being more antibonding than the former is bonding. Consequently, the total interaction is energetically unfavorable, in contrast to interaction in hydrocarbon polyenes. This would be equally true for both the cis and trans forms but, following Eyring’s argument, the delocalization is greater in the trans form.
  Consequently this form will be the less stable one.

Reference: Structure of 1,2-difluorocyclopropane and the "cis effect"


What are the fumes that form above liquid nitrogen?

The visible fog that forms when liquid nitrogen is poured into an open container is almost entirely water fog: the boiling nitrogen chills the air above it, causing the humidity in the air to condense into fog.

There may be some microdroplets of liquid nitrogen in the fog too, but the air, even after it's been chilled, is still much warmer than the boiling point of nitrogen, so there won't be much.  (You can condense air by running a continuous stream of liquid nitrogen through a heat exchanger and blowing air over it, but mostly you get liquid oxygen when you do that.)

Liquid nitrogen has a relatively small heat of vaporization, as these things go: 199.2 kJ/kg according to Air Liquide.  (Compare 572.2 kJ/kg for dry ice.)  That plus the Leidenfrost effect means it's actually rather difficult to freeze yourself with the stuff. There's a standard demo where you pour liquid nitrogen into an audience member's cupped hands. The Leidenfrost effect prevents it ever coming in contact with their skin, and it only leaves a mild chill behind.  Try this with dry ice and your "volunteer" will need emergency treatment for frostbite.

Gaseous nitrogen is second only to the noble gases in its inertness, and it makes up 78% (by volume) of the air you're breathing as you read this, so obviously the "fumes" aren't harmful in themselves.  However, as was mentioned in an earlier answer (which seems to have been deleted), if you let liquid nitrogen evaporate in an enclosed space, it can "displace" the oxygen that makes up another 21% of the air, at which point anyone in the area may asphyxiate.  The sensation of suffocation is triggered by having too much CO2 in your bloodstream, not by inadequate oxygen supply, so you might not even notice that anything was wrong until you collapsed.  That's the second most dangerous thing about evaporating liquid nitrogen.

The most dangerous thing is, if you let liquid nitrogen evaporate in a sealed container, it's liable to explode on you.


References for learning about acid and bases in organic chemistry

Concerning your first question: I found that the  book "Organic Chemistry" by Clayden, Warren, Wothers and Greeves and here especially the 8th chapter contains a very good and readable introduction to acidity and basicity. After you have worked through this chapter you should be able to explain why certain molecules are as basic/acidic as they are and you see a lot of typical $\mathrm{p}K_{\mathrm{a}}$ values for a lot of substance classes so that you get some knowledge about what $\mathrm{p}K_{\mathrm{a}}$ values you can expect for other substances. As for your second question: If you want to know why reactions occur in the way they do you'll have to take courses in Organic Chemistry. Apart from that, I wholeheartedly recommend the aforementioned book again. If you work through this book (this will take you a while: it has 1200 pages... but then again, that is 1200 very readable pages :)) you'll have a very solid basis in organic reaction mechanisms and organic chemistry in general. If you want to dig a little deeper this book by Bruckner is very thorough but a little dry.

One particular perspective that I found very helpful in explaining why reactions happen in the way they do was the frontier orbital theory. A very good and readable book about its use in organic chemistry is this one by Fleming. The book by Clayden et al. also uses this theory a lot in explaining reaction mechanisms.


Would graphite stick to basalt cooled from its molten state in a graphite mold?

No, it would not stick. In our lab we commonly use graphite dishes/cups/etc to contain small amounts of liquid rock. All you need to do then is to hold it upside down and the solidified rock (now glass mostly) falls out. Sometimes there's a bit of graphite dust on it, but it can be easily scraped off.

Doing it in a reducing (or oxygen-free) environment will make it even easier, because graphite slowly burns to carbon dioxide with time, on the scale of hours, from my own personal experience.


What household substances could be distilled

This is tricky. Besides water, nearly everything has some hazard associated with it.  And even water distillation involves high temperature (100C).  Be careful distilling alcohol: the azeotrope of distilled alcohol is 95% ethanol which is VERY flammable! Plus, technically speaking, distilling alcohol may be illegal where you live.

One thing you could do is distill water with dissolved substances, most likely table salt (NaCl).  This would help you test the principle that distillation is a technique for separating substances. In this case, distilling salt water should yield pure water. A more visual example would be brewed coffee or tea, which is a suspension and should separate into water and crud!

And after that, I'd be careful! For example, concentrating acetic acid (white vinegar) is dangerous, as concentrated acetic acid is both corrosive and has choking fumes.  I would not even consider trying something that is not food grade, given your requirements.


Calculation of total pressure given ratio of pressure equilibrium constants

The equilibrium expressions for these reactions are:

$$ K_{\text{p},1} = \frac{p_{\text{Y}}^2}{p_{\text{X}}}$$
$$ K_{\text{p},2} = \frac{p_{\text{P}}p_{\text{Q}}}{p_{\text{Z}}}$$

Since $p_\text{A} = x_\text{A}p_\text{T}$ for component $\text{A}$, we can rewrite them as:

$$ K_{\text{p},1} = \frac{(x_\text{Y}p_{\text{T},1})^2} {x_\text{X}p_{\text{T},1}} = \frac{x_{\text{Y}}^2p_{\text{T},1}}{x_\text{X}}$$
$$ K_{\text{p},2} = \frac{(x_\text{P}p_{\text{T},2})(x_\text{Q}p_{\text{T},2})}{x_\text{Z}p_{\text{T},2}} = \frac{x_{\text{P}}x_{\text{Q}}p_{\text{T},2}}{x_\text{Z}}$$

Taking the ratios:

$$ \frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = \frac{x_{\text{Y}}^2x_\text{Z}}{x_\text{X}x_\text{P}x_\text{Q}}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}\label{eqn}\tag{1}$$

We're asked the ratios of the total pressures, and we're given the ratios of the equilibrium constants, so we just need to find the molar fractions ratio in that equation. We know that the degree of dissociation, let's call it $\alpha$, is the same for both reactions.

Assume the first equilibrium starts with $m$ moles of $\text{X}$:

$$\begin{array}{cccc}
  &\ce{X(g) &<=> &2 Y(g)}\\ \hline
\text{I} &m & &0\\
\text{C} &-\alpha m & &+2 \alpha m\\
\text{E} &m - \alpha m & & 2 \alpha m
\end{array}
$$

Then, the molar fractions are:

$$x_\text{X} = \frac{n_\text{X}}{n_\text{T}} = \frac{m - \alpha m}{m - \alpha m + 2 \alpha m} = \frac{1 - \alpha}{1 + \alpha}$$
$$x_\text{Y} = \frac{n_\text{Y}}{n_\text{T}} = \frac{2\alpha m}{m - \alpha m + 2 \alpha m} = \frac{2\alpha}{1 + \alpha}$$

Similarly, for the second equilibrium, assuming we start with $r$ moles of $\text{Z}$:

$$\begin{array}{cccccc}
  &\ce{Z(g) &<=> & P(g) &+ &Q(g)}\\ \hline
\text{I} &r &    &0     &  & 0\\
\text{C} &-\alpha r &    &+\alpha r     &  &+\alpha r\\
\text{E} &r - \alpha r &    &\alpha r     &  & \alpha r
\end{array}
$$

And we calculate the mole fractions of each species:

$$x_\text{Z} = \frac{n_\text{Z}}{n_\text{T}} = \frac{r - \alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{1 - \alpha}{1 + \alpha}$$
$$x_\text{P} = \frac{n_\text{P}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$
$$x_\text{Q} = \frac{n_\text{Q}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$

Plugging the found molar fractions into $\ref{eqn}$, we have:

$$\frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = \frac{\left(\frac{2\alpha}{1 + \alpha}\right)^2 \left(\frac{1 - \alpha}{1 + \alpha}\right)}{\left(\frac{1 - \alpha}{1 + \alpha}\right)\left(\frac{\alpha}{1 + \alpha}\right)\left(\frac{\alpha}{1 + \alpha}\right)}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

Which simplifies to:
$$\frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

Since $\displaystyle \frac{ K_{\text{p},1} }{K_{\text{p},2}} = \frac{1}{9}$, we get:

$$\frac{1}{9}
 = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

And finally:

$$\frac{p_{\text{T},1}}{p_{\text{T},2}} = \frac{1}{36}$$

So the total pressures are on a ratio of $1:36$.


Conversion of methylcyclopentanol to cyclopentanone

If you leave it to E1, you will not get the selectivity you need.

The difference between the methyl group and the methylene unit is steric bulk. You can select for deprotonation of the methyl for elimination via a bulkier base in an E2 reaction.

In addition, you will need to convert the substrate so that it has a good leaving group. For an alcohol, this is typically accomplished by first converting to a tosylate or a mesylate.


Do electrons of opposite spin attract each other?

Electrons can sort of attract. Let's review magnetic dipoles. First, suppose we have a current loop $I$ rendered in blue at radius $a$ around the $z$-axis in the $xy$-plane and we want to compute the magnetic field at some point on the $z$-axis marked in red. See Figure $1$.

The source point vector is
$$\vec{r}_s=\langle a\cos\phi,a\sin\phi,0\rangle$$
So the element of arc length along the current path is
$$d\vec\ell=d\vec{r}_s=\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi$$
The field point vector is
$$\vec{r}_f=\langle0,0,z\rangle$$
So the vector from the source point to the field point is
$$\vec{r}=\vec{r}_f-\vec{r}_s=\langle-a\cos\phi,-a\sin\phi,z\rangle$$
And its magnitude
$$r=\left\lVert\vec{r}\right\rVert=\sqrt{a^2+z^2}$$
And a unit vector in its direction
$$\hat{r}=\frac{\vec{r}}{r}=\frac{\langle-a\cos\phi,-a\sin\phi,z\rangle}{\sqrt{a^2+z^2}}$$
So we can get the magnetic field at $\vec{r}_f$ via the Biot-Savart law
$$\begin{align}\vec B&=\frac{\mu_0}{4\pi}\int\frac{Id\vec\ell\times\hat{r}}{r^2}\\
&=\frac{\mu_0I}{4\pi}\int_0^{2\pi}\frac{\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\langle-a\cos\phi,-a\sin\phi,z\rangle}{(a^2+z^2)^{3/2}}\\
&=\frac{\mu_0I}{4\pi(a^2+z^2)^{3/2}}\int_0^{2\pi}\langle az\cos\phi,az\sin\phi,a^2\rangle\,d\phi\\
&=\frac{\mu_0}{4\pi}\frac{2\pi a^2I}{(a^2+z^2)^{3/2}}\hat k\approx\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k\end{align}$$
Where the magnetic dipole moment of the loop is $m=\pi a^2I$ and we are looking at the far field of the dipole so $a^2+z^2\approx r^2$. The magnetic scalar potential of a dipole, valid away from its source current, is
$$\Phi_m=\frac{\mu_0}{4\pi}\frac{mz}{r^3}$$
so
$$\begin{align}\vec B&=-\vec{\nabla}\Phi_m=-\frac{\mu_0m}{4\pi}\left\langle-\frac32\frac{2xz}{r^5},-\frac32\frac{2yz}{r^5},-\frac32\frac{2z^2}{r^5}+\frac{r^2}{r^5}\right\rangle\\
&=\frac{\mu_0m}{4\pi r^5}\langle3xz,3yz,3z^2-r^2\rangle\end{align}$$
Along the $z$-axis, $x=y=0$, $r=z$ so
$$\vec B=\frac{\mu_0}{4\pi}\frac{2m}{r^3}\hat k$$
So the scalar potential checks. Along the $x$-axis, $y=z=0$, $r=x$ and
$$\vec B=-\frac{\mu_0m}{4\pi x^3}\hat k$$
so
$$\frac{\partial B_z}{\partial x}=\frac{3\mu_0m}{4\pi r^4}$$
This is the same as for a scalar potential of
$$\Phi_m=-\frac{\partial B_z}{\partial x}\,xz$$
because then
$$\vec B=-\vec{\nabla}\Phi_m=\frac{\partial B_z}{\partial x}\langle z,0,x\rangle=\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle$$
For the field at our loop due to an identical loop at $\langle-r,0,0\rangle$. So then the force on our loop is
$$\begin{align}\vec F&=\int Id\vec\ell\times\vec B\\
&=\int_0^{2\pi}I\langle-a\sin\phi,a\cos\phi,0\rangle\,d\phi\times\frac{3\mu_0m}{4\pi r^4}\langle0,0,a\cos\phi\rangle\\
&=\frac{3\mu_0mI}{4\pi r^4}\int_0^{2\pi}\langle a^2\cos^2\phi,a^2\sin\phi\cos\phi,0\rangle\,d\phi\\
&=\frac{3\mu_0mI\pi a^2}{4\pi r^4}\hat i=\frac{3\mu_0m^2}{4\pi r^4}\hat i\end{align}$$
So in this configuration the force is repulsive if the dipoles are aligned but attractive if pointed opposite one another. You can see this if you have a neocube where oppositely aligned strings of magnet spheres attract at the equators.
For electrons things are more complicated. If we have little bits of charge $\Delta q$ and mass $\Delta m$ orbiting around an axis at distance $a$ and angular frequency $\omega$ then the charge passing a point in the orbit per unit time is $I=f\Delta q=\frac{\omega\Delta q}{2\pi}$ so the magnetic dipole moment is $m=I\pi a^2=\frac12\omega a^2\Delta q$. The angular momentum is $L=rp=a\Delta ma\omega$ so we would get
$$\frac mL=\frac12\frac{\Delta q}{\Delta m}$$
But life isn't so simple for a couple of reasons: the ratio is actually a factor of $g_e\approx2.002$. Also there is a question of whether we want to consider the total angular momentum of the electrons or the component at right angles to the vector from one to another. We choose the latter and this component is $L_z=\frac12\hbar$ so we get
$$m=\frac{g_eq_e\hbar}{4m_e}$$
So the magnetic force is
$$F=\frac{3\mu_0}{4\pi r^4}\left(\frac{g_eq_e\hbar}{4m_e}\right)^2$$
Comparing with the electrostatic force,
$$F=\frac{q_e^2}{4\pi\epsilon_0r^2}=\frac{\mu_0c^2q_e^2}{4\pi r^2}$$
We see that the magnetostatic force is greater when
$$r\lt\frac{\sqrt3g_e\hbar}{4m_ec}=\frac{\sqrt3(2.002)(1.055\times10^{-34}\text{J}\cdot\text{s})}{4(9.11\times10^{-31}\text{kg})(2.998\times10^8\text{m}/\text{s})}=3.35\times10^{-13}\text{m}$$
This is so close that it kind of negates the whole calculation though: the electrons are going to be moving pretty fast so there will be a magnetic force due to their relative motion and at these distances you really need quantum electrodynamics to analyze the problem. Still, it gives you an idea of how close you have to get so that the net force has at least a chance of being attractive.


Does rearrangement occur in Friedel Crafts alkylation or not?

When you conduct the alkylation of benzene with n-propyl chloride, the product you obtain is isopropylbenzene, so a rearrangement occurred by virtue of the fact that the primary carbocation that forms will undergo a 1,2-hydride shift to form a secondary one. Most alkylation ( aromatic ring and alkyl halides ) are conducted in the presence of aluminium chloride which is one of the strongest Lewis acids. The principle is that the alkyl halide reacts with the aluminium chloride to form a salt: a carbocation and a complex ion of tetrachloroaluminate. You can have methyl chloride react in this way because even if it forms an unstable carbocation, it will readily react with the aromatic ring ( which acts as a Lewis base ). You can have methyl carbocation in solution such as in alkylation reactions, but you can’t isolate a salt containing methyl carbocation in solid state. If you had methyl chloride and aluminium chloride in solid state ( in contact with each other, they would not react to form carbocations.


Is there an easy way to make PVC from PVA?

Choring Tablets have positivly charge cholrine atoms you want negative, so chlorine tablets will not work.  And no you won't get a dioxin compound unless you burn it, then you may ge some.  If you want to make PVC from PVA you can use Hydrochloric acid but you may not get a good yield.


I did a cursory search and found some information particularly from books of late 19th century(1,2):
Cinnabar and iodine

On heating, it (cinnabar) is readily decomposed with formation of small scarlet crystals of mercuric iodide.

Cinnabar and bromine

[...] After digesting the cinnabar with excess of bromine liquid (or bromine water containing 3% of bromine or solution of hydrochloric acid containing 13% of bromine) for a few days, the whole is dissolved, and also any metallic mercury present [...] The amount of bromine required is very considerable, 1 part of mercury is obtained and 3.5 parts of hydrobromic acid remains in the solution.

References

Chemical News and Journal of Industrial Science, Volumes 37-38, Chemical news office., 1878
Dictionary of Arts, Manufactures and Mines by Robert Hunt, Andrew Ure, 1878


How were the elements discovered before modern technology?

Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?

In many cases, they weren't - the ancient... maybe we can't call them scientists. Let's call them "natural philosophers." The ancient natural philosophers (up until about 1700) only knew about a handful of elements. This is part of the reason that it took so long to put the periodic table together.

At the time, an element was something that was not reducible to smaller components. The ancient philosophers tended to conflate properties of things with this idea of fundamental components, and as a result, things like earth, air, fire, and water were thought of as elements. When we say they "knew about" modern elements, what we mean is that they knew of the substances - they did not necessarily know that they were chemical elements in the sense that we define them today.

That modern concept of an element defined by the number of protons in the nucleus wasn't developed until around 1913. The wikipedia article on the history of chemical element definitions gives a good overview of how the definition has evolved over time.

The time period you are probably most interested in is that between the "ancient" era (pre-history up until the mid-1660's or so) to the beginnings of the modern era (say early 1900's). During this period, chemistry was evolving from alchemy into a modern science. The systematic investigation and communication of results was occurring, and a large body of known but poorly understood chemical "recipes" was available to chemists. Using a range of reactions and isolation techniques, they basically just randomly tried things (maybe not completely randomly) until they reached a point where it seemed like something could not be decomposed further, or until they identified a new substance. In many cases, they would discover the new element, but not be aware that it was an element - in other words, they could isolate and identify a substance, but because our collective understanding of chemistry was still very limited, they had no way of knowing that it was an elemental substance.

Wikipedia's list of elemental discoveries gives details about the discovery of each element - although you should keep in mind that "discovery" in this case means "identified as a new substance," not necessarily "identified as a chemical element."


Is there also a leveling effect for weak acids in water?

To answer your first question, weak acids don't get levelled in water. Each solvent has their acid-base equilibrium window, and water has one that is between 0-14, meaning that any solvent with a pKa within that range can be distinguished in water. There are other solvents with a ore narrow acid-base discrimination Window, such as methanol can acid, with a range between -10 to about 3, which will be unable to discriminate between weak acids. The reasoning behind the acid-base discrimination window is a bit long winded (and maybe a little confusing), so I will try to break it down as much as I can.

Say you have an acid in water, then the pKa + pKb = pKw
This is true for any acid or base in water. If pKa<0 or pKb<0 then then most of the compound will be dissociated in water, as an acid or base, respectively. This is undesirable, especially if the pKa/pKb<<0. As you have noted in your comments, when the pKa of a compound is less than 2 units compared to water, then in water only 1% of it will remain in the undissociated form. If pKa>pKw, you will obtain a negative pKb which will behave more like a base in water. It follows that compounds with a 0< pKa< pKw  will fulfill the condition where both pKa and pKb> 0. Now guess what pKw is? 14. Hopefully this might explain to you why water is able to distinguish between weak acids with a pKa between 0 to 14.


how to remove products from reaction space?(to apply LeChatelier's principle)

The scheme you ask about is effectively the same as $\ce{A <=>[k_1][k_{-1}] B ->[k_2] C}$ where $k_2$ measures the effect of removing B making C the removed product. The actual reaction is much more complicated as presumably mixing/diffusion has to be taken into account. Leaving aside these complications  as $k_2$ increases then the amount of B in the reaction reduces and the amount of C increases if measured at a fixed time. As long as $k_2$ is not zero eventually all B is removed, you can see this in the figure which shows the amount of C for different rate constants $k_2$  assuming the initial amount of A = $1$ and $k_1= 1, \, k_{-1}=3$. When $k_2=0$ (top figure) no product is formed and A and B come to equilibrium. The lower figure show the effects on forming C by removing B.


Meaning of electrochemical cell notation without a salt bridge

This is an example of an electrolytic concentration cell.

Both the cathode and the anode have the same electrolyte present and both are in same solution. Thus, there is no need to separate the electrolyte in two different solution and connect by a salt bridge.

The anode and cathode both are metal-sparingly soluble salt-salt anion electrode. The $\ce {KCl}$ solution here is the common electrolyte.

The anode reaction is:$$\ce  {Ag(\text{metal})(s) + Cl^-(\text{salt anion})(aq) ->  AgCl(\text{sparingly soluble salt})(s) + e^-}$$

The cathode reaction is also similar: $$\ce {Hg2Cl2(s) + 2e^- -> 2\ce{Hg}(l) + 2Cl^-(aq)}$$

Observe that the $\ce {Cl^-} $ is present in both the reactions as a common electrolyte's ion.


Why is the change in entropy is different for different interpretations?

From the perspective of Statistical Mechanics:

Multiplicities do depend on the external parameters that specify the energy level of the system.

For ideal gas of $N$ indistinguishable molecules, the multiplicity is given by $$\Omega= \frac{1}{N!}\frac{V^N}{h^{3N}}\,\frac{2\pi^{3N/2}}{(3N/2- 1)!}(\sqrt{2mE})^{3N-1}$$

So, indeed the multiplicity increases when the position space increases.

When the external parameter like volume of the container is increased, the energy levels come closer and then the particles can have more microstates for the same $E$.

Prior to the volume increase, the microstates of the system are $\Omega_{N,E,V_i}\;_;$ when the volume is increased, not will be there the microstates $\Omega_{N,E,V_i}$ still be available after the removal of the constraint, but also, there be addition of microstates. In general, $$\Omega_{V_f}\geq \Omega_{V_i}\;.$$

So, as the volume increases, the multiplicity increases and so does the entropy.

As requested by OP:

From the perspective of Thermodynamics:

Suppose the two dissimilar indistinguishable gases $A$ and $B$ having the same number of particles, energy, volume are partitioned from each other.

Now, when the partition is removed, the energy of the gas $A$ confined to a localised area gets dispersed over a greater volume.

Thus the entropy of gas $A$ increases and that is given by \begin{align}\Delta S_A &= Nk\,\ln \frac{V_f}{V_i}\\ &= Nk\,\ln \frac{2V}{V}\\ &= Nk\,\ln 2\;.\end{align}

Same happens for gas $B$ also.

Therefore, the change in entropy of the system is given by \begin{align}\Delta S_\textrm{system}&= \Delta S_A + \Delta S_B\\ &= 2Nk\ln 2 \gt 0\end{align}

Therefore, entropy increases when the partition is removed.


Safe endothermic reaction that gets below freezing?

The reaction of baking soda and citric acid (both are easy to come by and safe) in water does cool the water noticeably. I have tried it multiple times, and while it does not reach freezing, it will reliably reach $6\ ^\circ \mathrm{C}\ \left(43\ ^\circ \mathrm{F}\right)$, which is cold to the touch.

  Add 4 teaspoons each of sodium bicaronate (baking soda) and citric acid to a cup. Then add the smallest amount of water you can and start stirring!

You can tell a chemical reaction is occurring because a gas is produced:

$$\ce{3NaHCO3 + H3C6H5O7 -> Na3C6H5O7 + 3CO2(g) + 3H2O}$$

But, wait! We want a temperature below freezing! It might seem a little like cheating, but the reaction still works. Carbon dioxide bubbles are still produced, and the temprature gets down to around $-6\ ^\circ \mathrm{C}\ \left(21\ ^\circ \mathrm{F}\right)$.

If you do not have a thermometer, you can prove the second version is colder than the freezing point of water by doing the following:

Get a small, flat piece of wood and dampen the top surface of it.
Do the reaction in a glass, using 4 teaspoons each baking soda and citric acid and replacing the water with crushed ice.
As the temperature drops, the water between the wood and the glass will freeze, freezing the glass to the wood. If you lift the glass, the wood will stick to it.


This took a bit of thinking about but here is a viable route from commercially available bulk materials.

Start with nitroethane, deprotonate ($\ce{t-BuOK}$ in THF would be my choice, but $\ce{NaOEt/EtOH}$ should do it) and react with the commercially available dioxane protected 3-bromopropanal. Reduce the nitro group - alkyl nitro groups are tricky to reduce, I've found Hg/Al amalgam to be pretty good but others may have different conditions. There's a recent review of reductions$\ce{^{[1]}}$. The product of this will do the SNAr with 4,7-dichloroquinoline, I would try $\ce{K2CO3/MeCN}$ for this, but here are plenty of other possible conditions. This is  a good point to purify, then reveal the masked aldehyde by acid hydrolysis and do a reductive amination with the commercially available N-Ethylethanolamine to give the desired product.

Reference

Manuel Orlandi, Davide Brenna, Reentje Harms, Sonja Jost, and Maurizio Benaglia, Organic Process Research & Development 2018 22 (4), 430-445
DOI: 10.1021/acs.oprd.6b00205


What kind of plastic is (really) safe to store vegetable oil?

If you are concerned, I would suggest that high-density poly ethylene (HDPE) or poly propylene (PP) to be better options. HDPE is a heavy duty plastic resistant to most chemicals and physical deformation- it is often used for household products including bleach, detergent etc, but is also commonly used for milk bottles and the like.


First off, I would like to note that you wrongly interpreted the very first equation. Below I quote the relevant part of the paper:

  In the Hartree-Fock (HF) approximation for, say, the $\ce{Na}$ atom,
  the ground-state wave function has the form

  $$
\Psi = \mathcal{A} (\Phi_{\mathrm{core}} \phi_\nu X) \, , \tag{1}
$$

  where $\Phi_{\mathrm{core}}$ is a product of (ten) spatial orbitals
  very similar to the orbitals of $\ce{Na+}$ and $\phi_\nu$ is the
  valence orbital (the one removed in ionizing to $\ce{Na+}$). ($X$ is an
  appropriate product of spin functions.)

So, $\Psi$ is not just some function of spin orbitals designated by a weird letter as you said, it is, as usual, their antisymmetrized product, where I bet the calligraphic A symbol ($\mathcal{A}$) as usual stands for the antisymmetrizer. The only small difference is that small calligraphic a is used in the text rather then much more common capital one.

Now to you actual question. Both core and valence orbitals are thought in the derivation as the solutions of the Hartree-Fock equations, which have a very well-known feature: they are defined to within a unitary transformation (see, for instance, here). So, orbitals which are solutions of the Hartree-Fock equations are not unique: there are infinite different sets of orbitals that minimize the Hartree-Fock energy and you are free to choose any of these set for any further work. This is, essentially, what is said in the paper:

  It will be important in our later analysis to note that the valence
  orbital in (1) is not determined uniquely by the variational
  principle. One can modify $\phi_\nu$ by mixing in an arbitrary amount
  of any core orbital (doubly occupied. in $\Phi_{\mathrm{core}}$)
  without changing the energy.

Just note that this is true not only for the valence orbital: all orbitals (i.e. core as well) are not uniquely determined by the variational principle, as I mentioned above.

Now, if the goal is to have a unique valence orbital, then it can be achieved by restricting it to be orthogonal to core orbitals. In that case when transforming orbitals you can not mix any amount of any core orbital into the valence one since it will immediately make it non-orthogonal with this particular core orbital. And since mixing in core orbitals is the only way you can change the valence one, orthogonality of the valence orbital implies its uniqueness.

  In order to obtain unique solutions for $\phi_\nu$, one generally
  requires that $\phi_\nu$ be orthogonal to the other occupied orbitals.

And it is important to note (as it is done in the next sentence of the paper) that :

  Although this orthogonality restriction has the desired consequence of
  leading to a prescription for a unique valence orbital, there is no
  reason to consider the orthogonal valence orbital $\phi_\nu$, to have
  a special significance over any other combination of $\phi_\nu$, with
  the various core.

Note finally, that due to the orthogonality restriction you can not also mix any amount of the valence orbital into the core ones, but you can mix core orbitals with themselves anyway you like since it won't do any harm for their orthogonality with the valence orbital. Thus, while the valence orbital is uniquely defined by the orthogonality requirement, the core orbitals are still defined to within a unitary transformation.


The four lines in the visible spectrum (designated by α through δ) were the first observed by Balmer. Notice how the lines crowd together as they approach the ionization limit in the near-ultraviolet part of the spectrum. Once the electron has left the atom, it is in an unbound state and its energy is no longer quantized.
When such electrons return to the atom, they possess random amounts of kinetic energies over and above the binding energy. This reveals itself as the radiation at the short-wavelength end of the spectrum known as the continuum radiation.

Source: Chem1


Single-molecule magnet with electrically-controlled permeability: How does the Titan Shield from Deus Ex work?

Is there any existing examples of single-molecule magnets with permeability that can be controlled via altering current flow just like in case of electromagnets? If not, is there an explanation why it is not possible?

Short answers: no, and not yet.
Longer answer, and a warning that I'm using recent results from my group as references:

Yes, single molecule magnets exist, but... The field of single molecule magnets is hot, but the working temperatures of single molecule magnets are still below liquid nitrogen. Things seem to be improving, but they do not seem to be improving a lot. See the plot below for how the maximum temperature at which long-term magnetic memory has been reported has improved over the years (for single-ion magnets based on lanthanides, which are the best single molecule magnets right now). While there is no consensus that room-temperature single molecule magnetism is impossible, it has not happened yet, and there is no explanation why it could not happen in the future.

Maximum magnetic hysteresis temperature vs year when the Single Molecule Magnet was reported. Plot generated dynamically by the SIMDAVIS (AGPL-3.0) web app: https://rosaleny.shinyapps.io/simdavis_dashboard/ [1]

Yes, magnetism in molecules can be controlled via electricity, but... In this work, and in several related works, Wernsdorfer and coworkers show electric control (and electric read-out) of the magnetic state of a single molecule magnet. [2] However, this happens at temperatures below liquid helium, and the machinery needed is only able to do this for a single molecule at a time. Furthermore, the device is so delicate that hundreds of attempts are needed to trap the molecule in the right configuration. If the device heats up by accident, the molecule is lost forever and one needs to start from scratch. Again, there is no solid reason why this could not be achieved at some point in the far future. As promised, an alternate way of coupling the magnetic (spin) state of molecules to electric current, reported recently by my group.[3] In this approach, molecular helicity can be used to generate magnetic current (like a molecular electromagnet), and in principle this could eventually be employed to control the magnetic state of a single molecule magnet (again, at some hypothetical point in the far future).

To complete the answer, while neodymium-based alloys behave very well, neodymium-based single molecule magnets are, so far, greatly outperformed by dysprosium, terbium and even erbium (see figure below). Since there are fundamental reasons for this, it is more likely that any future technology resembling the one described in the question will be based on dysprosium (as of may 2021 holding the record for highest temperature single molecule magnet) or terbium (the ion employed in the cited examples about electrical control of magnetism in molecules).

Maximum magnetic hysteresis temperature vs lanthanide ion. Plot generated dynamically by the SIMDAVIS (AGPL-3.0) web app: https://rosaleny.shinyapps.io/simdavis_dashboard/ [1]
References

Duan et al. Data mining, dashboard and statistical analysis: a powerful framework for the chemical design of molecular nanomagnets. Preprint at https://arxiv.org/abs/2103.03199 (2021).
Godfrin et al., Operating Quantum States in Single Magnetic Molecules: Implementation of Grover’s Quantum Algorithm,  Phys. Rev. Lett. (2017) 119, 187702, https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.187702
Torres-Cavanillas et al., Reinforced Room-Temperature Spin Filtering in Chiral Paramagnetic Metallopeptides, J. Am. Chem. Soc. (2020), 142, 41, 17572–17580, https://pubs.acs.org/doi/abs/10.1021/jacs.0c07531]


I think I got this figured out.

Acidity and basicity are thermodynamic properties.

Nucleophilicity/electrophilicity are kinetic properties.

Lewis acids are electrophiles.

Lewis bases are nucleophiles.

We can use electrophile/nucleophile in describing Lewis acids/bases. But care should be taken to ensure we don't conflate kinetics with thermodynamics.

I think I finally got this issue resolved.


Will electric field make organic matters (e.g. glucose) ionize in water?

The short answer to your question is no, particularly for your example of glucose. Part of the key to your question is the statement: "[...] will the electric field ionize the organic matters that are not ionizable?".  If they are not ionizable, then they are not ionizable :)

You could electrolyze glucose to polyhydric alcohols as in this patent. But in this case you are not ionizing glucose, but creating an entirely different compound (which is still not strongly ionic).

There are compounds that can be electrolyzed to ionic compounds, but that is really a very different process than what you are asking about.  You wouldn't have ionization of the compound, you would be electrochemically converting a non-ionic compound into an ionic compound. See for example, this abstract1 in which the authors electrolyze ethanol to acetic acid, which would then ionize in the aqueous solution to some degree.

But, as in your example of gucose, the ethanol itself would not be ionized. You would simply be producing, via electrochemical reaction, an entirely different, ionic compound.

1) Journal of the Electrochemical Society, "Electrolytic Oxidation of Ethyl Alcohol to Acetic Acid", 1947, volume 92, issue 1, 335-342


Compare the boiling points of isobutane and 1-butyne

The boiling points of non-polar hydrocarbons are determined by the extent of van der Waals forces in between them. More specifically, molecules with a larger surface area have larger van der Waals forces of attraction between the molecules.

From the structures, it is obvious that but-1-yne has a larger surface area than isobutane, and hence has the higher boiling point.
Your idea of H-bonding in but-1-yne is plausible, according to this source1:

Quantum chemical calculations indicate hydrogen bond energies in the range $\pu{1–2.2 kcal mol^–1}$. Interconnected contacts such as $\ce{C#C–H ··· C#C–H ··· C#C–H}$ exhibit a pronounced cooperativity effect, with energy increases of $\approx \pu{0.5 kcal mol^–1}$ compared with isolated contacts.

However, while this bonding may be sufficient to explain a higher boiling point compared to isobutane, it is still a very weak bond compared to the H-bonds in hydrogen fluoride or water.
References:

Thomas Steiner, Evgeni B. Starikov, Ana M. Amado, José J. C. Teixeira-Dias, “Weak Hydrogen Bonding. Part 2. The Hydrogen Bonding Nature of Short $\ce{C–H ⋯\pi}$ Contacts: Crystallographic, Spectroscopic and Quantum Mechanical Studies of Some Terminal Alkynes.” J. Chem. Soc., Perkin Trans. 2 1995, (7), 1321–1326 (doi:10.1039/P29950001321).


Arterial blood CO2 content, division and partial pressure

Even when Douglas equation has been criticized is the best approach I know to calculate CO2 content. You can calculate plasma CO2 content and total blood CO2 content

Douglas 1988 (DOI: 10.1152/jappl.1988.65.1.473)


What inference can be made from the fact that there is no reversal in how three components eluate?

Question 1
Yes, of course! Something that is "relatively" non-polar would travel further with a non-polar mobile phase because it would have less attraction to the stationary phase. While a polar substance would not travel as far with the mobile phase because it would have a greater attraction to the stationary phase.
Question 2
Now, that depends. Are you also changing the stationary phase? Exactly how polar is the new polar-mobile phase? It would really depend on which phase the solutes have a greater attraction to.
Question 3
You are on your way. Quantitative HPLC is the bane of my existence. There's a million factors to consider, such as column temperature, flow rate of the solvent, particle size, polarity and charge. Usually, it takes a bit of experimentaton to reach a perfect combination that allows complete separation of whatever it is you're interested in. Usually, your best bet in identifying a compound at a particular point is to run a little bit of pure sample on its own. In the end, above all else, you need to find a combination that doesn't result in chemicals overlapping with each other.
In qualitative chromatography, as long as you can resolve all possible signals, you've done well enough.


Why doesn't beryllium, an elemental solid, obey Dulong and Petit's law?

Not a  'Law' a such it is more like a 'rule of thumb' based on a limited set of data measure at high (i.e. room) temperatures. The heat capacity is not constant but varies with temperature and for metals reaches a limiting value of approximately $3R$ at high temperatures. (The limiting value is higher for molecules due to the many translational, vibrational and rotational energy levels).

The heat capacity is the rate of change of the energy with temperature, i.e. the slope  of a plot of energy vs temperature.  At low temperatures only the lowest few energy levels are populated, so the rate of change vs temperature is almost zero. As the temperature is increased, more energy is added, more levels become populated and $C_V$ rapidly increases because the rate of change of energy is large. At high temperature, very many levels are now populated and the rate of increase of energy with temperature is constant, and therefore $C_V$ levels off becoming almost constant.

In some metals, those with large energy gaps between levels, room temperature is insufficient for them to reach the point where the slope of internal energy increase with temperature becomes constant and so the heat capacity is smaller than $3R$. The value at higher temperatures (say 1000 C) should reach the limiting value.


is it true that if I connect an aqueous oxidant to an aqueous
  reductant via a wire, will a current flow along the wire? In
  particular, does current flow in the following scenario?

No true. Nothing will happen as the circuit is incomplete. The oxidant or the reductant species have to see each other "face to face". Guess what is the most powerful reducing and oxidizing agent in the world? It is the cathode or the anode respectively. Consider an electrolytic cell which is calm and unstirred. The anode and the cathode are dipping in the solution. The moment you turn on the current, only the species which is next to the electrode gets oxidized or reduced. Nothing happens to the bulk right away, so physical promixity is important in electrochemistry.


How to determine source of air pollution

The main idea is the following: you will need multiple sampling points and record enough data to obtain a representative dataset. Then when drawing wind roses with high percentiles (let say P95 or above) you will be able to easily point out an unique source of a relevant and exclusive pollutant.

But of course, reality is much complex... And there are some adverse effects that you must address before pointing out the source with a reasonable confidence.

First you better be not so close to the source, in order to measure immission instead of emission. Then environment and local conditions will average out. And wind speed, wind direction and concentrations become meaningful. In this case high percentiles likely originate from your source.

Second, your analyte must be relevant and if possible exclusive from your source. Measuring CO2 is not a good choice when monitoring traffic (it is stable, widespread thus global and not exclusively produced by traffic), NO is a better one because it is local, reactive and strongly related to high compression ratio thermal engine combustion source.

Third, if there are multiple isotropic sources, you will get nothing useful using wind roses. And your problem becomes really tricky.


Is there a formula to tell how many conformers of a molecule to generate?

I haven't seen any rigorous benchmarking or guidelines.

Consider that the total number of conformers goes up like $3^n$, where $n$ is the number of rotatable bonds, and ~3 is the approximate number of symmetrically unique conformers for each rotatable bond.

For small molecules with <3-4 rotatable bonds, using 100 conformers seems to be enough (i.e, only a small deviation from the crystal structure geometry). There's a more complete discussion in Ebejer's paper "Freely available conformer generation methods: how good are they?": J Chem Inf Model. 2012 52(5) pp. 1146.

(Slide from Jean-Paul Ebejer)

Beyond 3-4 bonds, it's less clear if sampling more conformers is a clear win. I would guess the answer is "yes," but I have not seen any papers (yet) dealing with such flexible molecules. One may need better tools, and also better force fields or computational methods.


Hybridization of oxygen in Nitrate ion and the location/bond of nitrogen's lone pair

The actual structure of the nitrate ion is symmetrical. It forms a trigonal planar molecule.

This means that the bonds between oxygen and nitrogen are all the same. The resonance structures are a simple way to explain that structure using two-electron bonds. In reality the "extra" electron is delocalised across all three bonds (this will involve the nitrogen lone pair and nitrogen will be sp2 with no free lone pair).

The average bond order will be, as you calculate 4/3 but this doesn't, as you claim, make the molecule unstable.


Leaving group ability and solvent's effect

The solvent generally doesn't affect the leaving group ability of an anion. But coming to your Frinkelstein case, it a special one cause here some precipitate is being formed (yes, that's correct that it is due to lower solubility product of say $\ce{NaCl}$ in acetone). As soon as some ppt. is formed, those atoms lose their ionic identity, settle at the bottom and don't interfere in the solvent as solute and so their concentration decreases in the solvent. Due to Le Chatelier's principle, the reaction goes forward to a great extent. This is also the case for other super leaving groups like gases (for eg. $\ce{N2}$, $\ce{H2}$).


This is a classic oxidative addition. You're kind of outside the classic world of single and double bonds. You have 4 electrons spread of 3 carbon atoms that link stabilize a bonding interaction with the palladium. Two of those electrons came from the palladium, so we think of it as having been oxidized, hence the name of the process.

See reference for general $\mathrm{S}_{N}2$ type oxidative addition. The only difference from your example is that the electrophile has a double bond that serves as an additional ligand for the palladium.


Graphene in Lithium-ions batteries: why?

Lithium has a standard redox potential equal to -$3.04$ V. Coupled with any compound at the cathode, it could hardly produce cells yielding more than $4$ Volts, because part of the available energy is needed and lost to extract the external electron of the Lithium atom. The performance of a cell built with a Lithium anode could be more important if the Lithium was already ionized. That's what is done with graphene. When introducing Lithium atom in a graphene structure, its electron is not so strongly attracted by the Lithium nucleus. Lithium inserted between graphene layers behaves as if it was already partly ionized. That is why Lithium in graphene produces an anode which can deliver about $5$ Volts if coupled with a convenient cathode made of Lithium and Cobalt(III) oxides. Graphene plus Lithium may be considered as a mixture of Lithium ions $\ce{Li+}$ and "nearly free" electrons delocalized in the graphene structure.


Finding final pH of the buffer solution without applying Henderson equation

As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$
$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$
from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:
$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$
$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$
$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$


Problem digesting sample for Ti analysis using ICP-OES

If you want to stay with a sulfuric acid type of digestion, there is this(1) paper that describes a microwave digestion using copper(II) and hydrogen peroxide to facilitate the digestion.

From my personal experience, if your lab is properly equiped, my first choice would be an $\ce{HF}$ digestion or a nitric/perchloric acid digestion.

1)"A rapid and simple microwave-assisted digestion procedure for
spectrophotometric determination of titanium dioxide
photocatalyst on activated carbon", ScienceDirect, Talanta 71 (2007) 1867–1872


Synthesis of indigo from anthranilic acid - reaction not taking place

Trichloroacetic acid was definitely the wrong way to go.

The reason is the chloroacetic acid is alkylating your anthranilic acid, adding the missing carbons Ben Norris comments on.  Ttrichloroacetic acid is not a great alyklating agent, owing to steric bulk.

If you are dead set against working it up and separating your anthranilic acid, you might try just adding the chloroacetic acid.

Otherwise clean it up and start over.


Will there be a temp change in free expansion

In a real gas, the internal energy is the sum of the kinetic energy of the molecules and their potential energy of interaction. An ideal gas is the limit of real gas behavior in the limit of large molar volumes in which the potential energy of interaction of the molecules approaches zero. The kinetic energy of the molecules depends only on temperature, but the potential energy of interaction of the molecules depends on both temperature and specific volume. The temperature of an ideal gas does not change when subjected to free expansion, but it does change for a real gas.  So the deviation from ideal gas behavior caused by potential energy of interaction of the molecules is what causes the temperature of a real gas to change in free expansion.