Are ther known compounds of Argon? What is their molecular geometry and hybridisation?

Argon is a quite unreactive element, however very few compounds of it are known.

The most popular, and currently the only neutral compound, is argon hydroflouride, $\ce{HArF}$, which was reported by L. Khriachtchev et.al..[1]  In the same issue there is also a comment on the matter by my former supervisor G. Frenking.[2]
Another interesting read on the subject is the prediction of the metastable $\ce{HHeF}$, by M. W. Wong.[3] It generally treats all compounds of the $\ce{HNgF}$ family.

All calculations done so far indicate, that the molecule is linear and only kinetically stable. The highly accurate CCSD(T)/aug-cc-pV5Z level predicts, that the $\ce{HArF}$ molecule is $5.87~\mathrm{eV}$ higher in energy than its fragments $\ce{HF}$ and $\ce{Ar}$. The minimum barrier on this level of theory is a mere $0.35~\mathrm{eV}$ along the stretching coordinate of the hydrogen argon bond. The bending is predicted to be higher in energy. The reported bond distances for $\ce{H-Ar}$ and $\ce{Ar-F}$ are $133~\mathrm{pm}$ and $197~\mathrm{pm}$, respectively.
An NBO analysis on the MP2/aug-cc-pVTZ level of theory shows a strong ionic character of the molecule, it is reported, that there is a strong $\ce{(HAr)+F- }$ charge transfer in the electronic structure of the molecule.
Wong performed accurate calculations on the QCISD/6-311+G** level of theory and combined it with a Quantum Theory of Atoms in Molecules approach. He calculated a positive value of the Laplacian of the electron density $\nabla^2\rho_\mathrm{b}$ for the bond critical point of the argon fluoride bond, which indicates a mainly electrostatic interaction. The contrary is the case for the argon hydrogen bond, where the Laplacian is found to be negative, i.e. charge concentration, and therefore indicating a dominantly covalent interaction.
The following picture is a snapshot of this behaviour at the BP86/cc-pVTZ level of theory.[4]

Solid lines are regions of charge depletion, $\nabla^2\rho_\mathrm{b}>0$, while dashed lines are regions of charge concentration $\nabla^2\rho_\mathrm{b}<0$. Red lines are zero flux surfaces, where $\nabla\rho_\mathrm{b}=0$. And red spheres indicate bond critical points.

This molecule is quite interesting for many reasons, however, since it is only kinetically stable it has to be treated with fullest force of quantum chemical calculations. The calculations I did on a low level DFT method are not sufficient at all on its own, as has been investigated by J. Panek et.al..[5]]

Hybridisation is in a wider sense the mixing of s and p orbitals;[6] and upon further examination of the electron density this will be observed in this molecule. Most likely for the hydrogen argon covalent bond. As hybridisation is just a mathematical transformation of the original atomic orbitals there is no real interpretative value in this case. For example, the NBO analysis at the BP86/cc-pVTZ level of theory give only a negligible 3% s contribution of argon to the $\ce{H-Ar}$ bond.

The VSEPR theory is only a observational concept for well-behaved molecules. It basically lacks a solid scientific basis and should not be used any more, but especially in cases like this, it must not be used.

Literature and Notes

Leonid Khriachtchev, Mika Pettersson, Nino Runeberg, Jan Lundell, and Markku Räsänen, Nature, 2000, 406 874-876.
Gernot Frenking, Nature, 2000, 406, 836-837.
Ming Wah Wong, J. Am. Chem. Soc., 2000, 122 (26), 6289-6290.
To avoid copyright issues I performed a quick calculation to illustrate the above raised point.
Jarosław Panek, Zdzisław Latajka, and Jan Lundell, Phys. Chem. Chem. Phys., 2002, 4, 2504-2510.
Hybridisation is a concept that dates back to the beginnings of Valence Bond (VB) Theory. It was originally implemented in the strict fashion of creating $s^{1/(n+1)}p^{n/(n+1)}$ or $sp^n$ hybrid orbitals for $n = \{1,2,3\}$. H. Bent later went on suggesting fractional hybridisation schemes, see Henry A. Bent, Chem. Rev., 1961, 61 (3), 275–311. Nowadays the term s-p-mixing in Molecular Orbital theory is almost synonymous with the VB theory term hybridisation.


How to find the units of the rate constant?

You are doing it wrong in situation when concentration of B alone was doubled but half life did not change.

In this situation, the reaction order should be 1 according to B as Half life in first order reaction doesn't depend upon initial concentration of reactants.

$$t_{1/2} = \ln2 / k $$

So the final order will be $2$ and the answer of the question will be b.


That's supposed to be the sodium salt of salicylaldehyde, also known as sodium salicylaldehyde, sodium ortho-formylphenolate, 2-hydroxybenzaldehyde sodium salt or sodium 2-formylphenoxide. CAS number: 3116-83-4.

There's actually an error in the structure. The oxygen should have a negative charge if the sodium is written with a positive charge. The salt as a whole is neutral. What's written here is one lone pair, there should be 3 of those and a negative charge.


Describe the bonding in a tribromide anion, what is its structure?

In fact, tribromide ion has the generic formula $\ce{AX2E3}$ according to Lewis diagram. Where  $\ce{A}$ is the central atom, $\ce{X}$ an atom bonded to $\ce{A}$ and $\ce{E}$ a lone pair on $\ce{A}$.  Using Valence Shell Electron Pair Repulsion (VSEPR) theory,  the geometry  of this ion will be
 :

So, the molecular shape is linear, while the electron pair geometry is trigonal  bipyramid.
 And the bonds in the ion are covalent.


How do I build a home-made apparatus for identifying what minerals are present in water?

This probably not what you are looking for, but the cheapest and least time-consuming analysis is the one done by others for free.

Over here (Germany), local water suppliers are required by law (Trinkwasserverordnung - TrinkwV) to monitor the quality of potable water. This includes a large number of microbiological and chemical parameters. The details are given in the annexes 1-5 of the TrinkwV.

The results are published and are are available online at no cost.

Examples (mostly in German) are:

RheinEnergie (Cologne area)
RWW - Rheinisch-Westfälische Wasserwerksgesellschaft (one of the largest suppliers in Germany)


Are titration, recrystallisation, functional group analysis, etc. outdated?

It is true that modern techniques make it faster to elucidate the identity of compounds and isolate them from the reaction mixture, however to call the traditional methods like titration, recrystallization, or column chromatography (basis of HPLC) outdated is far too broad. You could probably get away with calling them outdated if all you wanted to do was to characterize compounds/final product. But when you're in synthesis, you still need the 'traditional' methods to prepare larger amounts of intermediate.

For example, HPLC is only able to separate small amounts of compounds (possibly enough for characterization) and if you wanted enough to go to the next step of your reaction (or if you wanted to make a lot), HPLC would be a highly inefficient way of doing this. Recrystallization, Distillation, and/or Column Chromatography (depending on the compound) would be much better for this purpose.


Steric hindrance of an unbonded electron pair

Since I just covered this material in depth in my physical organic chemistry class, I figured I'd write up an answer.

Electronegative elements lower the energies of all molecular orbitals to which they contribute, of which their low-lying $\sigma ^{*}$ orbitals are of particular importance. Electronegative atom lone pair MOs, which have little bonding character and are also (relatively) high-lying in energy, are very close in energy and will donate their electrons to the low lying empty orbitals, producing a stabilizing interaction. Molecular conformations are in part shaped by these interactions, which is known as the donor-acceptor effect.

Hydrogen peroxide is a classic example of this effect. In addition to sterics predicting that the hydrogens be anti-periplanar, this conformation also minimizes the molecule's net dipole (another stabilizing effect), yet its most stable conformation adopts anti-clinal geometry. This can essentially be thought of as a compromise between between sterics and dipole minimization, and the donor-acceptor effect:

$\hspace{4.3cm}$

The $\sigma ^{*} (\ce{O-H})$ is an excellent acceptor, and oxygen's highest energy unbonded pair is one of the strongest donating. The lower energy lone pair occupies a $\sigma\text{-out}$ orbital, and the higher energy lone pair occupies a pure $\mathrm{p}$ orbital. The mixing of this orbital with the $\sigma ^{*} (\ce{O-H})$ orbital favors a $90º$ dihedral angle, but due to steric and dipole effects, the actual angle is $\approx 120º$.

$\hspace{5.3cm}$

In hydrazine, the $- 2.5 \ \mathrm{kcal \ mol^{-1}}^{[1]}$ preference for the two $\mathrm{n}(\ce{N})$ orbitals being syn-clinal over anti-periplanar is also explained by the donor-acceptor effect. When the $\mathrm{n}(\ce{N})$ are syn-clinal, they are each eclipsed by the $\sigma ^{*} (\ce{N-H})$ orbitals, and orbital mixing occurs. The effects of this are further seen in the $\ce{N-N}$ bond length difference between the two conformations: $1.448\ \mathrm{Å}$ in the syn structure and $1.489\ \mathrm{Å}$ in the anti structure. The shortening of the bond length in the syn structure is a direct result of the multiple bond character introduced by the $\mathrm{n}(\ce{N}) \rightarrow \sigma ^{*} (\ce{N-H})$ interactions$^{[1]}$.

$^{[1]}$ Wilcox, C.; Bauer, S. Journal of Molecular Structure: THEOCHEM 2003, 625 (1-3), 1–8.


What is the role of various additives in a titration of vitamin C with N-bromosuccinimide

Thanks, Martin, for pointing out the relevant paper with all the answers!
Basic procedure
This titration was first described by Mohamed Zakibarakat, Mohamed Fathy Abd El-Wahab, and Mohamed Mahmoudel-Sadrin in 1954 (doi.org/10.1021/ac60100a013) as an alternative to using 2,6-dichlorophenol-indophenol as titrant. Here is the procedure for a solution of pure acsorbic acid:

Into a 50-m1. conical Erlenmeyer flask, a known volume of the unknown ascorbic acid solution is introduced — i.e., 1 ml. contains 1 mg. of ascorbic acid. Then 5 ml. of 4% potassium iodide solution, 2 ml. of 3% acetic acid, and 2 drops of starch solution as an indicator are added. The aqueous N-bromosuccinimide solution (0.1% weight per volume) is introduced into the microburet and is allowed to run drop by drop, into the ascorbic acid solution, with continuous shaking. The end point is reached when the last drop of the N-bromosuccinimide solution added produces a permanent blue color in the ascorbic acid solution.

For fresh fruit, there is an additional step to prepare the analyte:

Edible Fruits. The new method has been applied to the estimation of ascorbic acid in certain edible Egyptian fruits known to contain ascorbic acid—e.g., oranges, lemons, tomatoes, grapes, watermelons, and water cress. The fruits used in the estimation of ascorbic acid were almost ripe. The sampling and extraction of the material under examination must be carried out with the minimum delay, so that no significant change in ascorbic acid content takes place prior to analysis. Each sample is squeezed or bruised and the juice is directly received in a known volume of 20% trichloroacetic acid, so that the juice is diluted with the stabilizing acid at least twice, to avoid the oxidation of ascorbic acid and also to precipitate interfering substances such as protein and thereby facilitate subsequent clarification of the extract. The acidulated juice is fil-tered and the determination is carried out on a known volume of the clear filtrate as described previously. The N-bromo-succinimide solution used in the titration is 0.01%—i.e., 1 ml. contains 0.1 mg. of N-bromosuccinimide (see Tables XI and XII).

Oxalic acid
Note that they did not mention oxalic acid, but they did use fairly acidic conditions to inhibit redox enzymes and precipitate proteins.
Oxalic acid does not interfere in the determination:

Interfering Substances. The following substances which might interfere have no influence on the titration process: carbohydrates such as glucose, lactose, sucrose, and starch; diketogulonic acid; reductones and reductic acid; urea; uric acid and creatinine; alcohols; formaldehyde; acetone; ethyl acetate and acetoacetic ester; thiamine hydrochloride and ribo-flavine; oxalates, tartrates, and citrates; amino acids such as glycine, alanine, valine, and isoleucine; and ferrous and ferric salts. The only interfering substances that are also oxidized by N-bromosuccinimide before iodine is liberated from potassium iodide in acetic acid medium include sodium sulfite, sulfide, thiosulf ate, and thiourea.

Role of iodide
They nicely describe the two relevant redox reactions used in the procedure:

The fact that ascorbic acid reacts very rapidly with N-bromo-succinimide, whereas many of the interfering substances react more slowly or even do not react at all, provides a reliable titrimetric method for the determination of ascorbic acid. N-Bromosuccinimide is an oxidizing agent and thus can liberate iodine from potassium iodide in aqueous acetic acid medium, but it oxidizes ascorbic acid to dehydroascorbic acid preferentially. Until all the ascorbic acid present in the solution is oxidized, no iodine is liberated from potassium iodide.
The slightest excess of N-bromosuccinimide added, after all the ascorbic acid content has been oxidized, will liberate iodine from potassium iodide, which is easily detected by the blue color developed with a few drops of starch solution added at the beginning of the titration process. The end point is thus easily observed by the appearance of a blue color in the ascorbic acid solution or extract. The presence of other reducing substances does not interfere with the titration with N-bromosuccinimide, since iodine is selec-tively liberated from potassium iodide before reducing substances present, other than ascorbic acid, are oxidized by N-bromosuccinimide, and consequently the end point is easily determined in the presence of starch. The end point is definitely blue in the case of pure solutions and pharmaceuticals, but violet in the case of biologicals and fruits.

So it would seem that as long as there is sufficient iodide to give the color change for the endpoint, the exact amount does not change when the endpoint is observed.


To estimate reactivity solely based on the the octet rule is a crude, (semi-)empirical approach in the first place. Atoms and molecules don't know anything about their number of electrons in shells, but the formation of molecules are based on the minimization of (free-)energy of the system.
The octet rule gives a hint, for what kind of systems the potential energy may be small, but does not say anything about the relative energies of two systems, both fulfilling the octet rule.

To "exactly" predict the reactivity, you would have to solve the Schrödinger equation for both states of your systems, find the eigenvalue of the energies, consider entropic contributions. Furthermore you would need to "sample" all possible pathways of your reaction, to check if the energy barriers are small enough to be overcome at a given temperature.
(To be really exact, you still would need to estimate entropic contributions)

Yet still a chemists' intuition often can predict reactivity very well. Concerning your specific problem, one possible qualitative explanation would be, that the octet rule is not fulfilled for copper, but only for oxygen, before the reaction happens. Taking a look on the structure of Copper(II) oxide on the other hand, shows that it fulfils the octet rule for both, copper and oxygen.(Ignoring d-electrons for copper)

Furthermore, the reaction enriches the electron density in the strongly electronegative oxygen, while reducing the electron density in the "electropositive" copper, which also stabalizes the structure of copper oxide.


Do alkali metals react with ammonia to form hydrogen gas?

What you have written is right, but it is just the first step.
The release and solvation of an electron is fast. Reaction of an electron with ammonia to release hydrogen $$\ce{2 e-(solv) + 2 NH3 -> H2 + 2 NH2-(solv)}$$ is slow. Deep blue (diluted) or bronze (concentrated) "sodium electride" solution in liquid ammonia slowly converts itself to the colorless ammonia solution of $\ce{NaNH2}$.

For the water case, all is extremely fast, as water is a much stronger acid than liquid ammonia. Blue-like color can be catched by 10000 fps camera, what is commented in SE  Chem Q - Reaction of potassium with water
BBC article, referring the Nature article by Pavel Jungwirth and colleagues (ref.1),

ABSTRACT: Alkali metals can react explosively with water and it is textbook knowledge that this vigorous behaviour results from heat release, steam formation and ignition of the hydrogen gas that is produced. Here we suggest that the initial process enabling the alkali metal explosion in water is, however, of a completely different nature. High-speed camera imaging of liquid drops of a sodium/potassium alloy in water reveals submillisecond formation of metal spikes that protrude from the surface of the drop. Molecular dynamics simulations demonstrate that on immersion in water there is an almost immediate release of electrons from the metal surface. The system thus quickly reaches the Rayleigh instability limit, which leads to a ‘coulomb explosion’ of the alkali metal drop. Consequently, a new metal surface in contact with water is formed, which explains why the reaction does not become self-quenched by its products, but can rather lead to explosive behaviour.

with high speed (cca 10000 fps) camera photos shows the blue colour near 0.3 ms after the metal/water contact (ref. 2). It may be rather more toward violet, it seems to me. The abstract of ref. 2 explains the blue color:

Alkali metals in water are always at the brink of explosion. Herein, we show that this vigorous reaction can be kept in a non‐exploding regime, revealing a fascinating richness of hitherto unexplored chemical processes. A combination of high‐speed camera imaging and visible/near‐infrared/infrared spectroscopy allowed us to catch and characterize the system at each stage of the reaction. After gently placing a drop of a sodium/potassium alloy on water under an inert atmosphere, the production of solvated electrons became so strong that their characteristic blue color could be observed with the naked eye. The exoergic reaction leading to the formation of hydrogen and hydroxide eventually heated the alkali metal drop such that it became glowing red, and part of the metal evaporated. As a result of the reaction, a perfectly transparent drop consisting of molten hydroxide was temporarily stabilized on water through the Leidenfrost effect, bursting spectacularly after it had cooled sufficiently.

References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, “Coulomb explosion during the early stages of the reaction of alkali metals with water,” Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).
Philip E. Mason, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, “A Non‐Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide,” Angew. Chem., Internat. Edn. Engl. 2016, 55(42), 13019-13022 (https://doi.org/10.1002/anie.201605986).


Are risks (hexavalent chromium) during Chrome plating temporary, or remain with the product?

Multiple questions should be broken in multiple items, but anyway,

If your plating process is well done, and everything is thoroughly cleaned after, the level of Cr(III) or Cr(VI) remaining should be low enough to be inconsequential to health and the environment. You're correct stating that regardin chrome, thre will be only a passivated layer of metallic chromium and not the trivalent or hexavalent.

Please do remember that the main reason that chrome plating makes things resistant to corrosion is that as soon as chromium is exposed to air, it forms a very thin layer of chromium oxide attached to the surface of the metal, preventing further corrosion. So, in fact, there is some Cr(III) on a finished piece of chrome plated material.

We aren't considering the environment were the finished part will be exposed/used. If aggressive enough conditions are present, the chromium layer will be corroded away and will enter the environment as Cr(III)/Cr(VI).

Regarding Cr(VI) detection, there is a "simple" method involving a solution of diphenylcarbazide described at https://www2.mst.dk/udgiv/publications/2009/978-87-7052-987-7/html/kap04_eng.htm

However, one can easily buy test kits for Cr(VI) on Amazon, ready for use, starting at about (February 2019) USD 110.


Does aluminum oxide react with rubidium?

As suggested in the comments, rubidium aluminates do exist.
Aluminates such as $\ce{RbAlO2}$ and $\ce{Rb6Al2O6}$ are prepared at higher temperatures (above 550 °C [1]) or even from the melt.

However, alkali metal vapors at elevated temperatures act as reducing agent, so I wouldn't expect aluminates anyway.
On the other hand, aluminides such as $\ce{RbAl}$ (Zintl phase) are formed from the melt at much higher temperatures (above melting point of aluminium) and pressures exceeding atmospheric.

Since the corundum-coated plate is supposed to perform in vacuum under mild temperatures, I think it shouldn't be affected by rubidium vapors.
There is a possibility of coating embrittlement over time as rubidium diffusion occurs, but I wouldn't expect it to be an issue.

References

Schläger, M.; Hoppe, R. Darstellung und Kristallstruktur von $\ce{K6[Al2O6]}$ und $\ce{Rb6[Al2O6]}$. Zeitschrift für anorganische und allgemeine Chemie 1994, 620 (5), 882–887. https://doi.org/10.1002/zaac.19946200522.


How can enthalpy change of a system be negative while entropy change is positive?

For example, suppose you have a solid block of TNT.

It explodes and releases much energy. $\Delta H$ is negative.

Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.


You cannot determine individual ion's molar conductivity at 1 M. The ions are not independent enough. Kohlrausch's law of independent migration of ions does not apply, it only applies at a fictitious concentration of 0 M NaCl! This value is found mathematically by extrapolation to zero concentration

Is there any way to compute the molar conductivity of an ion without
any experiments?

Yes, if you look up the tables of molar conductivities at infinite dilution, then you can determine molar conductivities of individual ions.


Synthesis of 1-naphthaleneacetic acid: Methods

If you have read the reference 1 carefully, you would realized that the amounts of $\ce{KBr}$ and $\ce{Fe2O3}$, as well as reaction temperature and time is very important to the reaction yield. That's the reason I asked you to list your procedures so that we can analyzed them and suggest some changes. Well, as reference 1 listed, follow the following procedure exactly and see the results:

Naphthalene with 0.0010% iron $(\pu{57.6 g},\ \pu{0.449 mol})$, chloroacetic acid $(\pu{14.1 g},\ \pu{0.149 mol})$, ferric oxide $(\pu{87.6 mg},\ \pu{0.549 mmol})$, and potassium bromide $(\pu{420 mg},\ \pu{3.53 mmol})$ were placed in a long-necked round-bottom flask fitted with a rubber stopper carrying a thermometer and an air-cooled tube about a $\pu{1 m}$ long, and the mixture was boiled gently on a sand bath for twenty hours. The best yield was obtained when the heating was controlled according to a temperature-time curve in which the temperature of reactants attained $200^\circ$ after ten hours and $218^\circ$ after twenty hours. After the reaction, a greater part of the unreacted naphthalene $(\pu{43 g})$ was recovered by distillation. The residue was extracted with hot sodium hydroxide solution, cooled, and filtered. Acidification of the filtrate with hydrochloric acid yieldeed the brownish precipitate, $\pu{19.4 g}$ (70%), m.p. $108-113^\circ$ ....

It is noted that the melting point can be raised to $132^\circ$ by repeated crystallization. First crystallization the crude product with 250-fold distilled water has yielded $\pu{9.5 g}$ of product with m.p. $124-126^\circ$. This yield is 34% based on  chloroacetic acid used, which is the limiting reagent and has reacted completely, and 45% based on naphthalene reacted.
The condition described above is the optimum conditions regarding to temperature, time, increased amount of $\ce{KBr}$, and the amount ratio of naphthalene to chloroacetic acid. Also note that:

In the absence of $\ce{KBr}$ and $\ce{Fe2O3}$, no naphthaleneacetic acid was formed.
In the absence of $\ce{KBr}$ but the presence of $\ce{Fe2O3}$, only 3% of naphthaleneacetic acid was formed.
In the absence of $\ce{Fe2O3}$ but the presence of $\ce{KBr}$, only 4% of naphthaleneacetic acid was formed.
$\ce{KBr}$ is the most effective additive for the reaction. The promoting action of several other metallic halides such as $\ce{NaCl},$ $\ce{KCl},$ $\ce{KI},$ hydrated $\ce{AlCl3},$ hydrated $\ce{AlBr3},$ hydrated $\ce{FeBr3},$ and $\ce{HgBr2}$ when mixed with $\ce{Fe2O3}$ in place of $\ce{KBr}$ was also observed but they are less effective than $\ce{KBr}$.

References:

Yoshiro Ogata and Jiro Ishiguro, "Preparation of $\alpha$-Naphthaleneacetic Acid by the Condensation of Naphthalene with Chloroacetic Acid," J. Am. Chem. Soc. 1950, 72(9), 4302 (https://doi.org/10.1021/ja01165a533).


Is Alkanol an official part of chemical Nomenclature?

IUPAC gold book doesn't recognize it, that's a pretty solid answer. I've never heard anyone say 'alkanol' before either, I've only heard people refer to these as simply alcohols. However, there are people who still use the word 'ethylene' to describe the compound 'ethene' so it could just happen to be some common language I haven't been exposed to.

To be clear, no it's not official, but yes it might be real terminology used by some organic chemists.


Electroplating metal inside polymer foam with acetone

Electroplating might be possible, but getting good coverage of internal pores is tricky, even with aqueous electrolytes, as deposition will likely occur preferentially at the outer edge of the workpiece, leaving the internal passages lightly- or un-coated.

A more effective approach is likely electroless nickel deposition.  As long as you can find a suitable acetone-soluble reducing agent, it might work out.

I would also recommend testing other solvents for their ability to permeate the polymer structure -- if you could find a protic solvent that wets it (even something like 50/50 ethanol/water!), you'd have a much better chance of success at electrodeposition proper, and also at making electroless work for you, since (I think) most electroless recipes are aqueous-based.


From p. 327 of ChemDraw 18 User Guide:

  Importing

  You can import graphics or documents from other applications into ChemDraw.

  Files
  Files that you insert become part of the drawing. You can edit a file only if it is in a ChemDraw format.
  To insert a file:

  Navigate to Edit → Insert File.
  In the Open dialog, select the file type from the drop-down list.
  Select the file and click Open. The file is embedded in the drawing.

Further, there is a table of file formats ChemDraw is capable of importing and exporting, from which it is evident that only the following graphical formats can be imported:

Bitmap (BMP)
Graphic Image Format (GIF)
JPEG (JPG, JPEG)
Portable Network Graphics (PNG)
Windows Metafile (EMF, WMF)
Tagged Image File (TIF, TIFF)

PDF or PostScript files can be attached as objects, but I suppose it's not what you are after.
Either way, it looks like there is no way to add AI files directly, so you need to convert them first to either raster (e.g. TIFF or PNG) or vector (WMF) format.
I agree with @Ivan that crafting structures in ChemDraw in the first place and then post-processing them with Adobe Illustrator is arguably a better workflow.


Relationship between inhibition constant Ki and interaction energy?

One complication of enzymatic inhibition constants is that there a multiple mechanisms to inhibition, and the type of inhibition can change how the inhibition constant relates to binding energy.

On one side of the spectrum you have competitive inhibition, which means that the $EI$ complex is functionally inactive, and cannot bind substrate. (The easiest way of getting competitive inhibition of an enzyme is to have the inhibitor bind in the active site, competing with the substrate binding. But that's not the only way. You can also have allosteric inhibitors which exhibit "competitive inhibition", if they bind the apo enzyme and by doing so prevent substrate binding.) This is the sort of inhibition you're assuming in your question, but isn't the only type of inhibition.

On the other side of the spectrum, you have uncompetitive inhibition. (N.B. uncompetitive - noncompetitive inhibition is something different.) This means that the inhibitor requires the substrate to be bound to the enzyme prior to binding itself, and it's the $ESI$ complex which is functionally inactive.

And then you have a whole range of variation where you have a mix of effects (binding to both the apo and substrate-bound forms, with various affinities), of which noncompetitive inhibition is a particular type.

You're correct in that for pure competitive inhibition, the $K_i$ is equal to the $K_D$ of the inhibitor. The problem arises when you have one of those mixed effect or uncompetitive inhibitors, where catalytic steps and substrate concentrations get folded (partially) into the $K_i$.

Personally, I would recommend Enzyme Kinetics and Mechanism by Paul F. Cook and W.W. Cleland, which goes into great details about reaction mechanism, kinetics and inhibition constants. It should include derivations and relations for all the relevant scenarios.


Why are there two contradicting formulas for Enthalpy (Delta H):

This is my take on this

  1) Delta H = Bonds broken - Bonds Formed

  2) Delta H = $H_{products}−H_{reactants}$

Equation 2 is pretty straightforward, (change in enthalpy is the enthalpy of the products minus the enthalphy of the reactants).

Equation 1 is a bit less straightforward. Let's define the system as where you are having the reaction, and the surroundings as everything else. When you break bonds, you take energy from the surroundings into the system. Then, when bonds are formed, energy is released from the system into the surroundings. Therefore, if there is more energy released in the bonds formed than there is energy gained in the bonds broken, the system loses energy. How I see it is, you are looking from the surrounding's perspective at equation 1 and from the system's perspective from equation 2, that's why you got two opposite statements (from your POV).


Which tautomer of hexane-2,4-dione is more stable?

TL;DR
The question you are asking is not worth asking. Systems like these are hard to judge and the equilibrium between these structures is extremely fast. Considering any structure as more stable than another is certainly a mistake, because it completely ignores that the molecule is not static.

The double minimum of the tautomers of acetylacetone
 Long time it was believed, that for the equilibrium of acetylacetone (AcAc) the enol tautomer (4-hydroxypent-3-en-2-one) has a symmetric hydrogen bond and shows therefore only one minimum, like the following scheme by DMacks from Wikipedia suggests:

A quite recent publication by Howard et. al. nicely explores the potential energy surface of this molecule and manages to explain the transitional barrier.[1] They find the barrier to be around 10.5 kJ/mol (2.5 kcal/mol) which translates to a rate of 1011 s-1, which is ultrafast. At room temperature you are not able to distinguish between those structures.[2,3]

The equilibrium of hexane-2,4-dione
When we move away from the symmetrically substituted case, we can expect things to be similar, but a bit more complicated. First of all, we would expect a transition barrier to be roughly around the same size, so the transition from one form to the other is still ultrafast.[4]

I did a few, very rudimentary calculations on the subject and I have to admit, that they were far more complicated than I have anticipated. For reasons I will explain in the notes section, I choose to only discuss the electronic energies on the DF-BP86/def2-SVP level of theory.[5]

First of all it should be noted, that the dione open form 3 is more than 35 kJ/mol higher in energy and will play an insignificant role in the whole mixture. However, I chose this to be the reference state.
The (4Z)-5-hydroxyhex-4-en-3-one 1 is therefore 39.3 kJ/mol lower in energy, while (3Z)-4-hydroxyhex-3-en-2-one 2 is 37.2 kJ/mol lower in energy. This separates the two states by just 2.1 kJ/mol. At room temperature it is impossible to really distinguish those two. It is an incredibly shallow potential energy surface. The transition state 4 is at -36.5 kJ/mol, which makes the activation barrier less than one kJ/mol. Now we should not trust these values absolute, but they give us a rough idea about the whole process. The virtual mode in the transition state is depicted below.

The same mode in 1 looks like this:

And in 2 it looks like this:

You can see how these modes are all essentially the same. Again you will probably have to cool your system to below 50 K to actually see a separation of those.

One remarkable feature is that against the predictions of judging the +I effect, this level of theory would actually predict 1 being more stable than 2. This is not even surprising since the ethylene chain is also rotating out of plane in this structure.[6]

Conclusion
The tautomers of hexane-2,4-one are in a very fast equilibrium with each other. Based on simple models, like +I effect, it is impossible to judge which tautomer would be more stable, i.e. which tautomer has the lowest absolute energy. Not even higher level approximations like density functional theory are reliable enough to predict this.

Notes and References

Daryl L. Howard, Henrik G. Kjaergaard, Jing Huang, and Markus Meuwly, J. Phys. Chem. A 2015, 119 (29), 7980–7990.
Assuming that every transition leads to the formation of the associated structure, i.e. $\kappa =1$, which is reasonable for this small a barrier, we can approximate via transition state theory the rate for a reaction with a barrier of around 10 kJ/mol:
$$ k = \kappa\frac{k_\mathrm{B}T}{h}e^{\frac{- \Delta G^{\ddagger }}{RT}}\approx 1\times10^{11}~\mathrm{\frac{1}{s}}.$$
You might need to cool the system to about 50 K to actually see the formation of two minima. Of course for the symmetric case these two would be identical, so you would still need to mark one of the sides with an heavier isotope. This of course will influence the measurement, but I think you can get where this is going to.
If you are as confused as me about the correct IUPAC nomenclature of the compound see: How to name the tautomers of hexane-2,4-dione correctly according to IUPAC rules?
This system is incredibly hard to treat with DFT. It might be also very hard with more elaborate quantum chemical methods. The transition state is easily obtained, but a caluculation following the intrinsic reaction coordinate seems to be impossible. The reason for this is that the transition state itself is in very close proximity to the structure of 2. The position of the proton does differ by less than 0.2 Å. See a figure of the superimposed structures.
Accordingly all calculations involving frequencies will fail, since the harmonic treatment is simply not enough for this system. This is actually an interesting case, where the structures are in very close proximity, so that adding thermal contributions leads to completely nonsensical results. Here it is, that the activation Gibbs energy between the tautomers is negative.
Fast equilibria like these cannot any longer be described accurately in the Born-Oppenheimer approximation, hence molecular dynamics need to be involved, or at least high level geometry optimisations on the coupled-cluster level with a triple zeta basis set. (The last time I tried that, it already took forever.)
Reasons for this to happen are plenty. It is partially discussed in Why is the eclipsed conformer more stable than the bisected conformer in 1-butene? Also related to this is Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered?


What do the numerals on the top right corner of the cells in the periodic table represent?

The numbers give you the basic ground state electron configuration of the atoms by telling you how many electrons are in each shell.

An electron's 'position' in an atom can be specified by four quantum numbers:

Principal quantum number ($n$). This tells you which shell the electron is in and can take values $1,~2,~3,~4~...$. The maximum number of electrons in each shell is given by $2n^2$ so the first shell can hold 2 electrons, the second 8, the third 18, the fourth 32 etc. The numbers given in your periodic table are the total number of electrons in each shell in the ground state (lowest energy state) of that element.
Azimuthal quantum number ($l$).
Magnetic quantum number ($m_e$).
Spin quantum number ($m_s$).

See this wikipedia article for more detail on the meaning of quantum numbers.

Electron configurations can be written in more detail using $s,p,d,f$ notation. See this question for more detail.


Are coefficients in a reaction taken as the exponents in the rate law expression?

But when i researched on rate of expression i undertood that a, b and c are not coefficients of the reaction

You are correct. $a,b$ and $c$ summed together represent the order of the reaction. Note that this is very different from the molecularity of a reaction.

  But in most part of my book the coefficients are taken as the powers a b and c

Without knowing the reactions for which this is done, it is impossible to comment further. However, assuming that the book is correct, we can say, that in those specific cases, your book was dealing with elementary reactions.

Recall that molecularity of an elementary reaction is the same as its order. However, most natural reactions occur through complex mechanisms, and in those cases, molecularity is rarely equal to its order.

For more details on the difference between molecularity and order of a reaction, have a look at this question.


Your initial example showing derivatives of glucose and fructose might be misleading since the nomenclature of natural product parent compounds does not necessarily follow the usual nomenclature rules of general organic chemistry.

In particular,

  P-102.2.2 Numbering parent structures

  The carbon atoms of a monosaccharide are numbered consecutively in such a
  way that:

  (1) a (potential) aldehyde group receives the locant 1 (even if a more senior
  characteristic group is present);

  (…)

(This rule is probably easier to understand if you draw the monosaccharides in their acyclic form.)

Generally, however, nomenclature of organic chemistry has different rules for the numbering. The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Therefore, the numbering in your own example corresponds to the name 3-phosphinopropanoic acid since the acid group is the principal characteristic group of this compound.


Is Pauling electronegativity well-defined?

If you have a single, unique value of $\chi_i$ for each element $i$ then yes, the sum of the three values is always zero.

If you are assigning electronegativities that differ with oxidation state or chemical environment, the answer can be no (for example, if element B has a different electronegativity in an AB bond than it has in a BC bond).

The Pauling electronegativity difference is

$${\chi}_{a}-{\chi}_{b}=(1\ eV)^{-\frac{1}{2}}\,\sqrt{{E}_{d}\left( a,b\right) -\frac{{E}_{d}\left( a,a\right) +{E}_{d}\left( b,b\right) }{2}}$$

where $E_d(i,j)$ is the dissociation energy for a single bond between element $i$ and element $j$.
The definition assumes that you can consistently and accurately say that the square of the electronegativity difference between A and B is equal to the difference between the dissociation energy for AB and the average of the dissociation energies for AA and BB. If this isn't consistently true for each combination of three elements A, B, and C, your sum will be nonzero. Pauling fit his electronegativity values to the data to average out these inconsistencies.


How to get the efficiency of a heat engine which undergoes an elliptical cycle?

Let $(V_0,P_0)$ represent the coordinates of the center of the ellipse, and let $P_\mathrm{max}$ and $V_\mathrm{max}$ represent the maximum pressure and maximum volume, respectively over the cycle.  Then, the work is the area of the ellipse, which is given by:  $$W=\pi(P_\mathrm{max}-P_0)(V_\mathrm{max}-V_0)$$This is, $\pi$ times the product of the semi-major and semi-minor axes.  This differs from the result which you gave.

The shape of the ellipse can be represented parametrically in terms of the angle $\theta$ around the cycle, assuming $\theta$ is measured clockwise from the point $(V_0-(V_0-V_\mathrm{max}),P_0)$:
$$P=P_0+(P_\mathrm{max}-P_0)\sin{\theta}\tag{1a}$$
$$V=V_0-(V_\mathrm{max}-V_0)\cos{\theta}\tag{1b}$$
Application of the first law of thermodynamics to the working fluid over a differential portion of the cycle gives us:
$$dU=nC_vdT=dQ-PdV\tag{2}$$But from the ideal gas law, $$nRdT=d(PV)$$Substitution of this into Eqn. 2 yields:
$$\frac{C_v}{R}d(PV)=dQ-PdV$$
So the differential heat added during an arbitrary portion of the cycle is given by:$$dQ=\frac{1}{R}[(C_v+R)PdV+C_VVdP]\tag{3}$$So the differential heat is zero when dQ = 0, or, equivalently, when $d\ln{P}+ d\ln{V^{\gamma}}=0$, or, equivalently when $PV^{\gamma}=\mathrm{Const}$.  This is the equation for an isentropic line tangent to the ellipse.

There are two angles $\theta$ at which adiabats are tangent to the ellipse.  These two angles can be obtained by substituting Eqns. 1 into Eqn. 3 with dQ = 0.  Once these angles have been determined, the total positive heat added Q during the cycle can then be obtained by integrating Eqn. 3 with respect to $\theta$ between the two angles.


Negative Ohm reading from a Electrochemical reaction?

How could I use the negative Ohms reading?

You're using the meter the wrong way. A meter is set up to have a particular low voltage and measures how much current flows. Since your electrochemical cell has a voltage and creates its own current the meter reading goes whacko.

  Should I be collecting data in a different way?

If measure current that would give you some idea. In a better controlled analytical chemistry experiment a special electrode pair with two flat plates is used so that the geometric configuration stays exactly the same.


Unfortunately, the proportion you are using is not correct. Acid value $\mathrm{AN}$ determination procedure is not much different from an ordinary titration with indicator with the only difference that the mass of titrant is normalized to the mass of oil $m_\mathrm{oil}$:

$$\mathrm{AN} = \frac{T\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}$$

where $ΔV_\mathrm{eq}$ is equivalent volume (ideally you have to use a difference with an external standard ("spiking solution"), but you probably didn't do that); $T$ is a so-called titer, or basically mass concentration of the titrant:

$$T = CM$$

so that

$$
\begin{align}
\mathrm{AN} &= \frac{C(\ce{KOH})\cdot M(\ce{KOH})\cdot ΔV_\mathrm{eq}}{m_\mathrm{oil}}\\
  &= \frac{\pu{9.5e-3 mol L-1}\cdot\pu{56.106 g mol-1}\cdot\pu{7.4e-3 L}}{\pu{1.015 g}}\\
  &= \pu{3.89 mg g-1}
\end{align}
$$

Also, a couple of side notes:

Always carry units across all steps and never omit them.
I recommend not to do intermediate calculations if you don't have to. Instead, first find an algebraic equation and only then plug in all the numbers.
Mind significant figures. If the mass of oil is given with 4 significant figures, it's a good idea to use a molecular mass of the titrant with at least 4 sigfigs too.


Partial pressure of water in a submerged balloon

Excellent question. On the one hand, the vapor pressure of water (or any other volatile liquid) is a function of the temperature alone, but on the other, if you assume that the mole fraction is constant you find that the vapor pressure of the water changes even though the temperature is constant.

The short answer

The water vapor pressure stays constant because as the pressure in the balloon increases, the mole fraction of the water reduces proportionally to maintain the equilibrium it has established.

The longer answer

Let's dig into the former approach I mentioned in the introduction first. We can begin with the Clausius-Clapeyron relationship.

The Clausius-Clapeyron relationship (often is simplified to):

$$\ln \left(P_{vapor}\right) = -\frac{L}{R}\left(\frac{1}{T}\right) + C$$

Where $L$ and $C$ are constants and $R$ is the ideal gas constant.

Using this model, we would expect the vapor pressure of the water to remain constant and therefore the mole fraction of water in the gas phase to decrease.

Dalton's law tells us: $$X_{water}=\frac{n_{water}}{n_{tot}} = \frac{P_{water}}{P_{tot}}$$

Using this expression we can see that, as the total pressure of the gas increases and the vapor pressure of the water stays constant, the mole fraction of water in the gas phase must go down. If we assume, as a first approximation, that the total number of gas molecules in the gas phase is constant, then a reduction in mole fraction can only be accomplished by some of the molecules of water in the gas phase being reabsorbed into the liquid water phase. This consequence is in agreement with the Le Chatelier principle prediction. A decrease in volume available to the water molecules will shift the equilibrium below to the left:

$$H_2O\,(l) \rightleftharpoons H_2O\,(g)$$

Thank you for such a rich question! How interesting that there are so many topics it relates too! Excellent!

Other thoughts

When we calculate the initial pressure of the other gases in the volume, we should use:

$$P_{tot} = P_{water} + P_{all\,other\,gas}$$

$$P_{tot} = P_{external} = 760\,mm\,Hg$$

$$P_{all\,other\,gas} = P_{tot} - P_{water} = 760\,mm\,Hg - 24\,mm\,Hg = 736\,mm \,Hg$$

The dissolved gas in the water in the balloon will also increase at higher pressures. How will this affect our calculations?
We have not considered energy concepts explicitly in this argument about the pressures being equal (in the first bullet above), but we could have considered the elastic energy stored in the bonds of the balloon. How could we include this to get a better estimate of the pressure of the other gases in the balloon?


What is that process called in which we add water molecules to an ion to form a complex?

I believe it is normally just called "addition of water".
There is also the term "hydration" but this can also refer to the addition of water to alkenes or alkynes which forms alcohols (in the case of alkene; I am not sure about alkynes).
For example, $$\ce{CH2=CH2 \text{(ethene/alkene)} + H2O -> CH3CH2OH \text{(ethanol/alcohol)}}$$

I personally think "addition of water" is just fine to use and people will get what you are saying.


Photoionisation microscopy of hydrogen - where are the p orbitals?

These states aren't 1s, 2s, 3s, 4s, like you suggest. These are actually highly excited states of hydrogen. Note the negative energies in wavenumbers (cm$^{-1}$) in the rightmost column of the figure you provide. This is how much lower in energy the states are relative to the ionization limit, 13.6 eV, so we are talking about excitations ~13.57 eV in energy above the ground state. In fact, these energies are so high they refer to them as "quasibound" states.

The states the authors are looking at Stark states (the experiment is done in a field of 808 V/cm). So the analytical results they compare to are the solution of the hydrogen atom in parabolic, rather than spherical polar coordinates. This is why the quantum numbers given in the middle column are a bit funky. So the (0,29,0) does not mean $n=0$, $l=29$, $m_l=0$ (as I'm sure you can see why!) The quantum numbers define the state in parabolic coordinates, so the numbers are ($n_1$, $n_2$, $m$).

The reason they do this is that parabolic coordinates are still separable (so you can solve it analytically) when an external electric field is applied along some axis. Spherical coordinates, on the other hand, are not. So you don't see pure s, p, d orbitals (as nice as that would be!). My math gets a bit rusty here, so maybe some one else can add to this, but I believe the images you see on the left and the middle column are actually linear combinations of the spherical solutions we are used to as chemists. In fact, there is a line in the paper that in the experiment they prepare their states by exciting to a mixture of $s$ and $p$ states.

I believe you can connect the parabolic and spherical solutions in an electric field by a unitary transformation if the field is weak enough, since you can get the Stark effect in spherical coordinates by perturbation theory.

Hopefully that helps clear things up a little bit. I'm not an atomic physicist, so I'm often confused by some of the language and tools they use. There is a nice Wolfram Demo on the hydrogen atom in parabolic coordinates with and without an external electric field you may find interesting here:

http://demonstrations.wolfram.com/TheHydrogenAtomInParabolicCoordinates/


Is there a single or double bond between S and O in propanethial S-oxide?

The $\ce{S-O}$ bond in sulfoxides is best considered as a single covalent bond with positive charge on $\ce{S}$ and negative on oxygen. The oxygen is significantly nucleophilic (see for example the Swern oxidation mechanism 1) and the sulfur will stabilise an anion on the alpha carbon.
$\ce{C=S}$ has significant enol content and is better considered $\ce{HS-C=CR}$. The $\ce{S}$ is significantly nucleophilic and is easily oxidised to the ene-disulfide $\ce{RC=C-S-S-C=CR}$.


Are there any chemicals with this structural skeleton?

CCDC doesn't contain any molecules of such geometry. The closest topology observed is among pnictogens such as $\ce{P, As, Sb}$, which form tetrahedron units $[\ce{P4}]$, $[\ce{As4}]$, $[\ce{Sb4}]$ which are readily coordinating with Au, Rh, Ag and Cu by edge-sharing:

The first publication (1) tetraphosphido-ligand presents the following structure of $\ce{Rh}$-complex:

The most relevant structure best fitting the required connectivity matrix would probably be bis(tricyclo[$1.1.0.0^{2,4}$]tetraphosphane)-gold (2):

Thanks to @TAR86 comment mentioning exotic boranes, I also found a research (3) on ab initio calculations for $\ce{B4}$ cores obtained via the following proposed reaction:

They also obtained crystal structure for bis(tetrahydropyran)-(tetra(t-butyl)tetraborane)-lithium where $\ce{B4H}$ core is present:

(1) Lindsell, W. E.; McCullough, K. J.; Welch, A. J. J. Am. Chem. Soc. 1983, 105 (13), 4487–4489. DOI: 10.1021/ja00351a067
(2) Forfar, L. C.; Clark, T. J.; Green, M.; Mansell, S. M.; Russell, C. A.; Sanguramath, R. A.; Slattery, J. M. Chem. Commun. 2012, 48 (14), 1970–1972. DOI: 10.1039/C2CC15291F
(3) Neu, A.; Mennekes, T.; Englert, U.; Paetzold, P.; Hofmann, M.; von Ragué Schleyer, P. Angew. Chem. Int. Ed. Engl. 1997, 36 (19), 2117–2119. DOI: 10.1002/anie.199721171


How to compare the boiling point of water, ammonia and hydrogen fluoride?

Though ammonia has three H atoms, it is not capable of forming four H-bonds

An important difference in terms of hydrogen bonding between ammonia and water, lies in the ratio between how many partial positive hydrogen atoms and how many lone pairs of electrons each have.

A water molecule has two partial positive hydrogen atoms and two lone pairs of electrons located on the oxygen atom. This implies that each water molecule can potentially have both its hydrogen atoms and both its lone pairs involved in hydrogen bonding.

But in ammonia, there is a shortage of lone pairs on the central nitrogen atom.
Hence it is capable of forming only two H-bonds.

And the difference in electronegativity is higher in $\ce{HF}$ than in $\ce{NH3}$.

Hence boiling point order is $\ce{H2O} > \ce{HF} > \ce{NH3}$


The intuition behind carbon's covalent bonds

The most naive answer would guess that since every orbital wants two electrons and there are two orbitals with only 1 electron in each, then 2 more electrons are needed.

However carbon typically combines the one $2s$ orbital and three $2p$ orbitals to form four equivalent $2sp^3$ orbitals each with one electron. The four hybrid orbitals have a tetrahedral shape. So after hybridization, the carbon atom wants to create four bonds to fill the four electron shell vacancies.


GCMS Peak Identification: Online or Open Source / Free Programs?

Have you tried the Golm Metabolome Database?
It is a collection of GCMS mass spectra. The focus is on metabolites.

However, on their webpage is the entry "Decision Trees". If you follow the link, there is the option to enter your own mass spectra information and receive a prediction of functional groups in the measured substance.
This might be of help to identify your molecule.
There are more details in the Wikipedia entry.


Is the cyclopropenyl anion antiaromatic?

When you have a ring that has $4n$ π electrons it's anti-aromatic if you force it into a fully conjugated cycle.

Often such rings find a way to break the unfavorable (or at least, less favored) anti-aromatic conjugation.  Large ones like cyclooctatetraene are likely to bend so all ring atoms are no longer in the same plane. Small ones like the cyclopropenyl anion shown here can't bend the ring out of planarity easily (or do it at all with only three atoms), but they can still break the conjugation by using unequal bond lengths or moving a ligand out of the plane giving that ring atom a pyramidal bonding geometry.

The pyramidal bonding geometry seems to be what cyclopropenyl anion actually does.  A computational result is shown at http://comporgchem.com/blog/?p=2987; the top carbon atom has its bonds directed towards the observer indicating a pyramidal geometry.


There's no "steel" in that stuff. It is simply filled with inorganic particles, probably (partly?) iron dust. I'm using jbweld too. Quick test with a bar magnet showed the tube with the "steel" component as well as the cured epoxy are indeed slightly ferromagnetic.
Most importantly, the filler reduces macroscopic shrinking during curing. That's a general principle: The density of the filler particles stays constant, while that if the glue component goes down, remarkably. The filler material type is not really important, only its surface must be compatible with the epoxy.
Plus the filler particles impede the propagation of cracks, and reduce the permeability to water, oxygen, etc. Again, that's a general property of fillers, not specific to this one.
It's hard to tell what the specific advantage of using iron/steel powder is, instead of some inorganic oxide (vulgo "sand"). A metal might catch environmental oxygen that diffuses in, increasing the temperature stability of the epoxy. If that's the case I don't know, but this stuff is remarkably temperature-resistant.


NO, this is not a viable way to make amides.

Phthamide anion will react with acyl halides, but the problem arises with the second step, the removal of the phthalimido group. The usual conditions described here of hydrolysis with hydroxide or reaction with hydrazine will result in removal of the acyl group to reform phthalimide plus the carboxylic acid, in the case of OH- hydrolysis, or the acyl hydrazide if using hydrazine. This is because the acyl group is the more electrophilic site for the nucleophile.


Amber solvatebox: How do I remove water that has traveled away

This has nothing to do with box walls (there are no walls!) or restraining the waters. You do not want to do that!

Do not confuse periodic boundary conditions with wrapping to the central cell. Look at the amber tutorials, it is all explained there.

You either need to set iwrap=1 in your amber minor put your trajectory trough cpptraj and use autoimage.

Again, all in the tutorial and manuals.


Material that is flexible when cold but becomes brittle when hot

It's indeed not common, as the brittle/ductile transition is usually the other way around. You can, however, find counter-examples (in a limited range of temperature) by searching for “inverse brittle-to-ductile transition”.


What is the electronic configuration of La⁺?

During ionization only the electrons in last shell leaves.So its 6s


Which is more stable: Ortho or Para Hydrogen?

At room temperature, about $75$% of $\ce{H2}$ is $\ce{o-H2}$ and $25$% is $\ce{p-H2}$. As the temperature drops, the relative amounts of $\ce{p-H2}$ increases. The two forms are in a temperature-dependent equilibrium $\ce{o-H2<=> p-H2}$, $\Delta H = \pu{-1.66 kJ/mol}$.
Ref. M. Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, New York $1994$, 1202 p. ISBN 0-89925-4547-8.


The Fresnel equations, http://en.wikipedia.org/wiki/Fresnel_equations, are used to calculate reflectivity of transparent substances based on the index of refraction of the liquid vs. that of air or other medium. There are complications introduced by conductivity, and solids may display anisotropic reflectivity, see http://www.hindawi.com/journals/isrn/2012/732973/.

Dispersion staining is a technique for staining microscopic specimens depending on the change in the index of refraction with the color of light (dispersion), see http://en.wikipedia.org/wiki/Dispersion_staining. This uses the shininess of substances in a liquid, often an organic immersion oil, to help distinguish or identify them.

Note that metallic reflection is based on a "sea" of electrons bouncing back an inverted wave, and differs from refraction. Liquid metals such as mercury are shiny for that reason, but I do not know if there are sufficiently conductive organic liquids that would have metallic reflection.


Your first step is incorrect. Think about it this way:
$5$ people have $10$ eyes. This group of people has $14$ eyes. How many people in the group?
The correct solution is:
This group has $14/10 = 1.4$ times more eyes than the known group. Thus, they have $1.4$ times more people. $5$ people $\cdot$($14$ eyes/$10$ eyes) $ = 5$ people $ \times 1.4$ (unitless ratio) $= 7$ people. Here, the units make sense. This is correct.
In your calculation you do ($5$ people)/($10$ eyes $\times 14$ eyes)$= 1$ people/($28$ eyes$^2$). This unit people/eyes$^2$ makes no sense. This is incorrect.
First you find the amount of $\ce{Ni}$ in the final sample.
Your sample absorbs $0.410$ which is slightly less than $0.522$ ($\pu{6 ppm }$sample which). So your sample has $(0.410/0.522) \times \pu{6 ppm} = \pu{4.713 ppm}.$
Water solution has density of $\pu{1 kg/L}$, so $\pu{500 mL} \to \pu{500 g}$. $\pu{4.713 ppm}$ in $\frac{\pu{500 g}}{1\,000\,000} = \pu{0.5 mg}$. $\pu{0.5 mg} \cdot 4.713 = \pu{2.356 mg}$.
The total weight of the sample was $\pu{0.35 g} = \pu{350 mg}$
$$100\ \% \cdot \frac{\pu{2.356 mg}}{\pu{350mg}} = 1.003\ \%$$
You should round it down to $2$ digits, because that was the accuracy of other data. The answer is  $1.00\ \%$.


Can a battery powered by iron and air really provide a feasible power source for automobiles?

This is certainly possible.  In fact, there are a large number of metal-air batteries being developed all over the world.

It is, however, strange that such high energy density is achieved, because iron ion itself would not offer very high energy density.  My guess is that likely they did it by having an extremely high matter density.  So this battery may be small in size for its capacity, but at the same time it may also be much heavier than a conventional battery.

That could be why they brand it as a shuttle power supply(well, energy density isn't really relevant for a bus anyways).  A very heavy battery on a hand held device is never good.   Cost is the more important factor there, but it is hard to tell at the moment.  0 loss over 200 cycles seems like a fairy tale, but I will give them the benefit of the doubt at the moment.

This battery will also likely to have a considerable loss of charge when not in use.   Again, this might not be a huge problem for buses, where you keep driving non-stop during days then kept it in the charging station overnight.

Anyways, don't expect this to be on your iPhone 10 or your laptop.  Even if it works as presented, it might not be that ideal for portable electronic devices, where people worry much more about weight and stand-by time than if it occasionally explodes, sadly.


Determining absolute configuration when the lowest priority group is not facing backwards

You had the right idea when you said "if the hydrogen is coming out of the page (i.e. as a wedge) we swap our answer from R to S or vice versa (to get the hydrogen going into the page)".
The difficulty seems to have came with your manipulation of the drawing. The diagram you have implies that the hydrogen is coming out of the page towards us (i.e. anticlockwise = (R)).
You could also flip the entire molecule through by 180° to give you the chlorine pointing out of the page towards you.


How do I calculate the total volume when i mix 1000 kg of glucose with 5000 L of water

You could calculate it only if

you know the density of the substance being dissolved and of the solvent
there is no interaction between the 2 compounds, causing the deviation from the Raoult law and the shift of the final density.

If both assumptions are met, the final volume is
$$V_{\rm mix}=\frac{m_{\rm 1}}{\rho_{\rm 1}}+\frac{m_{\rm 1}}{\rho_{\rm 1}}$$

where m and $\rho$ are respective masses and densities.

On the other hand, if the mentioned deviation is in place, the calculation will be wrong.

E.g. if you mix $\pu{1 L}$ of ethanol with $\pu{1 L}$ of water, then after cooling down ( as the mixture would heat itself), the result will not be $\pu{2 L}$, but $\pu{1.9 L}$

In such cases, it can be determined experimentally, or from tabulated data of solution densities.

Or, the calculation above can be usually used as a good estimation, especially if there are no significant temperature changes during dissolution.


Just add your gas mix in a container with NaOH solution and shake well or bubble it continuously through the solution. SO2 dissolves in water to form H2SO3 (and H2SO4 in the atmosphere when it rains resulting in what we call acid rain) which will be neutalized by NaOH to give dissolved Na2SO3. Your air should be freed of most SO2. This is how it is done in the industry. There are many other ways as well but most rely in neuralizing the SO2. Google Flue-gas desulfurization for more details.


Calculating KC Given Equilibrium Concentrations

Those are the concentrations of $\mathbf{N_2O_4}$ and $\mathbf{NO_2}$ at equilibrium. They are just given to you as part of the question for you to use in calculating $K_c$.


After opening, when is a can of Diet Coke most acidic? (Based on a true story)

Coke (or any carbonated drink) is most acidic when it is just opened. The acidity comes primarily due to phosphoric acid. Citric acid if present also adds some acidity. Carbon dioxide (which is present as dissolved form) will increase acidity though by  small amount. Carbonic acid, formed upon dissolution of $CO_{2}$ is a weak acid. Its contribution to acidity, when phosphoric acid is present would be negligible.

Facts

Solubility of a gas is more in cold liquids.
Carbon dioxide dissolves in water. The dissolved $CO_{2}$ is now in equilibrium with carbonic acid. The so formed carbonic acid is much less in concentration than $CO_{2}$ itself.

Considering small effect on acidity from carbonic acid:
 Just opened coke can will have higher carbonic acid (whatever be its amount), due to higher (relatively) dissolved $CO_{2}$. As temperature increases $CO_{2}$  fizzles out to air, and dissolved $CO_{2}$ is reduced  so is carbonic acid.

PS- Microwaving coke can will be dangerous. There is a hazard of explosion and severely damaging your microwave oven of-course !

I am not sure if cleaning with coke will help in your situation.


For something to be an oxidiser. It must be capable of being reduced. The presence of oxygen is no particular indicator of the oxidising ability of a given species.

In potassium permanganate, manganese is in its +7 oxidation state (a very high oxidation state), consequently it is a good oxidant agent as it is able to gain electrons, causing oxidation of another species and itself being reduced.

In ethanol, there isn't the same possibility for reduction as the molecule is already in its lowest oxidation state. Even ethanal  (an oxidation level up) isn't particularly oxidising (despite being able to accept electrons and become formally reduced).


Bond Lengths: Equatorial vs Axial in Trigonal Bipyramidal

For trigonal bipyramidal structure, we know that lone pairs are preferred first to be positioned in the equatorial position. So in $\ce{SF2Cl2}$ lone pairs will be positioned in the equatorial position.
After placing lone pairs in the equatorial position, double bonds are then preferred in the equatorial position but as there are no double bonds in the structures we are considering, so we jump on to the next rule i.e. atoms having higher electronegativity is positioned at axial position and atoms having low electronegativity is positioned at equatorial position. You can also interpret this in another way. You can also say that atoms having higher atomic size is positioned in the equatorial plane and atoms having lower atomic size is positioned in the axial plane.
Now coming back to your structures, $\ce{PF2Cl3}$ and $\ce{SF2CL2}$ we will see structures in this manner:

$\ce{F}$ is more electronegative than $\ce{Cl}$, hence the $\ce{P-F}$ bonds are going to be shorter than $\ce{P-Cl}$ bonds and $\ce{S-F}$ bonds are going to be shorter than $\ce{S-Cl}$ bonds. As $\ce{F}$s (fluorines) are located in the axial position, hence these $\ce{P-F}$ or $\ce{S-F}$ bonds which are actually axial bonds are shorter than that of equatorial bonds (i.e. $\ce{P-Cl}$ or $\ce{S-Cl}$ bonds).
In $\ce{PCl5}$ all the atoms around $\ce{P}$ are basically the same i.e. $\ce{Cl}$. So the electronegativity factor of $\ce{Cl}$ doesn't play any role in reasoning out why axial bonds are longer than equatorial bonds in case $\ce{PCl5}$. The reason is - in $\ce{PCl5}$ the axial bonds suffer repulsion from the equatorial bonds so they become long. The equatorial bonds providing repulsion to axial bonds also happens in case of $\ce{PF3Cl3}$ and $\ce{SF2Cl2}$, however this factor is overlooked/ not considered in cases of $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ because there this factor/ effect is not dominating. In those cases as the type of atoms attached (i.e. electronegative atoms) are different, hence the effect of bond length shortening due to the electronegative atom attached is the dominating factor.


Is a 50/50 mix of Acetone and Water inert?

Acetone is notoriously aggressive towards the plastic, rubber, and other non-metal components of internal combustion engines (and associated parts like fuel pumps). If the concentration of acetone is strong enough to dissolve your jell (assuming that is some organic compound) it is very likely strong enough to attack components of the fuel pump.

As a bit of anecdotal evidence regarding what I said above, note that of the dozen-odd oxygenated additives allowed by the US EPA for the oxygenated fuels program, neither acetone nor any other ketone is on the allowed list. Although this is in part due to the high volatility of acetone, it is also largely due to the fact that acetone would attack engine components as described above.

In short, what you've described is a bad idea for the health of your fuel pumps, even at very low concentrations of acetone.


Can the thermodynamic predictions of redox reactions based on E and dG contradict each other?

$\Delta G = -n {\mathcal F} \Delta {\mathcal E} $

$\mathcal F$ is Faraday's constant, which physicists currently believe has been constant throughout the universe for all time.  $n$ is the number of electrons transfered in an electrochemical reaction.  For a given reaction, $n$ is also constant in all places and times.

Thus, $\Delta G$ determines $\Delta {\mathcal E}$ and vice versa.

  Are there any errors in above calculations?

Yes, there must be.

  According to E∘ values, reaction (2) is more feasible than reaction (1). Am I correct ?
  According to ΔG∘ values, reaction (1) is more feasible than reaction (2). Am I correct ?

Both of these cannot be correct.

  Is it useless to use E values to predict the feasibilities of redox reactions ?

No, it is just as good and exactly as informative as $\Delta G$.

  Should we always resort to ΔG ?

No.  Feel free to resort to $\Delta G$ if you like units like kJ/mol instead of volts, but other than personal preference there is no reason to prefer one over another.

  Is it possible for predictions based on E values and ΔG values to contradict ?

No.

  If not how can these results be explained ?

UPDATE: Arithmetic error.

The choice of $n$ is arbitrary.  Once you pick $n$ for a given half reaction, it determines the choice of $n$ for the other half reaction, but that's it.  If you then consider a separate cell with separate reactions, you could write that new cell's equation with an arbitrary $n$.  The $\textbf{mol}$ in $\Delta G$'s units of $\frac{\textrm{ kJ}}{\textbf {mol}}$ is $\textbf{mol}$ of reaction, and thus the value of $\Delta G$ depends on how you write the reaction.  For example you could write equation (b) + (c) as

$$\ce{10M^3+ + 2 H2O + 2 MnO4- -> 2 Mn^2+ + 10 MO^2+ +4 H+}$$

Now, $n$ is twice as large, so $\Delta G$ is twice as large.  But $\Delta \mathcal E$ is unchanged.  This is because $\Delta \mathcal E$ has a basis of "$\textbf{mol}$" of electrons, not of moles of a reaction of arbitrary scale.  In reality the "$\textbf{mol}$" of electrons is converted to a different units (Coulombs) to make things more confusing.  (Remember, a volt is a joule per coulomb.)

So contrary to what I said above, the sign of $\Delta G$ but not the magnitude of $\Delta G$ is fixed, meaning that whether a reaction is absolutely spontaneous can determined by either $\Delta G$ or $\Delta \mathcal E$, but to compare the relative spontaneity of two reactions, either the $n$ values of the two reactions must be the same and then $\Delta G$ values can be used, or else without worrying about $n$ you could just look at $\Delta \mathcal E$.


The Woodward-Hoffmann rules only tell you whether a reaction is thermally allowed or not, they do not give any indication of the relative rates of multiple allowed reactions.

I guess you're asking why the 4π component in a [4+2]-cycloaddition doesnt undergo a conrotatory 4π electrocyclic ring-closure? This is plausible, but I suppose even if it did happen the cyclobutene product is highly strained so the equilibrium would lie in favour of the ring-opened diene anyway.


In the reaction between salicylic acid and acetic anhydride to produce aspirin, why is it necessary to quench the reaction?

Acetic anhydride is contaminating your final product. Quench simply means to destroy excess reagent before continuing. It might not matter too much in this particularly reaction since you're about to purify via recrystallization anyway, but imagine if the next step involved adding a very expensive nucleophile. That nucleophile would add to acetic anhydride and be wasted.


Different solder composition affects the connection strength?

Good question!

The short answer: if you are soldering bulk silver to copper, 63% Sn/ 37% Pb eutectic would probably be OK, but for silver plating, you might be able to make a successful join using tin/lead, if heat is quickly applied and removed, but might need to use a silver-bearing solder for thin silver plate or if the base metal under the silver is not easily soldered.

BTW, "cold soldered" is perhaps better expressed as low-temperature soldering, though both indium and gold can be joined at room temperature.

A longer explanation: reasons to use a specific solder composition.

To some extent, solder dissolves the surface of the metals to be united. This can be problematic with plated parts, e.g. gold plated over nickel, where the solder can completely remove the plating, making it difficult to adhere to the base metal. For that purpose, solders containing the plating metal are made, but they have other liabilities: expense and brittleness. Similarly, copper-containing solders are used for copper-plated electronic parts.
Not only alloys form between soldered parts, but also intermetallic compounds form, such as $\ce{Ag3Sn}$, $\ce{Cu6Sn5}$ and $\ce{AuSn4}$, some of which are brittle. Changing solder composition may not be practical to prevent some intermetallic compounds from weakening a joint, so barrier coatings and specific temperature cycling and storage may be used to retard their formation.
Pure tin solder, as well as zinc, has the property of forming whiskers, which not only short-circuit electronics, but was rumored to have caught Napoleon's troops with their pants down.

For these and other reasons, soldering can be considered as much a craft as a science. You may have to make some tests of the solder joints if they are critical, as in aerospace or medical applications.


Chiral centres with stereoisomeric substituents

Let's address your concerns one at a time:

The structure 1 you have drawn is chiral with $\ce{C4}$ having four different groups attached to it. $\ce{C4}$ has the (S)-configuration with the CIP priorities Cl>(Z)-propenyl>(E)-propenyl>H.
The structure has an enantiomer of the (R)-configuration. Just switch any two groups: Cl, (Z)-propenyl, (E)-propenyl or H. Each enantiomer will have the same specific rotation but opposite in sign.
The two propenyl groups are different stereoisomers, just as different as the (Z)- and (E)-2-butenes.
If the two propenyl groups were changed to -CHClCH3, say, (2R,4R)-2,3,4-trichloropentane 2a ($\ce{CH3CHClCHClCHClCH3}$), then the central carbon, $\ce{C3}$, is chirotopic but non-stereogenic. That is to say, if the chlorine and hydrogen at $\ce{C3}$ were switched as in 2b, the compound would still be chiral but not a different stereoisomer. Rotate 2a by 180o about a vertical axis and you will obtain 2b. They are identical.


Predicting the product of a hydrogenation reaction

To answer and understand this, let’s first consider the products of the hydrogenation ($\ce{H2; Pd/C}$) of 1-methylidene-4-methylcyclohexane (the same reactant except the methyl group is in para-position). This compound is symmetric and therefore achiral. We can draw the methyl group pointing upwards iwthout loss of generalisation. The hydrogenation is syn-selective and it can occur from the top or the bottom. We expect a $1:1$ mixture of the former methylidene group pointing upwards (the syn-isomer) and pointing downwards (the anti-isomer). Both are achiral (the plane of symmetry is retained in the reaction) but not identical; therefore, they must be diastereomers.

Now let’s move the methyl group back into the 2-position. By moving the methyl group, the plane of symmetry is lost; it suddenly makes a difference whether the methyl group points upwards (S) or downwards (R). For each of the two isomers, we can again hydrogenate from both above and below; we can expect two anti and two syn isomers at first approximation: we enter one stereocentre into the reaction and we form an additional one making two stereocentres in total and thus 4 theoretically possible isomers.

In explicit forms, starting from the (2S) isomer the syn-product could be called (1R,2S) and the anti-isomer would be (1S,2S). From the (2R) isomer, another syn (1S,2R) and anti (1R,2R) isomer is expected. Only when taking a closer look at the products, it becomes obvious that the two syn isomers are in fact identical because they can be transformed into each other by rotation; and that they are achiral because they generated a new plane of symmetry. Thus, the syn isomer is a meso isomer.

Putting this together, the hydrogenation products are two enantiomeric anti products and a meso (syn) product. These two classes are diastereomers of each other, thus ‘a mixture of diastereomers’ is the only correct answers.


Note that the chromate ion $\ce{CrO4^{2-}}$ has chromium in the +6 oxidation state like dichromate $\ce{Cr2O7^{2-}}$. Chromate is yellow, dichromate is orange.

From the Wikipedia article on sodium chromate: sodium chromate is formed by reaction of dichromate with hydroxide:

$$\ce{Na2Cr2O7 + 2NaOH -> 2Na2CrO4 + H2O}$$

You may be making sodium dichromate, but the strongly basic conditions are converting it to sodium chromate.


What happens to the argon in the Haber-Bosch process?

Inert gaseous components such as methane and argon are indeed should be eliminated from the system in order not to lower the partial pressure of the reactants too much. Technically there is usually a standalone gas separation plant where the extraction of argon from the recycle gas is performed basically using modified Linde process [1 , pp. 428–430].

Suitable cryoprocesses [...] are available for production of noble gases from synthesis purge gas. Initially, extensive separation of nitrogen, argon, and methane is carried out by partial condensation. Purge argon is then recovered from the condensate in a two-stage condensation process. If helium is to be recovered it can be concentrated by liquefaction of the hydrogen in the hydrogen-rich gas phase, followed by purification. The heavier noble gases, krypton and xenon, pass into the methane fraction.
The purge gas at pressures of up to $\pu{70 bar}$ is ﬁrst introduced into the adsorber (a) where traces of water and ammonia are removed. It is then transferred to the heat exchangers (b1) and (b2) for cooling. The gas is then fed into separators (c1) and (c2) where separation of liquefied nitrogen, argon, and methane from gaseous hydrogen takes place, and dissolved hydrogen ﬂashes off. The liquid bottom product is fed into the fractionating column (d1) where methane (bottom) is separated from the nitrogen fraction (top).
A methane-free nitrogen – argon mixture (liquid) is withdrawn from the middle of this column and fed as reflux into the $\ce{Ar}$ purification column (d2), where nitrogen is separated from argon. The bottom product is argon of product purity and is transferred to a vacuum-insulated storage tank.
Both columns (d1) and (d2) are operated within an $\ce{N2}$ cycle in which cold is produced by expanding high- or medium-pressure nitrogen.
The $\ce{CH4}$ bottom stream from (d1), compressed by liquid pump (g), is evaporated against feed gas and normally led to battery limits as fuel gas. The higher boiling noble gases krypton and xenon are contained in this stream and, in principle, can be isolated.

Reference

Häussinger, P.; Glatthaar, R.; Rhode, W.; Kick, H.; Benkmann, C.; Weber, J.; Wunschel, H.-J.; Stenke, V.; Leicht, E.; Stenger, H. In Ullmann’s Encyclopedia of Industrial Chemistry; Wiley-VCH Verlag GmbH & Co. KGaA, Ed.: Weinheim, Germany, 2001. DOI: 10.1002/14356007.a17_485.


If d-electrons are such poor shielders, why do trends increase more gradually across the d-block than the s or p-block?

Your question is similar to what I had in mind like a week back.

I know some textbooks do state a contradiction — that d-electrons have a poor screening effect, but then quickly use the explanation that due to increasing screening effect of nuclear charge from left to right in d-block elements (due to more electrons coming in the d-orbitals), trends are gradual.

Let me clear your mind of this false statement. Remember: it is not the screening effect that leads to a gradual change in trends across d-block elements, but  rather the repulsion between d-electrons and outermost s-electrons.

The follwing example from the 1st series of d-block elements should make this clear to you.
     As nuclear charge increases, electrons come in the 3d-orbitals, which is not the outermost. The outermost is the 4s orbital, already filled up with 2 electrons.

Now, from left to right in this period, the increased nuclear charge will tend to hold on to the outermost 4s electrons tighter, but the d-electrons have an opposite effect. They repel these 4s electrons, and hence net force felt by the 4s electrons decreases. So, atomic size is increases more than expected, and hence it is easier to remove these electrons too (less ionisation energy)

Hence, the increase in ionization energy, reduction is size, etc. is much more gradual while traversing from left to right in a period.


The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

  (…)

According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. You have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the nitrile group ($\ce{-(C)N}$).

The next rule (P-44.3.2) is looking at the longest chain. Therefore, the parent structure is a butanenitrile.

The following rules (P-44.4.1.1 and P-44.4.1.2) would be looking at multiple bonds, which is not relevant in this case.

Next, Rule P-44.4.1.8 is looking for a lower locant for the attached group that is expressed as a suffix (here: nitrile); however, the nitrile group has the lowest locant ‘1’ in both suggested cases; therefore, this rule doesn't help to make a choice.

The next rule (P-44.4.1.10) again is not relevant in this case.

Finally, Rule P-45.2.1 is looking for the structure that has the maximum number of substituents cited as prefixes to the parent structure. Therefore, the preferred IUPAC name is 3-methyl-4-oxobutanenitrile (two substituents) rather than ‘3-formylbutanenitrile’ (only one substituent).

By way of comparison, if you add more substituents to the methyl group, the principal chain can switch, for example 4,4-dichloro-3-formylbutanenitrile (three substituents) rather than ‘3-(dichloromethyl)-4-oxobutanenitrile’ (two substituents).


Ethylene glycol is completely miscible with water in all proportions (as the commenters point out). Therefore, once mixed, the glycol will not separate from the water, ever.

The only exception is if it gets so cold the mixture starts to freeze. Freezing will cause some degree of separation as the water will solidify first. But if it is that cold, separation will be the least of your problems.


Heat transfer with surroundings: why considered reversible?

The surroundings are typically modelled in thermodynamics as an ideal (reversible) reservoir with infinite heat capacity and negligible temperature gradients (i.e., infinite thermal conductivity).  The assumption is that the temperature gradients are confined to the system, rather than the surroundings, and the entropy generation is confined to the system.  Of course, in real life, this is only an approximation.


Stabilization of ipso and para position in Birch reduction by EWG of aromatic ring

TL;DR: This question may not be readily answered with Lewis structures.

The problem is the electron is delocalized into an antibonding orbital.  By analogy, we can examine this concept with molecular oxygen.  The Lewis structure for molecular oxygen:

This clearly is a double bond, with all spins paired.  Now observe the molecular orbital diagram below:

The top energy levels clearly show oxygen with two unpaired electrons, which contradicts the Lewis structure completely.  There is no real way to simultaneously convey these two situations.

Similar diagrams exist and may be used for the aromatic system with an electron-withdrawing group exist, but they are rather complicated.


Why is carbon monoxide the product in reactions that produce high flame temperatures?

This is just a guess... But I wonder if this is due to entropic considerations. In the equation you have written, there are two molecules of gas getting converted to three molecules of gas. $\Delta S$ is positive, so it decreases $\Delta G$ as the reaction temperature increases. However, in the reaction $\ce{(CN)2 + 2 O2 = 2 CO2 + N2}$ you have an equal number of gas molecules, so the entropic component of $\Delta G$ will be small.

I'm not really sure why that would matter. At very high temps, the first reaction will presumably have a very large, negative free energy. Of course, the $\Delta G$ for the $\ce{CO2}$ reaction is most likely negative, so you might have to control the feed stoichiometry carefully to ensure that you are getting only the first reaction.

EDIT:

I got the $\Delta G$ values from the NIST-JANAF website, and it doesn't look like the free energy for these reactions is much different at 4000 K.

$\Delta_f G_{(CN)_2}^{0} = 136.036\ \mathrm{kJ\ mol^{-1}}$

$\Delta_f G_{CO}^{0} = -446.485\ \mathrm{kJ\ mol^{-1}}$

$\Delta_f G_{CO_2}^{0} = -393.183\ \mathrm{kJ\ mol^{-1}}$

$\Delta G_{rxn1}^{0} = -1029.005\ \mathrm{kJ\ mol^{-1}}$

$\Delta G_{rxn2}^{0} = -922.397\ \mathrm{kJ\ mol^{-1}}$

EDIT 2:

So, I just read one of the original papers on this type of flame, and it sounds like the thermal stability of the products is an issue as well. I take it that at these very high temperatures, some of the energy can be transferred to bond-breaking processes, which would lower the total flame temperature. $\ce{CO}$ and $\ce{N2}$ have very strong covalent bonds, so perhaps they do not readily dissociate at these temps.


How many carbons does an alkane need to have an enantiomer?

If we allow our substituents to be isotopes, then 2 is all that is really necessary:

$\hspace{6cm}$

If we exclude such compounds from consideration, then cyclopropanes will give you smallest number of carbons at 5:

$\hspace{5.5cm}$

As a side note, for a non-alkane, only 1 carbon is necessary:

$\hspace{5.7cm}$


Why is the thioester bond weaker than a regular ester bond?

Sulfur is a much bigger atom than oxygen or carbon. As a result, resonance donation from the sulfur into the carbonyl carbon is limited, and the C-S bond doesn't have as much multiple-bond character as the carbonyl carbon-oxygen bond in an ester.

  What are the contributing factors to the fact that thioesters are less stable than their oxygen counterparts?

Kinetically or thermodynamically?

Stable with respect to what?

If we're talking about thermodynamics, and stability with regard to hydrolysis, then yes, thioesters are generally less stable than esters.

It is more favorable for water to attack a thioester's carbonyl carbon because it has a greater partial positive charge than a corresponding ester's. This is due to the resonance issue discussed above. A thioester also has a more stable leaving group, as the larger size of sulfur allows negative charge to be distributed better.


How can the free amino acid B be enantioslectively synthesitzed?

I'm guessing this is a homework type question. Yet, I have some concern of this being a exam question, probably one already taken. So I give a step by step answer depicted in following diagram (courtesy of https://web.chemdoodle.com/demos/sketcher/):


How does bismuth's lone pair affect its properties as a metal and alloy?

If you look at the chemistry of lead(II) and the work by Tony F. Hill (and others) on platinum group metals you will see that bismuth is clearly not the only metal to have lone pairs.
What Tony Hill did was rather clever, normally metal-ligand coordination is all about the donation of lone pairs to the metal and sometimes the retrodonation of electron density onto pi acidic ligands such as carbon monoxide and cyanide.
What Tony did was to get a borane to bind to a lone pair on a low valent metal such as ruthenium, osmium or rhodium. Here is a diagram of an example of one of his compounds.

The synthesis of this rhodium compound can occur from $\ce{[RhCl2(Ph)(PPh3)2]}$, this rhodium complex can react with $\ce{Na[HB(mt)3]}$ (mt = methimazolyl) to form an intermidate $\ce{(mt)B(H)(mt)2Rh(Ph)Cl(PPh3)2}$ which then decomposes to lose a molecule of benzene to form $\ce{B(mt)3RhCl(PPh3)}$. (Chem. Comm., 2005, page 221).
I am sure that tin has a stereochemcially active lone pair, if we consider the strcuture of $\ce{Cp2Sn}$, this is not quite like ferrocene. Instead of the two Cp rings being parallel. They are not. So I think that bismuth is not the only metal to have lone pairs on it. Even the bis Cp* complex of lead(II) (bis(h5-Pentamethyl-cyclopentadienyl)-lead) is bent away from the ferrocene type strcutre suggesting that the lead has a stereochemically active lone pair.


What is a electronic configuration of scandium ion?

The electronic configuration of the cobalt is $\ce{[Ar] 3d^1 4s^2}$ the electrons with the higher energy will be snatched and the electronic configuration of Argon is very stable then with scandium you will have easely $\ce{Sc^{3+}}$.

For the Cobalt it is a bit more difficult, like you write the configuration you cannot find a correct answer. For all element first you do first the configuration using the Klechkovsky rule and after you put all orbitals by principal quantum number growing.

So the cobalt configuration is $\ce{[Ar] 3d^7 4s^2}$ then if you snatch two electrons from the $\ce{4s}$ orbital you have a stable configuration for the $\ce{Co(II)}$ ion. You cannot snatch them from the $\ce{3d}$ orbital (even if $\ce{[Ar] 3d^5 4s^2}$ looks stable because the $\ce{3d}$ orbital is half full then the spin is maximal) because its energy is lesser than the energy of the $\ce{4s}$ orbital.

You can find the Cobalt at different oxydation states from $\ce{+I}$ to $\ce{+IV}$ but it depends on what you have in your solution or in your gas if you have a gas.

NB :
Remember that the configuration of the elements are given if gas phase, then for example the more stable configuration of the copper iron is for $\ce{Cu^+}$ and not for $\ce{Cu^2+}$, $\ce{Cu^2+}$ is stable in water, so the answer may depends on the problem you have.

Explaination for the copper :

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^{2+}}$ ion has a greater charge density than the $\ce{Cu^+}$ ion and so forms much stronger bonds releasing more energy.

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu^+}$ ion loses an electron to become $\ce{Cu^{2+}}$ which can then release this hydration energy.

I hope it can help you !


In the addition of HBr to isoprene why does the less stable carbocation form?

The carbocation will be formed as per stability rules but as it is conjugated system the cabocation resonates and as the less stable carbocation is more reactive  it is attacked by Br-


Where the equibrium will shift when we add one mole to both reactants and products in PCl5 giving PCl3 and Cl2

In a reaction of the form $\ce{A <=> B + C }$ if $a_0$ is the initial amount of A (and B and C are initially zero) then if an amount x reacts the equilibrium constant $K_e = x_e^2/(a_0-x_e) $ where $x_e$ is the equilibrium amount reacted so equilibrium A is  $a_e = a_0-x_e$. Solving gives $x_e = (-K_e+\sqrt{K_e^2+4K_ea_0})/2$.

Now adding 1 mol to each side gives $K_e= (x_e'+1)^2/(a_0+1-x_e')$. Solving again gives $x_e' = (-K_e-2+\sqrt{K_e^2+4K_ea_0+8K_e})/2$. You can work out the ratio  $x_e'/x_e$ for different values of $k_e$ for simplicity, say, with $a_0=1$.

You can repeat the calculation for different initial amounts of B and C and the equations are far more messy but the final result similar.


How do you handle manganese nitrate hexahydrate?

Manufacturers of chemicals can make life interesting for first-time users. What a boon the CRC Handbook is! Without even listing $\ce{Mn(NO3)2.6H2O}$, the CRC Handbook can solve your problem. It gives the melting point of $\ce{Mn(NO3)2.4H2O}$ as $\pu{25.8^\circ C}$. Undoubtedly, the melting point of the hexahydrate is even lower, but maybe not by much. In hot water, the tetrahydrate is "infinitely" soluble.

Heat your jar of hexahydrate to 30C in warm water until all the crystals are melted and withdraw what you need with a pipette. You might plan ahead and fill several small bottles with the hexahydrate for later use. I doubt that the material needs to be stored cold.


Read this first: Why methyl group is 2,4-directing?

From that answer, we understand that methyl groups (or alkyl groups in general) are o/p directing because of hyperconjugation, and not inductive effects! In fact, as you may also know, inductive effects are generally overshadowed by hyperconjugative effects.

The aromatic ring on the left is more activated because it has two alpha hydrogens - implying more number of hyperconjugative structures - as opposed to only one for the right aromatic ring (a small difference, but good enough for elementary level questions). Thus nitration takes place at the para position of the left ring.


Preparing a solution of hydroxylamine hydrochloride in HCl, struggling

Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl?

No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.

I think the purpose of the hydrochloric acid is to ensure that the pH stays acid.  One way you could make this solution is by:

Decide on a volume of solution you want to use, say $X$ mL.
Prepare at least $X$ mL of 0.25M hydrochloric acid by dilution from a more concentrated stock solution.
Weigh out the amount of $\ce{NH2OH.HCl}$ that you need.  It has a formula weight of 69.49 g/mol, so for $X$ mL of a 0.25 M solution you will need:

$$0.25 \frac{\mathrm{mol}}{\mathrm{L}} \times \frac{69.49\mathrm{g}}{\mathrm{mol}} \times \frac{X\;\mathrm{mL}}{1} \times \frac{\mathrm{L}}{\mathrm{1000mL}}=\frac{69.49\;X}{4000}\mathrm{g\;of\;\ce{NH2OH.HCl}}$$

Dissolve the weighed out powder with your 0.25M $\ce{HCl}$ solution in a volumetric flask, first fully dissolving the powder and then slowly adding more HCl until the target volume is reached.

The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.