Which one is the strongest base among the following compounds?

A good way to compare basicity of basic compounds is to draw their corresponding conjugate acids and compare their stability. As you might figured out from drawing resonance forms of compound (2)'s conjugate acid (its protonated form ), the positive charge is going to be shared by the 3 nitrogens in the compound. On the other hand, in compound (4)'s conjugate acid (protonated form) the positive charge is on a single nitrogen. Of course, this gives compound (2)'s conjugate acid higher stability.

Your reasoning about the nitrogen being sp2 makes it less basic in comparison with sp3 nitrogen is correct,however; it only holds true if you were comparing two nitrogens with the only significant difference being the hybridization state, but in some cases, resonance will stabilize or destabilize the conjugate conjugate acid of the base, this is usually have greater impact than hybridization with regard to ranking baisicity of the compounds as in the question presented.
Electronegativity is an important factor in ranking acids/bases,but as I said, it should be used with compounds where the only significant different in the structures is about it (for example, 4 compounds with same structure but only difference is Cl, I, Br, OH at a specific position), in other cases, other factors such as resonance stabilization and aromaticity should be also considered alongside electronegativity.


"Cationic polymerization can occur with the use of strong or Lewis protic acids. The mechanism of this reaction is yet unknown." https://www.ukessays.com/essays/chemistry/polysiloxanes-synthesis-structure-2201.php

Thereby, we are freed to speculate!

It is more likely that silicon goes pentacoordinate rather than tricoordinate. If a proton attaches to an oxygen on the siloxane, the silicon goes positive, which will attract an oxygen from: 1) a water, 2) a methanol, or 3) a methyltrimethoxy silane. The second and third options can only reform the original reactants. If water attaches after protonation, forming $\ce{H2O-Si-OHCH3}$, methanol could leave, followed by a proton from the water, to give $\ce{CH3Si(OCH3)2OH}$. This becomes a fourth attacking option. If this attacks the silicon of a protonated siloxane thru the $\ce{Si-OH}$ to give $\ce{CH3O(H+)-Si-OHSiCH3(OCH3)2}$, the pentacoordinated silicon can lose methanol, then lose the proton from the oxygen between the silicons, to form $\ce{CH3(OCH3)2Si-O-SiCH3(OCH3)2}$, and that's the first step, a dimer.

Because there are two more alkoxy groups on the silicons which can leave, more reactions can occur, and likely will produce a three-dimensional gel, or even a solid, if you cook it.


Dinitrophenyl (DNP) derivatives of amino acids (specifically epsilon-DNP-lysine)

Does "epsilon-DNP-lysine" refer to a product where the DNP is bound only to the epsilon amino group (but not to the amino terminus), or does it refer to a product where DNP is bound simultaneously to the epsilon amino group and to the amino terminus?

Good question.  ε-DNP-lysine usually refers to a lysine in which the dinitrophenyl substituent is bound only to the ε atom, i.e., to the side-chain nitrogen.

The reason is that in lysine-focused derivatization methods, the reaction with DNP can be done prior to protein hydrolysis.  Thus, during DNP derivatization, the vast majority of lysine residues will have no free α-amino terminus because they are part of polypeptide chains.  Only the ε-amino groups are free to react with DNFB.  Derivatization forms DNP-ylated proteins where the peptide bonds are intact.  Later steps in the procedure hydrolyze all the peptide bonds.  During this step, free α-amino groups are created.  But no DNFB is used in these later steps.

I found a paper from 1963 which describes the derivatization procedure in detail.

  ...Care was taken to
  deposit the sample at the bottom of
  the flask. Ten milliliters of bicarbonate
  solution were added, care being taken
  that no meal adhered to the side of the
  flask. The contents were then thoroughly
  mixed by gentle swirling, and the
  suspension was permitted to stand for 10
  minutes. A solution of 0.3 ml. of
  DNFB in 10 ml. of ethanol was then
  added and the contents of the flask
  were thoroughly mixed by gentle
  swirling. The side of the flask was
  rinsed with 3 ml. of absolute ethanol
  and then the contents were shaken in
  subdued light for two hours on a shaker
  with a wrist-like motion. Previous
  work (10) has established that 2 hours
  reaction time is sufficient for substantially
  complete reaction between DNFB
  arid the free epsilon amino groups of
  lysine. alcohol and most of the
  water were removed by evaporation
  under an air stream, and the residue was extracted with four 50-ml. portions
  of anhydrous, peroxide-free diethyl
  ether. The ether in each case was
  removed from the residue by decantation
  and the residue in the flask was dried at ambient temperature by
  aeration.

The ether washings will remove the vast majority of unreacted DNFB.

  ... Two hundred milliliters of
  constant-boiling aqueous HCl nere
  added to the flask, and the resulting
  mixture was boiled overnight at the
  reflux temperature, cooled and then filtered through a sintered glass funnel
  directly into a 250-ml. volumetric flask.
  The filtrate, and washings from the
  filtrate, were made up to 250 mi. with
  distilled water.
  An aliquot (2.0 to 4.0 ml.) of the
  hydrolyzate containing from 0.1 to 0.3
  mg. of epsilon-dinitrophenyllysine...

Now for your other question:

  Also, can the same thing happen with other amino acids that have an amino group in the side chain? i.e. can there be epsilon-DNP-arginine, for instance, or is this behavior unique to lysine?

DNFB is not reactive with arginine.  The arginine side-chain is best viewed as a guanidine, not as an amino group.  Guanidine is considerably less nucleophilic than lysine.


Sign of work done by gas during isothermal expansion

If we imagine our proverbial, frictionless piston containing some ideal gas, when $P_{surr} > P_{syst}$, the system is compressed by the piston and work is done on the system by the surroundings (i.e. work is positive). If the reverse is true ($P_{syst} > P_{surr}$), the gas in the piston will expand and work is done on the surroundings by the system (work is negative)
If you think of it in terms of internal energy, it might help. When work is done on the system, the internal energy increases for the ideal gas in the system (work is positive, internal energy increases). When the system does work on the surroundings (expansion), its ability to do work decreases (internal energy decreases, work is negative).
Hope this helps.


This is a titration problem for quantitatively determining the concentration of a solution.

What reaction does occur?

   $\ce{AgNO3 (aq) + NaBr (aq) <=> AgBr v + Na+ (aq) + NO3- (aq)}$
 or essentially $\ce{Ag+ + Br- <=> AgBr v}$

Why is the solubility product important?

   The solubility product tells you about the extent of the reaction. In this particular case it tells you, that you reached equilibrium between ions in solution and the precipitated salt. It tells you exactly the product of concentrations in a saturated solution.

What can you tell about the equilibrium state at the point when the first precipitate falls?

   The solubility product is matched, hence $\ce{[Ag^+][Br^-]}<K_s=7.7\cdot10^{-13}~\mathrm{\left(\frac{mol}{L}\right)}^2$

What is the amount of bromine ions added to the solution?

   $n(\ce{Br^-}) = V(\ce{NaBr})\cdot c(\ce{NaBr}) = 0.020~\mathrm{mL}\cdot 0.001~\mathrm{\frac{mol}{L}} = 2\cdot10^{-5}~\mathrm{mol}$

What can you tell about the concentrations in the final mixture?
First, what is the concentration of bromide ions in this mixture?

   $V_0(\ce{AgNO3}) = 0.5~\mathrm{L}$, $V(\ce{NaBr}) = 0.02~\mathrm{L}$, $V_t = 0.52~\mathrm{L}$
 $c_t(\ce{Br^-}) = \frac{n(\ce{Br^-}}{V_t} \approx 3.8\cdot10^{-5}~\mathrm{\frac{mol}{L}}$

Second, what can you tell about the concentration of silver ions in the final mixture?

   $c(\ce{Ag^+}) = \frac{K_s}{c(\ce{Br^-})} = \frac{K_s}{\frac{V(\ce{NaBr})}{V_t}\cdot c(\ce{NaBr})} = \frac{K_s\cdot V_t}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 2\cdot10^{-7}~\mathrm{\frac{mol}{L}}$

What is the number of moles of silver ions in the final mixture?

   $n(\ce{Ag^+}) = c(\ce{Ag^+})\cdot V_t = \frac{K_s\cdot V_t^2}{V(\ce{NaBr})\cdot c(\ce{NaBr})} \approx 1\cdot10^{-7}~\mathrm{mol}$

What is the initial concentration of the silver nitrate solution?

   $c_0(\ce{AgNO3}) = \frac{n(\ce{Ag^+})}{V_0(\ce{AgNO3})} = \frac{K_s\cdot V_t^2}{V_0(\ce{AgNO3})\cdot V(\ce{NaBr})\cdot c(\ce{NaBr})} = 2.08\cdot10^{-8}~\mathrm{\frac{mol}{L}}$


Is the kilogram still consistent with the old definition?

The kilogram is likely to be redifined in the near future as a function of the number of atoms of a sphere of ultrapure $\ce{^{28}Si}$, as per the answer to this  Chemistry SE question.
That means that the standard will be the Avogadro number, and the mass will be a derived unit.

Then, any stored mass anywhere will just be an historical curiosity.
This does not mean that it has not been important.
We need exact measurements as Lavoisier taught to all future chemists.
At that time it was the best that was available.
Now we might go to a still better standard.


You have it a little backwards.  We don't observe charge, and then infer the existence of a force.  It's the other way around.  We observe forces between objects -- they repel or attract each other -- and then build models which explain which objects are attracted or repelled by which other objects, and by how much.  That leads us to invent the idea of charge and assign it to objects, and later still to connect charge to other properties of the objects.

If your question is why is there an electromagnetic force at all, the answer is probably that it is easier to imagine a universe in which there are forces between degrees of freedom than one in which there are not.  Our current understanding of the universe is that can best be described as a set of fields, like the surface of the sea, and everything we see is some kind of ripple or wave in that field.  It's easy to see why ripples on the surface of the ocean should interact with each other -- exert "forces" of some kind on each other -- and hard to see how they could just totally ignore each other.  That's how it is with nature.

Granted, it looks like there exist wholly separate bodies (particles, objects, things) and we wonder why they should interact with similar things far away.  But that may be because our perception is severely limited, and the underlying fields are much more complex than the surface of an ocean.  If you want to explore the topic further, start by googling "wave-particle duality in quantum mechanics."


Concentration of a species outside buffer region (polyprotic acid titration)

At the one side, you can always calculate the percentages of the acid or it's ions in solution:
$$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{a,j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{a,j}}$$
with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons.

But on the other hand there are some special points during titrations. One of those points did you choose, which is where $\mathrm{pH} = \mathrm{pK}_{a,i}$.

For $\ce{H3PO4}$ there exist three different acidity constants, which describe the three possible dissociation steps. For the first step, the reaction is as you described: $$\ce{H3PO4 + H2O \xrightleftharpoons{k_{a,1}} H2PO4- + H3O+}$$

A little work on it gives you the Henderson-Hasselbalch-equation, which usually is good approximation to buffer regions: $$\mathrm{pH} = \mathrm{pK}_a - \lg\left(\frac{[HA]}{[A^-]}\right)$$

If you enter the same values for pH and pKa you have to solve $$-\lg(x)=0$$ which is true for $x=1$. So you know that you have a 1:1 ratio at this point.

This means for phosphoric acid:
At $\mathrm{pH=pK}_{a1}$, you've got exactly $50~\%~\ce{H3PO4}$ and $50~\%~\ce{H2PO4-}$ (blue, yellow). The same procedure can be applied to the other dissociation steps. Therefor at $\mathrm{pH=pK}_{a2}$ you've got $50~\%~\ce{H2PO4-}$ and $50~\%~\ce{HPO4^2-}$ (yellow, green) and at the last possible $\mathrm{pH=pK}_{a3}$ there are $50~\%~\ce{HPO4^2-}$ and $50~\%~\ce{PO4^3-}$ (green, red).

This can be applied to most acid-base dissociation steps but you have to pay attention for close successive $\mathrm{pK}_a$ values, as if the gap gets too small, the 1:1 ratio doesn't hold.


Can alkyl-substituted phenols be reduced to alkyl-substituted benzenes by zinc dust?

This is the basic principle of functional groups. Organic chemists quickly noticed historically, that some compounds reacted in the same way when presented with the same chemicals; thus allowing them to deduce that it was only a certain functional group (‘radical’ in historic terminology) that was reacting while the remainder of the molecule remained untouched.

Unless conditions or reagents are known to also attack other functional groups, they will typically be unharmed in a reaction and whichever features your molecule has, it will still react in the desired manner. For example, if phenol gets reduced to benzene by distillation with zinc dust, we can be pretty sure that cresols (e.g. paracresol, 4-methylphenol) will react in the same way, only that the final product will be toluene rather than benzene.

As I am not aware of zinc dust reacting with a double bond — the only other functional group present in chavicol — I am willing to say that it will indeed lose the hydroxy group and be reduced to allylbenzene as you desire.


Is menthol an especially volatile molecule?

The term "volatile oils" is a traditional term.

From the 1916 book A TREATISE ON PHARMACY FOR STUDENTS AND PHARMACISTS 5th edition at page 217:

  Although the boiling points of volatile oils are considerably above that of water, the oils pass over rapidly with the vapor of boiling water, and in the leading establishments in this country [USA] and Europe volatile oils are now distilled by passing steam under pressure into stills which contain the material on a series of perforated trays...

So it means volatile to the degree that it could be steam distilled.


$\ce{H3O^+}$ is really just a "shorthand notation" for what really occurs in solution.  Typically higher hydrates of the proton, particularly $\ce{H5O2+}$ the "Zundel" cation and $\ce{H9O4+}$ the "Eigen" cation are thought to be the main players in the proton transfer process.  As permeakra has pointed out, Grotthuss was one of the early pioneers studying this process and his name is still used to describe the mechanism.  The mechanism is still an active area of research with considerable controversy remaining over the details of the mechanism.  This reference provides some diagrams of the general process.

Basically, just as electrons are said to use a "hopping" transport mechanism, the same has been said of proton transport.  The making and breaking of hydrogen bonds, accompanied by the twisting and turning of protons completes the process.  This proton hopping mechanism between these larger water aggregates is thought to be a key step at room temperature.  However proton tunneling plays a role as well, particularly as the temperature is decreased.


Changing Zinc acetate dihydrate to Zinc acetate

Wikipedia is your friend...Zinc Acetate

    Decomposes at 237 °C (459 °F; 510 K)
    (dihydrate loses water at 100 °C)

So heating to 120-130 °C for several hours would remove the water, but not be hot enough to decompose the salt. The dried salt will be hygroscopic of course so store it in a desiccator.


Where hydrogen belongs in the Periodic Table

This comes down to the fact that hydrogen has a $1s$ valence shell, capable of holding only two electrons.

In some sense, hydrogen is like the halogens, in that it can achieve the electronic configuration of a noble gas (namely, helium, which has a full $1s$ level) by gaining an electron. This reduction yields the hydride ion, $\ce{H-}$. It should be noted that hydride does not have the same stability as the halogen anions (e.g. hydrogen has an electron affinity of 0.75 eV vs. 3.40 eV for fluorine).

In another sense, hydrogen is like the alkali metals, as it only contains a single electron in its valence shell. Whilst the alkali metals can attain noble gas configurations by losing their valence electrons, it's a bit abstract to talk about hydrogen doing the same by losing its single electron to generate a proton ($\ce{H^{+}}$)(1). Nevertheless there is strong analogy between the electronic structure of hydrogen and the alkali metals which motivates hydrogen being often placed in this group.

Periodic table makers sometimes get away with duplicating hydrogen above the alkali metals and the halogens, or putting it in no group in particular.


Best way to dissolve and consume cysteine HCl monohydrate powder

I wouldn't worry at all about the oxidative dimerization of cysteine to cystine when you consume freshly prepared solutions.

In the lab, you'd use oxidants, such as $\ce{FeCl3}$ (in the presence of air) or hydrogen peroxide to achieve this reaction; both of which you're not likely to add to your amino acid diet!

As far as the article which you mentioned is concerned: The stabilization methods in question will only be relevant to maintain the composition of prefabricated and sterile-packed solutions over a long period of time.


Why are anti aromatic compounds unstable?

In conjugated-pi ring systems of N atoms, you typically have a single low energy orbital and then paired (degenerate) energy levels until you have N orbitals total. This has to do with the number of nodes in the conjugated pi system, and the number of ways you can lay them out with N atoms.

Now, when you add in electrons you add two electrons per orbital, so your 4n+2 electrons fill 2n+1 orbitals -- the single low energy orbital and then n complete sets of degenerate pairs.

If you only have 4n electrons, you can fill the single low energy orbital, then 2(n-1) complete sets of degenerate pairs ... but with two "extra" electrons which aren't paired up. This allows those two electrons to "spread" out between the pair of equal-energy orbitals. Those unfilled/partially filled orbitals mean that the anti-aromatic compound is quite reactive.

The other issue is that "stability" is not an absolute property. Stability is always in reference to some other state. The typical anti-aromatic systems you'll encounter tend to have those frontier orbitals as non-bonding or anti-bonding orbitals, meaning the electrons in the orbitals aren't really contributing to the molecular stability. Generally, if you build the energy level diagrams for alternate systems (e.g. a bent ring where you don't get localization) you'll find that - while the lowest energy orbital isn't as low energy as in the ring-conjugated system - the sum of the energies across all orbitals is indeed lower. And it's the total energy of the system which is optimized. (For example, in cyclobutadiene the ring-conjugated form would have two electrons in a low-energy bonding orbital and two in non-bonding orbitals. In contrast, linear butadiene bonding orbitals are not as low energy as for the ring-conjugated cyclo form, but there's two of them. And the sum of the energies from two linear bonding orbitals is lower than what you get from a single cyclo bonding orbital.)

(Final note: this is all high-level generalization. There's more sophisticated levels of quantum mechanical theory where things get more complex.)


On this site, Nasa give some closer information on the MOXIE system.
It splits electrochemically $\ce{2 CO2}$ into $\ce{2 CO}$ and $\ce{O2}$.


Aqua regia is a concentrated mixture of nitric and hydrochloric acids. The two acids react to generate soluble nitrosyl chloride, water and nascent chlorine: $$\ce{HNO3 + 3HCl -> NOCl + 2H2O + 2Cl}$$

Secondly, the gold is oxidised by nitric acid to gold(III) ions: $$\ce{Au <=> Au^{3+} + 3e-}$$

This reaction is normally highly unfavourable and so very few gold(III) ions are formed if you put gold into only concentrated nitric acid. However, the nascent chlorine is highly reactive as it has an unpaired electron in a high energy orbital and it is able to react with the gold(III) ions to form the soluble tetrachloroaurate(III) complex. This removes the gold(III) ions from the equilibrium, driving the reaction to completion.

Overall the reaction is: $$\ce{Au + 3HNO3 + 4HCl -> [AuCl4]- + 3NO2 + H3O+ + 2H2O}$$


You’re mistakenly conflating the bond dissociation energies between the diatomic components and the product molecule. If the bond dissociation energy of water is larger than the bond dissociation energy of the individual diatomic components, then the system is “lower energy” (preferred) in the $\ce{H2O}$ state and that “extra energy” in the atomic state will be released in the form of exothermic heating. Remember that elemental $\Delta H = 0$, and the water molecule has stronger bonds/lower enthalpy and would require more energy to dissociate than the elemental components (which is what your math shows), meaning water won’t spontaneously dissociate into the diatomic gases since that would be an endothermic process and require some external energy. As an aside, keep in mind that dissociation of bonds in a molecule like water won’t be consistent between atoms (after the first hydrogen bond has broken from an $\ce{H2O}$ molecule, the remaining hydroxyl group requires less energy to dissociate). In sum, if your product has a higher bond dissociation energy than your reactant, it’s at lower enthalpy and the net of the reaction will be a release of energy in the form of exothermic heating.


Examples of Lewis base nature of sulfur dioxide

You generally want to find a much stronger Lewis acid (typically forming a solvate/adduct) or make a complex with a transition metal: this way either unshared electron pair of oxygen or sulfur can be donated.
Canonical examples are given in Shriver & Atkins’ Inorganic Chemistry [1, p. 135]:

  To act as a Lewis base, the $\ce{SO2}$ molecule can donate either its $\ce{S}$ or its $\ce{O}$ lone pair to a Lewis acid. When $\ce{SbF5}$ is the acid, the $\ce{O}$ atom of $\ce{SO2}$ acts as the electron pair donor, but when $\ce{Ru(II)}$ is the acid, the $\ce{S}$ atom acts as the donor … $(\ce{[RuCl(NH3)4(SO2)]+}).$

$\qquad$

Figure 1. Left: Crystal structure of $\ce{SbF5 · SO2}$ (ICSD-67273 [2]). Right: Crystal structure of $\ce{[RuCl(NH3)4(SO2)]+}$ (ICSD-26061 [3]). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#90E050}{\Large\bullet}~\ce{F}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#248F8F}{\Large\bullet}~\ce{Ru}$; $\color{#9E63B5}{\Large\bullet}~\ce{Sb}$.

Reactions for the synthetic procedures adapted from the corresponding articles [2, 3]:

$$
\begin{align}
\ce{SbF5(s) + SO2(l) &->[\pu{-40 °C}] SbF5 · SO2(sln)} \tag{R1} \\
\ce{[Ru^{II}(NH3)4(HSO3)2](aq) + 2 HCl(aq) &->[reflux][15 min] [Ru^{II}(NH3)4(SO2)Cl]Cl(aq) + 2 H2O(l) + SO2(aq)} \tag{R2}
\end{align}
$$

References

Atkins, P. W. Shriver & Atkins’ Inorganic Chemistry; W.H. Freeman and Co.: New York, 2010. ISBN 978-1-4292-1820-7.
Minkwitz, R.; Molsbeck, W.; Preut, H. Kristallstruktur von $\ce{SbF5 · SO2}$ / Crystal Structure of $\ce{SbF5 · SO2}.$ Zeitschrift für Naturforschung B 1989, 44 (12), 1581–1583. DOI: 10.1515/znb-1989-1218. (Free Access)
Vogt, L. H.; Katz, J. L.; Wiberley, S. E. The Crystal and Molecular Structure of Ruthenium-Sulfur Dioxide Coordination Compounds. I. Chlorotetraammine(Sulfur Dioxide)Ruthenium(II) Chloride. Inorg. Chem. 1965, 4 (8), 1157–1163. DOI: 10.1021/ic50030a015.


Why are electrolysis cell plates turning a brown color?

Since the carbonate radical is covalently bonded, it probably is not being decomposed electrolytically. Much more likely is that iron in the stainless steel is being removed and oxidized, leaving a thin film of "rust" on the plates.

A few ways you can prove this:

Immerse the browned plates in oxalic acid, which will slowly dissolve iron oxides, but leave carbon untouched.
Substitute platinum electrodes -- admittedly expensive, but Pt is the "gold" standard for electrodes. They should not discolor.
Spray the browned plates with a solution of potassium ferricyanide. If free iron is present, a blue color ("Prussian blue") appear.


Why are d-d electronic transitions forbidden and weakly absorbing? Why do they occur at all?

In order for an electronic transition to be allowed (occur with strong intensity), certain "selection rules" must be obeyed.  You may already be familiar with the rule that the electron spin quantum number cannot change during a transition. It is this rule that forbids singlet-to-triplet absorption or emission (emission follows the same rules as absorption) and consequently makes phosphorescent emission so much weaker than fluorescent emission.

$\ce{\Delta S = 0}$ the Spin rule

Another rule, often referred to as the "LaPorte" rule or the "Orbital" rule, states that in molecules with a center of symmetry (centrosymmetric molecules), transitions within a subshell are forbidden when $\ce{\Delta l =0}$.

$\ce{\Delta l = \pm 1}$ the LaPorte rule

The orbital angular momentum quantum number "l" describes the type of orbital (s, p, d, f) within the subshell.  In centrosymmetric molecules, the various s, p, d, f orbitals cannot mix within a subshell.  For example, the 4d orbital cannot mix with the 4p orbital in a centrosymmetric molecule, such mixing being symmetry forbidden.  Octahedral complexes can be centrosymmetric.  In such a case 3p->3p, 3d->3d transitions would be forbidden by LaPorte's rule.  Tetrahedral molecules do not have a center of symmetry and p-d orbital mixing is allowed, so in the case of tetrahedral molecules 3p->3p and 4d->4d transitions may appear stronger because a small amount of another orbital may be mixed into the p or d orbital thereby removing the violation of LaPorte's rule.

If we can do something to remove the centrosymmetric nature of an octahedral complex, then we "relax" LaPortes rule and weak transitions can be observed.  For example, if a molecular vibration removes the molecular center of symmetry, then a d->d transition can occur if light is absorbed at that instant.  Things such as spin-orbit coupling, vibronic coupling, or ligands that can mix with the metal's d orbitals would all allow relaxation of LaPorte's rule and thereby permit weak transitions to be observed.


The latter amide definition in you question did not evolve from the former name of organic amides, but the process was rather opposite, or at least they have a common root. The name "amide" appears to be quite old. There is even a notion of amidogen, not used very often today, but it appears in the book Elements of Chemistry from 1840, along with some inorganic amides like "amide of potassium" or "amide of mercury". Later book Chemistry, Theoretical, Practical and Analytical as applied and relating to the Arts and Manufactures by James Sheridan Muspratt from cca 1860 besides inorganic amides mentions also a peculiar substance "margaramid" that was obtained as a byproduct of ammonium salts of fatty acids preparation by treatment of fats/oils with ammonia. In this time e.g. acetamide was known as well already.

And the definition in Cyclopaedia of Chemistry with Its Applications to Mineralogy, Physiology, and the Arts from 1854 states:

  Amide, Amidogen. The term amide is given to a class of substances which contain ammonia deprived of an atom hydrogen, (NH2.) Hence, when they are treated with potash, ammonia is disengaged, the third atom of hydrogen derived from water again uniting, so as to form the necessary quantity in ammonia. That such a substance as amide or amidogen exists, may be proved by passing dry ammonia over heated potassium or sodium. A green body is formed, as hydrogen is evolved, which is replaced by the metal, NH2H + Na, becoming NH2Na and H. Some of the most familiar of these amides are potassamide NH2K, sodamide NH2Na, oxamide NH2C2O2, benzamide NH2C14H5O2, sulphamide NH2SO2, carbamide NH2CO, platosamine NH2Pt. Some of these amides are formed by heating organic salts of ammonia, by which they lose water, and leave an amide. Thus oxalate of ammonia, when distilled, gives water and oxamide (NH3C2O3 = NH2C2O2 and HO). It is probable that many animal substances, as albumen, belong to this class.

(typos/mistakes included)

Please note that this is not an exhaustive historic research at all :)


I think that this is basically an accounting problem. The full electron configuration of $\ce{Ne}$ is: $1s^22s^22p_x^22p_y^22p_z^2$. So, what does one now define as core electrons? It could be: "all electrons not in the outermost shell", yielding the $1s$ electrons. An alternative is "all electrons in completely filled shells", yielding all 10 electrons as core electrons. (Note that these two definitions yield identical results for almost all non-noble gas elements, with some difficulties surrounding the $f$-block.) Both definitions have some merit, I would prefer the first one, giving an effective charge of $+8$, although I do not know what one does with that number next.

A word on effective nuclear charge
As noted on the wikipedia page linked in the question, to arrive at more meaningful effective nuclear charges (as might be helpful for the systematic creation of basis sets for computational/quantum chemistry), one has to take the effective shielding strength of the different orbital shapes into account. The $d$-block-contraction, as can be seen in the somewhat different chemistry of $\ce{As}$ compared to $\ce{P}$ and $\ce{Sb}$, is due to this. Thus, in the end, a pen-and-paper calculation as alluded to above can be useful to think about some concepts in a qualitative fashion, but more extensive calculation is needed for quantitative reasoning.


What you call a "gem-diol alkene" might also be called an "ene-diol" or "endiol" by analogy to the name enol.

Enols are a tautomeric form of aldehydes or ketones.  They have one double bond and one C-O single bond.  Since "ene-diols" have one double bond and two C-O bonds, the oxidation number of the O-bonded carbon is 2 higher than with enols.  This corresponds to the oxidation number of carboxylic acids.  Thus, "ene-diols" are (theoretically) tautomers of carboxylic acids.

Ene-diols are not stable because they would rapidly tautomerize the more stable carboxylic acid form.

Hydrogen bonds like the one you have drawn are unlikely to stabilize enols very much, for reasons that Jan noted in a related question here on chem.SE.


Why May K Expressions Be Manipulated As Such?

Your intuition is correct. The reason equilibrium constants from different reactions can be strung together and multiplied comes from a generalization of Hess' law which covers Gibbs' free energy changes (apparently also called Bordwell thermodynamic cycles). Enthalpy is a thermodynamic function with a special property; it is a state function. That is, enthalpy changes in cycles are determined solely by the initial and end states of a system, and not the trajectory in which the initial state travels to the end state (as happens for functions such as work $W$ and heat $q$). Any state function can be manipulated in the way enthalpy is used for Hess diagrams. Thus, by adding reactions, you can add their respective reaction free energy changes $\Delta_rG$.

Let us now take two reactions, with free energy changes equal to $\Delta_rG_1$ and $\Delta_rG_2$. There is a relationship between a reaction's free energy change and its reaction quotient $Q$, namely:

$$\Delta_rG = \Delta_rG^o + RT\ ln\ Q$$

, where $^o$ indicates standard conditions. For reactions 1 and 2, we therefore have:

$$\Delta_rG_1 = \Delta_rG^o_1 + RT\ ln\ Q_1$$
$$\Delta_rG_2 = \Delta_rG^o_2 + RT\ ln\ Q_2$$

Adding the reactions, we obtain a reaction 3 where the same formula applies:

$$\color{\red}{\Delta_rG_3} = \color{\navy}{\Delta_rG^o_3} + \color{\green}{RT\ ln\ Q_3}$$

However, since we can add the free energy equations for reactions 1 and 2, reaction 3 is also represented by:

$$\Delta_rG_1 + \Delta_rG_2 = \Delta_rG^o_1 + \Delta_rG^o_2 + RT\ ln\ Q_1 + RT\ ln\ Q_2$$
$$\color{\red}{\Delta_rG_1 + \Delta_rG_2} = \color{\navy}{\Delta_rG^o_1 + \Delta_rG^o_2} + \color{\green}{RT\ ln\ Q_1 Q_2}$$

Clearly, we have $Q_3 = Q_1 \times Q_2$. Considering the equilibrium conditions of all reactions ($\Delta_rG = 0$), then the reaction quotients are defined as the equilibrium constants $K$, and similarly one finds $K_3 = K_1 \times K_2$.


Oxidation numbers of phosphorus pentachloride

When considering the oxidation state of covalent compounds, we need to pretend they are ionic. When we do so, we assign all of the electrons to the more electronegative element. The electronegativity of phosphorous using the Pauling Scale is 2.19 and the electronegativity of chlorine is 3.16. Since chlorine is the more electronegative element, we assign all electrons in the bonds to the chlorine atoms. Thus $\ce{PCl5}$ is treated as the fictitious ionic compound comprised of $\ce{P^{5+}}$ ions and $\ce{Cl-}$ ions.


Can beta D-glucose be present hydrolysate after hydrolysis of cellulose and how can it be converted into alpha D-glucose?

In aqueous solution, $\alpha$-D-glucose and $\beta$-D-glucose equilibrate. So glucose derived from aqueous acid hydrolysis of cellulose will eventually become an equilibrium mixture of α and β.

A 2006 paper goes into great detail on the kinetics and thermodynamics of the interconversion (in water).  They say:

  At 30 °C, the rate constants for ring closing toward α-pyranose and β-pyranose are 0.044 s−1 and 0.072 s−1, respectively.

Keep in mind those figures don't include the effect of acid catalysis, and are incomplete without rate constants for the formation of the ring-open aldehyde from the two different pyranoses.  But they should give you a rough sense of how quickly the process occurs.  A later figure in the same paper shows that it takes roughly 12,000 seconds or 3.3 hrs for nearly complete (~90%) relaxation to equilibrium.

Also see chemistry.stackexchange.com/a/7311/13184 for more information.


Changes over time in sodium hydroxide concentration in a hot water bath used for pretzel manufacture

You need both.

Adding water, as water evaporates, but hydroxide does not.
Adding the hydroxide solution, as it is taken away by pretzels

But I suppose the evaporation is the major effect.

Probably the most convenient way is measuring density, with areometer $\pu{1.0-1.2 g / ml}$

Additionally, hydroxide gradually absorbs carbon dioxide, becoming sodium carbonate. pH paper strips may do good here, if solution looses it's alkality.


Why is the distance between the atoms shorter in the dicarbon anion compared to the cation?

Looking up in a current compilation of bond lengths in the International Tables for Crystallography (IT), apparently there is indeed not so much deviation in the distance between two adjacent carbon atoms in a $\ce{C(sp)#C(sp)}$ pattern:

(IT, 2006 edition)

Some hints to render this excerpt easier to access.  The IT indeed discern different bond types (13 for all combinations of $\ce{C-C}$,  and 625 for all organic / organometallic compounds), and consequently, the source offers separate listings for, for example, $\ce{C(sp^3)-C(sp^2)}$ or $\ce{C(sp^2)-C_{ar}}$, too (first column). The second column is about the structure around the pattern in question highlighted in bold. For the $IT$, $\ce{C_{ar}}$ explicitly represents an aryl C in an six-membered ring; $\ce{C^{\#}}$ any $\ce{C(sp^3)}$ - linearly substituted, or not.

The third, fourth and fifth column are the unweighted sample mean $d$ in [Å], sample median $m$, and sample standard deviation $\sigma$, followed by the lower and upper quartile, $q_l$ and $q_u$, as well as the number of observations $n$.  Overall, the compilation is based on a subset of 10 324 X-ray and neutron diffraction data found in the Cambridge Crystal Database by September 1985, judged to be "good enough" (the source offers additional insight about the criteria applied).


Correct Rate Equation for a Reversible Reaction

If all you want to find is the final concentration of all the species at equilibrium than you don't need to solve these differential equations.
To solve this types of questions easily in chemistry we define a quantity $K_{eq}$. Where $K_{eq}=\frac{K_f}{K_b}$ and $K_{eq}=\frac{\text{product of concentrations of products at equilibrium to the power of their stoichiometric coefficient}}{\text{product of concentrations of reactants at equilibrium to the power of their stoichiometric coefficient}}$.
So for your case we can set the equations as such:
$$
\begin{align}
\begin{array}{cccccc}
& \ce{A} & \ce{+} & \ce{B} & \ce{<=>} & \ce{C}\\
\text{Initial} & \pu{1 M} & & \pu{1 M} & & \pu{0 M}\\
\text{Final} & (1-x)\pu{M} & & (1-x)\pu{M} & & x\pu{M}
\end{array}
\end{align}
$$
$$
K_{eq}=\frac{K_f}{K_b}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\
2=\frac{x}{(1-x)^2}\\
2x^2-5x+2=0\\
(2x^2-4x)+(-x+2)=0\\
(2x-1)(x-2)=0\\
x=\{\pu{0.5M}, \pu{2M}\}\\
$$
But as $[\ce{B}]\leq \pu{1M}$, $x$ has to be equal to $\pu{0.5 M}$.
Which gives, $[\ce{A}]=[\ce{B}]=[\ce{C}]=\pu{0.5 M}$.
This also shows that your solving of differential equations was also correct!


Meanings of isopropyl, n-butyl, isobutyl, s-butyl and t-butyl and differences between them

(1) Yes, they're alkyl substituents.  But that doesn't mean they have to have been literally added to some other molecule.  There's nothing wrong with speaking about a part of a molecule as a "substituent," even if that exact piece was never actually literally substituted on the main chain in a chemical reaction.  Think of it as a hypothetical substituent.

(2) You get the names this way:

(a) Use the parent hydrocarbon and change "-ane" to "-yl."  So a 4 carbon substituent becomes "butyl" from "butane."  The 3-carbon substituent becomes "propyl" from "propane."

(b) There are two isomers of butane.  The one with 4 carbons in a row is called "n-butane" for "normal" butane.  The other has 3 carbons in a row, with the 4th branching off, this is called "isobutane" from a Greek prefix "iso" meaning "same as," which is the same prefix in "isomer."  Of course it's not the same molecule, but it has the same chemical formula.

(c) There are two ways to attach the group derived from n-butane to a main chain.  You could attach it from one of the end carbons, in which case we call it "n-butyl."  Or you could attach it from one of the inner carbons, in which case we call it "sec-butyl" because the attachment point is a secondary carbon (instead of a primary carbon as it is in n-butyl).

(d) There are also two ways to attach the group derived from isobutane. You can attach it from one of the end carbons, in which case we call it an "isobutyl" group, or you can attach it from the center carbon, in which case it's "t-buty" or "tert-butyl" for "tertiary" butyl because the attachment carbon is a tertiary carbon.

There are only two groups you can derive from propane: the group can either be attached from one of the end carbons, in which case it's a "propyl" group, or it can be attached from the middle carbon, in which case it's an "isopropyl" group.  (There is no isopropane parent, so this is a little inconsistent.)

(3) There are a few names for bigger alkyl substituents, but they aren't as often used I believe, because it starts to get even more complex and idiosyncratic, which is why...

(4) ..these are not the IUPAC names for these alkyl substituents.

You can certainly swear a vow to always use IUPAC names, and many of us have done so when we first learned organic.  But with time comes familiarity with these little guys, who are more compact and roll off the tongue faster than their systematic equivalents, and you should not be surprised if after some years (if you stick with it) you find yourself using them all the time just because it's convenient.


How to do complexometry titration of calcium oxalate?

Does it have to be complexometric?

You could acidify the sample and do a redox titration of the oxalate with permanganate or dichromate. And since the amount of oxalate is the same as calcium, bob's your uncle.


Lowest freezing point room temperature liquid?

After a bit of random searching through small molecules, I found this very large dataset with 28,645 experimentally measured melting points. Sorting them by increasing melting point and checking the first few entries individually, it seems that N,N-diethylmethylamine holds the record, with a melting point of -196°C (!) and a boiling point of 63-65°C.

I also mentioned another possibility for very low melting liquid substances at room temperature, namely ionic liquids. Even though the species present are much larger and have ionic interactions, presumably the ions pack so extremely poorly in the solid that freezing is very unfavourable.  Aldrich already offers multigram quantities of ionic liquids melting at -88°C, and ongoing research offers even lower melting alternatives, such as the ones in this article which purportedly do not crystallize until -150°C. This limit may well be improved upon in the next few years.

If mixtures are allowed as well as pure substances, then a combination of small organic molecules or ionic liquids will very likely reach even lower temperatures than the examples above without freezing.


What materials reversibly swell up slightly when in a moist environment?

A classic hygrometer consists of bundle of a dozen or so human hairs (blonde are supposedly best, because they're typically thinnest, i.e. react fastest to humidity changes), stretched between a metal spring with a pointing device and a fixed point.

A uniaxial effect is much easier to record precisely, and hair is thin (=fast) and far more mechanically reliable than a piece of swollen hydrogel.


Why isn't ammonium strong acid, since it's charged and ammonia is neutral?

As chemists we are given experimental observations, such as the difference in measured pKa between ammonium and hydronium ions, and we invent concepts such as "electronegativity" to explain this result in conjunction with many other similar results. If this concept is insufficient in your mind to explain the experimental data, you may be motivated to seek a more satisfying and self-consistent explanation. For me electronegativity is sufficient in this case. The fact that ammonia is "stable" and can be put in a bottle does not mean that it is somehow unstable when protonated to give ammonium ion. Many ammonium salts can be put in a bottle, even salts of weak acids like carbonic. Having a positive charge on an atom is not in itself an unstable state-note how stable salts like sodium chloride (with a full positive charge on sodium) are, as well as species like tetramethylammonium ion-- you can make the hydroxide and put it in a bottle.


Which mechanism should a reaction between the cyclopentylmethanol and the HBr should follow?

It's a tricky example, but the $S_N1$ process is favored because the solvent is protic and the resulting 6-member ring is more stable than a 5-member ring.

If ring expansion weren't possible, I believe your intuition is correct that the $S_N2$ mechanism would dominate. See this discussion. You're not alone!


Correct way of drawing D-configuration with this stereochemistry?

You don't have to memorize the whole structures. Easier is to memorize the general structure including the stereochemistry, and the residues, R (except from proline). The Fisher projection is useful, but on your depiction, it's not in the usual orientation. The carbon chain should be vertical, carboxy group at the upper side, residue at the bottom. Hydrogen and amino group (pointing towards observer) horizontal.

(non-ionized forms)

One can easily memorize that for L- amino acids, amino group is on the left, for D- amino acids the opposite, which is the case for your orientation as well (and don't forget to invert chiral residues for D- AA's, as Zhe has already noted in the discussion; it is the case for isoleucine and threonine).


What is the number of electrons being transferred in electrolysis of aluminum oxide?

I guess the misunderstanding arises from the fact that you treat $n$ as the same entity as the lowest common multiple (LCM) used to balance the net equation using half-reactions for the electrolysis of molten aluminium(III) oxide:
$$
\begin{align}
&\text{cathode:} &\ce{Al^3+(melt) + 3 e- &-> Al^0(l)} &|\cdot4\tag{red}\\
&\text{anode:} &\ce{2 O^2-(melt) &-> O2^0(g) + 4 e-} &|\cdot3\tag{ox}\\
\hline
& &\ce{4 Al^3+ + 6 O^2- &-> 4 Al^0 + 3 O2^0}\tag{redox}
\end{align}
$$
or, finally, with the smallest whole-number coefficients:
$$\ce{2 Al2O3(l) -> 4 Al(l) + 3 O2(g)}\label{rxn:R1}\tag{R1}$$
as well as normalized to a single formula unit:
$$\ce{Al2O3(l) -> 2 Al(l) + 3/2 O2(g)}\label{rxn:R2}\tag{R2}$$
In the equation
$$Δ_\mathrm rG = -nFE$$
$n$ is the number that matches the amount of electrons (in mol — since we are using molar convention for Gibbs free energy) transferred in the balanced equation from a reducing agent to an oxidizing agent.
From the balanced reaction \eqref{rxn:R2} it is evident that the number of electrons $n$ transferred from $\ce{O^2-}$ to $\ce{Al^3+}$ is $3 \cdot 2 = 6,$ where $3$ is the number of oxygens per formula unit, and $2$ is the number of electrons each oxygen donates.


Why is benzenediazonium fluoroborate water insoluble and stable at room temperature?

After a bit of research on benzenediazonium fluoroborate, I came up with an (complete?) answer although it is based on some approximations and/or speculations.

Benzenediazonium fluoborate is a versatile organic salt used to create various other organic compounds like by helping to introduce aryl, arylazo, arylhydrazono, or amino groups; forms fluoroarenes upon heating and are building block for heterocycles. They have a much higher stability than the corresponding chlorides and shock-insensitive. Generally decomposes at decomposes at 114-116 °C (Source).

So, what causes it much higher stability? Maybe it is their ion size. The benzenediazonium cation and fluoroborate anion are of same size resulting their stability[citation needed?]. Or maybe it is due to the resonance of benzenediazonium cation.

But when you apply heat to the salt, it is bound to decompose, maybe because of two reasons. One is that exceptionally stable nitrogen is formed on its decomposition (i.e nitrogen is formed from nitrogen cation) and second is the $\ce{N-F}$ bond is long and weak. From the information regarding structure of benzenediazonium fluorobotate, it is evident (Here):

  Benzenediazonium tetrafluoroborate, $\ce{C6H5N2+•BF4−}$, crystallizes
  in space group P21/a with unit cell dimensions a = 17.347(2),
  b = 8.396(1), c = 5.685(1) Å, β = 92.14(1)°, Z = 4. The structure was
  solved by direct phasing methods using the program SHELX 76. The
  parameters were refined by full-matrix least-squares to a final
  R = 0.063 for 1346 observed reflections. $\ce{C—N}$ and $\ce{N≡N}$
  bond lengths are 1.415(3) and 1.083(3) Å, respectively, and the bonds
  of the benzene ring vary from 1.371(5) to 1.383(4) Å. There are
  three $\ce{N-F}$ close contacts of ≤ 2.84 Å and the positive charge
  appears to be shared between the nitrogen atoms.

Below is the abstract of the mechanism of benzenediazonium tetrafluoroborate decomposition to give an insight (full pdf behind paywall is here):

  The thermolysis of benzenediazonium tetrafluoroborate was studied by
  thermogravimetry in dynamic mode. The decomposition of
  $\ce{[ArNN]+BF4−}$ in the solid state with the formation of
  $\ce{C6H5F, BF3, C6H6}$, and $\ce{N2}$ starts at T>348 K. The speed of
  the thermolysis was estimated gravimetrically and by infrared
  spectroscopy, considering the change of the intensity of the
  absorption band at 1498 cm−1, which corresponds to fluorobenzene. The
  maximal rate of thermolysis observes at the 366.5 K. A kinetic scheme,
  which includes the formation of a neutral complex
  $\ce{[C6H5δ+⋯BF4δ−]}$, is proposed for the thermolysis of
  arenediazonium tetrafluoroborate. The decomposition of the complex
  with the formation of free-radical intermediates explains the chain
  character of the thermolysis.

There is large hydrophobic part in diazonium ion and in $\ce{BF4-}$ ion, size is large enough that hardly any hydration energy would be released to cause the dissolution. Hence the salt is insoluble. (Here)


Calculating exposure time to inactivate coronavirus by UV light

yes, there are several errors in your calculation.
Germicidal lamps specify an effective power output in UVC. This lamp
https://www.osram.de/ecat/PURITEC%20HNS%20UV-C-UV-C%20lamps%20for%20purification-Ultraviolettstrahler-Industrie-Spezialbeleuchtung/de/de/GPS01_1028570/ZMP_1022255/
consumes 16W of electricity but has a UVC output of 4W. You need to calculate with the 4W value.
Then you need to calculate the intensity of your irradiation on the surface you want to inactivate the virus on. The distance from the lamp to the surface is key, as the 4W UVC power remains contant over the surface of the light cone that shines from it. In the lamp's data sheet it says that this is valid for up to 3m from the lamp. (after 3m it most likely decreases due to other effects.) So you need to calcuate the distance from your lamp, and you need to calculate your powerdistribution over the surface of the sphere/cylinder (depending on your geometry), with the lamp in the center. It does not do to calculate the surface of the room as a whole. Given the irradiation intensity on the surface of that sphere you can now calculate the time needed until the UVC output of the germicidal lamp manages to reach the 90% inactivation dose for a given virus. For corona viruses different values for that dose are given in different papers. Some say 10.3mJ/cm^2 for 99%, some quote 0.6mJ/cm^2 for 90% inactivation. This one gives a good overview: https://www.nature.com/articles/s41598-020-67211-2.pdf
I recommend to choose a simple geometry for calclation of the irradiation power. Living rooms are quite tricky. Spheres are better.


How do I purify the resulting compound after a nitro- to amine-group reduction?

It is very easy to purify the amino compounds from the reaction mixture of reduction of nitro- to amine- group. Firstly, we will do  workup after completion of the reaction by extraction method with organic solvents  either with DCM or with ehtylacetate, then organic layer wlii be separated, and then dried with anhydrous sodium sulphate and the solvent will be evaporated under reduced pressure by applying rotary evaporator. Then after making slurry by adding little amount of silica gel we will pack slurry in Column-chromatography silica gel (100-200 mesh) by using eluent as Ethylacetate:Hexane (20% to 60%),  the amino- group compond will be purify. This is so because nitro- group is less polar compared to anino- group. So, nitro- compound will come first from column chromatograhy and amino- compound will come letter in pure form.


Equation for the equilibrium constant for the hydration of carbonyls using UV spectroscopy

The law is not as you state although it often gets reported as this. The '$A$' you quote is the optical density. The Beer-Lambert law is $I_{tr}=I_0e^{-\epsilon_\lambda [C]\ell}$ where $I_{tr}$ is the intensity of transmitted light for a molecule at concentration $[C]$ at wavelength $\lambda$ and cell path length $\ell$, and $\epsilon_\lambda$ is the extinction coefficient at wavelength $\lambda$.

From the definition you can see where the expression you ask about comes from even with their different notation; it looks as if $w=-{\epsilon_\lambda [C]\ell}$ where ${\epsilon_\lambda [C]\ell}$ is the optical density.


Why do NMR spectrums have the higher chem shift values closer to x-axis?

This convention stems from historical reasons only. In the early days of NMR, there were a number of different ways to report NMR data; values for 'shielding constants' of nuclei was of most interest, rather than what we commonly refer to as chemical shifts. The two most common scales were the tau and delta scales . The tau scale had the chemical shifts running in the opposite direction, with increasing shielding constants running from left to right.

A third scale, the sigma scale was very commonly used, especially in heteronuclear analyses, and again had the shielding constant increasing to the right. This scale was favoured often by theorists, as it was usually a direct measurement of the shielding constant.

A mixture of tau and delta values was regularly seen in the literature up until IUPAC made their initial recommendations, in 1970. This was to adopt the delta scale, and report chemical shift relative to TMS at delta 0ppm, as values of chemical shift, which are dimensionless and, hence, field independent.


Application of LeChatelier's principle for reactions with solid products

I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.

Let's get back to the fundamentals of Le Chatelier's principle :

  When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract (partially) the effect of the applied change and a new equilibrium is established.

Let's assume you have a reaction $\ce{A(g) <=> 2B(g)}$. When lowering the pressure, the system will adjust itself to counteract this, i. e. increase the pressure. This can be achieved by displacing the equilibrium towards the right, as the direct reaction increases the amount of gas molecules.

In your case, however, there is no increase in the amount of gas compounds. Regarding the solid, as it does not participate to the overall pressure in the system (only gases do), it should not be taken into consideration here. Therefore, a change in pressure wouldn't alter the equilibrium state of your reaction.

That's for the math-less part.

You can also understand this concept using the equilibrium constant. For my example, it would be $K_c = \frac{[\ce{B}]^2}{[\ce{A}]}$. Lowering the pressure corresponds to an increase in volume, thus a decrease in both concentrations. This leads to a ratio that is different from the equilibrium constant, which will force the system to a new equilibrium.
$$K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{\left(\frac{n_\ce{B}}{V}\right)^2}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V}$$
When increasing $V$ to $V'$,
$$\frac{[\ce{B}]^2}{[\ce{A}]} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V'} < K_c$$
As this ratio is lower than $K_c$, the reaction will evolve towards the formation of products.

For your equation,
$$K_c = \frac{[\ce{B}]}{[\ce{A}]} = \frac{\frac{n_\ce{B}}{V}}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}}{n_\ce{A}}$$
there is no dependance on the volume, thus none either on the pressure.
C is not included as it is a pure solid ($a(\ce{C}) = 1$).

  Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not  really a hard limitation at all.

It is true that, if you find a way to remove products from the system, you can reach full conversion of your reagents. But that's actually the catch : it can be very hard to remove those products. Take as an example the Haber-Bosch process :
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
As it is a slow reaction, a high temperature is required to get the product in a reasonable time. How would you remove $\ce{NH3}$ from the system, leaving the reagents inside? We then use other ways of increasing the yield (pressure and temperature), but they do not allow a full conversion of the reagents.


Nanoparticle tracking analysis v.s Transmission Electron Microscope

You don't mention how homogeneous your solutions are or how much "considerably less" is. As summarized in this paper, "Under ideal conditions, particle size analysis by DLS and NTA both can be accurate to 2%, but heterogeneous samples can have errors an order of magnitude larger." (See references in that paper for details).
It's also important to keep in mind that you are comparing techniques which measure samples in very different ways and under very different conditions. TEM measures individual particles in a dried (and often treated) state in a vacuum, while NTA measures an ensemble of particles dissolved in solution. In each case, the size distribution is often just as important as the so-called "size". If the size-distribution of your measurements overlap, perhaps there's no conflict after all.
Also, if your particles have some sort of coating which Au nanoparticles, for example, often do, then you might expect differences in the apparent size measured in these two different states.
Using controls is a good way to get around some (but not all) of these problems, but then the same questions need to be raised about the control particles and any difference in their behavior from the particles under study.
So, as is usually the case in well-done science: "It's complicated..."


As pointed out already, the van der Waals (vdW) equation is also an empirical equation.  However, I think what you are saying is that the vdW equation is a physically motivated empirical equations where the empirical parameters in principle correspond to something "real" (e.g. molecular size and interaction strength) in contrast to a fit to an arbitrarily chosen fitting function such as a polynomial.

If you have lots of data points so that you can fit many parameters then a simple fit will be more accurate since it tends to have more fitting parameters than the physically motivated equation.  However, the simple fit will likely only be accurate for interpolation, and is still likely to be comparatively less accurate for extrapolation.


Why does the beryllium 2+ ion have a larger radius than a helium atom?

You are mixing apples and oranges.
Or, to be more precise, an ionic radius for $\ce{Be^2+}$ with coordination number (C.N.) 6 and van der Waals radius of $\ce{He}$.
To make it clear I compiled data for van der Waals and covalent radii [1, p. 9-58] as well as ionic radii [1, p. 12-13]:

$$
\begin{array}{lccccc}
\hline
\text{Element} & R_\mathrm{vdW}/\pu{Å} & R_\mathrm{cov}/\pu{Å} & \text{Ion} & \text{C.N.} & R_\mathrm{i}/\pu{Å} \\
\hline
\ce{Be} & 1.52 & 0.99 & \ce{Be^2+} & 4 & 0.27 \\
  &  &  &  & 6 & 0.45 \\
\ce{He} & 1.40 & 0.37 &  &  & \\
\hline
\end{array}
$$

Apparently, there is no experimental ionic radii for helium to compare with as helium, well, represents one of the most non-reactive elements.
Also, all radii set are determined using various methods and cannot be directly compared.
From the foreword [1. p. 9-58]:

  There are many scales of van der Waals radii, but they are not fully consistent with one another. The van der Waals radii determined by Bondi [...] from x-ray diffraction data, crystal densities, gas kinetic collision cross sections, critical densities, and liquid-
  state properties are the most widely used values. [...]

  The covalent radii tabulated here are recommendations for single covalent bonds,
  and they are based on a comprehensive evaluation of experimental data [...].

Ionic radii are estimated first and foremost from the experimental crystal structure data, averaged and supplemented by empirical and theoretical calculations (extrapolations, e.g. using Zachariasen method).

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


Positively charged ions wouldn't usually lose more electrons, they've already lost some.

The positively charge ions you have are called cations. These go to the cathode, which is negatively charged, as opposites attract. So, positive ions go to an area of negative charge and gain electrons, i.e. reduction takes place.

Now how can we negate any confusion between cathode and anode (it's very easy to get mixed up)?

Let's examine the word cathode a little more closely. Essentially there are two parts, 'cath' and 'ode'. The key part here is the ending of the word, which is 'ode'. The Greek root of this means, according to the Oxford English Dictionary, way. So as long as you can remember that 'cath' usually refers to a positive and 'an' to a negative, you can put the two together. Cathode meaning 'positive-way', as in positive things go this way.


First, note that the protonated amino group of an amino acid like glycine or aspartic acid has a pKa of about 9.6 - the number for glycine. At pH 1 it is indeed "fully protonated", but more correctly about 1 molecule in 10^(8.6) will have a free amino group at any given point in time (always in flux due to the rapidity of proton transfer). This amino group can react with a carboxyl group (from another amino acid) at the carbonyl carbon to give a tetrahedral intermediate, which in turn can eliminate water to give the peptide bond.

Under equilibrium at pH 0 or 1, the energetics typically favor the reverse of this reaction - hydrolysis of the peptide bonds to give the amino acids - this is what goes on in the stomach when you digest a protein - but it is possible that these conditions could produce a small amount of material from the reverse reaction - the amide-bond forming process.


Why does 1 cubic meter = 1000000 cubic centimeters?

Take a look at the Wikipedia article for SI prefixes, in particular, the following excerpt:

When units occur in exponentiation, for example, in square and cubic forms, the multiplication prefix must be considered part of the unit, and thus included in the exponentiation.

$1\ km^2$ means one square kilometre, or the area of a square of $1000\ m$ by $1000\ m$ and not 1000 square metres.

$2\ Mm^3$ means two cubic megametres, or the volume of two cubes of $1000000\ m$ by $1000000\ m$ by $1000000\ m$ or $2\times10^{18}\ m^3$, and not 2000000 cubic metres ($2\times10^{6}\ m^3$).

Thus, the relationship between the cubic metre and the cubic centimetre is as follows:
$$1\ m^3 = (1\ m)^3 = (100\ cm)^3 = 100\ cm \times 100\ cm \times 100\ cm =  1000000\ cm^3 = 10^6\ cm^3$$
$$1\ cm^3 = (1\ cm)^3 = (0.01\ m)^3 =  0.01\ m \times 0.01\ m \times 0.01\ m = 0.000001\ m^3 = 10^{-6}\ m^3$$
For a real case, converting $243.7\ cm^3$ to $m^3$ step by step:
$$243.7\ cm^3 = 243.7\times 1\ cm^3 = 243.7\times (1\ cm)^3 =243.7\times (0.01\ m)^3 = 243.7\times 0.01\ m \times 0.01\ m \times 0.01\ m = 243.7\times 0.000001\ m^3 = 243.7\times 10^{-6}\ m^3 = 2.437 \times 10^{-4}\ m^3$$
This is simply the way it was defined, and it is the standard everyone adheres to. More examples of SI prefix usage can be found on this page from one of the organizations which created the SI units, the BIPM.


Why isn't hybridization done in America?

In Russia hybridization of carbon learns at the beginning of the 10th grade, before the main organic chemistry. Hybridization in the atoms of metals (complexes) are studied in grade 11. These two topics are difficult for students to understand. But they can be on the exams.

Advantages:

It is easier to understand the spatial structure
Explanation of coordination numbers in complexes
For an explanation of the facts, when the atom forms a large number of bonds than the number of unpaired electrons in the ground state

Disadvantages:

Difficult to understand
There are many different theories
Rarely used in the study follow-up program


Activity Coefficients from Electrode Potentials

Sort of 'yes', sort of 'no'.  Consider the following points:

Activity coefficients are notorious for changing with $\mathrm{p}\ce{H}$, ionic strength, and concentration (among other things).  You might measure the the activity coefficient of $\ce{Cu^{2+}}$ in a $0.7~\mathrm{M}$ $\ce{CuSO4}$ solution and get one value, but try to measure the same thing in a low $\mathrm{p}\ce{H}$ and $1~\mathrm{\mu M}$ $\ce{CuCl2}$ solution and you may get a completely different value.  In practice, people usually end up measuring activity coefficients across a range of conditions that are relevant to their specific use-case.
The Standard Hydrogen Electrode is a theoretical device; and actually constructing something similar is difficult.  In practice, you would just want to use a commercially available double-junction silver-silver chloride or saturated calomel electrode. The potentials of those two are well-known and fairly stable.

See Chapter 2 of "Electrochemical Methods" 2nd Ed, by Bard and Faulkner for more information.


The repeated heatings were just to ensure that you had oxidized all of the copper carbonate to copper oxide. If you heated it a second time and the mass decreased, that would just mean that some of the copper carbonated had not been oxidized by the first heating. Then you would just keep heating until the mass quit decreasing, meaning that you had definitely oxidized all of it.

So the answer to you question is simply that the mass theoretically should not have decreased at all with a second heating, unless some of the copper carbonate had not been oxidized the first time.


Effect of protecting group on diastereoselectivity of LiAlH4 reduction of α-hydroxyketones

I think Jan did an excellent job of answering your question, however I am writing a slightly longer answer largely with more details concerning the Felkin–Anh rule, just so that someone who is not already familiar with them can learn something too.

Assuming that there isn’t an unusually electronegative atom on the carbon next to the carbonyl, the largest group prefers a conformation where it is perpendicular to the carbonyl moiety. This gives two relatively competitive lowest energy conformations, with the medium and smallest groups differing in their proximity to the carbonyl oxygen.

For the TBDPS-protected substrate shown above, the OTBDPS group would be considered the largest group. The medium and small groups are methyl and hydrogen respectively. This placement is also consistent with a modification to the Felkin–Anh model, which places electronegative substituents (such as –NR2 or –OR) perpendicular to the carbonyl (an explanation is given at the bottom of the answer). In any case, the two conformations of interest are the two shown below:

Moreover, the nucleophile doesn't prefer to attack the carbonyl carbon at a 90° angle; rather it prefers something closer to the tetrahedral angle (Bürgi–Dunitz trajectory). Naturally, the nucleophile prefers attack from the opposite side to the large group. This leaves two possibilities: for the nucleophile to attack over the small group, or for it to attack nearest to the medium group. The former is preferred, and the major diastereomer will be the syn diol:

On the other hand, Lewis basic groups such as $\ce{-OCH2OBn}$ are also capable of forming a chelate with the metal together with the carbonyl oxygen. This feature forces the $\ce{OR}$ group to be held parallel to the carbonyl. Nucleophilic attack still occurs along the Bürgi–Dunitz trajectory, and the nucleophile attacks opposite the larger of the two remaining groups. This will invert the stereochemical outcome as compared to the Felkin–Anh model previously described, and hence leads to the formation of the anti diol.

Now, to answer your question of why does the choice of protecting group matter. The previous discussion illustrates that the reason we get the Cram Chelate product is the chelation of the metal by the $\ce{OR}$ oxygen and the carbonyl. This chelation can take place easily if the $\ce{R}$ group is small, but is greatly hindered with a bulky alkyl/silyl substituent, in which case we get the product determined by the polar Felkin–Anh relationship.

The bulk of the TBDPS group does indeed prevent chelation, and you get the observed stereochemistry in accord with the Felkin–Anh rule. However, chelation is possible when BOM is the protecting group (Jan already offered an excellent explanation), and you end up with the Cram chelate product.

When there is an electronegative atom or group α to the carbonyl, there is a slight change to the Felkin–Anh model which may be applied. This ‘slight change’ is that this electronegative group becomes the equivalent of the ‘large’ group in the Felkin–Anh model.

The reason for this that the energy of the C–X σ* antibonding orbital is rather low, and so when the C–X bond is perpendicular to the carbonyl group, this can overlap with the π* MO of the carbonyl group to make a new, lower energy LUMO.

In "human" speak, this means that this conformational arrangement is more
reactive than any other. With this change in mind, the Felkin–Anh type addition now proceeds as before. However, the stereochemical outcome will generally be different from the original Felkin–Anh model.

For more information, see: How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?


Do the spectra of compound materials combine linearly with respect to the ratios of the constituent compound spectra?

Yes, you are right. If you have $n$ components in a mixture each with a spectrum $S_1$, $S_2$,...,$S_n$, the observed spectrum $S_{obsvd}$ would be the sum of $S_1$+ $S_2$+...+$S_n$.

In FTIR you either measure the transmittance or the absorbance. The term intensity is more or less used when you deal with molecular or atomic emission of light (fluorescence etc.) As long as there is no chemical reaction among the constituents, the absorbance should be linear with the concentration.


Is it possible to speed up radioactive decay?

It is possible to modify nuclear decay rates using chemistry, though it is rare and the effect is usually very small. Here I summarize the information available in this link. You may want to see the references within.

There is a type of nuclear decay called electron capture, where a nuclide directly captures an electron from the innermost electron shells and transforms a proton into a neutron. Therefore, there is coupling between the nucleus and the wavefunctions of the innermost electrons in this form of radioactive decay.

Usually the core electrons are very weakly disturbed by changes in chemical environment, but they are changed slightly. For especially light atoms, where the core electrons are very close to the valence shell, such as $\ce{^7_4 Be}$, this change results in a measurable difference in decay rate, varying by 0.1-1% compared to the isolated atom.

In a more dramatic case, there is a subset of beta decays called bound-state beta decay, where the electron released by the decaying neutron is immediately captured by the nucleus after decay. Apparently, if the parent atom is stripped completely bare of its electrons, and if the energy involved in the nuclear decay is comparatively low, once again there is a meaningful coupling between nuclear and electronic states. For the case of rhenium-187, the neutral atom $\ce{^187_75 Re}$ has a half-life of 42 billion years, but upon full ionization to $\ce{^187_75 Re^75+}$, the half-life is reduced to 32.9 years! For dysprosium-163, the neutral atom $\ce{^163_66 Dy}$ has half-life so high it has not been measured to decay, but when ionized to $\ce{^163_66 Dy^{66+}}$ its half-life is reduced to 47 days!


Why does the C-C bond break earlier than the C-F bond in the pyrolysis of perfluroethane?

I got this explanation from one of my seniors.

It is true that $\ce{C-F}$ bond is highly polar as fluorine is more electronegative than carbon. In fact it is this very difference in electronegativity that makes C-F bond very strong. This is because, in addition to the covalent bond strength between carbon and fluorine, there is an additional electrostatic attraction between positive carbon and negative fluorine.

We can think of it has the $\ce{C-F}$ bond has both covalent  and ionic character. This makes C-F bond extremely stable and hard to break.

In fact in organic chemistry we usually say that the $\ce{C-F}$ bond is not broken in reactions.

We can also verify this from their bond energies:

$\ce{C-F}$: $\pu{544 kJ mol-1}$
$\ce{C-C}$: $\pu{346 kJ mol-1}$


Entropy has two general precise definitions.  That which is most useful here is from Boltzmann, S(E) = k ln W(E) where S is the entropy, E is the energy of an isolated system (which is the easiest system to consider), k is Boltzmann's constant, and W(E) is the number of states of the system that have energy E.

For example, if the system has exactly one state with energy E, then S = 0.  If there are 2 states with energy E, then S = k ln 2.  And so on.

Now "organization" has no precise definition, but is rather an descriptive label we might apply to categorize states of the system.  Some of those states look "organized" to us, some look "disorganized."  For example, consider a model of atoms adsorbed on a crystal surface.  Let's model the surface as a square 3x3 array of sites and put 5 atoms on it.  A priori, there are 9!/(9-5)!4! = 126 ways of putting 5 atoms on 9 sites.  Now which of these states are "organized?"  That's to some extent a matter of opinion, but a common definition is "even" or "symmetrical."  So, for example, the state where there's an atom on every other site, in a checkerboard pattern, looks "organized" -- as if the sites formed a 2D crystal.  There's only one such state.

So now let us supposed we observe  this system, and we have some way of knowing, either from measurement or method of preparation, that it is in an "organized" (symmetric, checkerboard-like) or "unorganized" state.  There is only one "organized" state, so if we know the system is in an organized state then its entropy is k ln 1 = 0.  On the other hand, if we know the system is in an "unorganized" state there are 125 possible states it could be in, so its entropy is k ln 125, higher.

This is generally the case.  States we are willing to call "organized" generally have severe restrictions on the degrees of freedom: their values are distributed symmetrically in some way, e.g. the atoms are distributed uniformly, or in a regular array, the velocities are all the same, and so forth.  Because of these restrictions, it is almost always the case that there are fewer states of the system that satisfy the restrictions than that don't.  So the entropy of the system when it's known to be "organized" is less than when it is known to be not.

In short, it might help if when you hear "organized" you mentally translate that to "has some symmetry restrictions on how the degrees of freedom (e.g. atoms) are arranged."  Hopefully it is intuitive that there are far fewer ways of arranging atoms symmetrically than not, and therefore that the entropy of systems known to be symmetric is lower than the entropy of systems that are not.


I don't know whether liquid nitrogen can be used to cool a volume as large as 5 gallons safely (that is a lot to do at once in a kitchen or restaurant) but it can and has been used in cooking.

Here is an example from youtube.

But it can be dangerous if not done carefully. This BBC story has a warning of what happens if it isn't.


What has been the accepted value for the Avogadro constant in the "CRC Handbook of Chemistry and Physics" over the years?

I dislike offsite image links since they disappear over time. So below is Table 1 from andselisk's comment on the question.
The paper itself is behind a firewall and I can't see it.

See Table 1 in Becker, P. History and Progress in the Accurate Determination of the Avogadro Constant. Rep. Prog. Phys. 2001, 64 (12), 1945–2008. DOI: 10.1088/0034-4885/64/12/206. It covers year up to the year 2001 and is going to save a lot of time. – andselisk


melting and boiling ...really equilibrium?

I think what you are asking is this:  Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant).  By contrast, equilibria for phase transitions don't require specific ratios of products to reactants.   [For instance, at the phase transition between ice and water, ice and water can be in equilibrium with any relative amounts of ice and water.] Why is this?

Consider a gas-phase chemical reaction, say

$$\ce{A(g) + B(g) <=> C(g)}.$$

At equilibrium, the chemical potentials of all species are the same.  Associated with this is an equilibrium constant,

$$K_p(T) = \frac{\displaystyle\frac{p_\ce{C}}{p^\circ}}{\displaystyle\frac{p_\ce{A}}{p^\circ}\cdot\frac{p_\ce{B}}{p^\circ}},$$

which specifies a strict relationship between the partial pressures of the reactants and products.

Alternately, consider a solid-liquid phase transition,

$$\ce{A(s) -> A(l)}.$$

At the melting temperature, the chemical potentials of the two phases are the same.  However, by contrast with the example gas-phase reaction, this is true independent of the relative amounts of $\ce{A(s)}$ and $\ce{A(l)}.$

The question then arises: why the difference between the two?  The answer is that the particular mathematical form of the equilibrium constant—and the attendant constraint it imposes on the relative amounts of reactants and products—arises from the entropy of mixing, and there is no entropy of mixing in a phase transition!  At equilibrium, in the example gas-phase reaction, the entropy of mixing just balances the relative free energies of the reactants and products.  That's why we have an equilibrium.  If it weren't for the entropy of mixing, reactions would always go completely to the side whose pure components had the lower free energy (which is exactly what happens in a phase transition).

By contrast, there is no entropy of mixing term in a phase change, because the difference in phases keeps the two components separated.  And without an entropy of mixing term, the relative amounts of the two phases become irrelevant to their relative chemical potential, and thus irrelevant to the equilibrium state (by which I mean any relative amount of the two phases is allowed at the phase transition temperature).

This also explains why, when we are (say) somewhat above the melting temperature, we don't have a small amount of solid in equilibrium with a large amount of liquid (as would be the case if this behaved like a chemical reaction).  The reason for this is there's no entropy of mixing to favor this mixed state over the liquid alone, i.e., over the pure substance with the lower chemical potential (which, above the melting temperature, is the liquid state).  [Or more accurately, while there is an entropy of mixing, because the two phases aren't mixed at a molecular level—phase is a bulk property (i.e., a collective property of many molecules), not a molecular one—the entropy of mixing term (which is essentially an exchange entropy) is negligible in this case.]

N.B.: Just for clarity, let me emphasize that, as a substance is melting or boiling, it is not at equilibrium.  It's not at equilibrium during a phase transition until there is no macroscopic change in the system, i.e., until it settles into a fixed ratio of one phase to another (and that it can be at equilibrium at any ratio).

*Incidentally, this is also why, when we have a mixed-phase chemical reaction where one phase can't mix with the others, that phase isn't included in the equilibrium expression.  For instance, for

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)},$$ we find that

$$K_p(T) = \frac{p_\ce{CO2}}{p^\circ}$$


Why are cations ordinarily smaller than anions?

It isn't true that any cation will be smaller than any anion.

You intuition that atoms with larger electron counts from lower rows of the periodic table will be bigger is correct. Size for atoms is determined by the highest occupied electron shell or orbital as the interactions with other things that define size are, crudely, caused by interactions between the electrons in their outer orbitals.

So a Rb+ cation radius is about 166pM but an F- anion is smaller at about only 119pm. There is a good table of such values in the Wikipedia article on ionic radii which also explains some of the trends and some of the difficulties in actually defining size for ions.


What are the parent compounds of the following?

Find the longest chain by starting at one end of the molecule and counting the number of carbon atoms along each possible path.  I've attached a figure with numbering of the longest chain.  Hopefully it will answer your question and you will understand how to find and number the longest chain.  In the future, post your thoughts, so we can better understand how to help.


How can we directly add half cell potentials to measure the EMF of a galvanic cell?

You are writing $\Delta G_3=\Delta G_1+\Delta G_2$. But it should be $\Delta G_3=\Delta G_1+2\Delta G_2$ (I am considering  $\Delta G_2$ as free energy change for the conversion of $\ce{Ag+}$ to $\ce{Ag}$, which is considered 2 times in the final reaction) Because $\Delta G$ also depends on the number of moles of reactants and products, if you add two times the second reaction, the $\Delta G$ will also be used as twice its value. There you've made a mistake.

Thus the equation will be $$2E_3^0 = 2E_2^0 + 2E_1^0$$ So you will have $E_3 ^0 = E_1^0 + E_2^0$.


Why is the Debye Hückel Limiting Law called the "limiting law"?

The Huckel limiting law allows us to determine the activity coefficient of an ion in a dilute solution of a known ionic strength.

Here's an appealing answer which makes perfect sense to me that I found on EveryScience

  The name 'limiting law' is applied because in the limit of arbitrarily low molalities (ie as the concentration of the solute gets
  closer and closer to zero), all solutions are expected to behave in a
  manner consistent with the expression.


Testing capacitors is almost always done with a two electrode setup.
There are some details here about interpreting the data, but the experimental setup is that you just connect the counter and reference electrode inputs together to one half of the capacitor and the working electrode input to the other.
A three electrode setup is used when we only care about the interfacial potential of a single electrode (the working), and thus use a measurement near it by the reference electrode to control the counter electrode accordingly.
In a capacitor, we are generally interested in controlling the voltage across the entire device and observing its behaviour as a whole, so we don't need a reference electrode. (though depending on your specific device, it may be useful to observe only one electrode for any number of reasons)


Oxide surfaces develop a charge in H2O. Do we have any, even vague, handle on how long does an indivdual charge last on avg. before being neutralised?

Proton transfer reactions in water such as the ones you wrote down are diffusion controlled under ambient conditions (rate constant about 1e+12 per sec.) A snapshot in time will have a certain distribution of charges on the surface, but these are continually in flux, so any one charged site will have an exceedingly short lifetime. You could reasonably represent that system with a fixed set of charges on the surface, for some types of models (e.g., states at equilibrium).


Lifetime of a (reversible) hydrogen fuel cell system

Fuel cell manufacturers quote a range of different lifetimes which depend on many factors. Backup cells tend to have lower lifetimes as the start-stop process degrades them compared to continuous operation. On the other hand, use as engines in vehicles requires a reasonably long life.

Impure fuel will damage the platinum (CO for example will poison the catalytic surface). But platinum is recyclable so either poisoning or other factors won't necessarily be a fundamental barrier to use (some companies take used cells back at the end of life and reuse the platinum). Membranes and electrodes also degrade mechanically depending on duty cycle and any mechanical stabilisation in the design.

Nedstack quotes lifetimes ranging from 2,000 hours for backup cells (which presumably means years of actual operation as they only get used when other power is cut which isn't a routine happening) and 8,000-16,000 hours for continuous operation (which is two years). Tests of fuel cells in vehicles have reported lifetimes of 7,500 hours (which presumably relates to several years of use in practice), though they seem to need more maintenance than diesel engines.


Is preference given to long alkyl group or branched cycloalkane?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), two methods are possible for naming systems composed of rings and chains: either the ring or the chain can have seniority. To construct the preferred IUPAC name (PIN), the ring is selected as the parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the preferred IUPAC name (PIN) of the compound that is given in the question is 1-ethyl-2-hexylcyclopropane.

In general nomenclature, however, the context may favour the chain, so that the parent structure has the greater number of atoms in the principal chain. The corresponding systematic name for the compound given in the question is 1-(2-ethylcyclopropyl)hexane.

Note that according to the obsolete 1979 recommendations, a hydrocarbon containing a small ring attached to a long chain was generally named as a derivative of the acyclic hydrocarbon. Many books are based on these nomenclature rules and thus may expect an answer that favours the chain.


Intermolecular and Intramolecular Forces (Vapor pressure of n-butylamine and methanol)

What are intermolecular molecular forces?
The forces(electrostatic force) between molecules of a compound. These forces are responsible for intermolecular distances and how tightly the molecules are held together. If the intermolecular forces(van der Waals force) are large, it means the molecules are tightly held together in a group and they are fixed in their position.

Now what is boiling point? It's a temperature at which liquid state of a compound changes into its gas state. What's the difference between gaseous and liquid phase? Gaseous phase has very large intermolecular distance when compared to that in liquid phase. If a large force exists between the molecules and they are quite fixed in their position, then large amount energy or heat will be required to move them away from each other. Large heat means greater temperature and hence greater boiling point. So the lesson is that, the stronger the intermolecular forces, the higher the boiling point.
The similar concept in applicable in melting point. the stronger the intermolecular forces, the higher will be the melting point.

You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably(In case you may be wondering, how are they related. Click here). If the intermolecular forces are large, then molecule in liquid phase are not able to leave to liquid easily, hence lowering the escaping tendency and therefore lower vapour pressure.

Note: This is a very basic explanation.


Reversed size dependence in ion solubility

This can be explained with some crude thermodynamic models. The lattice enthalpy $\Delta H_\mathrm L$ of an alkali metal halide $\ce{MX}$ is given by

$$\Delta H_\mathrm L = \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

and the hydration enthalpies by

$$\Delta H_\mathrm S(\ce{M}) = -\frac{k_2}{r_\ce{M}}\qquad  \qquad \Delta H_\mathrm S(\ce{X}) = -\frac{k_2}{r_\ce{X}}$$

where $k_1$ and $k_2$ are some positive constants and $r_\ce{M}$ and $r_\ce{X}$ are the ionic radii of $\ce{M+}$ and $\ce{X-}$ respectively.

The solubility is therefore a balance between these two quantities: larger (magnitudes of) solvation enthalpies increase solubility, but larger lattice enthalpies reduce solubility. The solubility $S$ is thus related to the difference of these terms. We ignore the entropic contribution, or to be precise, we assume that $\Delta S$ does not vary too much between different salts $\ce{MX}$.

$$S \sim \frac{k_2}{r_\ce{M}} + \frac{k_2}{r_\ce{X}} - \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

Let us now assume that

$$r_\ce{M} = a+b \qquad \qquad r_\ce{X} = a - b$$

where $a,b$ are more positive constants. Here we have implicitly assumed that $r_\ce{M} \geq r_\ce{X}$, but it works the other way round too, so there is no loss of generality. Thus

$$\begin{align}
S &\sim \frac{k_2}{a+b} + \frac{k_2}{a-b} - \frac{k_1}{2a} \\
&= \frac{2ak_2}{a^2-b^2} - \frac{k_1}{2a}
\end{align}$$

This is clearly a minimum when $b = 0$, i.e. $r_\ce{M} = r_\ce{X}$. So, salts with similar cationic and anionic radii are less soluble, and vice versa.


How does bombarding a molecule with electrons cause ionization?

This type of ionization is known as electron ionization (there are many types of ionization methods used for MS). One possible reaction for EI is:
$$\ce{M +e- ->M^{•+} +2e-}$$ source

If the incoming electron has the correct energy to interact and passes close enough to the molecule, it's electric field can impart enough energy to one of the molecule's electrons for it to be ejected. What happens to the electrons after that isn't particularly important as far as ionization is concerned, but in most EI sources, electrons are emitted perpendicular to the axis of the sample source and are collected on the other side by an electrode, preventing charging of the source chamber and focusing the emitted electrons into a beam.


I cannot say as to your specific sensor, but most oxygen sensors are electrochemical sensors built around concentration cells.

Take a look at the following diagram of a simple zirconia-based oxygen probe from Wikipedia. This type of sensor is very like the one in automobiles.

In the middle of the sensor is a gas permeable zirconia $(\ce{ZrO2})$ membrane between two gas permeable electrodes. The membrane and electrodes form the electrochemical cell. In many cases, an electrochemical cell will use a reference electrode with a standard reaction going on for comparison. In this sensor, the standard reaction is the same one that is going on at the other electrode. The difference is that at the standard electrode, the partial pressure ("concentration") of oxygen is known and at the other electrode it is unknown.

  But wait, how does that work?

Let's take a generic electrochemical process, where $\ce{M}$ in the following equation represents something generic that we can oxidize:

$$\ce{M(s) + O2(g) -> MO2(s)}$$

This cell has a standard potential $E^\circ$. The actual measured potential, however, will depend on the partial pressure of oxygen $P_{\ce{O2}}$ because the standard potential is defined at $P_{\ce{O2}} = 1\ \text{atm}$. The relationship that gives us the actual electrode potential at some other oxygen level is the Nernst equation;

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

Where:

$R$ is the ideal gas constant
$T$ is the temperature
$n$ is the number of electrons transferred (4 in my example)
$F$ is the Faraday constant
$Q$ is the reaction quotient, in this case $Q=\frac{1}{P_{\ce{O2}}}$

If $P_{\ce{O2}}$ is large, then $E$ is small and vice versa.

Now, we have two electrodes, our unknown and our reference, each with their own potential:

$$E_{ref}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},ref}}}$$
$$E_{unk}=E^\circ - \frac{RT}{nF}\ln{\frac{1}{P_{\ce{O2},unk}}}$$

The voltmeter in the device measures the difference between these two potentials and does some maths to calculate back $P_{\ce{O2},unk}$ and then convert it to concentration or percent. Note that it does not even matter what $\ce{M}$ and $\ce{MO2}$ were, since they are not in the equation and even $E^\circ$ cancels out.

$$E_{unk}-E_{ref} = \frac{RT}{nF}\ln{\frac{P_{\ce{O2},ref}}{P_{\ce{O2},unk}}}$$

  What substance might this have been?

If your sensor was this kind of sensor, the crusty mess is probably zirconia.


Schrödinger equation and degeneracy of atomic orbitals

Degeneracy refers to states having the same energy.
If you have two or more coordinates that can be related by a symmetry operation then the system will contain degenerate states because swapping the coordinates associated with those degrees will result in a Hamiltonian with the same solutions.
This is the case in both the hydrogen atom and in the particle in a cube.


Small atom, big atom and small atom, small atom - shouldn't their
  overlaps be proportional

No.  Look at the figure below and let's consider s-orbital overlap, just to keep it simple.  Remember that these are spheres with the electron density distributed throughout the entire volume.  In the "small-small" case there is effective overlap of the electron density in both 1s orbitals; we would expect the resultant bond to be strong.  In the "small-big" case, the small 1s orbital involves a significant fraction of its electron density in overlap, but not much of the electron density in the large s orbital is involved in overlap.  In the "small-big" case the overlap of electron clouds is ineffective and any resultant bond would be weak.


This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure.

First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{C5H9NO}$:
$$u=\frac{5\cdot2+2+1-9}{2}=2$$

However, OP's suggested structure was incorrect since chemical shift of -$\ce{OCH3}$ is usually greater than $\pu{3 ppm}$. Because of the chemical shift of -$\ce{CH3}$ group in the given $\ce{^1H}$-$\mathrm{NMR}$ is about $\pu{2.0 ppm}$, one can conclude that it is either olefinic or one $\alpha$ to a cabonyl group. Nonetheless, OP's suggestion of proton resonance around $\pu{6.66 ppm}$ is due to $-\ce{NH}-$ group is correct, but due to its deshielding nature ($\pu{6.66 ppm}$), one can suggest that it is an amido $-\ce{NH}-$ group. However, OP correctly assign it next to a $-\ce{CH2}-$ group.

Since I suspect non cyclic nature of the unknown, I have assigned splitting patterns and calculate relevant coupling constants assuming the spectrum is taken in $\pu{400 MHz}$ machine. All splitting patterns of remaining resonances are $ddt$ as depicted in fallowing diagrams:

Accordingly, these patterns and chemical shifts well match with N-allyl acetamide (see insert in first diagram).


Are fractional oxidation states possible?

The "oxidation number" is a theoretical value used to do electron bookkeeping and is one way of comparing the number of electrons "owned" by an atom in a molecule or ion versus how many valence electrons present in the atom as depicted on the periodic table.

Your calculation is assuming that all of the carbon atoms in octane have the same oxidation state, which is incorrect. The two terminal carbons have an oxidation state of -3 while the inner 6 have oxidation states of -2. Taking the average of these values gives you an average oxidation state, which in your case is a fractional number.

It is doubtful that the oxidation state calculated in this manner is of any use, since it is based on the assumption that all the carbons in the molecule are the same.


This structure is impossible as either the molecule is planar and then the C-C bonds on the outer ring are extremely long (and therefore weak), or you distort from the sp2 geometry (120 degree bond angles) of the central ring carbons so that the outer ring is beneath the inner one, which would also require a very high energy state. The only stable circumscribed ring structures you'll find are macromolecular.