Vacuum distillation apparatus selection and configuration

Don't bother with a powered pump: Get a water aspirator (I recommend the $20 Humboldt H-12020, which uses 10mm or 3/8" inner diameter tubing just like your glass), connect to faucet, and using utility water (1-2 gallons/minute) it can pull over 500 torr, which should be adequate for your applications.  Sure, you're flushing away a lot of water, but you don't have to bother with traps or filters.

(Actually, the H-12020 does have a check valve, but to ensure you don't get even a little water sucked into the apparatus when you shut off the aspirator you might want to add a separate valve or simple trap, but that's easy compared to a cold trap.)

You should buy "thick-walled vacuum tubing" for all connections if you want to run full vacuum.

Finally, for safety you do need some bump countermeasure.  A "capillary bubbler" is just a tube that lets a tiny stream of bubbles into the boiling flask.  Yes, that is reducing the vacuum, but a proper capillary tube admits a trivial amount of gas into the system.  When possible, however, I prefer stirring.  Running a magnetic stirrer in the boiling flask also prevents bumping, and is better than a bubbler because it doesn't introduce potentially contaminating gases into the mix (or require the supply of a neutral gas).

A decent reference for practical distillation is Zubrick's Chapter 20 ("Distillation for Dummies") available online here.


Definition of major, minor, and trace elements in analytical chemistry

There is no definite classification of what are the major, minor and trace elements are. While the numbers you mentioned are indeed the rule of thumb, they are rather loosely applied. It depends on field (geochemistry, solid state physics, analytical chemistry). There are however several guidelines.

Major elements are the elements that define the material in question. Some examples are:

Gold and silver in electrum,
Yttrium, aluminum and oxygen in YAG,
Hydrogen and oxygen in the ocean.

If you change the major elements, you are essentially changing the material. If you replace the hydrogen with potassium then instead of water you will get potash, which is not the same material and has vastly different properties (although this example might be a bit extreme).
Major elements are usually measured in percentage (either mass or molar), and are commonly above 1% of the chemical composition of the material.
Trace elements are elements that occur in such small concentrations that they do not change the essence of what a material is. Some examples:

Copper and platinum in electrum. It's still a gold-silver alloy, just with some other metals. The properties (such as colour) may change slightly.
Neodymium, erbium, etc. in YAG. YAG is used to generate lasers, and doping it with trace amounts of other rare earth elements may change certain properties of the laser (wavelength). However, it's still a YAG.
Almost every other element other than H, C, Cl and Na in ocean water.

Trace elements usually have concentrations of below 0.1%. In this case it's easier to measure them in units of ppm (parts per million, this mostly refers to mass). Remember: 1% = 10000 ppm. In extreme cases ppb (parts per billion) may be the appropriate unit.
Minor elements are everything in between. Technically, this means things between 1% and 0.1%. You might think of sodium and chlorine in ocean water as an example of minor elements (even though they're both slightly above 1%). To be honest, other than a mention in introductory textbooks about analytical chemistry and geochemistry, I rare see this term used in the professional literature.


There are no $\ce{NaCl}$ molecules in the solid salt nor in solution.

In solid salt, there are $\ce{Na+}$ ions, surrounded by 6 $\ce{Cl-}$ ions, while each $\ce{Cl-}$ is similarly surrounded by 6 $\ce{Na+}$ ions. Here is the Wikipedia picture of the structure.

During the dissolution, water molecules reach for a particular ion, oriented toward it by the opposite partial charge of their electric dipole.

$\ce{Na+}$ ions are torn away and wrapped by $\ce{O}$ side of $\ce{H2O}$ molecules, which have a partial negative charge.

$\ce{Cl-}$ ions are torn away and wrapped by $\ce{2H}$ side of $\ce{H2O}$ molecules, which have a partial positive charge.

Ions end hydrated in solution like $[\ce{Na(OH2)}_n]^+$ and $[\ce{Cl(H2O)}_m]^-$


When non-polar particles interact through dispersion forces, how fast does a temporary dipole induce a dipole in the neighboring molecule?

The upper limit are typically optical light frequencies. That is where the very last plateau is reached in dielectric spectroscopy. The electron shell becomes purely elastic.

I wrote a bit more about DES in my answer here: Difference between relaxation and resonance leading to an absorption spectral feature?


Above a $1000\ \mathrm K$, the $\Delta_\mathrm fG$ of $\ce {CO}$ from $\ce C$ is more negative than that of $\ce{CO2}$ formation from $\ce C$. Therefore, during smelting, when coke ($\ce C$) reacts with $\ce {SnO2}$, the formation of $\ce {CO}$ rather than $\ce {CO2}$ is thermodynamically preferable.

Below that temperature, the reduction with carbon is not possible since $\Delta_\mathrm fG(\ce{SnO2})\lt\Delta_\mathrm fG(\ce{CO2})\lt\Delta_\mathrm fG(\ce{CO})$. Therefore at all temperatures where this reduction is possible, thermodynamically carbon monoxide formation would be more favourable as compared to carbon dioxide formation. To see for yourself, the variation of $\Delta_\mathrm fG$ of various compounds, have a look at Ellingham Diagrams, a link to a comprehensive set of which is Given Here.

$\Delta_\mathrm fG$ is the Gibbs free energy of formation.


How come diluting a cell decreases the voltage?

Your assertion the voltage in a cell is decreased by diluting the cell  is not necessarily true. According to the Nernst Equation:

$$E_{cell}=E^0-\frac{RT}{nF}\ln(Q)$$

Where $Q$ is the reaction quotient relating $\frac{[\mathrm{metal~ion~formed~by~oxidation}]}{[\mathrm{metal~ion~reduced}]}$.

The value of $Q$ can be increased by: increasing the concentration of the metal ion formed by oxidation, or by decreasing the concentration of the metal ion that is reduced. By increasing the value of $Q$, the magnitude of the term $-\frac{RT}{nF}\ln(Q)$ is increased. This means that a larger number is subtracted from the $E^0$ of the cell, and that the cell voltage decreases.

The value of $Q$ can be decreased by: decreasing the concentration of the metal ion formed by oxidation, or by increasing the concentration of the metal ion that is reduced. By decreasing the value of $Q$ to a fraction, the value of the term $-\frac{RT}{nF}\ln(Q)$ is made positive. This means that this term is added to the $E^0$ of the cell, and that the cell voltage increases.

This is consistent with the Le Châtelier's Theorem. Essentially, increasing the relative concentration of your reactants drives the reaction forward (higher $\Delta V$), and increasing the relative concentration of your products drives the reaction backwards (lower $\Delta V$).


How to you show a 3-center 2-electron bond in a Lewis structure?

What about the probably "not so good" approach might be the bent bond as sometimes seen for diborane?

(source)

The latest recommendation by IUPAC I know, exemplified in rule "GR-1.9 Multi-center bonds":

(Brecher, Pure Appl. Chem., 80, 277-410, 2008; doi 10.1351/pac200880020277, open access)

stating:

  "As a matter of convention, any such multi-center character is ignored
  when producing chemical structure diagrams, and regular bonds
  connecting pairs of atoms are used instead."

To accommodate Karsten Theis' comment on the initial form of this answer, I would like to add the perspective provided by Robert Grossmann's The Art of Writing Reasonable Organic Reaction Mechanisms.  Starting page 270 (2nd edition, 2003) the chapter introducing reactions with transition metals draws attention about "Conventions of Drawing Structures"; highlighting that there are field dependent conventions about what line and dash represent.  To quote:

  "The conventions for drawing organometallic and inorganic compounds
  differ in subtle ways from those used to draw “ordinary” organic
  compounds. The most important difference is the way in which bonds are
  drawn. In organic compounds, one does not use a line to connect a bond
  to an atom. In organometallic and inorganic compounds, however, a line
  sometimes connects an atom and a  $\sigma$ or $\pi$ bond. In this
  case, the line indicates that the pair of electrons in the $\sigma$ or
  $\pi$ bond is shared with the metal also."

(loc. cit. p. 271)

  "An even more confusing situation arises in complexes in which the
  electrons in a $\pi$ system spread over three or more atoms are used
  to make a bond to a metal. In this case, the usual convention is to
  use a curved line to indicate the $\pi$ system and a single line to
  connect the $\pi$ system to the metal, regardless of the number of
  electrons in the $\pi$ system (the organometallic chemists’
  convention). However, sometimes the curved line is omitted and single
  lines are used to connect the metal to each of the atoms in the $\pi$
  system (the crystallographers’ convention). The representation that
  would make the most sense to organic chemists, in which a single line
  represents a two-electron  bond and a dative bond is used to show
  two-electron bonds between each individual $\ce{C=C}$ $\pi$ bond and
  the metal, is simply never used."

(loc. cit., p. 271)

  "Formal charges are usually omitted in inorganic and organometallic
  complexes. Only the overall charge on the complex is indicated. For
  example, formal charges are usually not assigned in Lewis acid–base
  complexes involving transition metals. The acid–base bond is sometimes
  indicated by an arrow pointing from the ligand to the metal, but more
  often it is indicated by an ordinary line."

(loc. cit., p. 271)

This precedes just the section about electron counting (specifically in organometallic compounds with transition metals). I speculate, however, Robert Grossman aiming to highlight differences between "organic" and "organometallic" convention meant does while accidentally deploying does not in the first paragraph.


Products of decarboxylation of sodium carboxylate

The first reaction is:
$$\ce{CH3COONa + NaOH -> Na2CO3 + CH4}$$
According to Encyclopedia of Inorganic Chemistry vol. 2:

The obtained methane in this process is not pure and may contain as
much as 8 % hydrogen and 10% other hydrocarbons such as the chloride
(to remove  acetylene) and concentrated sulfuric acid (to remove
ethylene and moisture). The purified  methane still contains hydrogen,
which may be removed as water by boiling at 110 °C the  excess oxygen
is absorbed concentrated sulfuric acid.

The second reaction doesn't make much sense to me with $\ce{BaO}$. If this reaction is happening in an aqueous medium, then $\ce{BaO}$ would probably have reacted to form $\ce{Ba(OH)2}$*. So I think you could justify representing the second reaction as:
$$\ce{CH3COONa + Ba(OH)2 -> CH4 + NaCO3 + BaCO3}$$
This is also how the aforementioned book represents the reaction, stating:

Heating sodium acetate with barium hydroxide, instead of sodium hydroxide forms nearly pure methane:
Sodium Acetate + Barium Hydroxide $\rightarrow$ Methane + Sodium Carbonate + Barium Carbonate
$$\ce{CH3COONa + Ba(OH)2 -> 2CH4 + NaCO3 + BaCO3}$$

So I guess the increased purity, in this case, is just due to the different amount of other products produced during the reaction.


The impact temperature has on the pKa of strong acids are in practical terms non existent as the equilibrium at any temperature lays very far to the right.

Practically, yes.

The impact of temperature on pKa are more significant for weaker acids.

Yes.

The impact of temperature on pKw is only significant around the point of neutrality. At lower pH's the few hydronium ions produced by H2O + H2O are so few that it bares no meaning on the logarithmic pH scale. In other words, the pH-temperature relationship is important when dealing with neutralization processes, but not so much when having pH setpoints set to anything in the milder acidic region and below.

$\mathrm{p}K_\mathrm{w}$ changes apply to all $\ce{H3O+}$ and $\ce{OH-}$, not just to those from water autodisscociation. If $\ce{[H3O+]}$ ( [] is the convention for molar conceentration ) is determined by $\ce{[OH-]}$, or if some process depends directly on $\ce{[OH-]}$ and not on $\ce{[H3O+]}$, it is significant.

Compensating for the pH-temperature relationship caused by the temperature dependent pKa is only important when doing measurements on processes where the temperature of the process differs from the temperature of the final product and where the targeted pH level during process control has to match the targeted pH level of the end product (at another temperature).
For example: say I want to produce an end product with a pH at 5 at temperature X. In order to make this product the temperature of my fabrication process has to be something other than temperature X. In this case, temperature compensation to account for changes is pKa becomes important - otherwise the end product won't have the desired pH.

Yes. E.g. for acetic acid, $\mathrm{p}K_\mathrm{a, \pu{25^{\circ}C}}=4.75$. That is the $\mathrm{pH}$ of solution acetic acid / sodium acetate in molar ratio 1:1.  Let imagine for illustrative purposes $\mathrm{p}K_\mathrm{a, \pu{50^{\circ}C}}=4.5$. If you used at $t=\pu{50^{\circ}C}$ for the final process $\mathrm{pH}=4.75$, there would not be reached 1:1 ratio.

in a different scenario the pH setpoint of a process becomes a means to an end. It's the pH of the process that determines the chemical reactions that takes place in that particular process (perhaps the hydronium ions merely acts as catalysts). In this case, temperature compensation in no way should be applied. This because the pH changes casued by changes in pkA are real and if the process demands say pH 3 for an optimal chemical reaction then it doesn't matter what the pH would have been at another temperature. The important thing is what the temperature is in the process right here, right now.

Yes, if I understant the paragraph well.

while pH changes that comes from changes to pKa actually do mean that a solution becomes more or less acidic, changes to pKw doesn't result in a solution becoming more or less acidic even though pH changes. This is because an equal amount of hydroxide ions a produced. So only the point of neutrality is changed.

This depends on what you mean by acidity. If you mean by acidity the activity or concentration of $\ce{H3O+}$, it is affected by $\mathrm{p}K_\mathrm{w}$ changes.

If so, then my question now is: What happens when we move into the alkaline region? I mean, pKb is, according to Le Châteliers principle, temperature-dependent as well, right? When looking at this graph https://ibb.co/LRs7G5y (algorithm based on NEN6411 (now DS/EN ISO 10523:2012)) it seems as if alkaline solutions ALSO becomes more acidic with increases in temperature. Why is this? Does it mean that the equilibrium constant pKb runs in reverse of pKa when temperature increases (less OH-)? Or does it mean that pKw plays are much larger importance once we wander into the alkaline region? I don't, however, see how pKw play any importance on the temperature-pH relationship of alkaline solutions. I mean, if we have an alkaline solution of say pH 12, then any extra/or less hydronium ions produced by changes in pKw would be dwarfed by the proportional much larger amount of hydroxide ions.

For weak acids, the primary parameters is $\mathrm{p}K_\mathrm{a}$. Then for their conjugate bases, $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a}$.
For weak bases, the primary parameters is $\mathrm{p}K_\mathrm{b}$. Then for their conjugate acids, $\mathrm{p}K_\mathrm{a}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b}$.
For solutions of strong acids, $\mathrm{pH} \approx -\log{[\ce{H3O+}]}$,  $\mathrm{pOH} =  \mathrm{p}K_\mathrm{w} - \mathrm{pH}$.
For solutions of strong bases, $\mathrm{pOH} \approx -\log{[\ce{OH-}]}$,
$\mathrm{pH} =  \mathrm{p}K_\mathrm{w} - \mathrm{pOH}$.

The graph you mention is for strong electrolytes, so $\mathrm{p}K_\mathrm{a}$ is not applicable.


The issue is that unlike for the methods based on Hartree-Fock, there isn't a systematic path to improving DFT. Unlike for HF, where we can treat the correlation as a pertubation (MP-n) or include excited Slater determinants (CI and CC) to generally improve our description of a molecule, we still don't have a way of determining the true form of the density functional (specifically the part governing exchange correlation) in DFT, so for now we use educated guesses for the general form (Local Density Approximation, Generalized Gradient Approximation, etc) and optimize them for specific systems.

In principle, if the exact form of the functional were known, DFT would be exact. This is why some authors like to use a separate term DFA (Density Functional Approximation) to describe the instances of DFT in use today.


Formation of oximes by ketones and aldehydes

An oxime is formed by the reaction of hydroxylamine (NH2OH) with either an aldehyde or ketone. The oxime nitrogen has a double bond (sigma and pi) to the oxime carbon, a lone pair and a single (sigma) bond to the hydroxyl oxygen. If we use valence shell electron pair repulsion (VSEPR) theory to deal with the problem.

Then the oxime nitrogen will be trigonal planar.

If the two groups on a ketone are identical then the molecule will be the same which ever way around we put the lone pair and the hydroxyl. But if the two carbon groups are different then we can have two isomers.

Consider for a moment the 2-hydroxyoximes used as LIX extraction agents for copper. There is a paper from the 1970s on the separation of the isomers of LIX 63 (T.T.Tammi, Hydrometallurgy, 1977, Volume 2, Issue 4 (October), Pages 371-380) which you migh want to read.

To make it more clear we have diagrams here of the two isomers of LIX63.


Answer on question 2.

As a former chemistry student, I saw that most chemicals were ordered at:

Sigma Aldrich
Biosolve
Fisher Scientific
ThermoFisher Scientific
Local pharmacy (at least in Holland)

Equipment was often ordered directly at the manufacturer (like an HPLC) unless the device is pretty general, like a rotary evaporator.


Although you probably know the answer by now, here is the cyclic transition state. It is an oxy-ene reaction. The Lewis acid lowers the activation energy of the reaction.


Does this produce the anti-Markovnikov product because of the ROOR?

Yes, having a peroxide will change the regiochemistry to an anti-Markovnikov addition. The reason, as suggested, is because of a change in mechanism to a radical process (below). The homolytic decomposition of the peroxide initiates the radical chain, generating two hydroxyl radicals. The hydroxyl radical abstracts a hydrogen from HBr to give a bromine radical. This is where the cycle really begins, with bromine radical adding to the terminal end of the alkene, leaving a radical at the internal position. The stability of a tertiary radical relative to a primary radical drives this selectivity. The tertiary carbon radical abstracts a hydrogen from the next molecule of HBr, giving the product and a new bromine radical which reenters the radical chain.


carbon dioxide and its behaviour in acid base balance

The disolved carbon dioxide increases the acid content of the human body. This happens when a carbon dioxide molecule bumps into a water molecule. Carbon dioxide makes water break into hydrogen and hydroxide ion. Hydroxide merges with the carbon dioxide molecule creating bicarbonate ion. Sometimes the hydrogen ion can become part of increasing acidity. Sometimes the hydrogen ion combines with the bicarbonate molecule and make carbonic acid. Carbonic acid could turn into carbonate ion and two hydrogen ions. This is what makes the human body acid base balance. The amount of bicarbonate is staying the same because the bicarbonate that does get converted into carbonic acid or carbonate ion an equal amount of bicarbonate is created from water and carbon dioxide. This just results in a huge amount of hydrogen ions.


What makes teflon such a strong material?

Teflon (PTFE, polytetrafluoroethylene) is prepared by the polymerization of tetrafluoroethylene as shown below.  If we took polyethylene and replaced all of the hydrogens with fluorines, we would have PTFE.

Here is a Table that compares the strengths of some single bonds involving a methyl group.

\begin{array}\hline
Experimental ~Bond~ Enthalpies for~ CH_3 - X~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
I & 57.6    \\ \hline
Br & 72.1    \\ \hline
Cl & 83.7    \\ \hline
CH_{3} & 90.1    \\ \hline
H & 104.9    \\ \hline
F & 115.0    \\ \hline
\end{array}

The $\ce{C-F}$ bond is one of the strongest bonds to carbon known.  As you add additional fluorine atoms to the carbon, the bond strength increases even further.  While the $\ce{C-F}$ bond is a covalent bond, it has considerable ionic character making the carbon positively polarized and the fluorine negatively polarized.  This bond polarization results in a strong electrostatic attraction between carbon and fluorine which accounts for the high bond strength observed.  It is this very high bond strength that makes fluorocarbons so stable (unreactive).  There are not many conditions that can supply enough energy to break this very strong bond.


Derive expression for heat flow from oil bath to metal sphere

Newton's law of cooling is

$$q = A h (T_0 - T_S )$$

With the area of the sphere being

$$A = 4 \pi R^2$$


Disease predicting software for small molecules

Do you mean predicting health effects/toxicity of a given chemical? I haven't seen such software and I doubt it exists. Much easier problem (predict optimal synthetic route to compound X remains unsolved.

By hand your best way is to check toxicity of related compounds. A program that can predict health effects of compounds will revolutionize drug design (and cause job loss for 99% of RnD chemists in pharma).


The basic explanation that you've recited for the Kinetic Theory of gases is trying to present a simple complete and consistent model which explains the observed behavior. Thus to give a consistent perspective it is easiest to think of gas molecules as elastic spheres. Of course, the assumption about the negligible size of the molecules does indicate that the shape of the molecules is irrelevant to the volume of the gas. But if you consider the molecules as having edges, or different moments of inertia about the three axis then the "simple" model gets too complicated for "simple" mathematics. Think of the all the additional variables that would be needed.

The basic model neglects other factors of course. In the real world gravity plays a part. The atmosphere decreases in density as the altitude increases. So the basic model is trying to illustrate the key factors necessary to understand the kinetic theory of gases, not all the factors.

Edit - The basic gas law PV=nRT has been modified by a number of expressions where correction factors have been used to make the equation more accurate and to give the equation more dynamic range. See some of the equations in the following wikipedia article.


What gas has the highest refractive index?

The question has to be more specific to get a specific answer.

Even in the table linked in the OP, bromine gas (1.001132) has a higher index of refraction than 1.000773.

At zero degrees C, 5462 Angstroms, and atmospheric pressure, The Refractive Index Dispersion and Polarization of Gases (1936) reports:

  $\ce{Si2H6}$ 1.0016574

$\ce{Si2H6}$ , disilane, is a gas at room temperature and atmospheric pressure.

The index of refraction is extremely dependent upon pressure.

As an introduction, see INDEX OF REFRACTION AND DISPERSION OF SEVERAL GASES IN CERENKOV COUNTER USE where indexes of refraction of gases as high as 1.02963 (for 182.79 psig ethane) are measured.


I'd say the suggested reaction of benzene-1,4-diol (hydroquinone) with $\ce{PCl3}$ to get 1,4-dichlorobenzene would not work (see  Waylander's comment). However, you can buy 1,4-dichlorobenzene very cheap (e.g., USD 21.50 for $\pu{100 g}$ bottle from Sigma-Aldrich) so why you need this trouble?

The suggested reaction of 1,4-dichlorobenzene with $\ce{NH3}$ to get benzene-1,4-diamine (p-phenylenediamine) would work, however, because there is a literature evidence for that (Ref.1). To do so, you need to use copper(I) chloride catalyst ($\ce{CuCl}$) and use 28% $\ce{NH4OH}$ solution as $\ce{NH3}$ source. Also need to heat the reaction mixture above $\pu{200 ^\circ C}$ in a pressure bomb (you may need to read this reference carefully before you start the reaction).

References:

Armand J. Quick, "The Preparation of p-Phenylenediamine and Aniline from their Corresponding Chlorobenzenes," J. Am. Chem. Soc. 1920, 42(5), 1033-1042 (https://doi.org/10.1021/ja01450a018).


Mechanism of ring opening of amine combined with formylation

This reaction is very similar to Reimer-Tiemann reaction which also consists of the reagent $CHCl_{3}$ + $NaOH$. The mechanism of that reaction is believed to have dichlorocarbene intermediate too. A good point to remember is that the dichlorocarbene intermediate is often used to form cyclopropane or carbonyl groups.

This is the mechanism for the reimer tiemann reaction. We can see the similarity in the this reaction and the one you have mentioned previously.


Is it possible to predict the lines in the atomic emission spectrum of Na?

The origin of the Rydberg equation is the Bohr model's equation for the energy of a hydrogenic state:
$E_\text{B} = -\dfrac{R_{H}hc}{n^2}$

This is accurate for other hydrogenic atoms, with only one electron too (bearing in mind that $R_{X}$ is proportional to the atomic number squared).

When you get to non-hydrogenic atoms you need to model inter-electron repulsion as well. This can be done by introducing an empirical parameter $\delta$, the quantum defect: $E_\text{B} = -\dfrac{R_{X}hc}{(n-\delta_X)^2}$. For sodium the quantum defect is 1.37, allowing good approximation of the transition energies between n-states in the electronic spectrum.

I don't think there is a similar back-of-the-envelope method to calculate the differences in energies between multi-electron atom l-energy levels (like the 3s-3p). The theory of angular momentum tends to mathematical complexity. There are likely good computationally derived approximations though.


Can we combine two reactions and then calculate the equilibrium concentrations?

Karsten's insight is correct. Your mistake is in assuming that you have either $\ce{N2O5}$ or $\ce{N2O}$ and $\ce{O2}$.

There are four concentrations of interest: $\ce{[N2O5]}$, $\ce{[N2O3]}$, $\ce{[N2O]}$, and $\ce{[O2]}$.

How can you relate these variables?

Well, you can apply conservation of matter in terms of oxygen and nitrogen.

For nitrogen:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$

For oxygen:

$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2 \ce{[O2]} = 5\times(4\ \mathrm{M})\tag{2}$$

Then, there are the equilibrium expressions:

$$K_{1}=4.5=\frac{\ce{[N2O3][O2]}}{\ce{[N2O5]}}\tag{3}$$

$$K_{2}=\frac{\ce{[N2O][O2]}}{\ce{[N2O3]}}\tag{4}$$

If you substitute $\ce{[O2]} = 4.5\ \mathrm{M}$ into (2), (3), and (4), you end up with:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$
$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2\times(4.5\ \mathrm{M}) = 5\times(4\ \mathrm{M})\tag{2'}$$
$$4.5=\frac{\ce{[N2O3]\times(4.5\ \mathrm{M})}}{\ce{[N2O5]}}\tag{3'}$$
$$K_{2}=\frac{\ce{[N2O]\times(4.5\ \mathrm{M})}}{\ce{[N2O3]}}\tag{4'}$$

That's 4 equations and 4 unknowns, and it's all algebra from there.

Assuming I did that correctly, I got $$K_{2} = \frac{45}{7}$$


In an oxime, the syn or anti configuration is determined by the position of the lone pair on nitrogen with respect to the alkyl or aryl group. That is, syn configuration is one in which the lone pair on nitrogen and the alkyl or aryl group are in the same side, while, in anti configuration, they are on opposite sides.


Why does bromine react with iron(II) but not iodine?

All four of the ions are reducing agents. But usually halogen gases' tendency to undergo reduction to become halide ions decreases down the group (along with their reactivity). Out of the $4$ ions, only $\ce{Fe^{2+}}$ has a mild enough reducing power.

$$\ce{2Fe^{2+} +Br2 -> 2Br^- +2Fe^{3+}}$$

But $\ce{Fe^{2+}}$ is not strong enough to reduce $\ce{I2}$ as it is relatively stable.

In fact, $\ce{SO3^{2-}}$ has the strongest reducing power among the four options, as it can reduce both $\ce{I2}$ and $\ce{Br2}$.

$\ce{OH^{-}}$ is a mildly strong reducing agent, it can react with both $\ce{I2}$ and $\ce{Br2}$, and its typical reaction is

$$\ce{4OH^- -> 2H2O +O2 + 4e^-}$$

$\ce{Mn^{2+}}$ is the least reducing, because it is the product of a strong oxidizing agent that has been reduced:

$$\ce{MnO4^{-} + 8H+ +5e^- -> Mn^{2+} +4H2O}$$

Here, permanganate ion $\left(\ce{MnO4^-}\right)$ is a very strong oxidizing agent; the product thus would be weakly reducing.

Actually, a similar idea can be seen with $\ce{SO3^{2-}}$ (sulphite) ions. Since it is a strong reducing agent, we can expect the product formed after it is oxidized is a weak oxidizing agent.

$$\ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e^-}$$

Indeed, sulphate ions alone are weak oxidizers, and stable.

I suggest you refer to E.C.S. (electrochemical series) to know whether a reaction would occur. It is easy to find online; e.g., at Wikipedia.


For a long time it was taught in school and universities that the HOMO of carbon monoxide is anti-bonding. Without more context it was also often taught that the bond order in CO is three, since there are eight electrons in bonding orbitals and two in anti-bonding orbitals. $$\text{Bond order} = \frac12(\text{bonding} - \text{anti-bonding})$$
By assuming that the HOMO is anti-bonding (it is not!) and taking away one electron, the  bond order has to increase to 3.5. This is wrong.

When we have a look at the MO diagram, a calculated version can be found here, we know that the HOMO, i.e. 3σ, is a bonding orbital, while the anti-bonding orbital is the 2σ. Upon ionisation, we would indeed remove one bonding electron and therefore the bond order has to decrease to 2.5 as you suggested.
However, it is not that easy. Strictly speaking the below MO scheme is, as well as MO theory itself, an approximation, and only one possible configuration. While we do not have to use resonance structures with MO theory, we have to consider other configurations (analogous to excited states). So naturally the bond order of CO is not strictly 3. And removing an electron does not mean we are removing it from only one orbital, rather than decreasing the electron density. Therefore we cannot accurately predict the bond order with these simple considerations.
Experimental observations and theoretical calculations suggest that the bond indeed becomes stronger when removing an electron. See the linked question and Philipp's answer within for more detail. (Don't look at the other answers, they are as wrong as they could be.)

In short: The bond order of $\ce{CO}$ is not exactly 3 and removing an electron will not increase the bond order to 3.5. In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in $\ce{CO+}$.

An orbital with bonding character has no node perpendicular to the bond axis; an orbital with anti-bonding character has at least one node perpendicular to the bond axis (electron density is zero). Strictly speaking there are no non-bonding orbitals.


Is enthalpy the same thing as potential energy?

In any way, ignoring the definitions given in a textbook is a bad idea. As for the question, the main problem in your thoughts is that you mixed too many things from different fields of science. Your example (two particles, kinetic energy, potential energy etc) looks like classical mechanics. So if you want to explain it, please, use tools offered by classical mechanics. Enthalpy is a measure of total energy in thermodynamics. It is used to explain processes taking place in large systems, which can be characterized with volume, pressure, temperature etc. It is not used to describe one particle.
Maybe you can take it into account and revise your question. Make it clear exactly what you want to understand: change in the kinetic and potential energy in interactions of two particles or change in enthalpy in exothermic and endothermic reactions.


As you state, hexane has a low boiling point. If the bubbles are hexane vapor, as opposed to an air leak, then cool the fluid and apparatus more. You might pre-chill the hexane in a dry ice bath. The sublimation temperature of $\ce{CO2}$ at 1 atmosphere is about 195 K, and the melting point of a mix of hexane's various isomers is about 179 K, so there should be no problem with crystals blocking the flow.

Be careful with the cold -- frostbite is not painful at first, but is as damaging as a burn!


First of all, the compound is not M(aa)2 type.M(aa)2 shows no kind of isomerism (with normal ligands). Since the donor atoms on a ligand are different, it would be M(ab)2. And square planar structures show only geometrical isomerism and not optical isomers, as you think. Why?

Optical isomerism is when the compounds can have non super-imposable mirror image like structure. They should not have a plane (or axis) of symmetry. However, in a square complex, it is not possible. Since it will always have a plane of symmetry or a mirror image which can be superimposed on the previous compound, even if it is of type  M(abcd) (Try this out for yourself)

But, you are right, carbon on your compound is chiral, so it can have different configuration of methyl and hydrogen groups in space so as to remove that plane of symmetry factor and make them optical isomers.

Geometrical isomers, however are variations of a compound about the central metal atom atom or ring or a double bond, in which the arrangement of groups is different. So square planar complexes can show geometrical isomers. The one is your question can be depicted as -

If you include the configuration of the chiral carbon of the ligand, there can be a few more structures made.


What is the role of the slash ("/") in chemical reactions?

The ambiguity seems to come not so much from Chemistry, but from English punctuation. Slashes are often used in place of "and", "or", or the latin "cum" (meaning combined with). It seems that it unfortunately is just context dependent, though I would argue its use as a replacement for "or" is most consistent grammatically and defining this to be the meaning of the slash would eliminate ambiguity in these situations.


Why is it that all strong bases contain at least one hydroxide bond?

All strong bases you know contain at least one hydroxyl group because we live in a world dominated by water, almost like in that old movie with Kevin Costner. Water chemistry is of paramount importance to us, and acid/base reactions in water pretty often boil down to $\ce{H+ + OH- ->H2O}$.

There is nothing special about the hydroxyl group that lets it dissociate. Come to think of it, the group itself does not even dissociate in bases - I mean, $\ce{O}$ and $\ce{H}$ remain together, the bond between them does not break. (It does in acids, but that's another story.) If you mean the dissociation of metal hydroxides, there is nothing special here either. $\ce{NaOH}$ dissociates like any water-soluble ionic compound. On the other hand, $\ce{Fe(OH)3}$ refuses to dissociate, like any water-insoluble ionic compound, of which there are plenty.

In a wider chemical world outside, there are bases which are many orders of magnitude stronger than that, and they don't contain any hydroxyl groups whatsoever. $\ce{NaNH2}$, $\ce{NaH}$, $\ce{BuLi}$ come to mind. You don't see them much in everyday life, though, because they don't survive in our world. Water act as an acid on them, and instantly transforms them to mere $\ce{NaOH}$, etc.

So it goes.


Velocity of electrons ejected from Cl atom

It would be the $3s$ shell because less energy is required to bump it off course. Due to electron repulsion and the shielding effect, the $e-$ in the $3s$ is more prone to get dislodged. And of course because of this, it will have a higher velocity when it gets bumped off. Another concept to note: $e-$ will have more energy at higher energy levels.

A reason why it wouldn't be the $1s$ shell is because due to the strong pull on the $1s$ electron. Since it is on the first energy level, it is strongly held by the nucleus. Even it were to get dislodged it would move very slow because of the strong pull by the nucleus.


Chemical to remove nail polish but retain ink

It would depend mainly on the chemical composition of the ink (permanent marker? standard ballpoint?)

If it was signed with standard ballpoint, you have a problem, because the main film forming / thickening agent used for ballpoint pen ink is the same used in the majority of nail polishes - nitrocellulose. This means that a solvent that removes nail polish (ethyl acetate, for example) will also remove the ink.


Why do I have to use a 1:1 ethanol:water solution to hydrolyse aspirin?

Hydrolysis means to break something apart (e.g. chemical bonds) with water.

If there's no water, there's no hydrolysis.


ESI & MALDI Mass Spectrometry: particle mass vs. measurement error

The answer to your question is not an easy one, as multiple parameters need to be taken into account to give an answer. At the high masses that you quote, most of the litterature trends come to their limits in terms of extrapolation, as the major mass ranges in mass spectrometry covers an $m/z$ range 10 - 4 000.

As you indicate in your question, one should separate the question into two parts: how does mass accuracy change with high $m/z$ values and how does mass accuracy change with high mass values, considering a high number of charges, as can be achieved in ESI.

Dependence with ion charges

I will first tackle the second part of the question, because this will be needed to answer the first part. If one considers a mass analyzer in which measurement accuracy ($\Delta m/z$) is directly proportional to the measured $m/z$ result, we have:

$$\frac{\Delta m/z}{m/z} = Constant = \frac{\Delta m}{m}$$

Therefore for such a mass analyzer there is no direct advantage in using multiply charged ions (as produced in ESI) compared to singly charged ions as produced in MALDI. From this point, it comes that on this criterium alone, one should consider if measurement accuracy increases faster or slower than a proportional increase to decide whether multiply charged ions come as an advantage or a disadvantage.

Dependence with mass analyzers

Let us now consider a series of classic mass analyzers and consider how they would perform in terms of high mass measurement.

Quadrupole filters and quadrupole ion traps

I group these in the same subtopic, as when one is concerned about high masses they have fairly similar properties. Commercial instruments are usually limited to $m/z$ 4,000 (or 32,000 for the specially devised systems). So ESI is a requirement for high mass species.

Quadrupole mass filters are usually operated in the constant mass width mode ($\Delta m/z = C$, with $C$ on the order of 0.5 to 1.3 amu), which means that the mass error will be roughly only dependent on $z$ and that there is an advantage to work on the high end of the $m/z$ range, as the overall error will be lower than on the low end. One point worth notice though: if you consider a 2 MDa system at $m/z$ 1,000, it would mean $z = 2000$. Thus spacing between neighbouring charge states is of only 0.5, lower than quadrupole resolution, which means that a deconvolution of the charge state distribution will not be doable and that is will be globally not possible to determine the mass of the molecule with this technique.

Time-of-flight analyzers

These are probably the mostly used analyzers for high mass compounds, both in MALDI and ESI modes. In an ideal TOF, the mass accuracy would depend only on the digitizer (and thus scale as $1/\sqrt{m/z}$), but it also depends on the applied potentials and other factors leading to a mass accuracy that scales as $\sqrt{m/z}$. This still means that considering the above points, mass accuracy will be better achieved at low $z$ values. The mass range limit for TOF analyzers is in theory infinite, but it is in fact limited by the detector (see below).

Orbitrap mass analyzers

Technically an Orbitrap mass analyzer could benefit from working at high $m$, low $z$ conditions, as mass accuracy in an Orbitrap scales as $\sqrt(m/z)$. The instrument is nevertheless limited by the possibility to inject ions within the trap, which is currently limited to 4,000 (or maybe 8,000, I did not check the most recent developments in the field). Thus ESI is required for such high mass systems, but the highest mass accuracies would be achieved on the high end of the achievable mass range.

FT-ICR mass analyzers

In FT-ICR mass analyzers, mass accuracy scales as $m/z$. Thus there is no effect of the charge on mass accuracy. Technically, FT-ICR instruments always include a quadrupole (trap or filter) between the ion source and the FT-ICR mass analyzer and the $m/z$ range is thus limited by the range of this filter (4,000 - 8,000).

Other issues of high mass measurements

One of the difficulties that arises when using high mass ions is also ion detection: devices using a secondary electron emission (such as MCP's or conversion dynode) have efficiencies that depend on the linear velocity of the impacting ion. Therefore the standard detectors are not well adapted to high $m/z$ ions. Some detectors have been developped to circumvent this limitation, but this remains an issue for masses in the 2 - 10 MDa range that you quote in your question.

One other issue is ion desolvation: as size increases, adducts (with salt, solvent or matrix) tend to increase also, leading to changes in the total mass of the ions and a broadening of the peaks. Mass accuracy will also be reduced because there is no direct rule for removing the masses of these adducts from the measured mass to lead to the mass of the molecule.


Purification of water with ferric chloride

If I'd state that the pH drops, would you know why?

If you dissolve $\ce{FeCl3}$ in water, a hexaqua complex is formed. These complexes are known as cation acids:

$$\ce{[Fe(H2O)6]^{3+} + H2O -> [Fe(OH)(H2O)5]^{3+} + H3O+}$$


What topics to review before starting organic chemistry?

As @inɒzɘmɒЯ.A.M pointed out in his comment, the usual prerequisite for an organic chemistry course is that a general chemistry course is already taken and successfully passed. The general chemistry is, of course, a rather large topic, so I'll just name a few subtopics which I feel are absolutely essential to start a venture into the world of organic chemistry.

Structure of matter.

Modern atomic theory (Schrödinger equation, atomic orbitals, spin,  Pauli exclusion principle, electron configurations).
The periodic table (periodicity, group trends).
Chemical bonding (covalent and ionic bonding, Lewis structures, resonance, VSEPR theory , introductory Valence Bond Theory and Molecular Orbital Theory).

Chemical reactions.

Nomenclature & stoichiometry (empirical and molecular formulas, chemical equations, mole, limiting reagents, yield).
Solutions (solubility, concentration, properties of solutions).
Acid-base reactions (definitions of acids and bases, pH, strong/weak acids and bases, trends).
Oxidation-reduction reactions (oxidation numbers, balancing oxidation-reduction reactions).

Thermodynamics.

Basic concepts (energy, work, heat, enthalpy, entropy, Gibbs free energy ).
The first and the second laws of thermodynamics.
Hess's law.

Kinetics.

Reaction rates (temperature dependence, Arrhenius equation).
Reaction orders (first and second order reactions, molecularity, rate laws).
Catalysis.

Chemical equilibrium.

Basic principles (the notion of equilibrium, the law of mass action, equilibrium constants, the direction of a reaction).
Shifting the equilibrium (changing amounts, pressure, volume, temperature).


What is the mechanism by which lactones are hydrolyzed during a Wolff-Kishner reaction?

This is a general feature of the hydrazine reagent that is employed in Wolff-Kishner reductions. When attacking an ester’s (includes lactone’s) carbonyl function, the generated tetrahedric intermediate can break down to form a hydrazide as shown in the scheme below.

The carbonyl function is not reduced but the ester (or lactone) is hydrazinolysed.

One could assume that the displace alcoholate could reattack the carbonyl group to regenerate the ester/lactone. However, apart from hydrazine being much more nucleophilic than alcoholates, the resulting hydrazide is also stabilised much like an amide, and is therefore less reactive.


How often is the IUPAC Goldbook Updated?

As various users have contributed to the arrival of this answer, it goes to the  community.

Firstly, the IUPAC is the authority of chemistry, which decides on and accepts terminology for international compatibility so that information can be unambiguously expressed by researchers, scholars, educators and students. So, it can be finally concluded that unless it is IUPAC approved, it is not official terminology.

Secondly, since the IUPAC is an enormous organisation, dealing at a global level, it can be understood that changes will be a little delayed as the information has to be checked for authenticity and a common end has to be reached. For example, it took quite some time to decide on the newcomers of the Periodic Table(Elements of Atomic Number 113-118).

Thirdly, a lot of terminology has a very limited scope and can be at the very least left out. For example, the iupac tag was dropped due to its limited scope.(See comments above).

Lastly, I thank the community for the comments which have fabricated the "bonds" of this answer.


Carbocation Rearrangements in SN1 reactions

I think the solvolysis would involve rearrangement to the cyclopentyl carbocation as indicated in the figure below.  The driving force for the rearrangement is two-fold

without rearrangement, the cyclobutyl compound would generate a secondary carbocation; with rearrangement a more stable tertiary carbocation is generated instead
release of strain as the 4-membered ring opens to a 5-membered ring

The tertiary cyclopentyl carbocation ultimately captures a solvent molecule as indicated.


I'll be using the brief guide to the nomenclature of Organic Chemistry as a reference, because I could not gain access to the blue book. This source suffices for this nomenclature example.

Seniority of parent compounds
The systematic name is based on the name of the senior parent compound, which is chosen by applying the following criteria in the order described below and shown in Fig. 1, until a decision is reached. For a complete set of criteria see Ref. [8].

In the given example, both the chains contain the senior most group, which is the hydroxyl (alcohol) group. The smaller chain, however, contains 2 hydroxyl groups as compared to the larger chain, which contains one hydroxyl and one alkylhydroxyl group (which is a lower priority). Since this is the first point of difference, The smaller chain would be the parent compound.
After identifying the parent chain, it is easy to name the locants such that the seniormost groups get lower locants; As you have written, the name would finally be 2-ethylhexane-1,3-diol.
This image is possibly more detailed than this question requires, but the image and the linked reference are good resources, should you need more information on naming tricky compounds.


I can't tell you for sure what is happening, but I can tell you what could be happening.
Physical Adsorption
It is possible that the proteins are not chemically adsorbed, but rather physically adsorbed. The article you cited implies that this is the case.
Many proteins in the presence of nano-scale hydrophobic "features" will partially denature, and the hydrophobic core will physically adsorb (attach via van der Waals forces). This effect is amplified if water is present, since the hydrophilic parts of the protein will be stabilized in water, allowing more of the hydrophobic parts to attach to the carbon feature.
Now you are probably wondering, "Why doesn't it attach to the PEG surface as well, since that is hydrophobic?" There are two possible reasons.
PEG is not as hydrophobic as you think

The PEG chains have oxygen in each monomer, which would reduce the non-polar character of the chains. It is possible that the difference is enough to make adsorption of non-polar amino acids on PEG thermodynamically unfavorable, while on pure-carbon structures it would be favorable.
Shape matters
Another possibility is that the actual shape of the features makes the difference. In order for a protein to denature and "attach" to carbon feature via hydrophobic interactions, it has to first expose the hydrophobic groups in the core. For a flat surface, the amount of initial denaturization (maybe not a real word) would need to be higher than for a "sharp" feature. Sort of like how a needle can slip into your skin without doing much damage. So, the smaller the characteristic length of the features, the more likely the protein would adsorb. There are also protein-protein effects on the surface, which are again reduced by smaller-scale features. In this scenario, you could think of the proteins as kinetically trapped in solution when near a regular PEG surface, while the activation energy for adsorption to a small feature is smaller.
I can think of two ways to test this idea:

Use different proteins on the same type of surface features - the ones with less exposed hydrophobic groups should adsorb less easily
Use different sizes of features - there should be a feature size that gives a maximum adsorption for a given protein.

Chemical Bonding
The other possibility is chemical bonding between the carbon features and the protein. It is unlikely that this will happen if the carbon features are really pure carbon (graphene, for example). However, if they are creating them by ionizing PEG with an electron beam, then it is highly likely that they are functionalized with something. Carbonyl groups would be my guess, but depending on the conditions there could be nitro groups as well. In this case, there is a possibility that hydrogen bonds would form, and if the surface is reactive enough, chemical bonds might form as well.
To test this idea, I would do the same thing as before - only this time, if we are right:

Proteins with more exposed hydrophobic groups should be less likely to adsorb
Feature size should not matter at all

If this turned out to be the case, you could then systematically test polypeptides made up of different amino acids to verify it and determine whether hydrogen bonding or covalent bonding was more likely.


Comparison of rate of dehydration of primary, secondary & tertiary alcohols on cyclohexane

The rate of dehydration is related to the ease of formation of the carbocation and the energy of the carbocation intermediate.

The ease of formation of carbocation is tertiary>secondary>primary. That is the ease of abstraction of $\ce{OH2+}$ COULD BE be R>Q/S>P.

Let's look at a comparison of the energy of the carbocation intermediates in all these cases.

  A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation $\ce{CH2=CH–CH2+}$ and benzyl cation $\ce{C6H5–CH2+}$ are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the $\pi$ bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge - and, therefore, stabilizes the carbocation.

Let's look at them individually,

P - allylic cation, primary alcohol
Q - allylic cation, secondary alcohol
R - allylic cation, tertiary alcohol
S - NOT an allylic cation, secondary alcohol

$\ce{R}$ is clearly the winner.

Now we may conclude that the charge dispersal is better in resonance than it is in hyperconjugation (explanation)

Here are the 3D diagrams of the P carbocation. Hydrogen cation doesn't look like it has to face significant steric hindrance to perform protonation of the alcohol.


Mechanism for formation of Quinol from a Phenolic Ether

Here is a mechanism for this enol ether hydrolysis.


Ballistic behavior of molecules on potential energy surfaces

I think the description you lay out in your question has it basically right.

MEPs, IRCs, etc. all assume that the geometric rearrangements that occur over the course of a reaction strictly follow the gradient uphill from reactants to transition state, and then downhill from TS to products (or intermediates).  Energetically, this provides the "path of least resistance."

In the paper you cited, it looks like the MEP would require some atoms in the system to "take a right-angle turn." Quoting a broader excerpt around the first sentence you include (emphasis added):

  As can be seen in Figure 3, after TS‘ S$_0$ the MEP takes a turn on the PES:  the O−O‘ distance increasing and the torsion around the O−C−C‘−O‘ dihedron nearly stopping, while the increase of the C−C‘ distance comes into action.
  ...
  At this point we have to remember that the computed MEP path represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic.
  ...
  What would be the consequences if the reaction continued along the torsional mode? Section 3.2 already suggested that this is of interest—the torsional mode is the principal internal coordinate of the T$_1$ MEP.

Nuclei have mass (obviously), and thus have inertia. While Newton's relevance at the atomic scale can be fuzzy at times, classical treatment of atomic motion has sufficient validity that acceleration of nuclei can be considered to require a suitable accompanying force (viz., PES gradient).  If the system has enough "inertia" and the PES gradient provides too small an acceleration to overcome it, the progress of the reaction can absolutely move away from the MEP.

Alternatively, at least one additional mechanism can introduce  deviations from the MEP: molecules at finite temperature are always vibrating. Thus, even in a case where the key internal molecular coordinates (bond lengths, angles, dihedrals) for a given reaction strictly follow the IRC, the "uninvolved" vibrational motion is still occurring, and so the  system is actually always oscillating around the IRC in a high-dimensional sense.  If the energy of the system is high enough, these oscillations have the potential to be of sufficient magnitude to "bump" the reaction coordinate away from the gradient-following pathway.

Rotational motion may also contribute to this, but I would think in a more subtle way than vibrational motion.

For further reading: Steven Bachrach of the Computational Organic Chemistry blog has an entire tagged category, "Dynamics," dedicated to this phenomenon.  In no particular order, some posts of possible survey value include:

An approach towards identifying dynamic effect without trajectories
Insights into dynamic effects
Bifurcating organic reactions

The entirety of Chapter 8 of his book is also dedicated to this topic.

Addendum: FWIW, as a chemical engineer, the (imperfect) analogy that I like to use is to liken the MEP to the 'reversible process' of thermodynamics:

Start at the beginning state
Take an infinitesimal step
Let the system infinitesimally relax
Repeat 2 & 3 until you reach the end state

As I (very approximately) understand it, irreversibility comes from changing the state of a system "too much, too fast", leading to net generation of entropy as the system relaxes back to the reversible pathway, and it's "kinetic energy" in a qualitative, generalized sense that allows the system to depart from the reversible path.


Why wouldn't you consider the [H+] produce from the weaker acid in a mixture of two acid?

You don't have to consider the system's sub-components separately, but it makes the algebra much easier.

If pKa of $\ce{HF}$ is 3.14 and pKa of $\ce{PhOH}$ is 9.95 and the problem gives you concentrations to ONE SIGNIFICANT figure, it isn't clear to me why you think the difference between .00072 and .00072000011 is worth worrying about...It can be ignored, clearly. It's not a question of the weaker acid, its a question of whether the values are close enough so that the smaller is significant with regard to the larger. In this case, clearly not. $\ce{PhOH -> PhO- + H+}$ is the chemical equation for the 2nd part, so

$$
\ce{\frac{[PhO][H]}{[PhOH]}} = K_a
$$

is the equation you derive from it (ignoring difference between concentration and activity, and ignoring water).

The fact that $K_a$ for this is $10^{-10}$ and that $\ce{[H+]}$  is on the order of 0.0007 means you get $\frac{X(0.0007)}{(2-X)}=10^{-10}$ which if you have any done a minimal amount of algebra practice/drill immediately suggests that X is going to be on the order of 3E-7 and again is there really a difference between 2 and 2-0.0000003? You'll note I'm not asking the slightly different question about whether the difference between 2.0000000 and 2.0000000-0.0000003 = 1.9999997; 2 is not the same as 2.0000000 when significant figures are in play.


Why does Freundlich adsorption isotherm not have a term for surface area?

The ratio

$$\frac{x}{m}= k P ^{1/n}$$

can be re-expressed in terms of the area $A$, surface-to-volume ratio $\lambda$, and density $\rho$ (note that $m=A\rho/\lambda)$:

$$\frac{x}{A}= k' P ^{1/n}$$

where

$$k'=k\rho/\lambda$$

The advantage of using $k$ as opposed to $k'$ is that you don't need to determine $\lambda$ to know how much (as in, what mass) adsorbent to use in a particular application (see e.g. Table 4 in ref 1).

There are two types of adsorbent material you are usually interested in:

Porous material; in this case, breaking up the particles will not significantly increase $A$, and therefore not change k (since $\lambda$ will remain approx constant).
Small particles with high $A/V$ ratio and adsorption on an outer surface of the particle. Now breaking up particles into smaller ones matters a lot, but this would be reflected in a different value of $k$, since $\lambda$ increases and $k\propto\lambda$.

Parameter $k'$ is more fundamental than $k$, as it reflects how much surface area the adsorbed molecules require, keeping in mind the range of applicability of the equation, as it does not predict e.g. monolayer saturation. Note that in neither scenario would a reduction in the size of the particles be expected to change $k'$ by much - certainly not in the first case, and, in the second case, depending on the importance of changes in adsorbent curvature or edge/corner effects.

Reference

G. P. Jeppu, T. P. Clement, "A modified Langmuir-Freundlich isotherm model for simulating pH-dependent adsorption effects," Journal of Contaminant Hydrology 2012, 129–130, 46–53 (DOI: 10.1016/j.jconhyd.2011.12.001).


Typography of atomic orbital subscripts p_x

The subscripts that specify orbital axes are written in italic. This is mentioned in ACS style guide.$^{[1]}$ So for instance

$$\mathrm{d}_{x^2 - y^2}.$$

Be careful though. Other subscripts for orbitals are generally upright, e.g.,

$$\mathrm{t_{2g}}.$$

$[1]$ Anne M. Coghill, Lorrin R. Garson. ($2006$). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online). (p 256)


Numerical on change in degree of dissociation wrt pressure

The dissociation of $\ce{PCl5}$ is given by the equation:

$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$

with $\displaystyle K_\text{p} = \frac{p_{\ce{Cl2}}p_{\ce{PCl3}}}{p_{\ce{PCl5}}}$. Putting this in terms of total pressure, $\displaystyle K_\text{p} = \frac{x_{\ce{Cl2}}x_{\ce{PCl3}}}{x_{\ce{PCl5}}}p_\text{T}$.

In terms of the degree of dissociation, $\alpha$, assuming an initial amount $y$ of moles of $\ce{PCl5}$, at equilibrium, we have $n_{\ce{PCl5}} = y - \alpha y\,\text{mol}$, $n_{\ce{PCl3}} = \alpha y\,\text{mol}$, and $n_{\ce{Cl2}} =  \alpha y\,\text{mol}$. Thus, their molar fractions are $x_{\ce{PCl5}} = \frac{1 - \alpha}{1 + \alpha}$, $x_{\ce{PCl3}} = \frac{\alpha}{1 + \alpha}$, and $x_{\ce{Cl2}} =  \frac{\alpha}{1 + \alpha}$, so:

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

Since at $p_\text{T} = \pu{2 atm}$, $\alpha = 0.4$, we get $K_\text{p} = \pu{0.381 atm}$. Hence, for $\alpha = 0.6$, since $K_\text{p}$ doesn't change (i.e. assuming it's an isothermic system), we get $p_\text{T} = \pu{0.677 atm}$. This is in agreement with Le Chatelier's principle: reducing the total pressure will push the equilibrium towards the side with more gaseous moles, so the dissociation constant in this case will be higher.


Dipole Moment of Normal Water vs Heavy Water

There is no measurable difference.

HOH 1.8546 ± 0.0006 D

DOD 1.8558 ± 0.0021 D

Dipole moment of water from Stark measurements of H2O, HDO, and D2O J. Chem. Phys. 59, 2254 (1973)


What reaction is happening when a lithium polymer battery "is on fire"

I'll take a stab at this one, but I'm not a battery expert.

What is a lithium "polymer" battery?

Wikipedia says that "lithium polymer battery" is now commonly used to refer to nearly any lithium ion battery that is encased in a flexible polymer shell instead of a hard metal casing.  There is a chance for confusion here because it apparently used to mean a battery with a polymeric electrolyte instead of a purely small-molecule liquid-phase electrolyte.  I'm assuming you mean the former "in a polymer shell" definition for the purposes of my answer.

What type of anode, cathode, and electrolyte materials?

There are many types of lithium ion batteries.  Electrolyte composition, cathode materials, and anode materials can all vary.  Perhaps the most common is a lithium cobalt oxide cathode with a intercalated graphite anode.  Organic solvents must be used as electrolyte because water is not stable to the high voltages that lithium ion cells provide.  Solvents stable over a wider potential range must be used instead to dissolve salts that can carry electric current.

What happens when it's crushed:

So for lithium cobalt oxide batteries with liquid electrolyte, I was able to find a good source describing the problem of "thermal runaway":

  The temperature of the cell rises due to the
  chemical reactions between the organic solvent and electrode materials leading to cell failure/explosion. The chemical reaction is prompted because of the coexistence of 1)
  combustible organic solvent with inorganic salt electrolyte 2) lithiated graphite anode and 3)
  partially delithiated LixCoO2 cathode in the charged state. The graphite anode intercalate
  fairly well during the initial cycles. As the cycle proceeds, the available intercalating sites in
  the graphite slowly decreases and the lithium deposition over the surface of the anode
  increases during charging process. Such a situation prompts non-even lithium deposition or
  even dendrite growth over the graphite surface. When the battery is in the charged state the
  cathode remains in a delithiated state. Flow of current over and above the
  tolerable/standard charged state causes instability to the cathode which starts to release
  oxygen into the electrolyte. Thus released oxygen reacts exothermally with the lithium
  plated over the graphite anode and increases the temperature of the cell making the cathode
  to release oxygen further. The cell which was in the dormant state in the absence of oxygen
  and heat now becomes an explosive device in the event the cell is met with any abuse.

That describes what happens when the battery is overcharged and leaks oxygen into the electrolyte.  I suspect that when the batteries are crushed, even if they aren't overcharged, that at least two things can happen leading to similar thermal runaway: First, the lithiated graphite is air- and water- reactive, heating the now-crushed battery and eventually lighting the electrolyte solvent on fire.  Second, the cell may be shorted by the mechanical disruption, so that even without air or water contact, the voltage in the battery can discharge very quickly.  This would create large heating from ohmic resistance, also eventually igniting the elctrolyte.


When a Calcium atom loses two electrons, it attains the stable electronic configuration of the noble gas Argon. Thus, Calcium cannot have a $\ce{+3}$ charge in any ionic compound. So, the ions forming your salt are $\ce{Ca^2+}$ and $\ce{PO4^3-}$. We need to balance the charges to obtain a neutral salt. We can see that the least common multiple of the charges (2 and 3) is 6. A total charge of $\ce{+6}$ can be obtained by taking three Calcium cations, and this can be balanced by taking 2 Phosphate anions.

Calcium Phosphate is $\ce{Ca3(PO4)2}$


When finding hybridisation of atoms, why are double bonds and triple bonds only considered once?

The hybridised orbitals are almost always involved in $\sigma$-bonding, and in housing the lone pair(s) of electrons. This helps us determine the geometry of the molecule.

Consider acetylene, $\ce{C_2H_2}$.

The carbon is $\mathrm{sp}$-hybridised. So, it has 2 $\sigma$-bonds - one to a hydrogen, one to another carbon. All this is in the third shell. So, a total of 4 orbitals (1 $\mathrm{s}$, 3 $\mathrm{p}$) undergo hybridisation and form 2 $\mathrm{sp}$-orbitals and 2 $\mathrm{p}$-orbitals. The $\pi$-bonding occurs due to these unhybridised $\mathrm{p}$-orbitals.

Hope this clears your doubt!

PS: The only exception to the first statement that I know is https://en.wikipedia.org/wiki/Aryne.


What is the "resonance hybrid" of benzene?

It means that what we usually consider to be the 'structure' of benzene is not the actual structure.

Although there are three π bonds in the structure of benzene, there is actually resonance, or electron delocalization. This means that the electrons are not localised in π bonds between two specific carbons, but distributed throughout the ring. This is one of the most important characteristics of an aromatic system.

The "localised" structures 1a and 1b above are resonance forms of benzene: neither 1a nor 1b are a true representation of benzene on their own; they are hypothetical, or "theoretical", structures.

The resonance hybrid is a weighted average of the resonance forms 1a and 1b. In this case, both forms contribute equally to the hybrid. So, each C–C bond has a bond order of (1+2)/2 = 1.5, which we can represent using dotted lines in 2 (each dotted line represents a fraction of a bond, so one solid and one dotted line means a bond order between 1 and 2).

More symbolically, we can also use the depiction with an inscribed circle 3. This means that the six π-electrons are shared between all six of the carbon atoms.


Moeller's Inorganic Chemistry textbook from 1952 has a fairly long discussion of different theories of acids, citing a lot of primary literature. For Lewis acids, it cites some reviews by W. F. Luder from the 1940's and some of Lewis' original work. Specifically, it says that Lewis in a paper in 1938 (J Franklin Inst. 226:293) described "phenomonological criteria" for Lewis acids and bases, which include the ability to displace weaker Lewis acids and bases.

My impression from Moeller's description of Lewis' work is thus that he defines an acid as a molecule in which one atom has an incomplete valence shell of electrons. He further views a Bronsted acid as a complex between a Lewis acid ($\ce{H+}$) and a Lewis base (the conjugate base). In a reaction like $\ce{NH3 + AcOH -> NH4+ + AcO-}$, he would say that the stronger Lewis base $\ce{NH3}$ is displacing the weaker Lewis base acetate from the Lewis acid $\ce{H+}$.

If I have time, I'll try to dig up some of those primary sources. Here's a quick list if anyone else wants to go after them:

G. N. Lewis; Valence and the Structure of Atoms and Molecules. The chemical catalog co. (1923)
G. N. Lewis; J Franklin Inst 226:293 (1938)
W. F. Luder; Chem Rev 27:547 (1940)
W. F. Luder and S. Zuffanti; The Electronic Theory of Acids and Bases, Wiley and Sons (1946)
W. F. Luder, W. S. McGuire and S. Zuffanti; J. Chem. Ed. 20:344 (1943)
W. F. Luder; J. Chem. Ed. 22:301 (1945)

UPDATE: The original 1923 Lewis text is available for free from Google Books here:https://books.google.com/books?id=2hxDAAAAIAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

The most relevant passage is from p.142:


Some electrolysis experimets are best first performed and then an attempt  at explanation.
For example, the products of the electrolysis of aqueous nickel nitrate with graphite electrodes are reportedly per a source (with video) as:

Electrolysis of a nickel nitrate solution produces oxygen at the anode, and hydrogen and nickel at the cathode.

Looking at related metal electrolysis, for example, Cu(NO3)2, per this source with again graphite electrodes:

Electrolysis of a copper(II) nitrate solution produces oxygen at the anode and copper at the cathode.

Lastly, with the transition metal manganese, a source on the electrolysis of Mn(NO3)2, reports, to quote:

$\ce{Mn(NO3)2 + 2H2O ± 2e- -> H2 + MnO2 + 2HNO3}$

Based on all of the above, and that hydrogen gas is forming per the reactions:
$\ce{H+ + e- -> .H }$
$\ce{.H + .H  -> H2 (g)}$
I would expect similarly some transient nitric acid formation with Ni(NO3)2 per its potential interaction between the nitrate ion and the hydrogen atom radical as follows:
$\ce{ NO3-  + .H  -> .NO2  + OH-}$
$\ce{.NO2 + .NO2 <=> N2O4 }$
$\ce{N2O4 + H2O -> HNO2 + HNO3 }$
Note, my claimed introduction of HNO3 (and its likely subsequent removal by nickel metal) can explain the comment "the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure", and is also consistent with the electrolysis of Mn(NO3)2 resulting in the creation of nitric acid.
And lastly, you are correct, oxygen is produced at the anode.


The detergent 'Positive 170' and what exactly makes it so corrosive?

Hint: read the column in the centre.  The potential risks in damage of tissue, described on the left and on the right hand column increase by increase of the solution's concentration, and temperature (and 170 F [about 76 C] is considerable), and obviously by the time of exposure to this cleaner.

As an example of what may happen to skin under attack of strongly basic solutions (sometimes called lye) if you do not wear appropriate protective gear, see this example. -- Be told, such pictures may disturb you.


The chemistry you have posted is for non-hydraulic cement, which requires air drying for precisely the reasons you enumerated.  The kind of cement that sets underwater uses silicates and various oxides (primarily calcium, aluminum and iron oxides). The basic chemical process that is occurring is hydration, whereby the mixture of oxides and silicates incorporates several water molecules directly into the crystal lattice.  The fact that the setting process both uses water, and also produces a structure that is insoluble and robust to attack by water, is what makes underwater setting of hydraulic cement possible.

With regard to teaching your high school class, it's probably better to stick to the lime cycle for non-hydraulic cement as far as detailed chemical reactions are concerned. One interesting point that you can tell them about hydraulic cement is that its chemistry is still not completely understood, and even the molecular crystal structure was only very recently sorted out by MIT scientists in 2009.


Find atomic number given the group and period

This is the quickest way to do this.

Memorise the following table which contains the atomic number of all the noble gases. Noble gases are placed at the end of every period, so if you remember this table, then using a bit of common sense, we can easily locate the position of any element.

  1st period :  helium - 2

  2nd period :  neon-10

  3rd period :  Argon- 18

  4th period : Krypton -36

  5th period : Xenon -54

  6th period : Radon - 86

Now you are good to go.

Example 1) Say you want to find the atomic number of an element placed in group 4th and period 3.

Neon(10) is the last element of period 2. Come down to the next period ,i.e period 3. We need to reach four blocks to the right to get to group 4. Going each block to the right will increase the atomic number by 1. Therefore, going 4 blocks to the right will give us the atomic number = 10+4= 14.

Example 2) 4th period : 17th group

The last element of period 3 is Argon(18). To get to 4th period and 17th group, we need to come down to period 4 and go 17 blocks to the right. The required atomic number is 18+17=35.

Note : This also requires you to remember that there are 2 elements in the first period, 8 elements each in the second and third period, 18 elements each in third and fourth period.


Marx, D.; Hutter, J. Ab initio molecular dynamics: basic theory and advanced methods; Cambridge University Press: Cambridge, 2012. This has some info on implementation of algorithm in CPMD code including flow chart of algorithm.
Thijssen, J. Computational Physics, 2nd Ed.; Cambridge University Press: Cambridge, UK ; New York, 2007. Chapter 6 comes with a short computer program downloadable from companion site written by author of the book.

Maybe these two help a bit.


I think you are confused. Heterogenous and homogeneous are words that are used to describe mixtures (of substances). In homogenous mixtures, the substances exist only in one phase, while heterogeneous mixtures the substances exist in two or more phases.

A molecule is homonuclear if it consists of atoms of a single chemical element ($\ce{O2}$). A molecule is heteronuclear if it is composed of more than one element ($\ce{H2O}$). Note that any pure molecule is considered a substance, and it is the combinations of these substances that form mixtures, either homogeneous, heterogeneous, or colloidal.


It is a valid generalisation of the skeletal formula principle to draw chloromethane as you did; you may also see it drawn in the following way:

$$\ce{-Cl}$$

In a similar vein, ethane can be drawn as a simple horizontal (or $60^\circ$ ascending or descending) line, ethene as an equality sign and ethyne as a triple line.

However, the concept of skeletal formulae, which was primarily invented to simplify drawing large, complex molecules, does not extend well to these very small ones. A chlorine atom with a line can be mistaken as $\ce{Cl-}$, a horizontal line as a dash, a double line as an equality sign and so on. Thus, while formally valid, you will not see this representation used often for very small molecules. Typically, it would be used from three explicit atoms upwards — as is the case in your second example, dichloromethane.


Absolute configuration in stereochemistry: R/S nomenclature

Just to remember the rules to assign an R/S configuration:

In order to assign the configuration as R or S:

Identify each of the chirality centers (most commonly an sp3 C with 4 different groups attached) Then at each chirality center....
Assign the priority (high = 1 to low = 4) to each group attached to the chirality center based on atomic number.
Reposition the molecule so that the lowest priority group is away from you as if you were looking along the C-(4) σ bond. If you are using a model, grasp the lowest priority group in your fist.
Determine the relative direction of the priority order of the three higher priority groups (1 to 2 to 3)
If this is clockwise then it is the R-stereoisomer (Latin; rectus = right handed)
If this is counter-clockwise then it is the S-stereoisomer (Latin; sinister = left handed)
If there is more than one stereocenter, then the location needs to be included with the locant, e.g. (2R)-

You can identify these previous rules on this example:

Source


Is it possible to dissolve Raspberry ketone in warm water?

According to PubChem, 4-(4-hydroxyphenyl)-2-butanone (Raspberry ketone) is estimated to be partially soluble in water, ca. 25.1 g/L @ 25 celcius.

https://pubchem.ncbi.nlm.nih.gov/compound/4-4-Hydroxyphenyl-2-butanone#section=Experimental-Properties

My initial thought is what would one need with Raspberry ketones? Their health effects seem to lack any clinical evidence.


More stable products are lower in energy.  At first glance the bonding arrangements might seem equivalent in both cases in terms of number of $\ce{C-C}$ and $\ce{C-H}$ bonds from $\ce{sp^2}$ carbons, which would result in no net change.  However, the change in molecular structure leads to a change in the molecular orbitals and the resulting energies of the molecule for the aromatic/resonance component to the energy. Using the Frost diagram technique we want to compare the orbital energies, nicely pictured here:

The key point to notice is the filled orbitals for seven member ring are slightly lower in energy than the filled orbitals for the six-member ring.  This means there is an increase in resonance stabilization energy for making a fully aromatic system rather than just a cation in resonance with benzene ring.


IBrF2 - van der Waals repulsions - and number of unique bond angles

I suspect the molecule $\ce{IBrF2}$ does not exist, and is purely conjectural (theoretical) at this point.  If it ever had been synthesized or observed, it would have made the cover of Science or Nature.  "First neutral ternary interhalogen compound found!" (A handful of anionic  ternary interhalogen species are known.)

But my hunch is that, if the molecule did exist, the conformation your professor describes would be favored because of symmetry. Having the same atom in (roughly) opposite directions maximizes the amount of coupling because the opposite orbitals have same energy and can combine better than if it were two different atoms.  Hence greater stabilization.

  Side note: why are axial bonds longer than equatorial bonds?

This is true in binary compounds like $\ce{PCl5}$, and apparently a sufficient explanation is that each axial chlorine wants to avoid 3 other chlorines at 90$^\circ$ angles, while the equatorial chlorines have 2 at 90$^\circ$ but the other 2 are further away anyway, at 120$^\circ$.  In other words, if you just want to maximize the average Cl-Cl distance, subject to the constraint that the P-Cl distance is reasonable, you're better off with a slightly elongated bipyramid instead of one inscribed in a sphere.

In the hypothetical $\ce{IBrF2}$ of course the axial I-F bond would be shorter than the "equatorial" I-Br bond, since Br is a bigger atom.


What is the enthalpy of formation of 1 mol of sulfuric acid?

Since enthalpy is a state function, all that is needed is the $\Delta H_{\mathrm{f}}$ of each reactants that form $\ce{H2SO4}$. Multiply the equations through by the correct fractions (including the enthalpies) to result in the correct coefficients and enthalpy for the desired reaction:

$$\ce{SO3{(g)} +H2O{(g)} -> H2SO4{(g)}}~~~~~\Delta H_{\mathrm{f}}=-193.9~\mathrm{kcal~mol^{-1}}$$

$$\ce{H2{(g)} + \frac{1}{2}O2{(g)}->H2O{(g)}}~~~~~\Delta H_{\mathrm{f}}=-68.3~\mathrm{kcal~mol^{-1}}$$

$$\ce{SO2{(g)} + \frac{1}{2}O2{(g)} \xrightarrow{\ce{V2O5}} SO3{(g)}}~~~~~\Delta H_{\mathrm{f}}=-47.25~\mathrm{kcal~mol^{-1}}$$

$$\ce{\frac{1}{8}S8{(s)} +O2{(g)} -> SO2{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-8.8625~\mathrm{kcal~mol^{-1}}$$

Combining these equations yields:

$$\ce{\frac{1}{8}S8{(s)} +2O2{(g)} +H2{(g)} -> H2SO4{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-318~\mathrm{kcal~mol^{-1}}$$

(according to the equations and numerical values provided)


What does this curve over the C-B bond mean?

That ‘curve’ is a bracket, and the whole thing is a shorthand notation for ‘two of the same group’.

Thus, the molecule could be termed bis-(3-methylbut-2-yl)borane.


Predicting pressure inside a container based on temperature

The problem with answering this question is that propane is not remotely close to being a permanent gas at typical room temperature. This means, in practice, that it is easily liquefied at room temperature with a little pressure and you can't, therefore, use the simple gas law equation to work out the pressure exerted by a given amount of the gas in a vessel of fixed volume.

This means that the tank will fairly certainly contain some liquid (as a butane-filled cigarette lighter will at very modest pressures despite butane being a gas at normal temperatures and pressures).

You can, therefore, estimate the pressure inside the vessel as being the vapour pressure of propane at the given temperature (which ignores any other gas included in the vessel though this may be a good approximation as the filling process is likely to sweep out any other gas). The Wikipedia entry on propane has a convenient chart of the vapour pressure and you can look up the value for a given temperature. The amount of the vessel filled won't make much of a difference to this as long as there is some liquid and some gas in the vessel.

Because simple gas laws don't do a good job of predicting the vapour/gas equilibrium, this isn't a gas-law question and the best answer will always be empirical (see the Wikipedia page).


Calculating the Gibbs Free Energy of Mixing

Let’s start with the last question: Calculate $\Delta_\text{mix}G$ when the pressures are equal in both compartments before mixing. Most physical chemistry textbooks treat this case (including Atkins), and the expression of $\Delta_\text{mix}G$ is:
$$\Delta_\text{mix}G= nRT(x_1\ln x_1 +x_2\ln x_2)$$ In this exercise, $1$ denotes hydrogen, $2$ denotes nitrogen $n=6\ \text{mol}$, $x_1=\frac{2}{6}$ and $x_2=\frac{4}{6}$. So, by substituting these values in the above expression, we find: $$\Delta_\text{mix}G= 6\ \text{mol}\times 8.314\ \frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\frac{2}{6}\ln\frac{2}{6} +\frac{4}{6}\ln\frac{4}{6})= -9462\ \text{J}$$ As $\Delta_\text{mix}G<0$ at constant temperature and pressure, this means that the mixing of two ideal gases is a spontaneous process .

Let’s now treat the first question with two different pressure in the two compartments. The initial Gibbs energy is $$G_\text{i}=n_1(\mu_1^0 +RT\ln p_1)+n_2(\mu_2^0 +RT\ln p_2)$$ Where $n_1=2\ \text{mol}$, $n_2=4\ \text{mol}$, $p_1=2\ \text{atm}$, $p_2=3\ \text{atm}$. $$G_\text{i}=2\ \text{mol}\times(\mu_1^0 +RT\ln2)+4\ \text{mol}\times(\mu_2^0 +RT\ln3)$$
When the partition is removed, and each gas occupies the volume $V_1+V_2$.
The partial pressure of hydrogen is $$p^\prime_1=\frac{2\ \text{mol}\times RT}{V_1+V_2}$$
The partial pressure of nitrogen is $$p^\prime_2= \frac{4\ \text{mol}\times RT}{V_1+V_2} $$  On the other hand: $$V_1=\frac{2\ \text{mol}\times RT}{2\ \text{atm}}=1\frac{\text{mol}}{\text{atm}}\times RT$$ $$V_2=\frac{4\ \text{mol}\times RT}{3\ \text{atm}}=\frac{4}{3}\frac{\text{mol}}{\text{atm}}\times RT$$ i.e.  $$V_1+V_2=\frac{7}{3}\frac{\text{mol}}{\text{atm}}\times RT$$
So $$p^\prime_1=\frac{2\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{6}{7}\ \text{atm}$$ $$p^\prime_2= \frac{4\ \text{mol}\times RT}{(7/3\ \text{mol/atm})RT}=\frac{12}{7}\ \text{atm} $$ The final Gibbs energy is $$G_\text{f}=n_1(\mu_1^0 +RT\ln p^\prime_1)+n_2(\mu_2^0 +RT\ln p^\prime_2)$$  $$G_\text{f}=2\ \text{mol}\times(\mu_1^0 +RT\ln\frac{6}{7})+4\ \text{mol}\times(\mu_2^0 +RT\ln\frac{12}{7})$$ The Gibbs energy of mixing is the difference between $G_\text{f}$ and $G_\text{i}$
$$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6/7}{2} +4\ \text{mol}\times RT\ln\frac{12/7}{3}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times RT\ln\frac{6}{14} +4\ \text{mol}\times RT\ln\frac{12}{21}$$  $$\Delta_\text{mix}G= 2\ \text{mol}\times 8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\times 298\ \text{K}\times(\ln\frac{3}{7} +2\times\ln\frac{4}{7})= -9744\ \text{J}\approx{-9.7\ \text{kJ}}$$
In this case, the value of $\Delta_\text{mix}G$ is the sum of two contributions: the mixing itself and the changes of pressure of the two gases to their final total pressure.

I hope it’s clear now.


Qualitative Analysis- Distinguishing between two ions or two salts

Answering this question requires a preliminary discussion of the solubility product constants of three silver halides and the formation constants of three silver complexes. First, note that $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ are all insoluble in water, but insoluble is a relative term in the end. Their respective solubility equilibria and solubility product constants, i.e., equilibrium constants for their dissolution, are:

$$\ce{AgCl(s) <=> Ag^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} = \pu{1.8E-10}$              \tag 1}$$

$$\ce{AgBr(s) <=> Ag^+ (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} = \pu{5.4E-13}$              \tag 2}$$

$$\ce{AgI(s) <=> Ag^+ (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} = \pu{8.3E-17}$              \tag 3}$$

Silver ions also form complexes with ammonia, thiosulfate ion, and cyanide ion. The formation equilibria and associated equilibrium constants are as follows:

$$\ce{Ag^+ (aq) + 2 NH3 (aq) <=> [Ag(NH_3)_2]^+ (aq) \quad $K_\mathrm{f(4)} = \pu{1.6E7}$              \tag 4}$$

$$\ce{Ag^+ (aq) + 2 S2O_3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) \quad $K_\mathrm{f(5)} = \pu{2.0E13}$              \tag 5}$$

$$\ce{Ag^+ (aq) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) \quad $K_\mathrm{f(6)} = \pu{1.0E21}$              \tag 6}$$

Now consider the following sequence of aqueous solution additions:

An aqueous $\ce{NaCl}$ solution is added to an aqueous solution of $\ce{AgNO3}$. Then $\ce{AgCl}$ precipitates, as per equilibrium (1).
Next concentrated ammonia is added in excess, i.e., well above the 2 to 1 stoichiometry of equilibrium (4). Then the following equilibrium, obtained by adding equilibria (1) and (4), occurs:

$$\ce{AgCl (s) + 2 NH3 (aq) <=> [Ag(NH3)2]^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} K_\mathrm{f(4)}  = \pu{2.9E-3}$              \tag 7}$$

Although the equilibrium constant is less than unity, adding concentrated ammonia in excess (in a hood!) results in all of the $\ce{AgCl}$ dissolving: the equilibrium is driven to the right.

Next $\ce{NaBr}$ is added. This results in precipitation of $\ce{AgBr}$ via the following equilibrium:

$$\ce{[Ag(NH_3)_2]^+ (aq) + Br^- (aq) <=> AgBr (s) + 2 NH3 (aq) \quad $1/(K_\mathrm{sp(2)} K_\mathrm{f(4)})  = \pu{1.2E5}$              \tag 8}$$

This equilibrium is simply the reverse of the addition of equilibria (2) and (4).

Next excess sodium thiosulfate ($\ce{Na2S2O3}$) is added. Then $\ce{AgBr}$ dissolves as per the following equilibrium, obtained by adding equilibria (2) and (5):

$$\ce{AgBr (s) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} K_\mathrm{f(5)}  = \pu{10.8}$              \tag 9}$$

Next $\ce{KI}$ is added. This results in precipitation of $\ce{AgI}$ via the following equilibrium:

$$\ce{[Ag(S2O3)2]^3- (aq) + I^- (aq) <=> AgI (s) + 2 S2O3^2- (aq) \quad $1/(K_\mathrm{sp(3)} K_\mathrm{f(5)})  = \pu{6.0E2}$              \tag{10} }$$

This equilibrium is simply the reverse of the addition of equilibria (3) and (5).

Lastly, $\ce{KCN}$ is added: in a hood, with proper safety precautions! Then $\ce{AgI}$ dissolves as per the following equilibrium, obtained by adding equilibria (3) and (6):

$$\ce{AgI (s) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} K_\mathrm{f(6)}  = \pu{8.3E4}$              \tag{11} }$$

When I did this demo in lectures, I skipped the last step in order to avoid dealing with the hazardous waste issue that cyanide solutions present.

So now, at long last, the OP's questions:

Distinguish between $\ce{AgCl}$ and $\ce{AgI}$ by adding ammonia.

From equilibrium (7), $\ce{AgCl}$ can be dissolved if an excess of concentrated ammonia is added. Trust me: this should be done in a hood! But, comparing equilibria (1) and (7), $\ce{AgI}$ is more than a million times less soluble than $\ce{AgCl}$, so no amount of concentrated ammonia will significantly dissolve $\ce{AgI}$.

The OP's second question involves distinguishing between silver ion and the silver ammonia complex.

The OP's answer sheet claims that adding $\ce{NaCl}$ should result in only silver ion giving a precipitate. But equilibrium (7) shows that sufficiently high chloride ion concentrations should result in $\ce{AgCl}$ precipitation. But silver also forms $\ce{[AgCl2]^-}$, $\ce{[AgCl3]^2-}$, and $\ce{[AgCl4]^3-}$ , when the chloride concentration is high. So adding chloride will not cause AgCl to precipitate, in agreement with the OP's answer sheet. See here as well.

Last thought: I have done this demo in lecture, minus the last cyanide step, and I think the OP should not be expected to know all this stuff, especially if an exam is involved. The entire point of this kind of problem is to show that we can control equilibria to our advantage in, e.g., gold extraction via the cyanide process, and insoluble is a relative term.

Source of all solubility product constants and formation constants:

Daniel C. Harris, Appendix I In Quantitative Chemical Analysis; 7th Ed.; W. H. Freeman & Company: New York, NY, 2007 (ISBN: 0-7167-7041-5; ISBN-13: 9780716770411).


Chelation is an entropy driven reaction, i.e. complexes with polydentate ligands are more stable.

Consider what happens when you dissolve copper sulfate in water:
\begin{align}
\ce{CuSO4 + 6H2O <=> [Cu(H2O)]^2+ + SO4^2- (aq)}\\
\end{align}

This complex forms and with it comes some kind of ordering of the water molecules around it. Dissolution is often favoured because of the solvation energy. Now consider a bidentate ligand like ethylene diamine, $\ce{H2N-(CH2)2-NH2}$, short $\ce{en}$. Adding this ligand to a copper sulfate solution will result in a ligand exchange:
\begin{align}
\ce{[Cu(H2O)]^2+ + 3en <=> [Cu(en)3]^2+ + 6H2O}\\
\end{align}

In this reaction you increase the number of particles from four on the left side to seven on the right side. Taking into account that you also disturb the order of the surround water molecules, this reaction is very much favoured by entropy. In an attempt of zeroth order approximation, one could assume that entropy roughly doubles. Since the Gibbs energy is inversely correlated to the entropy, this means, that the reaction will be spontaneous. This is usually referred to as the chelate effect.$%edit$


Analytical Chemistry predicting the order of elution of compounds in a reverse phase HPLC

The logic is wrong. Lower $\mathrm{pH}$ (actually you mean $\mathrm{p} K_\mathrm{a}$ here) does not mean more polar.

It is more complicated to decide the polarity then you think. For reverse phase HPLC, the more polar and more soluble compound come out first. For reverse phase HPLC, it is common to have acid (TFA for example) in the eluent. It this case, caffeine can be protonated and become much more polar because it is ionic now. Solubility is another factor to consider. So if you want to predict the order, you have to provide more information.


Increase in Viscocity in Gases with Temperature

From the basis of the kinetic theory of gases, viscosity in gases increases with temperature because the increase in kinetic energy will increase the frequency of collisions between the molecules. This increased number of collisions decreases the ability of the molecules to diffuse in the desired direction (i.e. the gas doesn't flow down the pipe to a reactor).


Do magnetic fields generated from ion flow through a membrane channel have any physiological relevance?

For currents in biological systems the magnetic field is minuscule, and likely has no physiological effect.

That said, magnetoencephalography is based on measuring the tiny magnetic fields created when nerves fire. Detection requires an incredibly sensitive detector, a superconducting quantum interference device (SQUID), to detect the location of the currents. A SQUID can detect fields of just a few atoteslas (aT)!

Perhaps one could investigate how much ion flow in electric eel muscle is constrained by the induced magnetic field... It could lead to shocking revelations.


Voltage vs current density - why are both parameters important to control an electrochemical reaction?

Reaction rate in electrochemistry is limited by diffusion of reactants from solution towards electrode. Once all reactants in near proximity for given type of reactions are exhausted, and if voltage is still applied, other reactions (if voltage is strong enough) may occur. For example, consider positive electrode in $\ce{KCl}$ solution (leaving aside paired negative electrode). It is true, that $\ce{Cl^-}$ anion is oxidized easier then water. However, once all chloride near the electrode is oxidised, the next possible substrate to be oxidised is $\ce{OH^-}$ that is always present is water via reaction $\ce{2H_2O <=> H_3O^+ + OH^-}$. Thus, depending on the current density on the electrode you will have different major products.

In short: resistance of the electrochemical cell is a complex function of temperature, voltage, current and solution composition.

The current density in electrochemistry may be adjusted not only by adjusting voltage applied, but also by adjusting surface area of the electrodes used. Also, it is critical to maintain conditions for effective transportation of reactants towards electrodes.


Why does hybridization significantly matter for organic chemistry?

Orbital hybridization is mostly useful in the rationalisation of molecular geometry. In fact it was invented by Linus Pauling for this purpose: to rationalise structures of organic molecules using the language of atomic orbitals. Yes, hybridization is also used to rationalise bonding properties, as well as acidity or basicity, but its primary use is to rationalise molecular geometries.

  I had the impression hybridization regulates a standard bond among
  same species of atoms, e.g $\ce{C-H}$ bond in all alkanes are the same. But
  that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.

Well, $\ce{C-H}$ bonds, of course, differ in $\ce{CH4}$ and $\ce{CH3-CH3}$, but not too much; they are approximately the same. Are you naive enough to hope that a model as simple as the orbital hybridization one would perfectly predict the molecular geometry of each and every organic molecule? Surely, it won't. It can just give a crude (but simple) approximation.

  What I am unable to reason is, why hybridization is so important for
  organic chemistry than to any other (general, inorganic..., physical)?

Where did you get that? Hybridization is a concept of general chemistry, it is used in both organic and inorganic one. Its relatively big importance for organic chemistry is due to the fact that it is the only simple model which can explain (approximately) molecular geometry of organic compounds. The original valence bond theory couldn't do it, so it was extended.

On explanation vs rationalisation

Now I would like to emphasize that hybridization provides a rationalisation of a molecular geometry (as well as some other properties), but not an explanation, so it can not be used to reliable predict the properties. What is the key difference between an explanation and a rationalisation in these context?

Well, an explanation of some phenomenon describes it in way that we can observe and test the consequences of our claims and premises. We can then tell from the observations are our claims and premises true or not, or, in other words, does our explanation work or not. The whole point of an explanation is to distinguish what is true from what is not true, in a sense, what corresponds to the actual reality from what does not do so.

In a rationalisation things work differently. We insist on some premises being true without testing them against the reality. We do observations, but we do not use them to tell are our premises true or not, rather on the basis of the desired premises and the observable evidence we start to make the observed outcome seem consistent with the premise.

In psychology rationalisation is also known as making excuses. It is a defence mechanism in which a controversial (from the society viewpoint) behaviour is justified "in a seemingly rational or logical manner to avoid the true explanation". With respect to scientific theories, it can also be thought of as a defence mechanism: when the real explanation is too complex for beginners we avoid it to defend them. Instead we first provide some simple rationalisation and later in the course of study the explanation is introduced.

The way rationalisation works for all these general chemistry models is as follows:

We do know geometry or some other property of a particular molecule, say, from an experiment, and we are seeking for some rationalisation of why the things are the way they are.
The rationalisation usually has to be as simple as possible, so that, for the case of molecular geometry, we usually start from VSEPR theory.
If the simplest possible theory can not rationalise the observed property, we choose a more sophisticated model. For molecular geometry, for instance, we could then try to use the original valence bond (VB) theory.
But even the VB theory (in its original formulation) won't work for organic compounds, so it has to be extended by two concepts: resonance and orbital hybridization. Note that this concepts are also required in inorganic chemistry as well, but not always: for some inorganic molecules the original VB theory is enough.

But however deep you go into this hierarchy of chemical models, all them are just rationalisation models: they provide simplified (and in many cases oversimplified) descriptions of the reality which is approximate (and in many cases rather crude). And besides all this process of seeking for an rationalisation is a bit like cheating:

If some model doesn't work, let us try another model.
If none of them works, let us introduce a new concept with some funky name in one of them, the purpose of which will be to rationalise this particular thing which couldn't be rationalised by the original model.

As a result, the predictive power of such models is quite small. They can be used to rationalise what is already known, but any predictions based on such models will be unreliable. And yeah, the concept of orbital hybridization is just one of these cheats mentioned above.