What is the directive influence of the tert-butyl group in electrophilic aromatic substitution?

What you've drawn in your bottom figure is an example of carbon-carbon hyperconjugation. The t-butylbenzene and the charge separated structure drawn to its right are resonance structures (so you should use a double-headed resonance arrow to connect them).  Additional resonance structures with the negative charge delocalized into the ortho and para positions of the benzene ring can also be drawn, as well as additional structures using the other methyl groups attached to the central t-butyl carbon.  Based on the carbon-carbon hyperconjugation argument we would expect that

the t-butyl group would be an ortho-para director and
that the t-butyl group would accelerate the rate of electrophilic aromatic substitution compared to benzene itself.

Generally, carbon-carbon hyperconjugation is a small effect (smaller than carbon-hydrogen hyperconjugation) and not that commonly invoked.  The problem that I have with using carbon-carbon hyperconjugation arguments in this case is that our resonance structures involve a high energy methyl cation.

Backing up to toluene for a moment, there are two arguments why the methyl group is an activating ortho-para director. You’ve already outlined the hyperconjugative argument where electron density is shifted from a $\ce{C-H}$ bond to the aromatic ring using pi orbitals. However, in addition to this pi transmission argument there is also electron transfer from the methyl carbon to the aromatic ring inductively through the sigma system. The methyl group is roughly $\ce{sp^3}$ hybridized while the aromatic carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow through the sigma bonds from $\ce{sp^3}$ to $\ce{sp^2}$ carbons. So both inductive (sigma) and hyperconjugation (pi) arguments can be used to explain the electron donating ability of a methyl group.

Here is some data on the relative rates for the electrophilic nitration of several aromatic compounds:

benzene=1, toluene=24, t-butylbenzene=15.7 (reference, p. 1060)

We see that t-butylbenzene undergoes electrophilic aromatic nitration slightly faster than benzene and slightly slower than toluene - and the effect is very small.  Further, it is also reported (same reference) that nitration of t-butylbenzene produces ortho, meta and para products in a 12 : 8.5 : 79.5 ratio. Since the bulky t-butyl group will sterically block the ortho position to some extent, the fact that the para isomer predominates and there is more ortho than meta isomer is consistent with the t-butyl group acting as an activating ortho-para director.

Small effects are often difficult to explain, but I suspect that the carbon-carbon hyperconjugation effect (pi transmission) is a small contributor at best here.  Rather, the inductive sigma transmission effect (electron flow from the roughly $\ce{sp^3}$ central t-butyl carbon to the aromatic  $\ce{sp^2}$ carbon) is how the t-butyl group donates electron density and stabilizes the electrophilic aromatic substitution transition state.  In fact, we could even argue that since carbon is more electronegative than hydrogen, the 3 methyl groups surrounding the central t-butyl carbon pull some electron density away from it compared to the 3 hydrogens surrounding the methyl carbon.  This would make the t-butyl carbon slightly less electron releasing than a methyl group in accord with observed relative rates.


Why is the first ionization enthalpy of molecular O2 less than atomic O?

A cousin of mine asked me to share her answer here:

In the molecular orbital diagrams shown in textbooks, the highest occupied molecular orbitals of $\ce{O2}$ and of $\ce{F2}$ are shown as antibonding. If the molecules and their cations had the same bond length, and the electrons in a species didn't "talk to each other", and the orbital energies of an isolated atom were not influenced by the approach of another atom, you could predict that the ionization energy of the molecular species is lower than that of the atomic species. The argument would be that these HOMOs have higher energy than the corresponding atomic orbitals (see comments to question).
As these assumptions are not quite true, it is more complicated, but luckily the quick and dirty analysis yields the correct answer.
According to the MO diagrams, the bond order of $\ce{O2}$ is 2 and that of the cation $\ce{O2+}$ is 2.5 (after all, we are removing an antibonding electron). Indeed, the bond length of the cation is shown as smaller in this source. I did not find a bond length for the $\ce{F2+}$ species. This shows that you have to at least adjust the MO diagram for the change in bond length.
To get a rationale for the ionization energies in a quantitative manner, you could calculate the energies of the four relevant species ($\ce{O, O+, O2, O2+}$) and look at the differences.


Creating a Galvanic Cell

To simplify, let's forget galvanic cells for a moment. Instead, just picture an oxidation/reduction reaction in aqueous solution.  If we had a solution of $\ce{Cu^+}$ and $\ce{Co^{3+}}$, the metal ions would exchange an electron, and we would end up with $\ce{Cu^{2+}}$ and $\ce{Co^{2+}}$. This happens because of the relative stability of the two ions; the electron is at a lower energy with the cobalt than with the copper (at these oxidation states). Now, if we separate the two metals into two different solutions, insert an electrode into each, and connect the two with a conductive wire, we can still transfer an electron. The potential between the electrodes comes from the same source as the mixture: the chemical potential between the two ions.

A salt bridge is also necessary. The salt bridge balances charge in the two solution. Otherwise, we would have a net positive charge in the copper solution and a net negative charge in the cobalt solution. This would create a potential difference and the reaction would eventually and stop.

Galvanic Cell using Non-ionic Species

Galvanic cells do not have to use species that are ionic in the oxidized and reduced states (or either, for that matter). Consider the cell pictured below. In this example, zinc metal is being oxidized to $\ce{Zn^{2+}}$ and $\ce{Cu^{2+}}$ is being reduced to copper metal. In other words, the zinc electrode is getting dissolved, and the copper electrode is growing. The anodic salt solution is used to create what is called a poised electrode. If a zinc electrode were inserted into pure water, according to the Nernst equation, the system would have an infinite potential.

$$\mathcal E = \mathcal E^0 - \frac {RT} {nF} \ln 0 \rightarrow -\infty$$

As soon as even a few molecules per liter are in solution (which would happen very quickly) the potential drops to around 50 V, and would continue to drop as more zinc was oxidized (although there are certainly deviations from the Nernst equation in an observed system). The salt solution puts the system into a steady state. In terms of the cathode, the metal does not matter. Anelectrode made from any conductive material would behave the same as a copper electrode in a solution of $\ce{Cu^{2+}}$; copper would plate onto the electrode as it is reduced. The system potential is not altered because the activity of a pure solid is unity. It is normally just convention (and for learning purposes) that the electrode material is the same as the oxidant.


According to the source mentioned in the comments to your question, the first step is indeed nucleophilic substitution of the OH group by $\ce{I-}$, faciliated by protonation of the alcohol. For the second step ($\ce{HI}$ reduction), a radical species was found as an intermediate, and therefore a reduction by single electron transfer (SET) with oxidation of $\ce{I-}$ to $\ce{I2}$ is proposed. Red phosphorus is required for the regeneration of $\ce{HI}$ from $\ce{I2}$. $\ce{P_{red}}$ reacts with $\ce{I2}$ to the phosphorus iodides $\ce{PI3}$ and $\ce{PI5}$, which are subsequently hydrolyzed to $\ce{H3PO3}$/$\ce{H3PO4}$ and $\ce{HI}$, which is reused in further reduction steps. Catalytic amounts of $\ce{HI}$ are therefore sufficient to run the reaction in the presence of phosphorus. However, a too low concentration of $\ce{I-}$ will lead to elimination instead of substitution in the first reaction step.

(Image source)

The SET mechanism of the reduction is not described in detail in the article, so the following is a bit speculative. As a resonance-stabilized radical is involved, homolytic cleavage of the $\ce{C-I}$ bond is one of the first steps, which gives the benzylic radical $\ce{R.}$ and an iodine radical $\ce{I.}$. A single electron is transferred from $\ce{I-}$ to $\ce{R.}$, forming the carbanion $\ce{R-}$ and $\ce{I.}$, which recombines with another iodine radical to $\ce{I2}$. Protonation of $\ce{R-}$ yields the final product.


You have calculated correctly for parts $\bf{a}$ and $\bf{b}$. However, to answer part $\bf{c}$, you need to use the given plot, the interception and slope of which can be calculated as follows:

The slope is $\tan \alpha = - \frac{1}{K_d}$:
$$\tan \alpha = \frac{1-0.2}{0.1-0.8} = -1.14 \ \Rightarrow \ \therefore \ \frac{1}{K_d} = 1.14$$
According to the given Scatchard equation (where  each receptor has a single ligand binding site), interception of the plot is also equal to $\frac{1}{K_d} = 1.14$. Therefore, the Scatchard equation can be rewritten as:
$$\frac{r}{[A]} = 1.14(1-r) \tag1$$
where $[A]$ is actually free ligand concentration (Wikipedia). Therefore, when $[A]= \pu{3.2 \mu M}$, find the relevant $r$ using the equation $(1)$, which is the average amount of $A$ per $\pu{mol}$ of $P$ $(r = \pu{0.785 \mu M})$.
This means: Total number of binding site per receptor is $1$ and only $0.789$ is bound by $A$. Protein concentration in the dialysis tube $= \pu{2.0 \mu M}$
Therefore, $A$ bound concentration of $P$ $ = \pu{2.0 \mu M} \times 0.785 = \pu{1.57 \mu M}$


What happens when deodorants or perfumes expire

The meaning of expiry dates

The date of expiry is widely interpreted by consumers as the date after which a product is known to detiorate.  However, the meaning of expiry dates varies widely from product to product, and from country to country.  Often times, the date of expiry is not a "first known bad" date, but a "last known good" date.  In that latter case, the dates indicates that manufacturers know their product will last up until the expiration date, but after that, they don't know what happen.  Their product could last 1000s of years, completely stable and undeteriorated.  Or it could degrade instantaneously the day after expiry.

I don't know for sure what perfume manufacturers mean by their expiry dates.  You could try asking the manufacturer for clarification.  Ask for their data from their product stability studies.

Likely routes of perfume degradation

All of the above notwithstanding, let's suppose that perfumes do degrade over a time span of several years.  Subject to that assumption, what could be happening happening?  First, I agree with the commenters that ethanol oxidation to acetaldehyde or acetic acid is unlikely to be a major contributor to perfume degradation.

However, the same is not true of other perfume components.

Oxidation

Aldehydes such as octanal, benzaldehyde, citral, are an important ingredients in many perfurmes.  Citral degradation has been studied and is accelerated by acid and oxygen, with the main ultimate product being p-cymene.  Benzaldehyde is oxidized in air to benzoic acid, and octanal is oxidized to octanoic acid.
Terpenes are also important pefume components.   Limonene is a component of many perfume blends and is a representative terpene.  It is also oxidatively unstable, with products such as limonene oxide, carveols, and carvones being formed.  Some of these oxidation products can trigger allergic reactions, which while not precisely "burning the skin", could lead to such a perception.

Other possible degradations

Perfume, especially if opened frequently, could also change by absorbing water from the atmosphere, leading to dilution and an alteration of its olfactory properties.  Frequent opening or exposure to the atmosphere could also result in differential evaporation of the more volatile components, which would also change its olfactory properties.

The bottom line

Expired perfume or deodorant might be just as good as regular perfume.  Or it might be degraded.  The degradation products might possibly include compounds that have been reported to form rashes.  It is unlikely, however, that the chemical mechanism for this is the formation of large amounts of acid.


Converting cyclic compounds to linear compounds (possible ?)

Yes, this can be done.  Unsaturated cycloalkanes can be ring opened using fairly straightforward methods.  For example, heating cyclobutene will produce 1,3-butadiene in an orbital symmetry allowed reaction.  But since your example involved cyclohexane, I'm guessing that you are more interested in the ring opening of saturated cycloalkanes, so that's where I'll focus my answer.

Catalysts are needed to ring-open cycloalkanes.  The catalyst complexes with a cycolalkane bond and then in a second step, hydrogen is added across this bond.  Mechanistically, it is basically the reverse of the hydrogenation reaction (e.g. catalytic hydrogenation of cyclohexene to cyclohexane).  Some of these catalysts are very indiscriminate, so the initially formed linear alkane will rebind to the catalyst and undergo subsequent isomerization and fragmentation reactions.  More recently catalysts have been developed that can open cycloalkanes to the corresponding linear alkane in fairly high yield and relatively free of other hydrogenolysis products. These reactions are run in solution at temperatures slightly above ambient.  Here is a link to a full text article on one such process that uses Lanthanide\Aluminum complexes;  Terbium seems to be the preferred lanthanide.  Yields are typically in the 65-80% range (cyclohexane, 67%; cyclopentane, 77%; cyclodecane, 77%).


HPLC Lab Report: Why is weight/purity % sometimes over 100%?

By far the most likely reason for reported purities of over 100% when using HPLC is the "analytical uncertainty of the method", but let me illustrate exactly how this is likely to arrive.

A standard procedure for estimating purity is:

Run an exhaustive series of known standards with (assumed) exactly known purities, or weight percents, on the HPLC.
Plot this calibration data with peak area on one axis, and the known weight % on the other axis.
Fit a line to this data.  The slope of this line is called the "response factor" or sometimes just the "response".
Run some unknown samples and measure the peak area for the unknowns.
Using the response factor, calculate a weight percent for the unknown sample.

If there are analytical uncertainties in the calibration standards, then the response factor may be randomly off by small amounts from the "true" value of the response factor.  Similarly, if there are uncertainties in the peak area of the unknown sample, even using the "true" response factor may give a value that is slightly off.

Statistically, these errors should be random, so that sometimes when the "true" purity is 100.000 % the measured value will be 100.6%, and sometimes it will be 99.4%.

Chemically, what is the source of the errors?  Well, maybe the autosampler that injected the unknown sample was supposed to inject 2.0000 μL of sample for analysis, but due to instrument imperfections actually injected 2.012 μL instead.  Maybe there was a tiny power surge in the AC lines going to the UV detector that was measuring the peak right when the compound of interest eluted, leading to an ever-so-slightly greater signal at that moment...

The list of possible causes is nearly endless, but none of these things are really cause for concern, unless you really need to know the purity to three or more decimal places.  From the report you provided, you can safely assume that the uncertainty in their analytical method is at least 0.6%, but not much else.


Here you need to consider two factors.

The one which follows $4n+2$ rule.
The stability of the ring.

When we see the IIIrd one, the structure don't follow the $4n+2$ rule when the double bond breaks. That means, that the double bond will be reluctant towards breaking. Hence its stability will be least.

COMING TO NEXT ONE

The order of stability of the ring is in the order: 6−member > 5−member > 7−member > 4−member > 3−member.

So when we combine these facts we find the correct order is B>A>C.


Why does the carbonyl of carboxylic acids get protonated and not the hydroxyl group?

Because the protonated acid is resonance-stabilized when the carbonyl is protonated, but isn't if the hydroxyl is protonated:

So the first protonated form is much more energetically favourable than the second.

The same resonance in the non-protonated acid means that the lone pair on the alkoxide has reduced basicity, meaning that the carbonyl lone pair is the more basic since it isn't involved in any energy-lowering orbital interactions(/resonances).


What does the Clausius Inequality mean in thermodynamics? (entropy)

Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously.

We know that $$\mathrm{\delta w \geq \delta w_{rev} \Rightarrow \delta w-\delta w_{rev}\geq 0}$$

and according to the First Law of Thermodynamics

$$\mathrm{\delta U = \delta q + \delta w=\delta q_{rev}+\delta w_{rev}}$$

Therefore $$\mathrm{\delta q_{rev}-\delta q=\delta w-\delta w_{rev} \geq 0}$$

Now using the definition of entropy yields

$$\mathrm{\delta S=\cfrac{\delta q_{rev}}{T} \geq \cfrac{\delta q}{T}}$$

Taking $\mathrm{\delta q=0}$ (since we have an isolated system no heat can flow into it), produces

$$\mathrm{\cfrac{\delta q_{rev}} {T} \geq 0 \Rightarrow \delta S \geq 0}$$


1,4-diphenyl-1,3-butadiene (E,E) and (E,Z) Isomer Newman Projections

I don't know what tempts you to use the Newman projection for these compounds. Trying to draw them, you might have missed the existence of the actual conformers. (Not speaking of the confusing 120° angle representation of the planar system.)

Thanks to the totally conjugated system, the molecules tend to maintain a planar shape. The two conformations, ‘cisoid’ and ‘transoid’ (deprecated terms; recommended: s-cis, s-trans; discussed also here) resemble double bond configurations because of the stability of the planar conformation (sterical hindrance has to be taken into account as well, it can be significant for the s-cis E,Z).

For example, the s-trans and s-cis conformations of the E,E isomer can be represented as follows.

Trying to show the conformation at all the three single bonds at once wound be difficult, but could be done. And there's more missing info, so the two Newman projections would represent the (E,Z)'s conformers as well.

I am aware that excluding the hydrogens in the Newman projection is unorthodox. With or without H's, I don't think that the Newman projection is useful here, as for someone unaware of the planarity tendency, might consider e.g. torsional angle 90° much more favorable than the 0° one.


This should do it

http://www.sciencebuddies.org/science-fair-projects/project_ideas/Weather_p004.shtml


Difference between isopropyl and isobutyl

The IUPAC Organic Chemistry Nomenclature does not explicitly define the "iso-" prefix meaning, as it is not used in the systematic naming. However, it is mentioned on several places when listing acceptable semi-systematic names. In the section A-2 (1979 Blue book, some also in 1993 Recommendations, section R-9.1, table 19a) those are (not mentioning neo- compounds)

isobutane (CH3)2CH−CH3
isopentane (CH3)2CH−CH2−CH3
isohexane (CH3)2CH−CH2−CH2−CH3

And in the section A-2.25 (some also in R-9.1, table 19b), where it lists acceptable "isoalkyl group" names:

isopropyl (CH3)2CH−
isobutyl (CH3)2CH−CH2−
isopentyl (CH3)2CH−CH2−CH2−
isohexyl (CH3)2CH−CH2−CH2−CH2−

Some related isomeric alkyls mentioned in this section are

sec-butyl CH3−CH2−CH(CH3)−
tert-butyl CH3−CH(CH3)2−
tert-pentyl CH3−CH2−CH(CH3)2−

There are also some other, unsaturated "isoalkyl" group names mentioned:

isopropenyl CH2=C(CH3)−
isopropylidene (CH3)2C=

From the above we can induce that "iso-" prefix, when related to alkanes or alkyl groups, it means the simplest (n-)alkane/alkyl isomer branched at the end of the carbon chain.

Isopropyl happens to be secondary (sec) alkyl (there's lack of carbon atoms to create primary alkyl isomeric to the n-propyl), all higher isoalkyls are primary (no branching at the alkyl "attachment").


What is the chemical equation, along with the change in heat of the crystallization process of Sodium Acetate

There is no chemical reaction.

The heat is released as the latent heat of crystallisation of a supercooled solution of sodium acetate trihydrate; you can obtain similar effects with sodium sulfate decahydrate or sodium thiosulfate pentahydrate.

When heated up, the hydrate dissolves in its own crystal water, which absorbs thermal energy similar to melting; when cooled down, the solution becomes supercooled.

Their state is called the metastable state: this means that it is not thermodynamically stable, but conditionally stable for kinetic reasons, as the crystallization requires a trigger.

The triggering can be done by, for example: insertion of a small crystal of the same salt, bending a steel chess (used in pocket heaters) or vigorous shaking.


How to treat pressure in computational chemistry?

Old question, but still seems to attract some interest.
In general, the total pressure of the system is not taken into considerations for corrections. The pressure at which you do the corrections is the partial pressure / concentration of the molecule.
I recommend calculating the thermal correction for $G$ at the temperature of interest and the default pressure ($e.g.$ 1 atm for Gaussian). You can then add the correction for the isothermal expansion/compression to the actual concentration $c_f$ / partial pressure $p_f$:
$$\int_{p_i}^{p_f}\bigg(\frac{\partial G}{\partial p}\bigg)_T dp = RT\ln\bigg(\frac{p_f}{p_i}\bigg) = RT\ln\bigg(\frac{c_f RT}{p_i}\bigg)$$
where $p_i$ is the pressure for the thermal correction ($e.g.$ 1 atm). Note that these formulas are based on the ideal gas approximation, but that's what software like Gaussian does under the hood anyway. Hence, you could in principle directly set the correct pressure and should get the same result, but I find it more convenient to have the same route section for all calculations.
Regarding your multiphasic systems: As you noted, the reaction will most likely take place in the solution, so you have to use concentration of the gas in the solution.


Why is crystallography still used in some cases for small molecule structure determination?

Our group uses both methods a lot so here are some examples why you would use X-ray, in addition to NMR, in organic synthesis.

Your compound isn't soluble enough:

Colleagues produce very large aromatic systems for organic electronics which are basically insoluble in all solvents. With a good NMR machine (600MHz and up) it might still be enough to get spectra, but full structure elucidation might be difficult. Since the compounds crystallize well it's quite easy to get a X-ray structure.

Low number of carbon and hydrogen:

I've produced compounds with a C:H:N ratio of 2:3:5. The 1H is two singlets, the 13C is two signals. COSY, HSQC, HMBC,... won't tell you much either and it's much easier to use the nice crystals and get a structure that way than trying some correlated NMR experiments including nitrogen.

3-D orientation/stereochemistry:

Imagine this compound which is quite similar to something we produced:

Since you got two quaternary centers here it's really hard to tell the configuration on those centers relative to each other using NMR. There's no coupling to determine the stereochemistry and a NOE experiment isn't very conclusive here. With X-Ray it's no problem to find out.

Actually, these problems are definitely the ones we use X-ray the most.

Absolute Configuration:

Theoretically it's also possible to determine absolute configuration with single crystal X-ray diffraction, while it's extremely hard to do, if not impossible, with NMR.

Summary:

Single crystal X-ray diffraction gives you the configuration of your molecule, including relative configuration of stereocenters and possibly even absolute configuration, right away. If you only care about configuration (and not behaviour in solution) X-ray is definitely superior to NMR, there's basically no question on how your molecule looks.

On the other hand you need crystals and it's much more complicated to perform than NMR, so this definitely stays the number 1 method for synthetic chemists.


Whether for all zero order reactions rate is independent of concentration of reactants

I think the issue is not that you can't define overall order, but that it isn't a meaningful quantity. This gets at the point that Zhe was making that while we can define the total order of a reaction as the sum of the exponents, it doesn't tell us anything about how the reaction occurs.

Your question gives a very good example of why it isn't a useful quantity: you might expect a zeroth order reaction to have no dependence on the concentration of reactants, but your reaction clearly depends on the concentration of $\ce{A} \text{ and } \ce{B}$, so the total reaction order isn't actually telling us anything about the kinetics of the reaction.

If the order of a particular reactant is zero however, you can say that its concentration won't affect the rate of reaction (within certain limits, as no reaction can be entirely independent of the concentration of an involved reactant).


Can the regions of 1s and 2s subshells overlap?

Subshells are not a real physical objects in the sense that we can't measure them, can't designate them. There are just a mathematical abstraction. So, there are meaningless in terms of physics to ask about their overlap. It can lead you to misunderstanding.

Let's look at the region where 1s-electron can be found with 90% probability. There are non-zero probability of 2s-electron being found there. Even if you'd consider the region where 1s-electron can be found with 10% probability, there are still non-zero probability of 2s-electron being found there. So, at first you should understand that 2s (2p, 3s, 3p and so on) electron have non-zero probability to be found anywhere except where its probability density function equals to zero.

Now let's say we have performed infinite number of measurements of 1s and 2s electrons coordinates. Let's discard 10% outliers in both cases. Now we have clusters which contain 90% probability of electron being found there. Will these regions overlap? Maybe. The answer is one can calculate how much outliers one should "remove" to obtain "non-overlapping regions" (I can't calculate this, unfortunately). For 1s and 2s it may be 95%-regions, or may be 89%-regions that not overlapped. And theoretically, one can perform such calculation for any pair of orbitals he wants.


Would hydrogen bond form in these scenarios?

The molecule your example, 3-pentanone, actually has a water solubility of $\pu{50 g/L}$ at $\pu{20^oC}$. The water solubility of it's pure hydrocarbon analog, n-pentane, is $\pu{40 mg/L}$ at $\pu{20^oC}$. A simplistic interpretation of this comparison is that the addition of a single hydrogen-bond acceptor functionality to n-pentane increases it's water solubility by greater than three orders of magnitude! So the answer to your question is pretty clearly that yes, hydrogen bonding is happening in this scenario.


Yes, the statement in your book is very vague.  Molecules are often classified by the symmetry elements they contain.  For example, $\ce{BH3}$ contains a $C_3$ axis perpendicular to the plane containing the $\ce{BH3}$ molecule.  "$C_3$" means that if you rotate the molecule around this axis by $360^\circ/3 = 120^\circ$, you'll get a molecule that is indistinguishable from the molecule you started with.  $\ce{BH3}$ also contains 3 $\sigma_\mathrm{v}$ planes of symmetry.  These planes contain the $C_3$ axis and one of the $\ce{B-H}$ bonds.  $\ce{BH3}$ also contains one $\sigma_\mathrm{h}$ symmetry plane that is perpendicular to the $C_3$ axis and contains the $\ce{BH3}$ molecule.  In addition to these symmetry elements, a number of other symmetry elements exist as well.

It turns out that only molecules that belong to symmetry classes (point groups)

$C_n$ (the molecule only contains a $C_n$ axis
$C_{n\mathrm v}$ (the molecule contains a $C_n$ axis and $n$ total $\sigma_\mathrm{v}$ planes)
$C_{\mathrm s}$ (the molecule only has a plane of symmetry that contains
the entire molecule)

can have a permanent dipole moment.  If the molecule belongs to any other point group it cannot have a dipole moment.  Since $\ce{BH3}$ contains a $\sigma_\mathrm{h}$ plane of symmetry ($\ce{BH3}$ belongs to point group $D_{3\mathrm h}$) - it therefore does not belong to groups $C_n$, $C_{n\mathrm v}$ or $C_{\mathrm s}$ - and cannot have a dipole moment.


Why does Calcium have a higher ionization energy than Aluminium?

Ionization energy is the minimum amount of energy needed to remove most loosely bound electron (valence electron).
The atomic number of Calcium  is 20. So it's orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s$2 3p^6 4s^2}$. This is the arrangement of electrons in the shells. No free electron is present in the above configuration, all the electrons are bonded. So for removal of an electron from the outermost shell higher amount of energy is required.
In case of Aluminium, the atomic number is 13. It's orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^1}$. One single electron is present unbonded in outermost shell, so it can be easily removed as compared to Calcium.
So, ionization energy of Calcium is higher than the ionization energy of Aluminium.


Is there a clear distinction between the various intra- and intermolecular interactions?

No, there aren't clear distinctions, particularly between intramolecular and intermolecular interactions.

First, a caveat that we generally divide molecules along covalent bonds. So those are strictly intramolecular, in my opinion. (Network solids are a bit tricky to classify.)

Let's go through the rest one at a time.

Dipole-dipole: These are usually intermolecular, but certainly in larger, more flexible molecules, intramolecular polar-polar interactions can change the molecular shape and preferred conformation. (Think about tying together two bar magnets on one piece of string. Certainly they'll interact.)
Hydrogen bonds: Yes, many of these are intermolecular. However, there definitely are intramolecular hydrogen bonds:

Van der Waals: Again, these can be intermolecular. In even a moderate molecule, intramolecular non-bonded van der Waals interactions can change conformations and geometries. Consider axial/equitorial preferences in cyclohexanes or other types of "steric clashes" because of big bulky groups. These are the effects of van der Waals interactions.

OK, so all the intermolecular interactions are also intramolecular ones too.

Hydrophobic interactions: I usually think of these as intermolecular but similar considerations make sense as intramolecular for the reasons I outline under van der Waals interactions above.
Ionic: I'll just think of these as different partial charges in a molecule. Clearly electrostatic interactions occur both inside a molecule (e.g., electrostatic non-bonded interactions changing conformations and geometries) and as intermolecular forces, e.g., water-cation interactions.

So besides covalent interactions, which we use to divide "inside" and "outside" a molecule, all of these are both intermolecular and intramolecular interactions.

I'm not sure I've seen a great chart. Maybe someone else can comment.


How much time liquid mercury takes to evaporate?

I found this statement in a paper about $\ce{Hg}$ contamination from dental work:

  The evaporation rate of elemental mercury at room temperature ($\pu{20 ^\circ C}$) is approximately $\pu{50 \mu g\:cm-2h-1}$ (range of $\pu{40-60 \mu g\:cm-2h-1}$).

They cited the following work as a source for that number:

  Gary N. Bigham, Wanyu R. Chan, Manuel Dekermenjian, Ali Reza, "Indoor Concentrations of $\ce{Hg}$ Vapor Following Various Spill Scenarios," Environmental Forensics 2008, 9(2-3), 187-196 (https://doi.org/10.1080/15275920802121975)

German regulations (as an example) say that 0.035 µg/l of metallic mercury vapour in air pose no risk under lifelong exposition, and 0.35µg/l can be harmful under long (months?) exposition for some people. The math is full of imponderables, esp. the actual amount that was spilled and size of drops.

Swiss officials https://www.toxinfo.ch/quecksilberhaltige-fieberthermometer say that a single broken fever thermometer can lead to intoxications, but the risk can usually be averted by collecting as much as possible and ensuring proper, regular ventilation.

The following publication suggests that a properly cleaned up spill from a fever thermometer should become undetectable within several weeks, but the airborne Hg concentration can be significant in that time, and regular ventilation is highly recommendable.

  E. Martin Caravati et al. Elemental mercury exposure: An evidence-based consensus guideline for out-of-hospital management, Clinical Toxicology 2008, 46:1, 1-21, https://doi.org/10.1080/15563650701664731

But one thing is quite clear: A single, large blob that you miss (under furniture, wooden floorboard etc.) can stay for a very long time, and lead to potentially harmful concentrations in the air. There are a lot of reports about little mercury puddles that were found under floorboards during laboratory renovations, even decades after the last mercury instrument had been thrown out.


How does the mixing of two gases increase entropy?

Here is a reversible process to do what you are asking for (i.e., unmixing the gases reversibly).  You have a cylinder containing the two gases already mixed.  The cylinder has a piston.  Connected to this main cylinder are two other cylinders joined to it through semipermeable membranes.  Each membrane allows one of the gases to pass, but not the other.  You gradually advance the piston in the main cylinder while, at the same time moving the other pistons outward in the other two cylinders so that, at any time, the pressures of the pure gases in the attached cylinders are equal to their partial pressures in the main cylinder.  The change in entropy for this adiabatic reversible process is zero.  This is the first step.  Now, in the second step, we close off the semipermeable membranes, and compress each of the pure gases in the two cylinders isothermally and reversibly to the original total pressure that was present in the main chamber.  So, if $p_1$ was the partial pressure of one of the gases in the main chamber and $p_2$ was the partial pressure of the other gas in the main chamber, the total pressure of the pure gases in their chambers after completing step 2 will be $p_1+p_2$.  The change in entropy for the 2-step process we described is minus the entropy of mixing.  So, if the process is done in reverse, the entropy of mixing will be positive.


Discovery of optical activity of enantiomers - where did the idea of using polarized light came from?

It wasn’t a standard lab procedure to test the effects of polarised light on a certain compound/solution/whatnot. However, back in the day it was standard lab procedure to re-run a certain experiment with practically everything your lab had: water, acids, solvents, table salt, cat hair, human hair, dust, fabric, … You would note if you observed an interesting effect and, where applicable, if that interesting effect would only occur in some cases.

So once you are aware that something like Iceland spar produced interesting phenomena (even though you might not be able to explain these phenomena just yet) you would start trying out all sorts of things to test what this interesting optical phenomenon is capable of doing. By chance, you would run across something that you would be able to observe as cancelling of light, even though all components of the system would let light through — and then you would attempt at understanding what you just observed.

They say science is merely grown men playing around for a reason.


According to Pauling's famous The Nature of the Chemical Bond , 3rd edition, at page 73:

  In the hydrogen molecule a quantum-mechanical treatment has shown that the two ionic structures $\ce{H+H-}$ and $\ce{H- H+}$ enter into resonance with the extreme covalent structure $\ce{H-H}$ only to a small extent, each ionic structure contributing only about 2 percent to the normal state of the molecule.

He goes on to state that other diatomics such as chlorine should have even less ionic character.

So all molecules, even homonuclear diatoms, have some ionic character.

For further information on diatomic hydrogen's ionic character see section 1.5 of the book and The Normal State of the Hydrogen Molecule Journal of Chemical Physics vol. 1, pages 593-596.

See also Overlaps of Trial Functions for the Hydrogen Molecule. II. Covalent and Ionic Character of H2 J. Chem. Phys. 23, 176


How much heat (in J) is required to produce 2.0L of N2(g) at 77.36 K and 1.0 atm?

Since nitrogen's boiling point is 77.36 K , it appears that the question is asking how much heat is required to boil nitrogen to produce 2.0 L at 1 atm.


It relates to the Hückel rule for aromatics. With the electron pair of the oxygen participating in the $\pi$ system of the oxazol (5-membered ring)  we get 6 $\pi$-$\ce{e-}$. Doing the same for the oxazepin would get 8 $\pi$-$\ce{e-}$.

Hückels rule states that planar, circular $\pi$ systems with $4n+2$ electrons are aromatic, resulting in a lower overall energy if the system is conjugated. (with $n$ being a positive integer or zero)
The other statement that Hückels rule does is that $\pi$ systems with $4n$ electrons are anti-aromatic, which avoid conjugation as it would increase the overall energy of the system.

Therefore it is favoured for your oxazol to conjugate, which results in a sp$^2$-hybridization for the oyxgen, while the oyxgen in the oxazepin avoids conjugation, resulting in a sp$^3$-hybridization.

The Hückel rule in turn can be quantized by a Hückel method computation or  graphically utilizing a Frost-Musulin-diagram.


Writing atoms of functional group backwards

They mean the same thing. If the OH group on the left were written as OH, it would look like there was a carbon-hydrogen bond. By inverting the order, it makes it clear that the oxygen is bonded to the carbon and the hydrogen is bonded to the oxygen. $\ce{NH2}$ might also be written as $\ce{H2N}$ in a structure drawing if a bond was coming from the right, but the functional group is the same in both cases—the group is connected to the rest of the molecule by the nitrogen atom.


Can a long polymer chain interact with itself via van der Waals forces?

You have a possible answer to your question in proteins, an example which includes some long polymer chains.

Intramolecular interactions - while not necessarily the driving force for formation of a collapsed protein globule (usually argued to be due to the hydrophobic effect, requiring intermolecular interactions) - are the basis for higher order structure in proteins and even some forms of RNA. I won't mention DNA since most commonly this forms a bimolecular double helix.

All of the interactions you mention may play a role in stabilizing a particular protein structure or "fold". A good and basic example of protein structure acquisition due to intramolecular interactions in proteins is polyalanine, which acquires alpha-helical secondary structure stabilized by intramolecular hydrogen bonds, typically between amide groups separated by 2 amino acids along the chain. Other types of structure stabilized by intramolecular interactions (H-bonds, usually) in polypeptides include other types of helices, beta sheets and various turns.


Can oxygen gas in air react with Mn3O4 to form MnO2?

Manganese(IV) oxide decomposes around 500 °C (German Wikipedia lists $\pu{450 °C},$ [1, p. 396] lists $\pu{530–585 °C}):$
$$\ce{4 MnO2 ->[\sim\pu{500 °C}] 2 Mn2O3 + O2}\label{rxn:1}\tag{R1}$$
On the other hand, according to [2, p. 12], $\ce{Mn3O4}$ reacts with oxygen only above $\pu{500 °C}$ and forms manganese(III) oxide:
$$\ce{4 Mn3O4 + O2 ->[>\pu{500 °C}] 6 Mn2O3}\label{rxn:2}\tag{R2}$$
Upon further heating, the reaction in the opposite direction takes place [1, p. 396]:
$$\ce{6 Mn2O3 ->[>\pu{940–1090 °C}] 4 Mn3O4 + O2}\label{rxn:3}\tag{R3}$$
All in all, the third reaction you've proposed $(\ce{Mn3O4 + O2 -> 3 MnO2})$ is unlikely to occur and the product will be $\ce{Mn2O3},$ which unfortunately would break the catalytic cycle.
As for the first two reactions, they are described in the literature all right [1, pp. 397, 399]:
$$
\begin{align}
\ce{MnO2 + SO2 &->[\pu{450 °C}] MnSO4}\tag{1}\\
\ce{3 MnSO4 &->[\pu{850–1155 °C}] Mn3O4 + SO2 + 2 SO3}\tag{2}\\
\end{align}
$$
Reactions \eqref{rxn:1}–\eqref{rxn:3} are also described in Greenwood's Chemistry of the elements: [3, pp. 1048–1049]:

It is $\ce{MnO2},$ however, which is by far the most important oxide in this group, though it is not the most stable oxide of manganese, decomposing to $\ce{Mn2O3}$ above about $\pu{530 °C}$ and being a useful oxidizing agent. […]
$\ce{Mn3O4}$ is formed when any oxide of manganese is heated to about $\pu{1000 °C}$ in air […].
If $\ce{MnO2}$ is heated less strongly (say $,\pu{800 °C})$ than is required to produce $\ce{Mn3O4},$ then the more stable α-form of $\ce{Mn2O3}$ results […].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)
Turova, N. Inorganic Chemistry in Tables; Springer: Heidelberg ; New York, 2011. ISBN 978-3-642-20486-9
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.


Is a buffer solution really inactive beyond a range of pKa±1?

The 10:1 ratio is really just something convenient for the chemists.  The buffer still works outside this range.  True, its effectiveness against base addition decreases with the amount of available acid at high pH, and vice versa for acid addition at low pH.  But if enough capacity remains, you can still live with the buffer even outside the 10:1 range.

Er, actually you do.  Our main blood buffer, based on dissolved carbon dioxide as the acid and bicarbonate as the base, has about 20 times as much bicarbonate as carbon dioxide (https://en.m.wikipedia.org/wiki/Bicarbonate_buffer_system).  Factors other than what a laboratory chemist calls "optimal" should be given their due.  Metabolic processes commonly produce (weak) acids, and the solubility of carbon dioxide is more limited than that of bicarbonate salts.  Both of these factors make a base-rich buffer desirable.


Alkene reaction with acid via SN1 reaction

My suspicion is that the resonance structures accessible to the allylic carbocation provide a lower energy state than the formation of the C--H single bond and the isolated carbocation that your expected pathway would require.

In order of reactivities, you are right to think that the alcohol is the controlling factor because of its pKa. Under acidic conditions it will behave more as a base than the alkene (It is much more basic than an alkene, or think of the Oxygen in an alcohol as being generally a stronger nucleophile than the alkene). This was one main consideration my O-Chem professor stressed was for us to be aware of how other functionalities in our molecules would behave under the reaction conditions of other steps (Eg, if going into a strongly acidic/basic environment after installing a reactive group). I am not an organic chemist by training, but I suspect that if you wanted to access the alkene reactivity, you would want some kind of alcohol protecting group.

This page has a brief discussion about three classes of nucleophile (and the next page describes what makes a good nucleophile): https://www.masterorganicchemistry.com/2011/03/04/the-three-classes-of-nucleophiles/


Do Bases and Acids actually react with each other?

$$\ce{NaOH + HCl -> NaCl + H2O}$$

$$\ce{Na+ + Cl- + OH- + H+ -> Na+ + Cl- + H2O_{(l)}}$$
Both sides $\ce{Na+}$ ans $\ce{Cl-}$ cancel out remaining reaction

$$\ce{H+ + OH- <=> H2O_{(l)}}$$

Any strong acid and strong base reaction the remaining reaction is hydrolization of water.


What is the difference in mosM and mOsm units, if there is any at all?

I think the clinical chemistry notation has not been standardized but certainly it has been inspired by chemists.
The capital M in mosM emphasizes that it is milliosMolar. Chemists use capital M to denote molarity.
On the other hand, mosm indicates milliosmolal. Chemists use small m to symbolize molality.
This notation is not universal.
From this slightly dated article on Osmolality, which is behind pay-wall
Osmolality, Harry F. Weisberg Laboratory Medicine, Volume 12, Issue 2, 1 February 1981, Pages 81–85.

Terminology and Units
In 1934, Gamble, et al.,11 introduced the term ”osmillimole” which has been successively transformed into mosM (milliosmolar), mosm (milliosmolal), the noncommittal mOsm, and the present-day mosmol (milliosmole). Raoult’s law states that the lowering of the vapor pressure (one of the four ”colligative”
properties of dilute solutions) is proportional to the mole-fraction (more easily calculated from molality) of solute present in the mixture.


Why is 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) often considered a prototypical room temperature ionic liquid?

This is a very good question. These compounds have been studied for long enough that most people working with them "know" the answers to these questions, but the answers are buried in physical chemistry articles from 20 or more years ago. The imidazolium tetrafluoroborate salts are popular RTILs because 1) they have the "right" properties, and 2) they are easy to make.

Concerning these compounds having the "right" properties:
Most RTILs require two characteristics in the their constituent ions to minimize the intermolecular order which would cause them to solidify. First, they need a cation-anion pair that cannot associate strongly, and second they need to have lower symmetry. Immidazolium tetrafluoroborate (BF4-) salts have both of these. The aromaticity of imidazole is important. The positive charge of the imidazolium cation is distributed through the resonance of the aromatic pi system. The negative charge in BF4- is distributed inductively over the four fluorine atoms. Additionally, imidazolium is planar and BF4- is spherical. Both the diffuse charge distribution and poor shape match prevent these two ions from having a strong association. The two different alkyl groups (ethyl and methyl) on the imidazolium ring are enough to break the symmetry of the system and prevent more efficient packing, which would lead to the substance being solid.

Concerning their ease of synthesis:
The imidazolium salts are easy to prepare from imidazole and various alkyl halides, all commercially available and inexpensive. The predominance of the BF4- anion is also a result of the synthesis. The synthesis of an imidazolium salt RTIL begins with imidazole and and an alkyl halide, say chloromethane. The product of this step is 1-methylimidazolium chloride. This salt is reacted with another alkyl halide, say chloroethane, in the presence of a base, say triethylamine. The products of the this step are 1-ethyl-3-methylimidazolium chloride and triethylammonium chloride. 1-ethyl-3-methylimidazolium chloride is a solid, because the chloride ion is small, with concentrated charge. The tetrafluoroborate salt is prepared by salt metathesis. 1-ethyl-3-methylimidazolium chloride is dissolved in water and an aqueous solution of silver tetrafluoroborate (AgBF4) is added. Silver chloride, AgCl, is virtually insoluble in water (Ksp = 1.77 x 10-10) and precipitates. The solution is filtered, and the filtrate is concentrated to produce 1-ethyl-3-methylimidazolium tetrafluoroborate (EMIM BF4-). AgBF4 and AgPF6 are commercially available, so you will see those ions more frequently. The other ions, while they have good properties for ionic liquids, are not usually available as silver salts, so those reagents have to be prepared. You are already doing three reactions to get to EMIM BF4-. Who wants to do more synthesis just to make their solvents?


Ab initio means calculations that are strictly from the principles of quantum mechanics. This contrasts with a non ab initio method which can involve parameters that do not arise directly from principles (i.e. results from experiment and/or 'better' calculations). One typical example of something that is not ab initio is a functional for Density Functional Theory. Because the exact functional is unknown, different functionals are generated using large datasets of high level calculations. These are often good at reproducing experimental data in many situations however you cannot derive them directly from the postulates quantum mechanics.

Molecular Orbital Theory (which can be implemented with the Hartree-Fock method) is an example of an ab initio method. One can (very roughly) come up with MO theory by approximating each electron as only feeling the effect of the average positions of the other electrons. So, if an orbital has an electron in it all the other electrons will feel a repulsion from the space occupied by that orbital. This results in a set of coupled one particle problems the solutions (eigenfunctions) which can be solved to find the Molecular Orbitals for the system.

More generally, ab initio calculations also refer to the more accurate post Hartree-Fock methods such as Configuration Interaction (CI), and Coupled Cluster (CC).


Transition state optimisation on the surface of periclase

To study surface reactions I recommend the growing string method (GSM) for surfaces. Here is a nice quote from the paper which developed it (Ref. 1)

  GSM’s efficacy was confirmed by comparison with CI-NEB on
  an extensive set of reactions characteristic of modern surface chemistry studies. In these cases, GSM reduces the computational cost (in terms of gradient computations) by about 45% on average over CI-NEB. In addition to high efficiency, GSM has the advantage of operating in single-ended way to enable explorative study of chemical reactions.

Key to the success of GSM is the use of the efficient hybrid delocalized coordinate system which greatly improves optimization and interpolation steps in comparison to NEB which uses Cartesian coordinates.

The exact code to reproduce their work is not publicly available, although you may be able to request a copy. However you should be able to reproduce it and more, with this python version of the code that I have developed.

Also, as others have pointed out, you should consider a higher level of theory like periodic wavefunction DFT (e.g. as implemented in VASP) .

References:

Jafari, M, Zimmerman, P. M.. "Reliable and efficient reaction path and transition state finding for surface reactions with the growing string method", J. Comput. Chem. 2017, 38, 645– 658. https://doi.org/10.1002/jcc.24720


In a galvanic cell, why do electrons move through the wire from zinc to copper?

If we consider the classical Daniel cell
$$\ce{Zn|Zn^2+||Cu^2+|Cu|}\text{(wiring  to Zn)}$$

then due higher tendency of zinc ions to    leave the metal, getting hydrated in solution, the zinc electrode obtains more negative electrostatic potential due extra electrons, compared to cupper electrode.

As electrodes are galvanically connected, electrons flow from more negative potential of the $\ce{Zn}$ electrode toward more positive $\ce{Cu}$ electrode, increasing potential of the former and decreasing the potential of the latter.

The $\ce{Cu}$ electrode, with the potential lower then its equilibrium potential, reduces $\ce{Cu^2+}$ to $\ce{Cu}$.

The $\ce{Zn}$ electrode gets the higher than equilibrium potential, as electrodes leave it by the wire, so more $\ce{Zn^2+}$ ions are transferred to $\ce{Zn^2+}$ salt solution, to approach the equilibrium potential.

Ions in both solutions, connected by diaphragm or salt bridge, migrate along the potential and concentration gradients, caused by unequal potentials and electrode reactions.

The whole cycle is closed.


Liquifying metal and toxic fume exposure

You ask:

  Aside from burning yourself horrendously, what dangers are associated with the melting of (we can assume nearly pure) tin, particularly in a closed space? Is what he did really dangerous?

According to the SDS for tin metal from Sigma-Aldrich, tin metal (at the purity level for the product mentioned here, which is <= 100% and may not be that of what was used in the video you mentioned) is not a hazardous substance or mixture.

There is some danger specifically relating to inhalation of tin dust or aerosols (see sections 6.1 and 7.1 in the linked document):

  6.1 Personal precautions, protective equipment and emergency procedures
  Use personal protective equipment. Avoid dust formation. Avoid breathing vapours, mist or gas. Avoid breathing dust.

  7.1 Precautions for safe handling
  Avoid formation of dust and aerosols. Further processing of solid materials may result in the formation of combustible dusts. The potential for combustible dust formation should be taken into consideration before additional processing occurs. Provide appropriate exhaust ventilation at places where dust is formed.

In section 8.2 (Exposure controls) it is noted that:

  Respiratory protection
  Respiratory protection is not required. Where protection from nuisance levels of dusts are desired, use type N95 (US) or type P1 (EN 143) dust masks. Use respirators and components tested and approved under appropriate government standards such as NIOSH (US) or CEN (EU).

In section 11.1 (Information on toxicological effects, a subsection of Toxicological Information), information pertaining to toxicity can be found. Inhalation here will refer to tin dust or aerosols that can be inhaled:

  Acute toxicity

  LD50 Oral - Rat - male and female - > 2,000 mg/kg (OECD Test Guideline 423)

  LC50 Inhalation - Rat - male and female - 4 h - > 4.75 mg/l (OECD Test Guideline 403)

  Dermal: No data available

If the tin was/is at the near 100% purity level, the main risks associated with handling it are associated with the formation of dust (workplace inhalation limits are given in section 8). The HMIS (Hazardous Materials Identification System) ratings in section 16 are 1 for health hazard, 0 for flammability and physical hazard, and no value assigned for chronic health hazard.

Conclusion: Assuming the person in the video did not inhale or ingest tin dust (a proposition that largely becomes moot after it has melted, in any case), there appears to be little to no danger associated with the activity with respect to melting tin in a closed space (especially such a small amount of it).


deduce chemical properties of a compound from elements

is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments?

Not yet, assuming you mean extrapolating such information from the molecule's structure

Physical and chemical properties are hard enough to extrapolate from a compound's structure, let along its composition. The entire field of theoretical chemistry has been working on predicting the structure-property relationship for several decades. Great strides have been made, but accurate physical property prediction has not been achieved (though I would love to be proven wrong). In most cases, very accurate computation of properties of all but the smallest molecules takes more computing time than you would need to do the experiment.

Your question implies an interest in deriving such properties from the formula. As Canageek said, this is not possible, given the phenomenon of isomerism. One example I just used with my class is that there are 22 known molecules sharing the formula $\ce{C6H12}$, which is a relatively simple formula. Two of the more common are1-hexene and cyclohexane, which you can see have very different structures and properties (chemical and physical).


Nucleophilic substitution in polar protic/aprotic solvents

It depends whether the reaction progresses through SN1 or SN2. In both, the departing of a leaving group is involved in the rate determining step, but only in SN2 is the nucleophile included in the rate determining step. In the following mechanisms, both of the first steps are rate determining.
(Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction)

It is important to remember that a good leaving group is defined by poor basicity, which can be quantified by a low pKa of the conjugate acid of the leaving group. Intuitively, this means the conjugate acid can lose a proton and the resulting base - our leaving group - will be reluctant to re-react with the proton. Or in our case, re-react with the electrophilic carbon. This generally means that good leaving groups are large, low charge density species; good examples are neutral molecules or even charged species like iodide ions.
This idea of strong and weak nucleophiles is situational. Hard-soft acid-base theory becomes a very useful reference in a lot of situations. It generally states that low charge density 'soft' nucleophiles and electrophiles have an affinity for eachother, and high charge density 'hard' nucleophiles also have an affinity for eachother. You can read more about exactly why this happens here:
https://en.wikipedia.org/wiki/HSAB_theory
Generally, we consider a saturated carbon a 'soft' electrophile, which will react very well with low charge density nucleophiles like iodide ions. What this means, is that in SN2 reactions, species that are considered good leaving groups are also good nucleophiles. For instance, you may predict - correctly so - that iodide ions are good leaving groups as well as good nucleophiles for SN2; this is why KI can be used to catalyse SN2 reactions by acting as both (usually better than your actual nucleophile and leaving group)!
(Image from Clayden Organic Chemistry)

These reactions can end up reasonably reversible, and can be driven to one side by modifying conditions. For instance, the reaction below (from Clayden organic chemistry) exploits the insolubility of a byproduct, effectively removing it from the equilibrium, shifting it to the right:
(Image from Clayden Organic Chemistry)

In SN1, nucleophilicity is irrelevant, since the nucleophile doesn't appear in the rate determining step, but I'd imagine the equilibrium position can be roughly estimated by relative leaving group abilities.
In answer to your specific example, fluoride would actually be a poorer SN2 nucleophile than you might think, worse than the bromide leaving group would be. The polar protic solvent will even further inhibit the nucleophilicity of the fluoride due to hydrogen bonding, slowing it down dramatically (so your assumption is accurate for kinetic reasons); and once it has been substituted, the fluoride ion is not a particularly good leaving group. In DMF, you don't have the hydrogen bonding issue, and due to poor solvation of anionic species in DMF, the fluoride is made slightly more reactive than it would be 'on its own'. And leaving group ability is also reduced due to poor solvation. This seems like it would be clear-cut; the fluorine will replace the bromine, right? However the bromide ion is a good nucleophile in this situation, and has the potential to kick off the fluoride in either solvent.
As you can see, there are a lot of factors to juggle here, and the actual degree to which the reaction progresses is specific to the exact conditions of the reaction. At a first glance, it is very difficult to predict, and is not necessarily as simple as the 'better' nucleophile replacing the 'worse'.


carbon dioxide in liquid and gaseous phase with liquid water

At a given temperature a mixture of gaseous $\ce{CO2}$ and liquid $\ce{CO2}$ will have a particular pressure. Thus more liquid $\ce{CO2}$ doesn't mean that more $\ce{CO2}$ will dissolve in the water.

Obviously liquid $\ce{CO2}$ has a lot more matter per unit volume than gaseous $\ce{CO2}$. So a little liquid $\ce{CO2}$ makes a lot of $\ce{CO2}$ gas.


How does brewing time and brewing temperature affect the flavonoid concentration of green tea?

You might want to have a look at

J. Peterson et al., Major flavonoids in dry tea, Journal of Food Composition and Analysis, 2005, 18, 487--501, http://dx.doi.org/10.1016/j.jfca.2004.05.006

and the references therein, e.g.

Christiane Lakenbrink et al., Flavonoids and Other Polyphenols in Consumer Brews of Tea and Other Caffeinated Beverages, J. Agric. Food Chem., 2000, 48*, 2848--2852, http://dx.doi.org/10.1021/jf9908042

From the abstract of the latter:

  [...] At brew times up to 2 min the composition of the brew solids was for each product practically independent of brew time, with flavonoids again comprising the major proportion (86−88%) of the total phenolics. [...]

Longer extraction times (> 10 min) or continuous (Soxhlet) extraction lead to partial decomposition of the flavonoids, most likely due to phenolic oxidation, cf.

Won Jo Cheong et al., Determination of Catechin Compounds in Korean Green Tea Infusions under Various Extraction Conditions by High Performance Liquid Chromatography, Bull. Korean. Chem. Soc., 2005, 26, 747--754, http://dx.doi.org/10.5012/bkcs.2005.26.5.747

I hope that helps a bit.


What free software exists to view molecule pictures within a file explorer?

I found this tool named thumbfish (Windows only). It can display thumbnails for variety of formats and is free


Yield of possible products from acid catalysed hydration of an alkene with a phenyl group

A basic principle in OChem says that stability of intermediate is the key factor in deciding the major product, in general. Another factor is stability of final product.

Will a kinetically favoured (minor) product (c) form?

It will form, as far as it's intermediate is known to exist (and it exists). However, that will be present in very minor amount, as the intermediate of it will not be as stable as that of (b) and (d). Additionally, the stable intermediate of (c) has 7 α-H, while that of (a) has 2 α-H, so product (c) will be major than product (a).

Here, (d) may be the major product due to high temperature.

The intermediates of (b) and (d) are equally stable. So, here we use the second factor. By looking at the structures of product (b) and product (d), we can say that (d) is major product due to extended conjugation and hyperconjugation with 9 α-H. Really?
The answer is YES. It is known that acid-catalyzed hydration reaction is highly exothermic$1$, which is why it is favored at low temperatures! (In accordance with the "Le Chatelier's Principle", a system will consume heat when temperature of surrounding is raised).
Now, I think, it's pretty clear what the order of yield of products should be :)

References
$1$: https://en.wikipedia.org/wiki/Hydration_reaction#Alkenes


Fischer indole synthesis: significance of choice of acid catalyst

I'd like to take this opportunity to give a nutshell review of the Fischer indole synthesis for benifit of future readers. This highly useful synthetic tool, which was named after the famous chemist, Emil Fischer who discovered it in 1883 (Ref.1 & 2)), is considered to be one of the best methods for preparing indoles. Hence, needless to say, it is the most widely used and versatile method for indole synthesis. The method simply converts converts ayrlhydrazones ($\bf{1}$) of an aldehyde or a ketone into corresponding indole ($\bf{8}$) in the presence of an acid catalyst (Ref.3 & 4). The overall reaction is depicted in the top box in following image:

The proposed mechanism of The Fischer indole synthesis is also included for the convenience. As indicated, ayrlhydrazones ($\bf{1}$) are formed from a condensation reaction between an arylhydrazine and an aldehyde or ketone (reference to 2,4-dinitrophenylhydrazine test for aldehydes and ketones). In the presence of acid (Brønsted or Lewis), imine $\bf{1}$ is in equilibrium with $\bf{2}$ (protonated imine), $\bf{3}$ (enamine form), and $\bf{4}$ (protonated inamine). Once formed, $\bf{4}$ undergoes [3,3] sigmatropic rearrangement (ene-reaction; could visualize as azo-Cope rearrangement), forming a new $\ce{C-C}$ bond to give $\bf{5}$ (this is a key step of the mechanism). Imine $\bf{5}$ gains back aromaticity to form $\bf{6}$, which is the most stable isomer between two. The anilinic $\ce{NH2}$ of $\bf{6}$ attacks electron deficient imine group in the side chain ($\ce{C}$ of $\ce{CNH2+}$ group) intramolecularly to give the corresponding dihydroindole derivative, $\bf{7}$ (which is also another key step), which eventually release $\ce{NH4+}$, $\ce{NH3 + H+ <=> NH4+}$, to give stable indole derivative,  $\bf{8}$. Reference 3 has given excellent review on the mechanism and hence, for benefit of interested reader, the PDF has been attached.
The key feature to produce corresponding indole is need of a $\ce{CH2}$ group as noted (as indicated by a green circle; $\alpha$ to the original ketone or aldehyde). If this $\alpha$-$\ce{C}$ is $2^\circ$, then $\bf{7}$ would still release $\ce{NH3}$ to give corresponding cyclic imine derivative (Ref.5). For example, compound $\bf{10}$ was formed ( Ref.5) when $2^\circ$ $\alpha$-$\ce{C}$ is attacked selectively at $\bf{3}$ to $\bf{4}$ conversion (thermodynamically stable enamine):

However, except for the situation to make $\bf{10}$, and the intention is making indole derivative of interest, the stereo selectivity at sigma-tropic rearrangement to give $\bf{5}$ and thereafter would not matter since at the end, all of stereocenters involved would be diminished (see the scheme above). Therefore, the choice of acid to get selectivity in general synthesis of indoles is not as important as it was told elsewhere. In reality, a variety of catalysts (Table I in Ref.3) have been used to effect the indolization of arylhydrazones. According to the Ref.3 (an excellent review of the Fischer indole synthesis):

Generally speaking, it may be said that acids and many compounds capable of complex formation with organic ligands may act as catalysts.

In addition to the variety of catalysts listed in Table I of Ref.3, there are wide variety of other acid catalysts including organic acids such as tartaric acid (e.g., Ref.5) and p-toluenesulfonic acid (e.g., Ref.6). The bottom line is the indole formation by this method is independent of acid catalysis (e.g., Ref.7), although it is markedly accelerated by it.
It is also worth noting that this reaction has proven very useful for the pharmaceutical synthesis of antimigrane drugs (e.g., Ref.8 & 9).

References:

Emil Fischer, Friedrich Jourdan, "Ueber die Hydrazine der Brenztraubensäure," Berichte der deutschen chemischen Gesellschaft 1883, 16(2), 2241-2245 (https://doi.org/10.1002/cber.188301602141).
Emil Fischer,  Otto Hess, "Synthese von Indolderivaten," Berichte der deutschen chemischen Gesellschaft 1884, 17(1), 559-568 (https://doi.org/10.1002/cber.188401701155).
B. Robinson, “The Fischer Indole Synthesis,” Chem. Rev. 1963, 63(4), 373–401 (https://doi.org/10.1021/cr60224a003)(PDF).
B. Robinson, “Studies on the Fischer indole synthesis,” Chem. Rev. 1969, 69(2), 227–250 (https://doi.org/10.1021/cr60258a004).
Sangram Gore, Sundarababu Baskaran, Burkhard König, “Fischer Indole Synthesis in Low Melting Mixtures,” Org. Lett. 2012, 14(17), 4568–4571 (https://doi.org/10.1021/ol302034r).
Seble Wagaw, Bryant H. Yang, Stephen L. Buchwald, "A Palladium-Catalyzed Strategy for the Preparation of Indoles:  A Novel Entry into the Fischer Indole Synthesis," J. Am. Chem. Soc. 1998, 120(26), 6621-6622 (https://doi.org/10.1021/ja981045r).
Okiko Miyata, Yasuo Kimura, Kanami Muroya, Hajime Hiramatsu, Takeaki Naito, "Thermal cyclization of N-trifluoroacetyl enehydrazines under mild conditions: A novel entry into the fischer indole synthesis," Tetrahedron Lett. 1999, 40(18), 3601-3604 (https://doi.org/10.1016/S0040-4039(99)00583-3).
Paul R. Brodfuehrer, Bang-Chi Chen, Thomas R. Sattelberg, Patrick R. Smith, Jayachandra P. Reddy, Derron R. Stark, Sandra L. Quinlan, J. Gregory Reid, John K. Thottathil, Shaopeng Wang, “An Efficient Fischer Indole Synthesis of Avitriptan, a Potent 5-HT$_\mathrm{1D}$ Receptor Agonist,” J. Org. Chem. 1997, 62(26), 9192–9202 (https://doi.org/10.1021/jo971368q).
Majid M. Heravi, Sahar Rohani, Vahideh Zadsirjan, Nazli Zahedi, “Fischer indole synthesis applied to the total synthesis of natural products,” RSC Adv. 2017, 7, 52852-52887 (DOI: 10.1039/C7RA10716A).


For a given atomic kernel, the orbitals energies are not fixed like a set of shelves. Their energy depends on filling themselves and also on filling the other orbitals by electrons, because of kernel shielding and mutual repulsion.
The mutual energy is $4\mathrm{s} \lt 3\mathrm{d}$ when empty. But as kernel charge gets increased and $\mathrm{4s}$ and then $\mathrm{3d}$ are being filled, their energy order gets switched.
The typical case is the 3d energy drop $\ce{K -> Ca -> Sc}$, where the order is:
$$\mathrm{K: 4s \lt 4p \lt 3d}$$
$$\mathrm{Ca: 4s \lt 3d \lt 4p}$$
$$\mathrm{Sc: 3d \lt 4s \lt 4p}$$
well described on  charts in SE Chemistry Q what-is-d-orbital-collapse, which are related to the effective kernel charge acting upon electrons on various orbitals.


Is the rate of an SN1 reaction really independent of the type of nucleophile?

Nope, the rates wouldn't be affected, even if the nucleophiles used are different (though if you're dealing with highly concentrated reactants, it might be a different story...).

Judging from the way you framed the question, you probably already know a fair deal about both the Unimolecular and Bimolecular Nucleophilic Substitution (SN) reactions. So I won't start from scratch.

In the SN1 case, as the name would suggest, the rate depends on only one concentration factor, namely the substrate. You already (correctly) mentioned that a carbocation intermediate is formed. At this stage, its just a matter of whether the nucleophile performs a 'frontal attack' or a 'rear attack', and doesn't really involve the nucleophilicity or the size (at least not much in your example) of the nucleophile. This is different from the SN2 reaction, which requires species with high nucleophilicity  and appropriate size to react (rear attack) with the substrate, leading to an unstable transition state.

How about an analogy?

Suppose you've got a lot of magnets stuck on your fridge door (at least I sure do..). Now suppose you're standing at a distance from the fridge, and you want to place another magnet on the fridge by throwing it. Since you've already got a lot of magnets stuck there, its going to be really hard to make your magnet land there. The weaker your magnet, the lower the chances of you landing it on the fridge door. However, if you have this really strong neodymium magnet, this wouldn't be much of a task; one throw, and it'll land so hard, it'll send the other magnets flying. This is a lot like the SN2 scenario. The attacking group must be a stronger nucleophile that the leaving group.

Now suppose your fridge door is devoid of magnets [which is precisely what my mother requires of our fridge.... :( ]. Then clearly throwing a magnet and hoping it'll land is a very, very good possibility. In this case it doesn't really matter whether the magnet is square or circular, large or small, strong or weak (not too weak). Apart from how large your fridge door is, there's no other factor that would keep the magnet from sticking to the fridge (albeit you aim). This is a lot like your SN1 scenario. The nucleophile's characteristics don't really matter (at least not in most typical cases). As long as there's space on the door, the magnet will stick.


Why is o-halophenol more acidic than m-halophenol?

It's because the halogen is highly electronegative. Therefore, the closer the halogen is to the alkoxide, the more it helps in stabilizing its negative charge after it's been deprotonated. Conversely, the further away the halogen is, the less it is able to help stabilize the negative charge. The more stable the conjugate base, the stronger the acid.

This question is basically testing induction, which is covered pretty well here:

http://orgomadesimple.com/organic-chemistry-acids-and-bases/


Which among phenol and 1,2-dihydroxybenzene has the higher boiling point?

Your structures (grabbed from Wikimedia Commons)

Phenol:

Catechol (1,2-dihydroxybenzene)

The three main influences on boiling point are:

Shape - molecules with a more spherical shape have lower boiling points than molecules that are more cylindrical, all else as equal as possible. Spheres have the lowest surface area to volume ratio of any of the solids. This is where symmetry comes in. More symmetric molecules are more spherical. The shape/symmetry argument does not work here since the two compounds are not isomers. The issue of close packing is relevant to the melting points of solids, since in a solid, most particles do not have translational motion - they are locked in a crystal lattice. In a liquid, such random translational motion is common. In the liquid state, molecules of most compounds are so disorderly arranged that it does not matter how well they pack as a solid. In your case, both molecules are rather like flat disks with a little bit of polarness sticking off one side. Their shapes are not so different as far as surface area to volume goes. Boiling point will be dominated by mass and polarity.
Molecular mass - bigger molecules require more energy to vaporize because they have more mass, and $KE=\frac{1}{2}mv^2$, all else as equal as possible. Mass difference applies in your case because the two molecules are similar enough in shape and polarity, except one has an extra oxygen in its formula.
Polarity - polarity controls the types and number of intermolecular forces, including hydrogen bonding. The more and stronger intermolecular forces between two molecules, the higher the boiling point, all else as equal as possible. Since catechol has two hydroxy groups and phenol has one, you have your strong argument for increased intermolecular forces.

The other easy thing to do is use the internet to look up the boiling points, and then explain the difference using the three factors. Once you know what the boiling points are, it is easy to choose the structural features that support the difference.

Phenol - $181.7 \ ^\circ\text{C}$
Catechol - $245.5\ ^\circ\text{C}$


Why does sodium polyacrylate solidify, on addition of water?

Sodium polyacrylate is an example of a super-absorbing polymer. It is a cross-linked (network) polymer that contains sodium atoms. It absorbs water by a process called osmosis. When the (sodium-containing) polymer is placed in contact with water, there is a tendency for the sodium to distribute equally between the network and the water.

That means, some of the sodium atoms want to leave the network and move to the water. When these sodium atoms leave, they are replaced with water molecules. Water swells the polymer network to try to keep the sodium concentration balanced between the polymer and the water.

The cross-links that connect the chains together prevent them from dissolving/breaking apart in the water. The resulting product appears very similar to fake snow, as you mentioned, a solidified powder like substance.

Sodium polyacrylate can absorb it’s own weight 300x over making it an amazing chemical to aid in water cleanups.

The question was good.


Can fumes from using bleach seep through plastic packaging and glass jars

Disclaimer: This answer is provided for information only. Use it at your own risk.

  I mixed 1/4 bleach and 2 cups water. Applied it on all surfaces. Then
  applied dish soap diluted with water.

This was not smart. Bleach should never be mixed with other chemicals, cleaners, or detergents. Rinsing with plain water first would have been less risky.

With bleach it is important to ventilate the area. Bleach should be thoroughly rinsed if it gets on anyone. Gloves are highly recommended for this type of cleaning.

  Does this produce toxic fumes?

Possibly.

Bleach attacks double and triple bonds in organic molecules. This action not only inactivates most disease causing organic particles, but also destroys the chromophores which give organic material their color. This reaction can sometimes liberate malodorous organic fumes which is best not to breathe.

Most inexpensive commercial bleach is a diluted form of sodium hypochlorite. Sodium hypochlorite can release chlorine gas; however, under normal usage bleach, nor its smell is likely to be harmful. The odor threshold for common bleach is 0.002 parts per million (ppm) which is well below the concentration it will cause any harm.

If on the off chance the smell is extremely irritating or if coughing occurs then get people and pets out of the area and ventilate the gas. Wait till the odor ceases to be irritating. If the odor is just annoying then it is likely safe.

  And can the toxic fumes seep through food even if they are covered and
  placed about 3 meters away?

If ventilation is good over a relatively short period of time then seepage is unlikely. If ventilation is poor over several days then seepage could turn into a problem. If the food smells or tastes either odd or like bleach, it may be best to throw it out.

  The cabinet still smells like bleach. What should I do about it?

Ventilating the cabinet till the smell goes away is best. Rinsing thoroughly with plain water, getting the water up as best as possible, and then allowing the surfaces to dry may work. Only after thoroughly rinsing and ventilating would it be worth it to clean with dilute vinegar help with the smell. Do not allow vinegar to mix with bleach!

A better disinfectant choice for just one bulged can may have been to use 70% isopropyl rubbing alcohol soaked into a dry rag. No chlorine fumes, much less likely to discolor, while still effective with germs.  This should go without saying...  Do not allow alcohol to mix with bleach!


SCF (Roothan-Hall) Method, why Density Matrix?

When doing the SCF calculation taking the RMSD of the density matrix compared to taking the RMSD of the diagonal of the density matrix takes about the same with respect to the total calculation time. The comparrison of the density matrices is not the time consuming step in the SCF calcultion. When you take the RMSD of the total density matrix you also "keep an eye" on the off diagonal elements, this is a more rigid test for convergence.


Is it possible for a substance to absorb visible light but still appear white?

Your eyes are not sensitive to UV light so any absorption there will be invisible to you. The visible spectrum you give has fairly uniform absorbance across the visible. The whiteness you see is primarily due to scattering of white room lights or sunlight off the powdered zinc oxide. Light is scattered when there is a change in refractive index between two mediums, in this case air and zinc oxide. The effect is enhanced when the crystals are of a very irregular shape and so have many different scattering surfaces. Sugar and salt show the same effect when powdered  even though large crystals of each of them can be made that are clear and colourless. As the oxide has a more or less constant absorption no wavelength is very much more absorbed than any other, or alternatively, all wavelengths are equally scattered.

If you can find a solvent with similar refractive index to the oxide (and in which it does not dissolve) then the crystals will become far more transparent and effectively disappear. It is easier see this effect if you wet a piece of frosted glass with water as it will then become far more transparent. Alcohol or benzene would be better as they have refractive index closer to that of many types of glass. Zinc oxide has a value of 2 higher than any common solvent, such as diiodomethane approx 1.7.


Why are so many ionic compounds brittle?

Ionic crystals are hard because of tight packing lattices, say, the positive and negative ions are strongly attached among themselves.
So, if mechanical pressure is applied to an ionic crystal then ions of similar charges may be forced to get closer to each other.
Now, by doing so, the electrostatic repulsion can be enough to split or disorient completely the lattice infrastructure. Thus imparting the brittle character.


Difference between Gibbs Free Energy and Exergy

Exergy is equal to the Gibbs free energy if you happen to be in the isothermal--isobaric ensemble $(N, P, T)$, i.e. in a closed system if the pressure $P$ and temperature $T$ are constant. This is indicated in Eq. 7 of the Wikipedia page for exergy.


Dissociation of water into H+ and OH-: Does the law of mass action hold at nanoscale?

There is definitely a breakdown of mass action at very low numbers of molecules.  This can happen either because concentrations are tiny, or because the system of interest is tiny.

In either case, your idea to turn to probability is an excellent one, and is how these systems are modeled.  But the reason doesn't necessarily have to do with quantum mechanics.  It's just a consequence that matter is not really a continuum that can have any concentration, but is in fact granular.  At the scale of atoms, there can either be an atom, or not be an atom.  You can't have 0.5 of an atom.

The right boundary at which mass action breaks down depends a bit on the particular system, in particular, if reactions are occurring that depend on H+.  In the less complicated case of no reactions depending on $\ce{H+}$, then the boundary between mass action being valid and not is actually fairly easy to calculate. If you know the average or expected value of $\ce{H+}$ ions in your water sphere (which you do, since you calculated it) the actually observed number of ions will follow a Poisson distrubtion.  If the mean number of atoms is $N$, then the standard deviation of atoms is $\sqrt{N}$, so the relative uncertainty in the mean number of atoms is proportional to $\frac{\sqrt{N}}{N}=\frac{1}{\sqrt{N}}$.  According to your calculation, $N$ is less than one, so the relative uncertainty is greater than 100%!  But if you enlarged the sphere so the expected number of ions was $N=10$, then the uncertainty would be about 32%, still pretty bad.  If you enlarged the sphere so $N=10000$, then the relative uncertainty is only 1%, meaning that mass action predictions for this system are pretty close to accurate!  Getting $N$ tht high requires a sphere about 10,000-times bigger in volume, which means the radius would have to be about 38 times bigger.  But keep in mind this result is only true for the water dissociation problem.  So each chemical system has its own scale where mass action breaks down.

For systems where reactions are occurring that depend on $\ce{H+}$, this simple logic breaks down.  Complex probabilistic simulations need to be done.  Depending on the details of the reactions, the chemical master equation or the chemical Langevin equation would have to be used.

One time I calculated the concentration of frogs in an aquarium was about 1 zeptomolar.  I don't think we could assume frogs followed the law of mass action!


What kind of string to use for the ice fishing experiment for kids?

This resource has an explaination of the behavior, which relies on the salt-dissolved water wicking into the string. This would mean that you need a type of string which will absorb water. It's surprising to me that your nylon string doesn't work (as long as it's thick/low-density enough to absorb water, probably unlike fishing line), is it possible it has a hydrophobic coating?

I would recommend any cotton-based string or twine. Cooking twine and sewing thread is often cotton, if you have access to that. I've had no problems with these.

Edit: I just thought, that you might want to try this not in a cup of water first (like in the link above). You might potentially run into not-string-related problems if your water isn't cold enough, or if your ice gets too submerged when you put the salt on it. It's possible your nylon string is fine.


Allotropy in Nitrogen and Bismuth of Group 15

Bismuth certainly has other known crystal structures at elevated pressures and temperatures, at least 4 others in addition to the rhombohedral structure stable at room temperature and pressure.  One place to start would be an article from NIF on shock physics of bismuth. For a scholarly article, you might start with 'Phase Diagrams of Arsenic, Antimony, and Bismuth at Pressures up to 70 kbars', W. Klement et al., Phys Rev 131(2) 632-637 (1963).

As for nitrogen, it too would appear to exhibit a variety of solid diatomic phases under pressure.  A starting point to explore might be 'High-pressure amorphous nitrogen', E. Gergoryanz et al., Phys Rev. B64 052103 (2001), which mentions a variety of phases in addition to the titular amorphous phase they explore.

Both Physical Review articles should be available to the general public...


Are the units of rate constants affected if we have more than one substance in the rate law

The heart of your problem is that you're assuming that there exist different types of the unit mole.

Consider the unit of length one meter. Take two meters of cotton thread, and two meters of jute thread. Place them together and you get a nice square of $1\times1 \text{ <unit of area>}$. Now what should the unit of area be? Should it be $=\text{cotton meter} \times \text{jute meter}$?

Well, it should NOT be $\text{cotton meter} \times \text{jute meter}$ because a meter, being a unit, is same for all substances being measured. A meter of jute is exactly the same length as a meter of cotton. Hence, the unit of area should be $\text{meter}^2$, which it always is.

Similarly, in your solution, you are assuming there exist two different types of the unit mole, one for $x$ and one for $y$. However, there is only one fundamental unit, and it measures the same amount of substance for both $x$ and $y$.

Hope it helps!


What goes on the NFPA 704 for multiple chemicals?

According to Section 4.2.3.3 of NFPA 704:

  Where more than one chemical is present in a building or specific
  area, professional judgment shall be exercised to indicate ratings
  using the following methods:

  (1) Composite Method. Where many chemicals are present, a single sign
  shall summarize the maximum ratings contributed by the material(s) in
  each category and the special hazard category for the building and/or
  the area.

  (2) Individual Method.  Where only a few chemicals are present or
  where only a few chemicals are of concern to emergency responders
  (taking into account factors including physical form, hazard rating,
  and quantity), individual signs shall be displayed. The chemical name
  shall be displayed below each sign.

  (3) Composite–Individual Combined Method. A single sign shall be used
  to summarize the ratings via the Composite Method for buildings or
  other areas containing numerous chemicals. Signs based on the
  Individual Method shall be used for rooms or smaller areas within the
  building containing small numbers of chemicals.


How do you know when the first compound is completely distilled and how do you know when the next compound starts

There is no such point when the first solvent is completely distilled and then the next one starts. From the very beginning, both solvents boil together, though the vapor initially is relatively enriched with the lower-boiling one. The cutoff point is set somewhat arbitrarily. And yes, the answer surely has something to do with temperature, because temperature is the only parameter we can control.

Like NotNicolaou said, if the boiling points of your solvents are reasonably far away, you may separate the mixture almost completely by collecting the fraction that distills near the boiling point of the low-boiling one. Otherwise, you might not be able to get pure compounds, unless you use an industrial rectification column.


yes it is possible, for example by adding a seed crystal under the right conditions.

See the reference for further information: http://www.google.com/patents/US2233606


How to find the number of hydrogen atoms in 0.5 mol of hydrogen gas?

There is no need to use
$$n = \frac{m}{M},$$
since you already have been given the amount of substance. You would need to use the equation, if you were given the mass $m$, e.g. as $\pu{0.5 mg}$.

If you want to calculate the number of hydrogen atoms in $\pu{0.5 mol}$ hydrogen gas, then you should consider that they are diatomic molecules, $\ce{H2}$.
The unit mole is represented by the Avogadro constant and is
$$N_\mathrm{A} = \pu{6.02214076×10^23 mol−1}
\approx \pu{6.022×10^23 mol−1}$$
Each hydrogen molecule has two hydrogen atoms, therefore
\begin{align}
n(\ce{H}) &= 2 \cdot n(\ce{H2}),\\
N(\ce{H}) &= n(\ce{H}) \cdot N_\mathrm{A},\\
N(\ce{H}) &= 2 \cdot n(\ce{H2}) \cdot N_\mathrm{A},\\
N(\ce{H}) &= 2 \times \pu{0.5 mol} \times \pu{6.022×10^23 mol−1},\\
N(\ce{H}) &= \pu{6.022×10^23}.
\end{align}


Why is molality per mass solvent rather than per mass of solution?

Molality $b$ is defined its way because being proportional for the given solvent to the solute/solvent molecular ratio.

Molarity or molar/mass fractions or molar_amount/solution_mass are not proportional to it.

The boiling and melting point shifts are proportional via ebullioscopic/kryoscopic solvent constants to molality.

$$\Delta T_\mathrm{b}=K_\mathrm{e} \cdot b$$

$$\Delta T_\mathrm{m}=K_\mathrm{k} \cdot b$$

The number of solvent (s) molecules per 1 solute molecules

$$\frac{n_\mathrm{s}}{n}=\frac{m_\mathrm{s}}{M_\mathrm{s}\cdot n}=\frac 1{ M_\mathrm{s}\cdot b}$$

If we provide as the input the solute mass and solvent volume and density, the above molecule ratio is

$$\frac{n_\mathrm{s}}{n}= \frac{m_\mathrm{s}\cdot M}{M_\mathrm{s}\cdot V \cdot \rho}$$


Why oxides with the element in higher oxidation state are more acidic

It does not appear to be always true.  According to Wikipedia, copper(I) becomes acidic, forming an anionic hydroxide complex, at about 1.5 pH unit lower than copper (II).  The higher oxidation state of copper qualifies as less acidic.


How is "dietary cholesterol" processed in a human body?

Most lipids, such as cholesterol (and triglycerides), are nonpolar and therefore very hydrophobic molecules. In order to be transported in watery blood, such molecules first must be made more water-soluble by combining them with proteins produced by the liver and intestine.

Lipids and proteins associate noncovalently to form lipoproteins (globular micelle like particles that consist of a nonpolar core of triacylglycerols and cholesteryl esters surrounded by an amphiphilic coating of protein, phospholipid, and cholesterol):

Lipoproteins have been classified into five broad categories on the basis of their functional and physical properties:

1. Chylomicrons, which transport exogenous (dietary) triacylglycerols and cholesterol from the intestines to the tissues.

Chylomicrons therefore function to deliver dietary cholesterol and triacylglycerols to muscle and adipose tissue and dietary cholesterol to the liver.

2–4. Very low density lipoproteins (VLDL), intermediate density lipoproteins (IDL), and low density lipoproteins (LDL), a group of related particles that transport endogenous triacylglycerols and cholesterol from the liver to the tissues (the liver synthesizes triacylglycerols from excess carbohydrates).
5. High density lipoproteins (HDL), which transport endogenous cholesterol from the tissues to the liver.

Circulating HDL acquires its cholesterol by extracting it from cell-surface membranes and converts it to cholesteryl esters through the action of lecithin-cholesterol acyltransferase (LCAT), an enzyme that is activated by apoA-I. HDL therefore functions as a cholesterol scavenger.

This enzyme transfers a fatty acid residue from atom C2 of lecithin to cholesterol with the concomitant formation of lysolecithin

The cholesteryl esters are hydrolyzed by a lysosomal lipase to yield cholesterol, which is subsequently incorporated into the cell membranes. Any excess intracellular cholesterol is reesterified for storage within the cell through the action of acyl- CoA:cholesterol acyltransferase (ACAT).

References

Principles of Anatomy and Physiology 13e Tortora, Derrickson
Biochemistry Voet and Voet: Lipoproteins


"Proof" that the chemical potential is less than zero for irreversible material changes

Q1. You won't need the chemical potential terms $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha$ (because it equals zero) in using the differential relation $\text{d}G = -S\,\text{d}T+V\,\text{d}P$ in this case. Nevertheless, the chemical potentials are still relevant in establishing criteria for phase equilibrium: $\mu_i^\alpha = \mu_i^\beta$, etc.

  Why should $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha = 0$?

Reversible changes in composition imply that the process is in equilibrium, so that $\text{d}G = 0$. Assuming thermal and mechanical equilibrium and constant $T$ and $P$ we have $\text{d}T = \text{d}P = 0$, and the result follows by considering the (full) differential relation satisfied by $\text{d}G$.

Q2. Yes, you do need a reversible path. The expression on the right, if you follow its derivation carefully, does represent a reversible path. The inequality occurs because the chemical potential terms have not been accounted for. So the cancellation works out.

It seems, however, that we can derive the same result much more simply. Consider a system at thermal and mechanical equilibrium with constant $T$ and $P$, but not chemical equilibrium. Then $\text{d}G < 0$, since the Gibbs free energy is minimized at equilibrium. If $T$ and $P$ are constant $\text{d}T = \text{d}P = 0$, so all that's left in $\text{d}G$ are the chemical potential terms, which must thus be negative.


Why is caustic soda (NaOH) used to clean beer bottles?

Caustics such as $\ce{NaOH}$ and $\ce{KOH}$ are effective in converting insoluble fatty acids in organic fats and oils into soluble soaps in a proccess called saponification. The caustics also react with cell membranes, DNA and RNA, destroying undesired yeasts, bacteria and other microorganisms.

That said, caustics react with the glass itself, removing the surface and etching the glass. Use will weaken and damage containers.

Use caution! Caustics also destroy skin and clothing, and are particularly damaging to the eye.


Maximum metal-metal distance for antiferromagnetic coupling

The annoying answer is that this will be system dependent. I'll try my best to explain with a quick recap of magnetism.

Unpaired electrons produce a magnetic moment
One of the factors that determines the strength of the magnetic moment is the number of unpaired electrons.

If you have a specific system in mind, I would suggest performing a series of single-point (no ionic relaxations) calculations where you increase the distance between the ions. Contrast the energetic between FM and AFM configurations. If you have a license, VASP can provide information on the magnetic moments.


In the case of lead oxide and many other inorganic thermochromic materials, the color change is a result of a phase change that slightly alters the distance separating the inorganic ions.

In ordered materials, the energy levels present in atoms and molecules are replaced by energy bands.  Said differently, in materials with extended order there are so many energy levels clustered around one another at a certain energy point and separated from one another by such small amounts of energy, that the distinct energy levels of the molecules "fuzz" out into a band.  Instead of a highest occupied and lowest unoccupied molecular orbital (HOMO and LUMO), a valence band and a conduction band results.

Just as the HOMO-LUMO separation will determine where in the energy spectrum an organic molecule will absorb light, the energy separation between the valence band and the conduction band (called the band gap) will determine where a material with extended order will absorb light.

In the case of lead oxide and other crystalline materials, slight changes in the crystal structure (phase change) that can be brought about by heating or cooling the material, will change the size of the band gap.  This change in the energy separation between the valence band and conduction band will result in a change in the optical (and electronic) properties of the material.  If the change in optical absorption occurs in the visible region, then the thermochromic properties of the material are apparent to the eye.

Here is a link to a nice paper that starts simply, but then goes into greater depth, on this subject.