Calculate Ethanol content after heating for N minutes

For a pure substance, to be able to calculate how much has boiled off after $N$ minutes you will need the enthalpy of vaporization, $L$ (in kJ/kg) and the heat input, $Q$ (in kJ) which will give you:
$$\tag{1} m=\frac{\epsilon \, Q}{L}$$
where $m$ is the mass that has boiled off and $\epsilon$ is the fraction of the heat input that actually goes into the liquid. $1-\epsilon$ would be the fraction of heat lost to the surroundings.

The very crude story

An extremely crude calculation would be to take the mixture enthalpy of vaporization and assume that only ethanol vaporizes such that the fraction of ethanol can be directly related to the total mass loss. Note that this calculation will give you a lower bound for the time needed to get rid of the ethanol

We can calculate the mixture enthalpy of vaporization by averaging the enthalpies of vaporization of water and ethanol:
$$\tag{2} L_{mixture}=x_{\ce{EtOH}} L_{\ce{EtOH}}+x_{\ce{H2O}}L_{\ce{H2O}} $$

Now we need to figure out the heat input. A precise calculation is impossible here, but we can use some reasonable estimates. According to an estimate on this (dutch) website when we are cooking we use about $164$ l/h of methane. Assuming that that is for cooking with 2 pots this translates to a gas flow rate of $\dot{V}=1.4$ l/min for a single pot. With the heat of combustion of methane, $\Delta H_c$ and the density $\rho$ we can then calculate the amount of heat released per minute:
$$\tag{3} \dot{Q}=\rho \dot{V} \Delta H_c $$
plugging in the values for methane we find $\approx 50$kJ/min

The enthalpy of vaporization of the mixture will constantly change, because the composition of the mixture constantly changes over time during the boiling process. Therefore, instead of simply using Eq. 3 in Eq. 1 to find $\dot{m}$ and multiplying with $N$ we will need to do an integration:
$$\tag{4} m_{boil-off}=\int_0^N \dot{m}\; dt $$
where $\dot{m}=\frac{\epsilon \, \dot{Q}}{L(t)}$. Now this is a bit problematic because $L(t)$ depends on $m_{boil-off}$, but it can be solved numerically (I feel like this should also be possible analytically, but can't figure it out right now, anybody?!).

Your example question (with $\epsilon=0.5$) leads to the following curve for the fraction of ethanol in your mixture:

As you can see the ethanol is gone after about 7 minutes. To do this calculation for your pecan pie mixture you will need to know the heats of vaporization of all your ingredients and the mass of each.

The more precise story

The more precise story is that not only the ethanol, but also part of the water will vaporize therefore slowing down the vaporization rate of ethanol. If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid. In that case you can use the theory available for distillation columns and something called the Ponchon-Savarit Method (the complete diagram for water-ethanol can be found here), which I will not recap here, because it is basically a book chapters worth of undergrad chemical engineering.

Note that the equilibrium vapor assumption is probably not a very good one due to air flows over the pot which keep pushing the system out of equilibrium. For the evaporation of ethanol this is good news by the way, because it will speed up the process a bit. Also note that with a true equilibrium you would run into the azeotrope of ethanol-water mixtures.


Why are the three most simple alcohols completely miscible?

This happens because the intermolecular forces between different low molecular weight alcohols are very similar; as you say, they are dominated by hydrogen bonding. You can consider mixing on a molecular level to involve exchange of molecules from the two phases at their interface; if the energy changes involved are very small, mixing will occur as it is entropically favoured (mixed phase is more disordered) and the average thermal energy is far higher than the barrier to diffusion.


The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen.

The major reason carbon-based life can exist at ambient temperature in an oxygen atmosphere is purely kinetic, not thermodynamic. You, the tree outside and everything else made of carbon is right now undergoing continuous combustion. However, in the absence of catalyst, this process is too slow below a couple hundred degrees Celsius for it to be self-sustaining. More technically, combustion of organic matter is a highly exergonic process, but the activation energy is high. The meagre amount of heat generated by the handful of molecules reacting is too quickly diluted into the surroundings, and the reaction does not accelerate and spiral out of control (a fire, as described by the eternal Feynman).

Very luckily for us, Life figured out this vast untapped source of chemical energy held back by kinetics approximately three billion years ago and developed a whole metabolic process to extract this energy in a stepwise fashion using catalysis, which we call aerobic respiration. Without it, multicellular organisms could well never had evolved.


Why does hydride acidity increase across period and down group in periodic table?

Yes, but that is somewhat circular reasoning, as electronegativity is largely just putting a numerical value to the trends in properties you are observing.  A better answer is to compare the effective nuclear charge holding on to the valence electrons as you go across or down the Table.  Going across the table, each electron gets added to the same shell, which means it shields the other electrons from the added proton poorly.  Consequently, Z* (the positive charge effectively felt by the outermost electrons) increases, and these electrons are more tightly held.  At some point the atom holds onto not only its own valence electrons, but those of the H atom when the H atom leaves, and so the H atom leaves as H+.  Conversely, at the far left of the table, the Group 1A atoms hold on so loosely to their lone electron that the H atom is able to leave with the metal's electrons, as H-.

Going down the table, you add a whole electron shell between the nucleus and outermost electrons with each period.  That is counterbalanced a bit by the fact that the shells become more closely spaced, but nevertheless for main group elements the extra shielding of the extra shells usually matters more, and so effective nuclear charge falls going down a group.

Finally, bear in mind the "acidity" of a hydride is the tendency of the compound to break up with the H atom leaving as H+.  So the greater the Z* of the partner, the more likely the H atom leaves without its electron.


Does a tertiary carbocation rearrange to another tertiary carbocation?

Option B is the correct one. The process of dehydration of alcohols follow E1 elimination mechanism which involves two steps out of which the first one is slow ionisation of C-X bond( where X is any hetro atom) and the second step is  fast removal of $H^+$ ion .Since the first step is slow it provides enough time for the carbocation to rearrange and form a stable carbocation . The mechanism itself is self explanatory. Firstly as Oxygen possess lone pairs of electrons so the $H^+$ ion attacks the OH group as a result the $H_2O $ molecular gets eliminated resulting in a positive charge there.  Now hydride shift takes place as the so formed $ \textbf{Carbocation is more stable as it has 7 alpha hydrogens compared to the 5 alpha hydrogens of the previous carbocation}$. Now the C-C bond undergoes ionisation to form a three degree Carbocation and a carbanion . Now this carbanion forms a bond with the Carbocation formed earlier to form a six member ring . Now the resulting carbocation undergoes methyl shift which produces more stable three degree Carbocation. Now a $H^+$ ion from the neighbouring carbon atom is ejected as a result the pie bond is formed.


What is the relationship between the two definitions of the difference between basicity and nucleophilicity?

The second set of definitions is the correct one.

Basicity is a thermodynamic property. In essence, the basicity is measured by determining the equilibrium constant of the following (generic) reaction:
$$\ce{B- + H2O <=> B-H + OH-}\tag{1}$$

The proton transfer that is the gist of this is an extremely fast process — faster than diffusion control due to the Grotthuss mechanism. You may have witnessed this during acid-base titrations: equilibrium is reached extremely rapidly. Another way to observe this is by adding carbonate bases to an acidic solution; $\ce{CO2}$ evolution is basically instantaneous. Therefore, there is no way to observe just one step (forward or reverse); you basically always have them both.

Nucleophilicity is a property that is related to the rapidness of the following, again generic, reaction:
$$\ce{Nu- + R-LG -> Nu-R + LG-}\tag{2}$$

As similar as this may seem to the above, the lack of a reverse arrow may already show that this is tied to only one directiong; that it is not an equilibrium process but rather a kinetic one. Most of the time, there will be one species that is clearly a good nucleophile ($\ce{Nu-}$ in the scheme) and one that is only a good leaving group ($\ce{LG-}$ in the scheme). The reaction will be driven to completion and stay there.

To exemplify the difference, consider what would happen if you added hydroxide to the acid-base equilibrium $(1)$ versus additional leaving group to the nucleophilic substitution $(2)$. In $(1)$, the equilibrium would readjust; if the base is less basic than hydroxide it will be deprotonated, i.e. the reaction will be shifted back to the reactants. In $(2)$, typically nothing will happen, e.g. if you add $\ce{OPPh3}$ to a Mitsuobu reaction; showing the kinetic nature of the process.

Another way to look at the problem — and to separate the parameter basicity from the parameter nucleophilicity — is to look at the influence of steric effects. For example, consider ethanolate, propanolate and deprotonated isobutyl alcohol. In all three cases, the negative charge is unambiguously located on the oxygen atom. Thus, we would expect a thermodynamic property to be relatively equal for the threeof them. At the same time, there is a clear increase in steric hindrance from ethanolate to the isobutyl alcohol. This steric effect should manifest itself in reaction kinetics and lead to the latter being less nucleophilic than the former. Indeed, sterically hindered bases are often described as non-nucleophilic while the same structure with less hindrance often works fine as a nucleophile.

Finally, let’s work at debunking the claims laid out in the first set of definitions. First, basicity. After the Arrhenius and Brønsted-Lowry theories of acids and bases came the Lewis theory. In this, the reaction of a Lewis base with a Lewis acid — not necessarily a proton — is an acid-base reaction. One of many possible examples would be the formation of a transition metal complex such as $\ce{[Cu(NH3)4(H2O)2]^2+}$. Here, the ammine ligand is acting as a base, but it is donating to a non-hydrogen atom (namely copper). The claim that basicity only goes towards hydrogen is obviously false.

Unfortunately, it is not possible to fully debunk the second part since a nucleophilic attack on a proton is not really possible (that would indeed be the thermodynamic acid-base reaction). However, we may remember that most of the non-metals can act as the recipient of a nucleophilic attack; for example sulphur in a Swern oxidation, phosphorus in a Horner-Wadsworth-Emmons reaction or iodine in a Dess-Martin oxidation.


Is trans-1,2-dimethylcyclopropane optically active or not?

First off ...
There is a difference between "chiral" and "optically active".  Something that's optically active has to have chiral molecules.  However, something can be chiral but if you have equal amounts of the mirror-image isomers, or if molecular motion can make an equal number of isomers like an umbrella-like inversion of a lone pair or an exchange of protons, then you do not see optical activity.  All we are looking at here is whether the molecule is chiral, not whether a sample of the stuff is in general optically active.
With that cleared up:
As Weijun says in the comments you have to have a mirror plane, or an improper rotation axis, or else the molecule is chiral.  A (proper) rotation axis does nothing in this regard.
Each of the carbon atoms with a methyl group on it is, in fact, a chiral center.  The four different groups attached to it are:
No 4.  A hydrogen atom
No 3.  The methyl group that's attached outside the ring
And...
No 2, No 1.  The ring going clockwise and going counterclockwise, as viewed from "above" the C–H bond, are counted as two different attached groups, as if the ring were "cut off" in either direction before rejoining the carbon atom you're looking at.
If you look at the ring attachment that way, you see that the two different directions rank 1 and 2, in some order depending on where the second carbon with the methyl group is, hence my numbering above.  So you can figure out an R or S configuration for each chiral center.  In the cis isomer you end up with one R and one S (thus mirror-image configurations), and a mirror plane does go between them making a meso rather than a chiral molecule.  But the trans isomers give both R or both S, which never allows a mirror plane or any improper rotation axis, and so they are chiral.


Do azeotropic mixtures evaporate proportionally?

Azeotropic composition is a function of the boiling pressure and therefore of the boiling temperature. This can be seen in looking up the composition of the constant boiling $\ce{HCl}$ azeotrope with changing atmospheric pressure. However, information on this is hard to find and the effect seems to be small in most cases. Wikipedia does discuss this a bit. Just thinking, this might be a possible research if one had access to an FTIR with a ATR trough cell so evaporation could be observed in situ.


The formation reaction of CH4 is exothermic. The heat of formation is : -75 kJ/mol


Solution of very small particles with neutral buoyancy and distinguishable to the naked eye

I think that the cheapest and fastest way is that of working on natural products, like actual tea leaves, or ground coffee (I just tried it with coffee, and it can be done):

If you want only particles with neutral buoyancy, or, if I understand well, floating in the solvent, you can use said natural products, which contain particles of different shapes and densities, and discard by fractionation the unwanted ones.

What I did:

Place the ground coffee in a glass of water and stir a lot.
Let the suspension sit. You will notice that part of the parcticles are on a top layer, some are floating in water, and some are at the bottom.
Put the glass in a sink, and gently add water to it, so that the top layer eventually falls out of the glass, together with some water
Let the suspension sit: if needed repeat the last step, otherwise go on
To discard the bottom layer, gently pour the solution in another glass, by leaving some water in the first glass: the first glass will contain the heavier bottom layer, and the second one will contain only floating parcticles
If needed, repeat the last step


Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate?

The keto-enol tautomerization cannot take place without at least a trace of acid or base, meaning that there is no direct shift of proton from alpha-carbon to oxygen or vice versa.

The two most common enolation mechanisms are:

Base-catalyzed (with organolithium reagent such as LDA (secondary amine))

In the base-catalyzed enol formation, the first step, proton extraction from the alpha-carbon, is slow. (The subsequent enolate formation is fast (due to resonance)). The base is proposed to form a chair-like transition state (TS) with the ketone, extracting the alpha-carbon proton and at the same time donating the Li to the oxygen. The resulting product is the classical lithium enolate.

Based on the structure of TS, you can then predict whether the “side chain” of the base would form a unfavorable syn-pentane interaction.

https://en.wikipedia.org/wiki/File:Ireland_model_enolate1.svg

Acid-catalyzed

In the acid-catalyzed enol formnation, the first step, the protonation of oxygen and subsequent “charge re-localization” to the hydroxyl carbon, is fast. (The subsequent enol formation is slow). Since the protonation can occur from any free proton in the environment, the size of the acid plays little direct role in the stereochemistry of the resulting enolate. (However, the size of the “side chain” in the ketone reactant does).

However, there is a 3rd mechanism when the “base” is a tertiary amine (as you appear to be very interested in) but the reaction is acid-catalyzed. In this mechanism, the amine behaves like a nucleophile and a carbinolamine intermediate is detected.

See Bruice, J. Am. Chem. Soc. 1983, 105, 4982

http://pubs.acs.org/doi/abs/10.1021/ja00353a023

(Also see Bruice, J. Am. Chem. Soc. 1989, 111, 962 and 1990, 112, 7361)

In drawing this carbinolamine intermediate, you can predict a bulky tertiary amine to favor a cis-enolate due to steric hindrance. However, as suggested in the paper, "severely sterically hindered" tertiary amine can be too bulky to follow this 3rd mechanism and just catalyzes enolation via the general base-catalyzed mechanism.


As far as I understand, there has been very limited experimental investigation of the properties of roentgenium (Z=111), if at all, simply due to the fact that so few atoms have been produced and they decay so quickly. It is possible to do some chemistry with even single atoms in seconds, but as you might expect it's an extremely arduous endeavour and is riddled with massive error bars. For now, just the synthesis of the element seems to dominate experimental roentgenium research.

There has, however, been ample theoretical investigation of roentgenium's chemical properties, both by extrapolation of behaviour for group 11 elements and by ab initio computational methods, at varying levels of approximation. One difficulty is that roentgenium displays strong relativistic effects, which make accurate calculations harder. Of course, atomic properties are easier to predict than molecular properties, which are themselves much easier to predict than condensed phase properties. So while estimates for ionization energy for atoms may be relatively accurate, physical properties of the pure materials such as ductility or melting point are to some extent little more than informed guesses.

Here is a list of some predicted basic properties of the period 7 $d$-block elements, including roentgenium (source, p. 1691):

That source expands upon some of the general properties of roentgenium. Its electronic configuration is predicted to be $\mathrm{[Rn](5f)^{14} (6d)^9 (7s)^2}$, which is dissimilar to the configuration for gold $\mathrm{[Xe](4f)^{14} (5d)^{10} (6s)^1}$. This is because relativistic effects cause a strong stabilization of $s$ orbitals (now more accurately referred to as $s_{1/2}$ due to spin-orbit coupling) by pulling them closer to the nucleus, while simultaneously destabilising the $d$ orbitals (as well as splitting the $d$ subshell into two groups of degenerate orbitals, two $d_{3/2}$ orbitals and three $d_{5/2}$ orbitals) by pushing them away. This means complete population of the $s$ subshell becomes preferable as relativistic effects get stronger. For example, see the effect here (ibid, p. 1667) in the triad Nb/Ta/Db in group 5.

Roentgenium is expected to be a noble metal, like gold, and in fact from its predicted standard reduction potential, is more noble than gold ($\mathrm{Au^{3+}(aq) + 3\ e^{-} \longrightarrow Au^0(s), \ \ \Delta E^0=+1.52\ V}$, for comparison). That said, once the roentgenium atom is ionized to Rg(III), it can reach higher oxidation states more easily due to less stable filled $6d_{5/2}$ orbitals, and so Rg(V) compounds are expected to be more stable. Interestingly, though in the 6th period transition metals gold (group 11) shows the local "relativistic maximum" with respect to stabilization of the $6s$ subshell, in the 7th period the maximum stabilization of the $7s$ subshell shift to copernicium (group 12) rather than roentgenium. This means roentgenium is expected to be less noble than copernicium.

With respect to physical properties, not much is predicted about roentgenium. It will likely be a very dense metal, with good electrical conductivity. The question about its colour is an interesting one. Wikipedia mentions it is expected to be silvery, even though roentgenium displays stronger relativistic effects, which is likely the cause of the colour of gold. This may be related to the switch in electronic ground state between gold and roentgenium. Unfortunately the source for the claim on Wikipedia is no longer accessible.


Looking for information about possible ingredients of a fertilizer

well , maybe you should have had googled it or spoke to the retailer about it . i dont know why you have no but its constituents are 20% nitrogen (N) and 20% phosphorus (P2O5). Afterall there must be NP written above 20-20-0 and these both must must be written above the KOMPOZE and this all will be enclosed in a red disc.and below all this will be a trademark of toros.


What does acidity of para-alkyl substituted phenols depend on?

The reason provided by ML cannot explain the difference in acidities since the +I effect doesn't come into play normally for the para positions.
A more likely reason is hyperconjugation (or no bond resonance) where the number of hydrogens directly attached to the benzyl carbon decreases as it goes from P to S.
This means that the number of destabilizing structures that are produced via this resonance decreases as you go down the list. This leads to greater stability in the deprotonated form compared to each other.
Therefore the correct order becomes S, R, Q, P.


1.Why is it difficult to prepare it, and how to prepare it?

Prafulla Chandra Ray prepared mercurous nitrite, $\ce{Hg2(NO2)2}$ by reacting metallic mercury with cold, dilute nitrous and nitric acid. According to his paper(1):

  Mercurous nitrite is the product of the combined action of nitrous
  and nitric acids on mercury :
  $$\ce{2Hg + NO2.H + HO.NO2 -> Hg2(NO2)2 + H2O}$$
   Some of the nitrite is decomposed by the nitric acid, the quantity
  of nitrous acid thus rapidly growing, until mercurous nitrite and nitrate
  are accumulated in molecular proportions. After some, the nitrous acid remains constant in the solution. It has become the catalytic agent between the nitric acid and the mercury,which now interact in the following manner:
  $$\ce{4Hg + 4HNO3 -> Hg2(NO2)2 + Hg2(NO3)2 + 2HO}$$

It is difficult to prepare the compound because many mercury-nitrite species is generated in the solution like  $\ce{Hg(NO2)2}$, $\ce{Hg2(NO3)2.2H2O}$, $\ce{Hg2(NO3)2.OH.Hg(NO3)2}$ etc. These salts exist in the solution and it is difficult to separate the main salt i.e mercurous nitrite from these salts and isolate in pure solid salts as all salts have similar properties and they tend to impure the solid mercurous nitrite.

  2.Can it be obtained by simply mixing $\ce{Hg2Cl2}$ and $\ce{AgNO2}$, where $\ce{AgCl}$ will be precipitated?

Reaction between $\ce{HgCl2}$ and $\ce{AgNO2}$ to yield mercuric nitrite $\ce{Hg(NO2)2}$ has been reported (2) but reaction between $\ce{Hg2Cl2}$ and $\ce{AgNO2}$ to yield mercurous nitrite $\ce{Hg2(NO2)2}$ is yet to be reported.

References

http://pubs.rsc.org/en/content/articlepdf/1905/ct/ct9058700171
http://chemiday.com/en/reaction/3-1-0-5880
http://www.insa.nic.in/writereaddata/UpLoadedFiles/IJHS/Vol49_4_3_AChakravorty.pdf
http://nopr.niscair.res.in/bitstream/123456789/11011/1/IJCA%2050A(2)%20137-140.pdf
https://www.researchgate.net/publication/289285311_On_mercurous_nitrite_and_a_basic_mercurous_nitrate_derivative


Why is absorbance in UV-VIS spectrometer negative even after calibration?

Possible causes include:

If it is a dispersive instrument (simultaneous recording of blank and sample in two cells): is the logic about which one is sample and blank set correctly?  At least some spectrometers by PerkinElmer and Shimadzu allow to swap how front and rear path are managed by the software.
How long did you run the spectrometer to warm up?  Maybe the instrument wasn't yet stable, elder spectrometers need 10mn+ to warm up all their internals.
Check the lamp for its run time and compare with the specifications by the manufacture.  Depending on the setup, the instrument's software may tell you this, or / and the lamps may have a counter (the little box between the plug and the mount of the lamp shown in the picture below).  If the lamp's intensity over time is not stable any more while recording sequentially blanks and samples, you will have a hard time to record faithfully spectra.  This cause typically remains unnoticed with mono-beam instruments less well maintained and run until the lamp «breaks» (multiple times older than the recommended run time).

(image credit)

Check your instrument with a known standard.  This either may be a liquid solution of a known concentration (e.g., by SigmaAldrich) to purchase, or already sealed into a cell, e.g.

(image credit)

or solid samples of metals deposit on glass or quartz

(image credit)

Don't forget that groups active in materials science / spectroscpy / optics may have such references to calibrate their instruments regularly and may be happy to share with you how to use these accurately, too.


How does the catalyst in a polyester resin or epoxy work?

The degree of mixing is very important in "two part" epoxy binder systems for coatings and adhesives.

In most cases the two liquids are quite viscous and the liquids themselves are better described as suspensions or colloidal dispersions rather than solutions. Unlike say two simple aqueous solutions when combined our two prepolymer liquids will not readily homogenize in a reasonable amount of time unless they are mechanically mixed in some way.  Even if the liquids do combine very slowly they will often gel or vitrify by the molecules at the interfaces reacting before the liquids homogenize.

In many epoxy systems we have, as Beerhunter stated above, some oligomer or polymer with epoxy functionality in one part and some liquid with amine (or sometimes hydroxyl) functionality in the other. Every amine needs to "find" an oxirane ring to attack or the curing process simply can't occur. The curing process works by nucleiphillic attack from amines to open epoxied rings.

The presence of leftover unreacted amine in cured epoxy materials often results in a phenomenon known as amine blushing or amine blooming. leftover amines or polyamides migrate to the surface and may in some cases react with air to form carbamates. This can be disastrous for desired materials properties like tack, adhesion (for the epoxy as a substrate), or cohesive strength of the bulk material.

In summary you can't build a robust three dimensional polymer network without thoroughly mixing because all the reactants can't react. Always mix as thoroughly as possible to get complete conversion.

My experience with polyesters is limited so I will not comment on that.


I think 4 should me more basic than 3 since $ -CH_3 $ has $ +I $ effect due to which electron density on $N$ increases.


Symmetries of the v=0 and v=1 quantum harmonic oscillator wave functions and those of the Cartesian Coordinates (References)

In general you need to understand point group tables.

These always have the same form, which is to say that symmetry operations are on the top line, characters (1,-1, 0 etc) in the main body. The most left hand column the Mulliken symmetry labels (also called symmetry species) for each irreproducible representation (one row of characters) .

The right hand two columns, often without header labels contain the operators for different type of transitions. x,y, z are used to dipole transitions because a dipole transforms as x, y or z. (The z direction is usually that of the principal axis). Rx, Ry, Rz are used, for example, spin orbit operators (These are coupling between electron spin and orbital motion, Not rotations).

The last column contains product operators $x^2$, $x^2-y^2$ etc and are generally used to describe Raman transitions since the Raman transition operator depends on a molecules polarisability (a volume) which in projection is proportional to an area hence squared terms.

If you want more complicated operators say $x(x^2-y^3)$ then generally these can be made from simpler operations. When making more complex functions consider the x as a $p_x$ orbital with two lobes of different 'colour'. The subtraction as in $x^2-y^2$ then looks like the $d_{x^2-y^2}$ orbital with opposite lobes coloured in the same way but each pair of a different colour. You can the use the symmetry of this object to determine which irreproducible representation it belongs to. In $C_{2v}$ (below) it is $A_1$.

The C$_{2v}$ point group shows these features. The operators are E, $C_2$ etc,. The Mulliken symbols for the symmetry species, in your case vibrational normal mode symmetries, are $A_1$, $A_2$. The characters are 1, -1 etc.

A totally symmetric normal mode $A_1$ has characters 1,1,1,1, is active to transitions with a dipole on the z direction and is also Raman active, as it has at least one squared term, x$^2$ for example. The $A_2$ symmetry species in not ir active, but is Raman active, due to the xy operator. You can follow the rest.

If you want to work out what the symmetries of a wavefunction are in a given point group, then the easiest way is to sketch the wavefunction and apply the symmetry operations from that point group. Choose a set of axes, z vertical and x horizontal and y out of the page, perpendicular to the other two. Count 1 if the symmetry operation is present and -1 if not.

For example the $v$=0 wavefunction for the harmonic oscillator has a gaussian or bell shape. By applying in turn the symmetry operations in C$_{2v}$ this wavefunction belongs to the $A_1$ symmetry species. The v=1 wavefunction has a node at the origin with an inversion in the values between negative and positive x. In this case C$_2$ is -1, $\sigma _{xy} = 1$ and $\sigma _{yz} = -1$ which is the $B_1$ symmetry species.


Copper acetate synthesis without hydrogen peroxide

I placed 1tbs salt, 1L vinegar, and 200g solid copper in a jar. After 3 months in a dark place the solution was extremely dark blue - so blue I couldn't see through it. This is what happened (bottom answer).
How may copper acetate ligands be manipulated to change colors?

Heat may be able to substitute for time and quicken the formation of copper acetate...I really haven't tried. Peroxide is dirt cheap. I would have used it if I had known.


Why are the secondary alpha hydrogens more acidic than primary alpha hydrogens (keto-enol tautomerization)?

You're actually asking two questions. There is both kinetic and thermodynamic acidity. There are times when performing deprotonation at low temperature will yield the kinetically favored conjugate base, i.e., the proton that is most exposed. If you increase the temperature of this intermediate, you can get to the point where proton transfers happen in reaction vessel to create the more stable thermodynamically favored conjugate base. Usually, this is the more substituted one.


Stereochemistry: synthesis of an epoxide

"Is it correct that this reaction is SN2, and the inversion occurs?" Yes, and yes, so the product is the one on the top.

"Inversion" in this context doesn't mean that every substituent has to be flipped. Maybe it's clearer if you take away the intramolecular aspect of it. This is a typical SN2 reaction as introduced in many textbooks:

Would you agree, here, that the stereochemistry at carbon has undergone inversion? It is nothing to do with the methyl group or the hydrogen, but rather it is because of the fact that the chloride (that was originally pointing down) was replaced with an oxygen (that is now pointing up).

Now to make it an intramolecular reaction, just replace the R with a single bond to the OH.


How much NaOH required to change the pH of a weak acid solution by 2

RANT - I hate problems that don't use significant figures properly. So is the answer supposed to be +/- 1 ml, +/- 0.1 ml, or +/- 0.01 ml?
First your overall approach is right. There are three parts to solving the problem:

(1) Calculate pHs
(2) Calculate how many moles NaOH
(3) Calculate how many ml of 0.1 molar solution.

So the problem is that there are four phosphate species, $\ce{H3PO4, H2PO4^-, HPO4^{2-},}$  and $\ce{PO4^{3-}}$. The precision needed for the answer is somewhat of a mystery. So what simplifying assumptions are valid?
I'm just going to guess that the volume of NaOH should be accurate to 0.01 ml.
So let's look at the problem another way. I start off with 20.00 ml of 0.1000 molar solution of phosphoric acid and titrate with 0.01 ml at a time with a solution of 0.1000 molar sodium hydroxide. As every aliquot of NaOH is added the pH is measured. So:

(1) when I have added exactly 10.00 ml I have effectively made sodium dihydrogenphosphate. I add 2 to this pH to get the target pH.
(2) I look to see volume of NaOH that gives the target pH and subtract 10.00 ml.

SIDEBAR

Just looking at the problem I'm guessing a few ml. Let's say 5.00 ml. Measuring the NaOH to 0.01 would then mean 1 part in 500 precision (2 parts per thousand). Let's say that the initial $[\ce{H+}]$ is $3.47\times 10^{-5}$. 1 part in 347 is about right. But a pH of 4.46 only has two significant figures, that is 1 part in 46 which is too little precision for my guess of about 2 parts per thousand being needed. So the pH should be noted as 4.460.

Looking at Wikipedia we can see that for phosphoric acid that pKa1 = 2.148,
pKa2 = 7.198, and pKa3 = 12.319.
(1) Given these pKa's we can't do better than 2 ppt.
(2) Only a small part of the $\ce{H2PO4^-}$ will ionize to $\ce{HPO4^{2-}}$ and $\ce{H^+}$. Thus the initial pH will be lower than 7.198 but quite above 2.148.
(3) Given (2) enough $\ce{H3PO4}$ might form to lower the pH when calculating to 2 ppt. This is right on the edge for 2 ppt precision...
(4) The species $\ce{PO4^{3-}}$ can safely be ignored in all the calculations. i.e.  $\ce{[PO4^{3-}] << 0.1000 molar}$
1.0 Quick and dirty method...
To avoid solving a 3rd or 4th order polynomial...
1.1 Calculate pHs
The reaction of interest is:
$\ce{H2PO4- <=> H+ + HPO4^{2-}}\quad\quad 6.339 \times 10^{-8}$
We'll let $\ce{[H+] = [HPO4^{2-}]}$ and assume that $\ce{[H2PO4-]} = 0.1000$ so
$ 6.339 \times 10^{-8} = \dfrac{\ce{[H+]^2}}{0.1000}$
$ \ce{[H+]} = 7.96 \times 10^{-5}$
pH = 4.099
Therefore end pH = 4.099 + 2.000 = 6.099

Checks
$\ce{[HPO4^{2−}]}$ of $7.96\times10^{−5}$ is just less than $1\times10^{-4}$ so the assumption that $\ce{[HPO4^{2−}]}$ is not appreciable to  $\ce{[H2PO4−]} \approx 0.1000$ is just ok.
We've also assumed that an insignificant amount of $\ce{H3PO4}$ will form which isn't true...  $\dfrac{\ce{[H3PO4]}}{\ce{[H2PO4^−]}} = \dfrac{\ce{[H^+]}}{7.11\times 10^{-3}} = 0.0112$

Da fix
We have 3 species of phosphate:
0.0008 $\times \ce{[H2PO4^{−}]}$ = $\ce{[HPO4^{2−}]}$
1.0000 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H2PO4^{−}]}$
0.0112 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H3PO4]}$
$\text{--------}$
1.0120 total $\ce{H2PO4^{−}}$
Normalizing $\ce{H2PO4^{−}}$ to 1.0000 total we have for the concentrations
0.0001 = $[\ce{HPO4^{2−}}]$
0.0988 = $[\ce{H2PO4^{−}}]$
0.001107 = $[\ce{H3PO4}]$
We rearrange the equilibrium for the reaction
$\ce{H3PO4 <=> H^+ + HPO4^{2−}}$
to solve for $\ce{[H^+]}$
$\ce{[H^+]} = \dfrac{\text{K}_{a1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = 7.97 \times 10^{-5}$
so everything now checks...
1.2 Calculate how many moles NaOH
Given:
(1) $\ce{[H^+]} = 7.96\times 10^{-7}$
(2) 0.1 molar = $\ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$
Assume:
(1) $\ce{[H3PO4]} \approx 0$
(2) $\ce{[PO4^{3-}]} \approx 0$
(3) moles change in $\ce{H^+}(\text{aq})$ is insignificant
$6.339\times10^{−8} = \dfrac{\ce{[H+][HPO4^{2-}]}}{\ce{[H2PO4^-]}}$
or
$\ce{[HPO4^{2-}] =} \dfrac{6.339\times10^{−8}\ce{[H2PO4^-]}}{\ce{[H^+]}} = 0.07962\ce{[H2PO4^-]}$
We also know that
$0.1000 = \ce{[H2PO4^-] + [HPO4^{2-}] = 1.07962[H2PO4^-]}$
$\ce{[H2PO4^-] =0.09263}$
$\ce{[HPO4^{2-}] =0.00737}$
$\ce{H2PO4^- + OH^- ->  HPO4^{2-}}$ so
Moles NaOH = $(0.020 \text{ liters})[0.00737] = 1.47\times10^{-4}$
1.3 Calculate how many ml of 0.1 molar solution.
liters 0.1 molar NaOH = $\dfrac{1.47\times10^{-4}}{0.1000} = 1.47\times10^{-3}$
ml 0.1 molar NaOH = 1.47


Why is the auto ignition temperature of petrol higher than that of diesel?

Auto ignition temperature is defined like this by Wikipedia:

  The autoignition temperature or kindling point of a substance is the lowest temperature at which it spontaneously ignites in normal atmosphere without an external source of ignition, such as a flame or spark. This temperature is required to supply the activation energy needed for combustion.

Gasoline (or petrol for non-americans) has an auto-ignition temperature about 50 °C higher than diesel.

The answer boils down to explaining why the typical chemicals making up the two fuels have different activation energy barriers to combustion in air. This reaction basically involves the constituent molecules getting enough energy to react with oxygen.

We can get a crude (and probably oversimplistic) understanding by looking at the typical components of the two fuels. Diesel tends to be made up from linear saturated hydrocarbons; petrol is typically a mixture of branched and unsaturated hydrocarbons. So, a little simplistically, the typical component if diesel will have a larger surface area than a typical component of petrol (branched chain hydrocarbons are more "bunched up" than their linear counterparts with the same number of carbons). To put it another way, in linear hydrocarbons each carbon is has, at least, two hydrogens between it and any surrounding oxygen molecules from the air but in unsaturated or branched hydrocarbons the component carbon atoms are often buried inside because they are surrounded by other carbons (consider the molecule shapes of linear hexane and 2,2 dimethyl butane which have the same number of carbons).

This is a very simplistic way of looking at the question but give a little insight. In typical diesel compound there are simply more opportunities for oxygen to attack the molecule and kick off a reaction because the amount of the molecule exposed to the outside is higher. By being more compact (lots of carbons buried deep inside the molecule) there is less opportunity for this in the components of petrol.

For the reaction to kick off some of the molecules in the mix have to have enough thermal energy to react when they meet. And oxygen has to hit the hydrocarbon hard enough or there is no reaction. Once an oxidation reaction starts, it releases energy increasing the likelihood of another molecular reaction. At a certain point when there is enough energy from new reactions and the mix doesn't spread this out too quickly, this becomes a runaway process and the mixture goes bang. This process is–at the molecular level–easier for diesel than petrol because the chances of an oxygen causing a chemical changes when hitting a hydrocarbon is higher.

Of course a lot of other factors are involved but this crude picture gives at least some useful insight. This was discovered experimentally in the early days of engine design as the designers observed that different extracts from oil had different combustion properties in engines and refined the way oil was distilled to give them the behaviours they wanted in their engines.


Explanation for the trends in nucleophilicity—Orbital interactions or electrostatic attraction

This is a rather intellectually-stimulating question and one that is also very difficult to answer. You have constructed a very good case for why the nucleophilicity order would not be expected to reverse in polar aprotic solvents, i.e. the order of intrinsic nucleophilicities of the halide ions should be $\ce {I^- > Br^- > Cl^- > F^-}$, based on the concepts of frontier molecular orbital interactions, as well as electronegativity and polarisability of the nucleophilic atom.

Affirming your viewpoint

Admittedly, the saturated carbon atom is a soft centre and based on electronegativity considerations, the energy of the $\ce {HOMO}$ of the halide ions should increase from $\ce {F^-}$ to $\ce {I^-}$, i.e. the hardness of the the halide ions as nucleophiles decreases from $\ce {F^-}$ to $\ce {I^-}$. Based on the hard-soft acid-base (HSAB) principle, we would expect the the strength of the interaction between the halide ion and the carbon centre to increase from $\ce {F^-}$ to $\ce {I^-}$.

We can provide more quantitative justification for this using the Klopman-Salem equation$\ce {^1}$:

$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$

The first term is indicative of the strength of the Coulombic interaction while the second term is indicative of the strength of the orbital interactions. In the reaction with the soft saturated carbon, it is actually expected for the reactivity to increase from $\ce {F^-}$ to $\ce {I^-}$ (ref 1, p. 117).

At the moment, all the evidence presented points us towards the conclusion that you have posited. However, we have overlooked the importance of a thermodynamic factor - the strength of the $\ce {C-X}$ bond.

A not-particularly-relevant add-on

Actually, if you think about it the considerations made in the Klopman-Salem equation are essentially kinetic considerations. We cannot conclude anything about reaction energetics from such considerations. The strength of the $\ce {HOMO-LUMO}$ interaction or that of the Coulombic attraction tells us nothing about how the electrons would organise themselves subsequently, what would be the strength of the resultant bonds etc.

A comment made on this by Olmstead & Brauman (1977)

On this topic of explaining the relative order of intrinsic nucleophilicity of the halide ions, Olmstead & Brauman (1977) had this to say$\ce {^2}$:

  The intrinsic nucleophilicities follow the reverse order of the polarizabilities (e.g., $\ce {CH3O^-}$ > $\ce {CH3S^-}$ and $\ce {F^-}$ > $\ce {Cl^-}$ > $\ce {Br^-}$). This could be due to a stronger interaction between the more concentrated molecular orbitals of the anion with the carbon center. It could also be simply a reflection of the greater thermodynamic methyl cation affinities of the smaller anions.

It is important to note that Olmstead & Brauman are merely postulating the reasons. However, I do feel that the point on thermodynamics leads us on the right track to answering this question. On the other point of "more concentrated MOs", I am unclear of how it can be interpreted.

Let us focus now on this point.

How does thermodynamics even influence kinetics?

Well... It certainly does here. What are we concerned with when dealing with nucleophilicity? As it is a kinetic phenomenon, we are essentially concerned with the rate of the substitution reaction. This makes the consideration of the activation energy ($E_\mathrm{a}$) a particularly important one because the higher the $E_\mathrm{a}$, the lower the rate of the substitution reaction.

Here, the $E_\mathrm{a}$ can be taken to be the difference in energy between the reactants and the transition state. Thus, we can decrease the $E_\mathrm{a}$ by either raising the energy level of reactants or decreasing the energy level of the transition state. The reaction energy profile for an $\ce {S_N2}$ reaction is shown below, taken from ref. 3 (p. 394).

There are two ways to see how the strength of the $\ce {C-X}$ bond can affect the $E_\mathrm{a}$. I am not sure if they may be considered equivalent perspectives. Nonetheless, I will present both.

The first perspective is well-explained by Francis & Sundberg (2007) with regard to the $\ce {S_N2}$ reaction$\ce {^3}$:

  Because the $\ce {S_N2}$ process is concerted, the strength of the partially formed new bond is reflected in the TS. A stronger bond between the nucleophilic atom and carbon atom results in a more stable TS and a reduced activation energy.

Another perspective (which may be somewhat similar to the first, or even possibly the same) is that the transition state structure bears resemblance to the product structure, based on the Hammond's postulate. However, this would be more applicable for an endothermic $\ce {S_N2}$ reaction where the transition state structure does indeed bear more resemblance to the product. When the product is more stable, i.e. has bonds that form more exothermically, the energy of the transition state would thus be lowered, and the $E_\mathrm{a}$ would also decrease, although not necessarily proportionally.

Conclusion

The argument you have presented failed to consider the important thermodynamics factor. Thus, it resulted in the predicted reactivity of the halide ions in the reversed order. The takeaway here is that the application of the $\ce {HSAB}$ concept alone does not always produce the correct prediction.

Additional information

More examples are cited for which the HSAB concept fails. These include:

The favourable combination of $\ce {H^+}$, a hard acid, and $\ce {H^-}$, a soft base (discussed here as well) (ref. 4).
The fact that the rate of reaction between $\ce {Ag^+}$, soft acid, and an alkene, soft base, is slower than that between $\ce {Ag^+}$ and $\ce {OH^-}$, a hard base (ref. 1, p. 110)

References

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Student Edition). John Wiley & Sons, Ltd. United Kingdom, 2009.
Olmstead, W. N.; Brauman, J. I. Gas-phase nucleophilic displacement reactions. J. Am. Chem. Soc. 1977, 99(13), 4219–4228.
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.
Pearson, R. G.; Songstad, J. Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. J. Am. Chem. Soc. 1967, 89(8), 1827-1836.


From here:

  Some buffers (e.g., MOPS and HEPES) cannot be autoclaved because they degrade upon heating.

and from here:

  [HEPES] Solutions may be autoclaved under standard conditions.

I guess nobody knows the answer.


Does boron form compounds without covalent bonds?

Boron can form ions but there is some fine print.  You won't get monatomic cations like the metals below it.  Instead, ionic boron structures are formed from clusters where the ionic bonding is driven by the molecular orbital structures in these clusters, not by electronegativity (cf. This answer).
Such clusters are internally held together by covalent bonds between the boron atoms, so in this sense boron is still forming covalent bonds.  The ionic bonds would be with atoms of other elements outside the boron cluster.  Since the valence shells of a neutral boron atom are less than half filled the clusters will likely have low-energy, bonding orbitals that require electrons from outside atoms.  Thus the boron clusters will be anionic and the ionic bonds will be most likely formed with electropositive metals.  As suggested in the comments, magnesium diboride, $\ce{MgB2}$, is one of the most widely studied compounds containing such boron clusters. It has drawn much research interest because of its relatively high critical temperature (39 K) for superconductivity, which may be related to the impact of ionic magnesium-boron bonding on the eletronic interactions that lead to superconduction.
Magnesium diboride has a layered structure in which magnesium layers alternate with boron layers.  The latter are covalently bonded into a hexagonal honeycomb, resembling a carbon layer in graphite.  However, in the boron layers each atom supplies only three electrons per atom instead of four, so the layers may act as electron-accepting structures to form macro-anion having the formula $\ce{B^-}$.  An ionic model for the diboride would then have the empirical formula $\ce{Mg^{2+}(B^-)2}$.  Here I discuss two references I have examined, in which the bonding is examined and the results may be compared with this model.
De la Mora et al. [1] compare magnesium diboride with other $\ce{MeB2}$ diborides using early transition metals and aluminum (the latter might also be regarded as having early-transition-metal character, as there is no $d$ block separating this element from magnesium).  They find that while all the diborides have significant ionic character, this ionicity is enhanced in the magnesium compound.  Thereby the magnesium compound has increased electrical-conduction anisotropy as the valence electrons are strongly localized towards the boron layers.  Zirconium diboride, with less ionic bonding and less electron localization, is also superconducting, but its critical temperature according to this reference is only 5.5 K versus 39 K for the magnesium compound.  The authors also suggest that an isoelectronic, even more strongly ionic $\ce{Li(BC)}$ compound may offer an even further enhancement in superconductivity.
Nishibori et al. [2] found that at room temperature, magnesium is essentially fully ionized to $\ce{Mg^{2+}}$ while the boron remains neutral; the negative charge is associated with the interstitial regions as if to constitute metallic bonds.  This still represents two-thirds of the theoretical charge separation for an ionic model and in that sense, the bonding between magnesium and boron may be deemed predominantly ionic.  At 15 K the electrons become more localized so the boron now has a significant negative charge and the percentage of theoretical charge separation exceeds 80%.
Thus both references agree that in magnesium diboride, the combination of an electropositive electron source with a favorable molecular structure for electron acceptance leads to strongly ionic bonding between the magnesium and the boron.  This applies especially in the low-temperature superconducting state, even as the boron-boron bonding within the boron layers themselves remains covalent.
References
1.
Pablo de la Mora, Miguel Castro and Gustavo Tavizonb,
"Comparative study of the electronic structure of alkaline-earth  borides (MeB2; Me=Mg, Al, Zr, Nb, and Ta) and their normal-state conductivity", Journal of Solid State Chemistry 169 (2002) 168–175, https://doi.org/10.1016/S0022-4596(02)00045-2.
2.
Eiji Nishibori, Masaki Takata, Makoto Sakata, Hiroshi Tanaka, Takahiro Muranaka and Jun Akimitsu, "Bonding Nature in MgB2", Journal of the Physical Society of Japan 70:8 (2001), 2252-2254,
https://doi.org/10.1143/JPSJ.70.2252.


Actually the % error is calculated as:
$\frac{\text{experimental} - \text{actual}}{\text{actual}}\cdot100\%$
That way

if experimental < actual then the error is negative
if experimental > actual then the error is positive

So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?

Not exactly zero...

EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then, 0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs

Correct for subtraction you have used. Addition and subtraction round to the the last common significant decimal place of all the measurements. That would be the 8 in 10.98 in the following example.
So $10.9\bar8 + 0.6754 + 0.4795 = 12.1349 \approx 12.13 $
In actually you should have used the reverse 0.997 - 0.99717 = -0.000171

there are no significant figures so this would mean the calculated 0.0171% error would be zero correct.

Now this gets sticky. You could write that the error was 0% but that would indicate only one part in 100 precision. If you write 0.00% it isn't entirely clear if that is two or three significant figures. But you have claimed that the hundredth place in the % is significant, which it is not.
Remember that the minimum nonzero error possible was +/- 0.001 which yields an error of 0.1%.
So the error would be best written as 0.0% indicating that the error is 0 to 1 part in a thousand.

Reference: Significant Figures at Wikipedia
I'll leave the propagation of errors for a different exercise.


In the carbonate anion, which atoms gain the two electrons?

One can draw Lewis structures to see where the lone pairs of electrons may be.

Here you can see that you can draw the ion like this, with one of the oxygen atoms having a double bond with the carbon, and the other two having a single bond. This means that the double bonded oxygen atom has no charge. The single bonded oxygen atoms have a net charge of $-1$ each as they have gained a lone pair of electrons.
However this picture does not tell the whole story. In reality, the negative charges are not localised to individual oxygen atoms in the ion. We can draw a total of three structures (called canonical forms):

These can be drawn because in actual fact there is delocalisation of electrons between the carbon and each oxygen atom. The true structure is not changing/oscillating between these three forms, but is actually called a resonance hybrid of all of the canonical forms. This is shown below. Therefore you can see that the negative charges are not localised to single oxygen atoms, but are instead delocalised across the entire ion.

If you want to read more about the carbonate ion and its structure: https://en.wikipedia.org/wiki/Carbonate#Structure_and_bonding


Why does the aqueous solution of sodium peroxide turns red litmus into white?

Your analysis seems right and you remember correctly that colour change of litmus of litmus paper will indicate the pH value of the solutions into which it is dipped.

That is, unless the dye (litmus) is irreversible transformed to another compound which no longer shows a pH dependent colour.

As the experiment states:

  A piece of red litmus paper turns white [...]

With other words: Litmus was bleached.

This bleaching typically is the result of an oxidative transformation of a (former) dye.

Here, the only conceivable oxidant in solution is hydrogen peroxide. Consequently, the initial compound in question must have been $\ce{Na2O2}$.


Electrons and protons are charged particles. The electrons have negative charge, while protons have positive charge. A neutral atom is an atom where the charges of the electrons and the protons balance. Luckily, one electron has the same charge (with opposite sign) as a proton.

Example: Carbon has 6 protons. The neutral Carbon atom has 6 electrons. The atomic number is 6 since there are 6 protons.


How is it possible for a substance to have a high heat of vaporization but a low boiling point?

This seeming contradiction can be reconciled by examining the thermodynamic quantities involved.  First, per Wikipedia, the enthalpy of vaporization is "the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure."  Written symbolically:

$$
\Delta H_{\mathrm{vap}} = H_{\mathrm{gas}} - H_{\mathrm{liq}}
$$

For this situation, it is most helpful to consider $H_i$ as the enthalpy per mole ("molar enthalpy") of the species of interest in phase $i$.

Formally, again per Wikipedia, the boiling point for a substance is "the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapor."  This is not so helpful for answering this question, however.

Another way of defining the boiling point is the temperature at which the liquid and the vapor of the substance are in equilibrium, for a closed system (no mass can get in or out) containing only the pure substance of interest at a given pressure$^*$.  The equilibrium condition here can be expressed in terms of molar Gibbs free energies (free energy per mole$^{**}$) as:

$$
G_{\mathrm{liq}} = G_{\mathrm{gas}}
$$

By definition, the molar free energy of a phase is equal to the molar enthalpy minus the temperature times the molar entropy:

$$
G = H - TS
$$

So, the boiling point equilibrium condition can be rewritten as ($T_b$ is the boiling point):

$$
H_{\mathrm{liq}} - T_bS_{\mathrm{liq}} = H_{\mathrm{gas}} - T_bS_{\mathrm{gas}}
$$

Subtracting $H_{\mathrm{liq}}$ and $-T_bS_{\mathrm{gas}}$ from both sides gives:

$$
T_b\Delta S_{\mathrm{vap}} = T_b\left(S_{\mathrm{gas}} - S_{\mathrm{liq}}\right) = H_{\mathrm{gas}} - H_{\mathrm{liq}} = \Delta H_{\mathrm{vap}}
$$

Here, $\Delta S_{\mathrm{vap}}$ is the molar entropy of vaporization, which is analogous to $\Delta H_{\mathrm{vap}}$.  The boiling point is thus:

$$
T_b = \frac{\Delta H_{\mathrm{vap}}}{\Delta S_{\mathrm{vap}}}
$$

So, what does this result mean?

Qualitatively, as noted in answers to the linked questions, $\Delta H_{\mathrm{vap}}$ is a measure of how much energy must be added to one mole of molecules in the liquid to free them from their associations with the other liquid molecules.  The physical meaning of $\Delta S_{\mathrm{vap}}$ is somewhat harder to describe, but corresponds roughly to the increase in the 'freedom to wiggle' afforded by the transition to the gas phase from the liquid.

So, in order to have a large $\Delta H_{\mathrm{vap}}$ (lots of energy required to drive molecules into the gas phase) but also a low $T_b$ (low temperature at which molecules "like" to transition into the gas phase) a chemical must have a large $\Delta S_{\mathrm{vap}}$, which means that each molecule gains a significant amount of 'freedom of movement' from entering the gas phase.  This makes sense in the case of ethanol versus water in Dissenter's question: ethanol is bulkier than water, so even though they both have similar values of $\Delta H_{\mathrm{vap}}$, ethanol gains more 'freedom to wiggle' in the gas phase and thus has a lower $T_b$.

This also helps to provide a qualitative explanation for why $T_b$ is proportional to pressure: as the pressure of the gas phase increases, the 'freedom to wiggle' gained by entry to the gas phase decreases, because higher $P$ means more collisions with gas-phase molecules.  Thus, one can expect $\Delta S_{\mathrm{vap}}$ to decrease with increasing pressure ($S_{\mathrm{liq}}$ isn't affected nearly as much by pressure changes as is $S_{\mathrm{gas}}$), which translates directly to the experimentally observed increase in $T_b$.

More details on the relationship between boiling point and pressure can be found in the Wikipedia articles on the saturation temperature and the Clausius-Clapeyron relation.  More information on equilibrium and Gibbs free energy can be found on www.chem1.com.

$^*$ In this model system, the "knob" one has available (see Gibbs' phase rule) to change the equilibrium pressure and temperature is to add to or remove from the system some number of moles of the species of interest.

$^{**}$ As orthocresol rightly notes in a comment, the equilibrium must be expressed in terms of the chemical potentials of each species present. However, in this case there is only one species present and thus examining the total molar Gibbs free energy is sufficient.


Is the correct configuration assigned to the given pair of geometrical isomers of 1,2-dichlorocyclopropane?

Those drawings are rather confusing.  In the following drawing I've reproduced your "cis" drawing and then below, I've redrawn it more clearly.  We now see that your molecule is actually the "trans" isomer.

The dark wedge bonds should be either drawn both "up" or both "down".  The same applies to the dashed bonds, they too are either drawn both "up" or both "down".  Your drawing is different, it has one wedge bond "up" and the other "down".  This is confusing and makes interpretation somewhat difficult.  In my redrawn version both dark wedge bonds are "up" and both dashed bonds are "down" making interpretation easier.


Boric acid is a Lewis acid, therefore when in water it accepts the lone pair from oxygen and becomes $\ce{(OH)3B^{-}-OH2^{+}}$.
From this compound a proton can be donated and this is how boric acid behaves as a Brønsted acid.


What is the plastic material of the CD cases?

These are called slimline jewel cases.

From the Wikipedia article on Optical Disc Packaging:

  The bulk of slimline cases are made with translucent or transparent polystyrene, and are available in multiple colors. A stronger alternative is made from semi-opaque, semi-flexible polypropylene which is strong enough to protect the disc, but flexible enough not to break easily.

Here is an image of a polystyrene case to compare to yours:


What does it mean by degree of aromaticity of a molecule?

Aromaticity might be viewed as resonance stabilzation energy, NMR shift, or altered reactivity.  Here is a way of measuring its change among related systems,

Chem. Commun.  48, 8144 (2012)
J. Am. Chem. Soc. 131, 189 (2009)
J. Am. Chem. Soc. 127, 16251 (2005)
J. Org. Chem. 67, 8798 (2002)
J. Am. Chem. Soc., 124, 13495 (2002)
Chem. Rev. 101, 1301 (2001)
Tetrahedron 57, 3689 (2001)

http://nptel.ac.in/courses/IIT-MADRAS/Engineering_Chemistry_III/pdf/1_Aromaticity.pdf
A more practical approach derives from MO theory and Hund's rule of maximum multiplicity.

http://www.youtube.com/watch?v=U6FX5RfLjd0
http://www.youtube.com/watch?v=H0so_NJ2aVw
Frost circle


How to partially hydrolyze triglycerides?

Over the last decades, partial hydrolysis of triglycerides has been examined in various contexts.

Note that this isn't my field of work and that there might hundreds of better and/or more instructive recipes, but a cursory search yielded the following freely available references:

THE DIGESTION IN VITRO OF TRIGLYCERIDES BY PANCREATIC LIPASE,
F. H. Mattson, and L. W. Beck,
J. Biol. Chem., 1955, 214, 115-125 (PDF)

A stereospecific analysis of triglycerides,
H. Brockerhoff, Journal of Lipid Research, 1965, 6, 10-15
(PDF).

The products from lipase-catalysed hydrolysis of bovine milkfat kill
Helicobacter pylori in vitro,
Cynthia Q. Sun, Charmian J. O’Connor, Alastair K. H. MacGibbon, and Anthony M. Roberton, FEMS Immunology & Medical Microbiology, 2007, 49, 235-242
(DOI)

It seems that the digestion of triglycerides by lipases in buffered solutions is a viable approach.


What chemicals cause the smell of a septic tank?

According to the Wikipedia article Septic tanks, as septic tanks are generally anaerobic environments, pungent smells from the formation of hydrogen sulfide $\ce{H2S}$, which is sometimes referred to as 'rotten egg gas'. As you stated, ammonia $\ce{NH3}$ can be produced - also with a distinct smell.

Also, according to the Wikipedia article, fermentation could occur, producing the odourless methane and carbon dioxide as well as phosphates.

Incidentally, $\ce{H2SO4}$is the chemical formula for sulfuric acid.


What is the volume of the produced SO2 gas at STP?

You have an excess of oxygen, right...?

So the limiting reactant is the sulfur, which is $2.4$% of the coal.

Spoiler:
Step 1:

   So calculate the sulfur content, which is $3.1 \times 10^7$ kilograms multiplied by $2.4 \times 10^{-2}$ which is $7.44 \times 10^5$ kilograms of sulfur.

Step 2:

   The $\ce{SO_2}$ is $50$% sulfur by mass (32 amu of sulfur vs. 64 amu total) so there is $1.488 \times 10^6$ kilograms of $\ce{SO_2}$ produced.

Step 3 (for moles, optional):

   If you want moles, divide by $64$, which is $2.325 \times 10^4$ moles (someone check my math)

Then again, you emphasized STP, so proceed with caution (sulfur doesn't react at $20$ degrees Celsius).


How to determine a Reynolds number for mixture of 2 liquids

The Reynolds Number is the ratio of inertial to viscous forces and is calculated according to this equation: $$Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density, $\mu$ is the dynamic viscosity, $v$ is the fluid velocity and $L$ is a characteristic length that depends on the flow being modeled. When estimating the Reynolds number of a mixture, the term $v L$ would not change but we need to make the assumption that the term $\frac{\rho}{\mu}$ of the mixture can be derived from the properties of the pure components.  This assumption will be valid if

The components have similar structure
The components are non-polar
The components are not associated
The difference in the kinematic viscosities is small (~15 cP)

Some of these conditions could be relaxed if one species is in significant excess.

Densities are additive so long as the volume of the mixture equals the sum of the volumes of the pure solutions.  (Search for volume weighted densities for more information.)  Viscosities of mixtures is a much harder thing to estimate.  One method is the Grunberg-Nissan equation: $$Log\eta_m=x_1Log\eta_1+x_2Log\eta_2+x_1x_2d$$ Where $\eta_i$ is the dynamic viscosity of component $i$, $x_i$ is the mole fraction of component $i$ and $d$ is an interaction coefficient.  If you assume $d=0$ then you have the Arrhenius equation for the viscosity of a mixture.

So, at the end of the day, if you state you are working under the assumptions that volume weighted densities equal mass weighted densities and use the Arrhenius equation to determine the viscosity of the mixture then you can average the weighted properties of the pure components to get an estimate of the mixture.

Just a word of caution under these assumptions you should take the mass weighted average of the densities but the mole weighted average of the viscosities.

I relied on this reference for much of this answer.  It will require a subscription to the service to access the PDF, but some of it can be found on google books.


Balloon is not an ideal system to study pressure-volume relationship. Because, on expansion, the elastic skin also expands and there needs to be an additional pressure build-up on the inner side to counter that force, very similar to the extra pressure in a spherical bubble. (where the surface tension acts exactly similar to the elastic balloon skin)

But in your arguments, you are forgetting an important point. Blowing a balloon also involves increasing the amount of air in it. Boyle's law holds for only a fixed volume of gas. If you change the number of moles, Boyle's law no longer stays valid.


Your problem is that you only accounted for the first dissociation of $\ce{H2SO4}$, a polyprotic acid -- your book needed the extra specificity from the second dissociation. I will walk through the entire process, including the parts that you know already.

Begin by finding the molar mass of $\ce{H2SO4}$ in order to find out to how many moles one gram of it is equivalent. Then, convert to molarity (concentration) using the given volume of water.

$$\ce{MM_{H_2SO_4} = 2*1.01 g + 1*32.06 g + 4*16.00 g = 98.08 g}$$

$$\ce{\frac{1 g H2SO4}{1}\times\frac{1 mol H2SO4}{98.08 g H2SO4} = 1.0\times10^{-2} mol H2SO4}$$

$$\ce{\frac{1.0\times10^{-2} mol H2SO4}{1 L H2O} = 1.0\times10^{-2} M H2SO4}$$

Although the ICE-box is a formality for such a strong acid, it can still be shown.

\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 1.0\times10^{-2} &  & 0 & 0 \\ \hline
 & \ce{H2SO4} & \ce{H2O} & \ce{H3O+} & \ce{HSO4-}\\ \hline
\text{Change}: & -x &  & +x & +x \\ \hline
\text{Equilibrium}: & 0 &  & 1.0\times10^{-2} & 1.0\times10^{-2}\\ \hline
\end{array}

The second ICE-box is a good way of organizing the second dissociation. Transfer the equilibrium concentrations from the first table. All calculations up to the line are for finding the change (using $\ce{K_{a(2)} = 1.2\times10^{-2}}$). Note that after $y$ is found, it is used again in the second ICE-box to determine the equilibrium concentrations after the second dissociation. Also note that you cannot neglect the $y$ after the second equation because of the similar magnitudes of the molarity and the $K_a$ and must use the quadratic formula.

\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 1.0\times10^{-2} &  & 1.0\times10^{-2} & 0 \\ \hline
 & \ce{HSO4-} & \ce{H2O} & \ce{H3O+} & \ce{SO4^{2-}}\\ \hline
\text{Change}: & -y &  & +y & +y \\ \hline
\text{Equilibrium}: & 0.5\times10^{-2} &  & 1.5\times10^{-2} & 4.8\times10^{-3}\\ \hline
\end{array}

$$\ce{K_a = \frac{[H3O+][SO4^{2-}]}{[HSO4-]}}$$

$$\ce{1.2\times10^{-2} = \frac{(1.0\times10^{-2} + y)(y)}{1.0\times10^{-2} - y}}$$

$$\ce{1.2\times10^{-4} - (1.0\times10^{-2})y = (1.0\times10^{-2})y + y^2}$$

$$\ce{1.2\times10^{-4} = (2.0\times10^{-2})y + y^2}$$

$$\ce{0 = y^2 + (2.0\times10^{-2})y - 1.2\times10^{-4}}$$

\begin{split}
\ce{y} & = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& = \frac{-(2.0\times10^{-2})\pm\sqrt{(2.0\times10^{-2})^2-4(1)(-1.2\times10^{-4})}}{2(1)}\\
& = \frac{-2.0\times10^{-2}\pm\sqrt{4.0\times10^{-4}+4.8\times10^{-4}}}{2}\\
& = \frac{-2.0\times10^{-2}\pm\sqrt{8.8\times10^{-4}}}{2}\\
& \approx 4.8\times10^{-3}
\end{split}

Plug into the p function to determine the pH.

$$-\log(1.5\times10^{-2})=1.82$$

Note that $-\log(2\times10^{-2})=1.69$ so your book probably rounded to one significant figure (which would make sense given the way the problem is worded).


Are there Exchange integrals in Semiempirical methods?

Semiempirical quantum chemistry models do carry out proper Hartree-Fock calculations, including both Hartree (electron-electron electrostatics) and Fock exchange contributions to the total electronic energy. As noted in the question, the tensor of 4-center Coulomb integrals is heavily sparsified, approximated, and parameterized to simplify and accelerate semiempirical calculations. Even in the most approximate models (CNDO), the exchange interaction between molecular orbitals does not vanish in general. For example, the exchange integral between a pair of molecular orbitals $i$ and $j$ with real atomic-orbital coefficients $\phi_{i \mu}$ is simplified by the zero-differential-overlap (ZDO) approximation stated in the question as

$(ij|ji) = \sum_{\mu,\nu,\lambda,\sigma} \phi_{i \mu} \phi_{j \nu} \phi_{j \lambda} \phi_{i \sigma} (\mu \nu | \lambda \sigma) = \sum_{\mu,\lambda} \phi_{i \mu} \phi_{j \mu} \phi_{j \lambda} \phi_{i \lambda} (\mu \mu | \lambda \lambda)$.

This answer might be confusing precisely what the original question means by "exchange integral". For example, no explicit integrals are involved in calculating the Coulomb matrix elements used by most semiempirical models, and there may be some class of Coulomb matrix elements in the atomic-orbital basis that are considered to be "exchange-like" that are indeed neglected in most semiempirical models (I'm not familiar with all variants of this nomenclature). The neglect of diatomic differential overlap (NDDO) approximation that eventually superseded ZDO does allow for nonzero $(\mu \nu | \lambda \sigma)$ values when $\mu$ and $\nu$ (similarly, $\lambda$ and $\sigma$) are different atomic orbitals on the same atom.


Tap water leaving white residue after drying

You either have hard water or a water softener system.
If you have hard water, then the residue is a combination of calcium / magnesium carbonate / chloride.
If you have a water softener system, then the residue is from the salt used to remove the hard ions.


How to sloppy freeze a dihedral angle within a range in a Gaussian modredundant optimization?

TL;DR: It is impossible (=not documented if it were) with Gaussian 09 Rev. E.01 and older, it might be possible with Gaussian 16, but in all cases not via opt(modredundant).

I am unaware of any sloppy freezing possibility in pre-Gaussian 16, and for the latter one has to dig much deeper in the manual and start constructing it; but at least it works in theory.

Gaussian 03

Unfortunately, you have misinterpreted the commands you found in the manual. You can find the full description for the opt keyword in via The Internet Archive, but I'll share the important parts here for convenience.

The construct

[Type] N1 [N2 [N3 [N4]]] [[+=]value] [A | F] [[min] max]]

indeed exists for a opt(modredundant) input section, but it behaves not as a sloppy freeze, but as a range exclusion criterion, which you can find a little lower in the manual:

  [...]
  An asterisk (*) in the place of an atom number indicates a wildcard. Min and max then define a range (or maximum value if min is not given) for coordinate specifications containing wildcards. The action specified by the action code is taken only if the value of the coordinate is in the range.
  [...]

For example, the input

D * * * * F 175.0

freezes all defined dihedral angles that are smaller than 175.0°.

The input

D * 4  5 * F 160.0 175.0

freezes all defined dihedral angles around the bond between atom 4 and 5 in the interval 160.0 - 175.0°.

The input

D 3 4  5 6 F 160.0 175.0

is a syntax error because it does not contain a wildcard.

Gaussian 09

The option modredundant to the opt keyword still exists, but has a slightly modified input. Specifically the [[+=]value]] part has been removed. Again you can look up the full page on The Internet Archive.
The examples above should, however, still behave the same.

  [...]
  Lines in a ModRedundant input section use the following syntax:
      [Type] N1 [N2 [N3 [N4]]] [A | F] [[min] max]]
  [...]
  An asterisk (*) in the place of an atom number indicates a wildcard. Min and max only apply to coordinate specifications containing wildcards. The action then specified by the action code is taken only if the value of the coordinate is in the min–max range (or below maximum value if min is not given).
  [...]

Therefore you also still get the error "Invalid extra data found".

Gaussian 16

In this version, while opt(modredundant) is still maintained, a new feature is rolled out. As far as I can see, there are no changes to Gaussian 09 regarding the modredundant input section. You can find it live on the Gaussian homepage or a copy on The Internet Archive.

According to this, the above examples should still produce the same results.

The new feature is Generalized Internal Coordinates or GIC. See a description on the Gaussian homepage or on The Internet Archive. You can see the GIC in action in my answer on Constrain bond angle In Gaussian molecular structure optimization and on Gaussian: Relaxed scan with modredundant optimization and dummy atoms.

In theory you should be able to freeze a GIC if it meets certain criteria for example from the manual:

  The following example removes an angle coordinate generated by default if ≥179.9°, substituting a linear bend:

A(1,2,3) Remove Min=179.9        !  Remove angle coordinate if too large.
L(1,2,3,0,-1) Add IfNot A(1,2,3) !  Add linear bends only if
L(1,2,3,0,-2) Add IfNot A(1,2,3) !+ the angle coordinate not active.

I have prepared a small example, but I am really not sure if it worked as intended. For cis-butadien the dihedral distortions should be small, so there isn't much visible. There are no syntax errors, but I am not really sure whether the section actually did what it should. Try it yourself and investigate further. (! recognises comments)

#p PM6                ! Semiempirical method for fast testing
opt                   ! Optimise
geom(AddGIC)          ! Use GIC, can also be specified with opt(AddGIC)
scf(xqc,MaxCycle=250) ! Switch to quadratic conv. if necessary, more cycles
symmetry(none)        ! Do not use symmetry

distorted cis-Butadiene (semi-constrained)
optimisation run
[Free title card can have multiple lines]

0 1
 C     0.000000     0.000000     0.000000
 H     0.000000     0.000000     1.089000
 H     1.026720     0.000000    -0.362996
 H    -0.513358    -0.889166    -0.363001
 C    -0.754249     1.306394    -0.533333
 H    -0.738680     2.222526     0.055206
 C    -1.531373     1.266770    -1.931370
 H    -2.044734     2.155933    -2.294370
 C    -1.615181    -0.098364    -2.761662
 H    -2.540288    -0.117779    -3.335874
 H    -0.765588    -0.163525    -3.439797
 H    -1.596709    -0.942933    -2.074433

sloppyfix=D(1,5,7,9)               ! Define an alias for the dihedral
sloppyfix freeze min=5.0 max=355.0 ! freeze in intervall

! Empty line above is important.

That the setup somewhat works can be seen with the following example of dihydrogen. First the atoms are separated 400 pm (=4.00 Å), and the value should be frozen at 250 pm (=2.50 Å), the PM6 optimised value is 76 pm (=0.76 Å).

#p PM6 opt(AddGIC,MaxStep=5) scf(xqc,MaxCycle=250) symmetry(none)

Free title card

0 1
 H    -2.000000     0.000000     0.000000
 H     2.000000     0.000000     0.000000

theHHbond=R(1,2)
theHHbond freeze max=2.5

! Empty line above is important.

Note that it will be frozen after the condition triggers, so it will not be able to the value that triggers it. Here are the values and messages you'll find in the optimisation run:

                           ----------------------------
                           !    Initial Parameters    !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name      Definition          Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! theHHbond R(1,2)              4.0            estimate D2E/DX2                !
 --------------------------------------------------------------------------------

[...]

 Variable       Old X    -DE/DX   Delta X   Delta X   Delta X     New X
                                 (Linear)    (Quad)   (Total)
 theHHbon     4.97283  -0.03796   0.00000  -1.00000  -1.00000   3.97283
         Item               Value     Threshold  Converged?
 Maximum Force            0.037962     0.000450     NO
 RMS     Force            0.037962     0.000300     NO
 Maximum Displacement     0.500000     0.001800     NO
 RMS     Displacement     0.707107     0.001200     NO
 Predicted change in Energy=-4.653046D-02
 Lowest energy point so far.  Saving SCF results.
 Internal coordinate with a value of  3.9728 a.u. has been conditionally frozen
 because it is in Min/Max range.
 The coordinate's label:
 theHHbond
 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

[...]

 Variable       Old X    -DE/DX   Delta X   Delta X   Delta X     New X
                                 (Linear)    (Quad)   (Total)
 theHHbon     3.97283  -0.07253   0.00000   0.00000   0.00000   3.97283
         Item               Value     Threshold  Converged?
 Maximum Force            0.000000     0.000450     YES
 RMS     Force            0.000000     0.000300     YES
 Maximum Displacement     0.000000     0.001800     YES
 RMS     Displacement     0.000000     0.001200     YES
 Predicted change in Energy=-0.000000D+00
 Optimization completed.
    -- Stationary point found.
                           ----------------------------
                           !   Optimized Parameters   !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name      Definition          Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! theHHbond R(1,2)              2.1023         -DE/DX =   -0.0725              !
 --------------------------------------------------------------------------------
 Lowest energy point so far.  Saving SCF results.
 Largest change from initial coordinates is atom    1       0.949 Angstoms.
 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

Note that in this example there is only one coordinate, ergo the optimisation finishes when the freeze condition is triggered. In this case there is quite a significant deviation from the condition max=2.5 and the actual value when it is frozen 2.1023. I would expect it to behave better with larger systems.

I would assume that the variable remains frozen for the remainder of the optimisation, with no more changes possible.
Standard disclaimer: Geometry optimisations with frozen parameters are critical for production purposes. Always check if you have a well converged molecular structure (stationary point) with no imaginary modes.


There's actually two questions being asked here:

  What is holding ink on paper?

and

  Why do some inks take longer to dry?

Ink sticks to paper because it has a high affinity for the medium it's being applied to. To group things into simple categories we could say, there are hydrophobic inks (non-polar inks that don't like water) and hydrophilic inks (polar inks that do like water). As with the inks, surfaces can also fall into this category. When ink is thrown onto a surface that it doesn't like, it aggregates, clumping together to avoid all interaction with the surface.

A simple illustration showing why this is unfavorable can be seen with water interacting with lipids. In the picture below, water is forming ordered cages around the hydrophobic part. These cages cause a decrease in entropy, and by the 2nd law of thermodynamics, this is unfavorable. You can think of the ink as the water, and the hydrocarbon chain of the lipid as the surface.

When inks do favor their substrate (surface) they will attach via h-bonding, london dispersion forces, and/or other inter-molecular forces.

As for the second question, the answer relies more on what the ink is made out of. According to the wiki on inks:

  Ink can be a complex medium, composed of solvents, pigments, dyes, resins, lubricants, solubilizers, surfactants, particulate matter, fluoresces, and other materials.

Generally, the ink, which has become mobile in its solvent, will remain mobile until the solvent evaporates. So you'll find that the dry time is dependent upon the rate of evaporation. That's why the dry times can be longer for some oil-based pens.


Prove that internal energy is a state function

The three laws of thermodynamics (or four, depending on the source) are postulates that scientists have assembled in an attempt to describe the way the universe works. The First Law is a reformulation of the principle of conservation of energy. It says very simply that while energy is indestructible it can be exchanged in various ways. It also implies that a system can exist in different states, each with a set of clearly identifiable properties including a fixed amount of energy. Declaring the state a system is in also stipulates the energy of the system. We can write this mathematically as $U=f(\mathrm{state})$.
One also presumes that if the energy of a system changes, then energy has been transmitted in some measurable form by an identifiable mechanism. In other words, for some process you can always write
$$\Delta U = \sum_x \mathrm{energy~in~transit~via~mechanism~}x$$
where the sum goes over all the ways we imagine energy might be transferred.
Classical thermodynamics suggests partitioning the energy in transit into two basic categories, work and heat. Heat is transferred when objects are at different temperatures (this forms the basis for the Zeroth Law of thermodynamics). If you insulate a system, heat is no longer a viable way to transfer energy. That leaves all other possible ways of transferring energy, which we call work. Alternately, if all identifiable means to do work are inaccessible, then the system can exchange energy only as heat.
The point is that if the initial and final states are known then $\Delta U$ is fixed, but there might be different ways (paths) that change the energy of the system by that amount. For instance, work can be used to change the temperature of a substance just as putting the substance in contact with an external heat bath can.


The structure of ethanenitrile is:

$$\ce{H3C-C#N:}$$

Qualitatively, we can draw some conclusions about the molecular orbitals in the structure. If covalent bonds result in in the involved electrons being lower in energy, then which pair of electrons in the structure has the highest energy?

However, if you are looking at nitriles, you probably already know that both lone pairs and pi bonds are good proton acceptors.

Let's consider what the difference is if the lone pair is the base or if a pi bond is the base:

  Lone Pair

$$\ce{H3C-C#N + H+ -> H3C-C#N+-H}$$

  Pi Bond

Which atom will the proton end up attached to? Which of the two following carbocations is more stable?

$$\ce{H3C-C#N + H+ -> H3C-C+=NH}\ \  or\ \  \ce{H3C-CH=N+}$$

Now... what is the relationship between the two conjugate acids drawn?

$$\ce{H3C-C#N+-H}\ \ and\ \ \ce{H3C-C+=NH}$$

Where is the HOMO?


Unfortunately the sloppy depiction of sulfoxides will cause such misconceptions. Your analysis is quite good:

  The central sulfur atom is bound to 3 atoms and then has a lone pair, so surely it has to use sp3 hybridised orbitals?

The only thing one should keep in mind is that the model of hybridisation starts to break down for these compounds (third period and below).
In this case it also is not strictly necessary to employ hybridisation, but as a first (very rough) approximation it can give you an educated guess.
In any case, the sulfur is approximately tetragonal coordinated.

Both depictions are indeed missing the lone pair at sulfur.

As such the first one would has 10 electrons around sulfur, which used to be regarded as correct, but is now a deprecated viewpoint. Sulfur would need to use d orbitals, which has been proved wrong.
The charge separated structure is superior to the above, because it does not suggest the use of d-orbitals.

The issue whether or not these compounds are chiral has been answered by Jan (Can heteroatoms with lone pairs be chiral centres?):

  The sulfoxide case. Sulfoxides $\ce{R1−SO−R2}$ have a free electron pair and a trigonal pyramidal configuration on the sulphur atom. The lone pair is in an orbital with a high s-contribution meaning that the nitrogen inversion highlighted above is rather complicated. Thus, sulfoxides can be assumed to be chiral on sulphur (although in practice they are often generated racemicly).

You have asked more general though:

  Can a molecule be achiral whilst it's resonance form chiral?

It is very important to realise that resonance only affects the electronic structure. Therefore in all resonance contributors the position of the nuclei is the same. Since chirality is determined by the spatial arrangement of the atoms, the question itself is void. There cannot be a set of resonance contributors which has chiral and achiral structures. Please also see What is resonance, and are resonance structures real?


Are there still "prairie dog refineries"?

Any collection of 'hooch stills' that were soldered (?) together would probably have been uneconomical almost as soon as they were built, because welding together a series of serviceable fractional distillation towers is not that technologically advanced and much more efficient.  However, there are certainly a number of relatively small refineries.

For instance, about the largest refinery in the U.S. (for comparison) is Exxon's in Baytown, TX that has a capacity of 560,000 barrels per day.

A couple of the smallest ones are the 5000 bbl/day refinery in Wyoming (Silver Eagle, in Evanston) and the 2000 bbl/day Foreland Refinery in Ely, NV.  The latter (if run all day) is equivalent to around 1.5 barrels a minute.

Also (to respond to a comment) there are still some refineries that don't do catalytic cracking (probably just straight distillation of the petroleum into fractions, which is viable for small refiners because they can save on expensive (and often proprietary) cracking units and are small enough that they can sell all the fractions that they produce.  The recently closed Flint Hill refinery in Alaska used another approach, reinjecting unused fractions back into the TransAlaska pipeline.)

For a list of refineries, the EIA is always a good place to look (pdf warning).

A wiki on refineries is "A Barrel Full" which has detailed information on individual refineries.


Deviation of Ytterbium from general trend

To compliment Jon Custer's comment, we can likely trace this deviation to the electron configuration of ytterbium. Note that ytterbium is not the only deviation among the lanthanides - Europium deviate as well:

Why is that? Note the electron configurations of europium and ytterbium:

Europium - $\ce{[Xe]} 4f^7 6s^2$
Ytterbium - $\ce{[Xe]} 4f^{14} 6s^2$

In europium, the $4f$ subshell is half-filled, an arrangement that is particularly desirable from a quantum mechanics standpoint. In ytterbium the $4f$ subshell is completely filled, another 'happy' arrangement. These configurations lead to changes in bonding behavior, for example noting that most lanthanides have the hexagonal close pack or the double hexagonal close pack crystal structures. The only ones that are different are samarium, which is rhombohedral, europium which is body-centered cubic, and ytterbium with is face-centered cubic.

The link between electron configuration and physical properties exists elsewhere in the periodic table, but the trends are not as clear. However, transition elements with filled or half-filled $d$ subshells are likely to have lower melting points than their neighbors.


What pressure units should be used when calculating electrode potential with Nernst equation?

Just as you are supposed to use activity instead of concentration for dissolved components when applying Nernst equation, you are also supposed to use fugacity instead of the partial pressure for the gaseous components. The reason for that will become transparent if you follow the derivation of Nernst equation using chemical potential.

Fugacity is defined with a reference to the standard state with the pressure of $\pu{10^5 Pa}$ or $\pu{1 bar}$. At ordinary pressures fugacity is numerically approximately equal to the pressure expressed in bars or atmospheres ($\pu{1 bar} \approx \pu{0.987 atm}$), so the author of the book is right.


Why is acetone less acidic than cyclohexanol?

if you consider the structure of acetone, it has 2 $\ce{CH3}$ groups and a $\ce{C}$ double bond $\ce{O}$ ($\ce{C=O}$). Therefore, for the $\ce{H}$ to disassociate, it must be removed from the $\ce{CH3}$ group which leaves now a $\ce{C=O}$, $\ce{CH2}$, and a $\ce{CH3}$ group where the negative charge lies on $\ce{C}$ a part of the $\ce{CH2}$ group. Conversely, the cyclohexanol would have the negative charge lying on the $\ce{O}$ because the $\ce{H}$ a part of the $\ce{OH}$ group would be removed.
Just by simple electron negativity rules, $\ce{O}$ is much more favored to possess a (partial) negative charge than $\ce{C}$. Thus, cyclohexanol is a much more stable acid than acetone which is reflected in its $\mathrm{p}K_\mathrm{a}$ value.


Volume of Oxygen Produced By Electrolysis of Water

Thank you very much for your idea and help.
I think that your calculations are correct.
But I think you'd better to try other simpler choices to make oxygen.
One way is air separation. If you separate nitrogen from air, you gain oxygen.
This can be done by "Zeolite oxygen generators". You can read these site about them.
1- https://www.zeochem.com/our-applications/medical-oxygen-generation
2- https://www.jalonzeolite.com/product-item/molecular-sieve-jlox-500/
There are many important factors that you have to consider about electrolysis of water that I know some of them:
First water has low conductivity. So, you have to add a strong inert electrolyte to make charge transfer between electrodes. This electrolyte could be any soluble salt, an acid or any substance that solves in water and then dissociates to ions. But, the electrolyte must not interacts to the main reaction and should be inert and surely it should be absolutely safe and not harmful for patient. Sodium or potassium sulfate or nitrate can be used.

Next, you have to choose an inert electrode to conduct electricity to the solution and then the gas be produced on it. Platin and gold can be use. But, they are expensive. Another choice can be carbon electrode.
And also take care about Hydrogen that produced on the negative electrode (cathode), that is flammable and can be exploded by spark.
And, so many other considerations to make oxygen efficiently and, next purify it to be suitable for hospital usage.
I wish you health.


How have I to understand in what those quantum numbers refer for ortho and para water?

It's explained in the first link you gave in the previous paragraph:

"The rotational quantum states of the asymmetric rotor, i. e.,
water, can be classified by $J_{K_{a}K_{c}}M$, with the total angular
momentum quantum number $J$, the projection labels $K_{a}$
and $K_{c}$ onto the molecule-fixed a and c axes as defined in
Figure 1 respectively, and the projection quantum number
$M$ onto the space-fixed $Z$ axis."

Do you know how angular momentum in quantum mechanics works? You can look at the projection of the angular momentum onto a specific axis and the total angular momentum. Turns out the Heisenberg Uncertainty Principle prevents you from knowing the angular momentum projections along more than one axis at once...


How are poisons discovered? Does someone have to die/be poisoned from it first?

Alle Dinge sind Gift, und nichts ist ohne Gift, allein die Dosis macht dass ein Ding kein Gift ist (The dose makes the poison) - Paracelsus

Poisons (I'm going to use this as an umbrella term for "toxins" and "venom" as well. Bear in mind though, they are not the same thing) have been known since antiquity.

Back in the good old days, you figured out if something was poisonous or not by eating/touching it (or getting someone else to do it), i.e- "discovering" a poison was simply a matter of chance.

These chance encounters alone, led to the discovery of numerous poisons.

With the advent of Chemistry, the gents in white lab-coats figured out that compounds that bear resemblance to the already well-known poisons, in terms of their functional groups and structures, are also toxic (albeit, to vastly different degrees). Poisons could now be identified a priori (you could tell it would end badly if you were exposed to such a substance, but not how badly).

And no, you can't really quantify a poison's effects without testing it on something. Nor can you tell how much of something would be needed to kill or severely maim.

This is where we bring in the idea of a median lethal dose $\pu{LD_{50}}$ which is, simply put, the minimum amount of a substance required to kill off half of all the animals in a particular test group.

$\pu{LD_{50}}$ values for a particular substance depend on the animal used.

Of course, the only way to get an accurate $\pu{LD_{50}}$ (which itself, is really a "mean/average" value of sorts) for a human is to actually poison someone, which doesn't sound very nice. So you do the next best thing, you measure it for a rat/chimp, couple it with your knowledge of the poison's mechanism of action, and extrapolate the value to something that would kill a person.

Another way to establish degree of toxicity, which doesn't involve killing animals, would be by exposing a cell/tissue culture (and not the entire animal) to the potential poison .


How to determine if a compound is a strong, weak, or non-electrolyte?

It's best to start with basic definitions: electrolytes are compounds that allow a solvent to conduct an electric current. The mechanism is generally formation of ionic species, that is, solvated atoms or molecules carrying a net charge. As charge carriers, ions are the most important contributors to the electrical conductance of solutions.

Salts represent a reliable source of ions in solution. They tend to dissociate to some extent into their component ions in solution. It is usually ok to assume that in the presence of solvent the ions in a salt can exist either as part of the solid lattice or as free ions in solution. Only the free ions contribute significantly to conductance. Therefore solubility and conductance are directly related. A salt that dissociates to a large extent is referred to as a strong electrolyte. One that does not dissociate much is called a weak electrolyte. Generally speaking the degree of dissociation depends on the lattice free energy and the free energy of solvation of the salt in the particular solvent under given conditions of temperature and pressure. Degrees of dissociation for different salts or conditions are generally tabulated as solubilities and may be described qualitatively or quantitatively (say as solubility product $K_{sp}$). Qualitative trends are usually memorized by chemists and are taught at an early level to students. While detailed explanations can usually be found for differences in solubility, they are not usually the subject of introductory courses. Memorization (and mnemonics) tend to be the rule.

Some compounds that would not generally be regarded as salts (although are sometimes prepared in salt form) contain ionizable groups that can dissociate to form oppositely charged ions. Good examples are acidic groups, such as carboxylic groups that generate an conjugate anionic (negatively charged) molecule and hydronium (in water) upon dissociation, and basic groups, such as amines, that can accept a proton from a donor compound (such as water) to form a cation (positively charged ion). Generally speaking molecular organic compounds are regarded as weak electrolytes, since the degree of dissociation of the ionizable groups may be incomplete to a large extent. This is not universally true. Some groups readily dissociate, for instance sulfates. The degree of ionization depends inasmuch on the nature of the group in question as on the structure of the compound and the solvent conditions, including in particular identity of the solvent and temperature.

Compounds that are insoluble can obviously not contribute to the conductance and are therefore not electrolytes. Some compounds readily solubilize in the solvent (for instance nonionic polar compounds in water) but do not dissociate into ionic forms. Such compounds are not electrolytes.


What is the relationship between configuration & optical rotation?

There is no general correlation between the the rectus (R) and sinister (S) prefixes and dextro/levorotation. In some cases the (R)-enantiomer is the dextrorotary enantiomer (as in the case of D-alanine), and in other cases the (R)-enantiomer is the levorotary enantiomer (as in the case of L-cysteine). The relationship can only be determined on a case-by-case basis with experimental measurements or detailed computer modeling.

For a really detailed explanation of exactly how chiral molecules rotate planes of polarized light, I recommend you check out this physics post, but to answer your other question, no it doesn't make sense for planes of polarized light to rotate according the Cahn–Ingold–Prelog Priority Rules. This system only takes into consideration the atomic number and number of bonds to atoms of a higher atomic number. Even though $\ce{-Cl}$ might rank higher than $\ce{-C(CH)3}$, wouldn't it make sense for the two groups to have different effects on rotation, in the cases of both direction and magnitude? This is why dextro/levorotation must be determined experimentally.