If you chose to recrystallise by variation of temperature, you have to find a liquid that dissolves your product much better when heated (or even heated to reflux), than at room temperature.  A qualitative test may be performed with a small quantity of substance, for example a tip of a spatula, that you fill in a test tube, followed by a small amount of liquid.

Does your compound dissolve well in the liquid already at room temperature? This is then not a good choice of sole solvent for a recrystallisation, as much of the product will be retained by the liquid.  Intentionally I use liquid, because while you likely find a commonly used solvent suitable for recrystallisation, some other chemicals not typical perceived as "solvents" may be useful, too.

Depending on your equipment available, you may extend the accessible temperature range towards lower temperatures by use of an ice bath, or a freezer.  Be aware, however, some of the solvents that are liquid at room temperature may solidify then, too (glacial acetic acid, for example).

Often, you face a situation like "ethanol (as one solvent) has a too high solvent capacity for X, and water (as an other solvent) has a too low solvent capacity -- regardless if warmed, or heated to reflux".  Then it is time to use a mixture of mutually miscible liquids as solvent of recrystallisation, to tune the solvent properties.  This is why, for example, aqueous ethanol (ethanol with an intentionally added proportion of water) is used as solvent of recrystallisation, too.

There are some seemingly obvious choices of solvents of recrystallisation, like methanol and ethanol for not so polar organic compounds, and other somewhat surprisingly well working examples (acetic acid for less polar compounds); hence, recrystallisation offers a spectrum of science, craftsmanship, art, and obviously some luck, too.  Note that recrystallisation aiming for purification may be well a different one as aiming for nice crystals (for the purpose of single crystal diffraction analysis, for example).

If you made your choice, retain a small quantity of non-recrystallised compound which may be helpful later.  Mix your solid (or tar-like) product with the solvent / solvent mixture at room temperature and heat it up; if you want to heat up to reflux, consider a reflux condenser.  At high temperature, all visible solid should be dissolved.  If this is not the case, there is to little of solvent present and carefully add more.  After each addition of some new solvent, wait a little bit until the temperature of your mixture is the same as prior to this addition. (Caveat: If your product to recrystallise is a mixture, one component may dissolve better, than the other.  If so, consider a fractional recrystallisation or other means of purification first.)  Then allow the solution a slow cooling to room temperature, for example switch out the heater (and stirrer, too) of the oil bath yet keep the round flask in the oil bath.

Scratching the inner walls of the flask with a glass rod may be dangerous as you accidentally may pierce the wall (and consequently loose product to the oil bath).  A long Pasteur pipette is a better choice, too much pressure applied will break it, but not the flask.  But in general, it is one good method to generate crystal seeds for nice crystals.  (The optimal temperature for crystal seeding need not be the same as the optimal temperature for growing larger crystals.)

If possible (vide supra), complete the crystallisation by cooling below room temperature (ice bath) as this often lowers even more the solvent capacity.  If accessible, collect the crystals at this / near this low temperature, too.  If you want to rinse your crystals with a "bad solvent" (i.e. with low solvent capacity), equally cool it in advance by an ice bath.  If you crop your crystals by vacuum suction, prior cooling the equipment (Buchner filter, suction flask) by an ice bath / in a freezer may be helpful, too.

As a final note (here), do not throw away the mother liquor / filtrate of your recrystallisation too early.  It may contain a considerable amount of product.  Concentrate it (for example, at the rotary evaporator), and repeat the recrystallisation for a second crop.  Do not merge crops of subsequent crystallisation until an analysis reveals them as of sufficient similarity, like comparable purity.


The reactions are ongoing this way:

Relatively free electrons of  potassium reduce water:

$$\ce{2 e- + 2 H2O -> H2 + 2 OH-}\tag{1}$$

That leaves metal positively  charged.

Liquid ammonia, if exposed to alkali metal, reacts with electrons much slower than water, forming a dark blue solution of solvated electrons. As electrons progressively kick out protons from ammonia, forming hydrogen, the solution finally turns to a colourless solution of NaNH2.

But back to water.

The potassium ions get hydrated, decreasing the charge.....

$$\ce{K(s)^{n+} -> K(s)^{(n-m)+} + m K+} \tag{2}$$

forming $\ce{KOH}$ solution in form of mixture of hydrated ions $\ce{K+ + OH-}$

But heavy potassium ions cannot keep pace with light and fast electrons and the drop of melted metal progressively gains positive charge and finally ends up by - as authors call it -  Coulombic explosion.

The hydrogen gets eventually ignited by microsparcs due the charge instability  even before the explosion. As "Terminator T1000-like" spikes of liquid metal eventually pierce isolating vapour+hydrogen layer, getting into contact with ignitable hydrogen-air mixture.

It was recently theoretically predicted by quantum chemistry simulation for several dozens of alkali atoms by Czech chemist Pavel Jungwirth And col. Chemistryworld-Alkali metal explosion explained

They have experimentally verified it by high speed 10000 f/s camera, using sodium/potassium alloy forming an eutectic with low melting point.

I knew that from the popular science radio broadcast interview, finding backwards some reference for it.

See also their article in Nature ( which I forgot about and found later):

Coulomb explosion during the early stages of the reaction of alkali metals with water

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


entropy calculation in an irreversible process

Quantities such as heat and work can vary depending on the path (how you get from an initial to a final state). In contrast a defining aspect of a state function is that it is independent of the path.
Entropy is a state function, as a corollary therefore a path-independent property. The experimental determination of entropy involves measuring the heat transferred as a function of temperature during a reversible process and computing $\int \frac{dq}{T}$. That is another defining property of the entropy. It is not trivial to wrap your head around why it should be so, but entropy can be thought of as helping quantify a limiting or ideal property, in particular the maximum work that might be obtained from a process, or the least amount of work required to bring about a non-spontaneous change. It defines processes in which there are no losses (losses being consistent with irreversibility).
As an example, at constant T and p the Gibbs free energy change (which is related to the limiting work) is related to the entropy as follows:
$$-\frac{\Delta G}{T}=\Delta S_{system}+\Delta S_{surrounding}$$


How to set the memory for the Gaussian 09 cubegen utility?

By reading the information from Gaussian about cubegen, you would find out, that use of cubegen is like the following:

  cubegen nprocs kind fchkfile cubefile npts format cubefile2

They further elaborate about nprocs:

  nprocs

  Number of shared memory processors used for electrostatic potential calculations. A value of 0 is equivalent to 1 (it is the default). Note that this parameter must be included if other parameters are specified. Previously this parameter was used to specify the amount of memory to allocate. The GAUSS_MEMDEF environment variable should be used instead.

You can read about setting GAUSS_MEMDEF on the site about the default.route file.

  Environment Variables

  All of these directives can also be set via environment variables or UNIX command line arguments. The environment variable GAUSS_XDEF provides a line equivalent to -X- in the Default.Route file. Similarly, the command line argument below specifies the same setting:

  $ g09 -x="value"

  Similarly, all of the following have the equivalent effect:

  Default.Route file: -M- 4GB
  bash shell: export GAUSS_MDEF=4GB
  tcsh shell: setenv GAUSS_MDEF 4GB
  command line: g09 -m="4GB" …

Based on that, you could try the way by exporting GAUSS_MDEF to the amount that you want, or by using the commandline. But I have no idea if the latter works with cubegen instead of g09. Let me know. :)


What is involved in this firey demonstration?

This is liquid methane, which boils at a temperature of −161 °C. (Methane can be condensed in a cold trap that is cooled with liquid nitrogen.)


Does my data show quadratic regression in a calibration curve?

As you have correctly assumed, the relationship between absorbance and absorbant concentration is linear (law of Lambert-Beer):
$$ A = \varepsilon \, l \times c$$

This should give you a straight line through (0,0) and your data.

Now, for a graphical analysis of data, let's not look at the numbers themselves but on what kind of a graph it gives us. So after 2 minutes with gnuplot I get something like this:

Now, there are two things that I note here:

The absorbance of the $\mathrm{IC_{25}}$ value is very very low (nearly 0) which indicates some wrong measurement. Maybe the cells weren't in the pathway of the light beam, only the medium?
You have four data points.

My advice is the following:

Double check your measurements (the absorbance and the cell density). It would be even better if you did several measurements for the different inhibitor concentrations and then calculate the mean (and standard deviation if you so desire).
Generate more data. It is very wonky to try and justify a linear regression through these 4 data points (although mathematically allowed). However, if all intermittent points are equally chaotically distributed along a straight line the linear relationship still holds (with a fairly high $R^2$). What this does to the interpretation of the data is a completely different story.


Why do carbon dioxide and sodium hydroxide not form sodium oxide and carbonic acid?

Sodium oxide is not a base in either the Bronsted-Lowry or Arrhenius senses.  It is technically a base anhydride, meaning that it can be hydrated to yield a base:

$$\ce{Na2O + H2O -> 2NaOH}$$

Carbon dioxide can similarly be viewed as an acid anhydride, meaning it can be hydrated to form an acid.

$$\ce{CO2 + H2O <=> H2CO3}$$

The difference is that the hydration of sodium oxide is essentially irreversible.  One program I used estimated $K_{eq}=10^{39}$ for sodium oxide hydration.  In contrast the hydration of carbon dioxide is reversible, with $K_{eq}\approx1.5\times10^{-3}$.

The reaction that you are wondering about, $\ce{CO2 +2NaOH⟶Na2O +H2CO3}$, is essentially the dehydration of sodium hydroxide by $\ce{CO2}$ to form sodium oxide and carbonic acid, that is, it is reaction 2 minus reaction 1.  But the equilibrium constants I just mentioned show that sodium oxide has an "affinty" for water that is about 42 orders of magnitude greater than carbon dioxide's affinity for water.  So that reaction is very unfavorable.  The reverse reaction would be extremely favored: Sodium oxide would be a powerful reagent for $\ce{CO2}$ removal prior to air liquefaction, at least as powerful as sodium hydroxide.

If you are curious, here is the R code I used to estimate the equilibrium constant of sodium oxide hydration, using the package CHNOSZ:

> subcrt(c("Na2O","H2O","NaOH"),c(-1,-1,2),T=25)
info.character: found H2O(liq), also available in gas
subcrt: 3 species at 298.15 K and 1 bar (wet)
$reaction
     coeff         name formula state ispecies
2031    -1 sodium oxide    Na2O    cr     2031
1       -1        water     H2O   liq        1
1157     2         NaOH    NaOH    aq     1157

$out
      logK         G         H         S         V       Cp
1 39.01742 -53229.29 -57143.24 -13.24728 -36.04979 -40.8745


Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron?

Although the question is a bit old, I think it still hasn't been answered yet. And by looking at the given answers it seems like the discussion went into a different direction at some point.

So let's compare the two compounds, here I plotted the $\ce{Cu(II)}$ centers of $\ce{CuSO4.5H2O}$ and $\ce{CuSO4}$ from their crystal structure data.

As you can see, it changes from a $\ce{[Cu(H2O)4[SO4]2]^2-}$ to a let's call it $\ce{[Cu[SO4]6]^10-}$. So all effects that are typical for $\ce{Cu(II)}$ should remain, it is still an octahedron, there is still Jahn-Teller present, we only got the change from four $\ce{H2O}$ ligands to four $\ce{[SO4]^2-}$ ligands.

Those two are similar in their ligand field splitting, where the latter one is slightly weaker, so we expect a red-shift of the absorption energy. And that is also what Nagase et al. investigated on (Ref.1):

  As shown partly in Fig. 1, the $\mathrm{d–d}$ band maxima of the hydrated and anhydrous copper sulfates were observed at $\pu{13200 cm-1}$ ($\ce{CuSO4.5H2O}$), [...] and $\pu{12500 cm-1}$ ($\ce{CuSO4}$), respectively, in their reflectance spectra. This is in agreement with the expectation from the spectrochemical series, and indicates that, in a series of complexes of comparable tetragonalities, the substitution of a coordinated water molecule with a sulfate oxygen weakens the apparent ligand field strenght about $\pu{150-200 cm-1}$. In addition to this shift of the absorption maxima, the following spectral changes were observed on dehydration: (i) the width of the $\mathrm{d–d}$ band becomes gradually narrower with the advance of dehydration; (ii) a broad and very weak absorption appears near $\ce{20000 cm^{-1}}$ in the spectrum of $\ce{CuSO4.H2O}$, and, with somewhat increased intensity, in that of $\ce{CuSO4}$; and (iii) a strong absorption (probably charge-transfer band) appears in the ultraviolet, and shifts remarkably to the lower frequency side with the advance of dehydration, indicating that the $\ce{[SO4]^2-}$ ions are getting more and more strongly bonded, as more and more sulfate oxygens are getting coordinated in the $x-y$ plane of the complex and shared among many $\ce{Cu^2+}$ ions.
  [...]
  Both the slight red-shift and the narrowing of the $\mathrm{d–d}$ band contribute to the disappearance of the blue color [...]

Or, as it is summarized in the book by K. Sone and Y. Fukuda (Ref.2):

  Therefore, the disappearance of the blue color of $\ce{CuSO4.5H2O}$ caused by dehydration is only an apparent phenomenon, which is brought about by a relatively small shift of the $\mathrm{d–d}$ band to the red end of the visible spectrum (and partially into the near infrared) and perhaps also by the spontaneous pulverization of the dehydrated product; if we had eyes which could register radiation from near IR to near UV, we should certainly feel that $\ce{CuSO4}$ is much more darkly colored than $\ce{CuSO4.5H2O}$!

  Since the ligand field strengths of $\ce{H2O}$ and $\ce{[SO4]^2-}$ are seemingly only slightly different from each other, these changes will only bring about a small shift in the $\mathrm{d–d}$ band towards the red. On the other hand, the $\ce{[SO4]^2-}$ ions, which are weakly held in the axial positions of $\ce{CuSO4.5H2O}$, are gradually brought into the equatorial coordination sphere, so that their CT interaction with the $\ce{Cu^2+}$ ion will be strongly enhanced. The observed spectral changes thus come about. Somewhat similar spectral changes are also observed when the double sulfates $\ce{M2Cu[SO4]2.nH2O}$ (where $\ce{M = Na^+, K^+, Rb^+, Cs^+}$ and $\ce{[NH4]^+}$) are heated.

References:

K. Nagase, H. Yokobayashi, K. Sone, “Spectrophotometric and thermal analytical studies on the dehydration of copper(II) sulfate and its double salts,” Thermochimica Acta 1978, 23(2), 283-291 (https://doi.org/10.1016/0040-6031(78)85070-9).
Kozo Sone, Yutaka Fukuda, In Inorganic Thermochromism (Inorganic Chemistry Concepts, Volume 10); Christian K. Jørgensen, Michael F. Lappert, Stephen J. Lippard, John L. Margrave, Kurt Niedenzu, Heinrich Noth, Robert W. Parry, Hideo Yamatera, Eds.; Springer-Verlag: Berlin, Germany, 1987 (ISBN 978-3-642-51019-9).


what causes the smell of a room ionizer? If ozone, what are we actually smelling when ozone interacts with nasal mucous?

Regarding our sense of smell, from the Wikipedia:

  The perception of odors, or sense of smell, is mediated by the olfactory nerve. The olfactory receptor (OR) cells are neurons present in the olfactory epithelium, which is a small patch of tissue at the back of the nasal cavity. There are millions of olfactory receptor neurons that act as sensory signaling cells. Each neuron has cilia in direct contact with the air. Odorous molecules bind to receptor proteins extending from cilia and act as a chemical stimulus, initiating electric signals that travel along the olfactory nerve's axons to the brain.2

Regarding the odor of ozone, from the Wikipedia:

  Ozone (/ˈoʊzoʊn/), or trioxygen, is an inorganic molecule with the chemical formula O
  3. It is a pale blue gas with a distinctively pungent smell.

From this site:

  While oxygen is not detectable by the human nose, ozone has an extremely low odor threshold.  Humans have the ability to detect ozone [concentrations of ~10 ppb] before many expensive measuring devices.  Also, ozone in air at high levels (greater than 0.1 ppm) is hazardous to human health.

  Ozone is a powerful oxidant (far more so than dioxygen) and has many industrial and consumer applications related to oxidation. This same high oxidizing potential, however, causes ozone to damage mucous and respiratory tissues in animals, and also tissues in plants, above concentrations of about 0.1 ppm. While this makes ozone a potent respiratory hazard and pollutant near ground level [...]

There are quite a few articles available online that discuss the "odor of ozone". It is described by some as "metallic" or as "similar to chlorine" (although note that "chlorine smell" is often "pool smell" caused by chloramines).

Given the reactivity of ozone (it oxidizes just about anything containing carbon to $\ce{CO2}$, which explains why it can cause inflammation and is generally regarded as deleterious to health), it is an interesting question whether there exists a distinct odor receptor that senses ozone or whether the response is nonspecific or mediated by pain receptors and registered as the "pungent smell" of ozone. Or perhaps it is other molecules altogether rather than the interaction of ozone with nasal receptors that causes the smell.

All of this is made even more complicated by the apparent loss of olfactory perception caused by exposure to ozone:

  An acute exposure to 0.2 ppm ozone impairs odor detection threshold.
  The cause is not clear, different modes of action might play at role.
  Olfactory threshold seems sensitive to photochemical air pollutants’ effects.

The "The Curious Dr. Z" blog hypothesizes on the source of "ozone smell" (see also reference 1 cited in the blog):

  It turns out that our olfactory receptors might be able to bind free radicals, and, since free radicals and ozone frequently occur together, that’s what we’re actually smelling.

  In 1991, a group of patients undergoing irradiation for cancer all complained about a strong, pungent odor they later identified as ozone. This lead to a study where researchers discovered that humans can actually detect the irradiation of tissue by detecting free radicals trapped in mucus.

  Another, older theory, is that ozone acts as the signal for irradiation. Ozone binds to and reacts with olfactory receptors the just as other odorants do, and effectively acts as a warning of more damaging chemicals nearby.

  We don’t know too much beyond this, and as far as I can tell, ozone olfaction is not a very active area of research. Bummer.

  Since we are so sensitive to the smell of ozone, and we don’t smell much else when we do smell ozone, I suspect that ozone/free radicals have many potential binding sites (a lot of those “hands” can grab ozone). When the ozone/free radicals does bind, it does so with high affinity, meaning it easily binds to receptors. This persistent and overwhelming smell experience is similar to other smells that indicate something noxious, including smoke and rotting tissue.

References

Lloyd M. Beidler. Olfaction, Handbook of Sensory Physiology, Chemical Senses. Springer Science & Business Media, 2012, ISBN 3642651267, 9783642651267


Intermolecular forces of attraction between positional isomers of alcohols

It does seem that placing the hydroxyl group at the terminal carbon maximises the hydrogen bonding interactions between alcohol molecules and also the van der Waal forces between them. This is best illustrated with a crude diagram I have made:

Finer differences due to the position of the hydroxyl group along the chain, such as the differences between the 2- isomer and the 3- isomer, are not explained here. My explanation does explain the markedly higher boiling point of the 1- isomer.


Is it possible to calculate the concentration percentage using chromatography paper?

Yes, it can be done.

I have done it on silica plates to compare the content of a natural product in different samples of plants.
It is , obviously, better to use HPLC, but in that case the sample preparation was very tedious so we decided to do it through TLC.
The content in the different samples was between 0 and 2%. Standards equivalent to a content of 0.5, 1, 1.5 and 2% we're prepared.

What you need to have succes is to be very accurate in the quantity that you apply on the plate and to have a good developer.
The quantities applied must be all the same, preferably not too much because then the spots are very big and it is more difficult to compare them.
Our product was developed with $\ce{KMnO4}$ yielding an intense yellow spot on a purple background.


Are labile tetrahedral [Mabcd] complexes optically active?

In a labile complex, ligands are detaching and attaching to the metal centre at extremely high rates. This means the solution will racemise quickly.  Therefore, even if the complex is truly optically active, you have equal proportions of each enantiomer and thus equal rotations of the polarised light plane - in opposite directions.
The end result is that rotation cancels and so none is observed.


Carbocation formation and stability from bromo compounds

Yes, both your answers and reasoning are correct. For the second compound a slight ambiguity might arise, because the $\mathrm{sp^2}$-hybridized carbons show $\text{-I}$ effect which would destabilize the carbocation, however considering the fact that there are not only  $\ce{2 Me}$— groups attached which show $\text{+I}$, they are also closer than one of the $\mathrm{sp^2}$- carbons, it is quite natural that the net inductive effect would stabilize it.


There is a nice little article in the Journal of Chemical Eduction Negative pH Does Exist pointing out examples of negative pHs in water, for example pH = -3.6 in mine water in California.  So -1.7 is not the lowest possible pH in water. Keep in mind that pH is defined in terms of hydrogen ion activity rather than concentration.  This is emphasized in Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+].  At high acid concentation, hydrogen ion activity greatly deviates from hydrogen ion concentration.  For example, in 16 molar HCl, hydrogen ion activity is 678!

And see Negative pH and Extremely Acidic Mine Waters from Iron Mountain, California (no paywall), which describes pH values even below -4.

In Hydronium ion activity in strongly acidic media. Remarkable agreement between independent estimates Am. Chem. Soc. vol. 95, pp 3055–3057, 70% sulfuric acid is found to have a pH of -9.8 (negative 9.8)!

However, ignoring activity and the true definition of pH, considering that pure water has a concentration of 55M, if all the water molecules were instead hydronium ions, -log(55) = -1.7.  Then in articles like Who Knows the Ka Values of Water and the Hydronium Ion? this value is attributed to the pKa of hydronium, with the supposed justification that:
$K_a = \frac{[\ce{H3O+}][\ce{H2O}]}{[\ce{H3O+}]} = [\ce{H2O}]$

The article New point of view on the meaning and on the values of
Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water is a serious consideration of the true meaning of the pKa of hydronium, and points out that the -1.7 value for the pKa of hydronium has no justification.

In conclusion, pH can be below -1.7 and the pK of hydronium isn't really -1.7.


In 2-bromo cyclohexanone,why bromine prefers to occupy axial rather than equatorial position?

I think there must have been some misunderstanding. This stabilization is working for all 2‑halocyclohexanones (but does not always cause the axial conformer to predominate), where bromo- one is not a special case, but has been used just as an example.

There is a combination of various effects, but the dipolar one is supposed to cause a destabilization by dipolar repulsion in the equatorial-X conformer. See the difference of partial charges distribution:

Measured or calculated molar fractions (%) of the axial conformers in different phases/solvents (taken from Yoshinaga, Fabiana, et al. "Conformational analysis of 2-halocyclohexanones: an NMR, theoretical and solvation study." Journal of the Chemical Society, Perkin Transactions 2 9 (2002): 1494-1498. doi:10.1039/B204635K):

$$
\begin{array}{c*}\hline
& \text{vapor} & \text{liquid} & \ce{CCl4} & \ce{CHCl3} & \ce{MeCN} & \text{DMSO} \\
\hline
\ce{F}  & 64 &  5 & 28 & 13 &  2 &  2 \\
\ce{Cl} & 86 & 26 & 62 & 42 & 15 & 13 \\
\ce{Br} & 92 & 53 & 81 & 66 & 36 & 33 \\
\ce{I}  & 96 & 87 & 92 & 85 & 65 & 63\\
\end{array}
$$

Overall stabilization effects in axial-X conformation of 2‑halocyclohexanones clearly increases in the order of F < Cl < Br < I.

However, the ‘gauche effect’ participates, which is stabilizing the equatorial conformation in largest extent for F, also for Cl, has almost zero effect for Br, and has a negative effect for I.


How are electronic configurations determined experimentally?

There are lots of techniques you could use. Most useful would be X-ray absorption techniques such as XANES or EXAFS. X-ray emission techniques such as XPF or EDS. You could look at electron emission spectroscopies (ESCA) such as XPS or UPS or Auger techniques and EELS spectroscopy, ELNES and EXELFS.

The important point to remember is that a small perturbation to the chemical environment e.g. by a change in oxidation state or coordination number, etc. will change the signals observed, this can lead to much more than electron configurations.

Each has its own benefit and speciality, depending on the sample and the exact thing you are looking for but that should get you going :)


How should elements be colored? And how many bonds does an atom have?

For the coloring part I would suggest you look at the CPK coloring convention.

As for the decision whether a bond is a single or a double bond: You could make this decision based on the bond length between two atoms. But you will need to find typical values for single, double and triple bonds for every combination of elements which seems a bit too much trouble. On the other hand, the depiction of double and triple bonds is used mostly for bonds between carbon, oxygen, sulfur, phosphorus and nitrogen (for other elements like metals the bond order is not readily identifiable, so I would leave those as single bonds). So, the amount of data to be collected might not be that large.

A very naive way to make the decision how many bonds an atom has could be the following (though this might not be very efficient):

calculate the distances between the atoms
for each element set a maximum distance another atom can be away from an atom of the element in question to be considered a bonding partner
for each element set a maximum number of bonds $N_{\mathrm{max}}$ an atom of this element can have, e.g. 4 for carbon, 4 for nitrogen, 3 for oxygen, etc. (for metals this might be a bit difficult to decide)
for each atom in the system order the list of atoms within bonding distance from lowest to highest distance. From this list choose only the first to $N_{\mathrm{max}}$th atoms as bonding partners. Then look whether some of the bonding partners qualify for double or triple bonds - if so you have to kick out some of the bonding members from the end of your bond candidates so that you don't exceed your maximum bond number.

There will be better ways out there to do this but this one worked for me in the process of creating "static" pictures of molecules.

Update:

You asked what size you should use for your atoms. Again, there is no definite answer to this - use whatever looks best for you. Nevertheless, I found that the sizes calculated according to the following formula give rather good looking atom size ratios for the different elements (maybe you have to multiply the resultant values by a constant to give good default sizes):

\begin{equation}
  \text{size} = \frac{1}{2 \sqrt{2}} \cdot \sqrt[6 \,]{\text{atomic number}} \ .
\end{equation}

Furthermore, here is a nice source for some typical bond distances. Other good resources would be the CRC Handbook of Chemistry and Physics or the book by Holleman and Wiberg if you can lay your hands on them. Those might come in handy if you try to adapt my method and look for values for the maximum bond distances. Another advice concerning the maximum bond distances would be that typical values for the bond lengths of bonds containing (transition) metals are between 200 and 300 pm, so a value of 350 or 400 pm might be a good choice for the maximum bond distance for those species.

Concerning the maximum bond number each element can have: here, the typical rules for drawing Lewis structures apply. You can find good resources for this on this site, e.g. here, and in every introductory general chemistry text. But you should be mindful of some special cases. You should not forget the cationic forms of the elements, when thinking about their maximum bond number. For example, carbon usually has 4 bonds, but there are very rare cases where it comes with 5 bonds ($\ce{CH5^{+}}$) although those are practically not important and some not so rare cases where it has 3 bonds (carbenes, carbocations, carbanions), so you might consider a maximum bond number of 5 for carbon, although for nearly all practical purposes 4 might do. Another example is oxygen. It usually has 2 bonds but when it is protonated or in Meerwein's reagent ($\ce{Me3O^{+}BF4^{-}}$) it has 3 bonds. Transition metals are very flexible when it comes to the number of bonds they have - it can be everything from 1 to 6 sometimes even more, so you should choose a high number, e.g. 8, for them.


Relative humidity and vapor-pressure in equilibrium

What I understood:
Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x$ atm at a certain temperature. Now Relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature.
So what this means is that if RH = 100% then the maximum amount of water vapours is present in the air. Now if RH = 50% means that the amount of water vapours present in the air is only half of the maximum possible.

So, does this mean the water has not been fully evaporated up to the limit it can?

Less than 100% humidity would mean exactly that. Assuming that there is still liquid water left, then an equilibrium has not been reached.

Does this mean that the equilibrium has not yet been established, and the reaction tends to shift in the forward direction? If yes, then what is stopping the reaction to move forward?

Exactly. Assuming an isothermal system with more liquid water available, then yes, more water would evaporate. Remember when water evaporates it cools the remaining water.

Does 100% RH mean the water and it's vapour are in equilibrium?

Yes, assuming both phases exists, and that the gas and liquid phase are at the same temperature.
Some  extra points:

The saturated vapor pressure of water is a complex function of temperature. So it is $x_1$ at $T_1$ but $x_2$ at $T_2$, and so on.

Ice sublimes and thus will also have an equilibrium water vapor pressure. The equilibrium water vapor pressure again depends on the temperature of the ice.


Why does the equilibrium constant depend on the temperature but not on pressure and concentration?

Please have a look at this answer of mine. It contains the derivation of the defining formula of the equilibrium constant.

\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}

This formula shows the temperature dependence explicitely, so it is clear that $K$ must depend on $T$.

Apart from the temperature this formula also relates $K$ to the standard Gibbs free energy of reaction $\Delta G^{0}$.
This quantity is defined for specific standard conditions involving standard pressure and equilibrium concentrations.
Since there aren't any other quantities contained in the formula (apart from the constant $R$) it follows that $K$ is defined for this specific pressure and these specific concentrations too and so it has no pressure or concentration dependence.


Calculate your DBEs (double bond equivalents):

\[\mathrm{DBE} = \frac{2c-h+2}{2} = \frac{2\cdot 11 - 14 +2}{2} = 5\]
Note that a benzene has 4 DBEs!

Shifts at 7.5 ands 7.4 are definitely not alkine protons. Aromatic protons are more likely, aren't they?
Largest couplings between benzene protons are typically ortho couplings. How large are they in your case?
Two sets of doublets at 7.5 and 7.4 ppm might indicate a 1,4-disubstituted benzene. You might want to play with some increment rules here.
You have two triplets at 3.5 and 2.4 ppm with each 2 H. These might be two neighbouring methylene groups in $\ce{C_{aryl}-CH2-CH2-O-R}$.
R in (5) might be $\ce{CH3}$.


Proof that Equilibrium Constant is a Ratio of Rate Constants

For a reaction step, the assumption that the rate is proportional to the product of the concentration of reactants raised to their respective powers is valid.

However, generally, reactions are made up of many reaction steps (e.g. SN1 reaction). For such reactions, you cannot use the assumption above. You have to understand the reaction mechanisms in order to determine the rate equation.

(For reactions that involve decimal powers, usually the steady state approximation is applied, and some mathematics is done along with empirical findings in order to determine the rate.)

Addendum: Yes. A reaction step means an elementary reaction. From Wikipedia: "An elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.". Therefore, there is only one reaction step.

Likewise, for the equilibrium constant,$K$, in your ratio derivation $K_c = k_f/k_r$, the chemical species must also be from that particular reaction step, and not the global reaction step. It makes sense because the equilibrium constant, K, is related to the free energy change, $\Delta{G}$ of a reaction step. In a two step reaction (e.g. SN1), there are two $\Delta{G}$ and therefore two equilibrium constants.

However, in practice, people do use the overall equilibrium constant, $\beta{}$, which is the product of the equilibrium constants at each reaction step. $\beta{} = K_1K_2K_3...$ (E.g.If the first step gives you 70% yield, and the second step gives you 70% yield, the global step would be 49% yield).(http://en.wikipedia.org/wiki/Chemical_equilibrium#Multiple_equilibria)

In short, for your rate constant,$k$ and your equilibrium constant $K$, in $K_c = k_f/k_r$, they must be of that of a single reaction step.


This is a decomposition reaction.  Synthesis requires a minimum of two reactants.  Decomposition is where a reactant breaks down into two or more products.  Here, one reactant (MnO2) goes to two products.  It seems like it may not be decomposition because MnO2 is becoming MnO3, but it is still one reactant to two products.


In contrast to $\ce{O2}$ (or $\ce{CO}$), $\ce{N2O}$ does not bind to hemoglobin and it is not metabolized in the human body. It is taken up through the lungs, gets in the blood and leaves the body the same way being exhaled.

According to a 2004 study by P. Nagele in PNAS (DOI), $\ce{N2O}$ acts as an antagonist for the  $N$-methyl-$D$-aspartate (NMDA) receptor. Other infamous antagonists are drugs like ketamine or phencyclidine.

Further information on NDMA receptor antagonists can be found here.


You are plotting different functions. The figures have no reason to be more than vaguely reminiscent of each other.

Now, those nice puffy things above are the isosurfaces of $\psi$ function. This is a function of three variables, and it just so happens that it can be represented as a product:
$$\psi(r,\varphi,\vartheta)=R(r)\cdot Y_{\ell,m}(\varphi,\vartheta)$$
You fix a value, find all points in 3D where the function takes that value, and plot them. That's what an isosurface is.

(By now, you might have noticed that the radial part $R(r)$ never appears at all in your attempts, so they got to be producing something different. And so they do.)

OK, then what are your plots? These are the spherical graphs of the angular component alone. You stay at the origin, point at the direction $(\varphi,\vartheta)$ with a stick having length $Y_{\ell,m}(\varphi,\vartheta)$, and leave a dot there. You don't care at all about the value of $\psi$ at that location. Besides, when $Y_{\ell,m}$ is negative (which it is half of the times), you point in the opposite direction. This, BTW, is the reason for having only one lobe in your p-orbital-like plots. No wonder you've got both lobes after you've squared $Y_{\ell,m}$ and thus made it always positive.

It could have been worse, you know. Think of a cone with its tip at the origin. Is it an isosurface of any function? Of course it is. What function? For example, $\cos\vartheta$ would do. What is the spherical graph of $\cos\vartheta$? A sphere, as you already know. Does a sphere look like a cone? Er, well...


Aromatic ring substitution: how are MNT and DNT further nitrated if nitro groups deactivate benzene rings?

On a industrial scale by using highly concentrated nitric acid known as fuming nitric acid (usually > 90%, but in the last step from DNT to TNT may also be anhydrous) along with oleum, which is a mixture of $\ce{SO_3}$ and sulfuric acid. Preparation in the lab may be achieved by more or less ordinary nitration at elevated temperatures. In a micro scale experiment [1] they used two different types of nitrating acid:

Nitrating acid I contains concentrated nitric acid 19% v/v, concentrated sulfuric acid 58% v/v and water 23% v/v.
Nitrating acid II is a mixture of concentrated nitric acid 25% v/v and oleum (30% $\ce{SO_3}$) 75% v/v.

Nitrating acid I was used for $\ce{Toluene -> MNT}$ and $\ce{MNT -> DNT}$ (65 °C). Nitrating acid II was used for $\ce{DNT -> TNT}$ (110 °C).

1 The Stepwise Nitration of Toluene: A Multistep Microscale Synthesis Based on an Industrial Process
Richard A. Russell, Robert W. Swltzer, and Robert W. Langmwe
University College, University of N.S.W., Northcotl Drive, Campbell ACT. 2600, Australia


How to interpret this formula about hydrogen bond energy

These exponents $10$ and $12$ are a little bit arbitrary. They have been chosen so as to obtain an energy curve corresponding to the measured H-bond energies. Other attempts have been published, like the famous Morse curve, which is, for covalent bonds versus HO distance $d$ :  $$E(d) = D[1 - e^{-k(d - d_o)}]^2$$ where $D$ is the dissociation energy of the covalent H-O bond. Apparently the curve with the two exponents $10$ and $12$ gives a better fit.


Answering your question as it stands is impossible, as we readers do not have enough information about your actual experiment. What do you mean with 'enthalpy of 1-butanol'? You can only measure changes in enthalpy, so you need to specifiy what 'change' you're looking at.

You could mean the enthalpy of evaporation or the reaction enthalpy of some chemical reaction or hundreds of other possibilities. It's essential to clarify what process you're referring to, so one can associate the corresponding starting (e.g. liquid) and end state (-> gas) to this energy change.

However, you stated that 200 ml of solution were heated up by whatever your process was. At constant pressure, the enthalpy required to change the temperature by $\Delta T$ is

$$\Delta H = m c_p \Delta T$$

where $c_p$ is the isobaric heat capacity. Knowing the value of $c_p$, you can calculate this energy change. Now where did that energy come from? Presumably, from your process. At this point, assume that there are no energy losses (f.e. neglect the heat lost to the surrounding air). Then by conservation of energy, the enthalpy increase measured by the temperature increase must be equal in size to the energy released by the process.

Finally, note that by convention we say if a process releases energy, it's enthalpy change is negative (we could've just done it the other way around; think of it as the same idea as defining a sea level).


How is carbon-halogen bond broken during Finkelstein reaction?

Think of the overall process and you will see where the energy comes from

Looking at just the reaction of iodide with the alkyl chloride doesn't give you the complete picture of the reaction. Sure, in that part of the reaction the bond-breaking and bond formation don't balance out. But in reality this doesn't mean there is no reaction: it just means that the equilibrium of that step strongly favours the alkyl chloride (there is still some dynamic formation of the iodide, perhaps in small quantities).

But the driving force here isn't that initial reaction. The real driving force is the precipitation of the KCl (it is very insoluble in the conditions of the reaction). From a mechanistic point of view this precipitation virtually eliminates the presence of free chloride in solution meaning that, if any is formed, it will rapidly disappear and not be available to drive the equilibrium back to the more stable alkyl chloride (even if the equilibrium of the halide exchange strongly favours the chloride).

From an energy or thermodynamic point of view, the overall reaction gets enough energy from the precipitation of the KCl (which releases energy) to compensate for the energy lost by substituting iodine for chlorine. So, if you look at the overall mechanism, the second step (precipitation) makes up for the apparent unfavorability of the first step.


Boiling/melting points of cyclohexane and cycloheptane

There are not many reputable sources on the matter (as I could find) since it seems that cycloheptane isn't the most popular chemical around (poor guy) but I will present you with the best explanation/argument/insight into the situation as I can.

Note that it would be solely my own interpretation, and a case of explaining results that we see, rather than formulating an explanation to encompass natural phenomena without observing the results prior.

The key ingredient that I noticed while looking through charts and whatnot was bonding which is connected to sterics. More on that later.

Melting and Boiling Points of Cycloalkanes

Above is a chart of the melting and boiling points of cycloalkanes. As expected, boiling point increases linearly. That is lovely! Thus, we can explain boiling point with natural intuition, similar to regular alkanes. More is more! If only life was so simple.

Now let's take a look at melting points. HOLY! Cyclobutane? Why you so high? (it is actually an error by the book. Nonetheless, it is higher by a little than cyclopentane and that is the point) And wow, is cyclohexane a monster, past the melting point of our dearest water. Something is wrong with this universe. It must be the martians.

But no. It is not the martians. At least I do not believe so. The answer, I believe, is sterics.

Sterics? you ask. Are you trying to say that cyclobutane bonds more structured than cyclopentane and cyclohexane bonds more structured than cyclohexane?

Precisely.

Although you may not expect much difference from cyclobutane and cyclopentane, cyclobutane is very close to a right angle (perfect square) and this would allow tight packing. Cyclopentane, on the other hand, is normally as an "envelope" which allows less close packing.

Similarly, cyclohexane has  very close cystalline packing which may account for its low melting point.

Then, after its crystalline structure melts away, our intuition gives way to the boiling point expectancies.

Again, this is not definite. If possible, I would like to juxtapose the solid structures of each cycloalkane, but unfortunately data seems scarce, especially for cycloheptane, which is of concern.


Are certain substances in furniture truly toxic?

The polyurethane would have to be absorbed through your pores when you're in contact with the sofa. That's not going to happen easily?
Theres a few tangent articles about the safety of these so called "cheap" sofas and mattress on livestrong and other websites.
I'd say it's not a concern. Sofa on bro.


Thermodynamics effect of change of temperature on equilibrium

First of all, spontaneity of a reaction depends on the kinetics, not thermodynamics.

Second, I always imagine LeChatelier's principle to work like this: When you raise the temperature for an endothermic reaction, it has more temperature to work with so the equilibrium shifts to the right. And if you lower the temperature for an exothermic reaction, that means there's more room for the energy created by the reaction.

But if you want to use $\Delta G = \Delta H - T\Delta S$ to check how the equilibrium behaves, you will only see the effect the temperature has on the free energy. It relates to the equilibrium via $$ \Delta G = -RT\ln K \Longleftrightarrow K = \exp\left(-\frac{\Delta G}{RT}\right)$$
This means that if you raise the temperature the expression in the exponent gets closer to 0, which in turn means that the overall $K$ becomes larger. So the equilibrium shifts towards the products.

However, I have neglected the influence of the entropy above. The formula should be, when completely written out:
$$ K = \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) $$

And with this you can always see whereto the equilibrium will shift.


What are the products in this SN1 reaction?

OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$

So it's SN1, so the $\ce{I}$ is going to leave spontaneously and give you the carbocation: $\ce{(CH3CH2)2CH+}$

The oxygen is in the ethanol so that will attack the carbocation giving the oxonium ion: $\ce{(CH3CH2)2CH-O+} ~ \ce{H-CH2CH3}$

Which will then be deprotanated by something acting as a base. There's probably some water around, but I suppose saying protonated ethanol and iodide is fine. This gives the final product of: $\ce{(CH3CH2)2CH-O-CH2CH3}$

For ethyl iodide to form, one of two things would need to happen:

SN1 mechanism: $\ce{OH-}$ would have to spontaneously leave forming a primary carbocation. It is SUPER DISFAVORED, so much so that it straight up won't happen.
SN2 mechanism: $\ce{I-}$ would have to act as a nucleophile AND $\ce{OH-}$ would have to leave. Iodide is an OK nucleophile, but OH- is a pretty bad leaving group. Both things considered this pathway is unlikely.

This was a pretty weak question by your (I am assuming) professor. This reaction would be horribly slow, and probably would barely go. Now, if they used sodium ethoxide instead of ethanol you get a nice Williamson ether synthesis and form the same ether and NaI, much more favored thermodynamically.

I just realized you probably meant why doesn't $\ce{I-}$ perform an SN2 on the protonated ethanol and have water leave. There are a couple of halfway decent chemical explanations for this:

Assuming there is any water around (there is pretty much always water around in the real world (especially with ethanol), unless remarkably heroic measures are taken to avoid it), the protonated ethanol will quickly react with it and form hydronium, ethanol and $\ce{I-}$. (Protonated ethanol has a $\mathrm pK_\mathrm a$ of about -2.4 and hydronium has a $\mathrm pK_\mathrm a$ of about -1.7.)
Assuming there is not water around, you are right it will probably react and form small amounts of ethyl iodide, however this will generate water which will stop this from continuing for the above reason.
You could also write the mechanism where the iodide deprotonates to form the ether (and avoid this whole fuss). BUT $\ce{HI}$ has a $\mathrm pK_\mathrm a\approx10$ where as oxonium proton $\mathrm pK_\mathrm a$s are usually between -7 and -2, so this would NOT be an accurate way to write the mechanism (in my opinion), but some professors get a little hand wavy here and allow it.

I know those are kind of sketchy workaround reasons, but honestly it won't be formed in any real quantity. There is probably another (probably better) argument from a thermodynamic stand point, but I don't know the bond energies off hand.


Understanding the solubility of Ca(HCO3)2

Calcium bicarbonate $\ce{Ca(HCO3)2}$ simply does not exist. It is impossible to fill up any container with pure $\ce{Ca(HCO3)2}$. Nobody has ever been able to produce a powder containing $\ce{Ca(HCO3)2}$. It is possible and easy to produce a solution containing the ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$, by bubbling $\ce{CO2}$ into a solution of calcium hydroxyde $\ce{Ca(OH)2}$. The following reaction will happen:

$$\ce{Ca(OH)2 + 2CO2 -> Ca^{2+} + 2HCO3^-}$$

The result is exactly what could be expected if some $\ce{Ca(HCO3)2}$ had been dissolved in water. And of course, it makes sense to admit that by evaporating the solution, the calcium bicarbonate could be obtained dry. Unfortunately it is not the case. When trying to evaporate this solution, $\ce{Ca(HCO3)2}$ is always and totally decomposed according to the following equation:

$$\ce{Ca^{2+} + 2HCO3^- -> CaCO3 + H2O + CO2}$$

And all you obtain by trying to evaporate a solution containing he ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$ is a deposit of calcium carbonate $\ce{CaCO3}$, whatever the method used for carrying out this evaporation.

Of course you can calculate the maximum amount of Calcium and of $\ce{HCO3^-}$ ions that may exist simultaneously in such a solution. You may even calculate the solubility of $\ce{Ca(HCO3)2}$. But this does not prove that $\ce{Ca(HCO3)2}$ exists as a pure compound. $\ce{Ca(HCO3)2}$ is a compound that only exist in solutions.


The carbon phase diagram shows that liquid carbon is only achievable at high pressure (~100 atm) and high temperature (~4500 K). This means that liquid carbon probably has very limited applications, and not many will be researching it. Funding is hard to come by when the applications of the research are not apparent.

Image source


While electrolyzing concentrated aqueous sodium chloride, why is it that chlorine is discharged but not sodium?

Let's take a look at a table of standard electrode potentials and find the relevant half-reactions:

Reductions (there is also a half-reaction for water reduction that doesn't involve a proton at -0.83 V, but for the sake of simplicity, we can ignore it):

$$\begin{align*}
&\ce{Na+ + e- <=> Na} & &E°=-2.71\ \mathrm{V}\\
&\ce{2H+ + 2e- <=> H2} & &E°=0\ \mathrm{V}
\end{align*}$$

Oxidations (reduction reactions reversed):

$$\begin{align*}
&\ce{2H2O <=> O2 + 4H+ + 4e-} & &E°=-1.229\ \mathrm{V}\\
&\ce{2Cl- <=> Cl2 + 2e-} & &E°=-1.36\ \mathrm{V}
\end{align*}$$

Now, if you know the concentrations of the species involved, you can use the Nernst equation to figure out the actual potentials, but we can reason around it without knowing the exact numbers.

The big thing to notice is how close the potentials of each of the reductions and oxidations are to one another—the reduction reactions are separated by almost 3 V, but the oxidation reactions are very similar. This means that for reduction, it is far less energetically costly to produce hydrogen gas than to reduce sodium, and in fact, if any sodium were produced, it would instantly react with the water to produce hydrogen gas, so if there is any water around, that half-reaction will be favoured. In the real world, the exact potential hydrogen is produced depends a great deal on the electrode used (see overpotential), but it's more or less impossible to electrodeposit metals with much lower reduction potentials than zinc in aqueous solutions. To actually produce metallic sodium, electrolysis of molten NaCl without any water present is the most common method.

Looking at the oxidations, the story's quite different. It doesn't take much more energy to oxidize chloride than it does water (and if the chloride concentration is high, the oxidation potential will go up). This means that (subject to the effects of the electrodes on overpotential), some amount of both half-reactions will occur and because chlorine is a gas and therefore leaves the solution and the oxidations potentials are so close (leaving little energy to overcome kinetic barriers), the chlorine produced will not significantly oxidize the water.


Interchange of axial and equatorial groups in Me2NPF4

Obviously, in the structure you drew you have two sets of two fluorines which are non-identical: The axial ones that form a 4-electron-3-centre bond with phosphorus and the equatorial ones that are bound by the classical 2-electron-2-centre bond. And also obviously, at higher temperatures this difference vanishes so the fluorines must somehow interconvert into each other (that is the easier explanation when compared to a different structure and is thus preferred by Occam’s razor) rapidly on the NMR-timescale. You noticed this already.

The mechanism for this is called Berry pseudorotation, and it involes a square pyramidal transition state as shown in this image taken from the Wikipedia page:

(The image is drawn with pentacarbonyliron but just exchange iron with phosphorus, four carbonyl groups with fluorine and the fifth with dimethylamine and voilà.)

Since this mechanism involves a significant change in the binding situation (we no longer can write two clearly 2-e-2-c and one 4-e-3-c bond(s)), we need a certain activation energy — that is usually supplied thermally. At $-100~\mathrm{^\circ C}$, there is not enough thermal energy to induce this change; at room temperature there is.

Ben Norris mentioned another important observation in a comment: There is not $^3J_\ce{PH}$ coupling visible. I suspect that this is due to the spectrum actually being $\ce{^{31}P\{^1H\}}$ — a proton-decoupled phosphorus spectrum.


What is the molecularity of a reversible reaction?

For the reaction $$\ce {A <=> B + C}$$
its molecularity is unimolecular. If we were given $$\ce{B + C <=> A}$$
Its molecularity would be bimolecular. Thus we see molecularity is dependent on the number of reactant particles, even if it's in equilibrium.


What is exactly the pH value of distilled water?

keep in mind that water (distilled, deionized, or tap) is NOT "pure" (i.e., pH equal to 7). The moment it comes in contact with air, CO2 gas begins dissolving into it, forming carbonic acid. The actual pH, therefore, will often be slightly less than 7.


What metabolite intermediate will accumulate in a mammalian cell-free extracts capable of glycolysis if lactate dehydrogenase is inhibited?

As you probably know, lactate dehydrogenase catalyzes the interconversion of pyruvate and lactate, with electron transfer via NADH/NAD⁺. If lactate dehydrogenase is inhibited, the fate of pyruvate in the cell extract will depend on what cell structures as wells as enzymes and other compounds are available. In other words, the answer to your question will likely be strongly dependent on the organism, as well as the particular circumstances of the experiment you are performing.

To give you a speculative answer: Assuming that the metabolic pathways of the mammalian cell still function and that oxygen is available to accept electrons in respiration, pyruvate could be metabolized through the tricarboxylic acid cycle, forming energy-rich compounds and carbon dioxide. If other fermentative pathways are available, pyruvate could be funnelled into those.

In any case, predicting what compunds would accumulate would require more information than that which is given in your question.


The confusion arises because of a simplistic definition of aromaticity.

What matters for aromaticity is the planarity of the relevant part of the molecule. In triptycene there are three planar benzene groups joined together. Each is individually aromatic, but the whole molecule is not. Similarly, triphenyl-methane has three individually planar benzene groups, each of which is aromatic, but the molecule cannot be planar because of steric interactions between the groups. There are many other such examples of complex molecules containing an aromatic component attached to other things.


Covalency of metal oxides in high oxidation states

Metal oxides are not always basic.  In many cases where the bonding is strongly ionic, the oxide ions do act as a base.  However, if the metal is in a high oxidation state an ionic bond would require a prohibitively high charge on the metal ion; the ionization energy required to put seven positive charges on one manganese atom far exceeds what can be reclaimed from the electron affinity of oxygen and electrostatic attraction alone.  The energetically preferred option for $\ce{Mn2O7}$ is to form polar covalent bonds instead.  Such bonds are similar to those in $\ce{CO2, SO3, P4O10,}$, etc., and thus $\ce{Mn2O7}$ will act similarly to those oxides -- as an acid anhydride.


How to quantize the shape of a molecule?

There are lots of descriptors of molecular size and shape that come to mind:

Moments of inertia (e.g., how different are they)
Radii (e.g., enclosing the molecule in a best-fit box)
Surface area or polar surface area (there are existing group-based methods)
Volume (e.g., of the van der Waals surface)

Some people have also used 3D shape descriptors like "spectrophores".


How does de Broglie actually prove Bohr's postulates?

de Brogile explains why orbitals are quantised

Strictly speaking de Brogile doesn't prove Bohr's postulates which are mostly wrong. But he did provide an explanation for the most important of Bohr's ideas: electron orbitals are quantised.

Bohr's whole model starts with the classical idea that electrons "orbit" a nucleus. But this has several problems not least of which is why don't they just radiate away their energy and fall into the nucleus (which is what accelerating charges should do in classical electromagnetism). Bohr made the extra assumption that the electrons around a nucleus could only "orbit" in a small number of fixed paths or orbitals which explained the actual observations. He didn't have any good reason why this was true.

de Brogile provided a reason. According to his postulates things like electrons are not just classical particles but have some of the characteristics of waves. and the thing about waves is that they interfere with each other. Electrons orbiting a nucleus can only do so if their wave-like property fits into a specific orbit (the way the vibrations in a guitar string produce only specific notes made from standing waves of integer divisions of the string length). Imagine integer numbers of electron waves around a nucleus and you are a third of the way of getting the picture (electrons orbit in 3D so the compatible vibrations are spherical harmonics in 3D space not the easier to picture vibrations of a 1D guitar string).

de Brogile, in short, produced a justification for the Bohr assumption that electron orbits had to be quantised.

Those pictures you see of electron "orbitals" are, in essence, pictures of the possible "waves" electrons can form when orbiting a nucleus.

All the above simplifies things a bit but the key ideas were from de Brogile and were more fully mathematicised by Dirac and Schroedinger giving us the modern quantum picture of how electrons behave.


Would any alkene possibly aromatize with proper conditions?

Your thought is actually happening in real life. Petrolium industry uses the same concept except for polymerizing ethene (may be that also used to get higher alkanes). Fluid Catalytic Cracking (FCC) is one of the most important conversion processes used in petroleum refineries. FCC (see the schematic diagram below)) is widely used to convert the high-boiling, high-molecular weight hydrocarbon fractions of petroleum crude oils into more valuable gasoline, olefinic gases, and other products such as aromatics (Ref.1):

For example, a conversion reactor is used to carry out several reactions. A few of them are:
$$\ce{n-C26H54 -> C6H12 + C6H14 + C10H20 + C4H10}\tag1$$
$$\ce{C10H20 -> C6H12 + i-C4H8}\tag2$$
$$\ce{C6H12 + 3 i-C4H8 -> 3C4H10 + C6H6}\tag3$$
The equation $(1)$ (cracking reaction of paraffin) shows the formation of cyclohexane from a long-chain paraffin. The equation $(3)$ (hydrogen transfer reaction between cyclohexane and isobutene) shows the conversion of cyclohexane to benzene.
References:

A. Kayode Coker, In Petroleum Refining Design and Applications Handbook, Volume 1; John Wiley & Sons, Inc.: Hoboken, NJ, 2018 (ISBN: 978-1-118-23369-6).


What happens to the bleach after mopping?

Sodium hypochlorite decomposes by a couple of different pathways:

$\ce{NaOCl\leftrightharpoons Na^+ + OCl^-\\
2OCl^- \leftrightharpoons O_2 + 2 Cl^-}$

In that case, the oxygen escapes (if it didn't react with "things to bleach", which is of course how bleach works in the first place), and you are left with ordinary salt.

The other pathway is

$\ce{3 OCl^- -> ClO_3 + 2 Cl^-}$

In that case you get some sodium chlorate, and again some ordinary NaCl salt.

Both NaCl and NaClO3 are crystalline solids that may be blown around like fine dust; some NaOCl may also remain behind. Note that according to wikipedia most commercial bleach solutions contain a small amount of sodium hydroxide to slow down the decomposition along the pathways mentioned above.

Straight dope link


I was supposed to predict the shape of the urea molecule $[\ce{H2N(CO)NH2}]$ using Lewis and VSEPR.

Don't forget that there are assumptions in VSPER to simplify the model. From Wikipedia:

  Valence shell electron pair repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules...

Essentially VSPER is for molecules in the gas phase where the only electrostatic interactions are within the molecule. Once the molecule is in a condensed phase, the assumption that it does not interact with external forces breaks down.

  What causes this change in geometry when the molecule crystallizes?

For one let's not forget that urea is a resonant structure, and does not have fixed bonds which also goes against assumptions of VSPER.

Further you have neighboring molecules with the hydrogens of those molecule taking some electron density from the resonance structure which is shown in the unit cell representation shown below:

In summary you have resonance which will flatten out the molecule and other molecules taking some electron density that combined will contribute further causing the molecule to flatten out.

  Does the same thing go for other molecules?

Yes, other molecules can condense and form structures that deviate from VSPER. Water in ice for example has a $109.5 ^\circ$ bond angle (tetrahedral angle) where as VSPER would predict $104.5^\circ$ (bent angle).

  Would there have been a way of predicting this as I was solving the question?

No, you would have to know how urea crystallizes and many other factors, then used a computer simulation to calculate (estimate) such.


Is the notion of orbitals different in theoretical chemistry?

Unfortunately, it only gets more complicated the deeper you dig. There is some explanation here: What exactly is an orbital?, but you should bear in mind that electronic structure theory is something that the average undergraduate student only barely touches. A strong background in QM is IMO mandatory to understand what some of these things mean. I'll see how far I can get with some of these queries, though, and some of it will be necessarily simplified.

The traditional "orbitals" that introductory chemistry teaches resemble what theoreticians would call spatial orbitals. These are distinct from spin orbitals, i.e. one-electron wavefunctions which describe both the spatial and spin orientations of the electron. The spatial orbitals only describe the spatial component.
For every spatial orbital, there are two possible spin orbitals, because you can have one electron with up spin and one with down spin. The simplest possible way to represent an "orbital" being filled with two electrons is by taking the product of the two spin orbitals, i.e. the product of two one-electron wavefunctions. So, the "two-electron orbitals" from introductory chemistry still have a mathematical form in theoretical chemistry, although it does depend on exactly what theory you are looking at (there are cases where the notion of a two-electron spatial orbital breaks down).
The reason why you don't see "two-electron wavefunctions" being discussed is because they have no real meaning in electronic structure theory. The N-electron wavefunction is useful because that is the way we choose to describe the entire molecule, and we can use it to extract information such as the energy of the system. The one-electron wavefunctions are useful because (1) they are used to construct the N-electron wavefunction (most simply by direct multiplication, or more properly using Slater determinants), and (2) we have developed methods to obtain the best possible forms of these. Two-electron wavefunctions are somewhere in the middle: they're not useful for describing the molecule as a whole, but at the same time there's no mathematical motivation for using them.

As for basis sets, it's a different ball game: we are no longer talking about what orbitals theoretically are ("theoretical chemistry"), but rather what is the best way to model them for a computer to get accurate results ("computational chemistry"). AOs themselves have no intrinsic physical meaning in a molecule, they merely serve as useful building blocks for MOs. Also, we don't even use the actual AOs as basis functions: we use things that resemble AOs as basis functions, to increase computational efficiency. So it is not too much of a stretch to add "unphysical" functions such as 1p, 2d, ... all for the purposes of getting better and/or faster results.
Also, orbitals in DFT have an entirely different meaning from orbitals in HF, but that's a story I'm not qualified to tell.

The problem with chemistry is that these theories cannot be taught to young kids - they are far too complicated and require far too much existing knowledge. And so we resort to teaching simple things, such as

"Orbitals are trajectories of electrons" - no.
"Orbitals are a region in which you have a 90% (or x%) chance of finding the electron" - no. Orbitals may be visually depicted by an isosurface which encompasses 90% of the electron density, but the orbital is not that region - it is a wavefunction.

For most people (cough organic chemists cough), simplified models work just fine. However, unfortunately, it can sometimes be a constant process of finding out that what you have been taught before is... wrong, or at best, heavily simplified. To understand exactly what justifies the simplification, though, you need to delve into the theory: and as I mentioned at the start, it is a very long journey. There are some good books in the Resources post, but again, a good foundation in QM is needed.


How can I prevent my spin-coated silica thin film from cracking when I heat it up?

The reference you sited has used gold surface to make the thin film. If the method you wanted is the one in that reference, folloeing is how they descibed it in experimental section:

  The modification of a gold electrode array on microchip with a SH-FMS thin film consisted of two steps. First a mesoporous silica film was deposited onto the electrode surface by the spin-coating technique. Then the thiol functional groups were immobilized inside the mesopores of the film by self-assembly chemistry. To prepare the silica film solution, a 2.578 g quantity of Pluronic F-127 surfactant was dissolved in 3.430 g of de-ionized water and 8.846 g of ethyl alcohol. Then 0.108 g of concentrated HNO3 was added into the mixture. Finally, a 5.0 g quantity of TEOS was added to the solution. The molar ratio of TEOS : water : ethanol : HNO3 : Pluronic F-127 was 1 : 8 : 8 : 0.05 : 0.008. The solution was shaken for 24 h, then spin-coated at the speed of 3000 rpm onto the bare surface of a gold microelectrode array on microchip (part No. M1450110, Microsensor System, Inc.) for 30 s. Fig. 1a shows the schematic of the microelectrode array. The specific electrode array had 50 finger pairs (15 microns finger width and finger spacing) with gold as the electrode material and quartz as the substrate. The electrical connection for the array was covered with plastic tape before spin coating with the film solution. After the spin coating, the tape was removed and the film-coated electrode was calcined on a hot plate at 350 °C for 8 min in air.

Hope this is helpful.


Do bond energies differ for surface and bulk atoms in nanoparticles?

Yes, there is, as surface atoms and bulk atoms have different bond energies in nanoparticles. This is a reason why nano particles coagulate easily and special measures have to be taken to prevent coagulation.


why can't I get my absorbance values to come out right on my oxybenzone uv-vis assay test?

The method, and your result, both make sense to me: it indicates that the test sample is 75.57% pure, although as you point out the result is not consistent with the assumption that the sample should pass the requirement for acceptance. However, why would you run the test if you knew ahead of time that the material would pass?

I would recommend that you repeat your test 2 more times from scratch - that way you would know if there was an error in your technique. If you get the same result (within experimental error), then you would have to admit that it is hard to argue with the data.


What is the variation of atmospheric pressure at large heights?

From the given composition of the atmosphere you can calculate the mean molecular mass $M$ of air.

Since Earth's gravitational field and temperature are assumed to be constant you can use the barometric formula

$$p(h_1) = p(h_0)~\mathrm{e}^{- \frac{M g}{R T} \Delta h}$$

to calculate the pressure at the top of Mt. Everest.


How to clean out a contaminated aluminum automotive radiator

Well I actually tried a small jar of the Enzymatic product with a sample of the nesting material I already removed as well as small piece of .023 6061 aluminum... Overnight the aluminum was completely decomposed.  The nesting material seemed unfazed.  So much for that idea!


Are bones generally harder to corrode than human skin?

Acids and bases are considered corrosive in industrial chemical hygiene. While it is true on a technical level that harder tissue (like bone tissue) will be a bit harder to corrode compared to soft tissue (like skin), a corrosive substance that can cause damage to soft tissue will cause enough damage to harder tissue to be a medical emergency; this is certainly true when you have a highly concentrated substance. If in your world, things that are worn on the person are made from either soft or harder tissue, then the harder tissue would provide slightly better protection. But to be absolutely clear, both types of tissue will be eaten away, it's just a matter of how much.

Metal reactivity with acids and bases is nuanced because it isn't just about hydrogen ion reactivity. The alloy known as hastelloy (trademarked name) and tantalum alloys are metals used in industry for corrosive environments. Silver and gold are resistant to some common acids, but not all (e.g. nitric acid will dissolve silver).


Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide

Here is a mechanism for the dehydration of an acid to an anhydride using phosphorus pentoxide.

The $\ce{OH}$ group is not a very good leaving group.  Even in simpler reactions like the dehydration of an alcohol to an olefin we often first convert the hydroxyl group into a better leaving group by protonation with acid or conversion to an inorganic ester (for example using thionyl chloride or phosphorus pentoxide).  The same thing takes place here - we are converting the acid's hydroxyl group into a better leaving group.

Why is $\ce{HPO3}$ such a good leaving group? Look at all the resonance structures you can draw for the anion $\ce{PO3-}$.  It's stability makes the reaction exothermic and strongly drives the dehydration process to the product side.

For completeness, here is a link to a drawing and discussion of the mechanism for the dehydration of an amide to a nitrile using phosphorus pentoxide. Same mechanism and principles (good leaving group, exothermic) as discussed above.


Proposed mechanism for the catalyzed oxidation of CO by O3: Are NO and NO2 catalysts or intermediates?

My take on this problem is as follows. Suppose a flask contains an equimolar mixture of $\ce{CO}$ and $\ce{O_3}$. The net reaction between the two gases will be

$$\ce{CO (g) + O3 (g) -> CO2 (g) +  O2 (g)} \tag{1}$$

But it needs a catalyst. If $\ce{NO_2}$ gas is introduced, then the reaction proceeds via the two reactions given in the OP's question:

$$\ce{NO_2 (g) + CO (g) -> CO2 (g) +  NO (g)} \tag{2}$$
and $$\ce{NO (g) + O3 (g) -> NO2 (g) +  O2 (g)} \tag{3}$$

So the nitrogen oxides cycle around. On the other hand, if $\ce{NO}$ gas is introduced, then the reaction proceeds via 'flipped' equations:

$$\ce{NO (g) + O3 (g) -> NO2 (g) +  O2 (g)} \tag{4}$$
and $$\ce{NO_2 (g) + CO (g) -> CO2 (g) +  NO (g)} \tag{5}$$

and again the nitrogen oxides cycle around. So it does not really matter whether $\ce{NO_2}$ gas or $\ce{NO}$ gas is introduced first: the two nitrogen oxides are the overall catalytst and either could be considered to be an intermediate. Basically, this is a distinction without a difference.


The bonding between $\ce{NH4+}$ and $\ce{Cl-}$ is ionic.

The bonding between $\ce{NH3}$ and $\ce{H+}$ in $\ce{NH4+}$ is a dative bond. (Since $\ce{H+}$ ion has no electrons, 2 electrons from $\ce{N}$ atom in $\ce{NH3}$ form the coordinate bond with $\ce{H+}$, thus forming $\ce{NH4+}$)

The bonding between $\ce{N}$ atom and $\ce{H}$ atom in $\ce{NH3}$ is covalent.


What are the possible concentration pathways for a 3-species system?

Of course, we can't display every pathway there is. We have to select a wide variety of rate constants to get a "good map" of the parameter-space. The approach I took was this:

Construct the differential equations

$$\frac{\mathrm dA}{\mathrm dt} = -k_1 A + k_3 C$$
$$\frac{\mathrm dB}{\mathrm dt} = -k_2 B + k_1 A$$
$$\frac{\mathrm dC}{\mathrm dt} = -k_3 C + k_2 B$$

Solve them numerically for every possible combination of 0.1, 1 and 10 for each rate constant and the above stated initial conditions
This leads to 27 pathways, which are plotted in a ternary diagram

The title of the ternary diagram is $k_1:k_2:k_3$, which is 0.1 : 0.1 : 0.1 in the first one.

(click for larger image)


Why can't nitrate ion in aluminium nitrate solution oxidise copper when a copper spoon is used to stir the solution?

Minor correction: It is not the spectrochemical series but rather the electrode potential tables. Those tabulated values are given under standard conditions
You are indeed correct in thinking that nitrate ion should oxidize copper. It is a game of pH. If you have very low pH and nitrate ions, the following reaction takes place (in the absence of other ions)...
$$\ce{Cu(s) + 4 HNO3(aq) -> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}$$


Comparing rate of reaction of acetic acid and sodium bicarbonate at different temperatures

Yes, there is, it has been done by others.  Typical acid-base reactions that occur in solution are diffusion controlled reactions. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. The reaction rate is limited only by how fast the acid and base can diffuse and encounter each other. It is usually easy to identify diffusion controlled reactions because they are extremely dependent upon the rate of stirring or mixing.  They are also very sensitive to the solution viscosity. Rate constants for these reactions are typically $>\pu{10^11 sec^-1}$.  Here is a link to a full paper that provides considerable experimental detail on one way to measure the rate of diffusion controlled acid-base reactions.


Are there phosphorescent/fluorescent materials that absorb near infra-red waves and emit higher wavelengths?

All molecules emit in the microwave region. They all have rotational energy levels and they absorb and emit all the time as energy flows between these levels.  But this not what you are asking I suspect. If you blast thin Gallium Sulphide wafers with a femtosecond laser then it is possible to generate terahertz radiation.


To directly address where the phrase comes from:

  Water is called the "universal solvent" because it dissolves more substances than any other liquid. -USGS

What are ideal qualities of solvent?

The strength of a solvent can be attributed to the strength of its intermolecular forces like london forces, dipole-dipole forces, ion-induced dipole, and hydrogen bonding. These are forces of attraction and repulsion. Solvation occurs when a molecule is surrounded by the solvent, so when there are strong intermolecular forces, stronger solvation occurs.

The hydrogen bond

Water makes use of the hydrogen bond, a type of intermolecular force experienced when hydrogen is attracted to the electronegative atoms nitrogen, oxygen, or fluorine. Hydrogen bonding is the strongest intermolecular force.

Hydrophobia

Water is a great solvent for hydrophilic molecules but hydrophobic molecules are, by definition, not easily disturbed by water. This makes a clear exception to the "Universal Solvent" idea. I haven't read any literature claiming it to actually be the best solvent, but it does work well in many chemical situations.


As far as I know, while fractional distillation would be a suitable technique to obtain O-18 water (yet, in very low yelds and in a way that is hard to control), the most used method nowadays is the membrane distillation technique: basically, a semipermeable and hydrophobic membrane is employed during the distillation (tipically made in PTFE), and the isotopes are separated by their ability to permeate through the membrane: lighter isotopes (O-16 water) tend to flow through the membrane, while heavier isotopes (O-18 water) are trapped in the apparatus and recovered.

You can find many detailed descriptions online, for instance:

http://www.ichtj.waw.pl/ichtj/library/Production%20of%20stable%20isotopes%20by%20membrane%20method.pdf

https://www.google.com/patents/US5057225


Does light absorption result in a momentarily weaker bond?

it may happen but it is not always necessary as 1S antibonding molecular orbital is lower in energy as compared to 2S bonding MO If electron jumps from 1S ABMO to 2S BMO  then BOND ORDER willhence bond will become stronger increase σ(1s) <σ∗(1s) < σ(2s) <σ∗(2s) < π(2px) = π(2py)


Great question!

The Claisen reaction happens under equilibrium conditions, and the addition of the ester enolate is highly reversible. Your mechanism omits the very important end of this reaction, which is that the β-keto ester product is highly acidic and easily deprotonated to yield the corresponding enolate.

The enolate formation drives the reaction forward according to Le Châtelier's principle. Very importantly, your question about the α-hydrogen. The deprotonation of the β-keto ester is not possible without two hydrogens on the original ester that will serve as the nucleophile.


Why does the Birch reaction not yield fully saturated products?

The mechanism of the Birch reaction includes single-electron transfers from the sodium dissolved in ammonia to the aromatic compound. These transferred electrons will attempt to occupy the lowest possible orbitals.

In the case of benzene, we have a delocalised $\unicode[Times]{x3C0}$-system. Delocalisation can be understood here to mean higher occupied orbitals and lower unoccupied orbitals. Therefore, the electron will find an orbital at a lower energy and it will gladly hop in to create the radical-anion intermediat.

Once this is protonated, the remaining radical system can still be seen as a delocalised pentadienyl radical. Thus, the energy level a new electron will occupy is still rather low and another anion can be created. Further protonation yields the product of the Birch reaction, a cyclohexa-1,4-diene — most importantly, a non-conjugated system. This can be regarded as a pair of isolated double bonds.

The energy of the unoccupied $\unicode[Times]{x3C0}^*$ orbitals in an alkene seems to be just too unfavourable for a free electron to occupy them. I don’t have the facilities to do calculations (and neither am I firm enough to be able to do them) so I can’t provide you with values to compare, but it has to be the case otherwise the reaction would continue. If it were not so, there would be no reason for the reaction to stop.


Why doesn't ionization energy decrease from O to F or F to Ne?

It's a good question, and I agree that it can be quite confusing, but Chemguide is (at least partly*) correct about this.
I'd phrase it this way, which I hope is clearer. First, think about what would happen if the "paired electron" factor was not a factor at all. We would then expect that, going from B to Ne, there would be a roughly constant increase in ionisation energy. It would look something like this. Please don't take the exact shape or numbers too seriously; they don't mean anything, only the trend does.

This is obviously not the reality, because in O, F, and Ne there is the additional factor to consider, which is that the electron being removed is already paired in an orbital, making it easier to remove. This means that the ionisation energies of O, F, and Ne are each decreased by a roughly constant amount (red arrow).

You can see that we recover the observed trend in ionisation energies, and we have indeed accounted for the pairing energy in all of O, F, and Ne.
There are two alternative ways of looking at it which may "click" better for you. I'll describe these in text.

Going from N to O, the nuclear charge increases by 1 unit, which on its own should cause the ionisation energy to increase. The pairing energy consideration, however, overrides this. So far so good.
Going from N to F, the nuclear charge increases by 2 units, which on its own should cause the ionisation energy to increase roughly twice as much. Therefore, to make IE(F) smaller than IE(N), you would need two counts of the pairing energy. But we can't do that: you are only ionising one electron each time, so you can only subtract one count of the pairing energy.
The argument can be easily extended to N vs Ne.

Going from O to F, the pairing energy does decrease the ionisation energy of F. However, the pairing energy also decreases the ionisation energy of oxygen, so this factor on its own cannot be used to say that IE(F) should be lower than IE(O), as it affects both. The same is true of F and Ne.

* Exchange energy, which Chemguide does not cover, is also a consideration. However, that's a smaller effect, and it's not necessary to invoke that to answer this question.


How to arrange carboxylic acids (phenylacetic acid and acetic acid) by increasing acidity?

The $\mathrm pK_\mathrm a$ difference between phenylacetic acid and acetic acid is around 0.5 and the trends become more obvious when other arylalkanoic acids are included in the comparison. Numbers in red are the $\mathrm pK_\mathrm a$ values.

Apparently, phenylacetic acid is a stronger acid than acetic acid. Adding a further phenyl subsituent even increases the acid strength.
The effect levels off with the distance of the phenyl substituent and the carboxylate. Note that 4-phenylbutyric acid is just as strong as acetic acid. It's as if there isn't any phenyl group at all.

Would you think that there is a $-I$ effect of the phenyl substituent on the $\ce{-COOH}$ group?


Positive electrode of an electrochemical cell?

As the standard reduction potential of Ag electrode is more than that of Fe electrode, Ag electrode will act as cathode (where reduction or gain of electrons takes place). The Ag electrode will be a positive electrode because it is electron deficient and thats why reduction is taking place at this electrode.

The Fe electrode will act as anode (where oxidation or loss of electrons takes place) and it will be the negative electrode because there is excess of electrons on this electrode.

Ecell = (Ereduction)cathode - (Ereduction)anode

      = +0.80 - (+0.77)

      = 0.03V


What are the physical manifestations of radial nodes?

I realize that plotting the square of the radial eigenfunction for an orbital will give the probability density of finding an electron at a given distance from the nucleus.

This isn't quite right.  The square of the radial function will give the probability density for s-orbital, but this is not the probability of finding the electron at a given distance from the nucleus, because the number of points a given distance from the nucleus increase with $r^2$.  To get the plots in Fig. c, the radial function squared in multiplied by $r^2$.

For example, for 1s, the probability density is greatest at the nucleus, but by multiplying by $r^2$ when r=0, a radial probability density of zero is obtained.

Furthermore, as you can see from the graphs on pages 187 and 188 here, for 2s and 3s the most probable location (highest probability density) is at the nucleus (r=0). Again, only by multiplying by $r^2$ when r=0, is a radial probability density of zero is obtained.

So "Fig. a" in the OP is misleading, the central region of 2s and 3s should be shaded much more intensely than the outer region(s).

  So here is what Im really wondering (please correct me on anything stated previously if I am mistaken). Is the first (smaller) hump on the squared radial eigenfunction (probability density function) for the 2s orbital the probability that the electron is in the 1s orbital (I believe this would be called an excited state if im not mistaken)? This would make the most sense to me, as it would be saying that it is possible for a 2s electron to be in either the 1s orbital or 2s orbital, but not in between (where the wave function goes from +y to -y)

No, the 2s electron is only in the 2s orbital.


Cyclopentadienyl radical geometry and MO considerations

A typical definition outlining the Jahn-Teller effect is that for any non-linear molecule in a degenerate electronic state, there exists a displacement of the nuclei along at least one non-totally symmetric normal coordinate, that gives rise to a distortion of the molecular geometry with a concomitant lowering of the energy

The underlying cause of the Jahn–Teller effect is the presence of molecular orbitals which are both degenerate and open shell.

The cyclopentadienyl radical, has degenerate and unequally occupied molecular orbitals, thus one can expect to see JT distortions; and indeed there is plenty of evidence to support it, earliest of which comes from the work done by Thrush et. al in 1956.

Experimental data can be found here (along with a list of references to the original works).

A publication I found particularly instructive was this The potential energy surface given below has been taken from there.

This image presents a downward looking view of linear Jahn–Teller distorted lower PES appropriate for $\ce{C5J5}$ . The coordinates $Q_a$ and $Q_b$ denote the two unique dimensions along which motion resolves the degeneracy of the conical intersection corresponding to the $^2E^"_1$ state. These coordinates lie in the space spanned by the four pair of $e^"_2$
normal coordinates, $Q_{ia}$ and $Q_{ib}$. Around the perimeter of the figure are pictures of the distorted $\ce{C5H5}$ geometries for the corresponding value of the phase angle $\phi$.

The work, later goes on to summarise the geometric distortions in the table given below.
The authors also compute $\epsilon_i$ for each of the four modes, and confirm that their sum corresponds to the experimental value of $\epsilon_T$ (this is done for both $\ce{C5H5}$ and $\ce{C5D5}$)


Heat and water absorption by calcium chloride

Anhydrous calcium chloride is an effective desiccant, but the absorption of water by anhydrous calcium chloride is exothermic; i.e. the reaction releases heat.

$$\ce{CaCl2 + 6 H2O -> CaCl2 . 6 H2O}\quad\Delta H = -97\ \mathrm{kJ/mol}$$


Joule–Thomson effect: why does temperature decrease during expansion

In order to expand, the molecules may do work, such as pushing against a piston, pushing against the container (as in water rockets), spinning a turbine, etc. Energy is transferred from molecular motion to another form, so the temperature drops.

In free expansion, however,

  the gas does no work and absorbs no heat, so the internal energy is conserved. Expanded in this manner, the temperature of an ideal gas would remain constant, but the temperature of a real gas may either increase or decrease, depending on the initial temperature and pressure.

[from Wikipedia].


25.9 kJ/mol refers to one mole of $\ce{HI}$ produced while 51.8 kJ/mol corresponds to one mole of $\ce{H2}$ consumed. It is good practice to give the full equation, especially when dealing with reaction enthalpies/energies $\Delta_rX$. For formation quantities $\Delta_fX$, I would always assume one mole of product.


mechanism of mechanically triggered explosion

There are differing propagation mechanisms for an explosion:

Thermal: a deflagrating explosive such as gunpowder has a comparatively slowly propagating wave-front, depending on heat propagating through the bulk material to set fire to the next layer. This is similar to a candle slowly burning down.
Compression: some brissant (but opaque: see 3. below) explosives create a shock wave. "[High explosives] detonate to produce a defining supersonic over-pressurization shock wave."
Light: fuel-air mixtures, glyceryl trinitrate and nitrogen triiodide explosion can be initiated by light (auto manufacturers are looking at laser "spark plugs" to improve internal combustion engine efficiency). In these types of explosives, light from the explosion itself might propagate a supersonic blast front.

So initiating an explosion through heat alone may initiate just a deflagration, without enough over-pressure to form a blast wave.


I think you are right. Grimm's law in its original version (1925) should not apply to the $\ce{-CF3}$ , $\ce{-CN}$ pair. This "law" is very restrictive and has only historical significance. To be honest I could not recall its statement and I have never come across it in the last 10 years or so that I work in the field. The modern definition of classical bioisosteres (as opposed to non-classical bioisosteres like the $\ce{-CO2H}$, tetrazole pair) has been extended exactly for this reason, to include functional group pairs with similar physiochemical properties that do not follow Grimm's law or any other classification of less general scope.


Cationic micelles similar to your examples can and do exist.  For example cetyl trimethylammonium bromide (Growth of Cationic Micelles in the Presence of Organic Additives, P. A. Hassan and J. V. Yakhmi, Langmuir, 2000, 16 (18), pp 7187–7191, DOI: 10.1021/la000517o).

Additionally, there are cationic micelle detergents (Ionization of cationic micelles: Effect of the detergent structure, Journal of Colloid and Interface Science, Volume 78, Issue 2, December 1980, Pages 330-337).

The bottom line is that there are a variety of use cases for cationic micelles, in detergents and otherwise.


What is a "symmetrical product" of an irreducible representation with itself?

In your case you need the equations for the characters as there are symmetric and antisymmetric parts in the product. Your question gives just the symmetric part and the antisymmetric part is $A_2$ normally written as $[A_2]$ to indicate this.

The characters are found using $\chi^+ =\left([\chi(R)]^2+ \chi(R^2) \right)/2$ for the symmetrical part and $\chi^- =\left([\chi(R)]^2- \chi(R^2) \right)/2$ for the antisymmetric part. Using the point group table you need to work out the square of all the operations for $E_g$ this gives $\chi(R^2)$ (e.g. $(C_2)^2 \to E$) and then square the operations which gives $[\chi(R)]^2$, i.e. $[E]^2\to 4$. From the list of characters you then identify the symmetry species, either directly by inspection (as for the antisymmetric in this case), or by forming the irreps from the reducible representation.


How could you accomplish the following transformation in sufficient yield?

I'm not an organic chemist, but upon looking for similar systems I found a nice collection of various papers that all use selenium dioxide, so allylic oxidation as you mentioned it with various other functional groups. I copied the sections from the papers (therefore sorry for the quality and the Chinese in one paper) together with the preparations.

To me this looks like the most promising route here. According to that mechanism it should favor the allylic position here selectively and therefore and also due to sterics hopefully not oxidize the longer alkyl chain.


How does an acid prevents or discourges oxidation of iron?

This is largely due to the extreme insolubility of iron (III) hydroxide, i.e. $\ce{Fe(OH)3}$.

One can write the oxidation reaction, somewhat simplified, as:

$$\ce{Fe^2+(aq) + 6 H2O(l) <=> Fe(OH)3(s) + 3 H3O+(aq)} + e^-$$

The $K_{sp}$ (solubility product) value for $\ce{Fe(OH)3}$ is very small: $2.79 \times 10^{-39}$ (Wikipedia).

From the equilibrium reaction it's easy to understand that acidic conditions (high $[\ce{H3O+}]$) push the equilibrium to the left, thus preventing the oxidation.

It also explains why Mohr's Salt resists oxidation much better than simple ferrous sulphate, as the ammonium ion provides some acidity.

In alkaline conditions the oxidation of ferrous ions to ferric ions is very fast.


Is there a particular reason for the way these structures are represented?

There is no real difference. Sometimes this comes about from the drawing software people use or personal preference. In some texts the authors or typesetters are using LaTeX packages to draw organic molecules, this can be rather tedious and it is easier to use 90° angles.

The angled hydrogens are a bit closer to what a 2d drawing of the 3d object would look like, but not to the point where any information is conveyed. It can make visualizing the compound in 3 dimensions a bit easier for me when they are drawn this way.

For the double bonds the cis/trans or E/Z information is conveyed, although this is irrelevant in your example.


In the following diagram, which carbocation is more stable, A or B?

The answer is A because the positive charge on the positively charged carbon can be delocalized into the attached benzene ring (I've drawn one of the possible resonance structures to illustrate this).

The same thinking can be applied to your question. Look at resonance structures C and D (we can draw many more) for the attack of an electrophile on a benzene ring.  If there is a second benzene attached and it is attached ortho or para to where the electrophile attacks, then the positive charge can be delocalized into the second benzene ring.

This additional delocalization will stabilize the system. Hence, the second benzene is an ortho-para activating substituent.  However, since this stabilization involves electrons involved in pi bonds, the stabilization (activating effect) will be weaker than when a substituent with a "more readily available" lone pair of electrons (amino, hydroxyl, etc.) is involved.


If melting points and freezing points are the same, what state is pure equilibrium?

At the melting point there is a change of state. This means that solid is transformed in liquid or viceversa. For this process you need energy which is supplied or withdrawn in the form of heat. The whole process occurs at constant temperature, which is precisely the melting point.

Adding or removing temperature has no sense. It is heat what  is added or removed. The temperature is just a result of the state. When you have a mixture of solid and liquid in equilibrium, this temperature is defined, is the same in the whole system and is the melting point.

Since it is state change, both solid and liquid are present and the temperature will not change while both are there in any quantity and if the system is in equilibrium. The quantities of solid and liquid will not vary until there is some heat flow. However, on the microscopic level, the molecules are still moving and some will migrate from the solid to the liquid while others (the same amount) will go from the liquid to the solid. This results in that, even if the system is in equilibrium, the small crystals will dissolve while the bigger ones will grow bigger.