Find empirical formula when individual masses are known

An empirical formula is, according to IUPAC’s Gold Book definition:

  Formed by juxtaposition of the atomic symbols with their appropriate subscripts to give the simplest possible formula expressing the composition of a compound

Wikipedia goes even further:

  the simplest positive integer ratio of atoms of each element present in a compound

and then

  An empirical formula makes no reference to isomerism, structure, or absolute number of atoms.

So, NiF2 is an empirical formula, while Ni2F4 simply isn't. Another example is given by, again, Wikipedia:

  For example, formaldehyde, acetic acid and glucose have the same empirical formula, CH2O

Their molecular formulas (which account for the absolute number of atoms in a molecule) differ, though: formaldehyde CH2O, acetic acid C2H4O2, glucose C6H12O6.


Are chemical treated as ideal gases when trying to figure out heat of formation using calorimetry?

Internal energy of any gas is given by
$$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$
Last three aren't affected by ordinary heating. And $U_\text{intermolecular}$ for ideal gas is zero. That's why for ideal gas $U$ is only function of temperature. But during a chemical reaction (even if it involves an ideal gas)
$U_\text{bonds}$ comes into play. The formation and destruction of bonds results in a decrease of potential energy.
Therefore, for a chemical reaction, even if the temperature remains constant the internal energy changes.


How to determine the volume of a substance produced when only given the equation at STP?

I think you erred in multiplying 2 moles of water by 22.4 L. There are 55.5 moles of water per liter of water. Keeping this relationship in mind, use dimensional analysis to find the moles of water produced in your reaction.

You're confusing the number 22.4 with the fact that 1 mol of gas at STP occupies approximately 22.4 L.


The authors mention three residues that they consider "second shell", 12, 77, and 102. They are shown in the figure below in green:

Residues Trp50 and Glu101 are directly involved in ligand binding, shown as space-filling in yellow. As you can see, they are part of the central beta barrel. The residues in green are also part of the beta barrel, but the side chains are not involved in ligand binding because they point to the outside. For that reason, the effect of mutating these residues on ligand binding must be indirect by repositioning the residues that point inside.

  What is the second shell, and how many different shells are there?

This is not a well-defined concept. The term is more common for hydration of ions (first shell of water molecules makes direct contact, second shell contact first shell and so on), or in the context of coordination spheres (https://en.wikipedia.org/wiki/Coordination_sphere). When researcher write papers, they come up with descriptions that fit their structures. A famous example is to describe polymerase active site secondary structure as hands, with a palm, thumb and fingers, see e.g. https://marlin-prod.literatumonline.com/cms/attachment/c9a1e727-5b4b-45ce-81c8-8a24c1aea2ef/gr1.jpg. Some of these descriptions become common (such as the beta-barrel mentioned above) and some do not (my research advisor once described an enzyme as hat-shaped - it did not stick, maybe because hats come in many different shapes and forms, and barrels are more defined).


How to prepare KF/Al2O3 i.e. potassium fluoride on alumina (/on aluminum oxide)

In Bull. Chem. Soc. Jpn., 1983, 56, 1885-1886 (PDF), the authors (Takashi Ando and coworkers) give the following method:

  Alumina (30 g, neutral alumina, type 90 for column chromatography, activity I, 70-230 mesh, Merck Jpn.) was mixed with $\ce{KF}$ (20 g) in 200 mL of water and the water was removed at 50-60 °C in a rotary evaporator. The impregnated alumina was then dried in a vacuum drying oven at 75 °C for several hours.

In another publication, J. Chem. Soc., Perkin Trans 2, 1986, 1133-1139, by the same main author, which focuses on the properties of the $\ce{KF}/\ce{Al2O3}$ system, a "dry preparation" by grinding $\ce{KF}$ and $\ce{Al2O3}$ under a nitrogen atmosphere is mentioned.


Calculate the standard oxidation potential from given potentials

You can write down the three half reactions:

\begin{aligned}
 \ce{Fe^{2+} + 2e- &-> Fe}      & E_1&= x\,\mathrm{V}\\
 \ce{Fe^{3+} + e-  &-> Fe^{2+}} & E_2&= y\,\mathrm{V}\\
 \ce{Fe^{3+} + 3e- &-> Fe}      & E_3&= z\,\mathrm{V}
\end{aligned}

The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.

An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).

Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry

$x$ joules per coulomb, or
$96500x$ joules per mole of electrons, or
$2\times 96500x$ joules per mole of reaction.

In the same fashion, the second half reaction has the electrons carrying

$y$ joules per coulomb,
$96500y$ joules per mole of electrons,
the same $96500y$ joules per mole of reaction.

The third half reaction is the sum of the first two, so its electrons carry

$2\times 96500x + 96500y$ joules per mole of reaction, which is
$3\times 96500 z$ joules per mole of reaction, or
$96500z$ joules per mole of electrons, or
$z$ joules per coulomb.

Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.


Hydrolysis of starch by means of amylase to unclog bathtub drain

The problem you'd face using any chemical to dissolve the rice is that rice forms a thick gelatinous surface coating as it dissolves, so that it would take an extremely long time for the amylase to penetrate... perhaps months or years. You'd be better off using some of the standard mechanical drain-cleaning techniques:

Disassembly of the sink trap (and drain, if needed) and cleaning or replacing it
Drain auger
Drain snake
Water-pressure drain cleaner

Note that augers and snakes can scratch sinks or toilet bowls, and that water-pressure cleaners can cause pipes to come apart.

N.B. If you've already poured a potentially dangerous cleaner into the drain, remove it safely before trying anything else!


Ozone's ground state is a singlet.  All electrons therein are paired,

However,

http://link.springer.com/article/10.1134%2FS0018143906040060#page-1
http://link.springer.com/article/10.1134%2F1.1870062#page-1


Specific heat for tetraatomic gas molecules

The heat capacity comprises contributions from translational motion (3 terms for each of $x, y, z$) plus (whole body) rotational motion ( 3 axis directions). Each motion accounts for $(1/2)R$ by equipartition theorem making the $(3/2)R + (3/2)R$ above.

Then there is a contribution from the fact that the molecule is vibrating. Non-linear molecules have $3N-6=6$ vibrational modes. Each vibration has $(1/2)R$ contribution from its potential energy and the same from its kinetic energy which in total is the $6R$ in your question.


how long would it take for this super-material to convert to the stuff
  I scribble with?

No, despite the fact that James Bond said "Diamonds are Forever", that is not exactly the case. Although Bond's statement is a fair approximation of reality it is not a scientifically accurate description of reality.

As we will soon see, even though diamond is slightly less stable than graphite (by ~ 2.5 kJ/mol), it is kinetically protected by a large activation energy.

Here is a comparative representation of the structures of diamond and graphite.

(image source: Satyanarayana T, Rai R. Nanotechnology: The future. J Interdiscip Dentistry 2011;1:93-100)

(image source)

Note that diamond is composed of cyclohexane rings and each carbon is bonded to 2 more carbons external to the cyclohexane ring. On the other hand, graphite is comprised of benzene rings and each carbon is bonded to only 1 carbon external to the benzene ring. That means we need to break 6 sigma bonds in diamond and make about 2 pi bonds (remember it's an extended array of rings, don't double count) in graphite per 6-membered ring in order to convert diamond to graphite.

A typical aliphatic C–C bond strength is ~340 kJ/mol and a typical pi bond strength is ~260 kJ/mol.  So to break 6 sigma bonds and make 2 pi bonds would require ~((6*340)-(2*260)) ~ 1500 kJ/mol. If the transition state were exactly midway between diamond and carbon (with roughly equal bond breaking and bond making), then we might approximate the activation energy as being half that value or ~750 kJ/mol.  Since graphite is a bit more stable than diamond, we can refine our model and realize that the transition state will occur a bit before the mid-point. So our refined model would suggest an activation energy something less than 750 kJ/mol. Had we attempted to incorporate the effect of aromaticity in graphite our estimate would be even lower.  In any case, this is an extremely large activation energy, so, as we anticipated the reaction would be very slow.

An estimate (see p. 171) of the activation energy puts the reverse reaction (graphite to diamond; but since, as noted above, the energy difference between the two is very small the activation energy for the forward reaction is almost the same) at 367 kJ/mol. So at least our rough approximation was in the right ballpark, off by about a factor of 2. However, it appears that the transition state is even further from the midpoint (closer to starting material) than we might have guessed.

This activation energy tells us that at 25 °C, it would take well over a billion years to convert one cubic centimeter of diamond to graphite.

Note 04/17/20: As mentioned in a comment, the original "estimate" link became  defunct and was replaced today with a new estimate link.  However the original article and estimate can can still be seen on the Wayback Machine and it estimates the activation energy to be 538.45 kJ/mol, reasonably close to our estimate.


There is a general belief that plastics are impermeable, probably because many liquids are stored in plastic containers. This is, however, not correct. Plastics are permeable to water and also to gases. The permeability is low but is not zero and therefore, when very long periods of time are involved, a perceptible change occurs. If absolute impermeability is required, then metals should be used.

The process involved is diffusion and depends basically on the type of plastic. The ones containing fluorine (Teflon and the like) are the less permeable and the polyesters the most permeable, probably because they are more polar.

What has happened to the plastic balls is that some water has permeated out and in some cases some air has diffused in.


How to determine the reactivity of a set of compounds that may undergo nucleophilic substitution?

In question 1), you are wrong about the tendency of leaving groups. While you are correct that the carbon-chlorine bond is more polarised towards chlorine, iodide is still a better leaving group. This is because it is much larger and can thus stabilise the resulting negative charge much better.

You can observe this in $\mathrm{S_N2}$ reactions, where iodides are much more reactive than bromides which are more reactive than chlorides (and fluorides are basically unreactive) even though the same polarisation argument could have applied.

In question 2), the stabilisation that alkyl groups give carbocations, or the electron density they donate into a phenylic ring actually comes from $\ce{C-H}$ bonds. Thus, the para-xylene derivative can stabilise a cation better since there are three $\ce{C-H}$ bonds it can use for stabilisation. No matter which way the methyl is oriented, it can stabilise. para-ethylbenzyl bromide can only be stabilised in two out of three orientations (when a $\ce{C-H}$ bond is orthogonal to the phenyl ring) and para-isopropylbenzyl bromide only in one. Thus, the rate of hydrolysis is accordingly.


More than likely the case of your laptop became charged and was attracting the salt. Yes salt crystals can be coerced by an electrostatic field.

Here is a post on a teaching site that actually uses salt (and pepper) to demonstrate electrostatic attraction.

https://www.teachervision.com/electricity/lesson-plan/5787.html


Order of the reaction between hemoglobin and oxygen

You seem to be overthinking this a great deal. The order of a reaction is not defined by the units of the rate constant, but rather by the sum of the exponents in the empirical rate equation. From the IUPAC Gold Book:

  If the macroscopic (observed, empirical or phenomenological) rate of reaction ($v$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ [...] then the reaction is said to be of [...] (total or overall) order $n = \alpha + \beta + \cdots$

Considering that the partial pressure of $\ce{O2}$ is directly proportional to its concentration, the forward reaction has a rate equation

$$k'[\ce{Hb}][\ce{O2}]$$

where the proportionality factor between $P(\ce{O2})$ and $[\ce{O2}]$ is absorbed into the new rate constant $k'$. The order is simply 2.


Electrophilic Substitution with Ceric Ammonium Nitrate and Lithium Bromide as reagents

First of all, the linked paper clearly points out the following:

Hint: In second step (A $\rightarrow$ B), combination of lithium bromide and cerium(IV) ammonium nitrate (CAN) is used as a brominating reagent

From a problem solving perspective, this step is trivial for someone who has read the problem well (not me, unfortunately :)

Ceric Ammonium Nitrate is a very strong oxidizing agent and has the ability of oxidizing benzene compounds first to Nitrobenzene, then to quinone or quinoid derivatives1. Assuming this line of reasoning works, the quinoid derivative could then undergo nucleophilic addition with $\ce{LiBr}$, followed by workup with acid to give the p-subsituted benzene derivative. This could be a probable reaction mechanism, if the substitutent was $\ce{-OH}$ instead of $\ce{-OMe}$ and if an acid workup step was involved. Also, this mechanism can give the unwanted product 4-bromo-3-isopropylphenol, as shown below:

However, this reaction mechanism is very speculative, and it also won't hold good here, because the substituent is $\ce{-OMe}$. Some further research on the links provided in the comments2 yielded the following results:

Alkyl aromatics were brominated in the side chain using sodium bromide and CAN. Asakura et al. reported bromination of C-5 of uracil nucleosides using CAN and lithium bromide in acetonitrile.

This is also the reaction present in (3), which seems to have been lifted without a citation to Asakura et al.

Finally, Asakura et al. 4 provides a suitable mechanism for this reaction, which I correctly assumed arose from the oxidation of $\ce{Br-}$ to $\ce{Br2}$, and then to $\ce{Br+}$

References:

Nikishin, G. I., et al. “Reaction of Aromatic Hydrocarbons with Cerium Ammonium Nitrate.” Bulletin of the Academy of Sciences of the USSR Division of Chemical Science, vol. 24, no. 6, June 1975, pp. 1328–29. doi:10.1007/BF00922077.
Nair, Vijay, and Ani Deepthi. “Cerium(IV) Ammonium NitrateA Versatile Single-Electron Oxidant.” Chemical Reviews, vol. 107, no. 5, May 2007, pp. 1862–91. doi:10.1021/cr068408n.
Cerium (IV) Ammonium Nitrate mediated oxidative transformations of diaryl methylenecyclopropanes. Shodhganga link
Asakura, Junichi, and Morris J. Robins. “Cerium(IV)-Mediated Halogenation at C-5 of Uracil Derivatives.” The Journal of Organic Chemistry, vol. 55, no. 16, Aug. 1990, pp. 4928–33. doi:10.1021/jo00303a033.


In the strictest sense, the orbital energies are simply eigenvalues of the multi-electron Schrödinger equation; unfortunately, the multi-electron Hamiltonian takes a nasty form and is impossible for us to solve exactly, so we can't get any insight on the energies from that approach...
Instead we inadvertently resort to LCAO-based methods, whereby exploiting symmetry the interacting atomic (or symmetry) orbitals can form molecular orbitals of the correct form. The energy of these is another issue; for some back-of-the-envelope calculations, severe approximations such as Hückel MO theory can be useful for a rough idea, but more accurate estimates just require much more complicated computational methods (e.g. the acclaimed Hartree-Fock).
As these methods resort to a large basis of AOs (and not just the simplified 2 AOs mix in- and out-of-phase to give 2 MOs picture) and other complications such as electron correlation, I'm afraid there is no simple rule to give an accurate enough estimate unless thorough calculations are made. So, to forwardly answer your question, rather impractically we can't know a priori the ordering of the orbitals in the general case.
Certain semi-empirical rules (based on the experience from multiple common systems and "chemical intuition") exist, such as the (often useful) approximation formulated for organic molecules $\sigma < \pi < l.p. < \pi^* <\sigma^*$ though this is very rough and suitable for organic systems, usually.
In the case of ammonia (as given in the question) a particular ordering is found; it should be noted that the two types of bonding interactions are both $\sigma$-interactions (i.e. locally cylindrically symmetric; there is no nodal plane within each individual bond). Despite the double degeneracy, the $e$ set still constitutes a $\sigma$-type MO (with a node in the overall MO shape, rationalising a higher energy than the $a_1$); so the ordering in ammonia is $\sigma_{a_1} < \sigma_e < lp_{a_1} < \sigma_{a_1}^* < \sigma_e^*$ -not so counter-intuitive after all, just different types of $\sigma$ MOs possessing different energies. (As a sidenote, the doubly deg. $\sigma_e^*$ are those allowing ammonia to act as a weak $\pi$-acceptor, placing it higher than simple $\sigma$-donors in the spectrochemical series.)
Summary: MO energy calculation especially in polyatomic species is hard and done with complex computational methods. Unfortunately, then, we can't have a general rule for the energy ordering in advance. However, often we can make educated guesses in analogy to systems we have already seen.
(For a more detailed look into the methods used to obtain the energies see Atkins & Friedman Molecular Quantum Mechanics, 4th ed.; Chapters 8–9)


Why is change in volume of the solution zero in case of an ideal solution?

If you have two component mixture of A and B, most important reason for volume of mixing being zero is that when solution is ideal heterogeneous interactions (A - B) are the same as homogeneous (A - A and B - B). If interactions are the same than volume doesn't change when you mix components. Other reasons which contribute to non - ideality of mixing is size and shape of the molecules. If you mix water with some liquid polymer, you can expect that  due to big difference in size and shape of water and polymer molecules, volumes will not be additive because water molecules being much smaller than polymer molecules can easily fit between them in free volume of polymer liquid. Such packing/fitting of water molecule is quite different than if you just look at water molecules in pure water or if you look at pure polymer since bulky polymer molecules can't just fit easily in free volume of other polymer molecules like water can.


Can energy savings be estimated from activation energy differences?

The statement of this questions mixes a number of issues, so there is, in my opinion, no definite answer to this question.

First of all, one must take care that activation energies will have an influence on the kinetics (rate constant) of the reaction that is catalyzed. In a very general point of view, the overall energetic requirement for a chemical reaction does not directly depend on the activation energy (for instance the equilibrium constant is not changed).

Thus if two separate processes are considered, a different activation energy will mostly mean a reaction rate that will be slower, so overall a process that might take longer but not necessarily need a higher energy input.

Now if one would add to this problem a constraint that would be that the reaction rates of bith processes should be similar, the main method to achieve such a result would be to change the temperature at which the process takes place. This is related to the activation energy through the Arrhenius equation:

$$k = A e^{-\frac{E_{a}}{RT}}$$

If one assumes that the pre-exponential factor is the same for both processes (in effect meaning that the transition states in the catalyzed reaction have similar constraints) then one can reach some conclusions on both processes.

For instance, the reaction rates at constant temperature can be calculated. At 298 K, 3 kcal/mol difference would mean a ratio $k_{1}/k_{2}$ of about 150, so the process with lower activation energy would proceed 150 times faster than the other one. An other option would then be to increase the temperature of the slower process. In such a case, the temperature would depend on the ratio of the activation energies (and not the difference) as :

$$T_{2} = \frac{E_{a2}}{E_{a1}} T_{1}$$

If one takes the values of the qustion and considers a process that would take place at 298 K, one would need to heat at 373 K to reach the same rate constant. From this, one could consider the additional energy input required to heat the reaction medium up to this temperature.


Index of Hydrogen Deficiency for para-Toluenesulfonylchloride

You have to start from the general formula for the degree of unsaturation
$$DU = 1 + \frac{1}{2} \sum n_i(v_i-2),$$
where $n_i$ is the number of atoms with a valency of $v_i$.

For para-toluenesulfunylchloride ($\ce{C7H7SO2Cl}$) you will therefore arrive at
\begin{align}DU &= 1 + \frac{1}{2}\Big[n_\ce{C}(4-2) + n_\ce{H}(1-2) + n_\ce{S}(6-2) + n_\ce{O}(2-2) + n_\ce{Cl}(1-2)\Big]\\
&=1+\frac12\Big[7(4-2) + 7(1-2) + 1(6-2) + 2(2-2) +1(1-2)\\
&=1 + \frac12\Big[14 -7 +4 +0 -1\Big]\\
&=6,
\end{align}

because sulfur uses all six valence electrons in this case.


What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays?

An article by Snell and Pleasanton, 'The Atomic and Molecular Consequenses of Radioactive Decay', (J. Phys. Chem., 62 (11), pp 1377–1382, $1958$) supports Ben Norris's comment.

  It is clear ... that $\ce{^{14}CO2}$ remains predominantly bound as $\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\ce{^{14}CO2 -> NO2^+}$ dissociation yielding $\ce{NO+}$, $\ce{O+}$ and $\ce{N+}$ follows [in], respectively, $8.4$, $5.9$, and $3.6$% of the decays.

A table summarising the results is given.

$$\begin{array}{|c|c|}
\hline
\mathbf{Ion} & \mathbf{\%\  abundance} \\
\hline
\ce{NO2+} & 81.4(16) \\
\ce{NO+} & 8.4(4) \\
\ce{O+} & 5.9(6)  \\
\ce{N+} & 3.6(4) \\
\ce{NO2^{2+}} & 0.40(06)\\
\hline
\end{array}$$


Boundary lines in phase diagrams and the lever rule

Lots of questions wrapped in one, but they do all come down to understanding what is represented on a phase diagram.  Up front, I'll suggest Porter and Easterling's classic text on 'Phase Transformation in Metals and Alloys' - it has lots of good info for lovers of thermodynamics and phase transitions.

In a system like Cu-Zn that you have pictured above, there are a number of different phases observed.  You could imagine that at a given temperature you could construct a free energy curve for each phase as a function of composition. At each composition, only one will have the lowest free energy, and one might think you could identify the phase diagram that way. However, a straight line across the G vs. composition plot is special - it represents where the chemical potentials of both elements are constant.  So a line that is tangent to two free energy curves represents a range of compositions where the elements in the two phases have the same chemical potential.  This then is how you construct the phase diagram - find the minimum free energy across composition, then find the common tangents (if any). An example is shown here:
The top plot is the free energy curves for all the phases, the bottom plot is the phase boundaries determined by the common tangents.  The dashed vertical lines up top are where the common tangent points are, coming down to their position on the phase diagram.

Now the $\delta$, $\gamma$, $\epsilon$ are specific phases.  In regions with just one of them listed, that is the only phase stable at the temperature and concentration.  Regions such as $\delta + \gamma$ are regions where both are stable. There relative amount has to follow the lever rule, which is just a linear variation across the two-phase region from, say, all $\gamma$ right on the boundary of $\gamma$ with $\gamma + \delta$, to all $\delta$ on the other side.

The horizontal lines are there mostly to emphasize when a new phase appears or disappears.
So, at point E on you diagram above, there is one specific temperature and composition at which $\delta$, $\gamma$, and $\epsilon$ are in equilibrium.  Go up in temperature, and it wants to be all $\delta$; come down in temperature and you enter the two-phase $\gamma$-$\delta$ region with the relative abundance from the lever rule. At P, the horizontal line is to point out that $\epsilon$ disappears into the $\delta$-liquid region.

So, at the eutectoid E, there is one point where the three phases coexist (Gibbs phase rule). Moving away from that point moves into single or two-phase regions. Everywhere along the horizontal line through E is $\gamma$+$\epsilon$, except at exactly point E.

As for the lever rule, that follow directly from coming along the common tangent - the ratios of the two phases have to vary linearly or else you are deviating from the common tangent.


Why is equilibrium achieved at different stages of a reaction?

If you define early or end of reaction by the how much the concentration of reactants change from initial reaction to once equilibrium is reached, it is because the equilibrium constant itself is essentially a ratio of the forward and reverse rate constants (can be approximated by the Arrhenius equation posted by t.c.). A high ratio means "more" products (at least a higher concentration) are present at equilibrium than reactants (so in a sense near the "end of the reaction"), since the forward rate is much higher than the reverse, a lower concentration of reactants compared to products is needed to maintain equilibrium. A low ratio means the opposite.


How is Bent's rule consistent with LCAO MO theory?

Beyond the educational value, Bent's rule has had significant impact on how we understand wave functions of molecules. It's simplicity is almost unique in the world of quantum chemistry; almost everybody can immediately understand it and apply it. Together with the VSEPR model it can lead to quite accurate predictions on the back of an envelope.
Molecular Orbital Theory vs. Valence Bond Theory (MO vs. VB)
In their respective infinite treatment these theories are congruent. For more on this topic please read What is actually the difference between valence bond theory and molecular orbital theory?
Important for this post is that both theories are approximations for the wave function of a molecule, hence they do not describe bonding beyond the stationary point they are applied to, and they operate within the clamped nuclei (Born-Oppenheimer) approximation.
In MO theory it is common to express the molecular orbitals as linear combinations of atomic orbitals. Those MO are usually chosen to be orthonormal. The MO themselves are delocalised, but there are algorithms to transform them into more intuitive VB type orbitals. This will provide us with a Lewis-like bonding picture. Famously used in this regard is Natural Bond Orbital (NBO) theory.
Bent's rule
Unfortunately I cannot find my copy of Bent's paper, so this answer will be based purely on the definition of Bent's rule.
The IUPAC gold book defines Bent's rule (DOI: 10.1351/goldbook.BT07000):

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

Water is a good example to demonstrate Bent's rule and its interplay with the VSEPR model. In $\ce{H2O}$ you expect oxygen to follow the octet rule, hence there will be four electron pairs around it, and it predicts a generally tetrahedral structure of the molecule. The angle for these shapes is about $\newcommand{\degree}{^\circ}109\degree$. Since hydrogen is more electronegative than no ligand, Bent's rule predicts a smaller angle than that.
Probably due to its simplicity, Bent's rule can be applied within the MO and VB bonding pictures.
Coulson's Theorem
Coulson's theorem is the formal theory which applies to Bent's rule (see Wikipedia).
We choose a set of orthonormal AO, as would be consistent with the LCAO approximation.
$$
\langle\chi_i|\chi_j\rangle = S_{ij} = \delta_{ij}
\begin{cases}
0, & i \neq j\\
1, & i =    j
\end{cases}\tag{1}\label{orthonormalAO}
$$
We can transform these orbitals into sets of orthonormal hybrid orbitals.
\begin{align}
\varphi_a &= \sum_i \lambda_i \chi_i &
\langle\varphi_a|\varphi_b\rangle = S'_{ab} &= \delta_{ab}
\begin{cases}
0, & a \neq b\\
1, & b =    b
\end{cases}\tag{2}\label{orthonormalHybridAO}
\end{align}
Let's look at a special case, where we choose atomic orbitals to be $\chi_i$ with $i \in \mathrm{s}, \mathrm{p}_x, \mathrm{p}_y, \mathrm{p}_z$. We can construct following hybrid orbitals (also compare Mathematical form of four hybrid orbitals):
\begin{align}\tag{3}\label{orthonormalSP3}
  \varphi_{xyz}
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_x
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_y
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\
  \varphi_z
    &= \lambda_\mathrm{s}\chi_\mathrm{s}
     - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x}
     - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y}
     + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}
\end{align}
If all $\lambda_i$ are chosen to be $\frac{1}{2}$, then we would come to the famous $\mathrm{sp}^3$ hybrid orbitals.
Let's generalise this a bit and drop the normalisation:
\begin{align}\tag{4}\label{hybrid}
  \varphi_a
    &= N\cdot(\chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi_{\mathrm{p}_k}) &
  \leadsto
  \varphi_a
    &= \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k}
\end{align}
From \eqref{orthonormalHybridAO} and \eqref{hybrid}, and ignoring the trivial case:
\begin{align}\tag5
  \delta_{ab}
    &= \langle\varphi_a|\varphi_b\rangle\\
  0
    &= \langle
      \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k} |
      \chi_\mathrm{s} + \lambda_{\mathrm{p}_l} \chi'_{\mathrm{p}_l}
    \rangle\\
    &= \langle\chi_\mathrm{s}|\chi_\mathrm{s}\rangle
    +  \lambda_{\mathrm{p}_k} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_k}\rangle
    +  \lambda_{\mathrm{p}_l} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_l}\rangle
    +  \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}
       \langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle
\end{align}
Given \eqref{orthonormalAO} we can simplify this. Note that $\chi'$ is actually a linear combination of $\mathrm{p}$ orbitals, and therefore they don't necessarily have to be orthogonal. The angle between two such orbitals is given by the inner product, i.e. $\langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle = \cos \theta_{kl}$.
\begin{align}\tag6
  0 &= 1 + 0 + 0 + \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l} \cos \theta_{kl}
\end{align}
Therefore Coulson's Theorem is
\begin{align}\tag{7}\label{Coulson}
  \cos \theta_{kl} &= \frac{-1}{\lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}}
\end{align}
If you call $\lambda$ the hybridisation index, then you see that the angle between two hybrid orbitals is dependent on the this index. In other words, the angle between two ligands around a central atom determines the $\mathrm{p}$ character of the hybrid orbitals that ideally describe this bond.
What you can also derive from this formula is, given a constant angle $\theta_{kl}$, if you use a hybrid-orbital with increased $\mathrm{p}$ character, the other hybrid-orbital must have less $\mathrm{p}$ character, i.e. $\lambda_{\mathrm{p}_k} < \lambda_{\mathrm{p}_l}$.
This is in principle all that Bent's rule requires in the framework of LCAO-MO theory to be consistent; the argument about the electronegativity of a ligand is based on observation of molecular structure.

I hope the above answers the titular question sufficiently. In the second part I'll try to show where the original arguments fail.
First of all, you are basing your argumentation on even more handwavy mathematics then I did above. You are also not considering the actual molecular structure and the implications of that in terms of the LCAO-MO approximation of the wave function.

The orbitals contributing to the O-H bonds are the lower two in energy.

This is not quite correct. In LCAO-MO (or MO in general) theory all orbitals contribute to bonding.

Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen s orbitals with the oxygen 2s (giving 2a1) and 2py orbitals (giving 1b2). The lone pairs are a pure px orbital (1b1) and a combination of pz with a small contribution from the hydrogen s orbitals (3a1).

The shown MO scheme does not ignore orbital mixing, and neither do you in your argumentation. If the oxygen's $\mathrm{s}$ and $\mathrm{p}$ orbitals were far enough apart to not mix, then only one sort of these orbitals could interact with the hydrogen orbitals to lead to bonding. This would either lead to a situation where there is no oxygen $\mathrm{s}$-character in the bonds, or all of it.
This basically follows the argument that for the heavier homologues of water the oxygen valence $\mathrm{s}$-orbital becomes the lone pair.

Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and 50% p/50% s for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more s character.

I don't have an optimal LCAO-MO solution at hand, but I have the next best thing: a HF/STO-3G wave function with $C_\mathrm{2v}$ symmetry, $\angle(\ce{HOH}) = 100.0\degree$, and $d(\ce{OH}) = \pu{98.9 pm}$. That is off by plenty, but it's Hartree-Fock, so that was expected. The general tenor wouldn't change, if you were to run it on the experimental structure. This is the output of the wave function:
 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-20.252 is O1-s=1.0006
 Alpha occ 2 OE=-1.258 is O1-s=0.7899 H2-s=0.0928 H3-s=0.0928 O1-p=0.0245
 Alpha occ 3 OE=-0.594 is O1-p=0.5365 H3-s=0.2317 H2-s=0.2317
 Alpha occ 4 OE=-0.460 is O1-p=0.6808 O1-s=0.1328 H2-s=0.0932 H3-s=0.0932
 Alpha occ 5 OE=-0.393 is O1-p=1.0000
 Alpha vir 6 OE=0.582 is H3-s=0.3144 H2-s=0.3144 O1-p=0.2947 O1-s=0.0766
 Alpha vir 7 OE=0.693 is O1-p=0.4635 H2-s=0.2683 H3-s=0.2683

Unfortunately this doesn't tell us much, apart from that there is obviously one $\mathrm{p}$ lone pair, which was expected due to symmetry. (In MO theory though, this would still count as a bonding π orbital).
From the above table, one would derive that the majority of bonding comes from MO 3, which is a nearly pure oxygen $\mathrm{p}$ orbital.
I've run an NBO6 analysis on this calculation, which is basically a unitary transformation of the MO, and arrived at the following description:
     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. (2.00000) CR ( 1) O  1            s(100.00%)
                                         1.0000  0.0000  0.0000  0.0000  0.0000
   2. (2.00000) LP ( 1) O  1            s( 71.29%)p 0.40( 28.71%)
                                         0.0000  0.8443  0.0000  0.0000  0.5358
   3. (2.00000) LP ( 2) O  1            s(  0.00%)p 1.00(100.00%)
                                         0.0000  0.0000  1.0000  0.0000  0.0000
   4. (1.99926) BD ( 1) O  1- H  2
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000 -0.7071 -0.5970
               ( 40.76%)   0.6384* H  2 s(100.00%)
                                         1.0000
   5. (1.99926) BD ( 1) O  1- H  3
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000  0.7071 -0.5970
               ( 40.76%)   0.6384* H  3 s(100.00%)
                                         1.0000
 ---------------- non-Lewis ----------------------------------------------------
   6. (0.00074) BD*( 1) O  1- H  2
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000  0.7071  0.5970
               ( 59.24%)  -0.7697* H  2 s(100.00%)
                                        -1.0000
   7. (0.00074) BD*( 1) O  1- H  3
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000 -0.7071  0.5970
               ( 59.24%)  -0.7697* H  3 s(100.00%)
                                        -1.0000

This results in two $\mathrm{sp}^6$ bonding orbitals, one pure $\mathrm{p}$ lone pair and a $\mathrm{sp}^{0.4}$ lone pair. From this we would conclude that more $\mathrm{p}$ character is directed towards the hydrogen ligands.

If we properly account for orbital mixing, there is no effect on the 1b1 lone pair orbital, [...]

That is correct, this is due to symmetry.

[...] while the 3a1 lone pair does in fact get more s character, [...]

In your argumentation 3a1 would have been a pure $\mathrm{p}$ lone pair, with mixing it does gain $\mathrm{s}$ character, which is obviously more than none.

[...] but this increased s character is not expected to exceed 50%, since that would just mean a switch of label between the 2a1 and 3a1 orbitals.  50% s in the 3a1 represents the maximum mixed case.

The part with the switch is only true if the energy of these orbitals also switches and the contribution of the hydrogen were equal in both orbitals.
Does this argument also imply that an $\mathrm{sp}$ orbital was the maximum mixed orbital?
I'm very sorry, but I cannot follow the logic as I do not understand what is meant with maximum mixed case. In LCAO you are able to continuously mix $\mathrm{s}$ and $\mathrm{p}$ orbitals, if that actually produces a sensible approximation to the wave function is a completely different topic.

That gives a final result of a maximum of 25% s in the lone pairs (50% in the 3a1 and 0% in 2b1) and 25% in the O-H bonds (2a1 and 1b2), still contradicting Bent's Rule, which predicts that the lone pairs should have more s than the bonds, not the same amount.

This would be exactly the result which you were to expect from tetrahedral coordination: two $\mathrm{sp}^3$ bonds, one $\mathrm{sp}$ and one $\mathrm{p}$ lone pair.

[One a side note, we know that H2O does not represent the maximum mixing case, as SH2 has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]

I am really interested in that side note. I still want to understand what is meant by the maximum mixing case, and especially how a smaller bond angle is consistent with more mixing.


Can nitrogen dioxide react with plastics of any kind?

Nitrogen dioxide is a free radical but its energy is rather low so it will be unlikely to be able to abstract a hydrogen from most polymers. If we consider polypropylene as the "standard plastic" then I would not expect nitrogen dioxide to be able to corrode it by abstracting hydrogens from the plastic.

But if we choose something else such as an epoxy as the standard plastic then nitrogen dioxide when combined with water could start to chemically alter the plastic. It can react with amines to form things such as N-nitroso compounds.

When nitrogen dioxide dissolves in water it forms nitric and nitrous acids. It can be regarded as a mixed acid anhydride for these two nitrogen oxy acids. The nitric acid could also start to degrade and alter plastics such as epoxy plastics by nitration of electron rich rings.

The acid formed when nitrogen dioxide dissolves in water could degrade plastic by increasing the rate at which esters such as the phthalate diesters react with water. It is important to keep in mind that many plastics are a combination of a polymer (or several polymers) and some additives such as plasticisers. If you degrade the plasticisers some plastics will stop being "plastic" and become brittle.


Should the mean energy be written in angular brackets or with an overline?

Both the IUPAC 2008 Green Book (section 4.2) as well as the ACS Style Guide (specifically Chapter 11) agree that both methods of representing the mean value are acceptable. There might be some style guides / international standards that state otherwise, but for the average chemist these two are generally sufficient.

If ever in doubt, just state the meaning in text.


What makes Gold so different from Platinum?

That one electron is making sure that Gold has a complete d-orbital, and therefor is much more stable. Elements want to achieve noble gas configurations, and if they can't, then they want to have a half-filled orbitals. In the case of gold the d-orbital is filled completely and the s-orbital is half filled. Whereas Platinum doesn't have that complete d-orbital filled.

Chemical properties (such as reactivity) are influenced by electrons. The amount of neutrons will determine if it can be radioactive or not(,...) look at Tritium : that is a hydrogen atom with 2 neutrons in it and it is radioactive. Whilst protium (the most common isotope of Hydrogen with 1 proton and 0 neutrons) is not radioactive.


Ionic radii are not measured on individual ions by some sort of mini-micrometer, but bond distances can be measured and then radii are assigned to the individual atoms/ions. The fuzziness of the assignments can be seen by the variation in ionic radii as the coordination number changes: if you give the ion more room (greater coordination number), it will fill the space - if you assume that the ligands have the same size as usual (whatever that is).
(There are materials (ferroelectrics) in which the coordinated ion does not completely fill the space allotted. In $\ce{BaTiO3}$, the $\ce{Ti^{+4}}$ ion is so small that it can shift its position in the octahedral cage of oxygen anions in response to an external electric field.)
The data in the Shannon paper are extensive, certainly a credit to the Central Research Department of DuPont in 1976, but the trends reported seem to be much more valuable than any specific datum.
Nicolau mentioned covalency as one factor in determining radii. Purepani brought up negative ions not having an effect of negative radii. Assigning a radius to a proton other than zero suggests that you are allowing it to have some electron density surrounding it, able to repel an anionic ligand. Electrostatically, a positive ion will attract some electron body (anion or atom) and come closer; depending on your initial assignment of radii, you can adjust to keep your idea of proton radius constant, or adjust, even into the negative region (because we assume you have a good imagination). Negative ions have a larger size than protons, and are less hard, so they can dent a little. One reason to use negative radii for the cations is that the anion may be dented only at one site, but has a "normal" radius everywhere else around it.
My approach to handling a negative radius is to conclude that the proton (or other mini- cation) is so electrostatically demanding that it can begin to bury itself in the outer shell of the anionic ligand, if the coordination number allows such a deviation in distance to be measured. And since we don't want to redefine the anionic radius, a negative radius is assigned to the proton, and the mathematics works out OK.
I guess you could say atoms and ions are somewhat mushy, and mathematics can deal with it.


Why doesn't rubidium react with nitrogen gas?

Lithium is the only group 1 element that can form a stable alkali metal nitride (at standard conditions), the other group 1 elements (such as sodium, potassium, rubidium,...) will react (at different reaction conditions) to form their respective nitrides (in the case of rubidium, $\ce{Rb3N}$ (rubidium nitride)), but the compound will be unstable and decompose quickly.  The reactions of the alkali metals (other than lithium) with nitrogen would not release enough lattice energy and would thus be endothermic, so they do not form nitrides at standard conditions

Read more here : http://en.wikipedia.org/wiki/Alkali_metal#Nitrides_and_pnictides


The reactions you have shown could in fact occur (in the forward direction) to a slight degree in a solution of cesium bromide in water. But I would expect them to occur to a lesser degree than the autoprotolysis of water (formation of hydronium and hydroxide ions from two water molecules), so their effect on pH would be insignificant, and the pH would effectively be neutral.


Do not do it !! ( putting acidic, or rather any juice to copper bottles )

You are in danger of copper poisoning.

Generally, by food processing laws, copper is not allowed to be in direct contact with food, as there is danger of copper contamination.

Especially acidic liquids, like vinegar or citrus juices, directly slowly dissolve copper in presence of oxygen.

$$\ce{4 RCOOH + 2 Cu + O2 -> 2 (RCOO)2Cu + 2 H2O}$$

Note that cold liquids contain usually more oxygen than warm ones.

For similar reasons, even if there is not copper, it is very bad idea to keep the content of conserved fruit cans in the already  opened cans. You can otherwise see the juice corroding the can inner coating in air presence.

Soluble copper salts are usually blueish/greenish what with yellow/orange colour of juice gives green colour.

Long term usage of copper for keeping water is on the edge. It can cause chronical low dose sneaky poisoning, especially if water is treated by chlorine.

$$\ce{Cu + Cl2 -> Cu^2+ + 2 Cl-}$$

Oxidation can lead to partially soluble basic copper carbonates, soluble by traces of organic acids, related to minerals malachite and azurite.

$$\ce{2 Cu + CO2 + O2 + H2O -> CuCO3 \cdot Cu(OH)2}$$

$$\ce{3 Cu + 2 CO2 + 3/2 O2 + H2O -> 2 CuCO3 \cdot Cu(OH)2}$$

Plus, there is danger some people would want to store there other liquids than water as well, as the OP.


Conceptual tussle with kinetics of equilibrium and non-equilibrium processes

OK, I think I am correct about this. If not then I suppose somebody will come along and correct me anyway.

The author here is referring to the net rate being infinitesimal. To understand where the infinitesimal comes from, we need to figure out what is being meant by an "equilibrium process", or a "reversible process". On page 43, Levine describes a reversible process as

  one where the system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.

Often these are exemplified using thermodynamics processes, e.g. expansion/compression of a gas, but since we're talking about chemical kinetics, we could use an example where the temperature of the system is being increased in a reversible fashion. This means that the temperature is being increased in infinitesimal steps of $\mathrm{d}T$. The total change in temperature of the process is calculated via integration:

$$\Delta T = \int_{T_1}^{T_2}\mathrm{d}T = T_2 - T_1$$

and likewise for other properties. If something (e.g. the pressure $p$) depends on temperature, i.e. $p = p(T)$, then we can find the overall change in pressure

$$\Delta p = \int_{p(T_1)}^{p(T_2)} p(T)\,\mathrm{d}T$$

If the system is at equilibrium ($Q = K$), then an infinitesimal increase in temperature leads to an infinitesimal change in the equilibrium constant $K$, and hence an infinitesimal change in $Q$ (essentially this is due to Le Chatelier's principle, but more properly it is explained by the system moving in the direction of decreasing Gibbs free energy). This leads to an infinitesimal change in the concentrations of the reacting species, and hence the overall rate of reaction is infinitesimal.

If nothing is changing, the system would still be considered as being at equilibrium. The net rate in this case is, of course, zero. However, such a system is not particularly interesting, as there is no process actually occurring.

So, the bottom line is that for a system in equilibrium, its properties must be changing at most infinitesimally.


Demonstrating orthonormality for wave functions for ground state and excited state of He

First prove a well-known fact that any two spin orbitals $\psi_i$ and $\psi_j$ are orthogonal if the corresponding spatial orbitals from which they are constructed are themselves orthogonal.

For two spin orbitals constructed from two different spatial orbitals orthogonality follows from orthogonality of spatial orbitals regardless of spin components.
For two spin orbitals constructed from the same spatial orbital orthogonality follows from orthogonality of spin functions.

Then you can work with spin orbitals to (re)establish a well-known result that for two Slater determinants differ in just one spin orbital, the overlap integral $\langle \Psi_0 \vert \Psi_1 \rangle$ is zero.
$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) & \psi_2(1) \\
\psi_1(2) & \psi_2(2)
\end{vmatrix} \, ,
\quad
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) & \psi_3(1) \\
\psi_1(2) & \psi_3(2)
\end{vmatrix} \, ,
$$
where
$$
\psi_1(1) = \phi_1(1) \alpha(1) \, ,
\quad
\psi_2(1) = \phi_1(1) \beta(1) \, ,
\quad
\psi_3(1) = \phi_2(1) \beta(1) \, .
$$
Algebraically, it is very easy to show that,
$$
  \langle \Psi_0 \vert \Psi_1 \rangle =
  \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \langle \psi_1(1) \psi_2(2) - \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) - \psi_3(1) \psi_1(2) \rangle \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \psi_2(2) \vert \psi_1(1) \psi_3(2) \rangle
  - \langle \psi_1(1) \psi_2(2) \vert \psi_3(1) \psi_1(2) \rangle \\
  - \langle \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) \rangle
  + \langle \psi_2(1) \psi_1(2) \vert \psi_3(1) \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \vert \psi_1(1) \rangle
    \langle \psi_2(2) \vert \psi_3(2) \rangle
  - \langle \psi_1(1) \vert \psi_3(1) \rangle
    \langle \psi_2(2) \vert \psi_1(2) \rangle \\
  - \langle \psi_2(1) \vert \psi_1(1) \rangle
    \langle \psi_1(2) \vert \psi_3(2) \rangle
  + \langle \psi_2(1) \vert \psi_3(1) \rangle
    \langle \psi_1(2) \vert \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(c_1 \cdot 0 - 0 \cdot 0 - 0 \cdot 0 + 0 \cdot c_1 \Big) = 0 \, .
$$
In the case when spin orbitals are normalized (to one) $c_1 = 1$, but it is not required here: the overlap integral will vanish for spin orbitals that are just orthogonal and not necessarily normalized (to one) as well.


What measures of the dielectric constant mark whether the solution is ionic, polar and nonpolar?

Just compare the dielectric constant of known substances to non-polar, polar, metallic designations. For example, the dielectric constant of water is around 80.0 and water is polar. H2 has a very low constant value near 2 I believe and it is very non-polar. Metals have very large epsilon values moreover.  Rule of thumb, the larger the dielectric constant, the more polar the substance or material. In math terms, the potential felt between 2 charges is V= (q1 q2) / (4pi epsilon r) where q1 and q2 are different charges separated by r. The higher epsilon, the smaller V becomes. How charges interact with each other hence depends on the dielectric constant of the medium or material.


The phrase "retain those 2 electrons" is wrong.
Metal surface has positive potential in solution,
but this doesn't mean that lost electrons are concentrated somewhere.
Electrons don't form layers. You can think of electrons as much more "volatile" things than ions, they may discharge everywhere depending on potential, for example, to reduce Zn ions in solution.
So, surface is charged positively and hence first ionic layer is negative.


This reaction is a Wolff-Kishner reduction [Organic Chemistry Portal, Wikipedia] which accomplishes the exhaustive reduction from a ketone to an alkyl group. Usually Wolff-Kishner requires forcing conditions, but I think this substrate could react under more mild conditions. The mechanism explains why.

The mechanism begins with the condensation of hydrazine onto a ketone (details not shown) to give a hydrazone. Under basic conditions, this hydrazone is deprotonated at nitrogen to give an anionic intermediate. In this case, the negative charge can be delocalized onto oxygen, resulting in an enolate structure. Typically, the negative charge is only shared between a nitrogen and carbon, so this substrate gives a particularly stable intermediate. Protonation of the enolate at carbon gives the first C-H bond necessary to form the product. A second deprotonation at nitrogen gives a similar flow of electrons to form another enolate structure, this time with cleavage of the C-N bond and release of nitrogen gas. Another C-protonation gives the lactam precursor to diclofenac. Cleavage of the amide with hydroxide (details not shown) gives the target.


Unique Solubility Expression for Sulfide Salts

To summarize the comments, the sulfide anion is a strong base. This site (sourced from the CRC Handbook of Chemistry and Physics) lists the second $K_a$ of $\ce{H2S}$ as $1\times 10^{-19}$, which is less acidic than water ($1\times 10^{-15.7}$).

Thus the $K_b$ of $\ce{S^2-}$, which is the equilibrium constant for the reaction with water, is $1\times 10^5$.

$$\ce{S^2- + H2O -> HS- + OH-}\ K_b= 1\times 10^5$$

After wrangling the algebra, if you start with a $1\ \mathrm{M}$ solution of $\ce{S^2-}$, at equilibrium $[\ce{S^2-}]=3.16\times 10^{-10}\ \mathrm{M}$ - almost all of it is converted to $\ce{HS-}$

Compare for some other anions (assume starting with total concentration $[\ce{A-}]+[\ce{HA}]=1\ \mathrm{M}$

$$\begin{array}{c|c|c|c|c}
\ce{A-} & K_b & [\ce{A-}]_{eq} & [\ce{HA}]_{eq} & \text{% protonated}\\ \hline
\ce{S^2-} & 1\times 10^5 & 3.16 \times 10^{-10}\ \mathrm{M} & \sim1\ \mathrm{M} & \sim 100\% \\
\ce{CO3^2-} & 2.1\times 10^{-4} & 0.99\ \mathrm{M} & 1.45\times 10^{-2}\ \mathrm{M} & \sim 1\% \\
\ce{F-} & 3.13\times 10^{-11} & \sim1\ \mathrm{M} & 5.59\times 10^{-6}\ \mathrm{M} & \sim 0\%
\end{array}$$


Visually, we can only estimate the  $\mathrm{p}K_\mathrm{a}$ value to be near the middle of the indicator range, shifted to more intense colour.

The $\mathrm{p}K_\mathrm{a}$ of an indicator could be determined more precisely by  combining $\mathrm{pH}$ meter with photometry, where we would get the well known " round step function" of absorbance versus $\mathrm{pH}$.

E.g phenolphthalein has indicating range $\mathrm{pH}=8.2 - 10.0$ for reaction
$$\ce{HInd(clear) <=> H+ + Ind^-(violet)}$$

(I will not fight over the colour name. It may be magenta, carmine, purpur, dark pink -  you know the 16 colours of Windows jokes. )

Let suppose the phenolphthalein $\mathrm{p}K_\mathrm{a}$ is in the middle of the range, i.e. $\mathrm{p}K_\mathrm{a}=9.1$.
Let suppose the ratio of concentrations of clear and violet forms at $\mathrm{pH}=8.2$ is $\ce{[Ind-]/[HInd]}=X$.

The equation for indicates are the same as for any other weak acid:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind-]}}{\ce{[HInd]}}$$

If we consider the mentioned border $\mathrm{pH}$ values,
then
$$\begin{align}
8.2 &=9.1+\log X \\
10.0 &=9.1+\log \frac 1X \\
X&=0.126 \\
\end{align}$$

The problem is the relative eye sensibility to small additions  of one form of an indicator to the other form.
 We notice much easier a small amount of violet substance in clear substance, than a small amount of clear substance in violet substance.

$$\begin{align}
\mathrm{pH_{low}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{noticeble}]}}{\ce{[HInd_{abundance}]}} \\
\mathrm{pH_{high}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{abundance}]}}{\ce{[HInd_{noticeble}]}} \\
\end{align}$$

As the consequence, the $\mathrm{pH}=10.0$ is significantly closer to $\mathrm{p}K_\mathrm{a}$ than $\mathrm{pH}=8.2$, therefore

$$\mathrm{p}K_\mathrm{a} \gt \frac {8.2+10.0}{2}$$

Let suppose for now the real $\mathrm{p}K_\mathrm{a}=9.7$.

Then $\ce{[Ind-]/[HInd]}$ is

2 for pH 10.0, i.e full violet weakened by 1/3
0.03 for pH 8.2, i.e 3% of the full violet.

The similar visual shifts happen to other indicators as well.


Is there such a thing as a Boron-Carbon-Lithium alloy?

In the study Compound Formation in the Systems Lithium-Carbon and Lithium-Boron, Journal of the American Ceramic Society, Volume 50, Issue 10, October 1967, Pages 520–523

molten lithium metal was added to $\ce{B_{4.0}C}$ in proportions ranging from 7 to 71% lithium.

  a  substitution  reaction  evidently  occurs  between
  lithium  and  boron  carbide  which  leads  to  the  formation  of
  lithium  boride  and  lithium  carbide


Finding large databases of sphere packings

I have found a partial answer to finding a large database of sphere packings, thanks to the comment from permeakra!

So far I have been obtaining .cif files from the large databases of crystal structures online such as, http://www.crystallography.net/.  Once I have the .cif files I use openbabel to convert those files to .xyz files.  From here I have a table that describes a crystal's structure by the x,y, and z coordinates of each particular element for example,

\begin{array}{c|c c c}
\ce{Mg}      &   1.01054     &   1.01054   &     1.01054\\\hline
\ce{Al}      &   1.01054     & 1.01054     &   1.01054\\\hline
\ce{Al}      &   4.04217     &   4.04217   &     4.04217\\\hline
\ce{Mg}      &   4.04217     &   4.04217    &    4.04217\\\hline
\ce{O}       &   2.11390     &   2.11390    & 2.11390
\end{array}

But my problem from here is that babel has failed to return the ionic radius which I suppose can be retrieved online via wikipedia or something.  I posted another question that is more specific to .cif files which expands on this problem which can be found here, Extracting "Crystal Radii" from a CIF file.\

Limitations to this approach include that I can only find 3D sphere packings and it's a rather long and clunky procedure for obtaining lots of sphere packings.


Valency is a concept that describes the power of an element to bind to other elements. There is an official definition in the IUPAC goldbook:

  valence
  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

So in principle it is not possible to know the valency just from the electron configuration, but you have to also know the chemistry it performs. It is also noteworthy, that according to this definition all elements have only one value for it's valency.

In the first period, according to the definition, hydrogen is monovalent, i.e. $\ce{H2}$ or $\ce{HCl}$. Helium on the other hand is technically divalent, since $\ce{HHeF}$ was theoretically predicted, see wikipedia. But it is somewhat safe to assume helium is inert and therefore zerovalent.

In the second period, where the octet rule holds, it is the number of unpaired electrons in the valence shell, with the exception of carbon. Unfortunately this is also the period which behaves most different from the rest of the periodic table. So the best way here is to just learn them as exceptions: $\ce{NaCl}$ (mono-), $\ce{BeCl2}$ (di-), $\ce{BCl3}$ (tri-), $\ce{CCl4}$ (tetra-), $\ce{NCl3}$ (tri-),(1) $\ce{Cl2O}$ or $\ce{H2O}$ (di-), $\ce{FCl}$ or $\ce{F2}$, $\ce{Ne}$ is zerovalent.

Below the second row it becomes a little bit more predictable. For most main group elements it is the number of electrons in the valence shell. This obviously coincides with the highest stable oxidation state.
Some examples may illustrate that. For the first group all binary compositions of the alkaline element $\ce{A}$ of the composition $\ce{ACl}$ are known. This is also the maximum of chlorine atoms any of these elements can bear, hence they are all monovalent.
 The same applies to the earth-alkalines, $\ce{Ea}$. Here all binary chlorides of the form $\ce{EaCl2}$ are known, hence they are divalent.
For the triels $\ce{M^{(III)}}$, all of the chlorides of the form $\ce{M^{(III)}Cl3}$ are known. However not all of them are very stable, i.e. $\ce{TlCl3}$ (wikipedia). So they are trivalent.
For the tetrels, $\ce{M^{(IV)}}$, the trend continues. The binary chlorides $\ce{M^{(IV)}Cl4}$ are known, even $\ce{PbCl4}$ is quite stable below $0~^\circ\mathrm{C}$ (wikipedia).  For Flerovium only predictions have been made, but it's believed to be very unstable in its tetravalent state due to the inert pair effect.
For the pnictogens, $\ce{M^{(V)}}$, all binary chlorides up to and including antimony are known. However, also $\ce{BiF5}$ is known, hence all elements can be considered pentavalent.
For the chalcogens, $\ce{M^{(VI)}}$, all binary fluorides with coordination number six are known. Most prominent is sulfur hexafluoride, for which a chloride counterpart exists. The elements can therefore be considered hexavalent. The tetra chlorides are also stable for all elements of this group, in that case the elements can be considered tetravalent. However, more common are compounds where the chalcogens are divalent, i.e. in the respective hydrogen compounds, $\ce{H2M^{(VI)}}$.
The halogens $\ce{Hal}$ are just amazing, and once you have looked into interhalogen compounds you will most likely remember a lot of them. For chlorine, the interhalogens $\ce{ClF_{\{1,3,5\}}}$ have been reported, making it pentavalent. For bromine, the same holds true: $\ce{BrF_{\{1,3,5\}}}$. Only $\ce{BrCl}$ has been observed. However this makes bromine also pentavalent. For iodine the colourless gas $\ce{IF7}$ has been observed, making it the only heptavalent element of the series. For astatine only a few compounds are known, and the current believe is, that it is monovalent.
For the noble gases, $\ce{Ng}$, there are a few compounds known, and their valency increases throughout the group. While argon can be considered at most divalent due to the discovery of the highly unstable $\ce{HArF}$, the divalent nature of Krypton is undoubtedly proven by $\ce{KrF2}$. Xenon is the most reactive and potent noble gas and it can be considered hexavalent according to the above definition, as of $\ce{XeF6}$ does exist. However, also $\ce{XeO4}$ is known, which would make it octavalent according to a different definition.(2)

Transition metals are a whole different story and it is difficult to find a generalised rule or observation for it, since the IUPAC definition is very strict on the matter. All elements have to be looked at in detail. It will certainly also exceed the purpose of this answer.
Just to illustrate the problem have a look at manganese. There are various compounds known, such as $\ce{MnCl2}$ (divalent), $\ce{MnF3}$ (trivalent) and even $\ce{MnF4}$ (tetravalent). Binary hydrides have not yet been reported. However, there are oxides of various composition, ranging from divalent to heptavalent.

Notes

The molecule $\ce{NCl5}$, pentavalent, has been theoretically suggested, see wikipedia and references within.
A more modern interpretation of the valency concept allows elements to have more than only one value. It is therefore always depending on the compound in which it is looked at. A very brief summary can be found on wikipedia.

  The most characteristic chemical property of an element is its valency. There are numerous measures of valency each with its own area of usefulness and applicability. Simple definitions refer to the number of hydrogen atoms that can combine with an element in a binary hydride or to twice the number of oxygen atoms combining with an element in its oxide(s).

(N. N. Greenwood and A Earnshaw: Chemistry of the Elements. Second edition. Butterworth-Heinemann, 1997. Chapter 2, p. 27.)


Stability is a funny word. It gets tossed around all the time, but a molecule's stability really depends on the context you are considering it in. For example, benzene is totally unreactive towards electrophilic addition of $\ce{Br2}$ like most alkenes, but $\Delta_\mathrm{f}H^\circ(\text{benzene}) = 49.0\ \mathrm{kJ/mol}$, while $\Delta_\mathrm{f}H^\circ(\text{cyclohexane}) = -37.8\ \mathrm{kJ/mol}$.

So when we talk about resonance and stability, what are we really talking about? What this really refers to the stabilization energy associated with conjugated systems. For every p orbital that is added to $\pi$-system, it mixes in phase to create a new molecular orbital that is lower in energy than the two starting orbitals, or mixes out of phase to create an orbital that is higher in energy than the two starting orbitals, and in some instances doesn't mix at all. This is illustrated here:

You can't directly compare the relative energies of the molecular orbitals of two unlike molecules simply based on their number of resonance structures, though. Different atoms and substituents will cause deviations and make comparison meaningless, so it is really only useful to compare homologous compounds, like butane, but-1-ene, and buta-1,3-diene.


Why is UV-Vis Spectroscopy not like IR Spectroscopy

I'm collecting ideas from the excellent comments and adding some of my own:

The typical UV/Vis spectrum has a narrow range

On the low energy side, it ends where the visible spectrum ends, at about 800 nm. On the high energy side, it ends where water and quartz start absorbing too much, at about 220 nm. Compared to an IR spectrum, it has a narrow range.

Most functional groups have no electronic transitions in that range

As far as organic compounds are concerned, the most common functional group to absorb in the UV is the benzene ring, and extended conjugated systems (e.g. beta carotene) will have electronic transitions in the visible. On the other hand, most organic functional group will show vibrational transitions in the range of a typical IR spectrum.

Peak widths

The spectrum of an electronic transition is broad because it is coupled with vibrational and rotational transitions. An example is the electronic spectrum of iodine:

Source: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/13%3A_Molecular_Spectroscopy/13-06._Electronic_Spectra_Contain_Electronic%2C_Vibrational%2C_and_Rotational_Information

In solution, the bands are not resolved, so the spectrum appears as a single broad peak.


How does activated carbon bind the adsorbate?

Activated carbon works by enlarging the surface area of elemental carbon. Activated carbon can reach surface areas of several hundred square meters to several thousand square meters per gram. Also elemental carbon is fairly non reactive, non toxic and cheap.

As seen in the picture, activated carbon forms a micro structure with incredibly small pores, so that many molecules can be adsorbed (the forces involved decrease with a factor $1/r^{6}$ with distance). The adsorption itself is an application of the London dispersion force (one of the Van der Waals forces).

I don't believe chemisorption has anything to do with this.


Is 1,3 alkyl shift allowed in Cyclobutyldicyclopropylmethyl Cation?

There may be 1,3-alkyl shifts but they are not needed here. Cation 1 should expand the more strained cyclopropane ring first to form cyclobutyl cation 2. After a 1,2-hydride shift to form a tertiary cation 3, the cyclobutane ring in 3 expands to form spirocyclopentyl cation 4. Another cyclobutane ring expansion of 4 leads to cation 5 and eventually to alkene 6.


Reaction of perchloric acid with cyclopropenol

The first step is indeed 1,2-addition of the Grignard reagent to the keto group, and product A is the alcohol:

Treatment of the alcohol with the strong acid $\ce{HClO4}$ cannot yield the usual dehydration product (alkene) in this case, as the corresponding alkene would have two double bonds in an already highly strained three-membered ring, which is highly unfavorable. Instead, after protonation of the $\ce{OH}$ group and subsequent loss of $\ce{H2O}$, the carbocation:

is obtained as the final product B (with $\ce{ClO4-}$ being the counteranion). The planar cyclopropenyl cation is a Hückel aromatic compound with $(4 \cdot 0+2) = 2$ $\pi$ electrons, and delocalization of the positive charge over the two phenyl substituents can provide further stabilization.


Obtaining atomic form factor from electron density?

Here is a possible source: Modern X-ray analysis on single crystals: a practical guide by Peter Luger, De Gruyter, Inc. 2014.

The relevant excerpt describing how to estimate the density of an atom j, and with that, the scattering factor (really scattering function) is below.

So atomic scattering factors are derived by taking the Fourier transform of the computed electron density of a stationary atom (the density has spherical symmetry, so the scattering factor is a function that only depends on $s = \sin \theta / \lambda$).


Determining the rate of the kinetics of a reaction isn't done with only one measurement, or looking at only one graph.

What zero order kinetics means, is that the rate of the reaction is independent of the concentration of chemical species. To determine the kinetics of a process, you need to take a series of measurements while altering one variable, namely the concentration of one of the chemical species under investigation.

In zero order kinetics, you should observe that as you change the concentration of reactant, for example, the rate of the reaction remains constant.


Can aqua regia/royal water be produced with sources of chloride and nitrate other than hydrochloric acid and nitric acid?

Not even zinc would react with neutral nitrate + chloride, why should gold ?

Aqua regia must be strongly acidic for nitrates to have oxidative properties for oxidation of chlorides to chlorine and to oxidize gold to trace amounts of $\ce{Au^3+}$(what redox potential allows)

Mixing of acids leads to these reactions:

$$\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 H2O}$$
$$\ce{2 NOCl -> 2 NO + Cl2}$$
$$\ce{2 NO + O2 -> NO2 }$$

That also means aqua regia dissolution ability decreases with time.

The major oxidative reagent is free chlorine.  Hydrochloric acid saturated by chlorine dissolves gold as well.   Chlorides additionally help gold dissolution  by reaction $\ce{Au^3+ + 4 Cl- <=>> AuCl4^-}$, this decreases the redox potential $\ce{Au/Au^3+}$, making gold dissolution by chlorine and even by nitric acid easier.

$$\ce{2 Au + 3 Cl2 + 2 Cl- -> 2 AuCl4-}$$
$$\ce{Au + HNO3 + 4 HCl -> HAuCl4 + NO + 2 H2O}$$
$$\ce{Au + 3 HNO3 + 4 HCl -> HAuCl4 + 3 NO2 + 3 H2O}$$

See also https://en.wikipedia.org/wiki/Aqua_regia


Why does the rusting process start from the bent/squeezed portion of iron?

I think this question is same as wondering as to why a certain cancer patient has a cancer in the kidney but not on the skin? Nobody knows. In the same way, rusting is a random process as well. It can start anywhere. This iron sheet is not ultrapure iron sheet of homogeneous nature. One can see a lot of spots in troughs and crests of this sheet. Secondly you do not know the history of this sheet. Is this galvanized or not galvanized? This rusted spot may have a local impurity which can initiate a local electrochemical cell.

In formal atmospheric corrosion studies, my colleague uses flat sheets of test material and rusting pattern is totally random. They also do a scratch test by drawing a cross X on the flat coated sheet, in case they wish to test a coating. The purpose is to expose the "fresh" metallic surface. The rusting then starts from the X pattern. This image is an example of flat sheet used in ASTM coating. The rusting pattern is nothing but random design on the uncoated sheet (left most).

https://hardide.com/technical/astm-b117-corrosion-resistance/


The acid in the vinegar helps to a) dissolve substances or b) makes the substance easier to remove. The water is neutral (more or less) so won't really dissolve any stain or such


Calculating reaction enthalpy given other reactions

As enthalpy is a state function, so regardless the multiple stages of a reaction, the total enthalpy change for the reaction is the sum of all changes.This is Hess's law.
So, you can see clearly that the third reaction results from adding  the first reaction to the reverse  of the second reaction, and so is the enthalpy change.


Halogenation of alcohols in the presence of ZnCl2

You are right, $\ce{OH-}$ is a bad leaving group. But the $\ce{HCl}$ in the reaction mixture protonates it. Thus, the bad leaving group $\ce{OH-}$ is converted to the good leaving group $\ce{H2O}$. Anhydrous $\ce{ZnCl2}$ might additionally facilitate the reaction by acting as a Lewis acid and it binds the expelled water thus preventing the back reaction.


Hydrolysis of salts containing amphiprotic anion

If the degree of ionisation is equal to the degree of hydrolysis, then (assuming you take the same initial molar concentration of [HCO3-]):

Let's say the initial concentration was 'c' (with no [H2CO3] and no [CO32-]), the degree of dissociation to be α and the degree of hydrolysis to be η

At equilibrium:

[CO32-] = c (1 - α)

[H2CO3] = c (1 - η)

Since we know α = η, we can say that [CO32-] = [H2CO3]

Only because you made the assumption that degree of dissociation = degree of ionisation, do you get pH = 7.

But, since we assume water to always be in excess, even in a real-world situation, difference in pH (from 7) would not be significant.


The meaning of covalent bonds being directional is that atoms bonded covalently prefer specific orientations in space relative to one another. As a result, molecules in which atoms are bonded covalently have definite shapes.

The reason for this directionality is that covalent bonds are formed by sharing electrons between atoms, or, in other words, as you said, by overlapping the atomic orbitals of participant atoms. And usually only few patterns of overlap are possible, consequently, only few spatial arrangements of atoms are possible.

Ionic bonds are different: there is no electron sharing (or atomic orbital overlaps) and the number of anions surrounding a cation is limited by the charges of the ions, their sizes, and the efficiency of the lattice packing.


Does the outcome of aromatic nitrations depend on the concentrations of nitric acid and sulfuric acid?

Nitration occurs as electrophilic aromatic substitution and activity of substrate varies depending on exact structure of ring system and substitutes in the ring. In general electron acceptors (nitro-, acyl-, nitrile- and sulfogroups) deactivates the reaction, while electron donors (hydroxy-, aryl-, alkenyl- and to lesser extent alkylgroups) activates the reaction. As nitration occurs step by step, it will advance to dinitroarene only if mononitroarene is not deactivated too much by introduced nitrogroup, and same for other nitroarenes. However, different nitration agents have different nitration activity with 100% nitric acid in oleum being one of the strongest, usual nitration mixture (70% nitric acid + 98 % sulfuric acid) being strong, 70% nitric acid being moderate, diluted nitric acid being gentle and sodium nitrite in acidic solution being one of the gentlest. The picture is further complicated because strong nitration agents are also strong oxidizers, capable of breaking apart delicate molecules like phenols. Given that, nitration is often performed in several steps, with first 'gentle' nitration to introduce the first nitrogroup without breaking the molecule and then further nitration with stronger agents (see, for example, synthesis of styphnic acid).

Exhaustive review of the topic is beyond what a simple post can cover. For further insight into the reaction I recommend to consult a decent university-grade organic chemistry textbook, like 'March's Advanced Organic Chemistry'.


Why doesn't the carbocation form on the carbon that is attached to the ethyl group in this reaction?

1,2 -hydride shifting takes place after the formation of the tertiary carbocation. After the hydride shifting, the carbocation becomes resonance stabilised. And the chlorine adds up there(the first product that you have shown). Also, as I said the carbocation was resonance stabilised. So you can draw resonance structure of the carbocation intermediate and the positive charge goes to 4 position and then chlorine adds up there (the second product that you have shown) and thus we get the second product. Hope this helps.


A general term used in all analytical chemistry is the lower limit of detection (LoD), which is generally defined as 3 times the standard deviation of the baseline/blank. So if you calculate the $\sigma$ standard deviation of the baseline in question, and you have a signal that is over $3 \sigma$, you can generally accept that as a peak. This comes down to the so-called 68-95-99.7 rule, according to which there is only a 0.3% probability that a peak higher than the LoD is an artifact.


Transformation of acyl group to carboxyl group in biphenyl system with two acyl groups at para position

Is it correct that you are aiming to synthesize 4-(4-hydroxyphenyl)benzoic acid [58574-03-1]?

The target compound is available in one step from biphenyl using potassium peroxydisulfate $(\ce{K2S2O8})$ as an oxidant, Palladium(II)-trifluoroacetate, $\ce{Pd(OCOCF3)2}$, as a catalyst, and formic acid as the carbon source for the $\ce{-COOH}$ group. The synthesis was published in
 Org. Lett., 2004, 6, 2437-2439 (DOI).

EDIT

Limited resources and experience will probably rule out Stille-type coupling reactions too.

Alternatives pretty much depend on which 4,4'-disubstituted biphenyl are available.

A methoxyarene can be converted to the corresponding phenol by $\ce{BBr3}$
or the cheaper hydrobromic acid. The $\ce{-OCH3}$ group resists oxidation, is a +M substituent and is not acidic. All this might be helpful (see below).

In order to introduce the $\ce{-COOH}$ substituent, two approaches seem to be possible in an undergrade lab:

oxidation of a methyl group by permanganate
Grignard reaction of a bromoarene with carbon dioxide


When I took P-Chem, decades ago, one experiment we had was using a dropping mercury electrode, DME. I don't recall the details, but the set-up was fairly elaborate, especially compared to a Ag/AgCl electrode. The DME is fairly common (or was, before we decided that Hg was probably better avoided by the undergrad). The reason WHY we used it was it provided a CLEAN surface. Think about it. We wouldn't go to such great lengths if the problem of contamination wasn't significant.

All materials absorb contaminants, may chemically react with near-by chemicals (solvent, ions, gasses), and will form a charge polarization layer, especially in polar solvents like H2O. In fact some electrochemical methods rely on it. But for getting the "pure" ideal thermodynamic (reversible) numbers the large tendency of electrode surfaces to change is a real problem.

Overpotential is the term we use to describe the additional energy (voltage) we have to put in to these non-ideal systems (either at the anode or the cathode) to overcome these "side reactions". This energy is ultimately converted (mostly) to heat. Although maybe not an overpotential issue, surface area can be important, so smoothness does matter.

I can think of no area in chemistry (short of biochem) where you have to get into the weeds (details) so quickly as you do with electrochemistry. I wonder if its because chemistry is actually about electric interactions between atoms, molecules, ions, etc.? I should also mention that on other nice thing about the DME was that the surface was smooth - although I can't recall whether the surface was smooth at the atomic level, most solids are far from smooth, another problem in reproducibility.


What is the mechanism for the ozonolysis of a nitronate anion?

Probably very analogous to ozonolysis of a double bond.

A typical synthetic procedure is detailed in OrgSyn.  Note the addition of dimethylsulfide during work-up; this is common in ozonolysis reactions as it helps reduce the ozonide and also reduces any other dangerous, highly oxidized by-products.  The discussion section of the OrgSyn article says,

  "There are various methods available for transforming a nitro group
  into a carbonyl group, including the Nef reaction (strongly acidic),²
  permanganate oxidation of nitronate anions (basic, oxidative),³
  persulfate oxidation of nitronates (basic, oxidative),⁴ treatment with
  a mixture of organic and inorganic nitrite (neutral, oxidative),⁵ and
  treatment of either a free nitro compound or a nitronate anion with
  aqueous titanium(III) (neutral, reductive).⁶ Each method is limited,
  however, by poor yield, inconvenience, or lack of generality.

  With the proviso that the substrate not contain a reactive
  carbon-carbon double bond, the present ozonolysis procedure⁷ appears
  to provide a convenient and efficient method for carrying out the
  desired "

The Nef Reaction is mentioned above, it is worth taking a look at the linked article since the Nef reaction is complementary to the ozonolysis reaction of nitronates.  You may also find the "Reaction mechanism" section interesting.

  What gas is produced in this reaction?

According to the above mechanism, no gas would be generated, rather $\ce{NO3^{-}}$ is produced

  Is there any chemoselectivity in the presence of a double C=C bond in the molecule?

No, as mentioned in the OrgSyn reference, this approach can't be used when other double bonds are present.


Why doesn't natural gas combustion release only one-sixth as much carbon dioxide as coal burning?

When you do these calculations be sure to keep track of your units and the actual reactions you are considering.

Burning methane is not the same as burning the same amount of carbon plus twice the amount of dihydrogen. You need to check the specific heats of combustion and you need to make sure you keep track of the units used to measure the heat produced in the reaction (per mol, per kg or per mol of carbon dioxide?).

The heat of combustion (assuming all products are at STP) for carbon is 393 kJ/mol and for methane it is 889 kJ/mol. Both reactions produce the same amount of carbon dioxide if combustion is complete. So burning methane produces about 2.3 times as much heat per mole of carbon dioxide as burning pure carbon (anthracite coal will be close to this).


Can nucleophilic attack be faster than rearrangement?

You are asking what would happen if Something like 1-bromo-2-methylpropane were to undergo an $\mathrm{S_N1}$ reaction. The simple answer is: Your premise won’t happen.

The first step in an $\mathrm{S_N1}$ reaction is the formation of a carbocation by the leaving group leaving. A general equation for this would be:

$$\ce{R3C-Br <=> R3C+ + Br-}$$

Where this reaction has its equilibrium and thus what concentration of the carbocation can be expected, depends on the relative stabilities of reactant and carbocation (the bromide is always the same so it doesn’t matter). Now primary carbocations are so unstable (in a thermodynamic sense) that thermodynamics predicts an equilibrium far to the left for primary haloalkanes:

$$\ce{RCH2-Br <<=> RCH2+ + Br-}$$

(This reaction arrow doesn’t do it justice. It should be a humungous left-pointing arrow and a minute right-pointing one.)

For tertiary or otherwise stabilised carbocations, we can expect a significant concentration of the right-hand side of the equation at equilibrium. Thus, we have a potential carbocation concentration that can potentially react with another nucleophile in an $\mathrm{S_N1}$ reaction. But primary haloalkanes will not even allow the carbocation to form, or in a different view, will pull back the bromide to regenerate the haloalkane before any nucleophile has a chance of attacking the carbocation.

That’s why primary haloalkanes predominantly react via the $\mathrm{S_N2}$ mechanism: There simply isn’t any significant amount of carbocation present for $\mathrm{S_N1}$ or any type of Wagner-Meerwein rearrangement.


From the Institute of Minerals, Materials and Mining,

  Most aluminium foil is made from pure aluminium, but increasingly alloys are used to improve properties and reduce the thickness required.

If you want to make sure that the Aluminium foil you're using is pure, the only way would probably to ask the manufacturer. Other than that, it could be the case that the foil is either pure or alloyed, depending on how it's made.


Why does ethanol and water mixture show positive deviation from Raoult's Law?

In more generality, Raoult's law is stated as $$P_i = x_iP_\text{vap}$$ and read "the partial vapor pressure of species $i$ is its (pure) vapor pressure multiplied by its mole fraction in the liquid phase." It assumes an ideal-gas gas phase and an ideal-solution liquid phase; an ideal solution is one in which all intermolecular attractions are of equal strength. More explicitly, the $a{-}b$ interactions are of equal strength to the $a{-}a$ and $b{-}b$ interactions, where $a$ and $b$ represent different chemical species in solution.

A positive deviation from Raoult's law occurs when the partial vapor pressure is greater than would be expected from Raoult's law. From a microscopic perspective, this implies that the $a{-}b$ interactions are weaker than the $a{-}a$ and $b{-}b$ interactions.

You should see now that your teacher's statement is a general rule-of-thumb following from the concepts outlined above, but does not necessarily hold true for all situations. Here, water forms hydrogen bonds of greater strength and magnitude than does methanol, and one expects that a water-methanol mixture will therefore possess weaker intermolecular interactions than those in a solution of pure water.


Will heating diatomic oxygen enough break the O=O bonds?

Yes, adding at least 495 kJ/mol of kinetic energy one way or another (thermally, photochemically by irradiation with photons of that energy, sonication, etc.) will cause $\ce{O2}$ to dissociate into monatomic oxygen.

$$\ce{O2 ->C[energy]\ 2O}$$


Why does the graph of compressibility factor(z) vs pressure increase after reaching the minimum?

After a certain amount of pressure, the molecules start repelling each other and this leads to a rise in compressibility factor.

At high pressures, the van der Wall constant $b$ which represents the finite volume occupied by the gas molecules plays a greater role than the van der Wall constant $a$ which represents the attractive forces between the gas molecules.


Avogadro's gas Law doesn't make sense? (it does actually)

Oh it definitely makes sense since it allows us to describe a lot of systems with very simple formulas.

The problem is that this law assumes "ideal gases" (just like the ideal gas law) where atoms/molecules are "points" (no volume) and they don't interact with each other. This works well at rather low temperatures, low pressure and for gases without strong intermolecular interactions.

But you are correct, there are problems and the law doesn't work in a lot of cases. For example if you go to high pressures the volume of the indiviual atoms/molecules does play a role, since if they are bigger the available space will be smaller and the pressure will be higher than predicted by the ideal gas law.


The fallacy lies in the assumption that mole fraction and the number fraction are always the same.
For example, take $\ce{Na2SO4}$
what is the atom fraction of Na: 2/(2+1+4) = 2/7
What about the mole fraction of Na? It 2$N_A$/$N_A$(2+1+4)= 2/7
However, this "simple" formula of yours breaks down in the case of mixtures even if we mix them in 1:1 mol ratio. Calculate the mol fraction of Na ions in a mixture using your formula for $\ce{NaCl}$ and $\ce{Na2SO4}$ and see that you will not get the right answer.


Does the dipole moment increase or decrease by increasing the bond length?

Strictly speaking, you can neither decrease nor increase the bond length a slightest bit.$\dagger$ All molecules of a given compound are equal. It is not like you may grab one with tiny pincers of some sort, stretch it a little and see what happens.

You may, however, consider a molecule of a different compound; indeed, that's what you do in your question. But that's not just a different bond length; it is different atoms with different properties, and hence different everything.

Most importantly for the question at hand, iodine has much lower electronegativity than fluorine, so the effective charges on $\rm H$ and $\rm I$ in $\ce{HI}$ will be much lower than those on $\rm H$ and $\rm F$ in $\ce{HF}$. It just turns out that the increase in bond length can't compensate for this effect, hence the lower dipole moment.

$\dagger$ For those who might want to point out possible vibrational excitation or some other subtle (or not-so-subtle) effect: don't. Let's keep it simple.


What does the circle inscribed in a benzene ring represent?

The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true.

It does not switch; it just stays there, in some sense "halfway between" these structures. There is no movement of double bonds, and in fact there are no double bonds. (Also, there is no movement of electrons either. They just sit there on their molecular orbitals, which by itself is an approximation, though a pretty decent one). All $\ce{C-C}$ bonds in benzene are the same; they are neither double nor ordinary. Read the Wikipedia page on aromaticity, or just look at this picture from it:

That's what the circle stands for.


Why aren't there 8 double bonds in octasulfur (S8)?

An $\ce{S}$ atom in a molecule must have 2 bonds to be neutral and in the $\ce{S8}$ ring it needs both of these bonds to continue the chain/ring. If you introduce some double bonded S atoms into the ring each of these will have 3 bonds and will thus be positively charged. So, while reducing the number of free electron pairs you increase the charge of the compound which would be wrong because $\ce{S8}$ isn't charged. You can only use free electron pairs to form double bonds if you have some electron deficient atoms in the neighborhood which can "receive" them.


Why are isotopically pure diamonds 50% more thermally conductive than other diamonds?

From THERMAL  CONDUCTIVITY  OF  NATURAL  AND  ISOTOPICALLY  ENRICHED  DIAMOND  - EFFECT OF NEUTRON IRRADIATION, D.P. White, Department of Physics, Merrimack College, N. Andover, MA 01845:

  The thermal conductivity was calculated using the Callaway method.  This method takes into account  the  effects  of  intrinsic  three-phonon  normal  (N)  processes.    These  processes  do  not create thermal resistance directly but affect the thermal resistance by transferring phonons from frequencies where they are not scattered to frequencies where scattering is more efficient.

[...]

  In order to  explain  the  large  increase  (originally  much larger  than  expected)  in  terms  of  isotope scattering it was necessary to include the effects of N processes.  Subsequent analyses of the thermal  conductivity  of  enriched  diamond  have  included  N  processes.    Wei et.al.  used  the Callaway  expression  to  fit  experimental  thermal  conductivity  data  on  diamond  with  the  natural isotope  concentration  and  isotopically  enriched  (0.1% $\ce{^{13}C}$)  diamond  over  a  wide  temperature range (100-1000K).


This is a strange question - isn't it possible to form an aromatic enol in all 4 cases? #3 is the least obvious, but will still proceed:

Maybe I'm misunderstanding the question, vote me down if so!


What is the structure of 92 m/z fragment for aspirin

This spectrum is recorded using electron impact. That means that the molecule is positively charged by losing one electron.
Since that is a very unstable structure, the molecule starts fragmenting losing neutral fragments.
More processes occur, but these are the ones that we see in the spectrum.
The peaks are the positively charged fragments.

A fragment with m/z of 92, 93 or 94 is typical for phenolic compounds.
The aromatic $\ce{C-O}$ bond is very strong and therefore first all the rest is lost.
In this case the fragmentations are:

Loss of the acetyl group: $180-42=138$
Loss of water from the acid group: $138-18=120$
Loss of $\ce{CO}$ from the acid group: $120-28=92$
Loss of $\ce{CO}$ from the phenolic group: $92-28=64$

These are all typical fragmentations.
The height of the peaks indicate the relative stability of the positively charged fragments.


Is it possible to produce sulfuric acid at large scale by oxidizing and hydrating pyrite?

Per a source, there is an iron bacteria path acting on FeS2 to create sulfuric acid:

Thiobacillus spp. are participants of active oxidative corrosion. They oxidize a range of sulfur compounds to sulfuric acid, and the acid released from the cells may attack alloys.

My experience, however, suggests that the easiest way to produce H2SO4 is by the action of a concentrated solution of Oxalic acid (used in commercial products) on magnesium sulfate (FeSO4 may also work, but likely leads to an iron impurity).
Per a source (old Wikipedia commentary) to quote:

Relationship to oxalic acid
The dissociation of protons from oxalic acid proceeds in a stepwise manner as for other polyprotic acids. Loss of a single proton results in the monovalent hydrogenoxalate anion HC2O4−. A salt with this anion is sometimes called an acid oxalate, monobasic oxalate, or hydrogen oxalate."

As such expected reactions:
$\ce{MgSO4 + 2 H2C2O4 -> Mg(HC2O4)2 + H2SO4}$
$\ce{MgSO4 + Mg(HC2O4)2 -> 2 MgC2O4 (s) + H2SO4}$
Be careful, the created Sulfuric acid could be very strong depending on the level of hydration.
Warning: Attempting (in the presence of H2C2O4) to produce fuming sulfuric by boiling down the solution can lead to a spontaneous ejection of the acid (reported in an old thread of mine).


I believe the answer to your question is summarized well by the documentation of Psi4, an open source quantum chemistry package which can do similar calculations. I am fairly certain this is the problem you're running into but I could be wrong.

  It is important to know that PSI4, like any other quantum chemistry program, does not compute the usual enthalpies, entropies, or Gibbs free energies of formation provided by most reference books. Instead, quantum chemistry programs compute “absolute” thermodynamic properties relative to infinitely separated nuclei and electrons, not “formation” values relative to elements in their standard states. If you are computing thermodynamic differences, like a reaction enthalpy computed as the enthalpy of the products minus the enthalpy of the reactants, then these “absolute” enthalpies are perfectly valid and usable. However, they cannot be mixed and matched with enthalpies of formation from reference books, since the zero of energy is not the same. Additionally, the “thermal energies” reported in kcal/mol are the finite-temperature corrections to the electronic total energy, and not the overall thermal energies themselves. If in doubt, use the reported Total Energies in Hartree/particle.

Link to documentation

In other words, you're value may be perfectly reasonable once you use it to calculate a change in energy, but as an absolute energy both values are fairly meaningless because they both pick a reasonable, but arbitrary, zero of energy.

This is also why you're internal energy can be negative. This simply means it has a smaller internal energy than whatever the reference internal energy is. I didn't check your work, but it is possible to have a negative energy depending on the reference value.

Also, here is the very thorough documentation (with examples!) from Gaussian. They also note,

  This is generally not the same as the output from Gaussian for a calculation on an isolated gas-phase atom. These values are referenced to the standard states of the elements. For example, the value for hydrogen atom is 1.01 kcal/mol. This is ($H_{H_2}^{\circ}(298K)- H_{H_2}^{\circ}(0K))/2$, and not ($H_{H}^{\circ}(298K)- H_{H}^{\circ}(0K))/2$ , which is what Gaussian calculates.

So you're going to have the same problem. Read through there and I think it tells you how to make the correction. I think you just need to inclue some zero-point energies.


What is the name of the diagram in this question? How do you read it?

So, Bohr's diagram is a super-simplified version of this, almost to the point where it is arguably incorrect.

This diagram, which shows the orbitals of an electron, shows where an electron has a >90% chance to be. That means that if an electron is in any of the orbitals in your picture there, and you take many snapshots of the atom, it will show a higher concentration inside the orbital. You will find quite a few outside the orbital, but the biggest concentration will be inside those "bubbles".

Basically it shows where you are most likely to find the electrons around an atom.

If you google "different orbitals" you will find picture similar to that of yours. (I tried just now). Hope that clears most of it up, if not I really reccommend you to watch Khan Academy's videos on the subject. That really helped me.