Thickening and/or slowing evaporation of IPA-based ink

This is more of comment than a solution.

It has often been noted here that chemistry is an experimental science so often you just need to try things.

There are a number of considerations here. It seems like it takes on the order of 10 minutes to stamp a batch of "strips" (whatever that is...) and that:

You need a stamp with multiple "stamp faces" to stay wet that long.
You also want the ink solvent to "dry" so that the ink solvent doesn't leave a noticeable "watermark" on the paper.
You want to use a minimal amount of solvent so that the dye stamp doesn't bleed into the paper.

(1) The dye must be fairly polar to dissolve in a solution of 50/50 mix of isopropyl Alcohol (2-propanol)  and distilled water. You can look at a table of solvents with their polarities listed at this webpage. You can try other solvents that you have, or mixtures that you can make, to get a rough idea of how polar the solvent (mixture?) must be.

Reasonable to assume about 1 gram/ml for any of the solvents listed. Estimate polarity of a mixture by assuming ideal solution, ie that total solvent polarity is equal to molarity of a solvent times its mole fraction. (ie 50 wt % is not the same as 50 mole %.)
You can check to see if the ink dissolved by seeing if the solution is "clear" - in other words shine a bright light through it. If you see scattering then the dye didn't dissolve.
You can probably estimate from your work with glycerine how concentrated the dye must be to be invisible when dried, but visible under fluorescence. (I'm guess that the dry dye itself is green. So you don't want to "paint" the surface with particles of the dye. Rather you want the dry to be carried into the fibers of the paper as the solvent carrying the dye is adsorbed.)

(2) Look at other solvents that have been used for inks, for example on this webpage. Choose some solvents to test which have a polarity that works from step (1). Now look at solvents vs vapor pressure. The lower the vapor pressure the slower the solvent will be to evaporate.

Note that a mixture won't evaporate homogeneously. So with an acetone/glycerin mixture the acetone evaporates preferentially to the glycerin.
There is also an effect here in that the ink solvent is adsorbed into the paper not just sitting on the surface of the paper like a drop of such solvent on glass. So the correlation between vapor pressures and drying time would be "good" but certainly not perfect.


I know that we will get an output of concentration vs intensity but that's it.

Actually, that's not what you are getting. Typically, you run an emission spectrum first to determine the maximum of the fluorescence emission (or any other point of interest in the emission spectrum). In the excitation spectrum, you monitor this particular emission line and its intensity while varying the excitation wavelength.


Bond rotation energy comparison of aromatic compounds

Cleavage of the double bond will lead to both compounds having two rings that are aromatic, however the amount of resonance energy(hence stability) will not be the same.

Both systems will have a Cyclopentadienyl anion, but one will have a Cyclopropenium cation
while the other has a Tropylium cation . As the Tropylium cation has more resonance structures, it will have a higher resonance energy and hence greater stability than the Cyclopropenium cation. So (b) has a lower bond rotation energy.


High pressure modification to the van der Waals equation

The most rigorous way to derive corrections to the real gas equation is the virial expansion. At second order this gives Van der Waals law, but this derivation is completely general and can be pushed to higher order.

My answer assumes a basic knowledge of classical mechanics, statistical physics and analysis.

Hamiltonian of the system

We consider a system of identical particles described by the following Hamiltonian:
$$H=\sum_i\frac{\vec{p}_i^2}{2m}+\sum_{i\lt j}U(|\vec{r}_i-\vec{r}_j|)$$
If you don't know what a Hamiltonian is, you can consider it as the energy of the system.

Canonical partition function for the perfect gas

For a perfect gas, you consider that the particles are non-interacting, so that
$$U(|\vec{r}_i-\vec{r}_j|)=0,\quad\forall i,j.$$
In this particular case, life is easy! In fact you can easily compute the Canonical partition function (this is a definition)
$$Z_\mathrm C=\frac1{N!h^{3N}}\int\mathrm d^N\vec{p}\,\mathrm d^N\vec{r}\,\mathrm e^{-\beta H}$$
which, in the case of a non-interacting gas, is simply
$$Z_\mathrm{PG}=\frac1{N!h^{3N}}\int\mathrm d^N\vec{p}\,\mathrm d^N\vec{r}\,\mathrm e^{-\beta\sum_i\frac{\vec{p}_i^2}{2m}}=\frac{1}{N!}\left(\frac{2 \pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}{2}}$$
In order to compute this integral, you can simply use the definition of Gaussian integrals. This is the partition function of a perfect gas and you can now derive all the properties of this system.

Canonical partition function for an interacting gas

If you want to consider an interacting gas, your partition function becomes
$$Z_\mathrm I=\frac1{N!h^{3N}}\int\mathrm d^N\vec p\,\mathrm d^N\vec r\,\mathrm e^{-\beta\left(\sum_i \frac{\vec{p}_i^2}{2m}+\sum_{i\lt j} U(|\vec{r}_i-\vec{r}_j|)\right)}$$
The integration over momenta can be easily computed as before (using Gaussian integrals). In contrast, the integral
$$Q_N=\int\mathrm d^N\vec r\,\mathrm e^{-\beta\sum_{i\lt j}U(|\vec{r}_i-\vec{r}_j|)}$$
is particulary difficult (even for simple forms of the interacting potential $U$). We can write temporarily the partition function of the system as
$$Z_\mathrm I=Z_\mathrm{PG}Q_N=\frac1{N!}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}{2}}Q_N$$

Grand Canonical partition function for the interacting gas

Now, as you sais, the critical parameter is the density $\rho$: at low density you recover the perfect gas behavior. Therefore, it seems a good idea to develop our formalism in powers of $\rho$. It is easier to work in the Grand Canonical ensemble (you don't have to worry much about this), where the partition function is given by (this is a definition)
$$Z_\mathrm{GC}=\sum_N^\infty\mathrm e^{\alpha N}Z_\mathrm C$$
This is just a generalization of the Canonical partition function $Z_\mathrm C$. With this definition we have, for the interacting gas,
$$Z_\mathrm{II}=\sum_N^\infty\mathrm e^{\alpha N}\frac1{N!}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac{3N}2}Q_N=\sum_N^\infty\frac{z^N}{N!}Q_N,$$
where we defined for simplicity
$$z=\mathrm e^{\alpha}\left(\frac{2\pi mk_\mathrm BT}{h^2}\right)^{\frac32}$$

Relation between the Grand Canonical partition function and the equation of state

From $Z_\mathrm I$ in the Canonical ensemble we can now compute the ratio $p/(k_\mathrm BT)$. The Grand Canonical potential is defined by
$$\phi=-k_\mathrm BT\ln(Z_\mathrm{II})$$
and this Grand Canonical potential is linked to volume and pressure by
$$\phi=-pV,$$
so that you finally have a link between the Gran Canonical partition function and the pressure:
$$\frac p{k_\mathrm BT}=\frac{1}{V}\ln(Z_\mathrm{II}).$$

Since $Z_\mathrm{II}$ is a development in powers of $z$, the same is valid for $\ln(Z_\mathrm{II})/V$. This gives
$$\frac1V\ln(Z_\mathrm{II})=z+\frac{a_2}{2!}z^2+\frac{a_3}{3!}z^3+\cdots$$

Density

Using an easy argument that I will not show here (unless explicitly required), you can then write the density as
$$\rho=\frac{\langle N\rangle}V=z\frac{\partial}{\partial z}\left(\frac1V\ln(Z_\mathrm{II})\right),$$
where $\langle N\rangle$ is the average number of particles in your gas.
Therefore, for the density we can write
$$\rho=z+a_2z^2+\frac{a_3}{2!}z^3+\cdots$$

Developement of the state equation in powers of $\rho$

Since we said that $\rho$ is our critical parameter we can write
$$\frac p{k_\mathrm BT}=\rho+b_2\rho^2+b_3\rho^3+\cdots.$$
In fact the perfect gas law was the first-order expansion with respect to the density. For a diluite gas, $\rho \ll 1$ and therefore we simply get
$$\frac p{k_\mathrm BT}=\rho.$$

Virial coefficients

It is now easy to link the coefficients $a$ to the coefficients $b$, by equating the powers of $z$. In our case you find:
$$b_2=-\frac{a_2}2,\qquad b_3=-a_2^2-\frac{a_3}3.$$
It now remains to link these coefficients to $Q_N$. By definition of $Z_I$ we can write
$$\ln(Z_\mathrm I)=\ln\left(1+Q_1z+Q_2\frac{z^2}{2!}+Q_3\frac{z^3}{3!}+\cdots\right)$$
By Taylor expansion of the logarithm we get
$$\ln(Z_\mathrm I)=Q_1+Q_2\frac{z^2}{2!}+Q_3\frac{z^3}{3!}-\frac{1}{2}\left(Q_1z+Q_2\frac{z^2}{2}\right)^2+Q_1^3\frac{z^3}{3}+\cdots$$
For $a_1$ and $a_2$ we have
$$a_1=\frac{Q_1}V\qquad a_2=\frac1V\left(\frac{Q_2}2-\frac{Q_1^2}2\right)$$
which gives
$$b_2=\frac1{2V}\left(Q_1^2-Q_2\right).$$

Computation of $Q_i$

Now, if we compute $Q_1$ and $Q_2$ explicitly (which is much easier than computing $Q_N$ for a general $N$) we get
$$Q_1=\int\mathrm d^3\vec r=V$$
$$Q_2=\int\mathrm d^3\vec{r}_1\,\mathrm d^3\vec{r}_2\,\mathrm e^{-\beta U(|\vec{r}_1-\vec{r}_2|)}=Va+V^2$$

Van der Waals law

With the expression for the virial coefficients and the analytical expressions of $Q_1$ and $Q_2$ we finally write
$$b_2=-\frac a2$$

Therefore we finally get Van der Waals law
$$\frac p{k_\mathrm BT}=\rho-\frac a2\rho^2.$$

At the second order in $\rho$ we get Van der Waals law, but this procedure is completely general and can be used to compute higher order corrections to the perfect gas law. Computing the coefficients $Q_N$ is quite cumbersome, but can be performed using the cluster expansion.


Here is a picture and an equation that may add to the comment LordStryker made.  Molecules vibrate because they have energy. First the equation, as $T \to 0$ (absolute zero), the first term will vanish, but the molecule will still have an energy given by $h\nu/2$.  So even at absolute zero a molecule will have some vibrational energy, this is called the "zero point energy".  The parabola (representing the hydrogen well, for example, that two hydrogen atoms will "fall into" when they bond and become a stable hydrogen molecule) gives a pictorial representation of why this must be - energy levels are quantized.  Even the lowest level is above the bottom of the parabola, so even in the lowest energy level at absolute zero, the atom or molecule will still have a non-zero energy which will cause it to vibrate.

$\displaystyle\varepsilon = \frac{h\nu}{\mathrm e^{h\nu/kT} - 1} + \frac{h\nu}{2}$


DOSY-NMR: How to calculate the volume of the molecule from the diffusion coefficient?

You can try and use the Stokes-Einstein equation for the diffusion coefficient $$D=\frac{k_\mathrm{B}T}{\zeta}$$ where the friction term is $$\zeta = 6\pi\eta r$$ and $k_\mathrm{B}$ is the Boltzmann constant, $\eta$ the solvent viscosity $(1\ \mathrm{cP} = 10^{-3}\ \mathrm{kg~m^{-1}~s^{-1}} = 10^{-3}\ \mathrm{Pa~s})$ and r the molecule's radius . You only have one number in diffusion constant so you can only get radius or volume assuming the molecule is a sphere.

The model assumes that the molecules are larger than the solvent molecules, i.e. solvent is effectively continuous not molecular. There are slightly different expressions for the friction, 4 instead of 6 for example depending on whether 'slip' or 'stick' limits are assumed.


Why does water become less dense when it cools

Under normal conditions (meaning your regular everyday freezer, or a cold day) water forms a crystal lattice below its melting point (as shown below). This structure is held together by hydrogen bonds, which are no longer overcome by the internal energy of the molecules. The hexagonal arrangement of the water molecules in this structure leads to a lower density due to the spaces left in the center of each hexagon (you can now fit fewer molecules in the same volume).

From Wikipedia

There are many other solid phases of water (at least 16 on wikipedia), some which have densities higher than normal liquid water, though most are not naturally formed (at least on Earth).


If it is so, then why do we call glowing objects in the dark as fluoroscent?

I think this is just marketing: most people have heard the term fluorescence, while phosphorescence or luminescence are largely unknown terms.

While a delayed emission from a singlet state is possible, most glow-in-the-dark items do in fact show phosphorescence, i.e. emission from a tripet state.


Why does delocalization lead to stability?

Electrons "want" two thing: be close to a nucleus and have the freedom to move. Also, being fermions, they are "forbidden" by the Pauli exclusion principle to have exactly the same state (e.g. same set of quantum numbers in an atom) as another one.

The electron distribution in a molecule is a compromise of the two "wants" while not trying anything "forbidden". In the $\ce{H2}$ molecule, electrons get to enjoy the attraction of two nuclei while having more space available than when bound to an isolated hydrogen atom. Because of the Pauli exclusion principle, only two electrons get to enjoy that low energy.

Electrons associated with conjugated double bonds also have "more room" without being further away from the nuclei than electrons associated with isolated double bonds.

Sorry for the anthropomorphizing... for a more rigorous treatment, look up particle in a box in a quantum chemistry book.


How to disarm and correctly dispose of tear gas spray?

Do not open, do not try to neutralize it yourself! Bring it to Recycling center (DE: Wertstoffhof). If your city is 100k+ there should be one. That's the place where you bringing your old furniture, the rubbish after renovations, and so on. They have a part for disposal of hazardous waste (DE: Sondermüll). What usually comes in there are paints, old car batteries, old pesticides, etc.... Another possibility - if you have some friend at chemistry school of some local university, give it him. Each Chemistry faculty has some "disposal" facility.


When to use chromatographic separation vs vacuum-assisted rectification distillation?

The decision to use a particular separation technique industrially is based on multiple factors. To speak generally, this comes down to ease of separation and process economics. The simplest and cheapest techniques would be ones like biphasic separation or filtration.

The distillation you refer to bears more relation to bulk chemical manufacture such as oil refining.  Process HPLC and simulated moving bed (SMB) are more involved techniques,  requiring more sophisticated technology and hence, is a more expensive technique. This puts it into the realms of higher value products which would tend to be pharmaceuticals or ultra high purity display chemicals.

So, for process economics, if we think in terms of Input-Process-Output we have cost of input material,  the cost of processing and the cost of output material.  If Input + Processing are greater than the ultimate commercial cost that the product can bear, then you don't have a product or it would need a key differentiator (which could be purity), to justify the extra cost.

If the costs justify it, then using higher vacuum can offset degradation risk involved in a fractional distillation. Physicochemical factors come into play to select this technique (under the ease of separation banner). Industrial distillation would tend to be used for more complex mixtures of desired volatile and semivolatile components.  A good Quora discussion on the topic is here:

https://www.quora.com/How-are-rectification-and-distillation-related

The technique, obviously relies on boiling point and exploitation of potential synergies of compound (usually liquid/oil) mixtures.

Chromatographic separation relies on adsorption/desorption by judicious selection of stationary phase and solvent (mobile phase). For commercial processes, this needs to be repeated over and over in a completely reproducible way with minimal variation. If the stationary phase is cheap, then less material has to go through the material before being changed. If the stationary phase is expensive (decent spherical functionalise silica can easily be over £2000/kg, chiral phases over 10-20k per kg. Irregular silica can be £500/kg), then you want a faster,  cleaner separation with minimal fouling of the support. Over 200 injections onto the same stationary phase is not unheard of.

These injections would also be maximised on production scale vs the standard analytical technique.
For SMB,  typically four HPLC columns are used and a more sophisticated algorithm is required to inject the right amount of material onto the correct column at the right time.

As commented on, isomeric separation,  particularly if the materials are solids,  then  favours process HPLC rather than high vacuum distillation. But again, the process economics would help decide this.

Other factors are internal expertise in the relevant techniques and/or capability of outsourcing partners (contract manufacturing organisations) in more niche techniques. SMB would probably fall into this category i.e. most manufacturing companies would look to a specialist 3rd party unless they were already engaged with this technology.


Are phenyl and allyl groups in prop-2-enylbenzene coplanar?

Probably not
Part of the problem with the question is geometry. Any two points always form a plane. What I suspect you mean is "is the allyl double bond in the same plane as the benzene ring?" in which case the answer is probably not.
To make the situation clearer here is a different numbering of the atoms in the allyl benzene molecule:

The question I think you intended is whether the 2* and 3* bond is in the same plane as the benzene ring.
There is no chemical or geometric reason why it would be. the 1-1* bond and the 1-2 bonds will freely rotate so there is no bond-related force driving the orientation of the 2-3 double bond. to illustrate here is a 3D view of one possible configuration:

I say one possible configuration as it is likely that the barrier to rotation of those two bonds is small and, at room temperature in the liquid, the molecule will explore all the possible configurations (it might be different in a solid crystal where packing forces between units might lead to one being favoured).
In short: there is no reason to suppose the double bond and the benzene ring will be in the same plane because the two single bonds connecting it to the ring will freely rotate.


In the E2 reaction, the size of the base can affect the selectivity of the deprotonation, because the steric interaction of a large base with a hindered proton will go over a higher transition state than the interaction of a large base with a less-hindered proton. This is kinetic control.

I'm not aware of any practical case where the selectivity of an E1 reaction can be controlled by the size of the base. The reason is that the deprotonation of the carbocation intermediate will have a transition state that is very close in energy to the carbocation itself. I agree with you that the size of the base should have some influence on the transition state energy, but in practice, it seems that it has less influence than the stability of the resulting alkene.

Also note that often E1 is run under equilibrating conditions, where even if a less-substituted (less-stable) alkene is formed, it can be protonated to reform the carbocation, and the material will funnel to the most stable product. This is thermodynamic control.


Is there any law for pH conservation in chemical reactions?

No, there is no law requiring pH conservation. And such a law would be chemical nonsense. Why? Read on.

You need to recall what pH actually is. It is defined as the negative base-10 logarithm of the activity (similar enough to concentration for almost all practical purposes) of positively charged hydrogen ions in solution.† Or in equations:

$$\mathrm{pH} = -\lg [\ce{H+}]$$

Note that most conservation laws in physics somehow relate back to the conservation of energy or something related. However, the pH of a solution cannot be easily traced back to any state function or anything else that would warrant its conservation.

In fact, the only thing conserved is the atom itself. Hydrogen, if $\ce{H+}$ is somehow present in a reaction or solution, does not disappear, it needs to be put in on one side of the equation and turn up somewhere else. In reactions that liberate protons of any kind, usually bases are added to capture them. In those that require protons, acids are added. What (if anything) is conserved is the number of hydrogen atoms but if they are not active as $\ce{H+}$ in solution they do not count towards pH. Therefore, there is no sense in postulating pH conservation.

†: $\ce{H+}$ as is does not actually exist in solution. One can treat it as $\ce{H3O+}$ for all intents and purposes, but actually it is better to think of it as a proton being shuffled back and forth by multiple water molecules. I think I recall four being stated as the lowest count in undergrad education.


How could you predict degenerate pi molecular orbitals just knowing what symmetry point group a molecule belongs to?

Which orbital symmetries you can get in a molecule always depends on the molecule’s point group. Only for those point groups which have degenerate irreducible representations can you get degenerate orbitals; so out of your examples only the $D_{4h}$ molecule could. This is entirely independent of Hückel’s method to determine energy levels.

But just because your molecule has the irreducible representations, doesn’t mean that a $\pi$ orbital has to have that degeneracy. Instead, it again depends on what you get when following symmetry rules.

You can calculate whether you will get $\pi$ degeneracies by drawing your molecule of interest, including all the p atomic orbitals and determining the molecule’s point group. Once you have that, for each symmetry operation, check how many p orbitals would be transformed onto themselves by said operation (and note whether with or without a phase change). Add that up. Once you’ve done that, you basically have the representation of the p-orbitals of the molecule that you can reduce to irreducible representations, finding out how many discrete and degenerate energy levels you have (albeit not including energy values).


In quantum chemistry, probably the easiest way to understand the "exchange interaction" is within the context of the Hartree-Fock model.
$
\newcommand{\op}{\hat}
\newcommand{\el}{_\mathrm{e}}
\newcommand{\elel}{_\mathrm{ee}}
\newcommand{\elnuc}{_{\mathrm{en}}}
\newcommand{\core}{^{\mathrm{core}}}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
$

To reduce the complexity of the electronic Schrödinger equation
$$
    \op{H}\el \psi\el(\vec{q}\el) = E\el \psi\el(\vec{q}\el) \, ,
$$
where
$$
    \op{H}\el
    =
    \op{T}\el + \op{V}\elnuc + \op{V}\elel
    =
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
we would like to separate the electronic coordinates from each other by writing down the many-electron wave function as a product of one-electron ones. Unfortunately, such separation won't work due to the presence of $\op{V}\elel$ term in the electronic Hamiltonian. But if instead of $\op{V}\elel$ potential, which prevents the separation of electronic coordinates, a model potential of the form $\sum\nolimits_{i=1}^{n} v_{\mathrm{MF}}(\vec{r}_{i})$ had entered the electronic Schrödinger equation, it would be reduced to a set of $n$ one-electron Schrödinger equations with the many-electron wave function being just a simple product of their solutions, one-electron wave functions $\psi_{i}(\vec{r}_{i})$.

More precisely, since we have to take spin of electrons and the principle of antisymmetry of the electronic wave function into account, the electronic wave function would be an antisymmetric product of one-electron wave functions $\psi_{i}(\vec{q}_{i})$ termed the Slater determinant,
$$
    \Phi
    =
    \frac{1}{\sqrt{n!}}
    \begin{vmatrix}
        \psi_{1}(\vec{q}_{1})  &  \psi_{2}(\vec{q}_{1})  &  \cdots  &  \psi_{n}(\vec{q}_{1})  \\
        \psi_{1}(\vec{q}_{2})  &  \psi_{2}(\vec{q}_{2})  &  \cdots  &  \psi_{n}(\vec{q}_{2})  \\
        \vdots                 &  \vdots                 &  \ddots  &  \vdots                 \\
        \psi_{1}(\vec{q}_{n})  &  \psi_{2}(\vec{q}_{n})  &  \cdots  &  \psi_{n}(\vec{q}_{n})
    \end{vmatrix} \, ,
$$
where the one-electron wave functions $\psi_{i}$ of the joint spin-spatial coordinates of electrons $\vec{q}_{i} = \{ \vec{r}_{i}, m_{si} \}$ are referred to as spin orbitals. Physically the model potential of the above mentioned form represent the case when electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons and given by $v_{\mathrm{MF}}(\vec{r}_{i})$.

Minimization of the electronic energy functional
$$
E\el^{\mathrm{HF}} = \bracket{ \Phi }{ \op{H}\el }{ \Phi } = \bracket{ \Phi }{ \op{T}\el }{ \Phi } + \bracket{ \Phi }{ \op{V}\elnuc }{ \Phi } + \bracket{ \Phi }{ \op{V}\elel }{ \Phi }
$$
with respect to variations in spin orbitals subject to constraint that spin orbitals remain orthonormal eventually results in the set of the so-called canonical Hartree-Fock equations which define canonical spin orbitals $\psi_i$ together with the corresponding orbital energies $\varepsilon_i$
$$
    \op{F} \psi_{i}(\vec{q}_{1})
    =
    ε_{i} \psi_{i}(\vec{q}_{1}) \, ,
    \quad
    i = 1, \dotsc, n \, ,
$$
where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator with
$$
    \op{H}\core \psi_{i}(\vec{q}_1)
    =
    \Big(
        - \frac{1}{2} \nabla_{1}^2
        - \sum\limits_{\alpha=1}^{\nu} \frac{Z_{\alpha}}{r_{\alpha 1}}
    \Big)
    \psi_{i}(\vec{q}_1) \, , \\
    \op{J}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{j}(\vec{q}_2) } \psi_{i}(\vec{q}_1) \, , \\
    \op{K}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{i}(\vec{q}_2) } \psi_{j}(\vec{q}_1) \, .
$$

Here, the first part of the Fock operator, $\op{H}\core$, is known as the one-electron core Hamiltonian and it is simply the Hamiltonian operator for a system containing the same number of nuclei, but only one electron. The second part of the Fock operator, namely, $\sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$, plays the role of the mean-field potential $v_{\mathrm{MF}}$ that approximates the true potential $\op{V}\elel$ of interactions between the electrons. The first operator $\op{J}_{j}$ rewritten in the following form
$$
    \op{J}_{j}(\vec{q}_{1})
    =
    \int \psi_{j}^*(\vec{q}_{2}) r_{12}^{-1} \psi_{j}(\vec{q}_{2}) \,\mathrm{d} {\vec{q}_{2}}
    =
    \int \frac{ |\psi_{j}(\vec{q}_2)|^{2} }{ r_{12} } \mathrm{d} \vec{q}_2 \, ,
$$
can be clearly interpreted as the the Coulomb potential for electron-one at a particular point $\vec{r}_{1}$ in an electric field created by electron-two distributed over the space with the probability density $|ψ_{j}(\vec{q}_2)|^{2}$, and for this reason it is called the Coulomb operator.
The second operator $\op{K}_{j}$ has no simple physical interpretation, but it can be shown to arise entirely due to anti-symmetry requirement, i.e. if instead of a Slater determinant one uses a simple product of spin orbitals, termed the Hartree product, there will be no $\op{K}_{j}$ terms in the resulting equations (the so-called Hartree equations). It is for that reason that $\op{K}_{j}$ is called the exchange operator.

To quickly understand why exchange terms appear when using a Slater determinant instead of a simple product of spin orbitals look no further than at the Slater rules. For $\op{V}\elel$ which is a two-electron operator we have

$$
    \bracket{ \Phi }{ \op{V}\elel }{ \Phi }
    =
    \sum\limits_{i=1}^{n} \sum\limits_{j>i}^{n}
    \Big(
        \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{i}(1) \psi_{j}(2) }
        -
        \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{j}(1) \psi_{i}(2) }
    \Big) \, ,
$$
where the second (exchange) part would be absent if $\Phi$ were a simple product of spin orbitals, rather than a Slater determinant.1)

To the question on why do we interpret exchange terms in the Hartree-Fock as a manifestation of some sort of interaction. On the one hand, it must be clear from exchange terms $\op{K}_{j}$ being part of potential energy terms $v_{\mathrm{MF}}$, that they are indeed related to some type of interaction between electrons. However, it must be said that, unlike for the four fundamental interactions, there exists no force based on exchange interaction. Strictly speaking, I don't even think that the term "interaction" has a definite meaning in physics, unless it simply stands for the "fundamental interaction" in which case it is merely a synonym for "fundamental force".

Exchange interaction is not one of the four fundamental interactions, so the meaning of the word "interaction" here is a bit different than that for fundamental interactions. Oxford Dictionary defines interaction as follows,

  1.1 Physics A particular way in which matter, fields, and atomic and subatomic particles affect one another, e.g. through gravitation or electromagnetism.

And exchange interaction indeed is a way by which electrons (of the same spin) "affect one another". This can be readily seen as follows. Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events:

$r_1$ - the event of finding electron-one at a point $\vec{r}_{1}$;
$r_2$ - the event of finding electron-one at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by
$$
    \Pr(r_1 \cap r_2)
    =
    \Pr(r_1\,|\,r_2) \Pr(r_2)
    =
    \Pr(r_1) \Pr(r_2\,|\,r_1) \, ,
$$
where

$\Pr(r_1)$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
$\Pr(r_2)$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
$\Pr(r_1\,|\,r_2)$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
$\Pr(r_2\,|\,r_1)$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.

If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts
$$
    \Pr(r_1\,|\,r_2) = \Pr(r_1) \, , \quad
    \Pr(r_2\,|\,r_1) = \Pr(r_2) \, ,
$$
and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,
$$
    \Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2) \, .
$$
In reality, however, the events $\vec{r}_{1}$ and $\vec{r}_{2}$ are not independent of each other, because electrons "affect one another",
$$
    \Pr(r_1\,|\,r_2) \neq \Pr(r_1) \, , \quad
    \Pr(r_2\,|\,r_1) \neq \Pr(r_2) \, ,
$$
and consequently the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ is not equal to the product of unconditional probabilities,
$$
    \Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2) \, .
$$

Inequalities above hold true for two reasons, i.e. there are two ways electron "affect one another".

First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons $\Pr(r_1\,|\,r_2) < \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) < \Pr(r_2)$, while at large distances $\Pr(r_1\,|\,r_2) > \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) > \Pr(r_2)$.
In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.

Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.
This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the latter, it is still noticeable when electrons are close to each other and not just at the very same location in space.

Now, for a Hartree product wave function
$$
  \psi_{\mathrm{HP}}(\vec{q}_{1}, \vec{q}_{2}) = \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) \, ,
$$
one can show that $\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ if two electrons are in different spin states as well as when they are in the same spin state2). But for a Slater determinant
$$
    \Phi(\vec{q}_{1}, \vec{q}_{2}) = \frac{1}{\sqrt{2}}
    \Big(
    \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) -
    \psi_{1}(\vec{q}_{2}) \psi_{2}(\vec{q}_{1})
    \Big) \, ,
$$
$\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ will hold only for electrons of unlike spin, while when both electrons are in the same spin state $\Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2)$2). And this inequality is a clear indication of exchange interaction.

1) I leave it to OP as an exercise. Derivation for a simple product of spin orbitals is pretty trivial, while for the case of a Slater determinant OP can consult, for instance, Szabo & Ostlund, Modern Quantum Chemistry, Section 2.3.3.

2) I leave it as another exercise.


Electron Transfer in Organic Reactions: Why Does Brown Iron Oxide Improve Burn Rate of Rocket Candy?

First of all, you're thinking of one of the two definitions of oxidation—breaking bonds in molecular oxygen and rejoining them in a different way in another molecule.

We need the electron transfer definition of oxidation however.

First off, to see what is happening we need to look at how many electrons are being transferred. In glucose the oxidation state of carbon is $\ce{0}$, whilst in carbon dioxide it is $\ce{+4}$. So during our rocket burning process, each carbon gives up four electrons. But what is it giving them to?

If we use molecular oxygen as our accepting species—our oxidisation agent—we have the following:

$$ \ce {6O_2} + \ce {C6H12O6}  \ce{->} \ce{6CO2} + \ce{6H2O} $$

With oxygen here going from an oxidation state of $\ce{0}$ in $\ce{O_2}$ to $\ce{-2}$ in both $\ce{H2O}$ and $\ce{CO2}$.

So if each $\ce{O2}$ molecule  is picking off two electrons and this is all it's doing, then why can't we replace this with two lots of $\ce{Fe^3+}$ to form $\ce{Fe^2+}$?

Before I write this in equation form, it's worth noting that organic electrochemical equations approximate organic matter (OM) with formaldehyde, $\ce{CH2O}$ since OM is biologically equivalent to a polymer (albeit complex ones!) of $\ce{CH2O}$. Note that glucose is therefore equivalent to six lots of this monomer. This simplifies writing balanced chemical equations a lot:

$$ \ce{2Fe_2O_3} + \ce {CH2O}  \ce{->} 4\ce{FeO} + \ce{H2O} + \ce{CO2} $$

Now molecular oxygen is by far a better oxidisation agent than $\ce{Fe_2O_3}$ however in fine powdered form, these iron oxide molecules are in immediate contact with your OM; we do not need to wait for molecular oxygen to be used up and for more to diffuse in.

So I'm warranting that red iron oxide ($\ce{Fe_2O_3}$) can be used as well as brown iron oxide (I think this means $\ce{FeO}$). The difference here is that in the former reaction each $\ce{Fe^3+}$ only captures one electron from your OM to form $\ce{Fe^2+}$, whilst in the latter case you're moving from $\ce{Fe^2+}$ to $\ce{Fe}$ metal which captures two electrons—so you need less. Note that formation of iron from $\ce{Fe^2+}$ will need really high temperatures or pressures, which is what you note in your question.

This is a super question in terms of getting people thinking about redox not just as a process which happens with ions, but with entire organic species. Such organic reactions can be characterised by a reaction voltage, just like the other types of electrochemical reaction you've been looking at. This is actually what happens in the field of biogeochemistry, where scientists look at which types of oxidisation agent are responsible for degrading OM in different environments. More can be found about this in this paper here.


Why does the hydroxide ion have a negative charge?

A water molecule is charge neutral because there is the same number of positive
charges as there are negative charges.

In this diagram, called a Lewis structure, the dots represent electrons while the lines or dashes represent a covalent bond of two electrons.

When water ionizes one of the hydrogen atoms absconds with itself and leaves
it's electron behind, giving us the hydroxide ion. The extra electron gives hydroxide a net charge of -1.

The brackets indicate that this is an ion, charge is denoted at top right.

To go deeper down the rabbit hole on this one I recommend reading up on the Octet rule and Electronegativity.


This is an answer to a question of mine, which I posted a few days ago, Deprotonated form of phenolphthalein.

I found this paper (it's in Japanese, but the relevant things are legible) https://www.jstage.jst.go.jp/article/yakushi1947/117/10-11/117_10-11_764/_pdf, which says that phenolphthalein has 2 pKas, pKa1 = 9.05 and pKa2 = 9.50. It also says that the pink form is the twice deprotonated; the monoanionic form exists, but is colourless.


In the Nernst equation, does T refer to the temperature of the standard cell or the cell under non-standard conditions?

The Nernst equation is $\displaystyle E=E^\text{o}-\frac{RT}{nF}\ln(Q)$ where $Q$ is the ratio of activities, $a$, of products over reactants, e.g. for $aA+B\leftrightharpoons cC+D$ then $\displaystyle Q=\frac{a^c_Ca_D}{a^a_Aa_B}$. The emf of the cell is $E^\text{o}$ which is measured when the emf of the cell is zero, $E=0$, no current flows, and means that the reactants and products are at equilibrium. In this case $\displaystyle E^\text{o} =\frac{RT}{nF}\ln(K_e)$ where $K_e$ is the equilibrium constant.

The measured emf $E$ changes with temperature as indicated by the Nernst equation and this is a good way to measure thermodynamic properties; for example the reaction free energy is  $\Delta G = -nFE$ and because $\partial\Delta G/\partial T|_p=-\Delta S$ then $\displaystyle \Delta S = \frac{\partial E}{\partial T}\bigg|_p nF$ which is a convenient way to determine the entropy change in a reaction by measuring the reaction emf at two temperatures and letting the derivative be $\Delta E/\Delta T$, which will be accurate for a small change in temperature, and give $\Delta S$ at the mean of the two temperatures. Since $\Delta G = \Delta H -T\Delta S$ the change in enthalpy can also be calculated.


Synthesis of chloroquine from 4-amino-7-chloroquinoline

My intuition tells me that the nitrogen in the ring is more nucleophilic than the exocyclic amine. A quick-and-dirty electrostatic potential calculation using B88LYP/6-31G(d) on the PM3 geometry shows a large electron density (blue and violet) on the ring nitrogen and lower density (yellow) on the exocyclic amine:

Are there conditions that might work? Yes. Should you try it, if you need that molecule? Yes. Will it be difficult? Yes.


I would start with guessing which one represents the "off" state. I would say (D), since the temperature slopes evenly from one air interface to the other, and is fairly constant in the heater layer. Based on this diagram you can also conclude as you did that layer 3 has the lower conductivity. And the slope in layer 2 indicates that it has a finite (not infinite) conductivity higher than that of 1 or 3.

Based on observations made in (D), you can conclude also that (B) represents the "on" state. As you explain, (C) is not sensible unless heating originates at the interface between layers 1 and 2. (A) also does not make sense, because it suggests that there is a heat sink between 1 and 2 (where is that heat going?).


That's a good, concise statement of Bent's rule.  Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all.

Bent's rule speaks to the hybridization of the central atom ($\ce{A}$) in the molecule $\ce{X-A-Y}$.

$\ce{A}$ provides hybridized atomic orbitals that form $\ce{A}$'s part of its bond to $\ce{X}$ and to $\ce{Y}$.  Bent's rule says that as we change the electronegativity of $\ce{X}$ and \ or $\ce{Y}$, $\ce{A}$ will tend to rehybridize its orbitals such that more s character will placed in those orbitals directed towards the more electropositive substituent.

Let's examine how Bent's rule might be applied to your example of methyl fluoride.  In the $\ce{C-F}$ bond, the carbon hybrid orbital is directed towards the electronegative fluorine.  Bent's rule suggests that this carbon hybrid orbital will be richer in p character than we might otherwise have suspected.  Instead of the carbon hybrid orbital used in this bond being $\ce{sp^3}$ hybridized it will tend to have more p character and therefore move towards $\ce{sp^4}$ hybridization.

Why is this?  s orbitals are lower in energy than p orbitals.  Therefore electrons are more stable (lower energy) when they are in orbitals with more s character.  The two electrons in the $\ce{C-F}$ bond will spend more time around the electronegative fluorine and less time around carbon.  If that's the case (and it is), why "waste" precious, low-energy, s orbital character in a carbon hybrid orbital that doesn't have much electron density to stabilize.  Instead, save that s character for use in carbon hybrid orbitals that do have more electron density around carbon (like the $\ce{C-H}$ bonds).  So as a consequence of Bent's rule, we would expect more p character in the carbon hybrid orbital used to form the $\ce{C-F}$ bond, and more s-character in the carbon hybrid orbitals used to form the $\ce{C-H}$ bonds.

The physically observable result of all this is that we would expect an $\ce{H-C-H}$ angle larger than the tetrahedral angle of 109.5° (reflective of more s character) and an $\ce{H-C-F}$ angle slightly smaller than 109.5°  (reflective of more p character).  In terms of bond lengths, we would expect a shortening of the $\ce{C-H}$ bond (more s character) and a lengthening of the $\ce{C-F}$ bond (more p character).


Sodium Peroxoborate reacting with Sulfuric Acid and Potassium Permanganate

Well $\ce{KMnO4} $ and $\ce{H2O2}$ react as follows:

$$\ce{2KMnO4 +3H2O2 -> 2MnO2  + 2KOH + 2H2O + 3O2}$$

Adding the two reactions together I get:

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 ->Na2SO4 + 2H3BO3 + 3O2 + 2MnO2 + 2KOH}$$


The antibonding orbital doesn't necessarily tell you the number of unpaired electrons - it's coincidental that for O$_2$, the HOMO (highest occupied molecular orbital) is a (1$\pi_g)^2 $ orbital - or, if you prefer to write it like you have for N$_2$, ($\pi ^1 _{2px} = \pi ^1 _{2py}$).
$\pi$ orbitals are doubly degenerate, so the lowest energy configuration is for the 2 electrons to occupy different orbitals (one in each of the 2 $\pi$ orbitals). This means they're both unpaired, as shown in this MO diagram of O$_2$ from Wikipedia. For an example of an MO diagram where there are unpaired electrons in an orbital which is not antibonding, try drawing the B$_2$ MO diagram (remember that s-p mixing occurs).

In terms of what they do, antibonding electrons destabilise the bond, "cancelling out" the energetically favourable effect of bonding electrons - try drawing an MO diagram for He$_2$ or Ne$_2$. It should be apparent why they don't form stable molecules.

As far as N$_2$ goes, I think you may have neglected to take into account the effect of the $\sigma ^* _{2s} $ electrons. Remember that antibonding electrons cancel out bonding electrons.


I've searched for this same data.  My interest regards electronic proporties of zeolite-class media.  I've written this same kind of question on Quora.

I think the reason this is rarely answered, and why there are no 'catalogs', is several-fold:

There exist a wide range of applications, based on the composition
of the water / etc.
The treatment media usually varies in
composition, meaning that no directive for the use of a zeolite
product can be given by a sales rep (only on-staff engineers, who
know how to use it).
Some uses create by-products that are regulated, creating possible liabilities for the seller by indicating a fitness-for-use.
Regulations change; per year, per locale, etc.

I think the groups that know about these are environmental engineering firms, whom have a stable of engineers that provide end-to-end services for their clientele.  Many of of these firms have Chemists on-staff, or a consultancy relationship that suits the same.  They know the chemical properties of their media, and need a zeolite / catalysts that can effect something very specific.  As a result, they're out their hunting the journals for something that targets their needs.

Hard to out-sell that without already having a dog in the game.


How do I determine molality from mole fraction without knowing how much of the solution I have?

Solution 1

Let's assume there is 1 mole total of solvent and solute. With a mole fraction of 0.2 solute, there is 0.2 mole of solute, the rest (0.8 mole) is benzene. We can calculate the mass of 0.8 mole benzene from the molar mass, and express it in kilograms:

$$ m_\mathrm{benzene} = n_\mathrm{benzene} \cdot M_\mathrm{benzene}$$
$$= \pu{0.8 mol} \cdot \pu{78.11 g/mol} = \pu{62.49 g} = \pu{0.06249 kg}$$

The molality $b_\mathrm{solute}$ is defined as the amount of solute divided by the mass of solvent:

$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{0.2 mol}}{\pu{0.06249 kg}} = \pu{3.201 mol/kg}$$

Solution 2

Let's assume there is 1 kg of benzene. That corresponds to 12.80 moles of benzene. If the mole fraction of solute is 0.2 and that of the solvent benzene is 0.8, they are present in a molar ratio of 1 to 4 (for every four molecules of benzene, there is one molecule of solute). So the amount of solute in that sample is 3.201 mole. We are ready to calculate the molality:

$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{3.201 mol}}{\pu{1 kg}} = \pu{3.201 mol/kg}$$

Problems lacking extensive quantities

It is helpful to understand the difference between extensive and intensive quantities. Intensive quantities do not depend on the size of your sample (examples include density, concentration, molar mass, partial pressure, coefficients in the formula of a compound). On the other hand, extensive quantities do depend on how much you have (mass, volume, amount of substance). Often, you define intensive quantities by dividing one extensive quantity by another (example: density is mass divided by volume).

In your question, you want to determine an intensive quantity (here: molality) from intensive quantities only (here: mole fraction). There is a larger set of exercises of that type. If it helps, you can arbitrarily come up with an extensive quantity to solve the problem. You can set the mass of benzene to 1 kg or 2 kg. Then, you can solve the problem. The answer will not depend on the mass of benzene, it cancels out. Other options are setting the amount of solute or the amount of benzene.

There are common exercises asking you for intensive quantities given other intensive quantities:

What is the formula of an oxide given the mass percentages of elements?
What is the density of xenon at STP given its molar mass?
What is the concentration of chloride ions in a solution containing 0.4 mol/L $\ce{MgCl2}$?
What is the partial pressure of dioxygen at STP if an air sample contains 20% of dioxygen?

For all of these, you may assume an arbitrary amount, volume, or mass (but just one) if it helps to solve the problem. Again, the actual value does not matter because it will cancel out.

For more on intensive and extensive, see e.g. https://chemistry.stackexchange.com/a/6538


Naming of asymmetric organic molecules with two nitroxides

The structure can be redrawn somewhat more clearly:

Let's call the basic structure, where R1 = R2 = H, ‘the skeleton’.

If your name was, after fixing some enclosing parentheses to square or curly brackets, even the ‘preferred IUPAC name’, it's not a nice one, because it does not reflect the symmetry of the skeleton, and repeats the TEMPO and adamantane building block names.

(From now on, I'm gonna use the blockquote for IUPAC organic nomenclature rules 2013 generated names, or the book citations, if section name included.)

"Almost preferred" IUPAC name, substituted carbamoyl biphenyl dicarboxamide

The skeleton is named

  N4,N4′-dicarbamoyl[1,1′-biphenyl]-4,4′-dicarboxamide.

This preferred name is obviously based on the biphenyl dicarboxylic acid diamide

  [1,1′-biphenyl]-4,4′-dicarboxamide

substituted with carbamoyl ($\ce{NH2-CO\bond{-}}$) groups at each nitrogen.

Simple monosubstituted derivatives of the skeleton (there are only two kinds of them initially) are named

  N4,N4′-dicarbamoyl-N4-methyl[1,1′-biphenyl]-4,4′-dicarboxamide

and

  N4-carbamoyl-N4′-(methylcarbamoyl)[1,1′-biphenyl]-4,4′-dicarboxamide.

R2 (“TEMPO-4-yl”) can be named,

  2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl (group)

according to the rules of making the radical to be prefix (e.g. when there is yet more prioritized radical):

  P-71.5 Prefixes denoting radicals

  (…)

  Examples:

  (…)

  $\ce{-O^.}$

  oxyl (preselected prefix)
  ylooxidanyl
    (not ylohydroxy)

  (…)

  P-71.7.d.1 (…)

  (3-oxylbenzoyl)oxyl (PIN)
  [3-(ylooxidanyl)benzoyl]oxidanyl

however the rules do not make a good distinction between residues (prefixes) and radicals.

So the resulting name would be
N4-(adamantan-1-yl)-N4′-[(adamantan-1-yl)carbamoyl]-N4′-(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)-N4-[(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)carbamoyl][1,1′-biphenyl]-4,4′-dicarboxamide. But it still redundantly repeats some building block names.

Unfinished IUPAC name based on substituted "diallophanylated biphenyl"

  P-66.1.6.1.1.4 Carboxylic acid derivatives of urea

  Two carboxylic acids are related to urea; have been known as ‘allophanic acid’ for $\ce{H2N-CO—NH-COOH}$ and ‘hydantoic acid’ for $\ce{H2N-CO-NH-CH2-COOH}$. These names are no longer recommended. Preferred IUPAC names for these two acids and their derivatives are formed systematically.

  Examples:

  $\ce{H2N-CO—NH-COOH}$
  carbamoylcarbamic acid (PIN)
  (aminocarbonyl)carbamic acid

  $\ce{H2N-CO—NH-CO\bond{-}}$
  carbamoylcarbamoyl (preferred prefix)
  [(aminocarbonyl)amino]carbonyl

The allophanoyl- prefix can be found in the older IUPAC 1979 nomenclature, rule C-971.

The skeleton could be named

4,4′-diallophanoyl[1,1′-biphenyl], or
4,4′-bis(carbamoylcarbamoyl)[1,1′-biphenyl],

if you want.

But now we would have to repeat the two differently substituted allophanoyls in the name. Or derive a name from the symmetrical one, but how to number it?

Urea based IUPAC name

Closer look at your structure reveals that it in fact is two N,N′ asymmetrically substituted urea

  P-66.1.6.1.1 Urea and its substitutive derivatives

  P-66.1.6.1.1.1 The compound $\ce{H2N-CO-NH2}$ has the retained named ‘urea’, Which is the preferred IUPAC name, With locants N and N′, as shown above the structure below. The systematic name is ‘carbonic diamide’. The locants 1, 2, and 3 have been used in the past and may be used in general nomenclature.

  $$\begin{array}{ccc}
\ce{H2\overset{$N$}{\underset{1}{N}}-\underset{2}{C}O-\overset{$N'$}{\underset{3}{N}}H2} & \hskip{3em} &
\ce{H2\overset{$N$}{N}-CO-\overset{$N'$}{N}H2} \\
\text{urea (PIN)} & &
\text{carbonic diamide}
\end{array}
$$

  $$\bbox[0.5em,border:1px solid black]{\text{Numerical locants for urea are no longer used in the IUPAC preferred names.}}$$

molecules, N,the-other-one's-N′-interconnected with

  [1,1′-biphenyl]-4,4′-diyldicarbonyl (group).

And that's a correct name, used e.g. in

  2,2′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)dipropanoic acid.

(If these propanoic acids were interconnected by different carbons, 2,3′ (there's no other option), such multiplicative name would be no longer a preferred one; see e.g. P-15.3.2.1.1. But never mind, we're making a nice one.)

The skeleton now can be named ([1,1′-biphenyl]-4,4′-diyldicarbonyl)diurea. (More precisely N,N′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)diurea. Or N,N′′-…? And what about such multiplicative names of asymmetrical derivatives? There's no definitive answer, but this would be for another question.)

So we'll use the other urea numbering, the numerical, non-preferred one.

Now we get a nice name:
1,3′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)bis[1-(adamantan-1-yl)-3-(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)urea]

(But I would like similarly short and non-redundant name that would end like “…bis[(2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl]” yet more.)


There are different types of lattices. A lattice point is a point in any of these lattices. Lattices are used to describe  highly ordered systems such as crystals and some supersolids. I am unsure whether lattices can only describe periodic systems (see quasicrystals). However, I will be assuming that we are talking about periodic lattices. I will also be assuming that we are dealing with crystals.
Crystal lattice
The crystal lattice is used to describe the lattice of a real crystal. For example, in NaCl, a lattice point in a crystal lattice represents the position of a sodium ion or a chloride ion.
Bravais lattice
Bravais lattices are more mathematical and abstract than crystal lattices. They are pretty much the same as crystal lattices. Unlike the crystal lattice, however, lattice points in the Bravais lattice no longer represent a position of a particular atom. Instead a lattice point represents a position in which an atom can be placed. In other words, a lattice point in a Bravais lattice is a point, which is equal and indistinguishable from any other another point. What matters in a Bravais lattice are not the points themselves, but how they are arranged (i.e. symmetry).
Reciprocal lattice
The reciprocal lattice is the Fourier transform of either the crystal lattice or a Bravais lattice. More often than not, it is used to refer to the Fourier transform of the Bravais lattice. More of that is explained in Physics.SE. However, I personally think that this video explains it better. A lattice point here can represent an atom or a point, depending on the context.
Finally, as mentioned by other posts, these points can be vacant.


What is the industrial yield of paracetamol?

For a variety of explanations of industrial techniques as applied to acetylation,  including acetaminophen synthesis,  try here
https://www.google.com/patents/US20120065423
The examples are right at the bottom and are only scaled down lab versions of the industrial technique. Over 90% yield should be easily achievable if acetic anhydride is used (and is achieved in the industrial synthesis of ibuprofen).  The patent examples give yields for the use of acetic anhydride and acetic acid, as well as the technology employed e.g. continuous stirred tank reactor (cstr)  vs plug flow reactor (pfr or tubular reactor)  vs reactive distillation.  Enjoy.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  15. aldehydes
  16. ketones
  (…)

Thus, the principal characteristic group of the compound that is given in the question is the carboxylic acid group. This principal characteristic group is expressed at the end of a name by means of a suffix or a class name, i.e. ‘cyclohexanecarboxylic acid’.

The other groups of the given compound cannot be expressed as a suffix. The cyclohexanecarboxylic acid is substituted using the usual principles of substitutive nomenclature.

For the aldehyde group, the suffix ‘carbaldehyde’, which generally can be used when the $\ce{-CHO}$ group is attached to a ring as in cyclohexanecarbaldehyde, cannot be used in this case. Instead, the prefix ‘formyl’ can be used in the presence of a characteristic group having priority to be cited as a suffix or when the $\ce{-CHO}$ group is attached to a side chain.

Likewise, the suffix ‘one’, which generally can be used to form the name of ketones as in cyclohexanone, cannot be used in this case. Instead, the prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix.

Therefore, the correct name for the given compound is 4-formyl-2-oxocyclohexane-1-carboxylic acid.


Total number of stereoisomers of truxillic acid

At first glance, not only does it seem like there are no stereogenic centres in truxillic acid, it is also highly symmetrical. However, if we consider different conformations of the chemically identical units (H, COOH or Ph), which give rise to different isomers, some carbons at the 4-membered ring can possibly become stereogenic.

Stereogenic centres are labelled with an asterisk: ε-truxillic acid and peri-truxillic acid do not have stereogenic carbon centres as their 4-membered ring carbons have chemically and conformationally identical neighbours.
It took me a while to get this.Thanks to @andselisk and wikipedia.


What you are trying to do is called a redox reaction (from REDuction-OXydation) in which one molecule will be oxidized (i.e. give one electron) and another will be reduced (i.e. receive one electron). The propensity of two different molecules to exchange electrons is called "redox potential" and measured in Volts.

In the case of water, its dissociation into hydrogen and oxygen has a potential of 1.23V(1), so a 1.5V battery would be enough (however a rechargeable battery usually has a potential of 1.2V which would not be enough).

In your case, you apply a 240V potential, which will not only promote many other reactions but also heat the water up because of the Joule effect and make it boil.

(1) I purposely ignored the sign not to confuse you.


What is the meaning of parentheses in "NAD(P)H"?

If an enzyme is called $\mathrm{NAD(P)H}$-dependent, that means it is using both $\mathrm{NAD^+/NADH}$ and $\mathrm{NADP^+/NADPH}$ as its cofactors. For example, most oxidoreductases are $\mathrm{NAD(P)H}$-dependent, which are able to oxidize a relevant substrate by transferring a hydride ($\ce{H−}$) group to a nicotinamide adenine dinucleotide cofactor (either $\mathrm{NAD^+}$ or $\mathrm{NADP^+}$), resulting in their reduced forms, $\mathrm{NADH}$ or $\mathrm{NADPH}$ (Ref.1):

You can find a list of $\mathrm{NAD(P)H}$-dependent oxidoreductases within Reference 1.

Reference:

Lara Sellés Vidal, Ciarán L. Kelly, Paweł M. Mordaka, John T. Heap, "Review of NAD(P)H-dependent oxidoreductases: Properties, engineering and application," Biochimica et Biophysica Acta (BBA) - Proteins and Proteomics 2018, 1866(2), 327-347 (https://doi.org/10.1016/j.bbapap.2017.11.005).


Which molecule is more acidic: CHF3 or CHBr3?

The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen.

\begin{array}{|c|c|c|} \hline
\text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline
\ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline
\ce{CHCl3} & 15.7 & 3.16\\ \hline
\ce{CHBr3} & 13.7 & 2.96\\ \hline
\ce{CHI3} & \sim 25 & 2.66\\ \hline
\end{array}

Looking at the $\mathrm{p}K_\mathrm{a}$'s, the first thing we notice is that the two haloforms at the beginning and end of the series are weaker acids than the two haloforms in the middle of the series. This behavior is often seen when there are at least two factors affecting acidity and they are in opposition to each other.

Let's consider inductive, steric and resonance effects and examine their expected effect upon haloform acidity.

Inductive effects should correlate with electronegativity.  Based only on inductive effects we would expect fluoroform to ionize and produce the most inductively stabilized cabanion ($\ce{CF3^{-}}$), while iodoform would ionize to produce the least inductively stabilized carbanion ($\ce{CI3^{-}}$).
Steric effects are often discussed in terms of bulky groups coming too close to one another, but what we are really saying is that the electron clouds around the groups are coming too close to one another causing electron-electron repulsion. Iodoform is a very bulky molecule.  The iodine substituents are very large and there are many lone pair-lone pair repulsions.  Upon ionizing iodoform can relax from a crowded $\mathrm{sp^3}$-like geometry to a more open and spread out $\mathrm{sp^2}$-like geometry.  As the halogen gets smaller there is less of a steric \ electron repulsion driving force for haloform ionization.
Resonance effects are not a significant stabilizing factor in the haloform series. Significant low energy resonance structures cannot be drawn. Thanks to TanYongBoon for noting this.

Conclusion: Inductive effects and steric effects (steric meaning electron-electron repulsion) control acidity in the haloform series. The two effects are in opposition to one another.  The inductive effect strongly stabilizes the fluoroform anion and plays a lesser role as we move down the series.  Steric \ electron repulsion effects strongly stabilize the iodoform anion and plays a lesser role as we move up the series.  Said differently, only one effect stabilizes the fluoroform (inductive) and iododform (steric) anions, but both inductive and steric effects play a role in stabilizing the chloroform and bromoform anions making chloroform and bromoform the most acidic molecules in the series.


Difference between valency of nitrogen and phosphorus though both have same valence electrons

At your current stage, it's probably easiest to consider the octet rule which applies to nitrogen and the expanded octet rule which applies to phosphorus.

If nitrogen and phosphorus made five bonds, they would have 10 electrons total. This is inconsistent with the octet rule but valid under expanded octet.

In reality, the situation is governed by the fact that you need more orbitals to accommodate the extra electrons for the two extra bonds. Unfortunately, no orbitals of comparable energy to the $2s$ and $2p$ are available, so the stabilization of the electrons in the putative molecular orbitals would be poor in nitrogen. With phosphorus, you might more readily access the $4s$ and $3d$ orbitals which are closer in energy to the $3s$ and $3p$ orbitals.


Any examples of liquids volatile at room temp but non-flammable?

Yes, there are lots. Start with Wikipedia's list of refrigerants; while it doesn't call out flammability, any fully-halogenated compounds will be effectively non-flammable. Trichlorofluoromethane has a standard boiling point just above standard room temperature.

You didn't say anything about pressure in your question. If you don't mind pressures higher than one standard atmosphere, and don't mind toxicity, candidates like ammonia, sulfur dioxide, and chlorine come into play. If you don't mind low pressure, you could even use water, although it might freeze as it boils off.


I don't know what flex and link voltage are; however, here's the rest of it as per this site at UCLA and this homework solution key from Illinois State:

The $\ce{N-N}$ stretch is at $\pu{2200 cm-1}$, the $\ce{N-O}$ stretch is at $\pu{1290 cm-1}$, and the doubly-degenerate $\ce{N-N-O}$ bend is at $\pu{590 cm-1}$.


If salt crystals make a cube shape, what is this 'X' marking?

When salt (NaCl) crystals are grown under normal conditions, the edges of the cubes usually grow faster than the faces (because the edges have more contact with the saturated salt solution than the faces). So the crystals grow with a square-pyramidal indentation on each face. The "X" you observe is your view of the edges of that inverted pyramid. On the other hand, if the crystals are grown really slowly under controlled conditions, this does not occur and you get a more or less perfect cube.


preparation methods of nicotinoyl glycine

As a microscale laboratory method, it should be possible to react nicotinic acid with, e.g., thionyl chloride, yielding nicotinoyl chloride (the acid chloride of nicotinic acid). This can be reacted with glycine to yield the desired product. Unfortunately, glycine would exist predominantly in zwitterionic form at neutral pH, so I would guess that pre-treatment of the glycine with base would probably be required prior to the reaction with nicotinoyl chloride.


There is no theory, nor is there a need for one. To put in in other words: yes, there is a theory, and that's the one you are already familiar with. Antimatter elements are precisely the same as ordinary matter elements, only the other way around. There "are" (read: "could be", though some were actually detected) antihydrogen, antihelium (all with the same energy levels as their real counterparts), also anticarbon, antiwater, anti-$\ce{NaCl}$, antiglucose, and pretty much everything anty-else.

To compose a substance of particles and antiparticles together is an entirely different question. Yes, there is a compound of electron and positron; it is called positronium and has the lifetime of a few microseconds or something. Electrostatic forces attract the opposite particles, rather than repel them (otherwise it would not be a compound, BTW), so it annihilates pretty quickly. There are other similar compounds, equally or more short-lived. Matter and antimatter do not sit well with each other; there is no going around it.

As for the spots of the Universe composed of antimatter, there was once a mainstream view that there must be many of those, about as many as those made of matter. Then the global symmetry broke, and things went... complicated. I'd rather leave it at that. Go ask at Phys.SE for a more comprehensive take.

So it goes.


What is the reaction of ammonium nitrate with hypochlorous acid

Do not for goodness sake mix chlorine bleach with ammonium compounds or ammonia, it has a good chance of ending badly.

I am not sure of the curved arrow mechanism, but what I suspect will happen is that the ammonia will attack the HO-Cl using the OH- as a leaving group. So by making the system less basic it will favour the reaction.

This will convert ammonia into NH2Cl, which will then be converted into NHCl2 and finally into nitrogen trichloride (NCl3). When I was a PhD student one of the other people has to make NCl3, he did it with great care. He did the reaction in such a way that the NCl3 was diluted in an inert solvent.

The scary thing about nitrogen trichloride is that it is a very unstable and shock sensitive explosive. One of my uncles is a chemist like me, he once witnessed an accident in which a man lost several fingers playing around with nitrogen iodide.

But back to nitrogen trichloride, as well as being explosive it is also a harmful and irritating substance. I would suggest you have nothing to do with it !


The IUPAC goldbook defines a zwitterion as being overall neutral. Any other definition is thus contrary to the standard definition.


What does 'a' stand for in 'a-C' according to research paper lingo?

a for amorphous is totally correct. Also used in other publications like here.


Where do negative ions migrate from the salt bridge to which electrode in the voltaic cell?

Oxidation half reaction (happens at the anode)

$$\ce{Mg -> Mg^2+ + 2e-}$$

The electrons get removed via the wire, and the magnesium cations enter the aqueous solution, giving it a net positive charge.

Reduction half reaction (happens at the cathode)

$$\ce{Ag+ + e- -> Ag}$$

The electrons are supplied via the wire, and the silver cations leave the aqueous solution to be deposited on the cathode, giving the solution a net negative charge.

Salt bridge

The salt bridge provides anions (negative charged particles) to the solution of the oxidation half cell, and cations (positively charged particles) to the solution of the reduction half reaction.

Silly mnemonics

Oil rig: oxidation is loss, reduction is gain (of electrons)

Red cat: reduction occurs at the cathode.

An ox: oxidation occurs at the anode.

  Where do negative ions migrate from the salt bridge?

They don't migrate to any electrode (neither cathode nor anode). Instead, they migrate to the solution of the solution of the oxidation half cell.


Why does (2z²-x²-y², x²-y²) have -1 trace in the character table of Td for C3 rotation?

I think I know now.

Choose rotation axis that makes equal angles with the $x$-, $y$-, and $z$-axes so that the $C_3$ rotation essentially permutes the axes. Without loss of generality, I could choose a $C_3$ rotation that permutes $x$ to $z$, $z$ to $y$, and $y$ to $x$. Rotation in either direction will give the same trace with this $C_3$ rotation. With this rotation (permutation), $(2z^2-x^2-y^2,x^2-y^2)$ becomes $(2y^2-z^2-x^2,z^2-x^2)$. Multiplication of basis function vector a by $2\times 2$ matrix in this representation yields

$$\begin{pmatrix}
  -1/2 & -3/2\\
  1/2 & -1/2
\end{pmatrix}
\begin{pmatrix}
  2z^2-x^2-y^2 \\
  x^2-y^2
\end{pmatrix}
=
\begin{pmatrix}
  2y^2-z^2-x^2\\
  z^2-x^2
\end{pmatrix}
$$

The trace of this matrix is $-1$ as in the character table. Choosing any axis via orthogonal transformation of this matrix yields same trace.

The tricky thing about this example was that I needed to use a rotation in a three-dimensional vector space and project it into a rotation in a two-dimensional vector space.


Why does nitration of aromatic amines give meta directed nitro compounds?

A couple things here.  First, it's nitric acid, not nitrous acid, that is the nitrating reagent.  Otherwise, your own explanation is correct.  Orthocresol points out that phenols are sufficiently inert to protons, to avoid this situation.

Second, you can make ortho/para directed nitrated aromatic amines, if you are a little more clever and if you have a tertiary amine (http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-3-6.html).  Use nitrous (not nitric) acid, which is weaker and will react only with the unprotonated amine.  That gives you a nitroso $\ce {-NO}$ group which you can then oxidize to the nitro group.


As you mention, there are two main conformations that cyclopentane may adopt.

In the absence of substituents around the ring, there is actually only a $\pu{0.5 kcal/mol}$ energy difference between envelope and half-chair (envelope slightly lower in energy) meaning cyclopentane itself (computationally at least) is in rapid equilibration with no 'major' conformation as is observed with cyclohexane. (Note that the energy of the totally planar conformation is about $\pu{5 kcal/mol}$ higher than that of the both bent conformations, and so is not observed to any great degree).
With a single substituent on the ring however, such as a methyl group, the envelope conformation becomes slightly more favourable (to the tune of around $\pu{3.5 kcal/mol}$) giving a barrier to the ring flipping and meaning there is a dominant conformation.


Early history of Earth is a question to paleohydrology and paleochemistry. Briefly, water wasn't pure on the early stages of Earth.

Brief history of water on Earth. The planet was hot and seismoactive (lots of vulcanos) in the beginning. Water (as steam) escaped from magma, cooled down and precipitated on the Earth's surface. The surface already had salts on it, so it became saline from the very beginning. A better question is: how come we have fresh water in rivers and saline water in oceans? This separation is due to water cycle on our planet. Rivers bring dissolved salts to the ocean, then evaporates and precipitates over the land. Salts cannot evaporate (at any practical rate), so they stay in the ocean. This is similar to washing dishes. Consider this model: dish gets cleaner - that is a river; drain water gets dirtier - that is an ocean; drain water goes through purification and becomes clean before going to the supply pipe - this is evaporation of water from the surface of an ocean and precipitation over the land in our model.


Relationship between Lewis Theory and Redox theory?

Under the Usanovich definition of acids and bases, any substance which loses electrons is a base while any substance which gains electrons is an acid. Here, any redox reaction is also an acid-base reaction, whereas Lewis theory has nothing to state in these cases.

Regarding your second question, blanket statements are dangerous. Depending on exactly what you mean, I can provide a counter-example. Here is a strong oxidant ($\ce{HNO3}$) reacting as a Lewis base:

$$\ce{HNO3(l) + H2SO4(l) -> H2NO3^{+} + HSO4^{-}}$$


Finding orbit radius using the Bohr model and Rydberg equation

To answer the second part:
We know $M=208m_e$, $Z=3$, $\hbar=\frac{h}{2\pi}$.
Part one has a mistake, as it is
$$
\begin{align}
&&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\
&&Mvr&=n\hbar\\
\implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2}
\end{align}
$$
We also know the Bohr radius:
$$a_0= \mathcal{k_e}\cdot\frac{\hbar^2}{m_e\cdot\mathcal{e}^2} \approx 5{,}29 \cdot 10^{-11}\mathrm{m}$$
Therefore we can write and cancel:
$$
\begin{align}
&& r&=a_0\\
&& \frac{\color{\green}{\hbar^2}}{\color{\red}{\mathcal{k_e}}\cdot m_e\cdot\color{\navy}{\mathcal{e}^2}} &=\frac{n^2\color{\green}{\hbar^2}}{M\cdot\color{\red}{\mathcal{k_e}}\cdot Z \color{\navy}{\mathcal{e}^2}}\\
\therefore&& Z\frac{M}{m_e}&=n^2\\
\therefore&& n&=\sqrt{Z\cdot208}\approx25
\end{align}
$$
The third part:
The Rydberg Formula is given as
$$\frac{1}{\lambda_{\mathrm{vac}}} = \mathcal{R}Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
with the Rydberg $\mathcal{R}$ constant defined for a photon emitted by an electron. We'll assume that the mass of the nucleus is 7 atomic units (three protons + four neutrons). Taking into account that $m_p\approx 1836m_e$, we arrive at
$$\mathcal{R} =\frac{\mathcal{R}_\infty}{1+\frac{M}{T}} = \frac{\mathcal{R}_\infty}{1+\frac{208m_e}{7\cdot1836m_e}}$$
Now the Rydberg constant has to be modified to include the mass of the particle:
$$\mathcal{R}_\infty = \frac{M e^4}{8 c \varepsilon_0^2 h^3}=208\mathcal{R}_{m_e,\infty}$$
With $\mathcal{R}_{m_e,\infty} = 1.097\cdot 10^7~\mathrm{m^{-1}}$ (wikipedia), I got to $\lambda_\mathrm{vac}=55.6~\mathrm{pm}$.
Without taking the reduced mass into account, i.e. $\mathcal{R}\approx\mathcal{R}_\infty$ I arrived at $\lambda_\mathrm{vac}=54.8~\mathrm{pm}$.
Both values are reasonably close to the solution given.
(If the question was really about the muon, the more accurate weight ratio is 206.77 and the corresponding wavelengths 55.1 pm and 56.0 pm.)


Though PTFE is not attacked by the halogens, they may discolor PTFE (particularly iodine, and at elevated temperature), since the plastic is somewhat permeable to halogens (end even to water).


One could say that hydrogen peroxide decomposes because it is unstable. But that would be cheating because this is simply restating the fact that we observe it to decompose in fancier language.

We could use some statistics about the reaction, like stating that the decomposition is thermodynamically favourable (this doesn't explain why a catalyst works but it helps to explain why the reaction happens). But, again, this is just a sophisticated way to repackage the observed fact that the reaction is observed to happen.

However, if we think what is happening at the molecular level, we can get some extra insight.

All molecules above absolute zero have some kinetic energy (that is they are bouncing around and banging into each other with a range of speeds, some slow, some fast). Temperature summarises the average amount of energy present in the compound (but the individual molecules have a range of energies, some higher, some slower than the average).

In a solution of hydrogen peroxide (or in the pure liquid, though I don't recommend ever handling the pure liquid) the individual molecules will be bouncing off each other a lot. Each collision will exchange some energy between the colliding molecules. Sometimes a Collision will involve enough energy that one or more of the O-O or OH bonds will rupture. This will generate some reactive species. Then, sometimes, the reactive species will reform in a way that causes water and oxygen to be formed (the details of the reaction are probably fairly complicated). The products oxygen and water have much stronger bonds than the hydrogen peroxide (or at least eh O-O bond in it) so the probability that they will react to reform hydrogen peroxide is very low.

The probability that peroxide will decompose this way is still pretty low, though, at least in dilute and very pure water solutions. But many substances (including some present in slightly impure water) will catalyse the reaction. The presence of manganese dioxide, for example, provides a different route to the same products (that is pretty much the definition of a catalyst). If a peroxide molecule bumps into a manganese dioxide surface, there is a much higher probability that a reaction will happen than if it hits another peroxide molecule (it probably transfers an electron from the manganese oxidising it and this can be very fast and also can precipitate more reactions that break up the peroxide and, ultimately, yields water, oxygen and leaves manganese dioxide). Human blood contains an enzyme that is designed to do the same job as peroxides and related active species are generated by other biological reactions and you really, really don't want them to build up in the blood. Catalase uses oxidation reactions involving iron atoms to do an even better job of facilitating the reaction than manganese dioxide.

So the detail of the reaction is all about molecules bumping into each other and causing bonds to break and reform in a different way. The overall reaction releases energy (which is to say that the new bonds formed are stronger then the ones that get broken and the reverse reactions are therefore a lot less likely to happen). Catalysts proved an easier route to the products than the reaction where peroxide molecules just bump into each other.


In the first condition when both flasks are boiling, the temperature of the gases in both flasks are the same (100 °C).  Using the ideal gas law on the system that comprises both flasks, where $n$ is the number of moles initially in each flask and $V$ is the volume of one flask.

$$ P_{init} = \frac{(2 n) R\;T_{boil}}{2 V} = \mathrm{0.5~atm}$$

$$ 2n = \frac{P_{init}(2 V)}{R\;T_{boil}} $$

Now if Flask #2 is immersed in ice water, we should assume that the pressure in the system will be constant but that the flasks will have different temperature.  For the pressure to be constant at different temperatures, material will have to flow from the hot flask to the cold flask; now, $n_1$ and $n_2$, the moles of gas in flask #1 and flask #2, will not be the same, but since no gas leaves the system comprising both flasks, then we know $n_1 + n_2 = 2n$

$$ n_1 = \frac{P_{final}V}{R\;T_{boil}} $$
$$ n_2 = \frac{P_{final}V}{R\;T_{ice}} $$
$$ n_1 + n_2 = 2n$$

Substituting:

$$ \frac{P_{final}V}{R\;T_{boil}} + \frac{P_{final}V}{R\;T_{ice}} = \frac{P_{init}(2 V)}{R\;T_{boil}}$$

$$ P_{final}\frac{V}{R}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}V}{R\;T_{boil}}$$

$$ P_{final}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}}{T_{boil}}$$

$$ P_{final} = 2 P_{init}\frac{\frac{1}{T_{boil}}}{\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)} = 2 P_{init} \frac{\frac{1}{T_{boil}}}{\frac{T_{ice} + T_{boil}}{T_{ice}{T_{boil}}}}$$

$$ P_{final} = 2 P_{init} \frac{T_{ice}}{T_{ice} + T_{boil}}  $$

This answer makes sense in that (i) if the temperature $T_{ice}$ is actually the same as $T_{boil}$, then the final pressure would be the same as the initial pressure, and (ii) the answer doesn't depend on the volume or moles of gas in the flasks, just the temperatures and the initial pressure.

  Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry.

This way is incorrect because a flask immersed in boiling water ($T=$100°C) is unlikely to have a temperature of 50 °C.  Similarly for a flask immersed in ice.

  Way 2: The sum of the moles of gas in both the flasks would remain same.

This is the correct method.  It does require assuming that the rate of heat transfer from the boiling / ice baths to the flasks is much much greater than the rate of heat transfer between flasks via diffusion through the tube.  However since the tube is small, this assumption is probably OK.


Formation of halohydrin vs alkyl dihalide

The halohydrin is usually formed by means of a solvolysis reaction where water is used as the solvent. So, it is present in such high amounts that it overpowers the larger nucleophilic strength of the $\ce{X-}$ ion since the latter is only present in relatively minute quantities. Furthermore, water is a polar protic solvent and thus it has the effect of weakening the nucleophilic strength of $\ce{X-}$ in solution by "encasing" it in a large hydration shell.


Which species occupy the equatorial positions in VSEPR's trigonal bipyramidal and octahedral geometry

Trigonal planar is a molecular structure that requires five ‘substituents’ around a central atom (counting lone pairs as ‘substituents’). Assuming main-group compounds, that means, we have already left the realm of easily derived molecules and need to resort to 4-electron-3-centre bonds to properly explain the molecule and be in line with the octet rule.

This very good answer by Ron does a good job of explaining what the central atom’s electronic structure should be considered to be and what the 4-electron-3-centre bond is. In a nutshell, assume the central atom to be $\mathrm{sp^2}$ hybridised with an unhybridised $\mathrm{p}$ orbital. This additional $\mathrm{p}$ orbital then interacts with two electronegative substituents forming a 4-electron-3-centre bonds; this bond can be rationalised in a Lewis formalism by looking at these two resonance structures:

$$\ce{X-\overset{+}{A}\bond{...}\overset{-}{X} <-> \overset{-}{X}\bond{...}\overset{+}{A}-X}$$

The bond order of a 4-electron-3-centre bond is approximately ½, in good agreement with a single bond and a non-bond forming equivalent resonance structures for each $\ce{X\bond{...}A}$ fragment. Each outer atom has a partial negative charge while the central atom has a partial positive one.

This already means a few things:

a lone pair can never be in axial position as this position is formed by a $\mathrm{p}$ orbital. If a lone pair were in axial position, no second ligand can be there and the structure is actually better described as trigonal planar.
the more electronegative a substituent is, the more likely it is to occupy the axial positions as these positions require stabilising a partial negative charge. Hence, fluorine is most frequently found here.

This is also somewhat of an example for Bent’s rule — the more electropositive a substituent is, the higher the $\mathrm{s}$ orbital participation for the orbital forming the bond to said substituent. Conversely, the more electronegative the substitutent is, the higher the $\mathrm{p}$ content. Naturally, a pure $\mathrm{p}$ orbital is on one end of that spectrum, clearly preferring electronegative substituents.


How to calculate valence factor (n-factor) for a element?

Hint :

n-factor of a molecule/compound is defined as the change in oxidation state per molecule.
You have correctly calculated the change of one carbon atom as 4.
But how many carbon atoms are there in the glucose molecule?

Note:

The average oxidation state of carbon in glucose is zero while in reality the different carbons have different OS. (Reference)
n-factor of a reaction is not defined but it is defined for a single species participating in that reaction.
n-factor of glucose is not equal to the n-factor of the product formed in that reaction, that is, $\ce{CO_2}$.


This is not exactly an answer making strong use of references etc. but it does give a good hint. The most ‘energetic’ bonds in ATP are the phosphoric acid anhydride ones i.e. the $\ce{P-O-P}$, which, to hydrolyse ATP, must be attacked by water. The most logical path of attack is one according to the $\mathrm{S_N2}$ mechanism, where a lone pair of water attacks the rear side of a phosphorus-oxygen bond and liberates some phosphate species. ($\mathrm{S_N1}$ can be ruled out, since it would result in a phosphocation which is surrounded by three electron-withdrawing oxygens making the cation highly unstable.)

The problem in this mechanism is physiological pH. It is safe to say that the entire ATP molecule is at least a dianion, if not even a tri- or tetraanion. Now water’s lone pair, i.e. its partially most negatively charged area must approach this already highly negative molecule — but that would result in the repulsion of same charges leading to a high activation barrier.

This would be circumvented by protonation, and indeed even at equilibrium there is a certain probability that a higher number of protons attaches to ATP, making it less negative. This probably helps to account for the slow but steady decay of ATP.


Doping, not to confuse with alloying, introduces a low amount of impurities or more generally point defects in a crystal. The overall periodicity of the system is thus not changed and no shift should be expected in XRD. However doping introduces some disorder that makes XRD peaks broader.


Is the effect of hydroperoxyl radicals significant in the oxidation of organic compounds in the Fenton reaction? Why or why not?

The effect of the hydroperoxyl radical in the Fenton process can be largely ignored as the oxidising ability of the $\ce{^∙OOH}$ radical is not as effective as that of the hydroxyl radical.
According to Chang et al., the reaction rate coefficients of p-nitrophenol and $\ce{OH}$ and $\ce{HO2}$ were found to be $\pu{9.05 \times 10^6 M−1 s−1}$ and $\pu{1.50 \times 10^−6 M−1 s−1}$, respectively. Considering the orders of magnitude of these two rate coefficients, the reaction pathway involved the p-nitrophenol reaction with $\ce{^∙OOH}$ radicals can thus be neglected. This can therefore be extended to cover the oxidation of organic compounds by hydroperoxyl radicals through the Fenton process.
References

Chang, M., & Chern, J. (2010). Decolorization of peach red azo
dye, HF6 by Fenton reaction: Initial rate analysis. Journal of the
Taiwan Institute of Chemical Engineers, 41(2), 221-228.
DOI: 10.1016/j.jtice.2009.08.009


How do I calculate the isoelectric point of amino acids, each of which has more than two values of pKa?

Since the $\mathrm{pI}$ is the $\mathrm{pH}$ at which the amino acid has no overall net charge, you need to average the $\mathrm pK_\mathrm a$ values relevant to the protonation/deprotonation of the form with no net charge. Here are the acid-base equilibria for tyrosine:

The form with no net charge is in red (+1 and -1 cancel out to give no net charge). It is the $\mathrm pK_\mathrm a$ values on either side of this form (in blue) that matter, hence the $\mathrm{pI}$ of tyrosine is $5.66$ (the average of $2.20$ and $9.11$).

It just so happens that $2.20$ is the carboxyl $\mathrm pK_\mathrm a$ and $9.11$ is the amino $\mathrm pK_\mathrm a$. If the side chain $\mathrm pK_\mathrm a$ were lower than $9.11$, then you should average the carboxyl and side chain $\mathrm pK_\mathrm a$'s instead.

The same logic applies to cysteine (look up the $\mathrm pK_\mathrm a$ values and draw out the differently protonated forms). You'll find that since the side chain has a lower $\mathrm pK_\mathrm a$ than the amino group, you average the carboxyl and the side chain $\mathrm pK_\mathrm a$'s.

This procedure can of course be extended to the amino acids with acidic side chains (aspartic acid; glutamic acid) and those with basic side chains (lysine; arginine; histidine).


How to quickly cool a test tube without cracking it?

Rather than working on your technique, check whether other test tubes are available.

Apparently, a glass-like material with a very low thermal expansion coefficent would be a good choice.

The glassblowing shop at the University of Delaware provides a nice overview of the possible materials and their properties, and there are definitely additional online resources, e.g. by manufacturers like Schott, Corning, etc.

Here's a summary (coefficients are given in $cm/cm/°C$):

soda lime glass $89 \times 10^{-7}$

Pyrex, Duran $32.5 \times 10^{-7}$

Vycor  $7.5 \times 10^{-7}$

fused quarz $5.5 \times 10^{-7}$

Please note that quarz is both pretty expensive and demands special care! Alkaline solutions and quarz are a bad conbination!


What the context of aiming for a highly concentrated aqueous solution of epinephrine / adrenaline?  What would be a sufficient concentration?  Would -- assuming a chemical reaction is intended -- in sight of a low reaction rate a steady of epinephrine in low concentration be sufficient?

While this lists a typical concentration of saturation in water of about 180 mg/L, other conditions are provided, too.  The list equally includes a statement like "soluble in acetic acid" -- so perhaps you do not need to decrease the pH value down to 1 (by HCl), and maybe you were able to obtain a sufficient concentration of epinephrine in a mixture of water / acetic acid, too?

Are there restrictions, imposed by the intended application / compatibility  toward (biological?) materials to stay near neutral conditions?  One way to reply to this and to tailor solubility of pharmaceutical active ingredients is to refrain from their form as free acids or bases, and to use them as salt.  In this sight, would be epinephrine tartrate, or an acetate, an alternative option?


When you ionize an element, all the orbitals go down in energy but orbitals with higher angular momentum drop more than those with lower angular momentum. When you start ionizing Ni, the $d$-orbitals drop much more than the $s$-orbitals and so, in the ions, the highest energy orbitals are the $4s$.

Relatedly, I believe (and I just ran a quick calculation to confirm) your electron configuration for $\ce{Ni+}$ is incorrect and should be [Ar]$3d^9$ for the same reason.


The largest contribution to the entropy change will be the translational component, which is easy to evaluate using the Sackur-Tetrode equation
$$S^\circ_{\rm{trans}}=R\ln(bM^{3/2}T^{3/2})$$
where $b$ is a collection of constants with a value of 3.75 (mol/gK)$^{3/2}$ $M$ is the mass in g/mol, $T$ is temperature in Kelvin, and the standard state is 1 M.

Due to the slowly varying logarithmic dependence on $M$ and $T$ this contribution is usually on the order of 100's of J/molK.

While the translational contribution is the largest, rotational and vibrational contributions can be sizable so this will give you an order of magnitude estimation at best.

Derivation starting from the wikipedia equation
$$S^\circ=kN \left( \ln \left( \frac{V^\circ}{N} \left( \frac{4\pi mU}{3h^2N}\right)^{3/2} \right)+\frac{5}{2}\right) $$
$V^\circ$ = 1 L = 0.001 $m^3$, $N=N_A$, $U=\tfrac{3}{2}N_AkT$, $M=mN_A$
$$S^\circ=R  \ln \left( \left[e^{5/2}\frac{V^\circ}{N_A} \left( \frac{4\pi \tfrac{3}{2}k}{3h^2N_A}\right)^{3/2}\right]M^{3/2}T^{3/2} \right) $$
The term in square brackets corresponds to $b$ and using SI units $b=1.20\times10^5$ mol(kg/K)$^{3/2}$ which corresponds to 3.75 mol(g/K)$^{3/2}$


Effect of temperature on Hydrogen Bonding?

Yes. Temperature disrupts bond of all kinds. Heat up a protein hot enough and you can even disrupt its primary structure - the linear sequence of amino acids, and amino acids are held together through covalent bonds. This is why prion contaminated organisms and instruments must be heated to extremely, extremely high temperatures. The question is whether or not the bond will break; the question is simply what temperature.

Take for example nitrous acid, which exhibits two unique $\ce{O-N-O}$ bond angles. One of these bond angles (the smaller one) results from intramolecular hydrogen bonding between the hydrogen and the oxygen on the other end. This occurs because there free-rotation is permitted around the single bonds in the nitrous acid molecule. The energy difference between the two structures is calculated to be $0.45~\mathrm{kcal~mol^{-1}}$ or $1880~\mathrm{J~mol^{-1}}$.


Effect of +I on the alpha hydrogen in aldol reaction

Is the alpha hydrogen acidic enough to react with base?

The answer is yes. As you said +I effect of the methyl group stabilizes the carbocation. It is weakly acidic but not weak enough to not be taken by any base.

Will it proceed through aldol reaction or will Cannizzaro reaction occur?

It will proceed through Cannizzaro mechanism even though it has an alpha hydrogen. That is because if you take it through aldol, the attacking nucleophile in the second step is $\ce{-C(CH3)2(CHO)}$ which is a very bulky nucleophile and so it isn't able to attack on $\ce{CH(CH3)2(CHO)}$. Think, if this reaction undergoes than the product will have $\ce{CH3}$, $\ce{CH3}$, $\ce{CHO}$ and $\ce{-C(OH)(CH(CH3)2)}$ groups all on one carbon. The steric crowding will be so immense that the reaction becomes infeasible. That is why it undergoes Cannizzaro reaction which has $\ce{H-}$ as the attacking nucleophile (in the second step).
I think your other questions won’t arise if you think this way.


Conductivity of a semiconductor at high temperatures

A number of different things are going on in this scenario.

First, the carrier concentration does increase steeply, although for Si at its melt temperature it is still only about 10$^{20}$/$cm^{3}$, much lower than a metal.

In addition, the band gap is decreasing with temperature (not from carriers, but from changes to the lattice spacing). From a graph in Sze's Physics of Semiconductor Physics, Si goes from 1.12eV at room temperature to a little under 0.9eV at 900K. Extrapolation (probably a bad idea) means that somewhere above 2700K there will be little to no band gap. However, this assumes the physics extrapolates linearly, which it doesn't. (Well, liquid silicon is metallic. However, there is still a large increase in conductivity at the melt point).

Finally, the phonon population increases with temperature, which will increase scattering, decreasing the conductivity (just like a metal). This can be seen from a graph of carrier mobility (also in Sze) with temperature. The mobility drops roughly as T$^{-2.2}$ or so.

Getting the balance of all of these just right under extreme extrapolation, well, that is hard. It is actually easier to point to papers on the electrical characteristics of undercooled liquid silicon (since that has been measured), than to superheated solid silicon (which has not been observed). - for ev


How does sulfuric acid dehydrate sugars?

$\ce{H2SO4}$ protonates the alcohol to make $\ce{R-OH2+}$ and $\ce{HSO4-}$, the $\ce{HSO4-}$ then takes a hydrogen from the carbon atom adjacent to the alcohol, yielding an alkene and water.  Alcohol dehydration reactions using acid are just elimination reactions.

Additionally, at temperatures above 340 degrees Celsius, sulfur trioxide exists in equilibrium with sulfuric acid in significant concentration.  The presence of sulfur trioxide explains the presence of sulfur dioxide  (via dissociation), or some redox chemistry occurring with the sugar.


Will this reaction form magnesium carbonate?

If such a reaction takes place at all it is not very efficient because magnesium hydroxide is not very soluble.

And if you want to get the metallic magnesium neither the carbonate nor the hydroxide will really work.  Both decompose at a relatively low temperature to the oxide, which is then solid up to $\pu{2800\!^\circ C}$.  Convert the magnesium to the chloride and melt that, preferably as a eutectic with alkali metal chlorides.  Then electrolyze the molten salt.  An aqueous solution of magnesium chloride  (or anything else containing magnesium) does not work.


Calculation of proportion of zwitterion from this please.

(I shouldn't answer this because it is a homework question with no effort, but since someone posted a wrong answer I'm answering anyway)

Only the pK of 9.6 is relevant because the pH is 10.6; therefore, think of the situation as having just two species, zwitterion (acid) and negative (conjugate base).

Because pH is one unit above the pK, the ratio of acid to conjugate base is 1 to 10.

Ratio of acid to total is 1 to 11.

1/11 = 0.0909


According to IUPAC (source):

  2.6 - If chains of equal length are competing for selection as main chain in a saturated branched acyclic hydrocarbon, then the choice goes in series to:
  (a) The chain which has the greatest number of side chains.

Therefore the correct name is 3-ethyl-2,2-dimethylhexane as opposed to 3-tert-butylhexane because the former has more side chains.