Calclulation of MP2 first order wave function

The main problem with the last statement,

$$
\begin{align}
C^{MP2} &= c\cdot c^T \\
\Rightarrow P^{MP2}_{ij} &= \sum_{a}C^{MP2}_{ia}(C^{MP2}_{ja}),
\end{align}
$$

is that one never has "MP2 MO coefficients" in this way; you would have MP2 natural orbital (NO) coefficients that come from diagonalizing this density, so it's the other way around:

$$
P^{\text{MP2}}C^{\text{MP2 NO}} = n^{\text{MP2 NO}}C^{\text{MP2 NO}},
$$

where $n^{\text{MP2 NO}}$ is the diagonal matrix of natural orbital occupation numbers. This still leaves the question of how to calculate $P^{\text{MP2}}$. Conceptually, the way I view it is that the set of $\{c_{ijab}^{(1)}\}$ are the amplitudes describing how to correct the uncorrelated ground-state wavefunction; they don't specify the wavefunction on their own, so $C^{\text{SCF}}$ will probably have to appear in the total density.

All of the following assumed a restricted wavefunction. $i,j,k,\dots$ are occupied MO indices, $a,b,c,\dots$ are virtual/unoccupied MO indices, $p,q,r,\dots$ are general (any) MO indices, and $\mu,\nu,\dots$ are general AO indices. The 2nd-order correction to the density, specifically the occupied-occupied and virtual-virtual blocks are given by Eqn. (11a) in Trucks [1]:

$$
\begin{align}
P_{ij}^{\text{MP2}} &= -\frac{1}{2} \sum_{kab} c_{jk}^{ab} c_{ik}^{ab} \\
P_{ab}^{\text{MP2}} &= +\frac{1}{2} \sum_{ijc} c_{ij}^{ac} c_{ij}^{bc},
\end{align}
$$

where I switched to using your notation. This is indeed similar to your idea from above; accumulate over both indices of the opposite space, then do a matrix multiplication.

The total density is then [Eqn. (7)]

$$
\mathscr{P} = P^{\text{SCF}} + P^{\text{MP2}},
$$

where $P_{pq}^{\text{SCF}} = \delta_{pq}$ because so far we're working in the MO basis. The back-transformation to the AO basis will be the same as for any other 2-index MO-basis quantity:

$$
\begin{align}
P_{\mu\nu} &= \sum_{pq}^{\text{all MOs}} C_{\mu p} P_{pq} (C_{\nu q})^{T} \\
\Rightarrow \mathbf{P}^{\text{AO}} &= \mathbf{C} \cdot \mathbf{P}^{\text{MO}} \cdot \mathbf{C}^{T}.
\end{align}
$$

Then, we can evaluate molecular properties that are represented as energy derivatives [Eqn. (6)]:

$$
\left.\frac{\partial \mathscr{E}}{\partial \lambda}\right|_{\lambda=0} = \sum_{\mu\nu} \mathscr{P}_{\mu\nu} Q_{\mu\nu}^{\lambda}
$$

where $\mathscr{E} = E_0^{(0)} + E_0^{(2)}$ and $Q_{\mu\nu}^{\lambda}$ is the matrix representation of the external perturbation $\lambda$ in the AO basis. This is analogous to the SCF result:

$$
\left.\frac{\partial E^{\text{SCF}}}{\partial \lambda}\right|_{\lambda=0} = \sum_{\mu\nu} P_{\mu\nu}^{\text{SCF}} Q_{\mu\nu}^{\lambda}
$$

Unrelaxed vs. relaxed density

The above definition of the density is incomplete, since it describes the unrelaxed correlated density. Our usual formulation of correlated methods is to describe correlation as some to-be-determined mixture of the virtual orbitals with the occupied orbitals. There is no such mixing in the above equations. In reality, $P_{ia}^{\text{MP2}} \overset{!}{\neq} 0$, so there is something that looks like a "transition density" with non-zero occupied-virtual blocks, in order to correct for the fact that our $C^{\text{SCF}}$ has no correlation. Allowing this mixing (which corresponds to orbital response) leads to the relaxed density. However, one has to solve the coupled-perturbed Hartree-Fock (CPHF or CPSCF) equations.

See between Eqns. (8) and (9) in [1] for the CPHF equations and how to solve them for the occupied-virtual mixing coefficients and finally the $P_{ia}^{\text{MP2}}$ block. Generally, note the similarities to the CIS density equations.

CPHF equations and first-order properties

As stated, when differentiating the energy with respect to some perturbation $\lambda$ to calculate first-order properties, one needs to contract the density with the operator property matrix. If you transform Eqn. (6) from the AO into the MO basis,

$$
\begin{align}
\left.\frac{\partial \mathscr{E}}{\partial \lambda}\right|_{\lambda=0} &= \sum_{pq} \mathscr{P}_{pq} Q_{pq}^{\lambda} \\
&= \sum_{pq} P_{pq}^{\text{HF}} Q_{pq}^{\lambda} + \sum_{pq} P_{pq}^{\text{MP2}} Q_{pq}^{\lambda},
\end{align}
$$

the SCF contribution simplifies due to the nature of the SCF density matrix in the MO basis,

$$
P_{pq}^{\text{HF}} = \left\{
\begin{array}{lr}
2 & \text{if }p = q\text{, p is an occupied MO} \\
0 & \text{otherwise}
\end{array}
\right.,
$$

leading to the expression

$$
\begin{align}
\left<\hat{\lambda}\right> &= \sum_{pq} P_{pq}^{\text{HF}} Q_{pq}^{\lambda} \\
&= 2 \sum_{i} Q_{ii}^{\lambda} \\
&= \mathrm{tr}\left(\mathbf{P}^{\text{AO}}\mathbf{Q}^{\text{AO}}\right) \\
&= \mathrm{tr}\left(\mathbf{P}^{\text{MO}}\mathbf{Q}^{\text{MO}}\right)
\end{align}
$$

for one-electron operators and wavefunctions that obey the Hellmann-Feynman theorem. I absorbed the 2 into $\mathbf{Q}^{\text{MO}}$, but this depends on the program implementations and their definition of the density. The point is that this contribution is relatively simple, and illustrates how for one-electron operators (dipole, quadrupole, angular momentum, ...), evaluating them for a converged Hellmann-Feynman wavefunction is very fast: one the density and property integrals are in the same basis, either do a matrix multiply and then trace, or do an element-wise (Schur) product and accumulate all the elements, so it scales as $O(N^3)$. No solution of CPHF equations is required. The same goes for the gradient with respect to nuclear coordinates $\frac{\partial E^{\text{SCF}}}{\partial x}$; it isn't a one-electron operator, and requires a mathematical trick to make $\frac{\partial P^{\text{SCF}}}{\partial x}$ disappear, but the SCF gradient doesn't require solving the CPHF equations either. This isn't true for $\frac{\partial E^{\text{MP2}}}{\partial x}$, which requires $\frac{\partial P^{\text{MP2}}}{\partial x}$, where you need to solve the CPHF equations to even form the full $P^{\text{MP2}}$.

Consider the remaining MP2 part of the property evaluation; the total density contains contributions from the occupied-occupied (oo) block, the virtual-virtual (vv) block, and the occupied-virtual (ov) block. I'm going to switch from $P^{\text{MP2}}$ and $c$ to $D$ and $t$ to make typing easier.

$$
\sum_{pq} D_{pq} Q_{pq}^{\lambda} = \sum_{ij} D_{ij} Q_{ij}^{\lambda} + \sum_{ab} D_{ab} Q_{ab}^{\lambda} + 2\sum_{ia} D_{ia} Q_{ia}^{\lambda}
$$

The first two terms can already be calculated from the unrelaxed density; $D_{ia}$ is the term that must come from solving the CPHF equations.

In general, the form of solving CPHF equations looks like $\mathbf{AU} = -\mathbf{V}$, where $\mathbf{A}$ is the orbital Hessian (2nd derivative of the energy with respect to orbital rotations), $\mathbf{V}$ is the gradient of some perturbation $\hat{V}$. In [1], it's $\lambda$, but Neese [3,4] and the Scandinavians usually use $V$ or $Q$ when the perturbation is associated with an observable. This perturbation gradient, termed the right-hand side (or RHS), describes how the perturbation might cause occupation of virtual orbitals, and the orbital Hessian describes how the response of one MO to some perturbation is coupled to the other MOs [5,6].

Start by rewriting $\sum_{ia} D_{ia} Q_{ia}^{\lambda}$ as the combination of a perturbation-dependent response $\mathbf{U}$ and the perturbation gradient $\mathbf{X}$:

$$
2\sum_{ia} D_{ia} Q_{ia}^{\lambda} = 2\sum_{ia} X_{ai} U_{ai}^{\lambda}
$$

$\mathbf{X}$ serves the same purpose as the $\mathbf{V}$ mentioned above, but rather than coming from an external perturbation, it is an intermediate quantity that describes the MP2 wavefunction.

First, the perturbation-dependent response. This part is not specific to MP2, and is also important for higher-order SCF-based molecular response [3].

$$
\begin{align}
U_{ai}^{\lambda} &= \sum_{bj} (\mathbf{A}^{-1})_{ai,bj} Q_{bj}^{\lambda} (\varepsilon_{j} - \varepsilon_{b})^{-1} \\
A_{ai,bj} &= 1 + \left( \left<ab||ij\right> - \left<aj||ib\right> \right) (\varepsilon_{a} - \varepsilon_{b})^{-1}
\end{align}
$$

The $\varepsilon$ are MO energies. If we don't assume the use of canonical MOs, this energy denominator is actually $F_{ab}S_{ij} - F_{ij}S_{ab}$.

Jumping ahead, the relationship between the ov density, the orbital Hessian, and the MP2 RHS is [Eqn. (9)]

$$
\sum_{bj} D_{bj} (\varepsilon_{j} - \varepsilon_{b}) A_{bj,ai} = X_{ai}
$$

See how this is structurally similar to solving for $U$, but the RHS is different. To show the equivalence between all quantities,

$$
\begin{align}
2\sum_{ai} X_{ai} U_{ai}^{\lambda} &= 2\sum_{ai} X_{ai} \sum_{bj} (\mathbf{A}^{-1})_{ai,bj} Q_{bj}^{\lambda} (\varepsilon_{j} - \varepsilon_{b})^{-1} \\
&= 2\sum_{bj} D_{bj} Q_{bj}^{\lambda}.
\end{align}
$$

The explicit equation for $X_{ai}$ is [Eqn. (11b)]

$$
X_{ai} = \sum_{jk} D_{kj} \left<ki||ja\right> + \sum_{bc} D_{bc} \left<bi||ca\right> - \frac{1}{2} \left[ t_{kj}^{ab} \left<ib||kj\right> + t_{ij}^{cb} \left<cb||aj\right> \right],
$$

which is in fact the MP2 Lagrangian, $L_{ai}$ [2, Eqn. (11)]. Again, notice the similarity to the CIS Lagrangian. The Lagrangian is required because the wavefunction doesn't need to be relaxed in order to calculate the correlation energy, so you pay the price here. These equations must be evaluated for MP2 geometry optimizations (during the gradient part) or for fully-relaxed MP2 molecular properties. To be clear, if one is only interested in the density and not molecular properties, only $\mathbf{DA} = \mathbf{X}$ needs to be solved for, and not the equations involving $\mathbf{U}$. In terms of scaling, SCF-based CPHF is about $O(N^4)$ (after back-transforming the $\mathbf{A}^{-1}\mathbf{Q}$ terms to be AO-direct) but might require fewer iterations than the SCF itself. According to [2], the MP2 gradient is something like $O(ON^4) + O(O^{2}V^{3})$, but they can do similar tricks during iterations to cut down on the number of fifth-order steps. In my experience, MP2 gradient calculations are about as expensive as an SCF polarizability calculation.

References

Trucks, G. W.; Salter, E. A.; Sosa, C.; Bartlett, R. J. Theory and implementation of the MBPT density matrix. An application to one-electron properties. Chem. Phys. Lett. 1988, 147 (4), 359-366. DOI: 10.1016/0009-2614(88)80249-5
Frisch, M. J.; Head-Gordon, M.; Pople, J. A. A Direct MP2 gradient method. Chem. Phys. Lett. 1990, 166 (3), 275-280. DOI: 10.1016/0009-2614(90)80029-D
Neese, F. Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling. Coordination Chemistry Reviews 2009, 253 (5-6), 526-563. DOI: 10.1016/j.ccr.2008.05.014
I cannot recommend the above paper enough; it contains excellent descriptions of how electric and magnetic response are calculated with explicit equations that are actually usable. There, you can see how CPHF is used for molecular response properties and the connection with excitation energies, found by solving $\mathbf{AU} = \mathbf{0}$ (diagonalization rather than solving a set of linear equations).
Toulouse, J. Introduction to the calculation of molecular properties by response theory.
Parrish, R. M. Psi4 Development: Theory Supplement, CPHF for Dipole Polarizabilities.


You need both an empty LUMO for the electrophile with an in phase HOMO for the nucleophile at the site of attack AND the proper charge distributions.

The LUMOs are critical; you can't shove electron density into more electron density, and LUMOs represent a lack of electron density.  However, the electron density for a nucleophile will be attracted to the partial positive charges.

Organic chemistry normally requires you make judgments about the relative impacts of various phenomena, and this is another example.


Guidelines for Solubility in Water of Organic Compounds in Relation to Carbon Chain Length

To determine the solubility of some compound, a chemist would systematically throw it into water and see what happens. You may draw some correlations like "plus six for two hydroxyls, minus five for five carbons, should be soluble"; they are somewhat useful, but mostly qualitatively.

There is a whole hierarchy of computational methods for prediction of solubility, but these are highly complicated.

BTW, haloalkanes are almost as insoluble in water as alkanes themselves.


Why is polyethylene not called polymethylene?

According to the Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008, i.e. the ‘Purple Book’), three different types of names can be used for polymers:

traditional names,
structure-based names, and
source-based names.

When traditional names fit into the general pattern of systematic nomenclature, they are retained, in this case: polyethylene (PE)

The systematic name for a polymer requires the naming of a preferred constitutional repeating unit (CRU). If necessary, this basic name is then modified by prefixes. The systematic structure-based name for polyethylene according to the Purple Book is indeed poly(methylene).

Polymers can also be named as being derived from a monomer (or precursors), which is named according to IUPAC rules. Such names are referred to as source-based names. The systematic source-based name for polyethylene according to the Purple Book is polyethene.


Why does branching decrease density and melting point?

I'm not sure that your statement about density is always correct. Looking at octane and its isomers the density changes very little, for example

octane           $703~ \pu{kgm^{-3}}$

2-methylheptane  $698~ \pu{kgm^{-3}}$

2,5-dimethylhexane  $694~ \pu{kgm^{-3}}$

2,2,4-trimethylpentane  $692~ \pu{kgm^{-3}}$

although the melting points do change considerably. In the same order as above they are $-57, -112,-93,-107$ so octane is the exception here; (data from Wikipedia pages).

This can be explained by the fact that melting can be considered to be governed by the repulsive part of the intermolecular potential this having a greater influence than the attractive part when the molecule has an odd shape. Rod like octane has to have deeper intermolecular potential than the odd shaped isomers as it has a higher melting point.

In the isomers the interaction should be dominated by the repulsive part of the potential as the odd shapes must try to push methyl groups into to one another. The potential itself has the same overall geometry, i.e. similar intermolecular minimum distance, as density is the same in all molecules above, but is shallower in nature due to the extra repulsion. Hence the isomers have significantly lower boiling points than octane.


What is the bonding picture in Peroxomonosulfate?

The peroxomonosulfate is a hypercoordinated molecule, in which the sulfur appears to exceed Lewis' rule of two, better known as the octet rule. In former times it was believed that d orbitals would contribute to bonding. This is not the case.
Bonding is better represented by

three equivalent $\ce{S-O}$ single bonds
sulfur having a formal charge of +2
these three oxygens having a formal charge of -1
one $\ce{S-O}$ single bond that leads to the peroxyl oxygen
this oxygen has a formal charge of zero
one $\ce{O-O}$ single bond
and the terminal oxygen having a formal charge of -1

This bonding picture is much more in accordance with quantum chemical calculations and an NBO analysis on the DF-BP86/def2-SVP level of theory. The involvement of d orbitals is minimal (<1%).

It is noteworthy, that this molecule pretty much has the same bonding around the sulphur as in sulphate. It can therefore similarly be described. Have a look at my rant against hypervalency or the bonding picture of the $\ce{PO4^{3-}}$ anion, which is isoelectronic to sulphate.


Mechanism for oxidative cleavage of tertiary alcohols by NaOCl

Reaction of an ester with a Grignard reagent gives two products: (1) a tertiary alcohol coming from the carbonyl carbon and two equivalents of the Grignard reagent and (2) an alcohol liberated from the ester. I believe it's this liberated alcohol that is reacting with bleach, since a tertiary alcohol is typically resistant to oxidation. However, the liberated alcohol might be primary, secondary, or tertiary depending on the identity of R2.

The information given says that after treatment with bleach, acetic acid is formed. This tells us two things: (1) that liberated alcohol is a primary alcohol, since the oxidation goes all the way to an acid and (2) that liberated alcohol has two carbons, since acetic acid has two carbons. That implies that the liberated alcohol is ethanol. The starting ester has four carbons total, which leaves two carbons for the carboxy fragment. The original ester was ethyl acetate.


Two liquids that react to produce a gas less dense than air at room temperature

If the other products aren't significant, I'll recommend the all famous alkali metals' reaction with water:
$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$
Hydrogen gas is less dense than air, both at RTP and at STP, because the main components of air (by mole percent) are $\ce{N2}~(78\%)$, $\ce{O2}~(20.95\%)$ and $\ce{Ar}~(0.934\%)$ - all of which have a molar mass larger than hydrogen.


Are the vibrational modes of CO2 active at room temperature?

Yes, the vibrational modes are present at any temperature, including absolute zero where the lowest vibrational energy levels only are populated; the zero-point levels. The molecule is linear ( OCO ) so it has $3N-5=4$ vibrational modes for $N=3$ atoms. The symmetric stretch  corresponds to both CO bonds stretching in phase, and asymmetric stretch to one CO stretching while the other compresses (this has highest frequency), and there are two degenerate bends where the OCO angle changes (lowest frequency). If the z-axis is along the OCO bonds, bends are in the zx and in the zy planes. Wikipedia has lots of diagrams showing these vibrational normal modes.

I have calculated the heat capacity as shown in the figure. The translational value is 3R/2 at all but the lowest temperatures (a few K) and similarly for rotational levels, which have a value of R, again above about 50 Kelvin. (The rotational constant is 0.3902 wavenumbers). The vibrational heat capacity from the modes (1388, 667,667, and 2349 wavenumbers) gives the rising curve on the figure. At 300 K the value is about 3.4 R or 28 J/mol/K. It is clear that only a few vibrational quanta are populated at room temperature. The high temperature value of the heat capacity adds another 4R (each vibration counts R) and is not reached until about 4000 K. Clearly never obtained in practice.

Total heat capacity for CO2 vs temperature. The blue line is the total heat capacity, the lower horizontal line has a value R, the upper one R+3R/2. The top horizontal line has no particular importance and is the value at 300 K and is approximately 3.4R as mentioned in the comments.


Microfiltration for saturated solutions of highly hygroscopic compounds?

Filtration is a mechanical operation.  Thus, if you have a solution of $\ce{ZnCl2}$, which (assuming you use water as solvent) dissociated into ions of $\ce{Zn^{2+}}$ and $\ce{Cl^-}$, a typical filter will not retain these ions.  Instead of using some vacuum (potential danger to evaporate some solvent, thus altering the concentration of he solution), you may consider filtration under pressure (e.g., across a Schlenk fritt for larger scale, or a syringe filter at smaller scale [e.g., testing a spin coating]).

(credit)
If the solvent capacity for $\ce{ZnCl2}$ is not (this much) temperature dependent, you equally could perform the filtration near room temperature (e.g., $\pu{432.0 g}$ $\ce{ZnCl2}$ per $\pu{100 g}$ of water at $\pu{25 °C}$ listed here).


Can one relate the frequencies of vibrational modes within a molecule to the molecule's HOMO/LUMO energies?

No, I could not think of any relation between them. Orbital energies are solutions of the electronic Schrödinger equation, and as such, they describe the electronic state of a molecular system. Vibrational frequencies are solutions of the nuclear Schrödinger equation, and as such, they describe the nuclear state of a molecular system. The quantities that describe the electronic state are not related to the quantities that describe the nuclear one simply because they describe two different subsystems: the electronic and the nuclear one.

Note that usually we treat these two subsystems as being independent of each other (the Born-Oppenheimer approximation), but even if you go beyond that, what you'll see is that changes in the nuclear subsystem (say, changes in geometry during vibrations) might affect the electronic state. Consequently, molecular vibrations in principle can alter HOMO/LUMO (as well as any other orbital) energies, but I doubt this interdependence of electronic and nuclear states is the relation you're looking for.


Best way to raise pH without alkali metal?

Tetramethylammonium hydroxide

This is a strong base like NaOH.

See Behavior of Polyethylene Oxide Based Nonionic Surfactants
in Silicon Processing Using Alkaline Solutions J. Electrochem. Soc. 1995   volume 142, 621-627:

  alkaline cleaning solutions based on quarternary ammonium hydroxides such as choline ($pK_b = 0.1$) and tetramethyl ammonium hydroxide ($pK_b = 0$)

and Ceramic Nanomaterials and Nanotechnology II which says:

  In this research, tetramethylammonium hydroxide (TMAH) was used, instead of KOH, to adjust pH of the precursor solution to 14.0

Also, TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society (1905):

  An estimation of the strength of the base by means of the velocity
  of saponification of methyl acetate in N/80 solution showed that it was
  somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the
  strength of sodium hydroxide is represented as 100, that of tetramethylammonium
  hydroxide, under the above conditions, will be represented by 92.


Determining perovskite crystal structure

At Geoff's request I'll make comments into an answer:

Since it is a multi-component system you need to think of a crystal structure with a basis. Not surprisingly, the basis here is $\ce{ABO3}$. In your picture from Wikipedia, start with just the blue atoms - they form the simple cubic. Then each blue atom shares 6 red atoms with other blues, so pick three for a given blue. Then place the green atom at the cube center to go with the blue atom at the lower right front blue atom. This is now your basis, and placing that basis unit at each corner of the cube (repeating to infinity) gives you the crystal structure

To explain further: each red has two blue neighbors, so each red is 'half-owned' by a blue. For a basis, you want only whole atoms to keep your mind a bit saner, so pick three to be 'wholly-owned' by a blue, and leave the others to be owned by another blue. Taking the left lower center blue atom, I'd pick the red atoms in the +x, +y, and +z directions, then the green atom in the (x,y,z) direction to make a nice basis.

I'll also add some on deviations from cubic symmetry.  If the green atom is at (1/2,1/2,1/2) than it all remains simple cubic.  If the greens are big fat atoms compared to the blues, they may want a little more room, which leads to distortions.  Simply stretching along one dimension is pretty easy, since it leaves 2/3's of the blue-red pairs at the same interatomic distance, stretching the other 1/3.  This creates an orthorhombic crystal, still with the same $\ce{ABO3}$ unit as a basis, but now slightly elongated along one direction (commonly taken as z).


A 4.00 L sample of gas is cooled from 67 °C to a temperature at which its volume is 2.60 L

Hint: Use Ideal gas equation:

  $$\large PV=nRT$$

Given pressure is constant, can you tell the final temperature from this relation?


Triphenylphosphine vs. trimethylphosphine in reduction reactions

Yes it can. It also has the advantage that the trimethyl phosphine oxide produced is water soluble and thus easier to remove from a reaction mixture than triphenylphosphine oxide which is notoriously difficult to get rid of. I have used it for a Staudinger reduction of an alkyl azide when triphenyl phosphine was unreactive.

However, trimethyl phosphine is more expensive, pyrophoric, has an unpleasant odour and is toxic requiring full personal protective equipment. Safety data here: https://www.alfa.com/en/content/msds/english/30143.pdf


What is the density of compressed natural gas (CNG) used in vehicles?

The density of CNG at the pressure of 200 bars (which is optimized for use in vehicles) is 435 $\frac{kg}{m^3}$. However, it's (not usually) 175 $\frac{kg}{m^3}$ in some vehicles in some countries. Note that it differs according to your home country. Source 1 and source 2.


How exactly does the electrolysis of water work?

Electrolysis of pure water is very difficult, but adding only a small amount of ions make the process easily achieved. In most places, there are enough minerals in the water  that the ionic strength or conductivity of the water is  great enough for electrolysis to effectively occur without needing to add additional ions to the water

Water dissociates into H+ ions and $\ce{OH–}$ ions; the $\ce{H+}$ ions are attracted to the negative electrode (the cathode) and are converted (reduced) to a hydrogen atom ($\ce{H}$) (i.e. $\ce{e- + H+ -> H}$). This is highly unstable and immediately reacts with another hydrogen atom to produce $\ce{H2}$, molecular hydrogen gas.
At the other electrode (the anode), oxidation occurs.  The $\ce{OH-}$ ions are attracted to the positive electrode where they are oxidized to form oxygen gas ($\ce{O2}$) and hydrogen ions ($\ce{H+}$).  However, if chloride is present, it will oxidize (instead of the $\ce{OH–}$ ions) and form chlorine gas, which will then react with the water to form hypochlorous acid.

The cathode reaction is: $\ce{2H+ + 2e- (cathode) →H2 (g)}$
The  anode  reaction  is:  $\ce{2OH– → 4e-(anode) +O2 + 2H+}$
The overall reaction is: $\ce{2H2O → 2H2 (g) + O2 (g)}$


Can 4-aminophenol be used to produce 2,5-cyclohexadien-1-imine?

TL;DR

No

Explanation

As far as I understand, the reaction being suggested is as follows:

The first thing one should notice is that 2,5-Cyclohexadien-1-imine is not going to stay as it is. It will tautomerise to the more stable and aromatic aniline.

Let us go back to the starting material - 4-aminophenol. Both the oxygen and the nitrogen are directly attached to the aromatic $\pi$-system. Both can donate a lone pair and cause a +R effect. Their electronegativities would result in a -I effect, which is very small, compared to the +R effects. The point is, the starting material is electron-rich, i.e. nucleophilic. It would not leave a carbocation (if it does form) alone.

Also, the thing about the lack of a hydride donor...


pH of rain with and without thunderstorm

Though I've never heard of that statement before, it's possible that nitrogen oxides formed by electrostatic discharge (not only lightning, but even glow discharge) dissolve in the rainwater, producing nitrous and nitric acids.

There is a 1997 citation describing comparison of pH in rain accompanied (or not) by lightning.

That said, presence of fly-ash from coal burning, sulfur oxides from volcanic activity and carbon dioxide from natural as well as anthropogenic sources probably have greater influence on pH.


When a reaction occurs at constant pressure and no work other than expansion work  (such as electrochemical) is performed, then it can be shown, from the definition of the enthalpy and the first law of thermodynamics applied to a closed system, that the change in enthalpy is equal to the heat:
$$\begin{align}dH&=dU+d(pV)\\&=dw_{pV}+dw_{other}+dq+pdV+Vdp\\&=-p_{ext}dV+dw_{other}+dq+pdV+Vdp\\&=dq+dw_{other} \end{align}$$
The last equality follows when the pressure is constant and equal to the external pressure. When only pV work is done then clearly heat and enthalpy change are equal. Readily measured heats of reaction provide a very convenient way to determine values of the enthalpy state function which can be tabulated and employed in further thermodynamic computations.


What are the order of the reaction under the following conditions?

Let's compare the rate when $A$ and $B$ are present at their initial amounts, say $A_0$ and $B_0$, to when the reaction is 99% complete.

When $[\ce{A}]\gg[\ce{B}]$, the reaction will be 99% complete when 99% of $B$ is gone, i.e. when $B = 0.01B_0$.

At that point, $$\frac{dx}{dt} = k[A][B]^2 = k(A_0 - \frac{0.99}{2}B_0)(0.01B_0)^2 $$

Now since $A\gg B$ also means $A_0\gg B_0$, then it follows that $(A_0 - \frac{0.99}{2}B_0) \approx A_0$.  Thus, when the reaction is 99% complete and $B$ is limiting,

$$\frac{dx}{dt} \approx kA_0(0.01B_0)^2 = kA_0 (0.0001)B_0^2$$

If we lump $k$ and $A_0$ into a single parameter, $k_{app}$, then we can write $\frac{dx}{dt} = k_{app}B^2$, which shows that when $B$ is limiting, the reaction is 2nd order.

The case for $A$ being limiting shows the reverse.


How to calculate the pH of a propanoate buffer after addition of sodium hydroxide?

The addition of the NaOH will "use up" 0.01 mol of propanoic acid and "create" 0.01 mol of propionate. To make things easier, just work in molarity. Thus 0.02 M NaOH.

The pH of the final solution can be found by adjusting the equation you used in part b,

$pH = -log \left( 1.26 \times 10^{-5}\frac{0.1-0.02}{0.05+0.02} \right)=4.84$

The result agrees with our expectations. Since the solution is a buffer, the pH is not expected to drastically change with the addition of a strong base. It is buffered.


How to compute the NMR shift in Gaussian09 for only one specific atom in a bigger unit cell?

Short answer, you can't.

It might save a bit of time, but probably not much. The calculation would still need to go through the normal SCF on the entire system to calculate the NMR shifts.

In principal, in evaluating the property calculation, you could save a small amount of time by only computing the one atom, but for most systems, I would think the overall SCF calculation would be vastly greater than computing a few properties.


Help me proving Dalton's law of partial pressure formula

After having some discussion with @JohnRennie

Let $P_i=n_i \dfrac{RT}{V}$ then $P_i=\dfrac{n_i}{n_{\text{total}}}\dfrac{n_{\text{total}}RT}V$

$P_i=x_i P_{\text{total}}$

Where $x_i$ is mole fraction of $i^{\text{th}}$ particle.


Availabilty of magnesium ions in phosphate buffer

If you're still interested in this, fluorescence may be a good method to test the phosphate dependence on magnesium activity. First record a fluorescence spectrum of a solution of magnesium and fluorescent dye, then add small volumes of concentrated phosphate and note any changes in the fluorescence. Of course, you would have to do a control experiment to see if the phosphate is directly perturbing the fluorescence. This write-up describes some magnesium-chelating dyes that you could use, though if you don't want to spend a lot of money on a dye, you may already have some chemicals in your lab that will work. Phenol looks like promising alternative as it fluoresces, is decently water soluble (83 g/L), and forms a complex with magnesium which breaks planarity, which would certainly change the fluorescence spectrum.

It would only take a day to get accurate quantitative results if you have access to a fluorometer.


Effect on acidity of benzene due to electron withdrawing groups

Both the predictions are good. The $\ce{NO2}$ will actually be an effective electron withdrawing group, and the lost of that $\ce{H}$ will stabilize the compound (if there is and $\ce{NO2}$ on it). It is also important to rescue, that both effects will be stronger just on para or ortho position. You can make the different structures to check it.


Is this relation in case of a mixture of ideal gases valid?

This relation is not valid if the gases are assumed to be ideal. see Amagat's law of partial volumes
Moreover, if the gases are non ideal, the relations between the volumes of the combining gases and the final resultant volume is very complex and the given relation is too simple and includes too less variables to even begin describing the actual non-ideal gas mixing.
It, however unlikely, might represent a certain special case that can only be judged from the context wherein the aforementioned equation was taken from.


Does sulfur and phosphorus expand their octet?

Whether sulfur or phosphorous actually expand their octet is contested within the chemistry community. Another term for this octet expansion is "hypervalency." You can find many works of research regarding hypervalency. The consensus, according to Wikipedia, is that both can expand their octets, but not to a significant extent. In other words, d-orbital utilization is low at best.

Regarding sulfur trioxide, this article may be of interest.

  We find no evidence to support notions of p π –d π back-donation from oxygen to sulfur.

Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: $\ce{C, N, O}$, and $\ce{F}$. And note that for carbon, there are well known exceptions in organic chemistry - carbocations (which have less than an octet of electrons). Carbocations however tend to exist mainly as reactive intermediates. Therefore, a more nuanced statement might be $\ce{C, N, O}$, and $\ce{F}$, when involved in neutral and thermodynamically stable compounds, generally have 8 valence electrons.

  this makes the concept of octet somewhat incorrect.


Is it possible for a liquid chemical to have a pH value lower than its aqueous solution?

The concept of pH tends to break down when water is not greatly in excess of the solute (e.g. if your product is a highly concentrated amine base). The apparent pH of your product, when measured using a pH meter or an indicator, may be relatively low (seemingly less basic) if the solution has a less polar environment than that of pure water--because ionization will be less favorable.

If no water is present at all, the product might still give a pH reading with a meter or indicator, but the term "pH" in this case would be irrelevant.

It always helps to talk specifics with chemistry, so I am curious as to what the ingredients of the product are.


How to write the formula of a mixed metal salt?

"$\ce{KNaSO4}$" is correct, but not because of electronegativity order, but rather the alphabetical one.
From IUPAC "Red Book", section IR-2.15 ORDERING PRINCIPLES [1, p. 40]:

IR-2.15.2 Alphabetical order
Alphabetical order is used in formulae as follows.
(a) Within the group of cations and within the group of anions, respectively, in formulae of salts and double salts.
Deviations from this rule may be acceptable if it is desired to convey specific structural information, as in Example 5 below.
Examples:

$\ce{BiClO}$ (anions $\ce{Cl-}$ and $\ce{O^2-})$
$\ce{NaOCl}$ (the anion is $\ce{OCl-},$ cf. Section IR-4.4.3.1)
$\ce{KNa4Cl(SO4)2}$
$\ce{CaTiO3}$ (perovskite type)
$\ce{SrFeO3}$ (perovskite type)

Reference

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. (PDF)


How does the position of an electron donating group affect acidity in benzoic acid?

The strength of the acid is determined by strength of the positive charge that is on the carbonyl carbon so in case of the presence of an electron donating group you have also to check what is its inductive effect whether it has an electronegative part which would attract electrons towards it or not ....the best way is to draw the resonating structure for every case and you will reach how is the activity affected by the EDG in different positions..one last factor you have to bear in mind is that the presence of an OH group in ortho position to the COOH will form a hydrogen bond which makes it even more stable than other positions ..if we take salicylic acid as an example then the strength is ortho>meta>para


Propylene Glycol Degradation due to Sunlight

The thermal degradation of propylene glycol and its uses in e-cigarettes has been a discussed in a previous question here on Chem.SE.

  Upon complete oxidation and degradation, propylene glycol will fully
  decompose to carbon dioxide and oxygen. However, if there is partial
  oxidation, you will get some intermediate species - formaldehyde or
  acetaldehyde.(1st answer)

  Propylene glycol does not decompose on heating, it simply boils. In
  theory, under certain conditions (pressure/encapsulation/catalysts) it
  could dehydrate to either acetone or Propionaldehyde, probably the
  former.(2nd answer)

Also, this link shows the decomposition products of propylene glycol which are all acid based:

  The main degradation product of propylene glycol was lactic acid (2-hydroxypropionic
  acid), with formic, acetic, and oxalic acids as minor products.

So, mainly degradation of propylene glycol in sunlight would yield a bunch of acids and small amounts of acetone. Worrying about the odors is a personal medical question and is off-topic in Chem.SE.


Full Valence Shell (Octet Rule) vs Formal Charge

First, keep in mind that Lewis-structures and formal charges are huge simplifications.  In reality, at the quantum level, charges and bonds are spread out and it's somewhat unrealistic to say "Atom X has a charge of Q" or "Atoms X and Y have an Nth order bond between them".  There's merely a cloud of electron density, and it's somewhat arbitrary to say which parts of that cloud belong to a given atom or bond.

It's perfectly valid to draw a structure with a double bond.  And in reality, I'm sure there is hyperconjugation that leads to a small amount of double bond behavior.  However, such a resonance structure puts a negative charge on an electropositive element, and a positive charge on a very electronegative element, and thus such a structure has a smaller contribution to the overall structure.


The 21.0% of oxygen by volume in air is from dry air. Data for dry air from webpage of Columbia University's Department of Earth and Environmental Sciences.

Note that the "% by volume"/100 for dry air also corresponds to the partial pressure in atmospheres. For air saturated with water at 25 C, the vapor pressure of water, according to Wikipedia, is 0.0313 atm. The second column is calculated by multiplying all dry air values by (1-0.0313)=0.9687.

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Gas} &\hspace{-5em}\text{dry air}&\ce{H2O}\text{ sat. air} \\
 & \text{% by volume} & \text{% by volume} \\
\hline
\ce{N2} &  78.084    & 75.640 \\
\ce{O2}  & 20.947    & 20.291 \\
\ce{H2O} & 0.0      &  3.13 \\
\ce{Ar}  &  0.934    &  0.905 \\
\ce{CO2} & 0.035    &  0.034 \\
\ce{Ne}  &  0.001818 &  0.001761 \\
\ce{He}  &  0.000524 & 0.000508 \\
\ce{CH4} &  0.00017  &  0.00016 \\
\ce{Kr}  &  0.000114 &  0.000110 \\
\ce{H2}  &  0.000053 & 0.000053 \\
\ce{N2O} & 0.000031 &  0.000030 \\
\ce{Xe}  &  0.0000087&  0.0000084 \\
\hline
\text{TOTAL}& 100.0027187 & 100.0026304 \\
\hline
\end{array}
$$

Now correcting the data from Warneck and Williams (evidently Warneck,  P.  and  Williams,  J.:  The  Atmospheric  Chemist’s  Companion:  Numerical  Data  for  Use  in  the  Atmospheric  Sciences, Springer Verlag, 2012) for the value of water saturated air

$\ce{O2} = \dfrac{20.291}{21}\times 8.47 = 8.18$ mg/L

This is much better agreement with the Benson (1980) data.

The OP's value of 1.26e-3 Mol/L/atm for the Henry's constant for O2 in water at 25C was evidently take from Table 8.23 of Warneck and Williams. The value also has a list error bracket of +/- 0.02.

$$\dfrac{0.02}{1.26}\times 100\text{%} = 1.6 \text{ %}$$

For the two dissolved oxygen values:

$$\dfrac{8.26-8.18}{8.26}\times100\text{ %} = -0.97 \text{ %}$$

so the values are within the error limits for Henry's constant.


How does the presence of various NMR active nuclei in a compound affect coupling and signal multiplicity?

I'd like to expand on the other answer a bit (at least on the question of coupling).  An NMR active nuclei X can couple with (and detectably split peaks of) another NMR active nuclei Y if X is both not too rare (as discussed in the other answer) AND does not have an extremely short relaxation time.

I'll focus on the common elements that an organic chemist might work with:  $\ce{H}$, $\ce{C}$, $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$, $\ce{P}$, and toss in $\ce{Mg}$ (as mentioned in the question), $\ce{Li}$ and $\ce{Si}$.

"The good"

NMR active isotopes with high natural abundance and not extremely short relaxation times: in our list $\ce{^1H}$, $\ce{^19F}$, $\ce{^31P}$, and, marginally, $\ce{^29Si}$ and $\ce{^6Li}$.
These all are decent NMR nuclei and abundant, so they definitely split peaks of nearby NMR active nuclei.

$\ce{^1H}$:  99.98% natural abundance, spin 1/2 (so 0 quadrupolar moment), T1
around 10 seconds for the standard.
$\ce{^19F}$:  100% natural abundance, spin 1/2, sharp peaks (thus
relaxation times not too short).
$\ce{^29Si}$:  A bit marginal.  Around 5% natural abundance (so the 'side
peaks' caused by splitting come from only 5% of the molecules and
thus will be small, called satellites), spin 1/2.
$\ce{^6Li}$:
Around 8% natural abundance, spin 1, but is an honorary spin 1/2 as
it has a very small quadrupolar moment.  Relaxation times are long.

These nuclei will split nearby NMR active nuclei (not many bonds away).  This is either a bug or a feature.  The splitting pattern can tell you a lot about the molecule along with other advantages you can get from internuclear coupling (I have a colleague who uses this all the time on fluorocarbons).  If you don't want the effect, though, you will have to use a decoupling pulse on these nuclei.

For instance, in the text "Organic Structures from 2D NMR Spectra" (Field, Li, Magill) the examples include many uncoupled experiments (Note:  Y{X} means Y is observed while uncoupling X): $\ce{^13C}\{\ce{^1H}\}$ (often), $\ce{^19F}\{\ce{^1H}\}$, $\ce{^13C}\{\ce{^1H, ^19F}\}$, $\ce{^1H}\{\ce{^31P}\}$, $\ce{^31P}\{\ce{^1H}\}$.

"The bad"

There are nuclei that, if enriched, would split nearby peaks (having not-too-short relaxation times), but have a low natural abundance.  These can be analyzed quite well using NMR, and may have peaks split by other active nuclei, but won't usually cause visible splitting in other nuclei.  Out of our list:  $\ce{^13C}$, $\ce{^15N}$.

$\ce{^13C}$:  Only 1% natural abundance, spin 1/2.  While it is often
examined either alone or in conjunction with other active nuclei, it
won't cause anything but the tiniest satellites in other nuclei.
$\ce{^15N}$:  Even worse.  Spin 1/2 but only 0.4% natural abundance.

"The ugly"

These might or might not have high natural abundance, but they have short relaxation times.  When relaxation times are very short, the nuclei is, in effect, self-decoupling.  Nuclei with very short relaxation times have, though the Heisenberg Uncertainty Principal, broad lines and thus often are pretty hard to work with in any NMR context.  They certainly don't show visible splitting of the signal of nearby nuclei.  From our list, we have:  $\ce{^14N}$, $\ce{^33S}$, $\ce{^35Cl}$, $\ce{^37Cl}$, $\ce{^79Br}$, $\ce{^81Br}$, $\ce{^127I}$, $\ce{^17O}$ and $\ce{^25Mg}$.  At least you don't have to use decoupling pulses on these nuclei!

$\ce{^14N}$:  Too bad.  Even though it's 99.6% abundant (the rest is $\ce{^15N}$), it
has spin 1 and a reasonably large quadrupolar moment, thus a short
relaxation time (often around 20 ms).  It might broaden the peaks of
nearby nuclei, but won't show coupling in any but the most symmetric
environments.
$\ce{^33S}$:  Ouch.  0.8% natural abundance, spin 3/2, high quadrupolar
moment, only NMR active sulfur nuclei.
$\ce{^35Cl}$, $\ce{^37Cl}$:  Between these, 100% abundant.  Both are spin 3/2 with
VERY large quadrupolar moments and short relaxation times.
$\ce{^79Br}$, $\ce{^81Br}$:  Same as the sad fate of chlorine.
$\ce{^25Mg}$:  Only active NMR nuclei of Mg, 6% abundance, spin 5/2, VERY
large quadrupolar moment.
$\ce{^17O}$:  The only NMR active oxygen nuclei.  Very rare (0.04% natural
abundance), spin 5/2 with a fairly large quadrupolar moment.
$\ce{^127I}$:  100% natural abundance, but VERY large quadrupolar moment.
Maybe also prone to CSA relaxation (I'm not sure).

COMMENTS on short relaxation in nuclei

Nuclei with very short relaxation times often get these via two relaxation processes.  One is quadrupolar relaxation (only for spin > 1/2) and chemical shift anisotropy (CSA) relaxation.

For high spin (>1/2) nuclei, relaxation can occur because charge on the nuclei is not symmetric (thus a quadrupolar moment) so that energy can be exchanged with an environment that is not highly symmetric.  The effectiveness of this type of relaxation rises as the square of the quadrupolar moments times the square of the applied magnetic field.  Spin 1/2 nuclei have quadrpolar moment zero.  Quadrupolar moments range all the way from values such as 0.0028 ($\ce{^2H}$), -0.0008 ($\ce{^6Li}$) to values such as 0.31 ($\ce{^79Br}$).

CSA:  Also caused by tumbling in an asymmetric environment.  Larger for nuclei with large shift ranges, proportional to gyromagnetic ratio squared times magnetic field strength squared.  Not really predominant in protons or any carbon but Quaternary carbon.

These relaxation mechanisms don't work if the environment of the nuclei is very (especially tetrahedral or octahedral) symmetric, in which case relaxation times can be long.

The perverse thing about these relaxation mechanisms is that they get worse (more decoupling) with higher field strength.  Usually anyone working with NMR dreams night and day of getting a higher field spectrometer.  However, if you are interested in a nuclei prone to CSA relaxation (say, $\ce{^199Hg}$ or $\ce{^195Pt}$, which are spin 1/2 and not prone to quadrupolar relaxation), and you want to see splitting of peaks (coupling) in nearby nuclei, you might have to move to a LOWER field NMR.

References

A good site with lots of information on NMR active nuclei is the NMR periodic table.


Why is the value of specific rotation different in different cases?

A polarimeter is typically a long cylindrical tube with flat ends. I think the 25 cm was referring to the diameter of the tube. The other wrinkle here is that the angle measurement may be + or -. So 357.75 may in fact be -2.25.

Specific Rotation (in first case)=$$\frac{-2.25^{0}}{(\frac{4.5}{250})(1)}= - 125^{0}$$

Specific Rotation (in second case)=$$\frac{-4.50^{0}}{(\frac{4.5}{125})(1)}=-125^{0}$$


Calculating concentration of dissolved ions

You're right and the given answer is wrong. Here's the math:

$$\left(1\;\mathrm{mol}\;\ce{Fe2(SO4)3}\over {399.88\;\mathrm{g}\;\ce{Fe2(SO4)3}}\right)\left({2.0\;\mathrm{g}\;\ce{Fe2(SO4)3}\over 100\;\mathrm{cm}^{3}}\right)\left({3\;\mathrm{mol}\;\ce{SO4^2-}\over 1\;\mathrm{mol}\;\ce{Fe2(SO4)3}}\right)\left({10\;\mathrm{cm}\over 1\;\mathrm{dm}}\right)^{3} =\ce{[SO4^2- ]} $$

$$\ce{[SO4^2- ]}= 1.5\cdot10^{-1}\;\mathrm{mol}\;\mathrm{dm}^{-3}$$


I'm not a big fan of MOs when this can be explained using just AOs. For sure, MO theory gives a much better and profound explanation, and I'm surely waiting for someone to do that, but that's what I would answer if the kids asked me this.

Oxygen has 6 electrons on its outer shell, which has $\mathrm{2s}$ and $\mathrm{2p}$ orbitals. The arrangement of these electrons is such that $\mathrm{2s}$ sublevel is completed, and $\mathrm{2p}$ sublevel contains 4 electrons. Following Hund's rule,  two electrons occupy one of three $\mathrm{2p}$ orbitals and other two come by one to each of $\mathrm{2p}$ orbitals left.

When water forms, oxygen shares two non-bonded electrons with hydrogen atoms, forming a covalent bond with them.

$\ce{H3O+}$ forms via donor-acceptor bonding, when oxygen supplies its lone pair, which it has two, to form a bond with $\ce{H+}$ ion.

To form $\ce{H3O-}$, you'll need an interaction between water molecule and $\ce{H-}$ ion. Even if we ignore the fact that hydride is a powerful base ($\mathrm pK_\mathrm a$ is around $42$) and will react violently with water, it would be impossible. In this theoretical case neither of the species has an orbital suitable to accept a lone pair from another species. Each has a complete outer shell.

Edit: it really exists though. Wow.

Sources:

Probing the transition state via photoelectron and photodetachment spectroscopy of H3O−
Calculation of the photodetachment spectrum for H3O−


Is the correct zeroth-order wave function $\phi^{(0)}$ an eigenfunction of the perturbation Hamiltonian $H'$?

No, writing $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ will be incorrect. I will give you two explanations for the same, one intuitive and other mathematical.

Intuitive Explanation:
Assuming that you can write $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$. But, $\hat{H}_0\psi_3^{(0)} = E_3^{(0)} \psi_3^{(0)}$, and in an overall sense, you can write
$$\hat{H}\psi_3^{(0)} = (E_3^{(0)}+E_3^{(1)}) \psi_3^{(0)}$$
This essentially means that electron-electron repulsion is keeping the wavefunction same as that of single electron Hamiltonian. Now, since in single electron Hamiltonian, each electron does not "see" the other electron, and they are allowed to be in the same spatial position. Now, on bringing the $1/r_{12}$ term inside the Hamiltonian, the energy goes to $\infty$ for such a case, and hence a singularity is created in the system.
So, it means that the $\psi_3^{(0)}$ is not an eigenvector of total Hamiltonian, $\hat{H}$. Since $\psi_3^{(0)}$ is defined to be eigenvector of $\hat{H}_0$, so the initial assumption that $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ is incorrect.

Mathematical Explanation:
You have given $\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$, which is correct. Along with that, we know  $\left\langle \psi_{m}^{(0)} \middle|  \psi_n^{(0)} \right\rangle = \delta_{mn}$. However, you could write the given statement $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ provided $\psi_n^{(0)}$ would be a complete basis set (for explanation, see below). However, $\psi_n^{(0)}$ is not a complete basis set, since for a He atoms, there are infinite number of ways to put two electrons in an orbital, since there are infinite number of orbitals. Since you are treating the orbitals as hydrogenic, it is not guaranteed that the orbitals will be orthogonal to each other once the $1/r_{12}$ term is pulled in.
Even for a hypothetical case if we consider only 1s and 2s orbitals are present and are orthogonal to other orbitals even after the presence of the $1/r_{12}$ term, still they would not be a complete basis set. The reason is simple: You have considered only $1s^12s^1$ configurations, and not $1s^2$ and $2s^2$.

For complete basis, proof:
Assuming $\psi_n$ is an othronormal and complete basis set of dimension $M$, one can
write: $$\hat{H}'\psi_3^{(0)}=\sum_{i=1}^M c_i \psi_i^{(0)}$$ Now,
$$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)} \Rightarrow C_m=0
> \forall m\neq 3$$ $$\Rightarrow \hat{H}'\psi_3^{(0)}= c_3
> \psi_3^{(0)}$$
However, this would have been true if, along with other conditions,
$\psi_n^{(0)}$ would have been a complete basis set.

So, from the given information, it is impossible to conclude that $\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$ mathematically, and from a chemistry standpoint, $\hat{H}'\psi_3^{(0)} \neq E_3^{(1)} \psi_3^{(0)}$.


Why is the relative atomic mass of carbon not exactly 12?

Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.


Dissolution of Pentahydrate of Copper Sulfate

When copper(II) sulfate pentahydrade dissolves in water it actually just becomes further hydrated, in what is called an aquo complex (a coordination compound containing only water molecules as ligands). According to this Wikipedia article:

  The pentahydrate ($\ce{CuSO4·5H2O}$), the most commonly encountered salt, is bright blue. It exothermically dissolves in water to give the aquo complex $\ce{[Cu(H2O)6]^2+}$, which has octahedral molecular geometry. The structure of the solid pentahydrate reveals a polymeric structure wherein copper is again octahedral but bound to four water ligands. The $\ce{Cu(II)(H2O)4}$ centers are interconnected by sulfate anions to form chains.

The aquo complex maintains the blue coloration that is present in the pentahydrate solid.


Why isn't ΔG equal to zero for this equilibrium reaction?

You are mixing up two different quantities. The equation you are using describes the standard Gibbs free energy of reaction $\Delta G^{0}$, i.e. $\Delta G$ for standard conditions without any requirement of the reaction being in equilibrium.
And this quantity can very well be different from zero.
For non-standard conditions the Gibbs free energy of reaction is

\begin{equation}
 \Delta G = \Delta G^{0} + RT \log K
\end{equation}

and as you stated correctly this quantity must (per definition) be equal to zero for equilibrium conditions.

For a derivation see this answer of mine.


Are Group 2 metal fluoride salts less soluble in organic solvents than Group 1 fluorides?

Don't get your hopes up.  Wikipedia identifies magnesium fluoride as insoluble in ethanol.  Calcium fluoride fares no differently in acetone.  While these data are hardly exhaustive, they suggest strongly that Group 2 fluorides are not a good way to get fluoride ions into solution.  Of course, we would not expect Group 2 ionic compounds with their doubly charged ions to get vary far in organic solvents generally.
Getting fluoride ions into solution with organic solvents is typically done with bulky 1+ cations, such as N,N,N-trimethyl-(2,2-dimethyl-prop-1-yl)ammonium ion, $\ce{C_{8}H_{20}N^+}$.  See Reference [1] as an example.  The use of cryptand-complex alkali-metal complexes fits this paradigm because such ions are also bulky.
Reference
1.
Victoria K. Davis, Stephen Munoz, Jeongmin Kim, Christopher M. Bates, Nebojslš a Momčilović, Keith J. Billings, Thomas F. Miller III, Robert H. Grubbs and Simon C. Jones, "Fluoride-ion solvation in non-aqueous electrolyte solutions", Mater. Chem. Front.,
2019, 3, 2721


Can sodium dithionite reduce the nitro group of 2-nitrobenzoic acid?

Yes I believe it can. Dithionite is a reliable reducing agent for aromatic nitro groups. The reaction of dithionite produces an acid solution which will need to be neutralised before isolation. Anthranilic acid is not very water soluble so an extraction with e.g. chloroform should separate the product cleanly. Tetrahedron Lett. 2013, 54 (21), 2600–2603 discusses dithionite reduction of o-nitrobenzamides and refers to other papers that have used dithionite on similar substrates.


You are right in saying that in your question, the gas is an non-ideal (real) gas. Basically, the reason why we use the ideal gas equation despite knowing that it is a real gas is simply because that the equation for real gases is too complicated and contain more variables which are hard to find given limited data.

The reason for this complicated formula is that several assumptions that we make for ideals gases don't hold for real gases. Recall that an ideal gas is considered to be a point mass - a particle so small that the volume of that particle is negligible. A real gas particle does have real volume. For an ideal gas, the collisions between gas particles was said to be "elastic" - no attractive or repulsive forces exist, and thus, no energy is exchanged during collisions. For a real gas, collisions are non-elastic. So, the ideal gas law must be corrected for this extra forces. There are several real gas laws, one is the van der Waals equation:

$$\left[P + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$

Notice how "corrections" are being made to the pressure term and the volume term. Since collisions of real gases are non-elastic, the term $a(n/V)^2$ is correcting for the interactions of these particles. Since real gas particles have real volume, the $nb$ term is correcting for the excluded volume. The values of $a$ and $b$ are constants, and must be determined experimentally for each gas.

As you can see from this equation that first the values of $a$ and $b$ must be known then you must solve this equation which can be complicated. Even then, when you do finally solve it, you will probably get a result which is similar to the result obtained if you used the ideal gas equation. This is why the ideal gas equation is continued to be used as in most cases it provides quite accurate results while bypassing complicated formulas.


Latent heat isn't heat, it is energy released or absorbed during a constant temperature phase change. The kinetic theory of gasses has little to say about phase changes. (Which by definition involve liquids or solids and may or may not involve gasses, most often "not".) Claiming a phase change releases latent heat is false, an error of comprehension (or articulation). If I slowly accelerate a tank of gas, its kinetic energy increases with no increase in temperature. Temperature is a measure of kinetic energy, not the only measure of kinetic energy. (Which is obvious since you know that temperature doesn't increase as most solids at their melting point do not increase in temperature as heat energy is added to them.) In the broadest (most general) sense, latent energy is the energy difference between the two phases. So where it goes (or where it comes from) is in converting one phase to the other. The assumption here is one phase is lower in potential energy than the other. You may know that entropy is defined in terms of heat and temperature. In classical thermodynamics energy and entropy are two fundamental concepts, they are on equal footing. In statistical thermodynamics, energy is still fundamental, while entropy can be derived from the ensemble of possible states that a system can be in. Whoops! Too much! In the case of ice melting, the intermolecular bonds (mostly hydrogen bonding) are 'broken' (which means there are a lot fewer of them and they are (on average) weaker). The energy that was potential energy of the bonds becomes "kinetic" energy as chemists understand the term in this context. In this context (as contrasted with the kinetic energy of a bullet or a ballistic electron), the kinetic energy includes all of the energy of translation, rotation and vibration: the rotation of a molecule doesn't change its speed, nor do the movements of the two hydrogen atoms with respect to the oxygen atom - their bond to the central O atom can be looked at as a (coil) spring and they can stretch and bend that spring. Faster and greater spring movement occurs during melting. (The rotation of the H atoms also occurs but the energy required to spin a H atom about its bond is very small as bond energies go, so is generally neglected with little loss of accuracy; this wouldn't be true for bonds between larger groups, where rotation around the bond could require quite a bit of energy). So, in the isothermal melting of ice, the place where the energy is going is in these intramolecular movements. (intra = inside). And in the isothermal freezing of water, the place where the energy is coming from is these same intramolecular motions. There is no one unique definition of kinetic energy and another for potential energy. In a group of atoms or molecules, the 'potential' energy is 'internal' and kinetic is 'external' and yet if one focuses in on a single molecule, some of its potential energy is due to its movement (or movements of its parts), so what is and what is not k.e. or p.e. may change with the context. In fact, the basic laws of energy conservation may be K.E. + P.E = constant or K.E. = P.E. (which is often expressed as K.E. - P.E. = 0. These two approaches are embodied in Hamiltonian and LaGrangian Mechanics which are the two related but complimentary ways modern physics is expressed. The LaGrangian is L = T-V (where T is k.e. and V is p.e.) and the Hamiltonian is H = T + V. These are used in both classical Physics and in quantum Physics.


what would be p-v diagram for irreversible isothermal expansion

If you have constant external pressure $P_{ext}$, the work on the PV diagram will be minus the area of a rectangle that is $P_{ext}$ high and $V_{final} - V_{initial}$ long.

When you want to show work as an area, you must plot external pressure on the y-axis of a PV diagram, not gas pressure. In the special case of a reversible expansion or compression, the gas pressure $P$ is equal to the external pressure $P_{ext}$ at every point during the process, so you are still plotting external pressure if you're relating work to the area under the curve.

Why does the external pressure matter in the calculation of expansion work in irreversible processes, and not the gas pressure?

The reasoning goes like this:

expansion work = opposing force x distance moved

opposing force = external pressure x area

volume change = area x distance moved

so

expansion work  = -(external pressure) x (volume change)

The minus sign comes from our decision to make work negative when energy leaves the system in an expansion, and positive when energy enters the system from the surroundings in a compression. This is the convention used in chemistry; you'll find the opposite convention used in physics!


Why don't decomposition reactions form pure elements as products?

Using your example, you simply have not given the system enough energy to do so. Water and $\ce{CO2}$ are very stable compounds under ordinary conditions of temperature and pressure, and thus are the preferred products under standard conditions for that reaction.

If you were to heat the system further, further decomposition would occur. For example, at higher temperatures you will convert $\ce{Na2CO3}$ into $\ce{NaO}$ and more $\ce{CO2}$.  Under extreme conditions, i.e. thousands of degrees K, you will ultimately have enough energy to break all of the chemical bonds and then you will be left with the elements.


Is it possible to build or embed the SMILES representation of compounds in 3D?

SMILES is insufficient

SMILES strings do not encode 3D structure information.  They only convey atom type, connectivity and bond types.  InChI is like SMILES in this regard.

Thus, you will need either (a) an algorithm to infer or guess a plausible 3D conformation of a molecule or (b) a file type that has already specified the 3D arrangement of the molecule.

File types for storing, reading, and showing 3D conformations

Probably the most standard way to represent the 3D conformation of a molecules is with a *.mol file.  There are many tools to read such files.  You can read more about the format on Wikipedia.

Estimating a conformation from SMILES

You can also use computational tools to estimate a 3D conformation from a SMILES string. Note I say a conformation rather than the conformation; molecules can in general have many valid conformations.  Also, tools for generating conformations rely on molecular force fields, etc.  These have many implicit assumptions; there is no guarantee that a computationally generated conformation will be the real conformation of a real molecule in the real world.

Here is some code for generating a plausible conformation from a SMILES string using rdkit.

from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole

my_mol = Chem.MolFromSmiles('NC(=N)N1CCC[C@H]1Cc2onc(n2)c3ccc(Nc4nc(cs4)c5ccc(Br)cc5)cc3')

my_mol

my_mol_with_H=Chem.AddHs(my_mol)

AllChem.EmbedMolecule(my_mol_with_H)
AllChem.MMFFOptimizeMolecule(my_mol_with_H)

my_embedded_mol = Chem.RemoveHs(my_mol_with_H)

my_embedded_mol

print(Chem.MolToMolBlock(my_embedded_mol))

The printed result is:

    RDKit          3D

 33 37  0  0  0  0  0  0  0  0999 V2000
   -8.0789   -0.7261   -1.9565 N   0  0  0  0  0  0  0  0  0  0  0  0
   -8.3618   -0.9375   -0.6556 C   0  0  0  0  0  0  0  0  0  0  0  0
   -9.4453   -1.5737   -0.3799 N   0  0  0  0  0  0  0  0  0  0  0  0
   -7.4690   -0.4468    0.2422 N   0  0  0  0  0  0  0  0  0  0  0  0
   -7.8136   -0.1283    1.6244 C   0  0  0  0  0  0  0  0  0  0  0  0
   -6.7632    0.8908    2.0392 C   0  0  0  0  0  0  0  0  0  0  0  0
   -5.5246    0.3855    1.3227 C   0  0  0  0  0  0  0  0  0  0  0  0
   -6.0688   -0.0733   -0.0461 C   0  0  1  0  0  0  0  0  0  0  0  0
   -5.2554   -1.2432   -0.6177 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.8658   -0.8320   -0.9216 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.6647   -0.1417   -2.0770 O   0  0  0  0  0  0  0  0  0  0  0  0
   -2.3059    0.1587   -2.1237 N   0  0  0  0  0  0  0  0  0  0  0  0
   -1.8139   -0.3885   -1.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -2.7692   -1.0082   -0.2227 N   0  0  0  0  0  0  0  0  0  0  0  0
   -0.4078   -0.3427   -0.6136 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0488   -1.0902    0.4772 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.3984   -1.0569    0.8486 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.3307   -0.2688    0.1543 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.6731   -0.3282    0.5615 N   0  0  0  0  0  0  0  0  0  0  0  0
    4.8291   -0.0477   -0.0843 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.9334    0.1757    0.5968 N   0  0  0  0  0  0  0  0  0  0  0  0
    7.0123    0.4129   -0.2413 C   0  0  0  0  0  0  0  0  0  0  0  0
    6.7153    0.3213   -1.5854 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.0623   -0.0682   -1.7942 S   0  0  0  0  0  0  0  0  0  0  0  0
    8.3378    0.7031    0.3040 C   0  0  0  0  0  0  0  0  0  0  0  0
    9.3324    1.3464   -0.4485 C   0  0  0  0  0  0  0  0  0  0  0  0
   10.5913    1.6060    0.1057 C   0  0  0  0  0  0  0  0  0  0  0  0
   10.8633    1.2259    1.4171 C   0  0  0  0  0  0  0  0  0  0  0  0
   12.5638    1.5736    2.1593 Br  0  0  0  0  0  0  0  0  0  0  0  0
    9.8883    0.5951    2.1844 C   0  0  0  0  0  0  0  0  0  0  0  0
    8.6313    0.3380    1.6284 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.8659    0.4742   -0.9343 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.5160    0.4406   -1.3141 C   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0
  2  3  2  0
  2  4  1  0
  4  5  1  0
  5  6  1  0
  6  7  1  0
  7  8  1  0
  8  9  1  1
  9 10  1  0
 10 11  1  0
 11 12  1  0
 12 13  2  0
 13 14  1  0
 13 15  1  0
 15 16  2  0
 16 17  1  0
 17 18  2  0
 18 19  1  0
 19 20  1  0
 20 21  2  0
 21 22  1  0
 22 23  2  0
 23 24  1  0
 22 25  1  0
 25 26  2  0
 26 27  1  0
 27 28  2  0
 28 29  1  0
 28 30  1  0
 30 31  2  0
 18 32  1  0
 32 33  2  0
  8  4  1  0
 14 10  2  0
 33 15  1  0
 24 20  1  0
 31 25  1  0
M  END

A semi-interpretable 2D image of this 3D conformation, also generated by rdkit, is shown below.  For comparsion, the "un-embedded" molecule, optimized to look nice on a 2D display, is also shown.

From the admittedly not-great 2D depiction of the embedded molecule, you can at least tell that the various aromatic rings are not coplanar.  For better visualization of 3D conformations, you would want to use a tool like py3dmol.

RDKit is just one example of a software with these kinds of capabilities.  I used it here because it's the one I know.  OpenBabel is another one, as Martin rightly mentions in the comments.


Why doesn't the particle in a ring model have a potential energy term?

The short answer is that the particle in a ring is defined to have no potential energy term (at least, not beyond what Karsten Theis has pointed out). Once you introduce something else into the system, then it is no longer really the particle on a ring problem any more.

However, if you want to relate it to the hydrogen atom, then note that in the hydrogen atom, the Coulomb potential is a function of only $r$, i.e.

$$V(r) = -\frac{1}{r}$$

in atomic units. In the particle on a ring model system, the particle is already constrained to be at a particular value of $r$, so even if we were to hypothetically introduce a Coulomb-type potential $\propto 1/r$, this would just be a constant. The effect of having a constant potential energy is just to shift every eigenstate up in energy by the same amount; it has no real physical effect.

What are the differences in the mathematical solutions? Well, you already know it, surely; the eigenstates of the hydrogen atom are atomic orbitals, whereas the eigenstates of the particle on a ring are simply $\exp(\mathrm im\phi)$ with $m \in \mathbb{Z}$ to satisfy the boundary condition.

Of course, this $\exp(\mathrm im\phi)$ term is indeed part of the mathematical form of the atomic orbitals. This is because when you solve the Schrödinger equation for the hydrogen atom, you can successively separate out different degrees of freedom. Firstly you separate the bit which depends on $r$, which goes on to become the radial wavefunction $R(r)$. Then you need to solve for the angular wavefunction $Y(\theta,\phi)$, and the way of doing this is to again use separation of variables, i.e. assume $Y(\theta,\phi) = f(\theta)g(\phi)$. That gets you to a differential equation which looks like $(-1/g)(\mathrm d^2g/\mathrm d \phi^2) = m^2$, which has $g = \exp(\mathrm im\phi)$ as its solutions. I'm glossing over some details here, but that's the general idea; consult e.g. Griffiths' Introduction to Quantum Mechanics for the nitty-gritty bits.


Let's consider two buffer solutions of the same kind ($pK_a=4.75)$, but of different concentrations:

$\ce{CH3COOH}\,0.01\,\text{M}$ and $\ce{CH3COO-} 0.01\,\text{M}$
$\ce{CH3COOH}\,0.1\,\text{M}$ and $\ce{CH3COO-} 0.1\,\text{M}$

The starting buffer volume is the same for both and equal to $100\,\text{mL}$.

Now, add a strong base $\ce{NaOH}\,0.05\,\text{M}$, few $\text{mL}$ only: see below $pH$ as a function of $\ce{NaOH}$ added volume:

Have a look how "fast" the $pH$ changes and you will conclude that higher concentrations buffers are more effective. Differences can indeed be appreciated from the graph: they witness the evidence that larger concentrations better "buffer" the addition of the base.

I would suggest you to do some calculations (e.g. for $1\,\text{mL}$ of  $\ce{NaOH}\,0.05\,M$ added) and see whether you get the same values of the graph (hopefully I have done them right...). Remember that the addition of a strong base will provoke a decrease in concentration of the acid, $\ce{CH3COOH}$ and a consequent increase of its conjugated base, $\ce{CH3COO-}$.


In almost all respects, deuterium behaves like hydrogen, chemically. You can safely drink pure $\ce{D2O}$ in reasonable quantities (though not enough to replace more than ~25% of the $\ce{H2O}$ of the body). In fact, prokaryotes can continue to grow in ~100% deuterated media, albeit more slowly. It does prevent mitosis in eukaryotes in large quantities, though.

Given that similarity, there is no reason to assume any different behavior between deuterated chloroform and that with just protium. But don't knock yourself out trying the experiment, as chloroform is hepatotoxic.


Deliquescence/efflorescence: How can I convert between salt concentration and RH?

You can't. The two phase diagrams are for two different parts of the system. There isn't enough data to go between the two phase diagrams. You have to use the two diagrams together to estimate the missing data (Relative humidity as a function of temperature over wt% magnesium percolate).

The top figure (Chevrier et al., GRL, 2009) is for a salt in solution. Think of having a sealed bottle with air over solution. The phase diagram shows  the relative humidity of the gas as a function of temperature. So at about 240 degrees the equilibrium system has about 45 wt% magnesium percolate in solution.

The solid circles in the lower figure (Gough et al., EPSL, 2011.) show relative humidity as a function of temperature. So at 240 degrees the relative humidity is about 50% for the the equilibrium system.

So at a temperature of equilibrium 240 degrees a solution of magnesium percolate will have about wt 45% of salt and the relative humidity of the air over the solution will be about 50%. You have to combine data from the two phase diagrams to figure out the whole system.

The open circles in the bottom bottom figure (Gough et al., EPSL, 2011.) shows a solid salt in various hydration states (5 vs 6 waters). This sort of detector would be like a crystal oscillator with a coating. As the hydration in the coating changes, then mass of the oscillator changes which changes its fundamental frequency. The chemical reaction (between hydrates with 5 and 6 molecules of water) is not temperature dependent hence the near vertical line.


Why any other element except fluorine wouldn't take electrons from oxygen?

You can consider thinking about charge to volume density. Fluorine is very small and has highest charge to volume ratio among other group 17 elements. For that reason only fluorine is able to snatch electrons from oxygen.


Is there any mathematical or logical proof that carbon with 4 different groups would be chiral?

Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4.

So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical groups attached.

Chemistry

In a tetrahedral carbon, the groups can't swap places. If that happens, we will call it a chemical reaction, and we are not allowing any chemical reactions.

Geometry

Tetrahdral should be taken loosely: For the plane defined by L1, L2 and the carbon atom, L3 is on one side of the plane and L4 on the other. The exact angles don't matter.

Showing that the carbon is chiral

If we put a mirror in the plane defined by L1, L2 and the carbon atom, the mirror operation swaps L3 and L4 (L3' on the side of L4, and L4' on the side of L3 - not sure we need a formal proof for that). I cannot superimpose that mirror image with the original (L1, L2 and the central carbon are already in the same position as the mirror image. I would have to rotate the molecule to get L3 and L4 superposed with L3' and L4', but then L1 would move away from L1', and L2 away from L2' - not sure we need a formal proof for that). Thus, the carbon atom changes configuration when taking the mirror image.

3D illustration

see https://proteopedia.org/wiki/index.php/R/S_nomenclature (you have to turn the molecules a bit to have L1 and L2 point in the same direction as L1' and L2')

Is this trivial?

No, not at all. For trigonal bipyramids there is sometimes pseudorotation, and for ammonia derivatives the trigonal pyramid can "umbrella flip", so these are typically not chiral. If there are four ligands on a central atom and the geometry is square planar, there is no chirality. So a couple of things have to come together to get a chiral center.


By definition, stereoisomers are the compounds which have same connectivity of atoms but different spacial orientation of atoms and two stereoisomers do not have same orientation from any point of view.
There are different possibilities can a molecule of cylclopentane-1,2-diol can have:
For simplicity, I have made two hydroxyl ($\ce{OH}$) into two methyl groups.
But by definition you can cross out two structures because there exist a vantage point from which two molecules look the same (rotate molecule 180 out of plane and then rotate it slightly in plane, you get 2 pairs of identical molecules i.e 1,4 and 2,3. Hence, there are only two stereoisomers possible for cylopentane-1,2-diol.
Hope that helps.


I work at CAS and recently answered a similar question on Twitter (@CASChemistry). In fact, there are 8 pre-50-00-0 CAS Registry Numbers in our database:

35-66-5
35-67-6
36-51-1
36-88-4
37-71-8
37-82-1
37-87-6
38-26-6

So, the lowest CAS Registry Number on record is 35-66-5 for "Benzacridine, 7,9-dimethyl- (9CI)".


Iron can corrode faster in soil than in air, particularly if partly immersed in the soil. Moisture and electrolyte are retained in the soil, making it a good anode. Also, microbes in the soil can accelerate corrosion.


How do I perform a gravimetric analysis of Portland cement?

http://books.google.com/books?id=VeXQAAAAMAAJ&pg=PA2&lpg=PA2&dq=Gravimetric+Analysis+of+Portland+Cement+%22aluminum
Copy whole URL. Page 2 for clickable index, then page 18.
DOI: 10.1039/AN9709500051

Digest it in "strong" HCl (tough lab for employment, worse if you service the fume hoods).  Everything except silica comes out as soluble chlorides (hydrated complex ions).  That says something about Portland cement in acid environments and, unless modified, in seawater.


Why lanthanides and actinides are shown separate from standard Periodic Table layout?

The Periodic Table arranges elements in blocks as each type of orbital fills with electrons - $s,p,d,f,g,h$.  Alkali metals and alkaline earths are $s$-block filling (but could be one $s$-block slot).  $p$-block six electrons to fill are trelides, tetralides, pnticides, chalcogenides, halides, inert gases (but could be one $p$-block slot).  Transition metal $d$-block is ten elements (but could be one $d$-block slot).  Filling the $f$-block are 14 elements, lanthanoids and actinoids.  That gets sloppy to print and the elements are (or at least were) overall obscure.  They get condensed.

Representing the Periodic Table has become an an art form.  The plain vanilla variety is terse and useful.

http://en.wikipedia.org/wiki/Alternative_periodic_tables


Fortunately, the explanation is not as complex as it might seem. It is an oft-made analysis, but one must be very careful when determining reaction spontaneity by looking at the standard free energies $\Delta G_r^o$ alone, as most reactions are not performed in standard conditions and therefore the actual reaction free energy change $\Delta G_r$ may be quite different.

You are correct that, in standard conditions, $\Delta G_r = \Delta G^o_r$ and therefore that $\ce{NO}$ is the most stable reduction product of $\ce{NO_3-}$. However, standard conditions entail, among other things, that $\ce{[H+]\ = \ 1\ M}$ (at least approximately). This corresponds to quite an acidic aqueous solution, with pH zero. If one were to examine the reaction in a more dilute acidic medium, then the value of $\Delta G_r$ for each reduction reaction would change. The reaction that produces $\ce{NO_2}$ has a $\Delta G_r$ that depends on $\ce{[H+]^2}$, while for $\ce{NO}$ the $\Delta G_r$ varies with $\ce{[H+]^4}$. It is clear then that, as one goes to a more and more dilute acidic medium, $\Delta G_r$ decreases more strongly for the reaction that produces $\ce{NO}$, and at sufficiently weak acidic solutions the relative stabilities of the products actually switches, so then $\ce{NO_2}$ becomes the most stable product. A very similar argument can be made for the switch between $\ce{H+}$ and $\ce{SO^{2-}_4}$ reduction in media with sulfuric acid. For $\ce{MnO^{-}_4}$, note that the conversion to $\ce{Mn^{2+}}$ implies the consumption of eight protons, and hence if it can happen, it will likely do so in acidic media. As the acidity of the solution decreases, the conversion to $\ce{Mn^{2+}}$ is quickly suppressed, favouring reactions for which the $\Delta G_r$ depends less strongly on $\ce{[H+]}$, and so we find that $\ce{MnO^_2}$ becomes more favourable.


Samsung's S-Plasma Ion - scientific basis

if you can take a look at the main web site, then it seems they make $\ce{HO2}$ and not $\ce{HO2-}$. This a peroxy radical which exists at trace levels in the ambient atmosphere, usually from reactions of light with oxides of nitrogen and volatile organic compounds. Its really reactive and I don't know for sure how long it will last in the air of the room. It will however react with $\ce{OH}$ radical and make water and $\ce{O2}$. $\ce{HO2 + OH -> O2 + H2O}$. $\ce{HO2}$ will also react with $\ce{NO}$ to form $\ce{NO2}$ and then depending if there is sunlight could make ozone. One of the issues for the machine will be that $\ce{HO2}$ will need to made at some point and therefore one of the first molecules it will encounter will be itself and therefore make $\ce{H2O2}$ (hydrogen peroxide). I think getting one into a lab and testing it would be a start.


How to determine if a compound is likely to become a solid at room temperature?

You are pretty much on the mark with looking at the strength of intermolecular forces. If you wanted to make a general comparison, the tendency to form a solid can be arranged by the strength of the intermolecular forces.$$\text{London Forces}<\text{Dipole-Dipole}<\text{Hydrogen bonding}<\text{Metallic}<\text{Ionic}<\text{Network Covalent}$$
Then, to compare within a group, you need some way of measuring the strength of that type of intermolecular force.

A more mathematical way to look at this if you have empirical values for a compound is to look at the $\Delta G_\text{fusion}$ at room temperature. $\Delta G_\text{fusion}=\Delta H_\text{fusion}-T\Delta S_\text{fusion}$, where $\Delta H_\text{fusion}$ is negative (since it will free up energy in forming the solid) and $\Delta S_\text{fusion}$ is negative (since, whether it started as a gas or a liquid, forming a solid will reduce entropy). Letting $\Delta G_\text{fusion}=0$ (reaction at equilibrium), we find that $T_{\text{fusion}}=\frac{\Delta H_\text{fusion}}{\Delta S_\text{fusion}}$. This allows us to see that when a compound becomes solid is a trade-off between the enthalpy released from forming bonds and the energy needed to reduce entropy at a given temperature.


When to disconnect tubing in vacuum filtration?

If you were to turn off the pump while leaving the object attached then if you have an oil filled pump or a water pump then you will have a suck back which will either damage the pump or contaminate the thing you are filtering, or if you are very unlucky both.

If you have a house vacuum system then you should close the valve / tap between your filtration and the vacuum manifold before you remove the hose to your flask. If you do it the other way around then you will interfere with the use of the house vacuum by others.

I think you need to know if you have your own vacuum pump or if you are using a house vacuum system where a centralised pump serves many people.


Stereoselectivity of dihydroxylation with osmium tetroxide

Dihydroxylation responds to steric effects and proceeds at less hindered positions. What will be the major product in the following reaction if the reagent used is osmium tetroxide followed by sodium hydrogen sulfite?

The crux of the original question seems to be why standard Upjohn dihydroxylation conditions (cat. osmium tetroxide w/ NMO as the terminal oxidant) doesn't respond to hydrogen bonding based directing effects, whilst stoichiometric osmium tetroxide in combination with TMEDA does.

One of the earliest reports of diamine ligands on osmium (of which TMEDA is one) was by EJ Corey's group in 1996 (, who obtained a crystal structure of osmium tetroxide bound to, in this case, a chiral diamine in a bidentate fashion. (If you're familiar with the Sharpless Asymmetric Dihydroxylation you'll see why this was an important first step).

With the coordination of diamine ligands to the osmium tetroxide understood, several groups began to think about why this allowed for hydrogen bond based direction of the dihydroxylation. The most comprehensive reference I've found is a chapter in Organic Reactions by Donohoe et al. (DOI: 10.1002/0471264180.or076.01 sadly not widely available, but a .doc file missing the figures is available from the publishers website which may be somewhat helpful).

Donohoe's group (in collaboration with several others) also published a detailed study of the transition states of these directed reactions in 2002 (DOI: 10.1021/jo026161y hopefully more widely available).

The essential argument being made is that the TMEDA (or other chelating diamine) is acting as a Lewis-base, providing electron density to the osmium tetroxide, which in turn makes it more willing to hydrogen bond (more electron dense carbonyls are more willing to H-bond). This is perhaps a sensible conclusion given how electrophilic osmium tetroxide is, with 4 highly electronegative atoms all pulling for electron density.

In the paper Donohoe proposes two possible TS structures for the directing effect, with either the oxygen or the nitrogen being involved in coordination to the hydroxyl of the substrate. Through NMR and IR work it was eventually determined that TS-A was the true pathway in which the TMEDA is donating to the osmium tetroxide, with the carbonyl on osmium using its lone pair to coordinate to the hydroxyl of the substate.

To conclude. Standard Upjohn conditions give the product expected by a steric argument alone (opposite to the hydroxyl) whilst modified TMEDA conditions give the product formed through H-bond mediated direction.

  But won't the same reason apply for m-CPBA

This part of the question should maybe be another question entirely, but I think its fair to say that osmium tetroxide bears zero resemblance to mCPBA, the electronics of a metal centre vs the electronics of a peracid are completely different and as such their relative ability to hydrogen bond and hence be directed will be significantly different.


What are complete active space methods and how are such spaces defined for molecules?

Multiconfigurational self-consistent field (MCSCF) calculations can be further categorized. What you are referring to is a complete active space SCF (CASSCF) calculation.

In CASSCF, the 1-particle orbitals are partitioned into 3 groups, called spaces, that must be continuous and non-overlapping:

core or inactive orbitals; these must always hold two electrons
active orbitals; these are allowed to be partially occupied
virtual or external orbitals; these must always hold zero electrons

What "complete active space" means is that within the active orbitals, a full configuration interaction (FCI) calculations is performed, so electrons may occupy orbitals in any configuration as long as spin is conserved. The core/inactive orbitals are identical to occupied orbitals and the virtual/external orbitals are identical to unoccupied orbitals from Hartree-Fock (HF), by definition only; starting orbitals for any MCSCF theory don't necessarily need to be HF orbitals. You will see the notation CAS(n,m), where n is the number of active electrons and m is the number of active orbitals. Again, this is for a given total spin.

Here is a rather crude depiction of what a starting active space before optimization would look like; here, there are 8 (active) electrons in 8 (active) orbitals to give a CAS(8,8):

There is also restricted active space and generalized active space (RASSCF and GASSCF). For RASSCF, there are additional spaces, 1 below and 1 above the CAS, where electrons can only be doubly excited from the lower space to the upper space. By definition, RASSCF is more general than CASSCF. The Molcas manual has a good explanation. For GASSCF, all restrictions are lifted and an arbitrary number of active spaces can be chosen with arbitrary restrictions on excitations between them. I will defer mostly to the primary literature.

I'm not sure what is meant by "typically written as eight digits". If you're referring to the strings that are printed to the screen by programs, like 222ab0ba00a or 21102110 (those are made-up examples), they represent the orbital occupations (and potentially electron spins) of a single determinant that comprises a configuration state function (CSF). The length is entirely dependent on how large the chosen active space is in terms of number of orbitals, and all the numbers should add up to how many electrons have been chosen for the active space.

Choosing the active space for an atom or molecule is done so that the static or nondynamical correlation is qualitatively recovered in the wavefunction. In FCI, all possible electron correlation is captured within a given basis. As the excitation level is truncated from all possible excitations to all possible 1 through 6-fold excitations (CISDTQ56) to all possible single and double excitations (CISD) and so on, we make the artificial distinction between static and dynamic or dynamical correlation. Crudely put, static correlation is a result of near-degeneracies in the wavefunction, requiring a linear combination of Slater determinants for a proper qualitative representation. Dynamic correlation is a result of electrons interacting with each other instantaneously and explicitly, rather than in an average way such as the mean field SCF approaches. There are more detailed explanations here and here. Again, in the FCI limit, or even as you approach something like CISDTQ or CCSDTQ for simple systems, these are the same thing.

Because an FCI calculation is being performed within a subspace, and the computational cost of FCI in general grows factorially, this subspace is by necessity kept small. If no static correlation is present in a system, then it's rather pointless to perform any MCSCF calculations, and it's better to do a dynamical correlation calculation (perturbation theory, truncated CI, coupled cluster) on top of a single-configurational wavefunction from HF.

With regards to these specific cases,

there is no static correlation in the water molecule, so any energy lowering from MCSCF would actually be caused by dynamic correlation. Compare correlation energies (post-HF energy corrections) between a singlet CAS(2,6) calculation (just to pick an active space that will run quickly) with an MP2 calculation.
For the beryllium atom, it is well-known that the ground-state wavefunction from CASSCF should be ~90% $\mathrm{(1s)^{2}(2s)^{2}}$ and ~10% $\mathrm{(1s)^{2}(2p)^{2}}$ divided evenly over the 3 $p$-orbitals (they are degenerate in the absence of any fields), so a CAS(2,2) will qualitatively capture this effect.
My guess for $\ce{Be(OH)2}$ is that this effect is only qualitatively changed by the presence of two hydroxide groups breaking spatial symmetry, which will remove the $p$-orbital degeneracy and require more orbitals in the active space.
I am not familiar with nitrate, but looking at the Lewis structure, there are 3 resonance structures which clearly contribute equally, with 2 electrons moving between 3 bonds. Without looking at orbitals, I would try a CAS(2,3) and a CAS(2,6) to start.

I have neglected to discuss any of the subtleties of MCSCF calculations which are numerous and not always obvious. The one I most blatantly ignored above is I assume the starting orbitals are in the correct order. Unless you reorder the orbitals prior to starting MCSCF iterations, CAS(2,2) will pick the last 2 occupied orbitals and the first 2 unoccupied orbitals. These might not be the ones you want! Look at your starting orbitals! The other one I left out is technical, which is how the MO and CI coefficients are optimized during CASSCF iterations leading to final CSFs, and single-state versus state-averaged or multi-state MCSCF.

Image source: Chemgapedia


Here's a list of types of radioactive decay. The most notable types of decay that are not among the classic three involve the direct emission of a free proton or a neutron, the emission of atomic clusters other than helium nuclei (alpha particles), the absorption of one of the innermost shell electrons into the nucleus, or spontaneous fission of the unstable nucleus.

Regarding the question about gamma ray energies, it is first important to notice that "gamma ray" is a generic term that refers to any electromagnetic energy (photons) coming from a nucleus or subatomic particle decays, and has no direct reference to the energy of the photon (though in the past it did). For example, a nuclear excited state of thorium-229, $^{229m}\ce{Th}$, decays through the emission of gamma rays with an energy of 7.6 eV, which corresponds to photons in the ultraviolet region of the electromagnetic spectrum.

Also, the answer depends on exactly what you mean by "power". Assuming you're comparing kinetic energies, Wikipedia says gamma rays from nuclear decay rarely have energy above 10 MeV, and another source says electrons from beta decay usually have energies up to 4 MeV. The energies are all of the same order of magnitude, which makes sense because all the particles are coming from the same source (an unstable nucleus). Theoretically gamma rays can have a slightly higher kinetic energy than other decay particles because photons have zero rest mass, therefore all the decay energy is converted into kinetic energy, whilst for electrons 511 keV is "wasted" on creating the rest mass (and even higher "wastes" for heavier particles such as protons/neutrons/alpha particles etc). I think one can also consider that some energy is used up in "climbing out" of the electrical potential and residual strong nuclear potential wells of the nucleus, neither of which applies to gamma ray photons (though positively charged species would in fact gain energy, being electrically accelerated away from the nucleus by repulsion, but this contribution is probably smaller than the nuclear force one). However, different types of nucleus have different allowed decay paths, and depending on complicated details the most energetic beta particles or proton/neutron emissions might have a slightly higher energy than the most energetic nuclear decay gamma ray.

As a final note, it is also possible to create even more energetic photons than any that could be produced via nuclear decay by putting electrons in a sufficiently powerful particle accelerator. Any charged particle under acceleration/deceleration loses some energy in the form of electromagnetic radiation. For ultrarelativistic charged particles, the photons emitted can be of extremely high energy. These are a type of x-ray termed "hard x-rays" (more energetic than so called "soft x-rays").

Edit: Correction regarding energy lost in rest masses of particles from decay. Energy is used up in the creation of electrons, but protons/neutrons/clusters already exist inside the nucleus, and so are not created. These particles could never be produced directly as a single one would have upwards of 930 MeV, far above typical nuclear decay energies.


Why are super absorbent polymers not made from polysaccharide-rich wheat straw and chaff?

Until the 1920's, polysaccharides were a prime choice for producing absorbent materials. However, their water absorbancy was limited to about 10:1 by mass (i.e. 1g of materiel could absorb 10g of water).  This metric, along with absorbancy by volume, are the properties of interest when designing a super absorbent polymer (SAP).

Subsequently, for several decades hydrogels dominated the SAP landscape. According to this Wikipedia article:

  "Water-absorbing polymers, which are classified as hydrogels when
  cross-linked, absorb aqueous solutions through hydrogen bonding
  with water molecules. A SAP's ability to absorb water is a factor of
  the ionic concentration of the aqueous solution. In deionized and
  distilled water, a SAP may absorb 300 times its weight (from 30 to 60
  times its own volume) and can become up to 99.9% liquid...
  [...]
  Superabsorbent polymers are now commonly made from the polymerization
  of acrylic acid blended with sodium hydroxide in the presence of an
  initiator to form a poly-acrylic acid sodium salt (sometimes referred
  to as sodium polyacrylate). This polymer is the most common type of
  SAP made in the world today.

  Other materials are also used to make a superabsorbent polymer, such
  as polyacrylamide copolymer, ethylene maleic anhydride copolymer,
  cross-linked carboxymethylcellulose, polyvinyl alcohol copolymers,
  cross-linked polyethylene oxide, and starch grafted copolymer of
  polyacrylonitrile to name a few.

There is also ongoing research into improving the ability of natural products to perform as SAP's.  According to the same Wikipedia article quoted above:

  "Today´s research has proved the ability of natural materials, e.g.
  Polysaccharides and Proteins, to perform super absorbent properties in
  pure Water and saline solution (0,9%wt.) within the same range as
  synthetic polyacrylates do in current applications."

Summary, TL;DR:
Until the 1920's, polysaccharides were the prime choice for producing absorbent materials. Primarily due to their relatively low absorbancy (~10:1 by weight, and lost water when pressure was applied), they were subsequently replaced with hydrogels (absorbancies of 50:1 to 300:1, and maintained water when pressure was applied). However, by the 1960's, polysaccharide based super adsorbent polymers (SAPs) saw a resurgence in use as researchers (primarily at the US Department of Agriculture) were able to re-engineer polysacharide based SAPs to have absorbancies comparable to that of the synthetic hydrogels.

The qualities that make a SAP desirable are primarily it's absorbancy and ability to maintain absorbancy under pressure, but also cost and availability of materials. It is the latter of these that drives the research to develop better techniques for the use of polysacharide based SAP's.