text:
Is propanol 1-2 diol 178 the same as propylene glycol?
I clean vinyl records in an ultrasonic vinyl record cleaning machine the manual states it uses propanol 1-2 diol 178 as a surfactant.  I think this is the same chemical as propylene glycol. Is it?

@Thomas A Jones — It is probably a different compound. First, "propanol 1-2 diol" is not a proper chemical name. Also the "178" part is weird. It rather gives the idea of the compound being a polymer, which would fit its role as a surfactant. By definition, a surfactant needs a polar (electron-rich) part AND a lipophilic (carbon-rich) part.

Make sure you are asking about the right name. Also, online search is your friend. Could be 1,2-Propanediol, with n=178 the degree of polymerization, thus PG. Or 178 could be the avg molecular weight, then n is about 2. Maybe something else.

I bought propylene glycol to use it as surfactant for the ultrasonic vinyl record cleaning machine. It is not the same as this does not get in the grooves, it stays on the surface and does not pull out the "white toothpaste".
What ever it is the price for the surfactant is much to high as it is 98%destilled water.

Yes, “propanol 1-2 diol 178” is the same compound as propylene glycol and appears to be a fabricated name used in a manual for a proprietary cleaning system by Kirmuss Audio.
According to the article from Analog Corner magazine [1, p. 22] (emphasis on the footnote):

The Kirmuss Audio KA-RC-1 comes with accessories: 60 mL of an antibacterial/antistatic/antifungal spray; a goat hair brush applicator; an optical-quality microfiber cloth; a larger microfiber cloth (decorated with rabbits) to cover the work space; a circular felt mat to support the record during hand cleaning; a combination carbon fiber/parastatic felt brush; and a bottle of stylus cleaner.
All of this costs $870—far less than at least one competing system and only
slightly more than the iSonic machine and its inferior record-spinning mechanism.
Granted, the accessories are not costly, but no matter how you figure it, Kirmuss doesn’t seem to be in this for the money: He’s not looking to lose money, but clearly he’s not trying to make a killing.
From what I’ve witnessed, he’s more concerned with waging a worldwide war against fungus.
The antibacterial/antistatic spray is identified in the manual as a “98-99% distilled water, 1-2% propanol 1-2 diol 178 mix.” That last one is propylene glycol,4 an organic solvent used in, among other things, pharmaceutical preparations and as a food additive.

4 That chemical name seems made-up, and Kirmuss at first disputed that the ingredient was nothing more than propylene glycol, but a subsequent correspondence made it clear.

Reference

Michael Fremer. The Seven “M”s of the Charles Kirmuss Vinyl Restoration System. Analog Corner, July 2019. (PDF, accessed 2020-09-03)


text:
Why are the OV and VO blocks of the CIS 1RDM zero?
I know the reduced density matrix has the form:
$$
\gamma(\alpha,\beta) = \langle\Psi_N|c^{\dagger}_{\alpha} c_{\beta}|\Psi_n \rangle
$$
but when $\Psi_N$ is the configuration interaction singles (CIS) wavefunction, why are the occupied-virtual (OV) and virtual-occupied (VO) subblocks zero?  I see multiple references state that it is due to Brillouins Theorem, but I don't entirely see that connection.

Unless I'm missing something, I don't think that the OV and VO blocks of the CIS density matrix are zero. We can calculate the whole thing quite easily. I'm going to adopt the conventional notation in which $p,q,r,s$ are generic MO orbitals indices, $i,j,k,l$ enumerate occupied MOs, and $a,b,c,d$ run over the unoccupied MOs. To facilitate writing some of the expressions, further define step functions $\chi(p)$ and $\pi(p)$ which are equal to 1 when when $p$ belongs to the occupied and unoccupied orbitals, respectively, and 0 otherwise.
As per your definition, the one-particle reduced density matrix (1PRDM) is
$\gamma_{q}^{p} = \langle \Psi_{\mathrm{CIS}} | X_p^+ X_q | \Psi_{\mathrm{CIS}} \rangle$,
where the CIS wave function using intermediate normalization is defined as
$|\Psi_{\mathrm{CIS}} \rangle = (1 + C_1) | \Phi \rangle$
with
$C_1 = \sum_{ai} c_{a}^{i} N[X_p^+ X_q]$
being the one-body particle-hole excitation operator responsible for generating singles out of the reference determinant $|\Phi\rangle$, here assumed to be the Hartree-Fock state. I'm using the notation $N[...]$ to denote the normal-ordered operator product with respect to the reference determinant.
Plugging in this definition, we get 4 contributions to the 1PRDM,
$\gamma_{q}^{p} = \langle \Phi | X_p^+ X_q | \Phi \rangle + \langle \Phi | C_1^+ X_p^+ X_q | \Phi \rangle + \langle \Phi | X_p^+ X_q C_1 | \Phi \rangle + \langle \Phi | C_1^+ X_p^+ X_q C_1 | \Phi \rangle$.
The first term is the Hartree-Fock density matrix and is equal to $\delta_{pq} \chi(p)$ (it has 1's along the diagonal of the OO block and 0's elsewhere).
The next two terms are the OV and VO blocks, respectively. Using Wick's Theorem or the diagrammatic method,
$\langle \Phi | C_1^+ X_p^+ X_q | \Phi \rangle = \sum_{ai} c_{i}^{a} \langle \Phi | N[X_i^+ X_a] X_p^+ X_q | \Phi \rangle = c_{q}^{p} \pi(p) \chi(q)$
and
$\langle \Phi | X_p^+ X_q C_1 | \Phi \rangle = \sum_{ai} c_{a}^{i} \langle \Phi | X_p^+ X_q N[X_a^+ X_i] | \Phi = c_{q}^{p} \chi(p) \pi(q)$.
Clearly, the OV and VO blocks are equal as they should be ($c_{a}^{i} = c_{i}^{a}$ for real orbitals) and non-zero since the CI singles excitation amplitudes are non-zero.
The last term, when expanded, results in
$\langle \Phi | C_1^+ X_p^+ X_q C_1 | \Phi \rangle = \sum_{abij} c_{i}^{a} c_{b}^{j} \langle \Phi | N[X_i^+ X_a] X_p^+ X_q N[X_b^+ X_j] | \Phi \rangle$.
There are two distinct ways to carry out the full contraction for this term which results in the VV and OO blocks. In the end, this becomes
$\sum_{abij} c_{i}^{a} c_{b}^{j} \langle \Phi | N[X_i^+ X_a] X_p^+ X_q N[X_b^+ X_j] | \Phi \rangle = \sum_{i} c_{i}^{p} c_{q}^{i} \pi(p) \pi(q) -
\sum_{a} c_{q}^{a} c_{a}^{p} \chi(p) \chi(q)$,
where the first term is the VV block and the second term is the OO block.
Altogether,
$\gamma_{q}^{p} = \delta_{pq} \chi(p) + c_{q}^{p}(\pi(p)\chi(q) + \chi(p)\pi(q)) + \sum_{i} c_{i}^{p} c_{q}^{i} \pi(p)\pi(q) - \sum_{a} c_{q}^{a} c_{a}^{p} \chi(p) \chi(q)$,
which is a working formula for the 1PRDM matrix element for any given MO indices $p$ and $q$.
Furthermore, subtracting off the Hartree-Fock part results in the normal-ordered part of the 1PRDM, which is sometimes called $\Delta \gamma_{q}^{p}$, or
$(\gamma_{q}^{p})_N = \Delta \gamma_{q}^{p} = c_{q}^{p}(\pi(p)\chi(q) + \chi(p)\pi(q)) + \sum_{i} c_{i}^{p} c_{q}^{i} \pi(p)\pi(q) - \sum_{a} c_{q}^{a} c_{a}^{p} \chi(p) \chi(q)$.

Very nice answer! If you haven't come across it yet, you might also be interested in Matter Modeling SE which focuses on computational chemistry/materials.

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text:
Using a weak acid to prevent the precipitation of Mg(OH)2
I have this exercise in my chemistry book, and I have been staring at it for so long but unfortunately I have no single idea what to do and where to start. I will be so thankful for any help.
$0.05$ moles of NH3 are added to one liter of MgCl2 of concentration $0.02$ mol/L.
How many moles of NH4Cl do we need to add in order to prevent the precipitation of Mg(OH)2?
Given Ksp=8.4×10^(-12)  ; pKb=4.5
My teacher's approach:
The solution of Mg(OH)2 is saturated at pH=10.4 since:
Ksp=4s³
Then s=1.28×10^(-4) mol/L
with [HO–] = 2s
pH=14+log[HO–]
Afterwards, using Henderson–Hasselbalch equation, using the given and calculated values, we get
[NH4+]=6.29×10^-3 mol/L
I thought that the calculated value of $s$ is true when Mg(OH)2 is put in pure water and not in a solution containing a weak base which decreases its solubility by common ion effect, so this method is not valid. But i don't know how to solve the problem. Can you help me please?
Sorry for the not perfect format. I did my best.

It is unclear to what pKb refers to. As pKb1+pKb2 of Mg(OH)2 is 1.8 resp 2.56 ( http://www.periodensystem-online.de/index.php?show=list&id=acid&prop=pKb-Werte&sel=oz&el=92 ), it must be pKb of NH3, which is 4.75, so 4.5 is quite incorrect.

Consider (1) for $\ce{Mg(OH)2}$ dissolution equilibrium, (2) for ammonia acido-basic equilibrium.
Combining both, you get required concentration of $\ce{NH4+}$.
$$\begin{align}
\mathrm{p}K_\mathrm{sp} &= \mathrm{pMg} + 2 \cdot \mathrm{pOH}\tag{1}\\
\mathrm{pOH} &= \mathrm{p}K_\mathrm{b, NH3} + \mathrm{pNH3} - \mathrm{pNH4+} \tag{2}\\
\mathrm{p}K_\mathrm{sp} &= \mathrm{pMg} + 2 \cdot \left( \mathrm{p}K_\mathrm{b, NH3}+  \mathrm{pNH3} - \mathrm{pNH4+}\right) \tag{3}\\
  \mathrm{pNH4+} &= \frac 12 \left( \mathrm{pMg}  -  \mathrm{p}K_\mathrm{sp} \right)    +  \mathrm{p}K_\mathrm{b, NH3}+  \mathrm{pNH3} \tag{4}\\
\end{align}$$
where "$\mathrm{pX}$"  means "$-\log{X}$"
 resp. "$-\log{c_\mathrm{X}}$".

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text:
How does acetone help to make a liquid cleaner stronger?
Acetone, as a solvent, seems to behave somewhat different than other solvents. It seems to solve things other unpolar solvents do not.
For example, I have two cleaner sprays from the same brand, and the one with acetone seems to be "stronger" in some way. I think I have noticed that difference before elsewhere, so I assume at least part of what makes it stronger is the acetone.

Acetone is pretty polar on the scale of solvents. See eg https://people.chem.umass.edu/xray/solvent.html

It's not special at all.

At the same time, I don't see why this deserves downvotes.

Well, acetone is at the same time polar (so it likes water) and lipophilic (so it likes fat). As such, it is a very good solvent.

I would add that relative toxicity, price, and the right volatility (easy drying and removal as well relatively easy distillation) also contribute to make acetone so special. Plus one for the simple answer.

I changed the title significantly, but for your answer should be no change.


text:
Fat polymerization vs hydrolysis and temperature change
I have an experiment on fats that I have to do in a few hours and I am very lost between fat/oil (lipid) polymerization due to high temperatures and hydrolysis. It appears that both result in the breakdown of fats but in very different ways. Our class assistant suggested to do an experiment about different temperatures affect fat breakdown which results in acid creation that affects pH. I would track that ph change at different temperatures with different fats for the experiment. Please help as I am lost if that can even be done with simply a heating system, different fats and oils and pH strips??

As far as I know, fats and oils do not polymerize. How much do you know about the hydrolysis (under which conditions it happens, what the products are)? If you want to know what experiments you can do with fats, oils, heat and pH strips, that is a very broad question, probably too broad to be appropriate for this site.

You're only going to get hydrolysis if you have OH- present


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Psi4 gradient calculation with complete basis set?
I'm trying to set up a gradient calculation for psi4. In my input.dat file, I have tried the following keywords:
set basis cc-pVTZ
gradient('CCSD(T)')

which is not working. Does anyone have any recommendations on what key words to use at the end of my input file to run a gradient calculation with a CBS? Any suggestions welcome.

What do you mean "with a complete basis set"?

Can you give the entire input file and the error? It could be dying because you never tell it to do an RHF calculation. I can't remember if Psi4 has a problem with that or if it will just assume that's what you want for a reference. Also, the syntax for a direct call to the gradient is here: http://www.psicode.org/psi4manual/1.0.0/opt.html
You might need to type something like:
G, wfn = gradient('ccsd(t)', return_wfn=True)
wfn.gradient().print_out()
np.array(G)
This just mimics the manual example in which you take back the wavefunction object and print the associated gradient.

A triple zeta basis is far from complete, especially for coupled cluster methods. The entire input file would be necessary to reproduce your specific issue, and recommend further steps.


text:
Using a the particle in a box - one dimension, to calculate the zero point energy of an hydrogen atom
I have been tasked with using the one dimensional particle in a box model to calculate the zero point energy of an hydrogen atom with different sizes. Using the formula $E_1=h^2/(8mL^2)$ for a box with length 0.3nm and 1cm I receive results that are very huge, which made me question the validity of my approach.
Is the one dimensional rigid box model really valid for atoms?

This is a classic type of question to ask in semi-advanced QM courses, typically followed by using perturbation theory to improve the initial treatment with a particle in a box. Without more information, it is hard to tell for sure if you've gone wrong, but my memory is that using a particle in a box is a pretty bad approximation of the hydrogen atom. This makes sense as the hydrogen atom energy levels scale as $1/n^2$ while the PIB ones scale as $n$, so they are qualitatively dissimilar. Also, I think the PIB should give too large an energy as you impose an artificial barrier.

Very thankful for your answer.
What I've done is just plug in the mass for an hydrogen atom together with the length of by box(0.3nm and 1cm). I was to answer in Joules and eV but receive answers in the magnitude of 10^21 eV and 10^27 eV. Until now we haven't covered perturbation theory, so I'm not sure if this is what the teacher will request of us later on.

That looks to me like a unit error. Make sure you have a teeny-tiny mass for the hydrogen atoms if you are working in SI units.

After reviewing my calculations I found a mistake. Now for the box of length 0.3nm I get the result ~2.227 meV. This seems more correct to me, but I'm not sure. Thanks for your help.

@PerArne Sounds like you might have an answer to submit!


text:
Based on Molecular Orbital Theory , how is it decided that the double bond in O2 molecule has one sigma bond and one pi bond?
My Chemistry textbook says this in regard to a C2 molecule:

It is important to note that double bond
  in C2 consists of both pi bonds because of the
  presence of four electrons in two pi molecular
  orbitals.

Is it possible to account for the double bond in an Oxygen atom consisting of a sigma bond and a pi bond in the same way, based on MOT?

$\ce{O2}$ MO diagram
From the MO diagram, we can see that it has 2 electrons in a $\sigma$ orbital, 4 in $\pi$ orbitals, and 2 in $\pi$* orbitals. Taking this together gives a bond order 2, with 1 coming from the sigma electrons and another 1 being the net value for the $\pi$ and $\pi$* electrons.

Wow! Thanks for the brilliant answer! So we have to use this method for checking type of bonds are present between 2 atoms in a molecule? Didn't find it anywhere else...

It becomes a bit more complicated for anything beyond diatomic molecules and there are cases where it fails (e.g. CO) but this is how the concept of bond order is generally introduced at a general chemistry level.


text:
How do I name a substituent which is a cycloalkane which is attached by a chain of more than one carbon?

Above is a picture of an alkane. How would I name this using IUPAC rules? I understand how to name other substituents such as (1,1-dimethylethyl) and some other methyl groups, but I don't know how to name the cycloalkane substituents. What are the rules for naming cycloalkanes as substituents?

Without touching the actual case, things like cycloalkanes as substituents are cycloalkyls.

Note that "1,1-dimethylethyl" is not a preferred IUPAC name.

@Loong I thought that 1,1-dimethylethyl was the IUPAC name and that t-butyl was the common name. Where would the parent chain start for this molecule?


text:
Separate less volatile component from distilling a mixture
Is it a strict rule that when we boil a mixture of two liquids, with one being less volatile than the other, the more volatile vapour has to form the greater portion of the total vapours?
If it is so, how are we able to obtain a pure liquid of less volatility in the distillate and an azeotropic mixture by distilling a certain composition of a non-ideal mixture? Shouldn't we be getting ALWAYS the more volatile component in the higher plates and thus in the distillate?
And if it is not so, please enlighten me with a valid explanation. Thank you.

Greater volatility means that the fraction of vapour represented by the component will be greater than that expected on the basis of its mole fraction in the liquid, ie more volatile=mole fraction in the vapour > mole fraction in the liquid

The opening paragraph in this question suggests a need to for more effort, but the remainder of the question and the answer that it has triggered indicate that maybe this shouldn't have been put on hold.

The more volatile compound forms greater vapour portion, compared to its portion in the distilled liquid.
It forms e.g. 15% molar fraction of vapour, if there is 10% molar fraction in liquid.
In case of deviation from the Raoult law, when there is interaction between compounds, an azeotrope with minimal  ( ethanol-water ) or maximal ( nitric acid water ) boiling point may be formed. This azeotrope acts like if it were the 3rd compound.
Within rectification columns , for each distillation stage upwards, the composition is slightly enriched with more volatile fraction.
And vice versa,  for each distillation stage downwards, the composition is slightly enriched with less volatile fraction.
If we are interested just in more volatile fractions, the column bottom is attached to the vapour source input, what is typical for labs.
If we are interested in all fractions, what is typical in oil industry, the input is injected somewhere in the middle of column, where the equilibrium composition is the most similar to the input.
The more or less volatile fraction may be an azeotrop. Components in ratio forming the azeotropic ratio cannot be separated by destination, unless other components are added.
The typical example is the azeotrope ethanol water, denying to get more than about 95% ethanol by distillation, as this azeotrope has lower boiling point than pure ethanol.
Pure ethanol can be gained by adding benzene,  forming the ternary azeotrope  benzene - ethanol - water, with even lower boiling point.


text:
Global reactivity parameters from open shell DFT calculations using Koopman's theorem
Koopmans' theorem is a useful approach to calculate the global reactivity parameters from the HOMO-LUMO energies. My question is, does it apply to the open-shell systems where we get two sets of singly occupied (alpha and beta) HOMO-LUMO energies?
I am working on rare earth systems. Rare earth (RE) elements generally show +3 oxidation states which result in open shell configurations for $\ce{Ce^3+}$, $\ce{Eu^3+}$, $\ce{Gd^3+}$ etc. I am performing unrestricted open shell DFT calculations for these systems, as restricted open shell calculations for RE systems require larger computational time and cost. In unrestricted open shell calculations, we get two sets of alpha and beta HOMO and LUMO. My question is, can I apply Koopmans' theorem here to calculate the global reactivity parameters?
Could anyone help with some literature on the reactivity parameters for open shell systems? Thank you!

Regarding the literature, it might be worth having a look at the section Books about Computational Chemistry and Quantum Chemistry. I'm no specialist in this field, so probably someone else could help you with the precise textbook for you.

Could you walk us through an example you understand, and then give an example of an open-shell system that you are wondering about? That would make it easier to answer.

Hello @KarstenTheis, I am working on rare earth systems. Rare earth (RE) elements generally show +3 oxidation states which result in open shell configurations for Ce3+, Eu3+, Gd3+ etc. I am performing unrestricted open shell DFT calculations for these systems, as restricted open shell calculations for RE systems require larger computational time and cost. In unrestricted open shell calculations, we get two sets of alpha and beta HOMO and LUMO. My question is, can I apply Koopmans' theorem here to calculate the global reactivity parameters?

You (probably) cannot apply Koopmans' theorem. (Please also note that it is named after Tjalling Charles Koopmans, the s at the end is part of the name.)
Open shell systems are usually hard to describe and unrestricted density functional approximations might not even describe the system correct qualitatively. The problem usually comes down to the multireference character of the partially occupied orbitals. In these cases you need a multireference wave function to correctly describe the ground state, e.g CASSCF methods. In a first order approximation, unrestricted methods can probably yield reasonable results, because (in a practical sense) it is using more than one determinante. However, when you use these methods, there will be fractional occupancy of multiple orbitals, which make the approximations of Koopmans' theorem break down. I would suggest a thorough literature review on the best practises for such systems. In short, HOMO and LUMO are not defined in these systems.

Thank you so much for the clarification.


text:
Dipole moments of dibromoethene isomers
There was an exercise in my textbook asking to compare the dipole moments of 1,1-dibromoethene and 1,2-dibromoethene(Z). The answer in the textbook says that the dipole moment of 1,1-dibromoethene is greater than that of 1,2-dibromoethene.
This seems counter-intuitive as the angle between the dipole moments of the two $\ce{C-Br}$ bonds in 1,1-dibromoethene is $120°$ and that in 1,2-dibromoethene(Z) is $60°$. A little physics tells us that the net dipole moment of 1,2-dibromoethene should be greater than that of 1,1-dibromoethene which contradicts the answer in the textbook. I want to know whether I am correct or not.

There is also the dipole moment of C-C bond.

https://chemistry.stackexchange.com/questions/133337/which-has-greater-dipole-moment-cis-1-2-dichloroethylene-or-1-1-dichloroethylen


text:
What are the degrees of freedom that define the temperature of an ionic solid (such as sodium azide)?
Degrees of freedom describe the different ways atoms move in a sample. For a pure ideal gas made of non-linear molecules, there are 3N degrees of freedom (N is the number of atoms in the molecule), 3N-6 of which are vibrational, 3 rotational and 3 translational. So per atom, we have three degrees of freedom.
How many degrees of freedom does solid sodium azide have (per formula unit) and what are these degrees of freedom classified as?

For a solid metal, I know there are three degrees of freedom per atom, and they are related to lattice vibrations (phonons). I also know that black body radiation of solids is used as a "mise en pratique" to define the SI unit Kelvin. That makes me think that solids, no matter whether metallic or ionic, share some feature related to kinetic energy when they are at the same temperature.
I am less concerned with the nomenclature and more with how these degrees of freedom change with temperature, giving rise to the same black body radiation at the same temperature, and being at thermal equilibrium with other samples of the same temperature through collisions at the interface between them.

This question came up in the comments to this answer: https://chemistry.stackexchange.com/a/121228

Related: https://physics.stackexchange.com/questions/210685/do-solids-have-translational-energy

Temperature vs heat capacity is discussed for gas and solids (including metals) in http://people.virginia.edu/~lz2n/mse305/notes/HeatCapacity.pdf

The Wikipedia article for an Einstein solid suggests that the temperature of a solid can be directly computed from statistical mechanics assuming all atoms behave like simple harmonic oscillators.

Degrees of freedom describe the different ways atoms move in a sample.

The classical equations of motion for a system of N atoms describe the time evolution of 3N positions and 3N momentum coordinates for the atomic nuclei (ignoring electronic degrees of freedom). The total energy is a function of those 6N degrees of freedom. Sometimes some of those coordinates do not provide a significant way of storing energy so they can be ignored. For instance, in a dilute gas, translational momentum provides a means of storing energy, but the potential is essentially flat, so position does not.

How many degrees of freedom does solid sodium azide have (per formula unit) and what are these degrees of freedom classified as?

Sodium azide ($\ce{NaN3}$) has $4\times6\times \text{N}_\text{A}$ degrees of freedom per mole of substance. Half are associated with momentum coordinates, the other half with position coordinates (ignoring once again electronic degrees of freedom). Positions can be expressed in terms of 3N-6 internal and 6 global coordinates, and internal coordinates expressed approximately as harmonic oscillator modes. Each oscillator mode is a linear combination of the 3N-6 spatial coordinates, the entire set being orthonormal. If the solid does not move we can ignore global coordinates.

I am more [concerned] with how these degrees of freedom change with temperature, giving rise to the same black body radiation at the same temperature, and being at thermal equilibrium with other samples of the same temperature through collisions at the interface between them.

Ignoring electronic degrees of freedom (that is, focusing only on nuclear motions), each atom contributes 6N degrees of freedom. Unlike in a dilute gas, here both momentum and position provide a means of storing energy. The amount of energy that can be stored in  each degree of freedom is, following classical statistical mechanics (high T limit), equal to kT/2, for a total of 3RT for one mole. However, quantum mechanics limits occupation to certain states rather than a continuum, and this can reduce the amount of energy each degree of freedom can store (the heat capacity).
There are some concepts in your question that need to be disentangled:
1) Equipartition, a principle that states that energy is dispersed equally among all degrees of freedom. At low T quantization may determine to what extent different degrees of freedom are thermally available. Too cold and only a reduced number of states may be occupied, and there is no equipartition. This is well explained in the wikipedia. The implication for the example you present is that, when compared to a metallic solid with softer vibrational modes, a molecular solid, in which atoms form covalent bonds having large force constants, might be expected to have a lower heat capacity: degrees of freedom corresponding to high frequency vibrational modes that distort covalent bonds can accommodate less energy (on average) at lower temperatures. Such arguments justify some trends in the table below, showing the heat capacity of various solids on a per atom basis (computed from heat capacities from Ref. 1).  Sodium azide is a combination of our definitions of ionic solid and molecular solid. For solid sodium azide the heat capacity (on a per atom basis) is $\pu{19.2 J/molK}$, a value intermediate to that of metallic and molecular solids. Note also that $3\text{R} \approx \pu{24.9 J/mol K}$, the heat capacity of a solid (on a per atom basis) predicted by the partition theorem. This value excludes an estimate of an electronic contribution to the heat capacity.

2) Temperature, an equilibrium thermodynamic property: bodies that exchange energy as heat are not in thermal equilibrium, their temperatures differ. The concept of thermal equilibrium also applies to degrees of freedom. For instance, if you (somehow) freeze particular motions (translational motion in a gas, say) you lower the temperature associated with those particular modes. If the system is then allowed to return to equilibrium, heat flows from warmer into the (accessible) colder degrees of freedom.
3) Blackbody radiation, which is intuitively thought of as a "photon gas" in a cavity. The concept of thermal equilibrium can be applied to this "gas". You can consult the wikipedia for more information on these last aspects:

Just as the Maxwell–Boltzmann distribution is the unique maximum entropy energy distribution for a gas of material particles at thermal equilibrium, so is Planck's distribution for a gas of photons.[8][9] By contrast to a material gas where the masses and number of particles play a role, the spectral radiance, pressure and energy density of a photon gas at thermal equilibrium are entirely determined by the temperature.

References

Engineering ToolBox, (2003). Specific Heat of Solids . [online] Available at: https://www.engineeringtoolbox.com/specific-heat-solids-d_154.html [Accessed 23/9/2019]. The source does not state the measurement temperature but it is presumably ~room T.

Thanks for the edits. I really like the comparison of ideal gases with solids in terms of degrees of freedom.

Degrees of freedom is often used to refer to motions in which energy can be placed, but the definition makes it a little hard to understand the differences among gases and solid. Better to take a step back.


text:
Does space in between have any significance in IUPAC nomenclature?

The correct IUPAC name of the following compound is:

(a) 2,4,6-trimethyl cyclohexanone
(b) 1,3,5- trimethyl benzophenone
(c) trimethylcyclobutanone
(d) 2,4,6-trimethylcyclohexanone

I was confused between option (a) and (d). I opted for option (a), but our teacher told that ans is an option (d). The only difference I see between these two options is that there is space in (a). Does space have any significance in IUPAC nomenclature?

There is no space between substituent prefixes and the parent structure. That's all there is to it.

Compare "phenyl acetate" and "phenylacetate" – two different compounds.

How could putting space in the middle of a word not make a difference?!

$\boxed{\textbf{d}}$ is correct. It is the standard IUPAC name.
However, "2,4,6-trimethyl cyclohexanone" will often be understood as 2,4,6-trimethylcyclohexanone, similar to how, in theory, neopentane, 2,2-methylpropane and 1,1,1,1-methylmethane are all equivalent.


text:
Volatility comparison for isomers of phthalic acid
All the atoms of these compounds are in a plane.
o-phthalic acid can show extensive intra-molecular hydrogen bonding as well as inter-molecular hydrogen bonding.
m-phthalic acid can show intra-molecular hydrogen bonding (though, in lesser extent) as well as inter-molecular hydrogen bonding.
p-phthalic acid can not show intra-molecular hydrogen bonding but can show extensive (more than ortho & meta isomers, due to no steric hindrance) inter-molecular hydrogen bonding.
Now, intra-molecular hydrogen bonding increases volatility whereas, inter-molecular hydrogen bonding decreases volatility (I am not comparing boiling points because these molecules decompose at temperature below their melting point).
What can we say about the order of volatility of these compounds?

How does m-phthalic acid show intramolecular bonding? Seems unlikely.

Do you have a temperature in mind? Because all of three compounds are solid at room temperature.

@MathewMahindaratne Melting point: o-isomer 207 *C, m-isomer 300 *C (sublimes without melting) & p-isomer 345 *C


text:
Why mustn't the sample be dipped in the mobile phase in paper chromatography?
As the title asks, I am curious about why the sample (ie, an extract of plant's chlorophyl transferred onto paper) must not be dipped into the mobile phase (ie, the solvent which goes up the paper)?
My first guess is that it is because the sample would dissolve into the solvent and only a small portion of the sample would go up the paper, but I am really keen to know if there is more to it than just this.

Dipping sample to a mobile phase would cancel all the purpose of chromatography as a separation technique.
Imagine properly performed chromatography as the athletic race, where all runners start at the same point and time. The finish order is by their average speed.
Now imagine a crazy version of a New York marathone, where runners start in  long sequence of runner batches. But nobody would  think about registration of runner starts, so there would be no idea which runner was the fastest to be decorated as the winner.
Back to chromatography, instead of chromatographic spots ( PC, TLC) or peaks (HPLC, GC), there would be useless, smeared, overlapping long bands.

There is only one mode of chromatography in which the sample is dissolved in the mobile phase and it is continuously passed through the column. It is called frontal chromatography.

Yes, I am aware of it, but it is different case.


text:
Hydrogen Bonding by Carbenes
A hydrogen bond is formed between hydrogen attached to highly electronegative atoms (nitrogen, oxygen, and fluorine) which are small in size too and the non-bonding pair of electrons of another such atom that may be present in the same molecule or in another molecule but is not directly linked to it with a covalent bond. But when we talk about chlorine, it doesn't form hydrogen bond due to its large size. But what happens in the case of carbon more precisely 'carbenes'? It's small in size, has a lone pair of electrons and is somehow more electronegative too. Does it help form H-bond like N, O, and F do?

Wait... You're asking whether a hydrogen bond was found with C lone pair? I think it may be possible, but there's a lot of things you mention that aren't exactly true. Your thinking about these bonds is somewhat outdated - virtually any atom may be acceptor or donor, it's just that bond with donor should be significantly polarised or h-bond with acceptor will have negligible energy.

related https://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-does-it-really-need-fluorine-oxygen-or-n?rq=1 https://chemistry.stackexchange.com/questions/31591/why-is-hcl-not-considered-to-have-hydrogen-bonding?rq=1

I'm waiting for an organic chemist to comment or answer this question, but in the meantime my opinion is that a carbene is too reactive to do anything as subtle as form a stable H-bond. But perhaps that is not relevant to the question. It is perhaps more of a hypothetical question.

@BuckThorn There's a lot of stable carbenes, also radicals, cation/anion radicals etc. You name it, it can be stabilised. In particular NHC carbenes are often used as ligands, but their high basicity may be a problem for non-transient hydrogen bond.


text:
Oxidation of fructose with conc. HNO3
I recently studied carbohydrates. I was taught that fructose on oxidation with strong oxidizing agents like concentrated $\ce{HNO3}$ yields glycolic acid and tartaric acid along with trihydroxy glutaric acid. Though I am confused why glycolic acid is formed instead of oxalic acid (since the terminal $\ce{-OH}$ should also have been oxidized) I am also satisfied that oxidation of the keto group in fructose breaks it off into 2 carboxylic acids and yields a 2 carbon(glycolic) and a 4 carbon(tartaric) containing acid.
But I am very much unable to figure out the formation of trihydroxy glutaric acid by oxidizing fructose since its a 5 membered carboxylic acid and in no reaction I know did I ever come across such breakage of ketones in 6=5+1. Some sources even cited the formation of trihydroxy butyric acid along with oxalic acid where I couldn't understand why tartaric acid wasn't formed by oxidation of the terminal $\ce{-OH}$ group.

A reference to your sources would be useful. HNO3 would oxidize both C1 and C6 to form carboxylic acid groups. α-Ketoacids undergo oxidative decarboxylation. Hence, the trihydroxyglutaric acid. Be aware that not every molecule of fructose is oxidized in the same way. Some oxidation pathways presumably lead to glycolic acid.

Well here's a few links http://chem-brains.blogspot.com/2012/04/biomolecules-five-marks.html, https://image.slidesharecdn.com/structuralelucidationoffructose-140417063812-phpapp01/95/structural-elucidation-of-fructose-4-638.jpg?cb=1397716722, to name a few. Well can you elucidate the oxidative decarboxylation of alpha ketoacids ? What I saw at most sites talked about formation of tartaric and glycolic acid and hence I would like to ask why the terminal C doesn't get oxidized in glycolic acid to form oxalic acid.

Trihydroxyglutaric, tartaric & glycolic acids are some of the products isolated from nitric acid oxidation of fructose. https://books.google.com/books?id=WUxLAgAAQBAJ&pg=PA465&lpg=PA465&dq=fructose+trihydroxy+glutaric+acid&source=bl&ots=E7WpT7vcQ_&sig=ACfU3U29dsOHPe8eZH6F_k-iLwETncySFg&hl=en&sa=X&ved=2ahUKEwiJgfybjdbkAhXuV98KHdcWALEQ6AEwDHoECAgQAQ#v=onepage&q=fructose%20trihydroxy%20glutaric%20acid&f=false. Study the Krebs cycle for the oxidation of alpha-ketoacids. Oxalic acid may well be formed and oxidized to CO2.

Ok so there are many possible products, right? But can you post the mechanism for oxidative decarboxylation of alpha ketoacids? I can understand that the COO- group in a pyruvate (let's say) will cleave away giving out CO2, but that would leave behind an unstable carbanion(CH3CO-), is that formation favorable? Also it has to convert into a carboxylic acid(a required addition of OH+), does that occur by the attack of H2O? Anyways its hard for me to find convincing answers. Would be great if you could elaborate. Thank you.

I'll post an answer. It is not an acyl anion but rather a cation.

Sir Arthur7: From your comments you appear to know how to liberate CO2 and a proton by pushing arrows as illustrated by the blue arrows in generic α-oxoacid 1. You are correct in discrediting acyl anion 5 for two reasons. First, it is unstabilized and secondly, its protonation would provide an aldehyde, not a carboxylic acid. So why not push the red arrows in 1 to generate an acyl cation equivalent? The downside is that you don't want hydride as a leaving group. The solution is to replace the acidic hydrogen of 1 with a species that likes electrons, e.g., a high oxidation state metal or a hypervalent, non-metallic atom. Certainly, the nitrogen of HNO3 fits the bill where M = NO2 in structure 2. In this instance, nitrogen is n = 5 in structure 2 and 4. Bear in mind that for the purpose of electronic bookkeeping, transformation of the acyl anion 5 to a acyl cation 2 is a 2-electron oxidation while species M undergoes a 2-electron reduction. ---continued---

Moriarity, et. al.,1 have effected this oxidative decarboxylation with the hypervalent iodine species, iodosobenzene 7. The path of the red arrows within the brackets provides a step-by-step version of their mechanism using α-oxoglutaric acid 6.2 Notice that mixed "anhydride" 8 is the operational equivalent of species 2. The authors propose the formation of cyclic intermediate 9 that collapses to 12 with loss of CO2 and a proton. [Note: Structure 10 is drawn for clarity. Collapse of 9 to succinic acid 12 is proposed as concerted.] The collapse of structure 8 might be thought to follow the course of the green arrows but because the reaction is conducted in anhydrous dioxane, it is unlikely that an acylium ion is formed directly because it is apt to form succinic anhydride. See examples studied in the chart below.

The Moriarity paper describes the iodosobenzene oxidation of mandelic acid to benzaldehyde and CO2. See if you can now write a mechanism for this reaction.
1) R. M. Moriarty, S. C. Gupta, H. Hu, D. R. Berenschot, and K. B. White, J. Am. Chem. Soc., 1981, 103, 686. https://pubs.acs.org/doi/10.1021/ja00393a040
2) α-Oxoacid is preferred over α-ketoacid. The term keto (C=O) overcounts carbons. Oxo = (=O).

Thank you so very much. I got my concepts cleared, got to learn new things. Well as you mentioned, in the conversion of mandelic acid to benzaldehyde how does the -OH group get oxidized to -c=0? Does it occur in the presence of iodosobenzene itself? I could figure out the rest of the mechanism. Thanks a lot again.


text:
What is an example of an isopropyl alcohol resistant caulk?
What is a caulk (or bonding adhesive) that can be exposed permanently submerged in an isopropyl alcohol (>91%) bath?
The idea came from this previous question:
Silicone vs Alcohol
It seems from one of the answers there the alcohol would be resistant but the adhesion properties may be compromised.
The more readily available the product, the better.

The following chemical resistance list suggests isopropyl alcohol will not significantly disrupt silicone adhesion, provided the substrate is also resistant. http://www.warco.com/pdf/Rubber-Elastomer-Chemical-Resistance-Guide-by-TLARGI-and-WARCO-BILTRITE.pdf

Not an answer, but a strategy to get one:  I would start by googling 'alcohol-resistant caulk' or 'alcohol-resistant sealant'.  This will give you hits for chemical companies that produce such products.  For instance,
https://www.azom.com/article.aspx?ArticleID=5854
claims to have an epoxy ("UV18MED") that is alcohol-resistant and can be used as a sealant.
https://bowersindustrial.com/product-category/concrete-sealers-caulking/chemical-resistant-caulk/
claims to have caulks that are resistant to alcohol.
Since each of these companies are experts on the properties and performance of their respective products, you should call them, describe your specific application, and ask if they have anything that would work reliably.  That's the best way to be assured the product will do what you want.
One problem you might run into is that these may be B2B companies, i.e., ones that sell in quantities designed for businesses rather than consumers.  If that's a problem, try calling some manufacturers of consumer adhesives and asking if they have anything.
I know this is more work than simply posting a question on CSE (and there is of course nothing wrong with posting such a question), but it's only by speaking directly to the manufacturer that you can be sure the product is designed to work in your application (unless you have a product label that indicates this).
While you're speaking to them, you might also want to ask about leaching -- whether components of their product might dissolve into the alcohol; you would need to judge for yourself how much leaching your system could tolerate.  You'll also need to let them know what kind of mechanical stresses and temperature changes the product must tolerate (e.g., an adhesive that is sealing an aquarium and also holding it together needs to be a lot stronger than a caulk that is merely sealing the top of a sink). The more specific details you can give them about your system, the better.


text:
How to convert Mn2O3 to Mn3O4?
I was preparing $\ce{Mn3O4}$ but it was when characterized by XRD, a $\ce{Mn2O3}$ phase existed. Now how can I convert $\ce{Mn2O3}$ to $\ce{Mn3O4}$?

You may need a good lab oven.
Manganese(II,III) oxide on Wikipedia

Preparation
$\ce{Mn3O4}$ formed when any manganese oxide is heated in air above 1000 °C. Considerable research has centred on producing nanocrystalline $\ce{Mn3O4}$ and various syntheses that involve oxidation of $\ce{Mn^{II}}$ or reduction of $\ce{Mn^{VI}}.$

As @Poutnik has answered, heating manganese oxides at elevated temperature results in forming of $\ce{Mn3O4}$, research paper suggest that the reaction gets affected not only temperature but the partial pressure of oxygen and further heating at 1400 °C results in formation of $\ce{MnO}$. Here is an excerpt from a research paper [1] (emphasis mine):

TG and DTA curves of $\ce{γ-MnO2}$ have been obtained in nitrogen, air and oxygen. The reactions $\ce{MnO2 → Mn2O3}$ and $\ce{Mn2O3→ Mn3O4}$ were observed in the ranges 450° to 600° and 750° to 1100° respectively, and the decomposition temperatures are affected by the partial pressure of oxygen. The endotherm at ~1200° is unaffected by the atmosphere and is not accompanied by weight loss. It is, therefore, due to a polymorphic transformation of $\ce{Mn3O4}$ rather than formation of $\ce{MnO}$.

Another process is carbothermic reduction [2], i.e. reduction of manganese oxides with the help of carbon or carbon monoxide. It requires high carbon monoxide/carbon dioxide pressure and precise change of pressure w.r.t. to temperature. The following reactions depicts how the standard energy of formation varies with temperature:
$$
\begin{align}
\ce{3 Mn2O3 + C &->[$T = \pu{1100 °C}$] 2 Mn3O4 + CO} &\quad ∆G^\circ &= -0.25 - 0.17T\\
\ce{3 Mn2O3 + CO &->[$T = \pu{1100 °C}$] 2 Mn3O4 + CO2} &\quad ∆G^\circ &= -170.71 - 0.004T\\
\end{align}
$$
References

Tinsley, D. M.; Sharp, J. H. Thermal Analysis of Manganese Dioxide in Controlled Atmospheres. Journal of Thermal Analysis 1971, 3 (1), 43–48. https://doi.org/10/bvwj2v.
https://www.alliedacademies.org/articles/practical-reduction-of-manganese-oxide.pdf
Puerta, M. C.; Valerga, P. Thermal Decomposition of a Natural Manganese Dioxide: A Laboratory Experiment for Undergraduate Students. J. Chem. Educ. 1990, 67 (4), 344. https://doi.org/10/fqg3kp.


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Magnetic nature of tetraamminedichlorocobalt(III) chloride
I know how to predict magnetic nature when the compound contains only strong field or only weak field ligands. But in $\ce{[Co(NH3)4Cl2]Cl},$ $\ce{NH3}$ is a strong ligand whereas $\ce{Cl-}$ is a weak ligand.
So, how do we know if the ligands are causing the electrons to pair up or not? I understand VBT and CFT to predict the hybridisation and magnetic nature of compounds.

It is low spin. Nearly all d6 is low spin. See https://chemistry.stackexchange.com/questions/112675/is-conh%e2%82%83%e2%82%84cl%e2%82%82cl-paramagnetic-or-diamagnetic/115285#115285 which is essentially a duplicate of this

@IanBush Quite the opposite. Iron(II) is $\mathrm d^6$ but most of iron(II) is high spin.

Agreed Fe2+ is by far the most likely of d6 species to be high spin, but everything else is low spin with very, very few exceptions. And the question is about Cobalt.

The ammine ligand is not really a strong-field ligand. It is not a strong π acceptor which is characteristic of strong-field ligands such as the nitrosyl cation or carbonyl. Instead, it is a weak π donor. Quite unlike other strong-field ligands such as cyanide, there are a great number of high-spin complexes with ammine ligands. Ammine (and aqua) ligands are better termed intermediate field. They can stabilise low-spin complexes but they typically also support high-spin environments.
Understanding when ligands cause the electrons to pair is very non-trivial but essentially guesswork. This is exemplified by some complexes being isolated in both the high-spin and low-spin states as is the case in $\ce{[Fe(ptz)6]^2+}$ ($\ce{ptz}$ being 1-propyl-1H-tetrazol-κN4). Furthermore, the metal has even more influence on the question with higher oxidation states typically stabilising low-spin configurations while lower oxidation states often favour high-spin complexes. The only clear cases are 4d and 5d transition metals which can be safely assumed low-spin unless stated otherwise.
With that said, cobalt(III) complexes feature a combination of a high (-ish) oxidation state of the central metal combined with a $\mathrm d^6$ environment which generally favours low-spin a bit more as it allows for maximum electronic stabilisation. Therefore, cobalt(III) complexes can be guessed as low-spin if they are not entirely made up of weak-field ligands. Ultimately, however, the experiment or calculation will have to answer.


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Cupric and cuprous copper
Copper has two chlorides: $\ce{CuCl2}$ and $\ce{CuCl}.$
Copper reacts directly with chlorine to form a copper(II) chloride. Why doesn't it form copper(I) chloride? Is it because $\ce{CuCl2}$ is more stable than $\ce{CuCl}?$ I want a more foundational approach to answer this question.

Regarding stability of Cu+ see: https://chemistry.stackexchange.com/questions/42855/why-cu-is-unstable-in-aqueous-medium

and https://chemistry.stackexchange.com/questions/115674/why-do-cu%e2%81%ba-ions-spontaneously-form-copper-metal-and-cu%c2%b2%e2%81%ba-ions-in-solution

and who said copper(I) chloride is not formed? Wikipedia says that copper and chlorine directly reacts with each other to form copper(I) chloride at 450-900 C.

There are several ways to look at this question.
On one hand, the heat of formation of $\ce{CuCl2}$ $(\pu{49.2 kcal/mol})$ is greater than the heat of formation of $\ce{CuCl}$ $(\pu{32.5 kcal/mol}),$ so if there is excess $\ce{Cl2},$ the product would naturally be the dichloride. On a kinetic basis, even with just a stoichiometric amount of chlorine $(\ce{1/2 Cl2}$ plus $\ce{1 Cu}),$ gaseous $\ce{Cl2}$ might produce a little cuprous chloride from bulk metal, but the product would be more finely divided and more likely to react, so the reaction would tend toward $\ce{CuCl2},$ leaving half the copper unreacted.
The electrochemical potentials reflect the same thing: in water, if you have enough oxidizing power to produce $\ce{Cu+}$ $(\pu{-0.52 V}),$ there is enough to go the rest of the way to $\ce{Cu^2+}$ $(\pu{-0.1 V}$ for $\ce{Cu+ -> Cu^2+},$ or $\pu{-0.34 V}$ for $\ce{Cu -> Cu^2+}).$


text:
Why does the reaction between 2‐methylbut‐3‐en‐2‐ol and hydrogen bromide show allylic shift?
In the following reaction

shouldn't the minor product be the major one as the carbocation formed before attachment of bromine is 3°? Also, can someone tell me why the product includes only nucleophilic substitution as there is no other condition mentioned to judge whether it should show elimination or substitution.
Please mention if what favours allylic shift in such type of reactions.

Even thought the immediate carbocation formed is tertiary, the stability of the double bond should also be taken into account. A more substituted double bond is also favored/stable. In this instance it seems that it's a competition between the double bond's stability and the carbocation's stability. There could be other situations where this is not the case, but you would need experimental data to back up stability (there's no universal ranking system that I know of to say when a double bond outweighs a positive charge).
In short, I would say that the relative stability of a tertiary double bond (compared to a tertiary carbocation) is what would justify the major product being labeled as it is.

There isn't a distinct primary carbocation or a tertiary carbocation. There is only one carbocation with two major resonance forms. OP's question has the same issue.

Can u please elaborate what you mean? @orthocresol


text:
Amperage for the electrolysis of water?
I am electrolysing water using several different potential catalysts for my eight grade science fair project. The electrolysis will be running at 2 to 2.5
volts over nickel electrodes, as I do not need a long term solution. How much amperage should I put on my DC power supply? Thanks and I would appreciate simplified answer compared to what you would normally give. Yes, I do understand ohms law.
Edit: I did some calculation and if I assume the resistance of the electrolyte is lets say 5 ohms, then we can assume a current of 0.5A at 2.5 volts. Is this correct?

Not exactly. To use Ohms law on an electrolysis cell, you have to subtract the cell voltage. Only when you increase the voltage above that level, a current starts to flow.

In electrochemistry, the rate of electrolysis depends on rate of charge entering the cell. Second most important point is that one can either control potential or current but not both during electrolysis i.e. you cannot have both values set. If you are fixing potential at 2.5 V, the value of current is not in your hand. I don't think Ohm's law remains valid during electrolysis in general.
So the main question one should ask is: What rate of electrolysis do I need? For example, you may want 200 mL of $\ce{O2}$ in 60 min. From here you calculate the charge Q needed to collect this volume of oxygen. Current is charge per unit time (3600 s) and this is your required current above the decomposition voltage of current. See second figure: http://www1.lsbu.ac.uk/water/electrolysis.html
See the sample calculation here: https://www.quora.com/What-amount-of-current-is-required-to-produce-0-224-cc-STP-of-O_2-at-anode-during-the-electrolysis-of-water


text:
Is it possible for the kinetic energy integral to be negative?
Is it possible for the kinetic-energy integral, Tij, to be negative? I was messing around with some HF code and found that the integral became negative on some off-diagonal terms. (This could also be a bug with the code.)
If the kinetic energy integral is negative, what would the physical meaning be?

Page 126 of this book and the following pages suggest that negative Tij values are reasonable and shows them occuring for a few example calculations.

Generally, the off-diagonal elements of matrices, especially in physical systems, are interpreted as the coupling between whatever the $i$ and $j$ elements correspond to. So, in this case, a negative element of $T_{ij}$ corresponds to a negative kinetic coupling between atomic orbital basis functions. In plain language, this means that basis functions $i$ and $j$ tend to mutually lower the kinetic energy of an electron placed in one of those orbitals. I am not sure it really makes sense to give much of an interpretation to this, however, because what one is really interested in is diagonalizing the Fock matrix, of which the kinetic energy is only one part.
Also, it is always possible to choose a basis where the kinetic energy matrix is diagonal, but again, this would not get you anywhere as in solving the HF problem, you will diagonalize the Fock Matrix which will surely put you in a basis in which the kinetic energy matrix is non-diagonal. What might be more interesting would be to take the actual molecular orbitals which are part of the solutions of the Roothan-Hall equations and re-compute the kinetic energy matrix over these orbitals. Looking at the coupling between these orbitals may be more easily interpretable for e.g. aromatic $\pi$-systems.

@jheindel has given an insightful answer. I want to supplement it with some mathematical analysis. Szabo and Ostlund note that the kinetic energy integral over two primitive $s$-Gaussians $A,B$ is equal to$^{[1]}$
$$
\left(A \left| - \frac{1}{2} \nabla^2 \right|B\right) = \frac{\alpha\beta}{\alpha+\beta} \left[ 3 - \frac{2\alpha\beta}{\alpha+\beta} \left| \mathbf{R}_A - \mathbf{R}_B \right|^2 \right] \cdot S_{AB}
$$
where $\alpha, \beta$ are the Gaussian exponents, $\mathbf{R}_A, \mathbf{R}_B$ are the centers of the Gaussians, and $S_{AB}$ is the relevant overlap integral, which we know is greater or equal to zero. The term inside the brackets can become negative, rendering the entire integral negative.
For more insight, one may do the following:

Graph a one-dimensional Gaussian. Observe that it is positive everywhere. (The results translate to three dimensions.)
Apply the kinetic energy operator to the Gaussian from step 1 and graph it. Observe that there are negative regions.
Multiply the result from step 2 with a second Gaussian. Observe that depending on the center of the second Gaussian, one can amplify the negative sections and suppress the positive sections of the result of step 2. By taking the integral over all space, one can obtain a negative total value.

[1] A. Szabo, N. Ostlund: Modern Quantum Chemistry, Dover Publications, 1st. edition, revised, 1996, page 412.


text:
Gallium makes aluminum soft like a wet tissue. Any other metals or alloys that are vulnerable to similar damage?
Gallium can make aluminum soft and brittle as long as it bypasses the aluminum oxide normally formed on the surface of the aluminum (for example by scratching the surface).
Mercury can weaken aluminum as well.
Are there any other metals or alloys that can be "weakened" by coming in contact with gallium, mercury or any other substance?

Any? Better ask about ones then don't...

Amalgam#Liste_amalgam-bildender_Elemente Basically only Cr, Mo, W, and Mn, Fe, Co are safe.

Just found the term. It's called embrittlement.  It can occur even on steel (hydrogen embrittlement). Hopefully someone will post a comprehensive answer, otherwise i'll post one.

I want to add that you probably refer to simple contact. Otherwise there should be plenty of alloys and metallic solids that suffer of embrittlement. Some are well known and still useful.

@Alchimista You are right, I'll edit the question to highlight it.

Gallium and Mercury, the latter in particular are absolute pests and poisons to most metals. There are too many to mention that will suffer the same consequences. I will mention the few that don't. Iron, Platinum, Tungsten and Tantalum all tolerate them fairly well. There are certainly more examples, but the common rule is; most don't.


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Is it possible to blow out a candle flame using pure oxygen?
This sounds like a stupid question perhaps, but if a stream of pure $\ce{O2}$ was directed at the flame as a concentrated jet would it blow off the candle?

I think that the opposite will probably happen. Since flame is the direct result of a combustion and oxygen gas is a reactant on the reaction, adding more to the flame would instantly increase the rate at which the combustion will occur.

and https://www.quora.com/If-you-blow-a-candle-with-100-oxygen-will-you-be-able-to-light-off

Not possible. See this video

@Nilay Ghosh But, should not the same reasoning work for air too ? Perhaps for oxygen the needed flow is much higher. At some point. it would cool down the candle enough and decrease the partial pressure of parafine enough to extinguish the flame.

By other words, what happens if oxygen flow speed overtakes the speed of the flame propagation ? Also, note that the geometry of the  candle in the video  plays for keeping the flame. If the candle were a tall one with pointy end and exposed knot, it could be different.

@Poutnik Perhaps you are right. Your points got me to think critically. I don't know much on this topic but there could be many factors which would determine whether the flame would sustain(speed of air, composition, air outlet cross section etc.) At first glance, this question seems stupid, but there is more to it if thought critically. Perhaps you are the right candidate to answer this question. Your cross-questions in previous comments are spot-on :)

In principle it surely works as with air. @Poutnik mentioned the relevant points, temperature of the stream and cooling as well as reaction kinetics and stream speed. Also let me point out that not (necessarily) all mixtures lead to explosions, usually there is a window with lower and upper limits. This means a mixture doesn't explode but it doesn't sustain combustion at first. You can detach and thus extinguish a flame from a burner nozzle by simply rising the speed and varying the ratio of combustible and comburent. Just think of a Bunsen burner. But I wouldn't make experiments, of course!!!

It depends on the wick.   A flammable wick, exposed to oxygen, will burn, but
a non-flammable wick, saturated with liquid wax, could be chilled by the gas flow,
and when the evaporated wax is blown clear of the wick, the candle flame
could extinguish for lack of fuel.   Basically, the oxygen flow might
(at sufficiently high velocity) remove the vapor fuel so that the flame
becomes a detached fireball, and the wick and wax of the candle cool.
The oxygen flow just has to be faster than the flame propogation in the
fuel-oxygen mixture.
A solid flammable wick, though, won't be displaced, just burn faster,
and on discontinuing the oxygen, the candle will still be lit.

What is a solid flammable wick? A stream can cool down in principle every wick. At least it coukd extinguish the flame and perhaps the wick keeps being slowing and superficially oxidised.

@Alchimista  A solid (cotton string) wick in a gas stream holds a boundary layer of stationary gas at its surface, unless we're talking about highly turbulent (or supersonic) flow.   The combustion of the cotton (which makes the top of a candle wick blacken and turn to ash) is much slower than the combustion of the evaporated candle wax, but an oxygen flow can change that.

Sure. You mos likely help the burning of the boundary layer but still you could blow it out. The Wick burning is basically the same as wax burning. I would say the first part of your question is correct. Anyway we all more or less understand the same.  Blow enough fast and one might blow out a candle even with oxygen.

@Alchimista - in the case of a steel plate being cut by oxyacetylene, the cut, once started, can be completed on pure oxygen, turning the acetylene off.   The flow of oxygen impinging on hot burnable steel doesn't cool it, it reacts with the surface layer, oxidizing with enough waste heat to liquefy the oxide and preheat the layer just beneath it...

Yes yes but you can't blow out what you are cutting. Burning and cutting something is not a candle flame.

Fire triangle dictates that a fire requires fuel, oxygen, and heat.
Directing a jet of oxygen at the candle will not remove the fuel, will add oxygen... so the only way for it to extinguish the candle is to change the heat (reduce it) dramatically.
If the oxygen is cold enough - perhaps from being highly compressed and decompressed rapidly to produce the 'jet' -  it might cool the candle enough to make it less flammable.
But I suspect it's more likely the oxygen stream would ignite and the cylinder might explode.

How would the oxygen stream "explode" ? The fuel cannot flow against the oxygen stream, and oxygen without fuel cannot burn. The flames can only exist past the point of mixture.

Easily, just need enough velocity/volume to reduce the temperature. Similar to when lighting an oxy/acetylene torch, the acetylene is turned on and lit .Then the oxygen must be added/turned on very slowly to avoid blowing out the flame.


text:
Boiling point elevation does not depend on the solute?
I've recently carried out an experiment trying to measure the boiling point elevation of water after adding 5M of NaCl and KCl. Although for some reason, KCl solution started boiling around 90 degree Celsius, theoretically, both KCl and NaCl should have the same effect on the boiling point of water right?

Well, yes, both should have the same (per mole, that is) and more importantly, positive effect.

Unless you are at a very high altitude, the boiling point of KCl solution should have been higher. Was the thermometer near the place that appeared to boil, or far away? Were you observing dissolved air bubbling out??

@DrMoishePippik I was not at a very high altitude and the thermometer was dipped in the solution. Anyway, when can we say that the solution is "boiling" how many bubbles should appear? I repeated the experiment a few times. Does that affect the surrounding pressure in a way?

It may need to boil for a while to get dissolved air out. The temperature will not change appreciably unless a lot of water evaporates, increasing concentration. Measure the volume before and after, or boil in a graduated container.

Dissolving substances in water decreases water chemical potential
$$\mu=\frac {\mathrm{d}G} {\mathrm{d}n}$$
respectively activity $a$ :
$$\mu = \mu_9 + RT \ln {a}$$
As consequence, saturated water vapour pressure above the solution is lower than the saturate vapour pressure  over pure water at the same temperature.
This leads to the higher boiling point of solution, compared to water, at the same atmospheric pressure.
See also ebullioscopy and ebullioscopic constant, useable to determine the molecular mass of the solute from the boiling temperature change and solute (w/w) concentration.
Similar principles, affecting freezing point, lead to cryoscopy


text:
Why my boiling pot turned black?
Recently I ran out of dish washing detergent, so I decided to pick up one of the recipes online, to make my own "ecological" dish washing detergent alternative.
The recipe was:

Mix ingredients:

400ml of water
200mg of kitchen salt
100ml of vinegar
juice squeezed from 3 lemons

boil for 10 minutes

After I stopped boiling and left the mixture to cool down, the fluid turned dark gray. Bottom of my stainless steel pot turned black. Here is a photo of it (after I did some scrubbing, but you can still see black on the side):

What happened? (What chemical reaction?) Additionally, are any of the formed substances known to be toxic?

What is your pot made of?

I think it was just demonstrated that it is not made of stainless steel

I really do not want to sound sarcastic or offensive. These things are always discussed in another SE. Most of the time tha

@Alchimista which SE is the right one to ask this question?

@NilayGhosh label says "High Quality Stainless Steel"

I think it should not be posed.At least not for safety reasons. All things you mentioned are common ingredients why do you expect they turn something to be used for cooking in something toxic?. As far the colour is concerned it can be a thin film deposited onto. Colour of metal surfaces is quite complicated. Perhaps some oxides formed. Also the diy recipe is just something to fill in a Web page, though lemon and vinegar might help is hard to see why salt should be used. And the boiling part is even more obscure.

In no way would the chemicals used produce a detergent. Wherever you got this information should be considered an unreliable source of information.

@Alchimista why do I expect they turn something to be used for cooking in something toxic? Because they were used in pretty high dose (salt especially) and they reacted with the pot surface. So far no one was able to tell what reaction took place.

@user2449761 — Stainless steel is actually "stainless" only under some mild conditions, i.e. everyday conditions in a household: almost neutral pH, 25°C and a normal atmosphere.

So, as usual, some power users just didn't want us to answer to a valid question.What is wrong with this question?

I think there's interesting chemistry here -- mixing vinegar and/or lemon juice with salt will produce an acid environment with chloride ions, which attacks metals a lot more readily than the weak acids or the salt alone.

Are you sure it is stainless steel?  Or maybe aluminum instead?

You used a salt/acid mixture and, upon bringing it to a boil to clean a stainless-steel pot, parts turned black.
All stainless steel is not alike, and all are not “stainless” or non-reactive towards oxidation – given the circumstances.
Getting to your specific question:
The “black” you are seeing upon boiling is a Fe(II)/Fe(III) oxide, the same or similar to the black oxidative protection [the seasoning of a skillet] so valued by the fastidious owners of iron skillets.
One of many procedures for “seasoning” an iron skillet is similar to what you did.  I have done the following A) cleaning the skillet down to shinny metal, particularly important if you do not know the history of the skillet. B) rub a mixture of table salt dampened with an acid [vinegar] over the pan until you get the classic yellow-rust of iron we are so familiar with C) now add water to the skillet and bring it to a boil.  This step is critical.  The yellow rust will turn black upon boiling.  This is a protective Fe(II)/Fe(III) oxide coating.
Now repeat with the salt/vinegar rub and boil until you get a uniform blackish coating on the iron skillet.    Finally, fry up some bacon and eggs to give a carbon-based coating to the skillet.  The latter is more of a “paint” but with the black metal oxide coating as a base, your skillet is protected against “rust” and mild oxidation.
The “original Teflon ® coating” if you will.

I wonder if some of the grey color is chromium (III) citrate given the presence of the citric acid from the lemons. I wouldn't expect very much of the Fe(II)/Fe(III) oxide to go into solution. The heat and acid must first degrade the chromium oxide layer on the stainless steel before solubilizing iron ions.

@Andrew, Yes, the citric acid could be "the joker in the deck" here!  You make a good point.  However, citric acid is a 2-electron reducing agent and I suspect oxidation when I put an acid and a metal together.  Even with the cast iron seasoning recipes that have been around for years the nature of the iron oxide is less than clear.  For that matter the Fe(II)/Fe(III) chemistry of cyanotype process is all but settled. https://pubs.acs.org/doi/pdf/10.1021/ed078p311.2

I wasn't thinking of citrate as a reducing agent, just as a chelator. for the chromium ions released by attack of chloride and acid on the chromium oxide layer of the stainless steel


text:
Name of "the" benzodiazepine
Wikipedia has a thorough article on the benzodiazepines. Of course, it is written primarly  from the pharmacological point of view; but there is also the "Chemistry" section, which includes the IUPAC name. The parent hydride for the compound class is named 1H-benzo[e][1,4]diazepine. I do not understand why it is benzo[e][1,4]diazepine and not benzo[f][1,4]diazepine: shouldn't the bond alphabetic designation sequence go around the original diazepine ring, starting with the 1,2-bond (a), then continuing to 2,3 (b), 3,4 (c), 4,5 (d), 5,6 (e) and, finally, 6,7 (f), to which side the benzene ring is attached? Could please someone clarify the matter to me?

Perhaps it's the name of the second tautomer.

Ah, I see. Parent hydride stays parent hydride even if the indicated hydrogen then moves around. Clearly the tautomer which assignes to the bond letter closer to the beginning of the alphabet should be chosen as the base for fusion.

You should rewrite this as an answer! Thank You!

Related: https://chemistry.stackexchange.com/q/96439/7951

Let's look at the „[1,4]diazepine“ part more closely. (Standalone, it's correctly named 1,4-diazepine.) The „-epine“ in the Hantszch-Widman system stands for unsaturated seven-membered ring. More precisely, one with the maximum number of noncumulative double bonds. (IUPAC uses that term instead of „conjugated“ probably to avoid confusion with physical electronic conjugation, which might not be the case, as we're at the level of topology, not even geometry.)
The exact structure is specified with indicated hydrogen. There are four 1,4-diazepine structures, 1H-1,4-diazepine, 2H-1,4-diazepine, 5H-1,4-diazepine, and 6H-1,4-diazepine (not considering their stability/existence in tautomeric pair/multiple):

In the world of abstract organic chemistry nomenclature, 1,4-diazepine without indicated hydrogen can be represented as following structure (dotted bonds for unknown bond order, any arrangement of noncumulative dbl. bonds assumed; no explicit hydrogens):

When fusing with benzene, 1,4-diazepine part has to be numbered as follows (not the other way), to get lowest numbers/letters when starting with heteroatom:

… obtaining benzo[e][1,4]diazepine (new numbering of the resulting structure marked, partially; coincidentally we get the same set of numbers for nitrogens):

Indicate the hydrogen to get our 1H-benzo[e][1,4]diazepine:

(But note that fused ring nomenclature is quite long and complicated, generally the final numbering could also depend on the indicated hydrogen position.)
Finally, note that the new nomenclature has simplified the names of benzo-fused rings, the current preferred name is 1H-1,4-benzodiazepine.


text:
32 Degree Thermochomic dye
I am doing a school project which involves color changing based off of temperature, and I need a thermochromic dye that changes color at 32 degrees Fahrenheit. Does anyone know where I could find some or explain how it is made?

You might try a solution of cobalt(II) chloride, $\ce{CoCl2}$, in plain water (which might freeze) or in a mixture of water and ethanol or water and isopropyl alcohol. At some concentration, which you'd experimentally determine, it should turn from pink to blue on cooling to 273 K. See Flinn Scientific's site for more details.
BTW, you might as well use Réaumur as Fahrenheit; both are provincial.


text:
Why doesn't ethanol dissolve in sodium bicarbonate?
Today I experimented with different solvents and solutes and was a bit surprised by one result. When I added ethanol to dilute sodium bicarbonate and tapped the test tube against my palm to mix it, I saw the mixture swirling, then it turned cloudy.
I don't understand why this happened -- as far as my understanding goes, ethanol is both a weak acid and a weak base, and it dissolved with NaOH and HCl with no trouble. Does this have anything to with the fact that ethanol and sodium bicarbonate are both amphiprotic? Or is there a different explanation?
Thanks in advance.

You are asking the wrong question. You should be asking why sodium bicarbonate is less soluble in water/ethanol mixtures than it is in water. it isn't ethanol failing to mix with water that is making the result cloudy.

Firstly ethanol being an organic compound, tends to dissolve less in polar compounds. But it can form numerous hydrogen bonds in water and hence is soluble in most aqueous solutions.
NaOH is a often considered a moderate base in organic chemistry. If you dissolve ethanol in aq.NaOH , it tends to oxidize the alcohol into its aldehyde which further undergoes aldol reactions, giving it a yellowish brown color when left for many days.
That is not the case with NaHCO3 as its an even weaker base, so it rarely participates in any reaction with ethanol. Also there are lesser possible hydrogen bonds per molecule in it. So its already less soluble in polar solvents like H2O, not to talk about in organic solvents like ethanol.
Hope this helps.

What tends to oxidise alcohol? Certainly not NaOH...

Neither did it seem much convincing to me at first, though I couldn't have been much sure about reactions that take a days to occur, so I just knew it as a fact, though I am sure that NaOH cant oxidize ethanol. But it was what I learnt from a site while surfing the net, here goes the link--https://www.reddit.com/r/chemistry/comments/48c0ai/i_mixed_saturated_naoh_in_water_and_ethanol_why/  perks of not trying to take a second opinion.

Well, I guess OP wacky reasoning got you fooled. Acid base equilibria are largely irrelevant here. You can rewrite your post, for example using things like https://en.wikipedia.org/wiki/Solvent for help also solubility tables etc.

Yes you are right it seems, I have corrected my false beliefs now. But don't you feel in this age of technology and with so easy access to internet around the globe, sites like Wikipedia are the ones that anyone searches for at first and hence there's nearly no need to share links to that? For me links to other useful, legitimate, and less searched  docs is more appreciated

Sodium bicarbonate is a ionic compound with quite limited solubility in water. Ethanol is worse solvent to dissolve ionic compounds than water is, as it is less polar. Therefore salts have generally worse solubility in ethanol than in water.


text:
Diastereoselectivity of the Pauson Khand Reaction
I hope someone may help me. I don‘t know how to determine the diastereoselectivity of the product of the Pauson Khand reaction in generally.
In the following I’ve got 3 practice examples,however in the third one I’ ve got no solution. I know the product and the regioselectivity.

Do you know the putative mechanism for the Pauson-Khand? That should help you. Or if you are familiar with Heck-type chemistry, you might be able to guess.

You should render an educated rationale for the stereochemistry in #3.

Here's a useful link http://stoltz2.caltech.edu/seminars/2009_Bennett.pdf


text:
Does bicarbonation require water?
In an aqueous solution of $\ce{NaOH}$ exposed to the atmosphere one can probably expect some degree of $$\ce{NaOH(aq) + CO2(g) -> NaHCO3 (aq)}$$
But, what would happen if just a solid block of $\ce{NaOH(s)}$ was placed in some completely anhydrous environment, populated only by excess $\ce{CO2(g)}$ at STP? After all,
$$\ce{NaOH(s) + CO2(g) -> NaHCO3 (s)}$$
is technically a spontaneous process with $\Delta G^\circ = -76.91\ \mathrm{kJ/mol}$, but I can only think of three possibilities:

Nothing happens—no water (i.e., as a catalyst), no reaction!
A passivation layer of $\ce{NaHCO3(s)}$ forms on the surface, preventing the rest of the $\ce{NaOH(s)}$ inside from reacting^
No passivation layer—the $\ce{NaOH(s)}$ just keeps reacting away until it's all $\ce{NaHCO3(s)}$!

Why the downvote? I merely happened to skip the possibility of carbonate formation : / Never said a "4th possibility" could not exist! : )

Downvotes without any reason have no uses, it only serves the ego of the person doing it.

Rather
$$\ce{2 NaOH(aq) + CO2(g) -> Na2CO3 (aq) + H2O}$$
and analogically on the solid $\ce{NaOH}$.
Bicarbonate in aqueous solutions cannot survive the excess of hydroxide, forming carbonate.
$$\ce{HCO3^{-}\ (aq) + OH- (aq)<=>> CO3^{2-}(aq) + H2O}$$
The same is further enforced by dehydration effect of solid hydroxide.
$$\ce{NaHCO3(s) + NaOH(s) -> Na2CO3(s) + H2O}$$
Only in excess of $\ce{CO2}$ Is formed bicarbonate:
$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O <=>> 2 HCO3^- (aq) }$$

Yep, and the first reaction would also happen in a solid, thereby generating at least some amount of $\ce{H2O}$.

@ManRow Not exactly for that reason, but the thermal decomposition $\ce{2 NaHCO3(s) -> Na2CO3(s) + H2O + CO2(g)}$ occurs very slowly even at room temperature.


text:
Software to predict equilibrium products in the ideal gas phase
I'm interested in software packages that can answer the following kind of question:

If I mix 30mol% $N_2$ and 70mol% $H_2$ together at 2 bar and 2000C, what is the equilibrium composition after all reactions have reached equilibrium. Assume all reactants and products are ideal gases.

This is a relatively straightforward calculation. All that is required is knowledge of standard Gibbs Free Energies of formation and specific heat capacities for each species which may be formed. Of course, the challenge is guessing which species may be formed - here an obvious candidate is $NH_3$, but it is certainly possible that the formation of other species could lead to a lower value of the Gibbs Free Energy. A software package with a large database of standard Gibbs Free Energies and ideal gas heat capacities would be very useful here - you could simply select to search over all chemicals which you suspect may be formed, or instead you could minimise the Gibbs Free energy over 'all substances in the database' (at least those composed of the available chemical elements - $N$ and $H$ in this case.)
Does anyone know of any software packages (open source or otherwise) which could do these calculations?

You mean 2 thousand Celsius degrees? That's pretty high, but I don't think things like NH could be favorable at such pressure.

The CALPHAD community has various thermodynamics programs such as ThermCalc and Pandas that do this sort of thing. It mainly depends on what databases you have available.


text:
Why is the 2nd ionization energy of chloride similar to that of sulfur?
The first ionization energy of chloride is significantly greater than that of sulfur. I suggest that this is very expected as the effective nuclear charge of chloride is higher and additionally, sulfur has even more electron-electron repulsion due to advent of paired electrons in 2p orbitals. However I am not too sure on why the 2nd ionization energies are almost the same.

The second ionization of chlorine leaves $[\text{Ne}]3s^23p^3$ where the $3p$ subshell is half-filled with one electron per orbital and gives a favorable exchange energy.  With sulfur this relatively favorable configuration is achieved instead after the first ionization, the second ionization of sulfur breaks the half-subshell and is thus less favorable.
The effect of the half-filled subshell can be seen elsewhere, as well.  The chart below (reference) shows trends in first ionization energies.  Counting backwards from the third peak (argon) we see that phosphorus exceeds sulfur violating the usual rules based on effective nuclear charge.  The first ionization of sulfur gives the half-filled $3p$ subshell after removing the electron, whereas the first ionization of phosphorus would have to break that configuration.  The half-filled shell effect beats effective nuclear charge here.


text:
Can I reverse an iron redox reaction with potassium hydroxide and electricity?
For a school project, about clean energy, me and my team are trying to build an iron battery. We are using a piece of iron (anode), potassium hydroxide (electrolyte), carbon electrode (cathode) arranged as shown below:

As to generate electricity from the following chemical reactions:
$$
\begin{align}
\ce{Fe + 2 OH- &-> Fe(OH)2 + 2 e-}\tag{1}\\
\ce{3 Fe(OH)2 + 2 OH- &-> Fe3O4 + 4 H2O + 2 e-}\tag{2}\\
\end{align}
$$
Then:
\begin{align}
\ce{O2 + 2 H2O + 4 e- &-> 4 OH-}\tag{3}\\
\end{align}
And the overall reaction is:
\begin{align}
\ce{3 Fe + 2 O2 &-> Fe3O4}\tag{4}\\
\end{align}
Will this succeed in producing some electricity? And, can the redox reaction be reversed by having an external power source with sufficient voltage like a car battery, then hooking the negative terminal to the iron and the positive terminal to the carbon electrode?

A redox reaction in a battery can often be reversed by plugging an external power source; this is known as electrolysis. Now what is your reaction? I see two half-reactions, and both are on the same side. What is on the other side? Where do these electrons go?

@IvanNeretin Will the carbon electrode not take in the electrons given off by the iron redox reaction?

Two oxidation half-reactions are shown, but what is the reduction half-reaction? What is the overall redox equation? Where does the iron oxide enter in? This needs some re-working.

@user10560552 No.

@IvanNeretin Is this better?

@EdV Is this more in the right direction?

Replacement for previous comment. Looking at the standard reduction potentials in wikipedia, I think the proposed cell will work as a galvanic cell, but the iron oxide is ferric oxide, not magnetite. I doubt it can be recharged, but you can always try! Anyway, have a look at the standard potentials and good luck!

@EdV Thanks for the helpful information

Yes, this will do. I don't believe the product will be Fe3O4, but that's an unimportant detail. Also, there is not much O2 in water, so the current will be rather low.

One last comment. As per @IvanNeretin oxygen content will be low, so maybe gently sparge in some air near the carbon electrode. A simple glass frit aquarium bubbler will do. But wear eye protection, respect KOH, and do not splash KOH solution on yourself or anything else. If you do any sparging, use water as a test case first, to get a safe gentle flow rate.


text:
eV and redox potential
I am trying to estimate the redox potential of an excited state as described in this reference.
The paper says:

The reduction potentials associated with the excited states
  cannot be directly measured and are typically calculated from
  known cyclic voltammetry (CV) and spectroscopic data. As
  an approximation, the excited-state potentials of a catalyst are
  related to its ground state potentials and its zero−zero excitation energy ($E_{0,0}$). $E_{0,0}$ can be estimated by the difference in
  energy between *PC $^3$MLCT$_1$ and PC $^1$A$_1$ and can be approximated by the maximum emission of the catalyst (Figure 2). As an example, the emission maximum for $\ce{Ru(bpy)3Cl2}$ in MeCN
  has been measured to be $\pu{615 nm}$ at room temperature and $\pu{582
nm}$ at $\pu{77 K}$ in a frozen MeOH/EtOH mixture. This corresponds
  to an $E_{0,0}$ value of $\pu{2.02 eV}$ at room temperature and $\pu{2.13 eV}$ at $\pu{77 K}$. Given the ground-state reduction potentials, the excited-state
  reduction potentials can be calculated as shown in Figure 4.

Then Figure 4 shows the calculation for the room temperature case:
$\ce{Ru(bpy)3^{3+} + e^- -> \text{*}Ru(bpy)3^{2+}}$
$E_{red} = \pu{1.26 V} - \pu{2.02 V} = \pu{-0.76 V}$
My question is: Why is it valid to use the $\pu{2.02 eV}$ from the spectroscopic measurement as $\pu{2.02 V}$ in the redox potential calculation?
From my understanding, $\pu{2.02 eV}$ is the energy between the ground state and the excited state, but these are units of energy, not potential, so I'm having a hard time seeing how can this calculation be done.

E = qV fixes the dimensions

@Alchimista I'm not sure how this answers the question

@ralk912, I feel the calculation cannot be that trivial. In this case one would write to the corresponding author if he can tell you. I dislike such papers which do not bother to mention the details of their obscure calculations. Their reference 21, for this method from 1970, never bothers to make a clear connection between E(00) and the reduction potential.

"but these are units of energy, not potential, so I'm having a hard time seeing how can this calculation be done" I refer to what seems the question to me, by commenting on dimension discrepancy.

It boils down to allocate somewhere in absolute energy scale the levels involved. One assume the first level accessed by spectroscopy to be the one where the electron is received upon reduction. For determining where the el.chem. potential is see https://almantaspivrikas.blogspot.com/2013/07/how-to-convert-redox-potential-vs-nhe_15.html?m=1 for instance. Obviously this is prone to multiple error sources. But it does not seem related to the core question, which is just about dimensions discrepancy. Potential times electron charge gives energy. I must say that most of your formula are invisible

....to me. That's why I am commenting on the only point I have clear.


text:
Reaction of aqueous sodium carbonate with aluminum foil
Aqueous sodium hydroxide is known to react with aluminum foil in an exothermic, hydrogen-gas releasing reaction like $$\ce{2NaOH(aq) + 2Al(s) + 2H2O(\ell) → 2NaAlO2(aq) + 3H2(g)}$$
However, on youtube, it looks like someone was able to substitute NaOH with Na2CO3 and get a vigorous similar-looking reaction as well: https://www.youtube.com/watch?v=ggPWdwcsyqM
Which makes me wonder — did they also form $\ce{NaAl(CO3)2}$ as well? E.g., via
$$\ce{2 Na2CO3(aq) + 4Al(s) + 6H2O(\ell) -> NaAl(CO3)2(aq) + 3NaAlO2(aq) + 6H2(g)}$$
Or was sodium carbonate merely used as a source of NaOH, from $$\ce{Na2CO3(aq) + H2O(\ell) \rightleftharpoons NaHCO3(aq) + NaOH(aq) }$$

I don't think there is something like NaAl(CO3)2. The closest I can get is dihydroxy aluminium sodium carbonate. https://pubchem.ncbi.nlm.nih.gov/compound/Dihydroxyaluminum-sodium-carbonate

@Nilay Ghosh Indeed, mostly likely something like "sodium aluminum carbonate" would probably be rather unstable as well, especially in aqueous solution. Even the Wikipedia article https://en.wikipedia.org/wiki/Aluminium_carbonate is pretty doubtful of the existence of any (non-hydroxylated) "aluminum carbonates" too...

It is rather:
$$\ce{2 Al + 6 H2O + 2 OH- -> 2 [Al(OH)4]^- + 3 H2 ^}$$where $\ce{OH-}$ comes either from hydroxide dissociation, either from carbonate hydrolysis.
$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$
The reaction with carbonate would gradually slow down as the carbonate/bicarbonate buffer will kick in.
Initial $\mathrm{pH}$ is ( see notes below line ):
$$\mathrm{pH_{init}}=14 - 0.5 ( \mathrm{p} K_\mathrm{b2} - \log{ [  \ce{Na2CO3} ]} )\\=  7 + 0.5 ( \mathrm{p} K_\mathrm{a2} + \log{ [  \ce{Na2CO3} ]} ) $$
because $\mathrm{p} K_\mathrm{a2} +
  \mathrm{p} K_\mathrm{b2} =   \mathrm{p} K_\mathrm{w} = 14 $ ( at $\mathrm{ 25\ ^{°}C}$ )
As $ \mathrm{p} K_\mathrm{a2} = 10.329$:
$$\mathrm{pH_{init}}=12.17 + 0.5 \cdot  \log{ [  \ce{Na2CO3} ]} $$
..and when some $\ce{OH-}$ is spent and $\ce{HCO3^-}$ is formed:
$$\mathrm{pH}=10.329 + \log \frac {[  \ce{CO3^2-} ]} {[  \ce{HCO3^-} ]}$$

From Henderson–Hasselbalch equation
$$\ce{pH} = \ce{p}K_\ce{a} + \log \left( \frac{[\ce{A^-}]}{[\ce{HA}]} \right)  $$
If there is supposed all $[\ce{H+}]$ comes from acid dissociation and majority of acid is not dissociated, then  :
$$\ce{pH}   \overset{cca}=
\ce{p}K_\ce{a} -   \ce{pH} - \log   {[\ce{HA}]_\ce{total}}
$$
$$\ce{pH} \overset{cca}=  \frac12 (
\ce{p}K_\ce{a} - \log   {[\ce{HA}]_\ce{total}} )$$
Alternatively
$$\ce{pOH} = \ce{p}K_\ce{b} + \log \left( \frac{[\ce{BH+}]}{[\ce{B}]} \right) \\  \overset{cca}=
\ce{p}K_\ce{b} -   \ce{pOH} - \log   {[\ce{B}]_\ce{total}}
$$
Then
$$\ce{pOH} \overset{cca}=  \frac12 (
\ce{p}K_\ce{b} - \log   {[\ce{B}]_\ce{total}} )$$
$$14 - \ce{pH} \overset{cca}=  \frac12 ( 14 -
\ce{p}K_\ce{a} - \log   {[\ce{B}]_\ce{total}} )$$
$$\ce{pH} \overset{cca}=  7 + \frac12 ( \ce{p}K_\ce{a} + \log   {[\ce{B}]_\ce{total}} )$$

How to derive the following equation :
$$\mathrm{pH_{init}}=7 + 0.5 ( \mathrm{p} K_\mathrm{a2} + \log{ [  \ce{Na2CO3} ]} )$$

@Adnan AL-Amleh From analogous equation pH=0.5(pKa - log c ) for solution of a weak acid, that is derived with a simplification from definition of an acid dissociation constant. All is the basic stuff from chemical textbooks.

@Adnan AL-Amleh Perhaps you are looking for https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation ?


text:
Sources for files for testing a thermal ellipsoid (ORTEP) program
I am trying to reimplement the functionality of the classical program ORTEP for producing thermal ellipsoid plots of a crystal/molecule in a modern computer language. Unfortunately, I am having trouble finding usable sample files with the required thermal displacement parameters so that I can test my program. Is there any good source of thermal data for various molecules? Alternatively, if I have e.g. a MOL file of a molecule (or some other easily produced chemical file format), is there any way to produce reasonable(?) simulations of the required thermal parameters? Ideally I would want a sampling of molecules of various sizes.

I was able to compile and run ORTEP3 on a CentOS Linux machine, with a little effort.
libx11 and pgplot are required. you will also need gfortran, and apropriate shell variables set up.  from my .bashrc file:
export PGPLOT_DIR=/usr/local/pgplot
export PGPLOT_DEV=/Xserve I can point you to a compiled linux version if you would like. provided you have libx11, and pgplot, compiling will look like: gfortran -g -o ortep3 ortep.f -lpgplot -lX11


text:
Propylene Glycol to clean vinyl records
I clean vinyl records with an ultrasonic record cleaning machine. Would propylene glycol be good to use as a surfactant in a 1 - 2 % concentration with distilled water?  Would it harm the vinyl record?  Thanks for your help.

I know nothing about cleaning records. However propylene glycol seems a poor choice to me. Propylene glycol has a low volatility, which means that it will not evaporate readily. I'd worry about it leaving a "sticky" layer on the record.

Thanks for your reply! Yes I definitely would not want a residue that would be picked up by the needle!

http://www.audioheritage.org/vbulletin/showthread.php?19643-Record-Cleaning/page2 Suggests some formulations may have used propylene glycol. I’ll second that the Discwasher formula likely had it rather than isopropyl.

Hey MaxW, Had a thought (I do that every once in a while). The cleaning process calls for spraying the surfactant onto both sides of a record, brushing the surfactant into the record grooves, and then immediately placing the record in a 5 minutes ultrasonic bath.  Would the 5 minutes ultrasonic bath clean the sticky layer away from the record? Thanks again for your help. Thom Jones

Hey Jon Custer, Thanks for the link. Read it and there is talk of using Propylene glycol in cleaning records - as a solvent not as a surfactant. Would it work as a surfactant?  Thanks for your help. Thom Jones

Is there a reason to not use an actual surfactant, e.g., dishwashing detergent, in very dilute solution? I ask because it seems that cleaning the records is the main goal and propylene glycol has no necessary connection to that. Just curious!

Hi Ed, there is a solution  which is described as 98% distilled water and "1% - 2% Diol2 Propandiol 178, a type of propanol" but it seems fairly expensive. An article states this is the same as propylene glycol which is available on Amazon and much cheaper. The manufacturer states that 1% - 2% Diol2 Propandiol 178, a type of propanol is a surfactant.  The record cleaning solution is described as allowing the ultrasonic process to penetrate deep into the record grooves so that the microscopic explosions clean dirt  away much better than any other process.


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Molecular orbital energies prediction with ML algorithms
In the recent years, computational chemistry community has focused on Machine Learning algorithms to predict molecular properties. Unfortunately, many of the authors of such papers are not chemists, and hence, there are little explanation on chemical aspects of this type of works.
One well known problem is to predict different properties of QM9 dataset given their atom features. The properties include HOMO/LUMO orbital energies, atomization energies, etc. The inputs include type of heavy atoms in the molecule, type of their bonds, type of atom hybridization, etc. You can see a complete list in the following tables

The question is how the inputs are related to the outputs. For example, I understand that HOMO\LUMO energies are solutions of the electronic Schrödinger equation. But which one of the inputs features affects these? I appreciate it if someone can expand on which ones affect frontier molecular energies. Thanks!

The interpretation of the how the features lead to a given output is a problem with any machine learning prediction. We can make some sort of hand-wavy arguments about why certain features might influence orbital energies, how much a given feature influences an output is entirely dependent on the structure of the model. Particularly for neural networks, where there are a number of layers between features and output, it is difficult to tease out what features are most important.

Thanks for the comment, @Tyberius. Well said. The "hand-wavy arguments about why certain features might influence orbital energies" is what I'm looking for.

Fundamentally, molecular properties are determined by the atoms, their type, their positions (which includes the chemical environment of each atom) and the number of electrons (total charge). Most of your mentioned features actually try to describe the chemical environment by classifying different structural patterns (e.g. closest neighbors, bond type). Change any of that and you will get different orbital energies. Of course some changes have a larger influence than others. Answering this question is essentially what chemistry is all about. Which makes your question a very broad one ...

One a side note: orbital energies are not solutions to the electronic Schrödinger equation! They are more like intermediate results.

Thanks @Feodoran. By nature, I agree that the question is a broad one, but, I hope that I made it clear that I'm looking for one liner answers such as yours. (Not one-line, but one paragraph! :D)

As several comments have mentioned, interpreting a machine learning method can often be difficult. Outside the question of machine learning, I think your question centers around "what would I want to know to roughly predict HOMO / LUMO energies?"
Unfortunately QM9 isn't very diverse chemically, but if we think from a general qualitative molecular orbital picture (e.g., building up diatomic molecules), you want to know:

Number of valence electrons (i.e., which orbitals are filled)
Atom type (which tells you valence electrons and atomic orbital energies)
Type of bond (e.g., single sigma bonds clearly will yield different MO energies than pi bonds)
Conjugation - conjugated and aromatic systems will have different MO energies than plain sigma-bonded species
Bond distances (from the Coulomb repulsion which encodes a distance $1/r$)

Given the above information, it's fairly easy to sketch out MO diagrams for $\ce{O2}$ or $\ce{HOH}$ so one can imagine that they're important for a machine learning algorithm too.
Now how do you know if they actually matter to the ML model?
One easy way is to drop one of the descriptors and retrain the model. Does it get worse? How much?
There are statistical techniques to give you some insight as well. For example, one can use LASSO to drop factors from the input representation that don't have much effect. You can also look at simple statistical correlation between input factors. I'd guess that "Atom Type" and "Atomic Number" are probably highly correlated with each other and only one is truly necessary.. but that's just a guess.

Thanks, @Geoff Hutchison! Just a note for future readers that I found atom Hybridization to have the highest impact on the quality of prediction, followed by hydrogen neighbor count. Among the bond features also, Conjugation has high impact.

Makes a lot of sense - IMHO it's important to think critically about ML methods including "what would I want to know as a human" in addition to automated methods.

BTW, if you want a different data set, you might want to check PubChemQC http://pubchemqc.riken.jp and http://doi.org/10.1021/acs.jcim.7b00083


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The most acidic hydrogen of 4‐hydroxybutanenitrile

Identify the most acidic hydrogen atom in the compound below:
$$\ce{NC-CH2-CH2-CH2-OH}$$

My teacher said that the $\ce{H}$ connected to $\ce{O}$ is the most acidic since $\ce{O}$ is more electronegative than carbon.
However, I see that the $\ce{H}$ connected to the carbon adjacent to cyanide must be more acidic since the conjugate base is stabilized by resonance, contrarily to that connected to $\ce{O}$.
Isn't resonance the most significant criterion in determining the stability of a conjugate base? Or does the electronegativity of the donor atom matter more?

Electronegativity nearly always matters more, unless there is a lot of resonance (think malonates). My other suggestion is to stop thinking that one can simplify it into an "X matters more than Y" or "Z is the most important thing" situation that you can use to predict acidities. There will always be exceptions, and the only way to figure it out is to look up the data. In this case, alcohols have pKa ~15, and nitriles ~30. Thus alcohols are more acidic. You can figure out your own rules once you have enough experience with the hard numbers to know what's right and wrong. Not before.