Compound with pressure sensitive melting point at moderate temperature and pressure I'm in need for a compound with ideally all of the following properties:

A melting point of 5 °C or less at 1 atmosphere (or slightly lower pressure)
A melting point of 12 °C or more at less than 20 atmospheres.
A significant enthalpy of fusion

(long term stability, a boiling point > 50 °C, and less than a 3 digit \$ cost per kg would also be nice).
Seeking such a compound, a melting point range this wide appears to be unusual at such small pressure differences. So far, the best I have is cyclohexane, but the melting point is a bit too high at 6.5 °C at normal pressure, and it needs over 100 atmospheres before the melting point gets above 12 °C, which is impractical.

I'm not going to do your research for you (at least not for now—perhaps if the mood takes me I'll do the Mathematica programming at some later point), but I will give you guidance that will help you find the answer:

It's hard to find data on melting points at different pressures for a wide range of compounds (and you want to search through a wide range of compounds).  What is much more comprehensively available are data that can be used to calculate such pressure dependency, namely the specific densities of substances in their solid and liquid phases, and their enthalpies of fusion.  So:

According the Clapeyron equation,
$$\frac{dT}{dp}=\frac{\Delta V_m}{\Delta S_m}$$
where $T$ is the phase transition temperature, $\Delta S_m$ is the molar entropy change for the phase transition, and $\Delta V_m$ is the molar volume change for the phase transition.
For a phase transition, $\Delta S_m$ = $\frac{\Delta H_m}{T}$, where $T$ is in Kelvin.  And you can determine $V_m$ (in $\pu{\frac{{cm}^3}{mol}}$) from the molecular mass, $M_r$ (in $\pu{\frac{g}{mol}}$), divided by the specific density, $\rho$ (in $\pu{\frac{g}{cm^3}}$):
$$V_m = \frac{M_r}{\rho}\equiv \frac{\pu{\frac{g}{mol}}}{\pu{\frac{g}{cm^3}}} = \pu{\frac{{cm}^3}{mol}}$$
Thus:
$$\Delta V_{m, fus} = \frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}$$
The subscript "fus" means "fusion", indicating this is a solid-liquid phase transition.
Substituting, we obtain:
$$\frac{\Delta T_{fus}}{\Delta p} \approx \frac{dT_{fus}}{dp}=\frac{\Delta V_{m, fus}}{\Delta S_{m, fus}} =\frac{\frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}}{\Delta S_{m, fus}}=T_{fus}\frac{\frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}}{\Delta H_{m,fus}} $$
What you want is to identify the substances that have the largest values for the above expression, and then filter according to your other criteria.
Note that this expression only gives an exact value for $\frac{\Delta T_{fus}}{\Delta p}$ if you integrate with respect to $T_{fus}$ using $\Delta H_{fus} (T), \rho_{liq}(T) \text{ and } \rho_{solid}(T)$—i.e., if you account for the temperature-dependencies of each of these. However, since you are determining $\frac{\Delta T_{fus}}{\Delta p}$ over a range not far from $T_{fus}$ at 1 atm, and since the densities for the phases that are stable at room temperature are probably at $25 ^\circ C = 298 K$, which is close to your desired $T_{fus}$, and since they will not change much over your pressure range, this should give you a very good approximation. Once you narrow it down to the actual substance, you can apply corrections to get a more precise estimate.

You could do this search manually by plugging in values into an Excel spreadsheet, or semi-manually by finding downloadable data tables with these values and importing them into Excel, but the best approach would be to use a program like Mathematica*, which has access to large chemical databases, and ask it to search for all substances for which it has density data in the solid and liquid phases, and enthalpies of fusion, calcualate the values for the above expression, and sort by magnitude.  You can then apply additional filters (e.g., melting point and enthalpy of fusion), as desired.

[*This of course requires knowing how to use Mathematica, or learning to use it.  It's possible someone on MathematicaSE could help you write such a query. Alternately, it's possible the online CRC is also searchable in this way (certainly seems it should be), though I don't know myself. One way to get help might be to check the reference desk of your local library.  They might have the capability to query chemical databases in this way.]
Once you have this, you can then identify the best candidate compounds, and further filter by cost and stability.

Note: It may not be possible to find a compound whose melting point changes enough (a comprehensive chemical database search would help you determine this).  Finding a large pressure-induced change in melting point is challenging, because substances don't change volume that much in going from a solid to a liquid. Where you do get a large pressure-induced change in a phase transition temperature is when going from liquid to gas because, under normal conditions (room temp and 1 atm), gases have about 1000x the molar volume of liquids. That's why water boils at a notably lower temperature when you go into the mountains, even though the change in pressure there is a mere fraction of an atmosphere. [By "notably lower" I mean enough of a change to affect cooking times.]

Plus you make your search even more challenging by requiring a high $\Delta H_{fus}$, since a high $\Delta H_{fus}$ lowers $\frac{\Delta T_{fus}}{\Delta p}$ (everything else being equal).


What is the precise definition of Ramachandran angles? The Wikipedia article Ramachandran plot is very schematic and doesn't even say when these angles are considered to be zero. It seems to assume that the peptide chain has no end, and always shows the next carbon atom. What happens when the amino acid is in the terminal position? It also asserts that the $\omega$ angle is always 180°, while in the practical cases this isn't exactly true.

The definitions seem to rely on the adjacent bonds. Two on one side, and one on the other side. They rely on the fact that two bonds on one side are in the same plane as the rotating bond. How are these angles defined when relevant bonds don't form a plane?

What is the precise definition of $\psi-\phi$ angles? Are there formulas for them that take atom positions as arguments with amino acids being in both terminal and non-terminal positions? Are they defined for amino acids in terminal positions?

Additionally, an article by Zhou et al. [1] also talks about the bond angle $\tau$. $\tau$ seems to be a logical addition because it corresponds to an additional degree of freedom, not covered by $\psi-\phi$ angles.

References

Zhou, A. Q.; O’Hern, C. S.; Regan, L. Revisiting the Ramachandran Plot from a New Angle. Protein Science 2011, 20 (7), 1166–1171. https://doi.org/10.1002/pro.644.

Definition

Protein backbone dihedral angles are called φ (phi), involving the backbone atoms COn-1 - Nn - Cαn - COn , ψ (psi) involving the backbone atoms Nn - Cαn - COn - Nn+1) and ω (omega) involving the backbone atoms Cαn - COn - Nn+1 - Cαn+1). Thus, φ controls the COn-1 - COn distance, ψ controls the Nn - Nn+1 distance and ω controls the Cαn - Cαn+1 distance. Where n is the current atom for which you want to calculate the dihedral angles. Dihedral angle varies between -180 to +180. It is positive if rotation between atoms is clockwise and negative if rotation is anticlockwise.

Source: http://mebinfotalk.blogspot.com/2013/06/how-to-calculate-dihedral-angle-using.html
And here is an illustration from the same source:

To calculate, use the cross product to get vectors normal to the (v2, v1) and the (v2, v3) planes, and calculate angles between them (it is easy to get the direction wrong, so I would test with known cases).
In the question, the OP states:

They rely on the fact that two bonds on one side are in the same plane as the rotating bond.

No, the plane is between the rotating bond and the adjacent bond. Two bonds of one atom always form a plane. Here is an illustration: Dihedral angle of gaseous and crystalline HOOH
3D demonstration
A nice interactive tutorial can be found at Proteopedia: https://proteopedia.org/wiki/index.php/Tutorial:Ramachandran_principle_and_phi_psi_angles
Omega
Omega is the torsion angle of the peptide plane. It is most often near 180 (trans peptide) but sometimes near zero (cis peptide).
In a PNAS paper from 2012 (https://doi.org/10.1073/pnas.1107115108), the authors state in the abstract:

Analyses as a function of the φ,ψ-backbone dihedral angles show that the expected value deviates by ± 8° from planar as a systematic function of conformation, but that the large majority of variation in planarity depends on tertiary effects.

In protein structure determinations, the omega angle is often restrained, especially in low resolution structures. Looking at structures deposited in the protein data bank, it might therefore seem that the distribution of the omega torsion around 180 is more tight (less varied) than it is in reality.
First and last dihedral
The first amino acid has no phi angle and the last has no psi angle. Exceptions are when the N-terminus or C-terminus are modified.
Bond angle tau
Different from the torsion angles phi and psi, the angle tau is a bond angle (depends on three atom position, not four). It is the angle at the C-alpha carbon between the nitrogen and the carbonyl carbon. As an $sp^3$ carbon, the C-alpha carbon is expected to have an angle around $109^\circ$. The larger the angle, the more (phi, psi) combinations are possible without steric clashes, i.e. the larger the "allowed" regions in the Ramachandran plot.


Thermodynamic model of inclusion process of alpha-cyclodextrin I have a question about thermodynamic model of inclusion process of CD, published by Tabushi (DOI: 10.1021/ja00471a043). Here, in 3rd step, action is described as: "An apolar solute (guest molecule) is transferred from water to the ideal gaseous state leaving a structured cavity behind. The cavity then collapses with redistribution of the water molecules." I don´t understand this part where guest molecule is transferred to the gaseous state (why this state). Can somebody try to explain this in some simple way if it is possible, please? Thanks for all the answers!

This is a variation on the application of Hess's Law (or looks like it, not having read the article). The idea is that the individual steps add up to the complete reaction

cyclodextrin with included water + hydrated guest $\rightarrow$ hydrated cyclodextrin with included guest

According to Hess's Law you can break up this process into a sum of intermediate steps. One of those steps is dehydrating the guest molecules and shoving the waters into the bulk solvent. That last step is what is meant by collapse of the cavity  with redistribution of the water molecules..


Do halogens bond more commonly with each other or themselves? To me it seems like diatomic elements like $\ce{F2}$, $\ce{Cl2}$, etc. occur more prevalently than diatomic molecules with both, like $\ce{ClF}$.

Is this the case, and if so, what causes diatomic elements to be favored?

There's a lot of circumstantial evidence that heterodiatomics are more stable than homodiatomics. It is usually ascribed to the observation that atoms of different "electronegativity" tend to have stronger bonds than atoms of the same electronegativity.

In your particular example, consider the reaction:

$$\ce{Cl2 + F2 -> 2 ClF}$$

I don't have the tables handy, but I calculate with electronic structure theory that this reaction is exothermic by about 25 kcal/mol. That's quite a bit! (The equilibrium lies far to the right.)

This would suggest that a mixture of Chlorine and Fluorine, heated up to very high temperature and then cooled down, would tend to reform as ClF.


SN2 nucleophilic attack with alpha unsaturation The following two questions prompted me to ask this question.

  Determine the order of the following compounds according to reactivity in Williamson Ether Synthesis.
  a. $\ce{CH3CH2CH2Br}$
  b. $\ce{CH2=CHCH2Cl}$

  Which of these represent the ether($\ce{CH2=CHCH2OC3H7}$) on forming peroxide in air
  a. $\ce{CH2=CHCH2OCH(OOH)CH2CH3}$
  b. $\ce{CH2=CH-CH(OOH)OC3H7}$

In both the case, I think the question is the relative ease of nucleophilic attack on carbon with $\alpha$-unsaturation. In the former, the bromine is a better leaving group. But, the presence of $\alpha$-unsaturation in the chloro compound, changes the order to b>a. In the latter, the attack of oxygen depends on the most stable electrophilic centre.

My question:-

Is $\alpha$-unsaturation favourable for nucleophilic attack? Relevant factors would be the increased electrophilicity of the carbon attached to the $sp^2$ hybridised ethylinic carbon and the repulsion of the negative nucleophile by the pi electrons.

As Ben Norris said, the aspects of nucleophilic and electrophilic attacks are not relevant in the second case.

Talking about the compound given in question 1(b), the first step of nucleophilic attack is opposed by the conditions you mentioned, i.e., of the electron withdrawal effect of the double bond and replusion of nucleophile by $\pi$ electron cloud.

But talking in terms of yield of the product, it is necessary to consider two things:

Chlorine is a good leaving group.
The stability of the carbocation intermediate formed by $SN_1$ mechanism (steric factors don't favour $SN_2$). The carbocation formed will be very stable due to conjugation with the double bond. Hence, the yield of the product is not much low.

While solving the above question, we should look at the stability of carbocation intermediate formed,which is a more dominating effect. Though Bromine is a better leaving group, and there is decrease in tendency for nucleophilic attack due to $\alpha $-unsaturation; both factors which support higher nucleophilic attack for the first compound, stabilisation of carbocation by resonance (conjugation) is highly stabilising, and this shadows the opposing factors.


When volume decreases and the product has less moles than reactants, why is there no increase in the reaction quotient? Assume a reaction:

$$\ce{CO + Cl2 <=> COCl2}$$

My textbook says that a decrease in volume will result in the reaction quotient being less than the equilibrium constant. Why is that?

With a decrease in the volume, a forward reaction proceeds. Thus, more product and less reactant means there should be a higher reaction quotient.

Keep in mind that for reaction

$$\ce{r_{1}R_{1} + r_{2}R_{2} + \ldots + r_{n}R_{n} <=> p_{1}P_{1} +\ldots + p_{m}P_{m}}$$

the reaction quotient is:

$$Q = \frac{\prod_{i = 1}^{m} [P_{i}]^{p_{i}}}{\prod_{j = 1}^{n} [R_{j}]^{r_{j}}}$$

Specifically, here:
$$Q = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$

For your specific reaction, the reaction quotient is getting smaller because the denominator has a quadratic dependence on the pressure where as the numerator only has a linear dependence on pressure.

The goal as usual is to make $Q = K_{\mathrm{eq}}$, which is equilibrium.

Since now, $Q < K_{\mathrm{eq}}$, you want to push the reaction forwards. That will increase the concentrations of product and decrease the concentrations of reactant, making the reaction quotient bigger.

Intuitively, that makes sense in your example. There are two molecules combining to form one molecule. By decreasing the volume and increasing the pressure, you're squeezing them together into the product.


pH of baking soda in water? If I dissolve 1  g of sodium bicarbonate powder ($\ce{NaHCO3}$) into 400 ml of distilled water, what will the pH of the resulting solution be?

1 g of sodium bicarbonate in 400 mL of water will give an approximately 30 mM solution with a pH of ~8.4. The interesting thing about the pH of solutions of amphiprotic compounds is that so long as the concentration of the salt is significantly larger than its Ka and Kb, the solution pH only depends on the dissociation constants. Basically any amount of bicarbonate above 10 mM or so will have the same pH in solution, so pH is not a good indicator of how much acid you can neutralize with it.

I'm not really sure what good this information will do you. How much this solution will be able to affect the paper is going to depend on how much paper you use for a given volume of solution and how acidic the paper is. I don't know much about paper—it could be that it doesn't really matter if you go overboard on the bicarbonate. It might be as simple as making a concentrated solution, testing periodically that your bath hasn't become acidic, similar to photographic stop bath, which has a pH indicator that changes colour when the bath is not acidic anymore. If residual bicarbonate is a problem, rinsing with water afterwards is probably advisable.


Is corovin Polypropylene or flat polyester? I'm wanting to know what material 'corovin' is as described on some duvet covers.  Googling it, some places say it is just another name for 'Polypropylene' while other places say it is 'flat polyester' whatever that means.

Which is correct?  As far as I'm aware Polypropylene and polyester are different types of synthetic fibers.

Thanks

To my understanding, Corovin is a technology using to manufacture Nonwoven fabric. These fabric could either be polypropylens or polyesters, yet technique is first develop for polypropylene (Ref.1). Following is the phrase I borrowed from the book (p. 134):

  Companies such as Du Pont, Freudenberg and Corovin began their research and development work on generating spunbonded webs. In Europe, Freudenberg, Terram, and Corovin started production of polypropylene (PP) spunbonded webs at the beginning of the 1970s. Within these technologies, filaments are spun directly out of the polymer melt through small die heads arranged side by side, stretched down to a diameter of $\pu{10-30 \mu m}$ and laid down on a porous lay down and transport belt.

A recent patent has been assigned to Fiberweb Corovin GmbH for manufacture of polyethylene-based, soft nonwoven fabrics as well (Ref.2). The abstract states:

  The present invention concerns a nonwoven fabric the fibers of which have polyethylene on at least part of their surface, where the fibers are thermally bonded and the non-woven fabric exhibits an abrasion of less than $\pu{0.5 mg/cm2}$, in particular of less than $\pu{0.4 mg/cm2}$, and an embossed area fraction of less than 35%, in particular of less than 28%. Furthermore a device is made available for the manufacture of a non-woven fabric, using a polyethylene-containing polymer, with a take-down system underneath a spinning plate that causes the take-down of the polyethylene, where the spinning plate has an L/D ratio of between 4 and 9.

References:

H.-G. Geus, "Chapter 5: Developments in manufacturing techniques for technical nonwovens," In Woodhead Publishing Series in Textiles, Number 181: Advances in Technical Nonwovens; George Kellie, Ed.; Woodhead Publishing: Amesterdam, The Netherlands, 2016, pp. 133-153 (ISBN: 978-0-08-100575-0).
Markus Haberer, Henning Röttger, "Polyethylene-based, soft nonwoven fabric," US Patent: 8420557, 2013 (Filed: 2006).


Is “mass-action ratio” identical to “reaction quotient?” Mass-action ratio
Reaction quotient

According to their definitions expressed in the above Wikipedia articles, these two concepts seem to be totally identical. Is that true? If not, what is the difference?

The concepts are extremely closely related.  The difference is that the mass action ratio is a formally a ratio of concentrations, or, for gases, of partial pressures.  The mass action ratio, and the term "mass action" in general, refers to situations where the chemical potential of the reactants and products is directly proportional to the concentration, or pressure, or mass, of the species involved.

The reaction quotient is is a tiny bit more general; the implicit assumption of mass action is not necessarily made.  The reaction quotient is the ratio of chemical potentials of reactants and products.  It works even in systems that don't follow the law of mass action, i.e. where the activity coefficients vary during the equilibration.

For all intents and purposes, the two concepts are the same.  If you want to split hairs, you'd probably do it the way I did here.


How to evenly mix NaCl and Lactose Q:
You have 10 g NaCl (s) and 90 g Lactose (s) (both easily soluble in water.) You want to achieve a 100 g mixture of these two compounds in solid form where they are completely evenly mixed. (A 1 mg sample of this mixture will always have 0.1 mg NaCl and 0.9 mg Lactose no matter from where in the mixture the sample is taken.)

Attempted A:
Boil a litre of distilled water in a clean container. In a sterile container put the pulverized NaCl and Lactose. Wait until boiled water cools to at least 50 C because high temperature increases risk of hydrolyzing lactose. While stirring with a sterile glass stir rod add the hot water step-wise until just enough water has been added to completely dissolve all the powder. Leave the solution on a warm plate (temperature shouldn't exceed 40-50 Celsius) and use a fan or similar device for airflow to speed up evaporation. Once most of water has evaporated put in a dessicator chamber to finish drying.

What I'm not sure about:
Would this actually create a completely even mixture because lactose and NaCl have slightly different solubilities so as the amount of water drops wouldn't one start precipitating earlier? If this is true how could you prevent that and are there any other issues?

The final solid mass would be a powder of sort, right? If you grind it / mix it will be homogeneous irregardless of the selectivity of their precipitation in the prior solution.


How do I determine the number and order of ligand orbitals? I am trying to create interaction diagrams for transition metal species by reducing the representations for both $\sigma$ and $\pi$ ligand orbital sets and mixing them with the metal orbitals. The order of metal orbitals is obvious because I know I have my 5 degenerate $n\text{d}$ orbitals, then my $(n+1)\text{s}$ orbital, which is higher in energy due to its principal quantum number, then my $(n+1)\text{p}$ orbital, which is higher in energy due to penetration (it's a p orbital); and the number of metal orbitals is apparent because my character table lists the each representation's matching function, so I can tell $\mathrm{A}_{1g}$ will have s-symmetry and $\mathrm{E}_g$ will match my $\mathrm{d}_{x^2-y^2}$ and my $\mathrm{d}_{z^2}$ orbitals, for example.

I don't see exactly how I should do the same for my ligand orbitals, though. For example, I know that my ligand $\pi$ orbital irreducible representation is $\Gamma_{\pi} = \mathrm{T}_{1g} + \mathrm{T}_{2g} + \mathrm{T}_{1u} + \mathrm{T}_{2u}$, and I know from my table's matching function list that there will be three $\mathrm{T}_{1u}$ $(x,y,z)$ and three $\mathrm{T}_{2g}$ $(xy, xz, yz)$ orbitals, but I don't know how to identify that there are also three $\mathrm{T}_{1g}$ and $\mathrm{T}_{2u}$ orbitals, since my character table doesn't list the matching functions for those representations.

Also; my understanding of how to decide on the ligand orbitals' relative energy levels is dubious. Yves Jean's Molecular Orbitals of Transition Metal Complexes says "one must analyze the bonding or antibonding character" (p. 43), but I am not getting it right every time. Can anyone give me some tips?

T refers to triply degenerate set, E refers to doubly degenerate set, A and B are singly degenerate. The total number of orbitals should add up to the total number of atomic orbitals on your ligands. 1,2,g,u are symmetry labels (look up Mulliken symbols). One thing to remember when constructing you MO diagram is that only orbitals of similar energy and the exact same type and symmetry can mix. So, you can only mix A with A, B with B, E with E and T with T. Also, only g with g, u with u, 1 with 1 and 2 with 2.


Brown ring from the anion detection of NO3- Today in our practicals I observed that when I did the wet test for nitrate (Making a 1 ml solution of salt solution+$\ce{FeSO4}$ solution+hot concentrated $\ce{H2SO4}$);
The brown ring appeared for a very short time and then disappeared.

But later on I came to know that it was nitrate only. What may be the possible reason for this temporary appearance of a tinge of a brown colour and then its sudden disappearance?

The brown ring formed in the confirmatory test for the nitrate ion is due to the complex called nitroso ferrous sulphate, or pentaaqua nitrosyl iron (II) sulfate with the formula $\ce{[Fe(H2O)5NO]SO4}$.
The concentrated sulfuric acid decomposes this complex and brown ring disappers in a few seconds when $\ce{H2SO4}$ moves to the upper layer. This is why slow addition of the acid (through the sides of the test tube) is advised.


How is work converted to heat? My book says work can be converted into heat but I want to know how work affects kinetic energy of molecules which, when transferred, is called heat?

Work can be understood as a variation of energy during which the transporters (may they be molecules) are spatially aligned or bonded, meanwhile heat is a variation of energy in which the transporters aren't aligned (therefore heat dissipates energy).
With this in mind, any machine that uses an ordered energy transfer (like your hand waving) can induce heat in a fluidic means, as particles in fluids are initially moving randomly and are not likely to move far on the orientation of your hand.
In this example, the induced work is little and restricted to the analysis on the surface of the moving hand, but if the analysis is made from further on a wider system, we can assume you're doing work to generate heat alone.


Halogenation of Phenol Why does only one $\ce{Br+}$ ion get generated in a non-polar solvent (such as $\ce{CS2}$), while more $\ce{Br+}$ ions are generated in aqueous or polar solvent (such as $\ce{H2O}$)?

The reaction of phenol with bromine is known as bromination of phenol. Solvent has great influence on the reaction. In different solvents, different products are obtained. The action of bromine on phenol can be explained as:
Reaction with bromine in water:—-

Phenol reacts with bromine water to give 2,4,6-tribromophenol.
In water, ionisation is facilitated. Phenol gets ionised to form phenoxide ion, which is even better ortho,para-directing. Bromine also gets ionised to a larger extent to form a large number of bromonium ions. And Bromine ions are highly stabilized in ionic solvents. So the formation of strong o,p-directing group and stabilization of $\ce{Br+}$ enhances the formation of tribromophenol.
Reaction with bromine in $\ce{CS2}$:—-

Phenol reacts with bromine in a  presence of Carbon disulphide to form a mixture of o-bromophenol and p-bromophenol. Among which p-bromophenol predominates. In $\ce{CS2}$ ionisation is not facilitated that much. As it is a non-polar solvent it doesn't the $\ce{Br+}$ on. Also $\ce{-OH}$ group is moderately o,p-directing. So as usual para-product is more formed.


Why is the fluorescence spectrum of pyrene time-dependent? I've got a solution of pyrene and I excite it with a laser. If I measure the emission spectrum 1 ns after absorption, a well‐structured emission with a small Stoke’s shift is observed.
However, if the emission is measured with a delay of 100 ns, a broad, featureless spectrum with a much larger Stoke’s shift is obtained.

Is this purely because pyrene has had some time to undergo solvent reorganization and thus the individual excited pyrene molecules have larger differences or is there something else at play?

I think that you have rediscovered the emission of pyrene exciplexes :)

As you have correctly assigned, the structured fluorescence spectrum shortly after the pulse is the emission of pyrene monomers.

Given sufficient time (100 ns), an excited molecule of pyrene diffuses towards a pyrene molecule in its ground state and forms an exciplex.

Fluorescence emission from these exciplexes is known to be broad and unstructured.


Why does the size of hybrid orbitals vary as sp³>sp²>sp? In my text book it said that

  Size of hybrid orbitals vary as sp³>sp²>sp

Does this size variation in hybid orbitals means that S orbitals are smaller than P orbitals?

But we know that larger the quantum number "n" is, the larger the orbital. I guess it has nothing to do with type of orbital.

What could be the reason for this variation in sizes?

Looking at the shapes of atomic orbitals shown in many textbooks can be deceptive. The probability density of the radial part of the wavefunction ($r^2|R_{n,l}(r)|^2$)  shows that the maximum probability for 2s orbitals is noticeably greater than for 2p, and for d orbitals the same applies with p greater than d.  (As the principle quantum number increases so does the position of the maximum probability density of an orbital, i.e. increasing $n$ increases the size of the orbital.)

The hybrid orbitals form from  the atomic orbitals and are added together in different proportions. So the position of maximum probability is a type of average value of the s and p. The sp hybrid has $2s \pm 2p$, the $sp^2$ in the z direction $2s+\sqrt{2}.2p$ and the $sp^3$ also in the z direction just as for the sp orbital. It seems to be that the maximum position of probability taken along, say, the z direction is the same in each case.


Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides Problem

$\ce{Cu}$ reacts with $\ce{HNO3}$ according to the equation
$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2 + H2O}$$
If $\ce{NO}$ and $\ce{NO2}$ are formed in 2:3 ratio, what is coefficient of $\ce{Cu}$ when equation is balanced with simplest whole numbers?

Answer

 9

Question
I am getting the coefficient of $\ce{Cu}$ as $2,$ which contradicts the answer:
$$\ce{2 Cu + 9 HNO3 + 5 H+ -> 2 Cu(NO3)2 + 2 NO + 3 NO2 + 7 H2O}$$

The final arbiter of formal correctness of chemical reaction enumeration  are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That can be useful for educational purposes, or rather to implant terror and despair in eyes of young chemistry students.
Chemists do not go that way. They usually use various heuristics to speed up the process, enumerating equations in several steps. For redox reactions, the heuristic is inventory of exchanged electrons, which is tracked by formal oxidation numbers (ON). These numbers are rather just arbitrary convention, but extremely useful.
For enumeration of your reaction ( with expressed ONs of relevant atoms):
$$\ce{Cu^{0} + HN^{+V}O3 -> Cu^{+II}(N^{+V}O3)2 + N^{+II}O + N^{+IV}O2 + H2O}$$
the first step is matching components undergoing redox change.

Copper changes ON from 0 to +II.i.e. by 2.
Nitrogen changes ON from +V to +IV ( $\ce{NO2}$), or from +V to +II($\ce{NO}$), i.e. by 1 resp. 3.
There is given $\ce{NO/NO2}$ ratio 2:3, so there would be as products $\ce{2 NO + 3 NO2}$, or their multiples.
$\ce{2 NO}$ exchange $2 \cdot 3 = 6$ electrons.
$\ce{3 NO2}$ exchange $3 \cdot 1 = 3$ electrons.

For simplicity, we can take formally $\ce{2 NO + 3 NO2}$ as a single entity, created from $\ce{HNO3}$ by exchanging totally 9 electrons.
Now comes the "magical" formula $9 \cdot 2 = 2 \cdot 9$.
9 Cu exchanging 2 electrons each gives 18 electrons totally.
2 ( $\ce{2 NO + 3 NO2}$  ) exchanging 9 electrons each has accepted 18 electrons.
So the intermediate result is:
$$\ce{9 Cu + HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
This is enough for the question answer, but we have to finish the equation. It may sometimes happen some coefficients would get fractional, so all would get multiplied to have all integer.
We need supply enough nitrogen for the salt and oxides: $2\cdot 9 + 4 + 6 = 28$
$$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
We need to create water from hydrogens:
$$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + 14 H2O}$$
And we are done. We can formally check the item conservation:
$\ce{Cu}$: $9=9$
$\ce{H}$: $28=28$
$\ce{N}$: $28=28$
$\ce{O}$: $3\cdot 28=84=54+4+12+14=84$
charge: $0=0$
For a quick check,we may just check items we were not enumerating before, like just oxygen and charge.


What is the direction of charge transfer between Aluminum and aliphatic amines physisorbed on top? Is there dipole formation upon physisorption of an aliphatic amine to an aluminum substrate? What is the charge transfer direction? What if the aluminum substrate is clean, surface oxidized or covered with -OH groups?
What if aluminum is replaced by other types of metals, such as Au and Ag?

Amines could bond to the aluminum surface via a Lewis acid-base reaction between the electron rich amine nitrogen and the electron poor Al atom on the surface (perhaps formed by loss of a surface hydroxyl group as hydroxide ion). The direction of charge transfer would be from the amine nitrogen to an aluminum atom. The resulting "Lewis salt" would have a partial positive charge on nitrogen and a partial negative charge on aluminum (like an aluminate ion or an amine-aluminum chloride adduct).


Which will be oxidised between water and Copper? I have always read that the one having higher  discharge potential undergoes the reaction( oxidation or reduction). But In this example why is there also oxidation of water?

The question is not fully complete (not your mistake but someone who wrote this exercise). I am not surprised that you are confused. The question is indeed silent about the electrodes. You should have asked the question: What is the cathode made of? What is the anode made of before starting the question. The second trick in your question in your question is, which I liked, is that it is an example of exhaustive electrolyte. You passed a huge current which consumed all free metal ions and the solvent can also be reduced or oxidized.

So let us look at your problem:

At the cathode you have two explicit choices: (i) Reduction of Cu ions which are floating in the solution, (ii) Reduction of the solvent, which is water in your case to hydrogen! There is nothing else which can be reduced here as per your question.

At the anode you still have two implicit (hidden) choices: (i) Oxidation of water and (ii) Oxidation of the anode material.

Your solved questions bluntly assumes that the electrode is inert, say it is made of carbon or Pt or gold, whatever but inert. So are left with only one choice. Oxidation of water. However, let us say they actually told you that the anode is made of Copper:

Oxidation potential of Cu

Cu(s) ---------> Cu$^{2+}$(aq) + 2e$^{-1}$    -0.34 V

Oxidation potential of water

2 H$_2$O(l)---------> O$_2$(g) +  4 H$^+$(aq) + 4e$^{-1}$    -1.23 V

Which one do you think will get oxidized? The Cu electrode instead of water!

This is the process and principle behind copper plating.

When your examiners are quiet about electrode materials, assume the solvent is going to oxidize, i.e., water.


Which white powders form acids when mixed with water? I have a mystery acid which isn't citric, boric, or oxalic acid. When mixed with water, it smells pungent, like hydrochloric acid--though I haven't compared them side by side.

Which other white powders form a strong acid upon being combined with water? Is there any powder that decomposes into HCl when mixed with water?

(Note: the mystery acid was sold as part of a set, along with sodium chlorite, for generating chlorine dioxide for use in aquariums.)

From the pH and solubility information alone, I would propose sodium bisulfate: cheap, readily available, safe in the powerfully oxidizing solution. (Possibly also potassium bisulfate.)

One way of manufacturing $\ce{HCl}$ produces sodium bisulfate as a byproduct:
$$\ce{NaCl + H2SO4 -> HCl + NaHSO4}$$ The odor of $\ce{ HCl}$ could be a minor impurity in this mystery acid. The other impurities also suggest it is not a reagent grade product.


Co-precipitation particle size If we are creating a precipitate of (say) Barium Sulfate from a solution of Barium Chloride mixed with Sodium Sulfate, what determines the particle size?
I assume one of the factors will be concentration of reactants in the solutions, but what else?

Kinetic aspects:
Low concentrations of interacting ions decrease rate of precipitation, leading to bigger crystals.
A high ionic strength of spectator salts leads to lower activity coeeficients , and  decreased rate of precipitation, leading to bigger crystals.
Higher viscosity leads to lower collision rates, decreased rate of precipitation, and to bigger crystals.
Low temperaturee leads to bigger crystals, as rate of collisions leading to crystallisation is lower due higher viscosity, lower molecular speed and lower rates of interactions leading to precipitation.
Note that kinetic aspects may be hard to involve in the proces control, aside of concentrations.
Thermodynamic aspect:
Crystals of low sizes have high surface Gibbs energy and therefore bigger crystals are thermodynamically preferred.
This is initially overrun by the kinetics of the precipitation, but in long term there is ongoing recrystallization in favour of bigger ones. This can be extremely slow for very little soluble salts.
This process is speeded up by high temperature, what increases rates of the opposite processes of dissolution and crystallization. AFAIK, the gravimetric procedure to determinate sulphates as barium sulphate involved heating $\ce{BaSO4}$ suspension near below boiling point for I guess 1-2 hours to speed up recrystallization. As bigger crystals are easier to manage, espacially in this particularly case, as $\ce{BaSO4}$ is known to form too fine crystals.


Grignards and diatomic halogens Can Grignards react with diatomic halogens as to perform an substitution reaction?

I'd imagine that the first part of making the reaction work - polarizing the halogen as to prepare to pull it apart - would be easy for a Grignard. Hell, even alkene pi bonds can polarize halogens sufficiently; why can't a Grignard?

So in essence, can phenylmagnesiumbromide react with $\ce{Cl2}$ to make chlorobenzene? The other product of this reaction would be $\ce{Cl-}$ ... which I could see combining with the $\ce{MgBr+}$ in some way.

Can Grignards react with diatomic halogens as to perform an
  substitution reaction?

Yes, but it looks like a mixture of products results, including something explosive.  Here is a link to an abstract of some early work that reports,

"Chlorination of Grignard and Iotsich reagents, RMgX, (where X = Br, I) gives considerable quantities of RI and RBr together with RCl. Bromination of RMgI results in RI and RBr. Bromination of RMgCl leads to RBr, and iodination of RMgX (X =Cl, Br), to RI."

An "Iotsich" reagent is an acetylenic Grignard ($\ce{RC#CMgBr}$)

Judging by the fact that 1) the abstract suggests that product mixtures of the various possible alkyl\aryl halides are often formed and 2) no further reports claiming synthetic utlity have emerged, I suspect that the reaction is of limited synthetic value.  Nonetheless, your initial suspicion that reaction should occur seems supported.

One further note, many MSDS's and "Hazardous Laboratory Chemicals Disposal Guide, Third Edition" report that when phenylmagnesium bromide is reacted with chlorine, a solid is formed that explodes when shaken.  The reference provided is the same J. Organomet. Chem. paper cited above.  Given the lack of detail provided on the solid, my guess is that it wasn't characterized.


Band MO of Sodium Filling of electrons confusion I wanted to ask a question about metallic bond - band electron theory.

Consider the diagram of Sodium below:

Circled are orbitals both containing $2$ electrons each which combine using LCAO to give a set of bonding and antibonding electrons.

However, the $4$ electrons involved in this bonding  appear to only give $3$ electrons in the bonding and antibonding MO's produced.

This confuses me. Surely using $4$ electrons, I expect all $4$ of these electrons to also be present in the rectangle drawn.

Why are there only 3 electrons shown? Is one electron being used for the next addition of AOs?

Any answers would be greatful!

From left to right, it shows 1, 2, 3, and 4 atoms of sodium, with a total of 1, 2, 3, and 4 valence electrons. So this diagram is different from others which have the uncombined AO's on the left and right, and the combined MO's in the middle. Here, we are adding more and more atoms to go from discreet energy levels (left 4 columns, up to four atoms) to a band structure (rightmost column, many atoms in bulk metal).

The only column that shows the AO is the left-most because for that one, there is only one atom, which means that there can't be any MOs.

Maybe this diagram makes more sense to you (the full and empty circles on the right show how the AO's are combined to make the bonding, non-bonding and antibonding MO's):

Source: http://xbeams.chem.yale.edu/~batista/vaa/nabands.gif


Are all gases equal in their ability to carry with them lower volatility matter at a given temperature? As we all know water is a great solvent in it's liquid form relative to other substances. I wonder however how it performs as a gas relative to other stuff.
Does anyone know how well steam carries compounds with it compared to other gases at the same temperature and pressure?

It depends on how a solvent is defined. Regardless of semantics, supercritical steam 217.75 atm, 373.946 °C is a very good solvent. It will dissolve fats, hydrocarbons or plastics. It also becomes very corrosive to metals.


Something about the idea of Molarity is confusing me Okay, so molarity is defined as moles of solute per liter of solution. It is used to describe the concentration. So in the lab, when I see a bottle labeled something like $\mathrm{3.5~M}~\ce{H2SO4}$, is it saying that the only chemical in the bottle is this $\ce{H2SO4}$?

Because the way I think of it is "there are 3.5 moles of $\ce{H2SO4}$ per liter of solution". The word "solution" is kind of throwing me off because it makes me feel like they are saying that there is more than just $\ce{H2SO4}$ in the bottle. Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?

I don't know why I'm having such a hard time grasping this. But any help would be awesome!

Correct me if I'm wrong, but don't you need at least 2 solutes to make a solution?

You are, unfortunately, incorrect. In wet chemistry, you need just a solvent$^\dagger$ and at least one solute to make a solution.  While this does mean that you need at least two constituents to make up a solution, you do not need at least two solutes.

  ... is it saying that the ONLY chemical in the bottle is this $\ce{H2SO4}$?

It depends on your definition of a chemical:

If you consider water to be a chemical, then no: there are two chemicals in the bottle$^\ddagger$, $\ce{H2O}$ and $\ce{H2SO4}$.
If you are not considering water to be a chemical, then yes: the only chemical in the bottle$^\ddagger$ is $\ce{H2SO4}$.

$^\dagger$ If not specified, the solvent is almost always water.

$^\ddagger$ Except for small amounts of inevitable impurities.


What techniques can determine the concentration of ethyl acetate in a solution containing ethyl acetate, acetic acid, ethanol and water? I need to monitor the concentration of ethyl acetate in an Fischer esterification reaction. What techniques or spectroscopies can detect the concentration of the compound? And which one do you recommend based on precision and less sample preparation?

You might consider taking Aliquots of the reaction (or if you have the equipment directly sampling via a cannula) into a GC-FID or quantitative GC-MS. This way you could monitor the concentrations of all species in solution given the right GC method. Areas under each peak will give you relative amounts, you would need to use an appropriate known amount of an internal standard that remains constant to get absolute numbers.


Electrolysis of concentrated sulphuric acid In my textbook it is given that for electrolysis of dilute sulfuric acid  at anode following reactions can occur:

At moderate concentrations

$\ce{2H2O ->  O2 + H+ +4 e-}$

And for high concentrations

$\ce{2SO4- ->  S2O8^2- +2 e-}$

SRP value for first reaction is less than second and hence the first reaction should take place. But why does that not hold for high concentrations?

This is because in concentrated sulfuric acid solution equilibrium condition will favour more water molecules than oxidation of water into protons. So there will be an over potential required (to go against the equilibrium) , that is extra potential beyond the theoretical reduction potential derived from thermodynamics to complete the reaction. So the total potential required will be theoretical Plus overpotential. In case of oxidation of sulphate reduction potential will be much less that for water ,thus oxidation of sulphate happens.


Why fossil fuel is too dangerous? This seems stupid question, but from many years I am studying that burning/combustion fossil fuel cause pollution because it releases $\ce{CO2}$.

I have recently read about $\ce{CFC}$ here, which is more dangerous than $\ce{CO2}$ and seems it is not release by burning of fossil fuel then why are scientist more concerned about fossil fuel instead of $\ce{CFC}$?

I want proper explanation Why fossil fuel is too dangerous?

Why fossil fuel is too dangerous?

Combustion of fossil fuel does not only release $\ce{CO2}$ but many other harmful substances(pollutants) like oxides of sulphur ($\ce{SO_{x}}$), oxides of nitrogen($\ce{NO_{x}}$), Carbon Monoxide($\ce{CO}$), etc...

oxides of sulphur ($\ce{SO_{x}}$):

When sulphur containing fossil fuel is burnt, then oxides of sulphur is produced. Sulphur Dioxide($\ce{SO2}$) is generally produced.

Low concentration of $\ce{SO2}$ causes diseases like asthma, swelling and irritation of respiratory tract.

High concentration $\ce{SO2}$ leads to stiffness of flower buds, which eventually fall off!

Oxides of Nitrogen($\ce{NO_{x}}$):

When in automobile engine, fossil fuel is burnt at high temperature significant amount of nitric oxide($\ce{NO}$) and nitrogen dioxide($\ce{NO2}$) is released.

$\ce{NO2}$ is lung irritant. it is highly toxic to living tissue.

$\ce{NO2}$ decrease the rate of photosynthesis, cause respiratory diseases in children, etc.

Carbon monoxide($\ce{CO}$):

$\ce{CO}$ is produced due to incomplete combustion of carbon. It is major constituent in smoke of vehicles.

$\ce{CO}$ binds with haemoglobin and form carboxyhemoglobin, which is 300 times more stable than usual oxygen-haemoglobin! If concentration of carboxyhemoglobin reaches 3 to 4% then oxygen carrying capacity of blood reduces.

Few more!

$\ce{SO2}$ with smoke and fog is responsible for classical smog(London smog)

$\ce{NO2}$ and $\ce{NO}$ is responsible foe photochemical smog(Los-Angeles smog)

Update:

I was reading more about environmental chemistry and i discovered that $\ce{NO_{x}}$ is responsible for 40% depletion  of concentration of Ozone layer!

  why are scientist more concerned about fossil fuel instead of CFC?

As commented by LDC3 The production of the more hazardous CFC's has been banned.

There are 95 derivatives of chloro and bromo fluorocarbon which are listed as ODS(Ozone Depleting Substances).

On 16th September, 1987 93 countries sign Montreal Protocol and accept not to use ODS.

Reference: GSEB textbook pdf


Monomers and polymers At Wikipedia a monomer is defined to be

  a molecule that may bind chemically to other molecules to form a
  polymer.

A polymer is defined to be

  a molecule composed of many monomers.

These definitions are obviously circular.

Are there non-circular definitions of monomers and polymers?

The term "monomer" only has meaning in the context of discussing polymerization.

A polymer is a molecular chain where the links of the chain are repeated identical or a least similar units.  Examples where the units are idential would be polypropylene, polyvinylchloride and polysiloxane.  Example where the units are similar but not necessarily identical would be DNA and proteins.

A monomer is a molecule that can react to form a single unit of a polymer.

For example, propylene is a monomer of poylpropylene and amino acids are monomers of a protein.


Which organic compound's NMR and IR are these? Can't find out. (Data available: Melting Point ~ 116 °C) Which organic compound's NMR and IR are these? I have figured out that the compound is a 1,4 substituted benzene. And the substituents are probably not -COOH and -OH from the spectroscopy. Kindly help me in figuring the other groups out. (Data available: Melting Point ~ 116 °C, gives sooty flame)

The integrals and shifts in the very simple NMR spectrum are a giveaway that this is a simple compound with two different types of aromatic signal, and another non-aromatic signal from a substituent, at a 2:1 peak integral ratio, hence possibly a para-disubstituted benzene ring with one proton in one substituent, none on the other.

The IR spectrum is practically identical to that in the NIST webbook for p-nitrophenol.
Some other clues:

Oop Signal at $\pu{800 cm^-1}$: para substitution
$\pu{1400-1600cm^-1}$: aromatic cc stretch, matches nicely with typical appearance for para substitution
No peak at $\pu{1700cm^-1}$: no carbonyl
Weak signals at $\pu{3000cm^-1}$: sp2 stretch, no sp3
Signals at $\pu{1350 and 1500cm^-1}$ may be nitro

Note this is a little tricky - a search would show some variation in reported spectra.

The melting point also matches reported values.

This is well in agreement with many of the comments under the original post.


What is the specific functional group in MDMA that caused addictive effect on user? First of all, the reason I'm asking this is because I'd just done an experiment detecting MDMA in a urine sample with GC-MS for my internship program. I'm having difficulties figuring the functional group which cause the said effect as in my question. I've read two journal Shulgin and Nichols and I thought they explored more of the hallucinogen effect from N-methylation.

Firstly, MDMA is widely regarded not to be physically addictive, which simply means that the compound (actually the absence thereof) will not cause potentially life-threatening symptoms on discontinuation, as is the case with alcohol or morphine. Note however that this does not make any statement on the addictive potential of the substance, as e.g. cocaine also falls into this category of substances, which are called psychologically addictive.

In general psychological addiction (which is not necissarily associated to substances, but can also involve habits such as gambling or eating) is widely believed to be strongly associated with the dopamine system and the overstimulation that dopamine facilitates during the addictive actions.

Which brings me to the second point:
The active site of the molecule. In a way, looking at both structures you will notice a set of similarity between dopamine and MDMA. It is not that simple however. The effects of MDMA are quite complicated and due to its status as illegal substance research is kind of limited.

Known effects are:

Known and experimentally confirmed increase in dopamine levels, although the mechanism is not known.
Serotonine release from vesicles to synaptic cleft (has effects on dopamine levels through mediation/coupling)
MAO inhibition (decreases dopamine and serotonine degradiation - among others)

Other effects like oxytocin response are also proposed and may also have effect on dopamine levels. But in this case the effect could as well be a down-regulation.

Which structural component is responsible for all of this? All of them, as you can tell from the complexity of the physiological reaction on the substance, especially if compared to similar compounds with in some regards drastically different physiological effects. Think of (meth-)amphetamine(very high dopamine response, next to no serotonin effects) or mescaline(direct agonist on specific serotonin and dopamine receptors, as well as dopaminergic effects). (It would really blow the reach of this question, but taking a look at other compounds, such as 2-CB or 3,4-CTMP will make this picture even more complicated, but also more complete...)

These substances can give you at least some hints at two honorable mentions in terms of 'addiction-causing functional group':

(N-Me-)Phenethylamine
3,4-(+M)-substitution pattern on the aromatic ring

A small sidenote on the difference between physical and psychological addiction: I am not 100% sure about the current research in that topic, but afaik in both cases there is in overexpression of receptors that causes the addiction, only that in the case of a psychological addiction the effects of the receptor-agonist is not live-threatening. In that sense the terminology which kind of suggests that psychological addiction is not associated with any changes in brain structure is somewhat misleading.


How does ice formation glue things together? Today, I took two crumpets out of the freezer, which had become stuck to each other where moisture between them had frozen. They were two separate crumpets, the top one stuck fast to the bottom one, so they didn't go into the freezer attached to each other in any way.

And it made me wonder, what causes the crumpets to become stuck? Why don't you just have a crumpet, a thin sheet of ice, and then the other crumpet? What's going on at the chemical level that explains why these two crumpets became stuck to one another?

The crumpets you have frozen together are very porous and convoluted at a microscopic level. One would then imagine that this means that any water vapour condensing on them as they start to freeze will percolate somewhat into the surface of the material and then freeze. The interior will stay warmer longer than the surface so this will help percolation. (Imagine an ink drop spreading into paper and you get the idea of percolation). On freezing water crystals become connected and thus many of these stick the two parts together. Rather like a sort of very fine velcro. However, most all foodstuff already contains a lot of water surrounding lipids, proteins and carbohydrates. Thus hydrogen bonding of condensed water to these intrinsic waters may also be important at the molecular level.


Ring expansion from a given cyclic carbocation How will the cyclobutane ring behave in the case of cyclobutylmethylium (cyclobutylmethyl cation)?

I initially thought there would be ring expansion to a five membered ring so that there may be less angle strain and a secondary carbocation instead of a primary one. But I have also been told that there will be ring contraction for stability purposes. I would be highly obliged if someone could explain the mechanism involved here.

I think your friend is thinking of the cyclobutyl carbocation which does ring contract to the cycylopropyl carbinyl carbocation (and also equilibrates with the  methallyl carbocation).

However, just as you thought, the cyclobutyl carbinyl carbocation does ring open to the cyclopentyl carbocation (ref_1, ref_2, ref_3).  This rearrangement is driven by carbocation stability (primary to secondary) and relief of the steric strain present in the 4-membered ring, again, just as you suggested.

Note: this question has been asked previously on SE Chem, however I believe the accepted answer is incorrect, as it seems to primarily address the cyclopropyl carbinyl case, and what is said about the cyclobutyl carbinyl carbocation (little to no ring expansion) is incorrect.


How do I correctly typeset metastable radionuclide symbols? Does anyone know if there is an officially sanctioned way to typeset symbols like technetium-99m (99Tcm or 99mTc)? I have seen both, although in more recent publications, I think, the latter predominates.

For my part, it looks somewhat wrong to place the symbol for metastable in the position you'd normally expect the charge. How would you write the ion, 99Tcm+? That looks quite clumsy.

So does anyone know if there's some IUPAC (or similarly authoritative) guidance on this?

According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics, Amendment 1, a metastable nuclide is indicated by adding the letter m (in roman type) to the mass number of the nuclide, as in the following examples:

$$\mathrm{^{133m}Xe}$$
$$\mathrm{^{99m}Tc^+}$$


How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules? It is easy to compare two ionic compounds when one of the ions is same. However, how do we compare two compounds if one of the ions is the same element but just has different charge and the other ion is completely different?

Fajan's rules for the prediction of bond character can be summarized in two points:

Large cation with low positive charge + small anion $\rightarrow$ ionic
Small cation with high positive charge + large anion $\rightarrow$ covalent

In $\ce{SnF2}$, we have a large cation with low positive charge. $\ce{Sn^2+}$ is larger than $\ce{Sn^4+}$ because it has 2 more electrons in the $5s$ orbital. The anion $\ce{F^-}$ is smaller than its heavier group homolog $\ce{Cl^-}$. For this combination large cation + small anion, the bond character is predicted to be more ionic and less covalent.

On the other hand, $\ce{Sn^4+}$ is a small cation with high positive charge, and $\ce{Cl^-}$ is the large anion, as already stated. Therefore, according to the rules, $\ce{SnCl4}$ is the tin halide with more covalent bonds, and indeed, tin(IV) chloride consists of tetrahedral $\ce{SnCl4}$ molecules in the solid, liquid and gaseous state.

However, the predictions made with these rules do not always agree with the real bonding situation, and $\ce{SnF2}$ is an example. There is experimental evidence for covalent $\ce{Sn-F}$ bonds in tin(II) fluoride (references: 1, 2).


Protonating/basic site and Brønsted basicity orders of beta, gamma, and delta lactams Consider the three cyclic amides:

$\beta$-lactam
$\gamma$-lactam
$\delta$-lactam

My question
What is the basic site (site of protonation) for these three lactams? And what should be their Brønsted basicity order?
My approach
If you consider $\ce{O}$ as the protonating site for the lactams ($\ce{O}$ can act as the basic site as the lone pair on $\ce{N}$ is delocalising with the carbonyl carbon, which increases electron density on $\ce{O}$, which helps to donate its lone pair to $\ce{H+}$, and after forming the conjugate acid, its resonance stabilisation is still there, which can be lost if it gets protonated through $\ce{N}$ (due to non availablity of lone pair). Delocalisation is not at all possible in $\beta$-lactams, due to huge angle strain. But $\delta$-lactam is the ideal situation for delocalisation which will increase the electron density on $\ce{O}$ atom, and hence will be the most basic due to the most availability of electron density of $\ce{O}$.
So, basicity order will be $$\ce{\delta-lactam > \gamma-lactam > \beta-lactam (considering O as the protonating site)}$$
But if you consider N as the protonating site for these lactams, the delocalisation actually decreases the basicity, as delocalisation induces non-availability of electrons on $\ce{N}$ atom.
Going by this logic , basicity order should be, $$\ce{\beta-lactam > \gamma-lactam > \delta-lactam (considering N as protonating site)}$$
So, these two logics contradict each other and hence I am unable to decide the protonating sites for amides, as in some cases $\ce{N}$ is seen to be protonating site, and in some cases $\ce{O}$. Can anyone explain what should be the proper logic to deduce the Brønsted basicity of these lactams?

As the significance of the two contrasting factors cannot be easily evaluated, let us first look at some evidence. Wan, Modro & Yates (1980) reported the $\ce {pK_a}$ values of the conjugate acids of the various lactams, as shown in the table below:

\begin{array}{|c|c|c|c|} \hline
\text{Lactam} & pK_a\\ \hline
\ce{$\beta$, 4} &  -0.96\\ \hline
\ce{$\gamma$, 5} & -0.65 \\ \hline
\ce{$\delta$, 6} & -0.18 \\ \hline
\ce{$\epsilon$, 7} & -0.46 \\ \hline
\ce{$8$} & -0.38 \\ \hline
\ce{$9$} & -0.18 \\ \hline
\end{array}

As the ring size increases, from the 4-membered $\beta$-lactam ring to the 9-membered ring, the basicity of the lactam increases. This seems to suggest that the oxygen atom may indeed be the protonating site for these lactams, even for those with the smaller ring sizes. However, it seems that the 6-membered $\delta$-lactam ring exhibits rather anomalous basicity.

The authors explain the general trend in basicity as follows, using the two resonance structures shown:

  If the two canonical forms of the $\ce {O}$-protonated lactam are considered, the basicity will depend strongly on the ability of nitrogen to donate its pair of electrons
  to give a delocalized cation. Hence structure II will be more important in the conjugate acid than the analogous contributing structure in the neutral lactam. This will cause more serious ring strain the smaller the size of the lactam ring, and hence tend
  to reduce the basicity. As the rings become very large, such strain factors will become less important and the basicity would be expected to level off at a pK value similar to that of a typical secondary amide. Only the 6-lactam is unusual in that its base strength is much higher than expected. It is commonly observed (27) that for five-membered rings exo double bonds are preferred, while for six-membered rings endo double bonds are more stable, although these preferences are not easily rationalized. Therefore contributions from canonical form II in the protonated 6-lactam should be more important than in the $\gamma$-lactam case (and possibly the other lactams as well), leading to a higher basicity for the 6-lactam than its ring size would lead one to expect from the trends in Fig. 2.

Reference

Wan, P.; Madro, T. A. ; Yates, K. The kinetics and mechanism of acid catalysed hydrolysis of lactams. Can. J. Chem. 1980, 58, p. 2423-2432.


How are nuclei stable? We all know that the density of the nucleus is very high.

Nuclei are made up of protons and neutrons, and while protons have the same charge, they are closely packed in a nucleus. How does the repulsion between protons not break apart the nucleus?

Since the gravitational force between two protons is negligible there must be another force holding the nucleus together. This is the strong nuclear force, which as the name suggests is extremely strong but it is also extremely short range and so it's effects are only felt on the scale of nuclei and baryons. As you can see in the graph, if two protons approach each other they repel each other due to electrostatic repulsion but once they get within about $\mathrm{3~fm}$ of each other the strong nuclear force begins to become significant and at about $\mathrm{2~fm}$ it starts to outweigh the electrostatic repulsion and bind the two protons together.

It is also interesting to note that at even shorter ranges the strong nuclear force is repulsive and so there is an equilibrium position, indicated on the graph, where the strong nuclear attraction and the electrostatic repulsion are equal in magnitude and so the net force on the protons is zero.

Note that the graph shown is for two protons and so in nuclei containing more than two protons the numbers will be different but the principle is the same.


Why are oxyacids strong oxidizing agents? I saw that all the oxyacids are strong oxidizing agent.
I don't know why all the oxyacids are strong oxidizing agents?
What is the reason?

Your premise is incorrect. Some oxyacids are oxidising agents. Some are not. Take e.g. carbonic acid. You cannot oxidise anything with carbonic acid, really.

Also, the choice of words is bad. Sulphurous acid (or hydrogen sulphonic acid; it doesn’t really make a difference; both are structural isomers) is also an oxyacid ($\ce{H2SO3}$) however it is a pretty good reducing agent oxidising itself to sulphuric acid. Heck, technically $\ce{H4SiO4}$ is an oxyacid, too, but it won’t oxidise anything at all.

So reducing the question to where it is correct we arrive at:

  Why are some oxyacids where the central atom is in its highest possible oxidation state strong oxidising agents?

And it basically answers itself. Depending on the electronegativity of the central atom it will be more or less ‘happy’ to cede all its valence electrons to oxygens and be the centre of an oxyacid. For those that don’t like that lot in life, most notably nitric acid, they will often find species whose electrons they can take back (the atoms are electronegative after all) and thus oxidise them.


Stereocenter or not? Consider 2,3,4-pentanetriol.Is the middle carbon atom a stereocentre?I know how to count the stereoisomers, but is the middle carbon a stereocentre by definition?Let us say we have a 2R and 4S configuration.Now if we change the orientation of the OH group at the 3rd carbon, we do get a different compound,but both the compounds are optically inactive, so it doesn't seem it should be called a stereocentre.If we consider 2R,4R, or 2S,4S if we change the configuration at C3, we get the same compoound.So I feel the carbon at C3 should not be a stereocentre.But it seems that in the first case, we are changing the C3 configuration and getting a new diastereoisomer, so C3 might be a stereocentre after all.Which one is correct?If C3 is a stereocentre, what would be its R,S nomenclature?

The carbon atom in question is called pseudo-asymmetric. Check it out on the IUPAC page.


Photolysis of N-bromo compound in acidic conditions (radical cyclisation) For the proposed mechanism, I believe that the N-Br bond is firstly broken by the absorbed light hv, then the radical formed on the N will perform an intra-molecular attack on the end of the carbon chain to form a 5 membered ring. Then the sulphuric acid becomes deprotonated by the nitrogen lone pair? Just want to check whether this is correct.

This is the Hofmann-Löffler-Freytag reaction, which was later examined by E. J. Corey (J. Am. Chem. Soc., 1960, 82, 1657-1668,
DOI).

When the reaction is performed in sulfuric acid, it is conceivable that the nitrogen atom of the N-Bromoalkane is protonated. Homolytic cleavage of the $\ce{N-Br}$ thus probably leads to a radical cation.

The latter then abstracts a $\delta$-hydrogen atom, resulting in a radical centre in the side chain. It has been suggested that this centre might recombine with the bromo radical previously released. With other words: $\ce{H}$ and $\ce{Br}$ have changed their position. Deprotonation of the ammonium cation and a subsequent nucleophilic substitution, releasing $\ce{Br-}$ would then result in the formation of the indolizidinium skeleton.


How would one carboxylate CBG into its acid form, CBGA? I have a project that would require cannabigerolic acid (CBGA) as a synthase substrate, but the only available chemical I can find is cannabigerol (CBG). CBG is the product of CBGA through decarboxylation at about 105 °C. Is this reaction reversible? How can I carboxylate the CBG I have available into CBGA?

The substrate is a resorcinol derivative: a strongly activated aromatic ring, which upon reaction with a base delocalizes its negative charge on the ring, making it reactive towards acids (Eg. CO2).

Here (entries 130-132) good conversion is claimed (85%) in a high pressure vessel with magnesium methoxide and dry ice, or a lower conversion (40%) is reported at ambient pressure with magnesium methyl carbonate


How can I deduce the linearity of XeF2 from the IR spectrum? The vibrational modes of $\ce{XeF2}$ are at $555$, $515$, and $213~\mathrm{cm^{-1}}$ but only two are IR active.
Explain why this is consistent with $\ce{XeF2}$ having a linear structure.

From what I know, since it has $3$ atoms it should have $3n - 5$ vibrational modes, thus $3(3) - 5 = 4$ modes. Yet the problem tells me it only has $3$, which would suggest a non-linear shape. How to explain this?

Let's start with the rule of mutual exclusion, it will come in handy.  It states that,  for centrosymmetric molecules (molecules with a center of symmetry), vibrations that are IR active are Raman inactive, and vice versa.

A linear $\ce{AB2}$ molecule will have $(3n-5) = 4$ vibrational modes, but because it is linear 2 of the modes are degenerate.  $\ce{CO2}$ is an example of a linear $\ce{AB2}$ molecule, look at the following diagram showing the 4 vibrations for $\ce{CO2}$ and notice that the bending vibration ($v_2$) is degenerate.  That is, there is no difference between those 2 bending modes, they are identical.

Notice too that $v_1$ is the only vibrational mode that does not change the dipole moment of the molecule, therefore it is the Raman active mode. Since the linear molecule is centrosymmetric, the Raman band does not appear in the IR, nor do any of the IR bands appear in the Raman spectrum. Hence, for a linear $\ce{AB2}$ molecule we would expect 2 IR active bands and 1 Raman band. This is consistent with the spectral data you reported.

If the $\ce{AB2}$ molecule is bent rather than linear (water for example), then there are $(3n-6) = 3$ vibrational modes as pictured below.  All of these vibrations result in a change in the dipole moment.  Therefore, for a non-linear $\ce{AB2}$ molecule, all 3 of the bands will be IR active. Since the molecule is not centrosymmetric all 3 bands may also appear in the Raman spectrum (note: while a change in dipole moment is required for an IR-active band to appear, a change in polarizability is required for a Raman band to appear; looking at the water vibrations pictured, they should alter both the dipole moment and the polarizability of the molecule). Hence, for a non-linear $\ce{AB2}$ molecule we would expect 3 bands in the IR and 3 in the Raman spectrum. This is not consistent with the spectral data you reported above.

In summary, your spectral data is consistent with $\ce{XeF2}$ having a linear geometry.

Note: Thanks to @porphyrin for pointing out that bent $\ce{AB2}$ molecules are indeed not centrosymmetric.


Is this a possible explanation as to why sigma bonds are stronger than pi bonds? Sigma bonds form due to the axial overlap of two orbitals whereas pi bonds form due to their lateral overlapping. Since the former is more effective, we can say that sigma is stronger than pi. We mean that the orbitals overlap to a further extent in sigma bonds and not as much in pi bonds(...or at least that is what I infer). To me this sounds like a circular proof/explanation, so I thought up another way to understand it.

In a sigma bond, the electron cloud of one atom is attracted to the nucleus of the other atom's nucleus.This happens because of the electrostatic forces of attraction between the positively charged protons in the nucleus of one atom and the negatively charged electron cloud of the other atom. The attractive force causes the electron clouds to the atoms to overlap, despite the fact that the electron clouds should repel each other.  However, the electrons start to repel each other after a certain extent of overlapping has occurred. (Because the repulsive forces between the two nuclei and the electron clouds negate the attractive forces)

In a pi bond, there is no positively charged nucleus that would attract the electrons, so there must be another factor favoring the bond formation.
Since I haven't studied the quantum mechanics of bonding yet, my best guess is that the pi bond (or lateral overlapping) occurs because of the pairing of electrons.
So this means that sigma bonds are stronger because of electrostatic forces, which consequently causes more effective overlapping than pi bonds.

Is my reasoning correct to some (if any) extent?

This is a very short answer and it might as well have just been a comment. It is probably also tempting in it's simplicity. For starters, let's assume, that nuclei are point charges.

When two atoms form bonds, all the electrons experience the attractive force of the nuclei, of both of them. This is independent of the type of orbital the electrons occupy.

In a σ orbital, the path of maximum electron density is between the nuclei. It is also where the electronic field of the nuclei is strongest. For a π orbital the situation is different.

Let's take a step back and consider an atom. The s orbital is spherical symmetric. The maximum electron density is (formally) in the nucleus. For the p orbital, there is a node at the nucleus. The maximum electron density is further away from the opposite charge, causing attraction to be lower. The energy of the orbital will be higher.

Similarly we can see that for π orbitals. The node of these kind of orbitals is exactly where the field of the nuclei is strongest. Hence the attraction is lowered, the energy raised, because the maximum electronic density is further away from the nuclei.

In very short terms: In σ orbitals the maximum electron density is on the bonding axis, while in π (and δ) orbitals it is off the bonding axis.


Do I have to consider the molecular mass of the oxygen atom or the diatomic oxygen molecule when determining the empirical formula of an iron oxide? In determining the empirical formula of a compound which constitutes a diatomic molecule, should we calculate the molar mass of the divalent molecule by multiplying the molar mass of the element by 2, or should we leave it as it is. Because, I happened to come across a problem, where I had to calculate the empirical formula for an oxide of iron. A similar question has been answered before, but it doesn't seem to clear my doubt - Why not to consider hydrogen and oxygen moles to determine an empirical formula. Hence, I restate it:

  If an oxide of Iron has $69.9\%$ iron and $30.1\%$ dioxygen by mass, what is the empirical formula?

Now, if I assume the mass of compound to be $100~\mathrm{g}$, then it contains $69.9~\mathrm{g}$ of iron and $30.1~\mathrm{g}$ of oxygen. If I calculate the no. of moles,

$$\text{Moles of }\ce{Fe} = \frac{\text{given  mass}}{\text{molar mass}} = 69.9/58.5 = 1.25~\mathrm{mol}$$

Moles of oxygen - this is where the problem comes. If I take molar mass of oxygen as $32~\mathrm{g}$, as in $\ce{O2}$, I get $0.94~\mathrm{mol}$. It will not make any sense, as the ratio of iron to oxygen will not be in any reasonable quantity. However, if I take the  molar mass to be $16~\mathrm{g}$, I get $1.88~\mathrm{mol}$ of oxygen, and I can divide both the numbers by $1.25$ to get the ratio as $1.88/1.25: 1.88/1.88 = 1:1.5 = 2:3$,
hence the compound will be $\ce{Fe2O3}$, which makes perfect sense. But the problem is, why should I calculate the no. of moles from the molar mass of $\ce{O}$ rather than $\ce{O2}$? After all, its is the oxygen molecule that will exist, and no the oxygen atom right?

Also, the solution is given as $\ce{Fe2O3}$, so I need to take the molar mass of oxygen to calculate the number of moles as $16~\mathrm{g}$. The question is, Why? Why not $32~\mathrm{g}$?

Does it have something to with the fact that the question states "dioxygen" rather than "oxygen"? If yes, what is the difference between dioxygen being present in $30.1\%$ or oxygen present in $30.1\%$? I am in a real mess.

When determining the empirical formula of a compound, you only consider the molar mass of the element. In the compound itself, it is only the atom (or ion) that exists.

If you would assume the molar mass of dioxygen, then you would also assume, that the iron oxide is made out of iron and dioxygen molecules.

I do not know, and also do not understand, why the question states "dioxygen" as that is - in my opinion - false.


Vapor Pressure Paradox This is quite an interesting situation I had thought of. Consider two beakers, each containing $\pu{180 g}$ of water. Also, suppose one of the two has $\pu{1.8 g}$ of glucose dissolved in it. Let's call the one without glucose as A, and the one with glucose as B. I keep a small bell-jar that is large enough to just cover both of the two vessels. Now I will try to measure the vapor pressure inside the jar. Here's a little diagram to show what I mean:

I do not have access to bell-jars or manometers so I'll try to use Raoult's law to find the vapor pressure inside the vessel. The law in a simplified way would be:

$$P_{\text{solution}}=P_{\text{solvent}}x_{\text{solvent}}$$

Where $x$ denotes mole fraction, and solute is considered non-volatile. Keeping vessel A in mind, I get the vapor pressure as 23.8 mmHg, while applying Raoult's law for vessel B, I get the vapor pressure as:

$$P=23.8\left(\frac{10}{10.01}\right)=23.7 \text{mmHg}$$

Now, both of the values can't be true at the same time, and that's what I believe is the paradox.

How can this paradox be resolved?

My thoughts are that perhaps water from vessel A condenses back onto B, thus slowly increasing water content in B, increasing the vapor pressure. However, since theres solute already in B, the vapor pressure can never reach the value provided by the pure water.

What are your thoughts? Does the paradox ever resolve or does it remain in an unbalanced situation?

Although "paradox" is not quite the right term, what you have discussed is actually a simple, yet interesting and important phenomenon.

Given the ideal situation as you have presented, your thoughts on what would happen are correct. If the system were to achieve $\pu{100\%}$ humidity with respect to the pure water, that would always be slightly over $\pu{100\%}$ humidity with respect to solution B, so water would condense onto B. This of course would cause the humidity with respect to pure water to drop, so more would evaporate. All things being ideal, eventually all of the water in A would transfer to the solution in B (plus the gas phase). The rate that would happen would depend strongly on the concentration of the solute in B.

An example of this situation playing out in the environment can be seen in the growth of water and ice particles in clouds. The vapor pressure of one particle can differ from that of another due to solute concentration, particle size (small droplets have a greater vapor pressure than larger droplets) and phase (supercooled water droplets have a greater vapor pressure than ice particles at the same temperature). Regardless of the cause of the vapor pressure difference, whether it's solute concentration or one of the other factors, the situation is the same as depicted in your question. And the result is the same; the particles having greater vapor pressure will end up evaporating and then condensing onto the particles having lower vapor pressure (of course we're talking about vapor pressure of water/ice only, not any solute). This is generally referred to as the Wegener–Bergeron–Findeisen process, although this technically only refers to water-to-ice vapor transfer.


Will Ammonia react with Gasoline mixed with Ethanol? Recently the South Koreans invented a car that runs on an 70% ammonia 30% gasoline fuel mix.  (link) In America we already have instituted policies that require refueling stations to mix Ethanol produced from corn stills with gasoline to extend our fuel supply.

My question is, if you added a ammonia to an ethanol/gasoline fuel mix, would it cause a reaction?  If not, would it still burn and be able to be used as fuel?

A fuel that is mostly ammonia is an altogether different fuel formulation that would not reap the benefits of the oxygenated fuels program in the US for gasoline. In other words, there would be no reason to mix in ethanol or any of the other approved oxygenated species used in gasoline (the specification requires 1.5% - 3.5% oxygen by weight during the winter (state dependent) and is reported as ethanol, regardless of the oxygenated species actually used).

According to this US Environmental Protection Agency document:

  Oxygenates are fuel additives that contain oxygen, usually in the form of alcohol or ether. Oxygenates can enhance fuel combustion and thereby reduce exhaust emissions. Some oxygenates also boost gasoline octane. The Clean Air Act requires use of oxygenated gasoline in areas where winter time carbon monoxide levels exceed federal air quality standards. Without oxygenated gasoline, carbon monoxide emissions from gasoline-fueled vehicles tend to increase in cold weather. Winter oxygenated gasoline programs are implemented by the states.

The octane levels of ammonia fuels are already very high at around 120. There would seem to be little benefit in the form of carbon monoxide emission reductions by adding oxygenated hydrocarbons to a fuel that is 70% ammonia.

So, the bottom line is that there is no reason to expect ethanol and ammonia to be in the same fuel formulation. Furthermore, if there were ethanol in the ammonia based fuel, there is no reason to expect the two to react in any significant way.


Does ice melt faster when submerged in water? Suppose you have a container with a drain at the bottom full of frozen H2O which you leave a room temperature to melt. Will the ice melt faster if you drain the water as it melts?

I would expect that the water, being warmer than the ice, will speed the melting of the ice. However, I would also expect that draining the water would remove allow more air to enter the system and heat the ice faster. Which explanation is correct?

Related question: if the container is a cooler to fill with warm beverages, will the beverages cool faster in a wet or dry ice-filled cooler?

Assuming the water and air are the same temperature (and that the air is mostly stationary) then the ice will melt quicker in the water. This is due to water having a higher thermal conductivity and heat capacity. In fact, this is easy to test: place a cocktail stick through an ice cube and lay it on top of a glass filled right to the top with (room temperature) water. The submerged half will melt quicker than that on top.

Regarding the second question: A warm object will cool quicker in icy water than in just ice for the same reason above but also because of increased surface area contact. However, icy water will be warmer than just ice. So the item will potentially be able to achieve a lower temperature in just ice (slow and steady wins the race).


Is there a consensus on whether amber can seal in air bubbles for millions of years? I read about deriving the oxygen content in the atmosphere in the Cretaceous by analyzing the air trapped in bubbles in amber from that period, but I have also found papers disputing the study's methods and showing that air isn't actually trapped in the bubbles because it's capable of diffusing through the amber (the amber isn't a perfect sealant), and thus the air found in that amber isn't necessarily as ancient as the amber itself.

My problem is is that almost all the relevant papers are from the late 80's; all those I've found are pro- or con-, I don't have the background to tell who is right, I haven't found a source that describes an actual consensus the field would have come to since then, and I see the claim about the original study showing what oxygen levels were in the Cretaceous cited in various places, but I find it hard to tell if it's because the objections were dismissed or if it's because it's a zombie claim.

Some sources:

What seems like the first paper to make the claim:
Gas bubbles in fossil amber as possible indicators of the major gas composition of ancient air
This source seems to be more recent and find the same conclusions but it's hard to tell (looks like at least one person is the same as the 1988 source):
Air Bubbles, Amber, and Dinosaurs - USGS
The main disagreement:
Is the air in amber ancient?
Some back-and-forth:

Google Scholar: "is the air in amber ancient"
The relative contribution of adsorption to the overall sorption and transport of small molecules in amber

An unrelated recent paper that comes to a different conclusion on Cretaceous air content (based on amber as well but carbon isotope analysis thereof):
Stable carbon isotopes of C3 plant resins and ambers record changes in atmospheric oxygen since the Triassic

Does anyone know if a consensus was arrived at on this question, and/or if one side is clearly more right than the other?

There is at least a fair degree of consensus (there will always be naysayers) that the results that were performed back in the 80s and 90s by Berner and Landis were false positives that could not be confirmed by further investigations. A good historical account of this can be found for instance in Nick Lane's book "Oxygen" (first published in 2002). Quoting from this book:

  The overall consensus, which has now persisted for more than a decade,
  was that the air trapped in amber could not have been ancient [...]

That "could not have been" has an air of finality about it, and you can probably take this book's word as accurate. However, all of the amber in the world has not been examined, so I would say chances may be slim but not none that some will be found containing traces of an ancient atmosphere. Probably part on the answer depends on how old the amber is, and how it was preserved. A very different question is whether gases in the amber can be regarded as representative of the atmosphere of origin, or useful. Perhaps reactions with the amber (oxidation) during burial have altered it beyond recognition.


Volume of Oxygen Produced By Electrolysis of Water I am trying to estimate the volume of oxygen that will be produced by electrolysis of water at normal temperature and pressure with a current of $\pu{50 A}$. The following is my calculation:
The reaction is
$$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Therefore, to produce $\pu{1 mol}$ of $\ce{O2}$ ($\pu{24 L}$), we need to move $\pu{4 mol}$ of electrons through the circuit and this is $4F = 4 \times 9.65 \times 10^4 = \pu{3.86 \times 10^5 C}$ of charge. With a current of $\pu{50 A}$, we move $50 \times 60 = \pu{3000 C}$ of charge per minute. Therefore, we produce $\frac{24}{3.86 \times 10^5} \times 3000 = \pu{0.186 L}$ of $\ce{O2}$ per minute.
Is this calculation correct? If it is, is there any way to improve the yield of oxygen?
Disclaimer: I am not a chemist and it has been years since I studied physical chemistry. The reason for this computation is to decide if electrolysis is a quick and feasible method to produce oxygen to help mild COVID patients at home (since there is a severe shortage of oxygen and concentrators where I live).

Thank you very much for your idea and help.
I think that your calculations are correct.
But I think you'd better to try other simpler choices to make oxygen.
One way is air separation. If you separate nitrogen from air, you gain oxygen.
This can be done by "Zeolite oxygen generators". You can read these site about them.
1- https://www.zeochem.com/our-applications/medical-oxygen-generation
2- https://www.jalonzeolite.com/product-item/molecular-sieve-jlox-500/
There are many important factors that you have to consider about electrolysis of water that I know some of them:
First water has low conductivity. So, you have to add a strong inert electrolyte to make charge transfer between electrodes. This electrolyte could be any soluble salt, an acid or any substance that solves in water and then dissociates to ions. But, the electrolyte must not interacts to the main reaction and should be inert and surely it should be absolutely safe and not harmful for patient. Sodium or potassium sulfate or nitrate can be used.

Next, you have to choose an inert electrode to conduct electricity to the solution and then the gas be produced on it. Platin and gold can be use. But, they are expensive. Another choice can be carbon electrode.
And also take care about Hydrogen that produced on the negative electrode (cathode), that is flammable and can be exploded by spark.
And, so many other considerations to make oxygen efficiently and, next purify it to be suitable for hospital usage.
I wish you health.


How to determine the Bravais lattice and atom basis vectors from a CIF file? Say I have a CIF file describing some material in terms of its symmetry space group, lattice parameters and in-cell atom positions. A simple example might be,

data_global
_chemical_name 'Graphene'
_cell_length_a                 2.46
_cell_length_b                 2.46
_cell_length_c                 1
_cell_angle_alpha              90
_cell_angle_beta               90
_cell_angle_gamma              120
_symmetry_space_group_name_H-M 'P 3 m 1'
loop_
_atom_site_label
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
C  0.00000   0.00000   0.00000
C  0.33333   0.66667   0.00000

I would like to extract from this Bravais lattice vectors and atom basis vectors. For this specific example, this might look like,

 2.456   0.0     0.0
-1.228   2.126   0.0
C
 0.0      0.0      0.0
 0.0      1.418    0.0

where the first two lines give the two lattice vectors (in some basis), while the next two give the positions of the atoms in the unit cell.

I have tried using Open Babel for this, converting from the CIF format to a VASP one.  This works well for the example above, but fails for a material such as tin sulfide (SnS), which has a structure similar to black phosphorus---four atoms in a unit cell, only two of which are explicitly listed in the CIF file.  The positions of the other two are implied by the symmetry group, and programs such as Mercury correctly visualize the structure based on the CIF file.  However, converting the CIF into VASP only gives you two of the four unit cell atoms.  I tried all other formats in Open Babel, with no success.

Is there a reliable way to convert a CIF file into a description in terms of a Bravais lattice plus atom basis vectors?


Regarding Kw, titrations, and pH So I have need of performing a titration upon a fruit juice to determine acid concentration. I will be using 1M NaOH as my base. Seeing as the solution will be coloured, it would be difficult to use an indicator such as phenolpthalein. Thus, I would like to use a pH probe.

My Question:

While titrating, should I be aiming to reach a pH of 7 or should I be aiming to reach the ionic product constant of water (Kw) at the temperature that I am performing the titration at? (e.g.6.63 at 50 degreees celsius)

Thank you in advance for your time and knowledge.

When using a pH probe, instead of aiming to reach the equivalence point of the titration, you should be titrating beyond it to create a complete titration curve. With an indicator, the end point is clearly indicated by a color change; with a pH probe, however, the end point is indicated by a steep change in pH in a plot of the data, thus requiring the collection of data beyond the end point.

While in theory the pH at the equivalence point of the titration can be calculated (and, for the limiting case of a strong acid and strong base, is exactly the $K_w$ that corresponds to the temperature of your solution), calculating the pH at the equivalence point and aiming to reach that pH is (i) bad lab practice, since if you're able to calculate the pH at the equivalence point, you're presumably trying to experimentally verify theoretical calculations, and (ii) not applicable to your current experiment, because the organic acids present in fruit juice are weak acids, whose conjugate bases are weak bases that will raise the pH at the equivalence point (and render the pH impossible to calculate for your purposes).


Rate of hydrolysis of silicon tetrahalides Concise Inorganic Chemistry for JEE(Main and Advanced) by JD Lee(Third Edition) writes

  If we compare the rate of hydrolysis for $\ce{SiX4}$(where X= F,Cl,Br,I), it is in the same order as that of their Lewis acid strength order,i.e.
  $\ce{SiF4}$ > $\ce{SiCl4}$ > $\ce{SiBr4}$ > $\ce{SiI4}$
  .This is because the effect of π back bonding is less dominant over the negative inductive effect.

My question is, why is the π back bonding neglected here even though 3dπ-2pπ back bonding is strong enough to be evoked in many other cases? Also, Wikipedia lists the bond strengths of $\ce{Si-X}$ as $\ce{Si-F}$ > $\ce{Si-Cl}$ > $\ce{Si-Br}$ > $\ce{Si-I}$ , and I feel that the rate of hydrolysis should be inversely proportional to this order as it will become increasingly difficult to break the silicon-halide bond when bond strength increases,and consequently difficult to hydrolyse the molecule

We can justify the rate of hydrolysis of Silicon tetrahalides if we just think on the statement carefully. The statement claims that the rate of hydrolysis is in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$. So, we are mainly concerned about the rate of the reaction, not on the thermodynamic stability or any energetic criterion for this reaction( I have dealt with it in the next paragraph). If we recall the characteristics for rate of   a reaction, it depends on the slowest step of the reaction. In the hydrolysis reaction of silicon tetrahalides, the first step, i.e the attack of a $\ce{H2O}$ molecule to form the trigonal bipyramidal intermidiate is the rate determining step. Also note that, no cleavage of $\ce{Si-X}$ bond happens in the rate determining step, but it actually happens in the next step, which is a fast step, so it doesn't contribute to the rate of the reaction. So, basically strength of $\ce{Si-X }$ bond doesn't affect the rate of hydrolysis reaction (the situation is exactly same as Nuleophilic aromatic substitution reactions ($S_NAr_2$) also). But if the $\ce{Si}$ atom becomes more and more electrophilic, the attack by a water molecule becomes more easier which subsequently results in the increase of the rate of the reaction, So the electrophilicity of $\ce{Si}$ in the tetrahalides increase in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$, and so the rate of the hydrolysis also.

Here is the mechanism of the first out of the four subsequent substitutions in the hydrolysis reaction of Silicon tetrahalides.

Secondly, if we really want to judge the rate of the reaction by numerical calculations, observe that the ultimate product of hydrolysis reaction of all the tetrahalides are same. Also, all the reactions are exothermic in nature. So, $\Delta _r H^0$ (assuming the  hydrolysis reactions are done in standard conditions; it's not a strict assumption, it may not be at standard conditions also, but the changes in the heat of reaction will be almost similar for all the tetrahalides) will be more if $\Delta_f H^0$ is more for the reactants (note that, $\Delta _r H^0 = \Delta_f H^0_{products} -\Delta_f H^0_{reactants} $ and we are bothered about reactants. If $\Delta_f H^0_{reacants}$ are more $\Delta_r H^0$ will be more negative and hence energetically favourable). The values for $\Delta_f H^0$ have the order, $\ce{\Delta_fH^0_{SiF4} < \Delta_fH^0_{SiCl4} < \Delta_fH^0_{SiBr4} < \Delta_fH^0_{SiI4}}$. So, the hydrolysis reaction of $\ce{SiF4}$ will be the least exothermic and that of $\ce{SiI4}$ will be the most exothermic, which is expected also from the bond energy considerations. So, if you have less release of heat energy, i.e thermodynamically less stable product, it will have lesser activation energy and higher rate constant, and thus the rate of the reaction will be highest for $\ce{SiF4}$ and least for $\ce{SiI4}$ as we will get the product along with relatively more thermodynamic stability. So, by thermodynamic consideration also, rate of hydrolysis should be, $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$.

Lastly, the $\pi$-bonding is not important in case of large size difference between $\ce{Si}$ and $\ce{F}$ ($2p$ and $3d$ have course greater size difference though you may not admit). The fluorine atom makes the silicon more electrophilic by its electron pulling ability rather than  back-bonding.


What is the pH of 1M Glycerol? If the pKa of Glycerol is 14.15. How do you calculate the pH for it?

I assume that the Henderson-Hasselbalch derivative that works for weak acids and bases is not applicable here.

The answer is approximately 6.88. Therefore a $1\  M$ solution of glycerol in water will be ever so slightly acidic (considering that the hydroxyls are much weaker bases than acids, i.e. that the equilibrium constant for the reaction $\ce{C3H7O2OH + H2O <=> C3H7O2O^- + H3O^+}$ is much larger than the constant for $\ce{C3H7O2OH + H2O <=> C3H7O2OH2^+ + OH^-}$, which may or may not be true. Inclusion of the second equilibrium will drive the true pH even closer to 7)

For very weak or very dilute acids/bases, solving a dissociation problem is a bit more tricky because the self-dissociation equilibrium of water must be taken into account. It is often a valid approximation to forget about it, as water has only a small tendency to self-dissociate and so its effect on equilibrium concentrations of $\ce{H^+_{(aq)}}$ and $\ce{OH^{-}_{(aq)}}$ is often negligible, but this is not your case; glycerol in water seems to be about as weak an acid as water itself.

The best way to solve equilibrium problems is to set up a large equation including all factors by combining smaller equations. We need to make an equation that takes into account all the species in the medium ($\ce{H^+_{(aq)}}$, $\ce{OH^{-}_{(aq)}}$, $\ce{C3H7O2O^{-}_{(aq)}}$ and $\ce{C3H7O2OH_{(aq)}}$ {for convenience I shall label the latter two $\ce{A^-}$ and $\ce{HA}$}) and the related equilibrium constants ($K_a$ and $k_w$). The smaller equations are found by balancing charges (one equation) and the amount of matter of each substance (one or more equations depending on the problem).

Charge balance: $[H^+]=[OH^-]+[A^-]$ (I)

Matter balance: $[HA]+[A^-]=C_{acid}=1\ M$ (II)

The equations for the equilibrium constants are:

$K_a=\frac{[H^+][A^-]}{[HA]}=10^{-14.15}$ (III)

$k_w=[H^+][OH^-]=1\times 10^{-14}$ (IV)

To solve the problem, we need to play with the equalities until we get a single equation with a single variable. It's convenient to obtain an equation in $[H^+]$ because we are able to find the pH directly by applying $-log$ to the answer. Notice that

$[OH^-]=\frac{k_w}{[H^+]}$ (V)

$[HA]=\frac{[H^+][A^-]}{K_a}$ (VI)

Substituting (VI) in (II) yields, after some algebra:

$[A^-]=\frac{C_{acid}K_a}{[H^+]+K_a}$ (VII)

Now insert (V), (VI) and (VII) in (I) so that you get:

$[H^+]=\frac{k_w}{[H^+]}+\frac{C_{acid}K_a}{[H^+]+K_a}$

A little bit of persistence will get you to the following polynomial:

$[H^+]^3+K_a[H^+]^2-(C_{acid}K_a+k_w)[H^+]-K_ak_w=0$

Replacing the values of $K_a$, $k_w$ and $C_{acid}$, the only positive root for the equation is $[H^+]=1.30688\times 10^{-7}$, which after applying the antilogarithm results in $pH = 6.88$.