Evaporation Rate of Water In AP Chem we did a lab investigating the relationship between vapor pressure and water temperature. The relationship is linear and as the vapor pressure approaches atmospheric pressure, it begins to boil. Obviously warm water will evaporate faster than cold water, so it made me wonder how do you calculate the evaporation rate of water? Is there a relation between the evaporation rate and temperature just like there is a relation of temperature to vapor pressure? Of course the surface area needs to be taken into account, but I'm more focused on the relationship between temperature and the rate of evaporation.

It gets complicated. A drop of water (30 mg) will evaporate completely in several hours under standard conditions. You need to consider these factors:

(1) Temperature
(2) Relative humidity
(3) Wind
(4) Surface area
(5) Thermal flux to water (container walls -> water heat flux)

By all means you can do it, but it is a question for physics forums. The main idea would be to write to equations:

(a) Heat flux +walls_to_water + air_to_water -heat_of_evaporation = 0
(b) Diffusion balance. Water pressure at its surface, vs water pressure at the top of the flask is governed by diffusion laws.
(c) Material balance: amount_of_water(t)=amount_of_water(t$_o$)-evaporated_water.

Those are differential equations, so it would be hard to work with them without a solid understanding of calculus.


The importance of the cold chain in the food and the pharmaceutical industry There are these everyday things that one should know as a scientist and especially as a chemist, but which never come to light in an academic curriculum — at least not in mine.
One such thing is the purpose of imperative non-disruption of the cold chain. It is clear that there are plenty of reasons why it is crucial to transport food and also pharmaceuticals in a refrigerated condition. On the one hand, it reduces the risk of bacteria proliferating significantly, but it also prevents labile substances from decomposing or reacting with other ingredients. So far, so good. But why shouldn't you cool down products again once the cold chain has been interrupted?
I have read, for example, in the package insert of a pharmaceutical product that the drug should always be stored below 25 °C. In pharmacies, the product is kept in the refrigerator, and I, as the consumer, should continue storage in the refrigerator if possible. If this is not feasible for me, e.g., due to lack of space, the storage temperature should not exceed 25 °C. It is also pointed out that I should not keep the drug in the refrigerator after storage at room temperature. This reminded me of the wise words of my grandmother, who said that fish, for example, should only be frozen once because otherwise, it will go bad. With many other foods, similar claims are made. Since these indications can be found across countries and cultures, I assume that there are known scientific reasons for this, but I am not aware of them. I can't figure out why it should be bad to refreeze a defrosted piece of meat or why I shouldn't put my medicine in the refrigerator on hot days anymore after storing it at room temperature for a while.
Maybe a few food chemists and/or pharmacists can shed some light on this.

Edit: Poutnik and Julian, I wish I could accept both of your answers. I find it difficult to decide which answer is "the best", as they both address a different aspect of my question which they explain in a very understandable way. I have decided to mark Julian's answer as accepted since he — despite his long membership — has a much lower reputation than Poutnik.

Avoiding cooling  again after keeping it ( for non critical products ) at room temperature is mainly to prevent forgetting it was not cold all the time and that it may not last as long as expected.
Taking food from the fridge, you may not remember after a week or two that it was outside longer than it should have been. It can violate the safety by growth of microbes, which does not fully stop when it gets cold again.
For drugs or similar critical products, it violates stability and guarantee of usability, as thoroughly described by @Julian answer. I did  related analysis within drug stability studies in past as well.
Food freezing is another case. Industrial freezing of food is done by shock freezing using liquid nitrogen. This leads to quick forming of small ice crystals that do not puncture the cell walls, so after melting food before usage, it is OK.
But when refrozen back e.g in a freezer, freezing is slow, leading to growth of large ice crystals puncturing the plant/animal cells. After another melting, cell liquids leak out, cells lose their tonus and the food gets unpleasant mechanical and sensory state, affecting the final taste. Additionally, what was told above about the fridge applies here as well.
Presence of free ice along with frozen vegetables is a sign the content was refrozen.
Slow melting of previously shock-frozen stuff leads to growth of bigger crystals, similar to recrystallization, as bigger crystals are thermodynamically preferred due to lower surface energy. So some cell wall tearing and subsequent deterioration of the content happens even during melting and continues until content is frozen again. Also, shock freezing still induces some stress on the product, even if lower than slow freezing.
So the final effect of

fast freezing -> slow melting -> slow freezing -> melting

can be much worse than just

slow freezing -> melting

as the former has more time and opportunity to cause damage.


Atomic weight = expected weight? The atomic weight of an element, is it accurate to say that another way to think of it is the expected value of that element's weight if we were to sample one at random from the environment?

Are man-made versions counted as part of the weight? I'm new to chemistry but for example let's say we had a typical carbon-12 and we just started removing/adding neutrons at will, resulting in a whole bunch of various isotopes of carbon. I don't know if it's even possible to do that but say we had carbon-6, carbon-7, carbon-8, ... etc etc etc and so on until whatever isotope of carbon can't hold any more neutrons. Would those be counted?

Atomic weights in any periodic table are based on the natural abundance of different isotopes of the element. But sometimes the source matters and the results will be different.

The most important thing to note here is that creating new isotopes is hard. You can't just "add" neutrons to an atomic nucleus and often when you do it decays to a different element.

So most periodic tables quote atomic masses using the natural abundance of the elements as they are found in nature.

When there are exceptions, different atomic masses might be quoted for samples know to deviate from the natural abundance (or, more commonly, the specific ratios of the isotopes will be mentioned). Most elements abundance will be determined by the natural processes in stars (or the Big Bang) that created them in the first place. Occasionally natural nuclear processes will alter that ratio. For example, the radioactive isotope of carbon, 14C has a half-life of 5-6 thousand years but is produced naturally in the upper atmosphere. It is well distributed in the air and incorporated into plants. But if those plants die, the amount of the isotope will decay over time lowering the ratio of that isotope in the plant. This is both the basis of carbon dating and the reason why it is possible to tell whether alcohol is produced from fossil fuel sources (old enough that all the 14C has decayed) or from fermentation of sugars (produced from living things so incorporating the atmospheric level of the isotope).

It is even possible for nature to contrive conditions that mimic a nuclear reactor causing havoc with the natural abundance of heavy isotopes of uranium. In the uranium deposits in Oklo (in Gabon) the ratio of 235 was found to be far lower than the normal levels elsewhere because of a natural nuclear reactor that ran 2bn years ago.

But we don't change the numbers in the periodic table because of this. When we talk about specific samples of elements where the ratio is not normal, we specifically mention what the ratio is compared to the natural level.


What is antisymmetric exchange? What is J-strain? Where does it come from? I'm reading a paper1 by Sanakis, et al. that characterises the magnetic coupling in the $\ce{Fe3S4}$ clusters present in bacterial ferredoxin II and beef heart aconitase as arising through something called 'antisymmetric exchange'. This looks like it may be a result of the apparent magnetic frustration of the system (three high-spin $\ce{Fe^{3+}}$ ions, all in essentially identical environments, all antiferromagnetically coupled with respect to their neighbours). The paper says that the clusters have a S=1/2 ground state, which is incompatible with a broken-symmetry model.

The wikipedia article on antisymmetric exchange (all four sentences of it) is not particularly edifying. I'm hoping an expert can explain what antisymmetric exchange is and what causes it as I've not heard of it be before. The paper also notes that the exchange situation in the cluster gives rise to three slightly different $J_{ij}$ values (again, despite the symmetry of the ions) and terms this 'J-strain'. This makes me doubt that, for instance, 'antisymmetric' is being used as a synonym for 'antiferromagnetic'.

[Edit]

Here's an example of the general connectivity of this kind of $\ce{Fe3S4}$ cluster (based on D. Gigas FdII (PDB entry 1F2G)). The cluster comprises a hemicubane. $\ce{S_{Cys}}$ denotes a cysteinyl sulfur.

(1) Sanakis, Y.; Macedo, A.L.; Moura, I.; Moura, J.J.G.; Papaefthymiou, V.; Münck, E.; Evidence for Antisymmetric Exchange in Cuboidal $\ce{[3Fe-4S]^+}$ Clusters; J. Am. Chem. Soc.; 2000, 122, pp. 11855-11863

Maybe OP has already contacted the authors of this paper by now, but this was interesting. I'm no expert but I can Google things and this was too long for a comment.

Antisymmetric exchange:
At first I thought it was simply an exchange interaction where the wave function's sign is changed during exchange, now I don't think it's so simple.

Antisymmetric exchange is also known as DM-interaction (for Dzyaloshinskii-Moriya). I don't know exactly what it is, here is the original paper citation - can't find it anywhere though.

I.E. Dzyaloshinskii, Thermodynamic theory of weak ferromagnetism in antiferromagnetic substances, Sov. Phys. JETP 5(6), 1259-1272 (1957)

A definition

A slideshow, which contains some more info (slides 16-23)

J-Strain: Reading this, "J-strain" isn't the only "strain" that's referenced in this paper. Also referenced within is "G-Strain" which has a more elucidating Wikipedia article.

So, maybe it's safe to say, like G-Strain, J-strain is the broadening of the "distributed exchange coupling constants" (a.k.a. J-values). Inferred from a line in the paper: "Both proteins exhibit distributed J-values (J-strain)".

What did J-strain have to do with the experiment? "...The observed J-strain, which presumably represents a distribution of protein conformations “locked in” upon freezing the samples."

Where does the broadening of the J-values (exchange coupling constants) a.k.a. J-strain come from? Well apparently there's this thing called J-coupling that seems relevant. J-coupling is also known as indirect dipole-dipole coupling, and gives info about molecule interconnectivity. J-coupling relates nuclei to local electrons (All this is from the J-coupling article). The broadening of the lines may come from anisotropic interactions, which affect nuclear spin energy levels. Sure enough, the paper says it can't rule out anisotropy. An Alternative source? Non-homogeneous regions in the sample, which could go back to the frozen/locked conformations.

I still think it'd be great if someone doing physics research in this field could elucidate some more though.

Another J-coupling Webpage (U. Wisc)

Of course I should cite the original paper, "Evidence for Antisymmetric Exchange in Cuboidal [3Fe−4S]+ Clusters". Linked in question.


Pepper escaping soap on water Following a book of science experiments for children with my daughter, we did the following:

put some water in a shallow dish;
put some ground pepper floating on the water surface;
touch the water surface with the tip of a small stick previously dipped in liquid soap;
then the floating pepper escapes radially from the soap.

But I was at a loss to explain why. I mumbled something about “surface tension”, but didn't convince even myself.

What was going on?

I am pretty sure that the pepper in itself plays no role but showing the surface motion of water. Am I correct?

The "soap molecules" are so called amphiphiles: They've got one end that likes water ("polar" or "lipophobic") and the other ("lipophilic" or "fat-loving") doesn't.

These amphiphile molecules like to form a single-molecule thick film on water, all standing up in an orderly fashion (foot in the water), so if you put a drop of it on a water surface, it spreads very wide, until the film is monomolecular.

Or, until something pushes back. In your case the pepper, when its particles are pushed together so they also form a tightly packed surface. Waiting longer, the soap will likely act as a detergent, wrap the pepper particles in a bilayer, and drag them under the surface.

Search for, e.g., surfactants, detergents for some pictures.

As to your last question: Some water gets dragged along with the surfactant, but no, there is no large lateral movement of water on the surface. The spreading front you see is just the soap.


Why do different metal ions release different colors of light? For example, potassium chloride released a purple colored flame?
Does it have to do with the energy level of the valence electrons in the metal cation that is being burned?

You are right, it depends on the energy level of valance electrons, but more precisely it depends on the difference between the highest occupied and lowest unoccupied levels' energies. If the energy difference is $\Delta E$ then the light emitted by the excited atom has a frequency $\nu = \frac{\Delta E}{h}$ so a wavelength $\lambda = \frac{c}{\nu}=\frac{h c}{\Delta E}$


Reduction of hydrogen sulfide by silver Why can $\ce{H2S}$ be reduced by $\ce{Ag}$ when left in air:
$$\ce{2 Ag + H2S -> Ag2S + H2}$$
If silver is a noble metal that should not reduce hydrogen?

Check the electronegativity of hydrogen, silver and noble metals. In addition, though, consider that the product, $\ce{Ag2S}$, is a solid. The reaction is driven in one direction by precipitation, continually removing sulfur from the air.


About spontaneous combustion of methane I heard that spontaneous reaction happens if $\Delta G=\Delta H-T\Delta S$ is negative.
For combustion of methane, according to Chemguide:

$\Delta H=\pu{-891.1 kJ K^{-1} mol^{-1}}$
$\Delta S=\pu{-0.2422 kJ K^{-1} mol^{-1}}$

Thus, I would deduce that at normal temperature, $\Delta G<0$, combustion of methane is spontaneous, while at very high temperatures, it will stop.
But, from another reference, I deduce the opposite conclusion: if we consider the diagram below (taken from https://fr.wikipedia.org/wiki/R%C3%A9action_chimique),

I would think that in order to be able to go to the intermediate state, one needs to provide energy, thus that it could not be spontaneous for this reason. One needs to provide energy, for example by heating, that is increasing temperature.
How is that possible? Any comment?

"Spontaneous" means different things in different contexts
Your penultimate paragraph captures a key idea. The explanation for why this is right requires a recognition of the context of the term "spontaneous".
The context of the statement at the start of the question $\Delta G=\Delta H-T\Delta S$ is negative is thermodynamic stability. But this is somewhat at variance with the more natural use of the term which implies "things happen without being pushed". This idea is closer to the idea of kinetic stability in chemistry.
You correctly identify the need to add energy to get the reaction past the transition state. Even if the reaction overall releases energy (thermodynamically spontaneous) the reaction won't just happen if there is a huge barrier to getting over the transition state. There are big kinetic barriers that stop the reaction "just happening". Oxygen and gasoline will react to release energy but this doesn't happen without the push given by the spark plug in the engine of a car.
That barrier is so low in some cases a compound will react with nearly everything with little excuse (chlorine trifluoride will set fire to asbestos). That is a spontaneous reaction in any context. Luckily, few thermodynamically spontaneous reactions are also kinetically spontaneous or humans would catch fire in air.
So when you see the term "spontaneous" ask what is the context: thermodynamic or kinetic? And don't confuse them.


Why does baking soda and vinegar remove stains? To remove a stain on fabric, I often rub baking soda into the stain, and then sponge it with white vinegar. This causes a chemical reaction, involving some fizzing and frothing, and once the fabric has dried, the stain (hopefully) has disappeared. Why does this chemical reaction remove the stain?

From my basic knowledge of chemistry, I know this is a classic Acid + Base -> CO2 + H2O + Others. But this doesn't involve the chemicals which have caused the stain.

Is it that these chemicals also react, and turn into other colourless products? Or is it that the frothing action from the reaction just dislodges these chemicals, allowing them to more easily be wiped clean? Or is there another answer?

You might try a comparison test to see if baking soda and vinegar is any more effective than plain water, or water and one of the ingredients. There are some possible reasons why each might be effective:

Some dyes and stains are set by exposure to acid, such as vinegar, which makes them harder to remove. It is possible that alkaline baking soda and water have the opposite effect, removing the stain.
Some stains, such as from red cabbage and red wine, are indicators, and change color when exposed to alkali or acid. The natural dye alizarin changes from dark red to lighter yellow under acid conditions, as in vinegar.
A powder such as baking soda may act as a sponge or adsorbent because of its large surface area. Talcum powder, baking soda or fuller's earth are also used for that purpose.
There might be a physical effect from the bubbles, as you state. For that reason, perhaps, some use plain seltzer (carbonated water) for stain removal.

Note that there are many conflicting statements on the efficacy of these stain removal techniques. It would be of benefit to humanity if someone conducted controlled experiments on removing a variety of stains with these chemicals! In the meantime, I just put a bit of HE laundry detergent on a stain, or a commercial cleaner such Lestoil, on the spot, before laundering.


What elements don't observe the octet rule? Apart from hydrogen which forms a duet, which element's don't observe the octet rule?

Octet rule is more advisory than a rule; It's usually obeyed by the main group elements.
Exception are paramagnetic compounds (obvious reason); first for elements in periodic system (He-configuration), 3-valent boron compounds; depending on counting system boron clusters (f.e. dodecaborane with 6-coordianted boron); transition metal complexes (these usually follow 18e rule, however there are plenty of examples with 14, 16 or even 20 electron complexes).


Why doesn't rubidium react with nitrogen gas? I ran across a problem in which I was asked to predict the result of a reaction between $\ce{Rb}$ and $\ce{N2}$. I was frustrated until I found a note in my textbook which stated that out of all the alkali metals, only $\ce{Li}$ reacts with $\ce{N2}$. I am curious to know the reason for this.

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

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


Overall enthalpy balance on a mixing process Let's say I have a mixing process involving $\ce{A}$ and $\ce{B}$ at temperatures $T_\ce{A}$ and $T_\ce{B}$, respectively. The $\ce{AB}$ mix is at temperature $T_p$. Our $ΔH_\mathrm{mix}$ is defined at $\pu{25 °C}$ (i.e., our reactants/products are not at the reference temperature) and we know its value. What would be the overall balance?

It is my understanding that we would have 3 steps:

Use the amounts and heat capacities of $\ce{A}$ and $\ce{B}$ and find $ΔH$ (open system) from their specified temperatures to the reference temperature $\pu{25 °C}$. This is $ΔH_1.$
Use the amount and heat capacity of $\ce{AB}$ mix to find $ΔH$ from the reference temperature to $T_p$. This is $ΔH_2.$
Finally, $ΔH = ΔH_2 + ΔH_\mathrm{mix} + ΔH_1.$

That is to say, the overall $ΔH$ is the sum of all the steps required to get the initial state(s) to the final state.

I have been told by my lecturer, however, that it should be

$$ΔH = ΔH_2 - ΔH_\mathrm{mix} - ΔH_1$$

Why are $ΔH_\mathrm{mix}$ and $ΔH_1$ subtracted?

Let’s use a trivial case to test the two possible results. If we set $T_\ce{A}$, $T_\ce{B}$, and $T_p$ all to room temperature, the first and third term would be zero (sign does not matter), and you should simply get the enthalpy of mixing as a result. The result you arrived at does that, but the answer you were given changes the sign.

Unless there are some unexpected conventions about the signs of the terms, your result makes more sense.


Is a polycarbonate beer mug appropriate for serving alcohol? Are polycarbonate & acrylic beer mug, whiskey glasses safe for serving alcohol? I mean does it leach or dissolve? What are the harmful effects in long run if any?

It should be fine if used correctly! Most drinking alcohol wouldn't be strong enough to denature the polycarbonate which is very resistant anyway hence why they make whiskey/beer cups out of it. Some moonshine might warp it! Straight spirits may over time start to dissolve the polycarbonate. The harmful effects would be intoxication I'd say and maybe alcoholism !...


Pure Hydrogen and Dry Hydrogen Difference Do pure and dry hydrogen are the same? What does dry hydrogen actually means. And what does pure hydrogen means.

It depends what you (or your field) is considering what an impurity is for a given substance. For example, if you want really pure methane, ethane will became an impurity. However, if you want gaseous alkane(s), there is no real difference between methane and ethane, so ethane is no longer an impurity.

On the other side, moisture can be your enemy, so pure methane (without ethane) can be considered bad if it contains 3% water whereas a mixture of methane and ethane (50/50) could be considered better because it contains only 0.1% water.

In your case, "pure hydrogen" would mean that "it is composed almost entirely of hydrogen gas" (but it can contain water) whereas "dry hydrogen" would rather mean that the gas does not contain water (but it can contain other molecules as well, but not too much)


Lucas test for alcohols The Lucas test for alcohols is based on the way primary, secondary and tertiary alcohols react with Lucas reagent. If on adding Lucas reagent to an alcohol, the mixture immediately becomes turbid, then it is a tertiary alcohol.

I am aware that alcohols get converted to alkyl chlorides on reaction  with concentrated HCl, and that it the alkyl chloride that is insoluble and hence responsible for the turbidity.

But why do tertiary alcohols react the fastest with HCl?

The tertiary alcohol has more R groups compared to primary & secondary alcohols, there by removing the C-O Bond from a tertiary alcohol is more easier. (the tertiary carbocation is more stable)


Is there a way to turn alcohol into a solid gel? I'm wondering if there is a way to turn alcohol or another bacteria killing substance into a solid gel form, sort of like the gel you can use to clean a keyboard.

  Source: Aliexpress.com Computer Cleaning Compound

I'm wondering if you could put either alcohol or other sanitizer into this sort of gel, without the sanitizer having a "best before" date.

I'm a complete beginner when it comes to chemistry so forgive me if there's an obvious reason why this wouldn't work. If it does work, or partially, could you give me a run down on what I would need or need to test to achieve my goal?

If your alcohol is ethanol, calcium acetate works very well.  I use calcium acetate to make gel in petri dishes, and sprinkle metal salts on it for colored flame demonstrations.  Calcium acetate is also used in some canned alcohols used for warming/cooking foods.

3 grams of calcium acetate dissolved in 10 ml of water will make a saturated solution (some may not dissolve, which is OK).  This is mixed with 75 ml of 95% ethanol, which will then form a colloidal gel identical to that used in canned alcohol warmers.  I do not know if this gel is suitable for your purposes, but it works well for a heat source.  If used as a heat source, a pinch of salt should be added to the aqueous solution before mixing with alcohol to make the flame more visible.


Hybridization and excitation of electrons, how is it possible? When studying the structures of molecules by hybridization why we take into consideration of excitation of electron if the excited electron only stays in the upper shell by absorbing for only 10^-8 seconds{approximately }?

We don't really consider the excitation of electrons. If it appears like this, then this is probably due to an incorrect understanding of what an orbital actually is: The attempt to describe each single electron by its own wave function which is independent of all the other electrons (which it is not).

An orbital is defined as single electron wave function $\varphi(\vec r_i)$ where $\vec r_i$ are the coordinates of electron $i$. We then try to approximate the many electron wave function as an (anti-symmetrized) product of orbitals

\begin{equation}
   \Psi(\vec r_1, \vec r_2, \dots \vec r_N) \approx \prod_{i=1}^N \varphi(\vec r_i)
\end{equation}

But this is strictly speaking not true, because the electrons are correlated with each other, due to the Coulomb repulsion $V\propto \frac{1}{|r_i-r_j|}$. The product ansatz does not work for the kind of differential equation we need to solve. The Hartree-Fock method tries to do so anyway, and gives therefore wrong (or at least only approximate) result.

This means the whole concept of orbitals is an approximation, which may give a qualitative understanding, but are failing when trying to get quantitative results.

One approach to correct this approximation is expanding the many electron wave function in some basis and solve the resulting eigenvalue problem. This basis is not unique, but one possible choice is to use the Hartree-Fock determinant and the excited configurations you can generate from it. But these excited configurations are more of an abstract mathematical thing and have limited physical meaning, because they are based on the above approximation. The real physics/chemistry is in the solutions of such methods. And those solution cannot be describe by the above ansatz for $\Psi$.


What is hybridisation of XeF6 in solid state? According to the following formula:

Type of hybridisation/steric no. = (no. of sigma bonds + no. of lone pairs)

it should be $\ce{sp^3d^3}$, however according to my textbook it is $\ce{sp^3d^2}$ in solid state. That is how it was written(in my textbook), emphasising on the fact that "in solid state it is $\ce{sp^3d^2}$". How does solid state of a compound changes its hybridisation?

Forget about applying hybridization outside the second row, especially in 'hypervalent' compounds. I know, that it is common to use and sometimes works, but it is incorrect.
The $\ce{XeF6}$ molecule is a hard spot. While, indeed, experimental data suggest that it adopts distorted octahedral geometry in the gas phase, there is evidence that the minimum is very shallow.
The octahedral structure of $\ce{XeF6}$ (which is probably a local minimum, or at least found to be one in more than one calculation) can be pretty easily described in terms of three-center four-electron bonds.
There is no good agreement on the nature of the stereo-active electron pair in $\ce{XeF6}$. It seems, that repulsion with core d-shells is important, but the recent articles I could get my hands on has little to no rationalization on the fact. Still, it must be noted that the minimum is very shallow, meaning that rough qualitative theories like VSEPR and MO LCAO are too rough for a good rationalization anyway.
The solid state typically involves a lot more interaction, and, as result, compounds may adopt very different structures. The simplest one example I can think of, I could point to $\ce{SbF5}$ for example. A monomer adopts trigonal-bipyramidal structure. However, in the solid state, a tetramer with octahedrally coordinated antimony and four bridging fluorines is formed. In terms of VSEPR, it is a move from $\mathrm{sp^3d}$ to $\mathrm{sp^3d^2}$. Similarly, $\ce{XeF6}$ has a crystal phase (one of six) that involve bridging atoms.


Why do ionic solids change to a structure with a higher coordination number at high pressures? This seems counter intuitive to me since ionic radius increases with increasing coordination number. Surely the solid doesn't expand under high pressure? Can you explain?

Lets put some numbers on the phase transition of NaCl from the normal NaCl structure (based on fcc, 6-fold coordinated) at atmospheric pressure, to the CsCl (based on bcc, 8-fold coordinated) structure at high pressure.
From Compounds and Alloys Under High Pressure: A Handbook, by E. Yu Tonkov (google found it) one gets that the transition occurs at 27-29GPa, with a volume change of 5%.
The density of NaCl, taken from Wikipedia, is 2.165 gm/cc, or about 27cm$^3$/mol, or 27E-6 m$^3$/mol.
Multiply on through and you get, very roughly, the PV energy gained by converting from the less dense NaCl structure to the denser CsCl structure is 40kJ/mol - this is not an inconsequential energy gain for the Gibbs free energy.  Thus, the PV energy gained by a denser solid with the higher coordination plays off (and pays off) against the ionic unhappiness.

In the literature there are also indications that the CsCl, and/or higher pressure phases, may be metallic in nature, so the assumption of purely ionic bonding may be in jeopardy.


Determine standard enthalpy of formation of salt given other reactions I'm a little confused by this question, perhaps my understanding of enthalpy isn't as good as it should be.

  For the compound $\ce{KNO3}$, which of the following reactions is used to determine the standard enthalpy of formation, $\Delta H^\circ_\mathrm f$?

  $$\ce{K(s) + N (g) + O3(g) -> KNO3 (s)\\
2K + N2 + 2O3 -> 2KNO3\\
2K + N2 +3O2          -> 2KNO3\\
K  + 1/2 N2 + 3/2 O2  -> KNO3}$$

I know that the $\Delta H^\circ_\mathrm f(\ce{KNO3})= \pu{−494.6 kJ/mol}$ from reading a table, I'm just confused about how do I get from that number to determine the formation of a compound.

The standard enthalpy of formation $\Delta H_\mathrm f$ refers to the formation of 1 mole of product, with the educts and product being in their respective standard states. The fourth equation is therefore correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H_\mathrm f$. The first two equations can be ruled out because they do not contain the educts nitrogen and/or oxygen in their standard states, which are $\ce{N2}$ and $\ce{O2}$, respectively.


How many electrons with l = 1 does Si in its ground state have? I was solving practice problems for electron configuration and periodic table, and I got stuck through a question:

  An atom of silicon in its ground state has how many electrons with quantum number $l = 1$?
   a) 14
   b) 2
   c) 8
   d) 6
   c) 28

According to the solution the answer is c: 8 electrons, but that wasn't my answer.

My analysis:

Silicon is a chemical element with symbol $\ce{Si}$ and atomic number 14, this is its electron configuration:

$$\mathrm{1s^22s^22p^63s^23p^2}$$

The quantum number $l = 1$ corresponds to $\mathrm{p}$ levels only. So we have:

$$\mathrm{2p^63p^2}$$

The total number of orbitals is: $6 + 2 = 8$ orbitals.
Since each orbital has 2 electrons, the total number of electrons is $8 \times 2 = 16$ electrons.

What's wrong with my answer? Maybe the question has a wrong answer?

You're on the right track, but those superscripts in the electronic configuration don't refer to orbitals. They refer to the number of electrons in that type of orbitals.

For more information, you might want to look at this question: What are the maximum number of electrons in each shell?


Do quaternary sulfur dications exist? We know that sulfur can form sulfides $\ce{R2S}$, with two substituents bonded to it. The simplest example of this would be hydrogen sulfide.

However, sulfur can also form sulfonium ions $\ce{R3S+}$, where 3 substituents are attached to the sulfur atom and a negatively-charged counteranion is present.

What I am asking is whether there is such a thing as sulfur bonded to 4 substituents, with each bond being a single bond, with 2 counteranions (either $\ce{(R4S^2+)(X^2-)}$ or $\ce{(R4S^2+)(X^-)2}$). Is there such a thing as that or something similar?

Ogawa et al. [1] were first to report a crystal structure (CSD-YAFNOI) of a compound with quaternary sulfur, bis(2,2′-biphenylylene)sulfurane:

Figure 1. Molecular structure of bis(2,2'-biphenylene)sulfurane (CSD-YAFNOI). Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$.

  Compound 1 was synthesized as follows (Scheme 1). Dibenzothiophene 5-oxide (200 mg, 1.0 mmol) in anhydrous tetrahydrofuran (THF, 10 ml) was treated with trimethylsilyl trifluoromethanesulfonate (0.25 ml, 1.3 mmol) under an $\ce{N2}$ atmosphere at −78 °C. After stirring at 0 °C for 30 min, the
  mixture was cooled to −78 °C and was treated with $\pu{1.0 mol dm-3}$ 2,2'-dilithiobiphenyl (1.0 ml, 1.0 mmol) in diethyl ether solution. The whole mixture was stirred at −78 °C for 1 h and at 0 °C for 30 min under an $\ce{N2}$ atmosphere. After evaporation of the solvent, the residue was washed with anhydrous diethyl ether (10 ml) and was extracted with anhydrous benzene (10 ml) under an $\ce{N2}$ atmosphere. The solvent was removed under reduced pressure, and the crude product was recrystallized from anhydrous THF at −20 °C to give 1 as orange rods in 96% yield.

  Scheme 1 Reagents: i, trimethylsilyl trifluoromethansulfonate in THF;
  ii, 2,2'-dilithiobiphenyl in diethyl ether-THF

Further work by Sato et al. [2] resulted in a synthesis and crystal structure (CSD-NEDCEE) of bis(2,2′-biphenylylene)sulfuranyl bis(tetrafluoroborate).

Structurally, it's a similar compound with a greater, nearly 90° (in contrast to 60° twist angle in neutral bis(2,2′-biphenylylene)sulfurane), twist angle between 2,2′-biphenylylene ligands, however water molecules and $\ce{[BF4]}$-counterions appear heavily disordered:

Figure 2. Fragment of the molecular structure of bis(2,2′-biphenylylene)sulfuranyl bis(tetrafluoroborate) (CSD-NEDCEE) showing the bis(2,2′-biphenylylene)sulfuranyl cation. Oxygen atoms from water molecules as well as tetrafluoroborate anions are omitted for clarity. Color code: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$.

  Recently, we have succeeded in the first isolation and structural determination of bis(2,2′-biphenylylene)sulfurane [10-S-4(C4)] (1) as a stable sulfurane(IV) having only carbon ligands.[…] We considered that this sulfurane would be a suitable precursor to provide the desired dication. Therefore, we tried the reaction of bis(2,2′-biphenylylene)sulfurane (1) with xenon difluoride ($\ce{XeF2}$) in the presence of $\ce{BF3 * OEt2}$ and indeed obtained the bis(2,2′-biphenylylene)sulfurane dication, [8-S4(C4)]²⁺ (2) as an amazingly stable bis(tetrafluoroborate) salt.[…] Here, we communicate the first isolation and structural determination of bis(2,2′-biphenylylene)sulfurane dication (2) having only carbon ligands. […]

  The sulfurane 1 was reacted with 1 mol equiv of xenon difluoride in the presence of $\ce{BF3 * OEt2}$ in dry $\ce{CH3CN}$ at −40 °C (Scheme 1). After the solvent was removed at room temperature, the residue was washed with $\ce{CHCl3}$ at room temperature, and bis(2,2′-biphenylylene)sulfurane bis(tetrafluoroborate) (2) was isolated as a stable moisture-insensitive yellow powder in 62% yield.

  Scheme 1

Subsequently, hexacoordinated derivatives – bis(2,2′-biphenylylene)dimethyl- and diphenylpersulfuranes – were synthesized and their molecular structures were elucidated [3].

References

Ogawa, S.; Matsunaga, Y.; Sato, S.; Iida, I.; Furukawa, N. First Preparation of a Sulfurane with Four Carbon–Sulfur Bonds: Synthesis and Molecular Structure of Bis(2,2′-Biphenylylene)Sulfurane. J. Chem. Soc., Chem. Commun. 1992, 0 (16), 1141–1142. https://doi.org/10.1039/C39920001141.
Sato, S.; Ameta, H.; Horn, E.; Takahashi, O.; Furukawa, N. First Isolation and Molecular Structure of Bis(2,2′-Biphenylylene)Sulfuranyl Bis(Tetrafluoroborate) [8−S−4(C4)]²⁺. J. Am. Chem. Soc. 1997, 119 (50), 12374–12375. https://doi.org/10.1021/ja971336k.
Sato, S.; Matsunaga, K.; Horn, E.; Furukawa, N.; Nabeshima, T. Isolation and Molecular Structure of the Organo-Persulfuranes [12−S−6(C6)]. J. Am. Chem. Soc. 2006, 128 (21), 6778–6779. https://doi.org/10.1021/ja060497y.


Side reactions of NHS Chemistry N-Hydroxysuccinimide (NHS) chemistry is commonly used in chemical biology to react with primary amines like lysine and the N-terminus of proteins. I was curious how labile NHS esters are to other nucleophiles like secondary amines and sulfhydryls. This knowledge will be helpful to know the extent of side products when I do a bioconjugation experiment.

The reactivity of NHS activated esters has mostly been desribed quite anecdotically. They are susceptible to hydrolysis in water, the kinetics of which is dependent on the nature of the NHS activated group (ester vs carnonate vs carbamate) as well as on the pH of the reaction buffer1, which is one way to show that primary amines are not the only nucleophile which can react with NHS-activated esters.

As for the chemical groups that have been observed to react with NHS activated esters, I did find a couple of examples upon looking up this issue. For instance, the Zenobi group have attempted a quite systematic study on the groups that can react with doubly NHS activated cross-linkers. Citing the abstract2 :

  As soon as additional cross-linkers were attached or loops were formed, other amino acids were also involved in the reaction. In addition to the primary amino groups, serine, threonine and tyrosine showed significant reactivity due to the effect of neighboring amino acids by intermediate or permanent Type-1 cross-link formation. The reactivity is highly dependent on the pH and on adjacent amino acids.

In an earlier work, the reactivity of a biotinylation NHS-ester based reagent was assessed on a single peptide,  [D-Lys(6)]gonadotropin releasing hormone. A similar reactivity of -OH groups was observed, with in addition that of the arginine guanidinium group3:

  In addition to the O-acylation of Ser(4) and Tyr(5) in this peptide, we have also identified a novel biotinylation of the Arg(8) side chain.

Other groups have also observed some reactivity with cysteins (sulfhydryl groups), but in a situation where the reagent is included within a ligand of the target protein. One can assume, as do the authors, that specific positionning of the ligand within the binding site can lead to binding with a cystein group4,5.

From these earlier work, one cas conclude, in answer to your question that although primary amines are the most reactive group for NHS-ester derivatization, other groups present in peptide side-chains (-OH for tyrosine, serine, threonine, guanidinium for arginine and sulfhydryls for cysteins).

The last issue raised in the question remains that of the secondary amine groups. It is widely accepted that pH influences strongly the kinetics of NHS derivatization, mostly through protonation of the amine groups which cannot act as nucleophiles in protonated state. This has been put to use in controlling the buffer pH in order to favor N-terminal reaction vs lysine side chain reaction. Although, in my experience, it is not at all a straightforward experiment and it is highly dependent on the nature of the derivatization group. One could hypothesize that since secondary amines are more basic then primary amines, the reactivity of secondary amines towards NHS should be reduced compared to primary amines, but I have no experimental or litterature proof for this part of the answer.


Why do many hypervalant chemicals have non-constant bond angles? According to the VSEPR model, when a molecule in formed in the shape of $\ce{AB_{x}}$,

like $\ce{IF7}$, $\ce{SCl6}$, and $\ce{PCl5}$,

its structure is found by reducing the repulsions between peripheral atoms. This gives rise to many shapes, most of which have constant bond angles. Examples of those with constant bond angles include

$\ce{CH4}$ with 109.5° bond angles
$\ce{CO2}$ with 180° bond angles.

My question is then, why do some have non-constant bond angles?
The following do not

$\ce{IF7}$ with 72° and 90° bond angles
$\ce{PCl5}$ with 120° and 90° bond angles.

The $\ce{IF7}$ and $\ce{PCl5}$ molecules have 7 and 5 ligands, and pentagonal and trigonal bipyramidal geometries, respectively. These geometries happen

  simply because
  there is no geometrical arrangement which can result in five equally
  sized bond angles in three dimensions.

(Wikipedia)

There is no arrangement in which all the angles would be the same, except in a plane, where the repulsion would obviously be a lot higher than in the pyramids.


Carbon Oxides and stability If $\ce{C2O4}$ is very unstable no matter the configuration(that is the structure based on the bonds) and especially unstable in the form that has an oxyanion and a carbocation than how come other carbon oxides are much more stable despite the charges like $\ce{CO}$ for instance.

There are 2 possible structures with octets for this. They are

1) $\ce{C=O}$ with the carbon having 1 lone pair and the oxygen having 2 lone pairs. No charges here.

2) $\ce{C≡O}$ with $\ce{C}$ having 1 lone pair and $\ce{O}$ having 1 lone pair making the oxygen positive and the carbon negative.

This one is an important but very unstable structure.

Why is it so unstable? It is because while carbon isn't so determining as to what charge it prefers, Oxygen is and it really hates being positive because it wants to be either neutral or negative since it is more electronegative than everything else except fluorine. This is why $\ce{H3O+}$ isn't really $\ce{H3O+}$ but rather an ion in a complex of ions is because of how oxygen hates being positive.

So why are other carbon oxides more stable than $\ce{C2O4}$ when they too have very unstable structures like $\ce{CO2}$ having as a resonance structure a single bond to 1 oxygen making it negative, a triple bond to the other making it positive, and the carbon staying neutral? And if every carbon oxide is either unstable no matter the structure or has unstable resonance structures than why do carbon oxides with just carbon and oxygen exist?

And would adding hydrogens to a carbon oxide make the carbon oxide even more stable? By this I mean that with the fact that carbon oxides tend to be unstable and hydrocarbons tend to be stable would a carbon oxide with hydrogens added to it be less stable than a hydrocarbon but more stable than a carbon oxide?

Is this how come glucose is so stable and 6 $\ce{CO}$s bonded together would be extremely unstable just like $\ce{C2O4}$ is?

Sorry if it is too many questions but it is all about 1 thing, Carbon Oxides.

Why is $\ce{C2O4}$ unstable?

Have you tried drawing it?

I get four different kinds of structures for it:

$\ce{(O=)O^{.}C-CO^{.}(=O)}$
$\ce{(O^{.})2C-C(O^{.})2}$
$\ce{O#C-O-CO^{.}(=O)}$
$\ce{O#C-O-O-C#O}$

They are unstable for the following reasons:

Biradical
Tetraradical (This won't really exist anywhere near anything else.)
Radical
Peroxo bond, very easy to break homolytically.

All of these compounds can easily decompose to give (the thermodynamical Mariana trench) carbon dioxide:
$$\ce{C2O4 -> 2CO2}$$

That is why it is so unstable.

Reduce 'em!

  And would adding hydrogens to a carbon oxide make the carbon oxide even more stable?

Yes. If you start reducing the compounds above, you will end up with ketones and water (because geminal hydroxy groups tend to give water and ketones), or even alcohols (if you continue reducing).


Can magnetic fields affect a chemical reaction? This question was asked recently in an interview and I just said that "Yeah, like if the reaction involves ions or paramagnetic species". But the interviewer went on and asked me to elaborate on HOW will the reaction change, and I was stumped.

So what I am asking here is what will happen if I allow a chemical reaction to take place in a live MRI machine?

Here is a nice review paper on the subject: Rodgers, C. T. Magnetic field effects in chemical systems. Pure Appl. Chem. 2009, 81 (1), 19–43. DOI: 10.1351/PAC-CON-08-10-18.

From the abstract:

  Chemical reactions that involve radical intermediates can be influenced by magnetic
  fields, which act to alter their rate, yield, or product distribution. These effects have
  been studied extensively in liquids, solids, and constrained media such as micelles. They  may be interpreted using the radical pair mechanism (RPM). Such effects are central to the field
  of spin chemistry of which there have been several detailed and extensive reviews.


Static electricity with calcium carbonate particles We are working with small calcium carbonate particles (microfossils) but one of the problems is they stick to the plastic or metal tools we are using, presumably due to static charge. For example, small particles on plastic tray will stick to the surface even if it is upside down.

I'm guessing the static potential is related to how much calcium carbonate wants to gain or give away electronics?
What is this called in this context?
How do I choose a material that has a low potential for static electricity (i.e. sticking) with calcium carbonate particles?

Some links of a commercial equipment to avoid problems of static electricity in powders during weighing:

https://www.mt.com/us/en/home/products/Laboratory_Weighing_Solutions/Accessories/anti_static_devices.html

http://lab.mt.com/antistatic/

https://www.mt.com/dam/mt_ext_files/Editorial/Generic/6/Anti-static_0x000246700004f0770006d93b_files/ion_e.pdf


Will bridged compounds the undergo SN1 reaction? $\mathrm{S}_{\mathrm{N}}1$ reaction involves only one molecule in the rate determining step. So, the molecule which undergoes $\mathrm{S}_{\mathrm{N}}1$ reaction should be stable when it forms a positive ion. $\ce{(CH3)3C^+}$ is considered stable due to inductive effect from the three carbon atoms and also hyper-conjugation. But, what about bridged compounds like 1-bromobicyclo[2.2.1]heptane:

When $\ce{Br^-}$ is abstracted the positive ion formed will be stable and therefore this will undergo $\mathrm{S}_{\mathrm{N}}1$ reaction. Am I thinking right? Please help me correct myself.

Very interesting question… I would give two answers, at different levels:

At undergrad level, I think the carbenium ion formed at the bridge head is less stabilized that “regular” ternary carbenium ions, because it cannot adopt the planar trigonal geometry which would be ideal for it.
At higher level, this is much more complicated: the norboryl cation is an example of non-classical ion…


When an electron returns to it's ground state does it release the same wavelength of light that promoted it in the first place? If this is true, I fail to understand why we see colour at all -- wouldn't any wavelengths absorbed by electron promotion be inconsequential to colour because there would an equal amount, and identical wavelength of light emitted when the electrons returned to their ground state?

The only way this would make sense to me is if the energy of the photon absorbed by the electron is given off as heat or something besides an identical photon when the electron returns to its ground state.

Not necessarily... it can return to the ground state in multiple steps, releasing different wavelengths each time.

Also, the excitation could be caused by something other than light in the first place (heat, for example).

Also, relaxation isn't instantaneous, so the emitted photon wouldn't really be "cancelling out" the color of the absorbed one as some time has passed.


How does non-electrolytic tinning work? Having made diy pcb 's at home I have used things like tinnit which is a clear liquid . When the pcb is dropped into them , a layer of tin forms on the exposed copper regions on the pcb . I was wondering if someone could explain what is happening or if someone could give an equation .

Here is a video of the process on youtube https://www.youtube.com/watch?v=wOqkI-S-wdI

This is a good question: Tin is above copper in the galvanic series, so logically, copper should displace tin, not the other way around.

Electroless (autolytic) tin plating can be accomplished by disproportionation of a tin compound. $\ce{2Sn(OH)3- -> Sn + Sn(OH)2-6}$, where the less stable Sn(II) yields Sn(IV) + Sn(0), or metallic tin. The divalent tin is usually $\ce{SnCl2}$.

Electroless tin can also be deposited by reduction of a tin compound, e.g.  thiourea reducing $\ce{SnCl2}$ to metallic tin. In a similar fashion, silver nitrate can be reduced to metallic silver to make a mirror on a glass surface. Note that the silver is deposited without  any effect on the glass (other than a nice, shiny coating).


Why is the uncertainty of a digital instrument always the smallest scale division? For a mass balance that can give readings to 2 decimal places, the uncertainty is assumed to be $\pm 0.01$. Can someone explain the rationale for this, as I would think it more logical for the uncertainty to be given as $\pm 0.005$.

Migrating my comments to an answer:

The big issue with your line of reasoning is that the transducer in the scale does not produce decimal output. Like all digital devices, the signal is digital (i.e., binary), so the representation of the weight will also be a binary number.

Then you're in a scenario where the uncertainty is something like $\pm 0.0001_{2}$, which is $\pm 0.0625$. In this case, I would say that the error is in the first decimal place after the decimal point.

You can't give an error of $\pm 0.005$, partly for the reason I gave, but also because the error in measurement is not necessary an error in rounding. It's possible that the transducer is temperature sensitive, so within a reasonable range, it might over or under measure by $0.01$ units.


Why are there 2 equivalents of Ca in an example stated in Wikipedia? Applying what's been said in the Wikipedia page for equivalents, if $1\ \mathrm{mol}$ of $\ce{NaCl}$ and $1\ \mathrm{mol}$ of $\ce{CaCl2}$ are dissolved in a solution, there is $\rm1~eq~Na,~2~eq~Ca$, and $\rm3~eq~Cl$ in that solution.

Is this a misstatement in Wikipedia?

The Wikipedia article uses a funny defintion of equivalent that I have not heard yet even though I use the term extremely frequently in the lab. But they define it well so here goes:

One equivalent of ions is defined as $1~\mathrm{mol}$ of charge. So if we dissolve $1~\mathrm{mol}$ of $\ce{Na+}$, we have exactly $1~\mathrm{mol}$ of positive charges in solution, so $1~\mathrm{mol} \cdot 1~e = 1~\mathrm{eq}$ ($e$ being the elemental charge). The $1~\mathrm{mol}$ of $\ce{Cl-}$ contributes $1~\mathrm{eq}$ of negative charges, balancing the entire thing.

Calcium, however, has a $\mathrm{+II}$ oxidation state. So $1~\mathrm{mol}$ of $\ce{Ca^2+}$ gives us $1~\mathrm{mol} \cdot 2~e = 2~\mathrm{eq}$ of positive charge.


Why does acetonitrile have a larger dipole moment and boiling point than acetaldehyde? Experimentally, acetonitrile has a larger dipole moment than acetaldehyde, but I've never understood why.

I always thought that the charge separation between carbon/oxygen is larger than that of carbon/nitrogen, and since the $\ce{C=O}$ bond should be longer than the $\ce{C#N}$ bond, the dipole moment $\mu = \delta d$ would be larger with both larger charge separation and larger distance.

I assume geometry might have something to do with it but I can't really explain such a large boiling point difference.

Data: Dipole moment values:

acetonitrile: $\approx \pu{3.4D}-\pu{3.5D}$ (source)
acetaldehyde: $\approx \pu{2.7D}$ (sources: 1, 2)

Using molcalc.org to get a qualitative idea about what is going on, here are the charge distributions and the partial charge assigned to the electronegative N (-0.19) or O (-0.40):

Even with the quick cheap calculation done in the educational software, the calculated dipole moment of acetonitrile (4.05 D) is larger than that or acetaldehyde (3.16 D).

How do we rationalize a smaller dipole for acetaldehyde given the larger negative partial charge on the oxygen atom? You could consider the dipole of the methyl group separately from the dipole of the two functional groups. In acetonitrile, these dipoles line up, so they are additive. In acetaldehyde, they are at an angle, so the (vector) sum will be less.

If you look at all the calculated partial charges in acetonitrile, the hydrogen atoms are assigned most of the positive charge (3 x 0.08), and the nitrogen atom most of the negative charge. This results in a large charge separation, which influences the dipole moment (charge difference x distance). Surprisingly (to me), the carbonyl carbon is assigned a charge of -0.06.

If you look at all the calculated partial charges in acetaldehyde, the largest positive partial charge is assigned to the carbonyl carbon (+0.33), making for a shorter distance of the charge separation.

To get a more quantitative picture, one would have to perform a more serious quantum-chemical calculation. This would also be an opportunity to look at the orbital shapes and their relation to the charge distribution. This quick calculation, however, already points out some of the salient differences.


Thermal decomposition of mercury thiocyanate A video on YouTube shows the thermal decomposition of mercury thiocyanate. A strange thing I've observed is that $\ce{Hg(SCN)2}$, which is the reactant, is a white powder and the product is a yellow-brown solid serpent-like structure. How is this possible? What is the chemical reaction for this decomposition?

video link - https://www.youtube.com/watch?v=FaC4MqxUnFw

The reaction is the decomposition of mercury(II) thiocyanate:

  $\ce{2Hg(SCN)2 → 2HgS + CS2 + C3N4}$

Where $\ce {C3N4} $ is the compound that makes up the "snakes," and grow due to the evolution of gases produced by the decomposition of the other products:

  $\ce{CS2 + 3O2 → CO2 + 2SO2} $

  $\ce {2C3N4 → 3(CN)2 + N2} $

  $\ce {HgS + O2 → Hg + SO2}$

This is also known as the "Pharaoh's Serpent" in pyrotechnics. The resulting solid can range from dark graphite grey to light tan in color with the inside generally much darker than the outside.
Click here and here for reference.


Does adsorption occur between liquids? While reading a paper in a field other than chemistry, the author mentions a phenomena that he labelled "adsorption" and describes it as follows:

  ...I am told (though I do not know which liquids they are), there are
  distinct liquids such that pouring 1 litre of one into a container
  already containing 1 litre of the other does not result in there being
  2 litres of liquid in the container (adsorption).

Now, I know almost nothing about adsorption beyond what I just read on Wikipedia, but I didn't see an  example there where two liquids interacted in the way described. Instead, the examples seemed to be ones where a porous solid (e.g., activated charcoal) adsorbs some liquid.

Does this occur between two liquids? If so, what's an example and how does it work in the absence of pores?

The author of the article has named the phenomenon incorrectly. Adsorption can indeed result in a lower volume than might be expected, for example when a gas interacts with a high surface area solid. In the case of hydrogen storage, for example, the adsorbed phase is dense enough at low temperatures that much more hydrogen can be stored in a container filled with a high surface area material, and at lower pressures, than an empty container of the same volume.

However, the situation that is described actually is a result of solvation, not adsorption, and the volume-reducing effect has been discussed in some. other. questions, have a look around!

In essence, adsorption is only discussed in the context of the interface between two phases, such as gas-liquid, gas-solid, or two immiscible liquids.

If someone were feeling really pedantic, they could make an argument that the solvated molecules in the liquid constitute a really small "adsorbent" particle, and the solvent molecules in the first solvation shell are "adsorbates", and more power to them. Really, the forces that are at work in both cases (solvation and adsorption) are very similar.

It might be an interesting study to investigate where the crossover between adsorption and solvation happens as a function of molecule size. After all, graphene is just a really big molecule and its adsorption properties are pretty great. A brief search on adsorption solvation crossover reveals that there is some investigation that has been done on the process of going from a 3D gas $\ce{->}$ 2D adsorbate $\ce{->}$ aqueous solvate here.

tl;dr adsorption does occur at the interface between immiscible liquids, but the OP's article is talking about solvation


How can I perform calculations including the relativistic approximation SC-ZORA and solvent approximation COSMO in Gaussian 09? I am working on transition metal centered oganometallic complexes.

How can I apply or run the relativistic and non relativistic zero-order regular approximation (SC-ZORA) using Gaussian 09 in a DFT calculation?

I would also like to know about the process of performing COSMO analysis in Gaussian.

I have only available Gaussian 09 package for this purpose.

Gaussian can’t effectively treat relativistic effects (look up integral=dkh [1] and try its limited use. AFAIK you can only have 50 parameters in internal coordinates which is not very realistic for common molecules today.) and has no ZORA implemented. It is also not able to do COSMO PCM calculations. Instead one would use other PCM schemes, e.g. IEFPCM which is the default in Gaussian.

So can you perform your calculations with Gaussian 09? Unfortunately not.

Instead, you should change to other programs that fit your needs better. As it is free for academic use [2], you can have a look at ORCA and if it fits to your needs.

[1] http://gaussian.com/integral/
[2] read the licence carefully, it is quite strict


Chromyl chloride test side products I have read the test for chloride ions (chromyl chloride test). The reaction is:
$$\ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 + 2KHSO4 + 4NaHSO4 +3H2O}$$
My doubt here is if chlorine gas is evolved in a side reaction. Is it evolved in detectable amount? The side reaction is:
$$\ce{6Cl- + (Cr2O7)^{2-} + 14 H+  → 3Cl2 + 2Cr^{3+} + 7H2O}$$
Can someone please confirm this?

If you write the ionic equation, it seems that chlorine gas will produce. However, we see their half equation and standard potential.

$$\ce{ Cr2O7^2- + 14H+  + 6e-  ->  2Cr^3+  + 7H2O}$$
Standard potential: $E^\circ = \pu{+1.33V}$

$$\ce{ 6Cl-  ->  3Cl2 + 6e-}$$
Standard potential: $E^\circ = \pu{-1.36V}$

Thus, the standard potential cell for this ionic you say is equal to  $$\pu{+1.33V} + (\pu{-1.36V}) = \pu{-0.03V}$$

Because the standard potential cell is negative, so, it is call not energetically feasible or this reaction will not happen.

P.S.: The explanation only work under standard conditions.


The effect of pressure on boiling point? My textbook states the following:

Qualitatively, I understand why the boiling point of a substance increases when the pressure is increased. However, I learned that if the pressure is increased when the reaction system is at equilibrium, the volume of the system decreases and hence, the reaction will proceed in that direction in which the volume of the system increases. In the above scenario, however, it has been given that water in the gaseous form occupies more volume than its liquid form and so, shouldn't the given reaction proceed in the forward direction, hence reducing the boiling point?

Please note that I am convinced and I believe that it makes sense for the boiling point to increase when the pressure increases but can somebody please explain the flaw in my reasoning with regard to the given equilibrium?

Much thanks in advance :) Regards.

Your flaw in the logic lies in that you think that the volume of the system includes the liquid. Le Chatelier's principle on pressure equilibrium only applies to when the compound involved is in gaseous phase, in an ideal sense. Therefore if you increase the pressure (decreasing the volume), then the reaction will progress towards the direction where gas volume DECREASES, which is towards liquid phase. Overall, the entire apparatus increases the vapor pressure on the gas (gas pressure on the external pressure), which increases the boiling point.


Can Polyethylene Dissolve on 50-60 °C I know that polyethylene's melting point is between 115–135 °C, however I would like to know if can it start dissolving between 50-60 °C?

In a real example, if I put warm food (straight from the oven, I think it can be max. 50-60 °C) into a re-closable polyethylene plastic bag and close it, will any chemical reaction affect the food (from the polyethylene's side)? Or is polyethylene safe at that temperature, and if the bag is warm it doesn't mean that any chemical reaction is started? I didn't experienced any alteration, but I'm not an expert and don't know much about chemistry yet, so I would appreciate some guidance.

There may be better answers forthcoming, but a chemical does not need to melt (i.e. be in the liquid phase) to be reactive. For example, your food very likely has some liquid component, and that could be enough to react with some solid. Now polyethylene is used all over the place and is very nonreactive in many non-rigorous situations (like food service), so you are probably just fine. But I wanted to make the more general point that you don't have to be a liquid or gas to be reactive.


Is it possible to perform ROMP2 numerical optimizations in Gaussian09? I wish to perform an restricted open-shell MP2 (ROMP2) geometry optimization.

Gaussian09 currently only provides energies at ROMP2 level of theory.

It should be possible to perform an optimization by displacement of the structure followed by single-point-energy evaluations at the displaced geometries to obtain a gradient. In other quantum-chemical software packages this (slower) numerical optimization technique is readily available. I cannot seem to get it to work in the case of ROMP2 in Gaussian09. I have already tried the usual options, such as #p opt roMP2/cc-pVDZ scf=tight and so on. All of my approaches failed and yielded the following error:

 NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-
 NUMERICAL EIGENVECTOR FOLLOWING MINIMUM SEARCH
 INITIALIZATION PASS

 ************************************************
 ** ERROR IN INITNF. NUMBER OF VARIABLES (  0) **
 **   INCORRECT (SHOULD BE BETWEEN 1 AND 50)   **
 ************************************************

I am aware of this website: http://www.somewhereville.com/?p=1571
and I have tried the solution outlined therein, but it has failed in the same manner as well.

Any help or pointers towards other software that is capable of ROMP2 optimizations would be appreciated.

I've seen this error when I want to combine an optimization with an orbital deletion (NBODel) in Gaussian. It cannot handle an optimization with more than 50 degrees of freedom, so you need to freeze as much as you can, or pre-optimize your molecule with a more traditional method and then freeze whatever you can and do the ROMP2 optimization on the parameters of interest.


Mechanism of formation of 2-naphthol red dye (aka Sudan 1) As you can see in the image diazonium can attack at 2 different positions to form different products then why only the first product is correct?
As there are two ortho positions available why does it attack only one?

I tried to draw the intermediates see if they are useful and correct

is it because there are 7 intermediates in the first case?
or steric factors play some role here?

The most important thing to point out here is that your arrows are backwards. The diazonium is not the nucleophile. It is the electrophile. The diazonium ion has one lone pair. If it were a nucleophile, there's no way that you'd end up with the diazo compounds drawn, where each nitrogen has a lone pair.

This also means that drawing out all of your resonance structures doesn't mean anything because that is not the intermediate. The real question to ask here is which site on the naphthol ring is more nucleophilic.

Consider the naphthol as an enol for the purposes of the addition. Your two starting structures are exactly the right ones for focus one. The unsaturation in the enol in the top example is such that the nucleophilic site is benzylic while, in the other case, it is far from the ring.

The intermediates after attack by an electrophile are shown (technically, the carbonyls should be protonated still, but I'm too busy to fix that now). Only in the top case, where we substitute adjacent to the other ring, is aromaticity still preserved in the other ring. This is why the substitution is strongly preferred there as opposed to the other position, where we proceed via an intermediate that is not aromatic.


Why doesn't self condensation occur in acetoacetic ester synthesis So, I came across a quite versatile method to synthesize aldehydes and ketones, its called the Acetoacetic ester synthesis.

Suppose I need some 2-pentanone. So to synthesize it via this method I can do this

But I'm more worried about adding a base to Acetoacetic ester. Won't self condensation occur between two molecules of Acetoacetic ester like this?

If it occurs how do we minimize it? If it doesn't then why?

Any help will be appreciated.

Self-condensation is typically not an issue with 1,3-dicarbonyl compounds, such as the one in your question. Why? Due to their acidity ($\mathrm{p}K_\mathrm{a}$ around 9 to 11), most 1,3-dicarbonyls exist in their enolate form in solution. Therefore, there is very little of the original electrophilic 1,3-dicarbonyl which the enolate can attack. Hence, the amount of self-condensation taking place is relatively little and usually not a concern.


How does a spectrophotometer measure absorbance based on the intensity of detected light? We know that $\displaystyle A = -\log\frac{I_t}{I_0}$, where $I_0$ is the intensity of light without a sample absorbing, and $I_t$ with the sample. For determining the absorption $A$, we need both $I_t$ and $I_0$, but how does a spectrophometer detector measure $A$ knowing only $I_t$ ? (The detector is placed after the sample in the optical path, so it does not measure $I_0$ in front of it.)
Does the detector measure $I_t$ or the ratio $I_t/I_0$ ?

There are two possibilities:

If there is only one optical pathway, you record i) the blank sample for all wavelengths of interest and store the detector's intensity information in function of $\lambda$ in a working memory.  Light passing the blank is then assumed to equate $I_0$. Then you record ii) the wavelength dependent intensities $I_\mathrm{abs}$ with your sample of interest, and report the subsequently corrected data as spectrum $\mathrm{Abs} = f(\lambda{})$.

.or.

Your instrument permits the simultaneous recording of sample and blank to record both intensity data and reporting the already corrected absorption in one run.  Be sure, though, to assert which of the two paths (= two sample holders) is about your sample, and the blank (which may depend on spectrometer and software).


Final temperature after caustic dilution: Interpreting the graph I found the below graph in the Occidental Handbook on Caustic.

I'm confused, what may be the parameter noted on the four curves? 80-100-120-140 F?

The final Temperature is on the y-axis & initial water T is fixed as 70 F.

The labels on the curves are the initial temperatures of the 50% caustic soda solutions.


Technical name for the holes (missing electrons) in a p-type semi-conductor? When a pure silicon crystal (giant lattice) is doped with boron atoms one of the four covalent bonds made between one boron atom and four surrounding silicon atoms is missing an electron as boron has three valence electrons. Is there a technical name for these holes (lack of electron in a covalent bond).

Is it acceptable to say that these holes are indirectly the charge carriers as they are allowing the movement of the electrons?

Furthermore, could you say that these holes are positively charged?(Hence the name p-type semi-conductor?).

Yes, the technical name is, in fact, an electron hole (typically referred to as simply a hole). It is considered suitable to describe these holes as electric charge carriers with a positive charge that is equal in magnitude to the negative charge of an electron. So, you have actually answered each one of your questions!


Units for dissociation constant and relationship to Gibbs Free Energy How does one "convert" between an equilibrium constant calculated from the Gibbs free energy for a dissociation reaction and experimentally determined dissociation constants?

$$
\Delta G^\circ=-RT\log(K_{eq})
$$

I think I understand the derivation of this equation and why the equilibrium constant should be unitless if it's defined in terms of activities.
But I have a binding reaction that has an estimated free energy change of $-32~\mathrm{kcal/mol}$, which yields a dissociation constant on the order of $10^{-24}$ at $298~\mathrm{K}$.
Is it correct to just multiply the equilibrium constant by the standard concentration (1 molar) to get a more typical dissociation constant with units? A yoctomolar dissociation constant seems kind of strange to me, but I'm not sure where I've made a mistake.

free energy change of -32 kcal/mol, which yields something on the order of 10^-24 at 298K.

Seems the main mistake is a sign error.  Also "log" must mean "natural logarthim" the way the equation in the question is written to be correct.

If the free energy change is negative, the equillbrium constant is $>1$

Should be $10^{\frac{32\frac{kcal}{mol}}{2.303\times0.00199\frac{kcal}{Kmol}\times{298K}}} = 10^{23}$


Major products of the reaction of 2-bromo-2-methylpropane and propanoic acid Here's my attempt:

the electrophile is a tertiary carbocation, so that rules out sn2. I believe the nucleophile is a rather weak base since carboxylic acids are stable. However, it is polar, plus the temperature is low. Thus it seems likely that it proceeds by SN1? So would "I" be the correct answer?

The commenters are correct; this reaction, based on the steric hindrance of the tertiary alkyl halide and the propionic acid, I would expect a mixture of SN1 and E1, and minimal if any second-order reactions to the mechanism.

Specifically, the propionic acid would slowly approach the bromide substituent, and the acidic carboxyl proton (pKa ~ 5) would begin to associate with the bromide, which is a good leaving group. This slow formation of a tertiary carbocation (stabilized by hyperconjugation of the central empty p orbital through the C-H sigma bonds) then leads to the propionate binding to the carbocation with no rearrangements occurring.

Even though this reaction is at room temperature, E1 should not be ruled out, as there is no reason why a carboxylic acid cannot grab a proton from the 9 available ones on the alkyl halide. Due to the necessity for the proper orientation for beta-elimination though, it is plausible that E1 is less likely.

Regardless, I would project both SN1 and E1 to be the most likely reactions to occur. However, if you were to pick only one, I would pick SN1 based on the reaction temperature (and call E1 a "secondary" major product if that makes sense).


Why do cotton clothes take a longer time to dry as compared to synthetic clothes? This may be in relation to the $\ce{-O-H}$ group present in cellulose, but I would like to have a more specific answer.

Cotton fiber is primarily cellulose:

The red atoms are oxygen, and the white atoms are hydrogen. Whenever you have hydrogen attached to something extremely electronegative (such as oxygen), there is a potential for hydrogen bonding to occur:

As you can see, the electronegative oxygen atom draws electron density away from the more electropositive hydrogen atom, which results in a very polar bond. The positive charge on one molecule attracts the negative charge on another, and a very strong physical bond forms.

Now compare cellulose with something like nylon:

Since water can hydrogen bond to cellulose, it "sticks" much better - we say that it is hydrophilic. Since it can't form hydrogen bonds to nylon, we say nylon is more hydrophobic.

The same is true for most synthetic polymer fibers, and as a result, cotton absorbs water much more readily than other fibers, and tends to "hang on" to the molecules for longer - which means cotton fibers take longer to dry.


Gaussian parameters for IP and EA calculation I have already calculated ionization potentials (IP) and electron affinities (EA) for a whole bunch of organic molecules (from around 20 to 25 heavy atoms, mostly carbon atoms, just a few oxygen or nitrogen atoms here and there). A few molecules are actually a little smaller (6 heavy atoms). I have scrupulously checked the geometry of each and every molecule using the following DFT parameters in gaussian 09:

# freq=noraman cphf=noread b3lyp/6-31g(d) geom=connectivity
 integral=grid=ultrafine scf=maxcycle=1000

and made sure that I got the stationary point in every case.

Then I follwed a published paper in which they applied Koopman's thoerem approximation using HOMO and LUMO energies as IP and EA values, respectively. My results look pretty reasonable in terms of comparing the molecular structures between themselves, and also in terms of the obtained experimental results from a biological system. However, I understand that one can always question the accuracy of a chosen computational method...

I would like to ask you if it would make sense for me to double check the computational results I got with a potentially more accurate method (basis set, functionals), and which one would you recommend in this case? It just so happens that I will have access to a pretty decent cluster (12 cores, RAM 12 GB, HDD 500 GB) with like 1000 hrs/core computation time.

Thanks in advance for any help you can offer.

If you can afford to do so, it is always a good idea to compare calculations between at least two methods. Good results with B3LYP/6-31G(d) in particular are based upon favorable error cancellation, and B3LYP was not even parameterized against any data in particular (B3PW91 was parameterized for thermochemistry and the PW91 correlation functional was replaced with LYP without changing any other parameters). So, double-checking is wise!

Modern perturbation theory-based approaches have become much more affordable due to the density fitting (DF) and resolution of the identity (RI) approximations (same thing, different name), combined with spin-component scaling which may or may not improve things. The scaling is now roughly $O(N^4)$ and a small prefactor with basis set size; compare this with B3LYP at $O(N^3)$, exact MP2 at $O(N^5)$, CCSD at $O(N^6)$, and CCSD(T) at $O(N^7)$.

Energies are one of the most obvious improvements of range-separated or long-range-corrected (LRC) density functionals. Here the authors find that LC-$\omega$PBEh deviates little from $G_{0}W_{0}$, which is a workhorse for IE and EA calculations. Perhaps the most common new functional that is seeing quite a bit of use in the computational organic chemistry literature is M06-2X, which is shown here to perform even better than most approximate coupled cluster methods for this set of molecules (which are all closed shell!):

With regards to the basis set size, there is only one correct direction to go and that is up. You may be able to afford CCSD(T)/6-31G(d) calculations, but the results will be meaningless due to a severely deficient basis set for anything but the simplest calculations. See here for a recent discussion about the use of basis sets in DFT versus wavefunction methods. Wavefunction methods have always required larger basis sets for quantitative results, especially when not performing a basis set extrapolation. This would mean something like Dunning's correlation-consistent basis sets, augmented with diffuse and polarization functions, such as aug-cc-pVDZ or aug-cc-pVTZ if one can afford it. An interesting and effective truncation of these basis sets (they become very large) is given by Truhlar's "calendar" basis sets.

Less well-known are the requirements for LRC-DFT, which due to modern parameterization schemes are larger basis sets and larger integration grids for the XC contribution, often larger than what is default. Even Pople's triple-zeta sets based on 6-311G will offer some improvement, but the Ahlrichs def2 and Jensen's pcseg-n are probably better.

Finally, a word about the cost of LRC-DFT. Although its formal scaling cost is identical to traditional global hybrids such as B3LYP, in practice range-separated functionals may be 2-3x as expensive due to how the different contributions are calculated and then combined, depending upon the package being used.


R configuration for cysteine I think my prof made a mistake. Shouldn't the answer be 1, 4 and 5 instead of 2 and 3?

If the programmers did not make a mistake it seems that you are right - ChemDraw is on your side.

$\hskip.5in$


Which are more reactive lanthanides or aluminium? I am not sure whether to consider the standard reduction potential or ionisation enthalpy.Which would be more reactive out of the lanthanides and aluminium

Since the lanthanides are neither in the same period nor in the same group as aluminium, comparision is actually not feasible. Factors like atomic size and effective nuclear charge tend to predominate such comparisions, rather than ionization enthalpy or reduction potential.

Bear in mind that due to presence of electrons in f-orbital for lanthanides provide poor shielding of electrons farther away from the nucleus. This poor shielding is due to the fact that f-orbitals are quite diffused in nature. Yet, we cannot say anything about reactivity of lanthanides when compared to aluminium.
If you are still a high-school student, I suggest you do not ask such hypothetical questions.


Why enthalpy of adsorption is negative? I have already read Is adsorption exothermic, and if so, why?  but it didn't answered my query.

To illustrate, let's take a hypothetical vacuum space. In that space, there is a solid and a gas, the gas is getting adsorbed over the solid. In that case, if we'll use momentum conservation, we'll get that the net kinetic energy of system has increased due to positive work done by the internal forces, but heat exchanged from system to surrounding or vice-versa is zero( as there is no outside interaction present. Therefore, dU>0 as U is sum of kinetic and potential energy of system( note that I haven't made the internal force as due to a field with a assigned potential function). We know dH=dU+PdV for processes in which the pressure outside the system is constant( which we have assumed). Therefore, as dV=0( because I had taken a fixed volume of space), therefore dH>0. But in books it is written dH due to adsorption is negative. Also, if I had assigned a potential function for internal forces, what change would have taken place? Please help me out.

I will point out another equation:

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$$

For adsorption $\mathrm{d}S<0$, as molecules go more organized.

Therefore also $\mathrm{d}H < 0$ must be true, otherwise $\mathrm{d}G < 0$, needed for a spontaneous process, would not be achieved.


Separating Lipids from Milk I am trying to develop a procedure to test the percent mass of lipids in (update: homogenized) milk. It doesn't need to be very accurate, just an interesting application of chemical principles. As I was looking on traditional ways this is tested, I came across the Babcock method, which involves pouring concentrated sulfuric acid into the milk, heating, and centrifuging the sample. According to Wikipedia, "key to this process is that everything in milk except the fat dissolves in sulfuric acid."

The lipids of are surrounded by what is similar to a cell membrane. The fat is an emulsion with these surfacants holding it in the water. The other components of milk are water, colloidal micelles, which hold other proteins (of no interest to me). My question is, how does sulfuric acid dissolve "everything else"? In a biological procedure I know of to break cell membranes, isopropyl alcohol is used. The cell membrane in that case is broken when the alcohol behaves as a surfactant toward individual triglycerides. However, sulfuric acid does not have a nonpolar region, so how does it work to dissolve the phospholipid membrane?

Many Internet resources on this topic are not written from a chemical perspective. In addition, they seem to confuse the lipid bundles with the other colloidal particles.

Also, why not just heat the milk strongly (such as with a Bunsen burner) to break the protein, or add electrolytes to it, as is a common way to separate colloids?

Milk is mostly water, and diluting it (roughly speaking) with sulfuric acid to a concentration of ~50% milk and 50% sulfuric acid will result in a very acidic, but still aqueous, solution.  We can examine the likely behavior of other milk components using basic chemical principles.

Sugars.  Milk is rich a variety of di- and oligo-saccharides such as the eponymous lactose, as well as many other more complex molecules:

  [Free oligosaccharides are natural constituents of all placental mammals' milk. Human milk contains 7–12 g/L oligosaccharides, making the oligosaccharide fraction a major component of human milk. Compared with human milk, the concentration of oligosaccharides in the milk of the most relevant domestic mammals is smaller by a factor of 10 to 100.]

Fortunately, the effect of sulfuric acid on the sugars is easy to predict: they will simply stay dissolved in aqueous solution.  Hydrolysis of the sugars is possible, but will not happen to any great degree without heating or prolonged incubation.
Proteins.  Milk contains thousands of unique proteins.  However, by far the most predominant proteins are called caseins.  Caseins are relatively hydrophobic, proline-rich molecules and in milk are present as tiny aggregates -- a colloid -- of molecules.  Adding sulfuric acid will denature the casein.  As a result, the casein molecules are likely to stick to each other and form a centrifugable pellet that goes to the bottom of a centrifuge tube after spinning.  Proteins contain amino acids linked via amide bonds that are unlikely to hydrolyze in the acid, unless very high temperatures or very long incubation times are used.
Phospholipids.  Phospholipids in milk include glycerophospholipids and sphingolipids.  Polar lipids play an important role in stabilizing the triglyceride (fat) emulsions that keep fat in solution.  They do this by forming bilayers and other structures on the surface of triglyceride microdroplets.  To do this job, phospholipids have a phosphate head group attached to a fatty acid tail.  The fatty part of the phospholipid sticks to the triglyceride, and the polar head group interacts with aqueous solution.  The polarity of the head group is tunable by pH.  At neutral to mildly acidic pH, the phosphate head groups are (negatively) charged, greatly strengthening their propensity to interact with water and maintain emulsion stability.

Phospholipids may be partially hydrolyzed by the acid, but without very high temperatures or prolonged incubation, most molecules will remain intact.

The strong acid conditions that arise from adding $\ce{H2SO4}$ will protonate these head groups, neutralizing the negative charge and vastly weakening their emulsifying properties.
Triglycerides.  Triglycerides are the molecules that provide most of the fat content of milk.  They are triacylglycerols.  Unlike phospholipids, they have no polar head groups, and thus they do not dissolve in water.  They would "want" to form a separate, oily phase from the water (just like mixing vegetable oil and water), but the fact that they are usually bound in tiny droplets by phospholipids prevents this separation.  Because triglyceride oils are less dense than water, if they did ever get the chance to coalesce into large droplets, they would rise to the top of the milk.  This happens when cream naturally separates from fresh (raw) milk, but in homogenized milk, the droplets are artificially forced to much smaller sizes, effectively preventing the droplets from coalescing or rising.

Upond acidification, the loss of phospolipid surfactant activity breaks down the stability of the fatty droplets, allowing them to rapidly coalesce, so they rise to the top of the tube, a process that is only accelerated by centrifugation.

Hydrolysis of triglycerides in the acid is possible, but unlikely.  In addition to required prolonged incubation or high temperatures, keep in mind that the separated triglyceride phase is actually not exposed to the acidic solution very much at all.


Is HF the only method for trying to solve the many-electron Schrödinger equation? The Schrödinger equation can only be solved analytically for the smallest of "molecular" systems. The Hartree-Fock method is a method of obtaining approximate solutions to the many-electron Schrödinger equation, and a bunch of methods try to reduce the error associate with the HF method.

However, all ab initio electron correlated methods I know of and have read about are based on HF; that is, they all use the HF solution and adds to it. So basically there is only one way in use today of attempting to solve the many-electron Schrödinger equation?

Then of course there is density functional theory, but here you strictly don't try to solve the Schrödinger equation, you just try to obtain the solution, right?

So is this a correct assessment? Do we truly only use one method of trying to solve the many-electron Schrödinger equation? Is there any research being done on alternative methods (other than DFT)?

So is this a correct assessment? Do we truly only use one method of trying to solve the many-electron Schrödinger equation? Is there any research being done on alternative methods (other than DFT)?

No, it is not correct. Saying that Hartree-Fock(HF) theory is the only method is falling short. Hartree-Fock is only one of many existing methods. While it is true that some single reference methods have a dependence on the HF solution (because they build upon it) when they include a reduced number of determinants, it is not the case for other. Particular mention deserves Coupled Cluster since it is insensible to the reference determinant found by HF. The proof is given by Thouless theorem (see Joshua Goings' post, or via The Internet Archive).
Thus, to give you an overview of the grand landscape of possible approaches I tried to compile the families of methods better established nowadays. I did not intend to specify individual methods because there are so many ... and people continue to investigate new ones. These are:

Deterministic

Semi-empirical (non ab initio): Hückel1/Tight binding2, Hubbard, ...
Wavefunction methods [Trygve Helgaker, Poul Jørgensen, Jeppe Olsen:
Molecular Electronic‐Structure Theory. John Wiley and Sons, Ltd: 2000. DOI: 10.1002/9781119019572]

Hartree method. Caution: mathematicians use this name for a related method.
Hartree-Fock (HF): can be Restricted (RHF), Restricted Open (ROHF), Unrestricted (UHF) or Generalized (GHF).
Configuration Interaction (CI), up to order $N$, which is equivalent to Full-CI1/exact diagonalization2/direct variational approach3.
Multi-Configuration Self Consistent Field (MCSCF). [Chem. Rev. 2012, 112 (1), 108–181.]

Complete Active Space (CASSCF).
Two Configurations.
Manual selection of configurations or other.

Multi-Reference Configuration Interaction (MRCI)
CASPT2, NEVPT2, partially (PC) and strongly contracted (SC)
variants of the N-electron valence state second-order
perturbation theory
Multi-Reference Coupled Cluster methods (MRCC)
Many Body Perturbation Theory methods (MBPT) [Isaiah Shavitt, Rodney J. Bartlett: Many-Body Methods in Chemistry and Physics, MBPT and Coupled-Cluster Theory. Cambridge University Press: Cambridge, 2009. DOI: 10.1017/CBO9780511596834]

Coupled Cluster methods and its simplifications
Möller-Plesset perturbation theory of order $n$ (MPn)
Random Phase Approximation (RPA)
GW-Approximation

Explicit correlated methods R12/F12 [Chem. Rev. 2012, 112 (1), 75–107.]
Valence Bond Theory [Sason Shaik, Philippe C. Hiberty: A Chemist's Guide to Valence Bond Theory. John Wiley & Sons, Inc.: 2008. DOI: 10.1002/9780470192597]

Density Functional Theory (DFT). [Robert G. Parr, Yang Weitao: Density-Functional Theory of Atoms and Molecules. Oxford University Press: USA, 1994. ISBN: 9780195092769]

Thomas-Fermi theory. Orbital-free methods.
Kohn-Sham. Many density functional approximations developed, families are:

LSDA
GGA, hybrid GGA
MGGA, hybrid MGGA
OEP

Pair density functional theory [Comput. Theor. Chem. 2013, 1003, 91–96.]
Natural Orbital Functional Theory (NOFT) [S. Goedecker, C. J. Umrigar: Natural Orbital Functional Theory. In: J. Cioslowski (eds) Many-Electron Densities and Reduced Density Matrices. Mathematical and Computational Chemistry. Springer: 2000, Boston, MA. DOI: 10.1007/978-1-4615-4211-7_8]
Geminal Functional Theory [J. Chem. Phys. 2000, 112 (23), 10125–10130.]
k-particle density cummulants methods [Int. J. Quantum Chem. 2003, 95 (4-5), 404–423.]
Density Matrices methods

Density Matrix Renormalization Group (DMRG) [Annu. Rev. Phys. Chem. 2011, 62 (1), 465–481.]
Density Matrix Embedding Theory. Can be combined with several methods. [J. Chem. Theory Comput. 2016, 12 (6), 2706–2719.]
One-body Reduced Density Matrix Functional Theory (1-RDMFT) [J. Chem. Phys. 1985, 82 (12), 5604–5607.]
Variational 2-RDM Theory [Phys. Rev. Lett. 2012, 108 (26).]

Dynamical Mean Field Theory (DMFT). [J. Chem. Phys. 2011, 134 (9), 094115., arXiv:1012.3609 [cond-mat.str-el]] Starts with a Hubbard model of a lattice problem, then connects it to an impurity problem that can be solved with other methods.

Non-deterministic (stochastic)

Quantum Monte Carlo [Chem. Rev. 2012, 112 (1), 263–288.]

Variational Monte Carlo (VMC)
Fixed-Node Diffusion Monte Carlo (FNDMC).
LR-DMC, lattice-regularized diffusion Monte Carlo
Self-Healing Diffusion Monte Carlo
Green Function Monte Carlo (GFMC). GF2, self-consistent second-order Green’s function
Auxiliary Field Quantum Monte Carlo (AFQMC)
Reptation Quantum Monte Carlo
Full CI Quantum Monte Carlo (FCIQMC) [J. Chem. Phys. 2009, 131 (5), 054106.]
BDMC, bold diagrammatic Monte Carlo
Time-Dependent Quantum Monte Carlo

There are also mixtures of various methods.
The methods above have been tested with molecules by the quantum chemistry community but there are others that have been proven only for one dimension
or toy models. Operator methods are one example.

Many of them try to approximate the Full CI tensor that is described in a basis of determinant functions. It is impossible to treat such tensor for medium to large systems so the point is to take the minimum number of determinants that contribute significantly. The difference lies in how they choose those determinants. There are also other important distinctions among them, like if they are variational, perturbative or none, size-extensive, size-consistent, ...

In summary, each has its advantages and disadvantages. I recommend anyone interested to read
SOLVING THE SCHRÖDINGER EQUATION. Has Everything Been Tried? 2011, Imperial College Press. Ed.: Paul Popelier.

Furthermore, the analytic properties (read regularity) of the wavefunction can be exploited to devise new methods. As Harry Yseretant puts it in his book:

  Approximating the solutions is thus inordinately challenging, and it is conventionally believed that a reduction to simplified models, such as those of the Hartree-Fock method or density functional theory, is the only tenable approach. This book tries to convince the reader that this conventional wisdom need not be ironclad: the regularity of the
  solutions, which increases with the number of electrons, the decay behavior of their mixed derivatives, and the antisymmetry enforced by the Pauli principle contribute properties that allow these functions to be approximated with an order of complexity which comes arbitrarily close to that for a system of one or two electrons.

Also, I would like to note that there is a collaborative project under the hood of the Simons Foundation whose purpose is precisely finding better methods to solve the electronic structure problem. And, of course, there are many more people working on the same task.

Footnotes

Naming adopted by the chemistry   community
Naming adopted by the physics     community
Naming adopted by the mathematics community

References

Mario Motta, David M. Ceperley, Garnet Kin-Lic Chan, John A. Gomez, Emanuel Gull, Sheng Guo, Carlos A. Jiménez-Hoyos, Tran Nguyen Lan, Jia Li et al. Towards the Solution of the Many-Electron Problem in Real Materials: Equation
of State of the Hydrogen Chain with State-of-the-Art Many-Body Methods. Phys. Rev. X 2017, 7, 031059. DOI: 10.1103/PhysRevX.7.031059


Why does phosphine have a dipole moment and a higher boiling point than carbon tetrafluoride? Phosphine, PH3, and carbon tetrafluoride, CF4, are small molecules of a similar size and the same mass of 88 au. CF4 has a dipole moment of 0, which is unsurprising given its tetrahedral shape. However, I would expect P-H bonds to be non-polar (P and H have electronegativities of 2.19 and 2.2) and therefore the molecule to also have a dipole moment of about 0, despite being trigonal pyramidal and therefore asymmetrical. So I would expect it to exhibit similar intermolecular forces and a similar boiling point to CF4. Yet PH3's dipole moment is quoted as 0.58 D. And PH3 has a boiling point of -88 °C and CF4's is -128 °C.

Why does phosphine have such a higher dipole moment and boiling point than CF4 despite it having non-polar bonds?

The dipole moment of $PH_3$ can be attributed to the lone pair on P, which is directed away from all three P-H bonds. You can imagine that there is a higher electron density near the lone pair, while the electron density is more or less uniform near the nonpolar P-H bonds. This does not happen for $CF_4$, so it has no dipole moment.


Reaction of cinnamic acid with Br2/Na2CO3 The first reaction produces benzaldehyde, and the next one (perkin's condensation)produces Cinnamic acid.(X)
Now the treatment of X with $\ce{Br2/Na2CO3}$ is whats troubling me.  $\ce{Na2CO3}$ being a base, abstracts the hydrogen from the  $\ce{COOH}$ group.  $\ce{Br2}$ reacts with the alkene portion to yield a cyclic intermediate. What next?
The solution claims that somehow The  $\ce{CO2-}$ group leaves and a bromoalkene forms. The addition Of moist KOH in the next step results in  $\ce{E2}$ elimination to yield option (c).
I cant quite digest the $\ce{CO2-}$ group leaving. I thought of a mechanism, which is similar to the syn-elimination in esters: but that doesnt work out well..
Any hints/ insights will be appreciated. The correct answer is (c).

The sodium carbonate is just there to mop up any $\ce{HBr}$. The $\ce{Br2}$ adds across the double bond to give cinnamic acid dibromide.
The treatment of cinnamic acid dibromide with $\ce{KOH}$ at elevated temperature eliminates 2 eq. of $\ce{HBr}$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.