text:
How easily should 304 stainless steel dissolve in a salt and hydrogen peroxide solution?
This is a simple question, although I don't have any exact numbers.
I was looking to buy a food-safe stainless steel sieve, and one of the reviewers showed the sieve they purchased heavily rusted and "punctured" after testing it in "hydrogen peroxide with salt as a catalyst", claiming it thus cannot be 304 stainless steel as stated by the seller.
I don't know what salt concentration that person used (let's assume >1%) and for how long was the piece submerged (probably no longer than several hours), but from what I read online – while it should be hydrogen peroxide-resistant, 304 stainless steel is not considered salt-resistant.
So considering a sieve's large relative surface area, plus the presence of salt (whose corrosive function is augmented by the hydrogen peroxide or not?), is it safe to presume the reviewer did not use a suitable test to claim the material is not 304 stainless steel?

related https://chemistry.stackexchange.com/questions/139905/rust-on-stainless-steel-caused-by-salt-solution

Of course it's not a suitable test? To tell what a material is you need to test for composition, structure. People are imagining that stainless steel cannot corrode - they are wrong.

@mithoron I'm not saying that this test could equate to a molecular level of testing, but as a general assessment how incompatible are the results for 304 stainless steel verification? The salt alone, unless used in huge amounts, probably shouldn't corrode that much that quickly, but I wondered if combined with the hydrogen peroxide it does create a more prominent corrosion effect that's expected to dissolve a sieve very quickly. Also related, several reviewers mentioned it is quite magnetic, unlike 304, but I know 304 is usually slightly paramagnetic so could the threaded form increase that?

Go to a nice, local kitchen products store. Buy a good sieve. If it rusts, return it.

@joncuster I need something with a very fine mesh in order to smoothen liquid solutions, and all those available locally seem to have the regular >1mm holes.

Depends on specifics. The bulk of food processing facilities are 316 and 304. The 316 and 317 are more resistant to pitting but the equipment is cleaned and dried on various schedules. All these alloys will rust exposed to salts. I am unaware of many higher alloys being used for food as in the chemical industries but I have no direct experience in food industry. PS : a magnet will indicate the austenitic stainless like 304 and 316.

I understand they are more or less salt rusting-prone, but the issue is whether the extreme rusting and dissolving demonstrated in the reviewer's images could be due to the hydrogen peroxide reacting with the rust. Because otherwise, even if the solution had several dozen % of salt, how quickly should a 304 sieve dissolve? Also regarding (para)magnetism, I did mention in a comment a couple of reviews said it was uncharacteristically magnetic, but the sieve is probably made of drawn wires and it is written this could induce paramagnetism.

The austenitics are not ferromagnetic when annealed ( 316 may show a trace). They become increasingly ferromagnetic as they are cold worked/strain hardened. It is very difficult for me to imagine 304 dissolving in salts at  ambient temperatures.

The drawing process is a colf working one, no? The question is whether the high surface area of a sieve + the presence of hydrogen peroxide (I read it interacts with rust, but I'm not sure if it's true for 304) could cause quick dissolving.

Small particles may dissolve , but I am expecting a sieve to be about 1 mm thick ; it would only have surface rust.

Well, I've added the pictures added to that review, if that provides any information. I found several videos demonstrating instant rusting using hydrogen peroxide, salt and an acid of sort, but I didn't find a video demonstrating the effect without an acid.


text:
On the energies of molecular orbitals
I found a useful (though simplified) scheme illustrating the energies of atomic orbitals, in crescent order:

I wonder if there was an analogous (simplified) scheme or rule for molecular orbitals, at least for diatomic homonuclear molecules, for which the electronic state is given by the so-called molecular term symbol
$$\displaystyle {}^{2S+1}\!\Lambda _{\Omega ,(g/u)}^{(+/-)}$$
where
${\displaystyle S}$ is the total spin quantum number
${\displaystyle \Lambda }$ is the projection of the orbital angular momentum along the internuclear axis
${\displaystyle \Omega }$ is the projection of the total angular momentum along the internuclear axis
${\displaystyle g/u}$ indicates the symmetry or parity with respect to inversion through a centre of symmetry
${\displaystyle +/-}$ is the reflection symmetry along an arbitrary plane containing the internuclear axis
Such scheme or rule would justify for instance the following succession:
$$\sigma_g(1s)<\sigma_u(1s)<\sigma_g(2s)<\sigma_u(2s)<\pi_u(2p) <\sigma_g(2p)<\dots$$

I don't know if that's exactly what you meant. But, how about this from https://pressbooks.online.ucf.edu/:
Chemistry Fundamentals by Dr. Julie Donnelly, Dr. Nicole Lapeyrouse, and Dr. Matthew Rex

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Why do some hydrates need to be heated over 100 Celsius to evaporate the water?
If the boiling point of water is at 100 °C, why does a hydrate need to be heated higher than that to evaporate water from the crystals? Does it have anything to do with their intermolecular forces?

For the same reason that some hydrates lose their water below 100°C: They don't know what liquid water is, and they also don't resemble it in any way.

Hydrates are substances that have absorbed water into their crystal structure.  Much of this attraction is caused by interparticle forces (notice I didn't say 'intermolecular').  Since water has a very strong dipole, it is quite often attracted to not only other molecules, but ions inside ionic crystals.
In some cases, water can easily be attracted into the crystal structure so much that it makes the substance deliquescent.  Other times, crystals may only absorb some water, nor others.  This is because of the competition in attractions between an ionic crystals ion-ion attractions (ionic bond) and the ion-dipole attractions to water (interparticle force).
To finally discuss why we have to go above the boiling point of water $100^\circ C$:  If the attraction between the dipoles of water and ions of the salt are stronger than the standard interparticle forces of water, it will take more energy to break those forces and transition them to gaseous state.
Likewise, if a substance has quite weak ion-dipole forces, it would more readily release the water and dehydrate without much or no added heat.

I did indeed notice you didn't say "intermolecular". ... I have no clue why you felt it was critically important to scrupulously avoid using that term here, but (due in no small part to the presence of the vaguely worded parenthetical) I did notice.

@R.M. While water indeed is made out of molecules, I suppose the term intermolecular was avoided to avoid calling ions in a lattice molecules.

@R.M. It is probably a bit dogmatic of me to use that term and to be quite opinionated about it.  Teaching students forces of attraction reveals the blurry lines of bonds/not bonds or intermolecular/interparticle attractions tends to be a learning blocker.  Using more generic words, especially in cases like this question where the reader might be learning these topics for the first time helps them to understand attractions can occur between different materials/compounds and we shouldn't box in specifically this idea into "only molecules".

While it may be a useful simplification, treating water in crystal just as separate molecules isn't really correct. Sometimes it's indeed hydrogen bonded water, but usually it's coordinated to metal.

And in either case, the crystalline substance and it's hydrate are two different chemical compounds. Chladrathe compounds are a bit different, as they don't even have a loosely related base form of the matrix substance.

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text:
Can Vdp work be considered irreversible work (e.g., production of entropy)?
Since conduction across a finite temperature difference will produce entropy, will work done due to pressure difference generates entropy because system tends towards mechanical equilibrium which indicates a spontaneous process.
Unlike VdP work, PdV work does not generate entropy (reversible) since it is in mechanical equilibrium as there is no pressure difference. For instance a system with oscillating boundary.

Pressure work by a piston against a constant external pressure is not reversible.

VdP work doesn't need to generate entropy.  It can happen reversibly.

Yes, you describe one of many sources of "friction"
Friction is any irreversible process.
Friction generally takes on three forms:
"Dry" friction applies an (approximately) constant force regardless of velocity. Mechanical parts directly rubbing against each-other.
"Resistor" friction acts in proportional to velocity (or electric current). Such as sliding machinery lubricated by an oil film, a mammoth getting stuck in a tar pit, or electrical resistance in a wire. Heat transfer over a finite temperature is another example, but instead of energy lost to heat free energy is lost.
"Snow plough" friction is proportional to velocity squared. Energy is lost in pushing the mass of fluid out of your way high speed. The swirling masses left in your wake later dissipate energy as heat.
How your vdP example dissipates energy.
Mechanical friction in a piston means that we have to apply an extra dP to get/keep it moving, overcoming "dry" and/or "resistor" friction depending on oil levels. VdP is lost to heat at the seal. If the piston is expanding, the heat from the seal can recover some of this energy as work in expansion, but there is still a net loss.
Gas speed of sound if the piston expands very fast (such as half the speed of sound), the gas pressure will be lower at the piston head and less work is done. Conversely, more work would be needed for compression. In this case dP is the pressure difference between the head pressure and the ideal head pressure. VdP is lost to heat in the gas, cooling it off less upon expansion and heating it up more upon compression. This is a form of "snow-plough" friction, although it is complicated by transonic effects.

VdP describes the shaft work done in a continuous flow process for an open system if the process is occurring isentropically (i.e. reversibly).  PdV describes the P-V work occurring in a closed system process, and, if this process is reversible, P can be determined from the equation of state of the fluid; otherwise, P must be specified by way of external constraints.


text:
Does concentrated HNO3 oxidize aluminium to form a protective oxide layer, or is the oxide layer already present so HNO3 is unable to react?
Sources on the internet like this one only say that aluminium does not react with concentrated or dilute $\ce{HNO3}$.
I have seen other related questions on this site but none answers my question specifically. I have read that aluminium on contact with air forms a protective oxide layer, so this layer should protect it from conc $\ce{HNO3}$.
Does conc $\ce{HNO3}$ "further oxidize" aluminium to aluminium oxide?

Please use proper punctuation in the future.

Standard aluminum etch is on the order of 25:5:1 phosphoric-acetic-nitric, with additional nitric increasing the etch rate.

@SafdarFaisal This is definitely a duplicate, but it's a shame that the answer to the post you link to doesn't cite more authoritative references.

The aluminum oxide layer ($\ce{Al2O3}$) produced by air oxidation on metallic aluminium protects the metal against further oxidation by both air and nitric acid.


text:
The vapour density of $N_2O_4$ at certain temperature is 30.Calculate the percentage of dissociation of $N_2O_4$ at this temperature?
The vapour density of $N_2O_4$ at certain temperature is 30.Calculate the percentage of dissociation of $N_2O_4$ at this temperature.$\ce{N2O4_{(g)} <=> 2NO2_{(g)}}$?
I am unable to understand the concept behind vapour density of a the mixture.
Currently I understand that
2 x vapour density=molar mass.
vapour density =
mass of n molecules of gas ÷ mass of n molecules of hydrogen.
vapour density
= molar mass of gas ÷ molar mass of H2.
I am unable to apply the above formula because a mixture does not have molar mass.
And I am also not able to understand that if
2 x vapour density=molar mass
then in the question molar mass of $N_2O_4$   at the certain temperature given   would be 60 instead of 92.
The correct answer is 53.33%

The big mistake you made was assuming a mixture does not have a molar mass.
Molar mass for a mixture is calculated by using the mole fractions and molar masses of each constituent.
Let:
$A$ represent $\ce{N2O4}$
$C$ represent $\ce{NO2}$
Then the reaction becomes:
$$\ce{A<=>2C}$$
First, we calculate the molar mass of the mixture using the given vapor density:
$$M=2v=(30)(2)=60\;g/mol$$
Then, we can set up the following system of equations in terms of molar masses and mole fractions:
$$X_A\;M_A+X_C\;M_C=M$$
$$X_A+X_C=1$$
Substituting all known values, the system looks like this:
$$92\;X_A+46\;X_C=60$$
$$X_A+X_C=1$$
Solving this system, we get the equilibrium molar fractions:
$$X_A=0.3043$$
$$X_C=0.6957$$
Then, we can calculate $K_X$, the equilibrium constant in terms of molar fractions:
$$K_X=\frac{X_C^2}{X_A}=\frac{0.6957^2}{0.3043}=1.5905$$
Finally, we use the relationship between $K_X$ and dissociation fraction $\alpha$ for this reaction to calculate it:
$$K_X=\frac{(2\alpha)^2}{1-\alpha^2}=1.5905$$
Solving for $\alpha$:
$$\alpha=\sqrt{\frac{K_X}{4+K_x}}=\sqrt{\frac{1.5905}{4+1.5905}}$$
$$\alpha=0.5333$$

"First, we calculate the molar mass of the mixture using the given vapor density:
M=2v=(30)(2)=60g/mol".but the vapour density of value 30 was given for N2O4 and was not given for the mixture of N2O4 and NO2.so doesn't 60g/mol indicate the molar mass of N2O4 at equilibrium instead of the mixture. How did you conclude that 60g/mol indicate the molar mass of the mixture.

The molar mass of $\ce{N2O4}$ is 92, and 92≠60, so that means it has to be a mixture. Even if we assumed no $\ce{NO2}$ was initially present before decomposition, that would mean $X_A$=1, $X_C$=0 , and we would still get the contradiction: 92≠60. Also, there's no such thing as "molar mass at equilibrium". Molar masses are the same regardless of whether we're at equilibrium or not.

Another way to look at it is this: $\ce{N2O4}$ is 92, and $\ce{NO2}$ is 46, so it makes sense that for a mixture of them, the molar mass of the mixture (60) has to be in between 46 and 92.


text:
Can I make bleach for laundry from sodium dichloro-s-triazinetrione
According to The Organic Prepper, one can generate bleach from pool shock.

The good news is, you can make your own bleach with granular calcium hypochlorite, the primary component of typical pool shock. It’s essential to confirm that your pool shock is not contaminated by other chemicals, such as algicides. Only HTH is acceptable for purifying your water.

However, this uses calcium hypochlorite ($\ce{CaOCl}$). Upon research, I found out that there are three types of pool shock available - calcium hypochlorite, sodium hypochlorite and sodium dichloro-s-triazinetrione. The shock that I have is sodium dichloro-s-triazinetrione and not calcium hypochlorite.
Is it possible to make bleach (or something that does a similar action) using this sodium dichloro-s-triazinetrione?

Steps they used were adding 1/2 tsp pool shock to 1 gallon water. Did not know there are 3 types of shock.

If you simply want to clean clothes, you can use all these solutions. Sodium hypochlorite, calcium hypochlorite, dichloro-s-triazinetrione have all the same oxidizing effect in solution. They will all oxidize and destroy stains on dirty clothes.

Thank you Maurice!!! I have small dogs with small bladders and use belly bands (diaper for male dog peeing) for overnite, so not even important wash, and have enough pool shock to make $1000 worth of bleach then.


text:
Nichrome alloy corrosion at high temperature
We are currently using a nichrome (80/20) wire based heater to surround a sample chamber that we wish to heat by radiation. Maximum sample temperature is 400degC in an atmosphere of air that may be high humidity. That means the wires get red hot. Our specific problem is that there is corrosion occurring that alters the emissivity of the wires, making temperature control difficult.
Is there a better alloy, or is there some kind of coating we can apply which will not alter through thousands of hot/cold cycles?
The type of thing I am thinking about is (for example) platinum coating the wire. Alternatively, any suggestions for a solution?

Nichrome is what's generally used in space heaters.

Depending on what your sample chamber looks like, you may want to look into the heating tapes that are typically used for baking out vacuum systems. Very brief search gives you this, in the $100 range, rated up to 500 C if installed permanently (says it requires a controller as well but you could probably hack something together for less, depending on relative value of time and money to you) https://www.briskheat.com/applications/process-temperature-control-and-maintenance/vacuum-bake-out/heavy-insulated-heating-tapes-bih.html

@llama Thanks, but it must be non contact

Nichrome routinely operates at red heat (roughly 1500 °F) with very long life. 400 °C is no problem, moisture is no problem. Look for some contamination if you are getting corrosion.

Well, the wire is certainly not bright and shiny after a number of hours of use

Nichrome generally has a black surface ( presumably oxide). Such as an electric stove heating element. New stock from the mill is  dull grey; this alloy is not polished like a stainless.

Our wire is quite bright and polished before we insert it into the heater.

The starting material would be cleaned to avoid wear in the wire dies. Then in drawing the finished wire has a natural burnished surface .  And possibly the finished wire is polished. New stock we had was primarily sheet, no  wire.


text:
Why all commercial SPR instruments use angular interrogation?
Surface plasmonic resonance biosensor devices are used to measure analyte concentration, kinetics, affinity, and specificity without any need for labels. This technique involves the bonding of the analyte to a functionalized metal surface (typically gold) in contact with a dielectric medium (glass). The interaction of a light beam with the gold surface changes based on the presence or absence of the analyte which affects the plasmon resonance properties. These changes are then measured to provide quantitative data.
Almost all commercial SPR instruments use angular interrogation, mesuring the angle at which the light beam excites the plasmons. However, additional three SPR detection methods are possible: intensity interrogation, wavelength interrogation, and phase interrogation [PMID: 34615234].
Each method has its pros and cons, for example, phase interrogation has better SPR detection sensitivity but the refractive index change may easily exceed the detection range.
I would like to understand why angular interrogation strongly dominates the market. Is it historical? or are there specific properties that make devices that use angular interrogation commercially superior?

Off the top of my head, based on some surface plasmon work now 40 years ago, using a single laser wavelength and varying the angle is by far the easiest to do experimentally, most reproduceable, and likely the easiest to analyze.


text:
How do different x-ray tube emission profiles (i.e copper vs cobalt sources) affect XRD samples containing iron (results included)?
Background:
I'm a student and have recently done some XRD on some iron ore mine tailings which came from an Australian mine tailings storage dam. The redness in colour and industry source led me to reason that a significant amount of iron may still be present in the tailings. I currently have access to 2 X-Ray Diffractometers each with a different x-ray source (Copper vs Cobalt) and I've heard that there can be significant differences in diffractograms due to the scattering of copper vs cobalt x-rays by iron. I wanted to see this effect in action so I ran the sample on both XRDs.
Instruments:
Below are the XRD instrument details, I am using Match! for PhaseID and TOPAS5 for QPA.

Parameter
XRD1
XRD2

Instrument
Bruker D8 Advance
Bruker D8 D

Emission Profile
Cu Ka 5 Berger
Co Ka 7

Results:
Assuming I've read and analysed the bulk XRF analytics right, the sample seems to contain mostly phosphorus at 45.2 wt% and manganese at 23 wt% (which I presume comes from the clay and soil covering used to contain tailings etc.) with only 1.3 wt% coming from the iron itself.
Looking at the following diffractograms where the blue trend is from XRD1 (D8A) and the green trend is from XRD2 (D8D) I am surprised by the stark differences in pattern, namely the transpositions of the background (overall elevated intensities) and 2$\theta$ positions (there seems to be a lag between the patterns).

The Questions:

How do the different emission profiles affect the diffraction patterns for this sample and what is going on physically?

Does the difference in emission profiles explain the higher background, translated 2$\theta$ positions, both, neither, or only one of the differences seen between the patterns?

How could I know quantitatively if an XRD emission source is good/suitable or bad/unsuitable for analysing samples containing iron (or any other element) components?

Is it reasonable to presume that the instruments have been mixed up in the naming files because the blue trend (Co emission XRD2) is lower than the green trend (Cu emission XRD1)?

That is a really broad question. My wife just got back from a one-week intensive short course on XRF and there are a lot of details to pay attention to in order to get accurate composition analysis. You might want to ask if there is a standard (maybe silicon?) to compare the two instruments. A lag in 2-theta is expected because the wavelengths (energies) are different for the two systems. A table of K-alpha energies are at http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/kxray.html which may help as well.

A 10 wt% $CaF2$ standard was used in the sample which was loaded and analysed with both XRD instruments.

Fair enough. Now, take the Cu and Co x-ray energies (wavelengths) and calculate the 2-theta shift of a given lattice spacing.

You mean by just rearranging the $$2d(sin (\theta ) = n \lambda$$  Bragg equation for $$2\theta$$ and assuming n =1 right?

That is pretty much it (you may need to watch for n=2 peaks as well - it has been a long time since I though deeply about diffraction).

How would I know which peak is n=1,2,..., n?

Well, they would move differently, or pop up when diffraction conditions were right. If you can just stretch out the axis (appropriately) of the blue curve to overlay the green, there is nothing to worry about. If some peak is not matched, you check for other diffraction conditions.

1.) Assuming you use a cathode tube as source for the X-ray radiation, the target material emits bremsstrahlung what (on first approximation)* looks like a continuous emission (:

(image credit above Wikipedia article)
On top of this you have a number of sharp emissions which are characteristic of the element's relaxation of inner (non-valence) electrons.  This is why you read experiments described as using e.g., Cu $K\alpha$, or Mo $K \alpha$ radiation.  You don't need to use solid metals, indeed liquid metal target like molten Ga or In offer X-ray sources of high brilliance (though smaller in diameter than conventional tubes).
The underlying principle for the diffraction experiment is the Bragg equation, $n\lambda = 2d \sin(\theta)$ where $n$ is an integer, $\lambda$ the wavelength of the X-ray radiation, $\theta$ the glancing angle of the diffraction experiment, and $d$ the distance of the lattice planes characterized.  (It just so happens that these interplane distances are about of the wavelength of X-ray radiation; given a well suited repetitive pattern, diffraction equally may be observed with visible light (say scale of about $\ce{500 nm}$), or even sound.)
Both single crystal (XRD) as well as powder (PXRD) diffractometers go great lengths to pick and select only the characteristic wavelength for the experiment.  Again, depending on the target material of the X-ray tube and thus characteristic emission spectrum, both design and material for the monochromator varies (e.g., for Mo radiation, graphite) widely.
2.)  To record the diffraction signal, you either a point detector (similar to a Geiger counter which moves in respect to X-ray source and sample), or a fixed CCD array (similar to a blown-up camera, horseshoe like bent around your sample).  They differ in their sensitivity for X-ray radiation (which equally depends on the wavelength used in the experiment).  An other plausible reason that green and blue trace in your superimposed diffraction pattern appear moved against each other are different strategies to counter (as in exclude a priori by experimental design) e.g., by variation of the time of exposure to collect intensities, averaging multiple single recordings to cancel out noise in intensity while varying along the $2\theta$ scale. There equally are many mathematical methods to discern the pattern of noise from pattern of signals and dedicated programs (e.g., fityk) before one even starts to assign the diffraction peaks.  Though the blue trace looks like to be better in terms of signal/noise ratio than the green one, the two traces include differences (presence/absence of signals) which can not fully attributed to variation of Cu/Co only.  A potential pitfall is preferential orientation by sample inhomogenity (e.g., not well enough grinded crystals) when filling a capillary sample holder/transmission geometry (sample is between X-ray source and detector) vs. placing the powder sample on a rotating dish/reflection geometry.
3.) Of course, you select a brilliant X-ray source for your sample to obtain a good signal/noise ratio, however this equally should account for the sample's composition.  While inorganic samples are much less prone to decompose under X-ray radiation than organic/biologic samples, your incident X-ray radiation actually may initiate X-ray fluorescence in your sample.  Then, the metal atoms' of your sample equally may contribute to the signal recorded.  If I remember correctly from the colleagues of inorganic chemistry, samples with Fe are susceptible to this type of interaction with Cu $K \alpha$ radiation.  Check this.
4.) You did not show an emission spectrum (or two emission spectra), but diffractograms.  If peaks are narrow (or broad) has to be taken into account to characterize your sample as well as the peaks' intensity pattern.  At the end of performing a Rietveld refinement, there should be a good match of the diffraction pattern recorded with a hypothetical diffraction pattern based on the sample's composition -- both along $2\theta$ as well as the intensities.  This is why publications add a difference trace below the diffractogram recorded during the experiment:

(image credit to Wang et al, CrystEngComm, 2018, 20, 699-702)
Engaging a PXRD experiment and Rietveld refinement is suitable both for pure compounds, as well for the analysis of mixtures like alloys, ashes, clays, etc.  So you may reach your goal to characterize your sample, ideally, you validate the results with a technique independent of PXRD (since it is a  mineral e.g., by X-ray fluorescence).
*) Of course this is not true, electromagnetic radiation is discrete along the energy scale (Planck).

Thanks for the answer! So following on, (2) how do I minimise preferred orientation?, (3) what metric quantitatively can I use to measure whether a certain x-ray source is good or bad, the metal's x-ray fluoresence number?, (4) I've since done a rietveld refinement and can provide the picture if need be.

(2) E.g., don't let needle-shaped crystals stack like matches in the sample capillary (transmission geometry); don't let plates (e.g., mica) stack like a pile of newspapers on the rotating disc of the sample holder (reflection geometry); grind (not squeeze) the sample finely, uniformly in size (diameter-like) and dimensions (e.g., https://www.materials-talks.com/basics-to-powder-x-ray-diffraction-how-to-achieve-high-quality-xrd-patterns-qa/). 3) I'm unaware of a fluorescence number, yet one can compare the present setup with setups used for characterizations of similar samples (literature).

An example would be to consider the same X-ray source, but instead of K$\alpha$ to pick $K\beta$ instead (example X-Ray Powder Diffraction: Why Not Use CuKβ Radiation? by Otto, https://doi.org/10.4236/jasmi.2018.83004). Equally, some of the $R$ values of the model solved and refined are not only about the plausibility of the model (interatomar distances/angles), but account about the separation of (wanted) diffraction signal and (still recorded) background ($R_B$, or Bragg residual).  Eventually, not a single criterion, but multiple criteria describe the quality of the experiment.

Thanks for the great insights especially for (2) as I've only been using rotating discs for everything I do. Now I know to preferentially use that for needle-like samples and sample capillary holders for plate like materials. For (3), good points, another good way I guess is just to compare the background from the two XRDs I have access to.

The comparison of backgrounds were sensible with fixed incident radiation, if sample crystallinity similar enough (if not identical sample), the same detector and same detector parameters were used. E.g., the croissant-shaped image plates for PXRD allow a quick recording along $2\theta$, but with fixed resolution (in $2\theta$). PXRD's point detectors may move in different angular speed ($2\theta/t$), often in much greater angular resolution than an image plate (fixed by CCD pixels dimension & design) can offer. Some Empyrean PXRD (e.g,, https://www.cmca.uwa.edu.au/facilities/xrd/empyrean)

routinely may down to increments of 0.0001° for each step along $2\theta$ with an angular reproducibility of < 0.0002° (https://www.malvernpanalytical.com/en/products/product-range/empyrean-range/empyrean) and if one averages multiple scans, the signal/noise ratio may be lowered, too.  So it is better to compare normalized intensities (if at all), not the raw data as recorded by these machines.


text:
Electrode potential
In my book the definition of electrode potential was given when a metal rod is dipped into its own ion solution.there occurs either oxidation or reduction due to which a potential difference exists between electrode(metal rod) and solution on interface(on surface of metal) called electrode potential.. but is it necessary to dip the
metal rod  into its own ion solution. What will happen if we dip the metal into some other solution?will then the electrode potential of that metal be same if dipped this into some other solution?.

do you mean $E^\circ_\mathrm{red}$?

Both E° and E..

Standard hydrogen electrode does not use a dipped metallic hydrogen bar. Neither chinhydron electrode (used sonetimes for pH measurement before glass electrodes) uses a metal as the part of the redox system. Electrodes (as conductors) are sometimes just inert conductive materials, like platinum or glassy carbon, not participating in the redox process.

There are two possibilities.

If the metallic electrode M is dipped into pure water or into a solution containing non-reactive metallic ions, like alkaline metals, the electrode potential is theoretically infinite during a couple of nanoseconds. But this does not last. Immediately after this time, one, two, three or several atoms M are oxidized by air or water. This produces a tiny amount of $\ce{M^{z+}}$ ions. As a consequence, both the concentration of $\ce{M^{z+}}$ ions and the electrode potential get to a finite value. The potential depends on the concentration of the $\ce{M^{z+}}$ ion according to Nernst's law.

If the metallic electrode M is dipped into a solution containing metallic ions $\ce{Y^{y+}}$ having a strong positive redox potential, the metallic electrode M is oxidized by these ions to produce $\ce{M^{z+}}$ ions and a deposit of metallic $\ce{Y}$ metal. If z = y = 2, the reaction is ; $$\ce{M  +  Y^{2+} ->  M^{2+} + Y}$$ In this case, the concentration of the ion $\ce{M^{2+}}$ is continuous increasing. The electrode potential of the electrode M is changing during the reaction. At the end of the reaction, the electrode potential is governed by Nernst's law.

Edit : The potential of an electrode dipping in pure water is not defined if the "solution" is totally pure water. Another way of saying it is to say that the potential is infinite if there is absolutely no ion of this electrode in solution. But it is practically impossible to get a solution with zero ion of the corresponding metal in solution. I repeat here Nernst's law :
$$\pu{E = E° + ( 0.059/2) log[M^{2+}]}$$
If [$\ce{M^{2+}}] = 0$, its log and the potential both tend to - ∞.
But if there is only one(=$1$) ion $\ce{M^{2+}}$ in $1$ L water, its molar concentration is $\ce{[M^{2+}] = 1/(6·10^{23})}$, and the potential is not infinite as before. It becomes $$\ce{E = E° - (0.059/2)· log(6·10^{+ 23} ) = E° − 0.70 V}$$
If there are just $2$ ions $\ce{M^{2+}}$ in $1$ L water, its molar concentration is $\ce{[M^{2+}] = 2/(6·10^{23})}$, the potential becomes $$\ce{E = E° - (0.059/2)· log(3·10^{+ 23} ) = E° − 0.69 V}$$ This is not infinite !


text:
Is it possible to estimate how much Plutonium existed during the earth's formation?
Plutonium is a man-made element, but minute quantities of it exist on earth. It's probable the earth had much larger quantities of it during its formation as a planet, but most of it decayed due to its half-life. In the same vein (no pun intended) as carbon-14 dating, is it possible to determine how much of it existed by determining if some of the radionuclides on earth are remaining products of its decay chain?
In other words, can we study uranium, lead, thallium, bismuth, etc. samples on earth and determine if they were transmuted from plutonium?

It could be interesting to compare the amount of Pu thought to be present in the Earth at 'birth' and the amount produced during natural reactors such as Oklo (estimated to be about 2 tons of Pu-239).

So do radionuclides have any signature that can be used to determine their previous decay state? Interestingly, the common Pu-239 isn't the longest lived isotope.

I think this: https://en.wikipedia.org/wiki/Plutonium#Occurrence answers it.

The country of Gabon should make Oklo a nuclear tourism destination with a centrifuge amusement park ride and concession stands serving glow-in-the-dark Cherenkov drinks and hot yellow cakes. Great way to boost their economy. LOL!


text:
How are non-racemic drugs produced?
In many cases the L- or S- stereoisomer of organic compounds is vastly more bioactive than other enantiomers.  So we see pharmaceutical companies producing escitalopram (i.e., S-citalopram), esomeprazole (S-omeprazole), esketamine (S-ketamine).  (Granted some of this may be nothing more than a gambit to prolong patent protection on the underlying racemic drug, but leave that aside for now.)
What tricks enable such complex molecules to be produced or separated into a single, specific isomer in large quantities?

Asymmetric synthesis is quite a big deal.

Another way is applying separation techniques on racemic mixtures, using chiral counterparts.

Of the naturally-occurring amino acids it may be said that they are both L-(Fischer) and at the same S-(CIP) stereoisomers. But this correlation is not true for all stereoisomers. While you have found several active S-stereoisomers, an active site knows nothing about the CIP rules.

Small misconception: L is not S. Chirality is a part and parcel of life and there is no coincidence that a certain enantiomer is more biologically active than the other. You have already noted that. I don't think that patenting is the main driving force to develop chiral medicines. There is a very sad story, called the thalidomide tragedy which led to the birth of severely deformed babies who were crippled for life due to one of its enantiomers. Since 1992, chiral analysis of drugs is a must!
Approximately 63% of drugs approved by the Food and Drug Administration (FDA, USA) between 2015 and 2020 were chiral drugs, according to a survey conducted last year in J. Med. Chem. 2021, 64, 2339-2381). Coming to the main query, large scale preparation of a single enantiomer can come from:
(i) Asymmetric synthesis or enzymatic synthesis
(ii) Biological synthesis e.g., certain antibiotic are synthesized by microrganisms, whose complexity which can startle human minds. Nature is also a synthetic organic lab.
(iii) Chiral crystallization
(iv) Large scale (preparative) chiral chromatography (simulated moving bed etc.)
(v) Countercurrent chromatography, large scale isolation method.

The thalidomide story as often told is simplistic. The enantiomers readily convert under biological conditions in the body so, had the manufacturer produce a single enantiomer product, there would still have been a tragedy.

@matt_black, This is correct. Thalidomide is still used for other medicinal purposes but not for pregnant women anymore.


text:
Is there any process that converts water into hydrogen and oxygen by using light and a catalyst?
Can water be converted into hydrogen and oxygen using sunlight and a specific catalyst? I can imagine a electrolysis reaction using a catalyst that uses light to release electrons. The goal is to find a process which uses only sunlight as positive source of energy.

Yes, lots. Search "photochemical water splitting".

Have you seen how photosynthesis works?

https://chemistry.stackexchange.com/questions/28768/bandgap-of-a-semiconductor-for-photocatalytic-water-splitting https://chemistry.stackexchange.com/questions/71172/what-might-serve-as-an-initial-starting-photocatalyst-for-this-large-water-split

The US Department of Energy has a page on hydrogen production processes, including descriptions of research into  photoelectrochemical and photobiological processes for hydrogen production.

Probably several of them. The one I found at the top with a search for "photocatalytic water decomposition" is Ref. [1]:

The photocatalytic activity for water decomposition of alkaline earth metal and alkaline metal antimonates, $\ce{M2Sb2O7}$ (M=Ca, Sr), $\ce{CaSb2O6}$ and $\ce{NaSbO3}$, was investigated. These antimonates were photocatalytically active when combined with $\ce{RuO2}$: both $\ce{H2}$ and $\ce{O2}$ were produced from the initial stage of reaction under UV irradiation, and the photocatalytic activity became stable as the reaction proceeded. The photocatalytic properties are discussed based on the distorted structures of $\ce{SbO6}$ octahedra.

Reference

J. Sato, N. Saito, H. Nishiyama, Y. Inoue (2002). "Photocatalytic water decomposition by RuO2-loaded antimonates, M2Sb2O7 (M=Ca, Sr), CaSb2O6 and NaSbO3, with d10 configuration",
Journal of Photochemistry and Photobiology A: Chemistry,
148, 1–3, 85-89,
ISSN 1010-6030,
https://doi.org/10.1016/S1010-6030(02)00076-X.

Here you can find a second example of photocatalytic decomposition of water:
Ruinan Zhang, Song Liu, Hongyan Yuan, Dan Xiao, and Martin M. F. Cho (2012) "Nanosized TiO2 for Photocatalytic Water Splitting Studied by Oxygen Sensor and Data Logger", Journal of Chemical Education,89 , 1319-1322. https://doi.org/10.1021/ed1009283

For the following reaction:
$$\tag{1}
\ce{H2O ->[$h\nu$] 2H + O}
$$
you could shine light on the water with energy equal to the atomization energy of water: 917.8 kJ/mol (130.34 nm, or ultraviolet light). This is called "photodissociation" (among other names for it) and when dealing with a molecule that contains more than two atoms, we may call it "photoatomization".
For the following reaction:
$$\tag{1}
\ce{H2O ->[$h\nu$] H2 + O}
$$
you can shine light on the water with energy equal to the difference between the energy of the water molecule and the sum of the energies of the $\ce{H2}$ and $\ce{O}$ species. Similar reactions can be found in our atmosphere and in space.

However, if powered by sunlight this is not so useful for producing hydrogen and oxygen at the surface of the Earth.

@DDuck The title asked if it was possible with "light" (not just sunlight) and this answer immediately jumped at my mind. Some of the similar reactions mentioned in the link I provided, are in fact powered by sunlight in the atmosphere, but I wouldn't say this reaction involving H2O happens as often as the ones described in that link.

Interesting that 130 nm (short UV) may be considered the "uncatalyzed" energy requirement.

You could indirectly use sunlight to generate electricity with a common solar panel.  It only takes around 1.23 volts to split water into H2 at the cathode and O2 at the anode.  The gasses can then be collected in separate tubes for storage and use.
Interestingly, solar panels do not require higher energy UV radiation to generate electricity, only needing inexpensive and naturally available sunlight.  This would be an important consideration for large scale production.

technically this doesn't just use light and a catalyst.

@SafdarFaisal: …in a sufficiently broad sense, the solar panel and the electrolysis equipment could be considered catalysts. After all, they do facilitate the reaction but are not consumed in it.

@IlmariKaronen Well, if you're going that far, you might as well broaden it only slightly further and call a synthetic chemist a catalyst for all sorts of reactions... which is a notion that I rather like!


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What does it mean by "Standard state" when calculating for standard heat of formation
For example for this reaction:
Step 1: $\ce{CO(g) + 2H2(g) -> CH3OH(l)}$
Step 2: $\ce{CH3OH(l) -> CH4(g) + 0.5 O2(g)}$
for step 1, when computing the standard heat of formation
H = H($\ce{CH3OH}$) - H($\ce{CO}$)
I don't include hydrogen gas because its "element is in its standard state so H=0".
What does it mean by standard state and how would I know ?

Please format sentences properly, checking spelling and grammar.

This is the first time I see Phoenician alphabet used in physical chemistry. It's cool and all, but the text should be written from right to left, if you want to go full proto.


text:
What is the hybridization of the carbon atom in cyclopropenyl carbocation?
The hybridization of the carbocation in the cyclopropenyl carbocation (which is aromatic) should be sp as it as the + and also is in a ring. But the carbocation has a sp2 hybridization, why?
For example, a phenyl carbocation has a hybridization sp.
Thank you!

I have a feeling you'd better forget what you think you know about phenyl carbocation, and start learning about hybridization from beginning.

@Mithoron So I tried explaining it to myself via the vsepr theory, which gives me a sp2 configuration for the C with +ve charge in cyclopropenyl carbocation. Also tried with the phenyl carbocation and turns out it's between sp and sp2. I did not understand the reason for this, but I assume the phenyl is not sp then? It would really help if you could correct me wherever I am wrong. Thank you!

Your "hybridization" is pretty much a post factum rationalisation depending on geometry which is an actual observable. For sp there has to be 180 deg angle and for sp2 it's 120, otherwise that's not it and that's the situation in both carbocations.

Not wishing to add fuel to the fire here, but the C-C orbitals within cyclopentane are often said to have "sp5" hybridization. Assuming that this remains the case in the cation, and that the cation is a "pure" p-orbital (for reasons of conjugation/aromaticity) this would imply "s2p" hybridization in the remaining C-H bond...

The carbon with a positive charge on the cyclopropenyl carbocation is the carbon which is not involved in the double bond.
Hence, when not a carbocation, this carbon forms 4 sigma bonds, 2 with 2 different hydrogen atoms and one with each of the other two carbon atoms. It has no line pairs.
As a carbocation, this carbon atom remains bonded to the two carbons, so whatever reaction mechanism you’re looking at, the positive charge means that this carbon is now only bonded to one hydrogen.
Thus, the positive carbon forms three bonds and has 0 lone pairs, so it has a steric number of 3 (and requires three atomic orbitals).
These three atomic orbitals are thus sp2 hybridized.
Diagram explanation below:

Here's an answer based on how I was taught this theory.
The carbon atoms in benzene have sp2 hybridization because they are attached to three entities (two carbon atoms, one hydrogen atom, no lone pairs). Yank off an H– to create a phenyl carbocation. The carbon atom from which the H– was removed is now attached to two entities (two carbon atoms, no lone pairs) instead of three. Therefore this carbon now has an sp hybridization.
The carbon atom without a double bond in cyclopropene has sp3 hybridization because it is attached to four entities (two C atoms, two H atoms, no lone pairs). Yank off an H– from this atom to create a cyclopropenyl carbocation. It is now attached to three entities (two C atoms, one H atom, no lone pairs) instead of four. Therefore this carbon now has an sp2 hybridization.
I think teachers and students should avoid using the word "should" when describing predicted outcomes from conceptual models like orbital hybridization theory. (On the other hand, it's OK to use the word "should" when describing how language should be used!)

Your logic might work for simple tetraatomic molecules, but fails completely here, in molecules that can change the geometry of carbon skeleton only so slightly.


text:
How to get rid of iron phosphate coating on steel after rust convertion
I tried to get rid of some mild rust on a couple of steel elements (cases, screws and whatnot). My intention was to get rid of physical rust by manually scraping it off, perhaps sandblasting and then, with help of chemicals, to get rid of all traces difficult to reach by physical means.
Unfortunately, I haven't done my homework and I got myself a rust converter instead of rust remover. The former bases on phosphoric acid which turns rust into iron phosphate that results in black coating on steel elements that's supposed to be some form of rust protection, while the latter is supposed to physically decouple rust from steel.
To the point, now that I'm left with bunch of blackened elements, is there any solution in which I can soak those metal elements to get rid of this black iron phosphate coating (it not only leaves black traces on my fingers, it also stinks horribly) so that I can apply proper rust removal product (hopefully I'll get it right this time) and get those elements galvanised as I initially planned?

Try standard rust removers, e.g., those with oxalic acid and/or EDTA, which might attack through the phosphate layer. Sulfuric acid dissolves it,  but might not be good for the underlying steel (https://pubchem.ncbi.nlm.nih.gov/compound/Ferric-phosphate#section=Other-Experimental-Properties)


text:
Solubility of pyridine and pyrrole in water
Pyridine and pyrrole are two of the first aromatic heterocycles encountered in organic chemistry. Recently, I was surprised to find that their solubility properties are vastly different.
The website of American Chemical Society says

Pyridine is completely miscible with water...

While ChemicalBook says

Pyridine is hygroscopic and is miscible with H2O and organic solvents.

Pyrrole has only limited solubility in water but is miscible with many organic solvents.

The two molecules appear to be similar in size, and both have nitrogen. My first guess is that their dipole moments are different, since the lone pair in pyridine does not participate in aromatic resonance, but I can't seem to find relevant data on this. Also, it feels strange that a molecule like pyridine is miscible in water.

Sure both have a nitrogen and both are aromatic. But their structures are vastly different. For example, the lone pair of pyridine is readily available for H-bonding with water while that of pyrrole is blended into aromatic system so not available. Only a polar bond of $\ce{N-H}$ make it slightly soluble in water.

Difference in basicity is probably most important here.

@MathewMahindaratne While the lone pair on pyridine can participate in hydrogen bonding, can't the hydrogen on pyrrole also, with the oxygen in H2O? Or maybe there's a significant difference in their strength which I overlooked?


text:
Is oxyhydrogen lighter than air?
And, if so, how much lift does it produce per cubic meter?
(let's say at 290ish K and 150 meters above sea level)
For context, I'm taking an intro-level physics class. While I know what a joule is, I feel like I haven't really been able to internalize its meaning. Same goes for a couple of other SI units.
This will probably improve with time, but that takes too long. And, as helpful as the Wikipedia "orders of magnitude" series has been, sometimes the mind's eye just doesn't cut it.
So, I'm making a real-life version of it (that may or may not become a YouTube video, depending on my motivation and free time)!
(also, with different, fewer examples, and within legal-ish limits)
And I figured that one of the ways I could visualize a megajoule was by detonating a balloon filled with oxyhydrogen.
But I feel like setting it off in the air and viewing with a drone may be cooler and possibly less disruptive than doing so on the ground.
Of course, oxyhydrogen doesn't just explode when you ask it to. But I feel like my hobby-level background with electronics would help me out here. My plan, bear with me, is to use some random 8-pin SMD mcu and altitude sensor to short a small piece of metal across a tiny EDLC to bring it up to a high enough temperature to ignite the hydrogen. But, while I can design that while waiting for the stuff I ordered to arrive, I'd like to do so with some kind of weight constraint, if oxyhydrogen is lighter than air at all.
Feedback on the ignition plan is also welcome.

I'm sorry, but having a megajoule of an explosive gas mixture nearby is NOT something you want to do (particularly since it can be set off by a static discharge). For example, stored energies of ~10 Joules are often considered hazardous in OSHA-relevant settings. A megajoule is asking to be a Darwin award winner, which one should not aspire to.

Note: a pound of TNT is about 2 MJ.

This is all to understand how big a joule is?

This is a matter of simple calculation. If you don't know how to calculate it, then all this looks like just another reason not to anything like that.

You could eat a candy bar  instead. It is interesting that four candy bars oxidized to $\ce{CO2}$ release as much energy as a kilogram of TNT exploding. It is all about the speed of releasing the energy, I guess.

It's a bad idea to put a factor of 10^6 between your example and the thing you want to understand.

Here is how to illustrate one or a few joules in a relatively sane manner.
Start with a large glass or clear plastic bowl, which you place on the floor in front of you. Now take a beaker and add some water to it, preferably with a dye to make things easier to see -- 100 mL for each joule you want to demonstrate. Then dump the water from waist level into the bowl. The "blast" energy when the water hits is roughly one joule per 100 mL of water.

Thanks! For the megajoule thing I just ended up weighing out some wood pellets and burning them, then accelerated to 70 mph on a straight stretch of highway and slammed on the brakes. I am at once enlightened, very dizzy, and trying to rent a large passenger aircraft


text:
Question about Mohr's Method procedure
Just had a quick question on Mohr's method.
In my analytical chemistry course we are finding the concentration of chlorine among several samples. The first step is to add a small amount of silver nitrate is added to the indicator solution and is then filtered. Is this step not superfluous? What is the benefit of
adding a small amount since it is just using up indicator solution?

Well, the silver nitrate is the thing that reacts - the silver cations, actually - with the chloride ions in solution, so you are titrating with that and not just adding a small amount of it. Then, the excess $\ce{Ag^+}$ reacts with the chromate which turns reddish-brown: you back-out chloride from the amount of $\ce{AgNO_3}$ you added to get to that point. I could see filtering after the chloride is precipitated as $\ce{AgCl}$ in order to recover the silver.

The first step is to add a small amount of silver nitrate is added to
the indicator solution and is then filtered. Is this step not
superfluous?

This step is not unnecessary. This is to teach the concept of a blank titration. Your analyte is absent in the solution in the case of blank titration.
The chromate indicator consumes a very small but measurable amount of silver ions (the titrant, from the buret), so you need to distinguish how much is consumed by the indicator and how much by chloride ions. Record the buret volume, B. Once the blank titration is performed, now you proceed to the sample titration,S. The true volume of silver nitrate is then S-B. This subtraction is only valid, when a fixed amount of indicator is added in Mohr's titration to both solutions.


text:
Can someone explain the chemistry around adding Isopropyl Alcohol to water contaminated gasoline - how does the IPA 'remove' water from the gas?
So there are many products on the market sold as fuel additives; some (like 'Heet' brand) are nothing more than 99.9% isopropyl alcohol in a nice package.
When trying to understand how they work and what they actually do you hear vague things like "Removes water in fuel. The alcohol will bond with water and solubilize water into gasoline so it can be burned and pass out exhaust"
From what I understand alcohol and water don't 'react' -> there is no chemical formula change. They simply mix really well. But how does this make this new mixture 'mix' with gasoline better?
I know that if you take 50% isopro (other 50% is water) and light it on fire, it will burn. But I understand that it's only the 50% of alcohol molecules burning, not the water. So at the end you'll still end up with the water remaining (excluding some evaporation from the heat).
What's going on? Can someone explain the actual chemistry please of how IPA removes water from the gasoline?
UPDATE: found this technical paper from a chemical engineer on the subject: https://web.archive.org/web/20130427123941/http://www.epa.gov/OMS/regs/fuels/rfg/waterphs.pdf
However, it still doesn't spell out the exact chemistry but should add more context to my question.

You want the water evenly mixed throughout the gasoline (well, you really want it filtered out). You don't want a blob of water heading into the injectors to be sprayed into a cylinder and not ignite.

I don't see what "chemistry" you want. Water in fuel is more of engineering issue as Jon says. If you add enough alcohol you get one phase - no separate water phase - no problem. Maybe it's that you reasoned backwards? Alcohol dissolves also in fuel and if you add enough, water has enough solubility in resultant mixture.

Alcohols are composed by a hydrocarbon chain (R-) and a hydroxyl group (-OH). The bigger the hydrocarbon chain is, the more soluble other organic substances (like fuel) become, but water becomes less soluble. When there is a right balance between R size and OH in terms of total alcohol composition, both substances (fuel and water) become reasonably soluble to form a homogenized mixture. Isopropanol is an example of an alcohol that qualifies.

Here is a wrong sentence from the second paragraph : The alcohol will bond with water and solubilize water into gasoline so it can be burned. This is wrong.  Twice wrong. Alcohol will not solubilize water into gasoline, and water cannot be burned. A mixture $50$% ethanol or isopropanol + $50$% gasoline dissolves less than 1 % water.

@Maurice I also felt that sentence was wrong and way too vague. So what do you think is going on? Why even bother adding a tiny bottle of alcohol to a gas tank? Is this just snake oil and doesn't do anything meaningful?

@ wild coast. I have absolutely no idea about the reason of adding some alcohol to a gas tank.

There is another reason. Nobody wants water to freeze in gas piping or other bottlenecks. Even liquid water may be an obstacle.

Your paper just says exactly what I did - if you add some alcohol to fuel then water is gonna be a bit more soluble in it. Yeah, it's still less than 1%, but it can be enough to not have another phase - it's not like there's gonna be a glass of water in your tank, but small amount could even condense out of air.

Related but behind paywall: Ternary phase diagrams for gasoline-water-alcohol mixtures

Alcohol in the gasoline will dissolve water and the typical dose is about 250mL. If the amount of water is large it may require repeated dosed to get the engine running right again. We once bought some watered gas in Apalachicola Fl couldn't find any dry gas until Dade City hobbling across the panhandle about 25mph. It took a case of dry gas keeping the fuel level below 1/4 tank to get the car running OK, by that time we were halfway thru Georgia. this was an endemic problem in northern climates almost alleviated by todays gasoline with 10% ethanol. I still have some cans of dry gas, waiting

HEET used to be 100 % methanol; has it changed ?

Water separates from gasoline and should not be injected or carbureted in large quantities because it will destabilize engine performance.
Addition of isopropanol in quantities comparable to the water may not make a homogeneous solution in gasoline, but the water-iPA phase will have a surface tension so low (50/50 at 25$^0$ C: 24.26 mN/m vs 21.22 for pure iPA Ref 1)

that emulsification will be very easy (from shaking due to road conditions like potholes, sharp turns or braking, stopping, etc.).
This sort of behavior is evidenced in practice by formation of a foamy layer between two phases, or just a foamy mass, of tiny separate phase droplets that do not easily recoalesce into two distinct phases. This upsets lab workers who are trying to separate the phases, but the engine is much more tolerant (or could be if the amount of water is not too great).
The engine operates well with injected gasoline; less well with some emulsified water; poorly with too much water in the emulsion. But in general, small amounts of water-containing emulsion can be  injected or carbureted and burned without difficulty.
In other words, the "chemistry" behind the effectiveness of isopropanol's ability to counteract the effect of water in the gas tank does not require a total clarification and transformation into a single phase, but only an assistance so as to be able to minimize the amount of water in each cylinder-load of fuel.
I suppose a surfactant that really emulsified and stabilized a cloudy gasoline-water mixture would burn quite well, even at high concentrations of water.
This is a slightly expanded form of a comment I had posted yesterday.
Ref 1. pubs.acs.org/doi/pdf/10.1021/je00019a016


text:
What can be reasons for such anomalies in the IR spectrum of iodoform?
A few weeks ago I collected an IR spectrum of iodoform in KBr as part of a computational chemistry course. For some reason, there is consistently a trough right before every peak, and I am unsure of this mechanism. I have spoken to a spectroscopist who said he has never seen the phenomenon in anything other than in KBr and iodoform. Is there a mechanism for this that anyone is aware of?
More specifically, the iodoform was ground into KBr using a mortar and pestle in a 1:20 m/m ratio and the spectrum was collected using a PerkinElmer Spectrum 100 FT-IR spectrometer.

Looks like a phasing issue which is common in raw NMR data, but I've never seen it for IR.

These 'derivative ' type features do seem to occur at the three most intense iodoform transitions, so it might be possible that the at these particular wavelengths no light is being transmitted by the sample and the fourier transform  performed by the instrument is, for some unknown reason, responding as shown. You could try diluting the sample by, say, 10 times.

Prior to using an instrument, I always recommend benchmarking it. It would have been appropriate for the instructor to test this. A polystyrene film should be placed in the sample compartment and its FTIR spectrum should be collected. Does it appear to be a well-behaved spectrum? This test sheet is included with most FTIRs. If so, then the instrument is functioning properly. If the spectrum is not well-behaved, it is time for a servive. However, a good polystyrene spectrum this does not guarantee that your sample has been prepared properly.
Please keep your cell phone away from the instrument. Electromagnetic interference can lead to the formation of these fake derivative shapes. While the Michelson interferometer collects data while phones ring and chats take place, the infrared spectrum may exhibit these derivative-like spikes. The majority of instruments are shielded from electromagnetic interference, but it is possible for these things to occur.
People who perform microwave spectroscopy, for example, know that their detectors pick up the Wi-Fi signal in the room. Wi-Fi peaks contaminate the spectrum.
Briefly, the spectrum should be re-done to ensure that everything, including the KBr/sample, is bone dry. Handling spectroscopic instruments requires a great deal of respect. They demand a high-level of attention before their output can be trusted. Your overall transmittance is pretty high, so I do not think it is a concentration problem.

All other spectra are well-behaved. Every spectrum that has been collected by this instrument works as expected. This exclusively appears with Iodoform in KBr.

Acquire the spectrum again with a fresh sample with new background. Even try another bottle of iodoform. Collect at least 32 scans. There is nothing unusual with iodoform. Its FTIR spectra are available and they look perfectly normal.  https://www.chemicalbook.com/SpectrumEN_75-47-8_IR1.htm

Molecular absorbances change the bulk refractive index of a material in that derivative-like shape.  You have reflection and transmission going on in that KBR pellet and the absorbance peak is changing the refractive index as well as absorbing light. both of which changes how much radiation goes through. That's a great spectrum for physics, not what a chemist wants to see.
And yes, my email says PerkinElmer.com

Thanks for the answer! Are there any textbooks that you know of or could recommend I look in to for more info on these phenomena?


text:
LCAO-MO ansatz for arbitrary hydrogenic molecule term
I am struggling to understand the connection between molecular electronic term symbols and translating the symmetries to the underlying wavefunction.
In particular, my understanding of the LCAO-MO method tells me that I can write the molecular wavefunction as
$$ \Psi(r_1,r_2) = N \left( 1s_a(r_{1a})  1s_b(r_{2b}) + 1s_a(r_{1b})  1s_b(r_{2a}) \right) (\alpha(1)\beta(2)-\beta(2)\alpha(1)) \; ; $$
where the spatial part encompasses the spatial symmetry and the spin part ensures that the state is a singlet.
Independently of this, I can determine that the ground state of the hydrogen molecule has a term symbol of $^1\Sigma$: it is $\Sigma$ because both atoms contribute $l=0$ to the total $L$ quantum number, which in turn will be 0, and it is singlet because of the spin restriction.
I am fairly certain that the fact that there's a positive sign in the spatial part would be represented by a $_g$ subscript for gerade spatial symmetry - is this correct? And if I had a negative sign, I'd get the $^1\Sigma_u$ state, right?
However, if I look at the electronic states of the hydrogen molecule, by for example looking at a list here, I am not too sure what would be the LCAO-MO equivalent for the states listed.
For example what would be the ansatz for the state $^3\Sigma_u^+$ (I do not know how to account for the $^+$ for example) Or let's say for $^3\Delta_g$? And how would I treat "pre-factors" such as the term $b^3\Sigma_u^+$?

There's a pretty good summary on wikipedia. Check out the "Wigner Witmer Correlation rules" section: https://en.wikipedia.org/wiki/Molecular_term_symbol


text:
Reversible monomolecular reaction with two reverse rates
I think this is a simple problem but my kinetics are terrible and I was hoping for some assistance following up on the work of a previous student. I'm open to any form of assistance and I hope this question doesn't come off as too big of an ask.
I'm working on a solution to determining a rate of catalysis ($k_2$) in a reversible reaction:
$$\ce{A <=>[k1][k1 + k2] B}$$
where $\ce{A}$ and $\ce{B}$ are expressed as molfractions because it's known that $\ce{A + B = 1}$. Experimentally, it's been determined that $k_1$ is the same in both directions so I've excluded the normal +k and -k notations.
In a simpler experiment the rate of $k_1$ can be determined by the integrated rate expression derived from the reversible reaction:
$$\ce{A <=>[k1][k1] B}$$
$$ \begin{aligned} \frac{dB}{dt} &= k_1A -k_1B \\ &= k_1 (1-B) -k_1B \\ &= -k_1(2B -1) \end{aligned} $$
which leads to the integrated rate expression:
$$\int\frac{dB}{(2B -1)} = \int{-k_1 dt}$$
$$2B-1 = \exp(-2k_1 t)$$
This is a useful form of the equation because 1. it's in terms of one species only, and 2. the measurement we take originates from the square of the population difference between the species A and B and given that $A + B = 1$:
$$(A-B)^2 = (2B-1)^2$$
Experimentally, we just collect a signal over time and the decay of the signal is fit to the square of that integrated rate equation which is $(\exp(-2k_1 t))^2 = \exp(-4k_1 t)$. So in the end:
$$\text{Observed Signal} = (2B-1)^2 = \exp(-2k_1 t)$$
I am now studying the system described in the first equation where I want to introduce a species that catalyzes the back reaction, and given that I can determine the rate without the catalyst ($k_1$), I am pretty sure I can find the rate $k_2$ if I can figure out how to make this new integrated rate expression arranged in terms of $2B-1$. Essentially, can someone help me find an integrated rate expression which solves for $(2B-1)$ in terms of both $k_1$ and $k_2$ so that I can run one experiment without the catalyzed back reaction to determine $k_1$ and then run a second where I can then plug in $k_1$ as a parameter to determine $k_2$. I've been working on this for days and I feel like I'm just missing something fundamental in my understanding.
I felt this question might be relevant to others, as it seems like a good question on basic kinetics. If this is not the case, it's due to lack of understanding on my part and I do apologize. Even if there is any resources someone could point me to I would really appreciate it, as my searches have yet to turn up anything useful.

A true catalyst cannot affect the back reaction without also affecting the forward reaction, as the equilibrium must remain the same

@Andrew I don't believe I'm trying to make that argument, but if it would make it more correct the issue it could be thought of as three rate constants. k1 forward and k2 and k3 in reverse. That would however, remove my experimental capability to determine any of the rate constants I do believe. I have experimentally determined that in the simpler case the k1 and k2 are the same. Its my understanding that the rate is concentration dependent and therefore could be scaled simply by adding a second rate to the back reaction in the more complex case.

The first kinetics equation  is wrong. There must be a minus sign in front of $\pu{k_{1}B}$. This formula should be :  $\ce{dB/dt  =  k_1A - k_1B}$. The consequence is A = B.

@Maurice You are correct, and it's been edited to reflect those changes.

@Andrew I think I understand you statement a little better now. This reaction represents a simplified version of a flippase catalyzed lipid translocation. Rate 1 represents flip-flop due to thermal motion, rate two represents unidirectional lipid flip. I don't believe as they are traditionally modeled that catalyst rate would effect the separate native flip-flop. I omitted these details to try and make the exercise useful to a more general audience.  If I'm incorrect though about this assumption I wouldn't be surprised.

Got it. In that case, the k2 reaction is unidirectional because it is coupled to ATP hydrolysis. That complicates things a bit, and it is more properly notated with a second curved arrow from B to A with rate constant k2 and ATP -> ADP + Pi rather than adding k2 to the initial arrow with k1, since they are different reactions.

But the more important point is that as long as the ATP concentration is saturating and relatively constant, your simple kinetic model of k1 + k2 for reverse is fine. Just bear in mind that k2 contains an intrinsic [ATP] term.

@Andrew I guess I'm failing to see the difference between k1+k2 on the reverse side when it's known k1 is the same in both directions. To be clear I'm admitting I don't understand and not challenging your statement. This is also not the ATP consumption being considered but the actual flip of the lipid. The lipid is considered the reagent here on both sides. The ATP consumption and the flipping process are related but it's not known if they are 1:1, that's kinda the end goal with this process.

If ATP hydrolysis is not coupled to the flipping reaction, the flippase will necessarily increase the rate constant for both directions by k2. It cannot increase the rate constant for one direction and not the other, because the equilibrium must still be 1. With the ATP hydrolysis coupling, that is no longer the case, as k2 is the rate constant for a reaction that is fundamentally different from the reaction with rate constant k1 and has a different equilibrium constant, even though the observed effect (flipping of a lipid) is the same in both cases.

If you let the reaction from B to A have a rate constant $k_3$ and solve the equation with $B=0$ at $t=0$ then $\displaystyle B(t)=\frac{k_1}{k_1+k_3}(1-e^{-(k_1+k_3)t})$ which you can use by changing $k_3$ to be $k_1$ or $k_1+k_2$ etc.

@porphyrin That is more or less the equation that I seem to have derived. I have multiplied the entire thing by 2 and subtracted 1 on both sides to make it proportional to the signal intensity. I appreciate the confirmation that it's just a nasty looking outcome compared to the clean rate equations I am used to seeing. Thank you.

If you let $k_3=k_1$ then this gives the equation you have. Remember when solving equations it is always necessary to specify initial conditions, such as at $t=0, B=0$

Just to repeat what porphyrin said in the comments:
Simple first-order reversible reactions tend towards equilibrium with a characteristic time that is the sum of the forward and backward rate constants.
So for an equilibrium $\ce{A<=>[k_f][k_r]B}$, then the system will tend toward equilibrium with a rate $r \propto e^{-(k_r + k_f)t}$.
It doesn't matter if in fact you can mechanistically "decompose" say the back reaction into two different (first-order) pathways that have distinct rate constants.  If $k_r=k_{r1} + k_{r2}$, then $r \propto e^{-(k_{r1} + k_{r2} + k_f)t}$, but that doesn't help you, at least not if your measurements are just the overall kinetics of the system.  Measuring kinetics will just give you the value of the net rate constant, whether you call it $(k_r + k_f)$ or of $k_{r1} + k_{r2} + k_f$ or just $k_{net}$, with no way forward to get at the individual steps.
Explicit assumptions
You will need to introduce additional assumptions to make progress.  You've implicitly assumed, I think, that:

there exists a condition where $k_{r2}$ is zero, and in this condition, $k_r=k_{r1}$.  Let's call this condition $\aleph$.
also that in this condition $\aleph$, the forward and reverse rate constants are the same, which is equivalent to assuming the equilibrium constant (for condition $\aleph$ ) is 1.  This means that $k_f=k_r$ for condition $\aleph$.
that there is another condition $\beth$, where $k_{r2}$ is not zero, but that $k_{r1}$ is still equal to $k_f$.  Since the total reverse rate constant $k_r$ is now higher than in condition $\aleph$, this assumption means that that the equilibrium constant in condition $\beth$ is different (lower) than in condition $\aleph$.

What you'd need to measure
In order to determine a value for $k_{r2}$ from experimental data, you would need to measure the kinetics of the system under both condition $\aleph$ and condition $\beth$.

From measuring condition $\aleph$, you'd get the value of $k_r+k_f$, which by assumption 2. is equal to $2k_f$.
From measuring condition $\beth$, you'd get two pieces of information:

First, you'd get the value of $k_{r1} + k_{r2} + k_f$ by fitting your timecourse data.
Second, you should be able to measure the new equilibrium constant by observing the behavior of the system at very long times.  In condition $\aleph$ the system must approach $A=B$ at long times, because of assumption 2. But now in condition $\beth$, it should trend toward a different ratio between A and B, and this ratio should be $\frac{k_f}{k_{r1}+k_{r2}}$.

You should be able to estimate $k_{r1}$ and $k_{r2}$, subject to your assumptions, with either measurement on condition $\beth$.
Note that you don't even need kinetic information from condition $\beth$ to measure $k_{r2}$, you just need to measure the new equilibrium condition.  Of course, if you can make both measurements on condition $\beth$, you should get the same value (within experimental uncertainty) for $k_{r2}$.  If you don't, then one or more of the assumptions I've listed is violated.

I can follow your proposal but there is something that does not seems consistent. I maybe mistaken but If the reaction studied is a true reversible reaction, in the equilibrium $K_c = \frac{|B|{eq}}{{|A|{eq}}}$. Since the concentration ratio $ \frac{|B|{eq}}{{|A|{eq}}}$ depends on the values of the rate constants for the forward and reverse reactions, you would get different values of $K_c$ with and without catalyst. How can you get two different values of $K_c$ at the same temparature?

I intentionally didn't specify the nature of the two conditions.  Differences in pressure, solvent composition, voltage, magnetic fields, etc., could lead to differences in the apparent equilibrium constant.  For the purposes of my answer, I'm just assuming that the two separate conditions exist, and I guess that we are not really interested in the thermodynamic connection between the conditions.

Everything you said is correct, but largely not applicable here. The first problem is that his method for measuring reaction kinetics relies on the quantity 2B-1 (rather than just A or B), which doesn't factor out readily when the forward and reverse rate constants are different. Second, and more importantly, his reaction is not an approach to equilbrium (although he wrote it that way), as the k2 reaction involves an additional reactant (ATP), and so is better considered as a competing, functionally irreversible reaction.

Andrew: I think I might agree with you on your second point, but not your first.  Measuring 2B-1 is what happens after you (a) assume $k_r = k_f$, and (b) apply a mass balance on A and B.  It's perfectly consistent with what I wrote.  But yes, it might be better to model $k_r2$ as a separate reaction entirely.

This explanation, while maybe a little complicated, does indeed describe the system well. I would have to agree with @Andrew though in that it's not quite useful in this situation due to experimental parameters which we discussed. As they suggested, the fractional difference between the species is what yields signal. So while your explanation does describe the system, it doesn't shake out the needed parameters to conduct the experiment. What the both of you seem to have agreed on, and I also agree on, is that I didn't quite understand the nature of my question when I asked it.

@CurtF. it might be better that I type up the solution I arrived at after discussion with Andrew. The results of this discussion actually ended up changing some experimental priorities and those need addressed. If the results seem useful I will come back and write them up. It seems by intentionally trying to generalize my question to make it more applicable to others, I have made what I was looking for far too broad. Both of your answers are correct given what I have provided, but Andrew's happened to be more in the direction I was going. Apologize for the confusion I have created here.

That would be great @hotmaildotcom1.  Looking forward to seeing your solution.  I'm leaving my answer as-is, even though it apparently doesn't help you, because it answers your question as asked.  And it might help others.

@CurtF. I agree and I definitely learned something from it as well. Thanks for your help.


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How is geometry optimization of small molecules different from the protein folding problem?
My layman's understanding is that finding the ground state geometry of small molecules is "hard". I don't have a good sense of how hard though. I suppose there are applications where methods like VSEPR are good enough, and other situation where some sort of ab initio calculations are required. But isn't the problem of protein folding essentially the same problem, albeit at a much larger scale? Yet we have methods like Alphafold which can produce good results in many cases. So what am I missing here? Why can't something like Alphafold give us ground state geometries of small small molecules if it can do something similar for much larger molecules?

VSEPR is hardly a method in this sense. Getting geometry of small molecules calls for different methods than folding, but you got it backwards. You certainly could try using what is used for small molecule in a much bigger one. Trouble is that would make the calculation terribly long, that's about that.

@Mithoron And use lots of memory. Lots and lots of memory. That is a likely to be the limiting factor as time.

Questions about AlphaFold have also been well-received here (for example).

It's not entirely the same problem imo. Small-molecule geometries are often very easy to get approximately right (think molecular models), and quantum chemistry is used to fine-tune lengths/angles etc. On the other hand, the primary issue in protein folding is not precision, it's more about the overall large-scale, tertiary structure, which is not really a thing in small molecules (except in highly conformationally flexible molecules, where DFT does tend to suffer). That said, it doesn't mean you can't train ML models to do small molecules - search e.g. 'machine learning potentials'.

Well, if you keep adding more molecules, you'll have to consider interactions between many, many more particles. Like if you had two 'particles', it would be 1 interaction, but if you have 10 particles, it's not just 10 interactions, but 9 factorial. Although that is a simplified view, I think that's why.

tldr: For small rigid or semi-rigid molecules (e.g., 0-2 rotatable bonds) we can do a good job. For larger, more flexible molecules it's still hard.
One of the challenges with getting a good ground-state geometry is the accuracy of the methods to optimize the geometry.
AlphaFold and the various alternatives that have cropped up after its release, work in part from the vast amount of data in the Protein Data Bank (PDB) and similar databases. At the moment, there are almost 200,000 released structures. Moreover, because of the size of proteins, there's a lot of information that can be deduced from the secondary and tertiary structures.
There have also been numerous "blind tests" such as CASP, which started in 1994. The challenge is to predict the geometry / fold given the sequence. Such open challenges have definitely driven the field forward, since there's great prestige.
Add in that there has been a lot of work on bimolecular force field methods, such as AMBER, CHARMM, etc. Many groups around the world run molecular dynamics simulations with these methods and validate against experimental data, such as NMR.
One might expect that small molecules are easier – after all they are smaller!
Often that's true. For small molecules with zero "rotatable bonds" we have no conformers to predict and can often use quite accurate quantum mechanical methods to optimize the geometry. Bond lengths and angles are often very accurate when compared to experimental crystal structures, and the overall root mean square displacement (RMSD) of non-hydrogen atoms is often under 0.5Å. (Hydrogens are omitted because they usually are not available in crystal structures.)
For example, from our recent preprint we find the GFN2 semiempirical quantum method provides excellent agreement with crystal structures for zero or one rotatable bond (e.g., median RMSD ~0.2-0.3Å and mean RMSD ~0.5Å). That means for most of these molecules, atoms are only a small fraction of an Ångstrom out of place!

The challenge comes from a similar problem to protein structure prediction. There are lots of possible conformers for larger more flexible molecules. Many of these are not low-energy conformers and gas-phase optimizations neglect potential intermolecular stabilization in solution or a crystal structure. That's partly why the plot shows a slope for RMSD with increasing numbers of rotatable bonds. (Also, in large molecules, small errors in angles, etc. will add up.)
So why has AlphaFold done such a good job with protein structures, but small molecular conformers are still hard?
I can think of a few possible reasons:

More resources have been devoted to protein structure prediction. AlphaFold devoted a big team and lots of GPU/TPU time to attack it.
Proteins have fewer possible rotatable bonds. A recent paper by the Rarey group yielded a library of over 500 possible small molecule torsions with preferences.
Small molecule force fields are likely not as good as protein force fields, in my opinion. While faster approximate quantum methods like GFN2 (and some newer ML methods) exist, they're still fairly new and not as accurate as DFT on conformer energies.

There are many conformer sampling methods out there for small molecules. Unfortunately, very few support use with quantum methods.
Put that all together, and .. flexible small molecules are still tricky.

Oh, I completely forgot to mention ring sampling. That's usually ignored in protein folding, but ring flexibility, especially in macrocycles is still tricky (e.g., you can't just change one torsion angle, because the torsions in the ring are correlated).

The combinatorial complexity of large structures means that simpler methods can never solve the problem
Guessing the structure for small molecules can require a variety of techniques that work reasonably well for small molecules. Even some very simple techniques can do a good job for molecules where the bonding is not complex. Simple molecular mechanics techniques, for example, ar often good enough to give moderately good 3D structures for molecules with dozens of atoms especially if the bonds don't involve complex electronic interactions (so chemical drawing tools often apply simple force-field models to estimate plausible 3D structures).
These tools use simple information on bond-lengths and spatial interactions (atoms far apart in bonding terms can't overlap in 3D space and van der Walls interactions can explain that and reduce spatial overlap). They sometimes give results that are not ridiculous.
It might seem that protein folding could be "solved" with similar approaches. After all the constraints on folding are often not long term electronic interactions but simple spatial overlaps (I'm simplifying a lot as there are some important interactions with specific amino acid residues a long way apart in from each other on the protein backbone but this doesn't affect the argument). Don't we just have to try a bunch of bond rotations in the backbone of the protein and residues and see whether we can find plausible structures where the spatial overlaps are minimised?
But, even ignoring those complex interactions, this is an essentially impossible task. While testing the overall structure by running through many bond-rotations and testing the spatial overlaps that result works when the number of rotations are small (perhaps dozens), the number required to test protein structures is far larger. The difficulty rises exponentially with the number of bonds that have to be tested. Imagine we can explore the configurations of a molecule by testing 10 different rotations of each rotatable bond (and doing some simple calculations for each). For a molecule with just one rotatable bond this takes 10 calculations. For one with 10 rotatable bonds this takes 1010 calculations, which is a lot more. For 100 rotatable bonds (very small for a protein) this grows to 10100 calculations, which is not just a big number but an unimaginably big number vastly exceeding the number of particles in the known universe.
The combinatorial complexity grows exponentially with the number of bonds that need to be tested. This pretty rapidly exhausts the computer power of whatever computer you throw at the problem. Never mind that this is an extremely simplistic view of the complexity of the problem. We will never have the computer power to solve these problems.
What about Alphafold?
One might question that analysis by mentioning that Alphafold has made big progress on the protein folding problem so things can't be as bad as I suggest.
But this is to misunderstand how machine learning algorithms like Alphafold work (the saem applies to their similar achievement at Chess and Go both of which are intractable to brute force techniques largely because of their combinatorial complexity). Alphafold (and its equivalent for Chess and Go) does not attempt to calculate every possible configuration of the problem. It studies a very large number of known structures and tries to find common patterns that repeat across many structures. Machine learning is very good at spotting common motifs that are not very obvious to people in known structures. It then applies those to unknown protein sequences and tests whether they work. This is a far, far simpler approach than trying to work through all possible options and results in a higher, but still imperfect, guess as to the likely structure of the unknown protein.
What makes this work is that there are common features in many known structures that are very likely to repeat in others. And many of those are not easily spotted by people or other, existing, algorithms. But it is not a perfect method and does not solve the protein folding problem: it just does a much better job than other algorithms.
So the "success" of Alphafold does not undermine the argument that the fundamental barrier to using small molecule techniques to solver bigger problems is combinatorial complexity. The problem grows faster than increases our computing power as molecule size grows.


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How do longer chains and more cis-double bonds affect melting point and viscosity of fatty acids?
In fatty acids, longer chains increase the melting point (and viscosity) by increasing London dispersion forces. Inversely, more double bonds decrease the melting point.
What happens when both chain length and number of double bonds increase at the same time; how much are London dispersion forces affected by each factor?
Eicosapentaenoic acid has a lower melting point (probably lower viscocity too) compared to docosahexaenoic and stearidonic acids. One carbon atom was added to the chain and one double bond at the same time.

18:4 -> 20:5 decrease in melting point
20:5 -> 22:6 increase in melting point

Strange.
Intuitively bosseopentaenoic acid should have an even lower melting point (even lower viscosity). Conjugation would appear to make bonding between different molecules more difficult and therefore would also decrease London dispersion forces (and MP, BP, viscosity. etc with them)

The above are pictures of a space-filling model I found from google images supposedly generated from Accelrys DS Visualizer.

It's not necessarily that strange. Adding a carbon has less of an effect on how the chains pack than inserting a double bond, which completely alters the torsional freedom of the chain. You might also imagine that even numbers of double bonds allow a more linear conformation of a chain to be retained, whereas odd numbers inevitably result in a kink (edited: or vice-versa, depending on the conformation around the double bond?).

@BuckThorn It seems that EPA with 5 double bonds in more linear than DHA with 6 double bonds. How do the electrons in the molecules interact to bind together? How do DHA bind together more strongly?

What is the origin of the figures? Are they supposed to represent conformations in the solid? Could you explain this in the post?

You may want to look at related posts such as this one: https://chemistry.stackexchange.com/questions/46891/trying-to-understand-the-causes-and-implications-of-kinks-in-fatty-acid-chains

@BuckThorn The above are pictures of a space-filling model I found from google images supposedly generated from Accelrys DS Visualizer.

Yes, but pictures of what? And what are you trying to show with the images?

@BuckThorn That EPA's 5 double bonds allow a more linear conformation. The kink is in DHA.


text:
Rotating molecule representations with rdkit
I'm trying to display some molecules in rdkit and save them as png files. Simple enough. However, when I display them, most of them are drawn at a weird angle (shown below) or even completely upside down. I tried using this section to correct this but no luck. I've looked through similar issues people had but none of them have been properly resolved as far as I can tell.

Does anybody have a solution to this?
Thank you !

You can use SetPreferCoordGen(True)
from rdkit import Chem
from rdkit.Chem import rdDepictor
rdDepictor.SetPreferCoordGen(True)

mol = Chem.MolFromSmiles('Fc1ccc2ncccc2c1')
mol


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Is it possible to have an anhydrous solvent for virology, specifically in vitro phage assembly?
Typical protocols for lambda phage in vitro assembly call for you to dump all your viral components and buffer solutions into water. Water is a great solvent, drives extraction of energy from atp, and probably contributes structurally at various steps in the process. But I am curious to know: how anhydrous can you get and still have at least half the typical yield of fully formed phages? Is there an alternative solvent that can suspend and disperse the viral particles and buffer solution, and then water can be subsequently added as a solute to drive the reaction? Of course, I am googling away at this question. But maybe a person reading this knows or at least has heard of something like this? Thank you in advance!

By anhydrous you mean mixing the same concentration of viral proteins in a polar solvent that is not water? What would you suggest? Methanol?

@BuckThorn yes, some type of solvent that is not water, so that water can be subsequently added in varying degrees to drive assembly. Is there a paper written on this already? What other solvents are possible candidates besides methanol?

Proteins can sometimes be dissolved in solvents other than water, but there is a risk of denaturation that can then lead to aggregation. Proteins aren't like lego pieces that remain in one uniform shape regardless of solvent.

Good solvents for proteins are trifluoroethanol, phenol, and maybe HFIP?  I think most of the time solubilization will also involve denaturation though.

Also for lambda phage assembly you need to solvate the genomic DNA, too, right?  It's much harder to imagine DNA as very soluble in anything that isn't water...

After a lot of research, I have narrowed it down to a few candidates: cyclohexane, methyl formate, 1-dodecanethiol, acetate, 1-Ethyl-3-methylimidazolium chloride, ethylammonium nitrate (EAN), ethanolammonium nitrate (EtAN), glycerol, and good old methanol. The last four are the most promising. Please note however that none of this has been confirmed yet through experiments.

How on earth did you narrow it down without doing experiments?

Ummm.... exactly. I am in the design phase, narrowing down possible candidates before I go spending money. Any ideas?  Thanks!


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Does each liquid in a binary mixture boil at its boiling point?
I have been looking for an explanation and can't seem to find one, so I am asking my question on this forum. I appreciate your responses.
Consider a binary mixture of cyclohexane and toluene (50% mol:50% mol) and I want to separate them with simple distillation or fractional distillation. Each component, as a pure liquid, will boil at its boiling point. However, Raoult's Law shows that each component will have lower vapor pressure when in a mixture. Do cyclohexane and toluene boil at their boiling points or will each component boil at a different temperature? I am not concerned with the boiling point of the mixture, but with the boiling points of each component.
Cyclohexane has a boiling point of 81 °C
Toluene has a boiling point of 111 °C

What does the phase diagram look like?

I was not provided with one. I am in an Organic Chemistry I lab course and the experiment was about separating the binary mixture specified. I was asked to discuss how Raoult's Law and Dalton's Law are relevant to the lab, but I am confused about how those are relevant as the two laws are for mixtures.

A quick Google will turn up many resources showing the liquid-vapor phase diagram.

The apparatus that was used had a thermometer measuring the temperature of the vapors. Cyclohexane boiled at 49°C for simple distillation and
46°C for fractional distillation. Toluene boiled at 52 °C for simple distillation and 62 °C for fractional distillation.

From what I can tell, a phase diagram will not be useful as I am not asking for the boiling point of the mixture. I would like to know if each component boils at its respective boiling point or not.

The pure components boil at their standard boiling point. The mixture boils at some other temperature determined by the composition as can be seen on the phase diagram. So I’m a bit confused on what you are really asking here.

Does cyclohexane boil at 81 °C and toluene boil at 111 °C in the binary mixture specified? Will at 81°C cyclohexane begin to be collected in the vial? Will at 111 °C toluene begin to be collected in the vial? Or will cyclohexane and toluene boil at different temperatures when in a mixture? This is what I am confused on.

The mixture boils at the point on the liquidus. Asking what temperature a component of the liquid boils at does not make sense thermodynamically. The solution has a boiling point at that composition, not individual components.

I'm understanding it better now. Thank you!

I am not concerned with the boiling point of the mixture, but with the boiling points of each component.

Mixtures don't have a boiling point like pure liquids do. Typically, the composition of the vapor that boils off of a mixture is a mixture of the components as well, but with different composition than the liquid phase. As a consequence, the composition of the liquid phase changes continuously, also changing the temperature at which it boils.
The exception to this general concept is when the composition of the escaping vapor is the same as that of the liquid phase. In that case, you have an azeotrope, and the composition or boiling temperature does not change for the remainder of the distillation (and it is impossible to purify one or the other component by distillation under these circumstances).

Do cyclohexane and toluene boil at their boiling points or will each component boil at a different temperature?

Whether you can sustain a "rolling boil", i.e. bubbles forming within the liquid phase, depends on the sum of the vapor pressures of the components. This has to be higher (at a given temperature and composition) than the ambient pressure for boiling to occur.
The situation is easier if one of the components is not volatile at all. Confectioners work with sugars dissolved in water, and the boiling temperature increases as water is boiled off. The temperature is used to determine the sugar concentration of the concoction or confection.

Thank you! Your response was very helpful!


text:
Measuring Rare Earth Element Concentration in a Solution
I am interested in repeating the experiment as detailed in this paper: Study on Rare Earth Elements Leaching from Magnetic Coal Fly Ash by Citric Acid
In the paper, coal fly ash is leached by citric acid. A sample of the resultant mixture is centrifuged, and the liquid component is filtered and then analyzed to determine the concentration of rare earth elements.
When the solution containing the REEs is fully filtered, how would one measure the concentration of REEs without the use of an atomic spectrometer? I do not have access to expensive laboratory equipment, and was wondering if there is an alternative that would work in my case.

What instruments are available?

@AChem Not much, high school lab. Stirrer, hotplate, scale, oven.

No, then forget about this analysis. Think about a relatively easy project with your teacher. There are plenty of possibilities with simple ideas.

Rare Earths are notoriously difficult to separate for analysis so this probably cannot be done with high school lab equipment. Ion-exchange chromatography for separation and/or ICP-MS for elemental analysis are the generally accepted methods I believe.

Ok, thank you both for your help. I'll try and come up with another idea to pursue.

@rubpy32, What about a grating spectroscope in a box for capturing flame emission spectrum in a photograph? You can do wavelength calibration with a simple fluorescent light.

@AChem Great suggestion! Perhaps you can provide a nice understandable publication the OP might find helpful?

@EdV, If the OP searches the keywords "shoebox" spectroscope, he/she can find many useful articles. Starting articles could be Wakabayashi, F. Resolving Spectral Lines with a Periscope-Type DVD Spectroscope. J. Chem. Educ. 2008, 85 (6), 849,  DOI: 10.1021/ed085p849.

Another decent summary is Review of Student-Built Spectroscopy Instrumentation Projects, J. Chem. Educ. 2020, 97, 8, 2185–2195. Plenty of references including some of my own contributions.

@AChem Will take a look. Once again, thank you for your help.

If you have come up with a solution, feel free to self-answer it. Self-answers are appreciated and it helps future readers.