Which unit should I use for m/z? I've written m/z 300 - 1100 Thomson in my paper, but the reviewer said that the Th is already deprecated. So in this case, should I write m/z 300 - 1100 Da? or something else?

TL;DR: According to current IUPAC recommendations, $m/z$ is an abbreviation for a dimensionless quantity.
Use of thompson unit is indeed currently discouraged.

There is an overview provided in Definitions of terms relating to mass spectrometry (IUPAC Recommendations 2013) [1, p. 1516] (reference numbers were updated):

Labeling of mass spectra
The labeling of the $x$-axis of a mass spectrum engendered the most discussion during the creation of this document; however, in spite of a general desire for a better way to label the $x$-axis of mass spectra, there was no broad consensus for any of the proposed changes.
Therefore, this document continues the use of the definitions of the Gold Book [2] and the similar definitions in the Orange Book [3].
The Gold Book recommendation is for the use of $m/z$ as an abbreviation for mass-to-charge ratio, a dimensionless quantity obtained by dividing the mass number of an ion by its charge number [2].
The thomson unit, defined as the quotient of mass in units of u and the number of charges $(z),$ was proposed nearly two decades ago [4], but has not been widely adopted and is therefore not recommended.
Labeling the $x$-axis of a mass spectrum with any unit of mass such as dalton (Da), atomic mass unit (amu), or unified atomic mass unit (u) is strongly discouraged due to the confusion that would result when reporting spectra of multiply charged ions.
The quantity plotted on the $x$-axis of a mass spectrum is a function of both the mass and charge of the ion.
Furthermore, the use of amu in place of u is strongly discouraged in all cases; it has been used to denote atomic masses measured relative to the mass of a single atom of $\ce{^{16}O},$ or to the isotope-averaged mass of an oxygen atom, or to the mass of a single atom of $\ce{^{12}C}.$

Gold Book entry:

mass-to-charge ratio, $m/z$
The abbreviation $m/z$ is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although $m$ is not the ionic mass nor is $z$ a multiple or the elementary (electronic) charge, $e$. The abbreviation $m/e$ is, therefore, not recommended. Thus, for example, for the ion $\ce{C7H7^{2+}}$, $m/z$ equals $45.5$.

For historical purposes the aforementioned letter to the editor regarding the thompson unit [4]:

The ‘Thompson’. A Suggested Unit for Mass Spectroscopists
Recently, there has been a rapid increase in experiments in which multiply charged ions are generated in mass spectrometers. […]
This welcome development has permitted the measurement of molecular weights of compounds having masses in the tens and even hundreds of kilodalton.
These advances exacerbate the problems caused by a common imprecision,
namely, the terms mass measurement, mass range, etc are often used imprecisely when mass-to-charge ratio measurement, mass-to-charge range, etc are intended.
The implicit assumption that the charge state of ions is unity is no longer valid when very-high-charge states are so easily produced.
With this letter come two suggestions.
First, that the longer but more exact term be used whenever it is intended.
Second, that a unit of mass-to-charge ratio be adopted.
After all, this is the quantity which all mass spectrometers measure; mass is a derived quantity requiring an independent measurement or knowledge of charge.
Such a unit would be defined as the quotient of mass, in units of u* and the number of charges, $z.$
The number of charges could be positive or negative, depending on the sign of the charge.
The name Thomson suggests itself in view of J. J. Thomson’s contributions to measurement of this quantity and his preeminent role in the evolution of mass
spectrometry.
Using standard rules for abbreviation, we have $\pu{1 Th} ≡ \pu{1 u/\text{
atomic charge}}.$
For example, the molecular weight of the peptide myoglobin (isotopic average molecular weight $\pu{16950 u})$ can be deduced from measurement […] of a peak at $\pu{998.0 Th},$ provided it is known that the ion bears $17$ charges.
If this suggestion is accepted, other simplifications ensue.
For example, the benzoate anion, mass $\pu{121 u}$ and charge $-1$ atomic units, is $-121$ Thomson not $m/z~121.$
The latter is actually the mass-to-charge ratio of the corresponding (unstable) cation!

* $\pu{1 u} = m(\ce{^{12}C})/12 = \pu{1.660540E-27 kg}$
The unified atomic mass unit, $\pu{u},$ is also known by the name Dalton and
the symbol $\pu{Da}.$

References

Murray, K. K.; Boyd, R. K.; Eberlin, M. N.; Langley, G. J.; Li, L.; Naito, Y. Definitions of Terms Relating to Mass Spectrometry (IUPAC Recommendations 2013). Pure and Applied Chemistry 2013, 85 (7), 1515–1609. DOI: 10.1351/PAC-REC-06-04-06. (Free Access)
IUPAC. Compendium of Chemical Terminology (the “Gold Book”). Online Version (2019-) Created by S. J. Chalk, 2nd ed.; McNaught, A. D., Wilkinson, A., Eds.; IUPAC Recommendations; Blackwell Scientific Publications: Oxford, 1997.
IUPAC. Compendium of Analytical Nomenclature (the “Orange Book”), 3rd ed.; J. Inczédy, T. Lengyel, A. M. Ure, Eds.; IUPAC Recommendations; Blackwell Science: Oxford, 1998. https://media.iupac.org/publications/analytical_compendium/
Cooks, R. G.; Rockwood, A. L. The Thomson-A Suggested Unit for Mass Spectroscopists. Rapid Communications in Mass Spectrometry 1991, 5 (2), 93–93. DOI: 10.1002/rcm.1290050210.


Tap water leaving white residue after drying There is no heating or cooling in this process.

We recently move to a new house and something strange is happening with the water in every tap of the house.

First thing I noticed is when you fill a bucket with the garden hose the bubbles in the water lingered around longer than they should have, ie. normal senario fill bucket with water and bubbles pop immediately, this senario the bubbles take a few good long seconds befor they pop.

Second thing I noticed was after waking away from the garden hose and returning after everything was dry there was white residue all over anything that the water had been on, so I wipe it down and did it again and same thing happened, so I went and tested all the taps in the house with the cold water and the same thing happened for all the taps, so all the taps are using the same water source.

I have heard of white residue being left behind after heating from a source like a kettle but never heard of white residue being left behind from a non heated source.

Is there any way I can find out what this is and should I be worried about this (more than I already am)?
Or alternatively does anyone here know what this is?

You either have hard water or a water softener system.
If you have hard water, then the residue is a combination of calcium / magnesium carbonate / chloride.
If you have a water softener system, then the residue is from the salt used to remove the hard ions.


Making oxygen from iron oxide (aka rust) with sodium chloride? I like videogames, and I've been playing a lot of "Oxygen Not Included" recently. It's not a perfect physics simulation, but it does appear to be realistic-ish (both Specific Heat and Thermal Conductivity make up non-trivial parts of the gameplay; attempting to respect the tyranny of the rocket equation; etc.)

The latest update added in a new reaction though, that has me scratching my head a bit:

  Rust (Iron Oxide) + Salt (NaCl) -> O2 + Cl

Other reactions implemented in the game make sense based on my (limited) understanding of chemistry, e.g.  Iron + Carbon + Lime = Steel, or H2O+ Electrolysis = H2 and O2, so I'm wondering if there's a rational basis for the Rust -> O2 reaction as well, or if this is the devs handwaving some chemistry for gameplay purposes. (Such as, for instance, the apparent annihilation of the Sodium into heat, above)

This reaction isn't feasible and won't likely occur.

Why not?

Rust and Salt are very stable, and happy with their current state. You can see this better by looking at each product.

Rust forms from Iron and Oxygen

  4Fe + 3O2 → 2Fe2O3

Salt is formed from Sodium and Chlorine

  Na + Cl → NaCl

The formation of both of these products is very favorable because both molecules want to bond together to complete their valence shells.

They do not want to break their bonds, and doing so is possible but requires a lot of energy making it unfavorable. This is why we don't see salt or rust breaking down into its elements without a high amount of energy. This all applies for the overall reaction. All the reactants are already happy and don't want to break down making this reaction, possible, but not feasible due to the high amount of energy that it takes.


Why does capping after each coupling in Solid phase peptide synthesis improve the results? Why does capping after coupling improve results during solid phase peptide synthesis (SPPS)?

The Case: I have conducted a lab experiment where we coupled three Fmoc-protected amino acids, and did a final capping (acetylate) at the end of the synthesis. However, I was told that one can theoretically cap after every single amino acid coupling to ensure that the final product is correct. That is, one can "eliminate" or terminate the synthesis of peptides where the amino acids are innocently coupled (?).

I don't really understand why, because if you cap then you "block" all of the peptides that you are synthesizing (?). Therefore, how can you add further onto the chain if you cap after each amino acid? This seems counterintuitive, or I might be a bit confused. If you are able to de-protect the cap to further couple on more amino acids, then what is the point of capping?

Is there someway that capping only “caps” onto amino acids that are not coupled or that did not couple the previous amino acid somehow?

if you cap then you “block” all of the peptides that you are synthesizing.(?)

No, you don’t, the order of steps is important. First you try to couple your new amino acid to the solid phase. You will end up with uncoupled ends with free amino groups and coupled ends with a protected amino group. Then you cap, e.g. with phenoxyacetic acid anhydride. This will only cap the uncoupled ends with free amino groups. Then the Fmoc or whatever protective group is removed rendering free amino groups for new couplings.

Say you wanted to create the sequence $\ce{CHEMISTRY}$. And say at one specific spot on the resin, you failed to couple the serine residue; you have $\ce{H2N-TRY-CO-resin}$ at that one spot. If you now do not cap, you might generate the sequence $\ce{CHEMITRY}$ because isoleucine will couple to threonine’s free amino group etc. By capping, you essentially generate $\ce{Ac-NH-TRY-CO-resin}$ which won’t disturb you anymore.

And the benefits don’t stop there. Without capping, at the end you would end up with the sequences $\ce{CHEMISTRY}$ (the one you want), $\ce{HEMISTRY}$, $\ce{CEMISTRY}$, $\ce{CHMISTRY}$, $\ce{CHEISTRY}$, $\ce{CHEMSTRY}$, $\ce{CHEMITRY}$, $\ce{CHEMISRY}$, $\ce{CHEMISTY}$ and maybe even $\ce{CHEMISTR}$. These are all of similar length to your desired product and hard to purify away via HPLC. But if you cap, you can only get $\ce{HEMISTRY}$, $\ce{EMISTRY}$, $\ce{MISTRY}$, $\ce{ISTRY}$, $\ce{STRY}$, $\ce{TRY}$, $\ce{RY}$, and $\ce{Y}$ — most of these are significantly shorter and easier to purify away.

Also, assume a constant amount of amino acids always being added to the resin to couple. If only the so far correctly built sequences are available, these will ‘see’ a higher ‘concentration per site’ of their coupling partners maybe resulting in slightly increased yield, too. (Note that this paragraph used layman’s terminology.)


How to extract water from a solution in formamide I'd like to determine the water content in a sample without buying an expensive machine (like Karl Fischer titrator). So I'd like to measure the water volumetrically. I've chosen formamide to dissolve the sample because of its high polarity.

I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.

1) Let's assume there is 0,5 ml water dissolved in the formamide fraction, is it possible to add 30 ml solvent B and get so a fraction of 30,5 ml (30 ml + 0,5 ml)?

2) Instead of formamide, do you know another solvent A which dissolves the sample including the water (and the watercontent of approx. 20% in the glucose syrup)) and a solvent B which extracts the water phase from the solvent A?

I'd like to determine the water content in a sample without buying an expensive machine (like [a] Karl Fischer titrator).

Rather than try to extract the water and measure it volumetrically, I think you could try to measure its activity, or else perform your own Karl-Fischer titration, which does not necessarily require any equipment.

Water activity measurement

Practically this means hygrometers.  That may sounds like another expensive machine, but E-bay tells me that capacitive hygrometers are available for under $20.  If your sample consists of sucrose, glucose, and glycerol, those substances are unlikely to interfere with capacitive hygrometry measurements of water activity.

You might even get very crude accuracy by incubating a mirror at a known (chilled) temperature, and then waving it above your water/sucrose/glucose/glycerol sample, seeing if dew forms, and then repeating the process at several different temperatures.

Karl Fischer titration on your own

As an alternative, couldn't you just set up your own Karl Fischer titration?  Commercial titrators often use coulometric detection for better accuracy, but setting up a volumetric titration is pretty easy.  You need:

Methanol (or some other alcohol)
Imidazole (or some other amine base)
Sulfur dioxide
Iodine

Imidazole and sulfur dioxide should be combined in a 1:1 molar ratio in an excess of methanol.  Iodine should be dissolved in methanol.

A 2012 paper has a nice description of how to do KF with visual end-point determination.  You just need to buy a KF titration solution, or make your own:

  Commercially available [...] Test Kit from [...] was used for reference moisture determination. All the equipments needed were provided with the kit. It consisted of two-component ethanol-based reagents: [a] working medium containing [a mixture of] imidazole, sulphur dioxide and diethanolamine, and [separately a titrant solution] containing iodine. According to manufacturer's instructions for use, 20 mL of [working medium] was added to the titration vessel, which was then tightly sealed with a septum. The titrant was delivered through the septum with a syringe to react with the present water, until a color change from colorless to yellow occurred. A known amount of sample was added by a syringe to the titration vessel via septum, and a newly filled titration syringe was used for a new titration until the color change from colorless to yellow occurred. Consumption volume of the titrant was read off the titration syringe. The water content, in percent by volume, was calculated from the consumption of the titrant and the sample volume.

If you've diluted with formamide

  I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.

The nice part about both water activity measurement and Karl Fischer titration is that they are both sensitive and accurate enough to still work on your sample even after dilution or dissolution in formamide.


What pressure units should be used when calculating electrode potential with Nernst equation? Example 17: Calculate the electrode potential of given electrode
  $$\ce{Pt,Cl2 (\pu{1.5 bar}) | 2 Cl- (\pu{0.01 M})}\,; \quad E^\circ_\ce{Cl2/2Cl-} = \pu{1.36 V}$$

  Solution: The reaction of electrode is
  $$\ce{\underset{\pu{1.5 bar}}{Cl2 (g)} + 2 e- -> \underset{\pu{0.01 M}}{2 Cl-}}$$
  $$E = E^\circ - \frac{0.0591}{n}\log{\frac{[\ce{Cl-}]^2}{P_\ce{Cl2}}} = 1.36 - \frac{0.0591}{2}\log{\frac{(0.01)^2}{1.5}} = \pu{1.483 V}$$

In the solution, the author has directly used the pressure (which is in bars) in the Nernst equation without first converting it to its SI units. How can this be right?

Just as you are supposed to use activity instead of concentration for dissolved components when applying Nernst equation, you are also supposed to use fugacity instead of the partial pressure for the gaseous components. The reason for that will become transparent if you follow the derivation of Nernst equation using chemical potential.

Fugacity is defined with a reference to the standard state with the pressure of $\pu{10^5 Pa}$ or $\pu{1 bar}$. At ordinary pressures fugacity is numerically approximately equal to the pressure expressed in bars or atmospheres ($\pu{1 bar} \approx \pu{0.987 atm}$), so the author of the book is right.


What is activating effect? In the book "Organic Chemistry" by Morrison and Boyd,It is written that

  Sodium borohydride cannot reduce weaker electrophiles like esters though Lithium aluminum  hydride can reduce esters to alcohols due to stronger activating effect of Li+ ion than Na+

What is activating effect?

(Google search did not give appropriate results.)

Consider a carbonyl group. We know that, under acidic conditions, $\ce{H+}$ can protonate on carbonyl $\ce{O}$ and hence activate the carbonyl group to nucleophilic attack at $\ce{C}$, and we call such an effect an activating effect. The metal ions $\ce{Li^+}$ and $\ce{Na^+}$ serve the same purpose as $\ce{H+}$, though $\ce{Li^+}$ is a stronger activator due to its high charge density and hence stronger attraction to $\ce{O}$.


What does negative current reading in galvanic couple mean? I used a multimeter in a galvanic testing between two metals. I connect one to metal A and the another end to metal B. Then I immersed the 2 metals in a salt solution, securing that they are apart in the solution. It gave me a negative current reading, then when I switched the multimeter probes, it became positive. What does the sign of current indicates?

The negative sign is simply indicative of how you've inserted your multimeter probes (which are labelled for polarity, or should be) into the galvanic battery circuit. In one case, the multimeter "sees" electrons flowing from the positive terminal to the negative one (hence the negative sign); in the other case, the opposite is true.


What are t2g and eg in CFT? In the crystal field theory (CFT), when the splitting of the d-orbital occurs, it gets divided into two parts.
The upper part with higher energy is the  $\mathrm{e_g}$ and the lower part with lower energy is called the $\mathrm{t_{2g}}$ as in:

So can anyone please explain me what is this $\mathrm{t_2}$ or the $\mathrm{e}$ or the $\mathrm{_g}$ below.

As stated in both the links Geoff and Philipp have kindly commented (1, 2) they are to do with symmetry labels we chemists like to assign to orbitals. Knowing an orbitals symmetry class can lead to a lot of simplifications down the road when you use quantum mechanical calculations and even dictate the reactivity of which orbitals are "allowed" to interact together in reactions.

In this case the $\mathrm{t_2}$ groups three of the metal atom's d-orbitals into a certain class while two of the orbitals belong to the $\mathrm{e}$ class. The $\mathrm{t}$ means triply degenerate while the $\mathrm{e}$ means doubly degenerate (degenerate means have the same energy).

The $\mathrm{g}$ is not about how many energy levels are degenerate rather it is an indication of the answer to a certain operation we can perform on an orbital. It instead relates to how the orbitals behave if we hypothetically were to put a line bisecting the orbitals or "invert" them around a centre point (crudely put "going from one corner to the other". If the phase of the wavefunction (or perhaps more pragmatic) the "colour" of the orbital lobes changes while doing this we label the orbital with a $\mathrm{u}$ if we have no phase change we label it with $\mathrm{g}$. They indicate the orbitals response to an inversion about the centre of symmetry.

We can do lots more of these operations to investigate the symmetry of the orbitals which together lead us to assigning the molecule to a symmetry group. Knowing this is very important for a plethora of reasons.


What does thermal stability of compounds depend on? Does it somehow depend upon the lattice energy of the compound?
My textbook says that Lithium carbonate is not so stable to heat and forms more stable $\ce{Li2O}$ and $\ce{CO2}$.
Could it depend on the electropositive character? Because my textbook further sates that the stability of carbonates increases down the group.

Lattice energy might be important if the decomposition took place in the solid phase.
But Li2CO3 melts at 723 C and boils (with decomposition) at 1310 C. Na2CO3 melts at 851 C and decomposes before boiling, according to an older CRC Handbook and Wikipedia, which seems to be pretty current.  According to those numbers, thermal stability of sodium and lithium carbonates seems fairly similar, and decomposition does not take place because of solid issues.

K2CO3 melts at 891 C, decomposes before boiling. Rb2CO3 melts at 837 C, decomposes at 900 C. Cs2CO3 melts/decomposes at 610 C. Lattice energy might be important with Rb and Cs, but thermal stability seems to become less stable as you go down the group, not more so.

The first three seem fairly stable for several degrees above the melting point. We could consider the melt to be a mix of CO3-- anions with the metal cations, all oxygen-bonded (similar to hydrogen bonding!), but much looser than in a crystal. The thermal stability of the CO3-- anion does not seem to be very dependent on the ionic size of these three metals (Li+, Na+, K+).

So, I would ask the textbook author to substantiate his claims.


What kind of molecules do I smell when draining cooked potatoes or rice? When I drain potatoes after cooking them, they exhibit a nice smelling odor. It is not very strong. Now I know that some perfumes use natural molecules and one of them could come from potatoes (or rice.) But I forgot what the name was for this molecule, or perhaps someone knows some perfumes using these kinds of compounds.

It would be interesting to know what the functions of such molecules are in the organisms. Just an enzyme, or possibly an attracting force for some reason?

What a good question. But unanswerable by me and my prior knowledge and skill in google-fu. I have some general answers. This is a complicated area, since smell is both a physical issue and an individual and unmeasurable issue. Olfactory science is at a surprisingly basic level, given how far we are generally. I tried investigating what compounds in rice provide odour, the answer seems to be; way more than I could possibly have thought. Check out this article - while it is rough to digest a scientific articl unless you have basis knowledge just look at that list at page 505-506. There are many aromas in rice!

As for what functions they perform - looking at the list it is probably impossible to generalize down to what functions such a wide array of compounds do. Certainly, some compounds that are odorous has a primary function of being odorous, attracting pollinators and attracting mates. For rice, which has been around humans since time immemorial (at least 9k BC) they have probably been through a selection process, the most benevolent smelling rice plants being selectively bred, but this is speculation on my part. Other compounds have other functions and just happen to have a strong odour. Generally, to be odorous, the compound must have some volatility. What can also be said generally and very broadly is that alcohols and aldehydes / ketones are more often odorous than other compounds. There are also certain inorganic compounds that are (and often strongly so), Ammonia, Sulfide and the "garlic" series (S-Se-Te) for instance.

Good question. If you learn more, let us know.


Does 50% hydrogen peroxide dissolve gold? We have seen that the silver electrodes on a piezo element are being seemingly dissolved by exposure to our mixture of $50\,\%$ hydrogen peroxide. I was wondering if we were to switch to gold electrodes instead of silver could this mitigate the issue. I have not found much if gold is resilient to $\ce{H2O2}$ and was looking here to see if anyone has some expertise.

I have no expertise, but Pourbaix does.  Let's take a look at some of his diagrams which offer a hint to what is going on, with both silver and gold.  Or more accurately, why gold is indeed less vulnerable to attack by hydrogen peroxide than silver.
Begin with just the diagram for hydrogen and oxygen in aqueous solution.  Ross[1] gives a diagram that includes hydrogen peroxide as well as the stable species hydrogen, oxygen and water (figure from this reference):

Pourbaix diagrams typically include a pair of parallel lines, here labeled (a) and (b) representing water reduction to hydrogen and oxidation to elemental oxygen respectively.  We see that hydrogen peroxide requires a higher oxidizing potential to form than elemental oxygen, by roughly half a volt depending on concentration.  This is no surprise, of course, since elemental oxygen plus water would be more stable.  Hydrogen peroxide is, of course, only metastable.
Now let's see the diagram for silver.  There is some variation from different sources, but Wei et al. [2] show the following, which is typical.

The boundary between elemental silver and various oxidized species (essentially, dissolved silver ions in acid and silver oxides in basic solution) actually lies below the line for water oxidation to oxygen, so in principle even dissolved air could attack silver given enough time, let alone the more powerful oxidant hydrogen peroxide.  In fact, at the potentials where hydrogen peroxide appears in the diagram from Ref. 1, one may even see the oxide of silver(II)!  We are lucky, actually, that the kinetics is slow enough so the silver electrodes have some lifetime.]
Now it is time to go for the gold.  Huayhuas-Chipana et al. [3] adopted the diagram below from Ref. [4], showing that gold is not oxidized at all when water is oxidized to elemental oxygen (the higher of the two red lines) and that even hydrogen peroxide with its potential at half a volt higher is only on the borderline of beginning to oxidize gold.  More over, the soluble cation $\ceAu^{3+}}$ appears only in fairly string acidic solution, further hindering any dissolution of gold.  Thus gold would indeed be expected to hold up more strongly in your hydrogen peroxide solution.

References
1.
P. N. Ross Jr. (2010), "Oxygen reduction reaction on smooth single crystal electrodes", Handbook of Fuel Cells
Vol. 2.
2.
Weifeng Wei, Xuhui Mao, Luis A. Ortiz and  Donald R. Sadoway, "Oriented silver oxidenanostructures synthesized through a template-free electrochemical route", J. Mater. Chem., 2011, 21, 432-438.
3.
Bryan C. Huayhuas-Chipana, Juan Carlos Morales Gomero, Maria Del Pilar, Taboada Sotomayor, "Nanostructured Screen-Printed Electrodes Modified with Self-Assembled Monolayers for Determination of Metronidazole in Different Matrices", Journal of the Brazilian Chemical Society 25(9) (Sept. 2014).
4.
J. Van Muylder, M. Pourbaix, Atlas of Electrochemical
Equilibria in Aqueous Solutions, 2nd English ed.; Marcel
Pourbaix, National Association of Corrosion Engineers: New
York, 1974.


What to do with (large) imaginary frequencies for constrained minimum structures? I am performing DFT calculations using ORCA 4.0.1 on an enzyme active site model. The model contains 89 atoms including the substrate (see Animation 1), five of which are fixed in space (the spherical atoms in Animation 1). I am using dispersion-corrected B3LYP with 6-31g(d,p) basis set, with the "RIJCOSX" approximation, and with a CPCM model for the solvent (dielectric constant $\epsilon = 4)$. From what I have seen in the literature, this method is often applied to enzyme models with success, and this is the reason for using it. Once I start computing reaction paths, I plan to try different functionals and basis sets to see if this give drastically different results. My geometry optimization tolerances are

Energy change: 5e-6;
Max gradient: 2e-3;
MRS gradient: 5e-4;
Max displacement: 4e-3;
RMS displacement: 2e-3

The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient). Additionally, it has been absolutely necessary to compute a better guess for the initial Hessian for the optimization, done at the B3LYP/3-21G-rijcosx level.

The fixed atoms are kept fixed to maintain the orientation of the active site residues. Only atoms where the side chain or backbone has been truncated are kept fixed. Without fixing these atoms, the residues will most likely  deviate way too much from the crystal structure, and no longer represent the enzyme environment we try to simulate.

Now, due to the constrained atoms, I expected imaginary frequencies to occur. However, for one of my presumably converged structures, I have quite strong imaginary frequencies. The list of frequencies start as follows: -220.645934, -69.117105, -65.597557, -61.314719, -48.720497, -40.439949, -32.562415, -21.395791, -16.637113, 6.126145. The two strongest negative frequencies are mainly a sigma bond rotation of a methyl group (of which the carbon is fixed in space during the optimization).

It is difficult to find any good literature on what to do with and how to interpret such negative frequencies in constrained optimizations. Is it "safe" to disregard the frequencies that mainly involved fixed atoms? Should I converge my geometry tighter to come closer to the true minimum? That's not so easy, I find, as the molecule is quite flexible. It took me quite some time to come up with a method that converges, which involve computing a good guess for the initial Hessian used in the optimization (computed at the B3lYP/3-21G level). I hope the imaginary frequencies are artefacts of the constraints, and that I can safely disregard them. However, they seem to have quite higher frequencies than what other people report in the literature (many say that their imaginary frequencies are around 20 cm-1).

This has to be a common challenge for people doing electronic structure calculations on large and constrained systems, but very few address this issue specifically in their publications.

I did read through this question, but it addresses not quite the same issue that I face.

Animation 1. The active site model. A total of 89 atoms.

Animation 2. Imaginary mode with frequency -220.645934 cm$^{-1}$.

What I am trying right now

I have three jobs waiting in the queue right now:

I have made a small rotation of the methyl groups strongly involved in the strong imaginary mode
I have started another geometry optimization of the structure given above (exhibiting the imaginary frequencies), but by reading the computed Hessian (the "exact" one, from same level of theory as optimization)
I have restarted an identical optimization, but by computing the "exact" Hessian from the start, giving a much better guess of which direction to go in.

I can add here later what these attempts result in.

The first I noticed is you said:

  The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient).

At first sight, the structure of your molecule should not leads to serious problems, but the above is a bad signal, and turns the results unreliable (at least doubtful). The constrains must be considered in the calculation of vibrational frequencies for they make sense.

Maybe no one can give you a foolproof steps series to follow. My suggestions are below.

What to do/check

If comments are right, and Orca does not care about fixing atoms for
the calculation of the vibrational frequencies, then you need to
change the software for this task (because in this case results are not meaningful) (This also answer the question "Is it "safe" to disregard the frequencies that mainly involved fixed atoms?"). The simpler way to check this is by looking at the number of freedom degrees reported (or deduced form the output). I was referring directly to vibrational f.d. because low frequency modes may be treated like rotations. I doubt that Orca lacks of this basic functionality. In such case I bet that this is described their User Manual.
The step you are doing are reasonable. My guess is that maybe you are far from a true minimum, in some unstable geometry, and that is why you have problems with accuracy changes. You said: "Should I converge my geometry tighter to come closer to the true minimum? ". Yes, it is always advisable to get good frequencies. So I would use not only tighter geometry convergence criteria but the TIGHT convergence option for the SCF part.
In the same lines, you can try using modern semi-empirical methods and then make the fine tuning with DFT. Even using a simpler DFT functional.
I agree with comments that you are not using the best basis set out there, but I think that this is not a problem here.
If you want to discard weird behaviors you can try with another functional, but I do not think that this is the problem (because it is known that B3LYP performs well in compounds with the elements involved here).

In brief. I would check if Orca compute correctly the frequencies, if not: change the software. If yes: discard those results, try to converge the structure using cheaper methods (if you are lack of computational resources) from different initial conditions, store those results and start new geometry optimizations and vibrational frequencies calculations starting from those previous converged states. The last step using the tight convergence criteria for SCF and normal convergence criteria for geometries, and using the target basis set (for energy and its derivatives).

Good luck

Note:

  Once I start computing reaction paths, I plan to try different
  functionals and basis sets to see if this give drastically different
  results.

Why do you expect that something like this would happen? You can save time by using a little large basis set and a better functional once you get the B3LYP/6-31g(d,p) results. It should not take too much time.

EDIT DUE TO COMMENTS

This answers generated some discussion about B3LYP or not B3LYP?

In that respect, my viewpoint is as follows:

B3LYP is not the most accurate functional (there should be no doubt about that).
It is reasonably good for geometries in systems like yours. Errors in geometries can be larger due to small basis set than due to B3LYP, or at least of the same order.
While there are functional with better results for geometries, you also must take care of how large basis sets those functional needs to be a better option. I mean, maybe using 5$\zeta$ basis sets they are better, but not using a small basis set.  (and maybe not)
The above discussion can be very long, but in any case, you are using approximations that (in my opinion) are much stronger than the errors due to  B3LYP. For example, the permittivity value, or the geometry of the constained parts, etc..
Also, even if you could perform a FullCI/CBS you shoud notice that the simulated system is different to the real system, so extremely exacts results in the simulated system may no improve your knowledge about reality.
So, if the functionals are accurate enough, what it is important here is obtaining the corresponding results for the functional/basis set. The functionales can be more or less problematic. B3LYP should not be problematic for your system. My guess is what I said in the answer.
Because of the above, I think that B3LYP is reasonable and enough for your case.


Can we process waste just by decomposing it to homogenous mass and extracting the elements from it? I. e., what if we just burn trash completely or dissolve it in acid, get the resulting paste and extract pure chemical elements out of it by applying some physics and/or chemistry to it until no more can be extracted? Just curious.

EDIT

To be perfectly clear: my main concern is whether there are any means to extract any useful element from this bulk of burned/dissolved/blender-ed trash. For example, we burn with Thermite the whole contents of some waste disposal site, until it's only multicolored ash there. Or pour the HSbF6 over it. From there, can we get the pure carbon out of the resulting mass? Pure aluminium? Pure iron? Gases, maybe, like oxygen?

Sure, this absolutely can be done.  In particular, you cannot destroy elements using conventional means like burning, so all the initial elements are still in the mass.  The question is whether or not this is a good way to extract what you want.  There may be better chemical techniques to do the extraction that are short of burning.  The branch of chemistry which deals with the optimal conversion of elements and chemicals to other useful substances is chemical engineering, and figuring out which process is "best" (e.g. fastest, cheapest, highest yield) is crucial for industry.

There is a very funny story related to this idea on the Periodic Table of Videos.  A scientist was trusted with the entire UK supply of plutonium (10mg, a very small amount) during World War II.  He accidentally spilled it on to his lab bench.  He then took a saw, sawed a large circle out of bench, burned that wood, and recovered nearly all the plutonium!  So yes, this works.  Here's the video, and this story begins around 15:30: Plutonium Video

EDIT: As to your update, there are lots of methods available, but it depends greatly on the starting conditions and materials.  For example, burning is a great way to extract carbon: burning hydrocarbons or carbohydrates burns off the hydrogen or water (respectively) and leaves mostly pure carbon.

But extracting metals might take more work.  Often, metals exist as ionic compounds.  For example, silver nitrate, $\ce{AgNO3}$, contains siver $\ce{(Ag)}$ in the +1 state.  While it might be possible to dissociate the silver with high enough temperature, it is far easier to extract metallic silver by reducing silver in the presence of a more reactive metal.  "More reactive" simply means a metal that will more easily give up an electron to the less reactive sliver, changing the silver ion $\ce{Ag+}$ to metallic $\ce{Ag}$.  Silver nitrate is highly soluble in water, and if you simply drop in a piece of metallic copper $\ce{(Cu)}$ metallic silver will slowly come out of solution and plate the copper with silver, while copper dissolves and creates copper nitrate.  The equation is:

$\ce{2AgNO3 + Cu->2Ag + Cu(NO3)2}$

but the crucial part is the ionic equation:

$\ce{2Ag+ + Cu->2Ag + Cu^2+}$

which shows that the silver ion is reduced to silver, and the copper metal is oxidized to copper ion.

Again, Periodic Videos has the footage of this exact reaction: Silver Video

Now I picked this example precisely because of this footage, but it's not the best example because silver is so low in reactivity that it exists metallically in nature (i.e. you can find lumps of silver if you are lucky!)  But many metals are so reactive that it takes a huge amount of energy to extract them.  Examples are sodium or potassium, which are so reactive they explode in the presence of water (go search YouTube and have a field day!).  To extract them as metals commercially usually requires electrolysis, which uses electricity to force electron transfer which would not otherwise occur spontaneously.

And on and on with other methods.  There is no universal extraction method, at least not yet, and this is why we have chemcial engineers to figure out better and better procedures to use.


Where do I find datasheets of zeolites? I'm looking into uses of zeolites as water absorbers for heat storage and absorbtion cooling. However, I don't find hard numbers on the likes of absorption enthalpies at different temperatures, saturation moistures, etc.

I'm not an institution, chemical suppliers are traditionally reluctant to sell to private individuals. I found zeolitic absorbers in few catalogues, but only with a descriptor of the type, no further data. One company I asked (that sell zeolites in amounts of kg to private customers) was unwilling to share data on their product (They're an engineering company doing product developement with zeolitic absorbers for heating/cooling etc.).

I only found (sketchy) data on single zeolites in working papers, but no comprehensive source.

So: seeing that vendors don't share the info, where do I find datasheets of zeolites?

I've searched for this same data.  My interest regards electronic proporties of zeolite-class media.  I've written this same kind of question on Quora.

I think the reason this is rarely answered, and why there are no 'catalogs', is several-fold:

There exist a wide range of applications, based on the composition
of the water / etc.
The treatment media usually varies in
composition, meaning that no directive for the use of a zeolite
product can be given by a sales rep (only on-staff engineers, who
know how to use it).
Some uses create by-products that are regulated, creating possible liabilities for the seller by indicating a fitness-for-use.
Regulations change; per year, per locale, etc.

I think the groups that know about these are environmental engineering firms, whom have a stable of engineers that provide end-to-end services for their clientele.  Many of of these firms have Chemists on-staff, or a consultancy relationship that suits the same.  They know the chemical properties of their media, and need a zeolite / catalysts that can effect something very specific.  As a result, they're out their hunting the journals for something that targets their needs.

Hard to out-sell that without already having a dog in the game.


Why is the Van der Waals constant of water lesser than benzene? Van Der Waals constant of water and benzene ($a$) are $5.537$ and $18.82$ respectively.
I have read that the value of Van Der Waals constant indicates the extent of attractive forces.
By this, shouldn't it be greater for water instead because water has hydrogen bonding?
Also, what are the factors(with priority) on which the Van Der Waals constant depends on ?

The parameter $a$ from the van der Waals equation

$$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$

is not to be taken only as the measure of strength of intermolecular bonding.

Rather, it is overall measure of significance and frequency of intermolecular interaction.

It may be based on purely elastic electrostatic interactions, where only the weakest van der Waals force may ( or even may not ) be involved.

The probability 2 benzene molecules interact is bigger than for 2 water molecules.

Notes also that for n-hexane is $\pu{a = 24.71 L^2\cdot bar mol^{2-}}$. It has bigger value than benzene, in spite of the weaker interaction due the lack of $\pi$ bonding. But its linear molecule is stratched with all the hydrogen atoms in many directions, so collisions are more probable.


Explain the difference in stability of permanganate ions in acidic/ alkaline solutions? The question is such:

  Use the data to comment on the stability of $\ce{MnO4-}$ in acidic and basic solution. Suggest a reason for this difference.

  $$\begin{align}
&\text{acidic solution:}&\ \ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}\qquad &E^\circ&=+1.70\ \mathrm V\\[6pt]
&\text{basic solution}:&\ \ce{MnO4- + 2H2O + 3e- &-> MnO2 + 4OH-}\qquad &E^\circ&=+0.59\ \mathrm V
\end{align}$$

I know that the permanganate ion is more stable in alkaline solution than in acidic solution, as the more positive electrode potential in acidic solution means more negative Gibbs free energy, therefore the reaction is more feasible in acidic solution, therefore less stable...

However, is there a chemical reason for this?

Actually, yes.

As we know, $\ce{H2O = H+ +OH-}$ . Chemical equation can be threated as algebraic equations as long as the final result has chemical meaning (i.e. no negative coeffecients ... unless you understand their meaning), thus we can add $\ce{4H+}$ to both sides of the second equation. This gives us equation

$\ce{MnO4- +4H+ + 2H2O +3 e- \rightarrow MnO2 + 4H+ +4OH-}$

Obviously, we can collapse four pairs of $\ce{H+}$ and $\ce{OH-}$ into $\ce{4H2O}$

$\ce{MnO4- +4H+ + 2H2O  +3 e- \rightarrow MnO2 + 4H2O}$

and now we can drop two $\ce{H2O}$ from both sides

$\ce{MnO4- +4H+   +3 e- \rightarrow MnO2 + 2H2O}$

which is exactly as your first equation. So, in fact it is exactly the same reaction. However, WTF they are written differently and the corresponding potentials are different? This is because standard potentials are implied to be measured at concentration of all ions and molecules (except water) in corresponding half-reaction  1 mol/l. However, what happend when one concentration deviates from 1 mol/l ? The corresponding potential deviates from its value as well, in accordance with Le Chatelier's principle.

When we set up concentration of $\ce{OH-}$ at 1 mol/l, we set $\ce{H+}$ at roughly $10^{-14}$, since $\ce{[H+][OH^{-}] \approx 10^{-14}}$ (water autoprotolysis constant). According to Le Chatelier's principle the equilibrium should move to the left, which is observable as a change in electrode potential.

The quantitative expression of this kind of dependence is known as Nernst equation.


Boiling point elevation with multiple solutes I've encountered a problem where glucose and $\ce{NaCl}$ are mixed with water and $\Delta T$ is to be calculated.
I've calculated  $\Delta T$ for both glucose ($i = 1$) and $\ce{NaCl}$ ($i = 2$) separately and added them.
Is this procedure correct or should I have treated them as one solute and used $i = 3$ as the Van't Hoff factor?

When there is more than $1$ solute in the solution, the total elevation in boiling point ($\Delta T_\mathrm{b,total}$) or the depression in freezing point ($\Delta T_\mathrm{f,total}$) is defined as the sum of each individual value for each type of solute.
What is Van't Hoff factor? According to Chem Libretexts, the Van't Hoff factor is defined as:

$$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \tag{1} \label{1}$$

Now, why does this work? Let's take an example to see how it works.
We take a simple case as what you have mentioned in the question:

Find the elevation in boiling point of a solution of $\pu{1 m}$ glucose and $\pu{2 m}$ $\ce{NaCl}$ [$100$% dissociated] in water.
$\mathrm k_\mathrm {b,water}= \pu{0.52 ^\circ C kg mol-1}$

We can do this in two ways. One, where we take $i = 2$ for $\ce{NaCl}$. Second, we state that $\ce{NaCl}$ dissociates into $\ce{Na+}$ and $\ce{Cl-}$ and take the three types of solutes to be present instead of two [going from basic principles].
Method $1$
Here we would have two different solute particles. Finding the value of $\Delta T_\mathrm{b}$ for both glucose and $\ce{NaCl}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,NaCl} \\
&= \mathrm{k_b}[1\cdot1 \,+2\cdot2] \\
&= 5\cdot\mathrm{k_b} \\
\end{align}
Method $2$
Here we would have three different solute particles. Here, $\ce{Na+}$and $\ce{Cl-}$ would have a value of $i=1$ and their individual molality would be identical to that of $\ce{NaCl}$ since the salt was $100$% dissociated in solution
Finding the value of $\Delta T_\mathrm{b}$ for glucose, $\ce{Na+}$ and $\ce{Cl-}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,Cl^-} +\Delta T_\mathrm{b,Na^+} \\
&= \mathrm{k_b}[1\cdot1 \,+1\cdot2 + 1\cdot2] \\
&= 5\cdot\mathrm{k_b} \\
\end{align}
Therefore, both methods yield the same answer. The second method is actually the derivation for value of $i$ for a dissociating salt.
Now, you can apply $i_\text{total}=i_1+i_2$ only when the concentrations of the given solutes are the same. The proof is similar to how the second method proves the first above. In such a scenario, you can assume them to be one solute since colligative properties don't care about how many types of solute there are but rather care only about the amount of solute particles present in the solution.


Does brewing tea with hard water reduce the amount of bioavailable fluorine? I read that brewed tea contains a relatively large concentration of Fluorine.

Fluorine can have negative health effects such as Skeletal Fluorosis. EDIT: I understand some Fluorine is necessary and has benefits. This is just to inform why I have structured this hypothesis. I am interested in the chemistry not the health effects as that is a different topic.

Tap water and especially hard water contains lots of Calcium ions.

Would Calcium and Fluorine bond to form Calcium Fluoride? I'm obviously not a chemist, but I assume it would be a lower energy state than the soluble ion pair in the tea, correct? It forms Fluorapatite on tooth enamel.

This Calcium Fluoride SDS indicates that eating solid CaF2 straight won't immediately kill you, so I assume Calcium Fluoride does not decompose in HCl stomach acid, correct?

So with that logic would the following be true?

Mol bioavilable F-  =  Mol F- from brewed tea  -  2*(Mol of Ca2+ in hard water)

If that is the case, it brings up some other interesting questions, like if moderately hard municipal water actually has any Fluorine in solution or not...

EDIT: To select a best answer I need to know:

if Ca ions bond with F ions in solution to form CaF2
and to know the percent or time-rate solubility of CaF2 in stomach acid.

Thanks!

$\ce{CaF2}$ has a relatively low solubility in water, about 15 mg/L, with a $K_{sp}$ of about $4\times 10^{-11}$.  According to the paper you linked, the concentrations of fluoride in tea are around 5 mg/L, which corresponds to about $1.3\times 10^{-4}$ M. Thus, the concentration of $\ce{Ca^2+}$ required to precipitate the fluoride out is about 2.5 mM, which corresponds to 0.1 g/L. Hard water is defined as having a calcium concentration above about 0.6 mM, with very hard water being 1.8 mM calcium. Thus, it is unlikely that there would be high enough concentrations for $\ce{CaF2}$ to precipitate out of solution.

Because of the large solvation energy of fluoride ion, the molecules in solution are likely to be nearly completely ionized. That is, the fluoride anions and the calcium cation will each have their own solvation shell of water molecules. That means that they will not interfere with each other until the concentrations are high enough that solid $\ce{CaF2}$ begins to precipitate.

As long as the concentrations are below the solubility limit, the presence of calcium ions will therefore not affect the availability of the fluoride ions.

Furthermore, since HF is a weak acid, in the acidic environment of the stomach some of the fluoride ions will be protonated to HF, decreasing the concentration of free fluoride and increasing the apparent solubility of $\ce{CaF2}$.


Calculating amount of ice required for heat loss Question:

  Given $400.0\ \mathrm{g}$ of hot tea at $80.0 °\mathrm{C}$, what mass of ice at $0 °\mathrm{C}$ must
  be added to obtain iced tea at $10.0 °\mathrm{C}$? The specific heat of the
  tea is $4.18\ \mathrm{J}/(\mathrm{g}\ \cdot°\mathrm{C})$ , and $\Delta H_{\mathrm{fusion}}$ for ice is $+6.01\ \mathrm{kJ\ mol}^{-1}$ .

$q$ is the amount of heat transferred. Since the solution is decreasing heat, then $q_{\mathrm{ice\ tea}}=-q_{\mathrm{ice}}$ We can then solve q using the information given.

$q_{\mathrm{ice}\,\mathrm{tea}}=\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\cdot ^{\circ}\mathrm{C}}\right)(400\,\mathrm{g})(-70\,^{\circ}\mathrm{C})=-1.17\times 10^5\,\mathrm{J}$

The next part cofuses me, this is what the solution manual does.

$q_{\mathrm{ice}}=1.17\times 10^{5}\,\mathrm{J}=(6.01\,\mathrm{kJ\ mol}^
{-1})\left(\dfrac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}\right)\biggl(m_{\mathrm{ice}}\times\dfrac{1\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}}{18.02\,\mathrm{g}\,\mathrm{H}_{2}\mathrm{O}}\biggr)$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\ \cdot\ ^{\circ}\mathrm{C}}\right)(m_{\mathrm{ice}})(10.0\ ^{\circ}\mathrm{C})$

I'm not too sure where this equation is derived from or of its purpose. The top part is determining the the enthalpy of ice but why is it being added to the specific heat of ice?

HINT:

$\pu{4.18 J/g K}$ is the specific heat of water. Also question assumes that due to addition of tea leaves the specific heat doesn't change much and remains equal to that of water.

So if you mix ice and tea and want the final temperature at $\pu{10^\circ C}$ then what do you expect of ice. Shouldn't it must melt and reach to $\pu{10^\circ C}$ and that will require a further energy to raise it temperature from $\pu{0^\circ C}$ to $\pu{10^\circ C}$. Also the mass of ice will remain same even if it get converted to water


2017 Nobel Prize; What does Cryo-electron microscopy data actually look like? It was just announced that the 2017 Nobel Prize for Chemistry has been

  awarded to Jacques Dubochet, Joachim Frank and Richard Henderson "for developing cryo-electron microscopy for the high-resolution structure determination of biomolecules in solution".

What does the data that Cryo-electron microscopy produces when examining the structure of "biomolecules in solution" actually look like? Is it electron transmission images of the molecules, or a diffraction pattern produced by single molecules, or something different? Are the molecules still in liquid solution when they are measured, or are they actually frozen by the time the data is collected?

Including an example image or diffraction pattern or some representation of the type of data obtained from a biomolecule in solution would be greatly appreciated!

From this Nobel Prize tweet:

This is a summary of the account by Peter Brzezinski, Professor of Biochemistry, Stockholm University, Member of the Nobel Committee for Chemistry from the Scientific Background document.

  [Brzezinski:] Jacques Dubochet developed methods for preparation of samples for cryo-EM studies of biomolecules in water. Joachim Frank developed methods for structural determination of biomolecules from analyses of ensembles of particles in solution. Richard Henderson demonstrated that it is possible to obtain atomic resolution structures of biomolecules using cryo-EM.

To answer the questions of the OP, I will use some of the excellent figures in the document.

  [OP:] What does the data that Cryo-electron microscopy produces when examining the structure of "biomolecules in solution" actually look like? Is it electron transmission images of the molecules, or a diffraction pattern produced by single molecules, or something different?

For larger particles such as viruses, it looks like the actual objects. Here is an example:

As you can see, the virus particles are not oriented (not a crystal, different molecules rotated differently). Other studies used two-dimensional or three-dimensional crystals. No diffraction images are used (in contrast to single molecule X-ray diffraction using free electron lasers). When the objects are large and have high symmetry, orientation of the molecules is easier than for small objects lacking symmetry. Thus, success in structure determination started with highly symmetric virus particles (pioneered by Henderson) and progressed to smaller objects requiring more powerful algorithms for orienting images (developed by Frank).

  [OP:] Are the molecules still in liquid solution when they are measured, or are they actually frozen by the time the data is collected?

Samples are in vitrious ice. For the technique developed by Dubochet, no cryo-protectant is necessary (different from cryo-cooling crystals for X-ray diffraction):

Advances in detectors

The image posted by the OP mentions advances in detector technology. Traditionally, electron microscopy uses film or CCD detectors. In both cases, signal from electrons are "converted" into visible light, introducing noise. The new type of detectors are "constructed from Monolithic Active Pixel Sensors, based on Complementary Metal Oxide Semiconductors, known as CMOS technology". They directly detect the electrons, giving images with less noise.

Disclaimer

For brevity, I skimmed over the contributions of the laureates and did not mention any of the many other contributors. The article I reference above is excellent; I recommend it as a starting point to a more in-depth study of the topic.


Oxidation states of boron I have just been looking at      https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements and found that boron has a -5 oxidation state. I would like to know which boron compounds form this oxidation state.

As I remember I have read somewhere that $\ce{Al3BC}$ is konwn to have Boron in -5 oxidation state.

you can also reffer to this PDF page no. 139
PDF IN GERMAN


Stretching frequency of nitro and nitrito I got the values of the stretching frequency of nitro and nitrito ligands from this webpage.
I understand why stretching frequencies are close in nitro and well separated in nitrito. How to explain why nitrito has slightly higher stretching frequency?

First of all, in the webpage, I think there is some shuffling between Nitro and Nitrito first in the headline. But, according to the contents, Nitro has closer stretching frequency, but Nitrito has distinguished stretching frequency.Let's take that one as the correct data.

The stretching frequency of nitro ($\ce{<-NO_2}$ )corresponds to the stretching of the bonds between $\ce{N}$ and $\ce{O}$, which are in between double and single bond, due to resonance, and the bond order between these $\ce{N}$ and $\ce{O}$ becomes almost equivalent in both the two  bonds.

Thus the stretching frequency of $\ce{-NO_2 }$ corresponds to partial double bond character between N and O and almost equivalence of two bonds.  On, the other hand, the two streching frequency of nitrito ($\ce{<-O^ -N=O)}$ corresponds to one $\ce{N -O}$ pure single bond and  one $\ce{N=O}$ pure double bond as there is no delocalisation in this case. So, due to that pure double bond stretching frequency of $\ce{N=O}$, one streching frequency of nitrito is higher than Nitro, which is consisting of only partial double bonds.


Calculating the density of a saturated salt solution Is there a way to calculate the density of a saturated salt solution from the solubility limits or does it have to be experimentally determined?

First instinct, is to add the mass of the salt to mass of the water e.g. a solubility of $\pu{80 g}$ of salt in $\pu{100 mL}$ would have a solution density of $\pu{180 g/100 mL = 1.8 g/mL}$. However, it seems the salt should affect the volume of the solution. What additional information is needed or does this quantity just have to be empirically determined?

Often, solubility is given as mass per volume of solution (not solute). Nevertheless, unless you have the mass percentage (w) too, you are not able to calculate the density of the solution.


When precessing nuclei absorb electromagnetic radiations, am I supposed to consider the wave nature or particle nature of light? I decided to peek into the NMR section of my spectroscopy book, and I found out lots of interesting stuff. And along with the interesting stuff, I did come along some stuff that I didn't understand.

My book tells me this (not copied word-for-word):

  Under the presence of a magnetic field, the two spin states of the nuclei split up, and if a correct wavelength of light is incident upon the sample, the nuclei get excited to a higher state.

In the above scenario, I believe light behaves as particles, wherein a photon of light collides with a nuclei, and push it to higher spin states.

There's another thing my book says:

  Under the presence of a magnetic field, the nuclei begin to precess (wobble like  a spinning top). If an incoming electromagnetic radiation has the same frequency as the nuclear precession, resonance occurs and the light is absorbed.

In this case, light behaves as a wave. It's the oscillating electromagnetic fields that are resonating the nuclei, and causing them to absorb the energy.

Out of the two cases that I mentioned, which is the one that actually happens? Is it resonance or spin excitation?

The nuclear spin states in an atom are excited by the radio frequency radiation and one spin state moves to the other when a photon is absorbed.

The precession part is not so much about individual spins but the magnetisation of the sample which is the vectorial addition of all the nuclear magnetic dipoles; precession is motion about the permanent magnetic field. (Think of the magnetisation rather like a vertical rotating pendulum if you want.) As circular polarised rf is used the spins are excited coherently and all add up 'in phase', hence a large magnetisation is produced that then undergoes precession. Loosing this phase in the magnetisation corresponds the the T2 decay time and loosing population excess above equilibrium to the T1 time. It happens that the precession frequency is the same as the transition frequency.

(An individual spin angular momentum is a complicated nutation type motion as it experiences a force from the rf field which is at right-angles to he permanent magnetic field.)


Why does Cody use Litharge? In many of the precious metal refining videos Cody (Cody's Lab) takes the sample with the precious metal and puts Litharge into it. After heating it in the furnace, while pouring it out the molten lead comes out and Cody says that the precious metal is all inside the molten lead(Later extracting it using cupellation). How does this happen?
https://www.youtube.com/watch?v=vltBwXncjCI&t=369s
From 3:01
Edit: My doubt is specifically the processes involved during Litharge Smelting.(that result in the precious metal being alloyed with lead)Why is Cody confident that all of the silver has been alloyed with the lead? Why cant the silver form a separate mass? My speculation is that all the impurities except the precious metal melts, the litharge is reduced into lead, the molten lead being denser than other molten impurities sinks to the bottom ,the silver being the densest goes to the bottom to meet lead and they alloy. Is this possible?

Litharge is $\ce{PbO}$. It is probably introduced to save silver from compounds containing oxidized silver like $\ce{AgCl}$. If metallic silver powder is mixed with $\ce{AgCl}$, this compound will react with $\ce{PbO}$ according to $$\ce{2 AgCl + PbO -> PbCl2 + Ag2O}$$ And $\ce{Ag2O}$ is an instable compound being easily transformed into $\ce{Ag + O2}$ at high temperature. So all metallic silver and silver from silver compounds are finally dissolved into lead, because lead dissolves metallic silver and not silver compounds. Without litharge, silver compounds like $\ce{AgCl}$ are lost.


Why do we observe signal coalescence during dynamic NMR? At low temperature, the rotation around the N-C amide bond of dimethylformamide is slow, so the two methyl groups are in different chemical environments. The H NMR thus shows a peak for each of the methyl groups.

As the T is increased, we see that the peaks begin to coalesce, until eventually they coalesce into a single average peak. The basic explanation is that, since the methyl groups are exchanging their chemical environment at a rate faster than the NMR time scale, they are considered to be within the same average chemical environment.

I am not very comfortable with this explanation, because it could also be argued that 50% of the time the methyl will be emitting signal A, and 50% of the time it will be emitting signal B - and the FID signal should still contain two signals. Alternatively, if the rotation is continuous, then I could expect a very broad peak extending from A to B.

My guess is that there is something about the structure of the signal and the Fourrier transform that I do not understand. Could someone help me out understanding why this coalescence occurs?

Thank you!

The basic explanation is quite simple: a spectrum reports the rate of precession of a collection (ensemble) of nuclear spins. Precession can be pictured as orbital motion about a common axis, like earth rotating about the sun (merely for purposes of illustration). Now assume you have two groups of spins (equal number in each), spins in each group precessing at a fixed rate, that is, spins in each group precess coherently. Since the NMR signal is an average over all of the spins, and you have two populations with distinct precession rates, you see two independent peaks in the spectrum.

If on the other hand there is fast exchange (fast to be defined later) between the spins in the two populations, each spin ends up with an average precession rate, the mean of the starting values of the populations. In that case, the NMR spectrum contains only one peak. The spins have been mixed and each population is no longer pure, the original populations have lost coherence.

In the intermediate regime, when exchange rate is not so fast that the populations are thoroughly averaged, the spins will inhabit mostly one population but their rates will deviate from the pure initial rate due to sampling of the rate of the other population. This jumping between populations leads to a loss of coherence, and the NMR spectrum will show broadened peaks. These eventually coalesce into a broad central peak with a precession rate equal to the average over the two starting populations. The condition for coalescence (fast exchange) is that the exchange between populations happen rapidly relative to the precession rate about the axis, thereby blurring the precession rate of individual spins.


How to calculate the heat absorbed by a reversible process? Given a nonideal substance inside a box with diathermal walls, the pressure is increased reversibly. How do you calculate the heat absorbed?

I assume we use

dE = dq + dw = dq - pdV

=> dq = dE + pdV

But I'm not sure how to deal with the "dE" term.

I assume you mean the first law as

$$dU = \delta q + \delta w$$

where $dU$ is the (state) change in internal energy, $\delta q$ is the (path dependent) work, and $\delta w$ is the (path dependent) work.

Saying the walls are diathermal says something about temperature (as you should confirm). Then, you should find a relationship between temperature change and internal energy change for the substance in your system.


Cyclopentane conformations What is the energy minimum of cyclopentane conformational diagram: the halfchair or envelope? How does cyclopentane flip?

As you mention, there are two main conformations that cyclopentane may adopt.

In the absence of substituents around the ring, there is actually only a $\pu{0.5 kcal/mol}$ energy difference between envelope and half-chair (envelope slightly lower in energy) meaning cyclopentane itself (computationally at least) is in rapid equilibration with no 'major' conformation as is observed with cyclohexane. (Note that the energy of the totally planar conformation is about $\pu{5 kcal/mol}$ higher than that of the both bent conformations, and so is not observed to any great degree).
With a single substituent on the ring however, such as a methyl group, the envelope conformation becomes slightly more favourable (to the tune of around $\pu{3.5 kcal/mol}$) giving a barrier to the ring flipping and meaning there is a dominant conformation.


If phosphorus trioxide is present in the atmosphere of Venus, could it not react with water to form (sufficient) phosphorous acid? Sufficient phosphorous acid to be decomposed in the measured quantity of phosphine ?
From Phosphine gas in the cloud decks of Venus:

We also rule out the formation of phosphorous acid ($\ce{H3PO3}$). While phosphorous acid can disproportionate to $\ce{PH3}$ on heating, its formation under Venus temperatures and pressures would require quite unrealistic conditions, such as an atmosphere composed almost entirely of hydrogen.

But according to Vega mission results and chemical composition of Venusian clouds there is $\ce{P4O6}$ in the atmosphere which according to Wikipedia would react with water to form phosphorous acid.
However there is supplementary information in the chapter "Equilibrium thermodynamics in the atmosphere and surface" at the end of the article "Phosphine gas in the cloud decks of Venus":

As an example of our approach, we present a calculation for phosphorous acid ($\ce{H3PO3}$). This compound will spontaneously decompose on heating to form phosphoric acid  and phosphine; this is a standard laboratory method for making phosphine. Phosphorous acid is not stable in gas phase, but could in principle be formed in cloud droplets by reduction of phosphoric acid.

Edit for some additional informatiom.
Also from the supplementary information (page 13):

Reactions of $\ce{P4O6}$, $\ce{P4O10}$, $\ce{H3PO4}$ and $\ce{H3PO3}$ were considered (the last of these only in solution phase in the clouds),..

So not considered was the possible occurence of $\ce{H3PO3}$ below the clouds.
The Recent Evolution of Climate on Venus (page 23) states:

The evaporation of $\ce{H2SO4}$ occurs at about 48 km, the average cloud base.The vapor phase continues to exist down to 432 K (38 km), where it is thermally decomposed.

So below 38 km, and above 160 ⁰ C there would be water in the gas phase, free from sulfuric acid, that could react with $\ce{P4O6}$.
Could have that possibility been ruled out for some reason ?

The issue seems to be not the reaction between $\ce{P4O6}$ and water, but whether $\ce{P(III)}$ can form at all.  The OP's first reference says no whereas the second reference says yes.  Basically the two references used different thermodynamic assumptions.  The first one assumes an atmosphere in equilibrium while the second assumes only a partial equilibrium; in the latter case $\ce{CO}$ is allowed to form from nonequilibrium (radiation-driven) dissociation of $\ce{CO2}$, and minor species are then equilibrated with the $\ce{CO/CO2}$ couple.  In the case of phosphorous oxides this favors $\ce{P4O6}$ over $\ce{P4O_{10}}$, although the former could be oxidized by other species such as sulfuric acid.
The issue has become important because the recent discovery of $\ce{PH3}$ among the clouds of Venus has led to a hypothesis that this compound is generated by Venusian microbes, but if $\ce{P(III)}$ is available for a disproportionation reaction it could provide an alternative abiotic (non-life) source.  As noted above, the references disagree on this, and the question of whether this alternative to a biological source of $\ce{PH3}$ is viable awaits resolution.
Relevant to the question of whether $\ce{PH3}$ is a true biosignature is this question about organic compounds on Venus, which apparently spawned the current one.


Will ice cream freeze faster with frozen brine or adding salt to ice? I've been making some ice cream without a ice cream maker, using salt, ice, mixer and steel canister since I can't get a ice cream machine around here and its fun. It works pretty good but it seems slower than a simple ice cream machine that uses a freezer bowl. And the freezer wall volume seems smaller than the amount of ice I use.

Will using a frozen brine (if I can even freeze it) or liquid cool faster than
using frozen ice and then adding salt to melt it and lower the temperature? whats the difference? Is temperature all the matter? (liquid brine at -10c vs ice at 0)
any other factors that can contribute to the time it takes to freeze the mixture?
Any idea what coolant those bowls use? (seems to freeze at freezer temps).

If the brine melts at the same temperature you reach with the ice and salt, then it should take the same time to freeze the ice cream since the freezing time depends on heat transfer. The benefit of frozen brine is that there is more heat capacity in the frozen brine than in ice-salt cooled water, so the ice-salt water may become warmer than the frozen brine.

It may be that there is poor heat transfer from the cream to the ice so it takes longer than you think. Also, if you are the one turning the crank, time seems to stretch out since your muscles get tired and yet the ice cream is still not frozen.


What happens to potassium iodide in the "elephant Toothpaste" reaction? General Questions on "Elephant Toothpaste"

Equations:

1)$\ce{~2H2O2(aq)->2H2O(l) + O2(g)}$

Context:

I have a few questions on the reaction commonly referred to as "Elephant Toothpaste". You see, I am doing a basic demonstration of chemistry to little kids. I certainly don't think I am qualified to perform this, (Well, I can, but I want to know the details behind this, which is why I am asking for help.) but I have no choice. So Here we go!

Procedure:

Pour ~$\pu{50 mL}$ of hydrogen peroxide1 into graduated cylinder.
A "Squirt" (Not my wording) of Dish Soap into the graduated cylinder.
~$\pu{10 mL}$ of potassium iodide ($\ce{KI}$) Solution into Graduated Cylinder.

The result is...Foam, and lots of it.
130% concentration

So, here is my question:

What happens to the potassium iodide?

The potassium iodide serves as a catalyst for the decomposition of the hydrogen peroxide by facilitating oxygen removal from the solution so the reaction is carried to completion. The rapid decomposition allows you generate a lot of gas quickly and the long cylindrical shape of the graduated cylinder directs the momentum of the gas in a vertical direction to achieve the gushing effect. Since potassium iodide is a catalyst in the reaction, it is not consumed in the process of making the foam, but exists as the ions that come from dissolving the salt in the hydrogen peroxide solution.

Reaction mechanism on wikipedia

P.S. This reaction is exothermic, so if you go really large scale, expect your vessels and the foam to be hot to the touch.


Do all gases occupy same volume at equal temperature and pressure conditions? Why does one mole of all gases occupy 22.4 L at standard temperature and pressure? According to my thinking, it may be due to equal diffusion of all particles when at a particular temperature and  pressure.

So going by this, one mole of all gases should occupy the same $x~\mathrm{L}$ at some other temperature-pressure conditions. Is this true?

Why do all gases occupy 22.4 L [per mol] at STP?

The question is based on a false premise. Only ideal gases are guaranteed to occupy 22.4 L/mol at STP.  There are many gases that are not ideal.

  So going by this, all gases should occupy same $x$ L at some other temperature-pressure conditions. Is this true?

No, again this would be true only for ideal gases, not all gases.

At STP, many commonly encountered gases on Earth behave nearly ideally, but certainly not all.  For example, vapor of acetic acid and other simple carboxylic acids show strong deviations from ideality.

Wikipedia is a good place to read about the assumptions of the ideal gas law and the kinetic theory of gases.  The reason that ideal gases share a molar volume of 22.4 L at STP is not related to diffusion, but rather to the assumption of: (a) non-interacting gas particles that are (b) effectively zero-sized relative to the mean distance between them.

You can read about equations of state for non-ideal gases on Wikipedia as well.  Popular equations to describe non-ideal gases include the van der Waals equation, the Redlich-Kwong equation, and a whole slew of others listed at Wikipedia's real gas page.


Why the reduction potential of the reaction AgCl (aq) /Ag + Cl- is +0.22V The point is $\ce{AgCl_{(aq)}}$ is the same as $\ce{Ag+}$ + $\ce{Cl-}$ (again, in water)
So my reduction reaction would be
$$\ce{Ag+ + Cl- + e- -> Ag + Cl-}$$ then the equation is simplified by removing $\ce{Cl-}$ from both sides (Spectator Ion).

Now there is $\ce{Ag+ + e- -> Ag}$ which has a potential of 0.80V, so why the original equation is 0.22V.

PS: I know it takes more energy to break the bonds and thus Ag is more stable with Cl and less likely to reduce. But the subscribed (aq) doesn't mean the bonds have already been broken by water?

I don't know what your source is, but it should definitely be an $\ce{AgCl(s)}$ in your first equation. Wikipedia gets it right and if you find any textbook that doesn't explicitly state the phase of $\ce{AgCl}$, you can be pretty darn sure they meant solid because talking about aqueous $\ce{AgCl}$ makes little to no sense precisely because it is so insoluble in water.

So, essentially you've already answered your own question; precipitation of $\ce{Ag(I)}$ as $\ce{AgCl}$ stabilises it towards reduction and lowers the standard reduction potential relative to $\ce{Ag+/Ag}$.


How do Quantum Software Packages Work? If one wants to calculate a moderate size Alkane (with say 10-15 Carbons , assuming 100 electrons , with Restricted Hartree Fock based methods) we can simply say that electron-electron part will be $100^4 = 100$ million integrals. If we remove the non-unique ones it will become $\frac{100M}{8}= 12.5 M$ Integrals. If we do a pre-screening it may become around 6 million.  Even if we go for DFT it will be around $\frac{100^3}{8} =125,000$ .

I have calculated a small system like Methane with just 10 electrons (around 400 unique/pre-screened integrals) with an efficient C++ code  (based on Rys Integration which is among the most efficient methods) and it took around 1 hour. When 400 cycles takes 1 hour I can assume that 6 million cycles may take years! But software packages do that just in seconds!

My question is if I just want to do a simple math like 2+2 for 6 million times it takes way longer than what these packages do with such complicated integrals!!
Anyone knows what their trick is? Do they have pre-calculated database or some approximation methods?

UPDATE:
there are many good hints in this question and thanks to all for that. A combination of symmetry, finding negligible integrals and using optimized methods work fine. The only unanswered part is an algorithm for finding the symmetric situations.

There are many tricks. Since I'm not a developer, I'm not privy to all of them, but there are some simple things to start.

Some integrals are predictably close to zero: Even in fully first-principals calculations, there are parts of the code that prune out integrals that will clearly be negligible. This cuts out a lot of work. Consider, for example, overlap or Coulomb integrals between atoms that are very far apart.
Gaussian Orbitals: Using Gaussian-type orbitals are incredibly fast, because most quantum integrals can be derived as simple recurrence relationships. That is, you almost never need to do the integration. See, for example, Obara and Saika and the recent PRISM algorithm from Gill and Pople.

I believe there are some actual quantum developers on here and perhaps they'd be willing to share other tricks. But these are big - eliminate any integral you don't need, and don't actually take the integrals in the first place. ;-)

Yes, there are numeric tricks (using BLAS, LAPACK, etc.) and use of multi-core processing. But these are actually minor optimizations.

UPDATE

I realized that I forgot to talk about convergence of the self-consistent field which is also optimized. There are various convergence strategies, most notably DIIS, which helps to extrapolate the final orbital coefficients and electron density.


Why isn't my galvanic cell making enough voltage? I'm in kind of a pickle here. I made a galvanic cell of Lead and aluminium for a chemistry internal assessment but unfortunately, it didn't turn out so well and neither me nor my teacher has any clue why.
The E0 value of the cell is 1.53v. I got a maximum reading of 0.29v

If you don't have time to read here is what I think the sources of error could be:
1. I used filter paper as a salt bridge and I don't think it was thick enough to allow ions to flow through it.
2. I used around a 1M concentration of KCl to dip the salt bridge in.
3. I suspect either of the metals was not correct but this seems improbable as I got a voltage just not the correct one.
4. The solutions of AlNO3 and PbNO3 were kept in round-bottomed flasks with a glass cork in them for a few days before I decided to do the experiment.
5. I got a negative reading for the voltage which was weird.
6. I got a higher voltage reading only when I touched the aluminium rod to the salt bridge in the solution

I used the following procedure:

I made 0.01M, 0.1M, 0.25M and 0.5M solutions of aluminium nitrate and a 0.1M solution of lead nitrate (both well dissolved) from lab equipment.
I poured each of them into separate 50ml beakers. I then placed an aluminium rod (that was 50% submerged) into the AlNO3 beaker and a thin piece of lead (that was well sanded before each reading) that was fully submerged in PbNO3.
I connected each of the metals to a multimeter via alligator clips
I then prepared a salt bridge using 5x25cm strips of filter paper which were dipped in a solution of KCL of about 1M concentration. The filter strips were folded and stapled.
I then placed the salt bridge in both the beakers and conducted the experiments.

I'm not in school right now and I don't have time to conduct the experiment again as the assessment is due very soon.

Just keep in mind that Eo is determined under highly controlled conditions and concentrations. I sense two key problems here. The major problem is already stated above that the resistance of a wet filter paper must be high, and this is giving you lower voltage. Evaporation of water from the filter paper would make the situation worse. Since you are not repeating the experiment anyway, make sure to identify the causes of lower voltage. Secondly, both lead and aluminum nitrate tend to form hydroxides in water if you don't add little bit of acid in them. Did you notice that lead solutions were cloudy or they were clear and transparent?

As a side note, negative voltage reading of the battery just means that your connections of the lead are the other way round on the voltmeter. Test your voltmeter with a battery keeping the signs on the battery in your mind. Once you get a positive reading of a simple AAA cell you have correctly made the connections.


Why are peroxides explosive? Why do peroxides tend to explode with little provocation? In general, what are typical products of their explosion?

I don't like to speak in generalities if I can help it.

But we can talk, for instance, about hydrogen peroxide. One of the most familiar schemes is a disproportionation where,
$$2 \ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2}$$
and as you study this particular species, you will see this reaction more and more often. Actually in solutions of hydrogen peroxide you can take a 30% concentrated solution and add some Mn and subsequently measure a significant increase in oxygen concentration over time. Although you should know the solution itself decomposes slowly in practice when exposed to light or due to trace impurities, but this pales in comparison to the case involving a catalyst. This reaction is, in part, why researchers believe a peroxide-type species is important for the Oxygen Evolving Center of Photosystem II. Another fun demonstration of this process is often where foam is generated from a large graduation cylinder, see here, upon addition of KI.

Now 30% is weak, if instead you look at 80% solutions you will notice that the decomposition will evolve steam because it is so exothermic. In fact, the first liquid-fueled rocket in 1939 used an 80% concentrated solution and decomposed it with calcium permanganate. Later engines utilized another important property of hydrogen peroxide for its decomposition until around 1980: the fact that it is a stronger oxidizer than chlorine/chloride.

Thermodynamically we say the disproportionation is favorable since it has a fairly negative (~ -100 kJ/mol) enthalpy change and a considerable change in entropy ( ~ + 70 J/(mol K)). The basis of this is that it is highly favorable to form dioxygen and this is a large driving force of this reaction in terms of enthalpy and entropy. You'll also see this danger in species that are really close to forming dinitrogen too, see here and here. It only takes a catalyst for this down-hill process to take off. Which could be quite deadly if unintentionally concentrated and then introduced to a catalyst in a closed system. Perhaps the story is similar for other peroxides.


What do the molecular orbitals of a single water molecule look like? Some time ago, (a few years back) I heard that the orbitals of a single water molecule were solved analytically (or was it just numerically). What do they look like?

I am interested in the shape (and equations) of those orbitals. There should be an exact solution around and that is what I am looking for (no LCAO or similar please).

First things first, orbitals are not real: neither atomic, nor molecular. Except for, probably, one-electron atoms (also known as hydrogen-like atoms), where one might still argue that to get the orbitals you need to treat the nucleus as a classical point particle that just creates a spherically symmetric electric potential which can be already considered as an approximation. For many-electron systems, one have to necessarily make the so-called mean-field approximation (plus the Born-Oppenheimer approximation prior to that for the case of molecules).

Second, in principle, no further approximation is done in LCAO-MO procedure; it is a purely algebraic trick of expanding a function over a complete set. Such expansion is in principle exact, but the problem is that a complete set is infinite, so in real-world calculations we have to truncate the linear combination at some point which indeed is yet another approximation.

Thirdly, analytical expressions for orbitals are known for one-electron systems only (such as $\ce{H}$ atom, or $\ce{H2+}$ molecular ion) since the electronic Schrödinger equation was solved analytically only for such systems.

And finally, strictly speaking, it is electron density that was imaged with STM, not orbitals. The fact that it closely resembles the density corresponding to orbitals we get in our calculations justifies the quantum chemistry as a whole: at the end of the day, with all the above mentioned approximations we still get something that is close the reality!


What are angular and radial nodes? Nodes are the points in space around a nucleus where the probability of finding an electron is zero. However, I heard that there are two kinds of nodes, radial nodes and angular nodes. What are they and what information do they provide of an atom?

The accepted answer has nice pictures, but has a couple of small factual inaccuracies in the last paragraph. Here's a bit more maths to illustrate what's going on.
Atomic orbitals, which are one-electron wavefunctions, are split into two components: the radial and angular wavefunctions
$$\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)$$
so-called because they only have radial ($r$) and angular ($\theta$,$\phi$) dependencies respectively. If either of these two components are zero, the total wavefunction is zero and the probability density there (given by $\psi^*\psi$) is also zero.
A radial node occurs when the radial wavefunction is equal to zero. Since the radial wavefunction only depends on $r$, this means that each radial node corresponds to one particular value of $r$. (The radial wavefunction may be zero when $r = 0$ or $r \to \infty$, but these are not counted as radial nodes.)
Analogously, an angular node is simply a region where the angular wavefunction is zero.* In the case of the p-orbitals, this is a plane. However, angular nodes do not necessarily have to be planes, as we'll see later.
The number of radial and angular nodes is dictated by the forms of the wavefunctions, which are derived by solving the Schrodinger equation. For a given orbital with quantum numbers $(n,l)$, there are $n-l-1$ radial nodes and $l$ angular nodes, as previously described.
An example
Let's take a look at one of the hydrogen 3d wavefunctions with $(n,l) = (3,2)$. $a_0$ is the Bohr radius of the hydrogen atom. We would expect $3-2-1 = 0$ radial nodes and $2$ angular nodes.
$$\mathrm{3d}_{z^2} = \underbrace{\left[\frac{4}{81\sqrt{30}}a_0^{-3/2}\left(\frac{r}{a_0}\right)^2 \exp{\left(-\frac{r}{3a_0}\right)}\right]}_{\text{radial: }R_{32}} \underbrace{\left[\sqrt{\frac{5}{16\pi}}(3\cos^2\theta - 1)\right]}_{\text{angular: }Y_{20}}$$
All the individual terms in the radial wavefunction can never be zero (excluding the cases of $r = 0$ or $r \to \infty$ as I described earlier). Therefore, this orbital has no radial nodes. Surprise, surprise.
The angular nodes are more interesting. The angular wavefunction vanishes when $3 \cos^2\theta - 1 = 0$. Since $\theta$ takes values between $0^\circ$ and $180^\circ$, this corresponds to the two solutions $\theta = 54.7^\circ, 125.3^\circ$. Both of these solutions are angular nodes. This is how they look like:

The dotted lines are the angular nodes. They are not planes, but rather cones. They correspond to one particular value of $\theta$, which in spherical coordinates is the angle made with the positive $z$-axis; I have marked these angles on the diagram.
If you can obtain the forms of the wavefunctions, then it is easy to find the radial nodes. Mark Winter at Sheffield has a great website for this; just click on the orbital you want on the left, then "Equations" near the top-right corner.

If we stick to complex atomic orbitals, which are simultaneous eigenfunctions of $H$, $L^2$ and $L_z$, then the dependency on $\phi$ is always of the form $e^{\pm im\phi}$, which can never be zero, so angular nodes never arise due to $\phi$-dependency. However, radial and angular nodes are most commonly discussed in the context of real atomic orbitals, obtained by linear combination of the spherical harmonics. These do have angular nodes that depend on both $\theta$ and $\phi$, but they are often much simpler to express in terms of Cartesian coordinates $(x,y,z)$ (obvious examples being the $\mathrm{2p}_x$ and $\mathrm{2p}_y$ orbitals). The $\mathrm{3d}_{z^2}$ orbital is an exception that was deliberately chose as an example because its angular nodes are not planes.


How to determine fire hazard of heating ethylene glycol to dissolve borax and boric acid I am looking to replicate instructions provided in the last paragrph of this article that requires dissolving borax (Sodium Tetraborate decahydrate ($\ce{Na2B4O7·10H2O}$)) and boric acid in ethylene glycol, for reasons of practicality and cost efficiency supplied in the impure form of concentrated automotive anti-freeze.

For reasons out of scope of this question I've modified the ratio provided as follows:

5 kg anti-freeze
2.5 kg Sodium Tetraborate decahydrate
1 kg Boric acid

In order to dissolve the borax completely it is necessary to boil off the water of crystallisation plus any water impurities in the anti-freeze.

While I have water in the system, the temperature of the mix is expected to be stable at around 100 °C, but once the water is removed the temperature will rise to potentially dangerous levels if the heat source is not removed in time.

I would like to understand the risks of this operation. The flash point of (pure) ethylene glycol is, according to Wikipedia, at 111 °C. The autoignition temperature is listed as 410 °C and seems to be safely above the boiling point of ethylene glycol (it would boil off before it auto ignites). However this hazard sheet for ethylene glycol states it 'causes ignition' at 100 °C in combination with (amongst others) Sodium Chlorite (no more details given).

Can I assume that Borax and Boric acid are 'safe' as they are not listed on the hazard sheet?

Is there an (online) reference where I can find out how Borax and Boric acid modulate the flash point and auto ignition temperature of Ethylene Glycol?

Chlorates, perchlorates, permanganates and other strong oxidants conflagrate even at room temperature with ethylene glycol.

Boric acid and borax, however, are not considered strong oxidants and should probably be safe to dissolve below the boiling point of water. That said, I would test a small batch in a safe place, to be sure there is not a large gain in temperature on dissolving.

BTW, you might be interested in reading of the preservation of the Swedish warship Vasa in polyethylene glycol, and long-term issues with acidification. N.B. Polyethylene glycol is generally considered innocuous, but ethylene glycol is toxic if ingested.


Do catalysts undergo physical changes during a reaction? I know that catalysts don't undergo chemical changes during a reaction, but do they undergo physical changes? If they do, are there any examples?

I'll just give an example of a catalysis to show you that catalysts do really temporarily react with the starting product and eventually will be regenerated after the catalytic reaction.

An example is the catalytic decomposition of formic acid to carbon monoxide and water by using sulfuric acid as a catalyst. The sulfuric acid donates one proton ($\ce{H+}$ ion) which is then bonded to the central carbon atom in the formic acid molecule. Then next another hydrogen atom from the sulfuric acid reacts with the hydroxyl group present on the formic acid molecule to yield water. In the end the sulphate ion snatches the last two present hydrogen atoms from the formic acid molecule and yields carbon monoxide. The sulfuric acid is regenerated.

$\ce{HCOOH + H2SO4 -> H2COOH + HSO4- -> H2CO + SO4^{2-} + H2O -> CO + H2SO4}$

So catalysts do undergo chemical changes in the catalytic reaction but are regenerated after the catalytic reaction is finished.

This means that if you are using a catalyst, you can normally reuse it after your catalytic reaction because it has the same chemical (and physical) properties as before the catalytic reaction.

Also : when you use manganese dioxide as a catalyst to decompose hydrogen peroxide, you will see that the black solid compound is the same after the catalytic reaction.

There are also so called 'sacrificial catalysts' (which aren't regenerated and do physically/chemically change, but aren't real catalysts) -> see http://en.wikipedia.org/wiki/Catalytic_cycle


Are enzymes with superior catalytic properties more sensitive to temperature changes? I am wondering whether enzymes with superior catalytic properties are more or less temperature sensitive than average enzymes.

I know that enzymes lower the activation energy, Ea. So the more catalytic superior an enzyme is the lower the value of Ea. Then the Arrhenius equation can be used to describe how the rate constant of an enzymatic reaction changes with temperature. But how can I use this information to decide whether the more catalytic enzyme is more or less temperature dependent?

There is no obvious direct connection between catalytic properties and stability of an enzyme.

You are trying to make a connection, but let's consider your statement:

  Then the Arrhenius equation can be used to describe how the rate constant of an enzymatic reaction changes with temperature.

According to the Arrhenius equation, all reactions (catalyzed by an enzyme or not) become faster as temperature increases. The higher the activation energy, the more pronounced the increase. However, if the enzyme unfolds, it will no longer catalyze the reaction.

  But how can I use this information to decide whether the more catalytic enzyme is more or less temperature dependent?

The biggest temperature dependence occurs at the temperature where the enzyme unfolds. There is the idea that enzymes have to be flexible to be effective catalysts, so there might be a correlation between optimal temperature for the activity and the unfolding temperature. The best way to figure this out would be to compare enzymes from mesophiles and thermophiles in terms of optimal temperature for catalysis vs melting temperature.


What happens if you put a piece of copper into a solution containing Al3+ ions? This is a question in my studies, and I am a little bit confused! I would think it happens nothing because copper is below aluminium in the reactivity series, right?

You are right with copper does not react with aluminium ions.

But as a side reaction, copper may get slowly oxidized by oxygen and dissolve in mildly acidic solution of aluminium salt.

Such a thing may happen, if you decalcify copper heating spiral by vinegar and let it stay overnight.

$$\ce{ 2 Cu + 4 CH3COOH + O2 -> 2 (CH3COO)2Cu  + 2H2O  }$$

Something similar can happen in aluminium salt solutions as well, because of hydrolysis.

$$\begin{align}
\ce{ [Al(H2O)6]^3+ + H2O &<=> [Al(H2O)5(OH)]^2+ + H+}  \\
\ce{ 2 Cu  + O2 + 4 H+ &->  2 Cu^2+ + 2 H2O} \\
\ce{ 2 Cu  + O2 + 2 H2O &->  2 Cu(OH)2 v} \\
\ce{ 2 Cu  + O2 +  H2O + CO2 &->   Cu(OH)2\cdot CuCO3 v} \\
\end{align}$$