text:
Lattice vectors calculation
How can I find the crystal lattice vectors, given the crystal parameters?
For example, consider a crystal of rhombohedral symmetry, where a=b and c are known, and alpha=beta=90, gamma=120. How can the lattice vectors be found?

That's hexagonal, not rhombohedral.

You can find this using elementary geometry. Also, since the unit cell may be positioned in different ways with respect to the coordinate system, there can be many different answers, all equally valid.
One possible answer implies alignment of the cell vectors $\vec a$ and $\vec c$ with the coordinate axes X and Z, which leads to the following:
$$\left\{\begin{align}
\vec{\bf a} =& \Big(|a|,0,0\Big)\\
\vec{\bf b} =& \left(-{1\over2}|a|,{\sqrt3\over2}|a|,0\right)\\
\vec{\bf c} =& \Big(0,0,|c|\Big)
\end{align}\right.
$$

A lattice vector can be defined as $\bar V= u\bar a+ v\bar b+ w\bar c$ where $u,\;v,\;w$ are integers describing the next lattice point out from the origin (at $[0,0,0]$) with position $[ua, \;vb, \; wc]$ and other lattice points are at $[nua, \;nvb, \; nwc]$ for integer $n$. The base vectors of the lattice are $\bar a, \bar b,\bar c$
Notes:
Conventionally $[u \;v \; w]$ is used to define a lattice vector as a set of parallel point rows whereas  round brackets $(h \;k \; l)$ describes the Miller indices that define parallel planes  satisfying the equation $hx/a+ky/b+lz/c=n$ where $n$ is an integer.
A lattice is formally defined as 'an array of points in space such that each lattice point has exactly the same environment in the same orientation'.


text:
Breaking degeneracy in phosphorus pentachloride
Recently, in a class discussion, the following point was brought up:

Although the axial and equatorial bond lengths in $\ce {PCl5}$ are inequivalent, when we consider the time-averaged $\ce {P-Cl}$ bond length, there should only be one value sincethe molecule is symmetric and thus, all the chlorine atoms interact with the phosphorus atom to the same extent.

The idea of "time-averaged" means that we are considering the $\ce {P-Cl}$ bond over a long duration. This bond length varies over time since the atoms are constantly in motion. When this bond length is "time-averaged", it means that we are taking the average of the bond lengths measured at infinitely many instants in time.
The argument is that although at any particular instant, the molecule would adopt a configuration with axial bonds being longer than equatorial bonds, when the bonds are time-averaged, they should all be the same length.
This has caused much dispute among members of the class and it has also led me to consult a professor who is well-versed with quantum chemistry. The professor mentioned that the argument put forward is not valid because in $\ce {PCl5}$, there is a breaking of degeneracy due to perturbation. He cited the Jahn-Teller distortion as a situation similar to that in $\ce {PCl5}$, where degeneracy is broken.
I am very confused by this. What exactly is this perturbation? Why does it exist for $\ce {PCl5}$ and not for similar molecules, such as $\ce {SF6}$ (as seen in the comments for this question)?

The experimental data shows that the two axial bonds ($C_3$ axis) are significantly longer than the 3 equatorial ones i.e. outside expt. error different. So there is no time averaging going on here as that would predict similar bond lengths. Additionally,  electronegative atoms preferentially fill the axial positions e.g $\ce{PCl3F2}$ indicating a difference in bonding between axial and equatorial positions. The proposed trigonal bipyramidal  ($sp^3d$ bonding) is consistent with minimising electron pair repulsion; $sp^2$ along equatorial;  $pd$ along axial bonds.

@porphyrin I am completely aware of the 3c4e bonding scheme involved and I understand the rationale behind the difference in bond lengths and thus, I am not looking for this in the answer that I require

@porphyrin I would like to understand why the bonds are not equivalent by the symmetry argument

By this argument, the bonds are not equivalent because there is no way to arrange five points around the common center so that they will be equivalent (except the flat pentagonal configuration, which is obviously bad for steric reasons).

Rapid pseudorotation on the observational time scale is what makes the bonds euivalent.

That being said, the pseudorotation appears rapid to some physical methods and slow to the others.

The molecule will undergo a Berry rotation , but that does not make the bonds equivalent it effectively just swaps them around, rather like the effect of  rotating the axes.  The experimental evidence is that the two types of bonds have different lengths. A similar, but simpler, effect occurs in ammonia inversion or $\ce{CH2}$ wagging in cyclopentene or any number of molecules with small barriers to rotation/inversion.

I would guess he means a spontaneous symmetry breaking, in which case the "perturbation" is any old force from the environment.  This can happen in physical systems: it's possible for a system of high symmetry to spontaneously fall to a lower-energy state of lower symmetry.  Only the tiniest of perturbations is required, if you wait long enough.  A familiar classical example is crystallization, in which below a certain temperature a system of atoms spontaneously loses complete translational symmetry even in the absence of an external potential.  A quantum example is magnetization when a ferromagnet cools below the Curie temperature -- the spins lose rotational symmetry and pick some essentially random preferred direction, in response to the tiniest of perturbations from the environment.  In the case of PCl5, one would assume, very roughly (and it would take detailed calculation to verify) that if 3 of the Cls pull in closer to the nucleus and 2 move further away, you get a better balance of nuclear-electron attraction and nuclear-nuclear and electron-electron repulsion than if all 5 remain at the same distance.  So the system spontaneously breaks rotational symmetry, in response to any old tiny perturbation in the environment.
That said, in the absence of an external potential, the rotational quantum state of the molecule will be entirely symmetric, and you will have an equal chance of finding the long Cl-P-Cl axis pointing in any direction.

So if the PCl5 molecule isn't interacting with the environment, say it is in some vacuum, then it would adopt the symmetrical configuration?

Why does the same not occur for SF6?

The whole point of spontaneous symmetry breaking is that the perturbation required to achieve it becomes vanishingly small, so if you mean a real vacuum, such as interstellar space far form a nearby galaxy, no, because even in that lonely place there'd be a perturbation greater than strictly zero.  If you mean a hypothetical empty universe -- who knows?  There's certainly a theoretical answer, but you have to bear in mind our theories were inferred from experiment in a non-empty universe.  They need not give the correct answer for other (untestable) situations.

I don't know (why it doesn't happen for SF6).  One could do deep calculations and maybe find an answer why the balance is  better struck in that case by a purely symmetrical arrangement.  We do know symmetry is strongly preferred in a symmetric environment, so much so that a case like PCl5 is a surprise that needs to be explained.  Spontaneous symmetry breaking is rare, most of the time the symmetry of the ground state of a system matches the symmetry of its environment.

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text:
How can electrons manage to be at the same distance from the nucleus at all positions in p, d and f orbitals?
In quantum mechanics electrons are arranged in certain orbitals namely s, p, d and f orbitals.The position of the electrons are determined by 4 sets of quantum numbers n, l, m and s. Electrons can only be at a fixed distance from a nucleus which is determined by the quantum number n, but in p, d and f orbitals how  does it manage to be at a fixed distance at every point in the region of these orbitals?(here is an example of p orbital in image)

There is no fixed distance.

Hm, it's not very hard to guess what the OP's misunderstanding is: The average distance only depends on n in hydrogen-like atom, as the energy levels of all l,m,s quantum states are degenerate.

I think you're confusing what the orbital shapes represent. The shapes are probability distributions-which means that you're most likely to find the electron somewhere within the shape.

Electrons can only be at a fixed distance from a nucleus

This is false

which is determined by n quantum number

The n quantum number tells you how big the orbital is-not precisely how far away the electron is.
In reality, the electron is moving around the atom, and is probably somewhere within the orbital shape.

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What are the intermolecular forces between polychloroethylene?
Does polychloroethylene have only van der Waals/London forces between its molecules? Surely, if all the chlorine atoms are on one side, due to the fact that chlorine is more electronegative than carbon, shouldn’t it have permanent dipole-dipole forces?

Everything has London forces.

There are permanent dipoles there,so no idea why you'd think otherwise.

PVC tends to have a relatively high stereoregularity, for sth. made by radical polymerisation, but a local dipole moment perpendicular to the backbone would be there any way. Not a very large one, compare methylene chloride, chloroform, etc.

@Karl  dipole moment can be completely cancel out by C-C rotations...it follows of your reasoning...

@santimirandarp hm, C-C rotations are not so easy in a polymer backbone, especially if it's substituted with a bulky chlorine atom

@Karl chlorine has only hydrogens as neighbours...To be sure, we may have to look at conformers energies for 2-chloropropane, but I couldn't find them. And also... I don't see a reason not to be produced randomly...

In a polymer backbone, you need pairwise C-C rotations. And I don't get your last sentence, what is produced randomly?

No, but for polymers of any reasonable size, the dispersion (and excluded volume) forces completely overwhelm any dipole-dipole forces, since they grow with the size of the molecule and dipole-dipole forces do not.  Also, bear in mind that the chlorines aren't all on one side of the molecule first because the way it's manufactured tends to put Cls randomly on each side of the chain, and second because the molecule has considerable conformational flexibility.  That's true even in the solid state, which for polymers is usually substantially glassy with interspersed domains of crystal.

PVC has a rather high stereoregularity, not random atactic, and ist partially crystalline, ordinary polystyrene is just an amorphous glas, PE is partially crystalline and it's glass transition temperature is below -60°C. ?!?


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How to find the mass of the following solution?
What is the weight of a 1L of 0.5M solution of  NaCl? Is it the whole solution weight 29.22 grams?

You might want to reconsider your views on solutions. To begin with, water is also a part of your solution, and it also has weight.

Please edit your post and add how you would try to solve this question (i.e. add your calculation). That helps users to help you.

Your 29.22 g is the mass of the NaCl in your solution.
You're missing the mass of the water. This is probably around 1 kg, and if you don't need a very precise result you could just say that the total mass of your solution is about 1000 g + 29.22 g = 1029 g, give or take a couple percent.
If you need a more precise number, the question gets much more difficult. You don't know the exact mass of the water because you don't know the volume of the NaCl in solution. It's not necessarily the same as the volume of 29.22 g of solid NaCl because there are interactions between the water molecules and the Na+ and Cl- ions in solution.
So rather than trying to understand and quantify those intermolecular interactions, it might be easier to check if someone measured it. Googling around, I'm finding a paper from 1985, but it's behind a paywall. It looks like people have measured the density aqueous salt solution very carefully at many temperatures and concentrations, but I can't find a comprehensive, authoritative table with their results. What we get is this chart at EngineeringToolBox (the 3rd one). Zoom in a little, and you get 1.04 g/cm^3 for 0.5 M NaCl at 20°C. That would make your mass 1.04 kg.
Depending on your standards, 1.04 kg is almost the same as the 1029 g we got using the "stupid" approach above (it's off by only 1%) or it's very different (we thought a ~0.7 M solution should have that mass, not a 0.5 M solution).
If you need better precision, go have a look at this paper. Good luck!

You’d first have to calculate the mass of NaCl in the solution:
moles NaCl = 1*0.5 = 0.5 mol
Therefore mass NaCl = mass/Mr = 0.5/(23+35.5) = 1/117g
The density of water is 1g/cm^3 therefore 1L=1000cm^3
So the mass of the water = density * volume = 1*1000 = 1000g
So the total mass of solution is 1000 + 1/117 = 1000.008547g

Please notice that this site has the policy that questions should show some own efforts. Therefore, please do not encourage users who did not show these efforts by answering them (until they did).

Besides, you are doing it wrong anyway.


text:
What is the reaction of ammonium nitrate with hypochlorous acid
What would the product of ammonium nitrate and hypochlorous acid? Would there be any reaction? Would the HOCl get broken down? Trying to determine if the free chlorine content would be diminished upon mixing in an aqueous solution.

Hi, welcome! What have you done to try to solve this problem? It will help answerers by showing what you know. Thanks!

I am not a chemist, so I am just trying to get a general idea of what is going on, someone please help me :(

That's pretty much like https://chemistry.stackexchange.com/questions/47121/mixing-bleach-and-ammonia and https://chemistry.stackexchange.com/questions/73456/how-does-aqueous-ammonia-react-with-hypochlorous-acid-in-water

Do not for goodness sake mix chlorine bleach with ammonium compounds or ammonia, it has a good chance of ending badly.
I am not sure of the curved arrow mechanism, but what I suspect will happen is that the ammonia will attack the HO-Cl using the OH- as a leaving group. So by making the system less basic it will favour the reaction.
This will convert ammonia into NH2Cl, which will then be converted into NHCl2 and finally into nitrogen trichloride (NCl3). When I was a PhD student one of the other people has to make NCl3, he did it with great care. He did the reaction in such a way that the NCl3 was diluted in an inert solvent.
The scary thing about nitrogen trichloride is that it is a very unstable and shock sensitive explosive. One of my uncles is a chemist like me, he once witnessed an accident in which a man lost several fingers playing around with nitrogen iodide.
But back to nitrogen trichloride, as well as being explosive it is also a harmful and irritating substance. I would suggest you have nothing to do with it !

To add to the "don't do it" chorus, there was an industrial accident where hypochlorite solution was added to acidified urea*.  The resulting explosion killed one person and injured several others, probably through reaction of hypochlorite and ammonia to form nitrogen trichloride.
*-The Chemaxx site (link above) has descriptions of many chemistry-related industrial accidents.


text:
Failure to Locate Stationary Point in GAMESS
I'm currently trying to optimize the geometry of a complex using GAMESS at the B3LYP level. However, I keep failing to converge the SCF. I've already tried the following:

increasing the number of iterations up to 300
increasing the convergence criteria up from 0.0001 to 0.0005
Switching on DAMP
Switching DIIS on and off.
Switching off SOSCF

I'm out of ideas. What else can I do? It's interesting to note that I'm testing the ligand against 3 metals: Copper, Zinc and Iron. While Copper and Iron fail to converge, Zinc converged on my first try. Would it be fair to say that given the same ligand, the complex I've designed is "impossible"?
Thanks!

Lot of unknowns right now.  Firstly, check your charge and multiplicity.

@LordStryker Interesting. I'm using Cu(II) which has a multiplicity of 2 and a charge of +2. My ligand is neutral and has no unpaired electrons, so MULT=2 ICHARGE=2. For Fe(III), the same thing applies, so MULT=2 ICHARGE=3. Or am I mistaken somehow?

@BernardJudeGutierrez The Fe(III) could be high-spin, depending on the ligand, so your actual multiplicity might be 6. (Intermediate-spin, with MULT=4, is uncommon, but also possible.) There might well be static correlation junking up the works for both Fe(III) and Cu(II), though I've rarely had SCF convergence issues with the Cu(II) species I've computed. Zn(II) has a closed d-shell, so it's a much less complex electronic system to work with; not surprising it's behaving better.

@hBy2Py thanks for the input. I'll try a couple of iterations on the Iron. I'm not sure how the ligand will affect the spin of the metal since this is a novel complex anyway. Would you say I should also try the same for Copper?

What's your basis set? B3LYP is already suspect for a potentially multiconfigurational problem like Fe(III). SOSCF is most useful once the DIIS error goes down, which can be hard with a bad guess. Damping is not as useful as one would think. Something better is to converge SCF in a small basis, then project to the larger desired basis, but I don't know if GAMESS can do that.

You could, sure. Shouldn't take long to tell if it's going to work better. AFAIK copper is always MULT=2 in 'normal' complexes, but if your ligand is exotic enough in the right ways, it might converge at MULT=4? If it does converge at 4, then you should definitely see if there's a stable broken-symmetry wavefunction.

@pentavalentcarbon I'm using B3LYP //631G(d). I use this because I've successfully modelled the ligand using this basis set. I'll try to see if I can do it with PM3. I'll also try SOSCF.

@hBy2Py, I've tried every iteration of multiplicities and they all still fail to converge. Maybe there's something wrong with the basis set? PM3 returns "Gradient out of Range" for this ligand.

Let's move to chat.


text:
Write SMILES to SDF in RDKit
I have a file with smiles and I'd like to convert the SMILES to SDF and tag the properties of each molecule, so I can use the SD file in another tool. Here is my code:
pp = pd.read_csv('anti.smiles', names=['Smiles', 'BA'])
pp =  PandasTools.AddMoleculeColumnToFrame(pp,'Smiles','Molecule')
PandasTools.WriteSDF(pp, pp_out, molColName='Molecule', properties=list(pp.columns))

And the error I get is this:
NameError: name 'pp_out' is not defined

Kindly assist

As-is, this isn't a minimum working example. Where does PandasTools come from? Can you add your missing imports so your code is runnable?

You must give the output file a name: 'pp_out.sdf'
With a smiles-file like
c1ccccc1O,Phenol
CCO,Ethanol

this works for me
import pandas as pd
from rdkit.Chem import PandasTools
pp = pd.read_csv('anti.smiles', names=['Smiles', 'BA'])
PandasTools.AddMoleculeColumnToFrame(pp,'Smiles','Molecule') # pp = doesn't work for me
PandasTools.WriteSDF(pp, 'pp_out.sdf', molColName='Molecule', properties=list(pp.columns))

and gives me this SDF
     RDKit          2D

  7  7  0  0  0  0  0  0  0  0999 V2000
    1.5000    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.7500   -1.2990    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.7500   -1.2990    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.5000    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -0.7500    1.2990    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.7500    1.2990    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.5000    2.5981    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0
  2  3  1  0
  3  4  2  0
  4  5  1  0
  5  6  2  0
  6  7  1  0
  6  1  1  0
M  END
>  <Smiles>  (1)
c1ccccc1O

>  <BA>  (1)
Phenol

$$$$

     RDKit          2D

  3  2  0  0  0  0  0  0  0  0999 V2000
    0.0000    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.2990    0.7500    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.5981   -0.0000    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0
  2  3  1  0
M  END
>  <Smiles>  (2)
CCO

>  <BA>  (2)
Ethanol

$$$$


text:
How does the water pollutant sodium nitrate affect the concentrations of other ions in tap (dam) water?
Context
In my assignment I've been given two water samples from the hinze dam. The first sample has been untouched, however the second has been contaminated with an unknown ionic compound. To identify the unknown contaminant, multiple tests were done, namely ISE tests and a TDS test.
The results were as follows
Results
Nitrate test – Nitrate ISE with an accuracy of ±10% full scale (calibrated 1 to 100 mg/L) and a range of 1mg/L to 14,000 mg/L
Pre-contamination sample – 3mg/L
Post-contamination sample – 2000mg/L
Ammonium test – Ammonium ISE with an accuracy of ±10% full scale (calibrated 1 to 100 mg/L) and a range of 1mg/L to 18,000 mg/L
Pre-contamination sample – 0.1mg/L
Post-contamination sample – 0.2mg/L
Calcium test – Calcium ISE with an accuracy of ±10% full scale (calibrated 10 to 1000 mg/L) and a range of 1mg/L to 40,000 mg/L
Pre-contamination sample – 10.8mg/L
Post-contamination sample 2.8mg/L
Chloride test – Chloride ISE with an accuracy of ±10% full scale (calibrated 10 to 1000 mg/L) and a range of 2mg/L to 35,000 mg/L
Pre-contamination sample – 67mg/L
Post-contamination sample – 316mg/L
Phosphate test – colorimeter test using a reagent that turns the solution blue using a 10mg/L standard and a 0mg/L standard (distilled water)
Pre-contamination sample – 2.3mg/L
Post-contamination sample – 0.56mg/L
TDS test with 200mL of post and pre contamination solutions. Water was poured through filter paper into beakers and the beakers baked in an oven, then the weight of the new solids in the beakers weighed.
Pre-contamination sample – 0.02g
Post-contamination sample – 3.89g
Observations
Conducting a flame test on the post-contamination sample's TDS provided an orange flame, which indicated that part of the compound was sodium.
Conclusion
Since the TDS indicated an excess of sodium and the ISE tests indicated an excess of nitrate, the pollutant is likely sodium nitrate.
Question
If the pollutant is sodium nitrate, why did the ions ammonium (although negligible)and chloride increase after contamination, when the other ions calcium and phosphate decreased?
So far I believe that the calcium decreased as it bonded with the dissociated nitrate.
Also, i think that the phosphate decreased as the dissociated increased sodium levels bonded with it.
However, I believe the decrease in calcium should cause the chloride to increase as the presence of calcium chloride would have its equilibrium shift in a way as to create for chloride as calcium is reduced. Which isn’t supported by the data?
Does solubility affect how the ions change?

Nitrate won't form an insoluble salt with calcium. You have an analytical problem. You are changing ionic strength, which changes activity. Yet, you don't mention how you measured the individual ions. Your ions won't increase/decrease simply by spiking in sodium nitrate.  So, again, as mentioned in your previous post, you have an analytical problem - ie this is a poor analysis causing a confusing conclusion.

I've edited my post to add the methods for how I measured the other ions. But if I understand correctly you believe my data is erroneous because the only ions that should change are the contaminant's ions (nitrate)? Does this also mean there can't be any other explanation for why the ions change?

Just about every one of your samples are out of the calibration range and therefore inaccurate and cannot be believed. You WILL NOT form a calcium nitrate salt - look up the solubility. You cannot magically form chloride - your analysis is wrong or the "contaminant" is contaminated. Your ammonia didn't change, it is within analytical error. ISEs are easily interfered with by other ions, so you need to read the instructions first to determine what those are. "Does solubility affect how ions change" Of course, that's what solubility is.

Ion selective electrodes are known to have two major limitations:

Ion selective electrodes are not ion-specific

Most ion selective electrodes are known to have interfering ions. For example, the fluoride ion selective electrode has interference by hydroxide.
This website lists technical specification sheets for common ISE methods, including known interferences. Check out the interferences on the ISEs that you used. That might explain some of your issues.
http://www.nico2000.net/datasheets/electrodes.html#ionlist

The response of ion selective electrodes is dependent on ionic strength

The total ionic strength changes the activity of the ions, and ion selective electrodes measure equilibrium interactions across the membrane. These equilibria depend on the activities of the species.
If your two solutions were not equal ionic strength, your measurements could be off.
Here's a link to a guide on dealing with these two issues.
http://www.nico2000.net/Book/Guide7.html


text:
What is an uninegative ligand
I can't find a simple definition via google. All I get are research papers, that I don't understand.

I'm pretty sure it just means the charge of the ligand is -1.
https://en.wiktionary.org/wiki/unipositive
Cyanide qualifies, ETDA does not.
Anyway, the word is on its way out. "Monoanionic" seems to be more popular.


text:
Question on assumption in Michaelis-Menten kinetics
In Michaelis-Menten kinetics, it is assumed that the substrate concentration is much higher than the enzyme concentration. In which step of the derivation of the Michaelis-Menten rate law is this assumption used?
From the explanations given in different sources, I think this assumption is used in expressing the total enzyme concentration as the sum of free enzyme and enzyme-substrate complex. Is my understanding correct? What happens when the enzyme concentration is in the same range as that of substrate concentration?
Why is it important to make this assumption?
Could someone explain?

Well, that's certainly not the case near the end of the reaction...

Let us look at the derivation detail. I will try to note, all assumptions/approximations made. I hope I don't miss any. The basic model is as follows: First an enzyme substrate complex is formed. The substrate can either be released unchanged, or undergoes modification to form the product. This can be captured in the following two equations
$$\ce{E + S <=>[k_a][k'_{a}] ES } \tag{1}$$
$$\ce{ES ->[k_b] P } \tag{2}$$
First, assumption herein is that initially, so little product is formed that the reverse reaction (product combining with enzyme and re-forming substrate) can be ignored, and hence the unidirectional arrow.
Rate of formation of product is, $v = k_b \ce{[ES]}$
And we use the steady-state approximation to write,
 $$\frac{\mathrm{d}\ce{[ES]}}{\mathrm{d}t} = k_a\ce{[E][S]}-k'_a\ce{[ES]}-k_b\ce{[ES]} \approx 0$$
Re-arranging, $$ [ES] = \overbrace{\frac{k_a}{k'_a + k_b}}^{K_M}[E][S]$$
Now, let us re-cast this in terms of concentrations of enzymes, and substrate we started with.

From the explanations given in different sources, I think this assumption is used in expressing the total enzyme concentration as the sum of free enzyme and enzyme-substrate complex. Is my understanding correct?

$$\ce{[E]_0 = [E] + [ES]}$$
No. The equation given above holds regardless of the initial substrate concentration.
Substituting for $\ce{[ES]}$
$$\ce{[E]_0 = [E] + \frac{K_M [ES]}{[S]} = [ES]\left( 1+ \frac{K_M}{[S]}\right)}$$

In which step of the derivation of the Michaelis-Menten rate law is this assumption used?

Now, finally, the substrate concentration is so large, that $\ce{[S] \approx [S]_0}$
This, finally gives us
$$\ce{[ES] = \frac{[E]_0}{1+ K_M/[S]_0}} \tag{*}$$
Plugging this in the expression for $v$, we get the Michaelis-Menten equation
$$v = \frac{k_b\ce{[E]_0}}{1+ K_M/\ce{[S]_0}}$$

What happens when the enzyme concentration is in the same range as that of substrate concentration? Why is it important to make this assumption?

You don't get a nice expression for $\ce{[ES]}$ like equation $(*)$ which only involves constants on the right hand side.
$$\ce{[ES] = \frac{[E]_0}{1+ K_M/([S]_0-[ES])}} \tag{**}$$

Many computational models that are used to study biological pathways have been built using Michaelis-Menten equation. In a living cell, the concentration of the substrate wouldn't be much higher than the substrate. What is the reasoning behind using Michaelis-Menten kinetics to model biological processes where these assumptions don't hold always?@getafix @Dr.J.

@Natasha This can be a question in itself. Short answer, it has does fit experimental observations well. I’ll have to do more research because this domain is a bit foreign to me. But like I said, ask on the site, and perhaps other people can weigh in

@Natasha:I'm not sure I quite understand your question on this point. What are you saying about the living cell? Are you saying [S] is not >> [E]? I believe it is in most, if not all cases.

Yes, I wish to know whether [S] >> [E] in biological systems?  There are a few sources that state the enzyme concentration and substrate fall in the same range. Excuse me for the naive question. Since I am not able to find the enzyme concentration in the literature in which the substrate concentration is reported, I'm confused.@ Dr. J.

I think it is more about expressing[S]. If we assume the substrate concentration is much higher than the enzyme concentration, then we can ignore the substrate that is bound as ES in the derivation of the MM equation.  If the enzyme concentration is in the same range as the substrate (or if the enzyme affinity is very high), then the amount of substrate bound as ES cannot be ignored, and the derivation is not simplified. In this latter case, [S] in the equation can't be considered as the total, but must be corrected by the amount bound.

First we need to understand two things. (1) Since an Enzyme is a catalyst, we assume that the forward reaction is much, much, more favored than the reverse reaction. That is, k1 >> k-1. (2) It is impossible to measure the concentration of the Enzyme-Substrate Complex. We can only know how much Enzyme is present, how much Product is formed, and how much Substrate is consumed.
Having said that, we get to the heart of your question. The rate determining step is the dissociation of the Enzyme-Substrate Complex into the Product and the Free Enzyme. So essentially V = k2[ES]. But since we cannot know what [ES] is, we have to relate it in terms we know. So [ES] should be able to determined from [E] and [S]. In order to do this, [ES] must be at a steady state until most of the substrate is consumed. This is why it's important that [S] be greater than [E]. [S] will continue to decrease and [P] will continue to increase. This won't hold true if there is insufficient substrate to form [ES].

Catalysts don't change the equilibrium of a reaction, so I am not sure that your first point in the first paragraph is correct.


text:
Which is more stable Ethyl carbocation or Neopentyl carbocation?
As per my knowledge, hyperconjugation is more dominating over inductive effect, so due to three alpha hydrogen ethyl carbocation should be more stable. In some books, while solving questions I observed that answers can match if I take neopentyl carbocation as more stable.
so which one of them is comparatively more stable?

Watch out: the ethyl carbo cation is a nonclassical cation; under certain conditions it is better described as a protonated ethylene. Stability is also a relative concept, which can't be applied to molecular ions/entities in an absolute manner.

Can you give specific examples? It may be possible that the underlying concept in some questions is not just the stability of the carbonation, for example, a kinetically controlled reaction...

Ethyl carbocation is more stable than neopentyl carbocation as we know hyperconjugation is more with ethyl carbocation


text:
Where can I find the standard molar entropy for ammonium bicarbonate?
I'd like to know the standard molar entropy S0solid for ammonium bicarbonate to use in an example in class, to calculate ΔS0 for the reaction, NH4HCO3(s) → NH3(g) + H2O(g) + CO2(g).
Already checked: Wikipedia, NIST Chemistry WebBook, Google. I don't have access to the CRC handbook.
(Arthur and Chester Miller already found the standard enthalpy of formation for ammonium bicarbonate, in this forum, in 2016; and it's in the German Wikipedia.)

“Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. gives values for the standard molar enthalpy of formation $\Delta_\mathrm fH^\circ$, the standard molar Gibbs energy of formation $\Delta_\mathrm fG^\circ$, and the standard molar entropy $S^\circ$ (all at $T=298.15\ \mathrm K$ and $p=100\ \mathrm{kPa}=1\ \mathrm{bar}$) of crystalline ammonium hydrogen carbonate as follows.
$$\begin{align}
\Delta_\mathrm fH^\circ&=-849.4\ \mathrm{kJ/mol}\\
\Delta_\mathrm fG^\circ&=-665.9\ \mathrm{kJ/mol}\\
S^\circ&=120.9\ \mathrm{J/(mol\ K)}
\end{align}$$


text:
What mass of potassium lactate is needed to 300.0 mL of 0.238 M Lactic acid solution to make a solution with a pH of 4.00? pKa (lactic acid) =3.86
Upon attempt, I've yielded an answer of 11.91g. My calculations do not take 300 mL Lactic acid solution  into account, and now I somewhat know the error is at least partially due to the fact that my working out is independent on the the quantity of solution. All help is highly appreciated.
Here is my working out:
Note:

my molar mass of potassium lactate is incorrect in the final step, none the less the answer is still incorrect.
For those who will ask, the answer is 12.6 g

Using the Henderson Hasselbalch equation pH = pKa + log [base]/[acid].
4.0 = 3.86 + log [lactate]/[lactic acid].
log [lactate]/[lactic acid] = 0.14.
[lactate][lactic acid] = 1.38
moles lactic acid = 0.300 L x 0.238 mole/L = 0.0714 moles.
moles lactate = 1.38 x 0.0714 = 0.09853 moles
grams lactate = 0.09853 moles x 128.2 g/mole = 12.63 grams = 12.6 grams (to 3 significant figures)
NOTE:  this assumes no change in volume upon addition of the potassium lactate.

I find it incredible that there is a down vote with no explanation. It appears the answer is complete and accurate, and if not, some comment should appear with the down  vote. This site essentially SUCKS.

I am relatively new to the site and I agree with you. Your post was indeed accurate and helpful. Perhaps the community gets bothered by students asking questions they're genuinely having trouble with.


text:
About how long do polymer chains have to be for their physical properties to resemble the polymer more than the monomer? (Or are at least comparable)
For example: According to wikipedia, Styrene is a liquid at room temperature, whereas isotactic Polystyrene does not melt until the 200s celcius. This is a simple property but quite a large change. I am curious how long polymer chains have to be (how many monomer units are present) before you can just consider them to be functionally identical to the kind of Polystyrene generally used in practical situations?  I know that there are different forms of polystyrene such as puffed/extruded, films etc, this is just an example.
The purpose behind this question is that I am considering doing some molecular dynamics/computational work on polymers and the longer the chains are, the more computer time is needed. So I'm wondering how large the molecules should be to reflect "real behavior".
Note: I'm sure it's a bit of a spectrum. You can probably make a smooth function between chain length and melting point, but I doubt it's a linear function... there's probably a point where it basically stops changing.
Edit: While I cited melting point I am actually more interested in electronic/orbital properties in case that helps.

A thing of 2 units is pretty different from the monomer, but also very different from the polymer. 10 units is closer to the polymer than to the monomer, but not quite there yet. 100 units is a polymer all right. As for electronic properties, they barely change at all, unless you have a conductive polymer.

Great question. Regarding Ivan's comment, even if electronic properties don't change much past a certain number of units, what we would consider purely steric effects may still vary considerably, which should be relevant for most polymers.


text:
Product of an intramolecular aldol condensation
When nonane-2,8-dione is treated with base, the product 1-(2-methylcyclopent-1-en-1-yl)ethan-1-one forms according to the following scheme: The arrow denotes the hydrogen that is first attacked by the base. But why does the reaction not take place like this? —
Why is the arrow-marked hydrogen less acidic than the previous one?

Its a reversible process and the 6 ring is thermodynamically favoured over the 8

Somewhat relevant:https://chemistry.stackexchange.com/questions/89790/regioselectivity-in-claisen-condensation-and-aldol-reaction

There is an error in the reation scheme drawn - Nona-2,8- dione cannot give a [5] ring, it will be [6].

@Waylander, Yes, I have made a very big error. I will repost the images within a short time.

The two protons are of similar pKa and you can get deprotonation at both sites with equal ease, however it is important to understand that most intramolecular aldol reactions are reversible and thus give rise to the thermodynamic product i.e. the most stable product. [6]-ring formation is always preferred over [8]-ring formation 1, 2. If you prepare the [8] product by other means, and subject it to the aldol conditions you will get the [6] product. The only exception to this is when you set up the reaction to be under kinetic control e.g. by using LDA, then you can get a mixture.

It is not a [6] ring, it is a [5]- ring

You can't get a 5-membered ring from the starting material you have drawn

Maybe the hydrogen in the second case is the allylic carbon containing hydrogen, so it is more stable. More stable means less reactive, which indicates less acidic. Allylic carbon means sp2 carbon beside another carbon.

authors: Αντώνιος Κελεσίδης


text:
Why doesn't the carbon dioxide produced by a fire put out said fire?
Carbon dioxide ($\ce{CO2}$) is used in some fire extinguishers to put out fire, but a combustion reaction itself produces $\ce{CO2}$. Why do we get fire then?
$$\ce{CH4 + O2 -> CO2 + H2O}$$
Why doesn't this $\ce{CO2}$ put out the fire?

It does. $\mathstrut$

The fuel for the chemical reaction is the CH4 (in this case) and the oxygen.  Assuming the CH4 isn't going to run out any time soon, the chemical reaction will continue as long as there is enough oxygen accessible. The force behind fire extinguishers can replace the oxygen available to the fire in its immediate area with CO2, and without one of the reactants (oxygen) available, the reaction won't continue.
The CO2 produced by the fire in an open area isn't going to replace the oxygen of the surrounding area.  However, in a closed system (for example, if you put a lid on top of a burning candle), then the air inside of the candle will become deficient in oxygen and rich in CO2 after a certain amount of time and the flame will go out, even without blowing on it or forcing it out.


text:
What measures of the dielectric constant mark whether the solution is ionic, polar and nonpolar?
At what numerical values of ε, then what organic compounds if any mark these thresholds.  How is that only for a vacuum has ε=1?  This is absolute lowest limit?

This looks like two separate questions. What does the value of the vacuum have to do with ionic, polar, and non-polar compounds? Any limits on the dielectric constant aren't necessarily related to the classification of a molecule. (I'm not saying this because I know the answer.)

Related answer: https://chemistry.stackexchange.com/a/32862/7951

Just compare the dielectric constant of known substances to non-polar, polar, metallic designations. For example, the dielectric constant of water is around 80.0 and water is polar. H2 has a very low constant value near 2 I believe and it is very non-polar. Metals have very large epsilon values moreover.  Rule of thumb, the larger the dielectric constant, the more polar the substance or material. In math terms, the potential felt between 2 charges is V= (q1 q2) / (4pi epsilon r) where q1 and q2 are different charges separated by r. The higher epsilon, the smaller V becomes. How charges interact with each other hence depends on the dielectric constant of the medium or material.


text:
What is the IUPAC name of kerosene?
Is there a standard chemical name for kerosene? Would IUPAC define a name or terminology for it?

Kerosene is a mixture thus does not have an IUPAC name

What does “IUPAC supported” mean?

@JavaScriptCoder iupac affirmed

No I mean, what does “IUPAC affirmed” mean?

https://chemistry.stackexchange.com/questions/94824/what-is-the-empirical-formula-for-kerosene

@JavaScriptCoder https://iupac.org/what-we-do/recommendations/

Kerosine is not a pure compound so dosen't have a standard IUPAC name. In fact the composition will vary depending on the source.
The major components are long chain (~10-16 carbon) branched and straights chain alkanes plus a mixture of cyclic alkanes and some aromatic alkybenzenes plus naphthalenes. The common factor is they all boil between ~150 °C and 275 °C (which is how they are separated from other components of crude oil).
So, no standard name just a range of varying compositions.


text:
What MnO2 and graphite do I need for making a dry cell battery?
I'm playing around with making my own dry-cell battery and I want to make sure I understand the pieces before diving into it.
My plan is to make it like a layer cake: zinc sheet, paste of electrolyte, layer of $\ce{MnO2}$ powder, graphite sheet, and a metal sheet with a wire connected to it.
What I'd like to understand better is:

Could I replace the zinc sheet with Zn powder? (Also this is interesting, but I don't understand what's meant by a Zn sponge, is it Zn powder with $\ce{Ca(OH)2}$?)
How pure/fine does the $\ce{MnO2}$ powder need to be? Potters used $\ce{MnO2}$ to color clay, is this the same stuff as what's in batteries?  I tried emailing a few places but they didn't know the chemical properties and I'm not confident enough to go for it without double checking first :/
Could I use carbon filters or even crushed charcoal or does the electrode need to be a solid sheet of graphite?  I think carbon filters and graphite sheets have the same chemical properties, but I'm not sure.


text:
How does the Jahn-Teller distortion result in both elongation and compression?
Recently, I have been reading up on the Jahn-Teller effect and I have read articles and posts on it, such as What is the mathematical basis behind the Jahn-Teller effect?. However, many of these rigorously discuss the quantum mechanical basis of the effect (which is fundamentally what it is about). But I seek to understand it using the simpler more intuitive "chemistry" explanation, which is through invoking an understanding of the nature of the d orbitals. Perhaps, the two explanations, both QM and chemistry, are equivalent and they very likely are... But I would just like to focus on the latter.
The reference I have cited was one of the literature I have come across which discussed the chemistry explanation I have mentioned above:

When looking for example at the configuration of $\ce {d^{0}_{x^{2}−y^{2}}}$ $\ce {d^1_{z^{2}}}$ ($\ce {d^4}$) and $\ce {d^{1}_{x^{2}−y^{2}}}$ $\ce {d^2_{z^{2}}}$ ($\ce {d^9}$), the ligand(s) along the z axis are much more screened from the charge of the central metal ion than the other four ligands along the x and y axes, because most of the electron density will be concentrated in the $\ce {d_{z^2}}$ orbital between the metal and the two ligands on the z axis. Consequently, there will be greater electrostatic repulsion between the electron(s) in the $\ce {d_{z^2}}$ orbital and the ligands along the z axis than between the electrons in the $\ce {t_{2g}}$ set of orbitals pointing between the axes and the ligands on the x and y axes itself. The ligands on the z axis will thus move further away from the central metal ion and thus lower the symmetry (z-out or elongation Jahn−Teller distortion). This causes the $\ce {d_{z^{2}}}$ orbital to be more stable (lower in energy) than the $\ce {d_{x^{2}−y^{2}}}$ orbital, as is shown in Figure 2.

As can be read, their explanation points to the elongation of the axial bonds. However, they do not explain how does the other scenario (i.e. compression) result in detail and they simply stated what happens in the other scenario. From what I read from the paper, the argument presented can only lead to one conclusion, and that is the axial bonds elongate. Could only help to fill in the other half of the story? In other words, I would like to know how do we explain the Jahn-Teller effect for the compression of axial bonds using a chemistry-orbitals approach.
Reference
Freitag, R., & Conradie, J. (2013). Understanding the Jahn−Teller Effect in Octahedral Transition-Metal Complexes: A Molecular Orbital View of the Mn(β-diketonato)3 Complex. Journal of Chemical Education, 90, 1692-1696. doi:dx.doi.org/10.1021/ed400370p

I cannot access the article right now because the ACS website is acting up, but I don't actually like the explanation as quoted: it seems to be circular logic. For the compression case, though, you simply need to replace $z^2$ with $x^2-y^2$ and $z$-axis with $xy$-plane. As an aside, the Jahn–Teller theorem, by itself, does not predict whether elongation or compression is observed. It only predicts that a distortion from $O_\mathrm h$ to $D_\mathrm{4h}$ symmetry will occur. Both elongation and compression fulfil this requirement.

@orthocresol Thanks for the insightful comment. I can send you the paper if you could give me some mailing address. Also, have you actually heard of this explanation using the nature of d orbitals before?

@orthocresol So is there any way to predict axial elongation/compression? I have read that electron density measurements performed on the complex ion can help us determine the orbital occupation in the eg set, allowing us to decide on whether elongation
(when $\ce {d_{z^2}}$ is populated) or compression (when $\ce {d_{x^{2}-y^{2}}}$ is populated) occurs.

@orthocresol Actually, it isn't quite right to say that it is compression along the z axis but rather an elongation along the x and y axes. Is that right?

(1) looking at the geometry is the easiest way (2) it is both at the same time, but it gets really wordy


text:
Comparing relative negative inductive effect of substituents (dialkyl amine, amine, alkoxide, fluoride)

Which of the following is correct with respect to the negative inductive effect of the substituents?
  1. $\ce{-NR2 < -OR < -F}$
  2. $\ce{-NH2 < -OR < -F}$
  3. $\ce{-NR2 > -OR > -F}$
  4. $\ce{-NH2 > -OR > -F}$

Since it would be directly proportional to electronegativity, $\ce{-F}$ will be highest. That cancels out (3) and (4).
Now, to me both (1) and (2) seem correct. But the official answer (since this is from an entrance exam) is (2).
Are both (1) and (2) correct? Or is there something wrong with (1)?

The inductive effect is the polarization of σ (sigma) bonds within a chemical species caused usually by differences in electronegativity between the neighboring atoms. The higher the polarization it causes, the higher the inductive effect. In this case, the decreasing order of electronegativity is:
$\ce{-F > -OR > -NH2}$
So, for practical purposes, the correct answer should be option 2.
However, substituents far from the atom you are considering can also affect the inductive effect and an amino ($\ce{-NR2}$) group can be more withdrawing than an alkoxide ($\ce{-OR}$) group.
For example: $\ce{-N(-CF3)2 > -OCH3}$
This amino substituent is probably more electron-withdrawing than the alkoxide in the right, because the first is bound to two highly withdrawing groups ($\ce{-CF3}$) while the second is bound to an electron-donating group ($\ce{-CH3}$)

It was more of a trick question because whenever we think of $\ce{-R}$, we think of alkyl groups intuitively


text:
Hypothetically, can sufficiently powerful centrifugation separate normally miscible chemicals of differing densities?
Centrifugation is used to separate phases in suspensions by density. I read that this is not possible with colloids and solutions, but the distinction between the three seems to be optical properties and correlated ranges of particle size. Larger particles take less time and centrifugal force to be separated, so is it possible that a sufficiently powerful centrifuge can separate miscible chemicals close in size like water and ethanol just based on their difference in density, even if they remix once the centrifuge slows? I know a more powerful ultracentrifuge can fractionate soluble proteins, a sucrose gradient being used to slow diffusion enough for fractionation afterwards, and a solution of caesium chloride will form a density gradient under ultracentrifugation.

It can even separate isotopes.


text:
How much ionising radiation is released during an atomic blast?
I've been looking for how much ionising radiation is produced during an atomic blast (as a percentage).
I have seen a claim of 5% of the energy in an atomic blast is ionising energy (Wikipedia), but if it's so little why do we put so much emphasis on radioactivity protection when discussing a nuclear blast. It just doesn't add up to me.

5% of a nuclear explosion is most definitely a lot of energy.

We don't. If you are close enough to the blast site for radiation to be an immediate issue, you won't survive the blast anyway. It is only in the long run that the radiation becomes important, when all other effects of the blast are long gone.

The amount of energy in an atomic blast releases as radiation (gamma rays, beta particles, alpha particles and neutrons) depends on how the bond is designed but isn't usually a big proportion of the total energy released. But that's not why we worry about the radiation from a bomb.
The reason why we worry about radiation is the radioactive products created in the blast (eg from  fission of the large nuclei in a fission bomb or from the potentially radioactive nuclei created when the initial radiation blast interacts with other materials). Typically many dangerous isotopes are generated that continue to release radiation for days or years. Worse the explosive blast scatters many of these over a large area.
But not all bombs are the same. Neutron bombs are designed to release far more of their energy as radiation (in this case a large blast of high energy neutrons) with the express purpose of killing people in the immediate area while having a much smaller blast. The original idea was to be able to use them on invading Russians in Western Europe without doing too much damage to the infrastructure. Neutron bombs can release ~40% of their energy as neutron radiation which is far more than the 5% released in a bomb tuned to create a large physical explosion.

I would treat Wikiepdia with a lot of caution. The relative yields of energy as heat, blast, initial radiation and delayed radiation to a person on the ground will vary a lot depending on the situation.
At the moment of detonation a major energy transfer mode from the inside of the bomb to the surrounding environment is the emission of gamma and X-ray photons. These photons give up their energy in the air, water or rock around a bomb.
While many people regard infra-red as "heat rays" all photons are "heat rays" as they are able to increase the internal energy of the things which they strike.
If we were to put a bomb in a deep hole in the ground then the amount of heat (IR) radiation which would reach a person on the surface would be zero. Also almost not initial ionizing radiation will reach the person. If the hole stays sealed than very little radioactivity will reach the person. However this blast will be optimised for giving a large semsic shock.
If we were to put the bomb in the sea then it will greatly reduce the initial radiation, the IR light. But we will get greater mechanical effects against the hull of a ship. Also the explosion will churn up a lot of very nasty radioactive fallout.
If we were to burst the bomb high in the air, then we would optimise the burst for the delivery of heat (IR) to the ground and also the air blast will be enhanced. The delivery of initial radiation will be lower and also the delivery of delayed radiation (fallout) to the person will be lowered. While much of this question and answer is physics, the behaviour of fallout is controlled greatly by chemistry. The chemical form of the nuclides at about 20 seconds post detonation will control their behaviour. The involatiles at that point will go onto large particles while the volatiles will end up in fine particles.
The burst on the surface of the ground will deliver less heat to a person at a moderate distance and also more delayed radiation in the radioactive fallout. What will happen is more of the fission products will fall quickly to ground near the blast. The air burst will inject much of the radioactivity into the upper atmosphere where it will have plenty of time to decay before it falls back to earth.
If you burst the bomb in space then the blast effect will be minimal or even zero. Thus I think that it is impossible to give a fraction of how much energy in blast, "heat", fallout and initial radiation for bombs in general.


text:
Does boiling milk really make the nutrients in it less useful to the body when ingesting it?
Does “denaturing” nutrients in milk by boiling it make them less useful, or actually more useful, as they undergo “denaturation” anyway during digestion?
I've read online that boiling milk "denatures" nutrients such as protein, calcium, and others contained in milk, rendering the milk less nutritious.
I've also read that because the nutrients in milk (and other foods) undergo "denaturation" in our body during digestion anyway, boiling it will not make milk less nutritionally valuable, and will actually help the body to more readily absord the nutrients in it into the bloodstream right after ingestion.
What is the truth in this matter? Thanks for any knowledgeable answers!  I put the word "denature" in quotes because it's a technical term and I'm not really sure in which ways it can correctly be used.

The term "denature" is generally used for proteins, and it is true that boiling milk will denature the proteins that are present. Calcium and other such elements would not be "denatured" per se. In fact, I can think of no adverse effect of boiling on the chemical nature of calcium, magnesium, etc. I'm not entirely sure what other minerals and/or nutrients are present in homogenized milk.
As for material being "denatured" during digestion, this is probably an inappropriate term since we generally use the term "metabolized" instead of "denatured". When proteins and other "nutrients" are metabolized, they result in useful products, as opposed to when they are "denatured". So, in summary, I'd say that boiling milk is not the same a metabolizing milk, and I'd say that boiling would likely render milk less "nutritious".

Casiens as opposed to whey especially micellar casein take a long time for your body to digest. Whey that has been separated and the predigested using enzymes and filtered among other treatments of the milk including heat makes the whey able to pass through your gut without normal digestion giving the body a boost of nitrogen helping muscles recover right after a workout. Caseins should be consumed during the day and Micellar Casein before bed to provide a constant nitrogen source for your muscles. As far trace nutrients and lactose go, you are better taking a good quality vitamin and using a better source of carbohydrate to avoid insulin spikes and provide long-lasting and even energy for your body. One instance of having to "boil" or raise milk to 185 deg. F. is in the making of yogurt. If you don't the yogurt will not set properly, if filtered as Greek yogurt is, you will remove much of the lactose and some of the whey. Not a scientific answer, but a practical one. Most Milk is pasteurized using heat so unless you get raw milk the point is really moot. Lastly, what exactly is meant by "less nutritious?"


text:
Two-electron integral algorithm
I am writing a molecular integral code from scratch. Right now my code have the following structure:
loop over shellpair ab:
    loop over shellpair cd:
        loop over primitive ij in ab:
            calculate E_ij
            loop over primitive kl in cd:
                calculate E_kl
                calculate R_ij_kl
                calculate V_ab_cd_ij_kl
        contract V_ab_cd_ij_kl to V_ab_cd

The above pseudo-code have alot of implicit notation, so let me explain what it means before asking my question.
E_ij and E_kl is the expansions coefficients from the McMurchie Davidson scheme.
R_ij_kl is the Hermite integrals also used in the Mcmurchie Davidson scheme.
V_ab_cd_ij_kl is the cartesian integrals found by (note the equations below are not strictly correct):
$$V_{ab,cd,ij,kl}=\sum E_{kl}\sum E_{ij} R_{ij,kl}$$
And V_ab_cd is the contraction of the primitive to the basisfunction:
$$V_{ab,cd}=\sum_i\sum_j\sum_k\sum_l c_ic_jc_kc_lV_{ab,cd,ij,kl}$$
When calculating integrals of higher angular momentum (above $0$), I make sure to reuse common elements in E and R. I.e. when doing p-type integrals px, py and pz have the first element in common, and it is only calculated once in the recurrence relations of the McMurchie Davidson scheme.
For my implementation of the above pseudo-code even the (ss|ss) integrals are very slow. When doing benenze with 6-31G the (ss|ss) integrals uses 2.5 seconds, which is more than can be justified by the use of Python (accelerated with Numba) instead of a compiled language. I have thought about rearranging and saving more intermidiates in memory, i.e. doing something like:
 loop over shellpair ab:
    calculate E_ab (Here E_ab means all E_ij saved in memory)
    loop over shellpair cd:
        calculate E_cd (Here E_ab means all E_kl saved in memory)
        loop over primitive ij in ab:
            loop over primitive kl in cd:
                calculate R_ij_kl
                calculate V_ab_cd_ij_kl
        contract V_ab_cd_ij_kl to V_ab_cd

In the above by calculating E_ab and E_cd, recomputation of E_ij and E_kl is avoided. At this point I know that I lag some expertise in knowing what is worth it compared to memory usage.
My question is know, what is a proper way of implementing molecular integrals with regard of what intermediates to keep at which points. (I know thing like vertical recurrence relations exist, but these will for obvious reasons not speed on my (ss|ss) integrals).

Semi-related, have you thought about using a lookup table for the Boys function? Your solution is probably faster than the analytic one, but some experts once told me about 30% of the time in calculating (ss|ss) is spent there.

Thanks for your observation.
I have thought about making my Boys function faster through look up tabels as  you suggest. This is also the recommended method in Molecular Electronic-Structure Theory.
I think that right now I am mostly concerned about figuring out how I want the broad structure of the code.

I agree. As far as I know, the way the Boys function is incorporated never changes; it’s just called as a function, as you do, so any performance improvements are localized.

@user1271772 Being slow is not the intention, but probably poking fun at the fact that being fast enough for practical computation is very difficult. Avoiding recomputation of intermediates is the primary form of reducing the number of FLOPs, since you can't change the scaling complexity.

I was trying to be constructive @pentavalentcarbon. I believe that not repeating these integrals will be counter-productive towards the goal of making the program slow. Otherwise, would this be helpful? http://vergil.chemistry.gatech.edu/notes/permsymm/permsymm.pdf

@user1271772 In that case I apologize. That document is not actually helpful because it is too high-level; using an analogy, it's like comparing Python with C that includes assembly in some places. The starting place is usually https://rsc.anu.edu.au/~pgill/papers/045Review.pdf, and in particular the section on "drivers". This is difficult because your intermediate is stored in a fixed-dimension Fortran array meant to behave as a hash table, where the key is an integer formed from basis set and shell pair info. In Python, a dictionary avoids some of that.

@pentavalentcarbon: By "too high level" do you mean that the notes were talking about something too trivial (symmetry relations), whereas the OP is asking about some more niddy-griddy details to squeeze in even better performance?

@user1271772 Yes, exactly.

@pentavalentcarbon: I thought the purpose was to build a program that is slow. However, perhaps the part that I missed, is that he wants this particular part to be done at decent speed.

@pentavalentcarbon  rsc.anu.edu.au/~pgill/papers/045Review.pdf seem very useful, thanks for the suggestion.

Unfortunately writing an integral code with decent performance is rather difficult. I also did it once for a hobby project and I'm still not overall satisfied. What I found helpful back then was the following thesis https://www.duo.uio.no/handle/10852/12863 (chapter 4). A more general remark: Low angular momenta integrals like (ss|ss) but also (ps|ss) carry some overhead. There are many inner loops running basicly from 0 to 0 or 0 to 1, which slows down. Many QM codes therefore use special routines to handle this integral classes, where they unrolled this loops.

Also semi-related: One possible solution to avoid double evaluation is to wrap a cache around your functions.  This can be implemented very easily with @functools.lru_cache, for Python2 this is provided by functools32.

You may wish to read about the PRISM algorithm (I believe the original paper is with Pople and M. Head-Gordon). This is the algorithm which is commonly used in the big software packages as it offers a trade-off between storage of certain expensive integrals and reuse of the vertical recursion relations for cheaper (one-electron) integrals.

@ErikKjellgren this question seems related to a recent one: https://materials.stackexchange.com/q/62/5. Do you know anything about it?

@ErikKjellgren here's another one you should see: https://mattermodeling.stackexchange.com/q/4478/5

There are many things to consider and improve your code. Many components
in an ab initio electronic structure calculations need supervision to achieve a
good performance. I am no expert on those calculations, but I will point you to some ideas and references. On the theoretical side, you could calculate E_ab for all shell pairs first. For more clarification on the order of operations and their scaling look at
Figure 3 in Multi-electron integrals. The article is written by a well-recognized researcher in the field that wrote a monumental book where you can find all needed expressions.
On the computational side, I would say that maybe Numba is not able to optimize your code. Writing it with Numba is not a guarantee of speed. I just took a glimpse at the code and, without going deep in the code, I saw many if statements. Only a straightforward observation. As general advice, it is all about how data is moved
and the flow of operations. Avoid allocations, try to rewrite as much as possible.
Profile your code to find the bottlenecks. Try to write it as vector or matrix operations to be able to use the large registers of the processor. There are many tricks to do these operations fast, but sadly they are buried elsewhere.
I would like to end with a reference to the recent literature on the topic. I just came up with an article that I managed to find again now: Fast Evaluation of Two-Center Integrals over Gaussian Charge Distributions and Gaussian Orbitals with General Interaction Kernels.


text:
MCSCF Hessian transformation
I have tried doing the transformation of the Hessian in MCSCF following the notation that is used in Molecular Electronic-Structure Theory (MEST). Here I try to show what steps I have taken (to get the wrong result).
In MCSCF we have that the second order contribution to the electronic Hessian is given as:
$$\overline{\overline{K}}=\left(\begin{array}{cc}
^{\mathrm{cc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{co}}\overline{\overline{K}}^{(2)}\\
^{\mathrm{oc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{oo}}\overline{\overline{K}}^{(2)}
\end{array}\right)\tag{12.2.40, MEST}$$
With,
$$^{\mathrm{cc}}K_{i,j}^{(2)}=2\left\langle i\left|\hat{H}-E^{(0)}\right|j\right\rangle\tag{12.2.41, MEST}$$
and,
$$^{\mathrm{co}}K_{pq,i}^{(2)}=^{\mathrm{oc}}K_{pq,i}^{(2)}=2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle\tag{12.2.42, MEST}$$
and,
$$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \frac{1}{2}\left(1+P_{pq,rs}\right)\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle\tag{12.2.43, MEST}$$
For the last equation the permutation operator $P_{pq,rs}$ can be expanded out, giving:
$$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}P_{pq,rs}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle$$
$$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{rs}^{-},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle$$
$$^{\mathrm{oo}}K_{pq,rs}^{(2)} = \left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{rs}^{-},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle\tag{1}$$
Now for the transformation of the Hessian we can write:
$$\left(\begin{array}{c}
^{\mathrm{c}}\overline{\sigma}\\
^{\mathrm{o}}\overline{\sigma}
\end{array}\right)=\left(\begin{array}{cc}
^{\mathrm{cc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{co}}\overline{\overline{K}}^{(2)}\\
^{\mathrm{oc}}\overline{\overline{K}}^{(2)} & ^{\mathrm{oo}}\overline{\overline{K}}^{(2)}
\end{array}\right)\left(\begin{array}{c}
\overline{c}\\
\overline{\kappa}
\end{array}\right)=\left(\begin{array}{c}
^{\mathrm{cc}}\overline{\overline{K}}^{(2)}\overline{c}+{}^{\mathrm{co}}\overline{\overline{K}}^{(2)}\overline{\kappa}\\
^{\mathrm{oc}}\overline{\overline{K}}^{(2)}\overline{c}+{}^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}
\end{array}\right)\tag{12.5.17, MEST}$$
Now I want to determine the expression of the right hand side. The first three terms I seem to get correctly, only the last part gives me trouble. Look for equation (Problem). Lets start with:
$$\left[^{\mathrm{cc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{i}=\sum_{j}2\left\langle i\left|\hat{H}-E^{(0)}\right|j\right\rangle c_{j}$$
The definition:
$$\left|\overline{c}\right\rangle =\sum_{j}c_{j}\left|j\right\rangle$$
can be used to find:
$$\left[^{\mathrm{cc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{i}=2\left\langle i\left|\hat{H}-E^{(0)}\right|\overline{c}\right\rangle \tag{12.5.18, MEST}$$
Now the next term:
$$\begin{array}{ccc}
\left[^{\mathrm{co}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{i} & = & \sum_{p>q}2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle \kappa_{pq}\\
& = & 2\left\langle i\left|\sum_{p>q}\kappa_{pq}\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle
\end{array}$$
Now by the definition:
$$\hat{\kappa}=\sum_{p>q}\kappa_{pq}E_{pq}^{-}\tag{2}$$
Giving:
$$\begin{array}{ccc}
\left[^{\mathrm{co}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{i} & = & 2\left\langle i\left|\left[\hat{\kappa},\hat{H}\right]\right|0\right\rangle \\
& = & 2\left\langle i\left|\hat{H}_{\kappa}\right|0\right\rangle
\end{array}\tag{12.5.19, MEST}$$
with,
$$\hat{H}_{\kappa}=\left[\hat{\kappa},\hat{H}\right]\tag{12.5.22, MEST}$$
Now the next term:
$$\left[^{\mathrm{oc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{pq}=\sum_{i}2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle c_{i}$$
Since it can be written that:
$$2\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle =\left\langle i\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle +\left\langle 0\left|\left[E_{pq}^{-},\hat{H}\right]\right|i\right\rangle$$
This now leads to:
$$\left[^{\mathrm{oc}}\overline{\overline{K}}^{(2)}\overline{c}\right]_{pq}=\left\langle \overline{c}\left|\left[E_{pq}^{-},\hat{H}\right]\right|0\right\rangle +\left\langle 0\left|\left[E_{pq}^{-},\hat{H}\right]\right|\overline{c}\right\rangle\tag{12.5.21, MEST}$$
And the last term. This is here where I cannot figure it out, I tried the following, by first applying equation (1):
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\sum_{r>s}\left\langle 0\left|\left[E_{pq}^{-},\left[E_{rs}^{-},\hat{H}\right]\right]\right|0\right\rangle \kappa_{rs}+\frac{1}{2}\sum_{r>s}\left\langle 0\left|\left[E_{rs}^{-},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle \kappa_{rs}\tag{Problem}$$
This can be rewritten as, by applying equation (12.5.22, MEST) and (2):
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[\hat{\kappa},\left[E_{pq}^{-},\hat{H}\right]\right]\right|0\right\rangle$$
Now by using the Jacobi identity:
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle -\frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[\hat{H},\hat{\kappa}\right]\right]\right|0\right\rangle -\frac{1}{2}\left\langle 0\left|\left[\hat{H},\left[\hat{\kappa},E_{pq}^{-}\right]\right]\right|0\right\rangle$$
Now by using the relation, $\left[\hat{A},\left[\hat{B},\hat{C}\right]\right]=-\left[\left[\hat{B},\hat{C}\right],\hat{A}\right]$ and $\left[\hat{A},\hat{B}\right]=-\left[\hat{B},\hat{A}\right]$:
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[E_{pq}^{-},\left[\hat{\kappa},\hat{H}\right]\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[\left[\hat{\kappa},E_{pq}^{-}\right],\hat{H}\right]\right|0\right\rangle$$
Now:
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\frac{3}{2}\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left\langle 0\left|\left[\left[\hat{\kappa},E_{pq}^{-}\right],\hat{H}\right]\right|0\right\rangle$$
Now by $\left\langle 0\left|\left[\left[\hat{\kappa},E_{pq}^{-}\right],\hat{H}\right]\right|0\right\rangle =\left[^{\mathrm{o}}\overline{E}^{(1)},\overline{\kappa}\right]_{pq}$:
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\frac{3}{2}\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\frac{1}{2}\left[^{\mathrm{o}}\overline{E}^{(1)},\overline{\kappa}\right]_{pq}$$
But this is NOT the right result in MEST the right result is giving as:
$$\left[^{\mathrm{oo}}\overline{\overline{K}}^{(2)}\overline{\kappa}\right]_{pq}=\left\langle 0\left|\left[E_{pq}^{-},\hat{H}_{\kappa}\right]\right|0\right\rangle +\left[^{\mathrm{o}}\overline{E}^{(1)},\overline{\kappa}\right]_{pq}\tag{12.5.21, MEST}$$
I cannot seem to figure out where I did my maths wrong and ended up with a factor 1/2 and factor 3/2.
As a side note the book explain the above math as:

where te elements of $\overline{\overline{K}}^{(2)}$ are defined in (12.2.41)-(12.2.43). Inserting these expressions in (12.5.17) and carrying out some simple algebra, we obtain

Since it is supposed to be simple, I might have forgot about something basic.


text:
Could a non-spontaneous reaction occur "on its own"?
Does spontaneity really only apply to the net free energy change of a reaction?
I used to be taught that spontaneous reactions could occur on their own BUT without enzymes or a spark they might take thousands of years to occur.
So teachers would say that spontaneity is more describing whether a reaction will occur on its own. Rather than a statement of reaction speed.
Because there is still activation energy to deal with, so technically spontaneous reactions to need an input of energy.
So is a spontaneous reaction more a statement of the reaction energetics?
And could a non-spontaneous reaction occur on its own given enough time?

Thermodynamics is not concerned with time only equilibria and that  means starting and ending conditions. A huge activation barrier, between reactants and products  makes compounds stable on a human time scale, for example carbon reacting with oxygen means that a diamond is stable for millenia at room temperature, but heat it in a flame and the effective activation energy is less and reaction can be completed in short order. Reactions occur whether they are spontaneous or not, its just a matter of how much product vs reactant is present at equilibrium.

An analogy may help.  Imagine a rock perched at the top of a hill.  If given a small push, the rock will roll all the way down the hill without any further input of energy (in fact, releasing energy as it goes).  That's what we mean by a "spontaneous" reaction.  The reverse (pushing the rock up the hill) is a "nonspontaneous" reaction into which you have to continuously put energy for it to proceed.
But now let's consider that small initial push.  Just how small is it?  If the rock is small and perched on a very sharp hilltop, maybe a strong breeze or a squirrel jumping on it is enough.  But if the rock is big and perched on a rather broad hilltop, it may need an earthquake or big tree falling on it to get moving.  Of course, earthquakes and big trees falling are much rarer events than a squirrel jumping around or high wind.  So if the rock needs a tiny push, we expect it to come flying down the hill quite soon after it's put up there, and if it needs a big push, it will be quite a long time before it gets rolling.
This is what we mean by the kinetics of the reaction -- how fast it goes.  A fast reaction needs hardly any push to get started, a slow reaction needs a big push, and small pushes are much more common than big pushes.  (The "push" for a real chemical reaction is some fluctuation in energy in the environment, like the coincidence of a bunch of simultaneous high-speed collisions of molecules.)
Hopefully the analogy makes clear that the thermodynamics of a reaction (whether it is spontaneous or not) is completely separate from the kinetics (whether it is fast or not).
Finally, use caution when everyday words (like "spontaneous") are used in scientific language.  The implications may not be the same!  In everyday speech "spontaneous" often has the additional implication of "quick."  If I said "I spontaneously break into song during meetings" you would find it odd if I only did it every 25 years.  There's an implication, in everyday speech, of speed as well as no need for a driving force. But that implication does not carry over to the scientific use of the word!

Spontaneity has no relation to speed or rate of reaction. A reaction might be spontaneous and at the same time it might take a thousand years to complete. And it is true that spontaneous reactions occur on their own without the input of external energy.


text:
Find the concentration of proton after adding a acid to NaOH
I have this simple problem that I cannot figure out though, only the first part.

100 ml of 1.0 $\frac{mol}{L}$ $NaOH_(aq)$ contains ( a ) g of NaOH.
  After mixing 100 ml of 1.0 $\frac{mol}{L}$ $H_2SO_4(aq)$ with the
  first solution, the concentration of proton becomes ( b )
  $\frac{mol}{L}$

I got that there are 4 g of NaOH in the solution, but I don't have clear ideas about what should I do to find the concentration of proton. Could you advise me?
The answer is 0.50 $\frac{mol}{L}$

Remember that both protons of the H2SO4 are available to react with the NaOH

When you say "The answer is 0.50 mol/L", how sure are you about that?
Because I think it's wrong.  The 0.50 mol/L answer seems to assume that the H2SO4 donates both protons, but that's not the case.

The $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ of $\ce{H2SO4}$ is about $-3$ and $1.92$ ($K_\mathrm{a2} = 1.2 \times 10^{-3}$), respectively. To solve the second part of this question, at least $K_\mathrm{a2}$ should be given. Since mole ratio of $\ce{H2SO4}$ to $\ce{NaOH}$ is $1:1$, the acid-base reaction taken place in the mixture is:
$$\ce{NaOH(aq) + H2SO4(aq) -> NaHSO4(aq) + H2O (l)}$$
(a) You already have done part (a) but I'll show you how to do the unit conversion:
$$\text{amount of}~\ce {NaOH} = \frac{\pu{1.0 mol} \text{ of }\ce{NaOH}}{\pu{1.0 L} \text{ of solution}} \times \pu{0.100 L} \text{ of solution} \times \frac{\pu{40 g} \text{ of }  \ce {NaOH}}{\pu{1 mol} \text{ of } \ce {NaOH}}\\ = \pu{4.0 g }  \text{ of } \ce {NaOH}$$
(b) Assuming $\mathrm{p}K_\mathrm{a2}$ is given, to calculate $\ce{H3O+}$ in solution, we must consider following situations. Since $\mathrm{p}K_\mathrm{a1}$  of $\ce{H2SO4}$ is large, we can assume following ionization (complete dissociation):
$$\ce{H2SO4 (aq) + H2O -> HSO4- + H3O+ \tag1}$$
Since $\mathrm{p}K_\mathrm{a2}$  of $\ce{H2SO4}$ is very small ($\ce{HSO4-}$ is a weak acid), we can assume following ionization:
$$\ce{HSO4- (aq) + H2O <=> SO4^2- + H3O+ \tag2}$$
Now similar to the calculation in part (a), you can find out amount of $\ce {OH-}$ in $\pu{mol}$ in the original base solution ($\pu {0.10 mol}$) and the amount of $\ce {H3O+}$ in $\pu{mol}$ in the original acid solution (also $\pu{0.10 mol}$), assuming second ionization do not take place (hypothetically, if second ionization did not initiate, it is safe to assume that each $\pu{mol}$ of $\ce{H2SO4}$ would give $\pu{1 mol}$ of $\ce{H3O+}$ and $\pu{1 mol}$ of $\ce{HSO4-}$). When original solutions are mixed, each $\ce{OH-}$ ion react with one $\ce{H3O+}$ ion to give one $\ce {H2O}$ molecule and one $\ce{HSO4-}$ molecule remain unreacted, according to the equation $(1)$. At the end, $\pu{0.10 mol}$ of $\ce{HSO4-}$ ions remain unreacted, since not enough $\ce {OH-}$ ions have been added. Nonetheless, total volume now is $\pu{200 mL}$. Therefore, the initial $[\ce{HSO4-}]$ is:
$$\text{Initial concentration of }\ce {HSO4-} = \frac{\pu{0.10 mol} \text{ of }\ce{HSO4-}}{\pu{0.200 L} \text{ of solution}} = \pu{0.50 molL-1}  \text{ of } \ce {HSO4-}$$
Now, we can visualize the second ionization according to the equation $(1)$ wuold take place:
$$
\begin{array}{lccc}
\ce{& HSO4-(aq) &+ & H2O &<=> &SO4^2-(aq) & + & H3O+(aq)} \\
\text{I} & 0.5 \times 0.20 && \text{a lot} && 0 && 0 \\
\text{C} & -\alpha && \text{a lot} &&  +\alpha && +\alpha \\
\text{E} & 0.5 \times 0.20 - \alpha && \text{a lot} &&  \alpha && \alpha \\
\end{array}
$$
By definition,
$$\mathrm{p}K_\mathrm{a2} = \frac{[\ce{SO4^2-}][\ce{H3O+}]}{[\ce{HSO4-}]} = \frac{\alpha \times \alpha}{0.5 \times 0.20 - \alpha} $$
I assume you may able to calculate the final $\ce {H3O+}$ concentration ($\alpha$) now.

So, if I understood, when solution are mixed, 0.10 mol of $OH^-$ react with 0.10 mol of $H_3O^+$ so that only 0.10 mol of protons is in excess. Therefore, since the solution now is 200 mL I have to do $\frac{0.10 mol}{0.200 L} = 0.50 \frac{mol}{L}$. Thanks for your help!

Since nobody has pointed this out explicitly, I believe this solution is wrong because it assumes the H2SO4 will donate both protons.  But H2SO4 is only a strong acid with respect to the first proton.  To determine how many molecules donate the second proton, you have to use the second dissociation constant for H2SO4.  The funny thing is that the OP said "The answer is 0.50 mol/L" which I assume came from an instructor or an answer key, which means the instructor got it wrong as well.

@Bennett The answer is from an answer key. Probably the exercise assumed that H2SO4 donates both H+. Otherwise, it would be less as you pointed out.

@TheBarbarios thanks.  Yeah I figured that -- the point is that the assumption is wrong, and gives the wrong answer :)


text:
Origin of broad CO stretch in infrared spectrum of tricobalt cluster
I'm quite confident that the peaks around $\pu{2000 cm^{-1}}$ are $\ce{CO}$ stretches. But I cannot find any reference on that massive broad stretch following the $\ce{CO}$ stretches. Does it have anything to do with high symmetry?

If there is expected to a range if different CO frequencies in your molecules I would clearly expect to see a series of discrete bands not a broad featureless absorption so my first action, before worrying about what causes this, would be to purify the compounds again and check that some other instrumental artefact has not occurred.

I agree. For example, I found it easy to get spectra with poor lineshapes like this on a benchtop ATR; thin-film on salt plates was better, but KBr pellets were best. This is not the kind of broadening you would see in paramagnetic NMR, but probably has to do with sample preparation.

Yea, that's a good point. It looks like I got some solvent or starting materials stuck between the clusters. But unfortunately I did not have enough lab time to recrystallise it.

I found an reliable article (http://www.ilpi.com/organomet/carbonyl.html) describing wide spectrum of metal carbonyls. Since Chemistry SE won't open that website, I have attached scheme with a range of carbonyl stretching frequencies in the article for your benefits:

The IR stretching of carbonyl can be varied based on the structure. According to the website:

In the IR, typical stretching frequencies are: (i) Uncoordinated or "free" $\ce{CO}$: $\pu{2143 cm^{-1}}$; (ii) Terminal $\ce{M-CO}$: $\pu{2125}$ to $\pu{1850cm^{-1}}$; (iii) Doubly bridging ($\mu_2$): $\pu{1850}$ to $\pu{1750 cm^{-1}}$; (iv) Triply bridging ($\mu_3$): $\pu{1675}$ to $\pu{1600 cm^{-1}}$; and (v) Semibridging: somewhere between terminal and $\mu_2$.

In literature, most tricobalt clusters were discussed included with other metal cations. For example: (a) Rational synthesis of tricobalt-molybdenum and -tungsten butterfly clusters with alkyne ligands in Journal of Organometallic Chemistry, 2002, 659(1-2), 142-150 (https://doi.org/10.1016/S0022-328X(02)01720-5) and (b) Clusters as Ligands. 5. Tricobalt Cluster Alkoxycarboxylates of Titanium and Zirconium Exhibiting Novel Structures and Properties in Organometallics, 1997, 16(24), 5289–5301 (https://pubs.acs.org/doi/abs/10.1021/om970837b).
You can also read: A. S. Goldman, K. Krogh-Jespersen, Why Do Cationic Carbon Monoxide Complexes Have High $\ce{C-O}$ Stretching Force Constants and Short $\ce{C-O}$ Bonds? Electrostatic Effects, Not $\sigma$-Bonding, J. Am. Chem. Soc., 1996, 118, 12159-12166 (https://cdn-pubs.acs.org/doi/10.1021/ja960876z).
Since you have completely described preparation procedure of your tricobalt cluster, I assume the broasness of carbonyl absorptions are due to different structural aspects of carbonyl on $\ce{Co}$ and other metals in the cluster. It is also possibility that you have a mixture of clusters. Note that range of $\nu_\mathrm{CO}$ could be $\pu{2100}$ to $\pu{1300 cm^{-1}}$, based on stereo arrangement (see above references).

This is nice and all, but I'm not sure if you're addressing the main point of the question, which is to do with the unusually broad peak at $2000\text{–}1000~\mathrm{cm^{-1}}$.

@orthocresol: Please see added  information in late edition.


text:
Which invariant reactions are congruent?
I am studying the concept of incongruent melting and invariant reactions. I was wondering, would all the invariant reactions that I show below be incongruent? As I understand it, we can only have congruent melting when the liquid to which our solid melts to has the exact same composition, and by the definition of these invariant reactions, this seems not to be the case in any of them.

You are correct, none of the reactions are congruent reactions.
Per IUPAC GoldBook:

Congruent Transition: A transition in which the two-phase equilibrium of melting, vaporization or allotropism of a compound involves phases of the same composition.

The key term here is "phases of the same composition." Since all of the reactions have two phases transforming into one phase (or vise-versa) the reaction is not congruent.
See the phase diagram for silicon and titanium. Congruent melting occurs for $\ce{Ti5Si3}$ and $\ce{TiSi2}$ where you observe a $\ce{\color{red}{comp A (s)} -> \color{red}{comp A (l)}}$ transition, whereas at the eutectics you have $\ce{\color{blue}{comp B (s)} + \color{green}{comp C (s)} -> \color{red}{comp A (l)}} $ transition where $\mathrm{B \ne C \ne A \ne B}$. Thus the eutectics are not congruent transitions.


text:
Activation energy of reactions
When a reaction occurs where does the energy to cross the activation energy comes from?My book states that it might look like there is violation of conservation of energy but the physics of small particles is different.

Reactions occur in a heat bath of the solvent. The kinetic energy of the molecules vary, only is the average constant as determined by the temperature.  Very occasionally, say one in a million collisions or less, by random events,  is the collision of sufficient energy to overcome the activation barrier.

The energy to cross the activation energy doesn’t come out of nowhere. Many times it’s something like heat (which is basically how fast the particles are moving).  For example, paper doesn’t combust to ash automatically[citation needed]. But heat it to
218 or so degrees centigrade, (including Fahrenheit 451) and it will auto ignite. This is due to the energy the heat provides.

That's just the title of the book not actual data.

@Mithoron it’s about between 218-240 centigrade, which includes 233 Celsius

@JavaScriptCoder: No offence to late Ray Bradbury, it was his idea  when he named his famous book, Fahrenheit 451. However, the auto-ignition temperature of any solid material is a function of various factors including the time it expose to the high temperature.


text:
marijuana testing process
Does anyone know a bit about how costly and accurate are the current THC measurement techniques?
Please speculate what drives cost chemicals, consumables or instrument time.

This is a rather broad but interesting question, there are several different general methods for the anaylsis of drugs of abuse. You need to consider what are you trying to do in the lab. I would consider the testing of urine samples and the assay of lumps of resin / plant matter to be different problems.

I would imagine that LC-MS would produce a good analysis and the current generation of machines are fairly expensive, tens of thousands of dollars

In many states in the US, THC and CBD are recognized as medicinal, and there is a great need to certify these substances for purity and composition.

As a starting point, THC testing is advertised online for the following prices:
--Green Leaf Lab: $75
--Pharm Labs : $75
These companies both use HPLC. Without looking anything up, I'd guess that: The machine is around $100,000. It measures a sample every 30 minutes. (There might be modern marvels of engineering that run several samples in parallel and run way faster - idk.) Solvents and consumables are cheap. Maintaining the HPLC is expensive. You need a trained technician to prepare samples, run the analysis, and troubleshoot. That's expensive. You also need someone to deal with customers, which can be time-consuming as well.
You are not the only person thinking about THC testing. Google finds a handful of articles on what a business opportunity it is, and equipment manufacturers are ready to sell you stuff.
Good luck!


text:
What's the point of solubility products?
What is the advantage of solubility products, e.g.
LiF(s) ⇄ Li+(aq) + F-(aq)  |  K = [Li+(aq)] * [F-(aq)] = 0.00184 M2 at 25°C (source)
compared to just stating that
Solubility in water = 0.134 g/100 mL (25 °C) (source)
(There's a little difference between (0.00184 M2)½ = 1.11 g/L and the solubility of 1.34 g/L above; I take it that's because of the difference between activity and concentration?)
It seems to me that solubility products are just a roundabout, complicated way of expressing the concentration of a saturated solution. I suspect that I'm missing something though: otherwise why would people publish long lists of solubility products?
What I read: Wikipedia; all my high school textbooks.

It all starts making sense when you are dealing with a mixture of salts having a common ion.

Oh OK. In a situation like, how much LiF can I dissolve in a solution that already contains 30 g/L NaF; or what is the concentration of Na+, Li+ and F- in a saturated solution of LiF and NaF. (I'd have to think about how to solve the resulting equations...)

Yeah, that's exactly what is it good for.

Solubility products figure prominently in equilibria involving precipitated species.  Following is an example that may arise in a practical situation.
Problem:  We have 0.01 M ferrous ion in water and we propose adding a base to drive out the iron as hydroxide.  We do not want excess base dissolving into the water so we will try magnesium hydroxide.  How well will it work?
The proposed reaction is then
$\ce{Fe^{2+} + Mg(OH)2(s) <=> Mg^{2+} + Fe(OH)2(s)}$
$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}$
Compare this equilibrium constant with:
$K_{sp}(\ce{Mg(OH)2})=[\ce{Mg^{2+}}][\ce{OH^-}]^2=5.61×10^{-12}$ (source)
$K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^{2+}}][\ce{OH^-}]^2=4.87×10^{-17}$ (Ibid.)
So then
$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}=\frac{K_{sp}(\ce{Mg(OH)2})}{K_{sp}(\ce{Fe(OH)2})}=1.15×10^5$
Since the reaction stoichionmetry implies that the sum $[\ce{Mg^{2+}}]+[\ce{Fe^{2+}}]$ will be constant at the original iron concentration of 0.01 M, we find that the iron level remaining in the water may be reduced by five orders of magnitude, the equilibrium concentration is below $10^{-7}$ molar!
Magnesium hydroxide, despite being supposedly "weak" because of its limited solubility, is actually like a strong base here, displacing the weak iron hydroxide base almost quantitatively -- but, in effect, self-regulating because of that limited intrinsic solubility.  The same property applies with respect to most heavy metals commonly found in water; their hydroxide solubilities, like that of iron, are much less than magnesium's.  Thus magnesium hydroxide (or, in practice, magnesium oxide which becomes the hydroxide in situ) is a good agent for treating water to remove heavy metals.  See here, under "Applications".


text:
Find volume of sample such that it contains 100ng of DNA
Im struggling with the maths of concentrationsand volumes and things..Ill give the question, then my thoughts on it:
A DNA trimer strand, ab, has moleuclar weight of $9309ng/\mu mol$. I#m designing a page gel to test the weights of few, but i only need to do it with this one to understand the maths of it. I need to find the volume that contains 100ng of the DNA trimer strand. It has a concentration of $876.36 ng/\mu L$ and a volume of $20\mu L$.
So, what I did is: in the sample we have $876.36 ng/\mu L *20\mu L=17527.2ng$. For just $100ng$ of this, we divide $17527.2ng$ by 100 to find the factor, and we divide the volume by the same amount - which gives us $0.114 \mu L$ of the sample. Is this right? It seems very little to me, but Im still new to this lab calculation stuff and am constantly surprised by it.
Thanks!

Your answer is correct, but I've done the calculation using dimensional analysis, as follows:
876.36 ng/ul (x ul) = 100 ng
x ul = 100 ng/876.36 ng/ul = 0.114 ul
And it is a very small volume indeed.