text:
Energy of electrons in the same subshell
So I have this quote from wikia (it's similar to my lecture pdf file) but I don't fully understand the statement

Although it is commonly stated that all the electrons in a shell have the same energy, this is an approximation. However, the electrons in one subshell do have exactly the same level of energy,[6] with later subshells having more energy per electron than earlier ones. This effect is great enough that the energy ranges associated with shells can overlap

In this statement, does it mean for example:
1) "Each" electron in the subshell 1s and "each" electron in subshell 2s have the same amount of kinetic energy.
Or
2) It means :"All electrons 1s have the amount of Kinetic Energy & All electrons 2s have the same amount of Kinetic Energy. However each electron in 1s doesn't have the same amount of Kinetic Energy compared to one electron in 2s" ?

One thing is certain: the textbook statements refer to the sum of potential and kinetic energy, not just the latter.

It's important to remember (or to learn, if it hasn't been explained before) that the only atomic systems for which we can resolve the Schrödiger equation exactly are what we call hydrogenoid atoms - atoms with only one electron, such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$...
As soon as you add a second electron, atomic orbitals change and they aren't exactly the same as for the hydrogen atom (this is a form of the three-body problem that also comes up in other fields of physics; if you remember your Newtonian mechanics, that's why you cannot solve analytically a three-body gravitational problem). The source of this change is electron-electron interaction.
However, electron-electron interaction, in general, introduces only minor distortions in atomic orbitals - for the most part, the orbitals of poly-electronic atoms are very very similar to hydrogenoid atoms in terms of shape, size, etc, which is why we use the same labels for them as for hydrogenoid atoms ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p}$, etc).
There is one property of hydrogenoid atoms that is very importantly changed by electron-electron interactions, though: in hydrogenoid atoms, the energy of an orbital is only a function of its main quantum number, $\mathrm{n}$, which means that all the orbitals of the same shell will have the same energy. In a single-electron atom, an electron in a $\mathrm{2s}$ orbital will have exactly the same energy as an electron in a $\mathrm{2p}$ orbital, because they only interact with the nucleus through a Coulombic interaction - which has spherical symmetry (it's the same in all the directions of space; it only changes with distance to the nucleus).
If there are other electrons in the same atom, occupying different orbitals, this breaks down the energy equivalence of same-shell orbitals - because, for instance, an electron in a $\mathrm{1s}$ orbital interacts differently with an electron in a $\mathrm{2s}$ orbital or in a $\mathrm{2p}$ orbital - due to symmetry, overlap, and a number of factors (these interactions are very complex, can't be solved analytically, and to some extent aren't even completely understood). The energy corrections are generally small, but the orbitals belonging to the same shell are no longer equal in energy - that's why we can order subshells, and we know that $\mathrm{2s}$ orbitals fill up before $\mathrm{2p}$ orbitals.

Can I ask one more question ? For example: let's take C (Z=6): 1s2 2s2 2p2. Does the "2 electrons" in 2s2 are equal (if not discussing hydrogenoid atoms) ? And the same for "2 electrons" in 2p2.

Also, according to your statement, you meant that electrons in 1s2 and 2s2 are not of same energy levels ? Thank you. So sorry, I am just want to be sure I understand you correctly

a) Yes, the two electrons in a doubly-occupied orbital (say, $\mathrm{2s^\mathrm{2}}$, or $\mathrm{2p_\mathrm{x}^\mathrm{2}}$, or $\mathrm{3d_\mathrm{z^\mathrm{2}}^\mathrm{2}}$) have the same energy (they are identical except for the spin function).
b) That's exactly right: orbitals in different shells (the main quantum number) have different energy levels.

Let's be clear. The electrons in the same sub-shell have the same energy when not in an external electromagnetic field. Since electromagnetic fields are infinite in size, it is only an approximation to claim that same shell electrons all have identical energy. (It can be a good approximation, but it is an approximation nevertheless). For the same type of atom (same nuclear charge), the energy of all the electrons in the same sub-shell are (approx.) the same. I'm not sure how you jumped from that to thinking that all s or p or d or f or g or h or.... sub-shells were of the same energy. Of the captured electrons quantum numbers, the Principle quantum number affects energy the most. Keep in mind that the sub-shells are not necessarily relevant. For instance, hydrogen has a 1f subshell and it is possible that an excited H atom could have an electron in that subshell. But normally we speak about ground state configuration. It should come as no surprise if I were to tell you that the 1f subshell required more energy than, say the 2s subshell of a hydrogen atom. (Although I do not know which is actually higher energy, I'd have to either do the calculation or check the tables). So, it seems to me that the question isn't whether or not subshells with lower Principle Quantum Number can be higher energy than subshells with higher Principle QN, rather the question is do occupied (in the ground state) subshells ever show that property. The Aufbau Principle predicts the 4s is filled before the 3d, etc. https://en.wikipedia.org/wiki/Aufbau_principle which clearly answers that question. (Since aufbau says the lower energy level is filled first). (Note that there are well known exceptions to the rule, take a look at the ground state electron occupancies of Cr or Cu, they are not what you'd expect from a naive application of aufbau (and it gets worse as atomic number increases, "worse" in the sense of more exceptions)). To be clear, a Nx subshell electron will never have more energy than a (N+1)x subshell electron* (except possibly during the transition from one energy level to another, but thinking of an electron which has absorbed more energy than it can have in a sub-shell, thinking of that electron as still in that sub-shell, is a matter of definition and I wouldn't buy it) *Where N and N+1 are the Principle quantum no. and x from the set (spdfgh....). I'll also add the fact that only the s subshells are spherically symmetrical, so all of the other shells will be oriented with respect to any external electromagnetic field, hence various sub-subshells will have different energies because they're  differently oriented towards that field. I should for completeness also note that two electrons share the same sub-sub-shell only if they have opposite spins, (which again will cause slightly different energies if in an external field).

It is inaccurate to claim that a 1f subshell exists for the hydrogen atom because the the maximum value of the angular quantum number is (n-1).


text:
Vacuum filtration: will Polymer elements interfere with the vacuum?
I am doing an experiment with vacuum filtration, where a Nylon filter is used to create a membrane (marked in pink in the picture attached). When I use the membrane, it frequently breaks in the process of transferring the membrane (on the Nylon filter) to another location. So I thought to put the Nylon filter onto a rigid ring, so that I can move the entire ring (with Nylon filter and my own membrane on top) without introducing cracks into my membrane. I wanted to 3D print this disk using a polymer resin, but was unsure, whether the polymer would interfere with the vacuum filtration process in some ways, or whether it would be good to use. I marked up on the picture, where the polymer ring would go (between the glass beakers).
Here is the image I mentioned to show my setup

Keeping aside that there are different filter materials (nylon, PFTE, paper, glass/glasswool) and filter funnels (e.g., http://emsar-laboratory.ro/en/magazin/filters-filtration/filter-funnels-125-ml/), what are the media you intend to filter?  For example, some organic solvents actually may corrode a frame (similar to an example like https://3dprint.nih.gov/discover/3DPX-004082) if made from poly lactic acid.  A hint about your filtration: look for a vacuum hose, the tubing attached to the sucction flask on the photo is too flexible and consequently, will limit the efficieny of your device.

It depends on the solvent.  DMF can weaken some of these non-paper /non-glass fibre filters. Additionally,  the order of events of filtration can cause breakage.  Pulling vacuum prior to adding solvent is different to filling the funnel and then applying vacuum

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text:
Does aluminium carbonate exist?
I read in a west African chemistry textbook (New School chemistry by Osei-Yaw Ababio) that aluminium trioxocarbonate (IV) (aluminium carbonate) does not exist. I need verified confirmations or oppositions.

Trioxocarbonate(IV) = carbonate($\ce{CO3^{2-}}$). Aluminium carbonate($\ce{Al2(CO3)3}$) does not exist. However its basic form does exist in basic form as Dawsanite mineral. Check wikipedia.

It does not exist in nature, but you could make it in the lab by adding pressure.
From a traditional pencil and paper calculation I found the pressure needed for
Al2/3O (s) + CO2 (g) = Al2/3CO3 (s)
is about 10 GPa...
Or you could make an Al2(CO3)3 (aq) solution with concentration of $10^{-5}$ M.
Well, still not impossible. And there should be some special solvents and/or complex structures that easily allow this reaction (postulation).

Could you add more detail, such as what pressure is required? Is there a specific procedure?

I don't have the computational software I used to have. But you could calculate the free energy and find temperature/ CO2 pressure where the Al2(CO3)3 could be stable

The reason why aluminium carbonate cannot exist is because it is simply too unstable and would rather be broken down instead.
Firstly, the aluminium ion is highly polarising with a high positive charge density (as it is a small ion which is triply charged). Thus, distortion of the carbonate ion's electron cloud is very great.
This is a very crude way of putting it but how the reaction happens is in fact in this manner: Aluminium ion "pulls" one of the oxygen atoms, which has a single bond to the carbon, in carbonate, towards it and that single bond breaks,  allowing one of the other oxygens to form a double bond with the carbon. And this gives carbon dioxide.
Also, the reaction is favourable due to entropy considerations as well. The carbon dioxide formed is in the gaseous phase.
Thus, the reaction between aluminium and carbonate would gladly proceeds forward. This is why it "does not exist" as claimed by your textbook.
Hope this clarifies it.

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text:
Why is aniline less basic than methylamine?
The reason given for the question is that in aniline the lone pair on N atom is delocalized throughout the benzene ring, making it less available for donation.
But the conjugate acid of aniline should be more stable given the fact that the positive charge on N atom is also delocalized throughout the ring, hence making it more stable than the conjugate acid of methylamine which has +I effect. On that account, shouldn't aniline be more basic?

Are you sure the positive charge is delocalised throughout the ring? Have you tried writing the resonant forms of $\ce{C6H5NH3+}$ that are involved in that?

Okay. Now I got it. The positive charge cannot be delocalised because N can't form 5 covalent bonds. Since the C-N double bond can't be formed, there would be no charge dispersion.

Exactly. And the basic form, aniline, is actually stabilised by resonant structures in a way that alkyl amines can't.

The question premise is itself wrong.

N in aniline is forced into an $\ce{sp^2}$ configuration because only then can its p orbital interact to form aromatic MO’s. By contrast, methylamine has its N in the usual $\ce{sp^3}$ configuration. An $\ce{sp^2}$ lone pair has a great contribution from an s AO than an $\ce{sp^3}$ lone pair (33% vs 25%), and because an s AO is lower in energy than a p AO (due to a region of electronic density closer to the nucleus), it is also lower in energy. Therefore, the HOMO of aniline has a poorer energy match with the H 1s LUMO than the HOMO of methylamine, and the resonance energy stabilizing the protonated adduct is smaller. Hence, aniline is less basic.

It’s worthwhile to note that acidity is a thermodynamic, not kinetic effect. Yet, I can still consider HOMO and LUMO interactions here because the interaction between N and H is largely local, leaving all other MO’s largely unaffected, so the interaction between the FMO’s determine the stabilization of the product and not just the TS.

You are wrong when you say that positive charge on the anilinium ion is delocalised in the benzene ring because that would mean nitrogen having $5$ bonds (which is not possible since nitrogen cannot expand its octet). Hence, the aliphatic amines are generally more basic than aromatic amines.

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text:
Identifying spontaneous redox reactions

Which one of the processes is NOT spontaneous?
\begin{align}
\ce{Ni + Sn^2+ &-> Ni^2+ + Sn} \label{1} \tag{1} \\
\ce{Cu + Fe^2+ &-> Cu^2+ + Fe} \label{2} \tag{2} \\
\ce{Ni + 2Ag+ &-> Ni^2+ + 2Ag} \label{3} \tag{3}
\end{align}

I failed this class last semester so I'm redoing the tests to see if I can get this over with, however I'm confused with this question. To find out if a process is spontaneous I need to identify the oxidizing and reduction agents.
\eqref{1} $\ce{Ni}$ is the reduction agent and $\ce{Sn^2+}$ is the oxidizing agent;
 \eqref{2} $\ce{Cu}$ is the reduction agent and $\ce{Fe^2+}$ is the oxidizing agent;
 \eqref{3} $\ce{Ni}$ is the reduction agent and $\ce{Ag+}$ is the oxidizing agent.
For a redox reaction to be spontaneous the reduction potential of the oxidizing agent needs to be higher than the reduction potentials of the reduction agent, therefore \eqref{1} and \eqref{3} are not spontaneous, but only one of the reactions can be not spontaneous. Where did I go wrong?

For starters I recommend writing out all of the relevant half reactions and looking up the standard reduction potential in the table for each half reaction.

For starting purposes, you should write the half cells, either with their oxidation/reduction potentials.
Now if you want to compare the oxidizing species then you have to consider the oxidizing potentials of all the species and the one with highest value will be the spontaneous half cell and the reverse of the other one will occur.
Remember you should consider only one type of potential at a time(i.e. either oxidation or reduction) for comparison.

The answer is actually the opposite of what you've said.

(Source: VCAA)
From this diagram above we can see that for B, Silver (Ag) which is the oxidising agent is below the reducing agent (Iron II). This does not obey the rule meaning it will be non spontaneous.

In science we generally think about processes which produce more energy being more energetic. Free Energy, mostly (afaik) used in chemistry is a thermodynamic term and every student of chemistry should understand that
Δ G < 0  favoured reaction (Spontaneous)
Δ G = 0  Neither the forward nor the reverse reaction prevails (Equilibrium)
Δ G > 0  disfavoured reaction (Nonspontaneous)

You should understand the concepts of entropy, energy, enthalpy, equilibrium, Gibbs and Helmholtz Free Energy and learn what assumptions are required to use the various relationships. (Equilibrium, Isolated, Closed, Open systems, constant pressure, constant temperature, constant entropy, etc.). You also need some perspective: Thermodynamics was created long before we understood about the molecular/atomic nature of chemicals. The concept of equilibrium is vital to understand. Beginning chemistry deals with processes near equilibrium. Equilibrium ignores time. In this worldview, any process is understood to have both a thermodynamic aspect and a kinetic aspect - with the kinetics being what is concerned about the timing of events. Another way to put this is that thermodynamics deals with the starting state and the final state of a particular process, its equations deal with path independent quantities.
When I looked at your three equations, I immediately noticed the Cu++. Do you, just from a common sense view, really think iron metal is going to oxidize copper? So you'd rather have a pocket full of iron nails get wet than a pocket full of pennies? Because the Cu is more likely to react (be oxidized) than the iron? Well, no. Iron is much more likely to rust than copper is to oxidize (although green copper oxidation is pretty common). By looking, I also noticed Ag+ and I know that silver is, like copper, a common metal used for cups, silverware (hint, hint) and coins. In other words, it is not very susceptible to corrosion (oxidation). In other words, the metal Ag° is not prone to oxidation which means that the ion must be prone to reduction. So, even though I am not that familiar with Ni oxidation potentials, I know Cu° is more stable than Fe° (at pH 7) and I strongly suspect Ag° is more stable than Ni°. I'd have to check the Ni° & Sn° couple, but ever heard of tin cans? Yeah, again the products of both A and C are known (by common sense) to be the more stable metals while just the opposite is true for B. Now, it would be much better to actually check the electrochemical series https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page). And in a chemistry class, you must make that your standard practice, but you probably shouldn't even had to have done that to answer this question → IF you understood this stuff.
It's important to note here that the reason they're teaching you this electrochemistry, is not because it is all that important to learn how to plate copper or silver out of a solution, but to get you to learn the relationships between free energy and spontaneity (and enthalpy and entropy and energy) which are very common in the chemical literature. You need to carefully learn this stuff. The reason why they're carefully teaching it is because it isn't easy to learn but it's important to learn. The greater the Gibbs Free energy the more spontaneous ONLY makes sense if you understand they're talking about the larger negative quantity. This is undoubtedly confusing, and once you're burned a couple of times, you'll realize that you have to make sure you're on the same page as the person talking about "greater" or "more" when they aren't being specific about the sign of the quantity. A spontaneous process has a greater negative free energy.

For spontaneous reaction: $\Delta G \lt 0$
Since $\Delta G = -nFE$, it is safe to say that for spontaneous reaction: $E \gt 0$. Thus, to answer this question, you must have access to the electrochemical series, the one like Guest posted in his/her answer. Let's look at the first reaction:
$$\ce{Ni + Sn^2+ -> Ni^2+ + Sn} \tag1$$
The two relevant half-reactions are:
$$\ce{Ni^2+ + 2e-  -> Ni} \quad E^\circ = \pu{-0.257 V} \tag4$$
$$\ce{Sn^2+ + 2e-  -> Sn} \quad E^\circ = \pu{-0.138 V}  \tag5$$
From equations $(4)$ and $(5)$:
$$\ce{Sn^2+ + Ni  -> Sn + Ni^2+} \quad E^\circ_\text{Rxn} = \pu{+0.119 V}  \tag6$$
or
$$\ce{Sn + Ni^2+  -> Sn^2+ + Ni} \quad E^\circ_\text{Rxn} = \pu{-0.119 V}  \tag7$$
Accordingly, the equation $(6)$ (original equation $(1)$) is spontaneous since $E^\circ_\text{Rxn} \gt 0$.
Now, look at the second reaction:
$$\ce{Cu + Fe^2+ -> Cu^2+ + Fe} \tag2$$
The two relevant half-reactions are:
$$\ce{Cu^2+ + 2e-  -> Cu} \quad E^\circ = \pu{+0.342 V} \tag8$$
$$\ce{Fe^2+ + 2e-  -> Fe} \quad E^\circ = \pu{-0.447 V}  \tag9$$
From equations $(8)$ and $(9)$:
$$\ce{Cu^2+ + Fe  -> Cu + Fe^2+} \quad E^\circ_\text{Rxn} = \pu{+0.789 V}  \tag{10}$$
or
$$\ce{Cu + Fe^2+  -> Cu^2+ + Fe} \quad E^\circ_\text{Rxn} = \pu{-0.789 V}  \tag{11}$$
Accordingly, the equation $(11)$ (original equation $(2)$) is not spontaneous since $E^\circ_\text{Rxn} \lt 0$.
Now we have the answer. But, the reaction $(3)$ could not be spontaneous as well. To ensure it is spontaneous, we can check the reaction following the similar path:
$$\ce{Ni + 2Ag^+ -> Ni^2+ + 2Ag} \tag3$$
The two relevant half-reactions are:
$$\ce{Ni^2+ + 2e-  -> Ni} \quad E^\circ = \pu{-0.257 V} \tag4$$
$$\ce{Ag^+ + e-  -> Ag} \quad E^\circ = \pu{+0.7996 V}  \tag{12}$$
From equations $(4)$ and $(12)$:
$$\ce{Ni^2+ + 2Ag  -> Ni + 2Ag^+} \quad E^\circ_\text{Rxn} = \pu{-1.0566 V}  \tag{13}$$
or
$$\ce{Ni + 2Ag^+  -> Ni^2+ + 2Ag} \quad E^\circ_\text{Rxn} = \pu{+1.0566 V}  \tag{14}$$
Accordingly, the equation $(14)$ (original equation $(3)$) is spontaneous since $E^\circ_\text{Rxn} \gt 0$.
Thus, among three given reaction, only the reaction $(2)$ is not spontaneous while other two are spontaneous.


text:
Are there proteins with permanently bound magnesium?
Do you know something about the existence of a magnesium containing protein/enzyme in which magnesium is coordinated in the structure and the structure is stable? (like iron in haemoglobin or zinc in zinc fingers)

Chlorophyll? This is not hard to find :) https://en.wikipedia.org/w/index.php?title=Magnesium_in_biology&section=9#Biological_chemistry

@gilleain Chlorophyll isn't a protein.

@DavePhD : Right, I was thinking of the cofactor since OP mentioned haemoglobin but regardless - it's not exactly a rare metal to find in proteins.

From the look of things you may actually be referring to a lot of different proteins, however I managed to
grab two that I found fascinating:

Enolase

A glycolytic enzyme which catalyzes the reversible dehydration of 2-phosphoglycerate to phosphoenolpyruvate:

Credits: Lehninger Principles of Biochemistry page 213 ("Enzymes")

In the enzyme active site, 2-phosphoglycerate undergoes strong ionic interactions with two bound Mg2+ ions causing the resulting intermediate to be easily abstracted.

GTPase specifically  ARF1 (ARF for ADP-ribosylation factor). These are important in priming phase in the formation of Clathrin-coated vesicles (CCVs).

Clathrin itself is  a protein that forms a polyhedral framework around vesicles that transport TM, GPIlinked, and secreted proteins from the Golgi to the plasma
membrane.

X-ray structures of ARF1.GDP and ARF1.GTP, were determined by Dagmar Ringe and by Jonathan Goldberg:

Credits Voet and Voet Biochemistry 4th ed page 435 ("Membrane Assembly and Protein Targeting")
X-ray structures of (a) ARF1.GDP (left) and (b)ARF1.GMPPNP (right) (GMPPNP is a nonhydrolyzable GTP analog in which the O atom linking GTP’s beta- and gamma-phosphorus atoms is replaced by an NH group.)
The bound nucleotides are drawn in stick form in white with their phosphorus atoms magenta and their bound $\ce{Mg^2+}$ ions shown as lavender spheres.
In ARF1.GDP, the protein’s N-terminal helix (red) together with its
covalently linked myristoyl group (not present in the X-ray structures) are bound in a shallow hydrophobic groove on the
surface of the protein formed in part by the residues of loop gamma 3.

Thank you for your answer. As far as I am concerned, enolase or other glicolytic enzymes don't have prosthetic groups with magnesium, so the metal in there is just a cofactor, being much of an activator. However, I have never heard about ARF1 and I think that this is the example I'm looking for. Thank you for the hint :)

The metal is actually bound to the enzyme enolase :) Read more: Plant Biochemistry J Bonner page 216 > Enolase contains a bound magnesium ion as an essential component...

Anytime! glad I could help :)


text:
Column order, signs of MO coefficients, and density matrix construction in Hartree-Fock
I'm attempting to make a Hartree-Fock solver, using Python, following Daniel Crawford's guide.
I have been successful with the first four steps. However, I am having some trouble with this part in step 5:
$$
F^0_c C^{'}_0 = C^{'}_0 \epsilon_0
$$
I have the correct transformed Fock matrix (I can verify this as the site as step-by-step matrices that you can check against). My problem is that the MO coefficient matrix I get does not quite match the one on the site. I obtain it by multiplying the above equation by $C^{' -1}_0$ on the right:
$$
F^0_c = C^{'}_0 \epsilon_0 C^{' -1}_0
$$
Now it's in diagonalized form, which means $C^{'}_0$, the transformed MO coefficients, is the eigenvectors of $F^0_c$ and $\epsilon_0$ is a diagonal matrix of the eigenvalues. After I obtain $C^{'}_0$, I obtain the original untransformed MO coefficients $C_0$ using the next equation on the site:
$$
C_0 = S^{-1/2} C^{'}_0
$$
where $S^{-1/2}$ is the transformation matrix used to get the transformed Fock matrix from earlier. My transformation matrix, by the way, matches exactly the one on the site.
This is where my problem comes in. The $C_0$ matrix I get is this:

The one the website has as a reference is this:

If you look carefully, you will notice that mine and the reference one have the same columns; however, some are swapped around, and some have sign flips. This causes problems with the next steps, such as finding the density matrix.
I thought maybe the order should be such that $\epsilon_0$ is in order. The $\epsilon_0$ matrix I get is indeed out of order, and the same permutations and sign flips that will make it in order also makes my $C_0$ match the reference one in terms of where each column is, but NOT sign. Sadly the reference site does not have a $\epsilon_0$ matrix or $C{'}_0$ for me to compare with, only the final $C_0$, so I'm not sure if those are coming out of order also.
My first question is, will these sign flips matter? Is ordering the $C_0$ matrix in terms of getting $\epsilon_0$ the right thing to do?
I am carrying out these matrix operations with basic NumPy functions such as numpy.linalg.eig and numpy.dot. Nothing fancy there. It's worth noting that finding the transformation matrix, $S^{-1/2}$, requires diagonalizing in a similar way, but I didn't have any column switching problems there. It came out exactly as expected.
My second question relates to building the density matrix at the end of step 5. The website has this equation:
$$
D^0_{\mu\nu} = \sum_{m}^{occ} (C_0)^m_\mu (C_0)^m_\nu
$$
It says sum over the occupied MOs. The molecule being used in the example calculation is water, so two H's, and one O, so there are 10 electrons, so 5 orbitals. If I sum using the correct $C_0$ matrix I get on the reference website, from m = 1 to 5, I get the correct $D^0_{\mu\nu}$ result. However my textbook (Szabo and Ostlund, Modern Quantum Chemistry) has this equation (changing some of the notation to match):
$$
D^0_{\mu\nu} = 2 \sum_{m}^{N/2} (C_0)^m_\mu (C_0)^m_\nu
$$
Where N is the number of electrons. Well, 10/5 is 5, so I'm summing from 1 to 5 again, but now there's a 2 in front, so the sum will be doubled. Now, using the one on the website gets me the correct answer, but what could this 2 in the book mean?

Welcome to Chemistry! Take the tour to get familiar with this site. Rather than posting screenshots of your terminal, what you can do is take the matrix blocks and add 4 spaces to the front of each line to get monospace text.

Why do all these tutorials always provide the integrals? The most interesting (and expensive) part in programming a QC Program is the calculation of the integrals....

I agree in an academic sense, but practically, it's useful to have the values to make sure the rest of your code is working. When I'm done with the other parts, I'm going to make some actual integral-evaluating code.
http://folk.uio.no/helgaker/talks/SostrupIntegrals_10.pdf
This is something I have found helpful in evaluating the simpler one-electron integrals. For the two-electron integrals I may just link up to psi4 or a similar program, as that's an entire field in itself...

I use both Psi4 and pyscf in a project of mine to generate 1- and 2-electron integrals. The problem is the other side of the same coin: you can work on methods without even knowing about integral evaluation and vice-versa. Implementing an integral engine is tedious for the naive case and obtuse/very difficult for the optimized case, and slow integrals are really slow. On Github, I have an Obara-Saika program and Joshua Goings has a McMurchie-Davidson program you can look at in Python.

The naive integrals in PyQuante use the THO algorithm, which is easier to understand, IMO, because the recursion works a bit differently. There's a bunch of links here you might find useful.

I'm likely to go with psi4 because my PI is (or rather, was) one of the developers so he has a lot of experience with it, but yes, actually writing integral-evaluating functions yourself will probably not be very good, although I am going to do it anyway, just for the learning experience of how the integrals are evaluated, which may give me some insight into how they are better optimized, even if I don't implement those optimizations myself.

In that case, I definitely recommend Peter Gill's review article.

In the future, this should really be two separate questions, but I will try and tackle both.
MO coefficient column ordering and signs
The electronic energy is invariant to orbital swapping within subspaces. If you interchange one occupied orbital with another, the total energy of a molecular won't change. The same goes for virtual-virtual. Now, you know that the MO coefficients are by convention shaped as $C_{\mu i}$, where the rows are the AOs and the columns are the MOs. What this means is that for $\ce{H2O}$ in a STO-3G basis, where there are 7 AOs forming 7 MOs where 5 are doubly-occupied and 2 unoccupied, switching the order of the first 5 columns should not change the electronic energy. The same goes for the final 2 columns.
As you say, the $\epsilon_i$ follow the same ordering as the columns in $C_{\mu i}$; this is because they are the matching eigenvalues and eigenvectors from diagonalizing the Fock operator. The eigenvectors are not going to be affected by a sign change, which doesn't change how the space is spanned.
Regarding why the ordering is not what you expect, there is no guarantee made by NumPy or the underlying LAPACK routine DGEEV that the results will be sorted, just that the vector/value pairs will be in the same order. If you want to sort the results, you can use argsort:
idx = energies.argsort()
energies = energies[idx]
eigvecs = eigvecs[:, idx]

Make sure the ordering of the MO energies is maintained along with the ordering of the MO coefficient matrix columns.
Constructing the density matrix
This is a very irritating difference in convention among many sources. See later, for calculating the electronic part of the dipole moment:

$$
\left<\vec{\mu}\right> = 2 \sum_{\mu\nu} D_{\mu\nu} \left< \phi_{\mu} | \hat{\mu} | \phi_{\nu} \right>,
$$
  ...

The factor 2 appearing above arises because the definition of the density used in this project differs from that used in Szabo & Ostlund.

This can also have an effect on whether or not there is a 1/2 in front of the $2J - K$ portion of the Fock operator.

You're right, the factor of 2 was because of the notation difference. In the online guide, it was 2J -K, whereas in the book, it was J - K/2. I didn't even think to look at that before, but that accounts for the difference.

For your first question: since (if $A x = \lambda x$)
$$A (- x) = - A x = - (\lambda x) = \lambda (- x)\text{,}$$
an eigenvalue with sign flipped is still an eigenvalue.
(In fact, in linear algebra, any multiplicative constant would be fine, but since we are interested in normalised states this reduces to $\pm 1$).
So no problem not matching signs.
For your second question: you're mixing stuff.
Your first equation sums
$$\sum_m^{occ}$$
which means 10 terms in your case.
Since every $\alpha$ spin-orbital contributes the same as another spatially-matching $\beta$ spin-orbital, you end up summing 10 terms in which each two are numerically the same.
Your second equation sums
$$2 \sum_m^{N/2}$$
which simply avoids this double counting: you sum 5 terms, each being twice the values in the last sum.


text:
Recrystallizing Sodium Acetate from solution
I made sodium acetate using the following reaction:
$$\ce{NaHCO3 + CH3COOH ->CH3COONa + H2O + CO2}$$
The resulting solution of sodium acetate isn't pure, it has impurities. I want to crystallize the sodium acetate to extract the pure crystals. I crystallized some if it in the fridge, sodium acetate has a higher freezing point than water but I would like to know the technical side of it. Do temperature and saturation affect the crystallization and how can I crystallize sodium acetate reliably?

The Purification of Laboratory Chemicals by Perrin and Armarego states about sodium acetate this procedure:

Cryst[alise]d from acetic acid and pumped under vacuum for 10h at 120$^\circ$.
  Alternatively, cryst[alise]d from aqueous EtOH, as the trihydrate. This material can be converted to the anhydrous salt by heating slowly in a porcelain, nickel or iron dish, so that the salt liquefies. Steam is evolved
  and the mass again solidifies. Heating is now increased so that the salt melts again. (NB: if it is heated too strongly, the salt chars.) After several  minutes, the salt is allowed to solidify and cooled to a convenient
  temperature before being powdered and bottled (water content should now
  less than 0.02%).

This doesn't really help. I was asking about crystallization and I don't have a vacuum pump or a way to melt salt.

The quote from Perrin is offering several options.  1) Crystallisation of sodium acetate from acetic acid; subsequently the crystals obtained are dried -- which, because it is hygroscopic and in order to do it faster, is done at higher temperature and under reduced pressure.  Or 2) a crystallisation from aqueous ethanol, yielding crystals which incorporate three molecules of water per molecule of acetate (like the one at shown at wikipedia).  These (by way 2) obtained crystals could be molten to eventually remove the water incorporated, but it will destroy the crystals and yield a powder.

@TheTesseract'sShadow.. the book cited is the go-to place for finding out how to purify a compound, you should maybe [edit] your post with the equipment/resources you have available, but in general you're going to struggle to do a decent job without even basic lab resources.

Sodium Acetate is particularly hard to recrystallize due to its high tendency to super cool. What usually happens is the crystals crash out entombing the impurities. I am assuming your making it from vinegar and baking soda. I would recommend concentrating your solution down to saturation and then boiling off 25mL of water at a time and allowing the solution to cool fully and collecting the resulting crystals each time. These WILL probably still be fairly impure. Because these will be the trihydrate salt they will melt with heat (anhydrous will not). You can melt these crystals in a hot water bath (boiling) or cough a microwave cough. Melt most of these crystals leaving a very small amount on the bottom of your beaker to act as seed crystals and allow the sodium acetate to cool very slowly. Once about half of it has crystallized add a small amount of water to the sodium acetate, mix well and pour it onto a towel and pat the remaining crystals dry.


text:
Which rings of this heterocyclic compound are aromatic?
I am wondering which of the inner two rings of this compound are aromatic?

I know the rules for aromaticity. But this case is difficult for me.
Regarding the smaller ring, I guess the sulfur contributes one lone pair of its valence electrons into the system. This means from the nitrogen (in the 5-ring) only one electron is needed for the π-system (with 6 electrons). But this nitrogen atom has only one lone electron pair, thus one of the electrons would not be part of the  π-system of the smaller ring. However, this nitrogen atom in the 5 membered ring is also part of the 6-ring and there it could share both of its free electrons to this larger ring. Then the second nitrogen which is only in the 6-membered ring could contribute one electron only to the aromatic system, and thus having a happy free lone pair left.
What do you think of that? Which rules, logic or reasoning can confirm this? And isn't it a problem that the first nitrogen shares one electron with the 5-ring and the two with the 6-ring?
Also, is there software which can predict this reliably? I haven't found anything so far.

That is kinda interesting, your reasoning is rather muddled though. If not taking into account other mesomeric structures, both heterocyclic rings would be non aromatic, but those structures could be so big contributors that they could be aromatic nevertheless.

I think that all the four rings are aromatic.
If the top nitrogen donates it's pair of electrons to the ring, and oxygen takes it's pair, the three fused rings would have 14 electrons in the rings' peripheral, causing it to be aromatic:

Even the 3D drawings of this molecule confirm it's planarity of the 3 fused rings:

I think we can at least suspect all the rings in the molecule to be aromatic.

Thank you very much for your helpful answer.
The negative charge on the oxygen can be shared with the other oxygen right? (two tautomers)
Also, which software did you use for the 3D drawings?

Do I conclude correctly from your post that we don't have to consider each individual ring as aromatic (as I have done), but the entire peripheral ring (as you have done), to get the aromaticity of polycyclic compounds? And that peripheral ring has to satisfy Hückels rule I guess.

@Jadzia That's right, just the periphery. For the 3D structure I used this website: chemapps.stolaf.edu/jmol/jmol.php?model=[INSERT SMILES HERE] In the place of [INSERT SMILES HERE] input the SMILES of the compound. For your molecule the SMILES is: Oc3nc2sc1cc(Cl)ccc1[n+]2c([O-])c3Cc4ccccc4

Thank you for your answer. Also, can we assume that the negative charge is fully distributed among the two oxygens? This would mean that the compound is a strong Lewis base, easily accepting another proton at the other oxygen, thus becoming a positively charged molecule in water.

I'm not sure of the other oxygen, but I think it's not similar to the first one, that's in conjugation with the nitrogen, but the other one can't, for it will disrupt aromaticity.

Thanks for your answer. I just found out that for polycyclic aromatic compounds the Hückel rule does not need to be satisfied. https://en.wikipedia.org/wiki/H%C3%BCckel%27s_rule

@Jadzia Not in the usual way, but the peripheral does count as one ring, and should satisfy Huckel rule.

Let us continue this discussion in chat.


text:
What is substance that has high melting point in its powder form?
In laminated glass bending process 2 glass sheet will be superposed in a furnance.The seperation powder will be interposed between layer for avoid 2 glass sheet being bound together.
I'm trying to find substance that can be replaced baby powder as a lubricant in the process of glass-bending. It has to be cheaper, able to be found as a powder, and it should not be a human hazard.
Recently, I found magnesium sulfate, and I believe that it could be used for my purposes. However, it is so hard to find $\ce{MgSO4}$ in a compact and powdered form.
Does anybody know of any suitable replacements for baby powder in the process of glass-bending?

Magnesium sulfate (Epsom salt) is easily crushed in a mortar and pestle. If you're using it on molten glass, first dehydrate it (removing water of crystallization) or it will "pop" violently on hot glass.

Is it ok if i apply it in solution form insted of powder? Is it gonna pop in 630 celcius degree chamber?


text:
What are some ways to predict that a double replacement reaction will occur?
For instance, say:
$\ce{2KOH + Pb(NO3)2 -> 2KNO3 + Pb(OH)2}$
How would I even go about beginning to predict this?

To determine if a double replacement reaction will occur, you need to look at the products. You correctly balanced the equation and flipped the anions and cations, so let's take a look. A double replacement reaction will occur if one or both of the products is an insoluble precipitate, a liquid, or a gas. To check each of the products you use the solubility rules. You check each rule in order for each of the products. Here is a link to the rules I found online:
http://www.dorettaagostine.com/Solubility%20Rules.pdf
Look at the double replacement reaction chart. The rules are followed in order. Let's look at $\ce{KNO3}$ first. The first rule states that alkali metal salts are soluble. Potassium is an alkali metal, therefore potassium nitrate is soluble. The second product is $\ce{Pb(OH)2}$. Looking through the rules, none of them apply until the 5th rule which states that hydroxides are insoluble. Therefore lead hydroxide is insoluble.
As you can see, 1 of the products is insoluble, therefore the double replacement reaction will occur. I hope that helps! If you have anymore questions or clarifications just ask.
EDIT: How to find the net ionic equation.
First reduce the equation into it's respective ions.
$$\ce{2KOH(aq) + Pb(NO3)2(aq) -> 2KNO3(aq) + Pb(OH)2(s)}$$
...becomes...
$$\ce{2K+(aq) + 2OH-(aq) + Pb^2+(aq) + 2NO3-(aq) -> 2K+(aq) + 2NO3-(aq) + Pb(OH)2(s)}$$
However, keep in mind that $\ce{Pb(OH)2}$ does not break down into ions because it is insoluble, as we discovered earlier. Therefore it stays in it's full formula. At this point you do the "cancelling out" by removing duplicate ions on each side of the equation. The cancelled ions are also known as spectator ions because they do not actively participate in the synthesis of the product.
$$\ce{2OH-(aq) + Pb^2+(aq) -> Pb(OH)2(s)}$$
The potassium and nitrate ions are removed because they appear on each side of the equation in the same quantity. These cancel out leaving the net ionic equation show above.
The net ionic equation will show the creation of the insoluble, liquid, or gas precipitate that forms as a result of the reaction. I hope that helped!


text:
Is it possible to export a rotating .gif format picture from chemapps.stolaf.edu?
I mainly use chemapps.stolaf.edu to get the 3D structure of a molecule from the SMILES. It renders the molecules as:

I have the option to rotate the image by specifying the $x$, $y$, and $z$ spin rates:

However I can't seem to find a way to save a spinning image as such. I can only save .png files. I need spinning images to demonstrate 3D images. Can anyone help me get a spinning image?

You can download Jmol, the Java applet that is hosted on that website. With it you can export to an animated gif.

@ThomasJungers Thanks for your input, but what does "Java applet" mean(I'm not good with computer terminologies)? Can you point exactly where it's hosted?

A Java applet is basically a software for websites, written in Java. There is a standalone version of Jmol (that you can run directly on your computer) that you can download here : http://jmol.sourceforge.net

You should preferably use JSmol as its the latest version and does not need java as it runs in a web browser.  See also www.molecule-viewer.com for a large number of examples rotatable molecules comprising all point groups also written using JSmol via javascript.

As of now, Java applets only work with Firefox ESR 32-bit. Assuming that OP is using Chrome and Windows 10, in my opinion the best solution would be to use a screen capture software, like ScreenToGif, which is FOSS, has nice interface and is also quite intuitive:

And here is the result:

I must remind though that the major limitation of GIF is that you are stuck with 256 colors. Structures displayed with GIF will never look as slick as those demonstrated in browser with JavaScript.

Excellent! Where do I download this?

Ehmm... I thought I left a link to the GitHub repo. Anyway, here it is again: https://github.com/NickeManarin/ScreenToGif/releases

If you are using Chocolatey package manager on Windows, choco install screentogif and you are good to go.

Ok now I got it downloaded, and I got a gif ready (I think I did). But I can't seem to save it, it keeps telling "The output folder doesn't exist", even though it does.

Let me remind you that there is a great FAQ/Help section: https://github.com/NickeManarin/ScreenToGif/wiki/Help which I pointed to in my answer. I have no idea what the problem is, I've never faced this problem myself.


text:
Why does more mass not mean more yield?
I have a very vague knowledge of thermodynamics, but let's take a reversible reaction:
$$\ce{CaCO3 (s) <=> CaO(s) + CO2(g)}$$
Now if we increase the mass of the calcium carbonate, I understand that it would not increase the concentration by a very large approximation (since it's density does not increase by a very large approximation) so I understand that the equilibrium will not shift to the right, but the thing that confuses me is doesn't more reactant mean more product and a greater yield means a shift to the right? This also applies to a pure liquid reaction:
$$\ce{HCl(g) + H2O(l) <=> H3O+ + Cl-}$$
Let's assume that in the first reaction you have a mole ratio of the respective reactants in the form of 1:0.5  clearly water is a limiting reagent, however let's assume that we increase the mass of the pure liquid so that the mole ratio becomes 1:1 now we have more water available for the reaction to occur hence more product would occur, but according to le Chateliers principle this would not occur since a pure liquid does not cause a shift, rather if the reaction was at equilibrium it would remain as so. Why does this happen?

In your first example at constant temperature and in a vessel of fixed volume (to stop carbon dioxide escaping) there will be equilibrium as long as some solid carbonate is present. Adding more makes no difference, because, well, there is equilibrium. I don't follow you second example. You seem to have gaseous HCl above a solution of same?


text:
Why does the iron rust with less salt
For my school project I was going to measure the rate that salt has on the rusting of iron.
To do this I set up an experiment where I had 15 beakers containing 200 milliliters of distilled water. I poured into the beakers varying moles of salt $\ce{(NaCl)}$. I had $\pu{0mol}$, $\pu{0.25mol}$, $\pu{0.5mol}$, $\pu{0.75mol}$ and $\pu{1mol}$ of salt. So now there are three beakers for every concentration. I then put a iron nail in every beaker, put plastic wrap over them to stop some of the oxygen flow and left it be for 10 days.
What I was surprised to find was that the beakers with no salt in them had more rust than the one ones that had. It was exactly the opposite of what I expected would happen, and I can't find a website that explains how this occurs. Does anyone have an explanation?

um, that's not normal. Are you sure you didn't mix up the ordering of your beakers? Concentration of oxygen in the water is also quite important for rusting, so are you sure all conditions are exactly the same in all beakers? Also, what are your nails out of, do they have a coating?

The nails did not have any coating. However, I did find something that did explain what was going on in my experiment. https://chemistry.stackexchange.com/questions/22043/does-concentration-of-salt-increase-or-decrease-rate-of-rusting

My nails where completely submerged and the plastic wrap blocked off some of the oxygen, so I would imagine that this would be the explanation.

Triplicate results are recommended, so you should repeat the experiment two more times.

I wouldn't just throw the results away, they can be used to learn and understand the process maybe even better than the right results!

Corrosion is a balancing game. In this case, by covering the beakers, you have limited the driving force for oxygen dissolution by limiting the amount of oxygen available to dissolve into solution and then the higher salt concentrated solution is able to further minimize the amount of oxygen that can be dissolved and participate in the redox reaction (as explained in the linked question's answer).


text:
Mechanism of Fischer esterification: Does the carboxylic acid gives off OH- or H+?
For the reaction
$$\ce{R-OH + R'-COOH ->[H+] R-COO-R' +H2O}$$
the reaction mechanism is given here.
What seems unusual to me is that the acid appears to be donating $\ce{OH-}$ rather than $\ce{H^+}$.
Can the mechanism be written as follows instead?
$$\ce{R'-COOH <=> R'-COO^- + H^+}$$
$$\ce{HA-> H^+ + A^-}$$
The inorganic acid pushes the equilibrium for the organic acid to the left and it remains undissociated.
Then, $\ce{H^+}$ from the inorganic acid first attacks the $\ce{O}$ atom in $\ce{R-OH}$ resulting in the formation of a water molecule and a $\ce{R^+}$ ion.
Next, as the equilibrium for organic acid is pushed to the right, the bisulfate ion removes the $\ce{H^+}$ from the acid creating a carboxylate ion. This carboxylate ion then reacts with $\ce{R^+}$ ion to form the ester.
In this mechanism, the carboxylic acid donates the $\ce{H^+}$.
I know one can't prove reaction mechanisms to be correct; however, which one the mechanisms given above is more likely to occur?

The accepted mechanism is far more likely than your proposed alternate mechanism for a number of reasons.
First, let’s consider an esterification reaction employing a chiral secondary alcohol such as butan-2-ol. Protonation of the alcoholic oxygen is no difficulty and will happen in equilibrium. To generate the corresponding ester from the conjugate acid of butan-2-ol, we would require some nucleophilic attack mechanism: the acid would either attack according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. If we assume an $\mathrm{S_N2}$-like mechanism, the asymmetric carbon atom would be inverted. If we assume an $\mathrm{S_N1}$ mechanism, it would racemise. Experimental evidence shows, however, that the alcoholic carbon atom strictly retains is stereochemistry — and this principle is used in many total synthesis in which asymmetric alcohols are first generated stereochemically and then transformed into esters of the same stereochemistry. To give you just one of many examples, consider the total synthesis of aetheramide by Gerstmann and Kalesse.[1]

Scheme 1: Example esterification of (R)-butan-2-ol

Scheme 2: Esterification performed during the synthesis of aetheramide by Gerstmann and Kalesse.[1]
Second, and immediately tied to the scheme as referenced above, many alternate esterification methods exist in the literature. In principle, they all work by replacing the acidic particle ($\ce{H+}$) of a Fischer esterification with a less reactive and/or more specific compound. Practically all methods — Steglich or Yamaguchi to name just two — work by adding a compound that is known to interact with the carboxylic acid and not as much with alcohols. However, this is only a weak indication since different methods could follow different mechanisms — even though the quantity of the evidence does seem overwhelming.
Third and finally, the acidic hydrolysis of esters — the reverse reaction of the Fischer esterification — has been studied by $\ce{^18 O}$ marked water. Polanyi and Szabo could show that upon hydrolysis the heavy oxygen atom remains bound to the acid fragment and not to the alcohol.[2] This offers intriguing evidence for the mechanism of nucleophilic attacks on a carbonyl group. I had thought that a corresponding isotope-marking experiment had been performed for the Fischer esterification, too. Unfortunately, I was unable to find a reference. Do note, however, that using isotopically marked oxygen on either the acid or the alcohol would, by careful analysis of the product, rule out or confirm one mechanistic pathway.

Scheme 3: Mechanism of hydrolysis as confirmed by Polanyi and Szabo using $\ce{^18O}$ marked water.[2]

References:
[1]: L. Gerstmann, M. Kalesse, Chem. Eur. J. 2016, 22, 11210–11212. DOI: 10.1002/chem.201602682.
[2]: M. Polanyi, A. L. Szabo, Trans. Faraday Soc. 1934, 30, 508–12. DOI: 10.1039/tf9343000508.

"If we assume an SN1SN1 mechanism, it would racemise" Why? For alkyl shift happening aat the butane carbenium ion?

@Mockingbird360 because the $\mathrm{S_N1}$ mechanism requires a planar carbenium intermediate which has lost all stereochemical information and is therefore attacked from both sides equally


text:
Difference between valency of nitrogen and phosphorus though both have same valence electrons
Both nitrogen and phosphorus have 5 valence electrons, but nitrogen has valency of 3 while phosphorus has valency of 3 or 5. Why is this so?
Maybe phosphorus can gain 3 electrons or remove 5 electrons.But is it possible to remove 5 electrons? Or is it that phosphorus valence shell is farther away than nitrogen, so it is easier for phosphorus to also remove 5 electrons but nitrogen cannot.
I have searched but i need a concrete answer.

https://chemistry.stackexchange.com/questions/62218/why-can-phosphorus-have-5-bonds-while-nitrogen-cannot

Possible duplicate of Can an atom have more than 8 valence electrons? If not, why is 8 the limit?

The question is utterly different from the proposed duplicate, @Mithoron.

@Jan But it's standard dupe for hypervalent stuff. In this case lots of reasons would do. For example claim about valencies isn't true, they are identical for both elements.

At your current stage, it's probably easiest to consider the octet rule which applies to nitrogen and the expanded octet rule which applies to phosphorus.
If nitrogen and phosphorus made five bonds, they would have 10 electrons total. This is inconsistent with the octet rule but valid under expanded octet.
In reality, the situation is governed by the fact that you need more orbitals to accommodate the extra electrons for the two extra bonds. Unfortunately, no orbitals of comparable energy to the $2s$ and $2p$ are available, so the stabilization of the electrons in the putative molecular orbitals would be poor in nitrogen. With phosphorus, you might more readily access the $4s$ and $3d$ orbitals which are closer in energy to the $3s$ and $3p$ orbitals.

But it's not reality...

https://chemistry.stackexchange.com/a/42765/9961

Fair enough. Wrong choice of words. Maybe "a better model."


text:
In a molecular dynamics context, is the methyl rotation in propene a symmetric or asymmetric internal rotor?
In a prior question, I asked about the (a)symmetry of the potential energy surface of the methyl rotation of propene.  In that context, the kinetic energy of the nuclear motions is of course assumed to be negligible, and thus nuclear inertial effects are irrelevant.
Here, I want to ask about the (a)symmetry of this same internal rotation, but instead in the context of molecular dynamics (MD), where the Born-Oppenheimer approximation still applies but the kinetic energy of the nuclei is non-negligible compared to the variations in the potential energy over the course of the motion of the system. For the sake of argument, assume that the propene is in a sufficiently high rovibrational energy state for the methyl rotation that it behaves as a hindered internal rotor, and thus the nuclear motion is not "trapped" in one of the wells of the potential energy surface.

Depending on how one imagines the methyl rotation of propene to occur, it seems like it could behave as either a symmetric or an asymmetric internal rotor in an MD context.  If the $\ce{H2C=CH\! -}$ portion were held fixed, the rotating methyl would be inertially symmetric due to the local $\mathrm C_3$ axis:

On the other hand, if the methyl group were held fixed, the rotating $\ce{C=C}$ portion would be highly inertially asymmetric:

Obviously this rotation can't simultaneously be both inertially symmetric and inertially asymmetric. So which is it, and why?
My inclination is to assume that the internal rotor is asymmetrical, since the overall system does not have any axes of three-fold or higher symmetry. However, I don't know a good mathematical or group theoretic 'language' in which to define such internal rotor (a)symmetry, and so I'm unable to prove this to my own satisfaction.

The best qualitative argument I've thought of in favor of it being an asymmetric internal rotor is that the axis of the rotation (green bar) does not pass through the center of mass of the system (purple sphere / orange bar):

FWIW, I calculate the skew angle between the two colored bars in the above geometry to be about $18^\circ$. Anyways, as a consequence of this skewed orientiation, I think the methyl hydrogens would oscillate parallel to the axis of internal rotation while they rotate around the methyl carbon.  Simultaneously, I would expect a sort of corresponding "waggle" of the ethenyl fragment.
Thus, directly related to the "why" question above: If this qualitative argument is correct, how can I express it quantitatively?

I still think that you are convolving whole body motion with that of internal motion.  The two can be separated just as we separate out various normal modes from one another and from whole body translations. Sure, the centre of mass of the whole thing does not have to be placed at an atom or bond and can be 'in space' as you draw.

@porphyrin I agree, they can be fully separated -- that's what I mean by "internal rotation" -- the "true" internal methyl rotation occurs in the absence of net overall rotation of the system. It is this very constraint (zero net total system rotation) that makes me think that there is some sort of 'asymmetric' character to the methyl internal rotation here, intuitively since the nominal axis of internal rotation doesn't include the center of mass. I may be using the wrong language to frame the question, though.

As I wrote in reply to your other question the internal rotation occurs in a 3-fold potential; at low energy (cold molecules) motion is hindered and may be restricted to be within  the wells. At high temp it is more or less 'free' as there is enough energy to overcome the barrier.  I would not call the 3-fold potential 'asymmetric' though as it has some symmetry.

@porphyrin , and I'm interested here in the inertial dynamics of the "hot" molecule (ignoring electronic excitation effects), where the potential energy barrier is more or less irrelevant to the character of the motion.

Then have a look at the  J Chem. Educ. article in the comments your other question and put the potential to zero or so small that it becomes irrelevant.

@porphyrin That J Chem Educ article doesn't address my question at all. First, the ethane internal rotation and the cyclopentane pucker are inertially symmetric, to my intuition, unlike the methyl rotation propene, here. Second, the article only deals with the PES profile of the surface (even at the high energy of, e.g., $n=40$), and not the inertial symmetry of the dynamics of the motion of the nuclei.


text:
Which one is the stronger acid among these two compounds?

According to me, acidity for these two compounds depends primarily upon two factors:

inductive effect due to the ring
electronegativity of the atom bearing   the hydrogen atom

If  the  first factor predominates, second compound would be the most acidic otherwise the first compound would be the most acidic(as nitrogen bears a methyl group too).
So, which one is the most acidic?

You can also consider this question from the other direction. Which of these conjugate bases is more basic?

You can also look up the pKa values first, and then choose an explanation that fits the empirical data.


text:
Does liquid ammonia react with hydrogen gas?
This may seem like a basic question to educated chemists, but I failed to get this answer on Internet searches.  I keep getting more complicated stuff like the Haber process which is not relevant.
At this website it describes how the absorption cooling process works in a propane powered refrigerator.  The system uses ammonia, water, and hydrogen in closed loop system.  In the evaporator section, it says liquid ammonia reacts with hydrogen gas thus vaporizing it and this process of ammonia boiling in hydrogen absorbs heat from the fridge box. It doesn't mention what the end products of the ammonia and hydrogen reaction are.
Is there a chemical reaction between hydrogen and ammonia here?  Or is there no actual chemical reaction, just partial pressure vaporization of ammonia liquid in hydrogen gas?

No chemical reaction, just vaporization :)

Hydrogen then does not undergo a chemical reaction with ammonia but the interaction is physical. It would be interesting to know how the molecules do interact since hydrogen in non-polar while ammonia forms strong dipole-dipole (and H bonds). More clarification on what’s happening here would be helpful!

No, it does not. I do not know what they described but I am pretty sure liq. ammonia does not react with hydrogen gas. I believe they use hydrogen gas as a carrier but I found that to be questionable as well, since hydrogen in residential setting can be pretty dangerous.

Hydrogen absolutely does react and causes the partial pressure of ammonia to lower dramatically, causing the ammonia to  evaporate and absorb great quantities of heat for the operation of the refrigerator.  It does not involve strong compound forming bonds (which is the main focus of chemistry students) but weak bonds like Van der Waal forces, acting as a catalyst.

Van der Waal force is Temporary Dipole Induced Dipole interaction, not reaction.
The weak association causes absorption of heat.

That is London force. Van der Waals force is the superset to that, including also Debye force for permanent-induced dipoles and Keesom force for permanent-permanent dipoles.


text:
What would be a suitable non-combustible lubricating oil or compound for a very high pressure air compression system?
This question came up at Engineering SE but remains unanswered.  If we are compressing regular air, not nitrogen, what oil or compound can be used
that would not combust  with the oxygen in the air.  High pressures cause high temperature which would ignite hydrocarbon based oils.  What compressor or pump lubricant would be suitable in these conditions?
With Internet search, I found this compound, but it is a grease intead of oil and there is no mention of ability to withstand high pressure.

Could you not consider using an oil-free design for your compressor/pump?

@long yes of course. Is this possible?

Not sure what specifications you are after, but a search on oil-free compressors will give you plenty to go on. Try this or this for instance. I have used a number of oil free compressors, but not under the bonnet of a car!

@long  Thank you for the links.  However, the pressures we will be encountering in our setup are easily 1000psi or more.  The specs on the site you listed say that oil-free compressors operate at much higher temperatures compared to oil-injected ones at only 3.5bar (47psi).  For this reason the oil-free compressors where out faster than oil-injected ones.   A diesel engine cylinder at peak is around 1015 psi which is designed to ignite oil.  We need lubricating oil that can endure this pressure and temperature, still lubricate, and not ignite.


text:
What is the mechanism behind the formation of phosgene from tetrachloroethylene
Tetrachloroethylene used to be a popular automobile brake cleaner. Unfortunately, when applied to hot brake rotors, it can turn into phosgene. This has resulted in many injuries and even fatalities.

At temperatures over 315 °C (599 °F), such as in welding, tetrachloroethylene can be oxidized into phosgene, an extremely poisonous gas.

What is the mechanism behind the oxidation to phosgene? I know that these things probably happen--alkene bond is broken and oxidized--but is there a detailed mechanism?

Not an answer but the industry practice in welding enviroments is never to allow the UV generated by the arc impinge on Tetrachloroethylene. - No heat is mentioned in the AWS literature. There are also wives tales about Argon reacting with the Tetrachloroethylene but I find this absurd as Argon is inert. Is it true that UV radiation alone can decompose Tetrachloroethylene or is heat required? (Not that we intend to allow open containers of Tetrachloroethylene in the welding department)

Given that the reaction occurrs at high temperatures, the reactive species is likely singlet oxygen, which reacts much like an alkene$^{[1]}$. Electron rich alkenes without allylic hydrogen, like tetrachloroethylene, react with singlet oxygen to form 1,2-dioxetanes$^{[2]}$.

A $[_\pi2_s + _\pi2_a]$ cycloaddition, though a poor reaction geometry for two alkenes, is made more accessible by the small, linear oxygen molecule. This, as well as peroxides' predisposition to nonplanar conformations, makes the transition state hardly any more strained than the dioxetane itself. Once formed, the dioxetane quickly decomposes to give two molecules of phosgene. To give you an idea of their stability:

A small sample of the [dioxetane] product, on warming to room temperature, melted and exploded.$^{[3]}$

$^{[1]}$ Wikipedia, Singlet Oxygen, Reactions, Organic Chemistry
$^{[2]}$ Singlet Oxygen: Applications in Biosciences and Nanosciences Volume 1
$^{[3]}$ Stereospecific Formation of 1,2-Dioxetanes from cis- and trans Diethoxyethylenes by Singlet Oxygen


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Differentiating Al2O3 from SiO2 using chemical methods
I have granules of $\ce{Al2O3}$ and granules of $\ce{SiO2}$. They look almost identical and I have mixed them up. Is there a simple chemical test that would allow me to differentiate between these granules?

HF will etch the SiO2, but not harm the Al2O3 (unless heated). Note that HF is potentially nasty stuff. Optical measurements, including Raman, would do it as well. X-ray fluorescence would as well. Rutherford backscattering too...

I have some HF... I am thinking of trying this with proper precautions in a fume hood, etc. After a while at room temperature I could expect SiO2 granules should shoe some kind of disintegration? I was under the impression that Al2O3 would also react with HF.

Certainly sapphire is resistant to HF unless heated well above room temperature. SiO2 will dissolve pretty quickly.

What about separating based on density? For corundum, I see 3.95 g/cc, and quartz is 2.65 g/cc. Find a liquid with a density between those two and you're good to go.

@ToddMinehardt Yup, flotation is the way to go - unless the granules are small enough for stable dispersion to be an issue.

Although not a chemical method, using flotation in bromine to separate your mixture will work at room temperature. Unlike the chemical methods given in other answers, this physical method of separation does not destroy or otherwise degrade the components of your mixture, so you can recover them.
The density of bromine at 20 $^\circ$C is 3.1 g$\cdot$cm$^{-3}$; the densities of $\ce{Al2O3}$ and $\ce{SiO2}$ at the same temperature are 3.95 g$\cdot$cm$^{-3}$ and 2.65 g$\cdot$cm$^{-3}$, respectively. The $\ce{SiO2}$ will float and the $\ce{Al2O3}$ will sink.
As noted in the comment thread above, this method will work unless the particle size is too small.
And as noted in the comments below, sodium polytungstate solution (with a density of 3.1 g$\cdot$cm$^{-3}$) is an alternative to bromine.
Because there are likely a large number of substances with a density in the range required to separate your mixture, the suggestions here represent a small number of those choices.

Bromine is not very practical for sink-float analysis. I'd rather suggest sodium polytungstate solution. Sigma-Aldrich offers it with the same density as bromine. http://www.sigmaaldrich.com/catalog/product/sigma/71913

@aventurin - Thanks for the info, I've edited my answer to include it.

Bromine has the extra feature of the possibility of separating the separator from his life...

Acids can differentiate $\ce{Al2O3}$ and $\ce{SiO2}$. There is a famous reaction called etching of glass where $\ce{SiO2}$ reacts with hydrofluoric acid  $\ce{HF}$. $\ce{Al2O3}$ also reacts with $\ce{HF}$ but it requires heat. This test somehow differentiates them but we need to determine which substance is which by reacting with an acid which reacts with only one oxide. Strong acids like $\ce{HCl, H2SO4}$ or will suffice.
It reacts with acids to form corresponding aluminium salts(sulfate, chloride)
$$\ce{Al2O3 + 3H2SO4 ->[\Delta] Al2(SO4)3 + 3H2O}$$
$$\ce{Al2O3 + 6HCl ->[\Delta] 2AlCl3 + 3H2O}$$
Silicon(IV) oxide does not react with any of the above acid. Rather silicon(II) oxide reacts with sulfuric acid or nitric acid to produce silicon(IV) oxide.

$\ce{\alpha Al2O3}$ does neither react with sulphuric nor with hydrochloric acid.

@aventurin aluminum metal does not react with any of the acids due to passivation layer of Al2O3 but since  Al2O3 is present here in powdered form, surface area increases and reaction of acids is possible.

The question is not about aluminium metal. My comment was about alpha aluminium oxide which is known to be insoluble in sulphuric and hydrochloric acid. Btw, I'm not sure if granules is the same as powder.

Give a fine powdered mixture of your substance and $\ce{CaF2}$ a lead crucible. Add $1~\pu{ml}$ of concentrated sulfuric acid and warm gently.

If your substance is $\ce{SiO2}$ then gaseous $\ce{SiF4}$ will form:
$$\ce{2 F- + H2SO4 -> 2 HF + SO4^{2-}} \tag{1}$$
$$\ce{SiO2 + 4 HF -> SiF4 + 2 H2O} \tag{2}$$
A moistened black filter paper placed on the cap of the crucible will show white $\ce{SiO2}$ when dried (reaction (2) reverted).
As an alternative to the black filter paper you could place a water drop hanging at a black glass rod above the small hole in the crucible's cap. You will see silicic acid at the drop if $\ce{SiO2}$ is present.

I believe they are mixed in the same container.

A classical method is to digest the oxide in a sixfold excess of an eutectic mixture of sodium and potassium carbonate at a temperature of about 800°C in a nickel, iron or platinum crucible.
$$\ce{Al2O3 + Na2CO3 -> 2 NaAlO2 + CO2}$$
$$\ce{SiO2 + Na2CO3 -> Na2SiO3 + CO2}$$
$\ce{Al2O3}$ and $\ce{SiO2}$ are distinguished by extracting the cooled melt with concentrated hydrochloric acid. $\ce{AlCl3}$ gets dissolved while silicic acid is insoluble.
The procedure can also be used to separate $\ce{Al}$ and $\ce{Si}$ from a mixture of $\ce{Al2O3}$ and $\ce{SiO2}$.


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Converting units and sig figs (finding molarity)?
This may seem very basic, but I'm a little confused.
"What is the molarity of a solution 500 mL of which contains 0.500 moles of HCl?"
Being that 500 has one significant figure, would it be 500 mL = 0.5 L as opposed to 500 mL = 0.500 L? In which case the answer would be 1 mol/L instead of 1.00 mol/L?

500 ml and 0.500 moles corresponds to 1.00 mol/l solution.  The next decimal place is uncertain, i.e after the second zero.

In terms of significant figures, you always use the number with least amount of sig figs to determine how many sig figs your answer has. In this case, one number has 3 sig figs, and the other has 1. Therefore your answer should have 1 sig fig and be 1 mol/L, as you said. To make the zeroes in the 500 mL become significant it would need to have a period afterwards like this: 500. mL. Perhaps the writer forgot to do that, however as it is written the answer would be 1 sig fig: 1 mol/L

I didn't realize it was you again :P No problem, hope it helped!

By the way, on this forum, if someone successfully answered your question, you can press the check mark underneath the upvote and downvote arrows to indicate your question has been answered. That way it will be visible that this question has a correct answer.


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Can Pyrex laboratory grade glassware be heated on a stovetop?
I don't have access to a hot plate, I was wondering if Pyrex glass can be safely heated on a stove top. It isn't an open flame, it's a radiant stove top with an electric coil underneath a smooth counter. I believe the heating should be even because the coil heats the entire stove top evenly.

I think the main safety question here is what is inside the Pyrex container. Heating should, in general, not be a problem.

I agree with Zhe. I can't see any reason as to why using a stovetop would be any different than using a hotplate.

Rather than thermal shock, a bigger hazard might be that the stove surface is very hard. Borosilicate percolators came with a metal wire trivet to spread heat and prevent scratching against enameled burner tops; you might use  a bit of wire gauze for that purpose.


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What's a good measure of the Eigen-ness of a wavefunction?
Just wondering if someone can recommend an easy-to-compute measure for the Eigen-ness of a wave function, such as the wave functions computed using ab initio method?
For a wave function solved using a small basis set, the Eigen-ness should be smaller than that of wave function computed using larger basis set.
Naturally for arbitrary wavefunction $|\psi>$ one can compute :
$<\psi | (H/E-1)^2 | \psi>$,
where E is the expectation value, which is the dispersion of the energy. This, however, is too difficult to implement.

Can you clarify this a bit? I'm not sure I understand what you're asking.

As to a measure of Eigen-ness, it's meant to quantify how close a candidate wavefunction is to the actual eigen state of the Hamiltonian. For example, we know the exact solution of Hydrogen atom ground state, but if approximating it with a GTO, we measure how good the approximating function is by finding out the overlapping integral. But for general moleculare wavefunction, we don't know what the exact eignen state is, in this case, what's the best way of evaluating the candidate?

For variational methods, you can know for sure that your wavefunction is more Eigen if it has a lower energy, but AFAIK there's no way to know how close or far you are from the maximally Eigen function.

The energy converges with basis set size. So you can compare how much it changes e.g. from double to triple to quadruple zeta. You can even extrapolate the complete basis set limit from this.

$\hat{F}\psi_i=\varepsilon_i \psi_i \implies \langle\phi_k|\hat{F}|\psi_i \rangle = \varepsilon_i\langle\phi_k|\psi_i \rangle$.  So the Fock matrix $F_{ki}$ can be approximated by the product of orbital energy $\varepsilon_i$ and the overlap matrix $S_{ki}$. The closer these two values are, the higher the "Eigen-ness".
Of course $\phi_k$ and $\psi_i$ have to be non-orthogonal, so $\phi_k$ could be, for example, a basis function or an MO for another molecule.  I've used this to approximate intermolecular Fock matrix elements here.

could you use an auxiliary basis for the set of $k$?

In principle, any function that's non-orthogonal to the MOs.  Of course, different functions will probe different parts of the MO and each MO will have a different "Eigen-ness"

Thanks. Is there a way to evaluate the overall eigen-ness without looking at its projections to another set of basis functions? The dispersion of engergy levels, for example, <ψ|(H/E−1)2|ψ>, can measure it without additional outside references. But it's difficult, almost impossible, to compute.

I don't know of any right off hand. However, you don't need to use another set of basis functions. The basis functions that you use to describe $\psi_i$ are non-orthogonal to $\psi_i$ itself


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Heating a mixture of methanol and water?
I have a mixture methanol and water, at 25°C, and I want to take it to 250°C. I'll use two heat exchangers: one for taking the mixture from a liquid state (25°C) to a vapor state (X°C), and the second for taking it to the necessary temperature (250°C).
I know that methanol boiling temperature is 64 °C and water boiling temperature is 100°C, and it will be necessary to use two 'kinds' of heat, one for taking the component from 25°C to the boiling temperature, and another called the latent heat, for the phase change.
So, having methanol and water two different values of boiling temperature, what will be the value of X°C, approximately? 100°C, less or more?

H2O/CH3OH has no azeotrope so boiling points are in the middle between points for pure stuff.

You need to know the composition of your solution and then look at vapor liquid equilibrium data to find the boiling point of your particular composition solution. See here for boiling point as a function of composition.


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How to decide at which end to start the numbering of the parent chain when the options for locant sets are the same?
I would like to know whether the blue or purple way of numbering is preferable? Following the two possible ways, I get the following names:

4-(1,1-dichloro ethyl)-5-(1-chloro-2-bromo ethyl)-octane
4-(1-chloro-2-bromo ethyl)-5-(1,1-dichloro ethyl)-octane

In each case, the locant set is (4, 5), so from which side should I start numbering from?

Maybe this may help, but I can only guess. I took the liberty reworking your question, please check if I have introduced any errors.// I think that first you would need to order the substituents alphabetical, then apply the lowest set of locants, hence 4-(2-bromo-1-chloroethyl)-5-(1,1-dichloroethyl)octane, but I could be completely wrong. It should be noted, that all names are unambiguous, so while they are not the preferred IUPAC name (PIN) they are not false with respect to general nomenclature.

Thanks for the link again Martin. There in the link the substituents were single atoms like Br, Cl, I. But here the substituents are chain themselves, so how do i arrange them alphabetically? Also the answer you mentioned is correct as that's what my book says. Also is it okay to name the other way? Thanks for editing my question and making it better. I appreciate your help

@Martin-マーチン The rules of Section P-14, which are relevant in this case, are general rules. They shall be followed not only for preferred IUPAC names but also for other names used in general nomenclature.

If numbering from each end of the parent chain gives the same set of locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.
This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

For example ‘dibromo’ comes before ‘chloro’ since ‘bromo’ begins with ‘b’, ‘chloro’ begins with ‘c’, and the multiplicative prefix ‘di’ does not change the alphabetical order.
However, the example given in the question does not include simple prefixes but the compound substituents ‘1,1-dichloroethyl’ and ‘2-bromo-1-chloroethyl’. Such compound names are considered to begin with the first letter of their complete name.

P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, the correct name for the compound given in the question is 4-(2-bromo-1-chloroethyl)-5-(1,1-dichloroethyl)octane since ‘2-bromo-1-chloroethyl’ starts with ‘b’ and ‘1,1-dichloroethyl’ starts with ‘d’.


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Reaction mechanism (SN2 and SN1 reactions)
How to determine whether a reaction is an SN1 or SN2 reaction?
I am unable to determine whether a reaction would form a transition state or not,
I referred the NCERT textbook but didn't found anything that could clear my doubt.

I would like to suggest this book: Electron Flow in Organic Chemistry by Paul H. Scudder. Helped me a lot!

SN1 is characterized by the possibility of formation of a stable carbocation during the reaction, whereas in the case of SN2 there cannot be formed a stable carbocation, so the most stable "intermediate" structure would be a transitional state. Stable carbocations are formed via groups with strong inductive electronic effects (+I those that repell electrons) or they can be stabilized by conjugation (the case for this one is when you have a double, triple bond or an aromatic ring directly bound to the carbon atom where the substitution takes place)
so, for example, (CH3)3C-Cl would make for an SN1 when reactiong with alcohols. Also compounds like CH2=CH-CH2-Cl.
The remaining cases, for example CH3-Cl or CH3-CH2-Cl would work via SN2, because the carbocation cannot be stabilized.
Hope that's gonna help you :)

Sn2 will form a transition state and Sn1 will form an intermediate product.


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Liquid to gas phase exothermic hydrolysis reaction examples
I am working as a petroleum engineer and most of our crude oils are heavy oils (very dense). In reservoir, I want to start an exothermic liquid to gas phase hydrolysis reaction.
I am planning to inject my feed into reservoir from wells and the feed will make the reaction with reservoir brine (There is too much water in the reservoir with oil). With this reaction, I am planning to achieve the followings:

With heat, the viscosity of the crude will decrease and production will increase.
Liquid phase feed is easy to inject and gas phase product (especially $\ce{CO2}$) may be an additional force to increase production, which is called as gas flooding.

I found some reactions like the hydrolysis of $\ce{CaO}$. But here, the feed is in solid phase. The product in this reaction is an alkaline and it may enhance oil recovery by changing the wettability or reducing the interfacial tension between oil and water.
I wonder if there are any other reactions like that.

See, heating the reservoir, pressurizing it, and decreasing the surface tension are different things, they are normally achieved by different means, and that's for a reason. You are looking for a reaction that would do all three jobs at once. Guess if you'll find one, it will perform poorly in each.

Yes, you are exactly correct.  I know that heating in a way (Microwave, steam flooding etc.) is the best solution for our reservoirs. Also, heating will cause the viscosity to decrease  and that's the main aim for me. So, I can say that I am looking for an exothermic reaction and water soluble products.


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Calculating concentration of H+ Ions using Ka Value
I have this homework question that I am a bit stuck on for an answer. If anyone could help that would be great.
Question:

For a $\pu{0.2M}$ $\ce{HNO2}$ solution ($\pu{K_a= 4.5 \times 10^{-4}}$) calculate the concentration of the $\ce{H+}$ and $\ce{NO2-}$ ions, $\ce{HNO2}$ molecules and the $\ce{pH}$ of the solution.

If anyone could give any hints or answers that would be great.

Pinpoint exactly where your stuck, otherwise this question will get closed as no effort.

I'm mainly stuck on how to approach this question; I'm unsure how I can find the concentration of the H+ ion given this information. With this information I should be able to do this question, but I'm stumped on how I can find this value. Do you have any tips? :S

I'm afraid you lack a few basic concepts before even attempting this question. Have you read any textbook regarding ionic equilibrium? Do you know what $K_a$ means?

Isn't it the equilibrium constant for the reaction $H^+ + NO_2^-  <-> HNO_2$?

Precisely! Suppose if I give you a reaction: $$\ce{mA + nB <=>[K_a] pC + qD}$$ what would be the expression for $K_a$?

Would it be: $K_a = ([C^p] * [D^q])/([A^m]*[B^n])$? This is how I've been finding constant values when given the relevant info (Products/Reactants) but I'm stuck on how to do this is in the reverse (as I end up with two unknowns; the H+ and NO_2- concentrations)

Let us continue this discussion in chat.

Answer:
$K_a = 4.5*10^-4$
Concentration (From ICE Table) of products/reactants:
HNO2 = 0.2 - x
H+ = x
NO2 = x
Therefore:
$$4.5*10^-4 = x^2/(0.2-x)$$
Rearrange:
$$x^2 + x*(4.5*10^-4) - (0.2(4.5*10^-4)) = 0$$
Using quadratic formula:
$x \approx 0.009$
$$pH = -log(10)$$
$$pH \approx 2.05$$.
That should be correct!

In order to tackle the problem, some assumptions have to be made first as seen in the working below which I have drawn up:

In ideal circumstances, we assume that when one mole of nitrous acid dissociates it produces equimolar amounts of hydrogen ions and the anion of the acid. Thereby explaining assumption (I). Furthermore, as weak acids only dissociate slightly we can say that there will be negligible difference between the initial and equilibrium concentrations. It must be remembered that when dealing with these problems all concentrations must be from the solution at equilibrium. Because of assumption (I), we can represent the hydrogen ions and nitrous acid anions both as "x", which solves the problem you were having with the two unknowns. From then onwards all that has to be done is to solve for "x", which in the above solution is taken to two decimal places. But also keep in mind that the units have to be manipulated as well.
I believe you have forgot to add units for the K$_a$ value, as if you had then your concentration of "x" would come to mol$^{1/2}$dm$^{-3/2}$ which is not a valid value to be used in the pH equation as it requires an answer in moldm$^{-3}$. I hope this explanation furthers your progress.

Welcome to SE! Nice to see your contribution.  Try to avoid inserting picture texts, they're hard to search via googling. Also take a look at our guide to MathJax, to format math/chemistry expressions.


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Isolating KI3 from solution
Is it possible to isolate potassium triiodide from KI solution? Sources claim that this is possible because potassium is a sufficiently large cation and triiodide is a sufficiently large anion, therefore it is capable of forming a stable solid. This is an explanation based on HSAB, where triiodide is the soft base and potassium cation is the soft acid. Has anyone tried to do so?

You cannot obtain KI3 in solid form. KI3 is nothing but a solution of potassium iodide and iodine in distilled water.

@NilayGhosh By evaporating the water away, the ionic solids will be left in solution right? KI3 being an ionic solid may thus also be obtained in this way. Also, what is your reasoning in saying that it cannot exists in solid form?

You're assuming it's stable...

For reference: https://en.wikipedia.org/wiki/Triiodide  Note that "some" have been isolated. No one wants to isolate the thallium salt if they can isolate the potassium salt instead...

If you evaporate all the water, the solid left will be potassium iodide not potassium triiodide. Potassium iodide renders the iodine to be soluble in water by formation of triiodide ion($\ce{KI3}$). Without the iodine and water, the formation of triiodide is not possible. See- Lugol's iodine.

Well, @NilayGhosh already commented on stability of $\ce{KI3}$ in aqueous solution, also $\ce{K+}$ is not bulky enough. But as long as OP didn't mention explicitly that it must be water we isolate potassium triiodide from, there is a workaround, if one desperately want to get its crystals (in an altered form though).
From isopropanol one relatively easy obtains red diamond-shaped crystals of (18-Crown-6)-potassium tri-iodide (1) as the result of reaction between potassium isopropylxanthate and iodine:

There are also many other examples in the literature showing how one can man make potassium "bulkier", but I found this particular example the most straightforward.
Of course, there is always a fallback option to crystallize ammonium triiodide $\ce{NH4I3}$ (2) as ammonium is probably the closest cation resembling potassium for this case.
(1) Sievert, M.; Krenzel, V.; Bock, H. Zeitschrift für Kristallographie - Crystalline Materials 2010, 211 (11), 794–797. DOI: 10.1524/zkri.1996.211.11.794
(2) Cheesman, G. H.; Finney, A. J. T. Acta Cryst B, 1970, 26 (7), 904–906. DOI: 10.1107/S0567740870003357

Thanks. Although I was really hoping to get pure crystals, I am fine with getting these not so pure ones as well. I just want to know whether it is isolatable or not. Once again, thanks for your insightful response.


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Could we reach a pH of 4.9 by blowing through straws?
According to Wikipedia's article on Breathing:

The permanent gases in gas we exhale are 4% to 5% by volume more carbon dioxide and 4% to 5% by volume less oxygen than was inhaled.

If I started with a clean beaker of pure water and measured the pH carefully and quickly, I might get a value close to 7.0. If I left it out in the air for a while, far from concentrated sources of pollution, it would equilibrate with the local CO2 concentration which would be 300 to 500 ppm depending on local variables and natural sources (people, plants), and that would lead to a (carefully) measured pH below 6, and probably somewhere around 5.8 or 5.7. Is that about right so far?
Now suppose that I then took turns with lab mates and constantly blew bubbles into the water through long straws that contained some charcoal and other filters to absorb organics from my breath, leaving only the added 4 to 5% CO2 from our respiration.
If the average area of all the bubbles present was about the same or greater than the average area of the water exposed to the air, would I be roughly correct in using a number like 0.02 or 0.03 atmospheres for the CO2 partial pressure in the plot as displayed below, and getting a pH of about 4.9?
Is this what would happen, or are there significant effects that I've neglected?
related: answer to Formation of carbonic acid from breath?

Screenshots from Effect of Dissolved CO2 on the pH of Water, Byck, Harold T., Science  19 Feb 1932, Vol. 75, Issue 1938, pp. 224, DOI: 10.1126/science.75.1938.224.

You are adding as much as will dissolve, given its concentration in the exhaled air. The straw parameters are irrelevant.

Just curious. How long would it take for the CO2 to equilibrate with the water? It's OK if u don't have the answer.

@TanYongBoon I don't know, I'd guess it would take more than five minutes and less than a day to get most of the way there, so say split the difference on a log scale and call it 2 hours? That's why I said we!

You would reach the saturation point of $\ce{CO2}$ in water because the simplified scenario is that your are continuously sparging $\ce{CO2}$ into your beaker. What is being ignored is gas diffusion and bubble dynamics at the bubble-water interface.

@J.Ari can you make a statement about the resulting pH when saturated? The concentration of dissolved CO${}_2$ is only part of the problem. Are you agreeing or disagreeing with the calculations and conclusion of the attached paper?

If you follow the link below, there is another derivation presented that shows the ultimate pH as 5.65. I don't follow why you say the dissolved $\ce{CO2}$ is only part of the problem. In the simplified case, this is an equilibrium problem only so the dissolved content is the only major factor.
http://ion.chem.usu.edu/~sbialkow/Classes/3650/Carbonate/Carbonic%20Acid.html

@J.Ari the gas is 5% CO2, 95% O2+N2+Ar... It's not like there is a bottle of pure CO2 bubbling through the water. The complicating factor is that the water has one interface with 5% CO2 (0.05 atm partial pressure) and another interface with roughly 0.05% CO2. In any event, the solution would not be saturated with CO2. I'm thinking of simplifying the question and removing the dual-interface issue which is interesting but should probably be isolated and asked separately, unless you'd like to post an answer as-is?

Seeking clarity...in your second comment, you say you are looking at this is like an equilibrium problem. So, given the orders of magnitude difference between the concentrations of $\ce{CO2}$ in the breath vs the air that you quoted, the equilibrium case will be dominated by the breath both from a thermodynamics standpoint and a mass transport standpoint.

@J.Ari Please post that as an answer! It's the beauty of the stackexchange system that answers can be voted on. I don't think that statement is necessarily true. But if you include it as the basis of an answer, defend it with an equation or a link or a fundamental principle and several chemist up vote it, it will be more convincing to me. Right now it is hard for me to distinguish correct comments from incorrect comments on this subject. Go for it, don't be shy! Thanks!

See here for partial pressure of $\ce{CO2}$ from exhaled breath that I used (28 mmHg = 0.037 atm). I am assuming Henry's law is applicable to calculate:
${7.4*10^{-4} M} =  [\ce{CO2}_{(aq)}]$
Used Henry's Law Constant of $2*10^{-3} \frac{M}{atm} $
See here for equilibrium definition and equilibrium constant value ($K=4.45*10^{-7}{M}$) I used:
$$K = \frac{[H^+]^2}{[\ce{CO2}_{(aq)}]}$$
I then solved for pH from the above equation after rearranging to solve for $[H^+]$, and got a pH of 5.2; so I don't think we can achieve pH of 4.9 from only breathing air into the beaker.

Thanks for your post! Can you add the numerical value for the equilibrium constant you used here within your answer itself please? Links go dead all the time, and if that link breaks your answer is no longer useful, and people don't have to go opening up new windows and hunting for a number somewhere. Just type the number inside your answer and it will stay there more reliably for future users and be more easily findable for me too! Also it will be interesting to see if your calculation also produces about pH 5.7 for 350 or 400 ppm CO2.

@uhoh Did you have a look at https://en.wikipedia.org/wiki/Carbonic_acid?  pKa (first dissociation) = 3.6, overall (including CO2) 6.3; and according the lower part of the page & table the statement that fizzy bottled water (of 2.5 atm pressure) reaches even pH 3.7 ...

@Buttonwood update That's pretty surprising, I hadn't looked there but I have now! I think there is a lot I can learn about chemical equilibrium by just focusing on the CO2/H2O system. Thanks! You are right to guess that part of my interest is of the form how low can it go? but the other part is to ask about the complete experiment. Thus the title "Could we reach a pH of 4.9 by blowing through straws?" & discussion of charcoal etc. I'm going to go purée some red cabbage soon for indicator and go grab a bottle of fizzy water and have some fun! :) Mmmm... fizzy purple cabbage water.

Thanks for the updated answer, looks good! For this one let's just assume there's a cover over the glass so that the CO2 concentration in the air above the liquid is the same as the air in the bubbles, so there's no trouble with dual interfaces. I'll read some more about that problem, and ask a separate question if it's not answered here already. Thanks!