text:
How to calculate the equilibrium constant from the density of a nitrogen dioxide and dinitrogen tetroxide mixture?

The density of an equilibrium mixture of $\ce{N2O4}$ and $\ce{NO2}$ at $\pu{1 atm}$ and $\pu{373.5 K}$ is $\pu{2.0 g/L}$. Calculate $K_c$ for the dissociation reaction

I did this:

Let initial amount of substance be $a$.
Initial vapor density $=(\text{mass of }\ce{N2O4})/2 =46$
For final vapour density I find effective molecular mass of the mixture $\ce{N2O4}$ + $\ce{NO2}$ at equilibrium using ideal gas equation which comes out to be $61.33$ and thus final vapor density is $61.33/2=30.665$.
Now supposing degree of dissociation to be $x$ , total amount of substance at equilibrium is $a\times(1+x)$.
As $\text{Vapor density} \times \text{total amount of substance} = \text{constant}$ at same temperature, $46\times a=30.665\times a\times(1+x)$
$x=0.5$ Using this in the equation
$$K_p=\left[\frac{4x^2}{1-x^2}\right]\times P$$ for this reaction I get $K_p=\frac43\pu{ atm}$ and then $K_c=0.043$.
The answer is 2 but how?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please also have a look here. I have tried to make your post more readable, but I cannot be sure, since I do not understand in the slightest what you have tried there.

I confirm your answer.
The molar density of the mixture is
$$\frac{n}{V}=\frac{p}{RT} = \pu{0.0326 mol/L}$$
So, the average molecular weight of the mixture is
$$\frac{2}{0.0326} = \pu{61.35 g/mol}$$
If $x$ is the mole fraction of $\ce{NO2}$ and $(1-x)$ is the mole fraction of $\ce{N2O4}$, then the average molecular weight of the mixture is also
$$46x + 92 (1 - x) = 61.35$$
Solving for $x$ gives $x = 1/3$. So, $(1-x)=2/3$. These are also the partial pressures of $\ce{NO2}$ and $\ce{N2O4}$, respectively (in atm). This leads to the values of $K_p$ and $K_c$ that you calculated.

A very elegant solution! I also like how you used the "molar density" term,  not the "concentration". Kudos for that!

I don't entirely understand your way of solving, but I obtained pretty much the same results calculating $K_p$ and $K_C$

$K_p = \pu{1.21e5 Pa}$; $K_C = \pu{3.9e-2 mol L^{-1}}$

in the following way. I assumed the following dissociation process:
$$\ce{N2H4 (g) <=> 2NO2 (g)}$$
Further I denote $\ce{N2H4}$ as compound 1, and $\ce{NO2}$ as compound 2. Ideal gas equilibrium constants are linked via the equation:
$$K_C = \frac{K_p}{(RT)^{\Delta n}},$$
where $\Delta n = \sum{n_j(\text{products})} - \sum{n_i(\text{reactants})}$. In this case $\Delta n = 2 - 1 = 1$. Also at equilibrium $K_p$ is
$$K_p = \frac{p_2^2}{p_1}.$$
Density $d$ for the mixture can be written as:
$$d = \frac{m_1 + m_2}{V} = \frac{m_1}{V} + \frac{m_2}{V}.$$
These terms can be obtained from the ideal gas law:
$$\frac{m_i}{V} = \frac{M_i p_i}{RT},$$
which in combination with partial pressures brings us to the following system of equations:
$$\begin{cases} d = \frac{1}{RT}(M_1 p_1 + M_2 p_2) \\ p = p_1 + p_2 \end{cases}$$
Considering that $p_1 = p - p_2$ and $M_1 = 2M_2$ ($\ce{N2O4}$ has twice the molar mass of $\ce{NO2}$) we have:
$$d = \frac{1}{RT}(2M_2 (p - p_2) + M_2 p_2) = \frac{M_2}{RT}(2p - p_2)$$
and
$$p_2 = 2p - \frac{dRT}{M_2}$$
from where we can calculate $p_2$ using given values:
$$p_2 = \pu{2e5 Pa} - \frac{\pu{2 kg m^{-3}} \cdot \pu{8.31 J mol^{-1} K^{-1}} \cdot \pu{373.5 K}}{\pu{46e-3 kg mol^{-1}}} = \pu{6.5e4 Pa},$$
$$p_1 = \pu{1e5 Pa} - \pu{6.5e4 Pa} = \pu{3.5e4 Pa}.$$
Now we can calculate $K_p$:
$$K_p = \frac{(\pu{6.5e4 Pa})^2}{\pu{3.5e4 Pa}} = \pu{1.21e5 Pa},$$
which is about $\frac{4}{3}$ atm, and, finally, $K_C$:
$$K_C = \frac{K_p}{RT} = \frac{\pu{1.21e5 Pa}}{\pu{8.31 J mol^{-1} K^{-1}} \cdot \pu{373.5 K}} = \pu{39 mol m^{-3}} = \pu{3.9e-2 mol L^{-1}}.$$
I haven't practiced these problems for a long time, so my solution may not be the optimal one.


text:
Determining Kohn-Sham and Hartree Fock virtual orbitals: The underlying field
In Frank Jensen Introduction to Computational Chemistry from 2007, I stumbled upon this paragraph about whether to assign meaning to the KS orbitals or not

Another difference is that the unoccupied orbital energies in Hartree–Fock theory are determined in the field of N electrons and therefore correspond to adding an electron, i.e. the electron affinity. The virtual orbitals in density functional theory, on the other hand, are determined in the field of N − 1 electrons and therefore correspond to exciting an electron, i.e. unoccupied orbitals in DFT tend to be significantly lower in energy than the corresponding HF ones, and the highest occupied molecular orbital–lowest unoccupied molecular orbital (HOMO–LUMO) gaps are therefore much smaller with DFT methods than for HF. This also means that orbital energy differences in DFT are reasonable estimates of excitation energies, in contrast to HF methods where excitation energies involve additional Coulomb and exchange integrals.

I can't seem to understand why the Kohn-Sham equations are solved in the presence of N-1 electrons and not N. From my understanding the Kohn Sham and the Hartree Fock equations are almost identical. Where does this difference come from?

That is true: in the Kohn–Sham model electrons both in occupied and in virtual orbitals are "moving" in the field $n-1$ electrons, while in the Hartee-Fock model electrons in occupied orbitals are "moving" in the field $n-1$ electrons, but electrons in virtual orbitals are "moving" in the field all $n$ electrons. And this is simply "by construction": you just have to look at the very definitions of these models. So, in short:

In the Kohn–Sham model, it is the exchange-correlation hole that by definition "excludes" a single electron from equations describing each and every orbital.
If you look at the Kohn–Sham equations
$$
\left(-{\frac  {\hbar ^{2}}{2m}}\nabla ^{2}+v_{{{\rm {eff}}}}({\mathbf  r})\right)\phi _{{i}}({\mathbf  r})=\varepsilon _{{i}}\phi _{{i}}({\mathbf  r}) \, ,
$$
where
$$
v_{\rm {eff}}(\mathbf {r} )=v_{\rm {ext}}(\mathbf {r} )+e^{2}\int {\rho (\mathbf {r} ') \over |\mathbf {r} -\mathbf {r} '|}d\mathbf {r} '+v_{{{\rm {xc}}}} \, ,
$$
it is last term of the Kohn–Sham effective potential, the exchange-correlation potential,
$$
v_{{{\rm {xc}}}}({\mathbf  r})\equiv {\delta E_{{{\rm {xc}}}}[\rho ] \over \delta \rho ({\mathbf  r})} \, ,
$$
that causes the above mentioned feature of Kohn–Sham orbitals.

For the case of the Hartree-Fock equations
$$
  \hat{F} \phi_i({\mathbf r}) = \epsilon_i \phi_i({\mathbf  r}) \, ,
$$
the summation in the Fock operator by the very definition of the model runs over all occupied orbitals only,
$$
  \hat{F}
  =
  \hat{H}^{\text{core}}
  +
  \sum_{j=1}^{n}[\hat{J}_{j} - \hat{K}_{j}] \, .
$$
For occupied orbitals, in the Hartree–Fock equation defining some particular (occupied) $i$-th orbital the Coulomb and exchange contribution from this orbital itself perfectly cancel each other, i.e. $\hat{J}_{j} \phi_i = \hat{K}_{j} \phi_i$ when $j=i$. Thus, in the Hartree-Fock model an electron from any occupied orbital is also moving in the field $n-1$ electrons. And this is of course physically more than reasonable: at the end of the day electron should not interact with itself.
But for virtual orbitals the situation in the Hartree-Fock model is different from that in the Kohn-Sham one: for any virtual orbital $\phi_k$ the sum in the Hartree-Fock equation defining it still runs over occupied orbitals only, but none of the terms $\hat{J}_{i} \phi_k$ and $\hat{K}_{i} \phi_k$ this time cancel each other since $k$ is simply grater than $n$.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Why does my absorbance change over time?
I have a solution of black phosphorus quantum dots suspended in water that I prepared by sonicating black phosphorus powder in n-methyl-2-pyrrolidone and centrifuging to isolate the quantum dots. Repeat measurements of the absorbance of this solution over a period of days resulted in a decrease in the absorbance; I know this to be due to the instability of the black phosphorus when exposed to water, air, and light.
However, I also have a solution prepared in the same way using graphene nanoplatelets instead of black phosphorus. Testing the graphene solution over a period of days resulted in an increase in absorbance. Why might the absorbance be increasing in the case of the graphene in water?

As usual is difficult to propose an answer without looking at the whole spectra. When you say Abs, is the value at a given wavelength, is the entire spectrum shifting up (down), does it change in shape? In the specific case water is volatile enough to alter concentration. Especially in summer if it happens that you are in the northern emisphere. Else it could be some aggregation, but the spectrum would likely change in its appearance

Alchimista, the whole spectrum is shifting down but the shape does not change. I am in the northern hemisphere.

Then the absorbance is decreasing for graphene as well? If the spectrum is shifting down cannot be that your solution gets concentrated....

Oh, sorry, I miswrote that. The black phosphorus spectrum shifts down. The graphene spectrum shifts up.

Than look in the literature (no idea at what level you are working) because everything as dots nanoparticles could of course change upon time. As you depict the story, a simple evaporation of water can do that. With water I would be surprised if it take place while measuring. But it can happen at storage level. If you have a small amount of solution/suspension in a relatively large container and you do not shake it, it can even happen because some water evaporate and condense at the top. You take the sample with a pipette and Voilà, you have already a bigger concentration than supposed.....

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is II a resonating structure of I?
Is the structure II a resonating structure of I? If yes, then how? Isn't the lone  pair on N non conjugated so as not to participate in resonance?
I
II

The bottom structure has a carbon with 5 bonds in the bottom ring.

Actually there are carbon atoms with 5 bonds almost everywhere.

@Mithoron. Almost everywhere in the structures depicted in the figure posted by the OP. :)

Ah, yeah. I see poor skill with drawing program there.

@Zhe you know I am not hexaccordinate carbon for nothing. I like so much these structures with 5 bonds carbons :D

Note that both representations contain errors.
Under normal circumstances, the valence of uncharged carbon atoms in a molecule equal to four.
Hence, in the upper representation, the carbon bond to the nitrogen bears one valence to much.  Keeping the rest of the structure drawn as close to the original, your first drawing should look like the following:

In the lower representation, both the carbons binding towards oxygen are depicted as if carbon were pentavalent, which under normal circumstances does not occur.  In addition the central atom of nitrogen is drawn as if it were tetravalent.  There are instances of tetravalent nitrogen atoms in organic compounds, however these are positively charged (like the tetrabutylammonium ion, or the iminium cation) on nitrogen.  Indeed the second (assuming a motif of $\ce{C=N=C}$) would be something not frequently seen, but on paper could be represented for example as

There is a carbon atom in the second structure with 5 bonds. It is not a real structure nor a resonance structure of 1

I count more than one instance of "pentavalent carbon".

@Alchimista look at my username ^^

@Hexacoordinate-C. This is quite an  outstanding nick! :)

@Alchimista yeah but it really exists ! :)

I know but it is indeed outstanding. By the way it could have been a real question with mistake forced by some "new to the user" programme. It is good that Burtonwood made a clear picture


text:
Difference between Oxidation Potential and Ionization Potential
Ionization potential indicates a substance's tendency to loose electron and become a positive ion. So, a substance becomes ion more easily if it has a lower ionization potential (lower energy needed to make it an ion).
Oxidation potential indicates a substance's tendency to loose electron (oxidize) and become a positive ion. So, a substance should become oxidized more easily if it has a lower Oxidation potential. But it is NOT the case. A substance becomes oxidizied more easily if it has a higher oxidization potential.
Why is it so? Can anyone explain?
And again, Potassium becomes ionized more easily than Lithium according to the reactivity series for metals, but Lithium becomes oxidized more easily than Potassium according to Electrochemical series. Why?

You're confusing two related concepts.
Ionization energy is the amount of energy required to remove an electron.
Oxidation potential is the opposite of the reduction potential, which is electrical potential (i.e., voltage) derived from comparing the spontaneity of the reduction compared to reducing a standard hydrogen electrode.
The former is a measure of how easy it is to remove an electron in the absence of anything else.
The latter is used to determine, in conjugation with a reduction potential, the overall reaction spontaneity for electrochemistry.
There are two other issues with your question (which is probably several questions).
The first is that reaction spontaneity is not governed by energy alone. Ionization energy cannot determine whether or not a reaction takes place, especially in solution, where entropic effects are very relevant.
Secondly, it's also unclear if you are asking about kinetic effects (how fact the reaction goes), which is very different from asking how spontaneous the reaction is.


text:
How are the Laguerre polynomials and spherical harmonics related to the electronic quantum numbers?
How are the Laguerre polynomials and spherical harmonics related to the quantum numbers $n, l, m_l$ of an electron in an atom?

http://www.astro.caltech.edu/~srk/Ay126/Lectures/Lecture3/SchrodingerModel.pdf

I'd write an answer, but busy today. In the meantime you might want to consult a physics text on the QM model of the hydrogen atom - most textbooks should devote a section to solving the Schrodinger equation, and should explain what the spherical harmonics and (associated) Laguerre polynomials are. Or someone else might answer, of course.

there is an answer to this question https://chemistry.stackexchange.com/questions/80199/balmers-formula-and-lagurre-polynomials that covers most of the points asked here.


text:
Why must basis sets used in computations be non-orthogonal?
While constructing basis sets for calculations using Hartree Fock, the basis set must be non-orthogonal. And because of this, we go through many steps to put the overlap and density matrices in the 'right' basis. (See Modern Quantum Chemistry (Szabo and Ostlund) or David Sherill's notes online).
My question is: why must this basis be non-orthogonal?

Besides the existing answers, it's important to note that the 'many steps' you mention are there so you can handle non-orthogonal bases, but the formalism works perfectly fine if your basis happens to be orthogonal already.

It does not have to be non-orthogonal. The basis set can be arbitrary, and one usually uses the chemically motivated Ansatz with atomic orbitals (LCAO), which just happen to be non-orthogonal. An alternative would be to use a plane wave basis set, which are orthogonal.

The basis doesn't have to be non-orthogonal, but, in practice, it is more efficient to span the desired Hilbert space for each atom with Gaussians than with proper orbitals as Gaussian integrals can be computed relatively quickly. Moreover, if you want to have an atom-centered basis set (which is the way most chemists think about orbitals), these atom-centered basis functions will overlap if they are near each other regardless of what function type you choose, leading to non-orthogonality.


text:
What kind of water would Oxygen-15 and tritium create?
I've recently become interested in chemistry, and have some questions about water.
I've heard of semi-heavy, heavy, tritiated, and oxygen-heavy water. However, what about various combinations of isotopes? Let's say: Oxygen 15 + Tritium? How about Oxygen 18 + Deuterium?
Would these all have distinct properties from one another? Have all the stable combinations been extensively studied and lab tested?

Oxygen-15 has half-life of about 2 minutes...

@Mithoron That's true, but water Oxygen-15 is used to measure tissue or tumor blood flow/perfusion. I'm more curious about the physical differences between the different combinations of isotopes rather than the practicality of making it.

Yeah it's used in PET, but you can't get macroscopic amounts of its compounds.

Because of the dynamic equilibrium of autoprotolysis of water ($\ce{H2O <<=> H+ + OH-}$ the isolation of water with mixed isotopes of H (like pure HDO, or HTO) is hampered. From a probabilistic approach, the natural abundance of $\ce{^2H}$ equals to 0.02% and the one of $\ce{^{17}O}$ to 0.04%. In other words, only 0.000016% chance to encounter a molecule of $\ce{D2^{17}O}$ outside a scientific lab, and expensive  (http://www.medicalisotopes.com/product-category.php?id=2) For molecules consisting of short-living isotopes, you will need a rapid characterization, mass spectroscopy may be one.

They would vibrate differently (in terms of frequency) leading to slightly different boiling point and kinetic properties. That's kind of it.


text:
How would I determine the relative acidity of substances?

Which of the following salt will produce a neutral solution (If any)? A) $\ce{KI}$ (Potassium Iodine) or B) $\ce{SrS}$ (Strontium Sulfur)

I tried to solve this problem by using the "Relative Strengths of Brønsted-Lowry Acids and Base in aq solution at room temp table". From what I know each of those two substance will dissociate into ions. Having the ions $\ce{K^+, I^-,Sr^{2+}, S^{2-}}$
Since K does not hydrolyze and I is a product of HI which is a strong acid it won't hydrolyze either. I also know that the only cations that hydrolyze are Al,Fe and Cr hence Sr will not hydrolyze. However how do I know weather S will hydrolyze? The other question I had difficulties was "Order the following from most acidic to least. SO3, CO3 and CO2".

In my homework one of the question was which of the following salt will produce a neutral solution(If any)? A) KI (Potassium Iodine) or B) SrS (Strontium Sulfur)
For (A) both K+ and I- will essentially stay as the ions. So assuming the solution is neutral to start, it will remain neutral.
But for (B), in a neutral solution, $\ce{Sr^{2+}}$ will stay as the ion,  $\ce{S^{2-}}$ will react as:
$\ce{S^{2-} + H2O <=> HS^- + OH-}$
so the solution will become slightly more basic.
"Order the following from most acidic to least. $\ce{SO3}$, $\ce{CO3}$ and $\ce{CO2}$".
I can't really make sense of this question since $\ce{CO3}$ is highly unstable, and there are three isomers.
Bubbling  $\ce{SO3}$ and $\ce{CO2}$ gases into water would produce acidic solutioons via reactions
$\ce{SO3 + H2O -> H2SO4}$
$\ce{CO2 + H2O -> H2CO3}$

Just to clarify in the second question we evaluate the acidity by the acid that is produce as the final product not the acids that are formed as the intermediates such as HSO4 and HCO3?

There are really no such species as $\ce{HSO4}$ and $\ce{HCO3}$ in aqueous solution. The species would have to be $\ce{HSO4^-}$ and $\ce{HCO3^-}$ which would formed from acid reacting with $\ce{H2O}$. $\ce{H2SO4}$ is such a strong acid that there would be just a trace of $\ce{HSO4^-}$, with most of the S species being $\ce{SO4^{2-}}$.


text:
Ambimodal transition states in some organic reaction pathways
In many computational studies, mechanisms appear to have so-called ambimodal transition states, i.e. a transition state which can lead to multiple products, in conflict to the common undergraduate idea of a reaction pathway.
To give an example, Houk, in Mechanisms and Origins of Periselectivity of the Ambimodal [6 + 4] Cycloadditions of Tropone to Dimethylfulvene,[1] presents the following scheme:

Reaction pathway for [6 + 4] cycloadditions of Tropone to Dimethylfulvene. Taken from ref [1]
In the scheme, TS-1 may either give 7 (the 'expected' product) or go directly to 6, completely skipping TS-Cope-1 (which would allow 7 and 6 to interconvert) — Houk describes this as an ambimodal transition state, that is one transition state that can lead to two possible products.

Although what he says makes sense, it seems to me that, given 6 is the lower energy product out of the two, it would be favoured thermodynamically anyway.
Upon further reading, I discovered the concept of valley ridge inflection, which appears to describe the phenomenon observed above, in which we completely 'miss' the  intermediary transition state:

Why do these ambimodal transition states occur, and how does the situation differ from normal transition state theory?

Notes & References
[1]:Houk, K.N., J. Am. Chem. Soc. 2017, 139, 8251. DOI: 10.1021/jacs.7b02966

According to Houk (who cites Pierluigi Caramella here) it is one transition state that leads to two products. If you look at the second picture of yours you go through a saddle point and after that you end up on top of another saddle point where you got two direktions to go down. Technically this was one transition state. Do you know this talk: https://youtu.be/plYVZWBPCws ? It explains a lot of this in detail. The ambimodal cycloadditions start at 1:03, but the part about dynamics before is very helpful

I haven't seen that talk, some bedtime viewing perhaps!

Why do these ambimodal transition states occur?

I don't know that there is a general reason for that. Potential energy surfaces are complicated. Very often, there is an abundance of local minima, separated by potential energy barriers. It is similar to what happens with mountain chains and valleys, but with many dimensions.
One could argue that the opposite could be asked: why would there be a single minimum associated with each and every barrier? Why would there be exactly a single downhill valley behind each mountain chain, instead of, sometimes, two different valleys being similarly accessible, once you climb over the top of a hill?

How does the situation differ from normal transition state theory?

Transition state theory is a useful simplification, assuming a simple potential energy surface. It is useful because, at room temperature, the energy differences in the different possible paths are very often sufficient for a single reaction pathway to predominate, so that others effectively do not happen. When the temperature is comparable to the difference between the energies involved in different possible reaction pathways, the situation is more complicated.

ambimodal reaction path
The two different reaction trajectories from the 4+2 are both slightly lower in energy but they also are both energetically and geometrically relatively close to the Ea for the 4+2 reaction
For these pericyclic reactions ([4+2] vs [3,3],both pathways are "allowed".  The cope rearrangement [3,3] has a lower Ea than the Diels-Alder, and the tajectory from the 4+2 Ts to the bifurcation for product formation is slightly energetically down hill from the 4+2 TS.  Sigma bond formation is not complete, and thus its easy tor this reaction tajectory to slip down either side of this bifurcation.
It also interesting that if you execute the reverse reaction, you'll drop into the cope reaction, before you get the reverse Diels-Alder ( see Wooward-Katz)
RTP

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


text:
Naming a hydrocarbon chain with multiple substituents equidistant from both terminals
Does the substituent that comes alphabetically first take precedence, or do we look for the next nearest substituent, or is it a combination of both?

You mean from both ends?

Possible duplicate of How to decide at which end to start the numbering of the parent chain when the options for locant sets are the same?

@Mithoron yes I do

@jerepierre thanks for the link, but I didn't find an answer there

The substituents in the compound that is given in the questions are not equidistant from both ends.
Numbering from left to right yields the name 5-bromo-3-chloro-7,8-diethyl-2,9-dimethyldecane (i.e. the locant set ‘2,3,5,7,8,9’), whereas numbering from right to left yields the name 6-bromo-8-chloro-3,4-diethyl-2,9-dimethyldecane (i.e. the locant set ‘2,3,4,6,8,9’).
Concerning numbering, 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.
Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is named as 6-bromo-8-chloro-3,4-diethyl-2,9-dimethyldecane rather than 5-bromo-3-chloro-7,8-diethyl-2,9-dimethyldecane since the locant set ‘2,3,4,6,8,9’ is lower than ‘2,3,5,7,8,9’.


text:
Balmer's formula and lagurre polynomials
How can you explain the Balmer's formula using the properties of Laguerre polynomials?
It's a homework question but I really don't get it.
1.What is the property of the Lagurre polynomials in the first place?
2.How is Balmer's formula related to the Lagurre polynomials??

When solving Schrodinger equation different potentials are included depending on the problem at hand, for example, particle in a box, harmonic oscillator, H atom etc. Then boundary conditions are applied; for a particle in a box, for example, the wavefunction has to be zero at either end of the box and this ensures that the solutions are quantised, i.e the wavefunction depends both on position x and an integer n and the energy also depends on an integer.
The different solutions to different types of differential equations were worked out many many years ago, and often by French mathematicians consequently the polynomial solutions obtained often carry their names, Laguerre, Hermite, Legendre and Chebychev (although he was Russian). These polynomials all depend on position x and one or more integers. The Hermite polynomials, for example, which form solutions to the harmonic oscillator, are often labelled as $H_n(x)$ where different polynomials are found with  $n=0,1,2 ...$, e.g. $H_0(x)=1, H_1(x)=1, H_2(x)=4x^2-2 \cdots$.
In solving the H atom the radial and angular solutions are separated and each solved independently. The associated Laguerre polynomials ($L_n(x)$) are used in finding the radial solutions, and the angular solutions  depend on the spherical harmonic polynomials which themselves depend on the associated Legendre polynomials $P_l^m(\cos(\theta)$.
(Notes. The polynomials are often described by recursion formulae meaning that once the first couple are known the rest are easily found. The associated polynomials are found by repeated differentiation of the polynomial, via Rodrigues' formulae. The polynomials are orthogonal.)


text:
Calculating molar mass of an oxidant by iodometry
I am stuck on a question from the Australian Chemistry Olympiad involving molar mass, shown here:

BTW, this whole paper is available online for free, and is distributed by its owners for free. Just search up: Australian Chemistry Olympiad Papers and look at the past papers.
Anyways, in this question, I am able to solve part a and part b, but part c is a little tricky. I understand how to solve for molar mass: it is simply the mass of the substance divided by the number of moles in the same amount of substance. However, in the solution they use the number of moles of the metal cation M, as well as the subscript variable x. Can someone please describe in detail what this means. If there are any more questions, feel free to comment below. Thanks in advance.
P.S. The answers are not my own, but those from the answer sheet.

What you did is correct. Why do you have doubts?

The answers are not my own, but from the answer sheet

I don’t understand how the person got part c

@Abhigyan the answers are from the answer sheet

The molar mass of a substance, as you said earlier, is:

simply the mass of the substance divided by the number of moles in the same amount of substance

In this question, you can see that as per part b), we found that there were $7.553\times10^{-4} \text{ moles}$ of $\ce{M^{n+}}$ present.
Hence, the salt, i.e., $\ce{M_xA_y.ZH2O}$ contains $\ce{M_x}$ = $7.553\times 10^{-4} \text{ moles}$ $\ce{M^{n+}}$.
=> $\ce{M}$ = $ \frac{7.553\times 10^{-4}}{x} \text{ moles}$ of $\ce{M^{n+}}$
$\therefore$ taken moles of $\ce{M_xA_y.ZH2O}$ = $\frac{7.553\times 10^{-4}}{x} \text{ moles}$
But we know that the taken moles have a mass of $0.2642 g$.
Now, divide $0.2642 g$ by $\frac{7.553\times 10^{-4}}{x} \text{ moles}$ to get the molar mass, in terms of x, which is $349.8x$


text:
Proton leak during ATP synthesis
I was told there is a proton leak during ATP production. When going through the mechanisms we went over in class, I was unable to find when this occurs.
During what part of ATP production is there a proton leak and what are the ramifications of this?

Tbh, I think this is more suitable for [biology.se]. But it might work here, too.

Are you asking about how the protons move round and through the ATPase ?

This occurs during Oxidative phosphorylation and the "proton leak" process is also called Uncoupling of Oxidative Phosphorylation
The mitochondrial electron-transport chain
Electron transport, causes Complexes I, III, and IV to transport protons across the inner mitochondrial membrane from the matrix, a region of low [$\ce{H+}$] and negative electrical potential, to the intermembrane space (which is in contact with the cytosol), a region of high [$\ce{H+}$] and positive electrical potential.

Three of the four electron-transport complexes, Complexes I, III, and IV, are involved in proton translocation.
Two mechanisms have been put across that would couple the free energy of electron transport with the active transport of protons: the redox loop mechanism and the proton pump mechanism.
Uncoupling of Oxidative Phosphorylation
Electron transport (the oxidation of NADH and FADH2 by $\ce{O2}$) and oxidative phosphorylation (the synthesis of ATP) are normally tightly coupled due to the impermeability of the inner mitochondrial membrane to the passage of protons. Thus the only way for $\ce{H+}$ to re-enter the matrix is through the $\ce{F0}$ portion of the proton-translocating ATP synthase.

In the resting state, when oxidative phosphorylation is minimal, the
proton-motive force across the inner mitochondrial membrane builds up
to the extent that the free energy to pump additional protons is
greater than the electron-transport chain can muster, thereby
inhibiting further electron transport.
However, many compounds,including 2,4-dinitrophenol (DNP) and
carbonylcyanide-ptrifluoromethoxyphenylhydrazone (FCCP), have been
found to “uncouple” these processes. In a pH gradient, they bind
protons on the acidic side of the membrane, diffuse through, and
release them on the alkaline side, thereby dissipating the gradient.

The chemiosmotic hypothesis has provided a rationale for understanding the mechanism by which these uncouplers act.
Uncouplers act by dissipating the proton gradient across the inner mitochondrial membrane created by the electron-transport system.

As uncouplers, they function by carrying protons across the inner
membrane. Their tendency is to acquire protons on the cytosolic
surface of the membrane (where the proton concentration is high) and
carry them to the matrix side, thereby destroying the proton gradient
that couples electron transport and the ATP synthase.
In mitochondria treated with uncouplers, electron transport continues
and protons are driven out through the inner membrane. However, they
leak back in so rapidly via the uncouplers that ATP synthesis does not
occur. Instead, the energy released in electron transport is
dissipated as heat.

The presence in the inner mitochondrial membrane of an agent that renders it permeable to $\ce{H+}$ uncouples oxidative phosphorylation from electron transport by providing a route for the dissipation of the proton-motive force that does not require ATP synthesis. Uncoupling therefore allows electron transport to proceed unchecked even when ATP synthesis is inhibited.

Uncoupling of oxidative phosphorylation:  The proton-transporting ionophores DNP and FCCP uncouple oxidative phosphorylation from electron transport by discharging the electrochemical proton gradient generated by electron transport
References

Voet and Voet 4th ed. Biochemistry: Oxidative Phosphorylation
R.H Garret & C.M Grisham Biochemistry 4th ed : Electron Transport and Oxidative phosphorylation


text:
What is a "hydrogen-like" or "hydrogenic" atom?
I'm studying some chemistry on my own in anticipation of the new school year and in my book, I came across the Rydberg equation for the first time. I worked through some examples and everything was fine until I came across this comment on the question, "Calculate the wavelength of the radiation released when an electron moves from n=5 to n=2":

For future reference: the Rydberg formula only works for hydrogen-like
atoms.

What is meant by "hydrogen-like"? I've heard that solving for multi-electron systems is (near) impossible, so I understand why hydrogen is used here, but I don't understand what "hydrogen-like" is.

A hydrogen-like atom (or ion) is simply any particle with a nucleus and one electron.

That should be sufficient to answer the question at hand, but I thought I should say a bit more, as some of these answers are potentially confusing.
The historical reason why the Rydberg formula only works for hydrogen-like atoms is that it was originally formulated to explain the spectral lines of hydrogen. It was never intended to explain the spectra of multi-electron atoms.
The physical reason, though, is because the Rydberg formula uses energy levels that only depend on the principal quantum number $n$, which has to be a positive integer:
$$\bar{\nu} = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \qquad n_1,n_2 \in \mathbb{Z}^+$$
and nowadays we know that this is only true for hydrogen-like atoms;$^*$ the energy levels of multi-electron atoms depend on both $n$ and $l$.$^\dagger$

The $n$-dependency was later successfully rationalised by the Bohr model, but saying that "the Rydberg formula only works for hydrogen-like atoms because the Bohr model only works for them" is misleading and misses the point, as:

This implies that the Rydberg formula was derived from the Bohr model, which is not true; it was merely empirically determined, and the formula predated the Bohr model by 25 years.
The Bohr model simply does not work for hydrogen-like atoms. The fact that it reproduces the Rydberg formula should merely be considered a serendipity; Bohr arrived at the correct result by the wrong method.
It does not lend any real insight into the proper reason why the Rydberg formula does not apply to helium, etc. (which I briefly mentioned above).

$^*$ In fact, the energy levels of hydrogen are not only dependent on $n$ (due to various small effects such as – but not limited to – spin-orbit coupling, and hyperfine splitting). Wikipedia has a good overview of the topic here and most QM textbooks have a chapter on the hydrogen atom, where they discuss these perturbations to the Hamiltonian and their effects on the energies. Not surprisingly, the inability to explain this was one of the failures of the Bohr model.
$^\dagger$ Of course, there is also a series of approximations here. The energy levels of multi-electron atoms are only approximately described by sums of orbital energies, so the transition energies are only approximately equal to a difference in energy between two orbitals.

I like your answer, but I think you confuse Balmer's formula with Rydberg's formula. Rydberg was aware of Balmer's equation predicting the spectral positions of the hydrogen lines and Rydberg successfully(!) tried to find a similar expression for non-hydrogenic elements (see Z. Phys. Chem. 5, 227 (1890)). He did this by introducing an $\ell$ dependent quantum defect, which for H is equal to zero.

@Paul Thanks for your comment and sorry for my late reply. I didn't know that! Wikipedia isn't entirely clear on it, and the different form of the equations used back then don't help, but from what I can tell: (1) Balmer found the formula in my answer for the specific case of $n_1 = 2$; (2) Rydberg extended it to alkali metals introducing a quantum defect; (3) the formula above is a specific case of Rydberg's formula with quantum defects set to zero. Am I understanding it right? I just wanted to make sure before I edit.

sorry for my late reply as well, I was on holiday the last weeks... You're absolutely right.

In the spirit of notable University Of Nottingham chemist Sir Martyn Poliakoff's comment in the Periodic videos episode Helium in Disguise (also YouTube):

Again and again, for scientists, if they hear something surprising that makes them think in a different way, it’s really good.

I'll add to the other excellent answers by listing some less common hydrogen-like "atoms".
The video is about Muonic Heavy Hydrogen or so-called ${}^{4.1}H$ is discussed at the end of this list. Yes, that's pronounced "hydrogen four-point-one." See the New Scientist article Atomic disguise makes helium look like hydrogen, and the primary article published in Science Kinetic Isotope Effects for the Reactions of Muonic Helium and Muonium with H2 Fleming, D. G. et al. Science  28 Jan 2011: Vol. 331, Issue 6016, pp. 448-450, DOI: 10.1126/science.1199421

Positronium: (e+e−)

Positronium (Ps) is a system consisting of an electron and its anti-particle, a positron, bound together into an exotic atom, specifically an onium.

It has Rydberg states with $E_n \sim -(6.8eV)/n^2$ or half that of normal hydrogen because the reduced mass is half that for an electron bound to a much heavier item. The longer-lived triplet ${}^3S_1$ state has a mean lifetime of about 142ns, and decays by electron-positron annihilation into three gamma-ray photons. It is sometimes studied by slowing and stopping positrons in powdered MgO where they capture an electron and tend to remain relatively unperturbed by the other atoms.

Muonium: (μ+e−)

Muonium is an exotic atom made up of an antimuon and an electron, which was discovered in 1960 and is given the chemical symbol Mu. During the muon's 2.2 µs lifetime, muonium can enter into compounds such as muonium chloride (MuCl) or sodium muonide (NaMu). Due to the mass difference between the antimuon and the electron, muonium (μ+e−) is more similar to atomic hydrogen (p+e−) than positronium (e+e−). Its Bohr radius and ionization energy are within 0.5% of hydrogen, deuterium, and tritium, and thus it can usefully be considered as an exotic light isotope of hydrogen.

True Muonium: (μ+μ−)

True muonium or muononium is an exotic atom made up of an antimuon and a muon. It is yet to be observed, but it may have been generated in the collision of electron and positron beams.

Muonic hydrogen:

Negative muons can, however, form muonic atoms (previously called mu-mesic atoms), by replacing an electron in ordinary atoms. Muonic hydrogen atoms are much smaller than typical hydrogen atoms because the much larger mass of the muon gives it a much more localized ground-state wavefunction than is observed for the electron.

Muonic helium:

Muonic helium is created by substituting a muon for one of the electrons in helium-4. The muon orbits much closer to the nucleus, so muonic helium can therefore be regarded as an isotope of helium whose nucleus consists of two neutrons, two protons, and a muon, with a single electron outside. Colloquially, it could be called "helium 4.1", since the mass of the muon is slightly greater than 0.1 amu. Chemically, muonic helium, possessing an unpaired valence electron, can bond with other atoms, and behaves more like a hydrogen atom than an inert helium atom.
Muonic heavy hydrogen atoms with a negative muon may undergo nuclear fusion in the process of muon-catalyzed fusion, after which the muon may leave the new atom to induce fusion in another hydrogen molecule. This process continues until the negative muon is trapped by a helium atom, and cannot leave until it decays.

above: From New Scientist.

Hydrogen-like atoms are atoms with a single electron "orbiting" a nucleus that has more than one nucleon.  As @Xerxes pointed out in a comment, you can in principle have a nucleus made up of particles other than protons and neutrons (nucleons).  Positronium might be an extreme example of this.
Wikipedia actually has an entry about Hydrogen-like atoms which goes somewhat beyond what you asked.

There are technical issues with this answer: The core of a hydrogen-like atom need not be made of nucleons. Also, deuterium has more than one nucleon and it is not hydrogen-like, it is hydrogen.

And now it is no surprise what a helium-like ion is.

I'd argue that hydrogen counts as a hydrogen-like atom. But that's splitting hairs.

Hydrogen-like ions are ions that possess only one electron, just like a hydrogen atom.
The fact that these ions have only one ion in the outermost shell makes it simpler to analyze their radii and energies, as a simple electrostatic model can be used to describe them. Species having multiple electrons are difficult to study and are beyond the scope of the Bohr model. This is because the inter-electronic repulsions are hard to account for in electrodynamic interactions that make up the atom's bound system.

You don't actually need electrodynamics for multi-electron atoms more than you do for hydrogen atom. It's just that Schrödinger's equation for multi-electron atoms is vastly more difficult to solve — even in its purely electrostatic form.

The Bohr model is just a historical viewpoint, it really shouldn't be used to rationalise these things anymore. Electrodynamic interactions don't need to enter the Hamiltonian, as Wildcat explained (the full Hamiltonian is here). The difficulty arises from the electron-electron repulsions.

@orthocresol Yes, I am very much aware of that. However this is a question about something regarding Bohr's model and that the reason why Bohr's model doesn't work with multielectronic atoms is because Bohr's model uses electrostatic interactions.

@Ruslan Why are you involving the Schrodinger Equation to a question about Bohr's model? True, Schrodinger's equation is a better view of the atom, but that's not Bohr model. The question is just, "What are Hydrogen-like atoms?" Answering that doesn't need Schrodinger's equation, correct me if I'm wrong.

I don't see a single mentioning of Bohr's model neither in the OP nor in the post linked there. And as Bohr's model doesn't work correctly even for hydrogen atom (e.g. ground-state angular momentum), I see no reason to use it at all, at least in the context of current question.

On top of the previous comment, the Rydberg formula was only empirically determined, and is therefore not tied to any particular theory. It is not derived from the Bohr model; in fact, the Bohr model was developed to rationalise it. However, the use of the Bohr model is no longer necessary, as the QM model accounts for the Rydberg formula perfectly well. So this is not "a question about something regarding the Bohr model".

Single electron systems are known as hydrogenic/hydrogen-like species like He+, Li2+, Be3+, etc.

I think one key aspect that is overlooked in these replies is the Born-Oppenheimer approximation. The nucleus of any atom can be approximated as a point with very few relativistic corrections due to its mass.
An electron on the other hand is essentially massless (comparatively speaking.  I know it is still a fermion and has mass). Once you put more than one electron in any orbital those relativistic corrections have consequences. So "hydrogen-like" means any atom that does not need Pauli or Dirac's insight to explain the deviations in the spectra i.e. the Bohr model.

I am not sure if you don't confuse some things. For example you can straight forward analytically solve the hydrogen atom considering the electrons and protons mass. You simply switch to center of mass coordinates. Levels change by roughly 1+(1/(1+1836)). Thats actually beyond BO. Relativity you need for high "nuclear" effective charges $Z$. Thats quite different things.

That is a good point.  I guess I thought that Dirac applied relativity to explain the Pauli exclusion principle.  In the case of the hydrogen atom you can simply switch to the center of mass.  Is that the same as normal coordinates? I thought that is what the BO approximation did to limit the 3N degrees of freedom thus reducing the amount of parameters needed to approximate more complicated systems i.e anything past a two body problem.

When you solve the H-atom problem the "conventional way" you first use the BO approximation and then fix the proton to the origin. When you solve it in the way I have scetched you dont apply the BO and you don't fix the proton. Rather you decompose the problem in relative motion of the two wrt centre of mass and the motion of the centre of mass (which you don't care about since thats just plane wave). the relative motion problem is completely identical to the conventinal H-atom except you obtain a different effective mass, hence energy levels.

That makes sense.  Thanks for the clarification.

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


text:
Critical temperature for an ideal gas
What is the critical temperature for an ideal gas? It can't be found from Andrew's isotherm as it is for real gases and states critical temperature as $\frac{8a}{27 R \cdot b}$, but for ideal gases $a = b = 0$.

The point is that ideal gasses don't have a critical temperature; by being composed of infinitesimal, completely non-interacting point particles, they are not subject to intermolecular interactions and phase changes. That's what makes them ideal, i.e. simple.

Thus they cant be liquified?

Sure enough, they can't.


text:
Nucleophilic substitution of 2-chloropropanoyl chloride
Upon heating 2-chloropropanoyl chloride with concentrated ammonia in a sealed vessel, a primary amide is formed but the chlorine atom at the beta carbon does not get substituted with an amino group, that  is, the alkyl chloride portion of the molecule  fails to undergo nucleophilic substitution.
Why is this so?

I'm guessing that the α-carbon is much more eletcophilic than the β-carbon, due to the presence of oxygen and the chlorine, so that leads nitrogen, to transform the 2-chloropropanoyl chloride to the corresponding amide rather than amine.

authors: Αντώνιος Κελεσίδης


text:
why bond length of double bond is less than that of triple bond?
why bond length of double bond is less than that of triple bond ? I came across this question many a times. I am confused because there is a sigma bond and two phi bond in triple bond and a sigma and phi bond in double bond. Hence bond length of both of them should be same. But it is not so.
Ex :- Carbon Carbon triple bond length is shorter  than C=C

Before you ask "why", ask "if".

perhaps you can give some examples to support your question?

Pretty sure phi bonds don't figure in $C-C$ or $C=C$ bonds...

You need to make up your mind whether $\ce{C#C}$ or $\ce{C=C}$ bonds are shorter.

%s character of sp hybridised orbital is 50% and of sp2 it's 33%. More S-character means that the shared electron pair is closer to the carbon nucleus making it stronger and shorter.


text:
Is the compressibility factor smaller or greater than 1 at low temperature and high pressure?
The compressibility factor of a gas is defined as $Z = pV/(nRT)$. If attractive intermolecular forces dominate then $Z$ tends to be smaller than 1, and vice versa if repulsive forces dominate.
In the case where temperature is low, attractive forces increase due to less kinetic energy, but at the same time repulsive forces also increase due to high pressure. How can I determine which force will dominate?

If we use the Van-der-Waals equation then a plot of Z vs reduced pressure can be constructed. By using reduced values, i.e. pressure and temperature relative to the critical values then a general plot can be made on which data from many different molecules can be superimposed. Look under sections dealing with the Law of Corresponding states in your textbook. From such a plot then it is possible to determine whether the attractive or repulsive part of the potential is the dominant one.
The figure shows such a plot of Z vs. reduced pressure $P_R$ at different reduced temperatures. The reduced temperature is $T_R=T/T_c$ where T is the actual temperature and $T_c$ the temperature at the critical point.

From the figure when $T_R \lt 2$ and $P_R \lt 6.5$ then $Z \lt 1$ and this indicates that a real gas (in so far as such is described by the Van-der-Waals eqn) has a lower pressure than an ideal gas and so this indicates that the molecules are more influenced by the attractive part of the potential than its repulsive part, which is due to their finite molecular volume. If the temperature is higher, above $T_R \approx 2$ (not shown), then Z is positive for all pressures, and similarly above $Z \approx 7$ for all temperatures.  In these cases repulsion dominates over attraction.

Are these results for a specific gas at different temperatures or for different gases?

No the curves are found experimentally to  be general for many gasses as reduced values of temperature and pressure are used. Such gasses are  nitrogen, methane, ethane, water carbon dioxide and butane, etc. Have a look for 'compressibility factors' and see McQuarrie & Simon, 'Physical Chemistry' chapter 16; anf fig 16:10.


text:
Silicon organic chemistry
I've heard that in some alien planet there might be life based on Silicon chemistry, but have we ever discovered molecules that are similiar to carbon organic compounds? Something like chains or rings...
P.S. I'm not asking if silicon-based life is possible, i'm just asking if there actually are silicon organic compounds(in the sense that they are similiar to their carbon counterparts)

Yes there are chains and rings of silicon, though they are quite totally different from their carbon counterparts. All in all, the question is way too broad to be answered.

@Ivan The OP's asking if molecular silicon rings/chains have been synthesized (Yep). Though the first line in the post seems to be asking, "Is silicon-based life possible?", the OP doesn't explicitly request that. So...not all that broad ;)

Some silicon based molecules are similar to the carbon analogs. However, the full richness of organic chemistry that we observe as necessary for life on Earth is not expressed in silicon-based chemistry, so silicon-based life is extremely unlikely.

This should be well suited in worldbuilding.SE.


text:
Voltage needed for electrolysis of water
Reading the wiki for electrolysis it reads that a potential of 1.23V is needed to separate hydrogen from the oxygen. With that being said, how was that calculated? Does it change with the water volume and conductivity?

I'm assuming you mean this Wikipedia article on the hydrolysis of water in which the method of determining that potential difference is described farther down the page.
I'll address your other questions - what might cause this value to change.

Volume

No, except see below on concentration. The reaction potential for this electrolysis is independent of the amount of water since it is actually a thermodynamic function. Electrolysis may occur at different rates and take varying time depending on the amount of water, but the potential needed remains the same.

Conductivity

Conductivity is a measure of the ease of transmitting electrical current (electrons) through a substance. So, increased conductivity will increase the rate of electrolysis, but since it does not change the original or final substances in the chemical equation, does not change any thermodynamic values like the reaction potential.
However, there is a danger here. We typically increase conductivity by increasing electrolyte content. Some of these electrolytes could be electrolyzed at a reduced potential compared to water. For example, if you use sodium iodide as your electrolyte, the iodide ion is oxidized to the triiodide ion at -0.53 V:
$$\ce{3I- -> I3- + 2e-}\ \ E^\circ = -0.53\ \pu{V}$$

Concentration

The reaction potential -1.23 V is the standard reaction potential, which is determined when all applicable species are at 1 molar concentration and 1 bar pressure. The two half reactions in this case are:
$$\ce{2H2O -> O2 + 4H+ + 4e-}\\
\ce{4H+ + 4e- -> 2H2}$$
Thus, this potential is only valid at $[\ce{H+}] = 1\ \pu{M}$ or $\pu{pH}=1$. As the reaction proceeds, water is consumed and the volume decreases. The amount of protons remains constant, so the pH increases. As the pH increases, so too does the potential required and the electrolysis eventually reaches equilibrium and progresses no further unless you are providing an overpotential.
Note that -1.23 V is equally valid at $[\ce{OH-}] = 1\ \pu{M}$ or $\pu{pH}=14$ thanks to Hess's Law.

The -1.23 Volts also implies a very low current. Any appreciable current necessitates overpotential as well. // Current/Voltage also depends on size and type of electrodes.  // Practical considerations imply at least a crude relationship between cell volume and surface area of electrodes. In other words it is hard to have 1 m^2 area electrodes with 1 ml of cell volume.

@MaxW Let's take for example the electrodes I'll be using which will be made out of aluminum, how can you theoretically calculate overpotential?
Also, is it minus 1.23 Volts or close to that number?

@Tonakis2108 - Sorry, but you obviously are not getting the point. You will almost certainly need over-voltage. The needed over-voltage depends on a dozen nebulous factors.

@MaxW Sorry for my ignorance this is all new to me. I am trying to understand electrochemistry as much as I can since it's not the main subject in my college.I am sure I will need over voltage, the answer also stated that as pH increases. This far I know but can it be calculated also? Or do I give a random voltage over 1.23 V?


text:
Effect of relative strength on nature of solution
If a solution has equal no. of gram equivalents of a week acid (say acetic acid) and a strong base say NaOH , will the soltution be neutral or the relative strength has some role making solution slightly basic?

Note that the concept of gram equivalents was used in the 19th century. In general chemistry, it has been superseded by the concept of molar mass.

@Loong Wish you could get that point across to our education system...

The conjugate base of a weak acid is basic, whereas the conjugate acid of a strong base is not acidic. When a weak acid and strong base neutralize each other, the basicity of the conjugate base formed makes the solution basic.


text:
What can explain the more weight gain of aluminum coated steel than non-coated steel after NaCl immersion?
I have a 2 week experiment. I put aluminum coated and non-coated steel of the same size and shape in separate beakers with NaCl solution. Both are sealed. But, I am surprised that Al coated steel gained more weight than the non-coated steel after immersion. I know that the weight gain can be attributed to the formation of corrosion products. The measurements shows that the coated steel gained more weight, which be indicates more corrosion. But I'm sure that it doesn't since the liquid solution is still clear without any sign of rust. Unlike on the non-coated steel which has formation of rust that settles at the bottom.

I think it is galvanic corrosion where Al acts as sacrificial anode. So Al gets oxidised. This way you are passivating the underlying steel. In principle, the behaviour should revert. The originally Al coated sample is now protected by a mechanically strong layer of Al oxide should stop getting weight, and by waiting enough, the uncoated one should start losing weight. It is steel anyway, so its corrosion may take long.

In general, one experiment is not representative and you always want to do series of them to collect statistical data and calculate error margin, e.g. for the average mass deviation in your case. Also aluminium in direct contact with steel in seawater points to galvanic corrosion, the question is which steel it was and how concentrated solution was in order to estimate potentials for cathode-anode pair in solution. Most likely Al cathodically protected Fe, but I'm not sure. Staneless steel can also be a deal breaker.

But I have 20 replicates.

The nail coated steel  is heavier than the non-coated steel. But I can see that the non-coated steel is heavily corroded unlike the coated steel. Does it mean my gravimetric measurements is inconclusive?

Perhaps the non-coated steel corrosion products are either more soluble or less physically robust. So they go into solutions or fall off the uncoated nail.

Does it also mean that the aluminum in aluminum coated steel corrodes at a faster rate than steel alone?


text:
Explanation of Lemon Juice-Invisible Ink
A cute homegrown spy-trick we all know: lemon juice invisible ink. But there is no scholary article available in the internet about this phenomenon.
A rough explanation I found in Scientific American:

Lemon juice—and the juice of most fruits, for that matter—contains carbon compounds. These compounds are pretty much colorless at room temperature. But heat can break down these compounds, releasing the carbon. If the carbon comes in contact with the air, a process called oxidation occurs, and the substance turns light or dark brown.

But, this is not the type of explanation to be content with.
Question: What is the mechanism going on here? Please, support your answer with sufficient information and references, don't give a rough explanation.
Edit: I don't think my question is duplicate as I am asking for a well-referenced and informative answer which the other question doesn't have.

Surprisingly inaccurate explanation, indeed. I think it may be caramelisation reaction there.

@Mithoron Isn't that explanation suggesting caramelisation too?

This "explanation" is so vague and partially wrong that it's hard to tell if author really thought about caramelisation and dumbed it down inaptly or maybe about something else.

Lemon juice works as a sympathetic ink partially due to its caramelization, as @Mithoron profoundly noticed -- especially upon extreme heating.
Recent review of the degradation process of lemon juice [1] suggests that browning of lemon juice is associated with 3 processes:

ascorbic acid degradation;
Maillard reaction;
caramelization.

One of the first in-depth investigations revealed the major role of ascorbic acid degradation referring to the browning phenomena [2]:

Compared with other fruit products, lemon juice contains few natural pigments to mask.
...
Because of the high acidity (pH 2.5) of this product, it was unlikely that browning was due to sugar-amine condensation and the results showed that ascorbic acid was the main precursor. Browning of lemon juice and model systems was proportional to the level of ascorbic acid; the presence of amino-acids in model systems increased the intensity of browning.
The principal role of amino-acids in the non-enzymic browning of fruit products appears not to be in the initial reaction leading to the formation of reactive compounds, but to increase the browning potential after the oxidation of the ascorbic acid to reactive carbonyl compounds. Thus, the pathway for the build-up of carbonyls depends on the pH and nature of the product, but in all cases the formation of melanoidin complexes originates from the polymerisation of carbonyl and $\alpha$-amino-groups.

It was also confirmed that the $\alpha/\beta$-unsaturated carbonyls are potent browning agents and also that dicarbonyls of the glyoxal type make a contribution to browning in the early stages.
More recent investigation at elevated temperatures (up to $\pu{120^\circ C}$) also supports the major role of ascorbic acid degradation among other factors of browning [3]:

Browning of clariﬁed cashew apple juice was caused by the degradation of ascorbic acid. Changes in the absorbance at $\pu{420 nm}$ and the ascorbic acid levels during thermal treatment of clariﬁed cashew apple juice were
described by ﬁrst order kinetics, showing the correlation between ascorbic acid loss and colour formation (browning).

It is concluded in all mentioned sources that formation of intermediate undesirable carbonyl compounds (mostly furfural and 5-hydroxymethylfurfural (5-HMF)) and various products of their polymerization and combination with amino acids (furfural is an aldehyde) upon heating is one of the major factors of browning.

Maillard reaction is a chemical reaction between an amino acid and a reducing sugar, usually requiring the addition of heat. The reaction is initiated by a condensation between the free amino group of an amino acid, peptide, or protein and the carbonyl group of a reducing sugar, leading to the formation of Amadori compounds (various furoylmethyl derivatives in case of fruit juices)[4].

Caramelization generally occurs at high temperatures (above $\pu{140^\circ C}$), when the juice is dehydrated and has high sugar content. Basically it is a pyrolysis and a final step of decomposition.

To sum it up, these are the products of first 2 processes (ascorbic acid degradation and reaction with amino acids (Maillard process)) (from [3], referred to as browning markers):

Bibliography

Bharate, S. S.; Bharate, S. B. J Food Sci Technol 2014, 51 (10), 2271–2288. DOI 10.1007/s13197-012-0718-8.
Clegg, K. M. Journal of the Science of Food and Agriculture 1964, 15 (12), 878–885. DOI 10.1002/jsfa.2740151212.
Damasceno, L. F.; Fernandes, F. A. N.; Magalhães, M. M. A.; Brito, E. S. Food Chemistry 2008, 106 (1), 172–179. DOI 10.1016/j.foodchem.2007.05.063.
del Castillo, M. D.; Corzo, N.; Olano, A. Journal of Agricultural and Food Chemistry 1999, 47 (10), 4388–4390. DOI 10.1021/jf990150x.

What kind of carbonyl compound does ascorbic acid form?

@Mockingbird I slightly edited the answer. Long story short, it seems these are mainly furfural and its derivatives, which are also responsible for browning in baking process.


text:
Differential equation for the rate of unimolecular reaction
Given in my book is the following reaction mechanism for the unimolecular $\ce{A -> B + C}$,

$$\ce{A + M ->[k_1] A^* + M} \\ \ce{A^* + M ->[k_2] A + M} \\ \ce{A^* ->[k_3]  B + C}$$
Here $A^*$ is an $A$ molecule with enough vibrational energy to isomerize or decompose. In other words, part of kinetic energy of a bimolecular collision (between $A$ and $M$) has been used to raise an  $A$ molecule to higher vibrational energy.
Since $A^*$ is never present at very high concentration we can use steady state appoximination to obtain an expression for rate of reaction in terms of $[A]$ and $[M]$.
$$\dfrac{d[A^*]}{dt} = k_1[A][M] - [A^*](k_2[M] + k_3) = 0\tag{1}$$
$$-\dfrac{d[A]}{dt} = k_3[A^*]\tag{2}$$

I don't understand how the $(2)$ equation is obtained, in my opinion it should be $$-\dfrac{d[A]}{dt} = k_1 [A][M]- k_2[A^*][M].$$
I did not understand how we get $k_3$ instead of $k_1$ and why not $[M]$ in the rate law ?

Your equation for the rate of change of $[A]$ is incorrect. It should be
$$\frac{d[A]}{dt}=-k_{1}[A][M]+k_{2}[A^*][M].$$

To start with, the equation you wrote for the rate of change of $[A]$ is not correct. It should instead be
$$-\frac{d[A]}{dt}=k_{1}[A][M]-k_{2}[A^*][M].$$
Note that $[A]$ is consumed in the first reaction step and produced in the second. Now, rearrange equation $(1)$ to get
$$k_{1}[A][M]=[A^*](k_{2}[M]+k_{3}).$$
Plug in the above equation into the actual expression for $-d[A]/dt$ to get
$$-\frac{d[A]}{dt}=[A^*](k_{2}[M]+k_{3})-k_{2}[A^*][M].$$
This, of course, can be simplified to yield the desired equation $(2)$:
$$-\frac{d[A]}{dt}=k_{3}[A^*].$$

Sorry, that was a big typo on my part :).

I see you edited your original post. It's still not entirely accurate. Species $A$ is consuned when it reacts with $M$. As such, it should be $k_{1}[A][M]$ for that step if you're describing $-d[A]/dt$. Also, it is not usually suggested to change the content of your post in a way that changes the original question. If you want to update an equation you suggested, you can clearly note the change without overwriting the original. Just for future reference!

I will take care from next time. Since I think it does not affect the original question in any way I will leave as edited.


text:
Ease of reduction of carboxylic acid derivatives
Lithium aluminium hydride (LAH) is able to reduce aldehydes, ketones, carboxylic acids and esters. However, sodium borohyride ($\ce{NaBH4}$) is unable to reduce carboxylic acids and esters. Given that LAH is a stronger reducing agent than $\ce{NaBH4}$, it can be inferred that carboxylic acids and esters are more resistant to reduction than aldehydes and ketones. Why is this so?

Well, resonance comes to mind as basic rationalisation for this.

Mithoron, I'm not sure how resonance has a part to play here. Could you elaborate a bit further? Thank you.

Well, I guess otherwise you'd not ask... So called resonance is rationalisation for relatively low enthalpy of formation of compounds.  If compound is "lower on energy" then it is less reactive, because reaction provides also less energy.

So the reason why carboxylic acids and esters are less reactive is because of the stability on them conferred by resonance?

Less reactive in reactions, which break resonance.


text:
What makes for a good 'swimmer' in order see the filling level of Xenon gas in a vessel?
Most foams will probably get heavier due to diffusion. Is there glass balls this thin?

Even though Xe is 4.5 times heavier than air, won't it mix with it on a short time scale? In general, I suppose one can use emission spectroscopy to determine the amount of Xe in air (characteristic bands $\pu{467.13 nm}$ and $\pu{462.43 nm}$).

Polystyrene spheres? Small glass hollow spheres are also available.

I've seen balloons of ordinary air used in lab demonstrations to show the density of xenon, so that would likely work.

You can easily blow large, incredibly thin bubbles of soft glass with a propane torch or alcohol lamp -- so thin that interference patterns color them. Though fragile, it might be fun to make one as part of the demonstration. You'll need to experiment with surface-to-volume ratio to get it to float in Xe (you could also use halocarbon computer duster or $\ce{CO2}$, rather than expensive Xe). N.B. Use goggles, not just safety glasses, because the thin glass, when it breaks, lofts into the air... and possibly, your eye.
Rehearse, though, because removing the bubble from the tubing is likely to break it. After you get a large, thin bubble, aim the flame at the point where it connects to the tubing, and heat it until the tubing droops to a hair-thin tube. Do not melt this thin tube shut, because outside air-pressure will crush the bubble as it cools! Break that thin tube, leaving a short stem on the bubble. Yes, there will be some slow diffusion through the neck, but the bubble should last for weeks. If the demo is in a temperature-stable location, you might be able to seal the neck without the bubble busting.
There are videos showing lampwork here and here, but I suggest using soft glass (not borosilicate "Kimax" or "Pyrex") for your first bubbles, as it takes much less heat to melt.
BTW, a similar demonstration of density is the Cartesian diver, which also makes a good first student project in flamework:

With ~6 - 8 mm diameter soft glass, blow a bubble which can fit through the mouth of a plastic soda bottle.
Cut the bubble off with a triangular file, leaving a ~10 - 20 mm stem.
Optional: flame-polish the cut stem.
Dribble water into the (cooled) bubble until it just floats in a beaker of water.
Put the bubble in a filled plastic soda bottle, cap the bottle, and squeeze.

When I've had student 11-16 years old do this, it went exceedingly well (I actually had to remind them when lunch-break came that they needed to stop).

Do you think a swimmer could be made that would be held up by gas phase iodine at room temperature?

At room temperature? Perhaps a non-porous aerogel, created under near-vacuum... in a very hot room. The vapor pressure of iodine at 30 c is ~21 mm Hg. See https://en.wikipedia.org/wiki/Aerogel


text:
Why isn't hexaaquairon ion a hydrate?
What is the difference between hexaaqua(metal) ions $$\ce{Fe^{3+},Al^{3+}, Cr^{3+}}$$ and a hydrate? Since both of these absorb water into it own molecular structure. Would it be that hydrates have neutral charges where as these have positive charges? Why did we name Hexaaquairon ion it's name rather than Iron(3) Hexahydrate.

Maybe you meant hexaaquairon, not hexaauoiron? Also what do you mean by "absorb water into molecular structure"?

What andselisk says, and one is an ion, possibly in solution, and the other is a crystalline solid. I.e. hydrates are solids, your stuff would still need a counterion.

So it's a different name for a different state? But they are really the same thing.

No. Hexaaquaions are ions, hydrates are, well, hydrates of an ionic (usually ionic) compound.

So the difference lies in the charge. One of them is when the water molecules attach themselves to the metal ions where as the other it is attached to a molecule of neutral charge instead?

The point is that hydrates is a term that can be a little confusing. Hydrate can stand in chemistry as a material which contains water, and this can be merely formal, as in the case of carbohydrates.Even ions within their aqueous solvation cage are hydrated. Historically, and at least in my language it can still colloquially happen, many hydroxides were called hydrates. In our case, the confusion can only be made between water- coordinating ions, and salts containing crystallisation water, such as CaSO4 × 2 H2O. Properly hydrates refer to the latter case.


text:
Why only molecules with three or more atoms undergo unimolecular reaction?

Only molecules with three or more atoms undergo unimolecular reactions. A diatomic molecule cannot dissociate in this way because it has a single mode of vibrational freedom. If this mode is excited by an amount equal to dissociation energy, the molecule dissociates in $10^{-12}$ seconds.

What is the connection between the dissociation of diatomic molecules and vibrational degrees of freedom ?
I think this is a false statement because molecular hydrogen dissociates into hydrogen atoms, even its dissociation energy is given alongside other diatomic molecules. Am I missing something ?

So according to you, the photodissociation of dimolecular oxygen to yield atomar oxygen, which eventually offers access to ozone, in the upper parts of terrestial atmosphere, is not unimolecular?

Just consider that molecularity of a reaction equals to the sum of the stoichiometric coefficients of reactants...

I'm guessing this quote is given in the context of classical transition rate theory. Although the statement may seem very strong at first, it is generally true but has some important caveats.
Energy Landscapes
Lets consider a diatomic molecule.  It's geometry can entirely be described by one degree of freedom - the bond length.  As we move along that degree of freedom, stretching or compressing the bond we can develop a surface of the energy of the bond at different distances.  A common model used to approximate this surface is the Morse Potential.

These surfaces are often call "energy landscapes".  Minima in the energy landscape are stable structures - our starting materials and different products.  If you push them a bit they move back down the surface to where they started.
And maxima are our transition states between minima, push a little and they'll move down one side of the barrier or the other and into a minima.
Also, a system can easily explore the parts of the energy landscape with less energy than the system currently has - the "classically allowed region" - like a ball rolling up and down the side sides of a valley.
Over time a system tends to move along the energy landscape to positions of lower energy if allowed time to equilibrate and release energy to the outside - like a ball rolling to rest at the bottom of the valley, losing energy to friction.
So normally, below the dissociation energy, the system will roll around in the bottom of the energy landscape, vibrating the bond.
In classical transition state theory, a reaction happens when the atoms move over the energy landscape  along a particular degree of freedom or group of degrees of freedom.
In order for a diatomic molecule to dissociate, it needs to have enough energy moving along the bond vibration degree of freedom to explore parts of the energy landscape at large $r$.  Which means it has to have a lot of energy in the vibrational mode as mentioned in your quote, an amount about equal to the dissociation energy.  If it mode has that much kinetic energy though then, as mentioned above the molecule will fall apart incredibly quickly.
Changing Energy Landscapes
However, rather that adding thermal energy we can change the shape of the energy landscape.  Many different energy landscapes exist at different energies.  the Difference between different landscapes is the arrangement of the electrons in the molecule.
By shining high energy light at a molecule, you can excite the electrons in it into a different configuration of orbitals, leading to a different shape .  This is what happens in photodissociation.  You excite the molecule into a different electronic configuration, and onto different potential energy surface.  On the new surface the molecule is no longer sitting in a minimum, but near a maximum and the system releases energy by moving down the surface - this time towards extending the bond length and dissociating, rather than towards a stable minimum.

The statement is essentially correct. Unimolecular reactions refer to the Lindemann scheme and its improvements called RRKM theory. The Lindemann model is $\ce{A + M <=> A^*  +M ;\; A^*\rightarrow P}$ where $\ce{A^*}$ is an energised molecule and P product. M is an inert gas or another A type molecule that gives up some of its energy in a collision to form $\ce{A^*}$ and this energised molecule can only react by transferring its internal energy into another normal mode. A classic example is the reaction $\ce{CH3NC \rightarrow CH3CN}$.
In a diatomic molecule there is only one mode so when energised to $\ce{A^*}$ it dissociates directly and rapidly ($\lt 10^{-13}$ s)  without being able to loose its energy to reform A, thus there cannot be an equilibrium set up as is proposed by the scheme above.


text:
What is the reaction rate constant of an effective reaction?
Consider the reactions
$$\ce{A + B ->[$k_1$] C }$$ and
$$\ce{C ->[$k_2$] E + F}$$
with reaction rate constants $k_1$ and $k_2$. I know that the two reactions can be written as follows
$$\ce{A +B->E +F}$$
with a rate constant $k_3$. How does $k_1$ and $k_2$ relate to $k_3$? Does it matter how many species are in the reactants (such as if the first reaction was simply $\ce{A->C}$)?

It is not clear that they are related. It is simpler to consider the equivalent scheme $A \rightarrow C \rightarrow E$ with rate constnats $k_1,\, k_2$ then the decay of A is
$A_t/A_0= \exp(-k_1t)$ the amount of  C rises and falls; it is
$$\displaystyle C_t/A_0 = \frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$ and
E is
$$E/A_0 = 1-A_t-C_t=1-\exp(-k_1t)-\frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$
When $k_2 $ is very large and much bigger than $k_1$ then $\exp(-k_2t) \rightarrow 0 $ and $\displaystyle E/A_0= 1-(1+\frac{k_1}{k_2})\exp(-k_1t)$ and in the opposite limit $k_1 \gg k_2$ then $E/A_0= 1-\exp(-k_2t)$.
So you can choose which limit is important. In either case the amount of E rises as time proceeds but with different rate constants. In the intermediate case there seems to be no rate constant but only a slope to the curve which will depend on the time at which the measurement in made. You could for example find the half rise time and try to relate that to rate constants, but it looks messy.

Can you explain how do you come with the first equation which describes the concentration of $[C_t]$  in terms of $k_1$ and $k_2$ as a function of time?(it will be a complete answer you describe that too.)

Write down $dC_t/dt =k_1A_t-k_2C_t$ and substitute for $A_t$ as given and solve the equation. At $t=0,,C_0=0$.


text:
Molecular orbital diagram for BF3
I'm trying to build a molecular orbital diagram for BF3 and I'm running into problems with irreducible representations on the F side.
2s for B has an irreducible representation of A1. 2p for B has an irreducible representation of E' and A''2.
2s for F considered non bonding
2p (along the bond axis) for F has an irreducible representation of E' and A'
2p (z) for F Has an irreducible representation of E'' and A''2
2p (the remaining one) has an irreducible representation of E' and A'2
From this I built A1 bonding and antibonding orbitals, A'2 bonding and antibonding orbitals, 1 A'2 and 2 E''2 non bonding orbitals.
I'm now stuck because I have 4 E' on the F side and only 2 E' on the B side.
I found the following diagram for BF3 online but it doesn't generate the E' anti bonding and also doesn't generate enough molecular orbitals.

Does anyone know how to mix 4 E' with 2 E'

Related: http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf

'Full disclosure': this is quite nicely explained in Albright's Orbital Interactions in Chemistry (as are many similar questions).

I think you are most of the way to the answer, but I will start the process from scratch for the sake of a full explanation for future readers.
First, we determine the symmetry group of $\ce{BF3}$, which by quick inspection we can determine to be $D_{3h}$ (since it has a principle $C_3$ axis, $3\perp C_{2}$ axes, and a horizontal mirror plane of symmetry).

Knowing this, we can now make use of the $D_{3h}$ character table (thank you @orthocresol for compiling these in mathjax format) to determine the irreducible representation of the atomic orbitals (or their symmetry adapted linear combinations SALCS) within the group.
$$\begin{array}{|c|cccccc|c|c|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & \text{linear/rotations} & \text{quadratic}\\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy) \\
\mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$
Using this, we can quickly determine the irreps of boron's orbitals using the 2nd column from the right.

$\ce{B}$: $2s=a_1'$, $2p_x,2p_y=e'$, and $2p_z=a_2'$.

Next, we will determine the irreps for the $3\ce{F}$ atoms, which will require that I consider combinations of $s,p_x,p_y, \text{and } p_z$ orbitals.
Below, I show how the symmetry operations of the group affect each set of orbitals and show the orbitals that result (image source).
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_s & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_x} & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{12ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_y} & 3 & 0 & -1 & 3 & 0 & -1 \\ \hline
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_z} & 3 & 0 & -1 & -3 & 0 & 1 \\ \hline
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0  \\ \hline
\end{array}$$
$\hspace{21ex}$
These reductions to irreducible representations can be done via the reduction formula $$n_i=\frac{1}{h}\sum_c g_c \cdot \chi_{i_c} \cdot \chi_{r_c}$$ where $n_i$ is the number of times a particular irrep $i$ occurs in the reducible representation, $g_c$ is the number of symmetry operations $c$, $\chi_{i_c}$ is the character of the irrep $i$ for the symmetry operation $c$, and $\chi_{r_c}$ is the character of the reducible representation $r$ for the symmetry operation $c$.
This allows us to write the irreps of the three $\ce{F}$'s orbitals.

$3\ce{F}$: $a_1'+a_2'+a_2''+2e'+e''$

With all the formalities out of the way, we can construct the qualitative MO diagram for $\ce{BF3}$. Like much of this post, my source for the MO diagram is the pdf of lecture notes for a Dartmouth chemistry course located at http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf

Working our way up through this diagram, we see that we start by assuming that the SALCs of the $s$ orbitals of the three $\ce{F}$'s are assumed to be low enough in energy where they do not interact with the orbitals of $\ce{B}$.
The next point of interest is the $e' (\sigma)$ MO, which we form from the $p_x$ SALC of the $\ce{F}$ atoms and the $p_x$ and $p_y$ of the $\ce{B}$. We could additionally consider the interaction of the $e'$ $p_y$ SALC to form a $\pi$ bonding interaction, but we will assume that, with the $\sigma$ interaction already present, this $\pi$ bonding interaction will be weak and thus can be neglected.
This previous point leads us to have a set of nonbonding orbitals $a_2'+e'(y)+e''$. We then also have antibonding orbitals for each of the bonding orbitals below.
As a check on our work, we can compare this to the somewhat more quantitative diagram given in Principles of Inorganic Chemistry by Brian PFennig using the Student Editon of Spartan:

The only noticeable difference here is that the nonbonding MOs are now placed with respect to their computed energy rather than being lumped together. The classification of orbitals as bonding, nonbonding, and antibonding seems to fit with the diagram we have above, with the caveat that this more quantitative diagram does suggest some $e'(y)$ bonding contribution, as its energy is somewhat lower than the p orbitals it came from.
Now that we have the correct diagram, we can then compute the bond order for each $\ce{B-F}$ using: $$\frac{\text{Bonding electrons } -\text{Antibonding electrons}}{2\cdot\text{number of bonds}}$$ which for this diagram gives a value of $\frac{4}{3}$ per bond. This gives a sense of why $\ce{BF_3}$ is a strong Lewis acid, since it can accept a pair of electrons into its antibonding LUMO and still leave a compound with a BO of $1$ per bond.
As a final point, it is worth noting that this method of classifying orbitals is far from perfect and in general we should really look at individual bonds to classify if the orbital is anti/bonding with respect to that bond. However this energy based approach is what is commonly taught in school and happens to work well in this simple case; just be aware that it can fail for even seemingly simpler molecules like $\ce{CO}$.

That is a really good explanation. I personally dislike the classification of bonding, non-bonding, and anti-bonding according to energy, because it's not really that rigorous and sometimes leads to wrong conclusions (point in case CO), but it often works, as it's here the case. I personally would classify the 3e' orbitals as anti-bonding according to symmetry with respect to at least two bonds. The same would be the case for 4e', but that's a discussion for another time.


text:
Can I eat the lemons after I have used them as a battery?
After completing the lemon battery experiment, can I eat the lemons when I'm done?
More broadly, what will happen to the lemons? What are the chemical reactions, what chemicals might be present/missing after running my lemons as a zinc/copper battery? Would it be poisonous, or just unpleasant?

Hmm, a single down vote amongst the ups. Feedback greatly appreciated if available :)

Wasn't me but I'll bet somebody thought there's no way this could be a good idea. The first answer reveals it's nowhere near that cut and dried, so I don't care for the downvote either.

In a lemon battery, the zinc from the galvanized nail is giving up electrons and transitioning into an aqueous state:
$$\ce{Zn → Zn^2+ + 2e−}$$
While zinc is entering the electrolyte, two positively charged hydrogen ions ($\ce{H+}$) from the citric acid with two electrons at the copper electrode's surface and form an uncharged hydrogen molecule $\ce{H2}$:
$$\ce{2H+ + 2e− → H2}$$
While zinc is a mineral our bodies need, I wouldn't recommend using this as a source of zinc. Also, zinc toxicity is a thing, so without knowing the dose you're taking, you risk poisoning yourself.

Lemon Battery Wikipedia

Unless you manage to dissolve a significant part of your zinc electrodes into the lemons, zinc toxicity is probably not a serious concern. As the Wikipedia article you linked to notes, zinc supplements containing over 0.15 grams of zinc per day have been commonly used with no lasting harmful effects. That's equivalent to about half a square centimeter of 0.5 mm zinc sheet entirely dissolved into the lemon. Mind you, I probably still wouldn't eat the lemons myself; if nothing else, I suspect they may taste pretty nasty.


text:
What, if any, are the differences between hemiwicking and wicking?
I'm encountering these two terms, but their uses seem to be different, and there's little documentation out there defining the terms.
Here is an example definition I've found of Wicking:

Wicking is the spontaneous flow of a liquid in a porous substrate, driven by capillary forces.$^1$

Here is an example definition I've found of hemiwicking (notice little difference between the one above):

Hemiwicking refers to the spreading of a liquid on a rough hydrophilic surface driven by capillarity.$^2$

Below is a typical depiction showing Wenzel wetting (a), Cassie-Baxter wetting (B), and hemiwicking(c).$^3$ My assumption would be that Wenzel wetting would actually be an example of wicking?

http://journals.sagepub.com/doi/abs/10.1177/004051759606601008
http://fluids.snu.ac.kr/Publication/hemiwicking.pdf
http://jss.ecsdl.org/content/3/1/N3095/F3.expansion.html

I'm not especially familiar with the detailed differences between the two much beyond the definitions you've posted here, but it seems to me that:

Wicking occurs in the bulk volume of a porous substrate, not just on the surface; whereas
Hemiwicking occurs strictly on the surface of an otherwise impermeable substrate

As you note, in both cases the fluid movement is driven by capillarity, however.

My assumption would be that Wenzel wetting would actually be an example of wicking?

I disagree. All three panels of the figure you've posted look to me as though they are trying to illustrate strictly surface phenomena, and thus none of them are forms of wicking.  The main difference between (a) and (c) is that in (a) the structured surface is wetted only immediately adjacent to the liquid drop, whereas in (c) the surface is wetted in an area beyond the extent of the drop. Thus, the Wenzel wetting is neither wicking nor hemiwicking.


text:
How do reactant proportions affect the products of a chemical reaction?
Is it correct that if you alter the proportions of the reactants of many chemical reactions, especially step-wise, the products will change?
Here is an example, please tell me if it is right or not:
\begin{align}
\ce{KOH + H2SO4 &-> KHSO4 + H2O}\\
\ce{2KOH + H2SO4 &-> K2SO4 + 2H2O}
\end{align}

Just because you can write a different chemical reaction given different stoichiometric numbers does not mean it is more (or less) likely to occur if you provide a different concentration of the reagent(s). I could write the a completely ridiculous - but balanced - chemical equation. It doesn't mean it will ever happen. If you wish to know more about which reaction will predominantly occur, you need to know more information about the thermochemistry and kinetics!

Chemistry is much more complicated than just writing balanced chemical equations. In your case, the reaction is in equilibrium. It would be all much clearer if you took a couple classes of general chemistry.
But if you really want a clear cut answer: since this is an acid-base neutralisation, you would be right. With one equivalent of $\ce{KOH}$, you would find mainly $\ce{KHSO4}$ (with traces of $\ce{KOH}$, $\ce{H2SO4}$ and $\ce{K2SO4}$, which would be mainly dissociated ions anyway.) By adding more, you would shift the equilibrium toward $\ce{K2SO4}$.


text:
Does deuterated water slow down the overall metabolism of a cell?
Would deuterated water, being heavier, slow down the metabolic rate of the cell and subsequently the aging process?
edit: lets say I wanted to observe a cellular event, like the formation of the blsatopore or ameboid motion, could I soak a cell in a solution with D2O without killing it or severely damaging it ?

Kinetic isotope effect

Yes it would, by a few percent. It may or may not be a goal worth pursuing, but there is more to it.
Different reactions would be slowed down to a different extent. Tiny as they are, these discrepancies suffice to disrupt the delicate biochemical machinery of the living cell. No life form more advanced than bacteria is OK with that. Deuterated water in high concentrations is unpleasant for plants and downright lethal for animals, including us.
So much for your slowed aging.

wait, why would it be lethal for animals? its not like it admits radioactive decay?

@Confusedbyeverything, different reactions are slowed down by different amounts.  Bacteria can handle this; humans can't.

To add a bit of nuance, you need quite a bit of deuterated water to notice this. In the short term, drinking 5 liters of D2O is about as harmful as drinking 5 liters of H2O. You need to consume D2O over a longer period to replace a significant fraction of the H2O in your body before the specific effects become noticeable.

Out of curiosity, could you add some references for the effect on deuterated water on plants and animals? In particular, what is the typical cause of death?

Yes, the kinetic isotope effect is the main reason due to the differential lowering of the zero point energy in reactants and transition state, which has the effect of increasing, slightly, the activation energy. However, this effect is small, a few percent in a single reaction. The reason that deuteration has an effect overall is due to the fact that biochemical pathways involve many intermediate steps each of which has a small isotope effect. The total effect is multiplicative.


text:
Which is the stronger base?
I would like to know which of the two is the stronger base, tri-(n-butyl)amine or tri-(tert-butyl)amine?
I feel that the one with tert-butyl would be stronger as the inductive effect of tert-butyl on the nitrogen atom is stronger, making the nitrogen atom more electron-rich and better able to act as a base. This is because the carbon atoms in tri-(tert-butyl)amine are closer to the nitrogen atom, with the furthest carbon just 2 carbons away. However, for the other amine, the furthest carbon is 4 carbons away. I am not sure if my reasoning is flawed. Could someone clarify?
I am also not sure if steric hindrance is an important consideration here, considering the bulkiness of the tert-butyl substituent compared to the n-butyl substituent.
Also, pardon me if I had written the names of the chemicals wrongly.

Your reasoning is fair. This seems to list most of the amines and its dissociations with some useful info However don't forget any steric hindrance that could degrade the (alkyl contributed) electrons releasing ability of tert-alkyl amine's N (because alkyl is larger than H+ atom). Last but not least  solvation of ions is a must to consider due to the ability to form Hydrogen bonds. a bit like catch 22 case :P

Tri-(tert-butyl)amine has never been synthesized, so this is more of a theoretical study.

I learned that Hünig's base was the strongest one available. $\ce{N(tBu)3}$ is probably too crowded to even exist.

@Martin-マーチン Assuming if it exist. This question has actually been taken off an inorganic chemistry textbook. However, the answer was not provided. Therefore, I had to pose this question on this platform.

@Martin-マーチン A friend of mine mentions that steric hindrance only affects kinetics. Since I am only considering thermodynamics, in terms of equilibrium, can I safely put it out of the discussion? Or should I still consider it?


text:
Why does hydration of alkene takes place in presence of an acid such as H3PO4 or H2SO4?
Why acid is used? Can’t the hydration be done without acid? If not, why?

What do you know about the mechanism of the hydration?

The acid is used as a catalyst. As can be seen from the mechanism below, the electron-rich double bond first attacks the positively-charged acid. The positively-charged hydronium ion is important because it is rather electrophilic and allows the nucleophilic pi bond to attack it, breaking the pi bond. This serves to initiate the reaction. Without an acid catalyst, the neutral water molecule would not be electrophilic enough for the pi bond to want to attack it, thus the reaction would not proceed.
(from http://slideplayer.com/slide/11423361/)

Great answer! It's worth mentioning how this electrophilic addition (hydration) works under acid catalyst by reducing the activation energy and abiding by Markovnikov rule as well. (This could be useful in comparing, the stability of carbocations when looking at primary, sec and tert-alkenes' hydration reactions).

Simple answer : In the presence of an acid the water tends to behave as a base rendering hydrogen ions which can be attacked by the pi electrons. The rate of hydration would be way slower if acid is not used as there are lesser number of H+ ions.


text:
Acidity of aluminium trihalides

Would the acidity of the aluminium trihalides be the same as that for the boron trihalides?

The interesting thing about the acidity of the boron trihalides is that they actually increase in acidity from fluorine to iodine. This is due to the presence of stronger pi bonding in boron trifluoride compared to the other boron trihalides, which means that stronger partial pi bonds would have to be broken when boron accepts an electron pair from a base.

Would we observe this trend in acidity in the aluminium trihalides as well?
Based on the same reasoning, we should be able to right?

I have chanced upon the answer in an inorganic chemistry textbook (Essential Trends in Inorganic Chemistry by D. M. P. Mingos) I was reading. It turns out that this trend only holds for the boron trihalides and does not apply to the aluminium trihalides. However, the book did not provide the explanation for the exception of boron.
My thinking is that perhaps, backbonding is not present in the aluminium trihalides, possibly due to incompatibility of the p atomic orbitals for pi bonding between aluminium and the halogen atoms. Does anyone else have any insights to share?

Aluminium trichloride typically exists as a dimer or as a polymer (or so I've read). This would suggest that the orbitals to too "rigid(?)" to bond properly, and intermolecular bonding is preferred.


text:
Acidified sodium sulfite in the reduction of iron(III) to iron(II)
Why does sodium sulfite have to be acidified in the reduction of $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$?
Does the acid act simply as a catalyst or is it necessary for the reduction process?

You'd have some problem with reducing precipitate...

Please help me pls


text:
What defines the oxidizing/reducing power of a coordination compound?
For example, I came across a problem in which I was asked to determine whether the complex $\ce{[Cr(H2O)6]Cl2}$ is an oxidizing or reducing agent.
How should I proceed? My teacher told me its the EAN number that defines this. What are your views?

I was looking at some revision lately on this. My textbook/other refs indicated a bit about standard electrode potentials as a reference in determining a species as an oxidant or reductant. Given that there's Cl and a metal Cr, I would presume this is a reductant (note: reducing power decreases in this order: Na > Zn > Pb > H2 > Cu > Ag > Br- > Mn2+ > F-) and S-block metals are good reductants. Why not take a look at reactions (both acidic and basic) which involves this agent?

Yep your approach is also right. I was thinking that since the Cr here is in +2 oxidation state it has an electronic configuration - [Ar] 3d^4 so it can do two things -

It could gain an electron to achieve a stable [Ar] 3d^5 configuration or

It could lose an electron to have half filled t2g subshell as per crystal field theory

I just did a quick search online : coordination compounds redox, it looks like coordinated compounds' redox are of a different level altogether. I may not have sufficient knowledge to address that aspect at all. Keen to know an answer you get.

Pardon my ignorance, but what is EAN? I cannot imagine how this EAN has anything to do with redox processes. Is it effective atomic number $Z_\text{eff}$? But then there is "number number", so I'm lost.

See the logic with EAN here is the transition metal atoms tend to have the exact number of electrons as present in the noble gases of their respective periods. For example a 3d series atom  would like to have an EAN number of 36 since that is the number of electrons present in Krypton. In order to do so it could gain or lose electrons to make its Effective Atomic No. 36.