text:
How can a thermometer ever show the actual temperature of an object if the object loses heat to the thermometer?
We know that when two objects are placed in contact with each other, after a period of time, the two objects will have the same temperature. Thus, if a hot body comes into contact with a relatively cold body, it will lose its heat and its temperature will reduce, while the temperature of the 'cold' body will increase.
If I put a thermometer in an object to measure its temperature, the body will lose its heat (or gain the heat, if the thermometer is hotter) until it reaches the same temperature as the thermometer. Doesn't this mean the thermometer is not giving us the EXACT temperature of the object, i.e the temperature before the measurement was taken, and is in fact, showing us a lower/higher temperature?

Thermal reservoir

The obvious answer is that it doesn't because any observation will change the state of the system you are observing. The only way to observe the system is to interact with it. But good observations don't change the system much, so your measurement is still meaningful.

An IR thermometer (or pyrometer, or...) detects the infrared given off by a warm body. Since that energy is being radiated regardless of the thermometer being present, the temperature measurement has no effect on the item being measured (although that item is, indeed, losing energy through radiation).

Shades of Heisenbergs' uncertainty principle.

Related: https://chemistry.stackexchange.com/questions/121197/temperature-measurement/121228

Lengthy discussion on meta surrounding original intent of OP: https://chemistry.meta.stackexchange.com/questions/4759/yet-another-closure-question

Welcome to the Uncertainty principle, a fundamental principle of knowingness, not merely measurement.

I'll start by mentioning that there's no such thing as an exact measurement—there is always some measurement error.  The only observations that can be numerically exact are counted numbers of discrete objects (e.g., the number of electrons in a neutral carbon atom is exactly 6).  And I say "can be", because if the numbers are sufficiently large, even with counted numbers there can be errors (we see this with vote counting).
Having said that, your idea is correct—if the heat capacity of the temperature probe is significant relative to that of the object being measured, then the measurement can significantly change the temperature of your object.
This is a particular concern when measuring small objects.  For such applications, researchers can employ (as Jon Custer mentioned in his comment) non-contact thermometry.  See, for instance:
https://www.omega.co.uk/temperature/z/noncontacttm.html

I somewhat disagree with "only observations that can be numerically exact are counted numbers of discrete objects", sadly this is not true in spectroscopy. When you are counting photons, there is always an uncertainty, called the shot noise. It is a fundamental limit.

@M. Farooq.  The statement is correct, since I was explicit in explaining that being countable is a necessary but not suffiicent condition for exactness.  Indeed, I even gave votes as a specific example of countable items whose number is nevertheless uncertain.  Counting photons could be another example.

@theorist, Errors in counting votes could be human or machine error. They can be counted exactly. However counting photons by any detector is another story. It is limited by shot noise. This is a fundamental limit. Basically, when we start counting discrete events $N$, the more we count the larger is the error. The error grows as square root ($N$). Hence the signal to noise ratio improves with larger number of photons being counted.

@M.Farooq My statement is still correct:  Being countable is a necessary but not sufficient condition for an observation to yield an exact number, whether it's votes or photons.  Not sure why you're arguing otherwise.  It almost seems like it's not that you're disagreeing with my statement, but rather that you're not understanding it -- i.e., that you're not understanding what "necessary but not sufficient" means.

By the way, it is not entirely correct to assume that photons behave as discrete objects.  Counting the photons is essentially counting the number of times the detector has absorbed portions of the electromagnetic field energy sufficient to free a bound electron.  Assuming that each portion is carried by a separate particle is a somewhat convenient metaphor which is not necessarily always correct.

For the question to have a concrete answer you would have to explain in more detail how the temperature is to be measured and how the thermometer and sample interact with the surroundings. If you use a contact thermometer properly then it follows by the zeroth law of thermodynamics that the measurement recorded by the thermometer is the temperature of the sample. If equilibrium is established between thermometer and object or sample (let's call that the system), and in addition the system is thermally isolated (adiabatic) or thermally equilibrated with the surroundings, then the thermometer shows the exact temperature of the object.
A very simple example: if you measure your body temperature with a thermometer, will the temperature reading be affected by the heat capacity of the thermometer, or by equilibration? Is transfer of heat going to be a problem? Not if you wait for equilibration to complete, and for your body (the thermostat) to compensate for any transfer of heat to the thermometer and accompanying (if small) temperature change in your body. Many instruments and experiments aim to approximate thermal equilibrium as closely as possible by using insulation and/or a thermostat and/or large constant temperature heat reservoir.
In an equilibrium  scenario the effect of the thermometer depends on experimental design, for instance whether the sample plus thermometer represent an isothermal or adiabatic system. As succinctly suggested in a comment, for the thermometer to report the temperature of the sample without altering it, it suffices that the experiment be carried out isothermally. This in turn requires a thermostat (e.g. thermal reservoir or heat bath) at the desired temperature, in contact with the system (including thermometer). If the reservoir is sufficiently large (has a very large heat capacity) compared to the system, or has a regulated temperature (a thermostat), then heat transfer between system and reservoir will not significantly alter the temperature of the latter once equilibrium is established, and the temperature of the system can in turn be expected to be that of the bath (surroundings). The thermometer might significantly alter other thermal properties of the system, but by careful design it should be possible to keep the system at a fixed temperature such that the thermometer accurately reports an unaltered temperature of the system. If the system is properly thermostated one might instead rely on a measurement of the temperature of the heat bath rather than directly measuring the temperature of the system. However, measuring the temperature of the system can serve to verify that it is in thermal equilibrium.
You can of course attempt to estimate the temperature of an object in a steady state situation in which heat streams through a system (consisting of sample plus thermometer), that is, there is a temperature gradient in the system. That is not an uncommon experimental setup, but is not a situation of true thermodynamic equilibrium, and is accompanied by various problems, for instance the possibility that the temperature recorded by the thermometer differs from that of the sample.

Admittedly T measurement is not trivial. Equilibration can be a hassle. There is an introduction in the Wikipedia  to the situation when a clinical thermometer is used, referencing the following: https://academic.oup.com/qjmed/article/95/4/251/1562440

With temperature measurement, the entire system must be considered = that includes, the object whose temperature is to be determined, the measuring equipment, and the environment.  In extreme situations all of these must be considered.
With a radiation sensor, there are at least 3 components that can affect the indicated temperature - radiation between the target and the sensor, between the sensor and the surroundings, and between the target and the surroundings.  For example, when measuring  furnace tube temperature, the pyrometer reading will depend on the total energy incident on the measuring head, less the energy radiated by the head to the target and environment.  The incident energy will comprise an element directly emitted by the tubes and dependent on the tube surface emissivity, and also a component due to energy emitted by the heat source (the flame) and reflected from the tube surface - approximately (1 - emissivity).  There may also be some effect due to cooler parts of the system which are in the line of sight of the head. As an example, in one installation an indication of around $\pu{1000  ^{\circ}C}$ was about 40 deg high due to the impact of radiation.
See for example the section on "Reflected radiation" in http://www.iffcokandla.in/data/polopoly_fs/1.2498172.1438009753!/fileserver/file/514468/filename/Aiche-33-006.pdf.


text:
Do commuting operators always share the same eigenfunctions?
Do commuting operators always share the same eigenfunctions? Could you kindly explain to me if this is true or false and examples for this, as I am a little confused.

Related: https://math.stackexchange.com/questions/1227031/do-commuting-matrices-share-the-same-eigenvectors

i'm still confused?

as a quick summary, no. This is only true if both operators are diagonalizable (can be transformed such that they form a diagonal matrix of the eigenvalues). A common subcase of being diagonalizable is when there are no repeated eigenvalues. Some operators with repeated eigenvalues can be diagonalized, but this is a more involved proof.

If two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators.
However, if one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator.
As an example consider the states $|l,m\rangle = |1,1\rangle$ and $|l,m\rangle = |1,-1\rangle$. These are both eigenvectors of $\hat{L}^2$ and $\hat{L}_z$ (the angular momentum operators).
The eigenvalues of $\hat{L}_z$ are $\hbar$ and $-\hbar$, respectively, but the state
$$|1,1\rangle + |1,-1\rangle$$
is clearly not an eigenvector of $\hat{L}_z$, but it is still an eigenvector of $\hat{L}^2$. This is probably why we use the term "mutual" rather than "same". Finally, we form linear combinations of the $|1,m\rangle$ states to get the $l=1$ states that are eigenvectors of, say, $\hat{L}_y$.
As a straight-forward example, consider the following two matrices:
$$L = \left[ {\begin{array}{ccc}
0 & 0 & 0 \\    0 & 1 & 0 \\    0 & 0 & 1
\end{array}}
\right]$$
and
$$Z = \left[ {\begin{array}{ccc}
0 & 0 & 0 \\    0 & 1 & 0 \\    0 & 0 & -1
\end{array}}
\right].$$
The vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are eigenvectors of both operators. The latter two are eigenvectors of $L$ with the same eigenvalue (namely $1$), but they are eigenvectors of $Z$ with different eigenvalues (namely, $1$ and $-1$). If we instead use the vectors $(0,1,1)$ and $(0,1,-1)$, these are still eigenvectors of $L$ with eigenvalue $1$, but they are no longer eigenvectors of $Z$, as you can verify.

Let operators $\hat{A}$ and $\hat{B}$ commute, $[\hat{A},\hat{B}]=0$, and consider the eigenvalue equation for $\hat{A}$:
$$
\hat{A}|\psi\rangle=\lambda|\psi\rangle.
$$
The first thing to prove is that $\hat{B}|\psi\rangle$ is also an eigenstate of $\hat{A}$ with the same eigenvalue. To do this:
$$
\hat{A}(\hat{B}|\psi\rangle)=\hat{B}(\hat{A}|\psi\rangle)=\hat{B}(\lambda|\psi\rangle)=\lambda(\hat{B}|\psi\rangle).
$$
If $\lambda$ is a non-degenerate eigenvalue, this means that $\hat{B}|\psi\rangle\propto|\psi\rangle$ (as for non-degenerate $\lambda$ all eigenstates must be the same up to a constant), so we can write:
$$
\hat{B}|\psi\rangle=\mu|\psi\rangle,
$$
where $\mu$ is the proportionality constant. This shows that $|\psi\rangle$ is also an eigenstate of $\hat{B}$ (with eigenvalue $\mu$). So commuting observables share the same eigenstates if the associated eigenvalues are non-degenerate.
If $\lambda$ is degenerate, then $|\psi\rangle$ need no longer be an eigenstate of $\hat{B}$. To see this, let us have $g$ linearly independent eigenstates, $|\psi^i\rangle$ for $i=1,\ldots,g$, with the same eigenvalue $\lambda$. Construct an arbitrary linear superposition within the $g$-fold degenerate subspace:
$$
|\chi\rangle=\sum_{i=1}^gc_i|\psi^i\rangle
$$
for arbitrary coefficients $c_i$. Then,
$$
\hat{A}|\chi\rangle=\hat{A}\left(\sum_{i=1}^gc_i|\psi^i\rangle\right)=\sum_{i=1}^gc_i\hat{A}|\psi^i\rangle=\sum_{i=1}^gc_i\lambda|\psi^i\rangle=\lambda\left(\sum_{i=1}^gc_i|\psi^i\rangle\right)=\lambda|\chi\rangle.
$$
This means that any linear combination within the degenerate subspace is also an eigenstate of $\hat{A}$ in that subspace with the same eigenvalue $\lambda$. In turn, this implies that all we can say about $\hat{B}|\psi\rangle$ is that it belongs to this $g$-fold degenerate subspace. This means that $\hat{B}|\psi\rangle$ could be proportional to the result of one of these linear combinations rather than one of the original eigenstates, showing that for degenerate eigenvalues the eigenstates of $\hat{A}$ and $\hat{B}$ need not be the same.
However, as the action of $\hat{B}$ on $|\psi\rangle$ is restricted to the degenerate subspace, then you can always diagonalize $\hat{B}$ within that subspace to find the eigenstates of $\hat{B}$. These eigenstates will also be eigenstates of $\hat{A}$ because they are simply linear combinations of the degenerate eigenstates, and as we have seen any such linear combination is also an eigenstate of $\hat{A}$. This means that you can always choose an appropriate linear combination of the original eigenstates of $\hat{A}$ to create a new set of eigenstates that are eigenstates of $\hat{B}$ too.
This answer is adapted from one I wrote in the Physics Stack Exchange which asked specifically about the example in which $\hat{A}$ is the Hamiltonian and $\hat{B}$ the parity operator.

source: Stack Exchange

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text:
A question about net ionic equations
I have a question regarding net ionic equations. In a solution, sodium fluoride and hydrochloric acid are mixed together. The "correct" net ionic equation is shown below. However, how can this actually BE a net ionic equation, since none of the ions change state (all aqueous)?

There's no condition that there must be changes in state.

Oh, perhaps what's confusing you is that $$\ce{Na+(aq) + Cl-(aq) -> NaCl(aq)}$$ (for example) is not valid. Well, HF is not the same as NaCl. HF nearly always sticks together, even in water. NaCl doesn't like sticking together.

So, is it basically because it's a weak electrolyte?

Yup, that's right.

To answer your question it is important to know which compound dissociate completely in water when in aqueous solution. For example, consider following reaction:
$$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3_{(aq)}} \tag1$$
When ionic compound dissolves in water, each molecule separates to corresponding ions, which influenced by water. All of $\ce{AgNO3}$, $\ce{NaCl}$, and $\ce{NaNO3}$ are ionic compounds, which are soluble in water. $\ce{AgCl}$ is, onthe other hand, an ionic compound, but it is insoluble in water (refer to solubility rules).  Hence the reaction $(1)$ can be rewrite as follows:
$$\ce{Ag+_{(aq)} + NO3-_{(aq)} + Na+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} + Na+_{(aq)} + NO3-_{(aq)}} \tag2$$
Like in algebra, you can canceled out common ions from each side of the equation $(2)$, which leaves:
$$\ce{Ag+_{(aq)} + Cl-_{(aq)} -> AgCl_{(s)} } \tag3$$
We called the equation $(3)$ a net ionic equation.
Now, let's look at your equation:
$$\ce{HCl_{(aq)} + NaF_{(aq)} -> HF_{(aq)} + NaCl_{(aq)}} \tag4$$
Now, $\ce{NaF}$ and $\ce{NaCl}$ are ionic compounds, which are soluble in water, hence they separated into ions in aqueous medium. How about $\ce{HCl}$, and $\ce{HF}$? They are soluble in water, but are they ionic? Since both are acids, to answer this question, we should look into their respective $\mathrm{p}K_\mathrm{a}$ values (or $K_\mathrm{a}$ values). Ref.1 is listed followings:
$$
\begin{array}{c|ccc}
\text{acid/base (chemical formula)} & \mathrm{p}K_\mathrm{a} & K_\mathrm{a} & \\
\hline
\text{hydrogen iodide }(\ce{HI}) & -10.0 & 1.0 \times 10^{10} \\
\text{hydrogen bromide }(\ce{HBr}) & -9.0 & 1.0 \times 10^{9} \\
\text{hydrogen chloride }(\ce{HCl}) & -8.0 & 1.0 \times 10^{8} \\
\text{hydronium ion }(\ce{H3O+}) & -1.7 & 50.1 \\
\text{hydrogen fluoride }(\ce{HF}) & 3.2 & 6.3 \times 10^{-4} \\
\text{acetic acid }(\ce{CH3CO2H}) & 4.7 & 2.0 \times 10^{-5} \\
\hline
\end{array}
$$
Accordingly, $\ce{HCl}$ should be ionic in aqueous solutions and should dissolve in water according to following equation:
$$\ce{HCl_{(aq)} + H2O_{(l)} -> H3O+_{(aq)} + Cl-_{(aq)}} \tag5$$
I put forward arrow because it'd dissociate completely, since $K_\mathrm{a} = \frac{\ce{[H3O+][Cl-]}}{\ce{[HCl]}} = 1.0 \times 10^{8}$.
However, $\ce{HF}$ is a different story. Since its  $K_\mathrm{a} = 6.3 \times 10^{-4} \lt \lt 1$, its dissociation is not complete, and most of $\ce{HF_{(aq)}}$ molecules stay intact (keep its covalent nature):
$$\ce{HF_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + F-_{(aq)}} \tag6$$
Note that $K_\mathrm{a}^\ce{HF} = \frac{\ce{[H3O+][F-]}}{\ce{[HF]}} = 6.3 \times 10^{-4}$, and hence $\ce{[HF]} \approx 1.6 \times 10^{3}\ce{[H3O+][F-]}$. Like sugar molecules, which are soluble in water but did not dissociate, $\ce{HF}$ stays as $\ce{HF_{(aq)}}$ in water. Thus, the equation $(4)$ can be rewritten as:
$$\ce{H+_{(aq)} + Cl-_{(aq)} + Na+_{(aq)} + F-_{(aq)} -> HF_{(aq)} + Na+_{(aq)} + Cl-_{(aq)}} \tag7$$
When you made common ions cancelled out, the net ionic equation renains as:
$$\ce{H+_{(aq)} + F-_{(aq)} -> HF_{(aq)} }$$
References:

K. Peter C. Vollhardt, Neil E. Schore, In Organic Chemistry: Structure and Function; Fifth Edition; W. H. Freeman & Co.: New York, NY, 2007 (ISBN: 0-7167-9949-9).

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text:
Why does ring contraction take place in Wagner–Meerwein rearrangement?
An example of Wagner–Meerwein rearrangement given in my textbook is substitution by $\ce{AgNO2}$ in iodocyclohexane. One of the product, cyclopentylnitromethane is through ring contraction.
I don't understand how this took place. Can anyone explain the mechanism?

It would be a mistake to think that a five-member ring is very strained.

I know it isn't 'very strained'. I am editing the question. The main question is how is the rearrangement happening.

This reference from 1952 claims that the product of the reaction is nitrocyclohexane. No rearrangement. https://doi.org/10.1021/ja01132a037. A previous study had claimed additionally the formation of 1-methylnitrocyclopentane, which is not cyclopentylnitromethane. What is the name of your textbook?

@user55119, S.N.Sanyal - Reactions, Rearrangements and Reagents ; Publisher: Bharti Bhawan ; Page-185. You may find this book online. Also, in your link, it is clearly mentioned that Rosanow discovered rearranged products in 1915, i.e. Wagner-Meerwin Rearrangement is happening

Kornblum was unable to confirm Rosanow's observation of the tertiary nitro compound (see last paragraph in the text). Assuming that there is some rearrangement, how would 1-methylnitrocyclopentane be formed rather than cyclopentylnitromethane as your text claims?

I'll post a solution.

I would think the product mixture depends on conditions.

Methods of structural analysis have changed considerably since 1915. I think it unwise to accept uncritically the suggestion of the formation of cyclopentylnitromethane.

Kornblum1 demonstrated that nitrocyclohexane 3 was the major isolable product from the reaction of iodocyclohexane 1 and silver nitrite. Earlier work by Rosanow2 (1915) claimed the additional formation of tertiary nitro compound 7, a result that Kornblum did not confirm. If indeed nitrocyclopentane 7 did form, a ring contraction is required. Ring contraction of cyclohexyl carbocation 2 sacrifices the greater stability of a secondary carbocation for a primary carbocation (4) and presumably some loss in stability through ring contraction in forming cyclopentyl carbocation 6. There is no evidence that primary nitrocyclopentane 5 is formed. However, a hydride shift in 4 would afford the more stable tertiary carbocation 6 that would lead to 7.

1) N. Kornblum and C. Teitelbaum, J. Am. Chem. Soc., 1952, 71, 3076.
2) See reference in Kornblum paper.

Structure 5 is given in my textbook as one of the products.

Structure 5 is not present in the Kornblum paper. There maybe a problem in interpreting the names of 5 and 7. OTOH, some information in textbooks is gleaned from memory.


text:
Aquo metal complex structure

Draw the structure of the complex $\ce{Co(NO_3)_2*6H_2O}$ i) without and ii) with one of the nitrates being in the coordination sphere as a bidentate ligand and identify if the inversion center is present in the complexes. Inversion is a symmetry operation that translates every atom through the center to the opposite side.

The first part seemed pretty straightforward and I drew the following structure:

From the structure above, I concluded that there is indeed an inversion center at the cobalt ion.
I'm not so sure about the second structure though. Since the cobalt complex already has six ligands, how could there be a nitrate as a bidentate ligand as well?

Who said nitrate cannot be exchanged with water in the inner sphere?

You drew the arrows in the first structure backwards. Those should point from the water to the cobalt.

@Zhe There should not be any arrows. Using dative bonds for depicting coordination complexes is not acceptable. See https://doi.org/10.1351/pac200880020277, p. 292.

To condense the previous two comments: Yes, there shouldn’t be any arrows at all but if you’re going to defy the recommendation (and there can be a good reason for it; e.g. I refuse to use the chemically inaccurate recommendations for phosphorus/sulphur and related compounds), you should do it properly which involves the arrows pointing from the electron donor to the electron acceptor ($\ce{H2O\bond{->}Co}$).


text:
Why do research papers mention experimental details in such detail?
The paper Analysis of cyclic pyrolysis products formed from amino acid monomer has the following to say in it's experimental section:

Glycine, phenylalanine, tyrosine, tryptophan, serine, and valine were purchased from Daejung Chemicals & Metals Co. (Korea). Arginine, aspartic acid, glutamic acid, and proline were purchased from Samchun Pure Chemical Co. (Korea). Alanine, asparagine, glu- tamine, histidine, leucine, and threonine were purchased from Junsei Chemical Co. (Japan). Isoleucine and lysine were purchased from Acros Organics (USA), and cysteine was purchased from Merck Co. (Germany). Methionine was purchased from Yakuri Pure Chemicals Co. (Japan). Fumed silica was purchased from Merck Co. (Germany).
...
Pyrolysis–GC/MS was conducted with a CDS Pyroprobe 1500 heated filament pyrolyzer (Chemical Data System, Oxford, USA) coupled to an Agilent 6890 gas chromatograph equipped with a 5973 mass spectrometer of Agilent Technology Inc. (USA)

Why is there so much emphasis on the sources and country of origin of chemicals/instruments in these sections? This behaviour is not restricted to this paper; several other papers go into similar levels of detail regarding their procedures.

These details allow for reproducibility check and make quality control easier. Basics of any scientific method. I'd suspect there is something wrong with the paper if this data weren't presented. You don't think this is nationalistic or somehow discriminating, do you?

If quality control is so important, wouldn't it be better if these were presented in an appendix with more data (say, with their batch numbers)? I don't think this is nationalistic or discriminating, just unneccessary, although I'm not the one with a PhD ;)

Concerning the point you rise: in some journals the entire experimental section goes or can go to supplementary publicly available files. So apparently words counting wasn't a problem or the journal is/was fine with loaded exp parts.

Recently, related: https://academia.stackexchange.com/questions/149989

It’s not that much detail. They don’t even provide the batch/serial numbers or date of purchase, means of storage etc. I’ve also read studies on humans where they measured hormones throughout the year but didn’t even mention where the study took place or where the humans came from.

Different world - same lesson: Doing electronic product development in a factory in China I found that products behaved differently when solar charged batteries were cycled. Recollection is that boxes had batch numbers but there were many batches and it was not obvious which were associated. I obtained some "digital scales" and weighed boxes of batteries. There were two overlapping groups of weight. I then weighed all cells from a sample of boxes and found that one batch had a single peak weight distribution  and the others had a double peak weight distribution. And went from there ... .  ...

Batteries were the same brand and notionally functionally identical. IF we had had date codes (which MAY have been able to be obtained from the manufacturer) it may have helped.

Experimental details are very, very important: they are used to ensure that results are reproducible, or at least that is the aim.
Let's talk about this specific example of suppliers and sources. One might naively think that a chemical purchased from supplier X is the same as the chemical from supplier Y. In an ideal world, that would be how it is. Unfortunately, this isn't the case, due to various considerations e.g. method of preparation, packaging, and so on. Therefore, even though the majority of the compound may be the same, there can be different impurities. And unfortunately, these impurities can lead to reactions working or failing. Many experimental chemists will personally know of cases where a reaction doesn't work until a reagent from a different company is used, or a reaction that stops working after buying a different bottle of a reagent.
There are many examples of this in the literature, but I'll choose the first one that came to mind, which is a reasonably recent paper from my old group (open access):
Mekareeya, A.; Walker, P. R.; Couce-Rios, A.; Campbell, C. D.; Steven, A.; Paton, R. S.; Anderson, E. A. Mechanistic Insight into Palladium-Catalyzed Cycloisomerization: A Combined Experimental and Theoretical Study. J. Am. Chem. Soc. 2017, 139 (29), 10104–10114. DOI: 10.1021/jacs.7b05436.

During these investigations, we also uncovered a critical dependence of the stereochemical outcome of the reaction on the batch of Pd(OAc)2 employed as catalyst. This discovery was made through the chance purchase of Pd(OAc)2 from a different supplier, which led to an unexpected
  ratio of enamide alkene geometries ((Z):(E) = 80:20), rather than the typical ratio of ∼97:3.  Screening of further samples of Pd(OAc)2 gave variable results, the most extreme being a reversal of stereoselectivity to 40:60 in favor of the (E)-isomer.

The exact supplier isn't named, but 1H NMR spectra of various batches are given and the performance in the reaction can be correlated with the exact form of Pd(II) in the catalyst, which ensures that future readers are aware of what to look out for. This is very important for good science, because what's the point of publishing your fancy new method if nobody can replicate it?
Of course, this doesn't always happen. However, it's far better to include the information, just in case it is needed.

As mentioned by Waylander in the comments, commercial samples of samarium diiodide (SmI2) also have a reputation for being inconsistent: see e.g. this report by Szostak, M.; Spain, M.; Procter, D. J. J. Org. Chem. 2012, 77 (7), 3049–3059. DOI: 10.1021/jo300135v.

Great answer, I liked the real-life example. If QC was this important, wouldn't they publish the details in an appendix with more data? eg batch no, age of sample etc.

I have personal experience of the variabilitiy of Samarium Iodide solution from different suppliers. The lesson is always titrate a sample and don't trust the concentration on the label. Same with BuLi

@Aniruddha Deb going to details even further won't really be useful. At the end a specific batch in not forever. In addition to what the answer said (plus 1) these things also serve as a kind of warranty for the authors. Say that I'd notice a strong influence such as that narrated by orthocresol, I'd stress that point clearly in my paper. So that others don't doubt as soon they can't replicate exactly my yields. In most cases such a listing of suppliers is redundant but comforms to good details is better than less or no details. Without exaggeration, if nothing points to a specific necessity.

@Waylander SmI2 is a great example; there is this report from David Procter & group which I'll add to the answer. I never had the "pleasure" of using SmI2, but I did do plenty of BuLi titrations...

Another excellent example is the Nozaki–Hiyama–Kishi reaction, where careful investigation of suppliers and batch numbers eventually lead to the discovery that contamination of a sample of chromium(II) chloride with ~0.5% nickel was responsible for the results of the original report, three years earlier, and was necessary for reaction reproducibility.

Similar to Waylander's experience, I also have problem with trialkylaluminum. The efficiency was highly depends on the vendor. Same with acetone on ketal formation reaction with ascorbic acid.

@AniruddhaDeb, it's not always quality control.  Sometimes it's variations that you didn't think were important.  For example, Jet A-1 jet fuel and RP-1 rocket fuel both meet the ASTM D1655 standard for jet fuels, but RP-1 was developed after it was discovered that rocket engines are more sensitive to exactly which hydrocarbons are part of the fuel mix.

You have already seen a detailed response by orthocresol. I can add one more extreme example, but it is not only the chemicals but the apparatus can also play a role in certain reactions. At times, glassware can also alter the rate of reaction. I am not an organic chemist, but some time ago, I was studying the epimerization of a certain natural sugar for developing a analytical method for its determination. Quacks claim that that particular sugar cures cancer. If you boil that sugar in water, glassware from different companies showed different rates of conversion.
If you happen to analyze commercial "enantiomerically pure" compounds, you will see this if often not the case. The story of commercial polymers is also the same.
If you get a chance to read older papers you will see very long passages and very minute details were provided often with a schematic diagram of the reaction apparatus. Today this level of detail is not deemed necessary and journals often put a limit on page numbers. For example, American Chemical Society journal Analytical Chemistry often returns the paper if the length exceeds 8 journal pages.
Once you get into the habit of reading papers, there is an online section, called Supporting Information. Good authors always write a detailed Supporting Information.

I vaguely recall reading several years ago about some reaction which was eventually attributed to trace amounts of lanthanides present in regular borosilicate glassware. There was also a relatively recent article discussing the effects of reusing teflon-coated magnetic stirrers. Trace amounts of catalyst are left embedded in the plastic even after aggressive cleaning (e.g. aqua regia) and can be catalytically active in posterior (even completely different) reactions. Such findings cast doubt on some "metal-free" reactions, which should more correctly be called "no-metal-explicitly-added".

Just a note that that those vials were made for trace analysis work. I am sure it must be a very high quality glass. My opinion at that time (4 years ago) was that it is the alkaline detergents that remain during washing, because when I washed the glassware with a base bath, the epimerization rate slowed down. The reaction was catalyzed by bases.

wouldn't it be better if these were presented in an appendix with more data (say, with their batch numbers)?

It's certainly a matter of judgment how much detail is necessary. Sometimes the descriptions are kept very general if the authors think that is sufficient for correct reproduction. The cited text reports at the level of supplier which is a very common one (it also serves in a convenience way in telling readers at least one supplier where the reagents can be obtained). The next levels would be to give also

brand
grade/purity
article number
and finally the lot

(BTW, I put lot numbers into a manuscript just a few weeks ago.)

I don't think this is nationalistic or discriminating, just unneccessary, although I'm not the one with a PhD ;)

In my experience the amount of details given differs between fields.

Chemistry is among the picky ones in terms of very detailed reporting. Other answers already give examples where such details did matter. Personally I suspect that chemistry may be a field where actually a lot of reproduction happens compared to, say, sociological studies (at least that's my analytical chemist's  view on synthesis):
this reproduction of the paper is a side effect of doing the synthesis in order to obtain the product.
As PhD student, I attended a workshop with participants from all kinds of STEM fields and we had an excercise where e.g. chemists were reading physics papers and vice versa and then discussing them. One really prominent point in that discussion was that chemists kept complaining about the physics papers that they were of no use because the description of the experimental setups was not sufficiently detailed and clear to go to the lab and build such an apparatus.  The physicists complained that the chemistry papers were full of details (implying that it is dangerous to give that level of detail since any competitor could come and pretty much immediately get to your lab's level)

w.r.t. the physicists' complaints - it could be that they were unused to it because in their practice (in physics) such details on experimental setup were not usually found to result in reproducibility problems and they were unaware how sensitive chemistry results were to experimental setup.  (I don't know that it's the case that in physics, just that it is a possibiility.)

I would note that the precise list of reagents and descriptions of reaction conditions can be critical for select reactions to obtain a success and/or avoiding unfavorable side reactions. An extreme example of the latter can include explosions and/or the destruction of the created product of interest.
An illustrated example can be found in the field of advanced oxidation processes (AOPs) as commonly employed in environmental remediation and also disinfecting.
Apparently, for example, Oxone (active ingredient includes Potassium peroxymonosulfate) is widely employed as an oxidizing agent including for swimming pools. The action of KHSO5 in the presence of transition metals (like iron or some ions in even trace amounts, like cobalt) can produce powerful radicals. In fact, a trace of cobalt ions in the presence of HSO5- can produce the powerful sulfate radical anion (see "COBALT/PEROXYMONOSULFATE AND RELATED OXIDIZING REAGENTS FOR WATER TREATMENT" a thesis by Georgios P. Anipsitakis, available here). To quote from the dissertation:

It was found that when Co(II), Ru(III) and Fe(II) interact with KHSO5, freely diffusible sulfate radicals are the primary species formed.

Also:

$\ce{Co(ll)/Fe(ll) + HSO5- → Co(lll)/Fe(lll) + .SO4- + OH-}$        (Eq. 2.1 for Cobalt)

Note, if HSO5- is in excess:

$\ce{Co(lll)/Fe(ll) + HSO5- → Co(ll)/Fe(ll) + .SO5- + H+ }$       (Eq.1.11 & 3.8 for Fe]

And, particularly on the catalytic ability of cobalt ions:

It is also important to state that cobalt is needed only in small-catalytic amounts, thus its potential toxicity may be overcome. As low as 72.3 µg/L of dissolved cobalt were sufficient for the complete transformation of 50 mg/L of 2,4-DCP and approximately 30% organic carbon removal (Figure 2.9a). This value is between cobalt concentrations found in drinking water and below those tested in a recent study that reported the role of dissolved cobalt as a catalyst in an ozonation process (110).

My point here being, that transition metals, in even trace amounts, can particularly be significant, especially when engaging in cyclic reaction systems, producing pronounced impacts.
Bottom line, impurities should not always be assumed to be insignificant.

One point that has not been explicitly covered in previous answers is the thought of the reader of the future.
In 10, 20, 50 years time the reagent/starting material you take for granted now may well not be available commercially. The procedure you usually follow to make X may be unfamiliar to the first year PhD student reading your paper. So stating exactly where your starting materials/reagents came from gives the future reader a starting point.
There are often assumptions made by researchers that the materials they consider common are well known to everyone. orthocresol mentions Pd catalysts, I never worked in a lab where anyone bought these as finished products, all were made up from PdCl2 given the relative costs of doing the synthesis vs buying the finished catalyst, but the relevant syntheses are not cited as often as they should be! I've even written a paper myself on common half sandwich compounds it was non-trivial to find a good, published, large scale prep on Aust. J. Chem. 70 (1), 113-119 despite the fact that a reasonable number of labs use them heavily.
As an example of the other extreme I've come across an older paper which stated "[non trivial, not currently commercially available] compound X was made by the usual method". No citation though! (I have not been able to dig the reference up for this, been out of chemistry a few years now and a search of my Papers library didn't throw anything up.)


text:
How to get the right formulas for combustion (rocket chamber) or where to find them?
I am a space enthusiast and am intrigued mostly by rockets.
I recently found a book HOW to DESIGN, BUILD and TEST SMALL LIQUID-FUEL ROCKET ENGINES and have been fiddling around with rocket engine calculations and calculated a theoretical engine.
Now, this book/PDF has a lot of useful calculations and many other tips and tables in it. And the tables are what I want to know about.
In the document, there are calculated performances for many fuels at different pressures, mixture ratios, and many graphs of how the performance (or some "part" of it) is affected by other variables.
I also found many sites with this kind of graphs, like the site from Robert A. Braeunig and more precisely this site with these charts again.
So the question is: How can I calculate these graphs with my "totally" random variables or at least where could I find this knowledge. I'm basically interested just in the graphs shown on the  "more precise" link.
Thank you for your answers and for sharing your knowledge with me.
Disclaimer: I have no intention of building a real rocket engine as I do not have manufacturing knowledge nor tools to make it (so even if I wanted to I could literally).

I recommend starting here https://www.youtube.com/watch?v=yUYxk-y-tU8 .


text:
Why can't we oxidise allylic alcohol to aldehyde using pyridinium chlorochromate?
I read in my book that we cannot use pyridinium chlorochromate (PCC). Why is that?

Reflux over activated MnO2 in Chloroform is a good mild way of taking allylic alcohols to aldehyde. Otherwise look at hypervalent iodine compounds  https://www.organic-chemistry.org/chemicals/oxidations/hypervalent-iodine-compounds.shtm

Thanks.I actually edited my question as I got the answer for the reagent to be used.(Mno2 and Seo2)But can you explain why PCC wont work

Which book says that? I am unaware of a difficulty and can find examples in the literature of the oxidation of allylic alcohols with PCC (such as here https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alcohol2.htm) . The main reason for not using it is that it is a carcinogen and Chromium residues are toxic

There are indeed examples of $\mathrm{PCC}$ oxidation of allylic alcohols to $\alpha, \beta$-unsaturated aldehydes. The first example is from the original Corey-Suggs paper while the other two cases are from the work of Dauben and Michno.

References:

E. J. Corey, J. William Suggs, “Pyridinium chlorochromate. An efficient reagent for oxidation of primary and secondary alcohols to carbonyl compounds,” Tetrahedron Letters 1975, 16(31), 2647-2650 (https://doi.org/10.1016/S0040-4039(00)75204-X).
William G. Dauben, Drake M. Michno, “Direct oxidation of tertiary allylic alcohols. A simple and effective method for alkylative carbonyl transposition,” J. Org. Chem. 1977, 42(4), 682–685 (https://doi.org/10.1021/jo00424a023).

@Mathew: Thanks for catching the journal error.

The question:

Why can't we oxidise allylic alcohol to aldehyde using pyridinium chlorochromate?

The question did not specify what kind of allylic alcohol OP is talking about. It could be either a primary, or secondary, or tertiary allylic alcohol (didn't cite the book so we don't know for sure). Regardless, pyridinium chlorochromate $(\mathrm{PCC})$ has been used to oxidize tertiary allylic alcohols. However, in these reactions, the newly formed carbonyl group is at carbon $\beta$ to hydroxyl bearing carbon (original allylic carbon. For example, tertiary allylic alcohols generated by Grignard reaction of vinylmagnesium bromide with relevant saturated ketones can be oxidized to the corresponding $\alpha,\beta$-unsaturared aldehydes in good to excellent yields with $\mathrm{PCC}$ (Ref.1):
$$\ce{R2-C=O ->[a) CH2=CHMgBr/ether][b) H3O+/H2O] R2C(OH)CH=CH2 \\ ->[PCC/CH2Cl2] R2C=CH -CHO}$$
However, before these discoveries on tertiary allylic alcohols, J. R. Holum (1961; Ref.2) has studied thoroughly on chromium(IV) oxide-pyridine complex $\ce{CrO3.2C5H5N}$ (first synthesis of pyridinium complexes, Ref.3). The abstract of that study (Ref.2) states that:

The chromium(IV) oxide-pyridine complex was found to be a good oxidizing agent at room temperature for the conversion of primary benzylic and allylic alcohols to their corresponding aldehydes. Data for twenty-one compounds are presented. Simple aliphatic and aromatic secondary alcohols are converted to ketones in fair to good yields; data for six are presented.

Thus, it can be concluded that what OP's textbook said about $\mathrm{PCC}$ and allylic alcohols might not exactly be correct.
In different perspective, $\mathrm{PCC}$  has been shown to be of particular value in the allylic oxidation of compounds containing an activated methylene group, such as 2,6-dihydropyrans. Indeed, it has been claimed that $\mathrm{PCC}$ is the reagent of choice in the allylic oxidation of $\Delta^5$-steroids. For example, when the reaction of $\Delta^5$-cholestoryl benzoate ($\bf{1c}$, Figure $\bf{A}$) with $\mathrm{PCC}$ is carried out in refluxing benzene, the isolated yield of $\bf{2c}$ was 89% while it was 78% when the reaction was performed in $\mathrm{DMSO}$ at $\ce{100 ^\circ C}$. In contrast, if the condition was pyridinium dichromate $(\mathrm{PDC})$  in pyridine at $\ce{100 ^\circ C}$, the yield was only 64% (Ref.4). In addition, these solvent systems are claimed to be superior to the more usual methylene chloride $(\ce{CH2Cl2})$. The biggest drawback of this system, the requirement of large excess of oxidizing reagent, has been successfully solved by using t-buyl hydroperoxide-$\mathrm{PDC}$ $(\ce{t-BuOOH}/\mathrm{PDC})$ mixture as an oxidant (Figure $\bf{B}$; Ref.5). However, $\ce{t-BuOOH}/\mathrm{PCC}$ system is found to be as not efficient as $\ce{t-BuOOH}/\mathrm{PDC}$ system. A low conversion was obtained using $\mathrm{PCC}$ catalyst has claimed to be due to the fact that $\mathrm{PCC}$ having a mildly acidic character (Ref.6).

References:

William G. Dauben, Drake M. Michno, “Direct oxidation of tertiary allylic alcohols. A simple and effective method for alkylative carbonyl transposition,” J. Org. Chem. 1977, 42(4), 682–685 (https://doi.org/10.1021/jo00424a023).
John R. Holum, “Study of the Chromium(VI) Oxide-Pyridine Complex,” J. Org. Chem. 1961, 26(12), 4814–4816 (https://doi.org/10.1021/jo01070a009).
Harry H. Sisler, Jack D. Bush, Oliver E. Accountius, “Addition Compounds of Chromic Anhydride with Some Heterocyclic Nitrogen Bases,” J. Am. Chem. Soc. 1948, 70(11), 3827–3830 (https://doi.org/10.1021/ja01191a085).
S.  V. Ley, A. Madin, “Chapter 2.7: Oxidation Adjacent to Oxygen of Alcohol by Chromium Reagent,” In Comprehensive Organic Synthesis: Selectivity, Strategy, & Efficiency in Modern Organic Chemistry – Volume 7: Oxudation;  Steven  V. Ley, Volume Editor; Barry M. Trost & Ian Fleming, Editors-in-Chief; Fifth impression, Pergamon Press: Oxford, United Kingdom, 2005, pp. 251-290 (ISBN: 0-08-040898-3).
Manolis A. Fousteris, Anna I. Koutsourea, Sotiris S. Nikolaropoulos, Abdelkhalek Riahi, Jacques Muzart, “Improved chromium-catalyzed allylic oxidation of $\Delta^5$-steroids with t-butyl hydroperoxide,” * Journal of Molecular Catalysis A: Chemical * 2006, 250(1–2), 70-74 (https://doi.org/10.1016/j.molcata.2006.01.031).
E. J. Corey, J. William Suggs, “Pyridinium chlorochromate. An efficient reagent for oxidation of primary and secondary alcohols to carbonyl compounds,” Tetrahedron Letters 1975, 16(31), 2647-2650 (https://doi.org/10.1016/S0040-4039(00)75204-X).


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Determining the Energy Density of a Battery given the anode and cathode
everyone, I'm using nanohub's introduction to the material science of rechargeable batteries course to do some self-studying until I can go back into the lab. On one of the practice problems, it asks to calculate the energy density of a given battery system. For the first problem, it asks this for a lithium metal anode, LiCoO2 cathode system. We are asked to assume an electrode thickness of 100 micrometers and a porosity of 35%. I assumed no porosity contributions from other components such as binders. When I calculate the value I don't get something close to the answer given in the solution which is 453.6 J/kg.
I have tried calculating this and the other problems a few different ways with different units but I am not getting the correct answer. The formula I am using is
Energy density = (potential difference between anode and cathode) * cathode charge (capacity)* electrode thickness (this was given in the PS)*(1-porosity(this was also given)) below is an image of what the given information and potential answers looks like

Cathode and anode will have different specific charge capacities. These capacities can be made equal by tailoring the thickness of the electrodes. The question is asking you to make sure that the charge capacity ratio is equal to one, which means you will need to find what anode thickness you must have for the capacities match (they give you cathode thickness of 100 microns)
This is the equation to find the energy density of a battery: $$E=VC/m$$
$E$ is energy density in Wh/kg, $V$ is nominal voltage in volts, $C$ is capacity in Ah and $m$ is mass in kg. Why are you multiplying this equation with thickness and porosity?
You were given some electrode geometric parameters and you will have to find the others. Remember, you are looking for the electrodes masses so that you can compute energy density (hint: start by looking into densities of your materials).
Keep in mind that not all electrodes will be porous (for example for Li metal you can assume porosity is 0%). Also, you will have to report results in $J/kg$, so make sure you keep track of your units and convert if necessary.


text:
Can Gay-Lussac's law be observed in the reverse direction?
Gay-Lussac's law states that if you decrease the temperature of a gas at constant volume, the pressure decreases by a corresponding amount. This is easy to carry out experimentally. But in practice is it possible to perform this the other way round, i.e. decrease the pressure and thus cause the temperature to drop, while keeping volume constant?
It seems to me that in order to reduce the pressure, you would go about lowering the chamber's volume by removing air from it. However, this would also remove a proportional number of moles from the chamber, so the temperature would remain unchanged.

When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you measure the temperature and pressure of the gas and find, in agreement with Gay-Lussac's law, that both T and P have decreased. Then if you ask yourself, which did I lower first, you cannot say. They decreased simultaneously.
The pressure and temperature are coupled (dependent) properties of the ideal gas when the volume is held constant. This is a consequence of the thermodynamic state postulate that for a simple compressible system, the properties of the system are fixed once two independent ones have been specified. The fallacy in the question here is the assumption that one might be able to change the temperature and that a pressure change somehow follows after. The cooling experiment changes both simultaneously. You never actually choose to alter one variable rather than the other.

You cannot cool a gas by reducing the pressure at constant volume. How would you reduce this pressure ? It is the contrary : you cannot reduce the pressure at constant volume without cooling it. And removing a part of the gas cannot be done at constant volume.


text:
The dissociation of hydrogen on UO2 and PuO2 surfaces: Homolytic or heterolytic?
I have seen that $\ce{H2}$ can dissociate on a metal-oxide surface by two methods:

Heterolytically, forming a proton-$\ce{O}$ bond ($\ce{OH}$ group) and a hydride-metal bond ($\ce{M-H}$) on a metal-oxide surface.
Homolytically, where two $\ce{OH}$ groups are formed (adsorbed protons) and the two electrons of the $\ce{H2}$ molecule are transferred to low-lying $\mathrm{d}$ or $\mathrm{f}$ states of the cations. This leads to a change in oxidation state and a reduction of the oxide.

However, the reasons for the different behaviours are not obvious to me. Assuming two comparative MO compounds have the same crystal symmetry, how does the electronic structure of the metal impact which process dominates?

EDIT:
I am particularly interested in the example given by: Differences in the Existence States of Hydrogen in $\ce{UO2}$ and $\ce{PuO2}$ from DFT + U Calculations
Here, the preferential formation of a hydride ion $\ce{[(UO_{2})_{n}]^{+}H^{-}}$ or hydroxide ion $\ce{[Pu_{n}O_{2n-1}]^{+}[OH]^{-}}$ has been shown. The authors give the following explanation:

The differences in the existence states of atomic hydrogen in the two dioxides are proposed to be dependent on the nature of $\mathrm{5f}$ electrons of uranium and plutonium; that is, uranium $\mathrm{5f}$ electrons are more delocalized and more favorable to participate in chemical bonding than plutonium $\mathrm{5f}$ electrons.

Why should the higher degree of electron itinerary (which enables electrons to more easily form chemical bonds) favour hydride formation?
This seems to conflict with: Interaction of hydrogen with actinide
dioxide (111) surfaces.
In this paper and others, $\ce{PuO2}$ has the more reactive surface. Here, the formation of a hydroxide is more favourable on $\ce{PuO2}$, and less favourable on $\ce{UO2}$. Therefore, I am confused as to how this availability of electrons changes.

This reference$\ce{^{[1]}}$ shows what can happen even in a single oxide, in this case magnesia, with different metal environments.

Magnesia is, of course, strongly ionic, so the dissociation of hydrogen to form $\ce{O^{2-}-H^+}$ and $\ce{Mg^{2+}-H^-}$ on the surface might well be expected.  Putting the magnesia as a thin film on silver does not change this.  But on gold the hydrogen dissociates in a redox fashion, forming only $\ce{O^{2-}-H^+}$ as the electrons are injected into the gold.  The lowered Fermi energy of the gold causes this transfer as the electrons "fall" into the lower energy state.  It's an example of gold acting as an electronegative element, as it does in compounds such as $\ce{CsAu}$, ultimately caused by the effects of relativistic mechanics on the electron structure of gold.
In the case of bulk uranium and plutonium dioxides, electron donation or withdrawal must come from the oxide itself.  With uranium (IV), the 5f orbitals are at relatively high energy, and where they have electrons those electrons are readily available for bonding and for donation -- but conversely, empty 5f orbitals are relatively poor electron acceptors.  So we get metal hydride formation like the $\ce{Mg^{2+}-H^-}$ bonds in plain magnesia or magnesia/silver.
With plutonium in the same +4 oxidation state, enough increased effective charge and orbital contraction have occurred to make the plutonium 5f subshell a better electron acceptor, making the $\ce{O^{2-}-H^{+}}$ bonding mode more favorable.
References:

Hsin-Yi Tiffany Chen, Livia Giordano, and Gianfranco Pacchioni, "From Heterolytic to Homolytic H2 Dissociation on Nanostructured MgO(001) Films As a Function of the Metal Support", J. Phys. Chem. C, 2013, 117, 20, 10623–10629.

2.
Chemical Forms of Uranium". Argonne National Laboratory. Archived from the original on 22 September 2006. Retrieved 18 February 2007.  Link

Thank you @Oscar Lanzi for your insightful answer. I have now made my question more specific as you suggested; however, your input is still a valued answer.


text:
Atomic shielding 2s to 1s
How much is the effecting shielding of the 1s electrons from the 2s electrons? . Because the 2s subshell overlaps with the 1s subshell this will cause the 1s electrons to 'feel' less charge than the nucleus charge.
However due to the fact that the 2s subshell is much bigger than the 1s subshell the effect of shielding will be small correct?

The effect of "reverse shielding", as I call it, is on the order of one volt of potential difference in light elements where the $n=2$ electrons are the valence electrons.
Let's look at lithium.  From this data page the effective nuclear charge on the $1s$ electrons in the neutral atom is 2.691.  If we assume that the effective nuclear charge is related to the electronic energy level by
Energy$=(-13.6 eV)(Z_{eff}^2)$
then we infer that the orbital energy level is -98.5 eV for each electron.
Now what is that energy level if we take away the 2s electron? For this we need the electronic energy level of the $1s$ orbital in the lithium ion, which is obtained by averaging the second and third ionization energies of lithium given here.  That leads to -99.1 eV per electron.  The 2s electron has effectively increased the energy level of each 1s electron by 0.6 eV.
A similar calculation for beryllium gives a difference of about 1.1 eV per $1s$ electron (-184.7 eV from the effective charge on $1s$ in neutral Be vs -185.8 eV  from the third and fourth ionization energies after forming the di-cation).  The second 2s electron has basically doubled the shielding.
The electron volt differences determined above have some impact on chemical reactions, whose energies are on the order of one electron volt per electron transferred in a redox reaction.

Can we apply this to molecular orbitals as well?


text:
How does pressure influence the dehydration of salts?
It's well known that the boiling point of a substance increases with pressure. I was wondering if something similar happens with hydrated salts and the temperature at which they dehydrate.
I mean, if the pressure is increased, does the temperature at which a salt dehydrates rise or is it not affected by the pressure?

Yes - the pressure will impact the dehydration temperature.
You have correctly identified that temperature and pressure are related by the Gas laws.
High pressure can also force water molecules out, as water is loosely bonded in salts.


text:
How can one reclaim isotopic cadmium from a solution of acid, i.e. H2SO4 or HCl etc.?
I have dissolved isotopic cadmium-114 with $\ce{H2SO4}$ and or $\ce{HCl}$.  Now that I have the cadmium-114 in solution, precipitating out $\ce{CdSO4}$ and or $\ce{CdCl2}$, how can I reduce these simply to leave the pure cadmium metal isotope?  I have dissolved $\ce{^{114}Cd}$ with acid in very complex electronic tubes and I wish to reduce the precipitated compound(s) to pure metal for reuse.
$\ce{^{114}Cd}$ or is running about $ 1,500 per gram.  So extremely useful to reclaim it. Can anyone tell me a simple way to do this?

What is "isotopic" cadmium?

What "isotopic cadmium"? There are several isotopes of cadmium. Which one?

Also 1. Names of elements aren't capitalized 2.It's HCl not HCL 3.Cadmium chloride is CdCl2 not CdCL. I have edited those errors. Make sure to follow them in future.

Apparently the question has nothing to do with isotopes. In my opinion, it may be rewritten so : How to reduce cadmium compounds into metallic cadmium ? Am I wrong ?

Thank you for the responses.  Cd 114 to be exact, but I don't see how the isotope would affect the chemical reaction very much.  I admit, I did not bother to look up the exact chemical nomenclature, but you did understand what i was saying.  Finally, for Maurice.  Yes, in chemical terms it would be reducing, but for my purposes, it is reclaiming.  Using acid to dissolve Cd in very complex electonic tubes and then reclaiming the Cd for reuse.

@DaleHarder do you know what other metal ions are in the solution?

Should not be any other metals in the solution.  If any only trace amounts.

The usual way is electrolytic from a solution of the sulfate, Al as anode.

(btw. you mean monoisotopic. The adjective "isotopic" itself has no sensible meaning. )

Why bother turning it into a metal? You'll hardly ever need the metal. Keep the salt, and use it the next time you'll need it.

A solution of sodium dithionite should do it.

@IvanNeretin For starters, what does anybody do you do with 114Cd? ;-)

@ Dale Harder. Why do you want to use Cd-114 ? It is one of the sable cadmium isotopes. Ordinary cadmium contains 29% Cd-114. Is it necessary that your "very complex electronic tubes" contain this isotope and only this one ?

I'd guess that the point of Cd114 is to have a non-radioactive Cd in the tubes. $\ce{^{113}Cd}$ and $\ce{^{116}Cd}$ are very long life radioactive isotopes.

Again, Thank you everyone for your responses.
I am a Laser manufacture and also a plasma tube re-furbisher where possible. I make new tubes, however, they are very complex glass work. So, saving a tube when possible is economical. The Helium Cadmium Laser gains 30% more power by using a single isotope of Cadmium. Cd114 is the best option and produces the most usable power, especially on 325nm UV lines. 441.6nm lines are not as sensitive but do benefit. I have no need of Cd salts. Just pure metal. Cleaning the tubes for reuse is a standard process. So, dissolving the Cd out of the tube is on st

OK wow. ;-)  My first thought at your question was "now this is a really crazy scrap metal collector, how do we keep him from poisioning himself." My old inorganic chemistry bible "Holleman-Wiberg" says that very pure Cd is produced via electrolysis of a solution of the sulfate with Al and Pb electrodes. Nothing on recommended pH, concentration, voltage, unfortunately. There is no specific reference given in there, but I guess "Gmelin´s Handbook" should have the missing info.

I´m not sure that you will find a procedure in the literature that is well suited to your application. But you surely have a lot of contacts into academia. Talk to one who has a wire into inorganic chemistry, convince them to write a small grant application for a one-year project to devise a safe and efficient small scale electrolytic Cd recycling process. There are a lot of good buzzwords in the subject. ;-)

https://onlinelibrary.wiley.com/doi/abs/10.1002/zaac.18970130118 Notiz über die elektrolytische Reinigung des Cadmiums

Did you accidentally create a second account? Old: https://chemistry.stackexchange.com/users/94579/dale-harder New: https://chemistry.stackexchange.com/users/94599/dale-harder

Thank you everyone for the discussion.  Sorry, I have been away for a few days, so was unable to follow up.
Reading the ideas, it is a toss up on a procedure.  Scraping the Cd off of an Aluminum electrode might work, but Al contamination would be very bad for the laser oscillation.  It might work for awhile, but would die from contaminants in short order.
The Al oxide, though only a molecular layer, is extremely tough and can withstand a great deal of heat and abuse.  I make Aluminum edge-wound voice coils for my speaker company and we depend on the oxide layer to act as an insulator from winding to winding.  Our coils can withstand 600 degrees and will melt before they fail.
Stripping the oxide gives you only a few seconds before it reforms in the normal atmosphere.  Any other thoughts?
Perhaps reduction of the salt by heating?
Thanks
Dale

Perhaps the answer lies here...I understand a lot of it, but not everything.  Please look.                                      https://onlinelibrary.wiley.com/doi/pdf/10.1002/pip.402   Thanks, Dale.

The reference is paywalled. In the old days, electrolysis into a mercury cathode, followed by distilling the mercury off, might have been done. Of course, nowadays that would be something only for a specialist laboratory. But maybe use a glassy carbon cathode? Then no aluminum to contaminate the cadmium.

From a solution of an aqueous cadmium salt ($\ce{CdSO4}$ or $\ce{CdCl2}$), try inserting a metal plate of say Aluminum.
With time and warming of the solution, a coating of cadmium should appear on the metal sheet.
Try scrapping off the deposited cadmium metal, per the replacement reaction:
[EDIT]$$\ce{2 Al(s) + 3 CdSO4 (aq) -> 3 Cd (s) + Al2(SO4)3 (aq)}$$
Note: depending on your tools for scrapping, you may induce some metallic impurities.
[EDIT] Importantly, my standard chemistry approach (based on a single-displacement reaction) is different from an electrolysis or electrochemical cell approach, employing an Aluminum electrode, as I suspect, that the electrode may quickly become coated and effectively inoperative.  Also, an electrochemical cell with a large surface area anode (here Al) would require an even larger surface area cathode to operate efficiently. Further, the ability to facilely recover the Cd metal is key to the cadmium recycling process. My proposed process may fulfill these requirements.

@ Ajkoer 1) Your equation is wrong ! You would need 3 CdSO4 to balance the equation. And it would produce Al2(SO4)3. 2) Are you sure that this reaction works ? Usually aluminum is covered by a protective layer of aluminum oxide, which prevents the metal from touching the solution. If it does it does not produce aluminum sulfite...

Thanks, I worked on it late and was originally thinking of using CdCl2. Will edit. However, more fundamental does my path work in practice?

Why would cadmium ions not get reduced on a Cd cathode in electrochemical cell? And why should Al be used as a cathode in the first place?

Cadmium can be reduced by metals like zinc and magnesium. With aluminum, it will be theoretically possible, if the protective layer of aluminum oxide would be removed. This protective layer is transparent, continuous and nearly insoluble in diluted acids. That is why for example Aluminum is used as container for handling concentrated nitric acid. The alumina layer protects the metals from corrosion. I don't know how aluminum behaves with cadmium salts. But I think there will be no reaction, because of the protective layer of alumina. Anyway cadmium may be reduced by electrolysis.


text:
The meaning of permanent dipole moment

I was asked in one question that which of all given molecules have a permanent dipole moment?

And $\ce{XeF6}$ was given as a correct answer quoting that it is a distorted octahedral, and hence the lone pair results in the molecule having a dipole moment. But since we know that the lone pair goes at each face of the octahedral one by one, the net effect should cancel out. Hence, is it correct to say that the molecule has a PERMANENT dipole?
I am having similar doubt in the case of the very fundamental ammonia example. Since it undergoes inversion do we say that it has a permanent dipole?
Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me.

In the equilibrium (ie: lowest energy) situations, both NH3 and XeF6 have a permanent dipole moment. Since the molecule is very likely to be around its equilibrium geometry at any given time, then the general characteristics of the compound resemble those of the equilibrium situation. So, despite undergoing inversion via an apolar intermediate, ammonia is not an apolar molecule, and so isn't XeF6

Forget inversion. Molecules wiggle, bounce, and rotate all the time; how can we talk about the dipole moment? Now it points this way, and the next second it points that way. Is there any meaning in it at all? Well, yes, there is.

You might look at https://chemistry.stackexchange.com/questions/104468/why-is-ammonia-still-considered-polar-despite-undergoing-rapid-inversion/104629#104629 but I'm not quite sure if that is the question that is being asked (I think more than one this is being conflated here). For me for a molecule to have a permanent electric dipole means the expectation value of the electric dipole operator with the molecule in its equilibrium geometry is non-zero. The effect on the measured dipole due to motion is a separate issue.

@Ian Bush: Second part of the question is definitely related to: https://chemistry.stackexchange.com/questions/104468/why-is-ammonia-still-considered-polar-despite-undergoing-rapid-inversion/104629#104629.

Indeed, $\ce{XeF6}$ has a distorted tetrahedral structure in all three phases. Two of solid state crystal structure modifications ($\bf{A}$ and $\bf{B}$) are shown in following diagram (Ref.1):

The abstract of Ref.1 tells them all:

According to single crystal X-ray diffraction, neutron powder diffraction, solid state MAS NMR data, and differential scanning calorimetry, $\ce{XeF6}$ exists in at least six different modifications. Three of them are formed at temperatures above room temperature, one exists at room temperature, while two have been found at low temperatures. In the high temperature modifications $\ce{XeF6}$ forms a non-symmetric tetramer, better described as a cyclic trimer with a weakly associated monomer. The normal temperature modification is the previously described cubic phase IV, having disordered tetrameric and hexameric units. The low temperature modifications are regular tetramers. Only in presence of $\ce{HF}$ symmetric dimers are formed.

The tetramer $\bf{C}$ (ball and stick version of $\bf{A}$) represent the general version of all modifications, a cyclic trimer with a weakly associated monomer. The symmetric dimer $\bf{D}$ is the structure of $\ce{XeF6}$ in the presence of $\ce{HF}$ (highlighted by $\color{green}{\text{green}}$ ovals). Nonetheless, all of these modifications display distorted octahedral structure for individual $\ce{XeF6}$ molecule as represented by ball & stick 3D structure in the central box. None of $\ce{F-Xe-F}$ angles are equal to $90^\circ$, which is the normal angle of central atom with two adjacent ligands in regular octahedron. As a consequence, $\ce{XeF6}$ has net permanent dipole moment since all six $\ce{Xe^\delta+-F^\delta-}$ bonds are polar.
For your question about ammonia, look at Ivan Neretin's comment and here where there is an accepted answer.
References:

Sevim Hoyer, Thomas Emmler, Konrad Seppelt, “The structure of xenon hexafluoride in the solid state,” Journal of Fluorine Chemistry 2006, 127(10), 1415-1422 (https://doi.org/10.1016/j.jfluchem.2006.04.014).

Basically what is the meaning of a permanent dipole? I understand how it is different from induced dipole and instantaneous dipole, but these examples feel like instantaneous dipole examples to me.

Addressing your second question, permanent dipole moment experimentally means that a molecule will be deflected by an electric field. Consider water as an example, open your kitchen faucet slowly, so that a think thin stream is steadily flowing. Charge a comb and see its deflection. Such molecules have a permanent dipole moment.
If you had cyclohexane or heptane (a molecule without a permanent dipole moment) nothing will happen, i.e., their streams will not be deflected by an electric field.
Always focus on basic understanding first before being bogged down by complex cases of ammonia inversion or invoking operators for electric dipole moments etc.

Ok, thanks for the information. So would a liquid ammonia stream get deflected in an electric field or would the rapid inversions cancel out dipole moment of each other?

Ammonia has a dipole moment which is permanent. Whether it is an inverted state or or normal state, the individual bond dipole moments do not cancel. Hence liquid ammonia will be deflected by an electric field.

The distorted octahedral structure of $\ce{XeF6}$ produces a permanent dipole.
The key word is distorted structure, which results from the $\ce{Xe}$ lone pair).

If a structure is distorted, this results in a (lowering) breaking of symmetry.
If a molecule is unsymmetrical with strong polarising groups, there will be a local permanent dipole moment associated with it. In a general sense, you can think of bonds as vectors.
Now in your $\ce{XeF6}$ example:

Consider the direction of the bonds as vectors.
Then calculated (or visually consider) the sum of those vectors.
If the sum of all vectors is not equal to 0, then you have a dipole moment.

Now consider an perfectly octahedral molecule. You will find that the vector sum is now equal to zero, and therefore the molecule does not contain a permanent dipole.
Now for your $\ce{NH3}$ example. $\ce{NH3}$ has a distorted tetrahedral shape, where the bonds are forced away from the lone pair. The mistake is to assume bonds and lone-pairs are the same. The electron negative charge within the lone pair is held closer to the centre of the atom within the lone-pair orbital, and so repels the electrons in the bonds to a greater extent.
To contrast $\ce{:NH3}$ (here, $\ce{:}$ represent lone pair) with $\ce{CH4}$ (which doesn't have a permanent dipole). All bonded vectors in $\ce{CH4}$ are equal (same atom-bond type) and are evenly distributed.

Good as it is, your answer is from a different question.


text:
Rusting of iron is quicker in saline water than in ordinary water
I feel so because, saline water has more more electrons and because of that there will be oxidation will take place faster and hence the overall process will speed up. Is this explanation good enough or there's more to it?

Does this answer your question? Does concentration of salt increase or decrease rate of rusting?

@ManRow I read through it, but I still have some doubts. I can't seem to reply to the person who answered that question for some reason. Can you help me out?

You need 50 rep. to comment everywhere. BTW your explanation makes no sense :(

Leah - HINT In chemistry the using the right terms is vital. It isn't that "saline water has more more electrons" but that saline water has more more of something else...

@MaxW more of ions? Na+ and Cl- ?

Leah, yes that is it - ions, not electrons.

@MaxW Okay, but how do these ions contribute to speeding up the process of rusting?

To quote an educational source:

Water is the enabler of fast oxidation of iron so freshwater will also cause rust. However, salt water is a very good conductor (lots of dissociated ions) and so there are a number of electrolysis reactions that tremendously accelerate corrosion in salt water.

which means, in essence, that aqueous NaCl is a good electrolyte that can also promote so called galvanic corrosion reactions.


text:
Work done by one mole of gas during isobaric and isothermal expansion

If $100\ \mathrm{mol}$ of $\ce{H2O2}$ decomposes at $1\ \mathrm{bar}$ and $300\ \mathrm K$, the work done ($\mathrm{kJ}$) by one mole of $\ce{O2(g)}$ as it expands against $1\ \mathrm{bar}$ pressure is:
$$\ce{2H2O2(l) <=>2H2O(l) +O2(g)}$$

My attempt to solve the problem:
Since $2\ \mathrm{mol}$ of $\ce{H2O2}$ decomposes to give $1\ \mathrm{mol}$ of $\ce{O2}$, so $100\ \mathrm{mol}$ of $\ce{H2O2}$ will give $50\ \mathrm{mol}$ of $\ce{O2}$.
Now, The work done by $50\ \mathrm{mol}$ of $\ce{O2}$ as it expands is,
$$w_{\ce{O2}}=-w_\text{surr}=\int{p_\text{ext}\,\mathrm dV}=p_\text{ext}\,\Delta V=(\Delta n_\text{gas})RT=50\times8.3\times300\times10^{-3}\ \mathrm{kJ}=124.5\ \mathrm{kJ}$$
$124.5\ \mathrm{kJ}$ is in fact the answer given but we are asked to find work done by one mole of $\ce{O2}$.
Shouldn't the answer be $w={124.5\over50}\ \mathrm{kJ\ mol^{-1}}=2.49\ \mathrm{kJ\ mol^{-1}}$?
If $50\ \mathrm{moles}$ of gas expanded to a specific volume, say $V$, then $1\ \mathrm{mole}$ of gas should expand to ${V\over50}$. Then according to me work done by $1\ \mathrm{mole}$ of gas should be ${1\over50}$ of work done by $50\ \mathrm{moles}$ of gas. Does work add up like extensive properties of system?
Reference: This question appeared in JEE Main entrance examination held in India in 2016, the official answers and question papers are not publicly available, all I can find is this pdf https://www.resonance.ac.in/answer-key-solutions/JEE-Main/2016/papers/jee-main-2016-online-CBT-paper-Dt-10-04-2016.pdf#page=19 (Question no. 20 page 19) and the answers (also pdf) https://www.resonance.ac.in/answer-key-solutions/JEE-Main/2016/answer-key/jee-main-2016-online-CBT-solution-Chemistry-10-04-2016.pdf#page=7 (Question 20 Page 7)
$2.49\ \mathrm{kJ\ mol^{-1}}$ as calculated by me doesn't seem to be in the options at all.

I don't know why this question is getting closing votes. I took this question as an example to understand the concept of work in thermodynamics and to better formulate what I am asking.

Can you provide a reference (source) for the posted question?  It does look like an error in the official answer. I can't see an error in your calculation.

@BuckThorn Added references, I cannot find good one though.

It's clear from the official answer that there was an error in the question. It should not have said "per mole of O2".

@BuckThorn Strangely, as far as I know, no one challenged this question.

I might have overlooked something but I don't understand how the answer could be "per mole O2".

I’m voting to close this question because there might be an error in official answers and/or question.


text:
A Curious (definitely wrong) result of Absolute Entropy
The Gibbs free energy is defined as
$$G=H-TS$$
Differentiating both sides
$$dG=dH-TdS-SdT$$
For an adiabatic reversible expansion of an ideal gas, we would have $dG=0$, and $dS=0$.
Thus we can write, $dH=SdT$
Also, $dH=n C_p dT$.
Finally I reach the strange result that $S=nC_p$. How can I define absolute entropy so simply. Something is definitely very wrong here. Please correct me.

Integrate dT and add constant.

Double check your equations and assumptions. For instance for which kind of process can you assume dG=0? What happens to the temperature during an adiabatic reversible expansion of an ideal gas?

I am assuming dG=0 based on the fact that it is a reversible process and dG=0 in a reversible process. Does this not hold always?

No, for instance $dG \neq 0$  for a reversible isothermal expansion.

Ok.., how can I prove that dG≠0 in reversible isothermal expansion of an ideal gas?

BTW stumbled on this related post:  https://chemistry.stackexchange.com/questions/131179/is-constant-pressure-heat-capacity-entropy-during-adiabatic-reversible-expansi

So is dG≠0 in reversible expansion of ideal gas as well??...as is written in the answer of that post.

The Gibbs free energy change is zero in the case of reversible processes carried out at constant temperature and pressure, but that isn't the case if these conditions are not observed.
As a demonstration consider an isothermal reversible expansion of an ideal gas.
Since the temperature is constant, the free energy change is given by
$$\Delta G = \Delta H - T \Delta S$$
Because the gas is ideal and T constant, $\Delta U = 0$, and from the second law of thermodynamics it follows that
$$\Delta S = \frac{q_\text{rev}}{T}= -\frac{w_\text{rev}}{T} = \frac{\int_{V_1}^{V_2}p dV}{T} = \frac{\int_{V_1}^{V_2} \frac{nRT}{V} dV}{T} = nR\log \left( \frac{V_2}{V_1} \right)$$
It also follows that
$$\Delta H = \Delta U + \Delta (pV) = 0 + \Delta (nRT) = RT \Delta n = 0$$
since the system is closed.
But then
$$\Delta G = - nRT\log \left( \frac{V_2}{V_1} \right) \neq 0$$
The free energy change is not zero because the pressure changed during the process.
For the reversible adiabatic expansion of an ideal gas
$$\begin{align}  \Delta S &= 0  \\ \Delta H &= \int_{T_1}^{T_2} C_p dT  = C_p \Delta T \\ \Delta G &= \int_{P_1}^{P_2} V dP - \int_{T_1}^{T_2} S dT =  (V_1P_1^{C_V/C_P})\left(\frac{C_P}{C_P-C_V}\right)( P_2^{1-C_V/C_P}-P_1^{1-C_V/C_P}) - S\Delta T  \end{align}$$
The integral over P in the expression for $\Delta G$ can be evaluated after substituting the following expression for the volume in terms of pressure:
$$V=(V_1P_1^{C_V/C_P}) P^{-C_V/C_P}$$


text:
Definition of “solution”
In my textbook “solutions” are defined as follows:

Homogenous mixtures of two or more substances are known as solutions.

Should the two substances always be non-reacting?
The definition has no indication of this, buts it’s just something that came to my mind.
Won’t the solution still be homogeneous so shouldn’t we still call it a solution?

Your textbook definition is good enough to answer your question. If the substance reacts with the solvent yielding soluble product, then you got a solution. Of course, there is no definite answer as to wheter the product(s) will always be soluble, you need to look at the concrete example.

Please never use MathJax's math mode for emphasis.

@andselisk Thanks! I completely neglected the possibility of precipitate formation after the reaction! So yes...it should depend on the reaction taking place.

Sure will not use MathJax. Any particular reason by the way?

Ah...the formatting is much better without it.

The reasons against using MathJax's math mode for emphasis: 1. It's just wrong; 2. It's semantically incorrect to use math mode for highlighting the text; 3. Even if you were to use MathJax for emphasis you should've used $\textit{…}$ instead; 4. Every time MathJax is introduced, the web page loads corresponding JS scripts which are in general resource-heavy; 5. Did I mention it's wrong? Also, see Is it OK to abuse Mathjax for emphasis?

Example: Should bromine water be called a solution?

This is what Encyclopaedia Britannica definition of a solution in chemistry:

Solution, in chemistry, a homogeneous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility. The term solution is commonly applied to the liquid state of matter, but solutions of gases and solids are possible. Air, for example, is a solution consisting chiefly of oxygen and nitrogen with trace amounts of several other gases, and brass is a solution composed of copper and zinc.

A solution consists of solutes (at least one) and a solvent. The solute is define as the substance that is dissolved in the solvent. In other words, for solutions with components in the same phase, the substance(s) present in lower concentration are solutes while the major substance present in highest abundance is the solvent. For example, 190-proof alcohol is a liquid-liquid solution, which is the mixture of 95% ethanol and 5% water by volume. The major substance in this mixture is ethanol. Therefore, it is the solvent. The 5% water is the solute. Brass is a solid-solid solution of copper and zinc metals. General composition of brass is 65% copper and 35% zinc by weight. Hence, in brass, copper is the solvent and zinc is the solute. Air is a solution of multi-gases. In air, 72% nitrogen and 20% oxygen, and rest is other gases including $\ce{CO2}$ and $\ce{H2}$. Thus, the solvent of air is $\ce{N2}$ while $\ce{O2}$ is one of its solutes. Saline solution, on the other hand, is a solid-liquid solution. In a saline solution, solid salt ($\ce{NaCl}$) is the solute dissolved in water, which is the solvent (liquid).

A few days ago a senior member was arguing with me here that air is not a solution. I disagreed. Now Encyclopedia Britannica says the same that air $is$ a solution. Thanks for sharing.

@M.Farooq Na, sorry. A solution must be a condensed phase. Britannica is inconsequent in its definition, because it is per se impossible that two gases have any kind of solubility limit. You change temperature or pressure, the componentes in a gas mixture condense totally independent of each other. No interaction -> not a solution.

@Karl, It is ultimately semantics. There is no restriction on miscibility. ACN and water are miscible in all proportions. If we freeze a solution of water and NaCl, water freezes separately to some extent. Sorry, I don't buy this argument. Can you provide a solid reference which says that a solution must be a condensed phase? In microwave spectroscopy, there is a technique called chiral tagging where a enantiomer is exposed to a chiral alcohol in the gas phase and neon is used a "reaction medium". They form weak complexes in gas phase. Who says there can be no interaction in gases?

@M.Farooq "No interaction" was a bit of a badly chosen term. But gas mixtures do not have colligative properties, no solubility limits, nothing. Filing them under the same term just because of an indentical first half sentence in the definition is completely pointless. You learn nothing by that.  Your example with the chiral tagging is a proverbial exemption that in fact proves my rule: When you see gas mixtures show solution-like properties, you are just very close to condensation. Supercritical mixtures are a similar case.

@Karl, just a clarification, chiral tagging in microwave spectroscopy is done at millitorr pressures in a chamber. It is neither supercritical state nor a high pressure experiment. Do solid solutions have colligative properties? I am saying that this is all semantics. You must read the poem "Five Blind Men of Indostan" https://en.wikipedia.org/wiki/Blind_men_and_an_elephant

"And so these men of Indostan
Disputed loud and long,
Each in his own opinion
Exceeding stiff and strong,
Though each was partly in the right,
And all were in the wrong!"

@M.Farooq Got it. ;-)

@Karl, I appreciate when you are critical but sometimes it is splitting hairs. This is my favorite poem and I even show it in conferences when I am about show something where all of us do not agree :-)

I've done this myself more times than I should but I have to ask: why post an answer when you voted to close? I didn't think the question was at all ambiguous, at worst it was bound to lead to opinion-based answers (because the answer to the question depends on what one thinks the definition of solution should be). By the way, ever seen a submarine movie? Solution (as in "firing solution") has a very different meaning there.

@Buck Thorn: I wasn't the first one to do so. And, I wasn't sure vote came before or after my answer. Besides, these answers prevent other viewers asking the similar question.

That's a valid point. Perhaps worth bringing up in meta if no one has already.

Should the two substances always be non-reacting?

Short answer yes. If they two substances react, you cannot recover the original components which defies the definition of a mixture.
However, if you think deeply, what is the meaning of non-reacting? Sometimes words are not enough they can only be an approximation of a reality.
Let us take the example of a $\ce{I2}$ (iodine solid) and dissolve it in hexane or carbon tetrachloride, we get a purple solution. If I dissolve it in acetone, we get a brown solution. Chemists are okay to call it a solution. Thinking deeper, the purple color or the brown color indicates that the "species" of iodine is not the same into solvents. It means somehow the solvent is indeed interacting with iodine molecules. That interaction must be very weak, because the moment we evaporate the solution, we will get the exact amount of iodine and the solvent back!
So you can see there is no such situation where there is no interaction. Even there is some interaction between two strangers by gravitational forces, but it is extremely extremely weak.

You don't need to get into subtleties. Like every reaction you run in lab is in solution. Does it stop being solution when reaction starts? No, it doesn't.

From a physical chemist's perspective, those reaction mixture will no longer be a binary component solution, it will have multiple solutes (=reactants and products).

Chemists tend to loosely use the word solution e.g., dissolving HCl gas in water yields HCl solution. Is it a mixture of HCl gas and water- Nope!

Come on, no chemist says "HCl solution"! It´s hydrochloric acid, acide chlorhydrique, Salzsäure! :-)

@Karl, Yes, but the usage HCl solution for hydrochloric acid solutions is not uncommon. I just did quick Google Scholar search https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=%22+HCl-L%C3%B6sung%22&btnG=

@M.Farooq "HCl-Lösung" is an imo correct term for "solution of sth. in HCl". But yea, there is a lot of sloppy wordage used in scientific literature. ;)

@Karl, In analytical chemistry, it is okay to say "prepare a 0.1 M HCl solution", which means dilute pure hydrochloric acid with pure water so that its concentration is 0.1 M HCl.

@M.Farooq Can't argue with that!  :-)) Of course if one wanted to be nitpicky, they might insist you write "0.1M HCl·aq solution".

@Karl, Nobody can beat you! You can be a patent lawyer and earn a lot. :-)

@M.Farooq That badge goes to you! How can it not be a solution if it´s the result of diluting? wins not only this argument but also our other one about gas "solutions". Chapeau! :-))


text:
Are all corrosives either acids or bases?
Are all corrosives either acid or a base? From what I could remember from school, strong acids and bases can be corrosive, but can substances be corrosive from reasons other than that? I tried my best to check for the answer with my limited knowledge.
If it is not just acids and bases what is it then, how can we categorize them, if possible in simple terms?

Almost (if not all) all acids and bases are corrosives but all corrosives are not just acids and bases.

Wikipedia lists many corrosive compounds.  While a lot of them function as strong acids or bases, there are also these (not an exhaustive list):

Strong oxidizers including concentrated hydrogen peroxide
Fluorides (they say "fluoride ion", so salts as well as the acid are meant)
Organic compounds that can act as alkylating agents such as methyl sulfate

The moral of the story: Hazards can't be pigeonholed into specific types of chemicals.  Best to find and read the safety data sheet with any chemical.

I detect two downvotes, in both this answer and another one that previously had a positive score.  Oh boy are we having a bad day.  Anyone care to explain?

Besides seawater, as described below, I am unsure how the 3 mentioned examples act as corrosives? (:
Why are they corrosive?

@AndreasJørgensen, Look up the definition of corrosion provided by IUPAC. I think that you are thinking of corrosive nature from human safety point of view. That is not the typical meaning of corrosion in surface chemistry or material science.

Operatively, the most corrosive substance on planet earth is likely seawater (which has a pH of 7.5 to 8.4). Ordinary pure aqueous NaCl also does a very good job on corroding Aluminum alloys, being able to attack the protective Al2O3 coating.
Part of the explanation is likely due to the ability of NaCl to be an excellent electrolyte serving in a variety of galvanic corrosion reactions. The chloride anion is likely a factor given its ability in forming complexes and being able to act as a reducing agent.
Also, with respect to water per a source:

Water is the enabler of fast oxidation of iron so freshwater will also cause rust. However, salt water is a very good conductor (lots of dissociated ions) and so there are a number of electrolysis reactions that tremendously accelerate corrosion in salt water.

Also, more on water per another source, to quote:

This means the water is donating the hydrogen ion, which classifies it as an acid using the Brønsted concept. ... As these two reactions show, water can act as an acid or a base; molecules (or ions) that can do this are called amphiprotic. When an amphiprotic molecule (or ion) reacts with an acid, it acts as a base.

As to the question "Are all corrosives either acid or base?", considering the case of water, the 'or' part of the question may be problematic.

It is an open ended question. What is meant by corrosion to begin with? Corrosive is very general term which literally meant to "gnaw into pieces" like the worms.
Do you know pure supercritical water is quite a corrosive substance. It will attack many metals. Others have given you examples. Fluorine gas is corrosive. Who said that only acid and bases can corrode materials?
Formally, from IUPAC corrosion is "an irreversible interfacial reaction of a material (metal, ceramic, polymer) with its enviroment which results in consumption of the material or in dissolution into the material of a component of the environment. Often, but not necessarily, corrosion results in effects detrimental to the usage of the material considered. Exclusively physical or mechanical processes such as melting or evaporation, abrasion or mechanical fracture are not included in the term corrosion."


text:
Dispersity of synthetic DNA
Recently I was reading an article about DNA nanostructures and I was wondering--
I know that DNA in living systems is monodisperse, but when we make DNA synthetically, is it truly monodisperse or does the dispersity vary depending on how it is synthesized?
If it has the same molecular structure, why does the fact that it's made synthetically change the dispersity?
Thanks

Ten DNA strands of different lenght are still DNA. What do you want to do with it?

Two different cells will replicate DNA with different degrees of fidelity (dispersity, variability), depending on the capabilities and state of the DNA replication and mismatch repair machinery. In the same way, two labs will produce synthetic DNA of different quality depending on a lot of factors including the type of available equipment, chemicals, and  (not least) method of purification. Everything is error-prone, it's a fact of life. Check the message I just wrote and see if you can't find an error (despite my incredible innate error checking machinery). You probably will.


text:
Why are carbamates capable of inhibiting acetylcholinesterase?
Certain carbamate compounds, such as the insecticide carbaryl or the Edgewood carbamates such as EA-3990, can inhibit the enzyme acetylcholinesterase. Unlike organophosphorus compounds, however, they do so reversibly and are bound to the enzyme for minutes, as opposed to permanently. The carbamyl functional group differs from an acetyl group in that, instead of a methyl group being bonded to a carboxyl, a carbamate has amino group bonded to a carboxyl. Why exactly does this difference allow carbamates to inhibit AChE for minutes at a time, and why is cholinesterase that has been inhibited by a carbamate not susceptible to reactivation by oximes such as 2-PAM?


text:
Is the mole fraction used in Raoult's law calculated initially or at equilibrium?
Raoult’s law states that:

The partial pressure of any volatile constituent of a solution at a constant temperature is equal to the vapour pressure of pure constituent multiplied by the mole fraction of that constituent in the solution

My question:
Is the mole fraction of the constituent of the solution to be considered at the time when equilibrium between vapours dissolving into the solution and rate of evaporation has been achieved or at the beginning when the solution is just formed and evaporation has just begun change the mole fraction of the constituent which we have just calculated?
Also, experimentally, is there much difference in the mole fractions at equilibrium and at the beginning when there is apparently no vapour pressure (for zero vapour pressure in the beginning, I think  we can just close the lid of the container)?

Related: https://chemistry.stackexchange.com/questions/44773/raoults-law-and-mole-fraction

Application of Raoult's law assumes an equilibrium has been established between gas and liquid phases (otherwise the standard equations of thermodynamics, of which it forms a part, would not apply). For a mixture of components one can write a set of equations $\gamma_i p = \chi_i p_i^\circ$ that can be solved for a self-consistent composition. For instance, for two components one can write two equations in two unknowns:
$$\begin{align} \gamma_1 p &= \chi_1 p_1^\circ \\(1-\gamma_1) p &= (1-\chi_1) p_2^\circ \end{align}$$
with solution
$$\begin{align} \chi_1 &= \frac{p-p_2^\circ}{p_1^\circ-p_2^\circ} \\
\gamma_1 &= \left(\frac{p-p_2^\circ}{p_1^\circ-p_2^\circ}\right)\left(\frac{p_1^\circ}{p}\right) \end{align}$$
Therefore both the gas phase and solution composition are fixed provided the total pressure and vapor pressure of pure components are known.
As for your second question, it's impossible to answer because it depends on the solvent and details of the experimental setup (size of container relative to volume of solvent, temperature, etc)


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What are the rules for constructing Kohn-Sham orbitals?
I just recently began to study quantum chemistry and need some clarification for the construction of Kohn-Sham orbitals.
Consider a system of several atoms.  Let there be $N$ electrons in this system.
We can reduce the description of such a system to the $\psi_i$ single-electron wave functions describing the orbital, where the index $i$ takes a value from $1$ to $N$.
Now consider the same system in the framework of the  density functional theory.
The description converts to molecular wave-functions $\varphi_j$. Molecular orbitals are obtained from the Kohn-Sham equation:
\begin{align}
  \left(
    - \frac{\hbar^2}{2m} \nabla_j^2
    + \nu_\mathrm{eff}(\mathbf{r})
  \right) \varphi_j(\mathbf{r})
  &= \varepsilon_j\varphi_j(\mathbf{r}), &
  j &= 1,\dots, K.
\end{align}
How is it determined how many molecular orbitals are now? $K=N$?
How do the new molecular orbitals make up the overall wave function of the system?
I am a little confused by the fact that atomic orbitals form molecular ones according to very complex principles and it is unclear how this is taken into account.

Yes K = N. K-S functions are single-electron functions so you need one for each electron.

Thanks for the answer. If $\varphi_i$ is already molecular orbitals, it turns out they don’t have a strict binding to the electron of a particular atom? Moreover, the molecular orbital is a linear combination of atomic orbitals (if work within the framework of MO LCAO)

I'm not sure what you mean  by "strict binding to the electron of a particular atom". In a molecule, electrons aren't associated with specific atoms, nor even with specific orbitals, as all the electrons on the molecule are interchangeable. We use the concept of orbitals to create a mathematical description of the net electron density on the molecule. The mathematics of Kohn-Sham orbitals are such that each one represents the contribution of one electron.


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What is the diffusion limiting current density in comparison to rotating electrodes or fuel cells with pumps/stirrers?
If an electrolytic fuel cell operates in an acidic liquid electrolyte and electrons are supplied to a catalytic electrode so that the protons in the liquid solution will react and produce hydrogen gas, how much is the performance of the fuel cell limited by letting the protons simply diffuse onto the catalyst instead of rotating the electrode or stirring the liquid? I read about diffusion limiting current density but I wonder whether practically, fuel cells operating based on simple molecular passive diffusion are extremely inefficient. Can you give me a difference in magnitude between the current density of a fuel cell based on passive diffusion and a fuel cell based on rotating electrode or stirring? Is it 2 times higher or 2,000 times higher etc?

The question is too broad. These will depend on the type of electrolyte, its viscosity, ionic conductivity, characteristics of the electrode, the stirring rate that you use and so on.