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Non-conventional polymers
IUPAC defines a monomer as,

A molecule which can undergo polymerisation thereby contributing constitutional units to the essential structure of a macromolecule.

Bearing this in mind, I thought about molecules which exhibit hydrogen bonding. Take water for example: one water molecule bonds to another, and that one bonds to another, and another, etc. In the end, you'll have a long chain of water molecules bonded together. This process of water molecules coming together and hydrogen bonding sounds like they 'polymerise' in some way.
To explain my point, consider an infinite amount of water. Here, I'd imagine there to be an infinitely long chain of water molecules bonded together. This sounds like usual polymers I see, where there's an infinitely long chain of repeat units. However, the polymerisation is different for those.
Hence, I have two questions: could a single water molecule be described as a monomer? And, could a long chain of bonded water molecules be described as a polymer?

P.S. This doesn't just have to apply to water. The same could be said about a single alcohol molecule/a long chain of bonded alcohol molecules.

Hydrogen bonds are not actual bonds.

@a-cyclohexane-molecule Of course hydrogen bonds aren't actual bonds, and I never said they were. That's why I said 'it sounds like they polymerise in some way'. However, water molecules still interact in such a way that they 'bond' together, hence my thinking where they could form a 'bonded' chain.

I understand. Regardless, I pointed it out because there's an important distinction between bonds and strong intermolecular interactions. What you describe sounds like an unusual description of phase transitions; in your particular case, condensation of water from water vapor.

@a-cyclohexane-molecule Phase transitions aren't what I was trying to describe, but I agree it does sound a bit like that. However my question was simply whether water molecules could form a very long chain due to hydrogen bonding, and if so, whether you could describe this as being polymer-like.

There is no such thing as polywater but I'm old enough to remember the buzz.

There are actually two ways to answer this question: by lawyering or by trying to capture what a polymer is. I'll try both.
Answer by lawyer
The IUPAC defines the following terms as follows:

monomer: a molecule which can undergo polymerisation thereby contributing constitutional units to the essential structure of a macromolecule
macromolecule: A molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass.
molecule: An electrically neutral entity consisting of more than one atom ($n > 1$). Rigorously, a molecule, in which $n > 1$ must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

Using the data from this paper and a harmonic oscillator approximation, I estimate that the zero-point energy of a harmonic oscillator with two waters and a hydrogen bond is just over 45 kJ/mol.
Since the energy contributed by hydrogen bonding in liquid water is estimated to be between 2kJ/mol and 6kJ/mol, I conclude that the hydrogen bond of water cannot confine a vibrational state.  This means liquid water does not qualify as a single molecule.
Since a long chain of waters connected by hydrogen bonds does not qualify as a molecule, it cannot be a polymer, which means that water cannot be a monomer (at least, not in this sense).

A better answer
The real problem here isn't the technical stuff I wrote above (which, by the way, probably has a few holes large enough to sneak trucks through). It's that words have certain meanings, and we can't capture everything in a simple definition. For the same reason that you'd probably feel a bit weird calling "yellow" a "reddish-green", even though it is literally the result of mixing red and green light, it feels strange to call water a polymer.
Here are some common ideas that spring to mind when I see the word "polymer":

Made of organic monomers bonded together covalently
Structure of covalent bonds (e.g. branching) affects physical properties
Intermolecular forces affects physical properties

These are not strict, but they explain why quartz is not referred to as a polymer, even though it's made of covalently-bonded $\ce{SiO2}$ units: everything is covalently bonded and there aren't any intermolecular interactions to speak of! For another example, depending on the exact context, proteins are sometimes rarely thought of as polymers, even though they're made of repeating amino acids, because their structure is fixed (must be linear chain) and their intermolecular forces don't contribute much.
So the more sensible answer to this question that doesn't rely on lawyering is that liqid water just doesn't have the properties that most people think of when they hear the word "polymer" and so it isn't usually referred to as a polymer.
Supplemental Materials
The code I used to compute the harmonic oscillator energies
from math import pi,sqrt
# Compute the harmonic oscillator energy of two water molecules
# using a two-body QHO approximation
kg_per_amu = 1.66054e-27
watermass = 18.0 * kg_per_amu # Mass of water: close enough for now
reducedmass = watermass**2 / (2 * watermass)

hbar = 1.054571800e-34   # J-s
k = 845.4 # From https://www.princeton.edu/~fhs/paper79/paper79.pdf, N-m

E_zeropt = hbar * pi * sqrt(k / reducedmass) # for n = 0
E_zeropt_kjm = E_zeropt * 6.022e23 / 1000  # Scale from J/molecule to kJ/mol
print(E_zeropt_kjm)

Nice answer. Basically, it wouldn't be wrong per se to say that a chain of hydrogen bonded water molecules could technically be a described as a polymer, but instead it's simply a 'bit weird' and rarely done due to commonplace ideas of what a polymer should be?

@resno66 - Neither liquid water nor solid water is a polymer. Not at all. Stop trying to force fit water into some definition of as polymer. It just isn't.

@MaxW Stop being arrogant. I'm not trying to 'force fit' anything. All I've done is discussed my thoughts on how water sounds like* it could form some long chained 'polymer', and not once did I say that's how water actually behaves. It was simply an interesting thought which I shared. There's no force fitting of any kind. P.S. The read on 'polywater' was interesting - I was meant to mention this above earlier.

@resno66 The fundamental problem is that in a polymer, the bonds need to be covalent or near-covalent in nature, and hydrogen bonds just aren't strong enough to do that (in the "lawyer" section, I show that the bonds need to be 10x stronger to even come close to qualifying). So I'd say it's technically wrong to call a chain of water molecules a polymer...(cont)

I wrote the second part because there are a few issues in the technical section, and even if they turn out to be majorly problematic, I still would not call water a polymer, and those are the "more fuzzy" reasons why I would not do so.


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Effervescent in mineral spirits?
I'm shooting video of paint moving in mineral spirits.  To make for a little more action, is there a household granular/solid I could add to the mineral spirits that would gently effervesce ala alka-seltzer in water?

Are you looking for something that will effervesce in water or mineral spirits?

Something that would effervesce in mineral spirits.

A fish-tank bubbler with needle valve or regulator would not only provide bubbles but be controllable and would not need to be replenished, as would chemicals.
N.B. However you make bubbles in mineral spirits, it would help the highly-flammable liquid to evaporate, creating a potential explosion and fire hazard. You might get better images with less-miscible mineral or silicone oil, with less danger.

Thanks for the reply.  My thought is that a bubbler uses puffs, so it would mess with the effect, and the scale of what I'm doing is very small, a petri dish of spirits, so no hazard ;)

Those using motors do make noticeable puffs, but the vibrator type, especially through a fine stone, tend to make streams... which still might not meet your needs. Dry ice might work, if you could insulate it so that it doesn't sublime too rapidly.

source: Stack Exchange

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

sort: votes


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Meaning of electrochemical vs nonelectrochemical steps
Can someone please explain to me what is meant by an electrochemical reaction step and how it differs from a nonelectrochemical reaction step in the same overall reaction in the context of electrocatalysis?
A simple example with H2 or H2O would be sufficient.

$\ce{H^+ + OH^- <=> H2O}$ is a "nonelectrochemical" reaction step

And electrolysis of $\ce{H2SO4}$ in water is electrochemical. See http://www.bbc.co.uk/schools/gcsebitesize/science/add_gateway_pre_2011/periodictable/electrolysisrev2.shtml

@MaxW can you please elaborate why it is not an electrochemical step. To me it seems that two charged species interacted. This is the confusing part for me. Is there a definition one can use?

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


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Enzyme Kinetics two enzymes in complex with one substrate
Given an enzyme reaction where two enzymes are required to form the complex, for example $\ce{2E +S-> E3S->2E +P}$, how would I calculate complex concentration assuming steady state approximation ($\frac{d\ce{E2S}}{dot}=0$). Does the 2 in front of $\ce{E}$ multiply the rate or is it the exponent when looking at the rates going into the complex?

Welcome to Chem SE! If you haven't yet, take the [tour] and visit the [help]. As of right now, your question is a little unclear. I'm going to edit your question to show you some of the tools you can use on the site to write math/chem equations, but you should try to rephrase your question to better explain your problem. @Hellequin87

As this seems like a homework question (and may be closed), as an example consider the reaction  $\ce{H2 + I2 = 2HI}$. The rate, r is $\displaystyle r=\frac{1}{2}\frac{d\mathrm{[HI]}}{dt}=k_1\mathrm{[H_2][I_2]}-k_2\mathrm{[HI]}^2$ where the rate constants $k_1,\;k_2$ are for the forward and reverse reactions respectively. (This assumes that the reaction occurs as written.) As   steady state the intermediate or transient species has a rate of change of zero and you note in your question.
In the general case if the reaction is $aA+bB= dD+eE$ then $\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{d}\frac{d[D]}{dt}=-\frac{1}{e}\frac{d[E]}{dt}$


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Coordination number of spheres (all identical) in hexagonal close packing (HCP)
I was revising solid state chapter when I came across this question

What is the maximum coordination number of an atom in a hcp crystal
  structure of an element?

The word 'maximum' drew my attention. The answer provided was 12, without any explanation.

Now, it is pretty easy to visualise that the coordination number of the two spheres in the middle of the top and bottom layer is 12, but what about the other ones?
Update
After extending the lattice, it becomes clear that the coordination number of all spheres in layer A will be 12. The spheres in the original hexagon (layer A) can be seen as center points of some other hexagon.
Just confused about layer B.

The drawing only showed a small portion of the total layers of the spheres. There are zillions of spheres in each layer.

@MaxW The pictures show one unit cell. I guess that's the norm — to print only one unit cell in books to save space. Showing more than one would also make the diagram messy since the HCP unit cell is complex, compared to, say, simple cubic or BCC. Anyways, I can't figure out the layer B part (see my edit to the question). Any clues?

@MaxW Nevermind, I saw the layer by layer diagram and the doubt is clarified now. :) ... Turns out, ball and stick model was not the right one to use here.

There are no other ones. All spheres are equivalent.

Yes, in for example magnesium all atoms are the same and there is only one crystallographically distinguishable magnesium atom. It crystallizes in the space group $P6_3$/mmc having the Wyckoff-position of $2c$. If we look into the international tables of crystallography for this space group and the position we can see how the primitive unit cell of magnesium only consists of two atoms with the coordinates (1/3, 2/3, 1/4) and (2/3, 1/3, 3/4). As you can see each of them has a separate link. If we click that for the first atom (1/3, 2/3, 1/4) we can see that by applying the different symmetry operations for this space group and position we can construct 11 other atoms. In the picture below I added them around the magnesium atom with the corresponding coordinates. As you can see the 12th one is actually the other magnesium atom (2/3, 1/3, 3/4). Therefore I colored them in different colors. If you then draw in the polyhedron you can see that this would look the same for the other atoms as well and from the new ones you could go on constructing 11 other ones around it. And so they all have the same coordination number.


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Pseudo 3D skeletal structure style which emphasizes geometry
If one searches and navigates to a wiki page for, say, dichlorodifluoromethane one sees the 2D skeletal structure of the compound as below

This representation appears to be of a particular style that is not exactly what is rendered when one uses a rendering tool such as MolView or obtains data from PubChem and uses other converters/rendering tools.

The big difference may be that the former example shows bond angle information, stereospecificity and generally emphasizes the geometry of the molecule.
My question is whether there a database allowing access to data which are already in this style (e.g. bond angles and bond types included), or does one typically configure the finer details of the drawing / modifies the source data in order to achieve the specific style alluded to above?
Also, what is the most used name for the style of the former example (2D version which conveys 3D information)?

Most commonly known (I think) it is as Wedge-Dash Notation. Related in a way: How to wedge d/hash notation a molecule?, and Hash and Wedge nomenclature. I am not sure whether there are databases, but I think Chemdraw can construct these from a SMILES/InChI string, which in turn you could get from a database. Note that the representations usually use fixed angles and bond lengths, which do not necessarily have a 1:1 correspondence to the actual structure.

Apparently ChemSpider provides: CHFClBr. Please note that the Wedge-dash notation is not recommended for centres in molecules, which are not stereo-specific, i.e. for your example CF2Cl2.

I have not tried it, but the answer of Geoff might help: Free API to view 2D representation of Molecules || If you found something that works for you, I would appreciate that you share it.


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Bond angle in water vs in oxygen difluoride
My teacher taught that bond angle $\ce{H-O-H}$ is larger than $\ce{F-O-F}$ due to less repulsion in the latter's bond pairs.
I know that bond angle $\ce{H-O-H}$ is $104.5^\circ$, but I'm confused with that of $\ce{F-O-F}$ because variable angles are given in articles on websites and in books. Some are $109.47^\circ$, $101.55^\circ$, $103.1^\circ$.
I want to know which bond angle is greater and why?

We prefer to not use MathJax in the title field, see here for details.

Water has a bond angle of 105º and oxygen diflouride has a bond angle of 103.8º, which cannot be explained by repulsion since fluorine is larger than hydrogen$[1]$. This is phenomenon is, however, explained by Bent's rule:

Bent's rule is that in a molecule, a central atom bonded to multiple groups will hybridise so that orbitals with more s character are directed towards electropositive groups, while orbitals with more p character will be directed towards groups that are more electronegative.

p orbitals are oriented 90º from each other, so orbitals with more p character have smaller angles between neighboring atoms. The effects of this phenomenon can also demonstrated by comparing bond lengths.

$[1]$ Bent, H. A. Chemical Reviews 1961, 61 (3), 288.

Where did you get the value for oxygen difluoride from? I couldn't find anything reliable.

On the Wikipedia page for Bent's Rule. I'm trying to find another source now.

So Bent's original paper actually uses 103.8º for the bond angle of oxygen difluoride. He cites Ibers, J. A., and Schomaker, V.: J. Phys. Chem. 1953, 57, 699 .

On the DF-B97D3/def2-TZVPP level of theory the bond angle of water is $104.2^\circ$. Which matches up quite well, see what is the bond angle of water? The bond angle of oxygen difluoride is $104.9^\circ$. Unfortunately I couldn't find a good source for an experimental value. (Also the CCCBDB currently only produces errors.)
The values are too close together to actually compare them on this level of theory. According to Bent's rule (What is Bent's rule?) I would have expected the bond angle to be smaller in $\ce{OF2}$.

Maybe CCCBDB starts working again:

Oxygen difluoride: https://cccbdb.nist.gov/alldata2x.asp?casno=7783417&charge=0
Water:             https://cccbdb.nist.gov/alldata2x.asp?casno=7732185&charge=0


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Do milk and orange juice contain H2O molecules?
I'm asking as there is this woman who claims she's severely allergic to (you guessed it) water (H2O molecules). She says just a drop of water in her mouth sends her into anaphylactic shock and only adrenaline reverses this.
Now how's this relevant to this question? She says she survives on milk and orange juice, which ''because of their chemical composition'' bring her no harm. She's basically saying orange juice and milk are something 100% separate from water and don't contain free H2O molecules in them.

Here is some science for your own curiosity (full disclosure, it was my question) https://biology.stackexchange.com/questions/7905/what-is-the-smallest-molecule-that-can-present-as-an-antigen-to-the-immune-syste, but the direct answer to your question is that both contain significant amounts of water.

Off topic but lot of people say they are allergic to things just to call attention...might be particular of our times...

Most of the weight of a human body is water. Anyone allergic to water would die instantly. This sort of malarkey is why STEM education is necessary.

There is such a thing as "water allergy" but it's only a problem for getting water on your skin, not drinking it. https://en.wikipedia.org/wiki/Aquagenic_urticaria

@DSVA I'd just use quotations for the "allergy" bit (since it isn't technically an allergy) ;-) Great find by the way!

This popped up again. Besides some contaminate in water, I'd wonder if pure water on skin could allow some bacteria, fungus, or mold to thrive on the skin. So the biological contaminant is dormant when dry, but quickly becomes active when moist.

Both orange juice and milk are approx. 88% water, just like many other foods. The human body is about 60% water as well.
Therefore, she must be allergic to something her tap water contains other than water, such as chlorate from water treatment or some other contaminant which is not contained in orange juice or milk, assuming what is claimed is a fact and not just her imagination.
The solution is to install a good water filter in her faucet or drink bottled water in case there actually is some allergen in the tap water. In the worse case, she can try a drop of distilled water (which is pure water) just to confirm experimentally as well that it is not H2O that is the issue. Note: Drinking distilled water regularly is sometimes said to cause health problems as in the long term it deprives the body of minerals; see, for example, https://www.medicalnewstoday.com/articles/317698.php and https://van.physics.illinois.edu/qa/listing.php?id=28557. However, this is a controversial topic.

@Marzipanherz Edited accordingly, i have softened up the distilled water story as well as i made it sound more serious than it actually is.

We should not give medical advice here. The woman needs to see a doctor. What kind is the only question. The OP should not under any circumstances do medical experiments with her.

@MaxW Good point, sorted.

https://en.wikipedia.org/wiki/Aquagenic_urticaria technically it's not an allergy but it seems like it actually is the water and not something in it. But afaik these people do not have problems drinking water since you don't got skin inside of you.

Had a look too and, believe it or not, this is actually a thing indeed: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5438944/ Very controversial however and i am really not convinced: so you can bath in orange juice but not water? Rules out UK as a habitat then. Also, not a medical journal expert but the journals i found this published in seem very weak.


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What coefficients can I use in a Reaction-Diffusion Equation?
I am qualitatively analyzing an abstract chemical reaction
$A + B \rightarrow 2A$
where $A$ and $B$ are some liquid substances that undergo Reaction-Diffusion with a second order reaction (actually my model is much more complicated than that, but this is for simplicity). I have Reaction-Diffusion Equation that models this reaction:
$u = \left(\begin{array}{c}a\\b\end{array}\right), \quad u(0) = u_0, \quad d \in \mathbb{R}_+, k \in \mathbb{R}_+$,
$a_t = d \Delta a + k a b$,
$b_t = d \Delta b - k a b$.
However, I need some specific coefficients $d$ and $k$, and starting densities for $u$ to do my simulations on.
What would be a decent Fermi estimate for diffusion coefficient $d$, reaction rate $k$ and density $u_0$ for reactions present in reality?
In terms of speed, I'm interested in reactions like one of reactions present in the Belousov–Zhabotinsky Reaction-Diffusion model and I just need a very, very rough estimate of values (with physical units) to put in my model so that I keep studying behavior of reactions that might plausibly happen in reality for some specific substances. So far I used $k \approx 1 \frac{dm^3}{kg \cdotp s}$, $d \approx 1 \frac{dm^2}{s}$, and $u_0 \approx 1 \frac{kg}{dm^3}$, but I'm particularly unsure if this ratio of $k$ to $d$ is anywhere near realistic. I know I could just scale time, space, and mass in this equation, but I'm interested in specific value ranges or examples.

Too little information. For example diffusion coefficients tend to depend on if the system being studied is a gas, liquid or solid.

Thanks for the comment, I updated the question. So, let it be liquids. But is that of the order of 1 or more like 1 million? Would you be able to give an example of concrete coefficients for any R-D equation like that?

What you use depends on the specific reaction and then whether or not you use 'dimensionless' values. As an example of generating numerical version of equations see 'From Calculus to Chaos' by D. Acheson (publ. OUP) or for pattern generation  'Essential Mathematical Biology' by N. Britton (publ Springer) . As a start you might consider using concentration is units of, say, millimolar and times in millsec.

Diffusion coefficients for small molecules in water or other mobile solvents are around $10^{-9} \mathrm{dm^2s^{-1}}$, rate constants can vary hugely but say no more than  $ 10^6\mathrm{dm^3/mol/s}$ would probably be big enough and concentrations in the millimolar range.

Thank you for your comments, @porphyrin. If you could write an answer out of them, I'd accept it.


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Comparision of acidic strength of benzoic and peroxy benzoic acid
Out of the two which one is stronger  acid ?

Benzoic acid is 103.7 times stronger:
peroxybenzoic acid pKA = 7.9 [ref 1]
benzoic acid pKA = 4.2 [ref 2]
References:
[1]:  Ullmann's Encyclopedia of Industrial Chemistry, 5th ed. Vol. A19, Wiley, p. 206
[2]: Harris, Daniel (2010). Quantitative Chemical Analysis (8 ed.). New York: W. H. Freeman and Company. pp. AP12.


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Which chemical reactions don't involve bond breaking?
I recently came across a question in which an option is 'Breaking a chemical bond is the first step in any chemical reaction', which lead me to think of reactions which do not involve bond breaking. The only reaction I can think of is dimer formation, but a Google search reveals no more. Surely this cannot be the only example?

The reaction of a Lewis acid with a Lewis base results only in bond formation. Your question gets at the essence of chemistry, though. A chemical reaction is the rearrangement of electrons between atoms. So generally, most reactions will involve breaking bonds. Even for monomers (e.g. alkenes) to polymerize, $\pi$-bonds within the monomers have to be broken to form $\sigma$-bonds in the polymer. You seldom have one without the other.

Electron transfer is such an example; $\ce{D + A \to D^+ = A^-}$.

Precipitation reactions?  If you will grant me that an aqueous ion is free.  (But maybe you don't and consider the solvated ion bonded to nearby water.)

Per a source, to quote:

Like ozone, the predominant oxidation reaction mechanism for chlorine dioxide proceeds through a process known as free radical electrophilic (i.e. electron-attracting) abstraction rather than by oxidative substitution or addition (as in chlorinating agents such as chlorine or hypochlorite).

So, the reaction:
$\ce{.ClO2 + e- -> ClO2-}$


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How can I track the progress / read the output file of an ongoing calculation in GAMESS-LIBCCHEM from NVIDIA?
That is my question:

How can I track the progress / read the output file of an ongoing calculation in GAMESS-LIBCCHEM from NVIDIA?

The steps I have followed are in:

https://www.nvidia.com/en-us/data-center/gpu-accelerated-applications/gamess/

In there, there is a command line:
nvidia-docker run -v $(pwd):/results --rm nvcr.io/hpc/gamess:17.09-r2-libcchem -c "cd /workspace/examples && rungms cc-h2co.inp"

But, that command line print an output file only when the calculation has finished.
I want to read the output while the calculation is running and I do not know how.
I am running it on my laptop: i7 + geforce 940m + 12 gb of ram

You generally have much better luck with such highly specialized questions, if you contact either the developers themselves (nvidia-docker) or the user community.

This seems to relate to a specific program or programming issue, rather than an underlying chemical issue

Of course, you could always bring this up at chat. In matters pertaining to computational chem, I recommend asking Mod. @Martin about it ;-)


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Bromides limit test
I have a question in the monograph of sodium chloride. In the limit test of bromides the limit is 100 ppm but the concentration of the standard is much lower than the limit. So I want to know why.
Bromides
Maximum 100 ppm.
To 0.5 mL of solution S add 4.0 mL of water R, 2.0 mL of phenol red solution R2 and  1.0 mL of a 0.1 g/L solution of chloramine R and mix immediately. After exactly 2  min, add 0.15 mL of 0.1 M sodium thiosulfate, mix and dilute to 10.0 mL with water  R. The absorbance (2.2.25) of the solution measured at 590 nm, using water R as  the compensation liquid, is not greater than that of a standard prepared at the same  time and in the same manner, using 5.0 mL of a 3.0 mg/L solution of potassium  bromide R.

Without doing all the math I can hazard an educated guess.
First let's clear up the question.
For the sample its initial volume of 0.5 ml becomes 7.5 ml after all the reagents are added. So a 100 ppm max of $\ce{Br^-}$ becomes 6.67 ppm in the analyzed solution.
For the standard the solution is 67.1% $\ce{Br^-}$. I'm assuming it doesn't get diluted with the 4.5 ml of water, so the final volume would be 8 ml with the reagents. Thus the final solution would be 1.26 pmm.
So the difference is about a factor of 5.
Now typically one would create a calibration curve with multiple standards and calculate a best fit line. Around the best fit line the best precision will be at some point of line segment. Let's call that the "sweet spot" on the calibration curve.
Since absorbance is linear with concentration, the fitted line should go through the origin. Thus only one standard is really necessary. (Be careful here. Absorbance is linear with concentration, but the instrument is measuring transmission. So small differences in a lot of light are hard to measure, and small differences in very little light correspond to large differences in concentration.)
For all the error sources in the analysis the sweet spot must be around 1.26 ppm on the calibration curve.

I thought about the answer and I found it. Thanks for you efforts.

@abanoub - Was the answer something different? Bit of an explanation? Please add your own answer.


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Is there any difference between negative electron gain enthalpy and electron affinity?
Electron affinity is the amount of energy "released" during the addition of an electron in the valence shell of an isolated gaseous atom. The sign convention is opposite to that of thermodynamics convention i.e. positive means release of energy.
For many non-metals, the first electron affinity is exothermic, i.e. positive while all second electron affinities are endothermic, i.e. negative.
As we already know that electron affinity is energy released (the term "affinity" implies an attraction for the electron), then why we consider endothermic cases using this term?

https://chemistry.stackexchange.com/questions/64735/are-1st-electron-affinities-positive-or-negative

Well, energy is not always "released" when an electron is added to an atom in the ground state.
Noble gases for example, have a stable configuration and a high amount of energy is required to force the electron into a higher energy state and a less stable configuration, thus making the electron affinity negative.


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What compounds give two products on reacting with NH2OH?
This is a question from the chapter stereochemistry of my coaching module. The answer options include CH3CHO, DCHO, etc. How do I do this? I have an idea that if we take a racemic mixture of D and L isomers it is possible to get such products. Am I right? And is it possible? How do I do this question?

The oximes formed are not d/l, but E/Z so any assymmetric carbonyl compound will give rise to 2  products, DCHO for example will give 2 oximes while acetone will not.

@Waylander actually my book hasn't covered formation of oximes, it has only coveredmain definitions in stereochemistry like e/z, I'm not exactly sure what an oxime is, is there another way to get this answer?

Google "oxime formation". There is a good video explanation from the Khan Academy


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How can I prevent the effervescence of a solvent during evaporation?
How can I prevent the effervescence of a solvent like petroleum ether when I evaporate it on a water bath?

Break a ceramic utensil (mug, plate, any porous item will do) and add a few chips in the solution. This should lead to a smoother boil rather than the liquid jumping around.
This is a cheap alternative of anti bumping granules and works in a similar way by allowing the gas to accumulate on the rough surface and escape the solvent in a controlled fashion.


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Why Benzene is a good solvent even while it is not polar
Consider for a moment, ether, THF, benzene and hexane.
Why is the solubility of many organics in these solvents in the following order
THF > ether / benzene > hexane
What makes benzene a good solvent for many solutes when compared with hexane even while both solvents lack a permanent dipole moment.

It is not a good solvent because it is carcinogenic and banned from most good labs.

No benzene is a very good solvent, just a pitty it forms a carcinogenic metabolite in the human body.

A good solvent for non-polar molecules, organic or otherwise, but would be a bad solvent for ionic and highly polar  species.

@Abc Going by site-norms, it technically is homework :/ (Homework questions must demonstrate some effort to understand the underlying concepts - Else they'll be closed as HW)

I'd say that "demonstrate some effort to understand the underlying concepts" in this case would be equivalent to answering the problem, thus no point posting a question. On the other hand this apparently simple question has provided a very nice answer and could use others, so I vote to reopen.

Regarding "good" I think it is quite clear what is meant: the ability to provide a suitable solvent

Benzene does not have a permanent dipole, but it does have a quadrupole, which hexane does not. That is why it behaves as a more polar solvent.

Benzene is a good solvent for many organic compounds, the term good solvent to me does not mean a nice solvent with a good all round profile.
To me the perfect solvent would be non toxic, not flammable but combustable, very resistant to radiation, inert to nitric acid, be easy to distill, be cheap and easy to obtain. It would also have to be able to dissolve a useful range of different things.
Benzene is more able to dissolve many organic substances than a saturated hydrocarbon such as hexane, this is becuase the pi cloud of the benzene ring is able to interact with the pi clouds of various solutes. For example polystyrene will be solvated better by benzene than by hexane for this reason.
This good ability to dissolve things is why I regard benzene as a good solvent. It is a good solvent to dissolve things but often a bad thing to work with in the lab. Benzene has one special use, as a NMR solvent. One of my favourite tricks is if my student can not understand a proton NMR spectrum due to overlapping peaks. I tell them to repeat the measurement in d6 benzene. Often a lot of the chemical shifts for protons change, with a little luck the resulting spectrum is more easy to understand. Sometimes you need to look at both the chloroform and benzene solution NMR spectra to be able to understand enough of the spectrum to understand what your structure is.
While benzene is an unpopular solvent due to the health issues and flammability, there are plenty of aromatic solvents which have better health and fire profiles. For example Solvesso 150ND is an aromatic kerosene which is marketed by Exxon, this has different solvent properties to the aliphatic saturated kerosene marketed by Statoil under the name of "solvent 70".
This can have some advantages, if I want to dilute the ionic liquid aliquat 336 to 30 % in a diluent and then use this solution to extract metals such as copper, cobalt and iron from aqueous chloride media. Then with solvent 70 as the diluent (diluting agent) then the organic phase is very prone to separate into a layer of ionic liquid with metal complexes as the anions and a diluent rich layer. But when an aromatic solvent such as ethyl benzene or solvesso 150 is used then the organic phase can hold a lot more metal before this unwanted phase splitting occurs.
The aromatic solvent (diluents) tend to be more toxic than the saturated aliphatic (alkane) diluents. If you look at "A comparison of two methods of recovering cobalt from a deep eutectic solvent: Implications for battery recycling" by F.J. Albler et. al., Journal of Cleaner Production, 2017, volume 167, pages 806-814 it contains an assessment of leaving a puddle on the floor of an unventillated room is made for both solvent 70 and ethyl benzene / toluene is made. The solvent 70 level in the air is likely to pose a far smaller threat to health than the aromatics. But aromatics like solvesso 150 are likely to pose a far smaller carcinogenic threat than benzene.
It is important to not regard all alkanes or aromatics as being the same, hexane is a popular solvent in chemistry but it is very toxic. In the body it is converted to hexane2.5-dione which is very bad for you. But the other longer alkanes do not have the same harmful effect. You also need to consider the exposure route. For example a puddle of diesel fuel on the floor of a garage is far less harmful than exposure of a person's hand to a high pressure spray of diesel which injects it into the person's body.


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Can enamines be formed from α,β-unsaturated ketones?
This question was a result of discussion form here: α,β-unsaturated carbonyl compounds and alkyl addition
The reaction below (condensation of cyclohexenone with a secondary amine) does not yield any results on a literature search, even when I only search for the product by itself (unless my search was not good enough):

Can enamines actually be formed from α,β-unsaturated ketones? If not, what prevents it from being possible?

These two references concern the photoaddition of amines to benzene, not the addition to cyclohexenone. Enamine formation of cyclohexenone itself may be problematic but dienamines of enones have been reported. I'll put up an answer

@user55119 You are right, i made a daft mistake in placing the double bond. I corrected it now and removed the now irrelevant references.

Did find this one: http://pubs.rsc.org/en/Content/ArticleLanding/2016/RA/C6RA20163F#!divAbstract, but it is computational only not synthetic.

I'm not sure to what extent conjugate addition to the enone competes with enamine formation. At least, that's the first reaction I'd think of when I see amine + enone. Removing water will probably favour enamine formation if there's an equilibrium being established, which is probably why user55119's example has toluene+reflux (which I interpret to be a Dean-Stark setup). Morpholine boils at 129 °C which helps (toluene at 111 °C).

Stork and Birnbaum have reported (Tetrahedron Lett. 1961, 10, 313) the formation of dienamine 2 from octalone 1. Alkylation with methyl iodide leads to α-alkylation of the dienamine. Buffer conditions are required for hydrolysis of the dienamine to octalone 4.

This is a good answer but i think it fits really well with the original question (https://chemistry.stackexchange.com/questions/57494/α-β-unsaturated-carbonyl-compounds-and-alkyl-addition) as a special occasion where an α,β unsaturated ketone can actually form the enamine which then alkylates. I suggest to move this answer there. It doenst fit here that much since this question refers to formation (or not) of the imine 1 in the question. I also made a mistake in the question above, now corrected, which may have confused you, sorry.

You should correct the enamine reaction with methyl vinyl ketone  (MVK) at https://chemistry.stackexchange.com/questions/57494/%CE%B1-%CE%B2-unsaturated-carbonyl-compounds-and-alkyl-addition. The reaction continues to give  the dienamine of the aldol condensation product (Cf., 2 above). Perhaps using acrylonitrile rather than MVK would eliminate the issue of the aldol step. After addition, the less substituted enamine is formed.

Ok, i made the reaction more general. Also referenced your reply here since although the reaction doenst seem to be reported with pyrrolidine, in general it is possible to add alkyl groups in α,β-unsaturated carbonyl compounds by using secondary amines. Lastly, compounds like 1 above, namely 2 amido dienes can be formed but their synthesis seems more involved according to this: Beilstein J. Org. Chem. 2011, 7, 410–420. doi:10.3762/bjoc.7.53.  Not sure why though, which is the actual question!


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Water boiling with salt and without it
Does 1L of water with salt boil faster or slower than one with the same amount of water but without salt?
What I know is that since when you add salt to the water the temperature of boiling increase than the water with salt with take longer.

The rate of boiling is dependent on heat influx

What doe you mean with “boil faster”? E.g. reaching boiling temperature faster starting from room temperature, or evaporating faster to dryness?

Loong: reaching boiling temperature faster starting from room temperature

If the heat applied is equal than yes, 1 L of salt water boil slower than 1 L of unsalted water.
If you are intrested you can calculate the diference in boiling point with the formula $\Delta T = K\ b\ i$, where $i$ is the Van't Hoff factor, (2 for the salt NaCl), and $K$ is the ebuliocopic constant of water ($0.512\ \mathrm{^\circ C\ kg\ mol^{-1}}$), and $b$ is the molality of solution.

Thank you! Yes the same amount of heat is applied.


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Second order reaction cases
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Second-Order_Reactions
When you will read this webpage ,then you will find that they have made cases for different types of second order reaction .
In  $situation 2A $ of case 2 in  the article (eq 1.13 and 1.14) ,they have taken change in concentration of A to be dx/dt instead of
d(a-x)/dt .What is the reason for this ?
Also what I am also finding difficulty is how did they combine the reactants to make it $[A]^2$.......  it would mean that individually the concentration of a single reactant A  is following 1st order reaction. So the rate constant in disappearence of A should be first order instead of second so what is the reason for this .I t would have make sense for 2A to follow second order instead of just A

In linked article all the equations have numbers. Use the numbers to pin point exactly what equations baffle you. Why should we have to guess?!?

I can't see the equation numbers as they aren't shown in my opened webpage but still I have added more details in my question

If you are talking about the 1.13 and 1.14 equations, than whta he did was just a simple subistituction where $[A] = x,\ d[A] = dx$, if not, please identify the step.
And, yes, you are going to get diferent equations for $[A]_{(t)}$ depending on the order of reaction, $0^{th}$ order gives you a linear equation, $1^{st}$ order gives you an exponetial equation, and all the orders higher than 1 gives you a fraction diference like: $$\frac{1}{A_t^n}-\frac{1}{A_0^n} = n\ k\ t$$ where n is the order of reaction minus 2.
The step where they make $[A]^2$ is only because you have an elementary equation, it can not be assumed to be that if the equation was golbal or not elementary.


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Balancing the intermolecular redox reaction
The redox reaction is:
$\ce{SO2 + Na2CrO4 + H2SO4 ->  Na2SO4 + Cr2(SO4)3 + H2O}$
This is what I tried(tho I got stuck at one point which is indeed the exact reason why I've asked this question):
Oxidation half: $\ce{S^4+ -> S^6+ + 2e-}$
Reduction half: $\ce{2Cr^6+ + 6e- -> (Cr^3+)2}$
Net reaction: $\ce{3S^4+ + 2Cr^6+ -> 3S^6+ + (Cr^3+)2}$
Now how do I determine the distribution of $\ce{S^{6+}}$ ions among $\ce{Na_2SO_4}$ and $\ce{Cr_2(SO_4)_3}$??
A similar question to this is:
$\ce{C2H5OH + I2 + OH- -> CHI3 + HCO2- + I- + H2O}$
In the latter one, the distribution of $\ce{C^{2+}}$ among the products is also ambiguous.

I don't like the way you wrote the half cells. The overall notion is that when you add the half cells that you get the overall reaction. The point being you can't look up the half cells in a table the way you wrote them.

@MaxW I've always been taught to write them this way(even the books that I use for practice do it this way)...perhaps you can show me the correct way??

I'm used to using a standard electrode potential table as linked and the reduction reaction would be $$\ce{SO4^{2−} + 4 H+ + 2 e−  <=>  SO2(aq) + 2 H2O }\tag{EMF   +0.17}$$ to get oxidation reaction you of course flip equation and sign of EMF.

@MaxW oh okay...thank you..it is surely the more correct way to represent a half cell rxn..but in my scenario I'm just supposed to solve questions as fast as I can..most of the times its not even necessary to write those half cells while balancing a redox...so we skip most of the details...call it the flaw of education system or something else lol...but that's just how it is

With your way how do you tell the difference between $$\ce{Cr2O7^{2−} + 14H^+ +  6e− <=> 2Cr^{3+} + 7H2O}$$ and $$\ce{CrO4^{2-} + 4H2O + 3e− <=> Cr(OH)3 + 5 OH^−}$$

@MaxW for the first rxn..my way for writing it would be: $\ce{(Cr^6+)2 + 6e- \to 2Cr^3+}$ and for the second one: $\ce{Cr^6+ +3e- \to Cr^3+}$

Well, at first it seems a very complicated equation, but if we remove the soluble ions it gets easier to balance and than we can add then latter. So, this
$$\ce{SO2 + Na2CrO4 + H2SO4 -> Na2SO4 + Cr2(SO4)3 + H2O}$$
becomes this:
$$\ce{SO2 + CrO4^{-2} -> SO4^{-2} + Cr^{+3}}$$
The sodium ions where there to balance the charge, the sulfuric acid to identify the mean and the water will appear in solution.
Doing this we get that:
The reduction half: $\ce{CrO4^{-2} + 8H+ + 3e- -> Cr^{+3} + 4H2O}$
The oxidation half: $\ce{SO2 + 2H2O -> SO_4^{-2} + 4H+ + 2e-}$
Now multiply each half-cell by an integer necessary to balance the electrons and add the two half cells together.
$$\ce{3SO2 + 2CrO4^{-2} + 4H+ -> 3SO4^{-2} + 2Cr^{+3} + 2H2O}$$
Now, we need to add the spectator ions. So first we're going to add sulfate ions. In the process, we're going to join the already existing sulfate and chromium ions.
$$\ce{3SO2 + 2CrO4^{-2} + 2H2SO4 -> 2SO4^{-2} + Cr2(SO4)3 + 2H2O}$$
Now, is just add the sodium ions, and we're done:
$$\ce{3SO2 + 2Na2CrO4 + 2H2SO4 -> 2Na2SO4 + Cr2(SO4)3 + 2H2O}$$

I did not show the betwen steps, because I think you're capable of doing it your self with the simplified equation, but, if you need, I can edit and add the missing steps.
A tip that I give to you is balance the charges with electrons after every thing, so you dont need to find any NOX. After this is just a matter of summation of equations.

damn...that was simple as hell...thank you especially for that second last step...my book had a statement that really confused me: "three $S^{6+}$ ions are distributed in 2 molecules of $Na_2SO_4$ and 1 molecule of $\ce{Cr2(SO4)3}$."

In the oxidation half cell there should be two waters. // I'd also change "by this you can solve..." to "Now multiple each half-cell by an integer necessary to balance the electrons and add the two half cells together." // Where you have "Now, we need to neutralize the charges." that isn't right. The charges should already be balanced since the half cell charges are balanced. You need to add the "spectator" anions and cations.

MaxW I'm going to fix it. Thanks. As I'm not fluent speeker, a few things can be misspeled.
And in the neutralization step, I was talking about the charges present in both sides, and not any unbalanced one, but I'm going to change for something more comprehensive.

Nice. I was trying to tell @Carrick that he was making it hard on himself.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. The mhchem construct \ce{...} makes writing chemical equations much easier, please have a look here and here.

@Martin-マーチン Oh! Thank you. My fingertips have been painfull for two days.


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why do the complex [Cu(NH3)5NO2]SO4 show ionisation isomerism?
In ionisation isomerism, secondary valencies can be replaced by primary valencies on the condition that net charge going outside the coordination sphere must be equal to net charge coming inside it . Here NO2 ligand has charge - 1 on it . So how can it be replaced by SO4 which has -2 charge? Thanks!

You have two complexes.

$\ce{[Cu(NH3)5(NO2)]SO4^{-1}}$
$\ce{[Cu(NH3)5(SO4)]NO2^{-1}}$

Both $\ce{NO2^-}$ and $\ce{SO4^{2-}}$ are necessary for the overall complex to have a -1 charge overall. It just depends on which is attached directly to the copper ion.


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Special first order reactions

I am having trouble understanding the equations....how is concenteration of ester proportional to the volume of NaOH used  and how is initial concentration proportional to difference in the Volume of NaOH used at infinity and time t=0....How did we find this relation
The same thing happens during the hydrolysis of sucrose  or inversion of cane sugar where the reaction is monitored by polarimeter and the readings of the polarimeter are used .What is the common link in these types of reactions as both give equations in same type

Is the question alright now?? I didn't mean to ask this as a homework type format rather as one asking for the explanation for one of the steps here.

In a first order reaction it is only necessary to measure a quantity proportional to concentration. This could be a voltage or volume of reactant added etc. The reason this happens is because the ratio of quantities is all that is needed, thus if $X$ is some quantity at time $t$ and $X_0$ that initially there ($t=0$) then the equation for reaction is $\displaystyle X=X_0e^{-kt}$ where $k$ is the rate constant. Rearranging gives $\displaystyle \ln\left(\frac{X}{X_0}\right)=-kt$ and as you can see only the ratio is needed.
In your case, you can see from the equations you show, how the volume is used to be equivalent to something proportional to concentration $X$ and $X_0$ as used above. This is because of the way the titration works to neutralise acid with alkali (or vice versa) with equal amounts of each and for this an indicator is used to indicate the 'end point'.

But I don't undrrstand one thing ......if difference between the volume at t=infinite and initial volume is directly proportional to a then why can't we say difference between volumes at t=t  and volume at t=0 is proportional to a-x?.I am asking this silly doubt because I don't wanna cram the equation....instead learn how to use it

And what is the role of H+ ions here.Will they be come during rate lawv or are they ignored due to being in excess?


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Enol form of acetone is treated with D2O

Could find the mechanism in brief online. Still have few queries. What makes the D2O not attack the the central carbon (to which OH is attached) and choose other the carbon instead (both have a double bond). Again, in the 'repeat' part of mechanism, there's only one carbon with double-bond (carbonyl group). Why does D2O attack the second carbon again and not the carbonyl group one? I'm really confused.

While I would encourage the questioner to offer additional thoughts about the mechanism of this process, I don't think this question should be closed. There is a good question contained here that shows some understanding of the process (i.e., why doesn't water attack the carbonyl directly).

The oxygen of the water will attack the carbon of the C=O forming the hydrate, but this is in equilibrium and completely reversible and a dead end so of no consequence when considering the enol reaction.

Short answer - Electrphile seeks an electron rich site to attack
Explanation -
D2O contains partial charges on the atoms because of the difference in the electronegativity of D and O. Oxygen being more electronegative than Deuterium has partial negative charge and D has partial positive charge.
In the enol form the carbon atom directly attached to Oxygen also has partial positive charge (difference in electronegativity O being more electronegative than C). So, the already partially positive charged Deuterium can't attack a site which also has electron deficiency. So, it attacks the carbon next to it which is slightly electron rich when Oxygen lone pairs attack the adjacent carbon.


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Computing the turbulent flame length
There is a problem I found online for practice that I cannot solve:
Problem

Methane, $\ce{CH4}$, emerges from a $1\ \mathrm{cm}$ diameter nozzle
  at a rate of $120\ \mathrm{cm^3/s}$ and a mole fraction of $0.04$, and burns as a turbulent flame. The laminar flame speed of methane is $45\ \mathrm{cm/s}$ and the turbulent intensity is measured as $v_\mathrm{rms}=2\ \mathrm{m/s}$. Given the correlation for the turbulent flame speed $$S_\mathrm t = 0.35 S_\mathrm L\left(\frac{v'_\mathrm{rms}}{S_\mathrm L}\right)^{0.7}$$
  estimate the length of the flame. Assume that the flame is conical in shape

What I've tried

Compute the turbulent flame speed $S_\mathrm t=4.47\ \mathrm{m/s}$
From mass conservation, the mass flow rate exiting the pipe equal to the mass flow rate going through the flame. But there are still two unknowns before we can compute the area of the cone: the density of unburnt mixture and of the flame.
Assume $p=1\ \mathrm{atm}$ and $T=300\ \mathrm K$, compute density at the pipe exit using the Ideal gas Law $\rho = 1.15\ \mathrm{kg/m^3}$
Compute the adiabatic flame temperature at those conditions $T_\mathrm{ad}\approx1050\ \mathrm K$ (using $C_p$ at $300\ \mathrm K$)
Compute the resulting density at the flame $\rho_\mathrm f=0.323\ \mathrm{kg/m^3}$.
Compute the area of the cone and then the height of the cone

This method yield to the computation of the negative square root. What am I doing wrong? Even if there is a better way to approach this problem, I feel like my method should still give good results.

(1) think temperature is red herring (2) phrase "and the turbulent intensity is 0 measured as v rms = 2m/s." seems wrong.

@MaxW (2) was a typo but what about (1)? Problem does not mention temperature, that's just what I came up with.

I'd assume that flow out from nozzle was a stack of thin discs. So the cone is from a number of stacked discs. Each disc travels away from nozzle at 45 cm/s. Turbulent flow is due to oxygen diffusing from sides of disc which happens at 4.47 m/s. So it takes so time for oxygen to diffuse into center of disc.


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What is 'desorption hysteresis'?
I am currently reading a paper titled "Polystyrene Nanoplastics-Enhanced Contaminant Transport: Role of Irreversible Adsorption in Glassy Polymeric Domain" (link here).
In the paper, the authors mention the term 'desorption hysteresis' as the mechanism of the interaction between nanoplastics and contaminants. Here's the relevant excerpt (from the introduction section):

... at low nanoparticle (i.e., “carrier”) concentrations significant facilitated
    transport of organic contaminants requires not only strong adsorption of contaminants to the carriers, but also significant desorption hysteresis (a collective term referring to both slow desorption kinetics and thermodynamically irreversible adsorption) of contaminants from the carriers. Desorption hysteresis can either be due to the physical entrapment of contaminants in the complex matrices of the carriers, or strong specific adsorptive interactions that lead to irreversible binding of contaminants to the carriers.

I am having some difficulty in understanding what do they mean by adsorption hysteresis. From what I've read, hysteresis is how the system in its current state is related to some past state, but I don't understand this concept fully.
Could someone please explain what is desorption hysteresis? I couldn't even understand the explanation in parentheses (right after the term in bold above). Is it about the rate of desorption of contaminants from nanoparticles?

Hysteresis occurs when the state of the system depends on its past states---that is, the state of the system is memory-dependent.
The canonical example is a ferromagnet: if you apply and slowly ramp up an external magnetic field to a ferromagnet for some time, and then slowly ramp the magnetic field back down for the same amount of time, you might expect the ferromagnet to return to its original state, but it doesn't. Instead, if you track the magnetization of the magnet as a function of the applied magnetic field, you see these hysteresis loops. Each applied magnetic field corresponds, therefore, to two possible system states, and which state the system exists in will depend on its past history.
Desorption hysteresis is hysteresis applied not to magnetization but to adsorption and desorption, and your excerpt then goes on to describe the physical interactions that drive this phenomenon.

Thanks, but I still don't understand how the concept of hysteresis applies to desorption. Does it mean that whether the desorption of contaminants from nanoparticles occurs or not depends on the system's past? If so, what past state is that?

@Don_S, yes. The article mentions thermodynamically irreversible adsorption. If you repeatedly perform adsorption-desorption cycles, eventually the adsorption sites in your material will be clogged up by these irreversible adsorption events and your material won't be able to adsorb and desorb as much stuff anymore.

In the end of the Materials and Methods section, the authors calculate the Hysteresis Index (HI), and they say that "the higher the HI value, the greater degree of desorption hysteresis". From the HI calculation, it appears that the greater the observed concentration of adsorbed-phase (over the calculated concentration), the higher HI will be, but what does that mean? Why would the degree of desorption hysteresis be greater if the observed concentration will be larger than the calculated concentration? Also, what is the practical meaning of the desorption hysteresis degree being "greater"?

@Don_S, those questions are unrelated to your original question and to my answer, and I don't feel comfortable answering them. I recommend you post those questions as a separate question.

Are they truly unrelated? It all revolves around the concept of hysteresis which I've never encountered before, and so I have a hard time grasping. I thought that the hysteresis index is part of that concept, and that it would help me understand what they are talking about in the paper. I feel that opening a different post about it would be redundant, but I might be wrong. Let's see if anyone else prefers it that way too. Thanks for your help!

"Hysteresis" as a concept can most easily understood by describing it as phenomena that depend on the past history of the material, so that for the same external conditions, the system will behave differently depending on its past history. In the context of adsorption and desorption, this means that the amount of sorbate in a surface will be not just a function of the pressure of that sorbate in the surrounding medium, but also of the history of sorption - so, for instance, a surface will present a different amount of sorbate at the same pressure when its desorbing than when its adsorbing.

Numerically, this can be defined as the difference in adsorbed material, for the same pressure of sorbate, when pressure is increasing and when its decreasing - or, from the POV of the surface, when we're adding sorbate to a clean surface or when we're removing it from a saturated surface. If we plot sorbate vs. pressure, we'll get different lines going "up" and "down" - while ideally both ways should go along the same route. By analogy with magnetic hysteresis, the higher the difference between the adsorption and desorption load/pressure lines, the "higher" the hysteresis or hysteresis index.


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is Li ion less stable than Li atom?
so our teacher introduced the concept of ionization energy... heres my doubt :  if we remove an electron from , say, Lithium , we need to supply energy first and that will be very less compared to the energy we'll have to supply to , say, oxygen. now the Li ion will have more energy than the lithium atom. but doesnt more enegry = less stability? does that mean Li ion is less stable?
what i really want to ask is,  when we remove an electron from an atom, does the atom ' gain' energy or is the gain actually the energy supplied to the electron?

https://chemistry.stackexchange.com/questions/98109/why-is-dipositive-dilithium-more-stable-than-neutral-dilithium/98125

You can't compare a Li atom and a Li ion. There's an electron missing. And $E=mc^{2}$.

What do you mean by "stable"? In a vacuum, $\ce{Li+}$ is quite stable. In water, Li ionizes readily to $\ce{Li+}$, with $\ce{OH-}$ accepting that electron. Of course, in vacuo, if an $\ce{Li+}$ ion were to be placed near an $\ce{Li-}$ ion, they'd quickly redistribute the charge.
There is also a difference between chemical (thermodynamic) stability and kinetic stability. Lifting a ball up a hill, or lifting an electron to a higher level, gives it potential energy, but if the ball rests in a cup on the hill, or the electron is in a metastable state, it will not go back on its own without a push.
Thermoluminescence is a case in point. Over time, some materials such as ceramics absorb radiation from cosmic sources and nearby radioactive minerals, putting atoms into excited states. Upon reheating, electrons hop back to lower states, emitting light. This is used to date ancient pottery, since the metastable states are preserved for millenia.
Another example of metastability is diamond. It takes considerable pressure and a specific temperature range to convert carbon, e.g. graphite, to diamond -- but once that pressure is removed, graphite is the preferred state. However, once formed, diamond is stable at room temperature for eons.
See Difference between thermodynamic and kinetic stability, answered by thomij, for a much more complete answer.

Actually, this question about the difference in stabilities of the metal atom and the metal ion was discussed in my chemistry class before and I think the idea is not difficult to understand.
If we were to look at the $\ce {Li}$ atom and the $\ce{Li+}$ ion (and electron) in isolation (i.e. ignoring all the interactions with surrounding chemical species, then we can most definitely say that the $\ce {Li+}$ ion (and electron) is higher in energy and thus less thermodynamically stable than the neutral $\ce {Li}$ atom (and electron). This is simply because energy was invested in the ionisation of the $\ce {Li}$ atom so clearly the products of this ionisation, which us an endothermic process, must be higher in energy than the reactant (i.e. the $\ce {Li}$ atom).
However, in reality, things do not occur in isolation in most cases. For example, we lithium is reacted with water in a redox reaction. We see that the more thermodynamically unstable $\ce {Li+}$ ion quickly interacts via ion-dipole interactions with neighbouring water molecules. These interactions are exothermic and bring down the energy of the system drastically. Now, if we consider the lithium ion in this chemical environment versus the unreacted lithium atom, we can clearly conclude that the lithium ion, in its chemical environment with all the coordinating water molecules, must be lower in energy than the latter.
I hope I managed to clarify things for you.

We certainly have to supply energy to an atom to ionise it. Removing an electron from outermost shell of an atom requires energy known as ionization energy. Although, since Li is a metal, the value of its ionization energy is low. Also Li atom is more stable than Li+. However, in aqueous medium, due to extremely large hydration energy (energy evolved when a substance is hydrolysed), in aqueous medium, Li+ is more stable.


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What does atm•L mean?
In my chemistry book I always see a dot for units. For example, atm•L. Does this mean “1 atm per Liter?” By Atm I mean atmospheric pressure, not atom.

dah... Thanks for the correction! // Yes $\mathrm{atmospheres}\space\times\space\mathrm{liter}$

@MaxW How do you conceptualize that? Im just confused on that. Does this mean the unit of pressure in regard to one liter ? If you know what im saying?

101.33 Joules = 1 l atm // You have to bust "complex" units down to primitives. So a liter is 1000 cm^3 and so on. Then rearrange to something that makes more sense. The whole think is called dimensional analysis.

@MaxW okay. I know dimensional analysis but when there are two units it’s confusing. So really atm•L needs to be broken down to make sense, more or less? Is that kind of what you mean.

atm-l presumably showed up in some formula. You wouldn't want to willy-nilly convert atm-l to joules without doing a dimensional analysis on the rest of the formula.  The formula itself may be trying to solve for temperature.

R = V x P/T x n = 0.o821 liter x atmospheres/K x mol

A liter-atmosphere is the amount of work that must be done to move one Iiter of material against a pressure of one atmosphere.

$\mathrm{atm \cdot L}$ is a derived unit for energy. Pressure is defined as being force/area, and volume is distance cubed. If you multiply out the units of these quantities, you can see it is equivalent to force times distance, i.e. work, measured in energy.

Is there a particular context where atm⋅L (or pressure⋅volume more generally) is the "natural" way to express a quantity of energy?

@R.M. see the new link

@ringo What do you mean by “If you multiply out these units?”

I am talking about performing dimensional analysis. Consider measuring the area of a sheet of paper. You multiply its length by its width (both measured as distances) and you get its area in terms of distance squared. Read more about it here: https://physicsabout.com/dimensional-analysis-physics/


text:
What are the partial charges in an SN2 transition state?
What are the partial charges of on the atoms in a SN2 reaction?
The nucleophile is normally negatively charged and has a lone pair of electrons. This would make you think the transition state would have a negative charge on the nucleophile. The leaving group also is taking the electrons from the molecule with it. So what is the charge separation?
Here are conflicting images.
First image

Second image

Not quite sure what you mean by So what is the charge separation? For SN2 reaction of bromide with chloroethane, bromide has -1 charge as reaction "starts" and chloride has -1 charge when the reaction "ends."

I added some links for clarification.

? I don't see any "conflict" between the links. What do you think is different/wrong?

One has negative charges on both the nucleophile and leaving group. The other has a positive charge on the nucleophile and a negative charge on the leaving group.

That just isn't correct. For which of the images do you think that there is a positive charge on the nucleophile?!? For the second the leaving atom isn't even shown...

Take a closer look at the second image. There are partial negative charges on the nucleophile and the leaving group, and a partial positive charge on the carbon center. Your images are in accordance with one another.

The problem here is you're reading more into partial charge than you should. If ever you take 2 partial negatives and a partial positive, and you can tell the total charge, something is wrong. It's not that detailed.

So sorry, I linked the wrong image for the second link. https://qph.fs.quoracdn.net/main-qimg-5e8619e5b488fb01d32d2fb51c861fa8-c

Again, I don't see a problem....

Second image has a positive charge on the nucleophile. The first image has a negative charge on the nucleophile. What is correct?

Well they are both just diagrams and imperfect representations of what happens. In the second diagram the molecule being attacked is shown with a charge imbalance. I interpret the $\delta^+$ on the nucleophile in the middle image to mean that the positive charge is moving off the molecule being attacked and onto the nucleophile. Hence the nucleophile is donating an electron and the electron is going to the iodine atom which is shown with $\delta^-$.

I think this illustrates nicely why looking at multiple sources is good. Two different representations get you thinking about what is really going on...

I am more used to seeing the first image... It may depend on whether the nucleophile is initially charged. Maybe we may see a partial positive charge on the nucleophile in the TS if it was originally uncharged and we may see a partial negative charge if it was originally negatively charged.

I would regard the second image as wrong.

I suspect that the second image is a typo, and where there is delta+, it should read delta-.


text:
Water values and water hardness
We have drinking water with the following values:

Calcium mg/l 62.7
Magnesium mg/l 10.3
total hardness 7.) dH 11.1
Total hardness 7) mmol/l CaCO3 1,99
Carbonate hardness °dH 8.5
Hydrogen ion concentration (pH-value) - 7,8

To use the water for my plants, I reduce the pH with phosphoric acid to 6.8 PH.
If I now want to find out how many calcium and magnesium ions are in the water, do I only need to read out the amount of calcium / magnesium that are listed or must the calcium from the total hardness / carbonate hardness be added too, since it was present in the form of CaCO3 / Ca(HCO3)2 and not as ions before I brought the pH Down.

Looks like Houston TX water , not too bad. Some plants like it. Occasional use of rain water will help for house plants preferring neutral of acid pH. There are also acidic fertilizers., ( I use a lot of "starter" 9 -45 -15, but most of my plants like acid ).You may be able to find the chemistry of local municipalities and compare to what you have before you "fix" it.

What's your soil pH? No point titrating with H3PO4 unless soil pH is too high. If it is, powdered sulfur over winter is the usual way to handle it.


text:
Molar mass of a protein from osmotic pressure

The height of a column of liquid that can be supported by a given pressure is inversely proportional to its density. An aqueous solution
  of $0.010 g$ of a protein in $10 mL$ of water at $20^oC$ shows a rise of
  $5.22 cm$ in a mercury tube. Assume the density of the solution to be $0.998 g/cm^3$ and the density of the mercury to be
  $13.6 g/cm^3$. (a) What is the molar mass of the protein?

The answer key says the answer is $4.8*10^3 g/mol$, but I got $3.5*10^2 g/mol$. Below is my reasoning:
Osmotic pressure = pressure to raise mercury, so $MRT = rho * g * h$
$\frac{(0.010/M)}{(10.010/0.998/1000)} * 8.3145 * 293 * 10^3$ (pascals) = $13.6/1000*(100)^3 * 9.8 * 0.0522$ (pascals). Solving this equation for $M$ gets $3.49*10^2$, instead of 3.8*10^3.
Can someone please point out where I am wrong?
Please do not point this out as a duplicate, as I have already referenced the question here. I used their approach but still cannot get the correct answer.

Osmotic pressure is $cRT$, not $MRT$. You should put all dimension in your equation, and make sure they check out!

If you are going to use R in the SI units so, do the same for the rest, volume and mass.
As Karl said, putting all dimensions in the calculation helps you to identify the mistakes, because in the end it must give you a molar mass unit like $kg mol^{-1}$.
Please format your equation with the frac{}{} command and dont froget the \ on rho, if is convinient is better use two $ than just one for the solution.
Ah! and the link doesn't work, plese check it.

Just to make this clear, a 1 M solution has a concentation $c = 1 \rm{mol/l}$. "M" for "molar" is not a mathematical symbol. The capital $M$ designates molar mass, which is the number you want.

@liuzp While we're nitpicking, units should be given in roman, not italics, by using \rm{}  ;-)

@Karl. Since when does M designate molar mass and not moles/liter? I've seen it used to represent molar mass (MW), but most often it seems to be used to express concentration in moles/liter (molarity).

@Dr.J. Sorry, old misconception.  https://en.wikipedia.org/wiki/Molar_mass , and "molarity" M is not a mathematical symbol. It's a unit. https://en.wikipedia.org/wiki/Molar_concentration#Definition

Or let's say it's sort of a unit. It's not really clear what it is, mathematically, which is probably why IUPAC retired it.

@Karl roman? this is new to me, could you explain more?. And I'm going to fix the forgoten space among the kg and the mol.

Osmotic pressure of a solute in water is given by the formula
$$\Pi = i\rm{MRT}\tag{1}$$
Where:
$\Pi$ is the osmotic pressure
$i$ is the Van't Hoff factor
$\rm{M}$ is the molarity of the solute (mol/L)
$\rm{R}$ is the universal gas constant 62.36 ($\rm{L}\space \rm{Torr}\space$ $\rm{K}^{−1}$ $\rm{mol}^{−1}$)
$\rm{T}$ is the absolute temperature ($^\circ\rm{K}$)
We'll assume $i =1$.
Let's use $\rm{M_W}$ for the molecular weight of the protein. So the moles of the protein is equal to $\rm{g/M_W}$.
$$\dfrac{0.010\space\rm{g}/\rm{M_W}}{0.010\space\rm{L}}\times 62.36\space\rm{L}\space \rm{Torr}\space\rm{K}^{−1}\space\rm{mol}^{−1}\times293\space\rm{K}  = 52.2 \space \rm{Torr}\tag{2}$$
$$\rm{M_W} = \dfrac{0.010\times62.36\times293}{0.010\times52.2}\space\rm{g/mol}= 350\space\rm{g/mol}\tag{3} $$
Your mistake was believing the book answer.

If you need to check this sort of answer, then choose R with convenient units. It makes solving the problem much easier.

You (like the OP) are using M for molarity and molar mass. The former is a bit old fashioned, and it's all confusing.

@MaxW Thanks for your response. Don't you mean 0.010 L instead of 0.0010 L though?

@coder - Yes, you are correct. 10 ml = 0.010 Liters. blush... Thanks for pointing out the error.


text:
How to calculate the enthalpy of reaction using Gaussian 09?
I'm having a hard time trying to understand what's the correct way to calculate the enthalpy of reaction given that I've already calculated the frequencies of my compounds in Gaussian. Below I show the reaction of interest:

As you can see in the figure, experimentally, this reaction leads to the formation of ionic species. As for my calculations, I have all four species optimized (all of them separated, I'm not working with complexes) with their respective frequencies.
As for the enthalpy, I'm using the following formula:
$$
\Delta_\mathrm{R} H=
  \left[(\text{Iodide anion}) + (\text{cationic product})\right]
  - \left[(\ce{FeCpI(CO)2}) + (\text{phosphine reactant})\right]
$$
All terms are the sum of the electronic energy and the thermal enthalpy as given by Gaussian.
Am I doing it correctly? Because I'm not sure, if I have to consider both products together or separated as I'm currently doing. Can you give me some light on this?

In principle your approach is correct. There are a few pitfalls though.
Quantum chemistry is usually not well equipped handling ions, especially anions, within reactions, hence the evaluation of these reaction energies will probably contain some errors. This is mostly due to the fact that you are technically calculating everything in the gas phase, and apply thermal corrections later. In well behaved systems, e.g. organic compounds without charges, that does not matter much. But even then we try to exploit error compensation with (for example) isodesmic reactions.
If you are treating the ions separately, you are neglecting all possible interactions between them, too. That might lead to additional errors. One way to approximate this is to calculate the ion pair together. That might not always lead to the correct result since solvents will also play a role here. Obviously you could also include these.
Another problem for your system is probably conformational space. Especially the phosphine is somewhat flexible and will adopt a few conformations (isolated as well as complexed). Here it is also important to find the global minimum, to evaluate the energies correctly.
It is advisable to look for similar systems in the literature and see what they have done. You may also find benchmark studies that will give you more information on how well your methodology performs for such systems. There is no general recipe for success; it always involves a lot of hands-on work. Eventually you will want to compare your calculated results to experimental ones to verify your findings.

Thanks, @Martin! Your answer will surely help me a lot!
Concerning the error contained in my calculations, what if I only want to see trends in the Enthalpy among different Phospines? More specifically, I'm varying the stereochemistry of them. I'm not really interested in obtaining accurate values.
As for experimental results, I only have yields.

@Benjamin233 Try calculating isodesmic reactions: $\ce{[Fe(phos)]+ + phos' <=> [Fe(phos')]+ + phos}$ where $\ce{phos}$ is your reference phosphine, something like $\ce{PMe3}$, and $\ce{phos'}$ is the one you'd like to investigate, and $\ce{[Fe(X)]+}$ is short for $\ce{[Fe(Cp)(CO)2(X)]+}$... That should give you the thermodynamical order of your target compounds. Here it is most important, that you have the global minimum of all structures.

I have five different phosphines, and each one has different stereochemistry, so I compared their relative energies to see which one is the most stable. So, what would be the difference between just comparing their energies and calculating isodesmic reactions? I actually did it and I could observe reasonable results, but which approach is more reliable?

@Benjamin233 If you calculate relative energies of complexes which are stereoisomers, then then that is essentially the same as computing isodesmic reactions (you can show that by doing simple maths). You cannot compute relative energies of complexes which are not isomers, as they have no common point to be relative to; then isodesmic reactions are a way to use error cancellation. You can search this site for some examples.

@Benjamin233 This one. Ours. [Chemistry.se]. Or direct search link: isodesmic

Why is it that the energy difference between two complexes (they are isomers) is only 0.4 kcal/mol, but when I do the calculation for their respective isodesmic reactions, the energy difference is now 5 kcal/mol? I have the feel that calculating the reaction enthalpy for the isodesmic reaction is the right way because the result resembles more the experimental result lol, but I don't understand why such a big difference?


text:
what is the basic factor which determines the order of ionization energies of K,Li,Ca,S?
There was a question which was asking about the order of ionization energies of
K,Li,Ca,S.
The electronic configuration of all the elements were as follows

Li:${1s^2 2s^1}$
K :${1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$
Ca :${1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$
S :${1s^2 2s^2 2p^6 3s^2 3p^4}$

From the electronic configurations:

Potassium should have lowest ionization energy as it is very easy to remove electron from 4th shell.
Then it should be followed by Calcium as its nuclear charge increases from going across the period
Then it should be followed by sulphur as it is more easier to remove the electron from 3rd shell than from 2nd shell.
Then it should be followed by Lithium as it has lowest principle quantum number(n) so it is more difficult to remove an electron from 2nd shell than from 3rd and 4th shell.

So according to my calculation the order should be  K < Ca < S < Li.
But the correct answer in my textbook is K < Li < Ca < S. If I am wrong anywhere then what factor would help in determining the Ionization energies of the following elements, and what would be the exact order of I.E for these elements.

frankly I hate these sort of questions. Typically they call for some rationalization which only can be justified by some hand waving.

Looked up ionization energies (kJ/mol) from Wikipedia

K  = 418.8
Li = 520.2
Ca = 589.8
S = 999.6

so the book answer is correct...

Ok, here is best hand waving explanation that I can give. The general trend...

Li and K are both alkali metals. As the atomic number goes up the shells get further away from the nucleus and the inner electrons shield the valence electron. So in order of increasing ionization energy
Li > Na > K > Rb > Cs > Fr
So to start we easily get Li > K.
Ca and K are in the same period in the periodic table. Generally as you go up in atomic number in a period the ionization energy will increase. So Ca > K.

Sulfur is a non-metal in the upper right hand of the periodic table so it should have the highest ionization energy.
So we're down to either S > Ca > Li > K or S > Li > Ca > K.

To me giving either of these answers wouldn't be embarrassing.

If you go to the Bohr model of the hydrogen atom then energy of the $n$th shell is
$\mathrm{E}_n  \propto \dfrac{Z^2}{n^2} $
If we assume that the 2s and 3s electrons are perfectly shielded by all the electrons in the lower shells, then
$ \mathrm{E_{Li}} \propto \dfrac{1}{2^2} = 0.25$
The two Ca 3s electrons being equal the charge is 2
$ \mathrm{E_{Ca}} \propto \dfrac{2^2}{4^2} = 0.25$
So I would have guessed S > Li > Ca > K
Now I know that is wrong. I know that the "problem" is with the shielding of the electrons. Below is a section of the periodic table with ionization energies listed.
I just don't know how to explain the "right" answer without hand waving about shielding.
I don't know how you'd rationalize all the relationships between the group 1 and 2 elements.

If you assume the Bohr model as noted above then you end up with this which correctly predicts the trend down a group but overestimates shielding of the inner electrons.

I can't understand what are you trying to say about hand waving?

Hand waving - a pejorative label for attempting to be seen as effective – in word, reasoning, or deed – while actually doing nothing effective or substantial. It is most often applied to debate techniques that involve fallacies, misdirection and the glossing over of details.

As good an answer as one could come up with.

I think you are trying to say that  respective ionization energies of elements is just a matter of fact and we should see it just as a number, but anyway thanks for the answer.

@pranjalverma - Some of the tends in the ionization energies you can deduce from the position of the elements in the periodic chart. However some you just can't. // The answer is somewhat of a rant. These kinds of problems are posed many times to new students and then the answer is "justified" by what I believe to be some deceptive logic.   You can invoke shielding to explain Li/Ca, which is true, but I think you need a very detail calculation to prove it rather than some hand waving. In other words the hand waving for Li/Ca would also have to explain Na/Sr where the relative values reverse.


text:
Compatibility of Grignard reagents with aromatic nitro groups
Is it possible to form a Grignard reagent from 1-chloro-3-nitrobenzene, as shown below?

Specifically, will the nitro group react with the Grignard reagent formed?

No. Aromatic nitro groups react with grignards - see Bartoli reaction.

An organozinc would probably work

It really depends on what you're trying to accomplish with your organometallic reagent.

Aromatic nitro groups are not compatible with Grignards as they react with the nitro group. An example of this being useful is the Bartoli synthesis of indoles from vinyl Grignards (Tet. Letts. 1989, 30 2129, Curr. Org. Chem.. 2005, 9, 163). The first step of the Bartoli is addition of the vinyl Grignard to the O of the N=O of the nitro group so I presume this would be true for other Grignards and is the reason for the incompatibility.


text:
How can prepare ppb solution of sodium
How can prepare 1000ppb solution of sodium from sodium chloride salt in one litre.

Welcome to Chem SE! If you haven't yet, take the [tour] and visit the [help]. This question is likely to be closed, as we expect posters to include some of their own thoughts/attempts at solving a problem.

The title is misleading since you want to prepare a concentration in the order of ppm and not ppb, which would be a bit more difficult since sodium is ubiquitous. Also note that non-standard units such as ppm and ppb should actually be avoided; preferably use an unambiguous unit such as mg/kg.

@Loong - Like you point about sodium being ubiquitous. Not sure what special precautions might be necessary for ppb levels of Na. Obviously at least distilled water not deionized, and very clean glassware. Not sure if you could leach that much sodium from glassware or not.

Assume 1 liter of solution has a mass of 1 kg.
1000 ppb = 1 ppm = 1 mg/liter.
Sodium chloride is only 39.34 % sodium so you'd need 2.54 mg.
It is tough to weigh out that little so weigh 2.54 g and dissolve in 1.000 liter volumetric flask. Use a volumetric pipette and take 1.00 ml of that solution and dilute to 1.000 liters in a second volumetric flask.


text:
Cahn–Ingold–Prelog priority of CN vs CH2NH2
Why does $\ce{CH2NH2}$ have lower priority than $\ce{CN}$? The carbon on the first molecule is bonded to (N,H,H) while the carbon on the second molecule is just bonded to (N). I can't seem to find this scenario anywhere online.

Each multiple bond is considered to be equivalent to an individual bond. So $\ce{-CN}$ has the same priority as (N,N,N), making it higher than (N,H,H). Read more about it here.

Thanks for the quick reply. Also, I have one more question. In the case of C≡C-H vs C(CH3)3, why does the first one have higher priority than the second? They are both the same two branches out, but then the first one becomes (H) vs the second one which is (H,H,H,H,H,H,H,H,H). If two branches contain the same elements but just a different number of them, does the shorter one automatically get priority? Thanks.

Because the terminal carbon is considered as being equivalently bonded to (C,C,H), while the methyl groups are all (H,H,H).

Important to note that the additional N's on CN are phantom. Treat them with the same priority as N, but compared to a real N, it has lower priority.


text:
Why would a ~1 cm thick layer of argon be a significantly poorer conductor of heat than air?
In yesterday's new Periodic Video, Argon (new) - Periodic Table of Videos, after about 07 min 00 sec, Sir Martyn Poliakoff says:

The final, and I have to admit perhaps slightly boring application of argon, is in double glazing (of windows) to keep our houses warm. Because argon has a heavy atom, the atoms don’t move very fast in the gas phase, and therefore they’re bad at conducting heat (emphasis added).
So if you have double glazing (that’s two sheets of glass in your window) if you put argon between the glass, it is much more effective at insulating the inside of the house, keeping cold out, than if you use air, because air has a much higher conductivity of heat.
So, if you look out of the window, you may find that you are looking through argon gas.

At a given temperature or kinetic energy per molecule, velocity will roughly scale with the inverse square-root of the mass, so I estimate that the argon atoms would be moving
$$\sqrt{\frac{28}{40}} \approx 0.84 $$
only 16% slower than nitrogen molecules would be moving.
Why then would a trapped ~1 cm thick layer of argon gas be a significantly poorer conductor of heat than nitrogen gas or air?
I'm wondering if it has more to do with the fewer internal degrees of freedom of a monatomic gas, than it does with the mass of the molecule. There would be a similar collision rate, but each collision of an atom of Ar could transfer less energy to the glass surface than a molecule of N2 could transfer.

Ar has no molecular vibrations in which to store heat. Oxygen and nitrogen do. At impact diatomic molecules can transfer kinetic energy of linear motion as well as vibrational energy to the substrate.

@MaxW sounds like you agree with the last paragraph?

A problem with air insulation, as opposed to argon or even N2 - O2 mix, is that air contains water vapor, which efficiently passes heat, particularly over the temperature range where condensation and evaporation are possible (e.g. winter in temperate or colder climates).

$\mathrm{C_p[\ce{Ar}] ≈ \frac{3}{2}R}$.   $\mathrm{C_p[\ce{O2}] ≈ C_p[\ce{N2}] ≈ \frac{5}{2}R}$

The thermal conductivity coefficients of some gasses in units of W/(m•K) are helium 0.142, argon 0.016, air 0.026, methane 0.03, propane 0.015, bromine 0.004, and steam 0.018. These are all rather similar except for helium and bromine.
Using the kinetic theory of gasses the thermal conductivity coefficient is given by $ \kappa \sim n\bar c\lambda C_V$ where $n$ is concentration, $\bar c$ the mean speed, $\lambda$ the mean free path and $C_V$ the heat capacity. Using definitions $n\lambda \sim 1/\sigma^2$ where $\sigma$ is the diameter of a gas atom or molecule; $\bar c \sim 1/\sqrt{m}$ and the heat capacity is $C_V \sim sk_\mathrm B$ where $k_\mathrm B$ is Boltzmann constant and $s$ is a multiplier depending on the type of species, atom, diatomic molecule, etc.
The conductivity coefficient at constant temperature is therefore $\displaystyle \kappa \sim \frac{sk_\mathrm B}{\sigma^2\sqrt{m}} $.
Thus we can see that helium will have high $\kappa$ as it has a small diameter and mass, and bromine will have a small value of $\kappa$; large diameter and mass. The other gasses are similar to one another because although the heat capacity increases between an atom and molecule and increases further as the molecule has more vibrational modes, the diameter and mass also increase when the molecule has more atoms and these terms cancel out to some extent.
So it is not just about speed, but also size, mass and number of vibrational modes. Using argon over dry air would seem to give a small advantage, but it may not be cost effective, i.e. heat saved vs. initial cost and energy needed to use argon.
(The molecular heat capacity increases above that for translational motion with the number of squared terms in the energy, potential and kinetic, each term contributes $k_\mathrm BT/2$ to the energy and so $k_\mathrm B/2$ to the heat capacity. Each vibration populated thus contributes $2k_\mathrm B/2$ and rotations the same for a linear molecule, but $3k_\mathrm B/2$ for a not linear one.)

Thank you for your answer! I'm curious what the multiplier $s$ would be (approximately) for argon versus nitrogen.

When you say the thermal conductivity coefficients are all "rather similar" (argon .016, air .026) you may find people will disagree. Argon is 40% lower than air which I would call "significantly different".

@matt-black Of course it is a matter of opinion what is similar or not,  but compared to He and Br they are similar and each differ by a lot more to these gasses than to one another.

A monatomic gas has $ 3k_B/2$ ( 1/2 in each of x, y and z). A diatomic has 2 rotation directions because rotation along the bond does not change anything so rotations add another $2k_B/2=k_B$. Then vibrations add another $k_B/2$ because of their kinetic energy and the same for their potential energy  making an additional $k_B$ per vibrational. The vibrations only count if vibrational quanta are populated at the prevailing temperature. Same applies for rotations but these are usually populated at room temperature, so for a diatomic $5k_B/2$ or $7k_B/2$ if the vibration is included.


text:
Which natural orbitals are better for CASSCF calculations for organic diradicals?
I am trying to perform CASSCF calculations for a series of diradicals. As input orbitals I have used two types of natural orbitals. One is unrestricted natural orbitals and the other MP2-based natural orbitals. From both sets I am getting different trend for the properties I am trying to look into.
Can anyone suggest me which is the better way to know which results are reliable ?
I have used ORCA 4.0 version.

Since you have to reconverge the CAS orbitals anyway, it doesn't really matter which starting orbitals you choose, as long as they are sensible choices with respect to the spin state. They should give you the identical solution in the end, if you have chosen the same active space. The context of your question is not enough to actually help you with your problem. Highly related: Is it reasonable to use natural orbitals of an unrestricted HF calculation as a basis to start a CASSCF calculation for a radical

If you need more help with this topic, I suggest you [edit] in more information, like how you determine your active space, what kind of molecule, etc. (I know that is sometimes difficult, when involved in active research, but you can try with a more simple model compound that shows the same behaviour.)


text:
"Proof" that the chemical potential is less than zero for irreversible material changes
From Levine's Physical Chemistry 6th edition:

In a closed system, $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP$ not only applies to constant composition but also reversible changes in composition (page 116).
For an irreversible process: $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP + \mu\,\mathrm dN \lt -S\,\mathrm dT + V\,\mathrm dP$, thus $\mu\,\mathrm dN \lt 0$ (page 130).

For 1), why do we even bother with the chemical potential for closed systems if we can just find a reversible path in which 1) applies?
For 2), I thought that although $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP + \mu\,\mathrm dN$ can apply to irreversible processes, you had to calculate a reversible path. If so, that would mean that the $-S\,\mathrm dT$ on the left was calculated for a reversible path and the $-S\,\mathrm dT$ on the right was calculated for an irreversible path and they wouldn't cancel.

Q1. You won't need the chemical potential terms $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha$ (because it equals zero) in using the differential relation $\text{d}G = -S\,\text{d}T+V\,\text{d}P$ in this case. Nevertheless, the chemical potentials are still relevant in establishing criteria for phase equilibrium: $\mu_i^\alpha = \mu_i^\beta$, etc.

Why should $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha = 0$?

Reversible changes in composition imply that the process is in equilibrium, so that $\text{d}G = 0$. Assuming thermal and mechanical equilibrium and constant $T$ and $P$ we have $\text{d}T = \text{d}P = 0$, and the result follows by considering the (full) differential relation satisfied by $\text{d}G$.
Q2. Yes, you do need a reversible path. The expression on the right, if you follow its derivation carefully, does represent a reversible path. The inequality occurs because the chemical potential terms have not been accounted for. So the cancellation works out.
It seems, however, that we can derive the same result much more simply. Consider a system at thermal and mechanical equilibrium with constant $T$ and $P$, but not chemical equilibrium. Then $\text{d}G < 0$, since the Gibbs free energy is minimized at equilibrium. If $T$ and $P$ are constant $\text{d}T = \text{d}P = 0$, so all that's left in $\text{d}G$ are the chemical potential terms, which must thus be negative.

I thought that dG = 0 only for reversible processes involving constant T and P.

@Yuki, can you quote the part of my response that you're concerned about? There are lots of dG's in my answer and I'm not sure what you're referring to.

Thermal and mechanical equilibrium does not imply dT=dP=0 as both T and P can change reversibly. Also for 2), the expression on the right, -SdT+VdP, is definitely not a state function and thus for an irreversible process such that dG < -SdT+VdP it would not make sense to say that it is calculated for a reversible path.

I think I understand your answer now. Because thermal and mechanical equilibrium is assumed for all equations, -SdT + VdP is always reversible both in the right and left side of dG = -SdT + VdP + udN < -SdT + VdP. In the original derivation of dG = -SdT + VdP + udN, when Levine said that this equation was only valid if you used a reversible path, he meant that specifically -SdT + VdP had to have a reversible path, not udN (which is valid for irreversible paths).

@Yuki, (> Thermal and mechanical equilibrium does not imply dT=dP=0) sorry, you're right. I have updated my answer accordingly. And you've addressed the second part of your comment already.


text:
Electrode specification
What factors go into selecting an appropriate material for an inert electrode for electrolysis?
It looks to me like metal oxides or platinum/platinum-coated titanium are the preferred options. However, I'm not really clear why platinum should be a good option since it doesn't have a particularly high standard electrode potential (1.188V). Lead oxide, gold, and silver all have higher standard electrode potential and are all cheaper, so why are they not preferred?

These questions are hardly related. Also electrodes coated with oxides are a different type.

@Mithoron I've deleted the second question and I'll ask in separately. Can you please explain what you mean by "different type"?

The exact choice of electrode really depends on the goal of the experiment and what will be in your cell:
The most basic requirement is that the solvent or electrode material not decompose at the potential you want to work at.
For each solvent/electrode combination, there is a potential window in which you can safely work.
In the most common situation—working in aqueous solutions, the choice of working electrode mainly depends on whether you want to observe oxidation or reduction.
Electrodes like platinum or gold are commonly used for oxidation reductions because they have high overpotential for oxygen evolution in aqueous solutions (platinum in particular, performs poorly for reduction as it is rather good at catalysing the hydrogen evolution process).
Conversely, for reduction reactions, mercury and bismuth have potential windows that extend to quite negative potentials, but will themselves oxidize as you move towards positive potentials.
There are also some materials that work pretty well for both positive and negative potentials, such as glassy carbon and boron-doped diamond.
For analytical purposes, you can get away with (and often prefer) small electrodes, so cost is less of a concern, but if the application is mass electrolysis, large electrodes are required to have a decent reaction throughput and more expensive materials may be impractical.
Cheap materials like stainless steel and graphite can be used as electrodes in some cases, though they are more limited in the systems in which they may be used.
As far as the reason to choose platinum over lead oxide, gold, or silver is concerned, the standard electrode potential isn't a primary consideration. (Silver's is not higher than platinum's, incidentally)
The standard electrode potential reflects the thermodynamic situation of a particular half-reaction, but is not terribly informative of how the material will behave in a given cell as it does not account for the kinetics of the reaction or any other factors.
Silver is usually not a good choice for a working electrode as it oxidizes quite easily in aqueous solutions, gold is actually a fairly common choice as it has a wider potential window than platinum in aqueous solutions, but it is easily fouled by organic/biological contaminants.
Platinum doesn't have the widest potential window, but is generally resistant to fouling and being reacted itself (giving it a long lifetime).
In a nutshell, electrode choice is dictated by many competing factors: cost, lifetime, potential range, mechanical considerations, etc.
In an analytical lab, screen-printed gold electrodes are quite common because they are relatively inexpensive, suitable for many different reactions, and can be discarded and replaced when fouled.
In an aluminum refinery, you'll find carbon and coke electrodes used in the Hall-Héroult process, as they need to be very large and withstand the high temperatures of molten cryolite.
In the cloralkali process, the anode is usually made from titanium as it must resist the highly oxidizing conditions that produce chlorine, but the cathode can be made from much cheaper nickel as that half of the cell is only used to reduce water to hydroxide and hydrogen gas.