text:
What does isohydric solution mean?
While studying ionic equilibrium, I came across the term "Isohydric Solutions" which is defined as:

Solutions of electrolyte are said to be isohydric if the concentration of the common ion present in them is the same and on mixing such solutions, there occurs no change in the degree of dissociation of either of the electrolyte.

But as I understand from reading the statement, in the isohydric solutions, we have two solutions with a common ion whose concentration is the same in both the solutions.
So, when we mix the two solutions, the concentration of the common ion in the reaction quotient should increase (to be more precise, it should double) and then according to Le-Chatelier's principle, both the reactions will go backward which I think can only happen when there is a change in the dissociation constant of the 2 reactions (as equilibrium constant is constant for constant temperature).
So what exactly does the definition of isohydric solution mean?

Volume increases, not concentration.

Ah! I think i get what I was doing wrong. The concentration of both the acid and the anion will change by the same amounts, hence cancelling the effect. Thank you very much.


text:
Height of CCP Lattice
Are the height of the Cubic Close packing Lattice and the cubic Hexagonal Close Packing the same?
My understanding says that these unit cells are formed by different arrangement of atoms, ABC for CCP and ABAB for HCP and because they are made from two different layers their heights should be different, is it so? And if it is, how do I calculate the Height of a CCP lattice?
(I know how to calculate height for cubic HCP.)

You can't really compare cell heights because the cells differ geometrically, one is either a hexagonal or a rhombic prisms (depends on convention) and the other is a cube.

@OscarLanzi I am not comparing the two unit cells. At school, we have been though how to calculate the height of an HCP cubic unit cell. I want to know if there is a way to calculate the height for a CCP unit cell.

No, the CCP unit cell does not have a height.

@IvanNeretin umm, what?

You might have heard that a cubic unit cell is described with only one parameter.

@IvanNeretin yes

That's it, then. If you know that parameter, you don't want to know any other parameter that might or might not be called "height". There is no other parameter.

Also, you seem to have a wrong picture concerning the distances between layers.  In fact, they are related to the cell height in HCP, but not in CCP.

Face-centered cubic (FCC, or closest-packed cubic, CCP) and hexagonal close-packed spheres have the greatest (and equal) density of packed spheres, occupying 74.048% of the actual volume. Ref 1
However, because of the different interlaying of the hexagonal, tight packing in two dimensions, the repeating unit, or unit cell, is described differently for each crystal type. While each layer in a FCC crystal is identical to every other layer, and to each layer in a HCP crystal, the different layering makes it possible to distinguish between them, and to identify repeating units. The smallest repeating unit in a FCC crystal contains 4 atoms and can be described with one size parameter. The smallest repeating unit in a HCP crystal contains 6 atoms. Although its density is the same as that of a FCC crystal, it needs 6 atoms to describe its layering. Interestingly, body-centered cubic crystals can be described with only 2 atoms per unit cell, but have a density of only 68% (they are not close-packed). Ref 2
The construction of FCC and HCP crystals is identical for two layers, but the third is differently placed. HCP has only two positions, a and b, but FCC places a third layer c so that the atoms do not lie directly over any in the first or second layer. HCP places layers as a,b,a,b,a,b...; FCC does it this way: a,b,c,a,b,c,a,b,c...

FCC and HCP unit cells are drawn to show aspects of their packing, so interatomic distances (lattice parameters) specified are between differently placed atoms. Comparing heights isn't really comparing repeating units. If you measure the distance between two adjacent atoms in FCC, you will find the same interatomic distance in HCP. However, comparing the distance between more remote atoms can give you different results. It's a bit like comparing your vertical height at age 6 months with your horizontal dimension now. They're similar, but different because you are looking with a different viewpoint.
It's fascinating to consider that crystal structures are energetically distinguishable not just by next-door atoms, but by atoms down the street by one house. Or maybe more!
Ref 1. https://en.wikipedia.org/wiki/Close-packing_of_equal_spheres
Ref 2. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Solids/Crystal_Lattice/Closest_Pack_Structures

source: Stack Exchange

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text:
Atoms vs molecule when talking about the avogadro's number?
By far, my understanding is that a molecule is made up of atoms bonded together. For example, a molecule of water ($\ce{H2O}$) has 2 hydrogen atoms and one oxygen atom.
However, when it comes to Avogadro's number, I'm getting confused because it mixes the concept and treats it like it is the same.

Avogadro's number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance.

If I take 1 mole of water, with the molecules definition for the Avogadro's number, it has $ \mathrm{6.022×10^{23}} $ molecules of water. With each molecule having 3 atoms, it means that 1 mole of water has $\mathrm{18.066×10^{23}} $ atoms.
If I take 1 mole of water with the atoms definition for Avogadro's number, it has $\mathrm{6.022×10^{23}}$ atoms of water.
What is the right way to think about this?

A mole is always just a number. Out of context that means you don't know what you are counting. You always need to define exactly what is being counted. This resolves the ambiguity (one mole of water molecules contains one mole of oxygen atoms but also two moles of hydrogen atoms: there is no ambiguity if the thing being counted is specified.)

The Avogadro number is just a number, like 12. You can have 12 eggs, or 12 boxes of eggs.

$1$ mole of water has $\pu{6.022E23}$ molecules of water, but not $\pu{6.022E23}$ atoms of water, because the expression "atoms of water" has no sense. You are allowed to state that it contains $\pu{18.066E23}$ atoms. But you are not allowed to speak of "atoms of water". Water has not its own atoms of water.

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


text:
Differences in AChE inhibition kinetics between VX and VR
According to the referenced paper, VR has an AChE inhibition reaction rate constant almost 4 times that of VX. Interestingly, its reaction rate constant for aging of the enzyme is less than that of VX, despite the fact that, as acknowledged by the authors of the paper, aging tends to occur faster when the leaving group during aging is a branched alkoxy group. Finally, VR-inhibited AChE is also more prone to spontaneous reactivation compared to AChE inhibited by VX. What is the reason for these differences in reaction kinetics for the interactions of these two agents with AChE?
Reference:
Worek, F., Thiermann, H., Szinicz, L., & Eyer, P. (2004). Kinetic analysis of interactions between human acetylcholinesterase, structurally different organophosphorus compounds and oximes. Biochemical Pharmacology, 68(11), 2237–2248. doi:10.1016/j.bcp.2004.07.038


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ELI5: What is the effect of ultrafine Teflon (PTFE) particles in rats and mice
I am attempting to research the results of case studies regarding polytetrafluoroethylene [PTFE] (commercially known as Teflon). More specifically I am attempting to make sense of this Abstract on Acute pulmonary effects of ultrafine particles in rats and mice. Not being a chemist I'm having a hard time understanding the results and hoping someone can explain in laymens terms. More specifically, the abstract mentions;

We used ultrafine Teflon (polytetrafluoroethylene [PTFE]*) fumes (count median diameter [CMD] approximately 18 nm) generated by heating Teflon in a tube furnace to 486 degrees C to evaluate principles of ultrafine particle toxicity that might be helpful in understanding potential effects of ambient ultrafine particles. Teflon fumes at ultrafine particle concentrations of approximately 50 micrograms/m3 are extremely toxic to rats when inhaled for only 15 minutes.

That seems straight forward enough. They heated up Teflon in a tube, and made rats inhale it, who subsequently died. Then it goes on to say;

We found that neither the ultrafine Teflon particles alone when generated in argon nor the Teflon fume gas-phase constituents when generated in air were toxic after 25 minutes of exposure. Only the combination of both phases when generated in air caused high toxicity, suggesting the existence of either radicals on the particle surface or a carrier mechanism of the ultrafine particles for adsorbed gas-phase compounds.

This is the part thats unclear to me. The Teflon particles alone, I think means the compound before they heat it, but what is the 'fume gas-phase when generated in air'? How is that different then the fumes they made the rats inhale (from which they died after 15 minutes)? What does the 'combination of both phases when generated in air' mean?

Aging of the fresh Teflon fumes for 3.5 minutes led to a predicted coagulation resulting in particles greater than 100 nm that no longer caused toxicity in exposed animals.

How would coagulation affect the toxicity of the fumes? Does it change the chemical structure of what the animals inhale somehow making it less toxic? Naturally speaking, I would assume 'more' particles would result in higher levels of toxicity, not less.

Perhaps better on the Bio stack? Particle size, accompanying toxic gases, any number of things impact uptake and impact.

Im asking for an explanation of mostly chemistry related terms. Not about the actual impact on mice.

I think the authors' choice of words might have confused you a bit.

We found that neither the ultrafine Teflon particles alone when generated in argon nor the Teflon fume gas-phase constituents when generated in air were toxic after 25 minutes of exposure.

What the authors mean is that when they used only ultrafine Teflon particles on the rats it did not kill them, nor did the Teflon fume gas-phase constituents when the rats were breathing that air for 25 minutes.

Only the combination of both phases when generated in air caused high toxicity, suggesting the existence of either radicals on the particle surface or a carrier mechanism of the ultrafine particles for adsorbed gas-phase compounds.

The individual substances, that is, ultrafine Teflon particles and Teflon fume gas-phase constituents were non-toxic for the rats. When the two were combined together in the air and the rats were made to breathe in that air was the point that they died, meaning that when these substances were combined in the air, some radicals$^{[1]}$
were being produced or some form of carrier mechanism was taking place.

The Teflon particles alone, I think means the compound before they heat it.

No, they meant that the ultrafine Teflon particles after heating. Alone meant these Teflon particles, by themselves, are not toxic.

what is the 'fume gas-phase when generated in air'

Gas-phase is simply a way of referring to the gaseous physical state. See here$^{[2]}$ for the definition. The phrase fume gas-phase refers to the fumes that the author observed on heating the Teflon to the temperature at which it turned into gaseous phase (phase is the word for referring to the physical state of a substance) while producing fumes.

How would coagulation affect the toxicity of the fumes? Does it change the chemical structure of what the animals inhale somehow making it less toxic? Naturally speaking, I would assume 'more' particles would result in higher levels of toxicity, not less.

Its not about more or less here. After some time, coagulation takes place causing the Teflon to take a solid form in the whichever utensil the researchers were using. What they mean by the coagulated Teflon being non-toxic is that in whichever way they made the animals (they have not specified which animal though) ingest it, it did not kill them. The quantity of the Teflon, the form in which it was ingested and by which animal was it ingested has not been specified, at least I couldn't find any clear reference in the abstract and I don't know how to access the full paper.
References:-

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

I see you're edit quite a bit, but you're only making superficial edits, while editing new questions is for deciding if they are gonna get closed, and possibly deleted, or no. So edit should either substantially improve q. or may be a problem as it may interfere with reviewing. https://chemistry.meta.stackexchange.com/questions/3689/what-are-some-basic-guidelines-for-good-edits-on-chem-se You can also ask in chat if in doubt https://chat.stackexchange.com/rooms/3229/the-periodic-table

@Mithoron I really appreciate the help mate. I'll improve :) Thanks for reaching out.


text:
What is the difference between an Adiabatic System and an Isolated System?
In my school text book there is a passage that says:

The presence of reactants in thermos flask or any other closed insulated vessel is an example of an isolated system.

Then a few passages of text later it says:

we take a system containing some quantity of water in a thermos flask or in an insulated beaker. This would not allow exchange of heat between the system and surrounding through its boundary and we call this type of system as adiabatic.

Does that mean all adiabatic systems can be called isolated systems and vice versa?
This got me confused. Please provide some clarity on this.

Isolated system has zero energy interchange. Adiabatic system has just zero thermal energy interchange. E.g. raising air adiabatically expands, doing work and cooling itself down.

If something is hot to the touch, then it is not adiabatic. If you can push it, then it is not isolated.

An isolated system can exchange neither heat nor work with its surroundings.  An sdiabatic system can exchange work but not heat.

@IvanNeretin more specifically it would be, adiabatic if you don't feel heat but can push and isolated if you don't feel heat and can't push

There are three ways a system can interact with its surroundings:

heat flow across the system boundary

work flow across the system boundary

a flow of matter across the system boundary

An adiabatic system is one for which there is no heat flow into or out of the system.
A closed system is one for which there is no flow of matter into or out of the system.
An isolated system does not interact with its surroundings at all, so there is no heat flow into or out of the system, no work done on or by the system, and no flow of matter into or out of the system.
Thus all isolated systems are adiabatic, but not all adibatic systems are isolated.  To put it another way, being adiabatic is a necessary but not sufficient condition for being isolated.
Also, your textbook is not quite correct in its language when it says:  "This would not allow exchange of heat between the system and surrounding through its boundary and we call this type of system as adiabatic." [emphasis mine]   If you're a high school student using a high school textbook, this imprecision is understandable (they don't want to overly burden you with subtleties), so don't worry about it.
But, if you are interested in exploring the topic more deeply, read on:
Heat is not something that is exchanged between the system and surroundings.  Rather, it is thermal energy that is exchanged, and we call this exchange of thermal energy "heat".
More specifically: If systems actually contained "heat", then they could exchange heat with their surroundings.  But they don't.  Instead, what they contain (assuming $T > 0\,\mathrm{K}$) is thermal energy.  It is this thermal energy that can (if the system is not adiabatic) be exchanged with its surroundings.  And we call this exchange of thermal energy, which manifests itself only during a change in the system, and appears only at the system boundary, "heat".  I.e., heat is not a property of the system.  It is a property of the path through which a system undergoes change.
To give an analogy:  Thermal energy is like cars, and heat is like traffic.  Suppose cars are being moved from one parking lot to another along a connecting road.  The flow of cars from one parking lot to another is called traffic. The parking lots don't themselves contain traffic, nor are they exchanging traffic; they're exchanging cars.
This distinction between system properties (like internal energy) and path properties (heat and work) will become increasingly important if and when you continue your studies in thermodynamics.


text:
Molecular geometry of BaH2 according to VSEPR theory
I am asked to predict the correct molecular geometry of $\ce{BaH2}$ according to VSEPR theory.
I assumed that it is linear due to barium being an alkaline earth metal and therefore not having any lone pairs when bonded to two hydrogen atoms, but apparently the bond angle between the two hydrogen atoms is about 119°.
Why is that the case and how can one explain this observation with VSEPR theory?
$\hspace{3cm}$

Your logic is correct. VSEPR fails to predict the correct structure of BaH2. See https://pubs.acs.org/doi/10.1021/acs.jpca.9b10248

isnt BaH2  ionic? does VSEPR applies to ionic compounds?

There is a d-Orbital-Contributions to HOMO in d0 MX2-Complexes.
All higher‐level calculations show SrH2 and BaH2 to be bent with angles of ∼140° and 120°, respectively, while CaH2 is linear with a flat potential‐energy surface for the bending motion.
The use of a core‐polarization potential together with the 2‐valence‐electron pseudopotential approach allows an investigation of the relative importance of core‐polarization vs direct d‐orbital bonding participation as reasons for the bent structures. The calculations strongly suggest that both contribute to the bending in SrH2 and BaH2. Thus, the bent structure of [BaH2] may be considered to arise from a stabilization of the b2 HOMO upon bending by mixing in metal dyz character in second-order perturbation theory.
Reference:
https://core.ac.uk/download/pdf/232777714.pdf
http://www.chimdocet-inorganica.it/SITO_ESERCIZI/Complementi/COMP1/VSEPREccezioni.pdf
This observation has been rationalized successfully by an increased involvement of (n - 1)d-orbitals in the bonding as M gets larger,19-21
and by the influence of core-polarization interactions between the M and X sites.2 Elements in the d-block have relatively high atomic masses and they tend to have stereochemically inactive electron pairs.


text:
Prediction/computation of mass spectral ionisation intensity
There are quite a few approaches (ranging from empirical additive rules based on presence of functional groups to full blown in silico QM calculations) to predicting molecular properties such as UVmax absorption or IR vibrational frequencies. While not perfect, they give a general ballpark range of what to look for.
I’ve recently been doing some quantitative MS work which has involved a lot of dilutions and calibrations.
It got me curious  if there are approaches to predict peak intensity as it’s not really something I’ve ever read about (or even included in software similar to basic IR/NMR prediction). It obviously wouldn’t be useful for quantification, but could give a vague sense of how strongly something will be observed (if at all).
To give an example, imagine mixing 1 mmol of paracetamol and 1 mmol of aspirin in DMSO and recording an LCMS of this—although you have the same amount of both the intensity in the MS TIC will vary significantly.
It seems to me that the parameters are quite well controlled (voltage, temperature, $\mathrm{p}K_\mathrm{a}$ etc.)—is it possible?

There is the Competitive Fragmentation Modelling method, which attempts to predict MS via machine learning: https://link.springer.com/article/10.1007/s11306-014-0676-4

Looks like an interesting paper, but looks like it’s focus is predicting MS/MS fragmentation

I'm not sure how uniform LC-MS results are across platforms (unlike say NMR). How much variation can you expect if you perform the experiment with a different instrument? Will a few calibrations suffice to make intensities comparable? If the answer is "yes" then there is no good reason to suppose simulation (prediction) should be very difficult (or impossible).

For QM calculations of EI-MS this article is an interesting read: https://pubs.acs.org/doi/10.1021/acsomega.9b02011. The outlook also mentions potential for CID-MS.

The absolute response coefficient (i.e. peak intensity in counts per mM or similar units) will be way too strong a function of instrument-specific parameters like solvent composition, pH, one or two or three temperature variables specific to the ion source, one or or two or three more voltage parameters particular to the ion source, whether a given emitter is dirty or fresh, how big it is, etc.
However, it may be possible to predict relative response coefficients across a series of compounds.  Most approaches I've seen rely on prediction of gas-phase basicity values, or alternately proton affinity of gas-phase neutral species (the assumption being that you are predicting the ionization efficiency of singly-charged ions in positive ionization mode arising from neutral molecules).
Folks have been working on "machine learning" approaches in this area.  Here is one example...


text:
Can we balance a redox reaction in basic medium just like as acidic medium and add OH- ions on both sides to cancel out H+ Ions?
My teacher told me that the method (which I mentioned above)  is wrong but I find many references of saying the same method correct.

Provide an example please.

Probably the teacher would prefer introducing hydroxide ions directly rather than first using solvated hydrogen ions and then canceling them out.

Yes my teacher uses the same method as you mentioned here. I want to ask which is more correct or which is more accurate?

Related: https://chemistry.stackexchange.com/questions/140462/balancing-a-reaction-where-dichromate-ion-reacts-with-ethanol-to-give-criii-an/140478#140478

There are of course two ways of balancing redox half-equations in basic solution : writing it directly with $\ce{OH-}$ ions or starting to do it in acidic conditions and then destroying the $\ce{H+}$ ions by adding $\ce{OH-}$ ions. Let's compare these two approaches with chromate ions being reduced to chromium(III).
First method (acidic solution). The redox half-equation in acidic conditions is quickly obtained : $$\ce{CrO4^{2-} + 8 H+ + 3 e- -> Cr^{3+} + 4 H2O} \tag{1}$$ In basic solution, $\ce{8 OH-}$ have to be added on both sides to destroy the $\ce{8 H+}$ and get :$$\ce{CrO4^{2-} + 8 H2O + 3 e- -> Cr^{3+} + 4 H2O + 8 OH-} \tag{2}$$ which can be simplified thus : $$\ce{CrO4^{2-} + 4 H2O + 3 e- -> Cr^{3+} + 8 OH-} \tag{3}$$
Second method (basic solution). Balancing the redox half-equation from $\ce{CrO4^{2-}}$ to $\ce{Cr^{3+}}$ without $\ce{H+}$ is not so easy. Because the $4$ Oxygen atoms of the chromate ion are of course transformed  into $\ce{4OH-}$ ions, provided enough $\ce{H}$ are available. So this requires enough $\ce{H2O}$ on the left-hand-side to compensate for the $\ce{4 H}$ atoms included in these $\ce{4 OH-}$ : $\ce{2 H2O}$. But these $\ce{2 H2O}$ molecules bring new oxygen atoms. It is not obvious to see that, at the end, $\ce{4 H2O}$ (and not $2$) have to be added on the left-hand side. The final half-equation is :$$\ce{CrO4^{2-} + 4 H2O + 3 e- -> Cr^{3+} + 8 OH-} \tag{4}$$ This is equal to $(3)$. But it not so easy to obtain.
Final remarks. 1. Whatever the method used, it may be useful to state that the $\ce{Cr^{3+}}$ ion does not exist and makes a precipitate in basic solution so that the final equation should be written $$\ce{CrO4^{2-} + 4 H2O + 3 e- -> Cr(OH)3 + 5 OH-} \tag{5}$$
2. The same reasoning could have been done starting from the ion $\ce{Cr2O7^{2-}}$ in acidic conditions. With the same conclusion.


text:
Why is liquid mercury conductive if you do not have the lattice periodicity to describe the collective behavior of conducting electrons?
We have electric conductivity in metal because the regularity and interaction in a crystal allow for the creation of conduction bands. We should not find these bands in mercury so how could liquid mercury be an conductor?  How can we describe the electrical conductivty of mercury.

why wouldn't any metal be conductive in its liquid state?

Indeed, they all are. Regularity is simply not a factor.

You just can't us Bloch functions since there is no repeating units. That does not mean that there are not widely shared mobile electrons, we just can't give a nice $E$ vs $k$ picture for their energy and momentum.


text:
Are atom coordinates in molfile necessary to produce a valid InChI identifier
I took this molfile of benzoic acid:

  ACD/Labs0708041726

  9  9  0  0  0  0  0  0  0  0  1 V2000
   13.2076   -5.2994    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   14.3539   -4.6250    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   13.2076   -6.6480    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   15.5003   -5.2994    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   14.3539   -3.2763    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   14.3539   -7.3224    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   15.5003   -6.6480    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   13.2076   -2.6020    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
   15.5003   -2.6020    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0  0  0  0
  1  3  1  0  0  0  0
  2  4  1  0  0  0  0
  2  5  1  0  0  0  0
  3  6  2  0  0  0  0
  4  7  2  0  0  0  0
  5  8  1  0  0  0  0
  5  9  2  0  0  0  0
  6  7  1  0  0  0  0
M  END

and used inchi-1 program to produce this InChI identifier:
InChI=1S/C7H6O2/c8-7(9)6-4-2-1-3-5-6/h1-5H,(H,8,9)

When I edit molfile and make all atom coordinates identical, e.g. 13.2076   -5.2994    0.0000, inchi-1 produces the same InChI identifier.
Is this a fluke or a normal behavior? I see that atom coordinates are placed into AuxInfo=... string. Are there cases when correct atom coordinates are required in order to produce a valid InChI=1S... string?

Apparently, it is derived from the connectivity table and not from the coordinates.

It seems so. I found a definitive answer, which I will post.

For the interested (accidentally describing benzoic acid), the anatomy of a MOL file, although there is a relevant entry on Wikipedia, too.

Quoting from "InChI, the IUPAC International Chemical Identifier" article [Heller et al. Journal of Cheminformatics (2015) 7:23]:

Each atom is described by a number of properties: its
chemical element name; x,y,z- coordinates (all or any of
them may be zero);

When I edit all coordinates to zero, inchi-1 produces a valid InChI identifier:
InChI=1S/C7H6O2/c8-7(9)6-4-2-1-3-5-6/h1-5H,(H,8,9)

and auxiliary info, which reflect no valid atom coordinates available:
AuxInfo=1/1/N:6,3,7,1,4,2,5,8,9/E:(2,3)(4,5)(8,9)/rA:9nCCCCCCCOO/rB:d1;s1;s2;s2;d3;d4s6;s5;d5;/rC:;;;;;;;;;

I did a similar test earlier, but I didn't retain the strict 10 character column width for each coordinate, which produced an error:
Error 34 (no InChI; Cannot interpret atom block line:     0.0000  0.0000    0.0000

For a simple compound like benzoic acid, OpenBabel certainly accepts the SMILES string containing no coordinate information at all as sufficient input to reply to obabel -:'O=C(O)c1ccccc1' -oinchi  by InChI=1S/C7H6O2/c8-7(9)6-4-2-1-3-5-6/h1-5H,(H,8,9), too.  Of course SMILES simplify and thus possess limitations in representing a structure uniquely and unambiguously, thus the restraining «for a simple compound» at the beginning.


text:
Addition of 1 equivalent bromine to 1,3 pentadiene at temperature below 0 degree Celsius gives which compound?
As far as I know reactivity of alkenes towards addition with halogens is directly proportional to their nucleophilicity, so bromine in this particular case should attack the double bond at Carbon-3 but in my textbook answer Bromine has been made to react with double bond at carbon-1

I am confused. What are you trying to ask?

What is the solvent? At which pH and at which temperature is this?

Temperature below 0 degrees celsius and ph wasn't mentioned in the question


text:
What if kinetic energy of ejected electron = 0
When hf ≥ work function, Then the electron still comes out. So, if I say kinetic energy of ejected electron = 0, it should still come out. Right ?
Then, how does the electron even move out or gets ejected if its $v=0$?

Comments are not for extended discussion; this conversation has been moved to chat.

Rather than debating whether an electron with no kinetic energy is ejected consider how an experiment to measure the photoelectron effect might be conducted. A photosensitive electron emitter is irradiated and the kinetic energy of ejected electrons is measured. One way to do this experiment is described in MIT Physics Department lab course notes. The setup is described in the following schematic:

A photodiode is irradiated with the filtered output of a mercury lamp. Only one of the available narrow bands of the lamp's spectrum is selected by the filter. The light strikes the K coated surface of the cathode of the photodiode. Emitted electrons are collected at a ring-shaped anode separated from the cathode by a gap, setting up a small current. An adjustable retarding voltage can be applied between the cathode and the anode to suppress the current, and the minimal cutoff voltage V which completely suppresses the current is measured. This voltage is theoretically related to the work function $\phi$ of the material and to the energy $h\nu$ of the incident light as
$$V = \frac{h}{e}\nu - \frac{1}{e} \phi$$
Now note perhaps the most important point with regard to your question: only a limited number of light frequencies are available in the experiment, corresponding to the bands generated by the mercury lamp. In addition the motion of a collection of electrons (a current) is measured, not the velocity of a single emitted electron. While it is possible to perform single particle experiments, at no point in the MIT experiment is the kinetic energy of individual electrons directly measured.
The lab notes also provide a short history of the discovery of the photoelectric effect:

Crude though the early data were, the qualitative fact of  the  dependence  of  the  critical  cutoff  voltage  on  the wavelength of light emerged with sufficient clarity to induce the young Albert Einstein, working as a patent examiner in the Swiss Patent Office in 1905, to link the effect with the recent idea, introduced by Planck in 1900,that matter radiates its energy in quanta of energy hν.

and

It was not until 1912 that the technical problems of making  precision  measurements  of  the  photoelectric  effect were overcome [...]

In other words, 7 years passed between the time that Einstein postulated his theoretical description and a confirmation based on "clean" data.
Note also from the lab notes how inherently "messy" such an experiment can be:

It is not possible to precisely determine the work function for removing an electron, because the cathode surface interacts with the remaining gases in the photocell as a getter, so that the surface characteristics change  a  little  from  the  ideal  case.   It  is  important  to note that the electronic work function φ is a material constant which incorporates the different emission potentials of the cathode and anode.  The former is a difficult quantity to estimate due to the manufacturing process which makes  the  cathode  surface  inhomogeneous.   It  is  composed  of  a  mixture  of  potassium,  potassium  oxide  and oxidized  silver.   For  this  reason,  you  need  to  take  care that the same area is always illuminated.  The emission work function of the photoelectrons can vary locally!

There is the direct analogy to the Solar system.
Imagine an asteroid ejected from the Solar system with zero kinetic energy. It makes sense only if the asteroid is at infinite distance from Sun, otherwise it is not ejected yet. It would start long fall on the Sun.
Or, if other stellar systems are considered, than gravity of the Sun must be counteracted by gravity of the other stars, so there is zero potential gradient.
Now, replace stars by atoms and the asteroid by an electron. An electron with zero kinetic energy would be truly ejected only if there is mutually cancelled attraction toward neighbour atoms and ions, or if it is at infinite distance to them.
So the zero kinetic energy of ejected electron has sense and meaning rather merely mathematical as the limit point on the chart where electron energy as function of wavelength crosses the zero point.
Zero kinetic energy of electron has also quantum obstacles. Zero kinetic energy means zero speed and that would mean certainty of the electron position. Following Heisenberg uncertainty principle , momentum of electron could have any value.
Electron wave function, determining electron energy in an atom, is not function of time, so precision of predicted energy is not limited.

You might have a location between various star systems where the graviational attraction from the systems cancels. That could be used as a reference. I think the photoelectric effect experiment is useful as an introduction to quantum phenomena and it might be good to avoid invoking classical analogies.

@buckthorn You might have a location between various star systems where the graviational attraction from the systems cancels. You may have missed I have mentioned that. About avoiding classical analogies - it is case dependent. It applies here very well, while elsewhere is better to avoid it, if the essential is a quantum effect where there is no counterpart. But the limit case of zero kinetic energy at zero potential energy of the central force is where both worlds meet quite well.

That's true. I guess what the QM explanation does is reveal why the photon frequency has to be shorter than a given threshold for a current to be measured. The classical analogy would be that a planet can be freed from a given orbit by only one collision with an extrasolar obhect of sufficiently high energy (which is not entirely true but is the key takeaway for an introduction: light interacts as discrete wavepackets we call photons).

@BuckThorn Sure, but I have not been addressing the nature of the photoelectric effect, that has no classical solution. I was addressing ejected electron state with zero kinetic energy, from the electrostatic point of view, related to gravitational analogy with quadratic central force in both cases.

Yes, the field is treated classically.


text:
What are "Secondary Orbital Interactions" in Organic Chemistry?
We used these to analyze the stability of molecules in an Organic Chemistry Course.
My teacher analyses the stability of compounds on 3 levels:

Atom level by studying the hybridization of the main atom
Secondary Orbital Interactions: he first draws a good resonance structure, he identifies candidates that participate in the stabilization (Candidates are often a Pi or Sigma Bond) then
He draws the Energy Diagram of the Secondary Orbital Interaction.

But I can't see what Secondary Orbital Interactions are, and why do we draw their Energy Diagram?

Can you edit the question to include examples of some molecules or structures that your teacher used to describe the secondary orbital interactions?


text:
How can I Solidify / cure silicone caulk used to create tubing?
Curing and preventing silicon caulk from sticking to itself
Food Grade RTV Silicone Sealant Adhesive Clear 10.3oz when creating tubes.
Some videos show using hot water and some videos show using cold water and hydrogen peroxide Video link on using silicone to create tubes.  Anyone know of a more exact formula / ratios most people seem to just be guessing.
I'm not making a fuel line

I'm not making a fuel line but different size tubes which will be transporting water vapour / mist created using a ultrasonic transducer (used in pond foggers) to test with (testing a fogpoinc delivery system).
PS:  I know to use dark tubes to prevent algae growth in the tubes (this is just a delivery test setup).
I created / 3d printed different screw caps (extrusion molds) to create different size tubes to test with
See example:

Question:
Anyone know of a more exact formula / ratios curing and preventing silicon caulk from sticking to itself while creating the tubes shown in the video?  The video shows using water and hydrogen peroxide but doesn't specify the ratios to use.

What is the question  ?

Anyone know of a more exact formula / ratios curing and preventing silicon caulk from sticking to itself while creating the tubes shown in the video.  The video shows using water and hydrogen peroxide but doesn't specify the ratios to use.

It is not a question. It is a statement.

@RickT have you figured out a ratio that works for you? I was looking for the same thing and came here after doing some research.

@mvaneijgen unfortunately no, it just keeps sticking.


text:
Normalising Kinetic Isotope Effect Per Deuterium Atom
I've synthesised a protic and deuterated version of a specific molecule where there are two deuterium atoms in the latter. I then go about and measure the kinetic isotope effect (KIE) for a reaction involving these two molecules. However, in many papers where there are multiple deuteriums per molecule it is common to quote a KIE per deuterium in addition to the observed KIE.
This is usually achieved by taking the nth root of the observed KIE where n= no. of deuteriums (see attachment for example: 10.1039/ b405075d).
My questions are:

Why the nth root? What is the basis for this? No paper explains why they chose the nth root.
Should one use the observed KIE or the KIE per deuterium when discussing a mechanistic proposal?

It seems intuitive that if one deuterium changes the rate constant by $k_\ce{D}/k_\ce{H}$, $n$ deuteriums will change the rate constant by $(k_\ce{D}/k_\ce{H})^n$; but that seems way too simple to be true (it feels like there must be some more subtle considerations), and I'm interested to see an answer to this too.

@orthocresol I'm not sure it's intuitive (to me) as surely the location of said deuteriums will change their relevant contributions to the rate? Consider, the rate of Z-E thermal isomerisation of d2-stilbene: in one case you have both alkene C-Hs substit. for Ds but in another case one D is at the alkene position, another on the ring. Would it still be make sense to take the sqr. root and report the KIE as per D? I'm not sure...

Hmm, well, I was thinking about $\ce{CH3} \to \ce{CD3}$ substitutions such as those in the table you showed. I think in the situation you raised the $k_\ce{D}/k_\ce{H}$ for each position will in general be different, as you said.

@orthocresol As a separate question, for such an asymmetric case, are you aware of a way to partition out the contributions on a theoretical basis?

Nope, not at all. A speculative suggestion would be to deuterate one position at a time. You shouldn't take me too seriously, though: I'm not a phys org chemist in any way.

Unless the deuterium atoms are chemically equivalent, there doesn't seem to be any justification for the nth root rule.  And even atoms that might seem chemically equivalent may not be.  E.g. consider the oxidation of 1,1-d2-ethanol (CH3CD2OH) by the enzyme ethanol dehydrogenase.  Since the enzyme is chiral and the deuterium atoms are prochiral, I very much doubt that the KIE seen with 1,1-d2-ethanol will be the square of the KIE with 1-d1-ethanol.  (1-d1 ethanol i.e. CH3CDHOH is chiral, and the KIE of one enantiomer will be radically different than the other enantiomer.)


text:
Are the particles of a colloid "solid"?
I want to understand what are the particles of a colloid in the microscopic scale. According to Wikipedia:

A colloid is a phase separated mixture in which one substance of
microscopically dispersed insoluble or soluble particles is suspended
throughout another substance.

What I can't understand is how we can have a phase seperation when we have soluble particles? I looked for some pictures and I got more frustated. For example in the following picture Colloids:
the particles seem to be like ordinary atoms/molecules just scaled up. So how can this particles exhibit bulk properties and classified as "solid" particles? Is the picture a simplification? I mean should colloid particles be represented as clusters forming a phase (e.g. solid) that is invisible to human eye?

Colloid particles could either be individual particles or clusters of particles where the cluster is still small enough to not precipitate. However the point here is that there should be no active "clustering". If there is active clustering then the clusters will get large enough to precipitate.

So the term colloid "particle" should refer to either a "solute" like particle (with just greater size) or to a small cluster of particles so that Brownian motion prevent its precipitation (phase seperation)?

It can be one big protein molecule or thousand gold atoms or a million water molecules, whatever's kinda too big for "true" solution and too small for typical suspension.

@Mithoron Thousand gold atoms or million water molecules should form a distinct phase?

Op the cartoon just gives a qualitative comparison of size. Left the dots are ions or molecules, midlle the dots are small particles in a more everyday sense, eg small pieces of solids (they might differ from the same in bulk, at least they have a high surface / inner potential ratio) and right the dots are the same just bigger, or cluster of them.

@Alchimista If you can write it as an answer so I can accept it. Your comment clears my confusion.

@Anton you can write a nice answer trying to merge all comments and answers into one. Also https://goldbook.iupac.org/terms/view/C01172 and colloidal fluids (emulsions).

https://chemistry.stackexchange.com/questions/16671/how-many-molecules-does-it-take-to-have-a-phase

Colloidal solutions may contain particles made of liquids or of solids. The particles are liquid in substances like mayonnaise or moisturizer, whose particles are droplets of oil (in water) for mayonnaise or droplets of water (in oil) for moisturizer. They can be solid in substances like acrylic paints, where the particles are made of semi-solid paint dispersed in water. The great advantage is that these paints can be diluted with water. Fresh spots are easily removed. Colloidal gold does exist. The colloidal particles can even be gas in whipped cream, froth and foams.

Thank you for your answer. Because also the comments cleared up some of my confusion I will come up with an answer including your's and the comments above.


text:
What would be the physical interpretation of the equation dG = Vdp − SdT?
I was trying to understand the derivation of the formula $G=G^\circ+RT\ln (p/p^\circ)$ starting with the following relationships:
$$
\begin{align}
\mathrm dU&=T\,\mathrm dS-p\,\mathrm dV\tag{1}\\
\mathrm dH&=\mathrm dU+p\,\mathrm dV+V\,\mathrm dp=T\,\mathrm dS+V\,\mathrm dp\tag{2}\\
\mathrm dG&=\mathrm dH-T\,\mathrm dS-S\,\mathrm dT=V\,\mathrm dp-S\,\mathrm dT\tag{3}
\end{align}
$$
Under isothermal condition $\mathrm dG=V\,\mathrm dp$ $(\mathrm dT=0)$. Therefore, the change in $G$ from the standard state is given by:
$$
\begin{align}
\int\mathrm dG&=\int V\,\mathrm dp=\int\frac{nRT}p\,\mathrm dp\tag{4}\\
G-G^\circ&=\ln p-\ln p^\circ\tag{5}\\
G&=G^\circ+\ln\frac p{p^\circ}\tag{6}
\end{align}
$$
But I'm confused about the equation $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$. What does this physically mean?
I'm assuming $V\,\mathrm dp$ is some form of work done by the system, but that suggests $G$ of the system increases while it loses energy doing the work. I also have the same problem for $\mathrm dH=T\,\mathrm dS+V\,\mathrm dp.$ Can someone please explain what they really mean?

"I'm assuming   is some form of work done by the system, but that suggests  of the system increases while it loses energy doing the work."  You have it backwards. G decreases when the system does pV work isothermally:  If the system is doing pV work, p is decreasing, which means dp is negative.  And since V is always positive, VdP would likewise be negative.

@theorist The fact that V is always positive doesn't say much about the change in V, and the change in V should be considered, $\delta V$ can be positive or negative depending on the initial and final volumes of the process, here it looks like you are considering V constant which is not always the case for VdP changes. The overall change in the sign of VdP will depend on the V dependence wrt the P and dP together solving the integral W = - \int pdV for any given process.

@Ernek Nope. You're confusing the sign of the $p$-dependence of $V(p)$, namely $\frac{dV(p)}{dp}$, with the sign of $V(p)$ itself. If $dp<0$, then $V(p) dp<0$, always. This is independent of the sign of $\frac{dV(p)}{dp}$ (which, incidentally, must always be negative except for certain v. artifical cases). [Also, I am not considering $V$ constant. I simply didn't make the $p$-dependence explicit in my last comment, since it's obvious $V$ is $p$-dependent, and strongly so for the gas-based system described by the OP; I added the explicit $p$-dependence above to make it clearer for you.]

For a continuous flow system operating at steady state in contact with a constant temperature reservoir at the same temperature as that of the entering fluid, and with negligible change in kinetic energy and gravitational potential energy, the reversible shaft work per mole that the system does on the surrounding is equal to the integral of -vdP, where v is the fluid molar volume.  So $\Delta G$ per mole is equal to minus the shaft work per mole.

I see what I was misunderstanding. Just out of curiosity - does SdT also refer to some form of work?

I don’t think so.

The SdT term is normally associated with changes in heat ( dq ) . You can think of it as another way of transferring energy - similar to work (i.e also a mean of transfering energy) but with a different microscopic connotation.

The sign convention you use for work 'done by' and work 'done on' the system is not correct when you attempt to interpret the $\Delta G$ equation.
The first law is:
$$U = W + q$$
$$dU = dW + dq$$
$dW$ in this context is work done 'on the system' and is associated with an external pressure - (not the internal pressure of the gas)
[-> Consider an adiabatic process ($dq = 0$). Only by transfering energy from the surroundings to the system ($dW > 0$ ) there is an increase in the internal energy of the sytem $dU > 0 $]
Now, $PV$ work is associated with changes in $P$ and/or $V$ only which means:
$$ dW = \Delta(P \cdot V) $$
If, for example, the process is at constant Pressure, we get the common expressions:
$$dW_{on-the-system} = -\int P_{external}dV$$
$$dW_{by-the-system} = - W_{on-the-system} = \int P_{ext}dV$$
Only if the system is changing reversibly you can equal the magnitude of $ P_{ext} $ and $ p_{internal} $ in a process ($P_{ext} \sim p_{internal}$). Now, if the system is expanding (i.e gas pushing outwards), $dW_{by-the-system} = \int p_{intern}dV > 0 $ ( system doing work) and $dW_{on-the-system} = - \int p_{intern}dV < 0 \rightarrow dW < 0 $
Henceforth in $dG = Vdp - SdT$ , $\textbf{Vdp}$ is directly associated with work done 'on the system' ( and not 'by the system' as you state above) with pressure in $dp$ directly associated with the internal pressure (keeping track of the sign convention we used from the begining of the derivation).

Thus , if your system is at constant T and there is a positve change in $Vdp$ (i.e your equation for an ideal gas:  $\ln \frac{p}{p_0} > 0 $ for $p_f > p_0$ ) means that the positive work was done 'on the system',  energy was transfered from the surroundings to the system, hence why the system Gibbs Free energy increased.
If your $p_f$ would have been smaller than $p_0$ in $\ln \frac{p}{p_0}$ then $Vdp < 0$ and now is the system the one that is doing the work $\rightarrow \Delta G < 0 $ and "loosing" that energy
You can derive the same reasoning using the correct sign convention to interpret energy changes associated with $\Delta H$ changes.


text:
Why is zinc electrode at a higher potential than copper electrode in a galvanic cell?
I read somewhere that this happens because of better shielding of outer s electrons due to completely filled d orbitals of zinc, but I didn't understand it.
Since Zn has an electronic configuration of [Ar]4s² 3d¹⁰
And Cu has [Ar] 4s¹ 3d¹⁰
Both are having completely filled d orbitals so both of them should be having equal shielding.
Please explain how the above theory is right and if not right what comes into play?

For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell, which is what you specifically ask about, is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior.

It is the opposite - zinc is at lower potential than copper. IIRC, standard Redox potentials are -0.76 V for Zn/Zn^2+ and +0.34 V for Cu/Cu^2+.

It is not only about metals, but also about their ions.

I too have read this about "shielding" in the Libretext article on Standard Electrode Potentials, and thought it dubious.
The Schmidt-Rohr paper seems excellent. According to it, the main factor is that d orbitals in Cu can participate in metallic bonding (via promotion of a d electron to the unfilled s), whereas they can't in Zn as all the subshells are full. This leads to a lower "lattice" or cohesive energy in metallic Zn compared to Cu, which is reflected in properties such as melting point and hardness, as well as in greater tendency for an ion to escape from the metal in an electrochemical reaction.


text:
The role of PIDA in C-H activation reactions
PIDA ((Diacetoxyiodo)benzene) is a hypervalent iodine(III) compound that acts an attractive oxidizing agent in most of the transition metal catalyzed reactions. Can someone explain me the mechanism of PIDA in the hydroxilation of C-H bonds?

Do you have an example of the reaction you’re interested in? The mechanisms of these processes aren’t always well understood and often highly context dependent

"most of the transition metal catalyzed reactions" Really? Most of what group of reactions? It certainly isn't most C-H activation reactions.


text:
Net direction of dipole moment in oxygen difluoride
What is the net direction of dipole moment in $\ce{F2O}$ molecule?
In the case of $\ce{NH3}$ and $\ce{NF3}$ molecules both have net dipole moment with direction reversed due to strong electronegativity of fluorine atoms, whereas in the case of $\ce{H2O}$ the direction of net dipole moment is from oxygen atom towards lone pair with a value $\mu(\ce{H_2O}) = \pu{1.84 D}.$
But in my textbook the value of net dipole moment of $\ce{F_2O}$ molecule is given $(\mu(\ce{F2O}) = \pu{0.3 D}),$ but not its direction.
I have done a lot of research on google but I could not find any related information.

After going through a standard book on inorganic chemistry , I found the answer.
The direction of dipole moment in
$F_2O$ is just reversed as in the case of $NF_3$ i.e from oxygen atom towards fluorine atoms.
Reference: Concise Inorganic Chemistry 5th edition by J.D.LEE; Chapter "CHEMICAL BONDING"
topic "APPLICATION OF DIPOLE MOMENT".

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D)
On the other hand,

In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are acting in opposition resulting in a low dipole (0.3 D)
So, basically, the dipole moment is towards fluorine atoms (which are more electronegative than oxygen atom). The reason why H2O has a higher dipole moment is that the difference between the electronegativity of O and H is higher than that of O and F, so water has a higher dipole moment even tho both water and oxygen difluoride have a similar bent molecular geometry.

note that electronegativity in H, O and F are 2.1, 3.44 and 3.98 respectively. So the difference is 1.34 and 0.55 respectively.
Hope my answer helps :)


text:
Why are there only 7 periods in the periodic table?
Is seven a hard theoretical limit (there can’t be an eighth period) or is it there just because we haven’t discovered/synthesized anything beyond that?

It's an interesting and complex subject:  https://en.wikipedia.org/wiki/Extended_periodic_table

The  stability of heavy nuclei with too many protons has little to do with periodicity of patterns in electron  behaviour.

I don't have any trouble theorizing the 27th period. Synthesizing those elements is a "bit" more complicated.

https://chemistry.stackexchange.com/questions/2786/what-are-the-g-blocks-predicted-properties https://chemistry.stackexchange.com/questions/43144/how-long-the-block-starting-with-element-121-will-be

There is no theoretical limit to the number of rows, but...

On one hand, as Tom Lehrer states, there may be many others but they haven't been discovard.
On the other hand, after all s, p, d, f orbitals are filled in the seventh period, there might be a new row, or there might be an extension of the seventh row, as a new type of orbital (g? Anyone know it there's a name?). Each orbital has double an odd number (double because the Pauli Exclusion Principle allows pairs of electrons of opposite spins), so the next orbital would have 18 electrons. Is it possible? Inner electrons would be moving at very high speeds, and they'd also shield outer electrons from attraction by the positive nucleus.
On the gripping hand, transuranics may last a few seconds... or a few attoseconds, so getting spectra might be problematic.

The S, P, D and F orbitals aren't all filled. Each of the outer shells has more S, P, D, F and G orbitals with room for more electrons. It's just that the lower number elements we already know don't use those outer orbitals.

This is more of a few random comments then an answer :/ And you ask about thing you should know before you start writing!

@Mithoron, agreed, this is not intended as an answer, but as hints to the questioner where to look for info.

Meh, finding duplicates is way more useful.

On the other hand,  it will be difficult to synthetize elements heavier than Oganesson ($\pu{Z = 118}$) for lack of neutrons. When the number of protons increases through the periodic table, the ratio neutrons/protons is also increasing. In oxygen-$16$, this ratio $\ce{n/p = 8/8 = 1.0}$. In iron-$56$, this ratio is $\ce{30/26 = 1.15}$. In uranium-$238$, it is $\ce{146/92 = 1.59}$. For heavier atoms, this ratio keeps increasing.
One of the problems to be solved for creating the last elements (tenessine, oganesson) was to find a not too heavy nucleus having enough neutrons to be used as missile sent on a target already rich in neutrons. Calcium-$48$ is a rare isotope of calcium, but a good choice with $20$ protons and a ratio neutron/proton = $28/20 = 1.4$. This value is the highest possible value among the $30$ first elements. When $\ce{Ca-48}$ is sent on a target like uranium-$238$ ($92$ protons), no nucleus can be made with a number of protons greater than $92 + 20 = 112$, which is copernicium Cn. And this collision will produce a nucleus with a maximum of $28 + 146 = 174$ neutrons, which gives the isotope $\ce{Cn-286}$. This nucleus is already rather instable, as its half-life is shorter than $1$ minute. If a lighter isotope of calcium had been used as missile, there would not be enough neutrons to make any isotope of copernicium. Of course lighter isotopes of copernicium have also been found.
It is extremely difficult to produce a nucleus with more protons than Copernicium-$252$. For making heavier atoms, it is necessary to hit a nucleus having more protons than uranium. But these targets are more and more radioactive. For getting the last atom, oganesson ($\ce{Z = 118}$), it was necessary to send missiles made of $\ce{Ca-48}$ nucleus on a target made of Californium-$252$ ($\ce{98}$ protons and $154$ neutrons), whose half-life is $\pu{2.2 y}$. Used as a target hit by $\ce{Ca-48}$ nucleus, it may produce a nucleus with a maximum of $98 + 20 = 118$ protons, and $28 + 154 = 182$ neutrons, which is oganesson-$300$. This is a maximum formula. For some reason, some neutrons have been ejected in the collisions, and real oganesson nuclei have been obtained with less than $182$ neutrons.
Now the yield of this synthesis is wretched. A total of only $14$ atoms of oganesson has been produced in the first publication, and they last less than one second. And to obtain this result, it was necessary to bombard the target of californium with an intense bundle of calcium-$48$ non-stop for three months and registering non-stop all particles emitted by this bombardment. And those particles were mostly made of smaller objects than oganesson. $14$ atoms have been obtained in three months, during which the target was quickly decaying.
To produce heavier atoms than oganesson, one should find a missile heavier than $\ce{Ca-48}$ with a higher ratio neutron/proton, like $\ce{Se-82}$, where $\ce{n/p = 1.41}$, or like $\ce{Sn-124}$, with a ratio $\ce{n/p = 1.48}$. One should also use a target heavier than Californium. But heavier atoms have shorter and shorter half-lives.

Probably a very naive question, but could one send two streams of missiles at the target? Or are the chances of three atoms combining into one essentially zero?

Good question. I am not a specialist of this domain. But  the probability that one missile hits the target and produces a new nucleus is extremely low, I doubt the yield would be better in a triple collision. In order to produce 14 nuclei of oganesson, they had to bombard many millions calcium ions per second, and wait three months. About one nucleus par week !

One other possibility is that if we were to advance into an eighth period with the creation of new elements for which we can characterize chemical properties (we have begun to do so with hassium, atomic number 108 and copernicium, 112), we may find that the  periodic recurrences we have come to know from previous periods break down.  We may have the elements but not a good placement for them in the Periodic Table as it is currently rendered.
Two factors might be considered, one theoretical and one "experimental" (meaning "experiments" with computer calculations).  On the theoretical side, the eighth period would include elements with atomic numbers equal or greater than 137, this number being the reciprocal of the fine-structure constant.  The idea that such elements are impossible has become passe, but the theories that enable those elements also imply that relativistic effects will become a predominant factor in the electronic structure and properties, much more so than we currently see in Periods 6 and 7.  Such relativistic effects were not accessible when Mendeleev first conceived the Periodic Table and not prominent across known elements even after Einstein's theories came.  We should therefore expect the relativistic effects to scramble the periodic expectations based on lighter elements.
More concretely, computer calculations indicate that the electron-shell structure on which the Periodic Table is based is already breaking down in known superheavy elements.  According to Ref. [1] and as reported in Chemistry World, oganesson (atomic number 118) is approaching a non-shell, Fermi-gas electron structure, driven by those same relativistic effects  described in the previous paragraph.  Such a breakdown in the shell structure would imply a breakdown in the periodic evolution of chemical properties based on such a shell structure.  The Periodic Table may end in a fog.
Reference
1.
Paul Jerabek, Bastian Schuetrumpf, Peter Schwerdtfeger, and Witold Nazarewicz, "Electron and Nucleon Localization Functions of Oganesson: Approaching the Thomas-Fermi Limit", Phys. Rev. Lett. 120, 053001 – Published 31 January 2018.

Oh good I don't have to answer. I have slightly different information expecting the breakdown at 120 because the energy difference after putting in the relativistic corrections says that 120 should be 1s8 1p8 rather than 2s8 but the difference is guessing whose theory is better.

We want to see your take on this breakdown @joshua!

I don't have anything much to add. I show a breakdown at 120 because the orbital energies are wrong, another breakdown at 137 because electron orbital velocity is too high and another breakdown at 173 because the electric charge energy is enough to create electron-positron pairs. The result of these is relativistic corrections are no longer remotely negligable, which means it can't be periodic anymore.

The periodic system is based on chemical properties and electronic structure.
Even the seventh period does not have a relevant chemistry. The lifetime of the respective isotopes is so short an atom hardly has time to relax into its electronic ground state before the nucleus decays. Everything we could "know" about the electronics of the eigth period would be based on conjecture, speculation. Unscientific, because not only practically, but principally unfalsifiable.
Unless someone goes and finds stable isotopes out there ... ;)


text:
Is 2 ethyl 3 methyl oxirane optically active?
According to me this molecule does not have any any plane of symmetry nor centre of symmetry, but in my textbook it is given that this is optically inactive.

Then your textbook is wrong. Or maybe they mean a racemic mixture.

Please provide a reference for your textbook and some context. Also, the name in title doesn't match the structure (which is 2-ethyl-3-methyloxirane).

I am sorry, I have edited it. Thanks!
It was a question in the book "Advanced Problems in Organic Chemistry for JEE"
By Himanshu Pandey

You may draw the structure in a chemistry-aware search engine like Chemspider.  With title structure as sketched, omitting stereochemical information like here you see there is discern about the stereogenic centres to be either in RR, RS, or SS configuration.  One points to this noting chiral resolution.

Both oxirane carbon atoms are optically active, because both contain 4 different substituents. The left one has a methyl group, a hydrogen atom, the other oxirane carbon atom and the oxygen atom from the oxirane ring. The other oxirane carbon has the following substituents: a hydrogen, an ethyl group, the oxirane oxigen and the other oxirane carbon. I assume the author of the textbook made the mistake to count the oxirane ring as one functional group, and overlooked that this one functional group functions as two substituents.

Maybe the author of the book forgot the non visible H atoms.


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Confusion regarding orbital, electron and Quantum no’s
Now , In some textbook I have read that orbital is nothing but the shape of electron . s,p,d orbitals etc.
So , after knowing shape of an orbital . I got to know that inside the orbital is an electron and there is a probability of electron to be present in it at specific positions( So , please confirm me if we talk about only one electron inside any one orbital ).
Now , the i am getting infused from textbook after reading that azimuthal Q number does not talk shape of electron but talks about shape of orbital .
Now , it is because I also read that shape of an electron is same as shape of an orbital.
Also , L quantum number tells about at which place the probability electron is 0. That’s what L values tell us .
Does it also tell about shape of orbital ? Also , how does it tell us about sub energy level .
What do these lines here represent.

Please help me in correct here.

Find a better textbook.

@MaxW sure. Could you also please correct me?

Electrons don't come in s, p, d, and f "flavors."

Shape of orbital or shape of electron are two ways of saying about the same thing, although, strictly speaking, an electron has no shape.

@Maurice Ohk . Please help me with my other doubts as well

I simply don't understand your other question. A number does not "tell" something.

@Maurice Ohk. So , l = 1 means that probability of finding an electron at 1 angle is 0 for the orbital . That’s what I got to know.

@Maurice If I’m wrong. Please tell me then what does the L Quantum no convey,

@User883022. The first quantum number gives the number of nodal surfaces of the wave function. Or if you prefer the number of surfaces where the wave function is equal to zero. The second quantum number is the number of nodal planes, that is the number of planes where the wave function is equal to zero. It is smaller or equal to the first number. s orbitals have no nodal planes. p orbitals have one nodal plane

@Maurice why Are there two lines for s orbital then ?

See also https://chemistry.stackexchange.com/questions/117188/what-are-orbitals-actually/117191#117191

@User883022. Orbitals have no lines. Only spectra have lines.

The lines through the orbital depictions show the x, y, and z axes.

As soon as you draw something in three dimensions, orbitals or not, there are are three axes $x, y$ and $z$. It has nothing to do with $s$ or $p$ orbitals.

The atomic orbital is a mathematical model that tries to explain the behaviour of electrons around the nucleus of an atom. Like any other theory explaining the behaviours of sub-atomic particles, this is purely based on experimental data and mathematics. No one has actually seen how an electron looks like because it is simply not possible to 'see' sub-atomic particles like one might see cells of living organisms under a microscope.
The atomic orbitals are mathematical functions based on the theories and assumptions of the Wave Mechanic model of the atom.
Each atomic orbital features a set of unique values namely Principal Quantum Number ($n$), Azimuthal Quantum Number ($l$) and Magnetic Quantum Number ($m_l$).
The job of each of these quantum numbers is to give you information about the energy (you can estimate the position of the electron relative to the nucleus from this), angular momentum (you can estimate the region in which the probability of finding the electron is highest, and thus the 'shape of the orbital' from this) and magnetic moment of the electron, respectively.
All of these together - $n$, $l$ and $m_l$ (along with the Spin Magnetic Quantum number, $m_s$) give the unique quantum state of an electron. This is called its orbital. And according to Pauli's Exclusion Principle, no two electrons can have the same orbitals (the same value for all the four quantum numbers).
It is impossible to determine the actual shape of an electron. But, we can make mathematical models that help understand their behaviour and maybe use those to our benefits. Atomic orbitals or even the concept of electrons for that matter are mathematical models for explaining the experimental data.

Do we only talk about the probability of one electron in one orbital ?

Probability of finding an electron in a given space is all we can talk about. According to Heisenberg's uncertainty principle, it is impossible to find out the absolute position of am electron.

By Pauli exclusion principle , in every orbital. A maximum of 2n^2 electron are present .

An orbital is the region of space, in which an electron has a probability at least of 95% be present. The quantum numbers give the following information:

$n$ is the principal quantum number and it is responsible of the extension of an orbital (a 2s orbital is bigger than a 1s orbital)
$l$ is the secondary quantum number (Clebsh-Gordan series $l=0, \dots, n-1$) and it gives the shape of the orbital ($l=0$ implies a sphere)
$m$ is the magnetic quantum number and it gives the orientation of the orbital $m=-l, \dots, l$

Please, read some textbooks like the Atkins' Physical Chemistry.


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Can Gibbs free energy determine whether a reaction is endothermic or exothermic?
In the solution of a problem on thermodynamics, it is mentioned that:

$\therefore$ Reaction will be endothermic if $\Delta G^{o}>0$

In my knowledge, the Gibbs free energy is usually used for determining the spontaneity of a given reaction. So, this statement is new to me.
Please clarify how we can determine whether a reaction is endothermic or exothermic with Gibbs free energy.
Here is the link to the problem for anyone interested.

This sentence is wrong. A reaction is endothermic if and only if $\Delta H$ is positive. $\Delta G$ has nothing to do here. It can be positive or negative. If $\Delta G$ is positive, the reaction is endoergic !

@Maurice this is what I was thinking too. I checked the solution only because I got confused between options 1 and 2 (link is in the question).

@Maurice Can you make it an answer? It sounds trivial, but this is an error that many often confuses.

It can depend on the previous lines. If they were discussing a reaction that surely increases entropy then the only way for having DelG > 0 is to have an exothermic reaction. As it was a solution it is probably what was meant, and it is correct.

Just took a look at the link and that is way you should have take it. Whatever they say that is not a definition.

The standard heat of reaction is related to how $\Delta G^0$ varies with temperature:  $$T^2\frac{d(\Delta G^0/T)}{dT}=-\Delta H^0$$So if $\Delta G^0/T$ is increasing with temperature, the reaction is exothermic and if it is decreasing with temperature, it is endothermic.  And, from this it follows that, if the equilibrium constant is increasing with temperature, the reaction is endothermic and if the equilibrium constant is decreasing with temperature the reaction is exothermic.

This sentence is wrong. A reaction is endothermic if and only if $\Delta H$ is positive. $\Delta G$ has nothing to do here. It can be positive or negative. If $\Delta G$ is positive, the reaction is endoergic, and not endothermic.


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Adsorbable Organic Halides (AOXs) in bleach residue
I have a question regarding using household bleach, and wonder if someone knowledgeable can help, please?
I have used household bleach (sodium hypochlorite, sodium hydroxide, coco alkyldimethyl, amines, N-oxides, non-ionic surfactants) to bleach mold that had formed in my bathroom on the tiles. I probably used too much, undiluted, and the result has been large patches of salt residue which have formed on the bath tub.  I use the bath tub to shower, and to be honest, I hadn't removed some of the residue for a while, but when I have tried, it is quite stubborn.
I believe the breakdown of bleach results in harmless compounds, but I have read that a small percentage consists of adsorbable organic halides. My concern is, are these harmful, if they get in shower/bath water? Do they stick to the salt residue, or are they washed away quickly? The salt residue does not come off easily when rinsing, so are there no longer AOXs present, or are they potentially leaching into the water from the residue over time? I sometimes have cuts/dry skin so wonder about the possibility of absorption, and any risk this may pose.
Many thanks.


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Reasons you would not want to heat a crystal hydrate (Measurement of water content in hydrate)
I understand that some decompose under heat.
Some will form a different product, Such as CuSO4 becoming CuO if heated too much and we wouldn't recover all of the water.
Some, such as AlCl3 actually forms HCl instead of water and needs to be reacted with another compound first.
But what other reasons are there, specific to crystal hydrates? I need two reasons and I feel like I only have one so far as I would count both under decomposition.
Do some have a similar boiling point to water that I am not seeing? Would it be so low that we boil it off along with the water?
It looks like they are all stable and do not combust.
I'm very confused because all I see talks about decomposition. I don't know what I'm missing here.

This is a lab on crystal hydrates to find the % water and the formula of the Hydrate. IE our sample turned out to be MgSO4 7H2O. The lab question is specifically why you wouldn't want to use heat to find the percentage of water in hydrated crystals. The question is phrased that heating is not the recommended method, give two reasons why not. I believe I have one reason, decomposition, but can't find anything else that would tell me why you shouldn't use heat. I've torn up two books and the internet just tells me decomposition.

Another reason is that it is sometimes difficult to be sure that all water have been removed by heating. For example, copper sulfate $\ce{CuSO4·5H2O}$ loses $\ce{4 H2O}$ at $110°$C. To get rid of the last water molecule, it is necessary to overheat it to about $200°$C.

This is already mentioned in my post. Just not in detail.

The decompositionof anhydrous $\ce{CuSO4}$  into $\ce{CuO}$ does happen at much higher temperatures, after all water has been released. And you have not spoken about copper sulfate crystallized with five molecules water.


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Why Do Electrodes Develop Equilibrium Potential?
When we place metal electrode in solution of its ions, metal ions from metal have tendency to go into solution and ions from solution tend to enter the metal lattice. My question is what is a cause of this tendency for ions in metal lattice? Is it the same tendency which creates vapor pressure of liquids a.k.a Boltzmann distribution which allows some of metal ions (liquid molecules) that have enough energy to break metallic bonds and go into solution. If interactions with solvent or any other phase for that matter are favourable equilibrium position is more shifted towards solvation of ions (more negative electric potential of metal). Basically is Boltzmann distribution cause of tendency of ions in metal to move into solution and vice versa?

No. That has to do with electrons. Don't look it as an escape of ions. That is contingent to the ions being soluble. But it is not related to the electrochemical potential within the metals. Think of passivation, for instance. I know that this isn't satisfactory but I think it does orient you to the relevant property.

I wouldn't agree. Driving force of creating potential difference due to transfer of ions through interface is the fact that system is not in thermodynamic/chemical equilibrium. Atoms in metal lattice need to have sufficient energy to break metallic bonds, so that process can start to happen with some obersvable speed (Activation energy). Potential difference created on electrode/electrolyte difference depends on metal bond enthalpy, ionization energy, enthalpy of hydration of ions, activity of ions etc. Because of differences in chemical potential, difference in electric potential is made.


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What does radioactive decay look like?
Some element isotopes have half-lives that are as short as a couple minutes to an hour. Does that mean you can watch it (or catch it on video) turn into 2 different elements? What would something like that look like?

It does not look like a thing at all. One atom of it you can't see, and many atoms of it you can't have.

No ! Because elements having a short half-life are extremely radioactive. They emit such a huge amount of radiations per second that the camera trying to register it will be quickly destroyed.

There are people who know exactly what radioactive decay looks like. Unfortunately most of them are dead since high intensity radiation (from short lived isotopes) is really dangerous. Just look at the first responders to the Chernobyl accident.

@Maurice - and yet nuclear and particle physicists watch things decay just fine all the time. You build the detectors to measure what you are looking for without dying. Think CERN, or even simple time-of-flight detectors to measure in-flight decay.

Related https://now.tufts.edu/news-releases/scientists-are-first-see-elements-transform-atomic-scale

@Alchimista That's not just related, I think you have an answer right there :)

@BuckThorn I am under the impression that OP want to see a cube morphing into powder or at least another looking cube ;)

If you somehow magicked into existence a macroscopic solid cube of some radioisotope with a half-life in the range of minutes to hours, you would probably see the cube become blindingly white-hot then mostly vanish in less than a minute, before most of the sample has even had time to decay.
Large amounts of very radioactive isotopes release enough energy to vaporise themselves. This already happens to some extent with polonium-210. A single gram of Po-210 generates 140 W continuously just by existing, a fact which can be gainfully exploited in radioisotope thermoelectric generators. That in itself is kind of magic.
Due to the shorter half-life of your isotopes, a gram of these materials would pump out hundreds of kilowatts of thermal energy at the beginning. Even assuming most of this energy is not contained within the cube, it would heat up by hundreds of degrees every second until it reaches equilibrium at a few thousand degrees (due to black-body emission). Not only would it boil off as any metal would at these temperatures, but the radioactive decay also releases energetic particles which tear apart the solid as they shoot outwards, pulverising the cube at an atomic level from the inside.


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How to calculate non-stoichiometric ratio
I'm reading a paper "Photocatalysis-mediated drug-free sustainable
cancer therapy using nanocatalyst." Here, the authors synthetise nanoplates of $\ce{SnS_{1.68}}$ with nanodots of $\ce{WO_{2.41}}$.
The authors said the valence state for $\ce{Sn}$ are $\ce{Sn^{4+}:Sn^{2+}} = 0.681:0.319$ so the non-stoichiometric ratio of $\ce{Sn:S}$ is $1:1.68$.
For the $\ce{WO_{2.41}}$ they said, valence state for $\ce{W}$ are $\ce{W^{6+}:W^{5+}:W^{4+}} = 0.307:0.216:0.477 $ so the non-stoichiometric ratio of $\ce{W:O}$ is $1:2.41$.
I would like to know where the 1.68 and 2.41 numbers come from.
Thanks in advance.

Did you try averaging over the multiple valences?

@MaxW no, I don't have any idea how to get that values, if I average the valences what I should do next?

Huh? There's nothing to calculate - these are experimental values.

@Mithoron I get this information just from the experiment? But there is a way to get that proportion or the software to analyze the data do all the work?

1.68 and 2.41 may come from element analysis. I would be rather interested in where 0.307, 0.216 and 0.477 come from. As many combinations are possible with W : O 2.41.  With these 3 numbers given, the total W : O ratio is determined by them. Possibly X-ray analysis of different states of W atoms ? Or they mixed different W oxides ?

The numbers come from observation of the compounds, presumably by elemental analysis. They are experimentally observed results not theoretical calculations.

@Poutnik they used XPS to find the oxidation state of each one and find that proportion in their system

@matt_black thanks for the clarification

So W : O would be 0.307 * 3 + 0.216 * 2.5 + 0.477 * 2 = 2.415. Similarly Sn : S ratio is 0.681 * 2 + 0.319 * 1 = 1.681. XFS and elementary analysis are 2 independent ways how to get these numbers.


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What compound(s) contribute to bleach's smell and are they dangerous?
Just a few hours ago, I had to deal with a mold problem on some untreated wood and was told by a hardware store worker that bleach was an effective way to deal with the problem, following up with some wood filler paste. I did as I was told and left the room airing out with a window open and two fans running for multiple hours. In fact, they are still running right now; I was surprised to find that the smell of bleach was still lingering in the room. I also noticed that some bleach had gotten onto my hands and they smelled of bleach even after multiple washings with soap and water.
What chemical(s) are causing the smell to linger, in the air as well as on surfaces? Are they particularly dangerous, or just relatively mild product(s) from the degradation? Is there a way to eliminate the smell without producing harmful products?
I used a spray bottle with undiluted bleach and did not wipe anything up with damp paper towels. Even so, I assume that it isn't really too dangerous, given that it is unlikely I used more than 10 mL of the stuff, and there is no way most of that 10 mL is still aerosolized and/or inside the room. Regardless, I have looked on other questions in this StackExchange, and a lot of them talk about the breakdown of bleach, but I just want to be certain the room is safe enough to sleep in tonight. Thanks!
EDIT: After removing the cloth used to clean the wood (I don't know why I left it in there), the smell is less pungent but still noticeable. Figured this might be useful information to include.

Bleach is alkaline sodium hypochlorite solution, that eventually soaked inside the wood and is gradually releasing gaseous chlorine.
It is merely just disturbing and eventually irritating mucous tissues in longer term, not really dangerous. Some chlorinated public pools have strong chlorine smell as well, being smelled from swimming suit for long time until washed in the wash machine.
A countermeasure could by applying suitable diluted reducing solution, like ascorbic acid, pyrosulfite, thiosulphate ( big crystals from photo fixers ) or similar, that would react with hypochlorite and chlorine.
Questionable can be applicability on wood surface for estetic reasons. But if bleach was applied, this should not make things worse.

Thank you. It was quite irritating to sleep in but I am still alive so I take it you are correct. Gonna let it air out another day, though.


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Why can't the N- end of the nitrile anion attack the alkyl halide in an SN2 fashion during an alkylation reaction when OH- can?
I'm reading my Clayden textbook and something keeps nagging me. It says that when we're carrying out alkylation reactions with an enol attacking an alkyl halide with strong base catalysis (with hydroxide), we should keep the base separate from the alkyl halide electrophile to prevent it from performing the SN2 attack instead of the nitrile anion formed from deprotonating the nitrile at the alpha position.
But, if OH- can attack the alkyl halide as a "competing reaction", why can't the N- end as well instead of the nitrle anion carbon-carbon double bond? They both (OH- and N-) seem like equally good nucleophiles. Is it because when we have the option to choose from the "soft" end (the double bond in the nitrile anion) and the "hard" end (the N-) nucleophile to attack the soft alkyl halide electrophile, we'd much rather have a soft nucleophile to soft electrophile attack?

Yes, there's always a regioselectivity issue. Similar effects are seen with nucleophiles that can attack at multiple atoms, e.g. enolates (C vs O) and phenols/anilines (C vs O or N).

It can.
According to this JACS paper here steric effects are a major factor in whether the ketenimine is formed or the C-alklyated product - the more hindered the nitrile, the more N-alkylation.


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Does electron emit photon when it moves?
When the photoelectric effect happens or blackbody radiation happens or when an electron jumps from a ground to an excited state, we say light is emitted by electron.
Now, light is made up of photons. So, can we say that electron emits photons?

Comments are not for extended discussion; this conversation has been moved to chat.

Electrons bound in atoms and molecules emit a photon at change of their quantum-allowed energy level, saving the total energy, momentum and angular momentum.
Free electrons emit photons, if they are accelerated and the time derivative of their acceleration is nonzero. This emission is very low for low speeds and/or accelerations.(*)
This effect is observed from the space, when lightnings are accompanied by short bursts of gamma-ray-like photons, created by Bremsstrahlung ( deceleration) radiation. Their origin is in decelerated electrons from ionized air, that were accelerated by during the lightning by the electrostatic potential gradient.
Electrons are not the only quantum objects able to emit photons. Most of subatomic particles and atomic nuclei can emit photon to reach their lower and allowed energy level. E.g. distinguishing between X-ray and gamma ray photons is not by their energy ( they largely overlap ), but by their origin. The origin of the former is energy change of inner electrons of heavy atoms, th eorigin of the latter is energy change of the nucleus.

(*) An electron circularly orbiting in a cyclotron or synchrotron emits synchrotron radiation, but a linearly accelerated electron with constant acceleration does not emit this radiation ( or minimally, as conditions cannot be exactly met ). This is the reason to built linear accelerators. They are on one hand limited by the 1-pass only, on the other hand there is no radiative energy loss, which is significant for high particle energies. Things get more complicated with relativistic effects at high speeds, but you can still get the picture.


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How is graphene electrically conductive?
I've been doing some research into graphene for a project I'm working on, and have found that it can potentially be as strong as or even stronger than current superconductors. However, from my current understanding, a substance can only be electrically conductive if it has any free-flowing electrons or ions in its structure, and graphene doesn't have any. Its structure involves a carbon atom bonding with three other carbon atoms, though one of the bonds is a double bond meaning there are no free electrons. How does it conduct electricity then?
A google search just gives results on graphite and how it has a delocalised electron (due to its different bonding) that allows conductivity.
Any help is appreciated.
Thanks.

You have it all backwards. The bonding in graphene is no different from graphite, with its free-flowing electrons and all.

First refresh graphite. Then we might think of how/why a 2-D sheet of graphene is different from itself as constituent of graphite.

Thanks both of you. What I don't understand is why graphene shouldn't have a double bond as in this website: https://www.graphene-info.com/graphene-structure-and-shape.

Graphene and graphite have delocalised electrons
The root of your misunderstanding appears to be the assumptions that the bonding in structures like graphene can be accurately represented by drawing single and double bonds. While drawing such structures makes counting electrons easy (and drawing such structures easy), it does not explain what is happening with the electrons.
The idea of delocalisation is one you don't seem to have understood. So it is worth explaining the far simpler case of benzene which is where it started.
We normally draw benzene like this:

This allows counting of the electrons involved (each line allows two electrons so there are 6 in total in the double bonds). But this structure implies that the molecule contains two different types of bond. The bond between carbons 1 and 2 is a single bond the bond between 2 and 3 is a double bond. Unfortunately this is not the structure of benzene.
In reality, all the bonds in benzene are the same. And those bonds are about halfway between single and double bonds in length. The simple picture doesn't capture reality.
More sophisticated accounts can explain this but the only important point to note is that those 6 electrons are delocalised over all six carbons. Some chemists like to draw benzene like this to represent that feature,:

But this view makes it hard to count the electrons if you need to do that.
The important thing about the real structure is that it involves delocalised electrons which has big implications for the electronic behavior of benzene.
The reason why this matters for graphite and graphene (planes of graphite are exactly the same as graphene) is that the same feature occurs there. In both substances a single layer is like a bunch of fused benzene rings with the same delocalisation of the double bond electrons. But the delocalisation isn't limited to a single ring but occurs over the whole plane. We could draw a small part like this:

to emphasise the delocalisation. An this is, in one sense, better than the typical pictures of the structure containing localised single bonds. But again means we lose count of the total electrons.
The important thing is to not take the drawings with isolated single bonds too seriously. All the bonds are the same in the real structure and all have bond orders between those os single and double bonds. But, more importantly for the question, those "extra" electrons are completely delocalised over the whole plane of the flat structure. They are not tied to the notional location of the double bonds.
And the fact that the electrons are delocalised means that in that plane they can conduct electricity (not that conductivity perpendicular to the plane is orders of magnitude lower).

Graphene is simply one atomic layer of graphite – a layer of sp2 bonded carbon atoms arranged in a hexagonal or honeycomb lattice. Graphite is a commonly found mineral and is composed of many layers of graphene.
In graphene, each carbon atom is covalently bonded to 3 others. This atom contains free ‘delocalised’ electrons that can carry and pass on an electric charge.
Graphene does conduct electricity. This is because of its structure. In graphene, each carbon atom is covalently bonded to 3 others. There is always a fourth atom free atom. This atom contains free ‘delocalised’ electrons that can carry and pass on an electric charge.

I see. On this website, however, graphene is shown to have a double bond (unlike graphene). https://www.graphene-info.com/graphene-structure-and-shape
Also, from what I know, isn't the purpose of covalent bonds to achieve a full valence shell? In that sense, wouldn't one double bond and two single bonds between carbons make more sense for graphene?

@IAA do you know benzene and electron delocalisation? If so, in graphene you have something similar but in a 2D sheet instead of a single cell

Thanks @Alchimista. That's the biggest lead I have on it now. I don't know anything about benzene and electron delocalisation as I had thought due to it being stable there would be no delocalised electrons.

That's fine.. understand delocalisation concept through simple molecules like benzene then graphene would just be a piece of cake :)


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Confusions regarding oscillator in black body and value of one quantum
Q 1 By saying oscillator I.e atoms in the wall of black body. Do we mean electron ? Because boundary of any atom is electron right.
Q2 My textbook say that quantum is the smallest value of energy that can be emitted or absorbed. On web , it says charge of an electron i.e 1.6*10$^{-19}$C
so , is this the smallest value and there can be a value greater than this ?
“The value of one quantum is equal to the magnitude of the charge on an electron”

(1). No, they mean an atom. To oscillate, an electron must be a part of an atom. (2). That's charge, not energy. You can't compare them. If it is not clear, wait for 1 kilogram of time and read this again.

@IvanNeretin Also , they say smallest value of quantum is charge of electron . That’s the point. Yes,they are different things. I will paste the exact definition.

Please avoid asking multiple loosely correlated questions. Also, make sure you provide a complete citation for the sources you are referencing to (my textbook is meaningless) and write a concise title describing the question as precise as possible.

The "classical" smallest value of charge quantum ( as there are many quantised quantities ) is charge of electron. But it is not fully true, if subnucleon structures are considered. As the charge of u quark is +2/3 e and of d quark is -1/3 e.


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pH and acid value relation for an organic chemical
Would there be a general relation between the pH (say, as measured by a pH strip) and the titration-measured "acid value" of a organic chemical?
The acid value is typically defined as "the mass of potassium hydroxide (KOH) in milligrams that is required to neutralize one gram of chemical substance."
If you determine the acid value of something which technically does not have an acidic moeity e.g. (say) an oil or benzaldehyde I suppose what you are tracking is the residual acidity of some impurities?
Would pH then be a function of ill defined parameters like some ppm amount of residual water etc. (since pH would be strictly defined for an aqueous solution).
For context: The situation I have is an liquid organic chemical where the spec typically is an upper limit of the acid value. But someone started wondering whether there would be a pH strip correlation.

There is no strict relation between the two, even if we work in water and pH is well defined.

Thanks! That's what I wanted to confirm! Trying to make sure I didn't miss something.


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Forcing a steel pipe to rust completely
I understand rust.  Fe + H20 → OH- → Fe(OH)2 or 3 and/or FenOm
I understand rust prevention and steels that are resistant to rusting.  Trying to google this question gives thousands of answers on prevention.
But what I'm interested in is how to force steel to rust.  And not in an artsy way, so my steel stuff looks weathered and cool (these are the other thousands of search results for forcing iron/steel to rust).  I want to dispose of steel by turning it into iron oxide.  I don't actually really want to do this since it's a lot easier to just landfill or recycle, but I thought of it as an alternative and realized it's really not that easy.
The trouble seems to me that the passivation layer stops rust from completely going completely through the metal.  Can this be forced by running active corrosion circuit in reverse, or will the equation still slow to a crawl once a layer of rust builds up on the surface?

There is a huge variety of steels. Some are more quickly oxidized than others. But the rust formation is always a slow process, because four products must be present for rust formation : 1) water, 2) oxygen from the air, 3) $\ce{CO2}$ from the air, 4)  local impurities or spots (often carbon grains} on the surface of the metallic piece. If one of these requirements is lacking, no rust will be formed. Rust is an electrochemical process, that is not happening over all the surface. It only happens at points, where impurities are located.

A big aid to corrosion is chloride ion. Chloride will even assist in the corrosion of stainless steel.

As we in the trade say "rust never sleeps". Rust develops on the surface , rust is somewhat protective in some "weathering" steels such as Corten. Small additions of Cu ,Cr, and P , mostly provide limited slowing of rust . Otherwise there are many tables , based on conditions and location of exposure . The tables will give rust rates like 5 mpy ( mils per year ; mil is a thousandth of an inch , pardon my US units). The loss rates will be much higher on seacoasts or where rain often occurs. So with no other information ,you can guesstimate your steel will disappear in (THICKNESS divided by 0.005 ) years. Half that time for two sides exposed.

Chances that a weathering steel will be made into real pipe are slim and none.