text:
Why does transition state loses its ability to vibrate? (Transition State Theory)
In transition state theory, (according to my textbook) it is assumed that the the transition state loses its structure, and the ability to vibrate and rotate. If the transition state cannot vibrate, then my questions are

how does the previous bonds break and the newer bonds form?
Isn't the instablity and high potential energy of transition state
due to the  extra vibration of the atoms in the molecules?
Please guide me through this.
my book states:

"The transition state does not represent a real molecule, it is impossible to isolate a transition state. However it is assumed to possess properties common to real molecule, such as molecular weight, intermediate distances, a definite enthalpy, definite composition but loses its structure, and the ability to rotate and vibrate. The transition state may either return to the initial reactant or it proceeds to form products."

I am suspicious that your book makes this claim (exact quote would be helpful here). It would be more accurate to say that the transition state is a structure where the vibrations and rotations of the species will, in one direction, lead to product and, in the other direction, lead to the reactant.

Well the "transition state" is, by definition, a state where all the bonds are at their highest energy level. It implies that if any bond changes even a little bit, it would no longer be the "highest energy level".

Thx for the reply! I will quote the exact statement from my book.

One way to think about it is that the lifetime of the transition state is shorter than the fastest bond vibration or rotation. In order to break a bond, that bond is stretched out to beyond its maximum length. If the stretching mode has time to vibrate back to a shorter bond length, the energy goes back down.

Only some of the statements you quote are true, certainly a transition state cannot be isolated  since it lasts for less than a picosecond. In fact there is hardly any direct measurement of transition states, such as a spectrum or time profile or any structure. It is not that they are too fleeting to measure but that they occur very infrequently in any reaction so their population is always vanishingly small. This is a real experimental challenge.
What we do know is that the transition state is just like any other molecule with bonds vibrating/rotating  normally except for the bond(s) that are involved in reaction. They stretch and break then form new ones as the transition state is crossed. This pathway is the lowest point on the trajectory from reactants to products and is usually drawn at the top of the reaction energy profile as a 'saddle point'. (When at the saddle point looking at right angles to the reaction path the energy increases, looking forwards or backwards along the reaction path the energy decreases).  As the new bond has lots of energy it is initially vibrationally hot and rapidly loses energy either to solvent or into other vibrational modes in the product molecule or both.
In a complex molecule you can imagine the reaction starting as the species approach one another. The energy starts to increase as the atoms that are going to react  approach one another, but collisions with solvent can randomly knock them off course, or further on course for a while, or vibrations in the molecule may randomly add or remove some energy and so hinder or accelerate the reaction. One can then imagine the species  behaving somewhat as if they are diffusing along the reaction path, sometimes going forwards but sometimes being hindered. At any point, including the saddle point, they can diffuse around a bit and either return to reactants or proceed to products.
Even in simple atom-diatom reactions in the gas phase, such as $\ce{H + Cl_2}$ vibrations in the diatomic molecule  can inhibit reaction even when there is enough total energy to react; sometimes the energy is just in the wrong place and cannot aid the reaction. This can be the case because the diatomic is both rotating and vibrating and reaction will depend on the angle which it has to the line of approach ( atom to diatom) at the separation at which reaction starts.

Thnx for helping

My answer is probably unhelpful, but such is life.
===
To put it simply, transition states do not exist. They are artificial constructs. Thus, any attempts to work with them outside transition state theory are technically incorrect.
Molecules are quantum systems. We do not have a clear theory for interaction of quantum systems, only for isolated quantum system evolving in time, and even that is quite wonky. Thus any models about molecule interactions are very, very up in the air. It is actually very surprising that ANY of our models for chemical reactions work, since most of our models are quasi-classical in some sense.

Isn't the instablity and high potential energy of transition state due to the extra vibration of the atoms in the molecules?

No. Vibrations contribute very little to energy of any molecule. In fact, trying to ascribe high energy of transition state to something specific to molecules like vibration or strain energy is kinda questionable.
TS has high energy because it is chosen (defined) this way. It is, by definition, the highest point of the reaction path. The causes of the form of the reaction path are not considered in transition state theory. Actually, there are reactions that do not have a barrier and thus a transition state per se.

Your last two paragraphs make it sound like the energy of a transition state is arbitrary, defined as higher than reactants/products. If that is what you meant to say, that isn't correct. If it isn't what you meant, can you please rephrase to clarify?

@buckminst  to some extent. I rephrased a bit, is it more understandable now?

I think there is some confusion between removing a single degree of freedom (the vibration that changes the length of the bond that gets broken) and - in the text quoted in the question - freezing all degrees of vibrational freedom. I also don't understand the part about the transition state having no structure at all.
According to this paper's introduction, the transition state looks like the reactant except for a set of bond lengths of the vibrational mode that is frozen:

Transition state theory is a statistical mechanical theory of chemical reaction rates that may be derived from two fundamental assumptions. First one defines a reaction coordinate $s$ leading from reactants (negative $s$) to products (positive $s$) and a (generalized) transition state as a system partway between reactants and products with a fixed value for $s$ (thus the transition state is a system with one less degree of freedom than the reactants).

The hand-waving reason for doing this is that the degrees of freedom kept affect the entropy of the transition state (and can treated like a system at equilibrium), while the degree of freedom associated with making and breaking bonds has to be treated in a special way (it has to be frozen to be at the maximal energy). So if you think of the energy landscape around the transition state as a saddle point, the removed degree of freedom  is along the coordinate that shows a maximum, and the remaining degrees of freedom are at a minimum.
Just in case you are wondering about the second assumption:

The second assumption is that any system passing through the transition state does so only once (before the next collision or before it is stabilized or thermalized as reactant or product). These assumptions may be called the local-equilibrium and no-recrossing assumptions.


text:
Predict stereochemistry of product

When 3-methyl(1,1-2H2)pent-1-ene reacts with HCl, why do we obtain a racemic mixture?
The given reactant is optically pure.

That fully depends on whether the starting material is racemic.

I have made the edit.

Actually, I lied, it depends on more on the product you obtain. And looking at this again, there's a good chance that you will access an intermediate that scrambles the starting material's stereochemistry.

What is the product you expect? I think the CH group will form a new stereo center.
I think if you protonate your doublebond, you will end up with a CH+ carbocation. The Cl- has no steric preference wheather it comes from the top or the bottom. tBoth are equaly stericly hindered because your side chains can freely rotate.

@Inselino Depends on whether there is a hydride migration. That would racemize the current stereocenter.

If the reaction proceeds as @Inselino suggests, a secondary carbocation 2 would be formed. Addition of chloride at this point would give a mixture of optically active, diastereomeric chlorides. Further hydride migration forms the achiral carbocation 3, which leads to racemic chloride 4. If 1 and 2 are formed reversibly, deuterium may be washed out of the product.

source: Stack Exchange

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

sort: votes


text:
What happens if I draw too much current from a home made battery?
If I try to make a battery like one of these: https://www.wikihow.com/Make-a-Homemade-Battery And then I short the two terminals, what will happen?
Disregarding anything getting hot and burning up, I'm curious what happens to the chemical reaction. I assume there is a max speed at which the reaction can occur because it seems unlikely that shorting the wires would give "infinite" current and use up the reaction in an instant.
My assumption is that the current will max out and the voltage will drop to basically nothing. Is this true? And if so what is actually limiting the rate of the reaction? Is this the same chemical mechanism as a store bought battery or are they different somehow?

Totally correct. But ... of course the heating up might destroy or damage sth. long before the whole thing just ignites. Imagine decomposition of some part of the electrolyte, leading to a change in its composition and a precipitation of an insulating moiety on the electrode surface.

Yes, you're right. There will a maximum speed that is determined by the kinetics of several parts of the system.
Electronic Resistance: Pretty straight-forward. All the parts have some inherent resistance. Though this tends to be minial compared to some of the other causes.
Migration: At high currents, the ions become depleted (or in excess) at the electrode's surface. This means that the reaction will not run at the same potential until new ions diffuse from the bulk.
Adsorption: In order for the redox reaction to occur, there is often a step needed where the ions adsorbs to the metallic surface. This is a sort of chemical reaction, so has a rate constant associated with it.
Chemical Reaction: Some batteries can also involve a chemical reaction to form an activated reactant.
Electron Transfer: This is the actual redox part itself. Transferring an electron from the metal to the reductant, or vice-versa.
All of this will play a role, but I mostly think of migration effects when I think of kinetic limitations. Once that surface layer is depleted, all of the other processes stop too.
This is most likely to be worse for a home-made battery. The electrical resistance will probably be higher because there are longer wires, less reliable construction, etc. Distances between the electrodes is longer, so in the long run there's more distance for ions to travel to equalize a charge imbalance. Your electrode surfaces are likely to be dirtier, causing slower absorption.
The book Electrochemical Methods by Alan Bard and Larry Faulkner is a good resource if you want to do a deep dive.

This may be more of an electrical engineering question than a chemistry question.
A battery can be modeled as an ideal voltage source in series with its internal resistance.  If the load goes to zero ohms the voltage across the load would measure zero volts.  The current would be limited by the internal resistance of the battery.
Both the load and battery will have resistive heating.  The temperature will rise until the amount of heat generated equals the amount of heat dissipated.  This heat dissipation may cause anything from an imperceptible rise in temperature to the dangerous "thermal runaways" which are known to cause cell phones to explode.
Though the internal resistance of a battery is not usually discussed due to most load resistances being several orders of magnitude higher than the internal resistance of a battery, all practical batteries have an internal resistance.  What affects the internal resistance will differ greatly between the home-made vs store bought batteries. Store bought batteries are engineered to have a lower internal resistance.
Home-made batteries will likely have a very high internal resistance.  Even when shorted, those batteries may have limited current discharge.  The chemical mechanism on a homemade battery may not change much when a homemade battery is shorted.
(Disclaimer:: Shorting any kind of battery is dangerous and should not be tried at home.)

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


text:
Online list of water of crystallisation?
I was extremely happy to recently receive a copy of the Nuffield Revised Book of Data but was somewhat disappointed to find that it did not list data on the water of crystallisation for various compounds.
Does anybody know of any large online tables or databases of reliable water-of-crystallisation data?

Do you mean a list of compounds like $\ce{CaSO4.2H_2O, CaSO4.1/2H2O}$, etc. ?

Yeah, and perhaps the conditions under which they form? Ideally with excess water.

You can find some data in Wikipedia.

It's literally just a list of at most 20 compounds, which is hardly comprehensive nor does it list the conditions under which they are formed given that the same compound can form different hydrates under varying amounts of water.

What would be the scientific value of such a list? If you can answer that, it might lead to a place where it could be found, or an author who has compiled it.


text:
Bond-line structure of 2-methyl-3-phenyl-1-butanol

According to me the structure of 2-methyl-3-phenyl-1-butanol is the above one.
Is my answer correct?

only that the systematic name is 2-methyl-3-phenylbutan-1-ol

You are correct the bond-line structure of 2-methyl-3-phenyl-1-butanol is the same as you you draw in the question you can even confirm it on

PubChem
ChemSpider


text:
Quantum Chemistry
How to find the degeneracy of a given state in quantum mechanics? What is the formula?

Welcome to Chem.SE. It is a policy of the site that when posting a homework problem you should demonstrate your attempt to find a solution.

Sorry but I got my exams knocking on the door... It's a national level entrance exam for MSc, so I don't really have that much time to show the solution as well... really sorry... And it's not like I post a question without trying first.

The energy is normally written as $\displaystyle E=\ell(\ell+1)\frac{\hbar^2}{2I}$ where $\ell$ is angular momentum and $I$ the moment of inertia. The multiplicity is $2\ell+1$  and in the absence of any perturbation this is the degeneracy. You can see that 9 is the correct answer if $\ell=4$


text:
Distinguishing cis and trans positions

What I fail to understand is why does trans isomer contain $2$ $\ce{Cl}$ at axial positions but $2$ $\ce{NH3}$ are not taken to be so. Isn't the two $\ce{NH3}$ in the trans isomer in the picture cis to each other?

Trans and cis here are relative terms: the structures shown are with respect to Cl, while it can also be made with respect to NH3 as well

According to IUPAC nomenclature, in case of octahedral systems, the terms cis and trans are only used for octahedral centres of the form $\ce{[Ma2b4]}$, i.e. with four ligands of one kind and two of another.

The way I justify it relates to the main plane of the complex:
referring to your compounds, the plane I am talking about it the one which includes Co, the two NH3 and ethylenediamine.
If you want to "join together" with a line the two chlorides, you have to "cross that plane", therefore your compound is trans.
In the other case (the compound on the right), the plane includes Co, ethylenediamine, chloride and NH3. If you want to do the same thing, i.e. linking the two chlorides or the two ammonia together, you don't cross that plane. In the end, your compound is cis.
Also, take a look at this website as well.
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Coordination_Chemistry/Properties_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_Transition_Metal_Complexes_II
Feel free to add more questions about this!


text:
Molecule with a quadrupole moment in an electric field
How does an uncharged non-polar molecule that has a quadrupole moment (such as carbon dioxide) behave in an electric field? I know that in a homogeneous electric field, ions travel while dipoles orient along the field (rotate) and non-polar molecules are not affected.
What kind of electrical field, if any, would exert a force on a molecule with a quadrupole moment?

One application to selectively bind to carbon dioxide rather than dinitrogen (which does not have a strong quadrupole) is linked to in this report.

Different moments of a charge distribution couple to different components of the external electric field. In the case of the quadrupole moment, the coupling is to the gradient of the electric field (EFG). Such an interaction is for instance relevant in NMR of quadrupolar nuclei, NQR (not to be confused with naked quad run, according to the wikipedia - o tempora, o mores) and Mössbauer spectroscopy, although those techniques consider the nuclear quadrupole moment.
In the case of atoms and molecules without permanent electric monopoles or dipoles, EFG interactions with permanent quadrupoles are the leading field-multipole interaction energy term (ignoring dispersion terms, ie induced dipole-induced dipole). In some cases such interactions can be of particular importance, for instance in aromatic compounds (see Kocman et al. cited below).
The quadrupole moment has been measured in $\ce{CO_2}$, see eg Chetty cited below, which includes experimental methods.
References
Electric quadrupole moment of graphene and its effect on intermolecular
interactions
M. Kocman, M. Pykal and P. Jurecka
Physical Chemistry Chemical Physics
Vol. 16, 2014
N. Chetty and V.W. Couling
Measurement of the electric quadrupole moments of CO2 and OCS
Molecular Physics
Vol. 109 (5), 2011, 655–666

Not a bad answer except that it doesn't tell us what happens when there is coupling.

@matt_black I don't fully understand your question. The energy of the molecule is altered by the interaction of an EFG with the quadrupole moment, naturally. There is an associated force if the orientation of the EFG and quad tensors is not that of the minimum energy state.

IF a molecule with a dipole interacts with the strong electric field in, for example, a microwave oven, it ends up absorbing a lot of the radiation causing a lot of warming. What sort of effects do we see for radiation interacting with quadroploes? how big are they and what are the things we observe?

@matt_black Ok, I see your point. The question is what strength of EFG would lead to a non-negligible energy compared to kT. That will take some additional thought. The quad constant is provided in the reference I cite, so it's a question of finding the expression for the interaction with the EFG and solving.

The paper showed by Chetty and Couling showed a temperature-dependent effect. I'm suppose the field orients the molecule, and at low temperature it is more oriented than at higher temperatures. But that's just a guess from skimming the paper.

The origin of the T-behavior is explained in eg MOLECULAR PHYSICS, 1968, VOL. 14, NO. 1, 63-72, see equation 3.

Correction: Yes, figure 2 shows the temperature behavior of the Buckingham constant. According to equation 2 you expect a 1/T dependence, but the quadrupole moment is probably weakly T-dependent.

To determine the force on an arbitrary multipole moment, we first expand the E-field in a Taylor-series around the point $\vec{r}=0$:
$$\vec{E}(\vec{r})=\vec{E}_0+\vec{r}\cdot(\nabla\vec{E})_0+\frac{1}{2}\vec{r}\vec{r}:(\nabla\nabla\vec{E})_0+...$$
where the number of dots in the product denotes how many indices to contract over. We can then determine the force using this expansion, since it is just the charge multiplied by the E-field:
$$\vec{F}=q\vec{E}_0+\vec{\mu}\cdot(\nabla\vec{E})_0+\frac{1}{2}\mathbf{\Theta}:(\nabla\nabla\vec{E})_0+...$$
Typically, people use the traceless quadrupole, which would add an additional factor of $\frac{1}{3}$ in the third term, but I'm going to work with the basic definition throughout. So this shows that a quadrupole subject to a large electric field double gradient (not sure of the correct terminology for this) will experience a force even if it is charge neutral and has no net dipole.
For completeness, we can also write the energy since the force is just the gradient of the energy:
$$W=q\phi-\mu\cdot\vec{E}-\frac{1}{2}\mathbf{\Theta}:(\nabla\vec{E})+...$$
where we see, as TryHard said, that the E-field gradient interacts with the quadrupole.
As an aside, one can also obtain the torque (cross product of force and r) on an arbitrary multipole:
$$\vec{T}=\vec{\mu}\times\vec{E}_0+\frac{1}{2}\mathbf{\Theta}\dot{\times}(\nabla\vec{E})_0+...$$
(I'm using the nonstandard notation $A\dot{\times}B$ to mean $\sum_{ijkl}\epsilon_{jkl}e_iA_{jl}B_{lk}$, where $\epsilon_{jkl}$ is the Levi-Civita operator and the $e_i$ are a set of orthonormal unit vectors). This suggests that a quadrupole will experience a torque when interacting with a field gradient.
For more information on this, I would read chapter 3 of Jeanne McHale's Molecular Spectroscopy. There are some typos and notational quirks(e.g. in the equation for torque, McHale just writes a cross product for the 2nd term, which isn't defined between matrices), but its overall a good book to give background in classical electrostatics and a quantum mechanical description of spectroscopy.


text:
Why is Mg(OH)2 sparingly soluble in water but soluble in NH4Cl

Is $\ce{Mg(OH)2}$ sparingly soluble in water because it's hydration enthalpy is not greater than its lattice enthalpy?
Also, is it soluble in $\ce{NH4Cl}$ because it is a stronger base
than $\ce{NH4(OH)}$ hence it displaces Ammonia and since it is a gas
the equilibrium reaction goes forward as the gas is removed?

Ammonia is basically infinitely soluble in water, and magnesium hydroxide is more soluble in an acidic medium.

Probably it won't (fully) answer your question, but I spotted a mistake:
NH4OH is the common way you use to describe an aqueous solution of ammonia. Therefore, when you write your equations, you consider is as ammonia, not as ammonium hydroxide: it acts as a base. On the other hand, NH4Cl is ammonium chloride, therefore it acts as an acid.
This explains why Mg(OH)2 is more soluble in NH4Cl than NH4OH: in the first case, you mix a (weak) base and a (weak) acid, in the second one you mix two (weak) bases.
With respect to your first question, keep in mind that when the hydration (solvation) energy overcomes the lattice energy the compound is soluble in water (solvent). If they are about the same, then it is mildly soluble (not super sure about it, though). I hope this helps!

The ammonia does not come off as a gas.  The reaction lis driven forward, dissolving more magnesium hydroxide, even though the ammonia remaibsin solution.
Magnesium hydroxide comes off a bit strange when we try to classify bases as "strong" or "weak".  It looks like a weak base because it is only sparingly soluble in water, but the portion that does dissolve acts like a dilute solution of a strong base (see, for instance, the reaction of magnesium hydroxide with a typical transition metal ion discussed here).  So, like other strong bases, the magnesium hydroxide will tend to react with the ammonium ion according to the reaction we already know:
$\ce{NH4+ + OH- <=> NH3 + H2O}$
It is this removal of hydroxide ions, not any evolution of ammonia gas, that pulls the dissolution of magnesium hydroxide along.
Because magnesium hydroxide is sparingly soluble, the reaction proceeds only until the solubility equilibrium is re-established.  You will typically end up with a mixture of ammonia, ammonium salt and magnesium salt in a slightly alkaline solution.  But there will be a lot more magnesium ions in solution with this reaction than with adding magnesium hydroxide to pure water.


text:
Stability of Arsenic Pentafluoride
How is the above mentioned compound stable enough to occur naturally? Also, I've read that it's a covalent compound with a

Trigonal Bipyramidal

Structure. The Arsenic atom is $sp^3d$ hybridized. How does that work out? I don't see how it can attain such a hybridization state.

Look up Bonding in hypervalent molecule

One way to visualize this is to consider the As to have three $sp^2$ hybrid orbitals (using $p_x$ and $p_y$) that are involved in bonding to the equatorial ligands and two $pd$ hybrid orbitals (using $p_z$ and $d_{z^2}$) involved in bonding to the axial ligands. Overall, the As is $sp^3d$ hybridized in that 5 orbitals are being used, but that doesn't mean that you necessarily have five equivalent $sp^3d$ orbitals. The hybridization can be different because the axial and equatorial bonds are not equivalent.
The $sp^2$ hybridization of the equatorial ligand bonds gives the trigonal planar shape with 120 degree angles that we expect from $sp^2$ hybridization, and the $p_z$ and $d_{z^2}$ orbitals are both oriented along the $z$ axis to bond well with the axial ligands directly above and below the As, perpendicular to the plane of the equatorial ligands.
[Added clarification: Note that this is a deliberate oversimplification to help visualize. In reality, the s orbital will also be involved in bonding to the axial ligands and the $d_{z^2}$ will also interact with the equatorial ligands, so the hybridization will not be as distinct as I described.]

Thank you @Andrew, but I still have a bit of a confusion - If one looks at the electronic config of As -  $[Ar] 3d^{10}4s^24p^3$, one can see that the $p$-orbitals are half filled.  But how does this allow for the formation of $sp^3d$ orbital here?

It might help to imagine first moving a $4s$ electron to the $4d_{z^2}$ orbital so that each of the As orbitals involved in bonding has one electron to start, even though it's not actually in that excited state. This is called "promotion and hybridization". Another way to think of it is to start with $\ce{As^5+}$ (with all of the $4s$, $4p$ and $4d$ orbitals empty) and five $\ce{F-}$ ions that can each put an electron pair into an empty hybrid bonding orbital. Does that make sense?

It doesn't allow and your mixing of p and d orbitals is obsolete and even deprecated.

@mithoron - If this were PF5, I would agree that using d orbitals in bonding is an outdated model. However, for fourth row elements like As, this model is still commonly taught even if the participation of d orbitals is still questionable, and given that the OP asked specifically about sp3d hybridization, it seemed appropriate to work within that model. It provides a direct connection to the observed structure that is much less clear in a more complex model involving ionic interactions.

@Andrew Ohhh, so $sp^3d$ hybridization is achieved by the promotion of one $4s$ electron to a $4d$ orbital. I presumed that this hybridization was impossible because the $3d$ orbital was completely filled. I hadn't realized that electrons can only be promoted to a higher energy orbital in order to achieve hybridization. But why would such a hybridization model fail for $PF_5$? Thanks!

In $\ce{PF5}$ and to a large extent in $\ce{AsF5}$, the $d$ orbital is high enough in energy that involving it in bonding does not give a benefit. The "extra" electrons instead reside completely on F atoms. The simple picture is for the axial F- ligands to both have approximately 1/2-strength bonds to the $p_z$ orbital, with added strength coming from ionic interaction between the negative charge on F and positive charge on As. This is called a "3-center 4-electron bond". The F atomic orbitals are low enough in energy that the equatorial bonds also have a lot of ionic character.


text:
Confusion regarding the Shape of XeF4
I found on internet that the shape of $\ce{XeF4}$ is square planar but what if one place $2$ of $\ce{F}$ on the equitorial position and rest two on axial position and the two lone pair on the leftout axial position then its shape will not be square planar but will change to something else
Can someone please explain this to me.

There are no axial and equatorial positions here, to begin with.

The shape of $\ce{XeF4}$ is indeed square planar as you mentioned it is so because:

The lone pairs are kept in anti position to minimise repulsion between them and also from the electrons in fluorine atoms. Keeping the lone pair anti results in 4 empty positions which are filled with fluorine, giving rise to a square planar shape and octahedral geometry.

Please fell free to comment any doubt, and give the tick if you are satisfied.


text:
Carbanion Stability
I was wondering that whether a Cross-Conjugated or a Extended-Conjugated carbanion is stable I have sort of memorise that a Cross-Conjugated one is more stable For example,
Which of the Carbanion is more stable,

I figure it must be the left one since the chlorine withdraws electron density.
For reference let the Right one be 2 and left one be 1

As phrased, this question is not answerable. Carbanions 1 and 2, are resonance contributors of a delocalized carbanion. These two contributors and a third (3) are shown in the image below.

A common misconception about resonance contributors is that each structure exists and the species alternates through them. This is not the case. If this were the case, we would be able to detect all three contributors using spectroscopic methods (though 1 and 3 would likely be indistinguishable). Carbanions 1 and 2 (and 3) all have the same response to all forms of spectroscopy. Each resonance contributor is an approximation of the true structure of the delocalized carbanion.
Another way of looking at the structure of the delocalized carbanion is to draw a resonance hyrbid. This structure shows the delocalization. A similar representation uses partial charges to indicate the positions where the negative charge accumulates.

Sometimes when we ask the question about the comparative stability of two resonance contributors, what we really mean is which is more important for approximately the true structure of the resonance hybrid. In other words, which of the three carbons in the hybrid has the most negative charge. We can apply the same reasoning to answer this question as we would to answer the stability question.
Since the two structures are resonance contributors of the same hybrid, we can ignore resonance stabilization as a consideration. Both have the same degree of resonance. We then need to consider induction. There is an electronegative chlorine atom in the structure than can stabilize negative charge through induction. Inductive stabilization is through the sigma-bond network and decreases over distance. You should be able to use this information to judge which resonance contributor is more important (1/3 or 2).

On a side note, the question of stability of these anions is complicated by the ease of elimination of the chloride anion to produce benzene, which is an aromatic compound, and the chloride anion.

thanks sir, But Can I Just reprashe the question a which one in its unionised form will have greater acidity

sir if you have time please just check this question https://chemistry.stackexchange.com/questions/40913/elimination-of-hcl-from-isomeric-chlorocyclohexadienes/108573#108573

@AdvilSell - That is a slightly different question, but not the one asked here. Since both compounds share a conjugate base through resonance, the difference in acidity for the acids must be in the difference between those structures.


text:
Can equilibrium be calculated at any time of an acid base reaction and still be the same?
Should the equilibrium constant $K$ be calculated at the end of an acid-base titration or can we calculate it in the middle of the titration for example? If so, will the calculated equilibrium constant be equal to the equilibrium constant calculated from the final state of the titration?

Depends on what you mean by 'calculate'

@WilliamR.Ebenezer like put concentrations in order to get a value?

do you mean plug in the concentration values in something like [product]/[reactant]?

@WilliamR.Ebenezer Exactly what i meant

then you won't get K, I regret. you will get the reaction quotient(Q), which will be numerically smaller than K till equilibrium is established. At equilibrium, Q=K

@WilliamR.Ebenezer and how can i find if my equilibrium is established in a tetration reaction where i've got the graph of ph=f(v)?

it's not a good idea to talk about equilibrium in titrations. every drop added reacts almost instantly, and an equilibrium is established after each addition, with one of the quantities in excess. pH vs. V plot is quite unrelated to equilibria here.

Because proton transfers are very fast, the equation $K_a=\frac{[\text{conjugate base}][\ce{H+}]}{[\text{acid}]}$ will always hold as long as your solution is well mixed. So if you are able to measure those three concentrations, you can determine $K_a$ using any mixture. As the concentrations change during the titration, the value of $K_a$ will remain constant.
Be aware that the value of $K_a$ will vary with temperature, so you should maintain a constant temperature during the experiments.

You just have to measure one concentration (or the pH) if you know the composition of the solution (which you should for a titration) and the net reaction and $K_\mathrm{W}$. Then, you can calculate all the other concentrations and the equilibrium constant. For example, the pH vs. V(titrant) graph mentioned by the OP would be sufficient to calculate $K_\mathrm{a}$ and verify that it is constant throughout the titration (for ideal solutions).

I should have written "determine those three concentrations" rather than "measure. . . ".

Acid-base reactions are quite fast. Once you mix the reactants, you can generally assume equilibrium. However, for a general chemical reaction, you have to wait for the chemical kinetics to adjust the concentrations until equilibrium is reached, where the composition given by $K$-values are valid.

Yes, but my case i'm talking about my tetration so at different volumes of reactors does equilibrium change?


text:
Different formulas for copper pyrites and bauxite
In my book in the chapter on General Principles and Processes of Isolation of Elements, I found that the formula for copper pyrites was stated as $\ce{Cu2S.Fe2S3}$ at one place, $\ce{CuFeS2}$ at another place and $\ce{Cu2FeS2}$ at another place.
Are all these taken to be the formula for copper pyrites or are these printing mistakes. I have searched the internet but could not find any useful information regarding this.

Both $\ce{Cu2S.Fe2S3}$ and $\ce{CuFeS2}$ are the equivalent means to denote chalcopyrite. The first notation, $\ce{Cu2S.Fe2S3}$, commonly used a few decades ago, shows that two sulfides are not just a mechanical mix, but form a chemical compound (same as for crystallohydrates, e.g. $\ce{CuSO4 * 5 H2O}$). The second one, $\ce{CuFeS2}$, is a formula unit, a more universal and modern representation. Chalcopyrite is a mineral of $\ce{ABX2}$ type, crystallizes in $I\bar{4}2d$ space group.

Figure 1. Unit cell of chalcopyrite $\ce{CuFeS2}$. Color code: $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.
On the other hand, $\ce{Cu2FeS2}$ is a reduced formula of $\ce{Cu8Fe4S8}$, a superstructured bornite [1]. A compound of $\ce{AB2X2}$ type, crystallizes in $F\bar{4}3m$ space group.

Figure 2. Unit cell of superstructured bornite $\ce{Cu8Fe4S8}$. Color code: $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.
Structurally, chalcopyrite and superstructured bornite have very little in common. Unless there is a specific context given, I'd rather say that $\ce{Cu2FeS2}$ is an outlier among the three and is probably a typographic issue. Also, it's not a good practice to mix dot-notated formulas with formula unit representations unless one wants to underline some structural aspects (e.g. molecular assemblies/coordination polyhedra/domains etc.)
References

Ding, Y.; Veblen, D. R.; Prewitt, C. T. Possible $\ce{Fe/Cu}$ Ordering Schemes in the 2a Superstructure of Bornite ($\ce{Cu5FeS4}$). American Mineralogist 2005, 90 (8–9), 1265–1269. https://doi.org/10.2138/am.2005.1518.

I know of five copper iron sulfide minerals:
Bornite - $\ce{Cu5FeS4}$
Chalcopyrite - $\ce{CuFeS2}$
Cubanite - $\ce{CuFe2S3}$
Idaite - $\ce{Cu5FeS6}$
Isocubanite - $\ce{CuFe2S3}$

Here is some more: talnakhite $\ce{Cu18Fe16S32}$, valleriite $\ce{Cu2Fe4S7}$, fukuchilite $\ce{Cu3FeS8}$, nukundamite $\ce{Cu_{3.39}Fe_{0.61}S4}$, haycockite $\ce{Cu4Fe5S8}$, mooihoekite $\ce{Cu9Fe9S16}$.


text:
How can I use a solvent gradient in column chromatography?
I am working to use the column, there are a lot of videos which explain it, but I don't understand this statement in the procedure.

The residue was finally purified by column chromatography using 100 % dichloromethane to 1:1 ethyl acetate/petroleum ether to 100 % ethyl acetate as solvent gradient

Could you please explain?

Please give a background to this question. What residue, why are you doing this, what is your educational background, what kind of column, etc. And btw. I recommend economising on the exclamations marks.

Using a gradient in column chromatography means that the nature and composition of the solvent(s) used as eluent is changed during the process.
This can allow an easy and overall faster separation of compounds, either emphasising little difference in retention coefficients or taking advantage of the fact that some compounds are almost totally retained while others are eluted faster.
An extreme example is a mixture of a very polar compound and an unpolar one. In this case the separation will be easy anyway, but recovering both compounds would require an extremely long running time if the polarity of the eluent isn't varied on the go.
In your case the "column started" with dichloromethane until the elution of one or more compounds.
Then the eluent was switched to a polar one with a smooth transition ensured by 1:1 ethyl acetate/petroleum ether.
Note that besides retention considerations the switching must be performed gradually and not abruptly. Very rapid changes in polarity, meaning different interaction between the changing eluent and the stationary phase, can ruin the packing of the column and, in some cases, ignite the eluent if the process is exothermic enough.
Gradient is a common procedure in other chromatographic techniques as well, and can involve T as a parameter. The underlying principle remain the same.
See for instance http://www.chromedia.org/chromedia?waxtrapp=mxrorDsHqnOxmOlIEcCzBiDpC&subNav=curorDsHqnOxmOlIEcCzBiDpCbB


text:
Separation of sugar and fat in a column chromatography
I am chemist at organic and bioorganic chemistry,I want to know if we are separating sugar and fat in a column chromatography who will be down first ? and why ?
Thank you.

Column chromatography is unlikely to be the correct method for separation here.

Homework .... ;-)

Sugars are polar, fats are not. Thus if the mobile phase is polar and stationary phase isn't - then sugars will come out first. Otherwise (in case of reverse-phase chromatography) fats will come out first while sugars will be held by stationary phase.


text:
Can an acid have a high pH?
I am studying anesthetics and came across thiopental which is considered a weak acid. However it has a pH of 10.5. I always learned that greater than 7 is alkaline. I’m looking for some guru to help me outs thanks!

related: https://chemistry.stackexchange.com/q/7851/69977

Pure compounds such as thiopental don’t have a pH. The pH is a property of aqueous solutions.

What is the source of your statement that it has a pH of 10.5?

Are you talking about thiopental or thiopental sodium?

No, acids can't have pH greater than 7. By definition, at 25 °C, solutions with a pH less than 7 are acidic and solutions with a pH greater than 7 are basic.
In your question you say thiopental and pH = 10.5. Actually this is sodium thiopental with pH range of 10.2-11.2 for a 8% solution SOURCE. The solution is alkaline because of hydrolysis.
The pH value however can be less than 0 or greater than 14 for very strong acids and bases respectively. (Lim, Kieran F. (2006). "Negative pH Does Exist". Journal of Chemical Education. 83 (10): 1465.)

"Acid" is an ambiguous term. It can mean:

an acidic solution is an aqueous[a] solution with a concentration of H+ higher than pure water, i.e. [H+] > 10-7 mol/L, i.e. pH < 7, or
an acidic compound[b] is a compound that turns water into an acidic solution when the compound is dissolved, i.e. a compound that contains a hydrogen atom that can "come off"[c] as H+.

Thiopental is an acidic compound. Hydrogen ions can come off of the nitrogen atoms, leaving their electrons behind, like this. (You'd be left with H+ of course, not Na+) "Weak acid" means that some but not all of those H+ ions come off.
Once thiopental is dissolved in water you get an acidic solution. You can then add a base, say NaOH, until the pH is where you want it, say 10.5. At that point pedants will point out that your solution no longer contains thiopental but its conjugate base (=thiopental minus the H+). Everyone else will know what you mean when you say "thiopental solution, pH 10.5." This will be the same thing as sodium thiopental dissolved in water (see the link in Jakub Muda's answer).
[a]almost always
[b]or molecule
[c]dissociate


text:
Whats the difference between reversible process in thermodynamics and reversible chemical reactions?
We know that in thermodynamics, reversible process never occur since entropy should be conserved. However, in chemistry we do have a lot of reversed chemical reactions that forms reactants from products, previously formed by a direct reaction. So, what is the misunderstanding here?

Your first statement is not true. Overall entropy is not conserved and reversible processes do occur in thermodynamics.

Can you clarify? Are you asking about reverse reactions that happen simultaneously with the forward reaction (eg in an equilibrating system) or are you talking about converting products back to original reactants in a second process? If the latter, that second process is necessarily different from the reverse of the first process. Something must be done to make it favorable.

yes, I am talking about equilibrium

@porphyrin according to this answer and other sources, all real process are irreversible
https://physics.stackexchange.com/questions/111071/do-reversible-processes-exist-in-nature

There is no misunderstanding at all.
The second law of thermodynamics can be roughly expressed as: in a given process the entropy of the universe will always increase (defining universe as system + surroundings).
Reversible processes
For reversible processes the entropy of the universe remains constant. Most natural occurring processes are irreversibles.
A reversible process is a process where you can change its direction by inducing infinitesimal changes to the system. Throughout the entire process the system is at equilibrium, so from one state A to a state B you will not observe macroscopic changes.
The system goes from state A to state B in the following way: A + dA = B where B is infinitesimally different from A. These changes are applied over properties of the system like temperature and pressure by modifying its surroundings.
Reversible chemical reaction
In the case of a reversible chemical reaction it means that if you combine the products of that reaction you’ll obtain at least some of the reactants again. Usually these reactions require the same amount of energy that was consumed or released to obtain the products (This means that you must take into account the enthalpy of the forward reaction).
A typical example are lead batteries where lead reacts with sulfuric acid to form lead sulfate and $\ce{H+}$ ions while the battery is discharging and then you apply a direct current to revert the reaction and obtain lead, sulfuric acid and lead oxide again.
Even when a reaction is thermodynamically possible you must take into account the kinetics parameters. For example: it is thermodynamically possible to turn graphite into diamond but kinetically at standard conditions (1 atm, 25$^\circ$C) it will happen in millions of years.

Related: https://chemistry.stackexchange.com/questions/16917/relation-between-thermodynamic-reversible-process-and-reversible-reaction


text:
Why was a Plimsoll symbol chosen to indicate standard state?
Historically, the Plimsoll symbol (aka Plimsoll line) was created as hull mark that would serve as a ready indicator of whether a ship was overloaded and thus running too low in the water.  It was later adopted by chemists to indicate standard state (currently defined by IUPAC as 1 bar).
Why was the Plimsoll symbol chosen for this purpose?  According to Wikipedia,

The plimsoll symbol (⦵ or o) ... is used as a superscript in the notation of thermodynamics to indicate an arbitrarily chosen non-zero reference point ("standard state").

The language used by Wikipedia suggests it was chosen because it looks like a zero with a line through it, thus communicating "not zero".  Applying this to enthalpy or free energy seems redundant and unnecessary, though, since by definition there is no true zero reference point for either (among state functions, this exists only for entropy).  It's like having a special symbol attached to $ΔU$ to indicate that $U$ is conserved, when, of course, $U$ is always conserved.
Indeed, the problem with attaching a symbol to $U$ saying it's conserved is that it suggests that sometimes $U$ is not conserved (otherwise why would you need the symbol?).  Likewise, attaching a symbol to $U$, $H$, or $G$ saying that the reference state is non-zero suggests sometimes it could be zero.  I.e., I understand they needed a symbol for reference state.  But I'm puzzled why they chose one that communicated "non-zero" instead of just choosing something that meant "reference", period.
Nevertheless, I'm sure these people knew what they were doing, and had good reasons for what they did.  I just don't understand what that those reasons would be.  I'd thus be interested to hear from anyone out there with knowledge of the history of this.
Perhaps the answer (here I speculate) is that they wanted to communicate the obvious (that thermodynamic reference states can never be true zero reference states) — and that the reason they wanted to do this was because they anticipated tables of thermodynamic values would be used by a broad audience, some of whom might not understand the underlying thermodynamic theory.

I've always thought of it as a naught. This is the first time I hear of a nautical association. My concern is usually with choosing the right symbol. The fact that there are two choices (⦵ or o) seems a potential source of confusion. I've encountered the "naught" symbol more frequently, either empty or filled (but meaning different things). I think the idea was to employ a small, tidy and distinct symbol, the likeness to "zero" indicating a "ground" and thus reference state. Presumably the answer is buried in a grave or a library on the history of science. Also, check with IUPAC, as usual.

Sure enought, reading in IUPAC's notation guidelines (Pure & Appl.Chem.., Vol.54, No.6, pp. 1239—1250, 1982.), starting in section 2.2 and continuing in section 4, you see that use of the naught symbol is more common than that of plimsoll. That extra little line is bound to have cost the plimsoll in popularity. In any case, it's a really interesting origin story, would be good to see a complete answer.

@TryHard Note that in the citation, there was missing strike (horizontal bar) in the second symbol. (Now fixed.)

Plimsoll symbol was commonly used in the literature in XIX century to show the non-zero nature of standard values. It's ironic that due to typographical reasons that plimsoll symbol is often replaced by a zero sign in the superscript (ligatures for upper 0s were everywhere, but upper ⦵s were not so ubiquitous).

I find the argument that there is no true zero energy unconvincing. For instance, as the universe expands and exhausts entropy, isn't it also heading to a lower energy state? As you extrapolate, won't we reach a state that you can define as "zero energy". Isn't it just a matter of historical accident, of choosing reference states that are not true zeros.

The reference in Wikipedia for the history of this notation is "Prigogine, I. & Defay, R. (1954) Chemical thermodynamics, p. xxiv". Does anyone have a copy of this text? Does it cite a primary source?

Also, is there a geographical preference for the Plimsoll? I've never seen it in my US texts, which use the open circle "$^\circ$".

@Andrew This is a pure speculation, but probably it was either a political decision at that time (why would U.S. press promote a British MP by using his symbol?) or the lack of typographic equipment (Germany is/has been known as the major supplier of all sorts of advanced typographic machinery and ligatures; probably U.S. companies decided not to buy overseas and use what they had).

Atkins (Oxford U.) seems to use the plimsoll exclusively in his textbook.

According to Chemical Thermodynamics: An Introduction, by Ernö Keszei, IUPAC introduced the plimsoll in 1970 to denote the reference state. Since IUPAC was established in 1919, it's relevant that they waited so long to comment on this point (although perhaps they changed their opinion over time). My guess is that the two ecosystems coexisted peacefully but growing globalization required IUPAC eventually to pass judgement. Being exposed to and comfortable with different notations through journals it is likely that chemists never considered this question particularly important.

Prigogine's Chemical Thermodynamics gives a bit more context as to why a plimsoll symbol is somewhat more preferable and is indeed used to denote non-zero nature of standard values [1, pp. xxiv, 86]:

The properties of a pure substance are denoted by a superscript $^0$. The use of this symbol to denote "standard" thermodynamic quantities is then permissible only when these refer to the pure substance as standard state. Some other convention is needed to denote the more general "standard quantities" defined in equation (7.51), where the standard state may for example refer to an infinitely dilute solution. After much consideration the symbol $^⦵$ has been introduced. This symbol is based on the circle, and is thus closely related to the more common (but occasionally confusing) notation for standard quantities. It seems useful, however, to associate it with the plimsoll mark which, appropriately enough, refers to a reference state of loading of a ship; this use of an ideogram has, we believe, some value in keeping the essential nature of "standard states" in prominence.
Quantities denoted by the plimsoll symbol are still, in general,
  functions of temperature and pressure; the idea of "standardization"
  of temperature and pressure is not stressed. When necessary we denote
  quantities at the standard pressure (which for convenience is nearly
  always chosen as the unit of pressure) by the superscript $^†$, quantities at a standard volume (which for convenience is always chosen as the unit of volume) are denoted by $^‡$. Quantities referring to a (usually hypothetical) ideal system are denoted by $^\mathrm{id}$.
[...]
$$P = P^⦵(T,p) + P^M(T,p,x_1\ldots x_c) \tag{7.51}$$

It's also interesting that Lewis and Randall, known for defining the standard state in the beginning of XX century as we know it, used an upper circle [2, p.623]:

$G^\circ$ A property of a substance in its standard state.

However, they referred to it as to a standard reference state (distinguished from a standard state) throughout the textbook [2, p. 88, 99]:

While we cannot give the absolute magnitude of $\bar{H}$ for a substance in solution, we may ascertain how much greater or less this is than the heat content of the same substance in some chosen state. Thus, at any temperature, if we are dealing with the partial molal heat content of water in some solution in which water is the solvent, we may choose pure liquid water as the reference state, and denote its molal heat content by $H_1^\circ$.
[...]
It is highly convenient, especially for purposes of concise tabulation, to know the heats of reaction when various substances are formed from their elements. It is therefore desirable to choose some one standard reference state for each element. For this purpose, we shall, at all temperatures, take the element at a pressure of one atmosphere, and in that form which is most stable or most common at room temperature. Thus liquid mercury, gaseous oxygen, solid iodine, rhombic
  sulfur and graphitic carbon, all under a pressure of one atmosphere, will be considered to be in their standard reference state.

References

Prigogine, I.; Defay, R. Chemical Thermodynamics; Treatise On Thermodynamics; Longmans, Green & Co Ltd: Glasgow, 1954; Vol. 1.
Lewis, G. N.; Randall, M. Thermodynamics and the Free Energy of Chemical Substances, First Edition later Printing edition.; McGraw-Hill Book Company, 1923.

What's the difference between a "standard state" and a "standard reference state"? Isn't use of the word "reference" redundant in that context? To quote eg Atkins: "the standard state of a substance at a specified temperature is its pure form at 1 bar pressure."

Does that mean that "reference" implies a different stated pressure?

@TryHard The term "standard reference state" arose from the need to use some starting point to quantitatively describe thermodynamic functions of state such as heat of formation. Regarding the pressure, well, Lewis used 1 atm instead of these days' 1 bar ( I added additional quote on that).

I was trying to see how you did a plimsoll in MathJax, but it turned out you just pasted one in... :)

@Zhe Yep, I did, MathJax is very friendly towards Unicode:) I think we already had similar discussion and I suggested to use a compose key –  an efficient way to insert any character without the need to remember arcane Unicode sequences or waste time searching the internet:)

Summarizing what Prigogine wrote, it seems the open circle meant "pure substance", and the horizontal bar was then added to indicate a "standard quantity" of that pure substance, rather than standard pressure (the latter instead being designated with a single dagger).  Hence there was no notion of "not-zero" when the bar was originally introduced, and the Plimsoll was chosen simply because they needed to add something to the zero and, hey, there's already this good-looking compact symbol that does that and, in another well-known application, already has the connotation of "reference value".

And if my summary is correct, it must then be the case that the horizontal symbol's meaning was subsequently (between when Prigogine wrote this in 1954 and IUPAC's formal action in 1970) migrated from "standard quantity" to "standard state".  And perhaps at that time the "not-zero" association was introduced?


text:
Work done by gas in isothermal process

A $20\ \mathrm l$ steel cylinder containing $\ce{N2(g)}$ at $20\ \mathrm{atm}$ and $300\ \mathrm K$ is kept in a big jar containing air at $1\ \mathrm{atm}$ and $300\ \mathrm K$ and sealed. $\ce{N2(g)}$ starts leaking into the jar but reversibly and isothermally. When both
  mechanical and thermal equilibria are stabilized, pressure in the jar was $1.19\ \mathrm{atm}$. Determine work
  done by $\ce{N2(g)}$.

Since there is mechanical equilibrium, the pressures should be equal to $1.19\ \mathrm{atm}$ inside and outside the cylinder with nitrogen. And since temperatures are equal too, should I use this formula for work?
$$W=-nRT\ln\frac{V_{2}}{V_{1}}$$

This is a very poorly posed problem in my judgment.  The process you described does not seem to be reversible.  Did the problem mean to say "irreversible"?  Even so, once the N2 leaks into the air, it mixes with the air.  So how does it do work on the air?

@ChesterMiller I think they mean to say the work done against the pressure of the air around it, and that pressure at equilibrium, of N2 would be 0.19atm by partial pressure law. Then work will be -P∆V

So you are assuming that the N2 does not undergo mixing by diffusion into the air once it escapes?  If this is the calculation, then the first step is to determine the number of moles of air in the jar, such that the final pressure is 1.19 atm in both the cylinder and the jar.

Yes, and also I assume that the air does not enter the cylinder containing N2

The compression of the air outside the cylinder by the $\ce{N2}$ escaping from the cylinder takes place reversibly (as far as the air is concerned), but the process of $\ce{N2}$ leaking out of the cylinder through a valve or small orifice is definitely irreversible because of the Joule–Thompson throttling that takes place as the $\ce{N2}$ flows from a high pressure inside the cylinder to a much lower pressure outside the cylinder.  So to calculate the work done by the $\ce{N2}$, one cannot treat the expansion of the nitrogen as reversible.  Instead, one must focus on the reversible work required to compress the air.
Assuming that the $\ce{N2}$ does not diffuse into the air (i.e., the boundary between the air and the $\ce{N2}$ remains well-defined), the final volume occupied by the $\ce{N2}$ is $11 × 20/1.19 = 184.9$ liters.  That means that the volume of $\ce{N2}$ increased by $184.9 - 20 = 164.9$ liters.  This is the volume of $\ce{N2}$ outside the cylinder (in the jar) in the final state of the system.  If $V_0$ represents the volume in the jar outside the cylinder (and also the initial volume of air), the final volume of air is $V_0 - 164.9$.  So, the ratio of the initial volume of air to the final volume of air is equal to the pressure ratio of the air, and given by:
$$\frac{V_0}{V_0 - 164.9} = 1.19$$
Solving for $V_0$ then yields $V_0 = \pu{1033 L}$.
The work done by the air on the $\ce{N2}$ is given by
$$
\begin{align}
W_\mathrm{a} &= n_\mathrm{a}RT\ln{\left(\frac{1033 - 164.9}{1033}\right)} \\
  &= P_0V_0\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &= 1\cdot 1033\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &= \pu{-180 L atm}
\end{align}
$$
This is equal in magnitude and opposite in sign to the work done by the $\ce{N2}$ on the air.  So the work done by the $\ce{N2}$ on the air is 180 liter-atm.


text:
What's wrong in this?
While calculating pH of $10^{-8}$M HCl,  we take into account the $\ce{H^+}$ ions released by water. But water also releases $\ce{OH-}$ions.So she shouldn't they neutralise the $\ce{H+}$ ions?

There is a worked answer here  https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355 . and a more general one here https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566

Yes, water will release H+ and OH- ions and this is known as "self-ionization of water" and this is where the maximum pH in water (i.e. 14) comes from. But you should notice that one molecule of water will always give the same amount of H+ than OH- as H2O -> H+ + OH-
Whenever you add a source of H+, like HCl, then only the concentration of H+ will increase which causes an imbalance compared to pure water and this is where pH comes from. And the opposite is true for, e.g. KOH, which will release only OH- ions and no H+.

Yes, of course! H+ ions neutralize OH- ions, but they do so at a certain rate. At any given time, there are other, different water molecules generating the exact amount of ions being consumed.
This happens because the system reaches equilibrium, the condition that you assume when calculating pH. The conclusion is that while the H+ ions are constantly changing into water and viceversa, the concentration is always the same.


text:
Influence of the neighbouring molecules in a crystal at the example of XeF4
For reasons of symmetry I (without deeper chemical knowledge) supposed, that $\ce{XeF4}$ has a shape like $\ce{CH4}$. But according to Wikipedia its

crystalline structure was determined by both NMR spectroscopy and X-ray crystallography in 1963. The structure is square planar, as has been confirmed by neutron diffraction studies.

In a crystal the neighbor atoms have some influence on another and sometimes the chemical formula shows the numerical ratio between the elements but not draw the full picture of all interactions between the neighbor elements. Does the influence of the neighboring molecules in a $\ce{XeF4}$ crystal makes the square planar shap? Is it imaginable that an isolated $\ce{XeF4}$ molecule has a tetrahedral shape?
Additional question. What is the crystalline structure of $\ce{XeF4}$?

It seems to me that you imagine that some "square planar XeF4" will stick together in the same plane, but it is obviously not the case because fluorine atoms repel each others. In fact, if you take two "square-planar XeF4" on top of each other, one will be shifted so one of its fluorine atom is right above its neighbor Xe atom. And because of this arrangement, each Xe atom forms an octahedron with 6 surrounding fluorine atoms.

@SteffX Why not await two fluorine neigbors above and two below?

@SteffX Wow, andselisks answer indeed shows 2 F above and 2 F below :-)

There are two single crystal-structure investigations done at room temperature, both published in 1963. Both suggest that $\ce{XeF4}$ crystallizes in monoclinic system ($P 1 2_1/n 1$) and preserves square-planar geometry in solid phase; however, the angle $\ce{F-Xe-F}$ slightly deviates from $90°$.

Figure 1. Packing of $\ce{XeF4}$ molecules. Color code: $\color{#90E050}{\Large\bullet}~\ce{F}$;$\color{#429EB0}{\Large\bullet}~\ce{Xe}$.
Ibers and Hamilton [1] published a structure with an $R$-factor of $11\%$ (ICSD #27467)

The molecule has symmetry $D_\mathrm{4h}$ within the limits of error of this study, although such symmetry is not required by the space group. The angle $\ce{F1-Xe-F2}$ is $\pu{86 ± 30°}$ and the two independent distances $\ce{Xe}$ to $\ce{F}$ do not differ significantly and have an average value of $\pu{1.92 ± 0.03 Å}$ (Fig. 1). The shortest intermolecular $\ce{F...F}$ contact is $\pu{2.95 Å}$. This and all other intermolecular distances are reasonable; the molecule fills space well, providing further confirmation of the formula $\ce{XeF4}$. Since all of the intermolecular $\ce{F...F}$ contacts in this structure exceed slightly the expected van der Waals contact of $\pu{2.7 Å}$ it is not surprising that a second phase of $\ce{XeF4}$, with a density some $10$ percent higher than that of the present structure, exists (2). The structure of this denser phase is at present unknown.

Fig. 1. The structure of the $\ce{XeF4}$ molecule in the solid.

Templeton et al. [2] did a subsequent XRD experiment and presented improved crystallographic data (ICSD #26626) with $R$-factor of $8.6\%$ and stated that

The structure consists of a molecular packing of square planar molecules of $\ce{XeF4}$.

With a comment on [1]:

Ibers and Hamilton have deduced two structures by refinement of data with $h + k + l$ even. These data do not permit determination of the relative signs of the two $y$ coordinates. One of these two structures is in approximate agreement with our result.

References

Ibers, J. A.; Hamilton, W. C. Xenon Tetrafluoride: Crystal Structure. Science 1963, 139 (3550), 106–107. https://doi.org/10.1126/science.139.3550.106.
Templeton, D. H.; Zalkin, A.; Forrester, J. D.; Williamson, S. M. Crystal and Molecular Structure of Xenon Tetrafluoride. Journal of the American Chemical Society 1963, 85 (2), 242–242. https://doi.org/10.1021/ja00885a038.

This may be obvious, but since no one has stated it explicitly - a key difference between XeF4 and CH4 is that XeF4 has two additional lone pairs of electrons in its outer valence shell. These are positioned above and below the plane of the square planar shape. They are the reason that tetrahedral is unfavorable.

@Andrew I thought about this during the formulation of my question and for CH4 the same imagination is practicable. C gets the 4 electrons from H, alle H are in a plane and the 4 electrons from C form two pairs. But perhaps the hydrogen atoms are too big and have no enough space in the plane.

I think you've got a mistaken concept there. Each C-H bond needs two electrons, so the 4 from C and 4 total from H's (1 from each) only provide enough electrons (8) for the four bonds. There are no remaining lone pairs in the valence shell. (There are two additional electrons in the carbon 1s orbital, but it is inner shell.)

For the sake of completeness - Xe has 8 valence electrons, so adding one from each F, there are 12 total, accounting for 4 bonds (8 electrons total) and 2 extra pairs (4 more electrons).


text:
Thermodynamics at absolute temperature
Assume I have kept a single molecule in vacuum and in absolute zero (0 K) temperature. There is no energy left in the particle, and hence any type of motion is absent.
Now, what happens when I increase the temperature of that system? Does the molecule start to move? Does the energy increase?
What does temperature even mean in vacuum?
I don't know I'm asking a good question, but this troubles me a lot.

Temperature in vacuum is not a thing at all. To have temperature, you must have molecules in the first place. Then yes, as you heat them, they start to move faster.

The situation you described is not a vacuum because it has a single molecule in it, so what you're asking about is how the concept of temperature applies to an isolated molecule.
First, the idea that you can start at absolute zero even with a single molecule is not consistent because there will always be fluctuations/movement possible in the ground state of a quantum system (the zero point energy or ZPE). The cleanest example of this is the quantum harmonic oscillator: a particle in the ground state has an average position of zero ($\langle x \rangle = 0$), but the fluctuations are greater than zero ($\langle x^2 \rangle > 0$). Let's think about this in terms of the uncertainty principle. The uncertainty principle is given by $\Delta p\Delta x \geq \hbar /2$ where $\Delta x  = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$ and $\Delta p  = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$. Because $\langle x^2 \rangle >0$ and $\langle x \rangle = 0$, we can see that the uncertainty in the momentum is also non-zero. As we know, the kinetic energy is $p^2 / 2m$, so our energy uncertainty is non-zero as well.
Second, and onto the main point of your question, you can certainly define a concept of temperature with only one molecule. Consider that even a single molecule has many different energy levels (i.e. electronic, vibrational, and rotational degrees of freedom are all available). The partition function for our system is still defined as $Z = \sum_i \exp{\left(-\frac{E_i}{k_bT}\right)}$ where $E_i$ is the energy of the $i$th state.(See Note) So, if we start very near absolute zero we will see only contributions from the ground state, but as we increase the temperature of the system we will observe non-zero probabilities of populating excited states. Nothing is broken about the concept of temperature in this situation, the energy scales are just considerably smaller than those in an ideal gas.
Note:
Your initial reaction to applying thermodynamic principles to a single molecule might be to think that it doesn't apply because we only have one molecule and not an ensemble. However, within the density matrix formulation of statistical mechanics, the quantum uncertainty of a system is entirely indistinguishable from the statistical uncertainty of a system.


text:
How does lime prevent soda lime glass from dissolving in water?
How does lime addition as an ingredient prevent soda lime glass from dissolving in water? Does it form calcium silicate in the process of melting?

The addition of lime does indeed form calcium silicate, but it doesn't serve the purpose you mentioned. Silica of itself isn't water soluble, nor soda glass.
The difference is in the crystalline structure. Silica ($\ce{SiO2}$) consists of silicon atoms bridged by oxygen atoms, forming a periodic crystal with no net charges on any atom.
However when some alkali is introduced ($\ce{Na2O}$), the additional oxygen atoms remain charged, and cannot link silicon atoms together; those form silicate anions. The silica breaks and depolimerizes, and the remaining $\ce{Na+}$ fill holes in the structure to maintain charge balance. Thus the solid melts at lower temperature with reduced viscosity, it's "more fluid".
Lime contains calcium, which bonds more strongly to silicate anions than sodium. It also decreases viscosity of molten glass because of the non bridging oxygen, but in addition increases hardness and tensile strength of the glass in solid state.

Thank you for clarifying, it seems to be a common misconception that soda glass is soluble enough for it to be a concern. May I ask you an additional question here in the comments, since I do not want to spam the forum? Since neither calcium oxide or calcium carbonate is soluble in the melt (or am I wrong again?), how come the resulting glass is clear, instead of being cloudy due to a suspenion of solid CaO in it?

You probably mean in connection with glazing earthware. Lime (CaO) is added to the glass mixture as a flux agent, which helps lower silica's melting point (SiO2). Resulting in improving in hardness and adding resistance to leaching compounds from earthware.
https://digitalfire.com/oxide/cao


text:
Why doesn't self condensation occur in acetoacetic ester synthesis
So, I came across a quite versatile method to synthesize aldehydes and ketones, its called the Acetoacetic ester synthesis.
Suppose I need some 2-pentanone. So to synthesize it via this method I can do this

But I'm more worried about adding a base to Acetoacetic ester. Won't self condensation occur between two molecules of Acetoacetic ester like this?

If it occurs how do we minimize it? If it doesn't then why?
Any help will be appreciated.

Doesn't happen. You just get proton exchange.

Can you tell me the reason, please?

If the addition occurred at all AAE anion would prefer the ketone, a reaction that forms an intermediate alkoxide, whose conjugate acid has pKa ~15-18. The pKa of AAE is ~11. Seems to be an uphill battle.

@user55119 But I've learned that addition to ketones would require special conditions (i.e addition to ketones is mostly unfavourable due to steric hinderence).

I disagree. Addition of what to ketones? Grignard reagents? Stobbe condensation? Wittig reaction?

Self-condensation is typically not an issue with 1,3-dicarbonyl compounds, such as the one in your question. Why? Due to their acidity ($\mathrm{p}K_\mathrm{a}$ around 9 to 11), most 1,3-dicarbonyls exist in their enolate form in solution. Therefore, there is very little of the original electrophilic 1,3-dicarbonyl which the enolate can attack. Hence, the amount of self-condensation taking place is relatively little and usually not a concern.


text:
What are the possible magnetic quantum numbers of silver?
What are all the possible quantum numbers ml and ms of the electrons in the outermost s orbital of silver? I'm new to this concept. Can anyone walk me through this?

Silver has electronic configuration as [Kr] 4d10 5s1 .
The s orbital ( 5s ) has only one electron. And the s orbital here has n = 5 (as it is in the fifth period) and l = 0 .
The s orbital is symmetric in space (sphere) which has only one orientation and hence has no "different" orientations  each of which is represented by a particular value of ml. So ml can take only one value and that is 0.( If this was a p orbital which could take 3 different orientations, ml would take 3 values -> -1,0,1. In general an orbital with azimuthal quantum number l, ml would take 2l+1 values of ml  ranging from -l to +l )
The quantum number ms represents the spin. It can either be +1/2 or -1/2.
Hence ,  ml =  0 amd ms = +1/2 or -1/2.
Hope this helps!

There are 4 quantum numbers.
n - Principal quantum number with values 1,2,3,4... - in simple words look at the row in periodic table and this is the max value of n for a particular atom.
l - Azimuthal quantum number, has always values from 0 to (n-1). In simple words it representsthe type of orbital (s, p, d, f). Look at the following table.

ml - magnetic quantum number distinguishes the orbitals available within a subshell and has values from -l to +l. In simple words, each orbital wavefunction has a specific shape (photo above). There is 1 s orbital, 5 p orbitals, 10 d orbitals and so on. This numbers points to one of the orbitals.
ms - spin quantum number, represents the spin of an electron. Values +1/2 and -1/2.

Back to your question - quantum number of outermost s orbital of silver. Silver has an electron configuration [Kr] 4d10 5s1.
n = 5 as silver is in the 5th row (row is called period).
l = 0 for s orbital
ms = 0 as there is only 1 s orbital
ml = +1/2 or -1/2
I hope this explains your problem.

You have four quantum numbers: n, l, ml and ms.
The relationship between them is
n = 1, 2, 3... (theoretically speaking, up to infinite)
l = 0, 1, 2... n-1
-l <= ml <= l (that means, ml can assume all integer values between -l and l, including 0).
ms can be either 1/2 or -1/2.
l gives the "shape" of the orbital, and chemists give nicknames to the orbitals:
l = 0 is called s, l = 1 is called p, l = 2 is called d, l = 3 is called f, and so on (l = 4 is g, l = 5 is h, you got the idea)
Now, Silver is in the fifth period, therefore your n = 5.
Since we are only interested in the s orbitals, we have n = 5, l = 0 (for the reason given above). Since l = 0, then also ml is 0.
You have two electrons in the s orbital, therefore you have both 1/2 and  -1/2 for ms.
Therefore, your quantum numbers are n = 5, l = 0, ml = 0, ms = +1/2 and - 1/2.
I hope this helps!


text:
What would be a stable way to hold lithium hydride powder?
I currently have 100 grams of lithium hydride powder that needs to be held in a solid form. Initially, I was considering using cement, however, the moisture require converts the lithium hydride to lithium hydroxide. What would another moisture-free agent be that I could use for this task?
If it helps, the lithium hydride and the solidifying agent are to be used for neutron moderation.

How strong does the solid have to be, and what elements would be compatible, in terms of neutron cross-section?Paraffin wax has just H and C, but is not very strong. Seal it in polyethylene or polypropylene?

From what I remember, molten $\ce{LiH}$ is just poured under inert atmosphere into a container of the appropriate form, which is sealed later (or used in bulk – in aerospace industry). But it seems like you need some sort of a homogeneous solid matrix to keep $\ce{LiH}$ intact under ambient conditions, is it true?

Yes, you are correct.

If the LiH needs to be exposed to ambient conditions, it should be enclosed is a moisture-impermeable container, e.g., a glass ampoule which is closed off. Glass can be blown quite thin - but then you need to be quite careful. As DrMoishe suggested, PE or PP would be good, but I'm thinking of it as a closed tube, not as a binder.


text:
Why is a carbon to iodine pi-bond less stable than a carbon to fluorine pi-bond?
This question comes from a second semester undergraduate organic chemistry course. Please refer to the image below with regard to the question.
My understanding of the trend in section 4a is that "larger" halogens such as iodine by themselves are able to stabilize positive and negative charges better than "smaller" halogens like fluorine.
Moving into section 4b, my professor transitions with "HOWEVER", telling me that the trend in 4b is contradictory in some way to the trend in 4a.
The idea that my professor is attempting to convey still very much eludes my comprehension. So I ask:

(Main question) Why is iodine-containing molecule less stable than the fluorine-containing molecule?
What is he trying to show me with this trend comparison?
Is there a decent source that can elaborate on this section of my handout?

Thank you,
Brian

Could someone explain why such structures were used in 4b? This is a legitimate inquiry and I do not see how this question is preposterous.

If the bonds in 4b were sigma bonds and there was no charge, the order of stability would be reversed. I want to know why that is.

Well, you know what mesomeric stabilisation of cation is? That's exactly the context you're somehow missing.

To answer your question, I do not. Is it analogous to induction?

The main idea is that a cation can be stabilized by an adjacent atom mainly in two ways:

Distributing charge over pi bonds, easily visualized with resonance structures; the effect is called mesomeric stabilisation.
Distributing charge over sigma bonds; i.e. induction stabilisation.

As induction stabilization works over sigma bonds, it is stronger if the bond has the correct polarization; that is, if the other atom has low electronegativity, so it holds a positive charge easily. In your example iodine would be better as explained in a).
Mesomeric stabilization occurs through pi bonds, which are significantly weaker than sigma bonds. In this case the main issue is good "contact", or overlap, between atomic orbitals. As carbon is so small, it overlaps better with small elements, so in this case fluorine has the advantage, as explained in b).
In general, mesomeric stabilisation is the main mechanism for carbocations. This happens because not only is the charge effectively spread over more than one atom, but also the electrons compromised in the pi bond lower their energy compared to just staying in the valence shell of the donor atom.
However, note that this is not the case for all elements; take for example sulphonium ions. Sulphur is a bigger atom that forms double bonds poorly. Hence the main mechanism is induction.
A good book (albeit a bit old) for reference is Morrison's Organic Chemistry.

That was exactly the type of clarity I was looking for. You hit the nail on the head F Bat. Thank you. "Ah ha" moment achieved.


text:
Can a substance be extracted from solution by adding another substance with a higher solubility?
Suppose that I have a substance X dissolved in water. I then add a substance Y to the solution. Substance Y has a higher solubility in water than substance X, and more than enough of substance Y is added to saturate the water. Will substance X come out of solution? Why or why not?
For example, let's have X be sodium chloride and Y be sucrose.

Essentially, what you describe is a solting out process. Usually it's impossible to predict a priori whether it will or will not work, and how effective, so there is no uniform answer unless you provide a system to look into.

Related: https://chemistry.stackexchange.com/questions/79136/increasing-partition-coefficient-in-an-extraction


text:
Make paper from hair?
Hair is fibrous... Is there a good reason, why we don't recycle our own hair (and clothing, what have you) into paper-products? (rather than cutting down trees)?

Hair is protein, paper is polysaccharide.

The diameter of human hair is typically 100 microns (from 17 to 181 microns: https://hypertextbook.com/facts/1999/BrianLey.shtml ). The diameter of a cellulose fiber is about 10 microns ( https://en.wikipedia.org/wiki/Paper ), and a cellulose molecule is much smaller (up to about 10,000 glucose units long: https://en.wikipedia.org/wiki/Cellulose_fiber ).
While it should be possible to make a coarse papery product from hair, something is needed to make the hair fibers stick together without being sticky in the final product. Think how obnoxious it would be if you wash your hair, then dry it and it turns into cardboard!
Paper from cellulose fibers sticks together when it dries because there are so many hydroxyl groups on every cellulose molecule and fiber - no extra adhesive needed. It can be made quite thin - commercial paper is available as thin as 60 microns ( https://hypertextbook.com/facts/2001/JuliaSherlis.shtml ).
Hair, however, is long enough, and frequently curly enough, to be able to be woven into yarn or thread, and then into thick 2-dimensional products. The thickness of the yarn or thread gives the product more strength.


text:
What causes the turbidity when preparing Fenton's reagent?
New guy, please be gentle. I might ask questions not according to conventions, I am not a chemist nor do I study chemistry. Feel free to adapt the question and the tagging.

I am preparing a solution for Fenton's reagents according to [1], mixing 0.750 g $\ce{FeSO4 * 7 H2O}$ (blueish green, cyan almost) to 50 ml deionized (milli-q) $\ce{H2O}$ ([1] uses a slightly higher concentration, but we adapted it according to another publication). This solution needs to be adjusted to $\mathrm{pH}~3$, which, according to [1], can be done with $\ce{H2SO4}$. Now, even if I adjust the solution to well under $\mathrm{pH}~3$, let's say $\mathrm{pH}~1.5$, I still get fallout (presumably a ferrous(III) compound) and a generally turbid solution.
I don't understand what might be falling out and why the solution is turbid in general. What might be happening here? Is the solution somehow reacting with the surrounding air? Could it be further reacting with the water?

2. Fenton’s Reagent
This reagent has two components: $30\%$ (v/v) $\ce{H2O2}$ with an iron catalyst. The catalyst solution was composed of 20 g of iron(II) sulfate heptahydrate in 1 l of filtered RO water. Tagg et al.(34) tested this reagent within the context of extracting microplastics from wastewater. The authors identified this as the optimum concentration. The catalyst solution was adjusted to $\mathrm{pH}~3.0$ using concentrated sulfuric acid.

References

Hurley, R. R., Lusher, A. L., Olsen, M., & Nizzetto, L. Validation of a method for extracting microplastics from complex, organic-rich, environmental matrices. Environmental science & technology, 2018.

What water do you use? Is it distilled/deionized water, or just tap water? Also, what does the solid iron sulfate you are dissolving look like?

Deionized (milli-q). The iron sulfate is blueish green, cyan almost.

Looks about right; I'd probably just filter the solution and proceed further.

@andselisk hmm... I was instructed that the concentration with fallout, so the solution wouldn't be the same after filtration. I will discuss it with the local expert, though. Thx.

@andselisk thx for the edit. This will be of assistance next time I post a question on chem.se.

Yep, the concentration of $\ce{Fe(II)}$ might be a bit lower after filtration, but I presume you are not performing any quantitative analysis with Fenton's reagent anyway, so it shouldn't be a big deal. And you are very welcome (I included your comment regarding the precursors into the question); probably someone else here will provide you with better suggestion:)


text:
What is the name of this organic compound?

The organic compound in question is shown above. This is how I tried to name it:

Start from the left hanging  $\ce{H_3C}$.
$\ce{OH}$ gets the number 2 and $\ce{NH_2}$ gets 6. The lowest numbers possible for the functional groups keeping seniority/priority in mind.
There are two ethyl groups at 3 and 5.

Hence, the compound should be called 6-Amino-3,5-diethylhexan-2-ol.
I tried searching the web to confirm this, but nothing useful showed up.
Is this the correct IUPAC name of this organic compound?

The longest chain has 7 carbon atoms. So you would have an ethyl group in carbon 3 and the amino group will be a ramification in carbon 5. I would stand for:

5-(aminomethyl)-3-ethylheptan-2-ol


text:
Relationship between effective nuclear load and periodic properties
The effective nuclear charge is defined as the net positive charge experienced by an electron in a polyelectronic atom.  It can be calculated using the well-known Stars Rule.
Once I have calculated it for different chemical elements, I would like to know if there is (if there is any justification) any dependence between its value and the periodic property.
I remember hearing, for example, that the greater the value of the effective nuclear charge, the smaller the atomic radius (half the internuclear distance). And that the greater the effective nuclear charge, the greater the ionization potential.
However, I cannot justify these assumptions. Nor do I know whether this relationship exists in electronic affinity or electronegativity.

For the atomic radius, you may use a modified Bohr's formula: $r = \dfrac{53 n^2}{Z^*} \pu{pm}$

You can think of it in terms of electrostatic attraction.
If you model an atom as a positive point charge surrounded by orbiting electrons, the radius of each orbit is defined by the equiilibrium between the attraction from the nucleus and the repulsion from the other electrons. The effective nuclear charge would be the net force resulting from this competition.
As nuclear charge increases, the attractive force increases and the radius of the orbit decreases; hence the atomic radius decreases.
Because the attraction is higher, more energy is required to remove an electron from that low-energy state, which means the ionization potential is higher.
In the same fashion, a new electron added to a neutral atom will have lower energy with increased nuclear charge, which makes it easier to overcome the repulsive forces that the other valence-shell electrons exert on it. Thus the electron affinity is also higher.

In summary, one could say that: the higher the effective nuclear charge, the greater the ionization energy, the greater the electronic affinity and the smaller the radius, right? @F Bat

@user218559 Exactly. All those can be justified by this simple principle. Note though that I'm talking exclusively about the first electron affinity/ionization energy. Adding or removing many electrons is often more difficult and there are other considerations.


text:
Ability to produce hydrogen gas on reaction with Na/K
Out of a given set of compounds, I was asked to identify which would produce hydrogen gas on reaction with Na/K. One of the given compounds was:

I could not detect any acidic hydrogen in it, so concluded that it won't produce hydrogen. But the answer says the opposite.
So my questions are:

Why does the above compound produce hydrogen gas?
What can be the general set of rules to decide which hydrogen shall be acidic enough so that hydrogen gas is produced on reaction with an alkali metal?

This thing will produce a dianion, which according to Huckel feels quite well. As for the general set of rules, it constitutes much of organic chemistry.

What should be the ideal pKa value below which Na can produce hydrogen?

Actually in NaK it's rather potassium that is stronger reducing agent. Also it would be easier to say what doesn't react with such stuff, because most of organic compounds does.

This is not an acid-base reaction.

First of all, this is not an acid-base reaction. Alkali metals are not bases, rather strong reducing agents. For example, in this site, you may find a excellent explanation. Also remember, metallic $\ce{Li}$, $\ce{Na}$, or $\ce{K}$   does not abstract acidic hydrogen from organic molecules, not necessarily. For example,  metallic $\ce{Na}$ reacts with acetylene ($\mathrm {p}K_a$ of terminal alkyne-$\ce{H}$ is $25$) to give sodium acetylide, but it does not undergo aldol condensation with acetone with more acidic $\ce{H}$, $\mathrm {p}K_a$ of which is $19.3$ (https://owl.oit.umass.edu/departments/OrganicChemistry/appendix/pKaTable.html). The reaction of acetone and metallic $\ce{Na}$ rather undergoes dimerization (e.g., see here). Another example is the reaction of triphenylmethane (the $\mathrm {p}K_a$ of $\ce{H}$ on the central carbon is $33$) and metallic $\ce{K}$ in 1,2-dimethoxyethane, which has given solutions of triphenylmethyl potassium containing various cleavage and reduction products derived from triphenylmethane (Ref.1), but authors stated that no hydrogen was evolved in the course of the preparation.
However, the given compound has potential to undergo redox reaction with alkaline metal and release $\ce{H2}$ gas as a byproduct because the resultant dianion stabilized by resonance and show hint of aromaticity (Ref.2). Tentative mechanism is illustrated in following scheme:

References:

R. O. House, V. Kramar, “The Chemistry of Carbanions. I. The Reaction of Triphenylmethane with Potassium$^1$,” J. Org. Chem. 1962, 27(12), 4146-4149 (DOI: 10.1021/jo01059a007).
G. Boche, H. Etzrodt, M. Marsch, W. Thiel, “The Dianion 1,2,3,4‐Tetraphenylcyclobutadienediide,” Angew. Chem. Int. Ed. Engl. 1982, 21(2), 133-133 (https://doi.org/10.1002/anie.198201331).


text:
What's causing this dip in the FCS curve?
This is a fluorescence correlation spectroscopy measurement of light harvesting complexes. What could be causing the dip at 100 microseconds?

Edit:
The same thing was observed using fluorescence beads.
The only thing I have thought of is a problem reading in photon arrival times, or a problem with the laser.

It could be a saturation effect. Is it reproducible?

Can you provide more information about your experimental setup, and ideas about what might be causing this, and what you have changed in an attempt to troubleshoot this.

@TryHard Saturation of the fluorescence?

Are you the only person using the device? What is it's prior history and experience of other users? You likely have an instrumental artifact. Perhaps you can look at the raw (uncorrelated) data? The high laser power can cause problems such as saturation. Dips can occur if you have confined diffusion but this seems unlikely and you should be able to discount from the decay or a control such as very different samples.

http://www.fcsxpert.com/support/assays/troubleshooting.html

This is the first time FCS has been done on this system. It is normally used for the detection of immobilized molecules. The power used was not enough for saturation in immobilized experiments on the same sample.


text:
Aminoethyl or ethylamine?

I was trying to name the above compound which happens to be 3-(2-aminoethyl)-2-methylcyclohexan-1-ol.
On my first attempt, I had named it 3-(ethyl-2-amine)-2-methylcyclohexan-1-ol.
Which one is correct?

The IUPAC rules use "aminoethyl" for such a group. This is copied from the IUPAC "blue book" which can be found online here
Rule 811.3 - An -NH2 group, when not the principal group, is named by the prefix "amino-".
Your original name is very close to what would be used if the N were attached to the ring and the ethyl attached to the N. That would be "3-ethylamino-2-methyl. . ." according to this rule:
811.4 - Radicals RNH-, R2N-, and are named as substituted amino groups by changing "ine'' of the amine to "ino".

Is it also true for other groups with prefixes like, hydroxyethyl instead of ethylol?

Yes, the equivalent group with an HO would be hydroxyethyl-. If the oxygen were attached to the ring and had an alkyl group attached to it (an ether), it is called "alkyloxy-". For common groups like ethyl and methyl, the "yl" is dropped and it is just "methoxy-" or "ethoxy-".

Please note that your link points to obsolete IUPAC recommendations from 1979.

Yes, but I'm not aware of the newest version being available online, and these rules haven't changed. If you have a link to the newest rules, please post it.


text:
When a bond is formed between an electophile and a nucleophile, they both lose energy and become more stable right?

This is a section from clayden's organic chemistry textbook. I do not understand why it says energy gain and not energy loss

I agree, that doesn't make sense. If you looked at the enthalpy, I would say those bond-forming processes are exothermic, therefore an energy loss by the system (reactants->products).

In the context of the picture provided, it seems just a matter of language choice. When energy is lost, it is gained somewhere else because of conservation of energy. So a change in energy corresponding to a gain is at the same time a loss elsewhere. The author of that diagram could have used interchangeably energy loss without loss of meaning.


text:
How to write correct NMR integrals in a formula?
for my masterthesis I want to include the Equation how I calculate my conversion in a polymer reaction.
It is pretty easy, but I still want to write it down.
Basicly the equation is: (Integral A usally 4-5 and Integral B always 1 beacuse its one Proton, Integral B and C are equal)
$$ Conv=\frac{Integral_A-Integral_C}{Integral_A+Integral_B}=\frac{Integral_A-1}{Integral_A+1}
$$
But the formular with "Integral" in it looks not very nice while I thin the $\int{}{}$ sign seems to be a bit over the top.
If someone wants to know how I end up with the equation, here is the explanation:
It comes from the point, that my Monomer has 2 protons on each side of the double bond. When the polymer is formed they shift forming one broad signal. In the polymersolution I can identify one proton of the monomer which gets the Integral 1 while the other big peak contains the polymer as well as one proton from the monomer. Because both monomer protons have the same integral, both get a 1. My "clean" polymersignal is therefore the broad Integral -1. This has to be devided by the overall Integral which is the broad signal (which already includes one monomer proton)+ the other proton.

I’d just use $s_\ce{A}$ for signal.

And what for Conversion? Or just leave Conv?

I don't think there's a standard symbol. For what it's worth, in the context of kinetic resolutions, conversion is usually denoted with $c$ (e.g. https://doi.org/10.1002/1615-4169(20010129)343:1<5::AID-ADSC5>3.0.CO;2-I), which I suppose is intuitive. The bottom line is that you can choose whatever you want, as long as you define them in the text.

Assuming $Integral_C$ is the isolated peak, the denominator should be  $Integral_A+Integral_C$. I also have to assume you set the value of $Integral_C$ to $1.00$.

Integral B is the isolated peak but B and C are both 1 proton so it does not make a big difference

To me this opens a can of worms. Using $s_A$ is fine, but how is background handled? How are overlapping peaks resolved? The signal/noise ratio, and hence relative error, are critical in instrumental analysis. // Remember student in Chem 101 course who got 121% iron for a sample. me - "Didn't having more than 100% Fe  seem strange?" student - "That's what the calculator said."

Thanks but I think you are a bit over the top. This is not chemistry 101 this is a common method around polymer chemistry for a fast and cheap calculation of the conversion.
Maybe my explanation was not good, all signals are in one spectrum. Therefore the S/N Ratio should be the same unless it is not changing significantly with higher ppm....

You assume the total number of protons summed over reactants and products is constant during the course of the reaction, which is perfectly reasonable, and that all monomer protons are incorporated into the polymer (perfectly reasonably), but also that the total signal is a constant. Note that because polymer signal will relax more rapidly by $T_2$ relaxation, usually, signal ratios are not expected to be strictly proportional to concentration, but we'll ignore that point, assuming this is a solution experiment and the polymer is highly dynamic.
Therefore, you could write
$ f_{conv}= (s_A-s_C)/(s_A+s_C) = (s_{A,norm}-1)/(s_{A,norm}+1)$
where $s_C$ is the isolated signal from one type of monomer proton and  $s_B=s_C$, the other monomer signal, overlaps with the polymer signal to generate $s_A$. Then you define units for the area $s_B=s_C=1$ according to the expression on the right. This is explicit in the way I have written the equation. You should explain the meaning of the symbols in the text, of course.

Thanks for the answer how to write it. I did not really wanted to go into discussion about what I'm doing there. Maybe my expression was wrong. All three signals are in one spectrum of the crude polymer. So I'm not comparing 2 spectra with different concentration etc.
This method is very common in polymer history to determine the conversion in a crude solution as it is fast and cheap and when it is accurate around +-2-5% which is in range of the NMR Error its totaly fine.

@Inselino No problem, I understood how you intend to do the analysis from one spectrum. You can also write $f=(I_A−I_C)/(I_A+I_C)=(I_A−1)/(I_A+1)$, but then should mention that the areas are normalized to the area of one monomer peak.

OP I think that you got the way to handle this.  As far dynamics is concerned and just for non specialist readers: indeed the dynamics of the polymer can be so different that a signal from a polymer might be difficult to be even attained using the same conditions that readily give the monomer spectrum. When this is not the case than this method can be used, limitations taken into account as usual.