text:
Work Done by the gas in Isothermal Condition against constant external pressure
Question :
*

Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25L
against a constant external pressure of 2 bar . The work done by the
gas is( given that 1L.bar=100J)

Answer (from internet, that I always got) :

W=−Pext.ΔV=−2×0.15bar=−2×0.15×100J=−30J

I feel that in question they had asked the Work done by the gas, thus we do not have to use negative sign, as the volume is increase so, gas has done positive work. But, I am not able to find not a single solution that supported me, as it is the question and answer of NEET-2019 (A national level exam in India for Medical Students). So, please help to get the right answer, and suggest me if any correction is required within me.

Be aware you deal with 2 opposite sign conventions for work.
IUPAC and physical chemistry supports the convention Q and W have positive sign when done on the system, not by the system.  It means
$$\Delta U = Q + W,$$
with the positive work and heat increasing the system internal energy.
Historically, engineering and/or physics have been more inclined toward
$$\Delta U = Q - W,$$
i.e. positive work was the work done by the system, decreasing system internal energy.
Currently, at least physics in large extent shifts toward the former convention. There are many exceptions and there is a vaque boundary of convention domains. One has always to check or write, what convention was/is used.

If you use the IUPAC convention, then $W := - \int P ; dV$, but with the old convention  $ W := \int P ; dV$. If not, the 1st law will yield incorrect results.

@MetalStorm  I have not said otherwise. I would also add those integrals address only the "volume" or "PV" work, not the work not related to volume changes, e.g.work done to/by electrochemical cells.

Yes, of course. Just added because maybe the OP thinks that $ W = -P\Delta V $ or a similar equation still holds, regardless the convention.


text:
Can ionic and/or metallic bonding produce stable long chains? Like polymers, though not necessarily as useful
Question
Polymers are long chains ⛓ of covalent bonds.
Can similar structures exist for ionic and/or metallic bonding? They don't have to be as useful.

I know there can be polymers with ionic bonds in them, but I am talking about the entire structure (at least the base) being ionic and/or metallic.

Just in case, could they exist for other types of bonding too?

Would you consider a monocrystal of a ionic compound or of a metal as a ionic or metallic polymer?

@Poutnik Can it form a chain? Like a string.

They are 3D polymers. Like classical organic covalent linked polymers. But linked much more properly. Short linear chains are possible, but I would not call them polymers. They are hard to form and easy to break.

@Poutnik That would fit some definitions, but I was looking for something that forms a chain.

There would be the problem the forces along the assumed ionic/metallic chains would be the same as between the chains.

Google "ionomers"

In the case of ionic bonding, the answer would be that a stable structure is unlikely, although examples may exist among multimeric proteins. You may also check out the work of George Whitesides and others in the field of self-assembling soft matter.
However, to be clear, it is assumed that the arrangement in question is of alternating charges in a regular linear sequence (...+-+-+-+-+-+-...), that the attraction between nearest neighbors is isotropic, and that the arrangement is free of other forces (such as external fields). There could be a solvent but if it competes with the ionic interactions then stability is certainly impossible. Such an arrangement would be unstable in a vacuum as well except under exceptional circumstances. The reason is that electrostatic attraction operates over long distances, and charges will therefore experience attraction and repulsion from more than the nearest neighbors. In addition, since the bonding is assumed isotropic, there is little to interfere with bending  under the influence of these forces and in the presence of thermal agitation. If one monomer can bond to more than two (coordination number>2) then a sheet or a nonlinear 3D structure will ensue.
However, if one monomer can bond stably to only two others, then an entangled, reptating polymer-like structure would result.
With regard to metallic bonding, there is such a thing as a "molecular wire". These can form electrically conductive single macromolecules. However I can't at present comment on whether bonding in these would be regarded as metallic, at least in a way comparable to bonding in traditional metal solids.
You might ask at matter modelling SE whether such models have been studied in silico.

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sort: votes


text:
Is electrode reduction potential dependent on the nature of the solution(the reagents of the solution) it's dipped in?
I'm undergrad student majoring in materials science and engineering. I faced this question while studying for a course named "surface engineering of materials".
This slide is a part of my study material. It's discussing about the reduction potential of coating materials. But while listing them, there's no mention of the solution the materials are dipped in.
Isn't the electrode potential dependent on ln(Kp) here? and if the reagents are changed, won't
ln(Kp) and thus the electrode potential change?

These are standard reduction potential - you should already know what that means, and if you don't... https://en.wikipedia.org/wiki/Standard_electrode_potential

But while listing them, there's no mention of the solution the materials are dipped in.

Yes, you are correct. It depends on all the reactants and the solvent where these are. In my answer I expect that with solution you refer actually to the solvent. I will explain the dependencies of what you ask from:

The table you have shown displays the standard oxidation potential $E^\Theta$ of several half-reactions .
$E^\Theta$ will depend, easily said, on the standard chemical potential $\mu_i^\Theta$ of all the species participating.
For a solid species $\mu_i^\Theta$ will be a function of: (1) pressure (generally $\pu{1 bar}$), (2) specified temperature $T^\Theta$, and (3) the identity of the species itself. For two different solids $i$ and $j$, in general we have
$$ \mu_\pu{i}^\Theta(\pu{1 bar},T^\Theta) \neq
   \mu_\pu{j}^\Theta (\pu{1 bar},T^\Theta) $$
However, for all the solids in your list, these values are zero because they are elements in their most stable state, which happens to be the solid state.
For species in solution you need to specify a scale. Once you select it, $\mu_i^\Theta$ will be a function of: (1) pressure (generally $\pu{1 bar}$), (2) specified temperature $T^\Theta$, (3) the identity of the species itself, and (4) the identity of the solvent. For example if you choose the molality scale, then the standard state is a hypothetical ideal $1$ molal solution of this solute $ i $ in a certain solvent, at the pressure of $\pu{1 bar}$ and a defined temperature $T^\Theta$.

For the same solvent, lets say water, then two ions have different standard chemical potentials (so $E^\Theta$ will change)
$$ \mu_\ce{Na^+}^\Theta(\pu{1 bar},T^\Theta, \ce{water}) \neq
   \mu_\ce{Mg^2+}^\Theta (\pu{1 bar},T^\Theta, \ce{water}) $$
For different solvents, the same species will have different standard chemical potentials (so $E^\Theta$ will change)
$$ \mu_\ce{Na^+}^\Theta(\pu{1 bar},T^\Theta, \ce{solvent I}) \neq
   \mu_\ce{Na^+}^\Theta (\pu{1 bar},T^\Theta,  \ce{solvent II}) $$
This is why in the back of chemistry books, for charges species, they identify the identity of the solvent, e.g. putting between parentheses ($\ce{aq}$).

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text:
When starting a sentence with an IUPAC name that starts with a number, do you capitalize the first letter?
When you have an IUPAC name that starts with a number, i.e. "3,5-dinitrosalicylic acid", but it's at the start of the sentence would you capitalize the first letter in the name? For example:

3,5-Dinitrosalicylic acid can reduce sugars with aldehyde or ketone functionality.

or,

3,5-dinitrosalicylic acid can reduce sugars.

See: https://chemistry.meta.stackexchange.com/questions/4275/capitalization-of-titles-starting-with-chemical-name-with-noncapitalizable-prefi

According to the ACS style guide [1, p. 240]

Chemical names are not capitalized unless they are the first word of a sentence or
are part of a title or heading. Then, the first letter of the syllabic portion is capitalized, not the locant, stereoisomer descriptor, or positional prefix.

Therefore, the following capitalization is correct:

3,5-Dinitrosalicylic acid can reduce sugars with aldehyde or ketone functionality.

Reference

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. DOI: 10.1021/ed083p1603.1, ISBN 978-0-8412-3999-9.

I don't have access, but the new version should be here: https://doi.org/10.1021/acsguide

I used the previous reference for the third "offline" edition since the quoted part can be found there, and I don't have access to the latest ACS SG either. Added woke BS aside (i.e. Part 6: Inclusivity), I would suspect the content would be pretty much the same. If anyone who has access to the latest SG, please feel free to make a correction.


text:
Are there TLC stains for organo fluor compounds or anorg. fluoride that are useful for selecting the product containing fractions post purification
Are there TLC stains that work with fluoride or with organofluor compounds? I have a fluorinated phosphonic acid that does not stain with conventional means. I did research in many of the known and available staining resources, but I did not find something reliable yet. KMnO4, Phosphomolybdic Acid (PMA) Stain and Cerium Molybdate Stain, p-anisaldehyde as well as Vanilla stain did appear not to work. Do you have experiences here? For me I was not able to see an unambiguously contrast to the background. Maybe there is also a protocol that decomposes an organofluor compound beforehand and makes fluoride visible on TLC? I performed  many tests but nothing was really convincing until now. I know that a protocol for qualitative prove of F- uses chromic acid and the discoloration of alizarin dye - but this system is probably not applicable for TLCs.

What about TLC plates/sheets with UV254nm fluorescent indicator?

The product is UV inactive. When it would be UV active I would not try all these stains.

You do not provide relevant product properties to prevent not applicable suggestions in your questions. I have warned you what that leads to.

It seems that unfortunately it is not possible to stain fluoride containing compounds (C-F) bonds present in the compound.


text:
Detection of Nitrogen and Sulfur in Lassaigne's Test
In Lassaigne's test for the detection of nitrogen and/or sulfur in organic compounds, we prepare Lassaigne's extract. In this extract, NaCN is formed if nitrogen is present, and $\ce{Na2S}$ is formed when sulfur is present. NaCN on treatment with $\ce{FeSO4}$ and $\ce{FeCl3}$ gives a Prussian blue precipitate of $\ce{Fe4[Fe(CN)6]3}$, and $\ce{Na2S}$ on treatment with $\ce{Na2[Fe(CN)5NO]}$ (sodium nitroprusside) will give a violet color due to the formation of $\ce{Na4[Fe(CN)5NOS]}$. However, when both are present, NaCNS is formed. NaCNS can be detected by an $\ce{FeCl3}$ test, which gives a blood red color of  $\ce{Fe(CNS)3}$ (ferric sulfocyanide). My question is, when both nitrogen AND sulfur are present, will the individual tests for nitrogen and sulfur (Prussian blue and sodium nitroprusside test) be shown? Doesn't this depend on whether the sodium extract contains NaCN, NaCNS, or $\ce{Na2S}$? Is it possible to have a mixture of all 3 (NaCN, NaCNS, and $\ce{Na2S}$), in which case all the above mentioned tests will be answered?

Why do you want individual tests? It will give either NaCN or Na2S if either is present and NaSCN if both are present and you can distinguish NaCN, Na2S and NaSCN. Na2S is yellow solid and while NaCN and NaSCN are white, they can be distinguished by their solubility (NaCN is 63.7 g/100ml and NaSCN is 139 g/100ml at rt) and once you identified the sodium salt, you proceed with corresponding reaction. Tbh, Lassaigne test has become obsolete and nobody uses obviously due to its hazard factors.

Please note that nobody uses Lassaigne's test of the 18th century to test $\ce{S}$, $\ce{N}$ and halogens in organic compounds. It is unsafe and really obsolete method of qualitative analysis. I know that it is still taught in high schools in certain parts of the world. Just looking at the chemistry's point of view, your hypothesis is partially correct. One may see the formation of thiocyanate. However, all those products which you mentioned are dependent on the relative ratio of sodium metal and the organic compounds. If an excess of sodium metal is used, thiocyanate ion is destroyed during fusion and you get the regular products. In short it is possible to simultaneously detect $\ce{N}$ and $\ce{S}$ with the proper amount of compound to sodium ratio during fusion.
Reference:
Tucker, S. Horwood. "A lost centenary: Lassaigne's test for nitrogen. The identification of nitrogen, sulfur, and halogens in organic compounds." Journal of Chemical Education 1945, 22(5), 212-215 (DOI: https://doi.org/10.1021/ed022p212).

So it will show a positive test for all 3 tests (ferric sulfocyanide, sodium nitroprusside, and ferric ferrocyanide) if the correct ratio of sodium metal to organic compound is used?

I do not know the exact answer because a general answer cannot be stated, what if a molecule has two S for each N or any other ratio. The above reference suggests to vary the amount of sodium and see the effect for each compound.


text:
Why is the isenthalpic condition required in the Joule–Thomson effect?
I have been struggling to understand how the molecules behave during the Joule–Thomson effect. I would love to get some help on this concept. Here is what I got so far.
Under adiabatic and isenthalpic conditions, when a gas expand from high pressure to low pressure, depending on the initial pressure, initial temperature, and the molecular properties, the gas can either experience cooling or heating.
Specifically, if the gas was initially in the state where attractive intermolecular forces were dominant (typically at regions of moderate temperature and low pressure), as it expand, the kinetic energy of molecules are used to overcome the attraction reach greater separation. Also, the internal energy is overall decreased as the gas need to do work to expand. Thus the temperature of the gas decreases.
If repulsive forces were dominant (at regions of very high temperature or high pressure), as the gas expand, the repulsion kind of pushes(?) the molecules to reach greater distance. In other words, the potential energy is converted to kinetic energy here. And here although internal energy may be decreased to provide work for expansion, the molecule would have gained more kinetic energy from the repulsive forces such that the temperature of the molecules increases.
My question is, how does the system being isenthalpic play a role in this effect? I get that we need the adiabatic condition so that only the internal energy is being used to provide the work required for expansion. But what about enthalpy? How is isenthalpic + adiabatic expansion different from just adiabatic expansion?
Additionally, I would appreciate if someone could also explain how $pV$ from the enthalpy formula $H=U+pV$, changes during this expansion.

Are you familiar with the derivation of the open system (control volume) version of the 1st law of thermodynamics?  In  your assessment, you are omitting a viscous heating effect due to viscous friction (in the valve or porous plug) which offsets most of the effect of expansion cooling.

No I have never heard about that. I'm just following the textbook from Atkins which doesn't seem to explain much detail about this.

Get yourself a decent Thermo book like Smith and Van Ness or Moran et al and read up on their sections on the open system version of the 1st law of thermodynamics.

This is a throttling process due to a porous plug/constriction between the inlet and outlet. Constant flow is maintained by a pump and the pressure on either side of the plug is maintained at fixed values. The heat $q=0$ and the work done is $ W =\int_0^{V_f}pdV +\int_{V_i}^0pdV= p_fV_f-p_iV_i $ between initial and final states, therefore $0 = U_f-U_i+p_fV_f-p_iV_i$ which means that $H_i=H_f$ which  does not mean that enthalpy is constant.
In response to comment I have added some details
The coefficient used to quantify cooling or heating is
$$\displaystyle \mu=\left(\frac{\partial T}{\partial P} \right)_H$$
and this is obtained from the slope between points on a pressure vs temperature plot. These points are obtained by a throttling experiment. The initial pressure and temperature is decided upon and also the final pressure, which is always less than the initial one, and the final temperature is then measured. This produces one point on a $p$ vs $T$ plot. Next only the final pressure is changed and a second temperature point obtained and so on. A set of points that approximates to a curve is produced; the isenthalp. The slope of this  between two points is the JT coefficient $\mu$ and if this is positive cooling occurs and vice versa. The experiment is repeated many times with different initial $T_i$ and $p_i$ values and over the range of final pressures $p_f$. A plot is produced with many isenthalps  one above the other and by jointing the points where the slope is zero $\mu=0$ a curve is produced, the inversion curve, separating heating and cooling, see the figure.

Joule Thompson isenthalps and the inversion curve. The cooling area is to the left of the dashed red line. The blue lines are the isenthalps.The figure is from Wikipedia commons; (https://commons.wikimedia.org/wiki/File:Isenthalpic_contours_for_temperature_versus_pressure_showing_inversion_curve.svg)
The  JT coefficient can be related to heat capacity and volume for a non-ideal gas. It is always zero for an ideal gas since the the effect is dependent on intermolecular forces.
Using a Maxwell equation
$$\displaystyle dH=TdS+Vdp$$
We shall need to eliminate $TdS$ and can do this by using
$$\displaystyle TdS=C_p dT-T\left( \frac{\partial V}{\partial T}\right)_pdp$$
( this eqn. is more familiar for an ideal gas as $\displaystyle dS=\frac{C_p}{T} dT-R\frac{dp}{p}$ )
Rearranging gives
$$\displaystyle dT=\frac{1}{C_p}\left[ T\left( \frac{\partial V}{\partial T}\right)_p-V\right]dp+\frac{dH}{C_p}$$
and differentiating wrt $p$ and constant $H$ gives
$$\displaystyle \mu =\frac{1}{C_p}\left[ T\left( \frac{\partial V}{\partial T}\right)_p-V\right]$$
which for an ideal gas ($pV=RT$) is zero.

I'm guessing you meant that the enthalpy is constant. I did see this derivation on how enthalpy is proved to be constant throughout the throttling process. But what I'm still not getting is how constant enthalpy significantly contributes to the Joule Thomson effect. What would happen if the enthalpy isn't constant?

@asdfasdfasdf Not restricted by dH=0, gas could do pV work against external pressure. Even H2 or He that warm up at room T because of J-T effect, would cool down because of that work.

No, I meant that the enthalpy initially and finally is the same not that it does not change because we can only know the initial and final equilibrium points not any intermediates ones.

The substitution you have used
$$V = T\left( \frac{\partial V}{\partial T}\right)_p$$
implies
$$\left( \frac{\partial V}{V}\right)_p =\left( \frac{\partial T}{ T}\right)_p $$
which is valid only for ideal gases, while you try to derive expressions for non-ideal gases. It should be noted and justified.

@Poutnik The equation $\displaystyle TdS=C_p dT-T\left( \frac{\partial V}{\partial T}\right)_pdp$ is quite general and true of any pure substance and comes from assuming entropy is a function of $T$ and $p$ and using the Maxwell relationship $\displaystyle\left( \frac{\partial S}{\partial p}\right)_T=-\left( \frac{\partial V}{\partial T}\right)_p$

@porphyrin  I see. The trick must be dH <> Cp . dT // I admit I had to refresh my TD knowledge. It is 34 years since my graduation and then using these TD equations just occasionally, rather as a hobby.

been similar number of years since I taught this  :)


text:
What is a suitable process to eliminate an excess of pyridine from an aqueous solution?
I have a reaction in which pyridin is used as a reagent in excess. The resulting product is water soluble and an anion. Pyridinium there will act as a counter ion. Normally, when the product is soluble in organic solvents pyridine is removed by washing with HCl 1 M or with CuSO4 10%. What is the best way to achieve this when the product is water soluble in comparison? In the moment I would apply these strategies:

Applying vacuum and high vacuum until mass is constant (pyridinium salts can not be removed).

Setting the pH of the water phase slightly basic to pH 8-9 and (pyridinium pka = 5.2) and so multiple extractions with DCM/water.

Applying anionic exchange chromatography (Pyridin is washed through the column. Pyridinium is also replaced as a cation by a cation that is present in the mobile phase).

I am wondering if this is already the best possible approach or if you know of a means to remove pyridine more elegantly?

Do you have access to a freeze-drier?

What is the composition of the aqueous solution, apart from pyridinium ? I it acid ? basic ? Is there some ethanol ?`some ether ? some inorganic compound ?

@ Waylander: Access to liquid nitrogen, high vacuum (10^-3 mbar) and a lyophilization system (1 mbar).

@Maurice: The aqueous phase will contain the product, pyridinium as well as an excess of pyridine. No ethanol, no ether. Basic due to pyridin excess.

@raptorlane  I think freeze-drying under high vac is the easiest way to go if the pyridinium salt is an acceptable form.

In my case when performing the follow up reaction what seems to work is to extract most of the pyridine under neutral pH to the water phase and rotivap the organic phase. Subsequently applying high vac. For a water soluble compounds removing the excess pyridine takes longer because most goes to the aquos phase. The contamination is not of a problem on my case, but for yield determination it is obviously necessary to have a dry product.


text:
Nomenclature of Nonacyclo Compound
I can't understand which are the nine cycles in this compound. It seems to me that there are 11 faces and thus 11 cycles. And what is the numbering of the carbon?

Source: Advanced Organic Chemistry by Jerry March.

The number of rings is equal to the number of bond cuts necessary to transform the polycyclic system into an acyclic skeleton – not the number of faces.

The concise Comment of @Loong establishes the presence of nine rings in structure 1, which has the molecular formula $\ce{C20H24}$. The most saturated alkane with 20 carbons is $\ce{C20H42}$. If one subtracts the former formula from the latter one, the result is 18. What does this number mean? It is the number of hydrogen atoms (two) that must be removed from the acyclic alkane to form a new C-C bond (one). Thus, 18/2=9 rings (and/or multiple bonds).
In the first example, five arbitrarily selected bonds (in red) in 1 are "cut" and two hydrogen atoms added resulting in tetracycle 2. Again, four arbitrary bonds in 2 are cut to afford the $\ce{C20H42}$ acyclic alkane 3. Similarly, the sequence 1' $\Rightarrow$
4 $\Rightarrow$ 5 is equally applicable. Bonds must be selected to maintain the $\ce{C20}$ structure.
Other resources on this topic and links are available here on ChemSE.

Just as a doubt, I want to ask if there is a compound with formula C20H24 with 2 double bonds, then should it be a heptacyclo compound? I figured it out by using Double Bond Equivalent Concept

Yes, you are correct it would be a heptacyclo compound as long as there are no triple bonds which would have 2° of unsaturation.


text:
Effectivness factor for a cylinder
I would like to demonstrate the effectiveness factor of a porous cylinder, for diffusion-reaction (only radial diffusion, 1st order reaction, isothermal, diffusion coeff. is constant, steady-state),  such as
$$
\eta=\frac{2 \cdot I_1(\phi)}{\phi \cdot I_0(\phi)}
$$
using the info from this page integration of bessel function.
Also, I consider $$ dV = dr \cdot r \cdot d\theta dz $$
with:

V for volume
r for the radius, that ranges from $0$ to R,
$\theta$ for the azimuth, $0$ to $\pi$
z for the length, $0$ to h

So I have
$$
\frac{c(Y)}{c_{area}}=\frac{I_0(\phi Y)}{I_0(\phi)}
$$
with Y the dimensionless radius ($Y=r/l_c, l_c=R_p/2$, c and $c_{area}$ the concentration of the reactant in the media and at the surface of the media respectively, $\phi$ the Thiele modulus $\phi=R_p/2\cdot\sqrt{k/D}$, and
$$
r_p=\frac{1}{V} \iiint_{V}{}R(c)dV
$$
$$
r_p = \frac{1}{V}\int_{0}^{1}-k \cdot c(r) \cdot r \cdot dr \int_{0}^{\pi}d\theta \int_{0}^{h}dz
$$
such as
$$
\eta=\frac{r_p}{r_{area}}
$$
I would like some insight (or even clues) on how to proceed the integration of $\eta$. Because I might be doing something wrong and I can't seem to find the issue.

By definition, the effectiveness factor is the ratio between: (1) the rate of reaction integrated along the catalytic particle, and (2) the rate of reaction integrated along the catalytic particle as if all of it is covered at the surface concentration $C_A = C_{As} $. So, for an isothermal operation so that the rate constant may be taken out of the integral, we have
\begin{align}
 \eta &:= \dfrac{\displaystyle\int_V (-r_A')(C_A) \; dV}
               {\displaystyle\int_V (-r_A')(C_{As})\; dV}
       = \dfrac{\displaystyle\int_0^{2\pi}\int_0^{L}\int_0^{R} kC_A(r)
 \; rdrd\theta dz}
               {\displaystyle\int_0^{2\pi}\int_0^{L}\int_0^{R} kC_{As}
 \; rdrd\theta dz} \\
       &= \dfrac{2\pi Lk\displaystyle\int_0^{R} C_A(r) \; rdr}
               {2\pi LkC_{As}\displaystyle\int_0^{R} rdr}
        = \dfrac{\displaystyle\int_0^{R} C_A(r) \; rdr}
               {C_{As}\dfrac{R^2}{2}}
        = \dfrac{2}{R^2}\int_0^{R} \dfrac{C_A(r)}{C_{As}} \; rdr \tag{1}
\end{align}
Making the substitution $\lambda = r/R$ and $dr = Rd\lambda$, leads to the appearence of the dimensionless concentration $\psi(\lambda) = C_A(\lambda)/C_{As}$ ($\psi$ is for you $Y$). Thus, Eq. (1) leads to
\begin{align}
  \eta &= \dfrac{2}{R^2}\int_0^1 \psi(\lambda) R\lambda \; d(R\lambda) \\
  \eta &= \dfrac{2R^2}{R^2}\int_0^1 \lambda \psi(\lambda) \; d\lambda \\
  \eta &= 2\int_0^1 \lambda \psi(\lambda) \; d\lambda \tag{2} \\
\end{align}
As you have shown there, the dimensionless concentration profile that solves the difussion reaction process of a catalytic particle in cylindrical symmetry is
$$ \psi(\lambda) = \dfrac{I_0(\phi\lambda)}{I_0(\phi)} \tag{3} $$
Combining Eqs. (2) and (3)
\begin{align}
  \eta &= 2\int_0^1 \lambda \left(\dfrac{I_0(\phi \lambda)}{I_0(\phi)}\right) \; d\lambda \\
  \eta &= \dfrac{2}{I_0(\phi)}\int_0^1 \lambda I_0(\phi \lambda)\; d\lambda \\
  \eta &= \dfrac{2}{\phi^2I_0(\phi)}\int_0^1 (\phi\lambda) I_0(\phi \lambda)\; d(\phi\lambda) \tag{4} \\
\end{align}
Our mathematical friends from Stack Exchange informs us about the following relationship regarding Bessel functions, which we abuse
$$ \int_a^b u I_0(u) \; du = [uI_1(u)]\vert_a^b \tag{5} $$
And combining Eqs. (4) and (5) we have
\begin{align}
  \eta &= \dfrac{2}{\phi^2I_0(\phi)} [\phi\lambda I_1(\phi\lambda)]\vert_{\lambda = 0}^{\lambda = 1} \\
  \eta &= \dfrac{2}{\phi^2I_0(\phi)} [\phi I_1(\phi) - 0] \\
  \eta &= \dfrac{2}{\phi^2I_0(\phi)} \phi I_1(\phi) \rightarrow
  \boxed{\eta = \dfrac{2I_1(\phi)}{\phi I_0(\phi)}}
\end{align}


text:
What is the relationship between solutions and changes to states of matter?
For example, when liquid water evaporates, my instinct is to say that of course it's become gas, but I'm a bit unsure because, if I understand correctly, evaporation occurs because air dissolves the water, and it's not clear to me whether that counts as a phase change. Intuitively it seems like it should should water dissolves in the air or salt dissolved in water, etc. have different properties before and after being dissolved.  And yet, the way I've always heard it explained, phase changes are specifically due to changes in temperature and/or pressure, not due to chemical interactions with another another substance.
On the other hand, from what I remember from chem 101 and 102, when we considered chemical reactions occuring between solids dissolved in a liquid (usually acids and bases dissolved in water), we usually just labeled them as aqueous, meaning "in solution", whereas for non-aqueous substances, we'd label them with the relevant state of matter, solid, liquid, or gas.  Does that mean dissolved substances constitute their own state of matter?  Or that it's simply not meaningful to talk about the state of the solute independent of the solvent?
I also saw this thread, Is it appropriate to say "solid-in-gas solution" and "liquid-in-gas solution"?, where someone says, "Whenever there is only one phase, but there are two or more chemical species, then you have a solution", which heavily implies there's a fundamental relationship between phase changes and solutions, but it also doesn't seem quite right

It is usually helpful to look for a proper agreed upon definition such as in the IUPAC gold book https://goldbook.iupac.org/terms/view/S05746#:~:text=A%20liquid%20or%20solid%20phase,mole which states 'A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes......'

Water vapour in air is a solution, but not in the same sense as solutions in water.
Water vapour in air is solution in sense of homogenous mixture, where molecules move freely and independently.
Salts in water dissociate (are dissociated by water) to ions. Having a net charge, they form a nonhomogeneous electrostatic gradient.Ions are hydrated by water molecules that have an electric dipole and therefore attracted to the center of such a gradient. So here we see strong interaction and dependent motion.

Capacity for water vapour at given temperature is not a shared property of water and air, like solubility of table salt in water. It is a property of water alone. Water need not air at all to evaporate and to stay as vapour.
Presence of air has very tiny effect on the maximal absolute humidity in $\pu{g / m3}$, due secondary phenomena, that are significant only at conditions near critical conditions.
Like nitrous oxide versus it's 1:1 mixture with oxygen.
The former is at room temperature and pressure $\pu{100 atm}$ liquid, while it is gaseous at the same partial pressure $\pu{100 atm}$ in the latter.


text:
Chemical potential and condition for equilibrium in case of binary liquid reaction mixture
Suppose there is a liquid mixture with components A and B in equilibrium along with their vapours. Then with the Gibbs-Duhem equation we know that $$μ_A(\mathrm{liquid}) = μ_A(\mathrm{vapor})$$ and
$$μ_B(\mathrm{liquid}) = μ_B(\mathrm{vapor})$$
But then imagine a case where A(liquid) is getting converted to B(liquid), so for that case $$d G = - μ_A(\mathrm{liquid})\cdot dn+μ_B(\mathrm{liquid})\cdot dn$$ (for this reaction), so if $μ_A(\mathrm{liquid})$ is not equal to $μ_B(\mathrm{liquid})$, then this reaction will take place in one of the directions.
So doesn't it mean that $μ_A(\mathrm{liquid})$  must also be equal to $μ_B(\mathrm{liquid})$?

Reference for chem equation/expression formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

No, it doesn't. Are the liquids miscible?

The name is Duhem. Please use mathjax to edit math expressions and mhchem for chemical expressions. See the first comment and help section for more information.

Yes it does, if stoichiometric coefficients are the same.
For eventual reaction of 2 volatile miscible liquids $\ce{A}$ and $\ce{B}$
$$\ce{A <=> B}$$
all the respective chemical potentials must be equal, otherwise the system would not be at the Gibbs function minimum:
$$\mu(\ce{A(g)})=\mu(\ce{A(sol)})=\mu(\ce{B(sol)})=\mu(\ce{B(g)})$$

Then with gibbs durhem eqn we know that μA(liquid) = μA(vapor) μB(liquid) = μB(vapor).

That is not Gibbs/Duhem equation. Gibbs/Duhem equation states a restriction between activity coefficients at constant $P$ and $T$ must satisfy, this is, $\sum_j x_j d\ln(\gamma_j) = 0$. What you have written is the condition of chemical equilibrium of species between diferent phases.

But then imagine a case where A(liquid) is getting converted to B(liquid) so for that case Delta G = - μA(liquid)*dn+μB(liquid)*dn. (for this reaction),

We need to address how the chemical potentials are related between each other in this case. This is a bit more difficult than usual, because we have a reaction but two phases, i.e., a heterogeneous system. The main idea is that we can separate (1) the chemical reaction, and (2) the conventional chemical equilibria. This is not so straightforward to prove, but I will show what happens in this case.
First, lets apply the Gibbs phase rule for reacting systems, so we know how many relationships we need
\begin{align}
  F &= 2 - \pu{Num. of phases} + \pu{Num. of components} - \pu{Num. of chemical reactions} \\
  F &= 2 - 2 + 2 - 1 \rightarrow \boxed{F=1}
\end{align}
because we have two components ($A$ and $B$), two phases (liquid and vapor), and one chemical reaction.
The chemical reaction that you proposed, that we can generalize
$$ \ce{aA (l)} \rightarrow \ce{bB (l)} $$
gives us one condition that must be satisfied in the liquid phase ($a$ and $b$ are positive)
$$ \sum_j \nu_j \mu_j^L = 0 \rightarrow
    \boxed{-a\mu_A^L(P,T,x_A) + b\mu_B^L(P,T,x_B) = 0} \tag{1} $$
The other two, are simply the conditions of chemical equilibrium for species in a closed system
$$ \boxed{\mu_A^V(P,T,y_A) = \mu_A^L(P,T,x_A)} \hspace{1 cm}
   \boxed{\mu_B^V(P,T,y_B) = \mu_B^L(P,T,x_B)} \tag{2,3} $$
Eqs. (1-3) are the relationships we need to solve the system. The number of intensive variables are four:

The pressure $P$ [mechanical equilibrium implies $(P^V=P^L=P)$].
The temperature $T$ [thermal equilibrium implies $(T^V=T^L=T)$].
Two molar fractions, since the other one can be calculated with $y_B = 1 - y_A$ and $x_B = 1 - x_A$.

And the number of equations are three, so everything is fine.


text:
If Standard Reduction Potential and Standard Oxidation potential are equal in magnitude and opposite in sign?
If Standard Reduction Potential(SRP) and Standard Oxidation potential(SOP) are equal in magnitude and opposite in sign ,
Then as per this
As we know SRP of H2O is -0.83 V as per this SOP of H2O should be + 0.83 V but this is not the case the value is SOP of H2O is 1.23 V.
Why ? And how ?

Both SOP and SRO are related to a thermodynamic magnitude $\Delta G^0$. For the potentials to have equal magnitude but different sign, both processes must be the inverse of each other (i.e., one reaction and its inverse). In this case, you refer to two different electrochemical reactions that have a common reactant (H$_2$O) but are not the inverse of each other.

There is no such thing as the "standard reduction potential of $\ce{H2O}$". Reduction (or oxidation) potentials are not defined by a single species alone, they are defined by a half-reaction.

You're not talking about the same reaction, that's why!

As $$\Delta_\text{r} G = -nF E^{\circ}$$
and as
$$\Delta_\mathrm{r,forward} G = - \Delta_\mathrm{r,backward} G,$$
it should not be surprising that
$$E^{\circ}_\text{ox} = - E^{\circ}_\text{red}$$
Be aware that

The formal redox potential of a half reaction flips the sign with the reaction direction.
The above potentials are conventionally tabelizec for oxidation direction.
The physical value of measured or predicted electrode potential has just one sign, measured wrt SHE.

There is another way of taking care of the potentials in a galvanic cell, without changing the sign of the anodic half-cell. It is coming from the fact that in physics, all tensions are measured in volts. And they all can be considered as differences of potentials. I repeat : a DIFFERENCE of potential.
So when coupling two half-equations like in the Daniel cell, the two half-equations are : $$\ce{Zn -> Zn^{2+} + 2 e^-}$$ $$\ce{Cu -> Cu^{2+} + 2 e^-}$$The redox potentials of these half-equations are $\pu{E°_{Zn} = - 0.76}$ V for the $\ce{Zn}$ electrode, and $\pu{E°_{Cu} = +0.34}$ V for the $\ce{Cu}$ electrode. It is not necessary to change the sign of the $\ce{Zn}$ electrode (which gets oxidized) when considering the Daniel cell produced by the two half-equations.
The tension $\epsilon_{cell} $ of the cell formed by the two half-equations can be obtained by the DIFFERENCE of the two redox potentials of $\ce{Zn}$ and $\ce{Cu}$ : $$\epsilon_{cell} = \pu{E_o(red)- E_o(ox) = E°_{Cu} - E°_{Zn} = 0.34 \mathrm{V} - ( -0.76 \mathrm{V}) = 1.10 \mathrm{V}} $$ As the result must be positive, it is necessary to subtract the smallest potential (or the most negative - here $\pu{E°_{Zn}}$ ) from the most positive value ($\pu{E°_{Cu}}$), and not the contrary.
With this approach, it is not necessary to change the sign of redox potentials when a half-equation is reversed. Redox potentials are maintained always at the same numerical values, whatever their behavior (oxidation or reduction).
$$\pu{\epsilon_{cell} = E_o(red)- E_o(ox)}$$


text:
Finding chromium
If I were to go back in time to pre-industrial Earth, and I wanted to make stainless steel, would there be a way to find chromium and/or chromite? It doesn't matter what chemical form the chromium takes or what its purity is, as long as there is sufficient chromium.

See https://mineralseducationcoalition.org/minerals-database/chromium/

What are looking for ? A chromium mineral that does contain chromium without the scientist knowing it ?

Anything like post-apocalypse scenario when skills and knowledge is lost and remnants of people try to restart civilizationalmost from the scratch? Or just some reconstruction of medieval/early industrial procedures and findings?

@Poutnik Yes, a post-apocalypse scenario fits what I'm asking pretty well.

@Maurice In my hypothetical, there would be very few, if any, people that know what chromium is. In that scenario, how would you find it? I know my hypothetical is a little weird for this forum. I am writing a book, and doing research for it. Again, apologies if it doesn't fit the forum.

If they do not know what chromium is, why to search for it?

@Poutnik Because the person searching for it knows what it is, and wants to use it to make stainless steel.

I am sorry. I do not see what you are looking for. Are you looking for a man who knows what chromium is without knowing chromium ?

I would say making useable steel would be more impressive than finding chrome ore.

Are you looking for chemical tests for chromium? What kind of equipment or tests would be available or possible?

If the technology has been lost, meteoritic iron often has enough nickel and other elements to make it highly rust resistant: https://en.wikipedia.org/wiki/Meteoric_iron, https://www.atlasobscura.com/articles/before-iron-age-most-iron-came-from-space-meteorite-egyptian-bronze

This question might be more appropriate for earth sciences SE, depending on what you are focusing on....

Check out a show called "Dr. Stone". It will be perfect for you.

In some river systems chromite is a relatively common mineral. One could extract it in a similar manner to gold: panning, sluicing, etc. In certain rivers and creeks draining the right geology one could feasibly get anywhere between 10-100+ grams a day panning. The hard part would be reducing it and using it to make steel.
Native chromium samples have been found in strong natural reducing environments, notably kimberlite pipes, but they (both kimberlite pipes and native chromium) are very rare, while chromite is much more widespread.


text:
Is it possible to prevent elimination during Diels Alder reactions using Danishevsky's diene (or variants)?
Are there examples of keeping the methoxy group and then using it to perhaps do radical chemistry without eliminating to the enone?


text:
What material is used in coating these aluminum containers for the food industry?
What is the material used to coat (some might use the term 'lacquer') disposable aluminum containers in the food industry? And can it be removed in a safe way (for the environment) with 'home methods'?
I am not talking of the thin oxide layer that covers aluminum foil and most basic (shiny) containers; this layer can be tunneled through and a resistance measurement with an ordinary multimeter will show high conductivity even without scratching it.
I am in particular interested in thin, easily crushable containers like dessert cups (for example mildly acidic fruit mousse).
.
Source: Aliberico (see below)
The coating appears different from that used in aluminum cans, but maybe it is just a different thickness.
There is no conductivity no matter how much I scratch the surface, and only when I punch through the material with the probes I get aluminum conductivity.
The label just say Alu 41 and my google-fu has been found lacking when it comes to the coating.
I googlefound the websites of two manufacturers that produce the containers like the ones I am interested in, like Aliberico, but I cannot find the name of the coating material, just a tradename (Alucoat) and some vague marketing descriptions.
This is the 'datasheet' I have found for Alucoat: https://www.paroc.com/-/media/uploaded-product-docs/2023/01/16/15/15/alucoat-en-us.ashx?dmc=1&ts=20230409t1527075538 and there is no mention of the material.
So, let's focus the question to this particular instance: what is the coating of Alucoat made of? Can it be chemically removed, for example to create small pads?

'Alu41' is a recycling code and doesn't really help in identification of the coating shall there be one. I'm afraid there is no universal answer. First, and there might be no coating at all, see related question Which side of an aluminum foil should I put in contact with food? Second, rolls of foil are sometimes coated with wax (which annoys HPLC users), and the same might be happening with the containers. Third, there are numerous plastic-coated aluminium foil (e.g. Al-PP) types used industrially.

Metallic aluminum is always covered by an alumina layer (aluminium oxide) that is thin, continuous, adherent, transparent and waterproof. If a piece of aluminium is scratched so as to remove this layer, it is immediately reformed by the atmospheric oxygen. So it is not necessary to add a second layer of wax or lacquer to this oxide layer.

Aluminium can for storing fizzy drinks usually have an internal coat of plastics like PET to prevent reaction between the can and the drink. Disposable aluminium "containers" rarely do as they are usually intended for cooking where most coatings would burn or char. Disposable containers are not intended for long term storage so they don't need a coating (unlike drink cans).

@matt_black the reason I am asking is because there is a coating, unlike some basic disposable containers. I can tell from the conductivity: aluminum foil and basic "silver" lasagna container have a thin aluminum oxide layer that can be tunneled through or scratched by the probes and the material is all in all very conductive. The containers I am talking about are not: there is no scratching that will make them conductive and even with 24V applied to them there is no current flow. Also, there are companies that produce these coated containers. I will update my question with links.

Also, is there a better group where to ask this question? I asked it here because the process has to be chemical, not mechanical.

@Peltio I think this is a perfectly fine question for Chemistry.SE, especially after you you have provided additional info.

@Peltio I don't have any inside information here but the picture is compatible with the coating being anodisation (which is, essentially, a strengthened alumina surface). This might explain your conductivity observation.

You can experiment to find out what the colored coatings are. For example, to test for lacquers and similar organic coatings  (e.g., cellulose acetate butyrate, nitrocellulose, formvar, etc.), try various organic solvents (acetone, ethyl acetate, xylene, ethylene dichloride, etc.). N.B. Most of these solvents are highly flammable, and some are fairly toxic -- use in a lab hood with supervision. Formvar, for example, is not readily soluble in acetone, but is in ethylene dichloride.
Nitrocellulose lacquer can be identified by instability to heat, and the characteristic odor of nitrogen oxides as heat decomposes it.
If no solvent affects the coating, it might be colored anodized aluminum, though anodizing might be too expensive for disposable foil products. $\ce{NaOH}$ solution will dissolve the $\ce{Al2O3}$, but would not affect most organic materials in the same way (it might wrinkle a lacquer, rather than directly dissolve it).

Yes, the outside color of the containers I have reminds me of anodized aluminum, and so do all colors of the ones in the picture I posted. And it appears certain types of anodized aluminum do have exceptionally high resistivity:  https://www.finishing.com/69/03.shtml . I will try mechanical abrasion with fine grained sandpaper first, then acetone and then if I can do it safely the other solvents. Thanks for all the links.

@Peltio, you might report your results as an answer to your question, and help satisfy my curiosity, too!

So far I have managed to remove the yellowish exterior coating using NaOH, but the interior, which is silverish, appears to resist the hydroxide attack (I can see small bubbles only along a few, almost invisible, scratches on the surface). I am wondering if there is  an additional thin organic film on the inside or if the exterior was easier to attack due to diffused microscratches... I will report back when I find out.

FTIR analysis using a one bounce diamond cell will quickly identify the coating[s] on the aluminum.

Ok, after a few experiments and a more focused search, following Dr. Moishe Pippik's answer I have come to the conclusion that:
These cups are made of aluminum, covered with a layer of anodized aluminum and, at least on the interior, an additional epoxy film.
The yellowish exterior can be dissolved by NaOH so it seems it is just a plain anodized aluminum layer. After a few minutes of exposure to concentrated NaOH solution the thin layer dissolves (starting from microscopic scratches) and once the aluminum is exposed, the reaction takes off with generous bubbling and the formation of a white precipitate (which I assume is aluminum hydroxide).

The silverish interior is more resistant to NaOH, and that seems to validate the suggestion given in the accepted answer and the comment that some sort of protective film is added. I found confirmation on this other producer's website: https://www.ablpack.com/ where it shows the same type of packages and says that

...the epoxy coating on the surface protects the aluminum foil
material from direct contact with food and prevents human ingestion of
aluminum.

After I left the hydroxide inside the container for a few hours, it was able to erode the entire material at the perimeter and leak outside (in a plastic container!). This exposed the plastic film that appears to be on the inside only as can be seen in the following picture:

Note how the exterior (anodised aluminum layer) already started to dissolve.
Therefore, I believe the mystery has been solved (almost literally).
Creation of pads is better done by mechanical abrasion with fine sandpaper and then cleanup of the surface with sodium hydroxide.

Nice work! BTW, it seems you're trying to make electrical contacts or pads on aluminum. It's difficult to make long-lasting electrical connections because aluminum forms a very hard oxide coating immediately after being cleaned, and it is difficult to solder to copper wire. There are special fittings for making contact to Al wire, though:https://inspectapedia.com/aluminum/Aluminum_Wiring_Repair_Methods.php

@DrMoishePippik I have seen a method to solder on aluminum by covering it first with an alloy of tin and zinc, under silicon oil to avoid oxidation. Once that layer is formed, it is possible to solder normal tin or tin-silver solder. Not sure if it will be possible with these flimsy containers because it needs to get real hot... But for what I have in mind, simple temporary mechanical connections will suffice.


text:
Does effective nuclear charge decrease down the group?
I've read that the effective nuclear charge increases down the group.

This seems incorrect. As we go down the group the number of protons increases and the shielding constant also increases. We can approximate the shielding constant with the number of electrons between the nucleus and the valence electrons. Hence it seems that the effective nuclear charge is the same within a given group.
The effective nuclear charge doesn't depend on the distance ( Shown through the formula above ). So the valence electrons are further away from the nucleus and experience the same effective nuclear charge down the group. Hence they have less attraction to the nucleus and the atomic radius increases down the group.
Is my intuition correct? Or does the effective nuclear charge increase down the group?

The effective nuclear charge does not decrease along the groups.
It rather increases with the increase converging to zero. What usually decreases is the first ionisation energy, which depends also on the atomic radius:
$$E_\text{ion} \propto \frac{Z_\text{eff}}{r_\text{eff}}$$

Shielding constants $S$ in
$$Z_\text{eff}= Z - S$$
cannot be approximated by the number of electrons.
Electrons shield the nucleus just partially and differently, depending on orbital type and relation to the valence orbital.
For electrons much closer to nucleus, the constant for that 1 electron shielding contribution just converges to 1.
For more, see Slater's rules.

Imagine lithium atom with configuration $\mathrm{1s^2 2s^1}$. If $\mathrm{1s}$ and $\mathrm{2s}$ orbitals had not overlapped, there would have been valid $Z_\text{eff} =Z-2=1$, due the spherical symmetry of these orbitals and the Gauss law of electrostatics.
But as they do overlap, part of 1s electron shielding is close to zero or even negative, being behind the shielded 2s electron, effectively acting as additional nucleus charge.

The effective nuclear charge can increase or decrease down a group. As the nuclear charge is well known, the effective charge depends significantly on the shielding factor S which contains the effects of the electronic distribution (and not the electronic population : number of electrons).
For most transition metals groups, as the shielding of the d orbital is low (due to their symmetry) the effective nuclear charge tend to increase down the group compared to p or s block elements where the shielding is significant, the effective nuclear charge decreases down the group.
The shielding parameter is not trivial because it contains major electronic effects. Beyond this approximation, the shielding depends on the distance.


text:
Are there distinct phases of amorphous ice?
Amorphous ice is a solid phase of water that lacks crystal structure. It is stable below about 150K, above which it converts to crystalline ice.
There are four phases of amorphous ice: low medium, high, and very high density. Medium density is a very recent discovery and can be created by grinding ice at 77K.
Do these phases exist on a spectrum? There is no "in between" for different crystalline phases. But amorphous states do not have a precisely defined structure. If so there would be an amorphous ice with any density between 0.95 and 1.25 g/cm^3.

Given the different ice phases, it is conceivable that there could be amorphous ices with different radial distribution functions and a first order transition to a different radial distribution function.

According to Wikipedia, yes
Polyamorphism occurs when there is a first order phase transition between two glass phases. Water ice is one of the examples.
Glass phases are never thermodynamicly stable, the substance at the given T and P would rather be in a crystalline (or in very rare cases quaiscrystalline) phase.
A glass-glass phase transition only occurs when the molecules are kinetically free enough rearrange a bit but not free enough to find the crystal phase. In ice, which really wants to crystalize it's a minor miracle that such a transition can be observed.


text:
Why do my slow burning fuses light off?
So recently II have been making a few times some slow burning fuses with: 34 g potassium nitrate, 26 g sugar, 60 g of water and some wool yarn. However when I light it up it burn a few seconds and just lights off.
I have also dried it at 130 °C for 20 min. How can I make it last the length of the yarn?
edit: I was using cotton and not wool (sorry).

@Poutnik i have tried cotton yarn and some thick cardboard like yarn which were even worst

Why adding water ?

@Maurice I assume soaking KNO3 + sugar solution to wool thread + drying.

Have you evaluated the ideal theoretical or experimental ratio $\ce{KNO3}$ : sum of sugar and cotton?

Stoichiometric mass ratio $\ce{KNO3}$: sucrose is 3 : 1.
$$\ce{48 KNO3 +
 5 C12H22O11  -> \\
 60 CO2 + 55 H2O + 24 N2 + 24 K2O}$$
(A part of potassium oxide  and carbon dioxide may end as potassium carbonate, but it does not affect the ratio.)
The needed ratio is even higher as the cotton thread is being burnt as well. Your ratio is way too low. You may need to use hot solution, as $\ce{KNO3}$ solubility dramatically raises with temperature.

As info has been corrected, the stoichiometric ratio is for cotton similar as for sugar/sucrose, i.e. 3 : 1.
You can estimate the overall mass ratio from the nitrate/sugar ratio in solution and final/initial thread mass ratio.
If use use nitrate without sugar, the final thread mass should be 4 times bigger than its initial mass.
If you do use sugar and if the solution nitrate/sugar mass ratio is $a$ and the final/initial thread mass ratio is $b$, then the total nitrate/(sugar+cotton) ratio is
$$\frac{(b-1)a/(a+1)}{1+(b-1)1/(a+1)} = \\
\frac{(b-1)a}{a+b}$$
If we respect the desired 3 : 1 ratio and if we use the solution ratio $a$, then the final/initial tread mass ratio must be:
$$3 = \frac{(b-1)a}{a+b}$$
$$4a + 3b = ab$$
$$b=\frac{4a}{a-3}$$
I think it is worthy to try:

hot saturated $\ce{KNO3}$ solution, without sugar
$\ce{KNO3}$/sugar ratio 4-6/1, as correction on cotton.
as supporting procedure, accumulative deposion by repetitive soaking/drying.
to let (repeatedly) $\ce{KNO3}$ crystallize on the thread in saturated solution being cooled down.

(The original wool part had been left for completeness, but is not relevant any more.)
Estimated stoichiometric mass ratio $\ce{KNO3}$ : wool is 4.2 : 1.  (based on typical ranges of major elemental wool composition, represented by the model empirical formula $\ce{C15O5N4H20}$ ).
$$\ce{14 KNO3 + C15O5N4H20
-> \\
15 CO2 + 10 H2O + 9 N2 + 7 K2O}$$


text:
Reaction mechanism of enolate formation
In this reaction, KOtBu is supposed to be used to form an enolate first and then ethyliodide is added. I just can't figure out the reaction mechanism of this reaction, can someone help?

Looks like the enolate forms in the gamma position i.e. in conjugation with the double bond and then alkylates through the alpha position leaving the double in beta/gamma. Why does it not go the other side? Not sure, possibly due to the conjugation

Similar question here: https://chemistry.stackexchange.com/q/132890/16683 In my answer to that question, there's one additional step in that the double bond migrates back into conjugation with the carbonyl group. Of course, that's not possible here because the alpha carbon is quaternary. So it stops there. There are two interesting aspects: (1) why does it deprotonate there — it's probably because deprotonating at the gamma carbon leads to the more thermodynamically stable enolate. (2) why does the extended enolate react at alpha instead of gamma — not sure!

@orthocresol  what we're not seeing is the detailed breakdown of reaction products. There may be some gamma alkylation in the product mixture.

@orthocresol Thank you for your answer, I hope I understood correctly since my English isn't perfect. I edited the question to show the answer I ended up with, does it seem correct?

@roosaveera Yes, except that the first proton you're removing is from the wrong carbon. Also, you can submit that as an answer to your own question (and in fact that is encouraged!) :)

@orthocresol Okay thank you!


text:
Are elements made out of many atoms?
So im reading a chemistry book right now, and always thought an element is made out of ONE atom, with special properties that make it this element.
Amount of Protons, neutrons and electrons.
But this book now says, "most natural elements are made out of mixes of many isotopes; etc.etc. ..."
So, an Isotope is an atom with more or less neutrons than normal, right?
SO, to my main question, are element made out of many atoms then?
What about lets say, iridium. How many atoms does it know have?

Google is your friend: search "isotopes of Iridium".

In case of confusion over textbooks, learn the habit to consult the topic with available offline and online resources before asking. This way you prevent providing redundant information and save resources of other people to tell you, what is  already written in many ways on many places.

Let's take a better known element : Chlorine. It is similar to Iridium, as it is made of two isotopes.
In nature chlorine atoms are made of the two isotopes $\ce{^{35}Cl}$ ($17$ protons + $18$ neutrons) and $\ce{^{37}Cl}$ ($17$ protons + $20$ neutrons). On Earth, the chlorine element is made $75$% $\ce{^{35}Cl}$ and $25$% $\ce{^{37}Cl}$. It means that the chlorine molecules $\ce{Cl2}$ may have the three following compositions :
$9/16$ ($\ce{^{35}Cl-^{35}Cl}$) molecules (mass $70$ u), plus
$6/16$($\ce{^{35}Cl-^{37}Cl}$) molecules (mass $72$ u ), plus
$1/16$ ($\ce{^{37}Cl-^{37}Cl}$) molecules (mass $74$ u).
It means that pure chlorine $\ce{Cl2}$ is made of a mixture of three different molecules, in proportion $9/16, 6/16$ and $1/16$.


text:
Im confused over sulfur's ability to form 4 bonds
I'm a High School AP Chemistry student, and I am still confused on sulfur. I understand why it can form $2$ bonds to complete its outer shell, but when I was drawing a Lewis structure for Thiosulfite ($\ce{S2O2^{2-}}$), I drew it as (O-)-S-S-(O-). This seemed to work out perfectly, as the oxygens had each a minus 1 charge, and therefore only needed one bond. However, when I looked online, I saw a sulfur double bonded to another, and also bonded to two oxygen anions. This confounded me, as I saw the sulfur with 4 bonds! I did some research, and I found that the reason for this was that sulfur had vacant d orbitals. When I asked my teacher about this, she told me that it sulfur did NOT have vacant d orbitals. I went deeper down the rabbit hole, and found that there was a debate over vacant d orbitals even existing in the first place! What is the true answer, or the most likely answer to this question? Thank you!

See also https://en.wikipedia.org/wiki/Thiosulfurous_acid

You are drawing dead, at least for now. Thiosulfites and the parent thiosulfurous acid are unknown.

Sulphur has electronic configuration :
1s2 2s2 2p6 3s2 3p4
The 3 shell has 3 sub-shells namely s,p and d (having azimuthal quantum numbers 0,1 and 2)
Thus the more accurate configuration would be :
1s2 2s2 2p6 3s2 3p4 3d0
The energy difference between the 3s and 3d and the 3p and 3d sub-shell is low enough to allow 1 paired electron to ‘excite’ to the 3d sub-shell where it would become unpaired (and available for covalent bonding)
As there are two paired electrons (1 in 3s and 1 in 3p sub-shells), sulphur can form 4 and even 6 covalent bonds, as seen in SF6


text:
Why does the vapour pressure not depend on the surface area and volume of a liquid?
My teacher said that vapour pressure is independent of the surface area and volume of a liquid. But I think it should be dependent, because surface area is directly proportional to the evaporation rate and if I increase surface area of a liquid then more liquid will be evaporated and more vapours will form and hence vapour pressure will be increased. And similar approach for volume of liquid also. Where am I wrong?

I think that you forget about  molecules re-entering the liquid from the vapour. Eventually the vapour will come into equilibrium with its liquid and so the rate of leaving the liquid and the rate of colliding with the liquid surface and so re-entering, become equal. When this happens the vapour pressure becomes constant.

Sir, I kept thinking on it and got what you were trying to say, thank you sir for helping.

It's equilibrium vapor pressure. What actual partial pressure of H2O vapor is, in one place or another, depends on, like, everything.

Also I wouldn't be surprised if this is a dupe of half a dozen old q.

Vapour pressure is intensive property.
It depends for pure liquids (if we neglect minor secondary effects) only on temperature. Water molecules are not aware of the surface size nor the liquid volume to modify their behaviour.
Larger liquid surface areas accelerate the evaporation and condensation the same way. The system with larger surface area converges to equilibrium faster, but the equilibrium is the same regardless of the surface area or liquid volume if measured at the same temperature.
You can try an experiment at home. Fill 2 small containers with water. One wide and shallow, the other tall and narrow. Put them in an air tight container like the common clip boxes for food.
If your hypothesis were correct, water from shallow and wide container would progressively evaporate, condensing in the tall and narrow one.
If you observe it, let us know. It could lead to the effective way how to distill water without providing thermal energy.

@Poutnik Hello Poutnik. This has left me thinking. If you have a system where area effects cannot be neglected, you cannot postulate that free energy is only a function pressure, temperature, and composition of the system. In this case, the saturation pressure will not be a function of temperature, but also of surface energy, and thus will be different like, e.g., what you find in water tables. Do you agree with me?

@MetalStorm  I have intentionally avoided minor effects like for non-flat surfaces $p = p_0 + \frac{2\gamma}{r}$. // Like it does not make much sense to show students learning about impulse p=m.v, what it is like as 4-vector in the Theory of relativity.

@Poutnik But, isn't it the case that the OP is precisely asking for?

@MetalStorm If you read what he wrote here and in the related question about liquid volume, it is clear he is confused about basics of evaporation kinetics and thermodynamics.

@Poutnik Yes, he is confusing kinetics and thermodynamics, but well...

@MetalStorm  I always try to answer in context of assumed knowledge level of the asking user. More is often less. If a high school freshman asked you about the gravity law, you would not probably come with the Einstein tensor equation.

@Poutnik Seems fine then...

Very small droplets have higher vapor pressures as small crystals have higher solubilities. Whether this is strictly kinetics is open to discussion.

@jimchmst I have already mentioned it in the above comments. It is TD thing. And yes, while pressure or solubility are not saturated, evaporation or dissolution is faster. That is kinetics.

Your intuition confuses the long term equilibrium level with how fast it is reached
Your intuition that a larger surface area should increase evaporation rate is correct.
But we normally define "vapour pressure" as the equilibrium pressure. And this means the pressure that would be reached given infinite time. So, while a larger surface area in a closed vessel would reach that equilibrium pressure  far faster, this doesn't matter as we have as much time as we need to reach equilibrium even if the surface area were tiny. The rate of evaporation is irrelevant in both cases.
We don't often observe things that have reached thermodynamic equilibrium, which explains why intuition doesn't often match theory. But the idea of equilibrium if often important in chemistry and it is worth learning why it is independent on how fast things happen.

In which area or subject is this true? We normally define the vapour pressure (or tension) as the actual vapour partial pressure, which may or may not be equal to the saturation vapour pressure (tension of saturated vapour) or the equilibrium vapour pressure which are the pressures in some equilibria.

@VladimirFГероямслава The context where the question was asked. The context where the vapor pressure is constant. Had the teacher said the actual vapour pressure, they you would be right but the question would be meaningless as the actual vapour pressure is not something constant. We measure, for example, the humidity of air relative to the fixed, known capacity of air to hold water vapour which is the only constant relevant to an elementary chemistry class.

authors: Vladimir F Героям слава


text:
Formation of tertiary alcohol with Ester and ONLY 1 equivalent of Grignard reagent
I don't understand how 2 methyl groups can be added to an ester to form a tertiary alcohol when I only have 1 equivalent of MeMgBr. Doesn't 1 equivalent mean that I only have 1 Methyl group that I can add (so only a ketone with 1 methyl group would be formed)?
Thank you

It's one methyl group per molecule of ester starting material; but there's nothing saying that they react in a 1:1 ratio.

Then what does the concept of 1 equivalent imply?

1 equivalent just tells you how much you add to the reaction mixture; it doesn't tell you the ratio in which they react, either. To be clear, if you want the ester starting material to be completely consumed, you should probably add 2 equivalents (in fact, a real-life procedure will likely use even more than that, just to really make sure all of the ester reacts). But if you only add 1 equivalent of Grignard, then only half of the ester will react.

Thank you @orthocresol

Welcome to ChemSE!. In most instances Grignard reagents react faster with ketones than with esters at their respective carbonyl groups. The rate constant for the reaction of the reagent with a ketone is much larger than the rate constant for reaction with an ester. The rate of the reactions is also a function of concentration of the substrates and Grignard reagent.
Historically, Grignard reagents were prepared in situ. The substrates are added dropwise to the Grignard reagent, which is referred to as the normal mode of addition. As the reagent reacts with the incoming ester, a ketone is formed in the presence of a sea of Grignard reagent. Even though the concentration of ketone is low, the Grignard reagent concentration is very high and there is no way that the ketone will survive. That is why the reaction of Grignard reagents with esters usually uses two equivalents of reagent to form a tertiary alcohol having two identical alkyl groups. An alternative mode of addition is inverse addition.
An example from the chemical literature will help to appreciate the difference between these two modes of addition. N-Methylindole is metallated with n-butyllithium to form lithium reagent 1, which was prepared in situ. One equivalent of the reactive ester diethyl oxalate was added dropwise to one equivalent of the organolithium reagent with the expectation that α-keto ester 4 would be formed. However, the only products formed were α-diketone 2 and α-hydroxy ketone 3. Clearly, α-ketoester 4 was formed but it could not survive in the presence of an excess of reagent. In this particular case it is the ester group of 4 that reacts with the reagent and not the ketone. This ester is particularly reactive and the ketone may well be sterically hindered. It is apparent that α-diketone 2 is the genesis of the α-hydroxyketone 3.

When one equivalent of lithium reagent 1 was added dropwise to four equivalents of diethyl oxalate, the α-ketoester 4 was formed because there was always an excess of diethyl oxalate. Saponification of the crude ester produced α-keto acid 5 in 59% yield overall.


text:
Does the Hartree-Fock energy of a virtual orbital satisfy the virial theorem?
In calculating the ground state of atoms or molecules at the equilibrium geometry, the expectation values of the kinetic, $\langle T\rangle$, and potential, $\langle V\rangle$, energies relate to the total energy, $E$, according to the virial theorem:
$$
E = -\langle T\rangle=\tfrac{1}{2}\langle V\rangle.
$$
Since the solution of the Schrödinger equation at the Hartree Fock (HF) level is variational, the viral theorem holds for it. Also, the HF energy is the sum of the energies of occupied orbitals; therefore, these energies must also fulfill the virial conditions individually. Can the same be said about virtual orbitals?

Strictly speaking, virtual orbitals are not defined in HF. Additionally to the point that orbital energies so not sum to the total energy.

As you can easily see the mean field potential potential acting on every electron in atoms and molecules do not have this shape $V=Ar^\alpha$ so the virial theorem do not hold because of the repulsive electron repulsion in atoms and the other atomic potential in molecules.
If you perform a HF calculation you will see that the total energy is $E\approx -\langle T \rangle $. This means that the correlation effects do not deviate to much from this theorem because the atomic potential has a dominant contribution.
The total energy is not the sum of the occupied orbital energy. The total energy strictly is the negative value  sum of the ionization energies of all the electrons reason why there is a correction taking into account electron-electron interaction affecting every ionization.
The virtual and occupied orbitals are issued from the same eigenvalue equation and because of the potential, the virial theorem cannot be applied on them (or very approximately)


text:
Solutions which Disrupt Equilibrium of Ferric Chloride and Potassium Thiocyanate Reaction
Consider the reaction involving ferric chloride and potassium thiocyanate,
$$\ce{Fe^{3+}(aq) + SCN-(aq) <=> FeSCN^{2+}(aq) }$$
,where $Cl^-(aq)$ and $K^+(aq)$ are spectators. Here's my thinking on what could happen if disruptions arise.
If $I^-(aq)$ is added a redox reaction decreases $Fe^{3+}(aq)$ concentration so the reverse reaction is favored.
If $Cl^-(aq)$ is added then it will react with ferric ions and cause the reverse reaction to be favored. My understanding is that this reaction is always taking place but the addition of more chlorine ions speeds it up.
The addition of $K^+(aq)$ to the solution would not cause any reaction as it is unreactive with the other substances present.
If $CO_3^-(aq)$ is added a precipitate with the ferric ions causes the reverse reaction to be favored.
If $Pb^{2+}(aq)$ is added nothing would happen since plumbous thiocyanate is soluble.
Is this logic sound? Why would chlorine actually disrupt equilibrium when added if it's already present as a spectator and is there a general non experimental way to determine if additional ions will disrupt equilibrium?

See CH SE META post about policy for "Am I right?" questions

Can Pb^2+ form the complex with SCN- too, being a competitive reaction?

Actually, do not need to add anything to disrupt the reaction other than strong light, see "A practical FeCl3/HCl photocatalyst for versatile aliphatic C–H functionalization" at https://www.cell.com/chem-catalysis/pdf/S2667-1093(22)00165-8.pdf
, that notes "Efficient and cheap photocatalyst generated from FeCl3 and hydrochloric acid".  The ferric chloride photocatalyst is somewhat unique in liberating primarily solvated electrons reducing, in your particular equilibrium,  ferric to ferrous. Solutions with strong colors are generally prone to photocatalytic reactions.

@Poutnik I have edited the question to steer it away from solely being an "Am I right?" question.


text:
Explaining the low-lying electronic state of NO
The question asks the following:
'The molecule NO has a low-lying electronic state 0.015 eV above the ground electronic state. Explain why this is the case and state the electronic degeneracies of these two
states. [2]'
MO diagram of an NO molecule:

I am a bit confused by the question as I can't find anything on this in my textbook. The energy difference is small, so the electron configuration can't be very different to what's shown in the diagram - i.e. we can't just promote the electron into the σ* orbital as that would surely change the energy by a fair bit. I'm not sure what other actions can be taken to change the energy of an electronic state. Would appreciate it if someone could point me in the right direction.

The orbital diagram you show describes only the ground state. There is one unpaired electron and the multiplicity (degeneracy) is $2S+1$ where $S$ is electron spin. If the diagram is correct in its relative energies, the lowest electronic transition would be allowed from the partially filled doubly degenerate antibonding orbital to the next one.

There's spin–orbit coupling in NO, which isn't reflected in the MO diagram. The low-lying state has the same electron configuration but has a different total angular momentum... (can't remember the symbol for it in a molecule, I remember spin AM is $\Sigma$ and orbital AM is $\Lambda$... is it $\Omega$? It's been a while, sorry.)

But if you just google for 'spin–orbit coupling in nitric oxide' you'll find plenty of info.

@Orthocresol  the ground state is $^2\Pi_{1/2}$ and the first excited state at $\approx 121,\mathrm{cm^{-1}}$ is $^2\Pi_{3/2}$ which is hardly an excited state at all! (The next one is $^2\Sigma^+$ at $43965,\mathrm{cm^{-1}}$.)

@porphyrin Yup; that is the number that OP is looking for, though (121 cm-1 is 0.015 eV exactly, unless I miscounted?).

no miscount $8065 \times  0.015=121$


text:
How to find the added volume $V_e$
We have a volume $V_A$ of a solution of $\ce{CH3COOH}$ acid, its molar concentration is $C_A$ and it has $pH_0=3,4$, we add a volume of $V_e$ from distilled water to it, in this case it has $pH_0'=3,7$, then we titrate it with an aqueous solution of $\ce{(Na^+, OH^-)}$, its molar concentration is $C_B$, and its volume is $V_B$
let's consider that $\ce{CH3COOH}$ acid whose molar concentration $C_A'$ is weak and has a $pH$ that achieves: $pH=\frac{1}{2}(\mathrm{p}K_\mathrm{a}-logC_A')$ and we have $\mathrm{p}K_\mathrm{a}=4,8$
How can I prove that: $V_e=α.V_A$, where $α$ is a constant whose expression is required to be specified in terms of $pH_0$ and $pH_0'$
here is what i have tried:
\begin{equation}
\text{starting with these relations: $$}\begin{cases}
     *\frac{C_A}{C_A'}=F \\
     *\frac{V_A+V_e}{V_A}=F ⇒ V_e=V_A(F-1) \\
*C_A'=10^{\mathrm{p}K_\mathrm{a}-2pH_0'}
  \end{cases}
\end{equation}
I came to a conclusion That: $V_e=V_A(C_A.10^{2pH_0'-(pH_0+1,4)}-1)$ (from what's given we find that $pH_0+1.4=\mathrm{p}K_\mathrm{a}$)

The reaction of interest is:
$$\ce{HA (aq)} \rightarrow \ce{H+ (aq)} + \ce{A- (aq)}$$
First Equilibrium Let $X_1$ be the moles that have reacted of $\ce{HA}$ and let me name your $C_A$ by simply $C_0$. By stoichiometry, in equilibrium we will have
\begin{align}
  n_{HA} = C_0V_A - X_1 &\rightarrow C_{HA} = \frac{n_{HA}}{V_A} =
  \frac{C_0V_A - X_1}{V_A} = C_0 - Y_1  \\
  n_{H^+} = X_1 &\rightarrow C_{H^+} = \frac{X_1}{V_A} = Y_1  \\
  n_{A^-} = X_1 &\rightarrow C_{A^-} = \frac{X_1}{V_A}= Y_1
\end{align}
where we defined $ Y_1 := X_1/V_A $. However, this is data, because $Y_1 = 10^{-pH_1}$.
The law of mass action states that
$$ K_a = \frac{Y_1^2}{C_0 - Y_1} \rightarrow C_0 = \frac{Y_1^2}{K_a} + Y_1
        = \frac{Y_1^2 + K_aY_1}{K_a} \tag{1}$$
Second Equilibrium Upon adding water a new equilibrium will be reached. Since $pH_2 > pH_1$, the reaction has proceeded in reverse, which makes sense. Let $X_2$ be the moles that have been formed of $\ce{HA}$. By stoichiometry, in equilibrium we will have
\begin{align}
  n_{HA} = (C_0V_A - X_1) + X_2 &\rightarrow
  C_{HA} = \frac{(C_0V_A - X_1) + X_2}{V_A + V_e} = \frac{C_0V_A}{V_A + V_e} - Y_2 \\
  n_{H^+} = X_1 - X_2 &\rightarrow
  C_{H^+} = \frac{n_{H^+}}{V_A + V_e} = \frac{X_1 - X_2}{V_A + V_e} = Y_2  \\
  n_{A^-} = X_1 - X_2 &\rightarrow
  C_{A^-} = \frac{n_{A^-}}{V_A + V_e} = \frac{X_1 - X_2}{V_A + V_e} = Y_2
\end{align}
where we defined $ Y_2 := (X_1 - X_2)/(V_A + V_e) $. However, this is data, because $Y_2 = 10^{-pH_2}$.
The law of mass action states that
$$ K_a = \frac{Y_2^2}{\dfrac{C_0V_A}{V_A + V_e} - Y_2} \rightarrow
    \frac{C_0V_A}{V_A + V_e} = \dfrac{Y_2^2}{K_a} + Y_2 = \dfrac{Y_2^2 + K_aY_2}{K_a} \tag{2} $$
Solving for the asked ratio Now we divide, member by member, Eq. (1) and Eq. (2)
\begin{align}
  \dfrac{C_0}{\dfrac{C_0V_A}{V_A + V_e}} &= \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} \\
  \dfrac{V_A + V_e}{V_A} &= \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} \\
  1 + \dfrac{V_e}{V_A} &= \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} \\
  V_e &= \bigg( \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} - 1 \bigg) V_A
\end{align}
and going back to our definitions of $Y_1$ and $Y_2$
\begin{align}
  V_e &= \bigg( \frac{10^{-2pH_1} + 10^{-pKa} 10^{-pH_1}}{10^{-2pH_2} + 10^{-pKa}10^{-pH_2}} - 1 \bigg) V_A \rightarrow
  \boxed{\alpha = \frac{10^{-2pH_1} + 10^{-(pKa + pH_1)}}{10^{-2pH_2} + 10^{-(pKa + pH_2)}} - 1} \tag{3} \\
\end{align}
Notes:

I could get a result, but is still function of the acid equilibrium constant.
Not sure about the information given of the titration.
If $pH_2 = pH_1$, by Eq. (3)
\begin{equation}
  \alpha = \frac{10^{-2pH_1} + 10^{-(pKa + pH_1)}}{10^{-2pH_1} + 10^{-(pKa + pH_1)}} - 1 = 1 - 1 \rightarrow \alpha = 0
\end{equation}
which translates into $V_e = 0 $. This is fine, since if the $pH$ did not change, then we did not add any water after the first equilibrium.


text:
What are the direct reaction products of the urease reaction (H. pylori) in vivo? How do I know this?
In the literature, several options can be found:

Benini et al. 1999: urea + H2O --> carbamate (H2NCOO-) + NH3 [+ apparently one proton somewhere]
Mazzei et al. 2020 (review): urea + H2O --> carbamate + NH4+ [carbamate is then hydrolysed into bicarbonate and ammonia]; or: urea + OH- --> carbamate + NH3
Marcus et al. 2005: urea + H2O --> carbamic acid + NH3; carbamic acid is then hydrolysed to carbonic acid and another molecule of ammonia

Unfortunately, I don't have proper chemistry training and I can't find conclusive evidence for any of these options. Regarding carbamate/carbamic acid, a lot of other papers also write about the one, but draw the other in reaction schemes...
How can I know which reaction product of urease is the correct one?
I expect the reaction to take place at a pH of 7-8. I can't find the pKa of carbamic acid.

Without looking at the articles: are you considering products in vivo or in vitro? And under which conditions? There might be additional products that depend on the acidity of the solution and on other conditions (other enzymes etc).

It's really dependent on the pH of the solution that the reaction is being performed in, which changes over time due to the alkalizing nature of urease. Consider that H. pylori must counteract its acidic environment in order to survive, which it does by raising the pH of its surroundings.
I found a pKa value for carbamic acid to be 3.92 here.
Also, consider that this reaction is enzyme-catalyzed, so exact products may not be completely dependent on pH. However, the products will likely still exist in a protonated/deprotonated equilibrium dependent on the local pH. At lower pH values, where urease may not be as effective, you would mostly see carbamic acid, while at higher pH values you would probably see carbamate (deprotonated carbamic acid).
I am inclined to say that at relevant pH (near 7.5 - 8.5, the optimal pH for urease), you would see the deprotonated form of carbamic acid, carbamate, and the protonated form of ammonia, which is the ammonium ion. The ammonium ion has a pKa value of around 9.25, which I found in another paper on this topic (Wrobel et al., 2012). As such, it would likely be protonated. Carbamic acid would be mostly deprotonated as its pKa is far lower.

Thank you for your detailed answer! This makes sense from a chemical point of view. Now, I'm just a bit puzzled by how this reaction is meant to increase the pH in the periplasm of the bacterium, as this is usually ascribed to the generation of 2 NH3 molecules from one molecule of urea (e.g. Sachs et al, 2011, Fig. 1).

Ammonia is basic, so it follows that the pH would be increased.


text:
Are there any real dangers to using lead fishing sinkers to hold a shower curtain in place?
I am hoping someone can help me consider if there are any factors that would make the use of lead fishing sinkers such as these in the shower risky enough to stop doing it. Specifically, they are being used as weights to hold the bottom of a shower curtain in place in a bathroom with a very small shower whos curtain has a tendency to get blown inwards.
First of all, I am aware lead is not good for me or the environment, period. I am also aware this is not a health related stack exchange, and I am aware that this question may be asinine. The only people living in this house are 25+ y.o. and we will obviously not be drinking the shower water or going out of our way to play with them.
With all that out of the way, what I am specifically looking to know is if there is potential for chlorine or anything generally in city water (Pittsburgh, PA USA, if that helps) to react with the lead and form some compound that is more dangerous than simple contact with a lead ball infrequently. I am a computer engineer, so organic chemistry is far outside of my wheelhouse, but in my mind, there is also potential for this to be relatively safe as I have heard of city water using lead corrosion inhibitors to protect old lead pipes and mitigate their risks.
For some added context, I have added pictures of the sinkers after a few months in the shower and brand new. Looking forward to learning something!
Lead sinkers with opaque white coating after a few months of use:

Brand new lead sinkers:

I don't think lead is dangerous in these circumstances. Moreover, the brand new things do not even look like lead!

@IvanNeretin Probably just a result of the poor image quality, it had to be heavily cropped to fit in the 2MB upload limit, but the packaging/website states they are lead and from my little experience with lead, these definitely fit what I remember. Thank you for your input though!

They might be lead with some thin coating of a different metal.

As long as you don't have pets, or small children, or an adult exhibiting pica, that might swallow one, it might be OK. Lead can be coated with carbonate, reducing rate of reaction with water. Why not use nonlead weights, e.g., https://www.amazon.com/s?k=non-lead+split-shot+weight ? However, this seems to border on a personal medical question.

@DrMoishePippik availability and density are my two main considerations, to my knowledge nothing I can buy is going to be as effective for a similar size, and I already have to line the entire bottom edge of the curtain to keep it from blowing in as it is narrow and tall. It is indeed a borderline medical question, but as I mention my main concern is with any checmistry that I may be unaware of which could form a more dangerous lead compound than I would encounter with elemental lead in normal conditions. I would say that is more chemical than medical, although they go hand in hand here.

Bismuth/tin weights, 9.78 g/cm3, approach that of lead's density, 11.34 g/cm3. There are tungsten fishing weights, 19.28 g/cm³ for pure W, though less for an alloy -- but W has its own possible toxicity issues.


text:
Effect of inert gas on the rate of reaction
Is there a general effect on the rate of reaction of a dynamic equilibrium when an inert gas is introduced at a constant volume? I know that the position of equilibrium won't change, but much like a catalyst which increases the rate of both forward and reverse reactions, I'm wondering if the introduction of an inert gas will have a similar effect.
My preliminary thoughts were that the inert gas would 'block' successful collisions by the reactants and products, thereby decreasing both forward and reverse rates. However, these 'blocked' collisions could simply be the result of more collisions taking place since there are more molecules in the same volume so the rate doesn't actually change.
My other thought was that certain inert gases might increase the rate of reaction by acting as a heterogenous catalyst – a surface for the reaction to take place.
Clearly, my thoughts haven't led to a definite answer so I was wondering if anyone could clear up exactly what happens to the rate of reaction when an inert gas is introduced. Thanks!

At otherwise the same partial pressure of the reacting gas, an inert gas slows down heterogenous  reactions of a gas with condensed phases.
It is caused by local additional decreasing of partial pressure of the reacting gas at the boundary layer, where it gets partly depleted and is not being resupplied fast enough from the bulk volume.
The effect is the most obvious if products stay in the condensed phase. Like
$$\ce{O2(g) + A(s) ->[O2 0.21 atm vs air] B(s)}$$
Pure oxygen would have boundary pressure equal to bulk pressure. But with nitrogen and fast enough reaction, oxygen boundary partial pressure may drop significantly -- having near pure nitrogen there -- and so would the reaction rate.

This inert gas effect is in breathing or fuel burning context  empirically confirmed by better tolerance of lower oxygen  partial pressure due altitude, compared to lowered oxygen percentage at normal pressure.
The typical threshold range for burning organic/fossil fuel is $\pu{16-18 \%}$ of oxygen in air, with the partial pressure $\ce{0.16-0.18 atm}$ . But in mountains, locals burn wood at much lower oxygen pressure, due lower air pressure.

For homogenous reactions, the effect is more subtle, but an extra inert gas generally slows down reactions, by the following effects:

Its molecules take way part of energy released by the reactions, in macro scale via heat capacity and on molecular level. It decreases the probability of reactant collisions leading to the reaction.
It affects diffusion of limiting reactant, especially if the inert molecules are slow or reactants are not in the stoichiometric ratio.
If some reactions steps involves some intermittent reactants in excited state, it may cancel their excitation before they can react.

There is possibly more effects.

"Unimolecular" reactions are already well described by Porphyrin.

Do you know any information about what happens for non-hetereogenous reactions? In other words, just gases reacting like in the Haber process.

@PenandPaper But that IS a heterogenous reaction, gases reacting on surface of solid catalyst. If we added to H2 and N2 e.g. Ar, there wouldb be locally decreased partial pressures of N2 and H2 on catalyst surface. I mean even more than just due formed NH3.

Could you also just explain why the inert gas at the boundary layer stops the reacting gases from being replenished. Wouldn't this happen without the inert gas since the reagents are reacting anyway?

@PenandPaper see the question update.

In the gas phase in so called 'Unimolecular Reactions' an inert gas does play a vital part by collisionally exciting a potentially reactive molecule. The scheme is
$$\ce{A + M } \overset{k_2}{ \underset{k_{-2}}{\rightleftharpoons}} A^* + M \\ \ce{A^* \overset{k_1}\to product}$$
where $M$ is the inert gas and $A^*$ is vibrationally excited or 'energised' $A$. The rate law depends on $[M]$ and $[A]$ and can be found by applying steady state conditions to $[A^*]$. Methyl isocyanide $\ce{CH3NC}$ to acetonitrile $\ce{CH3CN}$ isomerisation is an example of such a reaction. The simple scheme (above) is due to Lindemann, a more advanced one is called RRKM theory.
Additional comment.
The rate of product formation with steady state on $A^*$is
$$\displaystyle \frac{dP}{dt}=\frac{k_1k_2[A][M]}{k_1+k_{-2}[M]}$$
so that the rate  is small and bi-molecular at low pressures of $[M]$ i.e. $(k_2[A][M])$ but becomes larger and unimolecular at high pressure when $k_{-2}[M]\gg k_1$ and is $k_1k_2[A]/k_{-2}$. Often a plot of $\displaystyle \frac{1}{[A]}\frac{dP}{dt}$ vs $[A]$ is made better to illustrate these effects.

So the inert has in this case increases the rate?

its not quite as simple as that,I've added a comment in my answer due to lack of space here.


text:
Feasibility of aromatic nucleophilic substitution of NO group by SH- nucleophile
(My approach)
In the first step, NO+ electrophile is generated which is substituted at the p position of the alkylated phenol. In the second step, HS- acts as a nucleophile and substitutes at p position with NO- as leaving group. However im doubtful about the feasibility of this reaction and NO- being a good leaving group. Also, the +M effect of -OH group lowers the possibility of nucleophilic substitution at para position. The only driving force for this reaction imo is HS- being a strong nucleophile but in that case, is it possible for it to substitute the OH group instead of NO??

The phenol will protonate the HS-   2. You need a significantly electron-withdrawing group p to the -NO for the displacement to work.


text:
Is it Rigorous to Derive the Arrhenius Exponential Term from the Boltzmann Distribution?

The Boltzmann distribution is a probability density function which expresses the probability of finding a particle in an energy state $\epsilon$ while in thermal equilibrium given a specific temperature. The Boltzmann distribution can be written as:
$$
p(\epsilon)=C\cdot\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}
$$
I have noticed that it is possible to derive the Arrhenius equation by slightly modifying the Boltzmann distribution. We simply have to express the Boltzmann Distribution in terms of the universal gas constant R by multiplying the Avogadro’s number.
$$
p\left(\epsilon\right)=C\cdot\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}=C\cdot\exp{\left(\frac{-\epsilon\cdot N_\mathrm a}{k_\mathrm BT\cdot N_\mathrm a}\right)}\iff p\left(E\right)=C\cdot\exp{\left(\frac{-E}{RT}\right)}
$$
Then normalize the function.
$$
C\int_0^\infty{\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}\,\mathrm dE=1}\Longleftrightarrow C=\frac1{RT}
$$
Finally, integrate the function from $E_a$ to $\infty$ to find the ratio of particles with sufficient energy.
$$
\frac1{RT}\int_{E_\mathrm a}^\infty\exp{\left(\frac{-E}{RT}\right)\,\mathrm dE}=\exp\left(\frac{-E_\mathrm a}{RT}\right)
$$
This gives the Arrhenius exponential term.
However, I’m rather unsure about the validity of this demonstration, as the Boltzmann distribution fails to model gas particle behaviour conceptually. Since the Boltzmann distribution is decreasing over its entire domain, this suggests that given any gas particle, it has the highest probability to have energy equal to zero. In other words, any gas particle will most likely be immobile.
Given this inconsistency, is the above proof rigorous or is it just mathematical luck? If it is luck, then what is a more rigorous way of deriving the exponential term in the Arrhenius equation?

Your BD interpretation fails in 2 aspects: // (1) - it predicts highest probability of velocity and Ek for 1 defree of freedom, in 1D.  For 3D, it directly switches to Maxwell-Boltzman D, with the nonzero highest probability of speed and energy. // (2) Highest probability of zero 1D velocity norm does not mean zero speed. similarly as highest probability density of electrons occurence does not mean the electron is supposed to be there.

By other words, as 1D velocity and BD apply in Arrhenius eq. context, the highest probability of v_x = 0 does NOT mean the probability of |v_x| > 0 is zero. It just means v_x = 0 m/s is more probable than e.g. v_x = 300 m/s.

Yes, I am aware that the BD. does not suggest that a given particle necessarily have velocity 0. However, it is still unlikely that a particle has highest possibility of not being in motion relative to a given axis.

You write 'it has the highest probability to have energy equal to zero.' but this is ok because most molecules will be found in the lowest energy level and fewer as the energy rises. The BD describes energy levels not a rate constant for reaction, to get to that you should consider transition state theory and start with partition functions etc. alternatively start from collision theory directly.

You have to consider the integral probability.
In fact,  for BD, the integral probability says probability of  v_x = 0 is zero, in spite of the fact the maximum of probability function is at v_x=0.
As $\lim_{\Delta v_x \to 0}{(p(v_x=0) \mathrm{d}v_x)}=0$

Normally one would start from the Maxwell-Boltzmann distribution (i.e. the velocity distribution) and using the reaction cross section obtain an expression similar to the Arrhenius one.

@Porphyrin I may be wrong, but I would say BD is directly applicable for A.E derivation, considering approaching 1D motion along reactant line.

@Poutnik I think you are correct, I have added an outline of the method as an answer

If the gas phase reaction is bimolecular $A + B \to C$ then we can observe the product formed by measuring  in a certain direction at angles $\theta,\phi$ with the solid angle ($\Omega$) relative the the colliding pair. The rate is a product of terms
$$\displaystyle \frac{dN_C}{dt}=s(\theta,\phi,v)vf_Af_Bn_An_BVd\Omega$$
where $s$ is the reaction cross section at the angles given and relative velocity $v=|v_A-v_B|$. The terms $f_A,f_B$ are the probability densities of the species having a velocity in the range $v\to v+\delta v$ so are functions of the velocity alone. $V$ is the volume defined by the intersection of A and B and $n_A,n_B$ are their number densities.
The total rate is that value integrated over all angles $\theta,\phi$ and means that we can replace the cross section by its integrated value which now depends on the relative velocity only, and the volume also is extended to the whole reaction vessel's volume. This produces
$$\displaystyle \frac{dn_C}{dt}=\int_0^\infty\int_0^\infty v\sigma(v)\,f_Af_B\,n_An_B\,4\pi v_A^2 4\pi v_B^2 dv_Adv_B$$
The phenomenological rate from chemical kinetics is
$$\displaystyle \frac{dn_C}{dt}=kn_An_B$$
and so the rate constant $k$ is
$$\displaystyle k=\int\int_0^\infty v\sigma(v)\,f_Af_B4\pi v_A^2 4\pi v_B^2 dv_Adv_B$$
We will assume that the velocity distributions $f$ are given by the Maxwell-Boltzmann distribution
$$\displaystyle f_A= \left(\frac{m_A}{2\pi k_BT}\right)^{3/2}e^{-m_Av_A^2/2k_BT}$$
where $m_A$ is the mass of  A etc. and $k_B$ the Boltzmann constant. Substituting for the $f_A,f_B$ gives, using the reduced mass $\mu$ and transforming to centre of mass coordinates,  (v. tricky) gives the rate constant at temperature $T$ as
$$\displaystyle k(T)= \left(\frac{\mu}{2\pi k_BT}\right)^{3/2}\int_0^\infty v\sigma(v) e^{-\mu v^2/2k_BT}4\pi v^2dv$$
and as the collision energy is $E=(1/2)\mu v^2$ the rate constant is
$$\displaystyle k(T)=\frac{1}{k_BT}\left(\frac{8}{\pi\mu k_BT}\right)^{1/2}\int_0^\infty E\sigma(E) e^{-E/k_BT} dE$$
The final step is to make a model of the cross section for the collision.  The hard sphere model in which only a fraction of the collision energy goes into reaction depending on the position of the collision, has a cross section of,
$$\displaystyle \sigma(E)=\pi d^2\left(1-\frac{E^*}{E}\right),\quad  \text{if}\quad E\ge E^* $$
where $E^*$ is the minimum energy for reaction, and otherwise $\sigma (E)=0$  when $E\lt E^*$. Substituting and integrating gives
$$\displaystyle k(T)=\pi d^2\left(\frac{8k_BT}{\pi\mu }\right)^{1/2} e^{-E^*/k_BT} $$
This shows how the exponential term initially from the velocity of the gas atoms /molecules when transformed into relative collision energy produces an Arrhenius type rate constant.
There are other ways of working out the rate constant, such as starting with the speed distribution rather than the velocity distribution, but here I have followed closely that in 'Chemical Kinetics and Dynamics' by Steinfeld, Francisco and Hase, (publ. Prentice Hall) where a much fuller description can be found.