text:
CVD MoS2 always get many particles/stuff on the surface, what is problem?
The $\ce{MoO3}$ and $\ce{SiO2}$ substrate is placed in the center of furnace and sulfur boat is in the upstream out of the furnace. In a general growth:

$\approx \pu{2 mg}$ $\ce{MoO3}$ and $\approx \pu{200 mg}$ sulfur is used (PTAS as promoter)
Gas: $\ce{Ar}$ $\pu{80 sccm}$ (sccm for standard cubic centimetre)
Growth time: $\pu{10 min}$
growth temp: $\pu{650 ^\circ{}\text{C}}$

Bellow are the images I got (both AFM and optical images), many stuff/particles on the $\ce{MoS2}$ surface.
What is problem and how should I change for set up/receipt? I have tried with different growth time (1,4,7,10 min), increase distance between $\ce{MoO3}$ source and $\ce{SiO2}$ substrate, change $\ce{MoO3}$ amount ($\pu{1 mg}$, $\pu{5 mg}$) but still I get the same results.

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Do these photos illustrate the situation prior/after the intended deposit?  Is your sort left to right, top to bottom? If so (or not), add labels and explain if they are taken in sequence in the question.  In micro photography, the use a scale of reference is very helpful, too; just to get an idea of the absolute and relative dimensions. What are the greater plates in the photo?  Not clearly stated.  So, you do worry about the almost black speckles in the top right photo? The black speckles are not evenly distributed on the small plates exposed.

Thanks @Buttonwood for the correction.


text:
Hazards of eating non-stick coating, particularly considering the coating primer
How unhealthy are teflon / non-stick coating chips that end up eaten? What role does the primer coating play?
Existing similar threads that don't go into enough detail: 1, 2
I understand that there is a lot of concern about PFAS and its analogs. However, most sources discuss the enviromental pollution inherent to the production of non-stick coatings; there is also the danger of them decomposing when overheated.
Both of these issues aside, how dangerous is it if you ingest non-stick coating? Particularly, how significant in this is the primer coating?

Teflon is pretty inert, so my thought is that it should just pass the digestive system (alleged EPA corroboration). Based on this, teflon is resistant to both the hydrochloric acid in the stomach and to the alkaline conditions in the intestine. I assume this holds true for all non-stick coatings, right?

As I understand it, coating metal with teflon can be done in various ways (none of them tremendously effective); this and this DuPont patent provides some overview. Basically, the metal surface is eroded and coated with a primer of PTFE, PFA $\ce{(C2F4)_n}$ and polyamide imide, which is applied at 5-10 um dry film thickness; it is then baked, and subsequent coats of PTFE are baked on top of it. Table 1 below shows the example composition from the patent.
Now, from what I can tell, the solvents used, like NMP, are basically irrelevant, as they are volatile and will all evaporate during baking. I was not able to find an SDS sheet or equivalent for the PAI referenced (AI-10 polyamide imide resin); Solvay states This document is not available online, please contact us to request documentation. They do mention in their general PAI presentation that The polymer, however, may be attacked by saturated steam, strong bases, and some high-temperature acid systems, but that's neither here nor there; still, assuming that human digestive tract pH doesn't rise above 7.5, PAI should be pretty inert as well.
So, long story short, would it be, then, safe to assume that non-stick coating flakes are not harmful when eaten, or is there some flawed logic here, or something that I have overlooked?

I understand that this forum is not meant to provide medical advice (actually, I would assume that, since medical advice needs some pretty rigorous published evidence to back it, in this case it really can't be had anyway, aside from general tips - studies and documentation on this issue are facing conflicts of interest at every turn). I am asking in terms of chemical logic and safety.
Table 1: 40% PFA/60% PTFE PRIMER (Tannenbaum / DuPont patent 1993)
 0.008  Zinc oxide
 0.050  "Afflair 153' titania coated mica from EM Industries
 6.936  Ultramarine Blue Pigment
 7.206  "T-30' PTFE from Du Pont
 1.038  "Ludox AM' colliodal silica from Du Pont
 4.930  Type 340 PFA from Du Pont
 4.959  Amoco AI-10 Polyamide imide resin from Anoco
65.38  Deionized Water
 0.47   "Triton X-100" octyl phenol polyether alcohol non-ionic surfactant from Rohm and Haas
 0.700  Diethylethanolamine
 1.399  Triethylamine
 3.859  Furfuryl Alcohol
 3.306  N-Methyl Pyrolidone

 100    Total

I closed your question using the "personal medical questions are not allowed" reason. Although your post does not violate this in an obvious way it's hard to know what kind of response you would be satisfied with, particularly when you ask "is it safe to assume that non-stick coating flakes are not harmful?" after writing that studies have shown it repeatedly. What could the answer be other than yes? On the other hand, the crowd at chem SE cannot be held accountable for decisions a visitor might make based on what they read here.

As you mention, the polymer coatings you refer to have been subjected to extensive experiments to check their toxicity under physiological conditions. It looks like you want information about degradation pathways that might lead to toxic compounds during the residence time of a polymer in your digestive tract. Rephrasing the question to indicate this more clearly would help readers.

You may find the following interesting: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4928218/

@BuckThorn I can see your reasoning; I guess I was reluctant to accept the mentioned studies at face value, so I appreciate the neutral perspective. Thanks for the link! This reminds me of Olestra; weirdly enough, teflon is probably the better choice...

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text:
Can a damaged neoprene (polychloropren) or rubber coating be cured (hardened) using a hot air gun and a curing agent?
Polymers can be hardened by a process known as vulcanization (curing of elastomers), a chemical process that involves a curing agent and heat.
According to Wikipedia, for the vulcanization of rubber, sulfur is used, whilst for neoprene (polychloroprene) metal oxides are used (MgO, ZnO and somtimes Pb3O4).
I wonder if a polymer coating (assumed neoprene) that was altered by cleaning alcohol (isopropanol) and became tacky/sticky, can be hardened again by applying a suitable hardening agent and heating with a simple temperature controlled hot air station.
If yes, which the temperature ranges and which precautions?
Envisaged application: restoring the coating of a laptop palmrest that became tacky/sticky.

Haha, yes, hypothetically. I recommend you split this in three questions: what is that polymer on laptop palmrests and similar applications that turns sticky? What happened to the same when it turned sticky? And lastly how can I harden it again?

Just buy a new palmrest...

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text:
Galvanic corrosion in heating system
I’m a training heating engineer and have am unsure on hydrogen build-up in heating systems. I understand that any free air and the water within a heating system will react with the iron content in steel radiators and cause a build-up of sludge (magnetite) as well as hydrogen as another by-product, is this correct?
Secondly is hydrogen produced as a by-product from galvanic corrosion in a heating system between two dissimilar metals? I understand you need two dissimilar metals, an electrolyte and a conductive path between the metals for corrosion to take place. Is oxygen required for galvanic corrosion and does the water break down during this process releasing hydrogen?


text:
why do we need to assign ΔH =0 for the formation of aqueous hydrogen ions?
Many books state that the enthalpy of formation of aqueous hydrogen ions is 0 BY DEFINITION, and hydrogen is chosen just for convenience. Once we do that, all other reactions have their enthalpies defined relative to this.
My question is, why do we even need to define it for a particular reaction? Shouldn't the enthalpy change be defined for every reaction, without the need to define one arbitrarily as 0?
we know that it takes 13.6 ev/atom to excite its electron. so why isnt this considered here?
Edit: There are similar questions with answers explaining why HYDROGEN is taken as 0. my question is why we must define zero at all

Reactions are a different matter altogether. I thought you were talking about ions.

Enthalpy of formation being zero by definition is for elements in their standard states,i.e. H2(g) in the hydrogen case.

@Poutnik This is about ions, not elements (or so it seems).

@IvanNeretin I know. But I do not remember if I have ever heard the formation enthalpy of hydrated H+ is 0, even by definition. It would mean total enthalpy change of atomisation, ionization and ion hydration is zero by definition too.

@Poutnik This definition does exist, though. You can't measure the enthalpy of one ion the same way you do for compounds, because no compound consists of just one ion. So we define (somewhat arbitrarily) DeltaH(H+) = 0 and work out the enthalpies of all other ions from there.

@IvanNeretin That makes sense.

@Poutnik To you, yes. To the OP, probably not so much. And now the discussion is too extended, and it has not even started. "My tale was heard and yet it was not told."

Heat effects accompanying half-reactions are impossible to measure. In order for heat effects to be determined, a reaction must be carried out, and two half-reactions have to be combined. So if a heat effect has to be given to any half-reaction, the heat effect of at least one of these half-reactions must be arbitrarily chosen.
The same reasoning is used for defining the redox potentials. Individual redox potentials cannot be measured. Only differences of potentials can. This is why one of these potentials must be arbitrarily defined. So the chemists fixed the redox potential of Hydrogen at $\pu{0 V}$. Hydrogen was chosen because it was already the reference for atomic masses in the beginning of the 20th century.

but why dont we consider the fact that I takes 13.6 ev per atom to remove an electron?

Heat of formation is defined through making the species from the elements. You can't make hydrogen ions from hydrogen without also getting an electron. For uncharged species, you can make the species (and just the species) from elements.

I repeat. The enthalpy change or the electric  potential of a half reaction is impossible to determine in solution. Of course 13.4 eV is needed to remove an electron from a H atom  in a vacuum, but not in solution. The heat of formation of $\ce{H+}$ can not be determined alone, because the electron is not emitted in a vacuum. It is emitted from a $\ce{H2}$ molecule. As soon as it has been removed from the $\ce{H2) molecule, it is absorbed by a foreign molecule or a foreign ion like $\ce{Ag^{+}$, without passing through an intermediate state where it is free.

Hydrogen while chosen by Dalton was never the accepted standard for atomic masses. Oxygen was chosen in 1903 and refined when its isotopes were discovered that resulted in two scales that were unified when C-12 was chosen to placate chemists. Today Avogadro's number is defined and all atomic masses are empirical altho C-12 comes closest to12. still. read about it here http://sites.science.oregonstate.edu/chemistry/courses/ch121-3s/ch121/Answers%20to%20interesting%20questions/history_of_the_atomic_mass_unit.htm I guess one had to live thru it all to understand what was happening.

@jimchemist. I perfectly know all what you try to explain to me. I have even taken part to the panel of discussion about the choice or refusal of Hydrogen, Carbon-12 or Oxygen in the years 1980.  I was the delegate of my country. It does not change the fact that in  order to decide which reduction potential would be accepted as a reference, The hydrogen was chosen for historical reasons, related to the old definition of atomic masses.


text:
Synthetic Autunite reaction with HCl
The mineral autunite, $\ce{Ca(UO2)2(PO4)2}$, is usually present in a hydrated form (i.e., $\ce{Ca(UO2)2(PO4)2 * {10}-{12} H2O}$). I'm interested in understanding the reaction with hydrochloric acid. I attempted to balance the reaction equation as follows:
$$\ce{Ca(UO2)2(PO4)2 + 4 HCl -> Ca(H2PO4)2 + 2 UO2 + 2 Cl2}$$
Will this practically happen in a weak HCl solution? Would it need to be heated? Would there be any additional side-products.

Is it pure autinite? Uranium ores typically contain only a little uranium-bearing minerals embedded in more common rock, and the surrounding rock is not in general highly soluble in HCl.

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@Oscar Lanzi The autunite is synthetically produced and has high purity.

In such a reaction, the oxidation number are not modified. This dissolution is not a redox reaction. It will not produce $\ce{Cl2}$ and $\ce{UO2}$ (as stated by Young), which would imply an oxidation of Chlorine by Uranium. To find the equation, consider that autunite is made of the following ions :
$$\ce{Ca(UO2)2(PO4)2 -> Ca^{2+} + 2 UO2^{2+} + 2 PO4^{3-}}$$ When adding $\ce{HCl}$ solution, the $\ce{H+}$ ions will react with the phosphate ion and it is the main change. If a very small amount of $\ce{HCl}$ is added, this addition will produce a precipitate of $\ce{UO2HPO4}$, and not $\ce{CaHPO4}$ as proposed as proposed by Young (there is too many phosphate ions in autunite - the only calcium phosphate which can be formed here is  $\ce{Ca(H2PO4)2}$ - see later on). The reaction is   :
$$\ce{Ca(UO2)2(PO4)2 + 2 HCl -> CaCl2(aq) + 2 UO2HPO4(s)}$$
If the solution is concentrated enough, and with adding more $\ce{HCl}$, the following reaction may happen, remembering that $\ce{Ca(H2PO4)2}$ is only weakly soluble into water :
$$\ce{Ca(UO2)2(PO4)2 + 4 HCl -> Ca(H2PO4)2 (s) + 2 UO2Cl2(aq)}$$
But, as soon as some excess of $\ce{HCl}$ is added, both precipitates are redissolved according to the following global equation : $$\ce{Ca(UO2)2(PO4)2 + 6 HCl -> CaCl2(aq) + 2 UO2Cl2(aq) + 2H3PO4(aq)}$$
As the final substances are all soluble, they are dissociated into ions, so that the last equation should be better written, without $\ce{Cl-}$ ions :
$$\ce{Ca(UO2)2(PO4)2 + 6 H+ -> Ca^{2+}(aq) + 2 UO2^{2+}(aq) + 2H3PO4(aq)}$$
Ref.: F.-P. Treadwell, Analytical Chemistry, Qualitative Analysis, Dunod 1924, p. 142 - 145.
P. S. : Today it is difficult to find documents about the chemistry of Uranium compounds and their analytical properties.


text:
An atom moving at its root-mean-square speed at 20 °C has a wavelength of 32.8 pm. Identify the atom

An atom moving at its root-mean-square speed at $\pu{20 °C}$ has a wavelength of $\pu{3.28E-11 m}$. Identify the atom.

Here's what I did:
$$u_\mathrm{rms} = \sqrt{\frac{3RT}{M}}$$
and
$$
\lambda = \frac{h}{mu_\mathrm{rms}}
\implies u_\mathrm{rms} = \frac{h}{m\lambda}
\implies \sqrt{\frac{3RT}{M}}=\frac{h}{m\lambda},
$$
where $M$ and $m$ are the masses of $1$ mole gas (in $\pu{kg}$) and $1$ atom of gas (in $\pu{kg}$), respectively. Here nothing is written about the gas, so I assume the molecule to be monoatomic only. Let the molar mass of the gas be $M$ (in $\pu{kg mol-1}$). So the mass of $1$ molecule is $M/N_\mathrm{A}$ (in $\pu{kg}$).
$$
\begin{align}
\sqrt{\frac{3RT}{M}} &= \frac{N_\mathrm{A}h}{M\lambda}\\
\implies \sqrt{\frac{3×0.0821×293.15}{M}} &= \frac{6.022×10^{23}×6.63×10^{-34}}{M×3.28×10^{-11}}\\
\implies \sqrt{\frac{72.2}{M}} &= \frac{4×10^{-10}}{M×3.28×10^{-11}}\\
\implies \sqrt{\frac{72.2}{M}} &= \frac{12.17}{M}\\
\implies \frac{72.2}{M} &= \frac{148.17}{M^2}\\
\implies {72.2} &= \frac{148.17}{M}\\
\implies M &= \frac{148.17}{72.2} \approx 2
\end{align}
$$
So $M=2$ and the molar mass of the gas is $\pu{2 kg mol-1}$. But there is no gas whose molecular mass is $\pu{2 kg mol-1}$.
So where am I wrong? Any other method is appreciated.

Why G? We not to keep M? // You have fast-forwarded to the result too fast. To say what you did wrong, it must be seen what you did.

I have not obtained 2. Check for numerical errors.

Why not to use $R=\pu{8.314 JK-1mol-1}$?

Oh yes, Thanks @Poutnik I messed up with the units

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The algebra/formulas looks correct, a tip would be to arrange
$$\sqrt{\frac{3RT}{M}}=\frac{N_\mathrm{A}h}{M\lambda}$$
to
$$M=\frac{N_\mathrm{A}^2h^2}{3RT\lambda^2},$$
then swapping in the numbers, it helps to avoid any errors propagating through if you miss write a number.
If you use the second equation for $M$, with the constants
\begin{align}
R &= \pu{8.314 J K-1 mol-1}\\
h &= \pu{6.626E-34 J s}\\
N_\mathrm{A} &= \pu{6.02E23 mol-1}\\
T &= \pu{293 K}\\
\lambda &= \pu{3.28E-11 m}
\end{align}
you get
$$M=\pu{2.02E-2 kg mol-1},$$
which is $\pu{20 g mol-1}$ which is the mass of neon.
A side note: if the mass was $\pu{2 g mol-1}$ then the gas could have been hydrogen as you don't need to assume it is monoatomic.

Try $\pu{8.314 JK-1mol-1}$ to get $\pu{8.314 JK-1mol-1}$ // $h=\pu{6.626E-34 Js}$ as  $h=\pu{6.626E-34 Js}$

See also upright vs italic

@nyra note that the problem says "an atom moving... identify the atom", so H2 would not work (though I guess technically deuterium would work).


text:
Why in calculating 25 mM of Tris HCl the answer in gram is automatically refer to the gram in 1000 ml
In the lab class,my teacher told me that the answer i get in grams is automatically the grams in 1000 ml.
Why is it possible without we doing anything the grams i get automatically the grams in 1000 ml?
For example
If i want to find how much Tris HCl powder i need for 25 mM
I did 25/1000 mol * 121.14 g/mol = 3.0285 grams
Here my teacher says it is 3.0285 grams in 1000 ml
Why is it automatically in 1000 ml
I have not done anything
Ps. Sorry if its a bit of stupid question but i just wonder

$\pu{25 mM}$ is not $\pu{0.025 mol}$ ( it would be $\pu{25 mmol}$), but $\pu{0.025 mol/L.}$ As $\mathrm{M}$ in this context means abbreviation for $\pu{mol/L}$ and $\pu{mM}$ for $\pu{mmol/L}$.
By multiplication with the molar mass, you get the mass concentration in $\pu{g/L}$.
$$\pu{\frac{25}{1000} mol L-1} \cdot \pu{121.14 g mol-1} = \pu{3.0285 g L-1}$$


text:
What exactly is giant covalent molecule?
What exactly is the definition of "giant covalent molecule"? All sources online don't give a definition but instead, allude to properties of examples of giant covalent molecules (such as diamond and graphite). What does the "giant" mean?
To this end, is table sugar (sucrose, $\ce{C12H22O11}$) a giant covalent molecule? I reckon that since it is a covalent compound with a crystalline structure, it should be a giant covalent molecule. Is my reasoning correct?

No. It is a small molecule. Ten times that big is still small. Pretty much all compounds have crystalline structure. This is irrelevant.

Some expressions do not have explicit and unique definition, like a "giant person". Stay with meanings of "giant", "covalent" and "molecule". BTW, it is not fructose, but sucrose.  Is sucrose molecule at least as big as molecules of polysacharides, proteins and DNA to be a giant candidate ?

@Poutnik My apologies, I meant sucrose; edited that part of the question. So the whole idea of "giant covalent molecule" is pretty much arbitrary?

@IvanNeretin I was under the (probably erroneous) assumption that since table salt is a crystalline lattice structure and is considered giant, table sugar would also fall into this camp since it is a crystal. Does it mean that being a crystalline lattice structure is not a sufficient condition in determining the size of the compound?

@Poutnik Oh ok, I get it now. Thank you for explaining! I'm quite new to chemistry so pardon my confusion over simple questions like this.

Comments moved to the answer.

Table salt is just a salt like a thousand others. There is nothing giant about it.

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The giant covalent molecule (GCM) in your context is a molecule ( 3D for diamonds, 2D for graphite ), which size is more or less (breaks are possible) limited just by the size of the particular solid phase region.
If sucrose molecules were interconnected by covalent bonds forming in the ideal case a single molecular structure of the size of the whole crystal, than it would be a GCM. But they are not. Sucrose crystals are formed by independent molecules. ( Table salt is a ionic compound, it does not form covalent molecules.)
Wikipedia.org - Covalent_bond -Covalent_structures says:

There are several types of structures for covalent substances, including individual molecules, molecular structures, macromolecular structures and giant covalent structures.
.......
Network covalent structures (or giant covalent structures) contain large numbers of atoms linked in sheets (such as graphite), or 3-dimensional structures (such as diamond and quartz). These substances have high melting and boiling points, are frequently brittle, and tend to have high electrical resistivity. Elements that have high electronegativity, and the ability to form three or four electron pair bonds, often form such large macromolecular structures.

I know table salt is an ionic compound; I was just thinking of "since table salt has a crystalline lattice structure and is considered to be giant, maybe 'crystalline lattice structure' is a sufficient condition for being a 'giant'" and extrapolated 'giant' to covalent molecules. I know now that it doesn't simply work that way. Thanks for taking the time to clarify.

@Tham Here, «crystalline lattice structure» merely refers to the regular spatial arrangement of the molecules: ideally, 1) it is possible to define a tiny parallelepiped (the unit cell) about the shape of the molecules, their orientation and distance to their closets neighbors. Then 2) any other location may be described by this unit cell, because the crystal is built from identical unit cells placed next to/on top of this reference unit cell by translation (similar to stacking shoe boxes), regardless if molecule(s) in this box is/are big, or small.


text:
How to calculate lower and higher calorific values using standard enthalpies?

Calculate the lower and higher calorific value of benzene, expressed in $\pu{kJ L^-1},$ if the standard enthalpy of formation is known:
$$
\begin{array}{lc}
\hline
\text{Compound} & \Delta_\mathrm{f}H^\circ/\pu{kJ mol^-1} \\
\hline
\ce{CO2(g)} & -393.52 \\
\ce{H2O(l)} & -285.5 \\
\ce{H2O(g)} & -241.6 \\
\ce{C6H6(l)} & 48.8 \\
\hline
\end{array}
$$
The density of benzene is $\pu{0.879 g cm^-3}.$

I know the following formula:
$$q = \frac{1000}{\mu}|\Delta H^\circ|$$
Is this formula correct? I took it from the internet because my teacher did not teach it to us. We also know that for benzene $\mu(\ce{C6H6}) = 78.$
I do not know how to use this formula and how to find the lower and higher calorific value of benzene. What is the correct formula for
lower and higher calorific values?

For better site experience, you can find useful How can I format math/chem expressions/equations/formulas on Chemistry SE. ( Not to be applied to titles )
See also upright vs italic

Ask yourself as the first step, how is calculated the reaction enthalpy from formation enthalpies of reactants and products. when you get it, you recalculate it per mass and then per volume.

What is your understanding of the terms "higher heating value" and "lower heating value?"


text:
Why one compartment in an osmotic container have different pressure?
I am reading about osmotic pressure and the derivation in Wikipedia, states:

Consider the system at the point when it has reached equilibrium. The condition for this is that the chemical potential of the solvent (since only it is free to flow toward equilibrium) on both sides of the membrane is equal. The compartment containing the pure solvent has a chemical potential of $\mu ^{0}(p)$, where $p$ is the pressure. On the other side, in the compartment containing the solute, the chemical potential of the solvent depends on the mole fraction of the solvent, $0<x_{v}<1$. Besides, this compartment can assume a different pressure, $p'$.

I can't understand why the other compartment can have a different pressure. Isn't pressure in the same height of a fluid the same everywhere? Also both compartments, if we consider the container as a U-shaped tube, are in contact with atmosphere. Isn't therefore the pressure the same in both compartments?

The pressure might be different because the fluid is different, or because its height is different on both sides, or because of some dude with a pump.

Take a look at the diagram at the upper right of the Wikipedia article you cite.  Look at the U-shaped tube on the right. The pressure (p) on the RHS of the membrane (at the bottom of the tube) is lower than that on the LHS. That's because the presence of solute on the LHS lowers the chemical potential of the solvent, causing solvent to move from L to R until the increase in chemical potential from the increased p just balances the decreased chemical potential from the presence of the solute. The p difference needed to equalize the solvent's chemical potential on the LHS&RHS is the osmotic p.


text:
Why measure both alkalinity and pH in pools if pH alone tells us how acidic or basic something is?
I am trying to better understand the chemistry of maintaining my pool so that I can use the least amount of chemicals to control algae growth and I'm confused by the need to measure both total alkalinity (recommended to be between 100 and 150 ppm) and pH. If pH is high then won't alkalinity always also be high, and vice versa? Or is there a specific need or advantage in checking for both pH and TA?

How is defined the total alkalinity ?

@Maurice parts per million

See Alkalinity.  Alkalinity is the strength of a buffer solution composed of weak acids and their conjugate bases. ....... Alkalinity is expressed in units of concentration, such as meq/L (milliequivalents per liter), μeq/kg (microequivalents per kilogram), or mg/L CaCO3 (milligrams per liter calcium carbonate). The latter can be well approximated as ppm.

The same pH can belong to various alkalinity. E.g. at the same pH=8, solution of NaOH would have much lower alkalinity than solution of Na2CO3 which would have much lower alkalinity than solution of NaHCO3.

Alkalinity is typically reported in terms of either bicarbonate (the dominant carbonate species from around $\pu{pH} \simeq$ 6.5 to 8.3) or carbonate. You can have a high $\pu{pH}$ and alkalinity (total carbonate) in the range you report. Basicity is frequently used as a synonym for alkalinity, which is not correct. You are introducing caustic alkalinity into the system by (likely) adding sodium hypochlorite to get chlorine into the pool, and there's typically lye ($\ce{NaOH}$) in that solution.
$\pu{pH}$ will give you a measure of acidity and basicity, while including alkalinity will add some information about the so-called buffer capacity (ability here to neutralize acid) of the solution. It's important for a pool to have buffer capacity to maintain the proper $\pu{pH}$ range, and the alkalinity part (total alkalinity will be carbonate + caustic) is needed to make that determination.
To do this you would likely need to combine a system of polyprotic acids and salts of those acids. I'm thinking about a combination of a phosphate and carbonate buffer, the mathematics of which I can't produce off the top of my head. But you would increase buffer capacity while maintaining that $\pu{pH}$. The relevant calculation/formulation of this sort of thing in general is represented in the Henderson–Hasselbalch equation.
Finally, I note that it appears to be commonplace to use soda ash and/or sodium bicarbonate to control $\pu{pH}$ in swimming pools, so my suggestion of the additional phosphate buffer might not relevant here: as it so happens, the carbonate/bicarbonate buffer system in our bodies (which tightly regulates blood $\pu{pH}$, among other things) would result from adding the two sodium salts in the appropriate quantities for the target $\pu{pH}$ range you specify.

Liquid bleach solutions (sodium hypochlorite) do not have NaOH in them.  Also commonly used is calcium hypochlorite in powdered form, also without NaOH.  And yes, usually just plain baking soda (sodium bicarbonate) is what's added to pools to increase the alkalinity.  pH is adjusted with HCl (down) or NaCO3 (up).  Stabilized chlorine (dichlor/trichlor) have cyanuric acid added, so people who use stabilized pucks tend to need to adjust pH up. Plain calcium hypochlorite tends to raise pH, so the pool usually needs downward adjustment of pH when chlorinating that way.

Really, the best tool a pool owner can get is a good drop-count test kit (Taylor, TF-100, etc).  These have adjustment tables and saturation index calculators that make this whole process much easier for non-technical pool owners.  It's not just TA and pH that need to be controlled in a pool - the total hardness also has a major impact on what the safe pH range for a pool is.

@J... Good points. Note that commercially available $\ce{NaOCl}$ does contain trace $\ce{NaOH}$, which is actually needed to stabilize the former. See this document from Oxy, one of the larger producers of industrial chemicals world-wide.

Fair point, but in the quantities present it's almost completely negligible as far as the pool chemistry is concerned.  Didn't know that, though, so hey, +1.


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Is the density of a mixture ever just the sum of the densities of each species in the mixture?
My current hypothesis is that this is true when the volume is finite and perhaps this would cause the densities to be additive.
For example, imagine there is a closed container, 2 L in size. And within this container, there is 5 g of O2 and 3 g of N. The density of O2 would then be 5/2 g/L while the density of N would be 3/2 g/L. Would the density of the mixture of O2 and N not be total mass of the mixture divided by the volume which is (5+3)/2 = 8/2 = 4 g/mol which is equal to the sum of the densities?
Also, when would you find mixture density using $$ \frac{1}{\rho} = \frac{x_{A}} {{\rho_{A}}} + \frac{x_{B}} {{\rho_{B}}}$$ vs. using $$ {\rho} = x_{A}\rho_{A} + x_{B}\rho_{B} $$
I've seen the first version using the inverse of the density in a few places, but I don't see why the second version isn't equally applicable as it makes sense to me intuitively. Or, perhaps the second version is only for ideal gases?
(This is all assuming the mixture is ideal, so there are no changes in volume from mixing (ex. water and ethanol).)

Note that the term density is used for pure substances or whole mixtures, not for mixture components. For the latter, it is mass concentration.

Density of mixture is always equal to sum of mass concentrations of all mixture components, as it is direct consequence of mass conservation law, with both mass sides divided by the shared volume. For ideal gases, these mass concentrations are equal to the particular gas density, if the gas was at given conditions alone in the volume.

Note that the term density is used for pure substances or whole mixtures, not for mixture components. For the latter, it is mass concentration.
Density of mixture is always equal to sum of mass concentrations of all mixture components, as it is direct consequence of mass conservation law, with both mass sides divided by the shared volume. For ideal gases, these mass concentrations are equal to the particular gas density, if the gas was at given conditions alone in the volume. Generally, it is not true and there are complicated relations of component densities and mixture density.
For mixing of ideal Raoult liquids with no volume change
$$V = V_1 + V_2$$
$$\rho = \frac {m_1 \cdot \rho_1 + m_2 \cdot \rho_2}{m}=w_1 \cdot \rho_1 + w_2 \cdot \rho_2$$
(  mass fraction is denoted $w$, while mole fraction is denoted $x$ )


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Why is the NMR result for my p-toluic acid product so odd?
I just started learning about NMR and did a lab session where we conducted Grignard rxn with CO2 to form p-toluic acid and my NMR result is:
1H NMR (400 MHz, CDCl3) δ 9.63 (s, 4H), 8.07 – 8.00 (m, 5H), 8.01 (s, 1H), 7.34 – 7.22 (m, 5H), 2.46 (s, 6H).
I should expect a peak around 12ppm which would signify the presence of a COOH group; however, I got a broad peak at 9.63ppm with the integration of 4H. It doesn't make sense and I did a melting point analysis to also check the identity of the product. The melting point for my product (175°C -178°C ) is quite close to the literature range(177°C -180°C ). The range is a little off but it could be due to the product not being dried enough. Ultimately, I can't explain why I'm not getting a peak around 12ppm. Could it be due to side products?

This is very clearly not p-toluic acid. Where is the Me group 3H singlet?

It seems that I made an error for the chemical shift. It should be 2.46ppm(s, 6H). My guess why its 6H and not 3H is that there might be side products and the chemical shifts for the methyl groups on the main and side products have around the same chemical shifts. I'm not sure about my reasoning. As for the 9.36ppm, it's lower than what I expected. However, from the melting point analysis, I'm certain I did manage to synthesize p-toluic acid, albeit being slightly impure.

What starting material did you use? p-bromo-toluene?

The shift of the acidic proton can be very variable, as can its integration. For this reason it is rarely diagnostic. One possible byproduct is the dimethyl biphenyl

I only know that I was given an aryl bromide with the formula C7H7Br to react with Mg to get the Grignard reagent and then react with CO2 to get the carboxylic acid portion. But I think it should p-bromo-toluene. My TA told me the final product should be p-toluic acid.

My guess is that some water could have entered during the formation of the Grignard reagent...that's the only reason I can think of to explain the results

ooo i see...dimethyl phenyl...could it be that some of the unreacted grignard reagent could've reacted with the starting reagent?

Grignards can dimerise under some conditions

If your dry ice isn't really dry (it might look dry, but there might be water on the surface) then that will give you a fair bit of protonation product. In this case that's toluene so you may be able to get rid of it with a good vacuum.

Part of the reason you're confused is that you're picking a bad signal to be your reference peak for integration. Integrations in NMR spectra are relative values - you always have to "define" the value for the integral of your first signal, and all other integrals are based relative to that. If you select a bad signal for your first integral, it'll mess up the rest of the analysis. Assuming there is only one clean sharp signal in the alkyl region and that it is a singlet, you should set that integral to 3.00 and see how the other integrals change.

If you took advantage of the base solubility of p-toluic acid during workup, you would have answered the question of whether or not you have a carboxylic acid or non-acidic product. If not, you should have.


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How do the assumptions of the Nernst equation impact its accuracy?
I was just wondering what impact the assumption that the equilibrium membrane potential for an ion is dependent on just the ion itself (if it is the only thing that is permeable). For instance, if I had a membrane that was only permeable to K+ but I had a lot of another cation on the same side as the K+ (which was impermeable), then according to the Nernst equation, I would still get the same equilibrium potential, Ek. However, I am confused as to how the fact that there might be a lot of positive charge on the same side as these K+ has no impact on the distribution of K+;is there a reason as to why it is not accounted for/ has no impact on the membrane potential or the electrochemical gradient of the potassium ions? (I realise that in reality there is no such large positive charge on the same side as K+, I am just asking this from a theoretical perspective).

It is rather issue of having a more complex system than the Nernst equation describes. Additionally, a lot of other ions would affect $a_{\ce{K+}}$ even at the same $\ce{[K+]}$

@Poutnik  So the presence of charged but impermeable ions does have an effect on the equilibriums potential of potassium? How would I quantify this/ what sort of model would I use in order to take these factors into account and is there a name for this effect? Thank you for your help :)

N.E. is written for ion activities which depend on ionic power which depends on summary ion concentrations and charges. // But there can be other influences due complexity of system, as cross-ion sensitivities. $$E=f(a_{\ce{K+}})=f([\ce{K+}],[ion2],[ion3],...,[ionN])$$

@Poutnik What is the name of the function that would take account of these? And are the ions that impact the equilibrium potential of K+ a major deviation from the results predicted by nernst/Goldman equations alone, and what physical reasoning is there behind these effects. Sorry for all these questions, but if you have any sort of books or text recommendations that I could read to find out it would be much appreciated :)

See membrane potential, Nernst equation, activity, activity coefficitent, Debye–Hückel equation.   General site resources. Great education source is chem.libretexts.org.

@Poutnik Thank you for these, I think what I was looking for were the Debye-Huckel equations. However, I struggle to really find their usage in membrane potential models and I was wondering whether these equations are actually applied to resting potential models etc? The wikipedia does not mention its use in this way and it itself is very theoretical (i.e it does not even go through the equation if there was asymmetrical ion distribution as in cell membranes)


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Kappa convention in additive nomenclature
I’ve recently enrolled in university and in my inorganic chemistry we were covering nomenclature as recommended by IUPAC of course. We never covered any examples where 2 different polyatomic groups which have 2 or more atoms that can coordinately bond, not talking about one ligand binding to central atom only once and without bridges. Any examples I find are either cyclic, bridging or have same previously described ligands. I have drawn an example picture below and here is my question:
Do you name it:
a) tetrachloridonitritothiocyanato-kN,kS-cobalt(II)
b) tertachlorido(nitrito-kN)(thiocyanato-kS)cobalt(II)
c) tertachloridonitriti-kN-thiocyanato-kS-cobalt(II)
d) as any of the above three with “-“ somewhere.
Or more generally when specifying with kappa do you write it after each name or at the end before central atom.
Quick disclaimer: I am not natively English, so sorry if i misspelled any of the names.
Thank you for the answer.

Some links for you: https://goldbook.iupac.org/terms/view/K03366 .... https://en.m.wikipedia.org/wiki/IUPAC_nomenclature_of_inorganic_chemistry_2005#:~:text=The%20kappa%20convention%20is%20used,bond%20to%20the%20central%20atom.

Thank you for the resources, but i have read red book before asking and didn’t not find the definitive answer to my dilemma as most examples there are either cyclic or equal groups using bis to simplify it.

Based on Red Book's 4-page summary (vide infra), I suggest to proceed with these steps:

What is the central ion? It is cobalt(IV).
What are the ligands: a nitrite (then to be called nitrido), four chlorides (then to be called chlorido) and a thiocyanate.  According to  your drawing, the later binds to $\ce{Co}$ via sulfur, thus named with the kappa notation as thiocyanato-$\kappa$S.
Red Book's four page excerpt (vide infra, 2.3) then instructs to sort ligands alphabetically.  In case of ligands occurring multiple times, prefixes are prepend which however do not alter the alphabetic sort established.

Thus, open to discussion tetrachloridonitridothiocyanato-$\kappa$S-cobalt(IV).
Disclaimers:

I'm not sure about the hyphen set just before cobalt(IV) and welcome critique on this.  Visually, I would perceive a form like tetrachloridonitridothiocyanato-$\kappa$Scobalt(IV) odd for the joining the italic S right prior to cobalt.

From perspective of organic chemistry, thiocyanato-$\kappa$S reads like a layer of redundant information.  There, if the group is not bound to the alkyl chain via sulfur, but via nitrogen (i.e.,  thiocyanato-$\kappa$N in parlance of Red Book), it would be called an isothiocyanate.  Possibly Red Book's rules want this to prevent ambiguity.

I know nitrosyl ligands ($\ce{NO}$) may bind either by oxygen, or nitrogen to transition metals in complexes. I'm not (yet) aware of experimental findings for an analogue isomerism for nitrite ($\ce{NO2}$) and thus assume one may skip the kappa notation for this ligand.

The references consulted for this preliminary answer are listed in the nomenclature section of chemistry.se's resources page; i.e., the brief guide and the 4-page summary of the Red Book.


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Reactivity with rubbing alcohol and PVC
Rubbing alcohol seems to make polyvinyl plastic more sticky; i.e. if I clean a tube connector with rubbing alcohol first and then insert it into the plastic tube, it supposedly secures the seal significantly better than attaching the connector without the rubbing alcohol. Why might that be chemically speaking?

It could be just that you clean plasticizers off of surface. Just guessing, but diffusion of plasticizers from PVC is quite quick


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is this compound meso? trans-1,4 dimethylcyclohexane
I had a question on a quiz that asked if the compound below was meso or not. I thought that it would be meso because there is a plane of symmetry going through the two methyls, but the answer was that it is not meso. I'm pretty sure that plane of symmetry is allowed to go through atoms, so I was wondering why it wasn't meso.

*If the picture isn't showing up, it is a trans-1,4 dimethylcyclohexane

What is a meso, come to think of it?

In meso compounds plane goes between atoms, otherwise you have identical substituents - no chiral center!

A meso compound must have an associated d,l pair, like the three tartaric acids.

I found the answer on a past post here:Does 1,4-dimethylcyclohexane have a chiral center?
seems that it is not a meso compound because it does not have a chiral center.

Minus one because you should have commented with the link or reported the duplicate.


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Is there a way to put helium or any other gas into food?
Kinda like pop rocks puts bubbles of carbon dioxide gas into their candy to make it pop, If someone put bubbles of helium into some sort of food light enough then would it float? And how do they even put gas into food in the first place?

Putting helium into food would not make it float in air. Even if you got helium into puffed up food, it would rapidly diffuse out.

Then what about incased food like hard candy (sugar, lactose, corn syrup, and flavors, frozen together)?

Helium will still diffuse out and air will diffuse in. And it will not float in air. You might be able to make a “molecular gastronomy” foam, using helium as the fill gas, then float that in a dense gas, but it would not last long.

See this for how they get carbon dioxide into Pop Rocks.

I see, is that just because helium is a different gas?

Helium consists of small atoms that can easily diffuse or effuse. Have you heard of ice cream made using liquid nitrogen? It is quite good and the nitrogen puffs it up a bit as it boils away. Replace the liquid nitrogen with liquid nitrous oxide (“laughing gas”) or liquid xenon (big bucks) and it would be more puffed up and way more fun!

Makes sense, finally are there any other gasses you know of that would be light enough to float, and be nontoxic?

Not really. You might look at this answer: https://chemistry.stackexchange.com/a/16059/79678 and the other answers.

See: https://chemistry.stackexchange.com/questions/135436/can-we-fill-potato-chips-bags-with-a-gas-other-than-nitrogen/135438

On the other hand, some of the more dense gases might float a sufficiently "fluffy" food (cheese puffs? popcorn? That's another question...). You could fill a bowl with perfluorobutane, halocarbon 610, with density around 11 kg/m3 at ~0 °C and see what comes up.
There are denser gases, such as uranium hexafluoride, $\ce{UF6}$,with density around 14 kg/m3 at ~50 °C, but that's nasty stuff with which to work, worse than perfluorobutane.


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Which temperature is more precise - Fahrenheit or Celsius?
The Celsius scale has 100 degrees between BP and FP of water;  there are 180 Fahrenheit degrees between the same points. Doesn't that make Fahrenheit a more precise scale? If so, then why is it always taught in Intro Chem that Celsius is a more accurate scale and was developed with more meticulous methods? Is it maybe that Celsius is more accurate even as Fahrenheit is more precise?

This is the common nonsense you will find on the web that Fahrenheit is more precise than Celsius because it has smaller divisions than Celsius between two points. Don't trust the web especially (quora, SE, Wikipedia etc.) and random people.

Precision refers to the measurement and is expressed using significant figures, and has nothing to do with the units or choice of scales as long as margins for errors reported correctly. See Wikipedia for further illustration.

@theorist I have also heard the strength of imperial-units-based bolts/nuts is higher than for equivalent size metric ones. If it is true, than not because of units.

@Poutnik   There are different grading categories for metric and Imperial fasteners, but I wouldn't say one type of fastener is broadly stronger than the other. Do you have a reference for that?  I'd be curious to read it.  Imperial sizes may be available in a wider range of grading categories, though I'm not sure.  E.g.:
https://www.americanfastener.com/astm-sae-and-iso-grade-markings-for-steel-fasteners/
vs.
https://www.engineeringtoolbox.com/steel-bolts-metric-grades-d_1428.html

@theorist No ref. just heard. But it may be just a prejudice, or perhaps slighly different geometry of the technical specifications, or accidental material advantage. I am not experienced here. :-)

"taught in Intro Chem that Celsius is a more accurate scale" the statement is true (the definition of Celsius scale is more exact), but you will hardly ever do a chemical experiment where the difference is important. We use Celsius because it is solidly linked to Kelvins, which is an absolute scale, that is you can properly do math with it.

@Karl  The current F scale definition is not less exact than C scale definition, as the former is defined by the latter  ( $\pu{0^{\circ} C} = \pu{32^{\circ} F},  \pu{100^{\circ} C}  = \pu{212^{\circ} F}$ ) ( saying that, I am a European metric guy, laughing at imperial units ).

As an aside, I wonder why these extra digits are necessary. 90-something percent of the time weather is reported - they quote the temperature within a 5-10 degree span. "In the 90s". Precision, my arse.

Reported as opinion-based :P

Celsius.
The preciseness of the scales are the same. Numbers are numbers. However, upon comparing the methods for obtaining the values, a judgement can be made.
Fahrenheit was initially set to zero with the eutectic temperature of a solution of a brine, a mixture of water and ammonium chloride in which water ice forms and is present. The secondary point, by which the scale was defined, was the body temperature of a human.
So, one scale was defined as the temperature water freezes and boils, the other defined in ice with smelling salt and then between the cheeks of the operator.
The historical fact that Fahrenheit has been retrofitted to be 180 degrees between boiling and freezing of water is a testament to which is more apt. Celsius.

I downvoted, for a quibble over presentation rather than content. The large bold Celsius heading at the top is, IMO, pretty misleading for two reasons. (1) If the Fahrenheit scale has ben redefined to match the Celsius scale, then the arguments you put forth are more of a historical quirk rather than a real modern consideration. The former is less relevant to users nowadays than the latter. (2) Given that that's the case, if a single-word summary were to be provided, then surely that should be Neither since precision is not defined by a scale but rather the measurement.

I'd almost certainly upvote instead if these nuances were made more clear. But, well, that's just my opinion!

@orthocresol If you want to downvote over quibbles then I'll leave you to it, you have a busy day ahead of you here on Chemistry! ;-) I do disagree with your quibble; IF the Fahrenheit scale is by definition a linear transformation of the Celsius scale - then the comparison is (even more) meaningless. The only way to actually decide would be to base on the definition(s) which are not a linear transformation of each other. And that race, completed some time in the past, was won by the Celsius scale. If I need to put something up my rectum to calibrate it, I'll go for the other option, thank you

Of course, Celsius himself defined his scale at having 100 degrees as the freezing point of water and 0 degrees as the boiling point! Another person, a few years later, flipped the two temperatures, to the eternal gratitude of those of us who have had to teach introductory chemistry courses!

Punch line of an old joke: “Rectum? It near killed ‘em!” ;-)

@StianYttervik I agree with you on all factual points, especially it being (even more) meaningless. I just think that that should be the very first point that's being made: the fact that it's meaningless, because (1) precision is a property of a measurement, which you wrote as "numbers are numbers"; and (2) F is nowadays defined as a linear shift of C. (3) Only after these considerations, does the historical definition of F and C come into play. The heading skips past 1 and 2, leading to what I perceived as an undue emphasis on 3. I'll leave it there, though; I'm happy to agree to disagree.

Can you check that last sentence? Who is apt to what?

@Karl Celsius is more apt (at measuring temperature.)

Accuracy and precision are determined by the quality of the measurement, not the units used. The units themselves don't limit accuracy or precision, since the measurement can be expressed to whatever number of significant figures are needed, e.g., $\pu{35.4848923 ^{\circ}C}$.
As another example, grams are not more accurate or precise than kilograms. If you had a measurement where the uncertainty was in the thousandths of grams, you would express it as, e.g., $\pu{34.308 g}$. In kilograms, that same measurement would be expressed, with the same uncertainty, as $\pu{0.034308 kg}$.
On the other hand, if you were limited to only expressing temperature in integers then, yes, you could express temperature more precisely in Farenheit than in Celsius, because the former's integer divisions are smaller. You might see this limitation on the TV news, if the weather reporters have decided that it's cleaner to express temperature to the nearest integer. But that's an external, artificial limitation, not a limitation in the scales themselves.


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Reaction of propanoic acid with chlorine used in wastewater treatment
I'm asked in a problem to write down the equation of the reaction of propanoic acid with chlorine used in wastewater treatment: $\ce{CH3-CH2-COOH +Cl2}$. The problem also states that, in this reaction, $\ce{CO2}$ and chlorides are produced.
I have searched about it and I have found that there is a reaction of propanoic acid with chlorine called chlorination process, but it does not fit with the products that the problem specifies should be produced...
What would this reaction be?

The main reaction is a free radical chlorination which is catalysed by light which splits the chlorine molecule to deliver 2 highly reactive Cl radicals. One of these adds to the 2 position of propanoic acid generating a hydrogen radical which combines with the 2nd chlorine radical to give HCl. The 2-chloropropanoic acid can also decarboxylate giving CO2 and chloroethane.

@Waylander. What do you consider the mechanism for the decarboxylation of 2-chloropropanoic acid  to CO2 and chloroethane?

@user55119  struggling to find one. Have you got any suggestions?

@Waylander. The information provided is rather vague. What chlorides are formed? At this time I would guess some form of radical process. Who knows?

@user55119 I would think it has to be. I cannot account for the production of CO2 any other way.

@Waylander: One might imagine additional beta-radical chlorination. Decarboxylation with loss of the beta-chlorine could lead to a number of perchloroethylenes. They sound like more of a pollution problem than propanoic acid.


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Question on theoretical existence (or lack thereof) of spherical symmetry of flerovium and/or oganesson
It is known that flerovium, with a [core]7s²7p² configuration, is predicted to be noble-gas-like; this is due to the "7p²" part of flerovium actually being (7p(1/2))², i.e. mathematically having an azimuthal number of 1/2 instead of 1, that splits well below the other four "7p" orbitals that become (7p(3/2))⁴.
However, I do not understand why the 7p(1/2) orbital would make flerovium noble-gas-like; the spherical harmonic corresponding to n=7 and l=1/2 is just mathematically not spherically symmetric.
Further complicating the matter is that oganesson, with a [core]7s²(7p(1/2))²(7p(3/2))⁴ configuration, should have a spherically symmetric "combination of the six 7p subshells" like the lighter group 18 elements(cf. Unsöld's theorem), but trying to plug in the mathematical 7p(1/2) and 7p(3/2) functions into Unsöld fails to give a spherically symmetric function as well.
Is there a way to mathematically "fix" this, or is this purely an artifact of relativistic effects in period 7?

Isn’t “theoretical evidence” an oxymoron?

Do we have information about the electronic structure of Flerovium ? I know its nuclei have been found. But their half-life is so short that they may disappear before having the chance of meeting enough lwctrons to become neutral atoms. Does anybody know ?

@Andrew It is and it is not. It is, if it would try to relate to facts about the observed reality. It is not if it is about to confirm/reject if some statement or  theoretical structure is or is not part of particular theoretical model. It is e.g. possible to find evidence that classical electrostatics says $F=\frac{q_1 \cdot q_2 }{ 4 \pi \epsilon r^2}$

@Maurice I'm sure five seconds of half-life time are enough to reach an electronic ground state, but you can never catch such a particle. You only know you had one after detecting it's decay.

In this answer the relativistic spin-orbit coupled probability density distributions as a function of angle are tabulated below for angular momentum states we encounter in known atoms:

The $p_{1/2}$ angular dependence is constant and thus already spherically symmetric with just the one orbital. For $p_{3/2}$ it is slightly more complicated. Of the two orbitals in this spinor one will have an azimuthal component of $3/2$ corresponding to the probability density function $(3/2)\sin^2\theta$ given in the table; the other will have an azimuthal component of $1/2$ and a probability density function $(1/2)(3\cos^2\theta+1)$. These two functions, applied to one $p_{3/2}$ orbital apiece, then add up to a constant $2$ as shown in right column of the table.
Thus, both spinors separately have spherical symmetry in their electronic probability distributions when fully occupied, leading to spherical symmetry in both the flerovium and oganesson valence-electron distributions.


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What is an example of an exothermic reaction with negative ∆S?
In high school we learn that when a reaction has both negative $\Delta{}S$ (entropy change) and negative $\Delta{}H$ (enthalpy change) it occurs spontaneously at lower temperatures, but becomes non-spontaneous at higher temperatures.
I know that water freezing below $\pu{293 K}$ is an example of such change, but I'm trying to find an example of a chemical reaction with such a property. Does it even exist?

Something like carbonation of oxides, eg CaO + CO2

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A very simple reaction which is readily accessible for study is the dimerisation of nitrogen dioxide:
$$\ce{2 NO2(g) -> N2O4 (g)}$$
This process is has a standard reaction enthalpy change of $\rm{\Delta_r H^o=-57.2\ kJ\ mol^{-1}}$ and a standard reaction entropy change of $\rm{\Delta_r S^o=-175.83\ J\ mol^{-1}\ K^{-1}}$. With these values, the standard reaction Gibbs free change $\rm{\Delta_r G^o}$ is very close to zero, which implies an equilibrium constant close to 1 in ambient conditions. Not only that, but nitrogen dioxide is intensely brown, whereas dinitrogen tetroxide is colourless, so the reversible shifting between the different species can be directly visualised.

Image source: Eframgoldberg, Nitrogen dioxide at different temperatures.jpg, commons.wikimedia.org. From left to right, the vial temperatures are -196 °C, 0 °C, 23 °C, 35 °C and 50 °C.

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There exist much simpler examples, like the synthesis of water according to
$$\ce{2 H2(g) + O2(g) -> 2 H2O(l)}$$
This reaction is exothermic. As the entropies of $\ce{H2},$ $\ce{O2}$ and $\ce{H2O(l)}$, in $\pu{J mol^{-1} K^{-1}}$, are respectively $131,$ $205$ and $70$, the reaction has a negative change of entropy.
A contrario, exothermic chemical reactions which have an increase of entropy are not so common. Apart from explosion of TNT or other explosives...

It is not exactly a reaction, but consider crystallization of $\ce{KClO3}$ solution. If you put together solid $\ce{KClO3}$ with its solution of particular concentration, the crystalization becomes spontaneous below particular temperature, where solution is the saturated solution for given temperature.
It is exothermic ($\Delta H \lt 0$), as the dissolution is endothermic,  with $\Delta S \lt 0$.
$$\Delta G = \Delta H - T \cdot \Delta S$$
$$T \lt \frac{\Delta H}{\Delta S} \implies \text{spontaneous}$$
$$T \gt \frac{\Delta H}{\Delta S} \implies \text{non spontaneous}$$

The nitridation/oxidation of transition elements is exothermic and decreases the entropy, since that one of the reactants ($\ce{O2}$, $\ce{N2}$) is gaseous, but at the end is dissolved in the crystal structure of the resulting product.
For example the formation of titanium nitride:
$$\ce{Ti(s) + 1/2 N2(g) -> TiN(s)}$$
It is higly exothermic and it evidently decreases the overall entropy.

A more general rule is any polymerization reaction. Since you chain molecules together you reduce the number of freely moving particles. Thus such reaction always leads to a decrease of entropy. This also means that any polymerization reaction needs to exothermic or it would not work. In some special cases like polymethylmethacrylate (PMMA, also known as plexiglas) the polymer can even be depolymerized into its monomers by heating it above the so-called ceiling temperature, causing the entropy penalty to outweigh the enthalpic benefit (most polymers decompose before reaching this temperature).


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Reversible compression and proving relation with state function
I am currently working on a question I am having problems with. I tried to solve it, but to no avail; it states that this expression is always valid for any three-variable dependent function:

It states that I need to prove that by considering the done volume work while doing a reversible compression on an ideal gas, expressed by:

and we set (x,y,z)=(p,V,T). The last derivative can be solved by using the ideal gas law.
Honestly, I don't know where to start. I don't understand how I should put up the differentials, because I don't see how w is dependent on p, the only thing that I thought of would be that the work is defined as -pdV, but I would not know how to work with the dV on itself that would be left.
I know that those are all partial derivatives and that the variable on the outside of the bracket is kept constant and that I need to formulate the function in such a way that both variables in the derivative expression appear on the right side, but I don't know the way to combine all three needed parameters. Any help would really be appreciated.

For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ).

$$pdV=\frac{nRT}{V}dV$$If temperature is constant, this is an exact differential.

For inspiration, use  $$\left(\frac{ \partial f} {\partial x}\right)_z=\left(\frac{ \partial f} {\partial x}\right)_y + \left(\frac{ \partial f} {\partial y}\right)_x \cdot \left(\frac{ \partial y} {\partial x}\right)_z$$ as $$\left(\frac{ \partial f} {\partial x}\right)_z=\left(\frac{ \partial f} {\partial x}\right)_y + \left(\frac{ \partial f} {\partial y}\right)_x \cdot \left(\frac{ \partial y} {\partial x}\right)_z$$

How should I start then?

@冰淇淋 You have been a contributor here for some time now, please follow the link that  Poutnik gave you. Also note you can copy-paste Poutnik's second comment content. Making an effort to improve readability will pay dividends.

https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange


text:
How is reversible expansion work a state function?
Work, as we know, is a path function. But between given initial and final states, there can be only be one reversible path. Is this true? If yes, then is reversible expansion work a state function? Can someone clarify this please?

A state function must be a function of a state, regardless of a path.

An expansion work, reversible or not, Is not a state function, because it depends on another state, the initial or the final state to where the expansion is carried out.

Given the initial and final states, it is not true that there can be only one reversible path.  There are an infinite number of reversible paths, and they have different amounts of reversible work (and heat).  But the difference between the reversible heat and the reversible work is the same for all the reversible paths.


text:
The effect of changing the pressure on an equilibrium
I was reading the section on Le Chatelier's principle (section 6.8) from Chemical Principles by Steven S. Zumdahl. On page 191 where I'm currently at, I do not understand why, to consider pressure changes after a system has reached equilibrium, instead I have to consider volume changes under constant pressure? I mean the pressure is changing, right?

When the volume of the container is changed, the concentrations (and
thus the partial pressures) of both reactants and products are changed. We could calculate Q and predict the direction of the shift. However, for systems involving gaseous components, there is an easier way: We focus on the volume. The central idea is that when the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system.
To see how this works, we can rearrange the ideal gas law to give $V = (R T/P) n$ or at constant T and P $$V \propto n$$
That is, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas present.
Suppose we have a mixture of gaseous nitrogen, hydrogen, and ammonia at equilibrium (a Fig. 6.9). If we suddenly reduce the volume, what will
happen to the equilibrium position? The reaction system can reduce its
volume by reducing the number of molecules present. Consequently, the
reaction
$$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$$
will shift to the right, since in this direction four molecules (one of nitrogen [...]

Since the pressure goes down after a rise (the system responds thus), the net effect is as if the pressure didn't change much at all, instead a reduction in volume caused the number of moles to reduce. $V \propto n$ under almost constant $P$. Sounds pretty wacky, but perhaps this is what the author is trying to tell?

There are tons of resources about the  principle. Try to search this site, Wikipedia,  chem.libretexts.org or general internet.

However, for systems involving gaseous components, there is an easier way: We focus on the volume.

This does not seem easier. First, it is about changing the pressure. If your system behaves like an ideal gas, it is clear that in order to change the pressure, something else must change. So the volume. But then the volume change is associated with a change in amount of substance $n$. That is not correct, the change in volume was at constant $n$, resulting in a change in pressure.
The example illustrates this. However, the example is wacky, too. It says when you reduce the volume, the reaction system can reduce its volume, too. There is only one volume; if you reduce it, that's the new volume.
It would be more helpful to say that the pressure goes up. If the reaction goes in the direction of fewer gas particles, that will lower the pressure again.

On page 191 where I'm currently at, I do not understand why, to consider pressure changes after a system has reached equilibrium, instead I have to consider volume changes under constant pressure? I mean the pressure is changing, right?

I don't understand either, and applying Le Chatelier's principle to situations where more than one thing changes (in this case, all the concentrations) is tricky, see e.g. here: https://doi.org/10.1021/ed086p514

Since the pressure goes down after a rise (the system responds thus), the net effect is as if the pressure didn't change much at all, instead a reduction in volume caused the number of moles to reduce. $V \propto n$ under almost constant $P$. Sounds pretty wacky, but perhaps this is what the author was trying to tell? What do you think?

The pressure will change. The reaction counteracts the initial change in conditions, but only partially.

It is apparently a problem of chemical  equilibrium. Suppose you are studying a mixture of $\ce{N2, H2}$ and $\ce{NH3}$ which is at equilibrium in a big syringe. It means that the concentrations of these three gases are not changing in time. Now you are to change the volume by pressing on the piston, or compressing this gaseous mixture. Both volume and pressure will change. Pressure will increase, and volume will decrease. As a consequence, the number of gaseous molecules in a given volume will increase.
But how will the equilibrium react ? There are two possibilities : the reaction may go to the right, or to the left-hand-side. Which one is chosen by nature ?
Well ! Nature chooses a way that produces an effect opposed to the cause of the disruption. Here the cause of the disruption is an increase of gaseous molecules par volume unit due to compression. The effect will be something that reduces this number of gaseous molecules per volume unit. Well ! How to do this chemically ? In the reaction, there are more molecules ($4$} on the left-hand-side ($\ce{N2 + 3 H2}$, than on the right-hand side ($\ce{2 NH3}$). So Mother nature will favorize this side and destroy the cumbersome  $\ce{N2 + H2}$ molecules in order to create more $\ce{NH3}$. This will help decrease the number of gaseous molecules per volume unit.
This is an application of the so-called Le Chatelier principle, which can be simply stated so : If a chemical equilibrium is disturbed by a disruption, Nature reacts by creating an effect which is opposed to the cause.


text:
Accidental hydrolysis of tert-butyl chloride
Yesterday, I performed the synthesis of tert-butyl chloride from tert-butyl alcohole in my practical organic chemistry class. However, I only obtained a relatively low yield of about 30 %. Since I used more water and sodium hydrogen carbonate solution for the washing steps than recommended (I misread the instruction), my guess is, that some of the product was hydrolysed back to tert-butyl alcohol.
Is this a reasonable explanation for my low yield?

Yes, hydrolysis or elimination is likely. Washing with water or NaHCO3 solution is OK for a primary chloride but not recommended for a _tertiary_halide as you have sadly discovered.

Thank you for your answer. In the instruction, the product (about 10 g of crude tert-butyl chloride) is washed 3 times with 5 ml of a saturated NaHCO3 solution and 3 times with 5 ml of water. Sadly, I misread the instruction and used 50 ml of each.

If you got your answer, feel free to self answer it. Self answer are appreciated and it also help future readers.

T-butyl chloride is known to be hydrolysed slowly by cold water. See the Org. Syn prep (1928) vol8 p50


text:
NMR terminology: "Transients"
I was reading a paper on hyphenating a gas chromatograph with an NMR. The authors describe a proof-of-concept type instrument. We have a couple of NMR experts here in SE, hoping to get some insight. The GC-NMR output is a plot of retention time (y-axis) vs. chemical shift (on the x-axis).
I wanted to confirm the meaning of the term "transient" as used in the figure caption of the following paper. The figure, taken from Analytical Chemistry (paper-behind paywall), shows how a GC is coupled to an NMR.

"Data acquisition parameters: 32 transients with 4k time domain points
and a spectral width of 5618 Hz were accumulated with a relaxation
delay of 500 ms."

Is transient just jargon for free induction decay signal? The complete term could be a transient signal/response. Also, if I understand correctly, 4k means that 4000 data points were collected in the FID after the NMR pulse, and the FID collection was delayed for 500 ms.
I gather this "transient" wording is borrowed from electronics, where it refers to a quick change in the current in a circuit.

How does one control the spectral width of observation during data acquisition, say, as in this figure, it is 5168 Hz?

Acquisition time of 28 s (of FIDs or transients?). This time seems too long. Is this reasonable?

Article in question: Grynbaum, M. D.; Kreidler, D.; Rehbein, J.; Purea, A.; Schuler, P.; Schaal, W.; Czesla, H.; Webb, A.; Schurig, V.; Albert K. Hyphenation of Gas Chromatography to Microcoil $\ce{^1H}$ Nuclear Magnetic Resonance Spectroscopy in Anal. Chem. 2007, 79, 2708-2713; doi 10.1021/ac0617767.

"Transient" for "decay" or whatever has a relatively short duration. Is a rather widespread term, in many techniques and fields. It can even be a pulse. Something Transient, in fact.

Yes, but "transients" is a jargon of NMR. Of course, it has ten other meanings.

"Data acquisition parameters: 32 transients with 4k time domain points and a spectral width of 5618 Hz were accumulated with a relaxation delay of 500 ms."

Is transient just jargon for free induction decay signal?

"Transients" refers to the number of scans, i.e. number of times the experiment is repeated in order to obtain better signal-to-noise. (The SNR scales as the square root of the number of transients, or number of scans.) So in a sense, yes, it refers to the number of FIDs recorded. I don't know where the term comes from, though.

Also, if I understand correctly, 4k means that 4000 data points were collected in the FID after the NMR pulse

Yes, although they probably meant 4096. It's common in NMR to loosely use terms like "1k", "2k", "4k" with the understanding that it actually means a power of two. (The original idea was that $2^n$ points is "most compatible" with the FFT algorithm, although nowadays that's not a problem and if you really wanted exactly 4000 points it's perfectly fine.) If you type "4k" into TopSpin (Bruker's NMR software), it will set it to 4096 automatically.
For what it's worth, I don't like this shorthand (typing 4096 is just two characters more than 4k and is far more precise), but well.

and the FID collection was delayed for 500 ms

No, the relaxation delay refers to the time between experiment repetitions, i.e. the time between the end of the FID and the beginning of the first pulse of the next experiment.

How does one control the spectral width of observation during data acquisition, say, as in this figure, it is 5168 Hz?

The FID is a series of discretely sampled points. The sampling frequency determines the spectral width as described by the Nyquist theorem. For a spectral width of $\pu{5168 Hz}$, one would need to sample real data points at a frequency of one point per $1/(2\cdot \pu{5168 Hz}) = \pu{96.75 µs}$.

Acquisition time of 28 s (of FIDs or transients?). This time seems too long. Is this reasonable?

Well, in one single FID $4096$ points are sampled at a rate of one point per $\pu{96.75 µs}$, so the total acquisition time is $4096 \cdot \pu{96.75 µs} = \pu{396 ms}$.
Of course, the entire experiment duration will take longer than that. One single repetition is roughly $\pu{900 ms}$ long ($\pu{400 ms}$ acquisition plus relaxation delay of $\pu{500 ms}$), and this thing is repeated 32 times, so without any further information, it seems sensible to assume that the entire experiment takes around 30 seconds. Indeed, on p. 2712 of the article the authors write "Averaging 32 transients (over $\pu{28 s}$)...".

Re: "I don't know where the term comes from, though" – I'm pretty sure it came about after the invention of Fourier NMR to distinguish collecting fids for a period of time, from averaging under a continuous wave experiment. The VnmrJ source code (now open source) has comments going back to 1986 or so, and some legacy hooks for CW experiments. You can see nt for "number of transients" clearly being used, even then  (e.g. https://github.com/OpenVnmrJ/OpenVnmrJ/blob/c306df2964c3a823588757772f9f5260722b46e1/src/xracq/lc.h)

@Landak, Thanks, this makes more sense. Transients in electronics has a specific meaning and now I am sure, it came from there. From OED, Electronics. A brief variation in current or voltage caused by a surge in power, the connection of a load, etc.

Since they also use transient power in the code you linked...the name is obviously of electrical origin.

Yes -- that's right. They're identical to [L]RC ringdown in electronics, but the link between electronics and NMR isn't very clear. I work in MRI – where "the bit of wire with capacitors in it wot collects your signal" is called an RF coil; in chemistry NMR it is a probe; to an electrical engineer it is a resonant (or "tank") circuit. Arguably it is a nearfield antenna – and the use of the word 'antenna' for NMR is itself incredibly controversial!

@Landak, Yes, I had asked an editor of magnetic resonance journal editor... the use of radiofrequency "absorption" and emission is highly controversial. He said this is the myth spread by MRI people.


text:
If a carbonic acid to bicarbonate reaction involves the release of H+, why does pH increase?
OK, I'm an engineer that took Aqueous Geochem 6 years ago, so please forgive this basic (get it?) question.  But why do the two reactions which start from dissolution of CO2 in water increase pH if they both release H+ ions?  I.e., the conversion of carbonic acid to bicarbonate releases a hydrogen proton, and the conversion from bicarbonate to carbonate releases a hydrogen proton.  Yet each reaction results in a higher pH.  Specifically, why don't the additions of H+ decrease pH?
$\ce{H2CO3 + 2 H2O ⇌ HCO3- + H3O+ + H2O ⇌ CO3^{2-} + 2 H3O+}$
Thanks, and apologies for the (probably simplistic) question.
Mike
Thank you all for your help with this.  I should clarify that what is really troubling me is the interpretation of the Bjerrum Plot for the carbonate system (aqueous geochem).  Looking, for example, at that plot here:
https://en.wikipedia.org/wiki/Bjerrum_plot#Bjerrum_plot_equations_for_carbonate_system
For HCO3- to transition to CO32-, a H+ ion is released (going rightward in the plot), but that is associated with higher pH (less acidity).  But if increasing H+ ions are the very definition of increasing acidity, why in the plot is that reaction associated with increasing pH?
Thanks again!
Mike

These reactions decrease pH. You may  confuse it with something else.

The reaction usually runs in the other direction. If a carbonate dissolves, it takes up a proton from the surounding water, becoming bicarbonate.

There are 2 ways, depending on what substances we start with. The forward way decreasing pH is $\ce{CO2}$/$\ce{H2CO3}$ releasing $\ce{H+}$, forming $\ce{HCO3-}$ and in alkalic pH also $\ce{CO3^2-}$. The backward way increasing pH is $\ce{CO3^2-}$, releasing $\ce{OH-}$ and forming $\ce{HCO3-}$, in acidic pH also $\ce{H2CO3}$/$\ce{CO2}$

"... if alkalinity of the tissues will produce tetany,
acidity may be expected to cure it. Unfortunately, one
could hardly try to cure a dying baby by shutting it up
in a room full of carbonic acid ..."                                           From J. B. S Haldane's book Possible Worlds (Harper and Brothers, 1928).

It seems you have misconception that creation of bicarbonates and/or carbonates causes pH increase.

There is sure to be a more technical (and more pleasing to the physical chemist) answer to this, but as a heuristic, don't consider what the carbonic acid is doing to solution, consider how the solution is acting on the bicarbonate.
The plot your referencing describes mole fractions of diprotic acids in solutions relative to pH. This isn't the same as the pH of the carbonic by itself. Of course, carbonic acid in a cup of water decreases the pH of the solution, but instead consider an equilibrium condition in which a happy amount of bicarbonate and hydronium exist. We're not concerned with what happens if we add more carbonic acid, we only care about what happens to the bicarbonate if we vary the concentration of our hydronium.
In a solution rich with hydronium such that there are many free protons to act on bicarbonate, equilibrium will shift towards the production of carbonic acid. Thus, at a lower pH, carbonic acid is the dominant species, as we can clearly see by our mole fraction of the acid hitting a maximum at lower pH's.
In a solution quite poor in hydronium, such that there are very few protons to act on bicarbonate, equilibrium will shift to produce carbonate. This is because we perhaps have free hydroxide ions in solution, which are perfect homes for bicarbonate protons. Therefore, we see a higher mole fraction of our carbonate ion at higher pH's.
Again, the dissociation of carbonic acid, which is really just dissolved carbon dioxide, is in fact a proton-producing process. But the plot your describing is a relationship that describes which species are dominant at different concentrations of hydronium, not the concentration of hydronium (pH) as carbonic acid dissociates.
To directly address the equilibrium you've set shown us- the real bump on the road is the dissociation of the bicarbonate. Remember that in this situation, we must have a very hydronium poor solution, such that it's likely that we actually have free hydroxide ions. So, what happens to the 2 protons on the far right side of your formula? Well, they pair up with the hydroxide! The issue is that you can't see that, because again the equilibrium formula you have doesn't intuitively show that there is an entire solution that impacts the equilibrium. So where do we get these hydroxide ions from? Take a look at one of the answers below!
The Le Chatelier way of saying this is if we decrease the concentration of hydronium relative to the bicarbonate, equilibrium will shift to try to increase it again. But clearly the best way to do that isn't by producing more bicarbonate, because the stoichiometry shows the fastest way to get more hydronium is by producing more carbonate. For every one mole of hydronium relative to bicarbonate, we have two moles of hydronium relative to carbonate. So in a weird way, by increasing our pH to become hydronium deficient, equilibrium shifts towards the intuitively lower pH side of the reaction to try and counteract this and we get carbonate as a byproduct. This may be the source of your confusion.
"Ask not what your acid can do for your solution, but what your solution can do for your acid"
- JFpKa (buh-dum-tss)

Many thanks, BrownBag -- that all helped considerably!!!

Glad to help!!!

No, addition of carbon dioxide to water will never cause an increase in pH. Think of the simplest scenario. Take pure water and dissolve carbon dioxide in it. The pH will decrease from 7 to ~ 5. Recall rainwater is slightly acidic due to $\ce{CO2}$.
You are most likely confused about carbonate-bicarbonate buffering system in geological settings? See here
http://butane.chem.uiuc.edu/pshapley/environmental/l24/3.html

It seems you have misconception that creation of bicarbonates and/or carbonates causes pH increase, because the plot has more of bicarbonate and carbonate at higher $\mathrm{pH}$ values. That is just a half of the truth.
The plot is about ratios of concentrations of respective carbonate forms in equilibrium.
There is equilibrium equation rewritten into logarithmic form, where square brackets are convention for molar concentration:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a1}}^{*} + \log {\frac {\ce{[HCO3-]}}{[\ce{CO2}]}}$$
When $\ce{CO2}$ is just dissolved in pure water with $\mathrm{pH}=7$, the right equation side does not match it, it is too low. So $\ce{CO2}$ converts to $\ce{HCO3-}$ until dropping left equation side and raising right equation side are equal, reaching equilibrium.
The similar equilibrium equation is for bicarbonate and carbonate:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a2}} + \log {\frac {\ce{[CO3^2-]}}{[\ce{HCO3-}]}}$$


text:
Isomerization of tertiary carbocations in cholesterol synthesis
The following reaction mechanism happens in cholesterol synthesis inside human body. I am unable to understand why on earth the hydride shift happens here. I mean its already a tertiary carbocation, by this hydride shift I am again going to another tertiary carbocation. Can someone throw some clue on this? This reaction happens inside human cell, i.e. pH 7.4, and room temperature.

The picture is from McMurry's book on Chemistry.

If the above transformation was depicted in a journal article, book, web site, add this as a reference.  For the educational benefit of future readers of your question, may you add the conditions (e.g., temperature, presence of reagents, reaction time, yield or percentage of conversion) leading to said reaction.


text:
What would the mechanism be for intermediate B to Supinidine, and what would by product C be?

I have managed to do the mechanism for the first step, but have no clue how 4 equivalents of MeLi are used for the second step of the reaction.

The first eq. of MeLi deprotonates the alcohol. This does not cyclise as it would form a 7-membered ring and that is not strongly favoured.
The subsequent eqs of MeLi remove the methyl carbamate group as described by Sawyer, et al., by consecutive additions of Me to give first the N-acetamide then the amine anion plus acetone. The N anion cyclises onto the chloride to give the product. The final eq. of MeLi adds to the acetone to give Li t-butoxide which gives t-butanol C on workup.
J. S. Sawyer, B. A. Narayanan, Synth. Comm., 1983, 13, 135.


text:
How are these two enantiomers?
I was wondering how these two chair conformations are enantiomers, because they do not seem to be mirror images of each other. The molecule on the left undergoes a ring flip to achieve the conformation of the molecule on the right. Also, I thought that in order for these two molecules to be enantiomers, the carbons would have to change both their R/S configuration, e.g. carbon 1 goes from R to S, and carbon 2 goes from S to R? I would really appreciate any insight!

Try rotating the one on the right clockwise looking from the top. might be easier if you build 3D models and compare them.

https://chemistry.stackexchange.com/questions/61107/is-cis-1-2-dimethylcyclohexane-a-meso-compound

https://chemistry.stackexchange.com/questions/61107/is-cis-1-2-dimethylcyclohexane-a-meso-compound

A flip of a ring alters the conformation of a cycloalkane, a reversible, gradually wiggle where all atoms remain connected with each other.  Passing from one enantiomer to the other enantiomer (e.g., R $\ce{->}$ S) is like driving either on the left, or on the right hand lane; or either light on vs. light off, binary: this is possible only if you break old bonds, and establish new bonds.


text:
Help determining Absorption Coefficient & Concentration from Uv-Vis reading
hope you're doing well.
I'm having some trouble determining the absorption coefficient and concentration values of a protein from its UV absorption results.
The experiment in question involves an ion exchange chromatogrpahy column to elute hemoglobin from an HB/BSA mixture, then quantitating the amount and concentration of hemoglobin in the samples and fractions. With the calculations that I'm currently using, I obtain an absurdly low yield, less than 2%, and I'm not sure where I'm going wrong.
My results are comparable to the sample results we were provided for the experiment, so I'm assumming that the error is a calculation error on my part. If someone could check over my approach, I would greatly appreciate it. The values below are completely made up, this is just to check my process.
Suppose I was working with a 2% Hemoglobin 0.1% BSA mixture.
Then, the (w/v) concentration of the haemoglobin should be 2 g/100 mL. If 0.3 mL of the mixture were added to the column, then a mass of $\frac{2\text{ g}}{100 \text{ mL}} \times 0.3 \text{ mL} = 0.006 \text{ g} $ of haemoglobin was added to the column.
Okay, say you collected a number of fractions from your column, pipetted them into a microtiter plate with buffer blanks and a 0.05% Hb standard solution as references, and measured the absorbance of the samples at a particular wavelength.
You want to determine the absorption coefficient for your 0.05 % (w/v) standard solution, so you do the following:
First, you average the buffer blank absorbance values, say its something like 0.035 and subtract this average from all absorbance values to obtain adjusted values.
Convert the 0.05% (w/v) concentration into mg/mL:
0.05 % (w/v) $\rightarrow$ 0.05 g/100 mL = 0.5 mg/mL.
Sample calculation:
Suppose this well has 50 uL of the 0.05% standard solution.
Then, the mass of Hb present in the well is equal to
m = (0.5 mg / mL) (50 uL x 1 mL/1000 uL) = 0.025 mg.
If the volume of solution in the well is 50 uL Hb standard + 150 uL buffer = 200 uL, then the concentration of Hb in the well is equal to the mass of HB present divided by the volume of solution in the well. That is,
[Hb] = $ \frac{0.025 \text{ mg}}{200 \text{ mL}} \times \frac{1000 \mu\text{L}}{1 \text{ mL}}= 0.125 \text{ mg/mL}$
And similarly for the other standard wells.
Then, by Beer’s law, the absorption of the absorbing species is the product of the absorption coefficient, the path length, and the concentration of the sample solution. Hence,
$$A=elc \implies e =A/lc.$$
If your plate has a path length of 0.7cm, and your sample has an adjusted absorbance value of 0.360, then the absorption coefficient e is given by
\begin{align*}
 e & = \frac{A}{lc} \\&= \frac{0.360}{(0.7 \text{ cm})(0.125 \text{ mg/mL})} \\
        &= 4.114 \,(\text{mg/mL})^{-1} \text{ cm}^{-1}\tag{1}
\end{align*}
Yes?
Then, suppose you perform the same calculation for two other standard solution wells, and get values of 4.13 and 4.09 for e respectively. You take the average of the three values for an absorption coefficient of 4.11 (mg/mL)^-1 cm^-1.
You then try to use this value to calculate the concentration of Hb present in the other wells:
For simplicity, say you ran a teeny tiny plate with only four wells, whose (adjusted) absorbances were
Well 3A: 0.203 (25 uL of Fraction 1, 175 uL buffer)
Well 3B: 0.651 (25 uL of Fraction 2, 175 uL buffer), and
Well 4A: 0.417 (50 uL of Fraction 1, 150 uL of buffer), and
Well 4B: 1.315 (50 uL of fraction 2, 150 uL of buffer).
You calculate the concentrations of Hb in each well as follows
3A:
By Beer’s Law, the concentration of Hb in solution is given by
\begin{align*}
c &= \frac{A}{el}\\
   &= \frac{0.203}{((4.114(\text{mg/mL})^-1 \text{cm}^-1)(0.7 \text{ cm})}\\
   &= 0.0705 \text{ mg/mL} \tag{2}
\end{align*}
This is the concentration of Hb in the well however, so to find the concentration of Hb in the fraction, you find the amount of Hb present in the well:
\begin{align*}
m &= c \times \text{ Volume of solution in well}\\
   & = \frac{0.0705 \text{ mg}}{\text{mL}} \times \frac{0.2}{ \text{ mL}}\\
   & = 0.014 \text{ mg }    \tag{3}
\end{align*}
Then divide it by the volume of the fraction solution in the well:
\begin{align*}
\text{Fraction concentration } & = \frac{0.014 \text{ mg}}{ 0.025 \text{ mL}}\\
                            & = 0.564 \text{mg/mL} \tag{4}
\end{align*}
The same calculation applies to the other wells. To calculate the total mass of haemoglobin recovered, you take the sum of the value determined in (3) for each well.
Although these values are made up, is this the correct approach?
I can’t find any examples showing how to do this, and we weren’t taught how to do so in the lab or in class (and it’s been a while since I’ve done any sort of stoichiometric calculations) so I’m hoping, but not sure whether this is actually correct.

Please explain your sentence (I copy) : Okay, say you collected a number of fractions from your column, pipetted them into a microtiter plate with buffer blanks and a 0.05% Hb standard solution as references, and measured the absorbance of the samples at a particular wavelength. ... Which volume do you pipette ? And how much standard do you pipette ?

Hi @Maurice, thank you for replying. The hypothetical arrangement here involves collecting fractions from the column in volumes of 4 mL. Each well of the microtitre plate has a total volume of 200 uL. The "blank" wells contain only 200 uL of your buffer, the Hemoglobin standard wells contain varying volumes (25, 50, 100 uL etc) of 0.05% Hb Standard solution with the remaining volume made up by buffer, and the sample wells each contain a particular volume from a given fraction (20 uL, 40 uL) with the remaining volume made up by buffer.

There are too many variable parameters. I am lost. Sorry !

It would be nice if you could choose a meaningful title.

A bit late, but this is my guess: the error is likely in the way the total yield was computed from the amount of protein determined for the individual wells. You should not just add up the amount of protein in each well, since this the sample volumes in each well represent only an aliquot of each fraction collected. You should instead scale those amounts by the ratio of the aliquot volume to that of the fraction. You may have done this after the fact, but that is not clear from your exposition.

Wait:
1)You're using several standards with the same concentration?
2)The samples (well, all but one) have higher absorbances than your standard?
3)Half of the samples have absorbances above 1?
I think I see at least part of the problem.
Higher absorbance means less light is getting through to the detector.  Due to limitations of the instrumentation (stray light, noise in electronics) this means that the instrument response will deviate from the expected linear behavior at higher concentrations.
Suggestions:

Prepare multiple standards with different concentrations, and plot absorbance vs concentration.  Use linear regression to check how well a straight line fits your data. The first time around, you'll want a lot of data points over a wide range of concentrations.
If your correlation coefficient is too low (I generally use r^2>=0.998 for UV-vis), try dropping data points one at a time from the high concentration end of the graph until you get a good linear fit to your data.
Dilute all your samples so they fall within the range of standards you decided to keep (preferably close to the middle, but it doesn't have to be perfect).


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How to separate manganese(IV) oxide and chromium(III) hydroxide?
I am trying to find a reagent that keeps either $\ce{MnO2}$ or $\ce{Cr(OH)3}$ solid and make other soluble so I can centrifuge or use filtration. I thought of $\ce{NaOH},$ but I could not find any information about how $\ce{MnO2}$ reacts with $\ce{NaOH}.$

Float separation? A solution of density ~ 4 g/ml would work.

Gut feeling: try (at small scale) if $\ce{Cr(OH)3}$ enters again into solution by addition of base (think like $\ce{[Cr(OH)6]^{3-}}$) while $\ce{MnO2}$ remains untouched; two small scale tests for each of the two in absence of the other, one small scale test where the two are present at the same time (low temperature, of course). Use then a centrifuge to separate $\ce{MnO2}$ (don't forget washing to remove the lye), and carefully precipitate $\ce{Cr(OH)3}$ again. All with caution and protection for yourself (heavy metals, chromium salts; lye). And report back to chemistry.se.  Good luck.

Add at least twice its volume of $\pu{2 M}$ $\ce{NaOH}$ and stir. The  $\ce{Cr(OH)3}$ precipitate will be dissolved if the solution is cold. It will make a solution of sodium chromite $\ce{Na[Cr(OH)4]},$ and $\ce{MnO2}$ will not react.
$$\ce{Cr(OH)3 + OH- <=> [Cr(OH)4]^-}$$
After filtration, $\ce{MnO2}$ is not easy to filtrate on paper. It can be centrifuged. The solution of chromite is then heated and diluted with a volume of water equal to four times the volume of the solution. The last equation is an equilibrium and will be reversed at high temperature, so that $\ce{Cr(OH)3}$ will precipitate again, and can be filtrated.


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The meaning of USFDA (in cosmetics)
The full meaning of USFDA base on the definition of cosmetic meaning and production

Aside of the particular answer below, the great reference for acronyms is https://www.acronymfinder.com

One likely possible meaning of USFDA is United States Food and Drug Administration, since

«The United States Food and Drug Administration (FDA or USFDA) is a federal agency of the Department of Health and Human Services. The FDA is responsible for protecting and promoting public health through the control and supervision of food safety, tobacco products, dietary supplements, prescription and over-the-counter pharmaceutical drugs (medications), vaccines, biopharmaceuticals, blood transfusions, medical devices, electromagnetic radiation emitting devices (ERED), cosmetics, animal foods & feed and veterinary products.»
source: Wikipedia (emphasis added ny mine)