text:
Difficulty in E-Z nomenclature and counting number of geometrical isomers

I have found above four molecules as the geometrical isomers just by drawing them and checking if they are superimposable. I'm not sure that these are the only ones.
I have tried using E and Z but couldn't go ahead with it because configuration around each double bond depends on configurations of all other double bonds and it gets circular.
How do I assign E and Z configurations to each of the four double bonds in this case to distinguish between all possible geometrical isomers?

E and Z are not of much help here (though you may assign them, of course). Your answer of 4 looks good to me.

Thanks, but I am also looking for the E and Z nomenclature of this thing.

Thanks a lot, but how do I assign E or Z to a particular stereoisomer of this  molecule ?

Geometrical isomerism is an outdated term no longer recommended to be used. Consider updating your textbook or getting a better one.

Related: Which alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol? and How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry. The explanation can be found in Rule P-92.4.2.2; in short: seqcis precedes seqtrans

https://chemistry.stackexchange.com/questions/68892/total-number-of-stereoisomers-of-cyclobutane-1-2-3-4-tetracarboxylic-acid

Rishi Shekher: You are correct that there are only four stereoisomers (A, B, C, D) of this tetraethylidenecyclobutane. Loong has given you a lead as to how to apply CIP rules to the assignment of the configurations of the double bonds. This method can appear confusing and it is certainly not intuitive. The configurations were generated with ChemDraw 21. I will show you how the CIP algorithm works.

Stereoisomer A: The configuration of each double bond must be determined independently. They are labeled in red in each of the digraphs 1-4. The digraph is constructed by following a path around the ring, CW or CCW, from the non-duplicate carbon (black dot) to the duplicate carbons (red dot), which are designated as being attached to three atoms of atomic number zero.
Focusing on digraph 1 (vide infra) and $\ce{C1}$ (black dot), the double bond immediately to its left ($\ce{C4}$) is assigned the temporary Z-configuration because the path "around the ring" to the right is longer, i.e., more carbons than the path leading to the left. This method is used to temporarily assign the five positions. The left hand chain has three Z's while the right hand chain has all E's. One proceeds out each chain from the black dot making a one-to-one comparison until a Z>E is achieved (CIP Rule 3). For $\ce{C1}$ this is accomplished at $\ce{C4}$ and $\ce{C2}$ where Z>E, respectively. Determinant double bonds are shown in blue.

Stereoisomer B: All positions in this isomer are equivalent. The double bonds are all of the E-configuration with Z>E.

Stereoisomers C and D: Stereoisomer C has a plane of symmetry. $\ce{C1}$ is equivalent to $\ce{C4}$ and $\ce{C2}$ is equivalent to $\ce{C3}$. Stereoisomer D has four equivalent double bonds owing to two planes of symmetry. Given that both  stereoisomers C and D have the Z-configuration, one would would be hard pressed to know which one to draw if asked to do so. This situation is unfortunate.

Thanks a lot for this much needed help. But, How can two different stereoisomers ( C and D ) in this case , have the  same absolute  configuration around all 4 double bonds? Shouldn't there be a way to distinguish between the two?

You have hit the nail on the head! There is a way to do it unambiguously using a technique applied to inositol-like\ (http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/inositol.html) (hexahydroxycyclohexane) structures. Adoption of this method would have to be applied specifically to structures like tetraethylidenecyclobutane. Otherwise, this method would jeopardize pre-existing CIP rules. Since your post appeared, I have been in touch with one guru in this area and I have contacted IUPAC about this issue. I may post this issue in the future  as a question.


text:
Does the equilibrium constant of an ideal liquid mixture reaction depend on pressure?
I want to calculate the equilibrium constant K[T] for a the following reaction:
$$\ce{CH3OH(l) + C4H8(l) -> C5H12O(l)}$$
It is assumed that the mixture is ideal and the reaction takes place at 14 bar pressure.
K[T] is defined as the product of the thermodynamic activities to the power of the stoichiometric coefficient:
$$\prod_j {a_j^{\nu_j}} = \prod_j {\left(  \frac{\hat f_j[T, p, \mathbf{x}] }{ f_j^{\mathrm{ref}}[T, p^{\mathrm{ref}}, \mathbf{x}^{\mathrm{ref}}]} \right)}^{\nu_j}$$
The fugacity is given by Raoult's law,
$$f_i = x_i \cdot p_i^{*}$$
where x is the molar fraction and p* the vapor pressure, which are given.
How can I determine the reference fugacity, and where in the calculations do I use the 14 bar pressure at which the reaction takes place?

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The equilibrium constant for a chemical reaction is only a function of temperature. However, if you want to calculate the equilibrium constant $K$ via an equilibrium calculation, e.g. using the equation you posted, then we need  to find the pressure dependence on the fugacity of the species $i$ as a liquid.

We write the relationship between the fugacity of a pure species $i$ with the Gibbs energy, for two different situations distinguished by a zero superscript
\begin{align}
  g_i   &= \Gamma(T) + RT\ln(f_i)   \tag{1} \\
  g_i^0 &= \Gamma(T) + RT\ln(f_i^0) \tag{2}
\end{align}
Subtracting Eq. (2) from Eq. (1)
$$ g_i - g_i^0 = RT\ln\bigg(\frac{f_i}{f_i^0}\bigg) \tag{3} $$
Equation (3) links two different states:

The fugacity of a pure species at pressure $p$ and temperature $T$.
The fugacity of a pure species at pressure $p^0$ and temperature $T$. The reference state for a pure liquid, in the context of this equation, is generally chosen as that where the substance at temperature $T$ is a saturated liquid, and thus $p^0 = p^\mathrm{sat}$.

Now we use the fundamental thermodynamic relation for pure substances
$$ \mathrm{d}g = v\mathrm{d}p - s\mathrm{d}T \tag{4} $$
The evolution $(p^\mathrm{sat},T)\to(p,T)$ at constant temperature, for a species labeled with $i$, is carried out by integrating Eq. (4)
$$ g_i - g_i^\mathrm{sat} = \int_{p_i^\mathrm{sat}}^p
   v_i\mathrm{d}p \tag{5} $$
The liquid molar volume will change between these two pressures. For a liquid, we make the assumption that if the pressure range is not that large, it will remain constant and equal to the liquid saturated volume at the saturation pressure $ v_i = v_i^\mathrm{sat} $. Hence, Eq. (5) turns into
$$ g_i - g_i^\mathrm{sat} = v_i^\mathrm{sat}(p - p^\mathrm{sat}) \tag{6} $$
Combining Eqs. (3) and (6)
\begin{equation}
  RT\ln\bigg(\frac{f_i}{f_i^\mathrm{sat}}\bigg) = v_i^\mathrm{sat}(p - p^\mathrm{sat})
  \to
  \frac{f_i}{f_i^\mathrm{sat}} = \exp\bigg[\frac{v_i^\mathrm{sat}
  (p - p_i^\mathrm{sat})}{RT}\bigg] \tag{7}
\end{equation}
The right-hand side of Eq. (7) is the famous Poynting factor.

We return to the equilibrium constant. As stated, the mixture is ideal. We deal with liquids, so the fugacity coefficient of species $ i $ in a mixture is $\hat{f}_i = x_i\gamma_if_i=  x_i f_i $, because $ \gamma_i = 1 $. Summarizing the reaction in the fashion $\ce{A + B -> C}$
\begin{equation}
  K = \dfrac{\dfrac{\hat{f}_C}{f_C^\mathrm{sat}}}
  {\left(\dfrac{\hat{f}_B}{f_B^\mathrm{sat}}\right)
   \left(\dfrac{\hat{f}_A}{f_A^\mathrm{sat}}\right)} =
  \dfrac{\dfrac{x_Cf_C}{f_C^\mathrm{sat}}}
  {\left(\dfrac{x_Bf_B}{f_B^\mathrm{sat}}\right)
   \left(\dfrac{x_Af_A}{f_A^\mathrm{sat}}\right)} \tag{8}
\end{equation}
Now we use Eq. (7) for the three ratios in Eq. (8)
\begin{equation}
  \boxed{K =
  \dfrac{x_C \exp\bigg[\dfrac{v_C^\mathrm{sat}
  (\color{blue}{p} - p_C^\mathrm{sat})}{RT}\bigg]}
  {x_A \exp\bigg[\dfrac{v_A^{sat}(\color{blue}{p} - p_A^{sat})}
  {RT}\bigg] \cdot
   x_B \exp\bigg[\dfrac{v_B^{sat}(\color{blue}{p} - p_B^{sat})}
  {RT}\bigg]}} \tag{9}
\end{equation}
In blue is emphasized the pressure dependence in the fugacity.
Eq. (9) shows that additional information is needed when we deal with liquid-phase reactions:

The saturation pressure $p_i^\mathrm{sat}$ for every species.
The saturated liquid molar volumes $v_i^\mathrm{sat}$ for every species, evaluated at $p_i^\mathrm{sat}$.

One book is this one in chapters $10-13$, and this one, chapter, $1$, $2$, and $7$.

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text:
What form of energy is produced by 2H2 + O2 -> 2H2O reaction?
I'm curious about are the details of exothermic nature of 2H2 + O2 -> 2H2O reaction. The explanations I've seen just use a term "energy release" without specifying what type of energy it is. Assume that we have a perfect mixture of H2 and O2 in a closed vessel and produce a spark, which will initiate the reaction. What is the mechanism, which increases the velocity (temperature) of the molecules participating in the reaction?

And here I'm so tempted to make a snarky comment... This question was already asked in some form, probably more than once. And in similarly confusing way.

If a molecule has surplus energy from reaction, its needs to be removed, lest reaction gets reversed. But it only takes a collision with another molecule - that's just what it takes. And all kinds of translational, vibrational energy, etc. are in it. Some light too as can be seen.

https://chemistry.stackexchange.com/questions/15250/how-do-breaking-chemical-bonds-turn-into-kinetic-energy https://chemistry.stackexchange.com/questions/63040/do-the-products-in-exothermic-reactions-have-higher-kinetic-energy-than-the-prod https://chemistry.stackexchange.com/questions/922/what-is-the-energy-package-released-to-the-outside-world https://chemistry.stackexchange.com/questions/44546/what-decides-whether-a-reaction-releases-light-or-heat

You're asking two separate questions.

What form of energy is produced when combining hydrogen and oxygen?
How does that become thermal energy?

The answer to 1. is that energy can be released in different forms, such as electricity in a hydrogen-oxygen fuel cell, or light. An efficient fuel cell produces little heat.
Question 2. is best considered through thermodynamics. Theoretically, it might be possible to predict the position and velocity of a few unrestrained molecules of $\ce{H2}$ and $\ce{O2}$ combining to form $\ce{H2O}$ (ignoring quantum uncertainty). However, en masse, this is effectively random. Ah... and what is the random motion of particles called?

Answer 1 does not describe the environment in my post. Thermodynamics says nothing about how the energy of chemical bonds is converted to the kinetic energy of H2O produced by this reaction and in what form. It models the gas as a bunch of balls undergoing elastic collisions. How is the kinetic energy of 2 H2 and 1 O2 set of balls suddenly increased after they meet and magically transmute into 2 H2O balls? Can elastic ball model describe this reaction without using a sudden magic dv for the H2O balls?

@PaulJurczak, what needs understanding? Electrons shift, the atoms to which they're attached move due to EMF.

That's better. So are we talking about H2O molecule vibrating and transferring its newly acquired energy via elastic collisions with other molecules?

@PaulJurczak Well, maybe instead of criticising, you should edit your question to actually ask what you want in a clear way? Using terminology like relaxation of intermediates instead of "form of energy" etc.

@Mithoron I have no idea what made you think I'm criticizing. My goal is to understand the physics underlying this reaction. I'm posting within the constraints of my knowledge. This is the first time I encountered the term relaxation of intermediates, so obviously I couldn't use it forming my question. I was hoping for a high school physics level explanation, Feynman style.

The form of energy depends on the reaction conditions but the molecular mechanisms are complex
When water is formed from hydrogen reacting with oxygen, a large amount of energy must be released (that's thermodynamics). How it gets released depends on the conditions of the reaction.
When oxygen and hydrogen are reacted in a controlled way on a catalytic surface inside a fuel cell, much of the energy can be extracted as electricity. But the detailed mechanism involves complex reactions on a specific catalytic surface.
When hydrogen and oxygen gas is ignited, the reaction is fairly uncontrolled and the energy emerges as both light and heat. How this happens is not easy to describe in a simple, single reaction, not least because in a gas there are many molecular collisions happening all the time and it isn't just about two molecules colliding and yielding the product. Plus, the reaction will usually have multiple steps.
For example, a hydrogen molecule might collide with an oxygen molecule to yield an H• radical and an HOO• radical (I'm making up possibilities for illustrative purposes rather than trying to give a realistic description). Both species might have more kinetic energy than the original molecules so will move faster, but they are also, both, highly reactive so further collisions will often yield more reactions, some of which will release more energy. In some cases, the reactive species generated will decay spontaneously, emitting light. In many other cases they will bang into other molecules distributing the excess energy as kinetic energy throughout the remaining gas molecules. This will continue until only stable molecules are left.
The point of this is that excess energy from the formation of new molecules can be released as light or can be distributed very rapidly as kinetic energy to other molecules in the gas (and more kinetic energy in the gas is heat). At a molecular level, the excess energy from a specific chemical reaction can appear as rotational energy or vibrational energy in bonds or as kinetic energy from faster movement of the molecules. But the specific amounts of energy will distribute very rapidly across all the types of energy because molecular collisions tend to redistribute energy very, very rapidly to an equilibrium across all the types on energy in all the gas molecules.
Ultimately, at least in uncontrolled reactions, this means we see the excess as heat.

rotational energy or vibrational energy in bonds this makes sens to me, but kinetic energy from faster movement of the molecules must be secondary to it. A single event of molecules A and B colliding and merging into a resultant molecule AB does not increase total kinetic energy of the system A and B, modeled with elastic balls. Only collisions of AB with other molecules in the system can transfer some of that newly acquired rotational and vibrational energy into kinetic energy. Is that accurate?

@PaulJurczak If two molecules collide but a bond breaks, releasing a lot of energy, then one of the resulting fragments may carry away a lot of kinetic energy (imagine a ball and spring model where the collision cause a highly compressed spring to unwind). But all the types of energy rapidly equilibrate anyway through further collisions.

If two molecules collide but a bond breaks, releasing a lot of energy, then one of the resulting fragments may carry away a lot of kinetic energy - this scenario is easy to model, but what I'm curious about is colliding molecules fusing into a single one. This scenario is not compatible with the unwinding spring model, since there is only one molecule and the other end of the spring has nothing to "push" against.

@PaulJurczak the "unwinding spring" is not a complete boding model but a very simple metaphor. It doesn't pretend to explain all possible reactions. BTW can you think of any strongly exothermic reactions involving two molecules just sticking together as the final step?

can you think of any strongly exothermic reactions involving two molecules just sticking together as the final step? I'm not a chemist, so I will not volunteer with an example here. I realize that many (most) chemical reactions proceed in forward and reverse directions simultaneously, converging towards an equilibrium. I would just like to zoom in from a large scale thermodynamics model to interactions between individual molecules leading to their increased kinetic energy. Trying to understand it from the first principles.

Thermal energy is released from the  reaction $\ce{2H2 + O2 -> 2H2O}$. Before the reaction takes place the system of $\ce{H2}$ and $\ce{O2}$ molecules are in a metastable state:

Given enough energy > activation energy the system gains enough energy to overcome the potential barrier and falls into a state of lower energy than the initial state.

Yes, but I'm asking about the physics of increasing the kinetic energy of H2, O2 and H2O molecules inside the vessel. What form of energy is being transformed into kinetic energy and how?

You can try to put your sample in a heat bath,due to the temperature difference between the heat bath and the initial temperature of the sample, thermal energy will be transferred from the heat bath to your sample enabling the reaction.

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text:
How can atoms get very close together even without direct covalent bonds between them?
Generally speaking, because of the Pauli exclusion principle, it’s difficult for the electron clouds of two atoms to overlap unless they are covalently bound together. The degree of overlap can be estimated with the van der Waals(vdw) radius. When the distance between two atoms is smaller than the sum of their vdw radii, the repulsion between them increases steeply.
However, atoms within the same molecules can be much closer than the vdw radii implied. For example, the half distance between the hydrogen atoms in CH4 is 0.89 angstrom, which is 74% of the vdw radius of hydrogen atoms. The half distances between the fluorine atoms and chlorine atoms in CF4 and CCl4 are 73% and 82% of their vdw radii, respectively (data sheet: C-H: 1.09, C-F: 1.32, C-Cl: 1.77, vdw H: 1.2, vdw F: 1.47, vdw Cl: 1.75). So how can these atoms get so close together despite the lack of covalent bonds between them? My guess is that vdw radius measures the distance between atoms of separate molecules. Because neutral molecules are bound together by the very weak dispersion force while in CX4 the four terminal atoms are tightly bound to the central carbon atom, they can overcome much larger repulsion between them.
My second question is whether there is a lower limit on how close atoms can get together. For example, C60F60 is much larger than CF4, which means the angles between adjacent C-F bonds is much smaller in C60F60. If the lengths of C-C bonds and C-F bonds are maintained, the distance between fluorine atoms may be unacceptably small, which may slightly stretch the C-C and C-F bonds to accommodate the fluorine atoms. Can we predict the stability of perhalogensted compounds (such as the instability of polytetrachloroethylene) by the atomic distances?

Consider the van der waals interaction equation for non ideal gases, there is a factor alpha and beta for molecule-molecule interactions and for their relative volume to their container.

Who told you that (I copy) ... "When the distance between two atoms is smaller than the sum of their vdw radii, the repulsion between them increases steeply" ?

I saw diagrams like these all the time https://p3test7.uni-freiburg.de/researchprojects/repulsive-van-der-waals-force

https://en.wikipedia.org/wiki/Buckminsterfullerene#Halogenation Not an answer but a comment on the effect of fluorination on buckyballs...


text:
Reaction of aluminium chloride with alkali metal
I wonder if we can get aluminium metal by adding alkali metal (ex. Li,Na) to aluminium chloride solution?
or
maybe what will happen if we mix aluminium chloride hexahydrate with sodium metal?

I am not sure about alkali metals but you can try manganese: https://patents.google.com/patent/US3900312A/en

If there is water in solution or in the hydrate, hydrogen would be released with priority.

Adding some sodium metal to any aqueous solution will only produce the reaction $\ce{2 Na + 2 H2O -> 2 NaOH + H2}$, whatever the nature of the solute (containing Al ions or not). The same reaction happens with aluminium chloride hexahydrate. To get an aluminium production with metallic sodium, the aluminium chloride has to be anhydrous. This was discovered by Davy in the years 1800 - 1810.


text:
Halogenation of organic acid derivatives
Chloropicrin can be produced through the reaction of sodium hypochlorite with nitromethane. On the surface, this seems to me to proceed through a mechanism analogous to the haloform reaction, with one notable difference being that, in the case of the generation of chloropicrin, the C-N bond is not cleaved.
On a related note, I found a paper that claims that acetate esters are also capable of undergoing the haloform reaction.
Because of these two facts, I am wondering if a process analogous to the creation of chloropicrin can occur with other organic acid derivatives. As an example, could methyl methanesulfonate react with bleach to give methyl trichloromethanesulfonate? If yes, are there any other organic acid derivatives for which this process can occur?
Source:
Nagata, D., & Nishiwaki, N. (2021). Are acetic acid derivatives really negative to the iodoform test? SN Applied Sciences, 3(9). doi:10.1007/s42452-021-04777-0

Mesylate? Not really, but any C-acid with nucleophilic conjugated base, and without electrophilic centers...

@Mithoron Could you explain why, preferably with something to back up your claim?


text:
Does aliquot matter for final concentration?
I am wondering if my calculation for getting the final concentration is right:
40 mL of solvent A was taken to extract a sample xy. From that 30 mL were taken and evaporated to dryness. Afterwards, 1 mL of a solvent was added for reconstitution. 100 uL were taken and were diluted to 1000 uL, analysis was performed with external calibration.
How to calculate the final concentration?
My thoughts were the following:
Take the obtained concentration from the calibration curve and multiply by the dilution factor 10 since 100 uL was diluted to 1000 uL. Now, I am struggling a bit. Is this already the final concentration or should I also consider the factor from taking 30 mL out of 40 mL?
Thank you very much for your ideas on this question!

It is always good to summarize and/or complement the verbal description by algebra. // Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. Advantage is: Focus on principles, Easier spotting of mistakes, Improved orientation, Q/A reusability and the permanent value.

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There is good chance that you will during formulating of expressions realize you need not to ask any more. BTW, you have skipped the concentrating step 30 mL -> 1 mL

Did you mean the original concentration of the sample? The final concentration of the aliquot is whatever you calculated using the curve.

Yes, I mean the original sample concentration @StanislavBashkyrtsev

$$c_\mathrm{r} = \frac{V_\mathrm{d} }{ V_\mathrm{i}} \cdot c_\ce{s}$$
$$c_\mathrm{e} = \frac{V_\mathrm{r} }{ V_\mathrm{a}} \cdot c_\mathrm{r}$$

$c_\mathrm{e}$ is analyte molar concentration in extract
$c_\mathrm{r}$ is analyte molar concentration in reconstituted extract
$c_\mathrm{s}$ is analyte molar concentration in the final sample,
being compared to the calibration curve.
$V_\mathrm{a}$ is volume of the extract aliquote
$V_\mathrm{r}$ is volume of the reconstituted extract
$V_\mathrm{i}$ is volume of the extract injected for final dilution
$V_\mathrm{d}$ is volume of the final diluted solvent

$$n_\mathrm{e} =  V_\mathrm{e}\cdot c_\mathrm{e}  = V_\mathrm{e} \left(\frac{V_\mathrm{r} }{ V_\mathrm{a}} \cdot \frac{V_\mathrm{d} }{ V_\mathrm{i}} \cdot c_\ce{s}\right)$$

$n_\mathrm{e}$ is the analyte molar amount in the whole extract
$V_\mathrm{e}$ is the total extract volume

All can be rearranged for mass concentrations and masses.
$$m_\mathrm{e} =  V_\mathrm{e}\cdot \rho_\mathrm{e}  = V_\mathrm{e} \left(\frac{V_\mathrm{r} }{ V_\mathrm{a}} \cdot \frac{V_\mathrm{d} }{ V_\mathrm{i}} \cdot \rho_\ce{s}\right)$$

$m_\mathrm{e}$ is the analyte mass in extract
$\rho_\mathrm{e}$ is the analyte mass concentration in extract
$\rho_\mathrm{s}$ is the analyte mass concentration
in the final sample, being compared to the calibration curve.

Thank you. When considering aliquot should there be also not a factor how much of the aliquot was taken, so for this example 3/4?

Some things matter for concentration but not amount, some vice versa. The fundamental law of mass/energy conservation helps a lot.

Is it also necessary to consider a third equation like:
$$c_{fe} = \frac{c_{e}* V_{e}}{V_{fe}}
$$
with $ c_{fe}$ as finale analyte molar concentration in extract
with $ V_{fe}$ as total original volume of the extract

$V_\mathrm{e} = V_\mathrm{fe}$ so $c_\mathrm{e} = c_\mathrm{fe}$

30 ml of extraction solution was dried and reconstituted to 1 ml, then diluted 100 ul to 1000ul so the concentration is now:
Extraction solution × 30 ml/1 ml × 100ul/1000 ul = Concentrated 3x
      ----> Sample run in analytical instrument ---->

What ever result you get from your analysis as grams analyte/ml must be divided by 3 to get the concentration of analyte in the extraction solution.
Now to find out total grams of analyte extracted in your 40 ml of extraction solution:

grams analyte/ ml × 40 ml = grams analyte extracted

Now, in order to determine the original concentration of whatever solution you extracted from (assuming 100% extraction efficiency$^1$), you take grams analyte and divide by volume of original solution before the 40 ml extraction solution was mixed with it (also assuming a 2 phase extraction with a separatory funnel).
So you need to know your original volume of solution.
$^1$ good luck with that, HPLC may be a better choice

It's good to keep in mind the distinction between amount and concentration.
Take the step where you go from 40 mL to 30 mL. Since you're not diluting, you're not changing the concentration of the solution, but you are changing the amount. The way you figure out the reduced amount of compound you have in the new sample is by looking at the ratio of volumes.
But take the 30 mL to 1 mL step. You (hopefully) haven't changed the amount of substance you have when you evaporated and reconstituted things, but you have changed the concentration. To figure out the new concentration, this is where you pull out the old c1 * v1 = c2 * v2 -- this isn't a magic invocation, it's just a representation of the fact that the amount of substance hasn't changed. (The amount of substance in a solution being the concentration times the volume - you do the calculation before and after and set them equal to each other, as the amount of substance hasn't changed.)
Your whole process can be thought of as a series of changes in the amount and concentration:

40 mL -> 30 mL -- Constant concentration; change of amount
30 mL -> 1 mL -- Constant amount; change of concentration
1 mL -> 100 uL -- Constant concentration; change of amount
100 uL -> 1000 uL -- Constant amount; change of concentration

So what you can do is take the value (the assay concentration) for the final solution and then work backwards through the steps, converting to concentration or total amount as necessary.
If you think about it stepwise, making sure you understand whether each step is a constant concentration step or a constant amount step, you can apply the relevant conversion factors backwards for each step until you reach the value you want. The benefit here is that this process is robust to (and forces you to think about) whether you want the amount of substance in the original sample or the concentration in the original sample (or if, for some reason, you need the amount in your 30 mL sample) - just add in an extra conversion at the appropriate location. It also generalizes to more complex sampling procedures.
(There's certainly potentials for shortcuts and rules of thumbs here to speed up calculations -- but if you're confused or uncertain, it's best to go back to first principles and work things out stepwise.)

Thanks a lot! So when I want to have a final analyte value in e.g. mg/g, it's necessary to use the molar amount multiply by analyte molar mass and then divide it by sample mass which was used in extraction?


text:
In which orbital is the positive charge of central nitrogen present in diazomethane?

Where is the positive charge of central nitrogen located?
It cannot be in the pure p orbital because in that case there will only s-orbital left for hybridization which is not possible. If it's present in the sp hybrid lobe which is also not possile because both sp hybrid orbital are being overlapped and formed sigma bonds with neighbouring atoms.
So in which orbital is this positive charge present?
More detail:
In ground state, valence shell electronic configuration of nitrogen was: 2s2 2p3. After internal excitation, it became 2s1 2p4. Two unpaired electrons made π bonds with neighours and the rest one 2s1 and one 2p2 orbitals hybridized to form two sp hybrid orbitals. Now one electron is lost from one of the sp hybrid orbital and so now both sp orbitals have one unpaired electron each. Now both  of these hybrid orbitals made sigma bonds with neighbours. After overlapping with neighbours the sp hybrid orbitals cannot contain positive charge.

"Positive charge being present in an orbital" is weird formulation. Orbitals have zero charge at the best if empty.
But as all 4 orbital pairs contributing to 4 bonds have 2 electrons, it cannot be even said a missing electron in one of them caused the positive charge.
It is rather that the middle N provided both electrons for one of its bonds and the other N forms another free electron pair.
By other words, the middle N uses 2 its electrons for the double bond with C and 3 its electrons for the double bond with the other N. The other N formally forms from its 5 electrons the 2 free electron pairs. The remaining unpaired electron and 3 electrons from the middle N form the double N=N bond. This all causes the charge shift.


text:
How strong (molarity) does NaOH solution need to be to saponify fats?
I have some large cast iron hob grilles of an awkward shape, and want to clean the burnt-on fat from them using NaOH solution. I have two options

Use a small amount of a strong solution, with a brush or turkey baster to keep them wet, while they sit on a plastic cement-mixing board with a raised edge, or
Use a large amount of a weak solution, in a plastic dustbin or suitable trough, to immerse them either fully, or partially and turn them over

The former would require a lot of hands-on time, and lots of opportunity for splashes, all the worse for being of a strong solution. The latter I could just leave to sit for 24 hours, as long as I knew the solution was strong enough to work, or at least work fast enough. I don't really want to buy, or dispose of, large amounts of NaOH.
I've search around but only found the SV, Saponification Value, for various fats and oils. This will at least give me a lower bound for how much I need, if I can estimate the amount that's there.
I have the gut feeling that there is a threshold pH below which saponification won't occur. However, the pH of even 0.01M NaOH is around 12. So my gut says concentration won't matter, and that any practical concentration of NaOH will work, albeit just affecting the speed of reaction.
I have found sources that suggest that 2.5M (10% by weight) will work in 10 minutes. I would expect the speed to vary simply as 1/concentration.
Are my suppositions on threshold and speed correct? I don't really want to spend a day on tens of litres of a weak solution that doesn't work.

There is a commercially available product called "Oven Mate" that is KOH in polyethylene glycol gel which is great for this. You paint it on, leave it then rinse off. I use it on oven trays and interiors. Works well.

And there is also the possibility of using a propane torch to complete the combustion process. Then just brush off the ash.

2.5 M might be a bit strong. 1.0 M is 40 g NaOH per liter of water, or around 4% solution.
Start with 1 % solution and leave it soak.  As with oven cleaners, a little heat should help move things along, like around 150 F.  No need to boil it.
I would check it every 2-3 hours.  It might be done by then.  A little dish soap, rubber gloves, and an abrasive pad should finish the job.
If you need to neutralize, some phosphoric acid from the local hardware store may be of use.  Please be careful and wear eye protection.  Caustic solution is very corrosive to skin as well.


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Strange intermediate product when benzaldehyde undergoes Cannizzaro reaction
In the school laboratory, a friend tried the famous Cannizzaro reaction for benzaldehyde. Some $\ce{NaOH}$ was added to benzaldehyde, and the solution was stirred for 30 minutes. What came as a surprise was that, after this time interval, a big, bright yellow, gooey-looking blob appeared in the flask.
As can be seen, the blob does not dissolve upon shaking. Whatever reaction occurred here was probably reversible, though, as the blob burst after the solution was transferred into a separation funnel, and the experiment went on to give normal products and reasonable yields, without any obvious anomaly.
Does anyone have any idea about what this strange intermediate product might be, or what might have caused it?

Robert DiGiovanni's answer comes close. Benzyl alcohol can indeed precipitate during the intermediate stage, but it is colorless. Because benzyl alcohol has some weak acidity, it can be partially deprotonated by a base as strong as sodium hydroxide, giving $\ce{C6H5-CH2-O^-Na^+}$. This indeed can impart a yellow color according to Sigma-Aldrich, which sells this salt as a solution in the alcohol. When the sodium hydroxide is neutralized upon workup, the benzyl alcohol fully resumes its neutral form $\ce{C6H5-CH2-OH}$, as if nothing untoward had happened.

@RobertDiGiovanni thanks for the correction.

pKa benzyl alcohol is 15.4 as read here.  But this is plausible.  Perhaps the O- would be on the surface of the blob, into the aqueous layer.

The O- from benzyl alkoxide does not conjugate with the ring. The Cannizzaro reaction involves a deprotonated acetal and hydride transfer. It is possible that a doubly deprotonated acetal with the negative ions effectively ion paired to Na+ could enable an electron pair from the ring to supply the impetus for the hydride transfer. This would result in a quinoid compound that would revert to benzoic or possibly hydroxybenzoic acid upon neutralization. This scenario fits the indicator story given above.

@jimchmst would such a quinoid compound be stable enough to form in this case? Would be nice if there are similar examples in the literature, I don't quite understand the mechanism described in the comment.

I missed the impact if the intervening CH2 group. Last paragraph removed.

I just tried to figure out a way to get a conjugated structure that would absorb blue light. the answer might be to buy some yellow benzylalkoxide  from Sigma-Aldrich and isolate the yellow color, Something is happening. It is possible that things happen that are not yet in the literature or are at least hard to find. The Cannizzaro Rx involves a hydride transfer.

Benzyl alcohol has a lower solubility (4 g/100 ml) than sodium benzoate
(63 g/100 ml) or benzaldehyde (7 gram/100 ml) in water.
The presence of NaOH contributed to the "salting out" of the more nonpolar benzyl alcohol, plus some other impurities, forming your "blob".
Benzaldedyde and benzyl alcohol are also (surprisingly) slightly denser than water.

Would a difference of 3g/ml (between benzyl alcohol and benzaldehyde) result in such a drastic change in colour and texture?

@jonohuang that certainly has to do with what was added to the flask and whether or not they reran it more dilute (solubility issues).  But I agree with Oscar Lanzi that there may be some deprotonation of benzyl alcohol.

Since benzyl alkoxide is colorless the yellow color must be an intermediate in the reaction or possibly something that can be formed to make the alkoxide yellow probably from oxidation. The colors are possibly related but might not be. Step one: repeat the reaction identically; same chemicals. Step two: Repeat the reaction with carefully purified benzaldehyde and NaOH. If it was an impurity then there is an additional product or an analysis of the original benzaldehyde is in order.

"In the school laboratory", yes, some additional optimization work may be in order, lest the "blob" escape the laboratory and ...


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Can it be said that an electron transfer between two neutral atoms comes from the neutral atomic electric field?
Let us consider the example of Na and Cl. Both atoms are called "neutral" simply because they have the same total number of protons as electrons, yet they have a very weak electric field which is hardly measurable. The electron transfer between two neutral atoms is ultimately driven by the electrostatic force.
Arguments like: This is because atoms and molecules are always trying to achieve the most stable, lowest energy state that they can don't really adress my problem, "lower energy" is not a fundamental force but the outcome of a process caused by electrostatic force.
So, in the end, can we say that this very weak electric field of a neutral atom is the main cause that it can attract electrons of other neutral atoms? Simply because the electrons are not classically like point charges in the same place as the protons?
Is there a more detailed book that deals with this in particular? Because I think this is fundamentally important.

While atoms do have zero net charge, the charge distribution is not generally spherically symmetric. And even it originally was for some atoms, like for 2 isolate H atoms, it stops to be, when 2 atoms approach. As the electrostatic potential stops to be spherically symmetric around their nuclei.

somewhat relevant: https://physics.stackexchange.com/questions/267371/is-there-an-electric-field-around-neutral-atoms

@DrMoishePippik what should "quantum force" even mean? Thats not a fundamental force.

@Poutnik so can i really say that it is ultimately initiated by the weak electric field around a neutral atom and the more symmetrical the distribution the weaker this (already weak) electric field becomes? You meantioned it would break symmetrie a bit if 2 atoms approach, but this is somehow circular, because one of the 2 atoms must have been asymmetric by itself from the beginning.

It may help if you stop thinking about atoms as natural balls and start thinking there are positive nuclei and probability clouds of negatively charged electrons.

@Poutnik That could be, I have a picture in my head where there is a nucleus and a diffuse structure which over time is a distribution of many locations of an electron in the form of its orbital shape.

@Poutnik Theoretically, I could imagine a perfect symmetric electron distribution around a proton as a gaussian sphere. The perfect negative distribution on the surface of the sphere would be quantitatively equal to a negative electron charge at the point of the proton and thus really perfectly cancel the electric field. Such an atom would of course never react with anything in its environment.

Only if the electrostatic potential around proton is perfectly spherically symmetric. Is it?

@Poutnik no, of course not. That was just a hypothetical thought.

When your body hits some obstacle, you can tell yourself it was very weak, hardly measurable electrostatic force of neutral atoms. It may help the pain.

At molecular level, quantum effects become as, or more, important than electrostatic forces. Consider that if only electrostatic forces were acting on a hydrogen electron, it would fall into the nucleus. If the electron were orbiting the nucleus, it would radiate energy (being accelerated) and then fall in. See https://www.eetimes.com/quantum-wave-functions-come-alive-may-the-bohr-model-rest-in-peace/

You don't need to consider quantum mechanics to arrive at a rudimentary understanding of how two neutral objects containing charges can attract, but it is important nonetheless to understand that quantum mechanical properties are very different from classical ones, as explained for instance in answers in this post. In what follows I start with the classical picture.
Two neutral objects can attract each other because the cancellation of attractive and repulsive contributions from the positive and negative charges is not exact, particularly at close distances, and varies with distance. At long distances the total field of one object becomes very similar to that of a neutral point particle, and is therefore very small and nearly invariant with distance between the objects. Therefore both field and force are negligible. At close distances those contributions don't cancel, in particular because the strength of the field falls with distance. This is why two (real) dipoles can attract (ideal dipoles can only induce mutual alignment). To look into this in more detail you can consult the EM textbook by Griffiths.
What Fritz London et al accomplished was to place these basic ideas from EM on a more solid QM footing by introducing concepts such as the atomic polarizability. This was in part necessary because in the absence of point particles you need a different way of computing the field generated by the electrons.


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Role of spin of electron in splitting of orbitals
What will happen if there are electrons with opposite spins present in the orbitals of ligand and metal atom during the splitting of orbitals in crystal field splitting?
Magnetic fields of two electrons with opposite spins should cancel out each other, thus its net magnetic field would be zero, there would be no repulsion between orbitals of the ligand and metal atom, and splitting of the orbitals $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ would not happen.

It's not the magnetic field due to spin that causes the splitting. It is the electrostatic field, coupled with the lowering of the symmetry from spherical to octahedral or whatever, that is behind the splitting of the d orbitals.

The repulsion between two electrons due to being in the same spin-state is only a part of the net repulsion between the electrons. Coulombic repulsion is present between pairs of electrons with same and opposite spins. The Hamiltonian includes the Exchange integral, the Coulombic integral, etc. In certain cases, it is even favorable for electrons to occupy parallel spins (see magnetism). The only condition in which Pauli exclusion principle applies is when the electrons are present in the same orbital, having all other quantum numbers identical.


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battery using zinc, copper, and vinegar
I filed one side of a penny to reveal the zinc; then I bent the coin and dropped it into a cup of white vinegar. Shortly afterward, many tiny bubbles were forming on the zinc side. Later on, I noticed a black residue on the zinc side. What might be the reaction that occurred? [I believe that the gas was either H2 or CO2 gas. I also believe that the black residue was either CuO  or carbon] {Zn + Cu + H ion + CH3COO ion --> ???}

The zinc reacted with the acetic acid in vinegar, with hydrogen gas being evolved. The black stuff is not carbon or CuO. Just some kind of crud.

Copper surface was probably a little bit oxidized into $\ce{CuO}$. Dipped into vinegar, this oxide is quickly dissolved by the acetic acid, producing $\ce{Cu^{2+}}$ ions in solution $$\ce{CuO  +  2 CH3COOH  -> Cu^{2+}  + 2 CH3COO^-}$$Then, zinc may react with these copper ions, producing copper metal which looks black if the dimensions of the grains are small enough.
$$\ce{Cu^{2+} + Zn ⟶ Cu + Zn^{2+}}$$

Thank you, Maurice.


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In gas law equations, if volume and temperature are directly proportional, why can't we write the equation as temperature divided by volume?
In gas law equations, volume, temperature, pressure and amount of substance may vary. In the simpler equations, like Charles' and Amonton's, temperature and pressure/volume are directly proportional. Why are the equations specifically with the temperature in the denominator, like this one with the reciprocal being untrue ( I'm asking a conceptual, not mathematical question)?
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

We can, and therefore there is a fallacy (a false assumption) in your question.

Interesting! This bothered me long time ago in elementary school. Anyway, later I learned elsewhere that the mathematical convention is to write the dependent variable first in ratio and proportion with a proportionality symbol. Such relations appeared all the time in chemistry and physics.  It is very sad that modern high-school algebra has eliminated entire chapter on ratio and proportion properties (esp. see invertendo property*).
So, $$ V \propto T $$ which means
$V$ is a function of $T$, not the other way round. Temperature dictates what the volume of gas will be.
$$ V =c  T $$
where $c$ is a constant of proportionality. Following the graphing convention, this means that one will plot $T$ on the x-axis and volume $V$ on the y-axis. This is nothing but a linear equation with zero intercept when $T$ is in Kelvins.
We can also write,
$$\frac{ V_1}{T_1} =c  $$
and
$$\frac{ V_2}{T_2} =c  $$
Since both fractions are equal to $c$,
$$\frac{ V_1}{T_1} =\frac{ V_2}{T_2}  $$
*Invertendo property of ratios tells us that a/b: c/d is the same as b/a:d/c.
$$\frac{ T_1}{V_1} =\frac{ T_2}{V_2}  $$
is also correct. It is not wrong algebraically. Here you don't care which one is the independent variable.
As another unrelated example is Ohm's law:
One can say, $$ I \propto Voltage $$ which means current depends on the applied voltage.
So, $$ I = c Voltage $$
$c$ turns out to be the inverse of resistance $R$.

"It is very sad that modern high-school algebra has eliminated ..." Has deletion of this topic from the curriculum been universal? I remember attending a weekend course that taught the topic. It was.... not the most interesting subject, but maybe useful. I would guess there is increasingly more to learn as more info is accumulated, and this is covered in passing with other topics. Teaching is riddled with topics that could be taught more deeply to great benefit but aren't, time being in short supply.

And I learned about rational numbers early on, as I would guess many do. That would be the basis for developing an understanding of ratios.

@BuckThorn, I think this chapter has been universal for high schools. I am talking about grade 10 (high school algebra). You must have noticed questions like these and dilution problems lately. It reminds of its need. These are all ratio and proportion problems. I you teach, ask your students if they recognize the proportionality symbol. They might not identify at all or call it alpha.

I suspect the OP might be confused as to why the ideal gas law cannot be derived as T/V = nR/p as opposed to V/T = nR/p . I am not sure about whether it is a question with grasp of algebra, but it might be.

I think this is functional notation, $y=f(x)$. Dependent variable(s) first. V is proportional to T and V is inversely proportional to P.

Ok, but I still have difficulty seeing how that aspect has anything to do with how you would relate the ratio V/T to the remaining variables. See also the comments under Poutnik's answer. But I suppose you might be right.

@BuckThorn, the OP writes "Why do we solve for V/T and not T/V if the volume and temperature are directly proportional? Answers will be obviously be very different". This is obviously incorrect.

I interpret that to mean that the OP thinks the ideal gas law would look differently. I agree that there is a problem there with understanding the algebraic solution of the problem, and that $V/T = c \implies T/V = 1/c $

Measuring the volume regardless of P and T defines the amount of gas. Adding P or T defines the state.  PV = nRT is actually a description of the internal energy of a [ideal] gas. It says that the energy is a function of the temperature and the amount of gas present.

@jimchmst, This is a good reasoning. I like it!

The questions in the title and the body ask different things.
We cound write $\frac{T}{V}$ instead of $\frac{V}{T}$, but there is, perhaps unwritten, convention using the cause as the denominator.
The equation $\frac{V_1}{T_1}=\frac{V_2}{T_2}(=\frac{nR}{p})$ is obviously valid only for ideal gases ($pV=nRT$) at isobaric conditions.

Actually, the question is one and the same. Why do we solve for V/T and not T/V  if the volume and temperature are directly proportional? Answers will be obviously be very different.

"Why are being untrue" gives the other question different sense.

@suse You are not solving for $\frac{V}{T}$, you're using three known variables to calculate an unknown fourth, or perhaps the volume ratio from the temperature ratio. Proportions don't care about fraction position as long as correct direct or inverse relationship between variables is preserved.


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Hydrogen bond in water vapor
As the temperature increases and liquid water changes to gas, are ALL the hydrogen bonds broken or they are just weaker?

Does this answer your question? Hydrogen Bonding in Water Phase Changes

Fun fact: The lack of hydrogen bonds explains why steam causes much worse burns that boiling water. Steam contains all the energy used to break the hydrogen bonds in water, so when steam hits your face you first absorb the energy the steam has taken up from breaking the hydrogen bonds it its liquid state. Then, in an exothermic reaction, steam is converted into liquid water and heat is released. This heat adds to the heat of boiling water as the steam condenses on your skin.

Under commonly encountered conditions in industrial applications, essentially all hydrogen bonds are broken. Only at high pressures is hydrogen bonding likely to occur.
What happens is something of a Catch-22. If hydrogen bonding in water is significant, then it tends to form cross-linked structures because each molecule can both donate and accept two bonds, so the hydrogen bonding tends to form cross-linked structures that condense into a liquid. So to generate steam at near-atmospheric pressure you need enough thermal energy (temperature) to essentially break all the hydrogen bonds at once. Only at high pressure, where the gas becomes denser and more like a liquid, can hydrogen bonding become evident in the gas phase.
We may compare this with hydrogen fluoride, where chainlike structures form instead of cross-linked ones because each molecule donates and accepts only one hydrogen bond. The chainlike structures can be supported in the gas phase, so hydrogen fluoride gas can show hydrogen-bonded oligomers at relatively low pressures. See here.


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Diffusion rate of drugs - Ficks model?
I have a question about diffusion rate of a slow release drug. The drug I've been taking has an extended-release formulation and contains beads encapsulated in a capsule.

The beads are coated with a polymer (ethylcelulose) and are of different diameters - between 0.8mm and 1.7mm. I wonder if the inconsistency in size is intended to help release the drug at different rates or if it's simply a quality issue.
I'm not sure if the Ficks model is applicable here. The model seems to suggest that an increased surface area increases the diffusion rate. Assuming the concentration and the coating thickness are constant across all beads, does that mean larger beads are supposed to be faster releasing and smaller beads slower releasing?

Are you sure it is diffusion limited, not dissolution?

@JonCuster Yes. According to the FDA "Drug release is controlled by diffusion through the coating membrane on the spheroids and is not pH dependent"


text:
Reaction rate derivation: amount of substance vs concentration
Let's say you have the reaction
$$A + B \; \rightarrow \; C.$$
I would like to understand why $\frac{\mathrm{d}[C]}{\mathrm{d}t}$ is proportional to $[A][B]$. I can see why
$$\frac{\mathrm{d}n_C}{\mathrm{d}t} \propto \frac{\text{number of collisions between } A \text{ and } B}{\text{time}} \propto [A][B],$$
but I cannot see why
$$\frac{\mathrm{d}[C]}{\mathrm{d}t} \propto \frac{\text{number of collisions between } A \text{ and } B}{\text{time}}$$
because wouldn't this be invalid if the reaction changes the volume?

If the reaction mixture is in a dilute solution, you would not expect volume changes. If it is in the gas phase at constant pressure, that's a different story. So this assumes the volume is constant (and the reaction is homogeneous, i.e. in a single phase, typically a solution).

For the elementary reaction $\ce{A + B -> C}$, the following rate expression
$$\frac{\mathrm{d}C_A}{\mathrm{d}t} \propto C_A\;C_B$$
is only valid when:

The reaction is performed at constant volume with ideal mixing.

The rate follows a power-law model.

No inflows or outflows present (batch).

The fundamental equation from which this expression can be obtained is
$$\dot{n}_{Ao}-\dot{n}_{A}+r_AV=\frac{\mathrm{d}n_A}{\mathrm{d}t}$$
Since no flows are present:
$\require{cancel}$
$$\cancel{\dot{n}_{Ao}}-\cancel{\dot{n}_{A}}+r_AV=\frac{\mathrm{d}n_A}{\mathrm{d}t}$$
Considering constant volume:
$$n_A=C_A\;V\implies r_A\cancel{V}=\frac{\mathrm{d}(C_A\cancel{V})}{\mathrm{d}t}\implies -r_A=-\frac{\mathrm{d}C_A}{\mathrm{d}t}$$
Considering a power law model:
$$-\frac{\mathrm{d}C_A}{\mathrm{d}t}\propto C_A\;C_B$$
Finally, using reaction stoichiometry to relate rates:
$$\frac{-r_A}{a}=\frac{r_C}{c}\implies r_C=-r_A\implies \frac{\mathrm{d}C_C}{\mathrm{d}t} = -\frac{\mathrm{d}C_A}{\mathrm{d}t}\implies \frac{\mathrm{d}C_C}{\mathrm{d}t}\propto C_A\;C_B$$

Thank you! How exactly is $r_A$ defined?

@Silenus $-r_A$ is defined as the amount of limiting reagent A that is consumed per unit volume, per unit time. It usually has $\pu{\frac{mol}{dm^3;s}}$ units, although constant volume allows for $\pu{\frac{M}{s}}$ units or equivalently, $\pu{\frac{mol}{L;s}}$.

Note that the sign next to "r" indicates whether the species in question is being formed ($+r_A$) or consumed ($-r_A$).

A chemical engineer's approach!

We are not extinct yet :)

$$\frac{\mathrm{d}n_\mathrm{C}}{\mathrm{d}t} \propto R_\mathrm{AB} \propto [\ce{A}][\ce{B}]V$$
$$\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{d}t} = \frac{\mathrm{d}n_\mathrm{C}}{V\mathrm{d}t} \propto \frac{R_\mathrm{AB}}{V} \propto [\ce{A}][\ce{B}]$$


text:
Does removing hexavalent chromium from a welded stainless steel part using phosphoric acid neutralise the CR(6) toxicity?
We had a meeting today and talked about solving some issues improving the quality of life of our worker.
We were talking about hexavalent chromium. We already deal with the welding fumes, but as I understand the red pigment that is on a stainless weld is one form of chromium oxide that is the toxic CR(6). So our metal polisher would be affected too from this form.
One way we remove those oxides from the stainless before polishing is using a surfox machine. Essentially, it is a mixture of phosphoric acid that is used as an electrolyte with the application of an electric current.
Now that you have the background, my question is: does the use of this method neutralise the toxicity of the CR(6) transforming it into something else that is not toxic or it simply creates something like a salt containing the same toxic property as the CR(6) itself?
It would help us know how to deal with the issue properly for our workers.
Thank you and have a nice day!

Chromium is present throughout the steel, and what you propose only impacts the surface. When you make a weld pool, you will bring chromium into the melt as you melt the steel.

Do you propose using the phosphoric acid bath after welding? In that case, the comment of @JonCuster does not apply -- the Cr deep inside the steel will not migrate to the surface after the weld is completed. Welders would still need protection from fumes, but not polishers, if the bath is effective in reducing Cr(VI). However, you'd need to have both the used bath and the metal surface analyzed to see how much Cr(VI) is present.

Yes we are using the phosphoric acid after the weld is done to remove the oxide layer produced when the metal was still hot and and the argon gaz protection is removed so the hot metal was in contact with oxygen. From my undertanding the hexavalent chromium can be produce by this oxidation reaction. The color differ depending on the temperature of the stainless parts when it come into contact with oxygen.

But the main issue i want to know is if the product that is created from the phosphoric acid when it dissolve the chromium oxide, what i presume is CR(6), is not toxic or if it created just another toxic salt in itself. This way we have to find a way to treat all that is around the machine as toxic too and when it dry it could produce a toxic dust that could go everywhere in the shop when we open doors.

You mentioned that your machine passes an electric current through the solution. Hexavalent chromium is reduced to chromium (III) by the application of an electric current (Sidenote, this is investigated as a method of treating Cr (VI)-containated wastewater). Sadly, I don't know how to calculate whether all of the chromium (VI) is reduced using the process you described or if a substantial amount still remains in solution.

An acid wash removes basic oxides such as Cr2O3 and Iron oxides and possibly CrO2 [it disproportionates] not chromic acid, CrO3, that happens to be soluble in water. H2PO4 does not reduce anything.

Ok so thank you all for your input. From what i've read in the last comments, the acid wash does not modify the CR(6) that is present and it stay in the residual solution. BUT passing a current through the solution could reduce the CR(6) to CR(3), but there is no way to quantify if CR(6) could remain. As CR(6) is a known carcinogen and have a pletora of health effect, CR(3) is safer, but is still dangerous. So the only option that is left to me is take a sample to a lab to measure the composition and ask for a proper way to neutralise/dispose of it. Thank you again for your help!


text:
synthetic procedure for 1,3,4 thiadiazole from acylhydrazide
I wanted to synthesize 1,3,4 thiadiazole-2- thiol substituted to  benzoic acid the reaction has to happen in such a way that the "acid hydrazide" group of benzoic acid hydrazide has to undergo cyclization in presence of carbon disulfide to form 1,3,4-thiadiazole is their any way to carry out the synthesize

2 steps:- react the acylhydrazide and carbon disulfide with KOH in EtOH then acidify the resulting potassium salt with sulfuric acid which causes spontaneous cyclisation to the required product as reported in this review from which I have taken the reaction scheme (top line applies).

i tried this procedure by stirring overnight and also tried heating with reflux but my acyl hydrazide derivative not dissolving in the first step

What solvents is it soluble in? THF, dioxane, iPrOH?

the basic nucleus is showing solubility in ethanol

@Dina obvious question, but did you go to the review, find the original paper it referenced and follow the procedure from there?


text:
How to interpret this pair of IR and 1H NMR spectra?
I need to figure out the structure of a $C_{8}H_{14}O_{2}$ molecule with this IR spectrum:

and these $^{1}$H NMR spectrum signals:

m means 4 or more spikes.
I worked out that the degree of unsaturation is 2 and I think the first IR trough is OH alcohol or C-H alkene, but the rest of the IR spectrum doesn't seem to correspond to anything. I can interpret some of the NMR signals, but it seems like there are a lot of possibilities that I don't know how to organize, especially when symmetry is involved.
Here are the IR and NMR reference tables I'm given:

Any help is much appreciated!

I was spoiled working with a FTIR with digital spectra and libraries. I no longer have access so I can give some bits and pieces. First there is no OH!  the blip at 3400 and strong peak at 1730 indicate ketone or ester the peaks at 1180 and 950 suggest ester that takes care of the 2 Os and 1 unsaturation. The nmr shows a typical ethanol ethyl group so likely an ethyl ester. That leaves C5H9 one unsaturation and the bond to the carbonyl There is a singlet methyl group down field, 2 methylenes and 2 vinyl. That leaves a -CH2CH2- bound to a carbonyl and a C=C The ethylene has 2 H's and one methyl attached. The peak at 940 suggests a trans double bond [ I forget what the trans coupling constants are]. The sharp peak at 3100 for the vinyl is troublesome possibly the symmetrical stretch is inactive. Another possibility is a methyl ester and ethyl group on the double bond. The chemical shifts will tell.


text:
Phase transfer catalyzed Claisen condensation using sodium hydroxide or carbonate. Why?
I was wondering if there was any obvious reason why I wouldn't be able to conduct a Claisen condensation between phenylacetonitrile
and an ester like methyl benzoate under phase transfer conditions (ie. aliquat 336), and sodium hydroxide or potassium carbonate and if there would be any impediments to that reaction taking. Is it necessary for the reaction to be conducted anhydrous? Is carbonate better than hydroxide?
Here are some refences.
Synthesis of Glycidic Esters in a Two-Phase Solid-Liquid System
https://sci-hub.hkvisa.net/doi/abs/10.1080/00397918209408010
REACTIONS OF PHENYLACETONITRILE DERIVATIVES WITH AROMATIC
NITROCOMPOUNDS IN BASIC MEDIA
https://sci-hub.hkvisa.net/https://doi.org/10.1016/S0040-4020(01)90658-1
https://sci-hub.hkvisa.net/https://doi.org/10.1016/S0960-894X(01)00443-7
https://sci-hub.hkvisa.net/doi.org/10.1039/B418156E

What solvent are you using for your organic phase?

Is toluene appropriate? either toluene or dimethylformamide, preferably the former.

Toluene should be fine. I have done phase transfer alkylations of phenylacetonitrile in two liquids phases (DCM/water) using TBAB/hydroxide which works well.

Waylander do you have confidence that If I use a alkyl ester instead of an aromatic ester, I would produce a ketone, under phase transfer conditions, I intend on using sodium hydroxide as an 50% aqueous solution perhaps,

The pKa of phenylacetonitrile in around 14. It it by so way the most acid site present so I see no great problem. The ketone product will readily deprotonate so some may go into the aq phase.


text:
Why sulfur dissolved in ethylenediamine turns green?
Davis, R. E., & Nakshbendi, H. F. (1962). Sulfur in Amine Solvents. Journal of the American Chemical Society, 84(11), 2085–2090. doi:10.1021/ja00870a017
I've also done this experiment. But what exactly forms and what really happens, I cannot find a recent text on the subject. Perhaps I'm searching the wrong terms?

Have you seen this? Spectroscopy of sulfur in ethylenediamine
Robert MacColl and Stanley Windwer J. Phys. Chem. 1970, 74, 6, 1261–1266.

And this one, Colored Sulfur Species in EPD-Solvents (Google the title, free access). Once you read these, you can perhaps write an answer. Interesting question. Recent paper does not necessarily mean the best or the most accurate version. Works from the 60-80s perhaps belong to the golden era, at least for spectroscopy and analytical chemistry in the USA.

two recent reviews on the topic: https://doi.org/10.1039/C3CS60119F and https://doi.org/10.1039/C8CS00826D

Very interesting question. I can't help but wonder if the green colour could be a combination of a large amount of weakly yellow neutral octasulfur plus a minute trace of the very strongly blue trisulfur radical monoanion.


text:
If change in enthalpy is defined for a reaction going to completion, how is it measured for reactions that don't?
I read that $\Delta H$ is the energy absorbed or released for a reaction going 100% to completion(here). However, what we would be measuring in a calorimeter would be the reaction going from initial products to equilibrium. So for reactions where K isn't very high, the heat released or absorbed wouldn't reflect $\Delta H$

"If change in enthalpy is defined for a reaction going to completion" who told you something like that?

@Mithoron I will link it here again https://chemistry.stackexchange.com/questions/112815/change-in-gibbs-free-energy-vs-change-in-enthalpy#:%7E:text=To%20start%20off%2C%20%CE%94G,change%20from%20reactants%20to%20products.

Additionally, chem.libretexts.org/Courses/University_of_California_Davis/… and khan academy (khanacademy.org/science/chemistry/thermodynamics-chemistry/…) define $\Delta G$ for a reaction as $Gproducts-Greactants$ meaning in the final state, only the products exist, which implies 100% conversion. Since $\Delta G = \Delta H - T\Delta S$, the initial and final states in that delta for all three quantities is going to be the same. So if $\Delta G$ for a reaction is defined for a particular final state(which those links imply), so would $\Delta H$

That was a rhetoric question, you know? I pointed your fallacy. You neither need an actual reaction happening nor its measurement, much less "going to completion". Such value is just an abstract.

For incomplete irreversible reactions, you have to use the conversion $X$ of the limiting reagent:
$$\Delta H=n_{Ao}\;X\;\Delta\overline{H}^{\;o}$$
For reversible reactions, you have to use the conversion at equilibrium $X_e$:
$$\Delta H=n_{Ao}\;X_{e}\;\Delta\overline{H}^{\;o}$$
In both cases, $\Delta\overline{H}^{\;o}$ is the standard molar enthalpy of the reaction, and $n_{Ao}$ is the initial amount of limiting reagent A present.
Note that for complete irreversible reactions (such as a combustion with excess oxygen), the conversion is implicitly omitted since its value for such case is 1 or 100%.
For a reversible reaction of the form:
$$\ce{aA + bB <=> cC + dD}$$
$X_e$ can be calculated from the following equation, given reaction conditions, initial concentrations, and equilibrium constant $K_C$ at reaction temperature $T$:
$$K_C=\left[\left(\frac{C_{Ao}}{1+\epsilon\;X_e}\right)\left(\frac{P}{P_o}\right)\left(\frac{T_o}{T}\right)\right]^{\Delta n}\;\frac{\left(\frac{C_{Co}}{C_{Ao}}+\frac{c}{a}X_e\right)^c\;\left(\frac{C_{Do}}{C_{Ao}}+\frac{d}{a}X_e\right)^d}{\left(1-X_e\right)^a\;\left(\frac{C_{Bo}}{C_{Ao}}-\frac{b}{a}X_e\right)^b}$$
If the reaction conditions are such that volume, pressure, and temperature all remain constant, only 1 reactant and 1 product are involved, only A is initially present, and the stoichiometric coefficients of A and C are both 1, the expression is simplified a lot, since we would have the reaction:
$$\ce{A<=>C}$$
With the expression:
$$K_C=\frac{X_e}{1-X_e}\implies X_e=\frac{K_C}{1+K_C}$$

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


text:
Which Gibbs' free energy is an activation energy diagram referring to?
In the case of activation energy diagrams in terms of potential energy, I understand that when we're talking about two individual molecules, the potential energy is highest in the transition state since it is the most unstable, and then falls.
When this activation energy diagram is expressed in terms of gibbs free energy:

As is pointed out by this answer,here, they indicate that for the individual molecule interpretation, macro properties such as quantities such as temperature and pressure are ill-defined.
However, if we treat this as representing the reaction macroscopically, that can't be the case also since that would violate the second law (gibbs free energy rising). You would instead expect something like this diagram:

In classical thermodynamics we always consider an ensemble, typically a mole of reactants. The activation energy 'free energy' is used  as a way to include entropy changes for an ensemble of molecules when using transition state theory from a thermodynamic viewpoint.  Even the activation energy is an ensemble measurement.

The x-axes in these plots is different. The overall $\Delta G$ plot uses the change in concentration. The plot with the activation energy uses a more abstract "reaction coordinate" which could be a spatial coordinate of the two interacting molecules, but also could be simply the time progression of the reaction of an individual molecule. So short answer - no it is not implying that the gibbs energy of the entire ensemble rises and falls in that manner

in my previous comment by 'ensemble' I mean that the average value taken over numerous molecules is used.

@Andrew have a look at the thread I linked. The accepted answer claims it's for the ensemble which is part of what's confusing me

It should be obvious from the time evolution of any reaction that the entire ensemble does not cross the transition state at the same time, not to mention that a macroscopic increase in G is a violation of the 2nd Law. It is true, however, that the value of $\Delta G^\ddagger$ is derived by treating the transition state as if there were a macroscopic population of that species.

@Andrew That is what I thought as well. That the Gibbs free energy cannot rise, and I have also only ever seen Gibbs free energy used in a macroscopic context, which is why I was confused and asked this question

Related: How exactly is activation energy defined?

I assume the activated complex phonomena of chemical reactions can be taken formally as 2 subsequent reactions, the first endergonic, the second exergonic,  both without the respective  own activated complex.

@xasthor Thanks for referencing my answer to a similar question. It is not an accepted answer though, and I'm not making a claim about the best way of labeling these diagrams. I did make a claim that some ways of labeling them don't make sense. Folks who are closer to or in the subdiscipline of physical chemistry probably can give a better answer to your question and to the linked question.


text:
Are there standard measures of adhesion?
I am looking for quantitative indicators of how well water displays adhesion with respect to other materials like glass, metal, wood...did not find anything immediately on a google search. I need only rough indicators- the values are not being used for any important application.
Thanks.

If you're looking for equipment to measure adhesion, there are commercial products, such as these from DeFelsko (which also has an inrtroduction to testing).

If your looking for standards for testing, there are many, such as ASTM D4541 Standard Test Method for Pull-Off Strength of Coatings Using Portable Adhesion Testers.

If you're looking for some lab exercises in measuring adhesion, there are online resources such as from the University of Akron,  the University of Hawaii and Rohm & Haas.

If you're designing your own experiment, consider different ways adhesion is measured, e.g. by peeling, by pulling perpendicularly, etc. Also consider the units of measurement.

You could make some simple test equipment, using a spring scale to measure the force needed to separate two plaques (e.g., squares of metal, microscope slides or wood).

Thank you Dr. Moishe- these are helpful links. Would there be a table of standard values somewhere? Is there a standard unit of measurement that specifies how well a liquid-solid would adhere? I am especially looking for values fo substances that would NOT be used as glue... like water, honey, oil etc.

the DeFelsk link above lists a number of testing standards for coatings. As for adhesion of water (and ice), this is of major importance in many areas, such as aircraft. See https://pubs.acs.org/doi/10.1021/am1006035, for example. A simple test can be performed with water and microscope slides. Note that measurements will be quite rate dependent for liquids; a quick pull will snap the glass slides before the water "lets go".


text:
Sigmatropic Rearrangement along with oxidation
I was going through an organic chemistry book I came along this question it's solution used sigmatropic rearrangement along with oxidation in Jones reagent, how does this rearrangement work??

As to the oxidative rearrangement, look here.

If you look at the mechanism of chromium trioxide oxidation (see below - from Wikipedia) you will see that it proceeds via the formation of a chromate ester which is then displaced by water. Since there is no solvent water present in the reaction system this gives the opportunity for the loss of the chromium species to proceed by an intramolecular reaction whereby one of the oxygens on the chromium does a sigmatropic rearragement with allylic alcohol pushing the double bond to the tertiary centre expelling the cyclic intermediate chromate species and leaving the chromate ester at the methylene. The oxidation then proceeds at this centre giving the unsaturated ketone product shown.


text:
Carbocation rearrangement with ring expansion
I was solving a particular question which landed me to a position where I have to decide whether the following ring expansion would take place or not by carbocation rearrangement:

If the expansion takes place, then Bromine(an electronegative element) would apply -I effect on carbocation and will make it unstable but also 5 membered ring is more stable than 4 membered ring due to less angle strain.
How to decide it then?

Stability is a combination of all factors, weighted towards their contribution to the overall $\Delta G$. In this case ring strain is a major issue, and one not necessarily likely to win out over the tertiary -> secondary carbocation transition you've drawn (the bromine has a lesser role to play as well, as you highlight). An equilibrium will form between two drawn structures. Absent modeling I can't give hard numbers, but my intuition is that the cyclopentane will be the more stable.
Ultimately though you're missing the greatest contributor to a beneficial $\Delta G$: quenching the carbocation. Whether or not you get a cyclopentane/cyclopentene is going to depend on how quickly a proton can eliminate from those methyl groups to form a terminal alkene. In most cases I think the 1,3-carbon shift is going to be faster, but not by much when there's six potential protons that can eliminate.
As with most carbocation systems you're going to get a mix of products.


text:
Why does "bis-tris propane" have two pKa values?
Wikipedia gives two pKa values for "Bis-tris propane".

What about bis-tris-propane gives this compound two pKa values where other compounds have only one? How am I to choose which pKa value to use in a specific scenario?

It has 2 secondary amino groups, which, both in protonized forms, formally form biprotic acid due 2 respective conjugate acids $\ce{R-
NH2^{+}-R.}$

And the two identical amino groups are sufficiently far apart to act almost independently - hence very similar pKa's. If both are involved in ionisation, you must include both in your equilibrium calculations.

@IanBush hm, similar enough to consider both e.g. for pH buffer calculations, but otherwise different enough. It has 6 bonds H-to-H.  Not having data for alkylendiamins, but for analogical dicarboxylic acids, HOOC-CH2-COOH has pKa 2.83, 5.69 with the same count  6 bonds HtoH .   HOOC-(CH2)2-COOH has pKa 4.2, 5.6 with 7 bonds HtoH .   HOOC-(CH2)4-COOH has pKa 4.43, 5.41 with 9 bonds HtoH .

NH2-CH2-CH2-NH2 has pKa  6.99, 10.08

I'm just offering a very simple handwaving model as to why the two pKas are similar. You can see it in polyphosphoric acids as well.  It's almost as good (IMO) as most of the models taught early on in chemistry, which because it is so difficult we need to handwave extensively.

In the protonated forms, I would guess the slight difference in acidity comes from there not being a destabilizing +/+ interaction and perhaps to some degree, inductive effect.

There are two ways I could read your question. Are you asking why some acids are polyprotic (because they have multiple protons they can lose) or why with two identical and symmetrically arranged functional groups, we have two equilibrium constants instead of a single (overlapping) one?

@Karsten, I'm more interested in the latter question: Why, with symmetric functional groups, are there two equilibrium constants?

The compound "bis-tris-propane"(1,3-bis(tris(hydroxymethyl)methylamino)propane, also known as BTP) has 2 secondary amino groups and can be written as
$$\ce{R-NH-R-NH-R}$$
which, both in protonized forms, formally forms a biprotic acid:
$$\ce{R-NH2^{+}-R-NH2^{+}-R}$$
dissociating in the schema:
\begin{align}
\ce{R-NH2^{+}-R-NH2^{+}-R(aq) &<=>[\mathrm{p}K_\mathrm{a1}] H+(aq) + R-NH-R-NH2^{+}-R(aq)}\\
\ce{R-NH-R-NH2^{+}-R(aq) &<=>[\mathrm{p}K_\mathrm{a2}] H+(aq) + R-NH-R-NH-R(aq)}
\end{align}
The acidity of the respective substituted ammonium depends on if the other nitrogen is protonated or not, therefore 2 different acidity constants:
$\mathrm{p}K_\mathrm{a1}=6.8$ and $\mathrm{p}K_\mathrm{a2}=9.0$.
The particular usage of one of $\mathrm{p}K_\mathrm{a}$, or both, depends on the context. Note that the $\mathrm{p}K_\mathrm{a}$ difference 2.2 is almost ideal for a broad pH buffering scope
$$\mathrm{pH} \approx 5.8 - 10.0.$$

(ported from my prior comments)
The compound has in its bi-protonated form 6 bonds between the respective acidic hydrogens.
Not having data for alkylendiamines, but for analogical dicarboxylic acids:

Compound
$\ce{H}$ to $\ce{H}$ distance
$\mathrm{p}K_\mathrm{a1}$
$\mathrm{p}K_\mathrm{a2}$
$\mathrm{p}K_\mathrm{a}$ difference

$\ce{HOOC-CH2-COOH}$
6
2.83
5.69
2.86

$\ce{HOOC-(CH2)2-COOH}$
7
4.2
5.6
1.4

$\ce{HOOC-(CH2)4-COOH}$
9
4.43
5.41
0.98

"Bis-tris-propane" $\\ \ce{R-NH-(CH2)3-NH-R} $
6
6.8
9.0
2.2

$\ce{NH2-CH2-CH2-NH2}$
5
6.99
10.08
3.09

Why, with symmetric functional groups, are there two equilibrium constants?

If the functional groups are close together, there will be electrostatic interactions (coupling). Adding the first proton (to the neutral molecule) is easier than adding the second proton (to the +1 cation). So adding the first proton (no matter on which side) will break the symmetry.
If the functional groups are spatially and electronically distant (uncoupled) from one another, you still expect distinct apparent association constants. Take the case where the pH equals the (microscopic) $\mathrm{p}K_\mathrm{a}$. Each site will be exactly 50% protonated. There are 4 microstates (not protonated, two cases of singly protonated, doubly protonated), and 3 macrostates (neutral, +1, +2). The +1 macrostate will be present at twice the concentration compared to the other two states, so the apparent equilibrium constant for the first protonation step and the second protonation step will be different by a factor of 4. As a consequence, the apparent $\mathrm{p}K_\mathrm{a}$ values will be different by $\log_{10} 4 = 0.602$.
For the related example of dicarboxylic acids mentioned in Poutnik's answer, you can see that the difference of $\mathrm{p}K_\mathrm{a}$ values does not approach zero, reflecting this statistical reason for a difference in the apparent equilibrium constants.

How am I to choose which pKa value to use in a specific scenario?

You first figure out what the major species are. At pH = 6.8, you have the +2 and +1 cations as major species, at pH = 9.0, the +1 cation and the neutral species. The compound is useful as buffer in the pH range of 5.8 to 10.0, but does not buffer well at around 7.9 (major species is the +1 cation, with very little, i.e. less than 10% each, of the other two species).


text:
Oxidation and Polymerization of drying oils
I am having trouble understanding the relationship between polymerization and oxidation of drying oils, specifically linseed oil used for fine art. In refining my own oil by filtering it then leaving it in the summer sun in a pan, the oil thickens and becomes strong and very fast drying. Similar thickened linseed oil from the art store is alkali-refined, and though thickened (therefore supposedly polymerized) still dries slowly. Why? So does polymerization not affect rate of curing? (Art store thickened oil is boiled in the absence of O2, so can polymerization happen without oxygen?) I can use an aquarium pump to blow air through some home-filtered oil for a number of days and rapidly thicken the oil. Is the thickening because it is starting to cure/oxidize, or is it because of polymerization, or both? I am trying to learn the relationship between these concepts, and how they affect cure rate and strength of the dried film.

Could you give us some ideas of drying times?  How are you applying these oils?  To what?

can polymerization happen without oxygen

Yes, under heat the Diels-Alder reaction can occur between double bonds of unsaturated oils without oxygen.  "Boiled oil" is partially polymerized. Solvent (mineral spirits) is used to thin it and "dryer" chemicals are added into the formula for faster drying.
Oxygen can also "cross-link" untreated oils, such as pure linseed, tung, or soya oil, as well as processed oils, but this usually takes longer.
However, drying rates of solvent based product can take markedly longer at lower temperatures simply because mineral spirits does not evaporate as well.  The solvent must evaporate before even "boiled" oil can dry.  "Store bought" product also may not be pure linseed oil$^1$.
In coatings jargon, there are also two degrees of dryness: set to touch and tack free.  The dryers generally help form a film on a surface faster.  So lightly dragging fingertips without sticking should happen sooner.  The whole coating being firm with finger pressure (tack free) can vary, especially if the solvent is slow to dry.
$^1$ "tung oil finish" is an example of this


text:
How to find the deprotonation percentage of an acid?

Given is an aqueous solution of mandelic acid (M = 152.2 g/mol; $pK_a$ = 3.4) with a molar concentration of 0.3 mol/ liter. A $\mathrm{pH}$ value of $2$ is measured for the aqueous mandelic acid solution. What percentage of the dissolved mandelic acid is deprotonated?

I know deprotonation means the removal of a $\ce{H+}$ from the acid. After googling , I found the deprotonation percentage can be obtained by dividing the $\ce{H+}$ concentration by the initial acid concentration. Deprotonation percentage shows what percent of $\ce{H+}$ is removed from the initial acid
However, I am unsure about how to proceed because it seems to me there are different ways that give a similar results, is this normal ? For example, we could just do
$$\mathrm{pH}=-\log_{10}([\ce{H+}])\rightarrow10^{-\mathrm{pH}}=10^{-2}=[\ce{H+}] \rightarrow[\ce{H+}]/0.3= 0.0333 = 3.33 \% $$
or we could do
$$K_\mathrm a=10^{-\mathrm pK_\mathrm a}=10^{-3.4}$$
$$K_\mathrm a=[\ce{H+}][\ce{A-}]/[\ce{HA}]\rightarrow K_\mathrm a=x^2/0.3\rightarrow\sqrt{0.3\times10^{-3.4}}=x=[\ce{H+}]$$
$$[\ce{H+}]/0.3 = 0.0364 = 3.64 \%$$
Are both ways correct ? Or is there another way to proceed ? I googled quite a bit but was unable to find a clear explanation

@wengen . The following information is absurd :$\ce{10^{−2}=[H+] → [H+]/0.3=3.33 }$

Is there any difference between "deprotonation percentage" and "dissociation degree $\ce{\alpha}$" ?

@Maurice IMHO,  the difference is just the proportionality constant 100 and the wording. I am not sure if I have ever seen the former until now. // Generally, if one searches A by searching its rarely used/unused synonym B, one may not find A.

pH is an extra info there. You can get the result directly from the provided pH, or you can calculate pH from the other data. Note that both values may differ.

@Poutnik By functions with dissociation degree, do you mean adding a coefficient in front of the concentration ? What would be the point of doing that ?

@Poutnik So it's normal both values differ ? I found this post where they say to do what I did, do you agree with them ? https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=72640

@Maurice You are from Lausanne ? So weird to meet a fellow Swiss here, I'm from Pully (east of Lausanne)

$$K_\mathrm a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{(c \alpha)(c \alpha)}{c(1-\alpha)} =  c \frac{\alpha^2 }{ 1-\alpha} $$

@Poutnik So this is simply algebra ? We simply use a coefficient to represent this "dissociation" ? That makes a lot of sense seen like that, in this case this is a quadratic formula we need to solve. If the percentage is really small like in our case, can we simply ignore the "$- \alpha$" in the denominator like we do with the $- x$ when finding the concentration $K_a = x^2 / (c - x)$ ?

@ wengen. Happy to meet a neighbor. Is there a private part of SE where we could have private discussions ?

@Maurice I don't think there is, but feel free to send me a chat request/ message on my Reddit account https://www.reddit.com/user/Lost_Woodpecker5064 , we could chat there

@Maurice If you have Quora or Discord, we could also chat there

As @Poutnik stated, one of the values you provided is extra. Let's try to make everything tidier.
Complex protonated entities aside, what chemical species are we really dealing with? $\ce{[HA], [A^-], [H_3O+]}$. The concentration of $\ce{H2O}$ shall remain constant. Right off the bat, we are dealing with 3 unknowns: the concentrations of the aforementioned species. Wait! - you protest - I know $\ce{[H3O+]}$, it's given in the problem as the $\mathrm{pH}$ value! Let's try to ignore that for the time being. Let's pretend we were given only the initial concentration (I shall denote it as $\ce{C^0_{HA}}$ where $\ce{HA}$ and $\ce{A-}$ are, of course, the acid and its conjugate base) and the $\mathrm{pK_a}$. Would one still be able to calculate this mystifying deprotonation percentage without being force-fed a $\mathrm{pH}$ value?
At any point, we could define the deprotonation percentage intuitively as:
$$p = 100\cdot \frac{\ce{[A-]}}{\ce{[HA]}+\ce{[A-]}} = \frac{\ce{[A-]}}{\ce{C^0_{HA}}}$$
This is not a formula worth memorizing, it could be readily derived intuitively. A deprotonation percentage entails determining... well, what percentage of your initial acid exists as the deprotonated form ($\ce{[A-]}$). When it comes to any problem of equilibrium, especially in the realm of simple proton-transfer equilibria, one should write down balances: mass balances, charge balances and the expression of the acidity constant(s).
$$
\ce{C^0_{HA} = [HA] + [A-]}
\tag{1 - mass balance}
$$
$$
\ce{[H3O+] = [HA-]}
\tag{2 - charge balance}
$$
$$
\mathrm{K_a} = \frac{\ce{[A-]\cdot [H3O+]}}{\ce{[HA]}}
\tag{3 - the acidity constant}
$$
3 unknowns, 3 equations! This should be more than possible to solve! But then, why are we given the $\mathrm{pH}$ supplementarily? If our system of equations is totally solvable, we should be able to calculate the $\mathrm{pH}$ on our own and, should it be different from the one in the problem statement, come up with an explanation, right?
Simple algebraic manipulation from the 3 equations alone leads to:
$$
\ce{ C^0_{HA} = [HA] + [A-] = [A-] \left(1 + \frac{[H3O+]}{\mathrm{K_a}} \right)  \Rightarrow [A-] = \frac{C^0_{HA}}{1+\frac{[H3O+]}{\mathrm{K_a}} } }
$$
$$
\ce{ [H3O+] = [A-] \Rightarrow [H3O+] = \frac{ C^0_{HA} }{ 1+\frac{ [H3O+] }{ \mathrm{K_a} } }  \Rightarrow C^0_{HA} = \frac{ [H3O+]^2 }{ \mathrm{K_a} } + [H3O+]} \Leftrightarrow
$$
$$\ce{ \Leftrightarrow [H3O+]^2 + [H3O+]\cdot \mathrm{K_a} - C^0_{HA}\mathrm{K_a} = 0}
$$
Upon solving this 2nd-degree polynomial, we arrive at a $\mathrm{pH}$ of $\sim \textbf{1.97}$. Not that far away from 2! But not exactly 2 either.
By plugging in this value in the identities above we arrive at:
$$
\ce{ [H3O+] = [A-] = 1.073 \cdot 10^{-2} \mathrm{M} \Rightarrow p \approx \textbf{3.58}\% \\ [HA] = 2.893 \cdot 10^{-1} \mathrm{M} }
$$
But this $\mathrm{pH}$ is still not 2. We could, for sure, accept the kindness of the authors of the problem of providing us with an approximation for the $\mathrm{pH}$, and then rejoice in our victory. Or, we could accept the $\mathrm{pH}$ value from the problem statement as a buffered one. On many occasions, analytical chemistry makes use of buffer solutions: mixtures of weak acid/conjugate base couples meant to keep the $\mathrm{pH}$ steady. In this sense, there is absolutely nothing stopping us from attaining a $\mathrm{pH}$ of absolute 2, with the assumptions made above still in place.
However, expect something to change! Well, the acidity constant won't change: it's constant! The mass balance won't change: after all, all of your mandelic/mandelate species come from the initial small quantity of mandelic acid we put in - $\ce{C^0_{HA}}$ - there's no mandelate or acid popping up from the ether. It's the charge balance that changes, as adding a buffer will add a new charged species into the equation (in this case, we want to make the $\mathrm{pH}$ more basic, from 1.97 to 2, we could add small quantities of $\ce{NH3/NH4Cl}$ which will add $\ce{NH4+}$ into the balance).
If you want your answer to be precise, as long as your $\mathrm{pH}$ is different from the 1.97 value we calculated earlier, you can no longer assume that pristine charge balance: your balance will be spoilt by the addition of whatever changes it from 1.97 to 2. Since a buffer is meant to keep the $\mathrm{pH}$ steady, we could determine $\ce{[A-]}$ by using the exact same equation as above:
$$
\ce{ [A-] = \frac{C^0_{HA}}{ 1+\frac{ [H3O+] }{ \mathrm{K_a} } } } = \frac{0.3}{ 1+\frac{ 10^{-2} }{ 10^{-3.4 } } } = 1.149 \cdot 10^{-2}\  \mathrm{M}
$$
Finally, $$ p = 100 \cdot \ce{ \frac{ [A-] }{ C^0_{HA} }} = 3.83\% $$

Wouah, it makes much more sense after reading this. Thanks a lot for the thorough explanation !


text:
Why is the hydrolysis constant of a salt of Weak Acid and a Weak Base so?
I read that the hydrolysis constant for a salt of a weak acid and a weak base is
Kh=Kw/(Ka*Kb)
Now I recall reading in the autoionization of water that Kw=Ka*Kb.
So, is there any relation between the two here? Can we write KH=kW/kW=1, does that even make any sense?
Or are they completely unrelated and I'm just randomly relating stuff?
Thank you.

In the 1st formula, Ka and Kb are unrelated, while in the 2nd one they belong to the conjugate acid/base pair.

@Poutnik Yes! That clears it since I know the derivations. Couldn't thank you more!


text:
Correctly converting the weight of compounds (e.g. Alpha GPC) to the weight of their FDA-recognized essential nutrients
Chemicals such as Magnesium L-threonate have to have a conversion ratio applied based on their molecular formula and molar mass to determine their nutritional value. In the case of Magnesium L-threonate, the common dose of 2000mg is equivalent to 169mg of dietary Magnesium.
Is there such a conversion for these common forms of choline to dietary choline and, if so, is my attempt to calculate it correct?

Alpha GPC (L-alpha-Glycerylphosphorylcholine)
CDP-Choline (Citicholine)
Choline Tartrate 1:1 (Choline Bitartrate)

Molar mass, Alpha GPC = $\pu{257.22 g mol^-1}$.
Molar mass, CDP-Choline = $\pu{488.32 g mol^-1}$.
Molar mass, Choline Tartrate 1:1 = $\pu{253.25 g mol^-1}$.
If I do a PubChem search on "choline" the returned result has a molar mass of $\pu{104.17 g mol^-1}$. Is that the correct result?
Can I just use the ratios? So...
(104.17/257.22)*100 = 40.49% for Alpha GPC
(104.17/488.32)*100 = 21.33% for CDP-Choline
(104.17/253.25)*100 = 41.14% for Choline Tartrate 1:1
Is this correct? As a non-chemist, part of my concern is that I'm unclear if just because it has the term "choline" embedded at the end of its name it means that it can really be counted as choline in this way (Alpha GPC, for example)? From a nutritional perspective Alpha GPC is highly related to choline and acetylcholine, but I thought it best to confirm that my understanding and calculations are correct.

One thing is equivalence in context of the same molar amount, other thing is equivalence in context of metabolism and biochemical paths. Your numbers are correct in the former context, but I cannot comment for the latter.

They may mean that, based on the prior question, you should be already able to answer the question on your own, at least in the previous former context. There is time to learn concepts of molar mass, molar amount and molar equivalence and reuse it for similar cases.


text:
How to draw L and D configuration for isoleucine and threonine?
I'm finding difficulties in drawing the D and L configurations of these two aminoacids, because, differently from the others, they have two chiral carbons (specifically α and β).
In particoular, in the slides of my professor (I'll share the picture), in the representation of L-isoleucine both the amino group and the methyl group are on the left side.
Instead, in the representation of L-threonine only the amino group is on the left, while the alcohol group is on the right (I expected it to be on the left side too).
Why L-threonine doens't have the OH on the left side?

Some reading here.  You are correct.  Threonine was one of the last amino acids discovered.

Yes. You get one chiral center "for free", but you have to memorize which one, and what the rule is, so it doesn't really help. See https://www.masterorganicchemistry.com/2017/05/24/d-and-l-sugars/

@Karsten Thanks again for your help

There are four stereoisomers of D-threonine. One is called L-threonine, and the other two are not called threonine (because they are diastereomers of threonine, so they have different physical properties).
So the D/L nomenclature is a way to distinguish enantiomers. For a history of how D or L were assigned, see https://chemistry.stackexchange.com/a/50561/72973.

I'm finding difficulties in drawing the D and L configurations of these two aminoacids

You can't draw these from first principles, you have to look up the stereochemistry. This is different from the R/S nomenclature (where the complete stereochemical information is available from the R or S assigned to the "chiral" carbons).
This is illustrated nicely for the sugars glucose and galactose. Both of them have a D and an L enantiomer, but that does not tell you the configuration of the other 4 stereocenters.

Thanks for your answer. In practice, I just have to learn the structure by heart.