text:
Forming conjugate acid of borate anion via protonation
I am a biologist so please forgive me as I likely won't be as rigorous as a professional chemist. I am using the RDKit package to extract organic molecules from salts, where one ion is organic and the other inorganic. My code is attached below.
My strategy is to first isolate the organic ion, and then add or subtract protons, $\ce{H+}$, as necessary. For organic anions I try to form the neutral molecule by protonating atoms with negative formal charge; for organic cations I try to remove $\ce{H+}$ from an atom with positive formal charge.
In the simple case of sodium acetate, this method works as expected to give acetic acid:

However, this approach fails with sodium trihydroxy(phenyl)borate:

since there is no way to add an extra proton to the boron atom which has negative formal charge.
I am guessing that it is not really possible to neutralize the trihydroxy(phenyl)borate anion by adding a proton.
Removing one of the $\ce{-OH}$ groups would work, but that is not acceptable for me, because that is not dependent on the pH.
Is this correct, or there is another explanation?
The code:
SMARTS_inorganic_ion="[Li,Na,K,Mg,Ca,F,Cl,Br,I]"

from rdkit.Chem import SaltRemover
remover = SaltRemover.SaltRemover(defnData=SMARTS_inorganic_ion)
SMILES_sodium_acetate="CC(=O)[O-].[Na+]"
mol_acetate = remover.StripMol(mol_sodium_acetate)

def neutralize_atoms(mol):
    # Recognize charged atoms
    pattern = Chem.MolFromSmarts(
        "[+1!h0!$([*]~[-1,-2,-3,-4]),-1!$([*]~[+1,+2,+3,+4])]")
    at_matches = mol.GetSubstructMatches(pattern)
    at_matches_list = [y[0] for y in at_matches]
    if len(at_matches_list) > 0:
        for at_idx in at_matches_list:
            atom = mol.GetAtomWithIdx(at_idx)
            chg = atom.GetFormalCharge()
            hcount = atom.GetTotalNumHs()
            atom.SetFormalCharge(0)
            # Adjut the number of hydrogen according to the change in charge
            atom.SetNumExplicitHs(hcount - chg)
            atom.UpdatePropertyCache()
    return mol

You basically found that boric acid is not a Brønsted-Lowry acid but a Lewis acid.

Side note: CCDC's Python API contains a function to test if an entry is organic (reference). Though to run their scripts, you need some special libraries behind a subscription.

Just add proton to oxygen - that's what happens.

Essentially, you are trying to determine the conjugate acid of the borate anion.
Yes, that approach is likely to fail with boric acid & borate anions. In fact if you want to form the conjugate acid of $\ce{RB(OH)3-}$ by protonation, instead of protonating the boron atom, you should protonate the oxygen atom, because that's the atom that possesses an actual lone pair that can accept the proton.

Formally speaking, this species in the middle would be the conjugate acid, because you have added a proton to the original base. In practice, the water molecule would "fall off" to reveal the original boric acid. [Of course, this is probably quite difficult to code in a foolproof way.]
The boric acid / borate pair is special:* it doesn't work via the usual "release $\ce{H+}$" like what acetic acid does.† Instead, boric acid accepts $\ce{OH-}$ to form borate. So, instead of adding and subtracting $\ce{H+}$, you need to give and take $\ce{OH-}$. Naturally, both of these two methods (producing $\ce{H+}$ and absorbing $\ce{OH-}$) have the same effect on the pH, that is, they both bring the pH down. Effectively, the following reaction is responsible for the acidity:
$$\ce{RB(OH)2 + H2O <=> RB(OH)3- + H+}$$
and it's not right to say that this is pH-independent. If you add more acid to the borate anion, the position of this equilibrium clearly shifts to the left, which regenerates the original boric acid. Conversely, if you add more base to this system, then the base will react with $\ce{H+}$ on the right, pulling the equilibrium position over to the right and generating more of the borate anion.

* Some might argue that borate is a problematic conjugate base because boric acid is a problematic acid, but I find that reasoning to be slightly circular. I think it's clearer to ascribe weirdness to the entire acid-base system.
† Actually, it's possible for boric acid to work as an acid by releasing $\ce{H+}$: $$\ce{RB(OH)2 <=> RB(OH)(O)- + H+}$$ but that's a different conjugate base, and your question focuses on $\ce{RB(OH)3-}$, so we can't go down that path.


text:
How does water sublimate at normal atmospheric pressures/conditions (i.e., without a phase boundary between solid and gas)?
I remember my high school chemistry teacher stating that water could sublime under normal atmospheric conditions that would exist in e.g., Lancaster County, PA, where Amish would hang clothes to dry in the winter in below-freezing temperatures.  I do not remember a similar discussion in my physical chemistry (thermodynamics) class in college, but the former example stuck with me.  If clothes do indeed sublime water at ca. 1 atm then that means there exists a vapor pressure of water (i.e., ice) below 0 °C at ca. 1 atm.
In looking into this, the USGS states that, "It is not easy to actually see sublimation occurring, at least not with ice. One way to see the results of sublimation is to hang a wet shirt outside on a below-freezing day. Eventually the ice in the shirt will disappear."
Yet, when I look at the phase diagram,

there is no path from solid to vapor at any temperature below 0 °C that is also anywhere near 1 atm.  The highest pressure at which a phase boundary exists between solid and gas is 612 Pa, or 0.006 atm.
The USGS article goes on to discuss the south face of Everest and the Chinook winds sublimating snow and ice suggesting that it does indeed occur, though at admittedly much higher elevations (and lower pressure) than e.g., Lancaster, PA, but these are nowhere near the pressures you would have in e.g., lyophilizers I've used.
The only way I can come up with an explanation is if the "pressure" in the above phase diagram is somehow influenced by the partial pressure of water (i.e., "humidity") as opposed to or in addition to the net pressure of the system in total.
So the question is, if you are above 0.006 atm but somewhere between −60 °C and 0 °C where a phase boundary does not exist between solid and vapor/gas, how can water sublimate at ca. 1 atm in the example of drying frozen laundry?  Additionally, is there a partial pressure of water at, e.g., −10 °C and 1 atm?
In partial answer to the latter question, this link shows vapor partial pressures down to −82 °C and seems to indicate that those partial pressures are irrespective of the net pressure of the system ("...if the pressure in the chamber exceeds the vapor pressure of the product while in the presence of ice, water may not be able to sublime.").

The phase diagram relates to a single component system only, with pressure equal to water vapor pressure.

And it assume the conditions are at equilibrium, which they are not in the example of drying clothes as water vapour is continually removed from the system.

There is confusion between vapor pressure (of ice) and atmospheric (ambient total) pressure. The link describes a freeze drying process, where a vacuum pump is used to reduce ambient pressure. The purpose of reduced pressure is to enable H2O vapor to be removed more quickly by removing air molecules which hinder diffusion. With regard to clothes drying, sublimation of ice will be hastened by a good wind, even if it is well below the freezing temperature. I suppose you could freeze dry with a blast of cold air instead of a vacuum, but it would cost more and probably introduce some dirt.

@JamesGaidis - I don't believe the removal of ambient pressure is particularly important for the freeze drying. The vacuum instead acts to keep the partial pressure of the water vapor below the equilibrium vapor pressure so that it continues to sublimate. Otherwise, only a small amount of solid would sublimate before equilibrium is reached. Since selective removal of water vapor is challenging, the easiest way to do this is to reduce the total pressure to below the equilibrium vapor pressure.

@Andrew: You are correct. Could we even determine a difference? The limiting factor is probably the amount of heat entering the solid ice. Too much, and the ice melts. Too little - no problem; the ice just sublimes a little slower; you will probably not even notice it. (You might complain it's too slow, but be careful if you add heat.) The easiest visual explanation is get the barrier out of the way so the H2O can go away, whether that barrier is air or water vapor.

@JamesGaidis - the big difference is that the water vapor can redeposit as ice, so the net rate of sublimation goes way down (obviously reaching zero when the vapor pressure matches eq. v. p.), whereas the air does not redeposit. So the important thing is inhibiting freezing not slowing down sublimation.

@Andrew: you are absolutely correct! Removing the water vapor increases the rate of sublimation.

If you put frozen water in a vacuum at a temperature below the normal freezing point, water will sublimate until the pressure in the vessel reaches the vapor pressure (which is below 1 atm at that temperature). At this point, water will deposit and sublimate at the same rate
However, if you take the same sample and put it outside, the direction of the process depends on the partial pressure of water in the air (on the humidity). If it is dry air, the sample will sublimate. If the partial pressure in the air is higher than the vapor pressure of the sample, additional water vapor will deposit in the form of ice.

The only way I can come up with an explanation is if the "pressure" in the above phase diagram is somehow influenced by the partial pressure of water (i.e., "humidity") as opposed to or in addition to the net pressure of the system in total.

Yes, the graph is for pure gas and solid phase, with no substances other than water. If the gas phase is air (i.e. a mixture), you can have water at equilibrium between liquid and gas even at temperatures below the boiling point. To figure out the partial pressure of water in the gas phase, you can read off the vapor pressure from the graph by going down vertically until you hit the phase separation line. You would also have to interpret the graph different than for pure substances if the liquid is not pure water, or the solid is not pure water.

THe statement "If the gas phase is air (i.e. a mixture), you will have to change the y-axis label to partial pressure of water in air." is incorrect. The phase diagram y-axis is total external pressure regardless of source. The partial pressure of H2O vapor is the P value at the phase boundary line for the given T.

@Andrew Point taken, I changed it.

For any solid or liquid phase, if there is available volume above it, there will be a non-zero vapor pressure at equilibrium. The presence of other gases has very little effect on the value of this vapor pressure.
On your phase diagram, the y-axis represents the total external pressure, which can be from a gas phase or mechanical (eg a piston pushing directly on the surface of the solid or liquid). To find the partial pressure of $\ce{H2O}$ vapor (ie equilibrium vapor pressure at the given T), move vertically downwards from the point representing current conditions (ie keeping T constant) until you reach the boundary line with the gas phase. The P value at this point is the equilibrium vapor pressure at this temp for any value of external pressure.
Because the vapor pressures are low at low T, the density of $\ce{H2O}$ in the gas phase is very low so the amount of solid that needs to sublimate before equilibrium is reached is very very small for a small volume of pas phase. Thus, in order to dry something out completely, you either need a very large volume of gas phase, or you need to continually remove water vapor from the gas phase to maintain a non-equilibrium condition.
The lyophilizer uses the latter approach, constantly pumping the gas phase out so that the partial pressure of water vapor (which in a closed vacuum system rapidly becomes equal to total pressure because other gases are not replenished) remains below the equilibrium vapor pressure at the temperature of the sample chamber.
The winter clothesline uses the former approach, since clothes hanging outside are effectively surrounded by an infinite gas phase volume. Good air circulation/breeze is helpful to keep the local vapor pressure from increasing very much relative to the wider area. Of course, this only works if the ambient air is not already saturated with water vapor, ie a relative humidity of 100% (which is not uncommon in the morning in the winter in Pennsylvania, but usually is not the case by mid-afternoon). The lower the relative humidity, the faster the sublimation will occur.

source: Stack Exchange

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text:
Chemical equilibrium graph

How can we determine which reaction belongs to which graph.
I mean in the above pic . We can observe two graphs and 2 equations. But there is no link between the two . Both the reactions proceed till completion . So both graphs must be a straight line . But that’s not the answer . I would appreciate if you could help me in finding the reason to the above problem Thank you for your help.
Source: problems in physical chemistry by narendra Awasthi

This is a graph without time on the x-axis. It describes how product and reactant concentrations are correlated. The axes are mislabeled, they should say amount (or maybe concentration) of reactant and product. Also, b) does not have a correct answer. Could you edit your question to include where you found the flawed execise?

Here is a textbook section with some context for the reactant vs product type of graph: https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.04%3A_Non-equilibrium_Conditions

@KarstenTheis I found it in a book used to prepare for IITJEE

@KarstenTheis but how did you figure out that 2 nd one is wrong and the first is right . I didn’t get the logic

@Integer Consider a generic reaction A+B->C, which has equilibrium constant K=[C]/[A][B]. What would a graph of product [C] vs. reactant [A] look like, holding concentration of B constant? Manipulate the law of mass action (equilibrium expression) and obtain [C]=(K[B])*[A] which is a linear equation of the form y=mx. Apply this logic to your examples, keeping in mind that only aqueous and gas phases are included in the equilibrium constant. For the iodine reaction (solvation), K=[I2(aq)], so there is not a relationship between the product and reactant to plot.

This actually is not a bad question IMO. If @Integer can clean this up and garner reopen votes, it would be a welcome addition I think.

@ToddMinehardt I would argue that the question from the OP is good and the exercise from the photo needs work.

According @electronpusher comment reaction(A) belongs to graph (Q).

I think that the graph is not accurate as the graph indicates that the concentration of the reactant is zero

@AdnanAL-Amleh I agree rxn A corresponds to graph Q. I'm not sure what your second comment means. Both graphs have a point where the reactant concentration is zero, and this implies that the product concentration will also be zero. This should be understandable, we are talking about equilibrium not conversion. Assuming not all reactant gets used up (conversion less than 100%), at equilibrium there must be at least some reactant in order for there to be any product.

@electronpusher"Both graphs have a point where the reactant concentration is zero, and this implies that the product concentration will also be zero" what about the K value ?

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


text:
Why is selenophene aromatic?
According to Wikipedia,

The selenophene molecule is flat and aromatic. Being aromatic, it undergoes electrophilic substitution reactions at the 2- or 2,5-positions. These reactions are slower than that of furan, but faster than thiophene.

Selenium is a bulky 4th period element and I think the size of selenium is too large to have conjugation with carbon atom and make the compound aromatic. Why is selenophene still aromatic?
Further, faster electrophilic aromatic substitution (EAS) than thiophene sounds suspicious. Why and how?

+1 I have edited your question a little bit, I wasn't sure what EAS meant. If the edit is not appropriate, feel free to roll it back. Also, adding a picture of selenophene to the question would be helpful.

The molecular single bond radius of Sulfur is 103 pm and Selenium 116 pm (source https://www.webelements.com/selenium/atom_sizes.html). Se is only 10% larger, I don't think your argument holds water.

Also, "aromatic" isn't a binary yes/no label (sometimes we like to think of it as being so, but really it isn't). It's an interesting question, I think there's a good discussion to be had.

The discussion to be had is going to be at least difficult and probably opinion-based. Unfortunately there is no objective way to quantify aromaticity and presentations like "It is a metallole with reduced aromatic character compared to thiophene." on Wikipedia are missing the reference framework. As long as the definition of aromaticity includes "behaves similar to benzene" I'm afraid there will be no straight answers to that. So to know more, one should read the publication which Wikipedia uses as a source for that quote (or not because it doesn't really say).

I have no access, but this looks promising: Pelkey, E. Selenophenes. Comprehensive Heterocyclic Chemistry III 2008, 975–1006. DOI: 10.1016/B978-008044992-0.00313-8.

related https://www.sciencedirect.com/topics/chemistry/selenophene

Reading a review by Hellwich et al. in MDPI's Molecules (doi 10.3390/molecules25245907) by 2020, the routes of synthesis bear similarity to the ones about thiophenes, too.

Selenium — like sulfur — is a chalkogen/member of group 16.  It is heavier than sulfur, and thus more metalloid. In this regard similar to the trend in the adjacent group 17 (the halogens) where iodine has more metal-like character than the other members, or group 15 (the tetrels) with carbon; yet silicon, germanium, tin, and lead.  Thus, formally, selenophene is isoelectronic to thiophene, a with six π-electrons over five ring atoms an electron enriched heteroaromatic compund (compared to benzene).
But, comparing pyrrole, furan, thiophene and selenophene, Yadav arguments

«The electron count [of π electrons for the Hueckel rule], however, is not enough. Along with the number of requisite electrons, all the bond lengths in the ring must be the same or very similar, as in benzene. In five-membered heterocylces such as pyrrole, furan, and thiophene, the size of the heteroatom lone pair orbital must be similar to that of the $p$ orbital on ring π bonds. Both the requirements are not fulfilled in thiophene because σC–S bond is significantly longer than σC–C bond and, in comparison to the p orbital on carbon, the size of lone pair orbital on sulfur is too large to allow effective overlap.»
(abstract)

Given the larger radius of selenium ($\pu{115 pm}$) vs. sulfur ($\pu{100 pm}$), this might be even more the valid for selenophene, than for thiophene.
The discussion however isn't a new one.  By older recordings, the aromaticity of tellurophen and selenophen were considered between the one of furan and thiophene:

(Fringuelli et al., 1974)
References:
Fringuelli, F.; Marino, G.; Taticchi, A.; Grandolini, G. A comparative study of the aromatic character of furan, thiophen, selenophen, and tellurophen. J. Chem. Soc., Perkin Trans. 2 1974, 332-337; doi 10.1039/P29740000332.
Yadav V.K. (2021) Relative Aromaticity of Pyrrole, Furan, Thiophene and Selenophene, and Their Diels–Alder Stereoselectivity. In: Steric and Stereoelectronic Effects in Organic Chemistry, 191-216. Springer, Cham. doi 10.1007/978-3-030-75622-2_9.

Note, Yadav's publication is behind a paywall for which our library has no subscription.  I only was able to access the abstract above.

@andselisk I hope the edits you made are back again. It looks possible that both you and me edited the attempt simultaneously.  For future reference, what is your argument to type π (apparently) directly, rather than by $\pi$?  Is entering the math mode a reason why you suggest to type π directly, to keep consistency with the text-mode around?

No worries, my edit is present. There isn't much to add to my editing comment: here, letter pi should be upright (not a variable), and, since MathJax doesn't offer an alternative to upgreek package, it's better to use a Unicode version that isn't slanted.

I think you follow IUPAC's Green Book here.  If I recall correctly, the 3rd edition (with Professor Quack on board) has/had a dedicated early section about when to use Roman, and when Italic fonts exactly on this point.  Like in his short version, don't mix µ (upright) with $\mu{}$ (italic).


text:
What is the rate law of a reaction: 2A + 3B → products?
Update: I found out that the coefficients of the chemical reaction are not the exponents in the rate law. It is actually the order of the species during the reaction.
It is known that in the reaction, first order in 2A + 3B → products, A was found to be a first order and B is second order. So, does it mean that the rate law for this reaction is R = k[A][B]2?
I have been taught that R = k[A]x[B]y
Where: R = Rate of the reaction; k = rate constant; A = Concentration of species A; B = Concentration of species B; x = order of reaction with respect to A; and y = order of reaction with respect to B
First question: So, does that mean for the particular reaction: 2A + 3B → products, will have a rate law of R = k[A]2[B]3?
Second Question: How do I determine the x and y? Like what does order of reaction mean?

R = k[A]^2[B]^3 would be true if the reaction was the result of a simultaneous collision of 5 molecules. We do not know such reactions.

If such a reaction "$\ce{2A + 3B -> }$ products" occurs, it is the sum of several more elementary reactions, which are taking place successively. Some may be fast, some slower. The rate of the global reaction is the rate of the slowest of these elementary reactions. It can be first or second order, or it can even be more complicated, depending on any catalytic effect. –

The rate determining step approximation is often not a very good approximation. For the kinetics of a reaction all steps just be considered. It is really only a useful approximation of there is exactly one step that is much, much slower than all of the other steps. In complex reactions, e.g. catalysis, this is hardly ever the case. The energy span model provides a more practical approach.

If $\ce{2A + 3B -> \text{product(s)}}$ describes the stoichiometry of a reaction, than you describe the balance of starting materials against products; this however does not state much about the kinetics, elemental (perhaps rate determining) steps, and overall reaction order described by kinetics.
Uni- and bimolecular elemental steps are the most frequently encountered.  The simultaneous collision of five atoms/molecules to yield in one step a product you might suggest however is of very low probability.

I remember that trimolecular elementary steps are already incredibly unlikely, so 'very low probability' might be an understatement.

@Martin-マーチン Newer research suggest that they're aren't that rare, especially if there's for ex. hydrogen bonding involved, afaik.


text:
Assuming 2O3 (g) → 3O2 (g), how much oxygen will be formed within 5.50 s for a reaction rate of 0.250 M/s?
Context: The rate of appearance of $\ce{O2}$ in the reaction: $\ce{2O3 (g) -> 3O2 (g)}$ is $\pu{0.250 M/s}$. So, over the first $\pu{5.50 s}$, will it form $\pu{1.38 M}$, or $\pu{4.13 M}$ of oxygen during the given time?
Which is right? Here are my solutions:

$\pu{0.250 M/s} \cdot \pu{5.50 s} = \pu{1.38 M}$

$\pu{0.250 M/s} \cdot \pu{5.50 s} \cdot 3 = \pu{4.13 M}$,
(3 in the calculation is the coefficient of the product in the reaction: $\ce{2 O3 (g) -> \mathbf{3} O2 (g)}$).

I feel like the first one is right? I am overthinking this, and I need to make sure that I am doing this right.

reaction_rate = abs(component_rate)/stoichiometric_coefficient

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

Hello, @Poutnik. Thank you for responding. May I ask what "abs" and "component_rate" refer? I am a bit confused since I have a different formula given by my instructor.

Thank you so much, @Buttonwood. I will use mhchem on my next posts.

abs is math function giving an absolute value. Component rate is the time derivative of the component(=reactant or product) concentration ( or partial pressure or total amount, depending on scenario). // Good practice is starting with symbolic algebraic formulas and keeping it this way until all formulas are ready to finally put there literal numbers.
It helps focusing on principles and improves orientation, spotting mistakes, the text reusability and its permanent value.


text:
Can the environment of a molecule affect its HOMO LUMO gap?
For example, if a particular molecule is inside a particular protein, can its absorption maxima change with respect to the isolated molecule? Also, within the same protein, can the molecule's HOMO-LUMO gap vary, depending on its location inside the protein?

Just a short remark: HOMO-LUMO is not always the most intense transition, therefore the absorption maxima

The answer is a definitive affirmative yes.
You don't need to reach out for proteins, organic dyes are known to suffer from solvatochromism in function of the polarity of the solvent they are dissolved.  It either may be a bathochromic shift (or, red shift) of the UV-Vis absorption recorded, or a hypsochromic shift (or blue shift).  In large proteins, you equally may have variation of pockets which are more lipophilic, and other strains which are more charged/polar, too.   With the right probing molecule, you may monitor these changes, too.
There are even (small molecule) organic dyes which are used as an indicator about the solvent polarity they are dissolved in:

(image credit to Wikipedia article)
e.g., of the Reichardt's dye family:

(image credit to Wikipedia article).
This not only is an issue for solutions as commonly perceived.  This extends, e.g., to dyes used in organic electronics, since their mutual presence in the solid state affects light emission (both efficiency, as well as wavelength) of OLEDs, as well as light absorption and efficiency of solar cells with organic dyes.
A vaste topic ... thus only two references:
Machado V. G., and Machado C. An Easy and Versatile Experiment to Demonstrate Solvent Polarity Using Solvatochromic Dyes. J. Chem. Educ. 2001, 78, 649-651; doi 10.1021/ed078p649.
Loving, G. S.; Sainlos, M.; Imperiali, B. Monitoring protein interactions and dynamics with solvatochromic fluorophores.    Trends Biotechnol. 2010, 38, 73-83; doi 10.1016/j.tibtech.2009.11.002.

For a case that is about a protein neighborhood, see retinal photoabsorption in the eye.


text:
Writing rate of disappearance and rate of appearance using rate law
I have studied that rate of disappearance and rate of appearance is the change in concentration of reactants and products (respectively) with respect to time.  Thus,
$$\text{ROD} = -\frac{\Delta c_R}{\Delta T}$$
and
$$\text{ROA} = \frac{\Delta c_P}{\Delta T}$$

and
$$\text{rate of reaction} = %
%
\frac{\text{ROD}} {\text{stoichiometric coefficient}} %
%
= \frac{\text{ROA}} {\text{stoichiometric coefficient} }$$
Then writing rate of reaction using rate law was taught.
And according to rate law rate of reaction can be expressed as:
$$\text{rate of reaction} = k[\ce{A}]^x[\ce{B}]^y$$
here $x + y = \text{overall order of reaction}$, with $\ce{A}$ and $\ce{B}$, the reactants.
Here is my confusion:  Now they wrote rate of disappearance and appearance using rate law. How using rate law they wrote expression for ROA and ROD.
For a reaction:  $\ce{aA -> bB}$ (elementary reaction), it was stated that:
$$\text{ROD}_\text{A} = k_d [\ce{A}]^a$$
So how ?
Using rate law we can write rate of reaction.. (OK) but how can we write ROA and ROD ??!!
Hope you understood what I meant to say.

How did you come to the last equation?

@PoutnikCan we relate rate law with ROD of  reactant A as this formula : Rate of reaction = $-\frac{1}{Stoichiometric\space Coefficient\space  of \space reactant A}\cdot\frac{\Delta[A]}{\Delta{t}} = k[A]^x[B]^y$

@Poutnik I have made amendment to the reaction. Sorry for caused ambiguity. Now can you please explain how we can write ROD and ROA using RATE LAW ?

IF ROD(A)=ROR . SC and if ROR= k . [A]^x . [B]^y then .......

If there is text in the equation, set this as text (cf. the revision). A $ROD = x $ means $R \cdot O \cdot D$ (a multiplication) yielding $x$ as a product; using $\text{ROD}$ = x$ however states $\text{ROD} = x$, i.e. the variable ROD equates to $x$.  And for the use of \mhchem syntax: have a look here. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

@Buttonwood What is this ? "Stoichiometric space Coefficient" ? Why have you inserted "space" word at some points ? I think that's a mistake in your edit

@VishalAnand Look at the revision history.  Because you did enter text in mathmode without clear pattern like $\text{my variable} = 12$, you used \space to separate the words.  In equations, I don't use \space commands; this is why the previous edit showed them as remains of your approach.

Does this answer your question? Do coefficients matter in rate law?

You overthink it.
It is just matter of eventual multiplication or division with respective stoichiometric coefficients.
If there is an elementary reaction
$$\ce{a A -> b B}$$
then for respective reaction, appearance and disappearance rates:
$$\begin{align}
R_\mathrm{r} &= k_\mathrm{r}[A]^a\\
R_\mathrm{a,B} = \frac {\mathrm{d[B]}}{\mathrm{d[t]}}=bR_\mathrm{r} &= bk_\mathrm{r}[A]^a = k_\mathrm{a,B}[A]^a\\
R_\mathrm{d,A} = -\frac {\mathrm{d[A]}}{\mathrm{d[t]}} = aR_\mathrm{r} &= ak_\mathrm{r}[A]^a = k_\mathrm{d,A}[A]^a
\end{align}$$
.. and backwards:
$$\begin{align}
R_\mathrm{r} = \frac{R_\mathrm{a,B} }{ b} &= \frac 1b  \frac {\mathrm{d[B]}}{ \mathrm{d[t]}}\\
R_\mathrm{r} = \frac{R_\mathrm{d,A} }{ a} &= -\frac 1a \frac {\mathrm{d[A]}}{\mathrm{d[t]}}
\end{align}$$

thanks .According to your solution can  we conclude the following expression :$-\frac 1a \frac {\mathrm{d[A]}}{\mathrm{d[t]}}= k_\mathrm{r}[A]^a$

@AdnanAL-Amleh If X=Y and X=Z then obviously Y=Z. :-)

Thanks . Can we teach the following expression for 11th grade . Is it correct?: $\text{average}R_\mathrm{r} = \frac{\text{average}R_\mathrm{d,A} }{ a} = -\frac 1a \frac {\mathrm{\Delta[A]}}{\mathrm{\Delta[t]}}= k_\mathrm{r}[A]^a$

@AdnanAL-Amleh Involving deltas and averages, you have to come with average for $\ce{[A]}^a$ too. If the delta is about 10% and a=3, it makes about 30% difference. It is a task for integration.

thanks  .
Is it useful to teach the following two expressions separately without reconciliation or connection  between them ,as the first expression about instantaneous rate and rate law but the second about average rate
1)$R_\mathrm{r} = -\frac 1a \frac {\mathrm{d[A]}}{\mathrm{d[t]}}= \frac 1b  \frac {\mathrm{d[B]}}{ \mathrm{d[t]}}= k_\mathrm{r}[A]^a $
2)$R_\mathrm{r} = -\frac 1a \frac {\mathrm{\Delta{[A]}}}{\mathrm{\Delta{t}}}= \frac 1b  \frac {\Delta\mathrm{[B]}}{\Delta\mathrm{t}}$


text:
Why are there two infrared vibrational frequencies for the terminal CO ligands of Fe2(CO)9?
I am struggling to rationalise why there are two infrared vibrational frequencies ($\pu{2082 cm^-1}$ and $\pu{2019 cm^-1}$) for the stretching of the terminal $\ce{CO}$ ligands of $\ce{Fe2(CO)9}$. From my understanding of the structure, all six terminal $\ce{CO}$ ligands are equivalent and thus I would expect only one vibrational frequency. Why there are two?
The structure is as follows:

The bridging ligands are accounted for by a vibrational frequency at $\pu{1829 cm^-1}$.

Do you have an illustration about the molecular structure?  Though about an hexacarbonyl iron complex, which was described earlier on chemistry.se here, degeneration of molecular symmetry may contribute to the presence of multiple IR absorption bands.

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

I have added the structure to the original post and accounted for the bridging ligands.

2082 cm^−1 = 63420 GHz = 63.420 THz. 2019 cm^−1 = 60530 GHz = 60.530 THz. If it was light, the wavelengths would approx. be 4.73 µm and 4.95 µm, respectively.

Despite the title of the publication, Fletcher et al. both present an IR spectrum of diiron nonacarbonyl in a matrix of ($\ce{Ar}$ + 10% $\ce{CO}$, recorded at $\pu{15 K}$) in the $\nu(\ce{C-O})$ region of about $2080 \dots \pu{1820 cm^-1}$ (left to right hand side) as below:

(composite of two illustrations by Fletcher et al.)
Given the data recorded, the authors' assignment is (emphasis in the original publication):

«The spectrum of $\ce{Fe2(CO)9}$ shows two striong bands in the terminal $\ce{C-O}$ stretching region, ~$\pu{2000 cm^-1}$, as expected ($a''_2$ + $e'$) for a $D_{3h}$ structure, 1.  In the bridging $\ce{C-O}$ stretching region, ~$\pu{1840 cm^-1}$, there is a single band, again as predicted, $e'$.»

Reference:
Fletcher, S. C.; Poliakoff, M.; Turner, J. J. Structure and reactions of octacarbonyldiiron: An IR spectroscopic study using carbon-13 monoxide, photolysis with plane-polarized light, and matrix isolation. Inorg. Chem. 1986, 25, 3597–3604; doi 10.1021/ic00240a014.

Since there is an entry about this compound in Wikipedia, you have access to the/a CAS registry number (here, 15321-51-4). If your school has the corresponding subscriptions, you may use this as a search criterion e.g., in Reaxys (by Elsevier) or SciFinder (by ACS) to identify more recent publications with an assigned IR spectrum of said compound.  You may complement the search with checking the publications citing the work by Fletcher et al. (landing page of the publication already lists 63 works known to the journal).
There is a good chance Martyn Poliakoff, co-author of the paper, and Sir Martyn Poliakoff, a chemist at University of Nottingham and still engaged in popularization of chemistry on youtube/The periodic videos, are the very same person.

It basically comes down to molecular symmetry and since Buttonwood provided an answer just seconds ago, I'll leave the following as visualisations for it.
An optimisation of $\ce{Fe2(CO)9}$ at the DF-B97D3/def2-SVP level of theory results in a D3h symmetric molecule.
This in term results in degenerate vibrations (E') for the bridging $\ce{CO}$ at $\pu{1922 cm-1}$.

There are degenerate vibrations (E') for the terminal ligands at $\pu{2059 cm-1}$

Furthermore, there is another vibration (A2'') at $\pu{2085 cm-1}$.

Please bear in mind that this is an instructional answer (if it really is one) and the level of theory chosen for it is simple enough to make some approximate observations.
Without further checking, the modes are surprisingly close to the values you have posted.

What a coincidence ... +1.

@Buttonwood I am very grateful for your answer, it saved me a lot of work! ;)

The publication is mentioned among the ones in the English edition of Wikipedia.  By title of the publication, I was on the rim to suggest the OP something in lines of «search by CAS number in Reaxys/Scifinder (if your school has access to them)» since the other ones (incl. the Russian edition) didn't tell much. Until I read the paper...

In other words, infrared spectra convey information about molecular vibrations, which are not necessarily well characterized as vibrations of individual ligands or bonds (though that impression can be taken away from some undergraduate courses).  That the six terminal CO ligands are expected to to be equivalent with respect to the average point symmetry of the molecule does not imply that there can be only one vibrational mode to which their stretches are a major contributor.

@JohnBollinger I'm not sure if isotopic labeling (on behalf of CO) may shed information if the terminal CO stay in place, or interchange with the µ CO bridging.  On the other hand, copy-paste benzaldehyde (c1ccccc1C(=O)C) into MolCalc, request optimization, calculate properties / tab «Vibrational Frequencies».  Already the one around $\pu{155 cm^-1}$ conveys the impression that wiggling the methyl group influences the relative orientation of carbonyl group and phenyl ring. Sadly, JSmol used for the visualization only reports steady state distances/angles here.

@JohnBollinger Sidenote, I filed a follow-up question on mattermodelling.se, which might be of interest for you in regard of teaching IR spectroscopy, too.  And the observation regarding JSmol likely will result an other one.


text:
is this the reason why time of flight mass spectrometers produce a mass spectagram with mass over charge on the x axis?
Is this the reason why time of flight mass spectrometers produce a mass spectagram with mass over charge on the x axis?
I haven't used one but from what I understand, they have an electron gun and it is meant to knock one electron off an atom, but on rare occasions it may knock more off(probably one more, so knocking two electrons off). The machine isn't aware of how many get knocked off but in its calculation, it assumes one got knocked off 'cos really only one is meant to be knocked off.  KE=1/2 M V^2  M=2KE/V^2  If  two get knocked off, giving a charge of 2+, then the kinetic energy is doubled, and likewise if three electrons get knocked off then it gives a charge of 3+ and the kinetic energy is tripled.  So on rare occasions where more than one electron is knocked off, mass ends up too much, by a factor of charge.
There's no complete way to always correct for that as we don't know the charge when the gun messed up. But my understanding is that we label the x axis as "mass over charge", so at least that is true to what has been actually plotted on the graph. We might adjust for any errors that look not right , if we can. And we look at it as mass. (though bearing in mind that some results could be off a bit due to the charge issue).
Is that the reason and thinking behind why the x axis is labelled as "mass over charge"?

What else but mass over charge would you expect there? You cannot distinguish twice the charge and half the mass.

mass over charge comes into play in many situations involving electrostatic accelerators and beam deflecting magnets. It is a fact of life, and how many things get multiply charged or not depends a lot on the particular experiment (thinking well beyond mass spec).

@Poutnik  I did not suggest that you'd get something other than mass over charge there.  But perhaps you could  if you had a way of detecting the number of electrons in an atom and used that method before and after the gun had hit the atom, then you would know whether the charge of the atom was  +1 or +2 or +3   Or, a way of measuring how much energy is produced by the gun accurately enough to know whether it'd knock one electron off or two.   Ways of knowing how many electrons will get knocked off an atom and how many atoms an electron has.

@Poutnik or this  alternative to a mass spectrometer  maybe this doesn't have that problem as doesn't use an electron gun   https://www.internetchemie.info/news/2008/jul08/nanoscale-mass-sensor.html

@barlop - there are, of course, ways of getting at charge and mass separately, they are more complex than a simple TOF mass spec and generally are not needed if you understand that the TOF system is giving you what it is giving you. As a fun aside, the inventor of the TOF mass spec was a nuclear physicist who built and ran Van de Graaff-type accelerators for photofission experiments.

Is this the reason why time of flight mass spectrometers produce a
mass spectagram with mass over charge on the x axis?

Because this is what the mass spectrometer is designed to measure - mass over charge of an ion. The term mass spectrometer is a misnomer. They do not measure mass. Initially, they were called parabola spectrographs. The reason arises from mathematics. If you look at the equation of the motion of ions in an electric field or magnetic field, you will always find mass to charge ratio somewhere. It can be difficult to calculate the motion of a charged particle in a field. Lots of differential equations.
Plotting of m/z on the x-axis has literally nothing to do how the molecules are ionized. Also, this has nothing to do with electron gun in a mass spectrometer's ionization source. When you remove sticky tape rapidly from it reel, this can also ionize molecules. Just for fun, some people have even shown emission of soft X-rays when you peel the tape off.


text:
Mapping of half-life decay for all elements/isotopes
Is there an open resource that helps illustrate all paths of decay for all isotopes (e.g. $\ce{^{99}Mo}$ decaying to $\ce{^\mathrm{99m}Tc})?$ I know the result state/transition map would be massive, but am curious if anyone knows if something exists in open domain.
Bonus: In addition to the nature decay process, I'm also hoping to see if there is any documentation regarding known transitions we can trigger from different types of cyclotrons.

Why the downvote? I would appreciate inputs, if the question is unclear.

Check the links in the comments here: https://chemistry.stackexchange.com/q/119782/79678. A lot of information available! Have fun!

This one is cool: http://www.nndc.bnl.gov/nudat2/.

Welcome to Chem.SE, VenomFangs. Don't mind the anonymous downvotes. If your question is clear and on-topic, it will definitely be answered.

It turns out there are two kinds of scientists who use this kind of data, those who design nuclear bombs and reactors and their products on Earth, and those that study them in space, i.e. Astrophysicists. I think that Astronomy SE might turn out to be a better place for this question than here, even though it's a reasonable nuclear chemistry question. There are also some nuclear engineers at Engineering SE as well.

Mentioning cyclotrons, I guess the OP may be focused on processing nuclear waste by nuclear reactions, leading to a faster way to end with stable isotopes.

The National Nuclear Data Center noted by @EdV has a number of different resources for nuclear physics/chemistry. In addition to NuDat you might also be interested in the Evaluated Nuclear Structure Data File (including searching By Decay), and the Evaluated Nuclear Data File which will give you data on neutron and gamma reactions.

My background is more software engineering, the example provided is just one common example I knew of when working on software tracking radioactive isotopes from birth to grave. As part of that process, I found out about different types of cyclotrons and that they can make many different isotopes. So I'm curious what we know they can make, what they decay to, etc. Really looking for a giant map showing states and ways we can transition them, need it be naturally or via processes involving cyclotrons.

Part of some of the medicine creation process sounded like they needed to go through a large series of steps to create something of use (the person running the cyclotron gave me an example of something that took 27 steps, but it wasn't feasible because of the short half lifes until they automated things more). So this questions is trying to get at the building blocks of transitions, which might be able to be used by people (or AI) to identify new processes we may be able to automate to create more useful medicines (or anything else for other fields). :)


text:
Can two hydrogen atoms with parallel electron spins form a covalent bond?
Is it possible for two hydrogen atoms with parallel electron spins (i.e. a pair of spin-up electrons, or a pair of spin-down electrons)
to form a covalent bond? I've seen some say there is no sigma bond formation in that situation due to the Pauli exclusion principle, whereas others say one of the two electrons undergoes a spin flip, with bond formation occurring naturally after that.
I would expect 50% of collisions in a gas of free hydrogen atoms to have up spins, and 50% to have down spins. So, on average, any given pair of hydrogen atoms are equally likely to either have parallel or antiparallel spins. Does this imply that only 50% of H–H collisions in that situation result in a bond, i.e. would this impact the yield of the reaction?
How would this look with other elements?

The mixed answers you get are actually describing the same picture. Initially there is no sigma bond in that situation, and then there occurs a spin flip.

@IvanNeretin After the flip has passed, however, a sigma orbital is created. Where does the energy for the flip come from? Shouldn't it come from the fact that some is released by the bond?

The flip does not need any energy.

@IvanNeretin So one electron reorients itself by the magnetic field of the other before the sigma bond happens so that both then have an antiparallel spin and can then transition into the sigma orbital?

When you say "parallel" and "antiparallel", what do you mean? As in, parallel to what? Normally when people use these words it's in the context of an external magnetic field being applied, is this the case here? If not, then you may want to just use "up" and "down" spin (and in this case, the spin states are degenerate, so indeed no energy is required to go from up to down spin or vice versa). Only when you have an external magnetic field do spin flips require energy (and even then, only in one direction; the reverse flip releases energy).

@orthocresol By parallel I mean: Up-Up or Down-Down, by antiparallel I mean: Up-Down or Down-Up from each electron spin of the hydrogen atom.

Aha, I get you. 50% parallel and 50% antiparallel is a bit awkward, though; technically, yes, if you pick any two hydrogen atoms you will get parallel spins 50% of the time and antiparallel 50% of the time. But it's probably clearer to say 50% up and 50% down.

@orthocresol But I wonder now, why does it need energy when you have an external magnetic field? For example, would the spin of electron in hydrogen change if I hold a bar magnet near hydrogen?

@jemeto7849 You just rediscovered Magnetization.

My guess is that the parallel and antiparallel spin states for an H2 would be indistinguishable when the H atoms are infinite distance from each other, so it's a bit like Schrodinger's cat(?) because you can't say for sure which spin which H atom has. Only when the H atoms come close to each other and start interacting, the two spin states would have different energies and would be distinguishable. I don't know the answer to your specific question though.

In the absence of any magnetic field or other distinguishing system, I would expect all of the atoms to exist as a superposition of "up" and "down" such that it is meaningless to discuss a difference between them. The formation of the bond results in a stable state that has net zero spin, but even in that state it is not possible to assign up or down spin specifically to either electron

I'll specifically answer this part of your question:

I would expect 50% of collisions in a gas of free hydrogen atoms to have up spins, and 50% to have down spins. So, on average, any given pair of hydrogen atoms are equally likely to either have parallel or antiparallel spins. Does this imply that only 50% of H–H collisions in that situation result in a bond, i.e. would this impact the yield of the reaction?

TL;DR: No, only 25% of collisions will be able to result in a bond.
In general, when you "measure" something in quantum mechanics, you force the system into the eigenvalues and eigenfunctions to some operator. In your case, you are looking at the electron spins, and once you measure them, your overall system is collapsed into an eigenstate of the spin operator. For a hydrogen molecule with its two electrons, those eigenstates are a singlet state $\alpha \beta - \beta \alpha$, and a triplet state with three "microstates" $\alpha \alpha$, $\beta \beta$ and $\alpha \beta + \beta \alpha$.
In the non-interacting case, we assume that each spin result ($\alpha$ or $\beta$) is measured with 50% probability on each of the free atoms.1 Then, for any pair of atoms, we measure each binary spin combination $\alpha \alpha$, $\alpha \beta$, $\beta \alpha$ and $\beta \beta$ with the same 25% probability. If those atom pairs now draw closer together and start to interact, then the $\alpha \beta$ and $\beta \alpha$ combinations must combine to form either the triplet or singlet microstates with equal probability; in other words, isolated atoms with opposite spins have equal chances of forming a singlet or a triplet state when they interact. However, the $\alpha \alpha$ and $\beta \beta$ combinations are also still possible, and since they belong to the triplet state, the latter still wins over the competing singlet state with a 3:1 ratio. And since we know that only the singlet state leads to formation of a bond, we expect 25% of atom collisions to be able to form a hydrogen molecule.
NOTE: As @Paul pointed out in the comments, the explanation here is solely regarding the probabilities of bond formation that come from the spin statistics. There are further effects from collision cross-sections and energy dissipation that I am not taking into account here. In reality, less than the 25% of atom-atom collisions stated above would actually result in the formation of a bonded molecule.

You may wonder: "If we measure eigenstates to the spin operator, how can the isolated atoms yield the specific spins combinations $\alpha \beta$ and $\beta \alpha$? The spin operator is still the same, so why wouldn't we expect to see the singlet and triplet states again? What happened to them?"
At this point, it's vital to remember one important rule: Any linear combination of eigenfunctions to the same eigenvalue is again an eigenfunction. 2 When two atoms are non-interacting, measuring the singlet or triplet microstates of $\alpha \beta \pm \beta \alpha$ is still perfectly possible; but the same is true for their linear combinations that form the "distinct" $\alpha \beta$ and $\beta \alpha$ states.3 What you measure now depends on what answer you seek, and each set of answers is as valid as the other. (But again, this is only true if there is no energy difference between the triplet and singlet states.)

Footnotes
1 Note that I'm treating this as the premise of your question. If your ensemble of atoms does not conform to this assumption, e.g., due to the presence of external fields, then the answer will also be different.
2 This is very easy to show. Assume $\hat{O} \Psi_1 = E_1 \Psi_1$ and $\hat{O} \Psi_2 = E_2 \Psi_2$, with $E_1 \ne E_2$. Then, $\hat{O} (a\Psi_1 + b\Psi_2) \ne (E_1 + E_2) (a\Psi_1 + b\Psi_2)$. However, if $E_1 = E_2 = E$, then $\hat{O} (a\Psi_1 + b\Psi_2) = E (a\Psi_1 + b\Psi_2)$.
3 Let $\alpha \beta + \beta \alpha = \Psi_1$ and $\alpha \beta - \beta \alpha = \Psi_2$, then $\frac{1}{2} (\Psi_1 + \Psi_2) = \alpha \beta$ and $\frac{1}{2} (\Psi_1 - \Psi_2) = \beta \alpha$.

your analysis is only based on spin statistics and ignores reaction cross sections. In addition, it is good to note that a collision between two atoms in the gas phase can only produce a bound molecule if it dissipates the bonding energy by release of a photon or by collision with a third body. Since radiative transitions in molecular hydrogen are dipole forbidden, three-body collisions are required to form molecules.

True. I'll add a disclaimer.

This is a good answer to the core question (whereas I've only challenged the premise), but I object to the statement "In the non-interacting case, we assume that 50% of our free hydrogen atoms have α and 50% β spin." If you haven't measured them, the correct statement is "we assume a 50% probability of measuring $\alpha$ and 50% probability of measureing $\beta$ spin."

Okay, I see what you mean. I'll incorporate that.

I think the premise of your question is flawed, which isn't your fault. It's the way students are introduced to quantum mechanics at the intro level. But if you could somehow generate a gas phase of free hydrogen atoms, it isn't correct to say that 50% would have "up" electron spin and 50% "down." What is correct is that if you measured the individual spins you would get 50% up and 50% down.
What's the difference? That's the weirdness of quantum mechanical systems. Individual electrons do not have defined spin until you measure it. Prior to that, they exist in a superposition of both possible spin states with a 50% chance of each result if you take a measurement.
So in your hypothetical, when two hydrogen atoms interact, there are no different cases to consider. Every pair of hydrogen atoms are the same. You have two electrons of undefined spin. When the atoms bond, the electrons become entangled so that the now diatomic system has net zero spin, but still neither electron has defined up or down spin. Only the net spin is defined.
If this doesn't make much sense, you're not alone. Quantum mechanics in many ways does not make sense. All we can say is that the empirical description we've come up so far does a very good job of predicting experimental outcomes. Why that description works is utterly unknown.

"when two hydrogen atoms interact, there are no different cases to consider. Every pair of hydrogen atoms are the same" - This seems wrong to me. If we measure their spins, we must get an eigenvalue to the total spin operator, and those eigenvalues will correspond to overall triplet or singlet states with their corresponding 3:1 weights.

@Antimon - Yes, if we measure their spins, they must collapse to a state represented by an eigenvalue. But prior to measurement, they have no defined spin. So if you somehow picked out a pair of atoms and measured their spins, you could create a pairing of up/up or down/down. But in a bulk mixture that hasn't been measured, all atoms are the same.

I'm still missing a connection between the bulk state of isolated atoms and the formation of a bond. I'll address this in an answer of my own, see if you can comment on that.

Yes -- if one or both electrons are in an excited state and they thereby occupy different orbitals. Of course, the excited state electron(s) would decay rapidly, so either the atoms blow apart from the resulting energy release or an electron spin-flips to preserve/strengthen the bond.


text:
Is water hydronium and hydroxide?
In our chemistry lesson when learning about the Bronsted-Lowry definition for acids and bases, we came across the reaction...
H2O + H2O -> H3O+ + OH-
...Where water is amphiprotic which means it acts as an acid and base. Does this mean that water is a combination of hydronium and hydroxide? How is it not harmful to drink then?

https://en.wikipedia.org/wiki/Self-ionization_of_water, or other Internet resources on the same topic, may help.

Numbers matter.

If there are arrows going both ways then that means it’s in equilibrium between the right side (products) and the left side (reactants).
It doesn’t mean that water is a mix of H3O+ and -OH, a vast majority of water will stay H2O, and the small amount of H3O+ or -OH wouldn’t be anywhere near a concentration to hurt you
I think the purpose of that was to show that water has the potential to form H3O+ and -OH in itself in an attempt to teach you about acid base equilibrium


text:
Equation describing electrons in a atom/ion
When not only 1 electron exist in the atom/ion the electrons of the atom/ion interact with each other and with the nucleus.
Do we solve this equations:

or this equation?

x is the position of a electron inside the atom/ion.
and we include the collective electric potential of the nucleus and all electrons inside the Hamiltonian?
Since the first equation will give different results than the second equation this will surely affect the shape of the orbitals and the chemical properties of the atom.
When do the different chemical properties become important enough?

It is complicated.

Bethe and Salpeter wrote a whole book on one- and two-electron atoms. A classic in quantum mechanics...

What exactly is the difference between the two equations you posted? What are $x_1$, $x_2$, ... $x_n$, and $x$, supposed to mean? Are the Hamiltonians different (this seems to be implied, but not clear)? Do the $\Psi$'s represent different types of wavefunctions (again, seems to be implied, but not clear)?

x1,x2...xn are the positions of all the electrons of the atom

OK, but what about $x$ without a subscript, and $H$, and $\Psi$? Like, it's obvious that $H$ is some Hamiltonian and $\Psi$ is some wavefunction, but it's not obvious what Hamiltonian and what wavefunction it is, like is it single-electron or multi-electron? I suppose by $x_1$ you really mean a vector $\vec{x}_1 = (x_1, y_1, z_1)$?

@orthocresol the Hamiltonians are different because they are for a different number of particles.

@orthocresol Exactly.

Sorry, but I asked a lot of questions to try and clarify what you're asking, and your responses don't entirely help. You need to explain all the terminology you're using, if your Hamiltonians (and wavefunctions) are different, then perhaps it's not a good idea to use the same symbol $H$ (and $\Psi$) in both equations? You should [edit] your question to include this detail, don't need to reply to me in the comments.

The Hamiltonian in the first equation is for  n particles while in the second equation it is for a single particle.

@orthocresol - I think what OP is saying is that the second version is similar to a DFT type approach - solving single electron equations while treating the rest of the electrons as a bulk potential field. Right?

Who's on first?

The short answer is that we'd like to do the first, but solving the multi-electron Schrodinger equation is impossible, so we use variations of your second option (ie considering many separate one-electron wavefunctions) to get something that matches experimental data as closely as possible, even though we don't ever get an explicit form of the true multi-electron wavefunction.
Two of the most common approaches are 1) Hartree-Focke self-consistent field theory and 2) density functional theory (which I think is closer to what you have in mind). Both are much longer topics to explain than a short SE answer, but it's easy to find information on both in chemistry texts or online.


text:
Why we use various layers in OLED (ETL,EIL,HTL,HIL)? How they increase efficiency OLED device?
Currently, I am studying the working of OLED devices! The device can work with just three layers which are Anode, Cathode and Emission layers. I know the fact that the work function is an important parameter in the device. I have read articles about Various layers which can we add in devices to increase the overall performance of the device.
The layers we add for improving efficiency are as follows

EIL (Electron Injection layer)
ETL (Electron transfer layer)
HIL (hole injection layer)
HTL (hole transfer layer)

I know the fact that To overcome losses in devices, a step-wise energy transport architecture is used, and the work function of layers are in between the emission layer and electrodes.
My question is How those layers can help in increasing efficiency? Why do we use them?
I need a detailed explanation of this question with proper work function values

The question is (perhaps too much) influenced by the view of cells built layer-by-layer (e.g., cathode, electron acceptor, electron donor, the transparent anode) where two in the middle total for about ~100 nm. But excitons in OLEDs die quickly, and must be split within ~10 nm of travel, one reason favouring ordered (stacked comb-like) hetero bulkjunction cells, or (simpler, e.g., spray coating produced) disordered hetero bulkjunction cells. The optimization of OSOLs isn't only optimization of the components (e.g., crystallinity vs. density of traps); a simple brief answer here won't fit.


text:
The use of far-infrared ceramics for water purification?
I am struggling to understand how exactly far infrared ceramic balls help purify the water. It is stated that active ceramic taormine balls help purify the water by forming a 2 micrometer wavelength electro-magnetic field (far infrared-red ray) along the water passageway.

Water molecules have a natural oscillation frequency of 2 micrometers. As the water passes through the filter, the FIR field will help in breaking up the water molecules (which are usually clumped together) into smaller pieces. This will help in maintaining a healthy body metabolism rate.

Far-infrared rays suppress the growth of excessive free radicals and shorten the cluster chain of water. The purified and activated water has smaller water clusters, which are more efficient at protecting cells from various stimulants.

Claims like these are made by companies. Is this feasible?

Claims about products associated with selling such products should not be generally believed until confirmed by independent sources not related to the selling profit. The last 2 paragraphs seems to me like tossing around scientific-like terms to overhelm ordinary people to believe there is big science behind.

I think this is a well-elaborated question that can be useful for the future reference. Probably it's self-evident for specialists, but for average consumer it might be not. Instead of criticizing the post, I would thoroughly express why and how these promotion materials are wrong so that the fraud becomes evident for everyone. The question got many triggered, so use the rage against the companies' shady practices, not a person's post who dared to question their legitimacy. This scam needs to be exposed, not hidden.

@andselisk Good point, but I wouldn't know where to start. These claims are so totally random ... :D

@Albert 1. Water has no oscillation frequency, 2µm or other. 2. ceramic balls at room temperature do not emit any more FIR radiation than any other substance. 3. water clusters (which exist!) break up and re-form randomly at a tremendous rate anyway. (so much for homeopathy). 4. FIR energies cannot make or inhibit radicals. Sorry. Who tries to sell this? https://www.osmiowater.co.uk/ Vitamin C shower filter, great. :D

This one gave me a good laugh, too: https://www.osmiowater.co.uk/water-filters/deuterium-depleted-water-ddw.html

Consider the gullibility factor. Some people even believe political campaign promises!

Suppose the balls would irradiate the water with infrared light, what would be their driving force doing so?  You can't generate energy ex nihilo (one reason there are (at least some) patent offices refusing grants on a perpetuum mobile).  And depending on your jurisdiction, treatment of water is regulated; the European (incl. UK) side of the pond actually a bit more than the US American, as Tom Scott once summarized for Coca-Cola's fail to market Dasani in Britain (youtube video).


text:
When doing free radical substitution, is the free radical a catalyst?
Consider a simple free radical substitution's propagation step for $\ce{CH4}$ to $\ce{CH3Cl}$:
$$
\begin{align}
\ce{CH4 + Cl^. &-> CH3^. + HCl}\tag{R1}\\
\ce{CH3^. + Cl2 &-> CH3Cl + Cl^.}\tag{R2}
\end{align}
$$
My understanding is that since the free radical is regenerated at the end of the reaction, it must therefore be a catalyst of some sort. Is this understanding correct or flawed?

You may find useful these links for text formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

Yes, $\ce{Cl^•}$ acts as a catalyst here.

@Poutnik Thanks for letting me know! I was trying to find the command for the last 5 mins but couldn't find it and thus went for the last resort!

Hmm,  $\ce{Cl^.}$ as $\ce{Cl^.}$ - nice!

@Poutnik No need for \bullet, \cdot or other macro or Unicode chars. Just use ^., as it's stated in manuals for MathJax-mhchem, $\mathrm\LaTeX$-mhchem, or another package chemformula.

See also IUPAC Gold book: Catalyst

@Poutnik thank you for the help with formatting in chemicals. I find it very useful.

@Poutnik Following the gold book, I know realize that this is a case of autocatalysis. Thanks

Your interpretation is interesting. As you state, the free radical $\ce{Cl·}$ is regenerated at the end of the reaction. It behaves like  Vanadium(V) which oxidizes $\ce{SO2}$ into $\ce{SO3}$. And once reduced into Vanadium(IV), it gets reoxidized by $\ce{O2}$ into Vanadium(V), according to the two equations $$\ce{V2O5 + 2 SO2 -> 2 VO2 + 2 SO3}$$
$$\ce{2 VO2 + O2 -> V2O5}$$ The main difference between the free radical substitution and the vanadium oxide catalysis is the fact that both vanadium oxides do exist as stable compounds. And the radical $\ce{Cl·}$ is not stable and cannot be isolated.

The question is, if we should consider transient, intermediate products in nonlinear complex reactions as catalysts. IUPAC Gold book seems to do so, but.....

Well, consider this argument on the undoubtedly actual expanded system, which would commence with the action of strong UV on a Chlorine gas in presence of Methane, and seemingly would proceed based on random kinetics, as follows:
\begin{align}
\ce{Cl2 + UV &-> Cl^.  + Cl^.}\\
\ce{CH4 + Cl^. &-> CH3^. + HCl}\\
\ce{CH3^. + Cl2 &-> CH3Cl + Cl^.}\\
\ce{Cl^.  + Cl^. &-> Cl2}\\
\ce{CH3^. + CH3^. &-> Products}\\
&\vdots
\end{align}
where not all reactions are equally likely again based on kinetics.
In the above more complex system approaching reality, I would claim the Chlorine presence (certainly in the form of the $\ce{Cl^.}$) has catalytic aspects. Note, both the chlorine radical and the methane radical are cycled within the system, albeit, the methane radical self-reaction is a termination step, whereas the chlorine radical self-reaction restarts the system. However, Chlorine is consumed in the system so not by definition a catalyst.

Good point that chlorine is consumed. Only the radicals regenerate each other. I don't see what you mean by "kinetics", likely the wrong word.

On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

@Karl won't the radicals be consumed if a chlorine radical hits a methyl radical?

MaguireFan123: Yes, indeed .CH3 + .Cl = CH3Cl  and the reverse may also be possible under UV light. Note: my use of dots to convene possible continuation of reactions.


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Hydrogen fuel cell purity / concentration
This is a chemistry/technology kind of question. Some PEM fuel cells work with a rated hydrogen purity of 99.99%.
What would happen if the fuel cell stack is supplied with a hydrogen purity/concentration below the rated value?
Would the lack of hydrogen damage some cells?
I guess that the fuel cell stack would have a lower efficiency and that would depend on the stack design and technology, but most important, would that low purity hydrogen accelerate the degradation of the stack? I am assuming that the hydrogen concentration is not monitored by a microcontroller that can shut down the fuel cell.

AFAIK hydrogen full cells require efficient catalysis which can be severely and irreversibly affected by impurities blocking active catalyst centers. The required purity is kind of trade-off decision between operational and lifetime cost.

Look up "catalyst poisoning". The difference inquantity between 99.90% and 99.99% is irrelevant. You might also see how a catalytic lighter works. https://en.wikipedia.org/wiki/Lighter#Catalytic_lighter

Thank you for the comments and answers.

It isn't only the concentration (and eventually, the amount) of impurity which has to be considered, but the chemical nature of the impurity.  Read the certificate of analysis by your supplier.

If you use transition metal electrodes which are sensitive to sulfides, and this is the main impurity of the gas used, you may alter the the electrodes locally (chemisorption), and eventually irreversibly (by reaction).

If on the other hand the hydrogen gas contains traces of e.g., argon not participating in the electrochemical reactions, then the efficiency of your fuel cell will alter only as long as the impurity is present.


text:
Finding dissolved amount of base from pH and solubility product

Calculate the amount of $\ce{Mg(OH)2}$ which is solubilized in $\pu{1.25 L}$ of a buffered solution at $\mathrm{pH} = 9.5$. $(K_\mathrm{sp}(\ce{Mg(OH)2}) = \pu{5.61E-12})$.

From $\mathrm{pH}$ I compute the $\mathrm{pOH}$, so $[\ce{OH-}] = \pu{3.16E-5}$. But how can I compute $[\ce{Mg(OH)2}]$?
$[\ce{Mg}^{2+}]=\frac{K_\mathrm{sp}}{[\ce{OH-}]^2}$ and then?
$[\ce{Mg(OH)2}]=[\ce{Mg^2+}]+[\ce{OH-}]^2$?

You compute [$\ce{Mg^2+}$] not [$\ce{Mg(OH)2}$] // You may find useful these links for text formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

I need $[\ce{Mg(OH)2}]$ for the solution, no?

If [$\ce{Mg^2+}$] is e.g. $\pu{1 mmol L-1}$, then $\pu{1 mmol}$ of $\ce{Mg(OH)2(s)}$ has been dissolved in 1 L of the buffer.

Thus $[\ce{Mg^2+}]=\frac{[\ce{OH-}]}{2}$, then $[\ce{Mg^2+}]=[\ce{Mg(OH)2}]$?

Neither, neither.

Let us continue this discussion in chat.


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How did we find relative atomic masses in 18th century or early 19th century?
The atomic masses of atoms or relative atomic masses were known in the 19th century.
But how did they actually calculate that relative atomic mass?
What I think is that if we have a sample of an element, let's say carbon, then we can find its atomic mass by dividing the total mass of the sample by total number of atoms in it.
But how could we find the number of carbon atoms in a sample?
(We are in early 19th century)
How could we do this job of finding number of atoms in a sample to find the atomic mass?

Aside of the referred question, eventually more relevant is the StackExchange site History of science and mathematics

Before the widespread use of spectroscopic methods, the preparation of derivatives for the qualitative and quantitative analysis was much more common. E.g., carefully executed reactions to solute ores, separate ions by adjustment of pH and eventual precipitation (gravimetric analysis). Changes of masses were monitored closely by precision balances. But you can't miss Morley still recognized today for his contributions in this field, especially to establish a reference on oxygen.

The reason i used the term early 19th century is to eliminate the answers that use atomic numbers and no. of neutrons in nucleus to calculate its mass. Also people may say that hydrogen is 12 times lighter than carbon-12 but my question is how do we know that?


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Why an atom is more stable when only sublevels s and p are full?
Supposedly when explaining electronegativity and stability of an element, they tell you that it is more stable if the last level is full. That works up to the third period, but after transition elements, they are full when the d sublevels are full, not when the p sublevels are full. Why is this? Is there any simple way to justify when one is teaching this to a high school level student?

do you some source articles that says what you are claiming?

Supposedly when explaining electronegativity and stability of an element, they tell you that it is more stable if the last level is full.

No "supposedly" about it.  That concept is definitely taught.

That works up to the third period, but after transition elements, they are full when the d sublevels are full, not when the p sublevels are full. Why is this?

You seem to be defining "level" as all the orbitals with a given primary quantum number.  Perhaps that comes from relying too heavily on the Bohr model of the (hydrogen) atom, which gets less and less adequate for elements with higher and higher atomic numbers.  In any case, that is not the definition of "level" that applies to the concept.
The definition of "level" (a.k.a. "shell") that applies is more along the lines of "all the orbitals populated in the ground state of any neutral element in the nth period but not in the ground state of any neutral element in the (n-1)th period".  I can appreciate that you may not find that definition very satisfying, but the whole concept is somewhat empirical.
In my experience, usually no descriptive definition is taught at all.  Instead, students are directly taught the pattern of which orbitals appear in which levels.  One could use this as an opportunity to point out the limitations of the Bohr model.

Is there any simple way to justify when one is teaching this to a highschool level student?

The fact that some levels comprise orbitals with a variety of primary quantum numbers is typically explained in terms of the relative energy of electrons in the various elements' various orbitals.  That is, for elements whose nd orbitals are occupied in the ground state, those are close in energy to the (n+1)s and (n+1)p, but comparatively far in energy from the ns and np.  An explanation of why the relative orbital energies follow the pattern they do would not be accessible to most high school chemistry students, plus it would constitute a diversion that available instruction time typically does not permit.

What you wrote might be a bit confused and you didn't include sources for what you said.. You may have misinterpreted your own material.
There is a pattern that breaks and i'll address that
In basic chemistry they can teach electronic configurations up to and including Calcium and stop there. And not go to Scandium. And they'd say(rightly) Calcium's electronic configuration is 2,8,8,2.
Things go funny after that because in basic chemistry they might say pretend that the third shell can take only 8 electrons. Really it can take 18 but pretend it can take 8.  So then you get as far as calcium(atomic number 20).
You mention transition metals. Scandium(atomic number 21)  is the first transition metal, where in basic chemistry they might say let's stop just before Scandium. Because the pattern up after 2,8,8,2 then changes.
What happens at Scandium is after the third shell gets 8 electrons, the 4th shell has to get two, (hence 2,8,8,2), but then, the third shell can start filling. And after the third shell is complete. 2,8,18,2 then the fourth shell can be added to e.g. 2,8,18,3 e.t.c.  Often that format of electronic configuration is used in basic chemistry and up to Calcium.  This online periodic table shows the electronic configuration in that format for each element.  https://ptable.com/#Properties/Series
It then becomes easier to use a model that is more complex, but says, there's each shell, 1,2,3,4,5,6   (maybe the order there is proximity to the nucleus). And each shell has subshells.
But the factor that controls the order that they fill up is "energy level". And the order of energy level goes a bit interesting around 3d(a subshell of the third shell), and 4s(a subshell of the fourth shell). And it's all simply subshells filling up.
They don't teach about the subshells at very basic chemistry.
The order of subshells filling up is, which is in ascending order of energy level, low to high, is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f ...
There is a diagram to help work it out by hand, and a graph showing energy levels too,  and the periodic table helps.
So you see 3p fills up, then 4s, then 3d, then some more 4.. So it's one subshell that fills up and then the next and then the next, in order of their energy levels.
It's not one shell that fills up and then the next e.t.c. because the shells don't divide completely by energy level as e.g. part of shell 4 could be higher than parts of shell 5.  It's that one subshell that fills up and then the next e.t.c.
And Where they teach that if the outter shell is full then it's more stable, I can't comment on whether that's true even pre calcium. But you can adapt that to the highest energy subshell. i.e. the last subshell to be filled.
You mention s,p,d.. so you might be aware that each subshell e.g. s,p,d,f..  contains orbitals and each orbital can contain a maximum of two electrons.
They tend to not teach the subshells and those electronic configurations in basic level chemistry but you seem to have researched it somewhat.
You can look at a periodic table where they block parts off, as 1s, 2s, 2p, e.t.c. 4s precedes 3d.  And that is helpful for predicting or determining electronic configurations. And there are "exceptions" like copper and chromium in the 3d section. And even more exceptions in 4d, as you see from that periodic table showing electron configuration https://ptable.com/#Properties/Series
One sees from a diagram showing energy levels of orbitals, https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/07%3A_The_Quantum-Mechanical_Model_of_the_Atom/7.06%3A_The_Shape_of_Atomic_Orbitals  "the energy levels become closer and closer together as the value of n increases".  If orbitals are close, there can be some competition.
An oddity re this is that in academia there seems to be some discussion that the order of orbitals in order of energy differs a bit, from element to element, and  debate on whether 4s<3d in the case of scandium. Eric Scerri argues that actually 3d<4s.  Whereas  Geoffrey Neuss https://www.thinkib.net/chemistry/page/37492/the-electron-configuration-of-scandium  mentions this paper of his freely available on springer.  And says that even though recent papers have argued that 3d<4s for scandium,  he argues in response to those recent papers, that we don't know if 3d<4s for scandium.
At a very high level they understand  some further things about orbitals and energy levels. https://pubs.acs.org/doi/pdf/10.1021/ed200673w
There is no dispute as to the electronic configurations of the elements. So you can see e.g. for potassium,calcium,scandium. A list of electron configurations of elements is here, in the notation that shows the subshells. https://sciencenotes.org/list-of-electron-configurations-of-elements/

I just read that the situation is actually different to what is commonly taught even re subshells. see https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium and  http://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html and  slightly relevant is some answers to https://archive.is/3kn7s

And https://chemistry.stackexchange.com/questions/162292/for-calcium-and-potassium-as-4s-is-then-lower-energy-than-3d-and-lower-means-o
mentions https://pubs.acs.org/doi/pdf/10.1021/ed200673w  or https://web.archive.org/save/https://www.docdroid.net/ingJp9G/ed200673w-pdf

No. Transition metals without any electrons in the 4p or 5p orbital are very reactive just like atoms without 2p or 3p orbitals fully filled.
The reason is quantum mechanical .Without any details if you solve the time independent Schrodinger equations for atoms which do not have fully filled the p orbitals by combining them with other orbitals of atoms the energy is reduced.


text:
How can boric acid forms pyroboric acid if it is a monobasic acid?
Pyro acids are prepared by removing one water molecule from two molecules of oxyacid. $\ce{H3BO3}$ has only one acidic hydrogen so two of that molecule will have two acidic hydrogens. If we remove that, then there won't be any acidic hydrogen and hence it shouldn't be able to form pyroboric acid or at least it should not be called an acid as it doesn't have $\ce{H+}$ ion to donate. This is the case with hypophosphorus acid, but why it isn't with boric acid?

$\ce{B(OH)3 + H2O <=> B(OH)4- + H+}$ // $\ce{HBO2}$ is a cyclical trimer with alternating O and BOH.

It is mono basic Lewis acid. But not an Arrhenius acid. So it doesn't have any acidic hydrogen.

@Infinite That isn't true. Each one of them can be abstracted, it's just like with any other protons.

@Mithoron I thought that boric acid is acidic because it abstracts HO- from water rather than because is adds protons to water which makes that possibility irrelevant in aqueous solution.

@matt_black It does both, to a similar extent.

From my answer:

The IUPAC goldbook states that:

"pyro" is used as a prefix designating compounds formed by heating a compound, usually with the elimination of water, carbon dioxide, or other simple molecule, e.g. pyroglutamic acid from glutamic acid.

Since Pyroboric acid is made by heating boric acid, it is prefixed as pyro. You can also call it tetraboric acid.
$$\ce{\underset{orthoboric acid}{4H3BO3} ->[373 K] \underset{metaboric acid}{4HBO2} ->[435 K] \underset{pyroboric acid}{H2B4O7} ->[red hot] B2O3  }$$

Hydrogen atoms need not bevunalterqbly acidic or non-acid for all time. Chemical reactions elsewhere is a molecule can change a hydrogen atoms acid-base reactivity
Suppose you were to react acetoacetic acid, $\ce{CH3-CO-\color{blue}{CH2}-CO-OH}$, with a strong base such as a Grignard reagent. The base extracts the catboxyl proton, and that's it. In the resulting anion no more hydrogen atoms, including those in the blue methylene group, have significant acidic properties.
Now suppose you esterofy the acid with ethanol, making what is commonly called acetoacetic ester, $\ce{CH3-CO-\color{blue}{CH2}-CO-OC2H5}$, and then apply the strong base. The base now deprotonates one of those hydrogen atoms in that methylene group. Removing the original acidic hydrogen from a different functional group has converted the methylene hydrogen from non-acidic to acidic.
Similarly, when boric acid polymerizes, it is essentially esterifying itself; but it's proton or Lewis acidity is not necessarily destroyed. It could be transferred to alternate sites.


text:
Should chemical names in figures be capitalised?
If I draw a molecule (for example glucose) and I want to name it, should I write glucose or Glucose? My aim is to draw a molecule, add the name below and then use this as a figure in a Word text.

It's really up to you. I usually stick with uncapitalised.

If you do this with the automatic function in ChemDraw ("Convert Structure to Name"), the label is not capitalized.

Treat chemical names as common nouns with uniform style. However, due to verbosity of certain chemical names, it might be a good idea to label each chemical structure with unique number and provide corresponding names in text. This is likely going to save a lot of hassle if you are asked to use PINs or fit your images into a column.

Stay with the rules set by IUPAC's colour books, e.g. Blue Book for organic compounds, and Red Book for inorganic compound, and  and don't capitalize the names if these names are used within an English phrase.  Descriptors like (E/Z) you always capitalize are a different thing, though.
If these documents are too long for you, you may consider their condensed versions (cf. chemistry.se resources page about material to learn chemistry, section nomenclature) for their brief guide and 4-page excerpts (guide, excerpt for organic; guide, excerpt for inorganic compounds).  Below the heading of the 4-page excerpts for both (which are published open access):

Both the brief guides, as well as the 4-page excerpts include flow charts how to determine the elements to consider (e.g., what are elements/substitutents; their number), how to name them, and how to sort them by priority. I still consider these and their intentional use of colour very helpful.


text:
How to calculate pH resulting from dissolving a salt in water?
The main question is how to calculate the changes in pH after dissolving a salt in water.
Suppose we dissolve 1 gram of magnesium carbonate (MgCO3) in 1 liter of Water. This will result in (approximately)

0.50 dGH
0.73 dKH
2.16 ppm Mg
7.84 ppm CO3

If we now add and dissolve 1 gram of calcium chloride (CaCl2.2H20) we get

0.39 dGH
0 dKH
2.73 ppm Ca
4.82 ppm Cl

(Assumptions: the water is pure distilled H20; the salt is fully dissolved; dissolution takes place at 20C and at sea-level (1 atm) pressure)
In total we will get dKH = 0.50 + 0.39, and dKH = 0.73 + 0.00
Since KH acts as a pH buffer due to its acid-neutralizing capacity, is it somehow possible to derive the change in pH from changes in KH and/or DH? If not, how can I calculate a change in pH from salt dissolution? Any background on how to do this would be welcome.

Welcome to Chemistry SE! MgCO3 solubility is 0.139 g/L so good luck in dissolving 1 g.  ( https://en.wikipedia.org/wiki/Magnesium_carbonate )

We have the policy which states that you should explicitly show your thoughts, effort and attempts to understand underlying principles and solve the question.  It'll make us certain that we aren't doing your homework for you. Otherwise, the  question may get closed and you may need to ask for  reopening when requirements are met.
As Homework is considered literal homework, exams, self-study questions, puzzles, worked examples etc.

dKH cannot be bigger than dGH, as the former represents Ca+Mg as bicarbonates, i.e. temporary hardness, the latter total Ca+Mg.

btw your question can be reopened once you edit it


text:
Cannabigerolic acid (CBGA) and cannabidiolic acid (CBDA): Solubility in ethanol
How much CBGA and CBDA can be dissolved in a given amount of ethanol at room temperature? I did a bit of research but could not find any source on solubility of these substances in ethanol.

For context, I'm looking to extract these from two different strains of ground cannabis flowers, one of which has a high concentration of CBGA and the other a high concentration of CBDA.

What makes this a fun experiment?

I'm not downvoting, but here are some suggestions about what is "wrong" with how the question was asked.  Questions on almost all Stack Exchange sites should be impersonal and objective, they should stick to the point, and they should ask only one question at a time.  For instance, the first and last paragraphs ("Sorry for …", and "Thank you …") are totally irrelevant, as is the personal "fun experiment" sentence. In general, Questions should avoid using "I", "me", "my", etc. Similarly this asks two independent questions, the minimum alcohol required, and measuring the success.

… As for the Question itself, it would be better along the lines of "Given specific weights and concentrations, what is the relationship between the amount of alcohol used and the amount of substance extracted?", and "How to determine the concentration of a substance in a tincture?".  These should be two separate Questions (and of course should contain more details than what I put in those examples).

@Karsten Theis: thanks for editing the title. I can totally understand it's fun only to me.

@Ray Butterworth: thank you for the detailed explanation, it helped me to understand how questions should be formulated here.

I did a bit of research but could not find any source on solubility of these substances in ethanol.

ChemSpider has some indirect prediction of the solubility of CBGA and GBDA in the form of a logP value. In a two phase system of water and octanol, substances with higher logP have a higher fraction in the octanol phase. Ethanol has a logP of −0.18, while the two compounds of interest have predicted values of about 7. This suggests ethanol would not be the best solvent, and water would be even worse. Something like Vitamin E acetate (infamous for vaping lung injuries) might be a better solvent.
The ChemSpider site also has information about toxicity, which typically increases with concentration or purity.

For context, I'm looking to extract these from two different strains of ground cannabis flowers, one of which has a high concentration of CBGA and the other a high concentration of CBDA.

The flowers probably have a pH near neutral (like other plant tissue). At neutral pH, both CBDA and CBGA would be deprotonated, making ethanol a poor solvent and water a good one.


text:
Organic Brønsted base that is not a Lewis base
Which non-nucleophilic organic bases get protonated by typical Brønsted acids (like $\ce{HCl}$, $\ce{H2SO4}$) but also don't form Lewis acid–base complexes with typical Lewis acids like $\ce{BCl3}$, $\ce{AlCl3}$, $\ce{ZnCl2}$?

I guess you ask about non-nucleophilic bases.

I am specifically not asking about non nucleophilic bases, since most of them form Lewis acid base complexes with typical Lewis acids. But of course all bases that do not form  Lewis-acid-base-complexes with typical Lewis Bases should be also non nucleophilic.

Better rethink that. Steric effects are like only thing that can create such effect. For example for https://en.wikipedia.org/wiki/2,6-Di-tert-butylpyridine

That is already very helpful. You can post it as an answer.

Meeting the condition you set is likely imposssible.
There are multiple approaches how to define an acid, and a base.  Despite differences between them, there are some which are related to each other.  For Brønsted and Lowry, a base is capable to attract and bind a proton; this already is an extension (mathematically speaking, a superset) of Arrhenius' focus on aqueous solutions.  But how do you do establish bond toward a proton? By extending a previously not engaged pair of electrons toward $\ce{H+}$; and this is the criterion of a base in Lewis' theory.
If you like a visual approach with Euler/Venn diagrams, than have a look at this illustration:

(image credit to Wikipedia).
And Lewis acids may, but needn't be an acid according to Brønsted and Lowry (e.g., $\ce{AlCl3}$).


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Molar conductivity calculation in a titration curve
At an exam for analytical chemistry, I was asked to calculate the molar conductivity of a solution for an ongoing titration. I just solved it the way our professor showed it to us but I have some doubts if it even is correct like this.
The question:
A diluted solution of hydrobromic acid is titrated with 2 equivalents of sodium hydroxide solution. Assume that the solution of $\ce{HBr}$ is diluted enough that the molar conductivity $\Lambda_0$ is the same as when at infinite dilution, and that volume change is negligible because $\ce{NaOH}$ is very concentrated. Calculate the values of molar conductivity at start of titration, equivalence point and 1 equivalent of $\ce{NaOH}$ beyond the equivalence point.
The given solution:
We just need to look at the tabulated molar conductivities. As there are only $\ce{H3O+}$ and $\ce{Br^-}$ ions, the molar conductivity at the start is $\pu{349.7 S cm^2/mol} + \pu{78.1 S cm^2/mol} = \pu{427.8 S cm^2/mol}$, respectively. At the equivalence point, as there are only sodium and bromide ions, the molar conductivity is $\pu{50.1 S cm^2/mol} + \pu{78.1 S cm^2/mol} = \pu{128.2 S cm^2/mol}$, and at $\pu{1 eq}$ above the equivalence point, we got one equivalent of $\ce{NaOH}$ and $\ce{NaBr}$, so $(50.1 + 50.1 + 78.1 + 198)\pu{S cm^2/mol} = \pu{376.3 S cm^2/mol}$.
My doubts:
First of all, if the value of Λ₀ is independent of concentration for our purposes, why do we need to take it times two for the end point of the titration? All what we are adding is another equivalent of $\ce{NaOH}$, but the solution already contains sodium ions. This would be comparable to a situation when we got $\ce{NaCl}$ and $\ce{KCl}$ dissolved near infinite dilution; would the chloride ions be counted twice?
And secondly, what does the $\Lambda_0$ (molar conductivity) even tell us? How would we use it to calculate the actual conductivity ($\kappa$) of our solution? All I know is that they are related with each other by
$$ \Lambda_0 = \frac{\kappa} {c^*} $$
Again, if we for example had a mixture of $\pu{1 mol/L}$ $\ce{NaCl}$ and $\pu{1 mol/l}$ $\ce{KCl}$ (same solution) and we calculated $\Lambda_0$, would it basically be $\Lambda_0 (\ce{NaCl}) + \Lambda_0 (\ce{KCl})$? And what concentration $c$ would we use, equally $\pu{1 mol/L}$, or $\pu{2 mol/L}$ as there is $\pu{1 mol/L}$ of 2 species?

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.  And with \pu{}, it is easier to type units consistently here, including a space which is retained, including in proximity to a line feed.

The addition of an aqueous solution of $\ce{NaCl}$ ($\pu{1 L}$ of $c = \pu{1 mol/L}$) to an aqueous solution of $\ce{KCl}$ ($\pu{1 L}$ of $c = \pu{1 mol/L}$) doesn't alter the concentration of $\ce{Cl^-}$, but halves both the concentration of $\ce{K+}$ and $\ce{Na+}$ because amount ($n$), concentration ($c$), and volume ($V$) relate by $n = c \cdot V$. Note the change in volume by mixing the two solutions.

Yeah what you say makes sennse, but I meant a solution which has both 1M NaCl and KCl in it

Note that 1 M concentration is quite high and there are significant nonlinear and interfering effects. Molar conductivity at 1 M is significantly different to the limit one at infinite dilution. i.e. $\Lambda_0$.  // See also wikipedia.org: Variation_of_molar_conductivity_with_dilution

Yeah that was an example to simplify the calculation, it should be more around 0.001M or around that

In a mixture of electrolytes, conductivities and conductances are additive.
Let's consider the beginning of the titration, with maybe $0.01 M ~\ce{HBr}$. Its molar conductivity $\Lambda_0$(HBr) is $350 + 78 = 428 ~ \mathrm{S~ cm^2/mol}$. Its conductivity is : $\sigma$(HBr) = $ \Lambda_0 $(HBr) c = $\Lambda_o$(HBr) $0.01 = 4.28·10^{-3}~ \mathrm{S/m}$ . Please observe the change of unit ! Here we will admit that $\Lambda_o = \Lambda$ for the sake of simplicity !
At the equivalence point, the solution contains no $\ce{HBr}$ any more. But it is made of $0.01$ M $\ce{NaBr}$, if the volume change is negligible. Its molar conductivity $\Lambda_0$(NaBr) is $50 + 78 = 128 ~ \mathrm{S~ cm^2/mol}$. Its conductivity $\sigma$ is : $\sigma$(NaBr) = $\Lambda_0$(NaBr) · $0.01 = 1.28·10^{-3}~ \mathrm{S/m}$
If the double amount of $\ce{NaOH}$ is added, the solution is $0.01$ M $\ce{NaBr}$ + $0.01$ M $\ce{NaOH}$. The two molar conductivities are : $\Lambda_0$(NaBr) is $50 + 78 = 128 ~ \mathrm{S~ cm^2/mol}$ and $\Lambda_0$(NaOH) is $50 + 198 = 248 ~ \mathrm{S~ cm^2/mol}$. The conductivities $\sigma$ of the two components are :

For $\ce{NaBr}$ : $\sigma$(NaBr) = $\pu{1.28 10^{-3}}$ S/m
For $\ce{NaOH}$ : $\sigma$(NaOH) = $\Lambda_0$(NaOH) · $0.01$ = $2.48~  10^{-3}$ S/m

As the conductivities are additive, the mixture containing both $\ce{NaBr}$ and $\ce{NaOH}$ has a conductivity equal to $\pu{1.28 10^{-3} + 2.48 10^{-3} = 3.76  10^{-3}}$ S/m


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Can anyone identify this instrument part and what instrument it comes from?
Can anyone identify this instrument part and what instrument it comes from?

A better picture might help.

Chances to keep a question like this active increase by provision of context.  E.g., found in an analytical lab performing soil analyses; the nozzle-like shape has a fine opening (or not), material is of (magnetic?) metal, and -- like for a crime investigation -- add a scale to the picture.  This could be a ruler (preferred), or (if small) look into your pocket for a well known coin to be put next to the unknown object. If the camera used possesses a macro function (i.e., yields a better focus for objects in proximity), activate it.

It is the inside of a centrifuge where the rotor gets attached (the top spins, and the two pins ensure proper seat of the rotor on the conical piece). Here is a better picture of the inside of a Sorvall RC-5C Plus Floor Centrifuge, with the hands serving as scale bar. This is partially disassembled to show how one would change the motor brushing.

Screenshot from https://youtu.be/rO1q2eHTyJ0?t=434
If this is for a trivia game, I want to be credited. If this is for some crazy exam question, I'm sorry I helped cheating.