text:
Are electrodes in an electrolysis unique for every electrolyte or can any random electrode be used with an electrolyte during electrolysis
For example, during the electrolysis of water, platinum electrodes are used, why can't I use any other electrode as long as they are conductors? why does some electrolysis require specific electrodes to work, instead of just any random electrode?

Electrodes - especially anodes - may corrode or even getting dissolved. E.g. neither Pt neither Au dissolve in HCl. But using them as anodes of HCl electrolysis would form [AuCl4]- and [PtCl6]^2-.

Electrolysis can be carried out with all sorts of electrodes, provided they are conductors. But the result depends on the choice of the electrodes and of the choice of electrolyte, because the electrodes can be corroded by the electrolyte and by the product of the electrolysis, as mentioned by Poutnik about $\ce{HCl}$ electrolysis.

Any electrode will oxidize if the potential is high enough and there is nothing present that is more easily oxidized or, if given enough time, potentials are close and conditions right. We had platinum electrodes in flow meters completely vanish in water over several weeks at 10 volts potential. Someone told the engineers that platinum was not reactive. Electrode choice, composition and even shape and configuration are important. In electroplating of large objects the resistance of the electrodes means the current densities are a function of distance from the connections and must be allowed for if plating is to be uniform.


text:
Is it possible for an appropriately vaporized crystal to have vapor properties in any way related to it's crystal?
Is it possible for an appropriately vaporized$^{1}$ crystal to have vapor properties in any way$^{2}$ related to it's crystal, or unique to the fact it was a crystal in it's previous phase?
The motivation for this question is regarding any potential applications as an analysis method.
$^{1}$ As a non-chemist its difficult to define this properly. Essentially meaning under appropriately controlled lab conditions.
$^{2}$ Stochastically or otherwise.

Depends how loosely you define vapor. You could probably spall nanocrystals off a bulk solid at relatively low energies and keep them airborne, though if the fragments are too small, they will no longer resemble the bulk crystal.

interesting thanks @NicolauSakerNeto would such airborne fragments constitute as a phase change?

Converting a homogeneous substance entirely into a vapour does not provide a simple way of obtaining information about the thermodynamic stability of the condensed phase from which it originated, since gas phases typically (almost by definition) do not store such information.
However, the equilibrium vapor pressure of a condensed substance reflects the chemical potential of the condensed phase. At equilibrium the chemical potential of the two phases should be equal. This principle can be exploited to investigate the thermodynamic properties of a substance in isolation or as part of a mixture. So if it is possible to establish equilibrium conditions between condensed and vapor phases and to avoid formation of competing crystal forms, then the answer would be yes, this property (the vapor pressure) is related to the crystal form.
In addition, some techniques probe phase transitions to learn about materials, for instance temperature scanning calorimetric techniques. These might also monitor the rate of mass loss from the condensed phase. The rate of conversion into the gas (or other) phase can inform on the structure of the condensed phase.

See also (e.g.) https://acp.copernicus.org/articles/18/3419/2018/ and https://link.springer.com/article/10.1023/A:1010188923713

many thanks for the description and references, it is exactly from the analytic perspective i was thinking. unsure why downvoted, a broad question yes. but as your references show, a legitimate one.

@antimony I'm happy the answer was useful. Your question is a bit vague about the objective, but it brings up interesting ideas. You might want to clarify why this is of interest to you (the objective).

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Predicting acid-base equilibria
I was looking at acid-base equilibria of weak acids and weak bases. To determine the equilibrium constant of a reaction of those, I found a formula on the libretexts page:

Where ΔpKa is the difference of the product acid's pKa minus the reactant acid's pKa. This formula makes sense to me, but how can one arrive at this formula, starting with the general definitions of pKa and law of mass action?
An example:

Now if we add the two reactions together (using NaOH, hydronium and hydroxide cancel):

But how can we get the equilibrium constant for this reaction, using only the K values listed above, and thus arriving at the supposed result of K=0.01 (pKa of water is 14)? I am kind of stuck, any help is appreciated.

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Suitable formatting can be done according to formatting math/chem expressions/equations.

Consider a generic acid dissociation reaction of the form:
$$\ce{HE <=> H^+ + E^-}$$
Let:
$A$ represent $HE$
$B$ represent $H^+$
$C$ represent $E^-$
So that the reaction becomes:
$$\ce{A <=> B + C}$$
We're mixing 2 different weak acids together, so each acid will have a different $K_a$ value. Assuming (1) represents the stronger acid (higher $K_a$), and (2) represents the weaker acid (lower $K_a$), we have:
$$\ce{A_1 <=> B_1 + C_1,   K_a_1}$$
$$\ce{A_2 <=> B_2 + C_2,   K_a_2}$$
Now, because (1) is stronger than (2), it will work as an acid and force (2) to work as a base.
This means we need to flip reaction (2) so that the products are the reactants and vice versa.
When we flip a reaction, the new $K$ value is the inverse of the old $K$ value, and when we add two reactions together, the resulting $K$ value is the product of each constant.
So, we get the following system:
$$\ce{A_1 <=> B_1 + C_1,   K_a_1}$$
$$\ce{B_2 + C_2 <=> A_2, \frac{1}{K_a_2}}$$
$$------------------$$
$$\ce{A_1 + C_2 <=> A_2 + C_1 + B_1 - B_2, \frac{K_a_1}{K_a_2}}$$
Where: $$K_{eq}=\frac{K_{a1}}{K_{a2}}$$
If we take the base 10 logarithm on both sides:
$$log(K_{eq}) = log\left(\frac{K_{a1}}{K_{a2}}\right) = log(K_{a1})-log(K_{a2})=pK_{a2} - pK_{a1}$$
We then define:
$$\Delta P_{ka}=pK_{a2} - pK_{a1}$$
So then:
$$log(K_{eq})=\Delta P_{ka}$$
Finally:
$$K_{eq}=10^{\Delta P_{ka}}$$

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How does alkaline pipe cleaner corrode plumbing?
The siphon of my bathroom sink was clogged up again, and I became curious on why it clogged up so quickly each time. So I decided to open the pipe and was shocked to see the plumbing like this:

Initially I thought it was mold, but the white stuff has is rather solid and it might be some sort of salt. Inside I found two three EUR cent coins which have a steel core and a copper coating. They were heavily corroded and seemed to have fused with the inner wall of the pipe. The copper coating was open and the steel core visible.

When I put it back together, the pipe cracked when I applied some torque with the wrench. The wall must have become really thin. The crash has a rust color, and the whole inside of the pipe also has this typical rust red.

I guess that the pipe is made from steel with a chromium coating.
A prior tenant has excessively used pipe cleaner, this highly alkaline granulate. So we have a steel pipe, a chromium coating, copper coins with a steel core, lots of alkaline granulate and hand soap.
My chemistry education unfortunately stopped in high school, but I would be really interested in the following things?

What could the stuff on the outside be? Some sodium iron chromium salt?
I've read that the pipe cleaner is really bad for plumbing (that's why I never use that stuff), but I don't understand how say NaOH would interact with solid Fe. How does that work?
Could the copper coins have made the rust process of the pipe worse?

Martin, The key problem is we do not know the composition of the pipe. It is certainly a low quality piping material. Also there is not guarantee that the previous tenant used a particular type of drain opener. Both acids and bases attack metals (slowly or rapidly). The situation is more complex when you have (unknown) alloys.

As @AChem states, it's not a simple question to answer. For example, did the drain cleaner contain aluminum, which releases hydrogen and causes embrittlement? Was the pipe aluminum or zinc "pot metal", both attacked by drain cleaner? Was there electrolytic corrosion? usw...

I see! Unfortunately I don't know what metal the pipe is. Local hardware stores sell these pipes in steel with chrome coating, my landlord mentioned aluminum. And I don't know the type of pipe cleaner, Then I guess in either case the corrosion is likely to happen, but the details depend on too many unknowns. Thanks for your time!


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Covalent character of bonds in uranium hexafluoride
In one of the questions in a qualifying exam it said that $\ce{UF6}$ is a "covalent compound". This fits the physical properties of the compound well, e.g. low boiling point, existence as molecules, etc. However, it still seemed a bit counter-intuitive to me as I was taught that covalent character is attributed to (1) the difference in electronegativity of the two atoms and (2) the oxidation state of the central atom.
A Wikipedia search gave a $\Delta E.N.$ of 2.60 between $\ce{U}$ and $\ce{F}$. In comparison, $\ce{MnO4^-}$, a compound with 4 electronegative atoms as ligands, a high central atom oxidation state, and covalent character, has a $\Delta E.N.$ of just 1.89. It seems that the $\ce{U-F}$ bond should possess ionic character.

U(VI) can be considered much more electron withdrawing than, say, U(III).

Probably the bonding in $\ce{UF6}$ has significant ionic character; as with many compounds neither the extreme of "all covalent" nor "all ionic" would really be accurate.
Ionic character in $\ce{UF6}$ need not imply the high melting and boiling points we commonly associate with ionic compounds. We have to reckon with the 6:1 stoichiometric ratio of proposed anions to cations -- such an extreme deviation from equality discourages forming a strongly stable lattice (whether ionic or covalent with delocalized bonds). An ionic model of $\ce{UF6}$ could be an ambient gas of neutral clusters similar to sodium chloride or lithium fluoride at higher temperatures.


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Colour of ferrous hydroxide crystal
My chemistry teacher said that pure $\ce{Fe(OH)2}$ is a "white" crystalline solid in anaerobic environments, and acquires the usual green tinge, the colour of $\ce{Fe(II)}$, when in contact with air. Wikipedia also agrees with this:

It is produced when iron(II) salts, from a compound such as iron(II) sulfate, are treated with hydroxide ions. Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge.

The above paragraph does not have any references. However, it suggests that the green colour is due to substances collectively known as "green rust", which is composed of hydroxides of $\ce{Fe(II) / Fe(III)}$ in different proportions. The following photo shows different colours of $\ce{Fe(OH)2}$.

I was puzzled over the mechanism behind this colour change, as $\ce{Fe(III)}$ has a yellow tinge, not green. My teacher suggested the possibility that the oxidation of $\ce{Fe(II)}$ to $\ce{Fe(III)}$ induces changes to the lattice structure, causing a shift in its spectrum which happens to coincide with the colour of the ferrous ion.
Is this correct? Are there any reliable sources that support or contradict with this explanation?

All your tubes show a strong air oxidation, as the stuff deposited in the higher part of the tubes above the liquid is a brown powder, made of $\ce{Fe(OH)3}$ and due to a complete oxidation of $\ce{Fe(OH)2}$. It can also be seen in the liquid of the second tube.
But this oxidation occurs in two steps : first formation of green-black $\ce{Fe3O4}$ and then into brown $\ce{Fe(OH)3}$ The nearly black deposit in the first tube at the left-hand side of the picture is due to a partial oxidation of $\ce{Fe(OH)2}$. It is made of magnetite $\ce{Fe3O4}$, containing a stoichiometric amount of ferrous and ferric ions ($\ce{Fe3O4}$ looks like $\ce{  1 FeO + 1 Fe2O3})$. When $\ce{Fe(OH)2}$ starts to become oxidized, as in the two last tubes on the right-hand side, it produces a few $\ce{Fe3O4}$ in much $\ce{Fe(OH)2}$, and this mixture is dark green, due to the dark green color of the diluted magnetite. At the end of the oxidation process, like in the second tube, $\ce{Fe3O4}$  gets more thoroughly oxidized, and the deposit becomes brown, but not so dark, and it is made of pure $\ce{Fe(OH)3}$. The successive reactions are :
$$\ce{6 Fe(OH)2  +  O2  ->  2 Fe3O4  +  6 H2O}$$
$$\ce{4 Fe3O4  +  O2  +  18 H2O  ->   12 Fe(OH)3}$$
As shown in the equations, the white precipitate of $\ce{Fe(OH)2}$ is rather sensitive to the tiniest amount of $\ce{O2}$, as one molecule of $\ce{O2}$ is able to oxidize a huge amount of $\ce{Fe(OH)2}$ (6 molecules $\ce{Fe(OH)2}$  - $6$ is a lot with respect to $1$) not magnetite. And even magnetite is still more sensitive to $\ce{O2}$. as one molecule $\ce{O2}$ is able to oxidize $12$ $\ce{Fe}$ atoms of magnetite, so as to produce $\ce{12}$ molecules $\ce{Fe(OH)3}$. $\ce{Fe(OH)2}$ is easily transformed in magnetite by oxidation. But dark magnetite is still more easily destroyed by $\ce{O2}$ into brown, but relatively pale $\ce{Fe(OH)3}$.


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Python implementation of the Lilly Medchem Rules code?
As the title suggests, I am looking for a Python implementation of the Lilly filter from the paper: https://pubs.acs.org/doi/full/10.1021/jm301008n.
The shared code on GitHub (https://github.com/IanAWatson/Lilly-Medchem-Rules) by the paper's authors is in C and C++. I guess because the article was published in 2012 and Python wasn't as popular back then.
I looked for a Python (the language I know) implementation but couldn't find any. Can anyone point me or provide me with a Python implementation? Thanks in advance.

Out of curiosity, what is your goal in finding a Python implementation? Is it just to understand better how they implemented these rules? If you want to write a Python script that used these rules, you could just make calls to this existing code.

I actually need it for a research paper I am writing on an AI-based de novo drug design. The reviewers of a journal requested that I should pass the AI-generated molecules through the Lily filter. I actually want to know how it works as well as use it. So, a Python implementation would facilitate this for me a lot.

Have you looked at the supplemental information? Between that and the paper, it's not hard to grasp what is going on. Also, python was unleashed to the world in 1991. The code you refer to was written in C/C++ because of speed - the UNIX implementation goes back to at least 1995.

This is not only borderline off-topic, but may be violating homework policy. Also I suggest learning C++ sooner rather than later.

C++ is not a requirement, but learning how to use multiple languages in your tool chain is in the modern world

I am the primary author of the Lilly Medchem Rules software.
Thanks to all the previous posters who have given really good replies.
Yes, python was not a big thing back when this was developed. This runs on
very large datasets and performace is very important.
The rules can be run as a pipeline from python using subprocess, see
https://github.com/IanAWatson/MedchemRulesPythonPipeline for an
example.
The output format is admittedly horrible, and will be fixed one day.
But doing this in a way that avoids external dependencies is complex.
If all you care about is which molecules pass, just grab the second
column of the output - which is the starting name.
As for running AI generated molecules through the rules, that is
absolutely a good idea, and is being done in many places. But there
is a word of caution. The Lilly Medchem Rules was formulated with
the input from expert Medicinal Chemists over many years. We
implemented rules as problematic molecules were encountered.
Nobody ever encountered something like a 1,4 bridged benzene ring,
c12ccc(C1CN)cc2. That would simply never have occurred. So
that ridiculoous molecule passes through the rules without
complaint - because there is no rule against it.
We could implement a rule to prevent that, but then we get into
a never ending game of whack-a-mole with the AI generators.
There are techniques for dealing with such unprecedented
molecules, and myself and colleagues are working on a publication
to outline some of the things we have done to make AI generated
molecules useful.
But even if a molecule satisifies structural precedent requirements,
it is still important to apply the Lilly Medchem Rules
to drive out precedented, but nevertheless undesirable, molecules.
You should thank your reviewer. Your paper will be better.
If people have questions/suggestions or problems with the software, I am happy to help.
Ian

Great to have you here, welcome to chem.SE. And thank you for your contribution(s)!

Thank you very much Ian. This was very helpful.


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How do you calculate the pH of 0.01 M H2S solution (hydrogen sulfide, an amphiprotic acid)?
I found a question in a physical chemistry book

What is the pH of $\pu{0.01 M}$ $\ce{H2S}$ solution? Given, $K_\mathrm {a_1} = 9×10^{-8}$ and $K_\mathrm{a_2} = 1.2×10^{-13}$.

I thought of applying the following formula,
$$\mathrm{pH} =\frac{\mathrm pK_\mathrm{a_1} + \mathrm pK_\mathrm {a_2}}{2}$$
However, I got confused by another formula which states
$$\ce{[H+]} = \sqrt{(K_\mathrm{a_1}× C + K_\mathrm{a_2}×C)}$$
Which formula should I use and how do I approach this type of question?

There is a lot of available online material about pH computation, including but not limited to this site and chem.libretexts.org. Be sure you have done proper review. // You can also derive the formula, using equations for inventory and equilibrium constants.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

I am asking about how do I determine which formula to use

Do not  use formulas like "black boxes" without understanding. Learn basics by studying, e.g. searching of pH polyprotic acids .  Asking should be used when other ways failed, to fill remaining gaps, not to lay foundations. // For this particular case, the 2nd ionization can be safely ignored and pH can be calculated like for monoprotic acid.

Note that the simplified formula [H+]=sqrt(Ka.c) for pH of monoprotic acid assumes [H2S]>>[HS-]=[H+]. Check, if deviations are acceptable, before accepting the result. Otherwise, in the worst case none of relations fits, you would have to use the general equation derived from the 2 equilibrium (acid, water) and 2 inventory (acid forms, charge) equations.

$\ce{H2S}$ is not an amphiprotic acid, it is a diprotic acid (meaning it can donate two protons rather than one). You could say e.g. dihydrogen phosphate is an amphiprotic acid because it can either donate (up to two) or accept protons.
The easiest way to deal with this is to ignore the second proton, calculate the pH using the first equilibrium, and compare the pH with the p$K_\mathrm{a}$ to check if there is appreciable dissociation of the second proton.
The second dissociation constant is disputed [1], but because the pH will be acidic, you can neglect dissociation of the second proton. It will have negligible effect on the pH.
Reference
Li, Q.; Lancaster, J. R. Chemical Foundations of Hydrogen Sulfide Biology. Nitric Oxide 2013, 0, 10.1016/j.niox.2013.07.001


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Is there any material that, subjected to electricity, extends in size in multiples without substantial change in heat for human use?
Is there any material that can molecularly restructure so as to expand its intra- or inter-molecular distance by no less than three-fold without substantially losing its structural integrity and remain solid without substantial thermal change to the touch, say, other than during the expansion process by the use of electricity, electromagnetism, ultraviolet light, radio frequency, easily reversible chemical exposure, or any other effect other than substantial heat loss or increase?
A lower scale example would be water where if talking about actual H2O, it will freeze with substantial discharge on audible wavelengths at some −40 °C (not at 0 °C, only contaminated water freezes at 0 °C), however, in that case, water is in a much less dense structure than when contaminated. Theoretically one could contaminate and decontaminate water (plus heat refreeze) and change its density.
I am aware of no other molecular procedures that could achieve the same with the material being at room temperature before and after whatever affect it is subjected, and seeking to learn if we are aware of any.

No. $\ce{   }$

@Mithoron Why don't you post that as an answer?

I think you have some misconceptions as to what inter- and intramolecular forces do, and how an external factors contribute to a change in macroscopic properties. Intermolecular forces govern what phase some substance is in with stronger intermolecular forces meaning a substance remains a solid at higher temperatures. Forces can't be expanded or contracted.
Applying a voltage to some sample could change its chemical properties, but only slightly. A voltage could theoretically induce a chemical change making the substance less dense, and a high enough voltage will turn basically any substance into a plasma.
Also, you're going to need some substantial evidence for your clam that

H2O will freeze with substantial discharge on audible wavelengths at some -40 C° (not at 0 C°, only contaminated water freezes at 0 C°)

Water freezing requires a nucleation point, like any crystal, but most anything -- including container walls -- can act as this nucleation site. And substantial discharge on audible wavelengths? I've taken many chemistry courses, but never heard of this.

My understanding about intra- and extramolecular forces are so limited, indeed, that I merely assumed the existence of these concepts based on my limited understanding of chemistry — hence I asked the question. Thank you for the answer!

“And substantial discharge on audible wavelengths?” I remember this back form about 2012-‘13 where what, as now I learn, is called “nucleation” takes place. It was described as this sort of freezing taking place with robust noise or potentially if you contaminate water while on the way to nucleation below zero degrees Celsius? My recollection may fail me — feel free to fix as necessary. As you mention “theoretically” the question was thought experiment for me although with the hope of real life use. Before you answer my own was, also, that likely the forces would at least linearly reduce[…]

in inversely proportionate to the increase in distance between molecules, but potentially exponentially so even if somehow something remained solid despite being subjected to any hypothetical affects to expand, the sturdiness would reduce to such extent that the application needed for would not be served to sufficiency.

At the molecular level, intermolecular forces are governed by the interactions of electrons a quantum mechanics. Yes, applying a voltage to some substance changes its properties (that's how electricity works), but I'd be hard pressed to say some material could operate in the way you described. If there was some clear relationship between applied voltage and material properties, it would take years, maybe even decades, of materials science research to engineer such a material.

https://en.wikipedia.org/wiki/Piezoelectricity is somewhat related...


text:
Energy values of steam reformation, water gas shift, hydrogen and methane burning seem to not add up correctly
I found the following reactions including the Energy-Delta all in Wikipedia and in at least one more source each:
(Delta with "-" are exothermic reaction which is generating heat)
Steam Reformation:
$$\ce{CH4 + H2O <=> 3H2 + CO}\quad\Delta H=+206\ \mathrm{kJ/mol}\tag1$$
Water gas-shift:
$$\ce{CO + H2O <=> CO2 + H2}\quad\Delta H=-42\ \mathrm{kJ/mol}\tag2$$
Hydrogen heat of combustion:
$$\ce{2H2 + O2 <=> 2H2O}\quad\Delta H=-572\ \mathrm{kJ/mol}\tag3$$
Methane heat of combustion:
$$\ce{CH4 + 2O2 <=> CO2 + 2H2O}\quad\Delta H=-891\ \mathrm{kJ/mol}\tag4$$
Now, if I add to (1) $\ce{H2O + 2O2}$ I may do (1),(2), and (3) in a row:
$$\ce{(1) CH4 +2H2O +2O2 <=> 3H2 + CO +H2O +2O2}\quad\Delta H=+206\ \mathrm{kJ/mol}$$
$$\ce{(2) <=> 4H2 + CO2 +2O2}\quad\Delta H=(+206-42)\ \mathrm{kJ/mol}$$
$$\ce{(3) <=> 4H2O + CO2}\quad\Delta H=(+206-42+2\times(-572))\ \mathrm{kJ/mol}$$
Subtracting $\ce{2H2O}$ on both ends of the equation I get:
$$\ce{CH4 +2O2 <=> 2H2O + CO2}\quad\Delta H=-980\ \mathrm{kJ/mol}\tag5$$
This almost looks like the reactions in (4) only with a $\Delta H$ which is $89\ \mathrm{kJ/mol}$ different.
From my understanding of the conservation of energy that shall not be possible.
What do I miss?

Make sure to account for whether H2O is liquid or gas

@Andrew As I understood it the energy difference between liquid or gas shall only make a difference for the lower heating value (lhv). The above number are all regarding the higher heating value (hhv). Or not?
Additionally the vaporation energy of water is 40,66 kj/mol. If that is relevant for 2 water molecules than there still would be an error of 8kj/mol.

@Andrew The only possible though improbable liquid hydrogen is in (1) the Water gas-shift which takes place between 90-230°. Even this would just meen a correction of 40.66 kj/mol. Though even if thats relevant there still would be an error of 48 kj/mol.

The heat of vaporization of water at 25 C is typically reported as approximately 44 kJ/mol. I'm not sure where you're getting 40.66.  The fact that your error is approx. 2 x 44 is not a coincidence.

All procedures occur between 90° and 900°. The value for 25° is irrelevant.

Hint: Just look for the states in every reaction. Irrespective of the temperatures at which the reactions happen

$\Delta H_r$ values are temperature dependent. You can only add them in the way that you are trying to do if they are all from the same temperature. The values you are using appear to be the standard enthalpy changes for these reactions, which by definition are the values for 25 C.  Whether or not you understand what standard enthalpy changes are, it is relatively easy to find the state of H2O in the reactions as defined for the values you have given. It is not the same in all of them.

Let's look at the 4 main reactions that we need to consider here. The reactions are as follows: Steam reformation, Water gas-shift, Combustion of hydrogen and the combustion of methane.
Similar to what is given in the question, I've listed the reactions with the states involved. [1]
\begin{align}
\ce{CH4(g) + H2O(g) &-> 3H2(g) + CO (g)} &\Delta H^°_{298} = \pu{205.9 kJ/mol} \label{steam-reform}  \tag{1} \\
\ce{CO(g) + H2O(g) &-> CO2(g) + H2(g)} &\Delta H^°_{298} = \pu{-41 kJ/mol} \label{water-gas} \tag{2} \\
\ce{2H2(g) + O2(g) &-> 2H2O(l)} &\Delta H^\circ_{298} = \pu{-572 kJ/mol} \label{hydrogen} \tag{3} \\
\ce{CH4(g) + 2O2(g) &-> CO2(g) + 2H2O(l)} &\Delta H^\circ_{298} = \pu{-891 kJ/mol} \label{methane} \tag{4}
\end{align}
Here, the goal is to get to \eqref{methane} from the other three equations. The path taken in the OP's solution produces the following result.
$$\ce{CH4(g) + 2H2O(g) + CO(g) + 4H2 + 2O2 -> 4H2(g) + CO(g) + CO2(g) + 4H2O(l)} $$
Simplifying the reaction by removing common factors, we get,
$$\ce{CH4(g) + 2H2O(g) + 2O2 -> CO2(g) + 4H2O(l)}$$
The $\Delta H^\circ_{298}$ value for the above mentioned reaction is correct to what the OP proposes and rightfully so. It comes out to be $\pu{979.1 kJ/mol}$. However, the more observant would notice. This isn't \eqref{methane}. We're missing one reaction.
$$\ce{H2O(g) -> H2O(l)}$$
The value of $\Delta H^\circ_{298}$ for this reaction is observed to be $\pu{44 kJ/mol}$. Using this in the equation we got previously and eliminating some of the water, we get \eqref{methane} with the right value of $\Delta H^\circ_{298}$.
References:

Subramani, V.; Sharma, P.; Zhang, L.; Liu, K. Catalytic Steam Reforming Technology for the Production of Hydrogen and Syngas. Hydrogen and Syngas Production and Purification Technologies 2009, 14–126. DOI: 10.1002/9780470561256.ch2.

To make this answer more generally useful, it would be good to explain why H2O(g) is used in those two reactions rather than H2O(l) (the standard state at 25 C). The reason (in case you don't already know it) is that phase changes are generally the largest sources of differences in $\Delta_r H$ between different temperatures. Otherwise the differences arise only from the difference in heat capacities between reactants and products, which is typically fairly small. By using H2O(g) at 25 C, the resulting $\Delta H$ is a closer approximation of what it would be at T>100 C.

I think your formula (4) is wrong. There shall be 2O2 and 2H2O.
Apart from that I did not found a source yet, that the energy-values are for a temperature of 298° Kelvin.
And apart from the formula (3) I did not find sources for the states (gas or liquid) for (1), (2), (4)

@ThomasR, the states for (4) come from the general definition of enthalpy of combustion. I'll add the source to the (1) and (2). It is behind a paywall though.

@ThomasR - It's actually rare to find $\Delta H$ values for temperatures other than 298 K, since that's the standard. If a source does not specify a temperature, it's almost always 298. But if you want to be sure, check authoritative sources like NIST or scientific journals, which will specify temp or use the proper notation $\Delta H^\circ$, where the $^\circ$ symbol indicates standard conditions of 298 K unless otherwise indicated in a subscript.


text:
Can two wavefunctions of 1s electrons in He atom have opposite phase?
Suppose we have a He atom. I understand that its wavefunction can be written as product of 2 hydrogenic orbitals:
$$\Psi(r_1,r_2) = \psi(r_1)\psi(r_2),$$
which in this case, are two 1s wavefunctions with paired spin. I want to know whether it is possible for the wavefunctions of each individual 1s electron to have opposite phase? If yes, can these wavefunctions interfere with each other constructively or destructively? If not, is it correct to say that both of these wavefunctions are orthogonal to each other?

What does "opposite phase" actually mean in this context? Could you add a definition of what you think it does? Thanks, and Welcome to Stack Exchange!

@uhoh By "opposite phase" I was imagining, whether the two 1 electron orbitals could be "out of phase" with respect to each other, similar to how we can write $1σ^*$ as linear combination of two out of phase 1s orbitals.

Do their wave functions have any angular dependence? If $n=1$ then $l=0$, so $Y_l^m(\theta, \psi)$ which has $\exp(im \phi)$ and $P_l^m(\cos \theta)$ dependence should be constant. I'm not very good at this, but I don't see any place where a phase can be defined. Unless their spins are opposite I don't think there's any other way that they can differ unless you excite one of them to $n > 1$.

I think the reason you're getting answers here that aren't directly addressing the question is because the question itself doesn't really make sense in a quantum world. The flaw underlying the question is reading too much into the use of the word "phase" (and maybe also "wave") and interpreting it/them as resembling their classical meanings.
In the quantum world, a wave function does not have a specific phase, nor do wave functions interfere constructively OR destructively. Rather, both phases coexist and wave functions interfere constructively AND destructively. For example, when two hydrogen atoms come together, molecular orbitals are formed from both the constructive (bonding) combination of atomic orbitals and the destructive (antibonding) combination.
So in your specific example of the He atom, since the 1s spatial orbital(s) do not have specific phase, they don't interfere with each other in that classical sense. As to whether they are orthogonal, the term "orthogonal" in reference to atomic and molecular orbitals is used only to refer to the spatial component, with the specific definition that the overlap integral over all space is zero. The spin component is not considered. In this treatment, spin is functionally considered as a fourth non-spatial dimension.
From a conceptual perspective, it is often most useful to consider that there is only one spatial orbital that contains two electrons of opposite spin rather than considering the two spin-spatial orbitals separately.

Well after pondering over your answer, I found a better 'intuition' for phase here: https://physics.stackexchange.com/questions/567289/can-someone-provide-a-physical-not-mathematical-intuition-for-the-phase-in?noredirect=1&lq=1

You can only do what you suggest (make product) if the electrostatic repulsion between the two electrons is ignored. This means that there are two single electron equations (just as for hydrogen but with $Z=2$) with energy given by the Rydberg formula. Thus the energy of either is $E_1=E_2=-4hcR_\infty=-54.4$ eV so $-109$ in total.
The perturbation caused by electron-electron repulsion is the expectation energy
$$\frac{e^2}{4\pi\epsilon_o}\int_0^\infty\int_0^\infty \psi^*\frac{1}{r_{12}}\psi r_1^2dr_1r_2^2dr_2=34\;\mathrm{eV}$$
and $\psi=R_{1s}(r_1)R_{1s}(r_2)/4\pi$ and $R$ is the 1s hydrogen radial wavefunction but with $Z=2$. This amount of energy is far from a perturbation and so we have to use the Variation Method to get a much better approximation to the energy.
The spin and spatial (radial/angular) parts of the wavefunction are treated as a product as electrostatic energies do not depend on spin. The ground state will be a singlet with spin paired as both electrons have the same spatial wavefunction so only a symmetric solution to the spatial wavefunction  exists. The total wavefunction is $\psi_{space}^A\psi_{spin}^S$ or $\psi_{space}^S\psi_{spin}^A$ where superscripts means symmetric or antisymmetric wrt electron-label exchange.

This means to complete previous answers.
The first problem I appreciate is that you are thinking of the electrons as distinguishible independent entities but in a helium atom you can not distinguish them, nor are they independent.
Besides, there is a fundamental flaw in the wavefunction you suggest: it is not antisymmetric with respect to the exchange of the coordinates of two electrons, it does not comply with the Pauli exclusion principle.
This is
$$\Psi(r_1,r_2) = \Psi(r_2,r_1) $$
and it should read
$$\Psi(r_1,r_2) = -\Psi(r_2,r_1) $$
and your are forgetting the spin-coordinates.
To construct a wavefunction that conforms with the Pauli exclusion principle you can use an Slater determinant
and for He you would obtain a wawefunction like:
$$\Psi = \frac{1}{\sqrt{2}} [\Psi_{1s}(r_1)\Psi_{1s}(r_2)] [\alpha(1)\beta(2) - \beta(2)\alpha(1)] $$
On this topic you may find helpful the book by D. A. McQuarrie and J. D. Simon, Physical Chemistry: A Molecular Approach, University Science Books (1997) or other book on Physcical Chemistry or Quantum Chemistry, or the following entrance of Chemistry LibreTexts on the topic: Antisymmetric Wavefunctions can be Represented by Slater Determinants


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Contaminants in Ion Exchange Chromatography significance?
I am a high school senior and decided to do a project on the essential amino acid composition of varying vegan protein mixtures. I have run into a problem concerning protein purification. One of the vegan protein powder mixtures I have chosen contains 1% fat (it does not really specify on the label but I assume it comes from the hemp seed protein; here is the link to the nutritional label: https://tinyurl.com/4fkymstf). My method of analysis is ion exchange chromatography and then light absorbance as seen here: https://tinyurl.com/4n3ukwa2. Can I just ignore the fat and continue to hydrolyze the proteins in 6.0 M HCl or does it actually matter? If it does matter I am going to cry because my only method of extraction is solubility-based (rice protein and hemp seed protein are insoluble in water compared to pea protein so I would have to do some alkaline solution BS to make them all soluble and then extract any residue left over from the alkaline additive which makes me sad). Note that I already found a way to extract the sodium on the label.
TLDR: For ion exchange chromatography in regards to amino acid analysis will a small amount of fat contaminant skew results to a significant degree/do I need a 100% pure protein sample?

Ideally, one should do chromatography after the crude product (hydrolyzed proteins in yours case) have been pre-cleaned. Fats do not dissolve in water, so I assume, you will carefully filter your sample before loading the ion-exchange resin. Also you can you do a solvent extraction in a separatory funnel with an appropriate solvent (ask your teacher) to remove fats before loading the ion-exchange resin. One good news is that sulfonated ion exchange resins will only retain positively charged compounds, the rest of the junk will just wash off if you are washing the resin with water after sample loading. However, resins are organic compounds, they can non-selectively bind tenacious/sticky molecules.
In short, you can continue your hydrolysis experiment as planned and then do fat extraction before loading the ion-exchange column. Alternatively, if you have not already satrted, you can do fat exraction by solvent exrtaction before the hydrolysis and then you will be assured that there no fats in your sample. The latter is preferrable.


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How are substances identified in the real world?
Recently in AP Chem I did a lab that involved doing various tests (melting on hot plate, flame test, etc.) to identify a metal out of a list of metals.
This had me wondering: how do chemists identify mystery substances in the real world? What tests are used?

There is a whole specialized field called "Analytical Chemistry". The test you list are very elementary and have no legal value. Modern and legal methods of identification require sophisticated instruments. If you have an example in your mind, then one can be specific and write a more detailed answer.

related https://chemistry.stackexchange.com/questions/86063/how-do-scientists-figure-out-the-chemical-structure-of-a-given-compound https://chemistry.stackexchange.com/questions/125879/how-can-an-organic-chemist-know-the-chemical-formula-of-a-natural-product-she-is

https://chemistry.stackexchange.com/questions/32240/how-can-an-inexperienced-chemist-determine-the-chemical-structure-of-a-molecule

Let's imagine aliens bestow upon us a block of some unknown pure substance. Guess we'd better figure it out!
One of the interesting things about this Universe is that, no matter what that material is, it must be composed of some combination of the (known) chemical elements, almost certainly the first ~100 or so in the periodic table (excepting some very unusual edge cases, such as stellar material, which isn't stable in conditions even vaguely close to ambient anyway). So the first order of business is to figure out which of these elements are present, and in what proportions. There are many methods to do this, but generally it's a good idea to use a procedure which does not significantly affect a sample (so-called "non-destructive analyses").
A reasonable option with no prior knowledge is to use X-ray fluorescence (XRF) spectroscopy. Nowadays there are small handheld "guns" that blast a sample with X-rays, which are absorbed by the inner-shell electrons of the atoms in a process called electronic excitation. When these excited electrons go back to their energetic ground state, they release X-ray photons of very specific energies depending on what atom they're attached to, and these photons can be measured. Here is a very nice video from Ben Krasnow of Applied Science which shows one of these XRF guns in action. He goes into the operation principle a bit and tears down the instrument.
There is another class of options, such as inductively coupled plasma mass spectrometry (ICP-MS), which is a destructive technique where you break a portion of a sample down into a plasma of its constituent atoms, and then fling these resulting ions through a magnetic field and measure their trajectory - the exact trajectory will depend on their mass-to-charge ratio, which can be measured very accurately and compared to the known elements. For these techniques (as well as many others), you typically have to "prepare" the sample for measurement, and unfortunately this can add some confounding factors.
Great, so now you know precisely which elements are in a sample and what their proportions are. Are you done? Not even close - the way those atoms are put together matters at least as much as the identity of the atoms. The number of different ways atoms can combine very quickly leads to dizzying complexity.
With a little bit of luck, your sample is composed of crystals. If the crystals are at least a few dozen micrometres across, you have a good chance of performing X-ray crystallography. The geometric regularity of crystals allows X-rays to penetrate the sample and mutually interfere as they bounce outwards, in a way that very sensitively depends on the exact placement and identity of atoms in a sample. This generates a so-called diffraction pattern, which can be very carefully analyzed with a lot of number-crunching, allowing back-calculation of the structure of the crystal, and therefore of the substance. This is perhaps the golden standard of chemical analysis. In the past 10-15 years or so, a similar technique called cryogenic electron microscopy (cryo-EM) has risen to importance, which allows a similar measurement of crystals by shooting electrons into a sample, instead of X-rays. Though there are significant additional complications, the main benefit here is that the crystals being measure can be much smaller, on the order of nanometres across.
Of course, the unknown sample may not be crystalline at all. From this point onwards, it becomes less clear what direction should be taken, and you get into a lot of case-by-case considerations. For example, nuclear magnetic resonance (NMR) spetroscopy is another very powerful and general technique which can provide information on the connectivity of atoms in a sample, but in a more indirect way than crystallography. Typically a sample is placed in a very intense magnetic field, often 10 000 or 100 000 times greater than Earth's magnetic field (enough to levitate small animals), which aligns the atomic nuclei of a sample in the direction of the magnetic field. The sample is then pulsed with radio waves, which knocks the nuclei over, and as they attempt to reorient themselves they emit photons of very specific energies, which once again can be used to infer the environment surrounding these atoms. This technique is typically very effective for organic substances composed largely of carbon and hydrogen, and less so for inorganic substances. There are many, many, many more techniques, each with their own strengths, weaknesses and regions of applicability. It takes a skilled chemist to figure out what the next move is.
Now, after all that, here's the kicker - virtually all real-world samples are not pure substances, but a mixture of several, sometimes even thousands, with their relative proportions varying by over a dozen orders of magnitude. The mere separation of these mixtures to allow accurate characterization of the individual components is its own huge field, full of complexities. It is not inconceivable for an unknown sample (especially one coming from a completely unknown environment) to admit only a superficial chemical analysis, even with the best techniques available today. Frequently, very special samples (such as space rocks) are analyzed to the best of our current ability, and then simply carefully stored for decades in the hope that that more advanced techniques can one day be found and allow an old sample to be further studied.

Whoops, the question was closed while I was in the middle of writing an answer. Hope that's not a problem...

Huh, I guess you bypassed this, by having old cache. These mechanisms are hardly perfect. While this q. is obviously too broad, and like a duplicate of a duplicate, your answer is valuable, so I guess I'm gonna link this with others.


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The difference between Zn and Al on acid and base reaction
Because it was written on my textbook that Zn and Al dissolves in acid and alkali, I tried.
When I put Al in NaOH solution it reacted immediately. But when I put Zn in the same concentration of NaOH solution it didn't react. I heated and added NaOH and at last it emitted some bubbles.This is what I observed when I reacted alkali and Al or Zn.On the contrary, when I put Zn into 9.5% HCl solution it reacted at once. But when I put Al into the same concentration of HCl solution, it took so long until it starts to emit bubbles.Could anyone explain why there is such a difference?

A "piece of aluminum" is not pure aluminum, but a sample of pure $\ce{Al}$ with a strong protective layer of $\ce{Al2O3}$. Interestingly, pure aluminum oxide forms complexes and reacts with NaOH, but $\ce{Al2O3}$ is only sluggishly attacked by $\ce{HCl}$. Here is a supporting reference:

In aqueous solutions of HCl acid aluminum wire does not dissolve due to presence of oxide film on surface.

And, a more detailed answer, per the same source:

Theoretically, the oxide film on Al is soluble in low pH solutions. However, in practice, the cathodic reaction, hydrogen reduction, results in a diffusion layer with a pH at the surface that is much higher than the bulk solution pH. This retards the rate of dissolution and can even bring it to a virtual stop if it enables a uniform passive film to form. In high pH solutions, the reverse is the case and dissolution is accelerated, instead of retarded, by the pH shift that accompanies hydrogen evolution. In less pure Al, the inclusions become dedicated sites for the hydrogen reduction reaction restricting the pH shift to their locations, and this allows dissolution to continue at anodic sites.

On the topic of zinc and HCl per an educational site:

Let’s examine the example of the interaction between zinc and hydrochloric acid. Zinc also reacts with HCl, releasing small bubbles of hydrogen and forming zinc chloride ZnCl₂.

But why small bubbles? My answer, there is also hydrogen embrittlement occurring on the zinc surface, as highlighted in recent research, "The Effects of Zinc Alloy Electroplating on the Hydrogen Embrittlement of High Strength Steels", to quote:

Hydrogen permeation measurements were performed and indicated that the quantity of hydrogen absorbed by the steel substrates is critically dependent on the composition of the initial deposit. Large quantities of hydrogen were trapped in the coatings and continued to diffuse into the steel over extended times after the plating process was completed.

I trust this helps.

Thank you very much for taking time to answer my question. I tried again with alminium foil and copper into HCl. This time bubbles formed on copper and aluminium dissolved very fast. Your answer was very helpful!


text:
What happens to the hydroxide ion concentration when you add caustic soda to ammonia?
Question.

Find the $[\ce{OH-}]$ and the pH of a $\pu{0.20 M}$ $\ce{NH3}$. If given $\pu{1 L}$ of the previous solution, when adding  $\pu{100 mL}$ of $\pu{0.2 M}$ $\ce{NaOH}$, what would be the new pH? Given $K_\mathrm b=1.8 \times 10^{-5}$

Attempt. From the info of the constant
$$K_\mathrm b=\frac{x^2}{0.2-x}\implies x=\pu{3.79 \times 10^-3 M}$$
Hence from the expression of the pH:
$$\mathrm{pH}=14+\log(x)=11.58$$
But for the second part I'm having issues to differ that when something is added, what I have to do is:

Since after the addition of a base, you are adding more concentration of $\ce{OH-}$, hence by Le Chateliers' principle, the system will move to the left, considering the formula is

$$\ce{NH3 +H2O -> NH4+ +OH-}$$
hence, the $[\ce{OH-}] = x-y$

Since it's just an addition, do $[\ce{OH-}]= x+y$, ignoring Le Chateliers' Principle

Pretty lost when it comes to when to use which bullet point, of course in both you'd use that $K_\mathrm b$ is constant after the addition of anything but still unsure.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

Convert it to a mathematical problem of a set of nonlinear equations. 2 equilibrium equations (ammonia, water ) and 3 inventory equations (molar concentrations of 2 bases + charge neutrality). Simplify where applicable.

This doesn't really help me at all, confused on what the equations should be and still doesn't answer my question about the bullet points, even though I appreciate it

You drive the dissociation equilibrium $\ce{NH3 + H2O <=> NH4^+ + OH^-}$ to the left with all those added hydroxide ions (Le Châtelier's Principle), so your hydroxide ion is little different from just having the $\ce{NaOH}$.

Note that Le Châtelier's Principle is qualitative, it does not help if you need quantification of the change.

The initial system contains $0.0038$ mol $\ce{OH-}$ (and $\ce{NH4^+}$) and $0.1962$ mol $\ce{NH3}$. Adding $0.02$ mol $\ce{NaOH}$ will consume $y$ mol of $\ce{NH4^+}$ and of $\ce{OH-}$ and produce the same $y$ moles of $\ce{NH3}$. As a result, the final amounts are :
$\ce{n(OH^-) = 0.02 mol + 0.0038 mol - y = 0.0238 mol - y}$ ;
$\ce{n(NH3) = 0.1962 mol + y}$ ;
$\ce{n(NH4^+) = 0.0038 mol - y}$.
Forgetting the units, the final expression for the equilibrium constant is $$\ce{K_b = \frac{[NH4^+][OH-]}{[NH3]}} = \frac{\frac{0.0038 - y}{1.1} ·\frac{0.0238 - y}{1.1}}{\frac{0.1962 + y}{1.1}} = \frac{(0.0038 - y)(0.0238 - y)}{(0.1962 + y)·1.1 }= {1.8·10^{-5}}$$ This is a $2$nd degree equation, and its reasonable solution is : $\ce{y = 0.0036}$. So $\ce{[OH^-] = 0.0202 M}$.
As a consequence, $\ce{[H^+] = 4.95 10^{-13}}$. And $\ce{pH = 12.305}$.
If we had ignored the presence of $\ce{NH3}$ in the initial solution, and added $100$ mL of a $0.02$ M $\ce{NaOH}$ in $1$ L pure water, we would have obtained nearly the same pH = $12.301$ ! Surprisingly enough, the presence of a large excess of $\ce{NH3}$ in solution has a negligible effect on the pH of the solution.

If you calculate addition of 100 mL of 0.2 M NaOH. to 1 L of water, pH would be 12.26. ( 14 + log ( 0.2 . 1/11 ). But I do agree additional pressence of NH3 does not change pH at first 2 decimal points.

Amazing! Just saw it.

I have added below the general solution, defining and solving a set of algebraic equations by the substitution method, as hinted in my comment before. But in this case, it is rather a cannon against a sparrow.
Enough is to replace in your equation $x^2$ by $x(x + [\ce{NaOH}\text{, final}]$, where $x$ is additional $[\ce{OH-}]=[\ce{NH4+}]$ due $\ce{NH3}$.

The general equation set approach:
There are often less standard situations, like this with $\ce{NaOH}$ addition, without handy equations to be used. In such a case, a generally applicable approach using known, generally valid equations can be applied.
Let perform the notation simplification to have easier writing:
$x=[\ce{H+}], y=[\ce{OH-}], z=[\ce{NH3}], r=[\ce{NH4+}], \\s=[\ce{Na+}], a=c(\ce{NH3},\mathrm{total}), b=c(\ce{NaOH}) $
2 equilibrium equations:
$$K_\mathrm{w}=xy$$
$$K_\mathrm{b}=\frac{yr}{z}$$
2 concentration inventories:
$$z + r = a$$
$$s = b$$
1 charge balance:
$$x + r + s = y$$

substitution for s:
$$K_\mathrm{w}=xy$$
$$K_\mathrm{b}=\frac{yr}{z}$$
$$z + r = a$$
$$x + r + b = y$$

substitution for x:
$$K_\mathrm{b}=\frac{yr}{z}$$
$$z + r = a$$
$$\frac{K_\mathrm{w}}{y}  + r + b = y$$

substitution for z:
$$r(\frac{y}{K_\mathrm{b}} + 1) = a$$
$$\frac{K_\mathrm{w}}{y}  + r + b = y$$

substitution for r:
$$\frac{K_\mathrm{w}}{y}  + \frac{a}{\frac{y}{K_\mathrm{b}} + 1} + b = y$$

The first term represent [H+] in charge balance equation. Its value is roughly $\pu{E-12}$ which can be neglected:
$$ \frac{a}{\frac{y}{K_\mathrm{b}} + 1} + b = y$$
$$ aK_\mathrm{b} + b(y + K_\mathrm{b}) = y(y + K_\mathrm{b})$$
$$ aK_\mathrm{b} + by + bK_\mathrm{b} = y^2 + yK_\mathrm{b}$$
$$ y^2 + y(K_\mathrm{b}-b)- K_\mathrm{b}(a + b) = 0$$
$$y = \frac{-K_\mathrm{b}+b \pm \sqrt{{(K_\mathrm{b}-b)}^2 +4K_\mathrm{b}(a + b)}}{2}$$

as $K_\mathrm{b} = \pu{1.8e-5}$, $a = 0.2 \frac {10}{11} \mathrm{M}$, $b=0.2 \frac {1}{11} \mathrm{M}$.....
$$y \approx \frac{-\pu{1.8e-5} + 0.2 \frac {1}{11}  \pm \sqrt{{(\pu{1.8e-5}-0.2 \frac {1}{11})}^2 +4 \cdot \pu{1.8e-5} \cdot 0.2 }}{2}$$
$$y \approx \frac{0.018164  \pm \sqrt{0.018164^2 + \pu{1.44e-5} }}{2}$$
$$y \approx 0.01836$$, therefore,
$$[\ce{OH-}] \approx \pu{0.01836 mol L-1}$$
$$\mathrm{pH} = 14 + \log{0.01836} \approx 12.26$$
If we compare it to $\ce{OH-}$ of $\ce{NaOH}$ origin:
$$[\ce{OH-}] = \pu{0.2 \frac {1}{11} mol L-1} \approx \pu{0.01818 mol L-1} $$,
$$\mathrm{pH} = 14 + \log{0.01818} \approx 12.26$$
(Without corrections for activity coefficients, it does not make sense for higher pH precission.)
so for $\ce{OH-}$ coming from $\ce{NH3}$:
$$[\ce{OH-}]\approx \pu{0.00018 mol L-1}$$

@Acedium20 Note that I added this as general solution, that is not needed here. Enough is to replace in your equation x^2 by x(x + c(NaOH, final)), where X is additional c(OH-) due NH3.


text:
How does bond formation affect the energetics of the bond?
I am an 11th grade student who has just started learning about bonds in compounds. My question pertains to a line written in JD Lee's Concise Inorganic Chemistry for JEE (Main and Advanced)
It was suggested that in all such case, a pair of electrons gets unpaired and one of the electrons is excited to the next available vacant orbital of slightly higher energy. The driving force for this process is the energy released in bond formation by unpaired electrons.
If we have an atom with its initial energy as 'x'J. and further add 'k'J of energy to it so as to excite the electron, then (according to this line), the energy released after bond formation is 'k'J, leading to a final energy of 'x'J.
We have been taught that the purpose of a bond is to lower the total energy. But here, the final and initial energies are same. Then what is the purpose of a bond? Perhaps I'm simply interpreting the line wrong, or is the bonding mechanism more complex than a simple "energy less"?

Have you looked at anti-bonding MOs and bonding MOs?

final and initial energies are the same

Now you have a perfect equilibrium between A + B <--> A-B
But this depends greatly on what A and B are.
Electrons do not have to be "excited" for bonding to occur.  It usually happens to complete the electron "shells", forming a more stable molecule.  More stable means lower energy.  Therefor, in that case the bonding reaction is favored.
"Covalent" bonds are so much more favored that for all practical purposes, the reaction of A + B is 100% A-B.  Weaker ionic bonds, such as salts, exhibit varying degrees of dissociation.  Here the "lower energy state" depends greatly on the "solvent effects".

Dissociation in water isn't nearly enough to say a bond is weak.


text:
Are acids effective in removing soap scum?
In my experience, most of the commercial products intended for removing soap scum from bathroom surfaces are ineffective. No matter how long you leave it on or how hard you scrub, the soap scum doesn't come off. The only successful way I've found is to scrub it with a stiff brush using elbow grease. The better cleaning products are have mild abrasives, which results in a dull surface finish.
I read some of the bathroom cleaning product labels and most contain primarily soaps and other high pH-level chemicals such as ammonia. Since soap has a high pH-level, shouldn't the soap scum respond better to an acid such as citric or acetic acid?
It's also common to add salt or hydrogen peroxide to an acid. Can they increase the strength of the acid, making it more effective? If so, how?

Soap is a combination of a metal (e.g. sodium) and a fatty acid (e.g., palmitic acid). If you use acid to break the metal off (e.g., $\ce{CH3(CH2)14COONa + HCl -> CH3(CH2)14COOH + Na Cl}$) you're left with salt water and a slimy fatty acid stuck to the surface. Sort of counterproductive.
Alkaline bathroom cleaners work by re-forming soap from fatty acids, which is soluble.
These cleaners may also contain chemicals to chelate or sequester metals that form poorly-soluble salts with fatty acids, such as magnesium and calcium in "hard" water.
Regrettably, they may take some time to loosen soap scum, and watery cleaners may drip off a vertical surface before they can be effective. To help the cleaner stay in place, wet a paper towel (or filter paper -- this is a site for chemists! ;-) with the cleaner and adhere it to the surface.
BTW, strong alkaline cleaners are more effective, but:

They're dangerous.
The alkali attacks glass (e.g., shower doors) and glazed porcelain, leaving the surface permanently damaged.

The white deposits may be limescale or soap scum, or a combination of both. Is there a high level of hardness in the water in your area? Therefore, one cannot generalize that acidic cleaners are ineffective. As long as the correct product is used, they are pretty effective. The soap scum is composed of calcium or magnesium salts of fatty acids. The substance is insoluble in water. Salts of sodium are soluble in water. As you may be aware, soap scum is greasy, and low quality acid cleaners simply bead up and flow downwards when sprayed on walls. A quality acid cleaner will contain wetting agents as well as a mixture of mild acids, such as lactic and gluconic acids. As a result, soap scum as well as limescale are wetted effectively while acid helps to dissolve the insoluble product. Calcium and iron are well chelated by glutonic acid. Lactic acid is a mild acid.

Soap scum is noticeably different from limescale which leaves a white dry haze or a in extreme cases a cement like deposit. The two almost always exist in combination and varies depending on the water source, so we should assume it's both with soap scum being the majority of its content. Your answer seems to disagree with the first one which claims soap scum will respond better to a base than an acid. Am I reading this wrong?  For the record, your answer is what I would have guessed, but science always has a surprise in store.

Btw, lye is the main active ingredient in commercial drain cleaners which dissolves  slimy concretions composed of organic matter, lime scale and soap scum. Lye is such a powerful base it is essentially a solvent with the benefit of not corroding pipes as would an equally power acid.

I don't think that soap scum is better removed by lye than acids alone.  One needs chelating agents plus surfactants and pH (low or high). The acids I mentioned such as lactic and gluconic acid along with surfactants are mild yet act on soap scum effectively. These acids do not corrode metals. One should not use strong acids or strong bases for scum removal or even lye. Both are damaging to the tiles and grit.

How about using an oxalic, acetic or citric acids?  They're relatively mild and while they can corrode metal, as long as they're not applied directly to metal surface, it should be OK. Can you recommend a good household surfactant?


text:
Why does chlorine displace bromide
Found that chlorine will displace bromide, and iodide from their ion forms. According to my textbooks, this is because chlorine is more electronegative than bromine and iodine. However, the chloride ion is less stable than the bromide and iodide ions due to its larger charge density.
However, based on the acidity of HCl, HBr, HI, HCl is actually the least acidic one hinting that shouldn't charge density be a more important factor in determining stability than electronegativity and hence Br2 should displace Cl- instead of Cl2 displacing Br-?

You are dealing with redox reactions, and chlorine is a stronger oxidizing agent than bromine.

I guess you are considering reactions in solution. In your analysis you are forgetting the strength of the interaction of the ions with the solvent molecules and how this affect the relative stability of the ions, and its reactivity. For this reactions it is better to use  reduction potentials and its relation to reaction free energy $\Delta G^0_r$ as a guide to its reactivity.

In the case of the acidity of HCl, HBr and HI, you must consider the differences in solvatation energy betewee the ions and the molecues (this are related to the pK$_a$).

Finally, looking at a table of ionic radii (Wikipedia), $r(Cl^-) < r(Br^-) < r(I^-)$.  How can you explain that the charge density smaller for the smaller anion if you assume that the charge distribution is spherical and all the anions have the same charge (-1e)?

The smaller charge density was a typo, I have already changed it. But what is the difference between solvation energy and the interactions of the ions with the solvent molecules. Aren't they the same thing?


text:
Intuitive explanation for enthalpy of hydration
In my chemistry course, we were discussing types of enthalpy and the enthalpy of hydration came up. My instructor defined this as

heat associated with adding water to an anhydrous salt.

However, I don't intuitively understand this. My instructor just used the example of copper sulfate pentahydrate — you put in heat to remove the water, thus when you put in water to the anhydrous form, you get heat. I do not find this satisfactory.
I was thinking that it could be possible adding water to the anhydrous salt causes bonds to dissociate within the salt, and similar to calcium chloride this generates heat, but this is just speculation.
By what mechanism is heat generated and how should I intuitively understand the enthalpy of hydration?

Are you familiar with the concept of bond enthalpy? Generally, there is always a change of enthalpy when a bond is created or broken, and this holds for all types of bonds (ionic, covalent, hydrogen, intermolecular ...). Enthalpy of hydration is therefore the change of enthalpy when particles (in your case ions) form bonds with the solvent (water). Usually, it is more instructive to consider the whole process of dissolving, constisting of ...

... first breaking the (ionic) bonds (lattice enthalpy), and then forming bonds between ions and water (enthalpy of hydration). Schematically depicted here.

The distinction between what's considered a chemical bond, an intermolecular attraction, and a coordination/dative bond is largely a manmade distinction. Nature doesn't care one way or another. All are mediated by photons and the physics of the fundamental electromagnetic force.
So, just as it takes energy to break chemical bonds and energy is released when chemical bonds form, it also takes energy to separate particles held together by intermolecular forces or coordination/dative bonds and energy is released when particles get together.
Hydration enthalpy, in the sense of the enthalpy of solvation when water is the solvent, is the enthalpy associated with adding or removing the ion-dipole intermolecular forces between the ions and water molecules. Hydration enthalpy, in the sense of the enthalpy of the creation of hydrates, is the enthalpy associated with creating or destroying the coordination/dative bonds in metal-aquo complexes.
Your instructor seems to be using the second sense of the word. Adding water to anhydrous copper sulfate allows coordination/dative bonds to form between the water molecules and the salt. Energy is released when new bonds form. Energy is required to break those bonds.

Hydration can have two meanings (IUPAC gold book), a chemical reaction that adds water (usually "across" a double bond), or a process where a particle from a gas phase moves into aqueous phase. Dissolution refers to dissolving a substance (usually a solid) in a solvent.
The enthalpy of dissolution (or of solution) can be artificially broken into two parts. First, you take the solid and remove all intermolecular interactions, effectively turning it into a gas (or maybe plasma for ionic compounds). Second, you take the isolated particles and place them into a solvent.
The enthalpy associated with the first part is called lattice energy (it is endothermic). The enthalpy associated with the second part is called hydration enthalpy (if the solvent is water) or solvation enthalpy (for any solvent). When you add the two, you get the enthalpy of dissolution.
Here is a conceptual diagram from a Libretext book:

Source

There is a difference between solvation and hydration. It looks to me that you are confusing both concepts. Your question maybe a duplicate of the following question. where a good explanation is provided, specifically for the Copper sulfate example, as it turns out.

This is decidedly not the same question. I am asking for an intuitive explanation of hydration enthalpy, not solvation enthalpy.

"I was thinking that it could be possible adding water to the anhydrous salt causes bonds to dissociate within the salt, and similar to calcium chloride this generates heat, but this is just speculation." It seems to me that you are referring to solvation here, not hydration. In any case the mentioned question's answer mostly refers to hydration.


text:
Is mass of solute included in calculation of enthalpy of solution?

$$
\begin{array}{lrrrr}
\hline
  & \ce{MgSO4} & \ce{K2SO4} & \ce{KNO3} & \ce{NH4NO3} \\
\hline
m/\pu{g} & 15 & 15 & 15 & 15 \\
M/\pu{g mol^-1} & 120.37 & 174.26 & 101.10 & 80.04 \\
n/\pu{mmol} & 124 & 90 & 148 & 187 \\
T_0/\pu{°C} & 28.0 & 28.0 & 28.0 & 28.0 \\
T_1/\pu{°C} & 74.1 & 22.3 & 0.0 & 0.0 \\
\hline
\end{array}
$$

To solve for the heat of solution of $\ce{MgSO4}$ we use the formula
$$q = -mc\,ΔT \tag{1}$$
What should be the value for mass $m$ given the data?
Should I add the mass of water $(\pu{30 g})$ and the mass of salt $(\pu{15 g}),$ or should it just be water?
$$
\begin{align}
q(\ce{MgSO4}) &= -(\pu{45 g})(\pu{4.184 J g^-1 °C^-1})(\pu{74.1 °C} - \pu{28.0 °C}) \\
  &= \pu{-8.68 kJ} \tag{2}
\end{align}
$$
$$
\begin{align}
\Delta_\mathrm{soln}H(\ce{MgSO4}) &= \frac{q(\ce{MgSO4})}{n(\ce{MgSO4})} \\
  &= \frac{\pu{-8.68 kJ}}{\pu{124 mmol}} \\
  & \approx \pu{70 kJ mol^-1} \tag{3}
\end{align}
$$

I don't we can ignore the the fact the percentage of salt is too high, you are adding 15 g of the salt to 30 mL of water. That is a lot! In this case, the heat capacity of the solution should be used not that of water and then total mass should be used. I don't know who makes up these questions in the textbooks/tests without thinking.

@AChem when you said total mass, you meant to add the molar mass of the 2 substances? MG and SO4?

@Red, I think it should be mass of magnesium sulfate in grams that you have added to the mass of water. The key issue is that you do not have specific heat capacity of that solution which is 50% MgSO4 by mass. It is a badly designed question.

@AChem Confused right now since my friend used 34.972 as m in the formula. I was wondering how he got it

@Red, Your calculations appear to be correct. Ignore your friend's random numbers. The only issue I see is that this is 50% by wt MgSO4, so water's heat capacity is an approximation.

@AChem, in other words i'll use 120.37 instead of 45?

I do not know how and where you are generating random numbers such as 120.37,  45, 33.972, that too without unit!  Your solution looks okay, as it is. Do you understand what is being done in the table and in the solution? When you solved it correctly, why are you still so confused?

@AChem, just so i would understand this what would be your value in m in the case of potassium sulfate?


text:
ΔU calculation for an ideal gas that starts at (P1, V1, T1) and goes to (P2, V2, T2)
To find ΔU for an ideal gas that starts at ($P_1$, $V_1$, $T_1$) and goes to ($P_2$, $V_2$, $T_2$), it's stated that there are two possible paths.

ΔU = ΔU(isothermal) + ΔU(isobaric)
ΔU = ΔU(isothermal) + ΔU(isochoric)

I don't understand why ΔU is not equal to ΔU(isothermal) + ΔU(isobaric) + ΔU(isochoric) since P, V, and T are all changing. For example, path 1 shows an isothermal path ($P_1$ to $P_2$ at constant T) followed by an isobaric path ($T_1$ to $T_2$ at constant P). From my understanding, the gas is now at $P_2$ and $T_2$, but this says nothing about getting the volume from $V_1$ to $V_2$.

From p, V, T, only 2 of 3 are independent. For p2 and T2, there is only 1 possible value of V2.

$pV = nRT \implies p=f_1(T,V), V=f_2(T,p), T=f_3(p,V)$
The isothermic + isobaric path:
$p_1$, $T_1$, $V_1$ $\overset{V=c/p}\rightarrow$ $p_2$, $T_1$, $V_3$ $\overset{V=cT}\rightarrow$
$p_2$, $T_2$, $V_2$
The isothermic + isochoric path:
$p_1$, $T_1$, $V_1$ $\overset{V=c/p}\rightarrow$ $p_3$, $T_1$, $V_2$ $\overset{p=cT}\rightarrow$
$p_2$, $T_2$, $V_2$


text:
Why does berkelium have two electronic configurations?
J.D. Lee Concise Inorganic Chemistry, Appendix E: Electronic Structures of the Elements [1, p. 601] provides two electronic configurations for berkelium:
$$
\ce{Bk}\quad
\begin{cases}
  [\ce{Rn}]~\mathrm{(5f)^9(6d)^0(7s)^2} \\
  [\ce{Rn}]~\mathrm{(5f)^8(6d)^1(7s)^2}
\end{cases}
$$
Why does Wikipedia — Berkelium lists only the following one?
$$\ce{Bk}\quad [\ce{Rn}]~\mathrm{(5f)^9(7s)^2}$$
Which electronic configuration is more appropriate and why does my textbook include both?
Reference

Sudarsan Guha. J.D. Lee Concise Inorganic Chemistry, for JEE (Main & Advanced), 4th ed.; Wiley India Pvt. Ltd.: India, 2017 (reprint 2020). ISBN 978-81-265-9114-5.

Electronic configurations of such heavy metals do not mean anything. Textbook writers can write whatever conjectures they want. What is the experimental evidence? Who investigated it? Without such info, this electron configuration exercises are just academic exercises.

Total quantities of Bk produced are in the gram range. Since the discovery of it. The largest batch I can find evidence for was 22mg to be used in heavier element production.

https://www.nature.com/articles/s41467-021-21189-1 has some interesting info on Cf and Bk electronic structure.

Why would there be only two?

Both of these configurations are appropriate and valid. $\ce{[Rn] {5f^96d^07s^2}}$ or simply $\ce{[Rn] {5f^97s^2}}$ is the ground state electronic configuration of berkelium while $\ce{[Rn] {5f^86d^17s^2}}$ is the electronic configuration of first excited state of berkelium. There is an energy difference of $\ce{0.92 eV}$ between these two states. The enthalpy of vaporization of trivalent Bk metal was calculated to be $\ce{2.99 eV (288 kJ mol^{–1})}$.
The Reference also states

A hybridized nondegenerate 6d and 5f virtual‐bound‐states model has been used to describe the properties of the actinide metals, including berkelium. It accounted for the occurrence of localized magnetism in Bk metal. Included was the relativistic energy band structure of fcc Bk metal, $\ce{[Rn] {5f^86d^17s^2}}$ and the conclusion was that berkelium is a rare earth‐like metal with localized (ionic) 5f electrons resulting from less hybridization with the 6d and 7s itinerant bands than occurs in the lighter actinides.

Reference : The Chemistry of the Actinide and Transactinide Elements, 3rd
edition, by Lester R. Morss, Norman M. Edenstein, Jean Fuger & Joseph J. Katz.

0.92 eV is a lot of energy (aprox. 0.03 Hartree) . I do not have access to your source, sure you meant a smaller amount. Please correct.

I rechecked my source and found that it really is written 0.92 eV.

Thanks. If I am not mistaken, in that case, one of the configurations would correspond to an excited configuration.

Yes that's exactly what the book says.


text:
Why phenolphthalein is not completely neutralized in basic solutions?
Phenolphthalein $(\ce{HIn})$ is slightly acidic. Why doesn't it completely break down into $\ce{H+}$ and $\ce{In-}$ and also reacting with all of the $\ce{OH-}$ ion present, unlike $\ce{NaOH}$ and $\ce{HCOOH}$ reacting completely to form $\ce{HCOONa}$ and $\ce{H2O}$ only?

Who told you that it is not neutralized in basic medium? It is deprotonated, indeed. The pink color is that of the anion. Look at the Wikipedia chart. It explains it beautifully. https://en.wikipedia.org/wiki/Phenolphthalein

@AChem unless you go too far, and the phenolphthalein overreaction and loses its color. This is also explained in Wikipedia. We can't trust sodium hydroxide, bring on milk of magnesia!

what would happen with the other half? @Poutnik

HPh is weaker. the basic medium is stronger. now why wont all of them react altogether unlike 30 ml 1M HCOOH with 30 ml 1M NaOH completely neutralize system. we dont use pKa in the last case maybe. @Poutnik

so it is theoretically possible for more amount of strong base to react with less amount of weak acid to take all of its H+ away and leave only the anions in the system but the exact same scenario wont be for phenolphthalein practically for some other reason?

Depending of the weak acid $\mathrm{p}K_\mathrm{a}$, the degree of the acid neutralization $\alpha$ depends on $\mathrm{pH}$, following famous equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)}$$
related to the chemical equilibrium reaction:
$$\ce{HA(aq) + OH-(aq) <=> A-(aq) + H2O}$$

$\mathrm{pH}$
$\alpha$

$\mathrm{p}K_\mathrm{a} - 3$
$\approx 0.001$

$\mathrm{p}K_\mathrm{a} - 2$
$\approx 0.01$

$\mathrm{p}K_\mathrm{a} - 1$
$\frac {1}{11}$

$\mathrm{p}K_\mathrm{a}$
$0.5$

$\mathrm{p}K_\mathrm{a} + 1$
$\frac {10}{11}$

$\mathrm{p}K_\mathrm{a} + 2 $
$\approx 0.99$

$\mathrm{p}K_\mathrm{a} + 3 $
$\approx 0.999$

For HCOOH, $\mathrm{p}K_\mathrm{a}=3.75$
For phenolphthaleinum, $\mathrm{p}K_\mathrm{a} \approx 9$
Note that the colour of ionized phenolphthaleinum disappears at very high $\mathrm{pH}$ by slow transition due nucleophilic $\ce{OH-}$ addition.

Not sure that last change is irreversible. My reading is it's just slow.

@OscarLanzi I have always thought until recently it is the second transition,  but I have recently (not handy) read it is some structural change. So, perhaps it is both, I have not done any testing.

Wikipedia says the structural change is that at high pH, a hydroxide ion adds as a nucleophile instead of extracting another proton; such a nucleophilic addition is in principle reversible but expected to be slower than proton transfer.

@OscarLanzi You may be right.


text:
How to remove hydrofluorosilicic acid from water?
I'd like to come as close as possible to pure water. Would a KF filter remove hydrofluorosilicic acid? If not, what would?
I could not find the concentration of hydrofluorosilicic acid, but would like to get this as close to zero as possible. The average reported chloride level is 12.4 mg/L. Otherwise, I don't know of any other additives. However, since this is municipal and treated water, it likely has whatever gets flushed yet not removed.

The best way to produce pure water out of any aqueous solution is distillation. It is also possible to use anionic and cationic resins : this will remove all dissolved ions. It will not remove bacterias.

@Maurice. Thanks for your reply. Is it possible for a regular person to buy one of these cationic or anionic resins? Considering the chloramines (I just learned from amonia), I'm not worried about bacterias.

Yes. These resins are on sale. You have to choose anionic resins, and not the more familiar cationic resins. You can find a lot of different offers on Google by typing "Anionic resins". I know the resins made by Dow Co. They are called Dowex. I can mention Dowex 1X4 (also called Amberchrom 1X4), which costs 130€ for 100 g. or Dowex 21K chloride, which costs 80 € for 100 g. Or Dowex Marathon MSA chloride, which costs 52 € for 100 g. There are usually several degree of mesh. 100 mesh correspond to beads 0.15 mm diameter. 200 mesh is 0.075 mm diameter. 50 mesh is 0.3 mm diameter

I think that almost answers the question. Can I assume KF from Berkey does not remove the HF acid? @Maurice, please reply as an answer so I can give you credit for your help.

I don't know what is "KF from Berkey". Never heard of this ...


text:
What's the maximum temperature of quicklime slaking?
Normally the maximum temperature of lime slaking is listed at about 100 C. I've read that using water with high salt concentrations will catalyze the slaking reaction and increase the boiling point thereby leading to higher slaking temperatures.
If you performed the slaking operation in a pressurized chamber this should increase the boiling point. Would it also lead to higher slaking temperatures

The purpose of "slaking" is to re-hydrate CaO, which was made from CaCO3 by heating.  Why would one want to raise the temperature of this (already exothermic) step?

@RobertDiGiovanni, perhaps to provide heat?


text:
When it comes to Intermolecular Forces, when the chain length of a molecule increases, is there more or stronger dispersion forces?
I have been told by my highschool teacher that there are more dispersion forces rather than stronger.However, reading up on articles, I have discovered that some explain the increase in boiling point due to stronger dispersion forces. Please clarify if this is an important distinction, or the same thing.

Could you expand a little on what you mean by "more"? Let's say we have two chains and somehow measure the force(s) between them. We now add one atom to each chain. In what sense do you mean there are "more" dispersion forces now?

Or to put it another way when you say "more" are you breaking down the total force between the chains into a sum over contributions from atom pairs

If by adding atoms you change the size of  any dipole, or change the type of atom, and so change the polarisability, then the size of the dispersion force can increase or decrease. However, if neither of this happens  and you add more of the same type of atom,  the dispersion will increase just because there are more atoms.

The confusion here, I suspect, is between thinking about the total force between molecules and the strength of the force contributed by each part of the molecule. In bigger molecules, the individual components don't get stronger, but there are more components, making the total force larger.


text:
Why is nickel anticorrosive?
I've been looking into corrosion and common ways to prevent it. Pure nickel seems to be the main anticorrosive material to use, but I can't find a decent answer as to what it has/does that other metals don't.
One answer I saw is that nickel oxide forms a thin layer and acts as a barrier for corrosion.
Does nickel do this before the corrosion process happens? Or is nickel oxide different from other metal oxides? To my knowledge, metal oxides are the product of galvanic corrosion (for the metal being corroded).
Thanks!

You have to think about the context. Corrosion under what conditions? Corrosion in the sea? In water? In air... different situation.

Nickel possibly owes its corrosion resistance in neutral water to its lower oxidation potential compared to iron; Ni to Ni++ is +0.25v; Fe to Fe++ is +0.44v; water to H2 at pH 7 is -0.41v. Under neutral conditions iron will corrode and nickel will not if only hydrogen ion is involved. Both should react with O2, 0.82v, at one atmosphere pressure. Dissolved oxygen is low, that inhibits the reaction. The air-liquid interface could change that. There is another caveat involving nickel. Nickel alloys such as high nickel stainless steels [such as 316], Hastelloy C, or Monel metal [ Cu-Ni alloy] Can be cathodic accelerators to corrosion of a more active metal, such as iron, electrically connected to it. Nickel, possibly more so in divided form, lowers the hydrogen overvoltage, hastening remote corrosion of the more active metal. I have seen corrosion problems with 316, Hastelloy C, and Monel where the nickel was cathodic. I have no information on the effect of pure nickel.
Care must be taken when dissimilar metals are in contact if one is capable of reacting in the environment.


text:
Is there any book for chemistry which has all the reactions in it ? A book of just reactions
Is there any book for chemistry which has all the reactions in it ? A book of just reactions. I have gone through way too many but each just explains the types of reaction and to get all the reactions collected at once place will take too long. So there any book which has all the reactions listed?

Not. We don't have enough trees

Any such book would be out of date before it was even printed with current rate of publication by the world's chemists.

@Waylander, There should another book highlighting the fake organic yields. Too much junk in the synthetic literature. Biochemistry / medical reproducibility research needs another treatise.

The ongoing Compendium of Organic Synthetic Methods is the closest thing I know of, but still far short of all reactions

There is no such heavenly book that contains all the reactions known to humanity; however, two databases are worth mentioning which summarize most of the published chemical reactions. One is Reaxys of Elsevier, and the other is SciFinder.
Reaxys includes not only the recent literature, but the nice thing is that it has data from the monstrous Gmelin Handbuch der Anorganischen Chemie (Handbook of Inorganic Chemistry), Beilstein's Handbuch der Organischen Chemie (Handbook of Organic Chemistry), all the reaction summaries published in organic/inorganic literarture, and patents. One can only be awed by the German contributions to chemical encyclopedias-like collections and writings which started in the early 1800s. In the previous century, the French tried translating Beilstein to keep up with the organic synthetic literature but gave it up as a hopeless exercise. Who would translate thousands of pages every year? As a result, many English-speaking organic chemists had to learn German until recently. By recent, I mean until the 1990s.
The second database is SciFinder of the Chemical Abstract Service (USA). It contains a relatively recent set of reactions (limited to the 1900s, if I remember correctly) from all published literature and world patents. Finally, however, another service called SciFinder${^n}$. It includes hundreds of thousands of abstracts from Chemische Zentralblatt, the German abstracting service from pre-1900 works. But, of course, these databases consist of millions of pages if you were to compile a book and are beyond the range of personal purchases. Moreover, SciFinder summarizes from more than ten languages.
Many Russian books summarize inorganic reactions; these books were discussed here in SE Chemistry some time ago (please add the name).

These databases are, of course, not free.

Organic Chemistry: 100 Must-Know Mechanisms by Roman Valiulin is the closest thing I can think of. Otherwise, pick up Clayden's Organic Chemistry or Bruice's Organic Chemistry and make your own list.
What is this for ? There's also databases of organic transformations for cheminformatics purpouses but nothing is even close to being complete (if it will ever be).


text:
accessing the ZINC15 and ZINC20 databases
I'm trying to access the ZINC15 and ZINC20 databases but I've been at it the whole morning and I can not figure out how to do it. I am not that experienced with programming so maybe there's something obvious I'm missing.
If I try to download larger chunks of data from here, I get all these weird file formats I don't know what to do with. And their website doesn't really have a clear documentation for the API.
Any tips, please ? Thank you in advance !
Update with answer: Nobody really said anything for a while here and in the meantime I found this YouTube video that has a method of downloading the data through powershell. I'm not entirely sure how "formal" thi is but it worked. Hope this helps any programming newbies out there looking for answers.

You can certainly add your own answer to your question with the powershell version.

It's not too hard. One of the "formats" from ZINC is a script to use curl or wget or Windows PowerShell to download chunks (tranches) of the molecules in the formats you specify.
If you go to ZINC 20 Tranches you can configure what you want along the top (e.g., I chose 3D representations, standard reactivity, wait OK, .. etc.) You can also select or deselect from the table to pick molecular weight or logP fractions.
Then, when you've picked what you want, click on the download button.

This brings up a new screen with the list of tranches and a place to pick:

the molecular format you want (e.g., SMILES, AutoDock, Mol2, or SDF) in gzip-compressed
the download type (e.g. curl, wget, or PowerShell commands, etc.)

This gives a file like this (in curl syntax):
curl --remote-time --fail --create-dirs -o AA/AAML/AAAAML.xaa.mol2.gz http://files.docking.org/3D/AA/AAML/AAAAML.xaa.mol2.gz
curl --remote-time --fail --create-dirs -o AA/AAMM/AAAAMM.xaa.mol2.gz http://files.docking.org/3D/AA/AAMM/AAAAMM.xaa.mol2.gz
curl --remote-time --fail --create-dirs -o AA/AAMN/AAAAMN.xaa.mol2.gz http://files.docking.org/3D/AA/AAMN/AAAAMN.xaa.mol2.gz
# …

There's more on the ZINC wiki although some may be out-of-date.

Thanks for the answer ! I updated the post just now with a Youtube video I found that shows the powershell way of doing this. I found it recently by complete accident while looking up something else.
Now I have two ways of doing it ! Excellent !


text:
Why is the d subshell only stable when all of its orbitals are filled?
I was learning about the Aufbau principle and quantum electronic configuration. Here are some of the yet-unanswered queries that I wish to know the answers to.

I discovered that the d subshell is only stable when each of its orbitals is either filled with a single electron or a pair of electrons. Why is it so?

What causes electrons to first fill up each orbital rather than to exist in the same orbital with opposite spins (in subshells other than s, obviously)?

Thirdly, if the 3d subshell has more energy than the 4s subshell, isn't it plausible that electrons are shared/transferred from the 3d subshell first, and then from the 4s subshell? Why doesn't it happen?

Mathjax is discouraged in titles, also subshell notations need to be upright. Andselisk was just correcting that.

Ad 3: Order of electrons energies in particular orbitals is not fixed, but depends on electron configuration, which affects electron mutual shielding and repulsion. The typical case are early actinoids, behaving rather like d elements, instead of f element behaviour in the prior period.

What do you mean by "stable" in your first question? What behaviour distinguishes "unstable" d shell fillings from stable ones?

Doesn't your first question apply to s and p orbitals too? Aren't p orvitals also unstable if they ate not completely or half filled?

I'll try #2

What causes electrons to first fill up each orbital rather than to exist in the same orbital with opposite spins (in subshell other than s, obviously)?

This is known as Hund's rule of maximum multiplicity: For a given electron configuration, the lowest energy is the one with the greatest value of spin multiplicity.
According to Ira Levine's Quantum Chemistry, the explanation is conceptually related to the Pauli exclusion principle. Pauli's exclusion principle forbids identical quantum numbers for two electrons around an atom. Applied to this situation, the principle is that electrons with the same spin in the same shell and subshell must be in orbitals with different magnetic quantum numbers m. The related concept is that when every quantum number is the same except m, a kind of spatial Pauli repulsion exists between such similar electrons.
The traditional explanation was that electrons with the same spin avoid each other, and that leads to a lower energy due to less electron-electron repulsion. Calculations in the 1970s showed that this was wrong. In some simple cases, the average distance between the two electrons was actually a little less (with more repulsion) for same-spin electrons than opposite-spin electrons.
The actual reason isn't distance between the electrons but a similar spatial concept. If you create an angle with the nucleus at the vertex and two electrons determining the rays, electrons with same spins create a greater angle on average than electrons with different spins. This reduces the chance that one electron will screen the other and allows the electrons to get closer to the nucleus on average (with more attraction) for same-spin electrons than opposite-spin electrons.

I can tackle #3, which is a bit of a subtle question.
In the transition elements of the fourth period, the $3d$ subshell lies at higher energy than the $4s$ subshell -- when the atoms are neutral. But the electronic energy levels change when you remove typically two or three electrons to form an ion of any of these elements in their most common salts.
Think of an ion with some number of protons in the nucleus and just one electron. The energy levels of the electron (though not the electron as such) follow the model constructed by Bohr, whereby the single $1s$ orbital is at one energy level; the $2s$ and $2p$ orbitals, which Bohr's model just called the second shell, are at another higher, single energy level; the third shell is at one still higher level, and so on. Thus the electron in what we now call a $3d$ orbital, part of the third shell, would be at lower energy than it would be in the $4s$ orbital in the fourth shell.
Real atoms and ions in ordinary chemistry, apart from hydrogen, are more complicated. They have multiple electrons which repel one another even as they are all attracted to the nucleus, and the quantum mechanical theories that succeeded the Bohr model show that the shells are no longer all at one energy level. What Bohr rendered as "the" energy level for the second shell ends up becoming a lower level for the $2s$ orbital plus a higher level for the $2p$ orbital; the interactions of the electrons with each other as well as the nucleus cause the higher angular momentum states (one unit for $p$, zero for $s$) states to have higher energy. Higher angular momentum states basically get higher in energy by avoiding the regions closest to the nucleus and intermingling more with other electrons.
When we get to the fourth period transition elements, the neutral atoms generally have just enough electrons for the higher angular momentum ($l=2$) $3d$ subshell to sneak above the higher-shell but lower angular momentum ($l=0$) $4s$ subshell. But taking (generally) two or three of those outer electrons away to form the commonly encountered ions means the interelectronic repulsions are reduced, and with that there is less separation of subshell energy levels within a single shell. The $3d$ subshell drops towards its lower angular momentum shell-mates and below the $4s$ orbital, so that the most stable form of the ion now has its outer electrons all in $3d$.

But I believe a part of that question has still not been answered - why are $4s$ electrons lost first if $3d$ electrons have more energy, albeit they owe it to $l$ (before losing electrons, the atom is neutral)?

The $3d$ electrons might be lost, but them when the $3d$ level drops it can swipe a $4s$ electron into a more stable state. We would not becable to tell this experimentally because according to quantum mechanics, electrons do not carry little tags saying which subshell of an atom they came from.


text:
Why are the phosphate ion and hydrogen phosphate not conjugate acid/base pairs of each other?

(e) Identify in the following reaction the Bronsted-Lowry acid and its conjugate base
$$\ce{PO4^3- + HNO3 -> NO3- + HPO4^2-}$$

Both $\ce{PO4^3-}$ and $\ce{HPO4^2-}$ differ by a $\ce{H+}$ ion so why aren't they conjugate acid base pairs?.

$\ce{PO4^_3-}$ is not an acid. It is a base. So the only acid in this reaction is $\ce{HNO3}$. Its conjugate base is the ion$\ce{NO3^-}$. The ion $\ce{HPO4^-} $ is the conjugated acid of the base $\ce{PO4^{3–}}$. It is not the acid creating the rection.

This query seems to be a product of misunderstanding the problem in question.
The question asks the following,

(e) Identify in the following reaction the Bronsted-Lowry acid and its conjugate base
$$\ce{PO4^3- + HNO3 -> NO3- + HPO4^2-}$$

Now, your reasoning is correct. $\ce{PO4^3-}$ and $\ce{HPO4^2-}$ are in fact conjugate acid/base pairs. However, the question doesn't ask for that. Rather it asks for the Bronsted-Lowry acid (in this reaction specifically).
According to The IUPAC Compendium of Chemical Terminology 2014, a conjugate acid-base pair is defined as follows.

The Brønsted acid $\ce{BH+}$ formed on protonation of a base B is called the conjugate acid of B, and B is the conjugate base of $\ce{BH+}$. (The conjugate acid always carries one unit of positive charge more than the base, but the absolute charges of the species are immaterial to the definition.) For example: the Brønsted acid $\ce{HCl}$ and its conjugate base $\ce{Cl−}$ constitute a conjugate acid–base pair.
(emphasis mine)

Now, a Bronsted acid is defined within the same book as the following,

A molecular entity capable of donating a hydron (proton) to a base, (i.e. a 'hydron donor') or the corresponding chemical species. For example: $\ce{H2O}$, $\ce{H3O+}$, $\ce{CH3CO2H}$, $\ce{H2SO4}$, $\ce{HSO4−}$, $\ce{HCl}$, $\ce{CH3OH}$, $\ce{NH3}$

In this specific question, the hydron donor is $\ce{HNO3}$ and therefore, it is the Bronsted acid. Both pairs - $\ce{PO4^3-}$ and $\ce{HPO4^2-}$ as well as $\ce{HNO3}$ and $\ce{NO3-}$ - are conjugate acid/base pairs though.


text:
Why is the 8N (octet) rule not applicable in the case of boron triflouride?
In my chemistry class we learned the so-called "8N rule" (very similar to the Octet rule) which helps one determine the number of bonding electron pairs and thus the number of bonds in certain molecules:
$$\text{No: of bonding electrons} = 2\cdot\text{No: of hydrogen atoms} + 8\cdot \text{No: of other atoms} - \text{Total no: of valence electrons}$$
This tends to work for most molecules I've tried. The lecture notes don't really mention any exceptions or limitations. Therefore, I assumed that it would work for a wide variety of molecules, especially ones that involve p-block atoms only.
However, the rule doesn't work with $\ce{BF3}$. Here I get $2\cdot0 + 8\cdot 4 - (3+3\cdot7) = 8$ bonding electrons which should translate to $8/2 = 4$ bonds (or one of the $\ce{B-F}$ bonds is a double bond). But this is apparently not the case - there are three single bonds or 6 bonding electrons. So where is my mistake?
I should say that this is not the first time I've encountered a rather simple molecule where the rule doesn't apply which is what gives confidence that it's not because of a mistake I made but because of a limitation of this rule that I don't know.

Octect rule is more than 100 years old. It does not apply to many compounds. People have come up with better explanations and bonding theories.

Then again, the bonds in BF3 are at least somewhat double.

In that case, is there a better and yet as handy alternative for calculating the number of bonds or the bond order?

The octet rule is a somewhat surprising model. It holds quite well for molecules as long as you are reading the 8 as an upper limit. Since it had never been formalised, this role of thumb is very open for interpretation.

Boron is one of the exceptions to the octet rule in which the octet is incomplete, and thus it only has 6 valence electrons instead of 8 to become stable. To put it simply, because Boron has 3 valence electrons, they react to only form 3 bonds which stabilises the atom with 6 electrons. I assume there is a more complex way to explain each facet of this incomplete octet that I don't know of, though.

The octet rule has a lot of exceptions. Examples :  $\ce{B2H6}$, $\ce{NO}$, $\ce{NO2}$, $\ce{SF6}$, $\ce{H2SiF6}$, plus the majority of compounds containing transition metals.


text:
Confusion related to exception in Fajan's rule (Regarding prediction in solubility of compounds in polar solvents)
My teacher taught us that higher the ionic character in a compound, higher is its solubility in polar solvents like water.
To illustrate this statement, he also gave us the following example-
LiOH< NaOH< KOH< RbOH< CsOH (solublity in water)
According to this example, if the anion is same in all compounds then ionic character increases as we go down the group and thus the compounds become more soluble.
But then our teacher also gave us an exception. He said that if we another example with alkaline earth metals, then ideally according to Fajan's rule BaSO4 should be more soluble than MgSO4. But this is not the case because as we go down the group the decrease in hydration energy is greater than the decrease in lattice energy and thus solublity decreases.
My question is that shouldn"t this explanation also apply for the alkali metals? Shouldn't the decrease in hydration energy be greater than the decrease in lattice energy here as well?

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

@Poutnik yes, I did mean that. Have edited that bit.

@Poutnik Also,  'Keep CH SE Q titles in plain text.' - the title had been typed in plain text at my end. Does it show something else at your end?

Note that barium hydroxide IS much more soluble than magnesium hydroxide, which is practically insoluble (and the exact opposite for sulfates).

@Poutnik may I ask why it is the opposite for sulphates. Also, I have edited my question. I checked my notebook and indeed, the exception was related to sulphates and not hydroxides. Sorry.