text:
Why do we have to prevent the hydrolysis of iron(III) nitrate?
I want to make particular concentration of ferric ions from $\ce{Fe(NO3)3.9H2O}$, and then I found this video
It says that we have to add nitric acid to prevent iron from hydrolysis, what does it mean? Why do we have to prevent it? If not adding it, does it affect our concentrate solution later?

Why not to get familiar with chemistry of iron(III) compounds before working with them? BTW, it is not a molecule and there is no $\ce{H20}$.

The iron becomes bound to hydroxyl groups in water.

Iron(III) nitrate dissolves in water as ions: $\ce{Fe(NO3)3 <=> Fe^{3+} + 3NO3-}$
The ferric ions forms a complex with water: $\ce{Fe^{3+} + 6H2O <=> Fe(H2O)6^{3+}}$
The iron(III) pentaaquahydroxy complex ion forms: $\ce{Fe(H2O)6^{3+} + H2O <=> [Fe(H2O)5(OH)]^{2+} + H+}$

Adding $\ce{HNO3}$ pushes the equilibrium back towards $\ce{Fe^{3+} + 3NO3-}$.
BTW, heating $\ce{Fe(NO3)3}$ solution to dryness drives off the nitric acid, leaving iron oxide.

The ferric ions is already hydrated when the salt is being dissolved, Anhydrous ferric salts usually exothermically hydrolyze, ferric chloride with $\ce{HCl}$ fumes, similarly as anhydrous aluminium salts.

If not adding it (nitric acid), does it affect our concentrate solution later?

Yes it does because slow hydrolysis with water, some $\ce{Fe^3+}$ would separate out from the solution as solid particles (Ref.1). This review focuses on studies on the hydrolysis and precipitation of $\ce{Fe^3+}$ from aqueous solutions of its inorganic salts, chiefly the nitrate, perchlorate, chloride, and sulfate. Note that the $\ce{Fe^3+}$ ion has been shown to exist as the octahedral $\ce{[Fe(OH2)6]^3+}$ complex in aqueous solutions of $\ce{Fe(NO3)3.9H2O}$ (Ref.2).

The hydrolysis reactions of these inorganic salt solutions, most of which occur slowly, consists of several steps: (1) formation of low-molecular-weight species; (2) formation of a red cationic polymer; (3) aging of the polymer, with eventual conversion to oxide phases; and (4) precipitation of oxide phases directly from low-molecular-weight precursors.

The characterized crystalline iron(III) oxides and hydrous oxides are $\ce{Fe2O3}$ and $\ce{FeO(OH)}$, each of which is polymorphic and very insoluable in water. For example, the $\ce{\alpha-Fe2O3}$ (hematite) has a $\mathrm{p}K_\mathrm{sp}$ of $41.7$ while that of $\alpha-, \beta-,$ and $\gamma-$ phases of $\ce{FeO(OH)}$ ((goethite, akaganeite, and lepidocrocite) are $41.7, 36,$ and $38.7$, respectively. Thus, one can expect after preparing in aqueous solutions, $\ce{Fe^3+}$ can lost significant amount to hydrolysis.
The first step of the hydrolysis of inorganic Fe(III) solutions is the formation of low-molecular-weight species. Two such reactions are given in Ref.1 as:
$$\ce{Fe(OH2)6]^3+ + H2O <=> Fe(OH2)5(OH)]^2+ + H3O+} \qquad K_\mathrm{eq} = 2.2 \tag1$$
$$\ce{Fe(OH2)5(OH)]^2+ + H2O <=> Fe(OH2)4(OH)2]^+ + H3O+} \qquad K_\mathrm{eq} = 3.5 \tag2$$
These formation reactions of low-molecular-weight species can be revised by adding appropriate strong acid (here it is nitric acid since we used $\ce{Fe(NO3)3}$) before they react further to make polymeric iron salts. Note that Hydrolyzed Fe(III) solutions containing the red polymer are not at equilibrium (Ref.1).
The polymer separated from hydrolysis has been studied to find the rate of their degradation by $\ce{HNO3}$, which results in immediate precipitation of an undetermined fraction of the iron. Thus, it should be advised that addition of the nitric acid should be immediate after the preparation of iron solution.

References:

Charles M. Flynn Jr., "Hydrolysis of inorganic iron (III) salts," Chem. Rev. 1984, 84(1), 31-41 (DOI: https://doi.org/10.1021/cr00059a003).
Neil J. Hair and James K. Beattie, "Structure of hexaaquairon(III) nitrate trihydrate. Comparison of iron(II) and iron(III) bond lengths in high-spin octahedral environments," Inorg. Chem. 1977, 16(2), 245-250 (DOI: https://doi.org/10.1021/ic50168a006).
Barbara A. Sommer, Dale W. Margerum, John Renner, Paul Saltman, and Thomas G. Spiro, "Reactivity and aging in hydroxy-iron(III) polymers, analogs of ferritin cores," Bioinorganic Chemistry 1973, 2(4), 295-309 (DOI: https://doi.org/10.1016/S0006-3061(00)80202-1).

The Gist
As mentioned in the comments, we have to prevent the formation of aqua complexes of $\ce{Fe^{3+}}$ as these tend to hydrolyze:
$$\begin{multline}
\ce{[Fe(H2O)_{2n}(NO3)_{3-n}]^{n+}.(NO3^-)_n.(H2O)_{9-2n} + H2O <=>} \\
\ce{[Fe(H2O)_{2n-1}(OH)(NO3)_{3-n}]^{n+}.(NO3^-)_n.(H2O)_{9-2n}  + H3O+}
\end{multline}$$
Note: In the above equilibrium, it was assumed that the denticity of $\ce{NO3^-}$ is 2. This may or may not be the case. I have avoided this ambiguity in the similar equation in the Remarks section.
Now, to prevent formation of aqua complexes, there are (at least) two ways to do this:

Decrease the pH to shift the above equilibrium to the reactants side in accordance with Le Chatelier's Principle.
Increase the concentration of $\ce{NO3^-}$ in the solution to prevent formation of aqua complexes in accordance with Le Chatelier's Principle, as explained in the section Hydration Isomerism of $\ce{Fe(NO3)3}$ below.

Your suggestion to add $\ce{HNO3}$ does both simultaneously, decreasing the pH and increasing the concentration of $\ce{NO3^-}$.
Hydration Isomerism
For a hydrated complex $\ce{[ML_y].xH2O}$, generally, assuming maximum valency of $\ce{M}$ is $y$, $y+1$ hydration isomers are possible given by the formulae:
$$
\ce{[ML_{y-n}(H2O)_n].L_n.($x-n$)H2O} %nasty workaround, possibly mhchem bug
$$
Note: This formula is most accurate for crystalline ($\ce{c}$) phase and the behavior is more complex in aqueous ($\ce{aq}$); however, for simplicity, we will consider this formula for our analysis.
Example
Consider the example of $\ce{[MX2].2H2O}$ where $\ce{M}$ has a maximum valency of 2:
$$
\ce{\underset{\text{isomer 1}}{[MX2].2H2O} <=> \underset{\text{isomer 2}}{[MX(H2O)].X^-.H2O} <=> \underset{\text{isomer 3}}{[M(H2O)2].(X^-)2}}
$$
Hydration Isomerism of $\ce{Fe(NO3)3}$
$\ce{Fe(NO3)3}$ has several hydration isomers and is commonly represented as $\ce{[Fe(NO3)3].xH2O}$, where $x$ is usually 9, which we will consider.
The usual coordination number and geometry of $\ce{Fe}$ are $6$ and octahedral, respectively, commonly forming six-coordinated complexes.
Thus, in $\ce{aq}$ phase, you are likely to observe (assuming that the denticity of $\ce{NO3^-}$ maybe $\pu{1}$ or $\pu{2}$), among many other isomers, the following complexes.

$\ce{[Fe(H2O)_6].(NO3^-)3.3H2O}$,
$\ce{[Fe(H2O)_4(NO3)_2].NO3^-.5H_2O}$, and
(very) low amounts of $\ce{[Fe(NO3)3].9H2O}$.

Note: These formulae are not quite accurate, as this Wikipedia article states.
This happens because, in the spectrochemical series, $\ce{H2O}$ is a stronger field ligand than $\ce{NO3^-}$, forming more stable complexes with $\ce{Fe^3+}$. We could represent this as:
$$
\ce{\underset{\text{isomer 1}}{[Fe(NO3)3].9H2O} <=> \underset{\text{isomer 2}}{[Fe(H2O)_4(NO3)_2].NO3^-.5H_2O} <=> \underset{\text{isomer 3}}{[Fe(H2O)_6].(NO3^-)3}}
$$
where, in aqueous phase, isomer 3 (hydrolyzed and otherwise) is present in abundance, and isomer 1 rarely observed. We don't want aqua complexes such as isomer 2 and isomer 3 as, reasoned in The Gist section, these tend to hydrolyze:
$$\begin{multline}
\ce{[Fe(H2O)_x(NO3)_y]^{n+}.(NO3^-)_n.(H2O)_{9-x} + H2O <=> }\\ \ce{[Fe(H2O)_{x-1}(OH)(NO3)_{3-n}]^{n+}.(NO3^-)_n.(H2O)_{9-x} + H3O+}
\end{multline}
$$
Shifting the Equilibrium
As pointed out in other answer, increasing the amount of $\ce{NO3^-}$ in the solution will, in accordance with Le Chatelier principle, shift the equilibrium towards the right. You are essentially carrying out the following reaction:
$$
\ce{\underset{will hydrolyze}{[Fe(H2O)_6].(NO3^-)3.(H2O)3(aq)} ->[\ce{H+NO3^-}][high concentration] \underset{won't hydrolyze}{[Fe(NO3)3]\cdot 9H2O}(aq)}
$$
Remarks
Denticity of Nitrate Anion
Denticity of the nitrate anion maybe 1 or 2. Therefore, in:
$$
\ce{[Fe(H2O)_x(NO3)_y]^{n+}.(NO3^-)_n.(H2O)_{9-x}}
$$
$x+y \in [3,6]; x>0, y>0$.
Two-Forked Action of Addition of Nitric Acid
Addition of $\ce{HNO3}$ is a two-forked strategy, decreasing the pH and increasing the concentration of $\ce{NO3^-}$, both avoiding formation of aqua complexes of ferric cation.

It seems not to address ferric ions hydrolysis, but hydration isomerism. Sure, you could formally replace NO3- by OH-.

@Poutnik I couldn't find many sources regarding hydrolysis of ferric nitrate. I assumed it meant hydration of ferric nitrate. Sometimes the water of hydration does deprotonate and ligate as hydroxide. Could you help a little here?

$\ce{[Fe(H2O){6-n}(OH)n]^{3-n}(aq) + H2O(l) <=> [Fe(H2O){5-n}(OH){n+1}]^{2-n}(aq) + H3O+(aq)}$ for $n=0..2$
$\ce{2 Fe(H2O)3(OH)3 <=> Fe(H2O)3(OH)2-O-Fe(H2O)3(OH)2 + H2O(l)}$ and further condensation... // the actual mechanism is complex and particular structures may be guesses. there would be also multiple parallel ways of simultaneous hydrolysis and condensation.

@Poutnik sorry for the bother, but could you provide a citation... I want to know the $\mathrm{pH}$ at which these reactions occur, and if at all, mixing $\ce{Fe(NO3)3}\cdot 9H2O$ would form hydroxide complexes at a neutral $\mathrm{pH}$. $\ce{HO-}$ is a weaker field ligand than $\ce{H2O}$.

No cit. available. Ferric ion start to hydrolyze even near pH 3-4. Hydroxide reportedly coagulates at pH 4.5 // Try searching for ferric ion hydrolysis. I do not have access to paywalled articles.  // https://www.sciencedirect.com/topics/earth-and-planetary-sciences/ferric-ion //  https://pubs.acs.org/doi/10.1021/jp980229p //  https://pubs.acs.org/doi/10.1021/ja01324a018

@Poutnik thank you. I have incorporated your suggestions.

@Martin-マーチン thank you for the revisions. I will keep these in mind when writing future posts. I also plan to edit my previous posts too, and I will keep these revisions in mind then as well.

I've gotten out the computer and tried to fix the MathJax. I'm pretty sure I didn't find everything though and there might be a bug with mhchem. Oh well… that's just life. Good luck with these kinds of posts. Btw. there is plenty of info on MthaJax on [meta].

Come to think of it: https://goldbook.iupac.org/terms/view/V06588 Valence for iron is not 6! That is the coordination number (most common). But that whole concept was stretched beyond the useful and is usually taught wrong in schools and textbooks… You might want to fix that, too.

@Martin-マーチン how could we forget about $\ce{[Fe(CO)5]}$? This does show an example of $\ce{Fe}$ having a valency of 5, but the oxidation state of $\ce{Fe}$ is zero here. We need to consider the valency of $\ce{Fe^{3+}}$.

@Martin-マーチン $\ce{[Fe(OH2)_6]^{3+}}$ is a 17 electron complexes and, theoretically, should be (somewhat) oxidizing in nature.

I find your comment a bit insulting. Did you actually have a look at the definition I have linked to? There are only a few possibilities here: You did not, you did not understand it, or you disregarded it. Honestly I do not understand how your comments address any of the concern I raised.

@Martin-マーチン sorry for the misunderstanding, I did look at it, but maybe I did not understand it fully. "The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted." I also noticed that ferric hydride is unstable, but $\ce{FeCl3.xH2O}$ points to a valency of 3 for $\ce{Fe}$.


text:
Is water required for metal to rust, or is only oxygen necessary?
What causes rusting? Some sources say ferrous metals don't rust without being exposed to humidity or water, whole others say they do rust as long as there is oxygen present, just more slowly. Which is correct? For example, if a vehicle were left in a bunker for a very long time, such as 100k years, with it dry the entire time, would it rust? And if so, would the rusting completely destroy it?

Cite your sources.

You have been supposed to find the provided link before asking.  // Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, misunderstanding, objections, question down-voting or even question closure.

$\Delta_\mathrm{f} G^\circ$ (free energies of formation) at $\pu{298.15 k}$ of $\ce{FeO}$ and $\ce{Fe2O3}$ are$^1$ $\pu{-101 5.227 kJ mol-1}$ and $\pu{-742.294 kJ mol-1}$, both negative, implying both these reactions are spontaneous at ambient conditions.
However, we wouldn't call it rusting, because rust is $\ce{Fe2O3.xH2O}$, the hydrated form of $\ce{Fe2O3}$. Thus, an $\ce{Fe}$ sample would corrode in an $\ce{O2}$ chamber but not rust. You may observe black/red powder which will be easily distinguishable from rust by examination of color.
References

Thermodynamic Data of Pure Substances. Ihsan Barin, Gregor Platzki. Third edition. VCH Verlagsgesellschaft mbH, Weinheim (Federal Republic of Germany), VCH Publishers, Inc., New York, NY (USA)

source: Stack Exchange

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

sort: votes


text:
Powder left everywhere after solution was left to evaporate. What could it be?
I dissolved a coin made of silver and copper and probably other metals in nitric acid. I neutralized the solution with some sodium hydroxide and then added salt(NaCl) to make Silver Chloride. A white percipitate formed and was filtered off. Some insoluable copper hydroxide was also mixed with the silver chloride. I washed the mixture with dilute hydrochloric acid to form soluable copper chloride, quite a lot of dilute acid was needed.
The waste solutions were left to evaporate in a fume hood so that they did not take up so much space. After a few weeks when i went to see if they had evaporated down i noticed a white powdery film that coated everything in the fumehood.
Why did this happen and what could this powder be?
I thought salts didn't evaporate with the solutions.

It is probably NH4Cl(s), slowly covering everything nearby long term situated $\ce{HCl}$ solutions, especially if traces of ammonia occur in air.  ( $\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$ )

I tested this theory by scraping of some dust and doing a silver nitrate test on it which yielded a positive result for chloride ions. Then i added sodium hydroxide solution to the powder in a test tube and used a wet ph paper to test for ammonia gas which also gave a positive result. Turns out that the Acid-base waste jug's cap was loose and that was what probabily released ammonia.

It is quite common knowledge among chemists with some lab experience.

The salts don't evaporate, but fume hoods have air flow going through them to enable the removal of the toxic fumes through the hood. That air flow can pick up droplets of solution, which then evaporate in the space of the hood — rather than staying in your original container.

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


text:
What do we get when we joint two α- glucoses via an α1,4 bond?
What do we get when we joint two α- glucoses via an α1,4 bond?

Is it alpha maltose or beta maltose?
In other words, if we cleave amylopectin with amylase (α-amylase or β-amylase), which form of maltose is the ultimate result?

I think Maltose, which has an axial anomeric hydroxy group, will be the answer to (1) and (2), because α-glucose has has axial anomeric hydroxy group; am I right? However, I could not identify which form of maltose has axial anomeric hydroxy group.

The anomeric carbon of of the second glucose unit of maltose interconverts between the alpha and the beta form in aqueous solution (just like glucose). The stereochemistry of the anomeric carbon in the glycosidic bond, on the other hand, is fixed.
For some sugars, you can isolate one or the other stereoisomer in the solid form. Glucose, for example, shows a large difference in solubility between the alpha and the beta form. If you crystallize glucose at room temperature, you get crystals of alpha glucose (details). In this crystal structure of maltose, the main form is the alpha isomer, with about 20% in the beta form. In solution, the ratio of alpha and beta will be different (and there will be some open form species as well), and will depend on the temperature.

Thank you!! So, is it correct to understand that for (1), both α-maltose and β-maltose can be made, and for (2), both α-maltose and β-maltose can be made?

@BlueVarious Yes, and usually you don't have a choice. Enzymes might make one rather than the other, but then they interconvert anyway and reach an equilibrium mixture (of alpha, beta and open form) after a while.

Thank you. So am I right in understanding that if the anomeric hydroxy group of α-glucose and the OH at position 4  of β-glucose are joined by an α1,4 bond, α-maltose and β-maltose would still be in similar chemical equilibrium?

I would be glad to have your answer to this question as well.　https://chemistry.stackexchange.com/questions/173748/are-the-anomeric-hydroxy-groups-of-%ce%b1-maltose-and-%ce%b2-maltose-axial-or-equatorial


text:
Does resonance stablization of carbocation increase ease of $\ce{S_N1}$ reaction or decrease it due to delocalization?
 (A)
 (B)
In both of the cases when a nucleophile approaches for $\ce{S_N1}$ reaction a carbocation is formed in place of $\ce{Cl}$ atom. I know that the rate of $\ce{S_N1}$ increases with increase in carbocation stability, and in (B) the carbocation will be resonance stabilized. But at the same time won't the positive charge be delocalized as well, leading to decreased effect of positive charge at the leaving group position? Won't that negatively affect $\ce{S_N2}$ reaction?

(A) Yes, rate of $\mathrm{S_N1}$ increases with the stability of the intermediate as the energy of transition state (partially positive charged at $\alpha$-$\ce{C}$) also decreases due to stabilization by delocalization.
(B) Since the rate-determining step (RDS) $\mathrm{S_N1}$ is the formation of the carbocation, the second step (which does slow down due to delocalization, but doesn't become as slow as the RDS), and thus, the rate is dependent on how fast the carbocation forms and not how fast the nucleophile approaches the positively charged intermediate; therefore, the overall reaction will speed-up.
Points to consider

$\ce{1-chloropropane}$ may undergo $\mathrm{S_N2}$ because the $\alpha$-$\ce{C}$ is primary, for which $\mathrm{S_N1}$ rates are usually slow. Some $\mathrm{S_N1}$ is always observed at primary $\ce{C}$ atoms as well, but rate of $\mathrm{S_N2}$ is usually much larger.

For $\ce{3-chloroprop-1-ene}$, nucleophile can approach at both the $\alpha$ and $\gamma$ positions. Here the product is the same, but it won't be the same for $\ce{4-chloroprop-2-ene}$.

Where is it proven that the energy of the transition state is dependent on its connected intermediates? That's a bit far fetched and hand wavy. Also, the RDS approximation works only for linear, non equilibrium reaction mechanisms. It is quite a terrible approximation.

I'm pretty certain I have asked you this before: please do not use MathJax as a way to format text.

@Martin-マーチン do you mean for me to change $\ce{3-chloroprop-1-ene}$ to 3-chloroprop-1-ene?


text:
Are the anomeric hydroxy groups of α-maltose and β-maltose, axial or equatorial, respectively?
This Q&A is a continuation from the following Q&A;
What do we get when we joint two α- glucoses via an α1,4 bond?
More straightforwardly, the question may focus on whether the bond represented by the following straight line means equatorial or axial　(Find the OH circled in blue in the following figure).

https://www.chemspider.com/Chemical-Structure.388329.html
Above figure is quoted from chemspide. If I understand correctly, the bond or solid line may indicates the anomeric hydroxy group of the alpha-maltose is equatorial, am I right(See the blue circle of the figure, quoted above)?
The chemspide also has figure which may indicates the anomeric hydroxy group of alpha-maltose has axical anomeric hydroxy group, shown as follows. (See the blue circle of the figure, quoted below)

https://www.chemspider.com/Chemical-Structure.6019.html
However, Wikipedia has figures which indicate the anomeric hydroxy group of α-maltose is axial and β-maltose has equatorial one.

https://en.wikipedia.org/wiki/Maltose
So, I' very confused.
On the other hand, PubChem has following descriptions, quoted below; this article
is titled simply "maltose," but this article also has the description which the  "maltose" is synonymous with the "beta-maltose".

However, although the article for the α-maltose titled  "α-maltose" but the article also　has the description  which indicate the "α-maltose  is also synonymous with maltose".
https://pubchem.ncbi.nlm.nih.gov/compound/alpha-Maltose#section=3D-Conformer
If I understand correctly, the structural formulas on PubChem indicates;

the anomeric hydroxy group of α-maltose　is axal but　directed Upward
the anomeric hydroxy group of β-maltose　is also axial but directed downward.

anomeric hydroxy groups of both A and B in this appear to be axial.

https://pubchem.ncbi.nlm.nih.gov/compound/alpha-Maltose#section=Structures

Maltose [4-O-(α-D-glucopyranosyl)-D-glucopyranose] always has an α-1-O-4 linkage. The bridging oxygen at $\ce{C1}$ is axial. For maltose, there are two anomers. The β-anomer (shown below) has the anomeric hydroxyl equatorial while the α-anomer is axial. These two anomers are interconvertible in aqueous solution.
On the other hand, cellobiose [4-O-(β-D-glucopyranosyl)-D-glucopyranose] always has a β-1-O-4 linkage. It also exists as α- and β- (shown below) anomers, axial and equatorial, respectively.
The diagrams are from here. A good read..


text:
Why does this lysozyme powder not look crystalline?
I am very new to protein crystallography and visualization. I have been asked to take photos through an optical microscope of the lysozyme powder directly as-is from the supplier. I suspended some of the powder in both paraffin and glycerol (10 mg/mL & 20 mg/mL for each) to observe the crystal form without solubilizing it.
I'm not sure what it is I'm looking at, however, in all of the suspensions I prepared, the particles look very different from literature images of both lysozyme and salt crystals. The image I included is a 10 mg/mL suspension in paraffin but is essentially what I'm seeing in all cases. A mixture of large chunks with some smaller flakes.
I am wondering if anyone with protein crystal experience can tell me what this form I'm seeing is? And perhaps any further suggestions on how to visualize the sample without any alteration. Thank you

They do not look like single crystal pieces, but I'm not sure that they are not polycrystalline in nature.

The typical protein crystal does not withstand mechanical stress equally well than e.g., the ones of table salt, or refined sucrose.  Hence the varying shapes of grains.
Photos about nice protein crystals by shape (habit) & extinction (check if you can use linearly polarized light / "crossed Nicols") often are recorded while they still are in their mother liquor (to grow them e.g., by sitting drop, hanging drop, or in capillary [see e.g. here] to mention some techniques) typically grown for in-house/beamline X-ray diffraction analysis.  Here, you want them to be representative of your sample (e.g., in chemical composition, and crystalline phase), and yet good enough for subsequent data collection and data processing.
The commercial (perhaps microcrystalline) powder you have access to, is prepared with the primary intent to offer you access to a sample chemically (by molecular composition) pure enough.  That is, concentrations of remaining solvent of reaction / purification, unwanted ions of heavy metals are below an (often arbitrary) threshold set.  If there once were crystals, they now are broken in a more or less random way while passing sieves, being scrapped-off from filters, dried (think about solvent molecules like water which have to leave the solid), filled into a bottle which upon transport is shaken / crystals bounced against each other.  This reduces your chances to meet a crystal with prominent cleavage planes.  Instead, you find rough and conchoidal surfaces. In addition, the size of the grains may follow a more or less broad grain distribution, too.

Your first sentence isn't clear to me, do you mean protein crystals are just as bad at withstanding mechanical stress as salt/sugar are, or that protein crystals are worse at withstanding mechanical stress?

Generally speaking, @llama, protein crystals are much more fragile than crystals of ionic salts or small molecules.  They also invariably contain a large proportion of water, so protein crystals generally cannot endure being very long removed from their mother liquor to a dry environment.

It often is easier to damage a protein crystal, than the ones of small molecules. It then is more difficult to record their shape/dimensions, perform an absorption correction; mechanically damaged crystals often feature broader/more often partially split/overlapping individual X-ray diffraction peaks which can render them unsuitable for single crystal XRD. You still have powder XRD as a complement which by today very well includes proteins (beside the sample a question of equipment & accessing a source brilliant enough). See e.g. Margiolaki (https://doi.org/10.1524/9783486992540-004, OA).

Sometimes, protein gets sold in a freeze-dried formulation. In that case, you don't expect crystals, and sometimes the protein will denature (which is find for some applications, such as protein standards in SDS PAGE). See e.g. Roy and Gupta


text:
Do all p orbitals really have the same energy?
If we solve the time-independent Schrödinger equation for any atom by considering only the electrostatic potential, an electron has the same probability of occupying the $p_{x}$ orbital as it does the $p_{y}$ and $p_{z}$ orbitals, because the electrostatic force is a central / spherically symmetric force which depends only on $r$, and the wavefunctions of the $p_{x}$ , $p_{y}$ , $p_{z}$ orbitals only differ in terms of the angular coordinates $(\theta,\phi)$.
However, a nucleus has a quantum spin which introduces a magnetic field, and the symmetry is broken. If we take into account the magnetic field of the nucleus, then the electromagnetic force which the nucleus exerts on electrons occupying the $p_{x}$, $p_{y}$, and $p_{z}$ orbitals is no longer central. So there should be a difference between the energies of the wavefunctions of the $p_{x}$, $p_{y}$, and $p_{z}$ orbitals.
How significant is this effect? In atoms with a small atomic number or with a small nucleus spin the effect is negligible and can be only observed near absolute zero. However shouldn't it become significant for atoms with bigger atomic number or with a large nuclear spin?

The degeneracy is only valid for one-electron atoms/ions.

As I understand it, when you start taking interactions like these into account (what you're describing here is hyperfine splitting), it's no longer valid to refer to $p_x$, $p_y$, $p_z$ orbitals anymore (which, as you note, are solutions to the Schrodinger equation when you neglect such factors). But yes, my understanding is that there is a lifting of degeneracy. (See the linked article.) || @ananta That's not true, what you're describing sounds more like degeneracy within a given $n$, and OP is asking about degeneracy within a given $\ell$.

Should we be asking questions about $p$ orbitals when all we have in one-electron system is an $s$ orbital?

You can excite the electron of a one-electron atoms into a $p$ orbital or any orbital at all. We can distinguish all the subshell states in the hydrogen spectrum when fine splitting (relativistic, nuclear spin interactions, etc) is taken into account.


text:
Ductless fume hood for dissolving aluminum foil in hydrochloric acid
I work at a food manufacturing plant and we use Tetra Pak packaging for our liquid foods. There is a layer of aluminum foil within the packaging material that we need to dissolve out of the package in order to examine the sealing of the inner plastic layer. We will be using 30% HCl for the reaction. We have a ductless fume hood with an acid gas filter for this process. I am concerned that the hydrogen gas released in the reaction could be an issue in the room as it is unlikely to be filtered out by the hood. We will be doing, at most, 16 packages a day (all at the same time) and the reaction would likely last for 45-90 minutes. Also, it would not be the entire package being dissolved but only areas near the seal being examined. Is there a reason to be concerned about the hydrogen gas released in this reaction when using a ductless hood? The lab is not particularly small and reasonably ventilated. If necessary we could locate the hood to an area outside the lab with better ventilation. Any input is greatly appreciated and I can provide more details if needed.

You seem to be afraid of something. What ? Hydrogen is a light gas. It is not considered as toxic. Its density is rather low. So it will leave the lower part of the hood. It burns or explode in a reaction with oxygen. Is it what you are afraid of ?

My concern may be unjustified. I am setting up this procedure and may have some nervousness that I’m not overlooking anything. My concern would be outside of the hood and if the hydrogen gas could be enough to combust. Our lab has a lot of other equipment that may be running such as an autoclave, water distiller, solid analyzer (basically microwave), and drying oven. My background is in microbiology so I feel slightly out of my comfort zone.

Hydrogen safety states, "hydrogen has no rating for innate hazard for reactivity or toxicity," so the primary concern is flammability. However, without measurement, it is impossible to give a definite answer.

Measure the mass of the aluminum in the foil. Either weigh the foil directly, or, if attached to a plastic layer, measure the mass of the whole before and after aluminum is removed to find the difference, i.e., mass of Al.

Calculate how much hydrogen that releases over 45 minutes, the worst case scenario. Given ventilation in the hood (calculate the rate of air exchange, either from the hood's specs, or using an anemometer), could the volume of hydrogen compared to the volume of air moved over that period of time approach the flammability limit of 4% hydrogen:air in the hood?

Calculate how much hydrogen could be released from 16 tests over a work-day (8 hours?). Given ventilation in the room (information that should be available from physical plant staff, since there would have been design specs for HVAC air exchange in the facility), could the volume of hydrogen compared to rate of air exchange in the room over that period of time approach the flammability limit of 4% hydrogen:air?

[Now, for my totally unjustified assumption: you're likely dealing with milligrams of aluminum, releasing a few milliliters of $\ce{H2}$ for each test. It seems unlikely, if you could capture all the gas from one test in a balloon, that you'd hear a pop! if you lit it.]

Your assumption is likely correct. I will plan to do some measurements to figure out the amount of foil that will be dissolved and if that can result in dangerous levels of hydrogen in the air at any point during the process. We would ideally be doing this with all the test packages at one time so the concern would be high levels at the peak time of the reaction. Thank you for the information.

Then, as you state, just multiply the amount of hydrogen by 16. BTW, if the amount of H2 is an issue, it could be "flared off" by leaving a small pilot light (e.g., an alcohol lamp) directly above the reaction vessel, so as to burn it off before it accumulates. A hot platinum-metal (Pt, Ir, Pd, etc., but not Os) wire could serve the same purpose. Again, it would seem unlikely much H2 would be released, unless the Tetra Pak is quite large.


text:
Schottky Anomaly
Simple question about the definitions, but I can't find it explicitly stated anywhere.
If you have a two level system described by $n_i = \frac{N}{Z}e^{-\beta E_i}$ and calculate the heat capacity
$$ c_v = \bigg( \frac{\partial U}{\partial T} \bigg)_{N,V} = \frac{\partial}{\partial T} NE₁\frac{e^{-\beta E_1}}{1 + e^{-\beta E_1}}  = \frac{1}{Z²} Nk_B(\beta E_1)² e^{-\beta E_1}$$
Where:
$ T $ is temperature
$ \beta = \frac{1}{k_BT}$
$n_i$  is the number of particles in energy level $i$.
$ Z $ is the partition function
You get a peak at low temperature.  In fact, I think this is true (I graphed it) for any system with a finite number of states of discrete energy.  But I don't understand how a system with finite states can have a temperature.  I understand temperature as the average translational kinetic energy per particle.  When there are other degrees of freedom available, addition energy added to the system is partitioned among this translational energy and other modes like vibration and rotation.  So it takes more total energy added to increase the average translational $E_K$ by the same amount.
If you have translational motion (around the centre of mass frame), then there are an infinite number of energy levels available, unless the velocities are bounded.
So by "two-level system" do we actually mean two states other than the translational energies?  Like a cloud of electrons (non-interacting... neutrons?) with spin up and down in a magnetic field.

Is this the temperature def. where you can get wacky results, like going to negative infinity? Well, physics can make whatever def. is useful for description of real systems, even is results might be unorthodox. BTW I think in the future asking stuff like that on Physics.se could be a good idea.

@FurrierTransform Hello. You are integrating the population fraction of a two-level system that will be found in the second level, i.e., $e^{-\beta \varepsilon}/(1 + e^{-\beta \varepsilon})$. However, that is not the way to relate $C_\mathrm{V}$ to the molecular partition function. Note even that the units don't make sense (you are integrating something unitless). Try to use perform $C_\mathrm{V} = 2kT \left(\dfrac{\partial \ln q}{\partial T}\right)\mathrm{V} + kT^2 \left(\dfrac{\partial ^2 \ln q}{\partial T^2}\right)\mathrm{V} $.

@FurrierTransform BTW, the anomaly that you refer to is another thing. It has to do with the decay to zero of $C_\mathrm{V}$ as $T\rightarrow \infty$, that will happen in that system. This is, you can heat something easier even if $T$ is really high (surprise!). If the math doesn't close to you, let me know in the comments and I can post something. (edit: in the last comment, I meant differentiate, not integrate, oops)

Related: Can somebody explain the low temperature Schottky anomaly for two level systems?

I understand temperature as the average translational kinetic energy per particle. - rather T is classically  proportional to average energy per a degree of freedom, later somewhat complicated by energy quantisation steps. There are no translational energy modes in solids.

@Metal Storm I forgot to scale the middle expression by $NE_1$, thank you.  The units in this term are now correct.  However this does not affect my question about whether the definition of "two-level system" means two levels in addition to a continuum of allowed translational energy levels.
This is the definition of (constant volume) heat capacity in Atkins Physical Chemistry (p43 and p560).

@Mithoron I was also wondering if it is from this that one derived those negative temperatures (I think so). However, my question is not on negative temperatures, though it might offer insight towards them.  I posted the question here rather than physics because it was in from my chemistry class using my chemistry textbook.

#ananta I did see that question, and I understand (I think) the answer there, however my question is less "Why does the heat capacity go down again?" as much as that that behaviour only makes sense if our classification of a "two-level system" does not disallow multiple levels of translational kinetic energy (around the centre of mass frame).  So maybe by "two-level system" we mean something with two spin states {up, down} in a magnetic field, AND an infinite number of energy levels associated with translation motion.

To answer 'But I don't understand how a system with finite states can have a temperature', we can start with the thermodynamic definition of temperature as
$$\displaystyle T=\left(\frac{\partial U}{\partial S}\right)_V$$
where as usual $S$ is entropy and $U$ internal energy.
In a system such as a mole of an ideal gas or a crystal, for example, there are an infinite number of energy levels, and increasing the energy ($\Delta U $ is positive) increases the number of ways these energy levels can be populated and so the entropy increases. The slope $dU/dS$ is therefore positive and so is the temperature. From the viewpoint of the Boltzmann distribution the population of any two levels is, where $E_2>E_1$,
$$\displaystyle \frac{n_1}{n_2}=e^{-(E_2-E_1)/k_BT}$$
If there are an infinite number of energy levels then the temperature is always positive because the upper level is always populated less than any one below it. To make the populations equal would need an infinite temperature and so infinite energy.
What about a system with $n$ particles but with a finite number of energy levels? Suppose there are two levels, for example, with energy $0$ and $E$. In this case a plot of entropy vs. energy has the form of an inverted semicircle. At zero energy all the particles are in the lowest level and so the entropy is zero. Similarly, when the total energy is $nE$ all particles are in the highest levels and again the entropy is zero. At energy $nE/2$ the two levels are equally populated the entropy has its maximum value. In the energy range from $0\to nE/2$ the slope $dU/dS >0$ and the temperature is positive but in the range $nE/2\to nE$, the slope of  $dU/dS$ is negative and so this corresponds to a negative temperature. The midpoint energy corresponds to infinite temperature and both levels are equally populated.
Negative temperatures are not uncommon, often called a 'population inversion' in a laser and occur also in many types of nmr experiments.

Things get interesting if the disturbed energy distribution for some degree of freedom (laser electron excitation, IR laser/maser vibration/rotation one) are not in equilibrium with other degrees of freedom. In a way one could say the system have different TD temperatures for different DoF.. In other way, it has none.

Yes, in jet cooled molecules, for example, there is a different vibrational and rotational temperature and v. low translational temperature because although the molecules are moving quickly their speed distribution can be tiny. It is also possible to produce molecules in the vapour phase in an excited state that is cooler than the ground state and which cools the gas, (but only by a bit) as it warms up by collisions. This takes a few tens of nanoseconds typically.

Thanks for this clear and well-written answer, however I fear that I did not phrase my question as clearly.  I understand why there is a Schottky anomaly, however I would like to clarify that by a "two-level system" we mean two states other than the translational kinetic energy levels.  Like an ideal gas of particles with two spin states in a magnetic field.  Because if there were only two level for each particle, that would be like all the particles having only two velocities: slow and fast.

yes states other than translation.

I understand temperature as the average translational kinetic energy
per particle.

Temperature isn't just related to the kinetic energy of particles, it also changes the potential energy. The potential energies are incorporated into the energy levels $E_1$ and $E_2$. The kinetic energy require consideration other than the equations you have considered.

But I don't understand how a system with finite states can have a
temperature.

All systems have a defined scaler temperature field (except for particles in high vacuum with long free flight where some particles absorb radiation (photons), for which it may not be defined, as pointed out in the comments). Now, to address the confusion, according to you, I think, a system with finite states shouldn't have a concept of temperature, because if we keep adding heat to the system, at some point, all particles in the lower-energy state will rise to the higher-energy state. What if we add more heat? This shouldn't be a problem because we can always add heat to the system, but the system has no more higher energy states to occupy. What now?

So by "two-level system" do we actually mean two states other than the
translational energies?

Yes, when you say a two-level system, you mean something other than just kinetic energy. It is the potential energy that is represented by the energy levels, $E_1$ and $E_2$, which are quantized and finite here. For a purely kinetic energetic model, consider the case of ideal gases.

Remarks
The additional heat simply increases the kinetic energy of the system once maximum potential energy (all particles in $E_2$ state) is achieved, which, actually, never is because there is always some probability to detransition.
Related: Can somebody explain the low temperature Schottky anomaly for two level systems?.

All systems have a temperature. Only those that are in thermal equilibrium. Particles in high vacuum may not be.

@Poutnik very interesting! I am curious to read more about it. Would it be more accurate to say 'all systems have a defined temperature field?'

If you mean a scalar T field or vector grad T field, sure, for macroscopic systems with dense enough matter. But for high enough vacuum with long enough mean free flight distance, eventually when some particles selectively absorb photons, there is no temperature.

@Poutnik I have incorporated your suggestions in the answer. Could you please provide some reference, I searched for a while and couldn't find what I was looking for.

Hmm, references to knowledge gathered over more than 40 years are hard to make, to dark ages when everybody was offline and very few books were available in EN language that I had not even known yet. //  Try to pickups some sources from searching site:stackexchange.com OR site:libretexts.org temperature particle equilibrium vacuum

@Poutnik Hello Poutnik. One possible (and easiest) definition of temperature, from statistical thermodynamics, is the thermodynamic property that is equaled when two objects are in contact sufficient time (relaxation time). I am not a chemist, so I would be interested in your point of view as a chemist: (1) a fluid changing its temperature in a tube at high velocity, (2) a temperature profile in cartesian symmetry, like a plane from $x=0$ to $x=L$, according to Fourier's law. Both systems are not in thermal equilibrium, so for you the notion of temperature is not applicable here?

@MetalStorm Well, there is temperature defined at least locally, in context of temperature gradient. I have meant in my earlier comments thermal equilibrium at particle levels. either between particles, either between energy modes/degress of freedom //  But imagine  for illustrative purposes CO2 gas, that has translational kinetic energy distribution of 300 K, but vibrational energy distribution of 400 K. // BTW, if not a chemist, what is/was you major focus in science? I would guess from SE profile some chemical engineering?

@Poutnik I understood now. Yes, how did you know? Right now I am working on mathematical modelling of lithium-sulfur batteries, but I have worked on computational chemistry regarding chemical equilibria.. So I had to study quantum mechanics and statistical thermodynamics. Also some electron paramagnetic resonance in some compounds (also computationally). I'd like to know more about inorganic or organic chemistry, though. Maybe I am a frustrated chemist, but I just don't like the industry/chemical process design (yes theoretically).


text:
Proton Exchange Membrane H2 generator transports too much water to output
I am using a PEM cell to generate H2 (eventually D2).  The PEM cell is fed water on the cathode (O2 side) only, but it transports about 3.5x as much water over to the cathode (H2 side) as it actually consumes in electrolysis.
Ultimately I want to have dry gas and want to minimize wastage since heavy water to get D2 is expensive.  The equations for PEM electrolysis suggest that there is no liquid water transported, so I could dry it with desiccant. Actual case seems like I need to separate water from H2 output and recycle it. I only need a small volume so I am looking for simplicity.
Is this the nature of PEM cells or do I have the wrong kind of cell?

Can osmosis be the reason? As membranes usually have some residual unwanted features. Like membranes for reverse osmosis have residual or being worn out permeability and PEM can manifest something similar.

I have a little experience PEM \ce{O2}/\ce{H2} cells from teaching in the first year lab. In my experience, you feed the cell with oxygen and hydrogen directly (in out case humid as it come from a previous electrolysis of water) not with water. As a product of the cell reaction you got a small quantity of water but it accumulates if you run the cell for enough time.

From the structure of the PEM membrane I would not expect migration of water through the cell since it only transports protons but I maybe mistaken.

I tried applying some back pressure (inflating a balloon, but that had no effect.


text:
Do complexes of a metal distribute in between the water and an organic compound layer?
My original question is one that is stated in the title, and I also have some side questions too.
I came across this question where you have a CuSO4 aqueous solution. Then you take an Ammonia aqueous solution and mix those two solutions. Then you take some Trichloromethane, and you mix it with the above mixed solution. The objective of the question is to determine the ratio of Cu2+ : NH3 in the formed complex.
So, for example, my assumption is that when CuSO4 and NH3 aqueous solutions are mixed, the Cu2+ ions form complex with Ammonia as [Cu(NH3)4(H2O)2]2+. According to the question there is excess Ammonia so, when you put Trichloromethane, Ammonia distributes between Trichloromethane and Water.
My first question is, does Cu2+ ions also distribute in between Trichloromethane and water? If it is, are there [Cu(NH3)4(H2O)2]2+ distributed between these two layers? (more precisely, are there [Cu(NH3)4(H2O)2]2+ or Cu2+ ions in Trichloromethane layer?)
For my second question consider following equilibrium.
\begin{align}
\ce{[Cu(H2O)_6]^2+ + 2NH3 &<=> [Cu(OH)_2(H2O)_6] + 2NH4^+}\\
\ce{[Cu(OH)_2(H2O)_4] + 4NH3 &<=> [Cu(NH3)_4(H2O)_2]^2+ +2OH- +2H2O}\\
\end{align}
In the question, it uses titration to measure the NH3 amount dissolved in each layer. It uses HCl for the titration. So, as these are equilibriums, is it OK to use this titration, because when HCl react with NH3, the above equilibriums reverse bias their equilibrium to release more Ammonia into the system?
A guide with the thought process about this question would be helpful.
Sorry in advance if my English is hard to understand. I couldn't find the English version of this question. I will a post photo of the original question, which is written in Sinhala, so if you need you can translate it from Google.
Also, below is a slightly modified accurate translation than Google translate that I made.

25.0 cm3 of an aqueous ammonia solution was mixed with 25.0 cm3 of a solution of 0.10 mol dm-3
CuSO4. The resulting deep blue solution was stirred well with about 50 cm3 of trichloromethane and allowed
to equilibrate at a certain temperature. 31.9 cm3 of 0.02 mol dm-3 HCl solution was required to
neutralize the ammonia at 25.0 cm3 of the organic layer. 20.8 cm3 of 1.00 mol dm-3 HCl solution was required to
neutralize 25.0 cm3 of the dark blue aqueous ammonia layer. Calculate the Cu2+: NH3 ratio of the complex formed by molecular association
of NH3 and Cu2+.

The distribution constant for NH3, between water and organic layer is 25.

I would like to understand the stuff. You state that the blue complex of copper and ammonia contains $4$ $\ce{NH3}$ for one copper atom. So why do you try to calculate the ratio $\ce{NH3/Cu^{2+}}$ by titration if you know that this ratio is $4$ ?

@Maurice OP doesn't really get the relevant equilibrium - there won't be any charged complex in the organic phase.

@Maurice sorry for not clarifying. I used it as an example to show there is an equilibrium. The ratio 4, was taken as an example as this is the most common equilibrium.

By ligands, there are meant molecules or ions attached to the central ion/atom, not the whole complex ion.

The diffusion coefficient  Diffusion??

@Poutnik I translated it with google. Sorry if I got it wrong. Is it called distribution constant?

Yes, it is called a distribution constant. (for given entity). You mean NH3 there, right?

@Poutnik yes, $K_\mathrm{D}$ is for $\ce{NH3}$.

You may use $\ce{H2O}$ for $\ce{H2O}$ and $K_\mathrm{D}$ for $K_\mathrm{D}$

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I think it is an interesting question which relates to the solvent extraction of metals by extraction agents such as acetylacetone. But I doubt if the ammonia complexes of copper will ever enter the chloroform layer. You would be better off considering the extraction of a metal by a larger and more lipophilic organonitrogen like a BTP.

[OP] My first question is, does $\ce{Cu^2+}$ ions also distribute in between Trichloromethane and water?

No, the charged species will stay in the aqueous layer.

[OP] So, as these are equilibriums, is it OK to use this titration, because when $\ce{HCl}$ react with $\ce{NH3}$, the above equilibriums reverse bias their equilibrium to release more Ammonia into the system?

Yes, that is the point of the titration of the aqueous layer. There is some free ammonia, and some ammonia released from the complex as you titrate with HCl.

[Instructions] The distribution constant for NH3, between water and organic layer is 25.

The titration of the organic layer with HCl gives you the concentration of ammonia in the organic layer. The concentration of (free) ammonia in water will be 25-times higher. You have to subtract the amount of free ammonia in water from the measured amount in the aqueous layer to figure out the amount that was complexed with copper.

I approached the question with the wrong path. I thought titration was used to measure free ammonia in aqueous layer. Thanks for the explanation and could you also clarify what "[OP]" in your answer means. Thank you.

@donthababakka OP is original poster, so that’s you. A bit of Stackexchange jargon.


text:
Creating a Simplified Model for Temperature Change in Auto-Ignition of A4 Office Paper
When paper reached a certain temperature, auto-ignition occurs, and then there is a process that makes the temperature rise to around 1000 to 1500 Celsius degrees. How can I make a simplified model for this process. the purpose of the model is the temperature change over time. the paper i am using in this paper is a normal A4 office paper. Furthermore you can assume the auto-ignition temperature is constant and that the heat transfer is neglectable.

Well, if there was no heat transfer, then maybe it could get so hot, but not in real world.

@Mithoron Maybe he means that the source that gives rise to the heat transfer directs all its energy to the paper, this is, if the source gives $\pu{5 J}$ to the system, then $\pu{0.1 J}$ won't drift away as a loss.

@Mithoron You can assume the object was initially created in that temperature. In no heat transfer, I meant there is no heat transfer to his surrounding air or other objects.

Too many variables undeclared: rate of heating, surface exposed (Crumpled? on a hot plate? suspended?), air movement (confined? open to still air? wind?)...

If there are gaseous reactants(O2), products(H2O, CO2) and inerts(N2), the heat transfer is among the major factors that determine the temperature.

A4 isn't a type of paper: it is a standard for the size of the paper. apiece of asbestos could be "A4". So the designation is useless in understanding the properties of the object.

The simplest way to start is to model an unsteady heat transfer in two dimensions. The heat equation is a partial differential equation that, under no flow condition, takes the following form
$$ \rho c_\mathrm{p} \frac{\partial T}{\partial t} - \kappa
   \left(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2}\right)
   = h(T' - T) \tag{1}
$$
where:

$\rho$ is the density of the paper.
$c_\mathrm{p}$ is the specific heat capacity at constant pressure of the paper.
$\kappa$ is the thermal conductivity of the paper.
$h$ is the volumetric heat transfer coefficient between the source and the paper.
$T'$ is the temperature of the source.

Consider the rectangle defined as the set $D = \{(x,y) \in \mathbb{R}^2 : 0 \leq x \leq L_x \; \wedge 0 \leq y \leq L_y\} $. The initial condition is that $T(x,y,0) = \pu{298.15 K}$ $\forall (x,y)\in D$. The boundary conditions are an important choice, for illustration purposes, I considered the following:

Three Dirichlet's boundary conditions over the sides of a rectangle, with the exception of the lower side:
\begin{align}
  T(x,L_y,t) &= 298.15 \; \pu{K} \hspace{1 cm} x \in [0,L_x] \\
  T(0,y,t)   &= 298.15 \; \pu{K} \hspace{1 cm} y \in [0,L_y] \\
  T(L_x,y,t) &= 298.15 \; \pu{K} \hspace{1 cm} y \in [0,L_y] \\
\end{align}
A Neumann's condition in the lower side, this is, a no flux condition
\begin{equation}
  \underline{\nabla} T(x,0,t) \cdot \hat{\underline{n}} = 0 \hspace{1 cm} x \in [0,L_x] \\
\end{equation}
In the following figures, this side is the one where you "enter" into the carvern.

Some recommendations:

Consider, when first treating a problem like this, that all the model parameters are constant. Thermodynamic and transport properties will considerably change from ambient temperature to $T \approx 2000 \; \pu{K}$, so I would advise to obtain these magnitudes as a function of temperature. But first, try to obtain some sense of what are the influence of the parameters in the model (some are obvious, of course).

You can model more realistically the source of the ignition, represented in the term $h(T' - T)$, considering it as a function of time. As an example, I will consider that the heat transfer coefficient is a linear function of the time, i.e. $h(t)=h_0t$.

I did some runs using the properties of wood, since it was not easy to find those of paper. I display the size of a real A4 paper; $\pu{210 mm}$ (in the $x$ direction) and $\pu{290 mm}$ (in the $y$ direction):

Some final points:

The paper of course will burn, and we will lose its mass, so our grid of integration should appear as a hole. You can regard this model as an approximation for the first instants of time, when the paper is still not obliterated from existence.
In Wikipedia I found that the autoignition of paper is around $\pu{220 °C}$ so you would want to double-check your values.
You may use different boundary conditions (I don't know the situation of the paper), specially for the Neumann's condition. Here, the source is explicitly in the partial differential equation, but you can come with another way of reflecting additional behavior changing the ones I have selected.


text:
Why is n=1 at a low pressure in the Freundlich Isotherm Equation?
On the topic of adsorption of a gas over a solid, the Freundlich isotherm explains this behaviour in an approximate manner via the following equation
$$ \frac{x}{m} = kP^{1/n} \tag{1} $$
where $x$ is the mass adsorbed, $m$ is the mass of the adsorbent, $P$ is the equilibrium pressure of the adsorbate, and $k$ and $n$ are the constants of the model.
The factor $1/n$ can have values between $0$ and $1$ (probable range from $0.1$ to $0.5$). What is the reasoning, that for a low pressure of gas, then we may set $1/n = 1$ or $n=1$?


text:
Valence, Variable Valence, and Valency
As per the latest IUPAC Compendium of Chemical Terminology, informally known as the "Gold Book", the terms valency or variable valence are not defined; however, valence is defined as:

The maximum number of univalent atoms (originally hydrogen or
chlorine atoms) that may combine with an atom of the element under
consideration, or with a fragment, or for which an atom of this
element can be substituted.

This definition was published in the article Glossary of Terms Used in Physical Organic Chemistry: IUPAC Recommendations 1994$^1$ by Muller, P., Chairman, The Commission of on Nomenclature of Organic Chemistry, Organic Chemistry Devision, IUPAC during the preparation of these recommendations (1987, 1993).
This glossary of terms is also available online. According to this definition, valence should be a fixed number.
However, lambda ($\lambda$) nomenclature for variable valence was developed in the article Treatment of Variable Valence in Organic Nomenclature (Lambda Convention): Recommendations 1983$^2$ by Powell, W. H., Associate Member, The Commission of on Nomenclature of Organic Chemistry, Organic Chemistry Devision, IUPAC during the preparation of these recommendations (1975, 1983).
I have many questions regarding this:

Has the concept of variable valence been discarded from the IUPAC recommendations for organic nomenclature between 1983 and 1994?

Is the nomenclature different for physical organic chemistry and organic chemistry?

Is the $\lambda$ nomenclature valid anymore?

References

Glossary of Terms Used in Physical Organic Chemistry: IUPAC Recommendations 1994. Muller, P. 10.1351/pac199466051077
Variable Valence in Organic Nomenclature (Lambda Convention): Recommendations 1983, Powell, W. 10.1351/pac198456060769

If valence is fixed (it totally is according to the definition we use), it can't be variable. However, valence isn't a useful concept (it never was). But that makes variable valence also no useful concept. And nomenclature is not the same as a chemical concept. The constant struggle about people saying one thing while meaning another.

@Martin-マーチン in certain cases valence does prove useful: carbon forms a maximum of four bonds.

First define what a bond is, then see whether that fits valency. There are examples where carbon is bonded to 5 partners. With this example alone, the concept has lost all meaning.


text:
Is there any evidence to show that chemistry does not represent a mass to energy exchange?
As per Einstein relationship,  $E=mc^2$ or its more correct form $E=\sqrt{(mc^2)^2 + (pc)^2}$. The mass loss would be extremely tiny given how little energy is exchanged in chemical reactions. But is there definite evidence to show that the energy contained in chemical bonds does not increase mass?

Do not mix different concepts, e- is stable, normally there is no mass loss. In heavy atoms, a similar expression holds to take into account  relativistic effects : the mass increases and the wave function is contracted. Another example, the effective mass in solid-state as an emergent phenomenon.

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At measurement of two compared quantities, there could be evidence their difference is statistically significant ( The mass has changed. ) or insignificant ( The mass does not change more than the limit the method can distinguish.) // By other words, there can be evidence they are not equal, there cannot be evidence they are equal.

You are correct that the energy difference would be tiny, however you are incorrect that the mass would increase. The formation of bonds would release energy, therefore the mass would actually decrease instead of increase.
Let us consider hydrogen molecule with bonding energy of $436\ \mathrm{kJ/mol}$, which is approximately $7.24\times10^{-19}\ \mathrm J$ for one molecule. In term of mass, that would be
$$\Delta m=\frac E{c^2}=\frac{7.24\times10^{-19}\ \mathrm J}{8.99\times10^{16}\ \mathrm{m/s}}=8.05\times10^{-36}\ \mathrm{kg}$$
Mass of a hydrogen molecule ($\ce{H2}$) would be $3.34\times10^{-27}\ \mathrm{kg}$, which mean the mass lost is approximately $2.4\ \mathrm{ppb}$, which is pretty insignificant. And I believe is beyond the sensitivity of most of balance in chemistry lab.
So yes, there is a mass difference, and yes it is small, and – as far as I know – beyond the current capability of most our balance in chemistry lab to measure.

https://youtu.be/Xo232kyTsO0

Depends on what they form from  say Carbon and hydrogen to ethyne.

It could be easier to calculate the relative mass change at molar, rather than molecular level.


text:
Decomposition of aliphatic diazo compounds
The oxidation of hydrazones produce a diazo compound: orgsyn.org - DIAZOPROPANE.
An intermediate in the Bamford-Stevens reaction is a diazo moiety.

When 2-diazopropane decomposes, does it produce 1-propene?
Generally speaking, do diazo compounds spontaneously decompose?
Is the oxidation of hydrazones used to access alkenes?

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This should be 3 questions

Most aliphatic diazo compounds are exceptionally unstable and they generally decompose as soon as they are formed, giving alkyl carbenium ions. Note that some compounds are fairly (usefully, but not indefinitely) stable in solution in an inert solvent, like diazomethane in diethyl ether solution. The carbenium ions formed on decomposition quickly rearrange and/or react with whatever molecule/ion is nearby (solvent, dissolved ions, etc.) to produce any number of products. Alkenes can be formed, but it depends greatly on reaction conditions.
After the diazo compound decomposes with loss of a nitrogen molecule, the free carbenium ion can expel a proton and become an alkene. It could also react with water to form an alcohol. If large amounts of acid are present, it can combine with the anion to form an ester, e.g. alkyl halide formation with concentrated hydrohalic acids. Finally, if decomposition is conducted in anhydrous alcohol, ethers may be formed.
So to explicitly answer your questions, 2-diazopropane can produce 1-propene upon decomposition, but it would depend on reaction conditions. Yes, aliphatic diazo compounds generally decompose spontaneously, but some can be held long enough to be synthetically useful. Because of rearrangements and the formation of multiple products, the synthesis of alkenes through a diazo intermediate (by oxidation of hydrazones, for example) would probably not be an advantageous route to access alkenes.
For more information, check out Söll in "Methoden der Organischen Chemie", 4th Edition, Volume 11/2, page 133 (1958).


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Can ΔG = ΔG° + RT ln Q be used to calculate K at different temperatures?
Specifically, can $\Delta G=\Delta G^{\circ}+RT \ln Q$ be used to calculate reaction quotients if we choose T to be a different temperature than the one provided by ΔG°?
For example, suppose that we are given ΔG° at 25°C and we are tasked with finding the equilibrium constant at 30°C. Do these equations hold?
$$0 \stackrel{?}{=} \Delta G^{\circ}+RT \ln K$$
$$K \stackrel{?}{=} e^\frac{-\Delta G^{\circ}}{RT}$$
I believe this is illegal, as it contradicts the fact that some equilibria are less favored at higher temperatures. However, I'm having trouble finding confirmation online, especially since the above equation has no easily searchable name.

Delta G is temperature dependent. See also van't Hoff equation : $\frac{\mathrm{d}(ln K)}{\mathrm{d}T} = \frac{\Delta_r H)}{RT^2}$ //  Try to learn more flexible searching, like site:stackexchange.com OR site:libretexts.org equilibrium constant temperature dependance   //
I am not aware of any law that forbids writing false equality statements, so it is probably legal, at least in most countries.

Related: https://chemistry.stackexchange.com/q/127313/72973. In both cases, temperature is an explicit part of the mathematical relationship but other quantities in it are highly temperature-dependent as well.

As mentioned in the comments, $K$ is itself temperature dependent. That is to say, $K = K(T)$ is a function of temperature, so
$$\Delta_\mathrm{r} G^\circ = -RT\ln K(T)$$
and the temperature dependence in $\Delta_\mathrm{r} G^\circ$ not only arises from the $RT$ term but also the $K$ term.
The dependence of $K$ on temperature is described by the van 't Hoff equation. I've written about it here (a while ago!).
$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = \frac{\Delta_\mathrm{r} H^\circ}{RT^2}$$


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Dilution formula for NH3 solution
I need this for an experiment:
ca. 3 mL concentrated (15 M) aqueous ammonia solution, NH3 (from a dropping bottle.)
I have the NH4OH pictured below. How can I dilute it to get what I need? What is the formula?

The label is a shame, I would not buy products from such a company for serious lab work.

A solution with a mass fraction of $25\ \%$ ammonia corresponds to a concentration of $c_1=13.32\ \mathrm{mol\ l^{-1}}$. So obviously you cannot dilute this solution to reach a higher concentration of $c_2=15\ \mathrm{mol\ l^{-1}}$, which corresponds to a mass fraction of $28.5\ \%$.
Depending on the experiment, however, you could simply use more of the available solution according to
$$\begin{align}n_1&=n_2\\[3pt]
c_1V_1&=c_2V_2\\[3pt]
V_1&=\frac{c_2V_2}{c_1}\\[3pt]
&=\frac{15\ \mathrm{mol\ l^{-1}}\times 3\ \mathrm{ml}}{13.32\ \mathrm{mol\ l^{-1}}}\\[3pt]
&=3.4\ \mathrm{ml}\end{align}$$
Also note that the fictitious substance “ammonium hydroxide” $(\ce{NH4OH})$ with a molar mass of $35.05\ \mathrm{g\ mol^{-1}}$ (as shown on the label) does not exist.

Makes sense, but can you please show me how you got 13.32 mol l−1?


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How to proceed with iodometric titration to determine caffeine concentration in liquid coffee?
Context
I'm a high school student who decided to investigate the relationship between caffeine concentration and brewing temperature of coffee out of their love for the substance. For my method, I opted for iodometric back titration, as detailed in this paper and shown in the image below. I've verified the method is viable with a known concentration of caffeine obtained by using caffeine tablets, but I am struggling to repeat the results on brewed coffee.

Problem
When performing iodometric back titration on brewed coffee (as opposed to pure caffeine solution) with starch as an indicator, the end point isn't clearly defined because it is no longer colorless due to the natural brown pigments in brewed coffee.
I've looked into several potential routes that could help me mitigate the issue (detailed below), but I'm looking for advice on the most convenient (requiring the least time, resources, and skills) way to proceed. Of course, I'd like to maintain a certain level of quality, but it isn't as vital.
Paths I've considered
1. Titration with spectroscopy
Carrying on with titration as previously, but using a colorimeter to detect the end point. From my limited understanding of the technology, I presume that I would be able to detect when the concentration of a certain compound (e.g. iodine-starch complex) would reach zero. A limitation I can foresee is that I would need to test a sample of the solution every time I add a drop of the titrant, which would become very time-consuming on many trials.
2. Extracting caffeine
Using an organic solvent to isolate caffeine and then perform the titration on that substance. The problem is that I'm not sure if the titration reaction (described in the paper linked above) will still hold. Furthermore, I am limited by the availability of organic solvents (can't use chloroform and likely not DCM either). Some other solvents are addressed in this question.
3. Filtrating the color of coffee away
Practically the inverse of the above, using a filter (activated carbon) to produce colorless liquid by removing pigments (melanoidins?). The problem is that information on this is quite limited when it comes to coffee and from my research both caffeine and iodine would be adsorbed by activated carbon. Perhaps I could mitigate this by pre-saturating the activated carbon with caffeine or iodine?
4. Switching to green tea
My supervisor actually advised this in the beginning, but I insisted on coffee (I know, I know). I wouldn't prefer this path because of the costs already sunk into preparing coffee samples, but if there is no other alternative...

All in all, these are just some of the possibilities I've thought of, but I am sure there are some missing. What would be your advice on how to proceed with iodometric back titration, considering the limitations in time, resources and skills?

Maybe take an aliquot after the precipitation reaction and dilute several times before titration to make the solution translucent enough to be able to see the blue-colourless change.

Classical volumetric titrations are not selective. This means if there are multiple components in the mixture, they will react with the reagents used in the titration. Since coffee and tea are real-world samples with hundreds of other compounds besides caffeine, there will be other oxidizable materials, which may/will react with iodine. One would suspect that tannins would also be oxidized by iodine. Your results will always be overestimated. What is the solution? A real sample titration requires sample preparation. Ask your teacher to help with extracting caffeine from tea or coffee with a suitable solvent in a separatory funnel. After extraction, you will have to evaporate the solvent. Those colors would not be extracted in a non-polar solvents. I will let you research more on what solvents starting from here Caffeine extraction. After that one may proceed normally as suggested in the Olympiad manual.