text:
How does the molecule consisting of two bridged aromatic rings look in 3D?
I can't understand how the following compound with bridged benzene rings looks like:

This is a cyclophane, namely 4,8-dibromo[2.2]paracyclophane.
Lonsdale et al. [1] first structurally characterized [2.2]paracyclophane in 1960:

The molecule consists of two facing benzene rings linked in para-positions by two $\ce{(CH2)2}$ bridges.
As a result of steric hindrance, the benzene rings are each bent into a boat-shape, but even so, their maximum distance apart in the molecule is only 3.09 Å, whereas the normal van der Waals distance is upwards of 3.4 Å.
[…]
The angle of bending of the benzene ring (14°) does not change
with temperature, but the aliphatic $\ce{CH2-CH2}$ bond appears to be unusually long at room temperatures* and has decreased by 0.07 Å at 93 °K. Moreover, although the distance between the two benzene rings does not change, there is a decrease of 0.03 Å in the overall length of the molecule from 291 to 93 °K.
[…]

Figure 4. Diagram showing projection of molecules at $0, 0, 0;$ $1/2, 1/2, 1/2$ and $0, 0, 1$ on to $(100).$

* The $\ce{CH2-CH2}$ bond length is 1.63 Å, which is above average 1.54 Å expected for the aliphatic $\ce{C-C}$ bond.
See CCDC entry DXYLEN for more details and an interactive 3D image.
Several related polysubstituted [2.2]paracyclophanes have been reported:

5,7-dibromo-4-hydroxy[2.2]paracyclophane [2] (CCDC: FULNUX);
4,8,12,16-tetrabromo[2.2]paracyclophane [3] (CCDC: SAWDIE);
4,5,7,8,12,13,15,16-octabromo[2.2]paracyclophane [3] (CCDC: SAWDEA).

References

Lonsdale, K. Y.; Milledge, H. J.; Rao, K. K. V. Studies of the structure, thermal expansion and molecular vibrations of di-p-xylylene, $\ce{C16H16},$ at 93 and 291 °K. Proc. R. Soc. Lond. A 1960, 255 (1280), 82–100. DOI: 10.1098/rspa.1960.0055.
Atman Vorontsova, N. V.; Zhuravsky, R. P.; Antonov, D. Yu.; Sergeeva, E. V.; Godovikov, I. A.; Starikova, Z. A.; Vologzhanina, A. V. Synthesis of Bromo-Substituted 4-Hydroxy[2.2]Paracyclophanes and [2.2]Paracyclophane-4,7-Quinones as Versatile Chiral Building Blocks: [2.2]Paracyclophane Derivatives as Versatile Chiral Building Blocks. Eur. J. Org. Chem. 2015(2), 325–330. DOI: 10.1002/ejoc.201403316.
Chow, H.-F.; Low, K.-H.; Wong, K. Y. An Improved Method for the Regiospecific Synthesis of Polysubstituted [2.2]Paracyclophanes. Synlett 2005, No. 14, 2130–2134. DOI: 10.1055/s-2005-872270.


text:
Why is copper discharged in the electrolysis of copper (II) sulphate solution instead of hydrogen?
In the electrolysis of copper (II) sulphate with graphite electrodes, I am confused why copper ions are discharged and form copper metal at the cathode. As far as I know hydrogen ions have a lower position than copper ions in the electrochemical series, and so technically shouldn't hydrogen be discharged due to its stronger oxidizing power compared to copper?

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

Should be aluminium discharged  before hydrogen?

No? Because aluminum is a weaker oxidizing agent?

Metallic aluminium is stronger reducer than hydrogen, while aluminium ions are weaker oxidizer than hydrogen ions.

So as copper ions are have stronger reducing power , it has less tendency to gain electrons compared to hydrogen, so why is copper discharged?

They do not. To what would you oxidize copper(II) ions? You should get your knowledge base straight before making conclusions.

The truth is the contrary of what you state. $\ce{H2}$ has a stronger reducing power than $\ce{Cu}$ (and not a stronger oxidizing power)

source: Stack Exchange

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

sort: votes


text:
What kind of rearrangement happens between 2,2,5-trimethylcyclopentanol and concentrated sulfuric acid?
Question:

My attempt:
After getting the carbocation after protonation, in order to form a more stable alkene, it rearranged the methyl group on the adjacent carbon so as to get the alkene in the option (b). I am confused that whether it should be "1,2 hydride shift" or should it be "1,2 methyl shift" because the migratory aptitude of $\ce{H-}$ group is more than $\ce{CH3-}$ but here if carbocation formation is the RDS (considering E1) should we not rearrange it to a more stable species?

try considering the fact that 'b' is more substituted as compared to 'a'

Have a look into this https://chemistry.stackexchange.com/q/161776/115228

Does this answer your question? Hydride shift or Methyl shift

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


text:
How does the amount of Calcium Carbonate in blackboard chalk influence its quality
Does the amount of calcium carbonate in chalk influence its charecteristics ?
If so please mention them

Welcome to the chemistry site of the SE network.
The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.
Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.

Amorphous pure calcium carbonate $\ce{CaCO3}$ is a white, common and rather soft mineral, called chalk or lime. It can be cut down in cylinders as thick as a finger. Using it on a blackboard will make a broad and thick line, and your fingers (and trousers) will quickly be full of white marks. It is too soft to be useful in a classroom. On the other hand, the same cylindrical pieces can also be made with plaster, which is hydrated calcium sulfate. But the obtained piece is too hard to be used on a blackboard. It will scratch the board. But if you make the appropriate mixture of plaster and lime, you can produce a piece of chalk useful on a blackboard.
So to produce a piece of chalk useful in a classroom, lime is reduced to powder, mixed with plaster ($\ce{CaSO4·\frac{1}{2} H2O}$). The mixture is then got wet. This will transform the plaster $\ce{CaSO4·\frac{1}{2}H2O}$ into gypsum $\ce{CaSO4·2H2O}$. This operation also transforms the mixture of two powders into a solid block, ready to be used for writing on a blackboard. Neither too soft, nor too hard.


text:
name of these compounds?
I'd call c. 6- (3 aminocyclohexyl- methylamino) cyclohexa-2,4-dien-1-one and d. 1r, 2s- 2 (2-formyl- 4-methoxycarbonyl-but-3-en-1yl) cyclobutanecarboxylic acid

How do you see two amino groups in c? What you call a "4-methoxycarbonyl-but-3-en-1yl" substituent actually is 5-methoxy-5-oxopent-3-en-1-yl.

You're right. In c it's 6- (3 hydroxycyclohexyl- methylamino) cyclohexa-2,4-dien-1-one. Thank you

Are you sure you correctly labeled the second molecule's stereocenters?


text:
Why are neodymium magnets called neodymium magnets?
Neodymium magnets have a formula of Nd2Fe14B, but why are they called neodymium magnets? There are more iron atoms, and iron makes up over half of the mass of a neodymium magnet. Why isn't it called something like ferroneodymium magnet instead?

'A rose by any other name would smell as sweet' - Shakespeare. Names of materials can be arbitrary. Metal organic frameworks (another class of compounds) have names after universities.

Putting iron or ferro into names would bring very little information. Neodymium is the essential part at functionality and information level.

If there were common neodymium magnets without iron...

Consider also "Chromoly" steel, which is more iron than chrome or molybenum, but the chromium and molybenum content is primarily what distinguishes it from other steels.

Don't take industrial designations too literally. Here is a doozy: to refine uranium they make an intermediate product, yellowcake, by adding "alkali" to uranium-bearing solutions. The alkali could be sodium or potassium hydroxide. Or it could be magnesia, which is sort-of an alakli -- or ammonia, which is totally out of line with chemistry!

Previously asked in Physics.SE: https://physics.stackexchange.com/questions/83810/how-did-neodymium-magnets-get-their-name

As Poutink already stated in the comments that Neodymium is the essential functionality. And quoting from Wikipedia,

The neodymium atom can have a large magnetic dipole moment because it has 4 unpaired electrons in its electron structure as opposed to (on average) 3 in iron.

This actually suggests that neodymium plays an important role in magnetic properties of these magnets and probably they are called neodymium magnets.

I can give a clear intuition on this if you have some basic knowledge in organic chemistry. Consider the compound $\ce{C2H5OH}$,ethyl alcohol. In this compound it has lesser percentage of oxygen than hydrogen and carbon. So why don't we say it simply a hydrocarbon rather than naming alcohol? This is because of the functional group $\ce{-OH}$, which characterizes the molecule's physical and chemical properties.

I sort of understand now, so do you mean it is named after the element(s) that give it its function, instead of naming it by element composition?

Yes in this case. If we had named according to the percentages then there will be no difference between this magnet and an ordinary iron one.

One thing to be discussed is we don't always name them simply by the functional substance. Sometimes we will observe an unexpected naming. See the Oscar's suggestion in the comments. And another simple example of such naming is "lead pencils" which actually have no lead in it!

The example "lead pencils" is inaccurate because pencil leads used to be out of lead.

That's what I am saying. It is an example for exceptional naming. Re read my comment.


text:
Negative Fehling's Test for glyoxal and glyoxylic acid
While I understand why aromatic aldehydes and ketones do not give positive Fehling's test, I do not know why Glyoxal and Glyoxylic Acid give a negative test too. Nothing has been written about these two compounds in my study material and I was hoping someone could help me out with the reasoning. I figure it's something to do with presence of alpha-H but I'm not sure.

See: https://chemistry.stackexchange.com/questions/90778/does-acrolein-give-fehlings-test/136835

I have seen that question, and the answer did not satisfy the user who asked nor did it answer my question.

There is a good diagram of the mechanism here that I have reproduced below.
The first step is the formation of an enolate anion of the aldehyde by removal of the alpha proton by hydroxide. The enolate then binds to Cu2+. If the enolate cannot be formed then the reaction cannot proceed. Glyoxal and Glyoxylic Acid cannot form enolates so they give a negative test.
The only proton that can be removed from glyoxal to form an enolate is the proton on a -CHO group. They are far less acidic than the alpha protons of an aldehyde. Glyoxalic acid obviously deprotonates at the carboxylic acid group, so no enolate forms.

Why can't glyoxal form the enolate?

@DhruvKaushik I have edited to answer this


text:
In regards to the Gibbs free energy equation, it seems that the two terms oppose one another, Enthalpy and -TS. Can you explain this?
So for example, if H is endothermic, then the enthalpy term is positive, meaning energy is absorbed by system overall, meaning net bonds broken, meaning the equation leans towards nonspontaneity. If entropy is negative, then -T∆S is positive, meaning net bonds formed, meaning equation leans towards nonspontaneity. It appears that they oppose one another in regards to bond formation...

Endothermic does not necessarily mean that there is net bond breakage. For example, you can break one strong bond and form many weak ones, and the number of bonds will increase even though the reaction is endothermic.

It is not about opposing terms but from experimental necessity. In the early days before the second law was understood it was assumed that the most efficient utilisation of chemical energy consisted of converting all the heat into work. But experimentally this proved not to be the case.
Lewis and Randall (Thermodynamics) put it very clearly

It is true that according to the first law, the external work performed must be equal to the loss of heat content of a system, unless some heat is given to or taken from the surroundings, but this is precisely the point first seen by Gibbs. When an isothermal reaction runs reversibly $T\Delta S$ is the heat absorbed from the surroundings, and if this is positive, the work done will be even greater than the heat of reaction.

Thus entropy and so the free energy had to be invented to understand the experimental data.

discovered and explained not invented

The combined effect of $ \Delta H-T\Delta S $ is a measure of spontaneity. The individual terms (i.e., $ \Delta H$ and $T\Delta S $)  are not the actual measure of spontaneity. But in general we say that exothermic reactions are spontaneous (it is when $T \Delta S $ term is negligible in comparison with $ \Delta H$) and sometimes if $\Delta S>0$ we say that the process is spontaneous (if $\Delta H$ term is negligible in comparison with $T\Delta S$ term).

Itepends on the direction the system is displaced from equilibrium. the differences determine where the equilibrium lies

At equilibrium, $$\Delta G = 0 = \Delta H-T \Delta S$$ $$\Delta H = T\Delta S$$  This means that a system at equilibrium is stuck there unless matter [chemical potential] or energy is changed to perturb the equilibrium.
Ice + heat = water; add some heat; $T \Delta S$ increases; $\Delta H$ must now increase; this means that heat must be removed and ice melts to maintain equilibrium. Spontaneity direction depends on which side the reaction is displaced from equilibrium. Both directions are spontaneous. One way is energy driven, the other way is entropy driven.
Energy and entropy really do work as a team.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Why do the lanthanides usually use the +3 oxidation state?
This might seem like a silly question, but why do all of the lanthanides use the +3 oxidation state? I know that some of them can use the +2 (europium, ytterbium) and the +4 (terbium, cerium, praseodymium) but their main oxidation states are +3.

I would say it is an experimental result !

It's basically down to ionisation energies, see my answer here: https://chemistry.stackexchange.com/q/53712/16683

Thanks! I couldn't find any other post, but it seems like there are.

Also see: https://chemistry.stackexchange.com/questions/53605/predominance-of-iii-oxidation-state-for-lanthanides

https://chemistry.stackexchange.com/questions/87162/europium-stability-in-2-and-3-state/87169


text:
Formula of what substance is shown in the clip Disappointed by Electronic?
I'd like to know the substance (if any) the model of which is shown in the clip Disappointed by Electronic.
The following screenshot is taken from the video electronicofficial —
Electronic - Disappointed (YouTube):

Just a cartoon.

Disappointed, @AChem, thought that maybe they were trying to convey some hidden meaning )

Blue is an oxygen atom; red is a carbon atom; green is a hydrogen atom. It is a cyclohexane ring where 4 carbon atoms are connected to 1 H and 1 OH. The two remaining carbon atoms are difficult to understand. One carbon atom is not connected to any atom out of the ring.(which is absurd). And the last carbon atom is connected to a group looking like -OH-OH (which is also absurd).

Thanks, @Maurice, oh,oh may be just an encoding for a sad interjection ?

I did a structure search within CCDC 2016 database (should be good enough for the 1992 video) for the connectivity graph with complete disregard of the used color scheme:

Here, X stands for any atom of the periodic table and T1 to T4 refer to the number of bonded atoms for each X.
The only molecule that comes up is glucose alongsode with related aldohexose stereoisomers (allose, galactose, mannose, talopyranose, talose and more).
I wouldn't try to pick the correct stereoisomer since the structure is being unrealistically bent and mirrored virtually all the time.
I tried my best to align the real glucose structure (for example, CCDC 1500503) to match the one from your screenshot:

I'm fairly certain the color code for atoms is meaningless and was an arbitrary artistic choice.
There are several atoms that give this away (labeled with yellow arrows).
The green color for terminal atoms holds pretty well for hydrogen atoms, but the single red bridging atom having only two neighbors cannot be assigned for carbon (this would imply linearity of $\ce{C=C=C}$ fragment).
The two blue atoms connected with each other have vastly different environments and hardly can be assigned to a meaningful functional group (at least in organic chemistry).


text:
What are observationally stable elements?
According to Wikipedia, a lot of the elements that have higher atomic numbers than dysprosium have isotopes that say "Observationally stable" instead of stable, for example in Isotopes of holmium. Why does this happen and what are observationally stable elements anyway?

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

See https://en.m.wikipedia.org/wiki/Stable_nuclide#Still-unobserved_decay

Theoretical calculations predicts some stable isotopes like calcium-40 to be radioactive. But nobody has ever measured that this isotope emits a radiation. It may be assumed that the half-life of this isotope is so large that not even one Ca-40 atom is destroyed in any sample of this element during one life (~100 y). So the Ca-40 isotope is considered as "observationally stable". Another possibility is that the calculation of the nuclear stability has to be re-examined. Who knows ?

Isotopes that are theoretically believed to be unstable but have not been observed to decay are termed as observationally stable. Currently there are 162 theoretically unstable isotopes, 45 of which have been observed in detail with no sign of decay, the lightest in any case being $\ce{^{36}Ar}$

I'll take this opportunity to shamelessly plug a closely related question on Physics.SE that I answered a couple of years ago.

"Observationally stable" does not refer to an element. Rather, it refers to  isotopes of a particular element.
Let us consider a hypothetical example. The intelligent beings on Planet Cartorze have measured nuclear masses, done energetic calculations, and concluded that carbon-14 should decay to nitrogen-14; but their experimental techniques to detect this decay are fairly crude and they can detect the decay only if half a typical laboratory specimen decays within 1000 years (the years on Cartorze are the same as on Earth). Of course it actually takes several times that long for carbon-14 to half-decay, so on Cartorze carbon-14 is observationally stable.
On Earth we are better at detecting radioactive decay, and we can render the decay of carbon-14 with reasonable precision. But when a calculated decay takes a half-life of billions of billions of years or maybe even more, we may not have experimentally verified it, and we have observational stability. Wikipedia reports that (at least to Earthlings) argon-36 is the lightest observationally stable isotope.

An element can be observationally stable, if all its "stable" isotopes are observationally stable. Bismuth with the only bismuth-209 was such a case until recently when the predicted decay has been observed. Holmium still is.

Or rather, ....   if some of its natural "stable" isotopes are observationally stable.  .....

Very generally, anything "observationally stable" is unstable thermodynamically, but stable kinetically ("frozen"), without observable change (or below change threshold).
For example, diamonds at room conditions. Or a car parking on a slope, but with the activated parking break. Or a brick standing on the smallest side. Diamonds are not observed turning to graphite, nor the car is observed going downhill, nor a standing brick spontaneously tips over and falls down.
In the context of radioactive isotopes, observationally stable are those with a predicted decay, but with the decay half-time above detection limits. Note that some detections of the slowest decays have been done indirectly, finding evidence of consequences of long term decaying. Typically, it is finding otherwise unexpected elements formed by the decay or disturbed isotope ratios. Typical reasons for slow decay rates is too high activation energy, or low probability conditions due quantum physics.
Some of such cases belong to the class of double beta decays, where the decay path goes via less stable in-the-middle nuclide (with higher energy) by quantum tunneling. An example of double beta decay is tellurium-128 with the longest known half-time $\pu{2.2E24 y}$. Having 1 mol of tellurium-128, the typical decay rate would be about 1 nucleus per almost 4 years.
Strictly speaking, TD stable states are literally "observationally stable" too, but the term is not used for these conditions.


text:
What is the Chemical Reaction that Explains why Rubbing Ascorbic Acid on Rust Stains Removes Them?
I am a pool maintenance technician. In the industry, we pour bags of Vitamin C (ascorbic acid) into the pool to remove rust stains from the fiberglass surface. I am trying to determine the chemical equation behind this phenomenon as it works really well but was unable to find anything online. Any help would be appreciated.

Why do vinegar and other acids remove rust?; Acetic(vinegar) and citric acid as the rust remover

@andselisk - I don't think this is just an acid reaction here. Ascorbic acid can also reduce Fe(III) to Fe(II), which is more soluble and can wash away.

And ascorbic acid also forms stable complexes with iron: https://www.amazon.com/Adapter-Desktop-ALINKVITE-Adapters-Wireless/dp/B09JBZZ8VF/ref=sr_1_25, https://pubmed.ncbi.nlm.nih.gov/6514543/ . This chelation of iron can be used to enhance its absorption in the body, https://pubmed.ncbi.nlm.nih.gov/15743017/

@Andrew I agree 100 %, that's why I dropped my dupe-close-vote and just left some links to related posts.

Conceptually, it's something like $$\ce{\text{ascorbic acid} + 2Fe^3+ -> \text{dehydroascorbic acid} + 2Fe^2+ + 2H+}$$
I say "conceptually" because the ferric iron will be some form of oxide rather than a free ion, and the ferrous iron might also be an oxide, depending on pH. The ferrous ion also can be chelated by ascorbic acid. The low pH of the ascorbic acid solution could also slightly solubilize some of the ferric iron without it needing to be reduced to ferrous.
But regardless of the details, the result is that the ferric iron that gives the orange stain color is solubilized in some form and washes away.


text:
Why is it necessary for the standard enthalpy of fusion to change at the transition temperature?
I have a table here:

For argon, for example, shouldn't the standard enthalpy of fusion @ 83.81K be 0 J/mol?
After all, it is already at the respective transition temperature.

The question in title does not make much sense. At fusion transition T, what is changing is enthalpy, not enthalpy of fusion (which is not zero)

It is not the enthalpy change that goes to zero at phase equilibrium, but the free energy change. And that involves entropy as well as enthalpy.
What happens is this: the "higher" phase -- liquid over solid, vapour over liquid -- embodies more disorder and thus more entropy, so the Second Law allows it to form with a loss of entropy in the surroundings -- meaning heat energy flows from the surroundings into the melting or boiling material. Then the First Law dictates that at constant pressure, which is the case in an equilibrium process, enthalpy goes up. Ergo a positive enthalpy change with melting or boiling even as the free energy change is zero.


text:
Should sample masses of various samples undergoing Thermogravimetric analysis (TGA) be same?
I want to analyze the thermal degradation phenomena for 3 different samples using TGA. One of these samples comprises Polymer only, while the other two are loaded with two different percentages of fillers. Should the test weight of all three samples be the same prior to TGA, or degradation trends would hold regardless of differences in the three sample weights?

As long as the weight limits are within the manufacturer's specs I don't think there is a problem. The y-axis of TGA is % weight loss, so it takes original weight into account.

@AChem, what if the TGA results are to be compared to gauge if the addition of fillers has fortified the resistance to thermal degradation of the polymer these are added to in various percentages? If it takes original weight into account, then should not all weights match? For instance, if we have 3 samples (2 filler and one polymer-only), should not all three starting sample weights be the same?


text:
How is the intermediate compound 2-chloropropan-1,3-diol is formed when hypochlorous acid is added to allyl alcohol in the preparation of glycerol?
The expected product is 3-chloropropan-1,2-diol since secondary carbocation is more stable than primary but it is not formed.
I have surfed the internet a lot regarding this and none seems to be a simple, reasonable answer. Can anyone please  explain this mechanism clearly.

Read this https://doi.org/10.1039/JR9500001282

Yes thank you Waylander, ive read articles 260 and 261. It seems to be very complicated though. Is the reason really a tricky one or is there a simpler one? I'm only a high school student so I would appreciate a simpler answer? Nevertheless, thank you. -GW

The answer seems to be pretty simple, surprisingly.
I've asked my chemistry sir and this is what he said.

In the underlined step, 2° carbocation is supposed to be formed.
But there is -I effect from OH group which makes the carbocation close to it, that is, 2° carbocation less stable.
Hence to avoid that, 1° carbocation is formed away from OH group and then the nucleophile 'OH minus' from HOCl attacks the 1° carbocation so formed.
Thus, 2-chloropropan-1,3-diol is formed as an intermediate instead of 3-chloropropan-1,2 diol.


text:
hot pack experiment: calcium chloride vs magnesium chloride
why is calcium chloride preferred in hot packs even though magnesium chloride produces higher negative enthalpy? Experiment performed at school shows higher temperature change and more quicker increase in temperature with calcium chloride than with magnesium chloride. However, on a theoretical basis, magnesium chloride has a higher negative enthalpy. How to explain this?

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.

Please use proper punctuation.

The enthalpy of formation has nothing to do with the enthalpy that you have measured. Both are independent parameters. One may be big and the other tiny, or the contrary. What you have measured is the enthalpy of hydration (or of dissolution). It is not the enthalpy of formation.

Thanks. Is the higher negative enthalpy for MgCl2 an enthalpy of hydration or formation? Why is calcium chloride preferred in hot packs? Is there more factors such as availability or cost in play than just an effective chemical process?

$\ce{MgCl2}$ formation enthalpy is for $\ce{Mg(s) + Cl2(g) -> MgCl2(s)}$.
Its hydration enthalpy is for $\ce{Mg^2+(g) + 2 Cl-(g) ->[H2O]Mg^2+(aq) + 2 Cl-(aq)}$
Its lattice formation enthalpy is for $\ce{Mg^2+(g) + 2 Cl-(g) ->MgCl2(s)}$
Its dissolution enthalpy is for $\ce{MgCl2(s) -> Mg^2+(g) + 2 Cl-(g) ->[H2O]Mg^2+(aq) + 2 Cl-(aq)}$
For the heat pack, the dissolution enthalpy is what you need.
Highly negative salt formation enthalpy means they is highly negative lattice formation enthalpy too, what increases value of salt dissolution enthalpy and decreases released heat.


text:
Why do amphiprotic substances react to completion in acidic and basic conditions?
I was under the impression that amphiprotic substances were usually weak acids/bases. If this is the case, how could their reactions go to completion in more acidic/basic environments?
e.g. my textbook says that
$\ce{H2PO4-(aq) + OH-(aq) -> HPO4^2- (aq) + H2O (l)}$
Wouldn't this in all other cases be an equilibrium? Or is it still an equilibrium, just so much shifted to the RHS that we consider it fully forward due to the excess of OH provided by the basic solution?
Any responses appreciated :)

It is an equilibrium all right, just shifted to one side.

Thats what I thought. Textbook probably simplifying it to as to not have to explain LCP etc. Anyhow, I think only considering the forward reaction is fine for simple purposes. Thanks!

The appropriate form of a chemical reaction is determined by the context or purpose the reaction is describing or illustrating.

Unidirectional reactions for titrations, synthesis, reaction schemas
Bidirectional reactions to point out equilibrium, e.g. for pH or concentration calculations
Substance-level reactions if focus is on used reactants and formed products.
Ion level reactions, if focus is on involved particles, excluding bystanders, mostly for redox or precipitation reactions.


text:
Effect of degree of freedom on entropy change during a reaction
In the reaction $$\ce{C_{(s)} + O2_{(g)} -> CO2_{(g)}} $$
one mole gaseous reactant gives one mole gaseous product, so there is no significant change in entropy. It can be verified by line of $\ce{C}$ gives $\ce{CO2}$ in Ellingham diagram being almost horizontal.
But oxygen has a degree of freedom 5 and at higher temperature $\ce{CO2}$ has 9. So there should be an increase in extent of random motion, leading to significant positive entropy change. Why is it not the case ? Does degree of freedom not affect entropy change of a reaction greatly ? Why so ?

What about the change in chemical potential in going from C and O2 to CO2?

Yes number of dof is important, and the entropy will be similar but not identical for both gasses. Using the Sakur-Tetrode eqn. for translational entropy, this depends on $\ln(mass)$ and rotational entropy on $\ln(\text{moment of inertia})$ so small changes between O2 and CO2. Vibrational contribution from oxygen is negligible at room temperature and larger but still small for CO2, this by population of vibration from Boltzmann eqn. Btw, try not to think of energy as random motion but as number of positions a molecule can take or the number of energy levels it can possibly be in.

@porphyrin thanks, although I don't know the equations you have mentioned . So the effect is small in this reaction... but in some other case, can there be a significant effect of dof ?

And you meant entropy , not energy , in last sentence, right ?

yes my typo it should be 'try not to think of entropy as random motion'. You will find the equations in most  phys. chem. textbooks, usually in sections dealing with statistical mechanics. You will also know that the heat capacity of O2 and CO2 differ and that $S(T)=S_0+\int_0^T C_p/T dT$ so that way also you expect the entropy to differ. To use this eqn. you have to know how $C_P$ varies with temperature .


text:
How to achieve desired response of a non Newtonian fluid prepared from silica nanoparticles and polyethylene glycol (PEG 400)?
I am working on non-newtonian fluid project. I have experimented with cornstarch and water and achieved the consistency desired with 55% cornstarch and 45% water.
Now I want to create similar outcome using silica nanoparticles and polyethylene glycol (PEG 400). However, I am not able to reach a proper result. I used the following ratio: 60:40 (polyethylene and silica), but I got a paste-like material instead of a shear thickening fluid.
And ideas on how to find an accurate ratio for desired result? Is the problem regarding the ratio or the mixing steps? Do I need specific equipment for the process?

I remember reading an identical question with writing identical comment: By polyethylene, you must mean PEG. If you are sure it is right mixture, experiment with the ratio.  Silika differs and may need variable ratio.

Silica particles do show non-Newtonian fluid behavior but you need a much more concentrated suspension than simple cornstarch. Secondly, the solvent must wet the particles. The trick is to use a very small amount of solvent. Add a few drops of a solvent at a time, mix well (ideally sonicate) and then test it with a spatula. If it suddenly becomes hard upon fast motion of a spatula you have the right ratio.
If you wish to go a little deeper: Shear thickening is usually observed with dispersed suspensions. Look at a very dilute suspension of silica NPs and PEG under a microscope. If the particles are sticking together like grape-bunches, do not expect a shear thickening behavior in this solvent. Instead, such a suspension might be shear thinning like ketchup (which is difficult to test without a rheometer).
You can add this predictive behavior to your project if your school has an optical microscope. People might appreciate it.

Thank you very much your answer is very helpful.
Is there other ways to properly mix materials since I do not have access to sonication?

Just mix with the help of a spatula.

I used a normal blender but did not help. I got a slime-like material instead of a non newtonian fluid.

Blender cannot work with very concentrated suspensions. Try optical microscopy first. As I mentioned above not all suspensions are shear thickening. What is this school project about?


text:
How is reaction enthalpy equal to heat in a coffee cup calorimeter?
In explaining how to evaluate data from a coffee cup calorimeter, textbooks often claim that
$$ \Delta H = q $$
This makes sense for an isothermic calorimetry experiment, where the temperature of the reaction mixture is kept constant by heating or cooling. It also makes sense for a bomb calorimeter, where there is heat exchange between the bomb and the surrounding water bath. How do you rationalize this for a coffee cup calorimeter, where ideally there is no heat exchange between inside and outside of the cup?

$\Delta H = q = 0$ and   $\Delta H = n \cdot \Delta H_\mathrm{r} + C_{p, \mathrm{fin}} \Delta T$

@Poutnik How is the sign of $\Delta T$ defined with respect to an exothermic reaction, for example? Does it refer the the change in temperature during the reaction, or the change in temperature to return to the original temperature?

For exothermic reaction, the 1st term in negative and the 2nd one is positive, the content warms up and $\Delta T$ is obviously positive. I.e. the change from initial to final temperature.

@Poutnik So this is the view that the product mixture is the environment at the same time as it is the system (the reactants turn into product, giving off heat to the solvent and "to itself"). Or in your mind, you let the system cool down (it releases all of the enthalpy in the form of heat, and you need $q = C_{p, \mathrm{fin}} \Delta T$ to heat it up again, see my answer.

If I got it correctly and the scenario is an isolated system, there is no cooling down. If reaction enthalpy is zero, temperature does not charge. With nonzero  enthalpy, temperature changes. Negatively taken reaction enthalpy, properly scaled, is then equal to heat, that would be needed to warm up the final system by the same T difference.

@Poutnik Yes, an isolated system as best as possible. I now follow the argument you made with the two equations you posted in the comments. There is a implicit third state (products at initial temperature), and I would argue textbooks and lab manuals should make it explicit (to remind folks that reaction enthalpy is defined with equal initial and final temperature, and that the enthalpy change in an isolated system is zero).

First, a remark about notation and dimensions. $\Delta H$ has dimensions of energy. I will use $\Delta_r H$ for the enthalpy of reaction, and it has dimensions of energy per amount of substance. The enthalpy of reaction is the enthalpy change of a system (where initial and final temperature are equal) when a reaction happens, divided by the amount of reaction. This means it refers to a specific reaction equation, and its value changes when you multiply coefficients by a constant factor.
Back to the question.

How do you rationalize this for a coffee cup calorimeter, where ideally there is no heat exchange between inside and outside of the cup?

You want to compare the state before the reaction to the state after the reaction at the same temperature. In the scheme below, this would be state 1 and state 3. However, experimentally we went from state 1 to state 2.

Enthalpy is a state function, so we can complete the path to state 3 by going from state 2 to state 3 computationally, and adding up the changes along the path. To get back to room temperature, we have to change the temperature by negative $\Delta T$. If we have good estimates of the specific heat capacity and mass of the cup's contents (and assume the heat capacity of the container is negligible), we can estimate the heat exchange necessary to get back to room temperature, as shown in the image. If the cup is insulated well, we are done because the enthalpy change between state 1 and 2 is zero, ideally. If not, we have to somehow estimate the heat loss or gain in that step.
The entire calculation assumes the absence of non-PV work and constant pressure.
There is a nice discussion of the conceptual difficulties in calculating the enthalpy change from state 2 to state 3 here. They conclude that using the heat capacity of water as an estimate for the heat capacity of the solution often tricks students into using the mass of water instead of the mass of the solution (which will typically be larger for the same volume). They also point out that the specific heat capacity of 1 mol/L saline (the product in one of the common GenChem1 reactions explored by calorimetry) is significantly smaller than that of water, so it might make sense to provide that value, both for accuracy and for better understanding. Relevant data is here and there. For example, the heat capacity of 1 mol/L aqueous sodium chloride is about 3.8 J/(g K) compared to 4.182 J/(g K) for pure water.


text:
Buffer when testing soda with DCPIP
We are testing a range of sugary drinks (coca cola, orange juice, and chocolate milk) with DCPIP to detect vitamin C. The coca cola has given a false positive and we were wondering if this is due to low pH or if there is a buffer solution to prevent this from happening?
Many thanks!


text:
Why does alkylamines have a higher boiling point than ammonia?
Here are the boiling points of ammonia and some amines:
$$ \begin{array}{|c|c|}
\hline
\textrm{Substance} & \textrm{Boiling Point} \\
\hline
\textrm{Ammonia} & \pu{-33.34^\circ C} \\
\textrm{Methylamine} & \pu{-6.3^\circ C}  \\
\textrm{Dimethylamine} & \pu{7^\circ C}  \\
\textrm{Trimethylamine} & \pu{2.9^\circ C}  \\
\textrm{Ethylamine} & \pu{16.6^\circ C}  \\
\textrm{Diethylamine} & \pu{55.5^\circ C}  \\
\textrm{Triethylamine} & \pu{89.28^\circ C}  \\
\hline
\end{array}$$
My question is, why every time you remove a hydrogen from ammonia and replace it with an ethyl group the resultant molecule has a higher boiling point? Since the hydrogens can make hydrogen bonds and the ethyl groups can only interact through London dispersion force, I would expect that the resultant molecule would had a lower boiling point.
Also, why is trimethylamine an exception to this trend?

Trimethylamine's the only example here where your idea works.

"Because it's heavier" would probably be the over-arching answer here. Also, trimethylamine doesn't have any hydrogens that do hydrogen bonding.

That's the point, hydrogen bonding is a stronger intermolecular force than London Dispersion forces. Yet, a molecule like ammonia (which has 3 hydrogen so it can make H-bonds) has a LOWER B.P. than methylamine which has only 2 hydrogens. I don't get why this happens, for me everytime you take an hydrogen out and replace it with a apolar carbon chain you should get a LOWER B.P. and not HIGHER.

And why is it that trimethylamine differs from the other examples?

When you compare ammonia to alkylamines you have to consider that the molar mass of your compound is also changing; methylamine has almost double the mass of ammonia and this will obscure the effect you're trying to look for. Other things (intermolecular forces, etc.) being equal, a compound with a higher mass will have a higher boiling point. Let's compare three alkylamines with the same molar mass.
$$\begin{array}{cr}
\text{Compound} & \text{BP (}^{\circ}\text{C)}\\ \hline
\text{Trimethylamine} & 2 \\
\text{Methylethylamine} & 36 \\
\text{n-Propylamine} & 49
\end{array}$$
From the table it's clear that changing a hydrogen-bondable hydrogen to an alkyl group reduces the boiling point, consistent with your suggestion that the stronger hydrogen bonding is being replaced with dispersion forces.


text:
What are the reaction rate constants, or coefficients, of coal combustion at high temperature?
Say I have coal that I will combust with $\ce{O2}$ at a high temperature of $\pu{1600-1800 K}$, what are the reaction rate constants, or coefficients, of the main reactions?
I have searched around literature, but it seems the constant increases exponentially depending on the temperature. However, I cannot find any research on some experimentally found rate constant at a temperature above 500 kelvin.

And you found some for lower temperatures?

Heterogenous reaction rates are very, very tricky. Surface geometry and matter transport is not something I would want to theoretically analyze.

@Mithoron Yes, see:  Liu et. al., 2022: Oxidation reaction constants for coal spontaneous combustion under inert gas environments: An experimental investigation

@Poutnik Why is that so? I want to model the dynamics of the reactions in a differential equation, and therefore I need rate constants for the dynamics of concentrations. Would you otherwise know some simplification or assumption for this?

The curse of solving differential equations is defining proper initial border and spatialconditions. Meteorologists could tell sagas about solving Richardson's partial diff. equations to predict synoptic situation by the most powerful available computers.

To get true rate constants from the rates, one has to know the true surface area and the true local boundary partial oxygen pressure across this area, which varies with local rates of mass transport. That leads rather to empirical, effective rate constants limited to particular scenarios. Even knowledge of the true rate constant would not be applicable for real scenarios due remaining unknowns.


text:
How did this paper mathematically express field driven water autoionization?
I recently read a paper on using an electric field to drive water autoionizaton. I'm trying to figure out how to use the Laplace equation on pg 9; 4th paragraph; to solve for voltage. I'm also interested in how this equation would change if I replaced the hemispherical tip with a parallel plate. Anyone strongly familiar with this subject matter? Any help is greatly appreciated!
https://www.sciencedirect.com/science/article/abs/pii/S0009261411011511
pg 8-9, 5.1. Conditions for ionization

Write the equations and add more details. Many people do not have access to papers.

Welcome to Chemistry StackExchange! It is rather hard to give you some help without knowing what is your background or where to begin the explanation. Poisson's and Laplace's equation from the article are usually introduced very early in any undergraduate physics course on electromagnetism (example 1, example 2). Note that the equations in article are written in spherical coordinate system, and a small peculiarity in the first one is the variable permittivity, $\epsilon$.

Instead of posting a screenshot, it would be better if you write the relevant  part as text.


text:
Why does changing extracellular concentration of a highly-permeable solute affect membrane potential (e.g. hyperkalemia)
Okay, this is an embarrassing question to ask and expose just how little I understand membrane potential given my field, but I've been trying to understand on my own for 3 years at this point and it's clearly going nowhere.
So, when you're first introduced to membrane potential, it's essentially explained as "with more positive charges inside the cell relative to outside, membrane potential increases". This causes confusion once the Nernst equation is introduced and it seems to be opposite, but the way I've explained it to myself is "the resting membrane potential for a given extracellular and intracellular concentration of an ion is the membrane potential needed to hold it in equilibrium at those concentrations, which is established by that ion plus all other ions' net charges. Not the membrane potential due to that ion's concentrations" i.e. to hold a positive ion mostly intracellular, you need a negative intracellular-extracellular electrical difference, so you say the resting membrane potential for those concentrations is negative.
Where I lose my absolute marbles is trying to understand how in vivo changing the concentration of a highly-permeable ion changes the equilibrium potential difference. Take hyperkalemia. Yeah, to hold the concentration gradient of potassium at less of a gradient (same intracellular concentration as normal, higher extracellular concentration), you would not need as negative of a potential difference. But I can't understand how the actual potential difference changes in that situation. Assuming there's no permeability to other ions (obviously incorrect, but their permeability is much lower than potassium, so I feel making that assumption should at least lead you in the right direction despite not giving you the exact correct calculation for potential difference, right?), the distribution of electrical charges has no reason to change to maintain this new concentration gradient. Instead, the highly-permeable ion should flow along its electrochemical gradient to reach its equilibrium concentrations. Why isn't this what happens?


text:
Hydrofluoric acid
At work we have a large tank that is comprised mostly of hydrofluoric acid, I don't know the dilution factor as I'm not the one making the bath, but I have a question, when emptying the bath we tried to check if acid really is heavier then water, when we insert the pH stick, it comes up "acid", but the stick itself only goes about 2cm into the bath, where theoretically the water would be, we emptied half of the bath so the acid would be emptied and tried again, same result, is the acid mixed completely with water or is it just stronger near the bottom of the tank?

I advise you to stay as far away as possible from the tank of HF.

Read Derek Lowe's comment on HF: https://www.science.org/content/blog-post/things-i-won-t-touch-1

Why on earth do you think that tank would we HF at the bottom and water at the top? That's not like oil and water!! Alcohol is lighter than water, but doesn't float at the top of your beer bootle. Aside, a company that has a large tank with HF and doesn't know what exactly is in it is a very bad health and safety risk for their employees and the general public.

All these answers are correct. But they are not very useful to the poor Alex

Liquids with unlimited miscibility with water, like ethanol, sulfuric acid and hydrofluoric acid do have different densities than water. This may lead to layered state, if one liquid is intentionally and carefully layered over or under water, or if there was just insufficient mixing.
If mixed properly, what takes quite short time with proper mixing, the solution of such liquid gets homogeneous, without separation tendency. Therefore, there is no reason to expect variation of the acid concentration, unless it was locally and significantly spent due the bath purpose and not mixed afterwards.
If there is such local acid spending then the result depends on how the liquid density changes by the process. If it decreases, it would have tendency to trigger convective raising stream and vice versa.
BTW, people without sufficient knowledge about dangerous chemicals should not be around them. A sudden splash of hydrofluoric acid on skin can painfully kill a person within not many hours. A part of danger is the acid takes its time for skin penetration before the person starts to observe dooming symptoms. Chemistry of fluorine had already too many victims, especially in first decades after the element discovery.

Thanks, the workplace is very strict with out protective gear and respirators, we work with the knowledge that protects ourselves and the only contact we have with HF acid is through a mounted crane, my question was purely of speculative matter and not dangerous curiosity

@Alex I had in mind rather assumed chemical background knowledge that should provide the answer.


text:
What is the structure of hydrated beryllium chloride
Is there water of crystallization or does water act as ligand? Also Be2+ is smaller than Li 2+ then why Li2+ has only 2 water molecules as ligands

$\ce{Li^2+}$ does not have 2 water ligands.It would oxidize water immediately.

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

It is $\ce{[Be(H2O)4]Cl2}$

As others have already mentioned it's more of a  $\ce{[Be(H2O)4]Cl2}$ if you want to write it like this. It becomes much more visible if we move to the solid state structure. I'm adding a picture of the unit cell of $\ce{BeCl2 x 2 H2O}$ here as well. From this you can see that each beryllium cation is tetrahedrally surrounded by four water molecules. And that's pretty much it. This 'unit' has some chloride around it but there is no direct contact between the beryllium and the chloride.
Calcium is another famous candidate for this behaviour. For  $\ce{CaCl2}$ hydrates we have, if I remember correctly trigonal prisms of calcium that are surrounded by water. Those prisms stack along their triangular faces. The chloride anions are always only in the second coordination sphere.
Now imagine you add a little heat to this. You already have your oxides close to cations that from really stable bonds with oxygen. Instead of water leaving it will destabilize the  $\ce{O-H}$ bond in  $\ce{H2O}$ and release  $\ce{HCl}$ rather than  $\ce{H2O}$. That's how those basic chlorides and related compounds form and the reason why you cannot make their anhydrous form from heat directly.
For your comparison with  $\ce{LiCl}$ (hydrated). Lithium is $\ce{Li^+}$ while beryllium is  $\ce{Be^2+}$. This by HSAB theory there will be a much stronger and shorter bond between beryllium and oxygen than for beryllium and chloride while for lithium this is much weaker. At this point we have to take the covalency into consideration as well but for pure comparison I rather keep it ionic.
If we compare the distances for both compounds we get d( $\ce{Li-Cl}$) = 264 pm and d( $\ce{Li-O}$) = 204 pm which means we can consider a $\ce{Li-Cl}$ bond here (that gives us an octahedron of the composition $\ce{[Li(H2O)2Cl4]^3-}$). For the beryllium compound we get d( $\ce{Be-O}$) = 160 pm and d( $\ce{Be...O}$) = 378 pm.
$\ce{BeCl2 x 2 H2O}$" />


text:
Are carbon nanotubes and graphene very flammable?
The reason I ask is because carbon is quite flammable in virtually every shape we encounter in nature. Even diamonds burn as a matter of fact. And after I just saw some steel wool burn easily, I started wondering about graphene and the likes. I mean, it is still a sheet, but it doesn't get thinner than a single atom, right? So in terms of surface area, this would be almost ideal for any form of oxidation. And since it is only 1 atom thick, I suspect it all wants to burn up rather quickly.
Does anyone know about this? Have experiments been conducted?
Edit: Steel Wool.

It depends on what you mean by very flammable. It is much more flammable than carbon forms with less surface/volume ratio, but much less flammable like many powdered organic matter able to form flammable gases.

Since graphene is usually adhered to a surface, that would greatly change its properties, including flammability.  BTW, "steel wool" is likely the metal wire "sponge" to which you referred. Even more flammable: cotton wool, used as an applicator for cosmetics. Pull some apart to increase surface area, and when ignited and dropped, most will be consumed before hitting the floor. [Of course, do this in a concrete or ceramic tile floored room.)

@DrMoishePippik Pippik. Thanks for the 'Steel Wool' part. Updated post accordingly. And regarding the substrate onto which graphene may be applied; yes, sure this is a completely different scenario. It's like saying Hydrogen is not dangerous in a closed box without oxygen, if you're getting my drift.


text:
How Amorphous solids are converted to Crystalline?
I know about a process of Annealing where a crystalline solid is heated to its melting point and then cooled rapidly so as to form amorphous solid which has random crystallite in uts structure.
But how is the reverse done. How amorphous is converted to Crystalline. Does it have to do anything with the fact that glass panes look milky after long time. Almost all Indian textbooks have this fact written in one of the chapters.

I'm not sure if it happens over time but we observe something similar here when we make glass ampoules from quartz glass tubes. Once we use the big oxyhydrogen torch to melt the thick-walled tubes you can see both, SiO2 gas formation as well as surfaces becoming rough and opaque. We always assumed it has to do with the glass melting and crystallizing again under the conditions. But many glasses also contain additives to avoid any far-range network formation.

The crystal nucleates and grows. The kinetics are slower, of course, but it happens just like it would in a melt. And, generally speaking, the quote from the textbook is, well, wrong - fused silica will remain fused silica at near-human temperatures for longer than we care to watch. (Now, when your intense pulsed laser self focuses inside a fused silica optic and rapidly nucleates crystal, well, so much for that optic...

@SamyakMarathe Comments are not supposed to provide answers.

It is only a question of time. Crystalline substances are more stable than amorphous substances. So when heated to near its melting point, any amorphous substance has from time to time the possibility of rearranging its atoms and molecules in a crystalline order. The barrier separating random order and crystal state can, contrary to the general practice, be exceeded for a small minority of energetic atoms and molecules, according to Maxwell theory. And once a crystalline arrangement is started at some point of the solid it is more stable. So it will not go back as easily to random structure. The crystalline structure will be growing and growing. Of course, rather slowly. This is how quartz crystal have been formed from silica in nature.

Well, the time may be a significant fraction of the age of the universe. Nobody worries much about their diamonds turning to graphite.

@Jon Custer. The activation energy between diamond and graphite is huge, because both phases are highly ordered, and one is cubic, and the second is hexagonal. The transition diamond-graphite will never be obtained within even the age of the universe. But the transition from amorphous silica to ordered silica (quartz) has a much lower activation energy. This transition may occur within a couple of centuries.

indeed, diamond to graphite is harder. But network glasses such as fused silica aren't particularly easy, and the more pure they are the slower they are. Highly pure fused silica tubes are used in furnaces operating at 1100C for years without nucleation. Even obsidian, which is not pure by any means, will last millions of years. Your modern window glass going crystalline? Not in your lifetime or that of generations to come.


text:
Can denticity and hapticity be defined simulaneously for a ligand?
I am freshly studying coordination chemistry.
What I understand from denticity and hapticity is:
Denticity - no. of electron donating sites in a ligand.
Hapticity - no. of electron donating atoms in a pi-bond conjugated ligand.
The use of the terms "sites" in denticity and "atoms" in hapticity makes it possible for a compound to have different denticity and hapticity at the same time. For example: ethylene can should be monodentate and η2. So, my final questions are,

Can we define both denticity and hapticity for a pi-bond conjugated ligand?
Are they supposed to have the same values?
Why was hapticity needed in the first place when one can count the nummber of pi-bonds and define denticity accordingly? Is it because more electrons can be donated by a conjugated system?

Thanks in advance.


text:
Can two cations bond?
I want to know if you forced two cations together (maybe by heating it up in a plasma or something) would they bond and share atoms like in a covalent bond? What’s more, instead of taking just the first valence electron from each cation, could you take half of the electrons from each atom and force them to bond so that the two atoms have a bond that’s 4 or 5 times stronger than it would be if you only took 1 electron?

Related: https://chemistry.stackexchange.com/questions/139526/is-it-possible-for-already-charged-metal-ions-to-form-a-metallic-bond/139527#139527

The mercury(I) ion is an exact example- it exists as a dimer, single-bonded using each atom's (relativistically stabilised) 6s orbital.


text:
Confusion regarding order of reaction
The following points are given in my textbook about order and molecularity:-
(i) Order of a reaction is an experimental quantity. It can be zero and even a fraction but molecularity cannot be zero or a non-integer. Its values are limited to 1,2 or 3.
(ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions.For a complex reaction, molecularity has no meaning.
(iii) Order is given by the slowest step(also known as RDS) and molecularity of the slowest step is same as the order of the overall reaction.
(iv) Molecularity is equal to the sum of stoichiometric coefficients of reactants in the elementary reaction.
However, I have observed an inconsistency in the above  points and would like a clarification.
Let us consider 2 cases:-
Case 1: Complex Reaction
By (iii)-order of overall reaction=molecularity of RDS. Since molecularity is 1,2 or 3, order must also be 1, 2 or 3.
Case 2:Elementary Reaction
By (iii)-order of overall reaction=molecularity of the reaction(since there is only 1 step so that must be the RDS). Since molecularity is 1,2 or 3, order must also be 1,2 or 3.
Now, combining these two cases, order =1,2 or 3. However, (i) mentions that the order can be fractional or zero. How is this possible?

Authors of the rules have not thought through all possible cases. E.g. nonlinear reaction schemes with a reaction feedback, regenerating reactants of some prior reaction steps.

@Poutnik Can you elaborate on that in an answer?

No resources for that now, but see e.g. https://www.colby.edu/chemistry/PChem/notes/ChainMech.pdf

The combustion of a candle is of order zero, in a first approximation. The amount of paraffin burned in a second is practically the same from the beginning to the end of the combustion.

Note that many processes have their pseudo-order lowered, if one component content is about  invariant within the process. E.g. A + B -> C with  [A] >> [B] is reaction of pseudo-order 1, with [A] implicitly involved in reaction rate constant. This applies to the above candle example too.

Kinetics of most reaction have little to do molecularity. BTW there's many issues with this post, see for example: Are you confused by confusing question titles including the word "Confusion"?, not to mention punctuation

@Mithoron punctuation edited. Could you suggest a better title? I'm unable to frame one without it being too long.

Well, earlier one was better. Your post is kinda "what is this kinetics all about", though.

@Mithoron How, I specifically asked what my question was in my post? How can it be improved?

Some reactions have no order. For example, the reaction $\ce{H2 + Br2 -> 2 HBr}$ has a reaction rate given by the formula : $$\ce{\frac{d[HBr]}{dt} = \frac{k[[H2][Br2]^{1/2}}{1 + k'[HBr]/[Br2]}}$$ This reaction has no order ! Ref. Arthur A. Frost, Ralph G. Pearson, Kinetics and Mechanism, John Wiley & Sons. New York, 1961, p. 236.

Here is some clarification.

You are correct that order of reaction is an experimental quantity.
Just like order of reaction is is an experimental quantity, molecularity is purely theoretical quantity and as of today, I have not found anyone who can observe it. And, it only work for elementary (1-step) reaction.
Order of reactions do only measurable for RDS. But just by stating that there exist an RDS, you are also stating that the reaction is not elementary, thus the idea of molecularity simply does not apply for a reaction with more than one step.
Molecularity have to be positive whole number the same reason you cannot have less than one atom, it is simply not make sense.

I hope this clear things up.


text:
Why does the Shielding Constant of Orbitals in Slater's rule get higher, the further away we are from the electron we are considering?
My text material says:

$\mathrm{Z_{eff}= Z-S}$ where $\mathrm{Z_{eff}}$ is the effective nuclear change on that specified electron; $\mathrm{Z}$ is the atomic number and $\mathrm{S}$ is the shielding constant or screening constant which is the measure of net repulsion on that electron caused by other electrons in the same shell and inner shells.

So I infer that the Effective Nuclear Force on the electron of our interest (supposing it lies in the outermost orbit) is the resultant of the Force of the Nucleus minus the repulsion of the other electrons. Am I correct in assuming that? Can "charge" and "force" be analogous in here?
If so, by Coulomb's law, shouldn't the closer orbitals repulse the electron more than the ones that are further away? Or in other words,  shouldn't the Shielding constant be getting higher as we approach the outermost orbit?
But the values in Slater's rule shows the opposite:

the shielding experienced by an $\mathrm{s-}$ or $\mathrm{p-}$ electron, electrons within same group shield $\mathrm{0.35}$, except the $\mathrm{1s}$ which shield $\mathrm{0.30}$ electrons within the $\mathrm{n-1}$ group shield $\mathrm{0.85}$ electrons within the $\mathrm{n-2}$ or lower groups shield $\mathrm{1.00}$ the shielding experienced by $\mathrm{nd}$ or $\mathrm{nf}$ valence electrons electrons within same group shield $\mathrm{0.35}$ electrons within the lower groups shield $\mathrm{1.00}$

Implying that the further we move from the outermost electron, the Shielding or the Repulsion increases, which sounds the opposite of Coulomb's law to me.
What am I missing here?
Source:

Chemistry LibreText - Slater's rules
Concise Inorganic Chemistry by J.D. Lee

There is important knowledge from classical electrostatics (going as far as to Newton classical gravity and his invention of integral calculus, as it is common for quadratic forces):

Two distributed non-overlapping charges with spherical distribution symmetries have on each other the same force effect as if they were point charges, located at the distributed charge center.

If we consider inner s orbitals, or superposition of all 3 filled p orbitals, they have spherically symmetric electron probability density.
Therefore, their shielding coefficients would be the same (1.00) as if located in the nucleus, if not decreased by overlapping of the shielding and shielded orbital. As the overlapping brings partial negative shielding.
Because, when a shielding electron is outside the shielded electron(wrt the nucleus), repulsion acts toward the nucleus, effectively being like an additional nucleus charge (= "negative shielding" ). If a point charge is overlapped by spherically symmetric charge distribution, it is attracted/repulsed only by the inner part of this distribution.
The more inner shielding orbital is, the smaller is overlapping and the better is shielding of the shielded orbital. Breaking spherical symmetry decreases the shielding even more.
See also Slater's rules topic on Wikipedia and Libretexts


text:
Estimating the energy required for dissociation from visible absorption spectra
I have been having trouble solving the following problem.
The question calls for an estimation of the energy (in kj/mol) required to dissociate ClO in its excited state when it is excited from the v = 0 of the ground state.
My first approach in solving this problem was selecting two transitions (v" = 2 to v' = 7 and v" = 3 to v' = 8).
I feel as though my error is in this part of the problem as the question asks for a transition from v = 0 but as far as I can see in the diagram there is no peak for the v = 0 transition.
Following this, I converted the wavelengths for each transition into wavenumbers and then subtracted the wavenumber corresponding to each transition:
$$\nu_{7'0"} = \frac{10^7}{287\ \mathrm{nm}} \approx 34843.205\  \mathrm{cm}^{-1}$$
Then, subtracted $308\ \mathrm{nm}$ in wavenumbers from the above value to give $2375.68\ \mathrm{cm}^{-1}$ for the transition.
Then using the above numbers I used the following formula to find $\omega_e$ and $\omega_ex_e$.
$$G\left(v'\right) = \left(v' + \frac{1}{2}\right)\omega_e - \left(v' + \frac{1}{2}\right)^2 \omega_e x_e$$
Then with $\omega_e$ and $\omega_ex_e$ I used the following formula to find the dissociation energy, $D_e$,
$$D_e = {\omega^2 \over 4\omega_ex_e}$$
Then subtracting the zero-point energy from $D_e$ to get $D_0$ and converting this value into kj/mol just to get an incorrect answer.
If anyone could point out where I'm going wrong that would be greatly appreciated.

Quiz 3 has been annoying.
From the graph, we know that the molecule dissociates at a wavelength of ~262.5 nm. We know this as this is when there are no more transitions, as indicated by the completely flat line.
Using the equation
E = (h * c)/ν and substituting in the relevant values, we get E = 7.57E-19 J. Multiplying by Avogadro's number nets us ~456000 J/mol = 456 kJ/mol.


text:
Are there tools or algorithms to detect organic molecules from 3D crystal structures?
The task I'm thinking of is to detect entire molecules from structures, such as in a crystallographic information file (CIF) or from the XYZ file generated in a molecular dynamics simulation. Crystallography itself doesn't help identify the molecular identity (such as represented in IUPAC name or as a SMILES string), but requires human interpretation or computational modelling.
Three example use cases:

Identify the spatial orientation of all the moieties of molecule X in a molecular crystal of X.
Identify locations of solvent molecules trapped in crystals
Identify parts of a molecule that has multiple copies in the crystal.

Does anyone know of such an algorithm/software?

If you can get the structure why would you not know the name beforehand?

I'm looking for part of a structure that is still molecular. In other scenarios, if you have multiple copies of the same molecule, their spatial orientation (such as packing) would be interesting to understand. Neither of these would be possible without identifying the molecular entity in 3D (e.g. as a grouping)

Its not clear what you want as the cif/sdf/pdb etc. data (obtained usually from x-ray crystallography) contains all the information, i.e. xyz coords and connection table to work out where the bonds are,  so you can draw the molecule at a pinch by hand from the connection table. If you don't have a connection table you can use known bond lengths and atom positions to work one out by writing some simple code.

I updated the questions with use cases, perhaps it will explain things better

Sorry, I don’t think that helps much. Perhaps if you have an example CIF in which you think this is necessary? The space group, for example will give you all the symmetric copies and orientations of a molecule in the unit cell. Finding a solvent isn’t very hard either, because you can look through the bonded components to find solvents or co-crystals.

Load the file into a 3D browser like JSmol. It will show bonds where they make sense, allowing you to distinguish between distinct molecules in the structure. Some cases might be more complex and you would need to know what is considered a molecule in that context.

Indeed, pretty much any program that handles CIF .. CCDC Mercury, JSmol, etc.

@GeoffHutchison, the problem is not with one or two CIFs, which can definitely be solved using these classic approaches you mentioned. I would need to do the same for hundreds to thousands of structures, which will not be possible without an algorithm. Perhaps defining a bond length constraint then gradually iterate over the atomic combinations can work.

@QGent - I think you misunderstand. Basically anything that handles a CIF can do what you describe. Most (e.g. Open Babel, CCDC Python interface) can be scripted to automate the analysis.

Your question has a few parts, but multiple solutions. Let's start with "XYZ from molecular dynamics"
3
XYZ file generated by Avogadro.
O     -2.86063    1.04260   -0.00000
H     -2.52287    1.38117    0.82085
H     -3.53528    0.39764    0.17713

This is trivial. Open Babel will read this, perceive the bonds through inter-atomic distances, and generate water. Most programs do this. (RDKit should use the xyz2mol package.)
Problems come when you want to perceive bond orders .. because it's a non-trivial task. Both Open Babel and xyz2mol have algorithms..
You ask about solvent molecules. Again, both Open Babel and RDKit have methods to find and/or remove fragments, e.g. "An open source chemical structure curation pipeline using RDKit"
Then you ask about multiple copies (e.g., in a CIF)
Here's part of an aspirin entry in the COD:
O1 O 0.62354(11) 0.1419(2) 0.61266(11) 0.0203(3) Uani 1 1 d .
H1 H 0.577(2) 0.025(5) 0.602(2) 0.060(8) Uiso 1 1 d .
O2 O 0.50970(11) 0.1879(2) 0.40933(11) 0.0212(3) Uani 1 1 d .
O3 O 0.78878(10) 0.41876(19) 0.72945(10) 0.0171(3) Uani 1 1 d .
O4 O 0.90669(11) 0.2189(2) 0.66325(11) 0.0215(3) Uani 1 1 d .
C1 C 0.65418(15) 0.4419(3) 0.50750(16) 0.0154(4) Uani 1 1 d .
C2 C 0.74806(15) 0.5203(3) 0.61181(16) 0.0162(4) Uani 1 1 d .
C3 C 0.80069(16) 0.7061(3) 0.60519(17) 0.0196(4) Uani 1 1 d .
H3 H 0.8632 0.7580 0.6776 0.024 Uiso 1 1 calc R
C4 C 0.76267(16) 0.8174(3) 0.49303(18) 0.0215(4) Uani 1 1 d .
H4 H 0.7997 0.9449 0.4882 0.026 Uiso 1 1 calc R
C5 C 0.67053(16) 0.7425(3) 0.38781(17) 0.0195(4) Uani 1 1 d .
H5 H 0.6444 0.8182 0.3106 0.023 Uiso 1 1 calc R
C6 C 0.61699(15) 0.5578(3) 0.39559(16) 0.0173(4) Uani 1 1 d .
H6 H 0.5534 0.5081 0.3233 0.021 Uiso 1 1 calc R
C7 C 0.59016(15) 0.2458(3) 0.50601(15) 0.0151(4) Uani 1 1 d .
C8 C 0.86823(15) 0.2623(3) 0.74362(16) 0.0166(4) Uani 1 1 d .
C9 C 0.89840(17) 0.1611(3) 0.86922(17) 0.0249(5) Uani 1 1 d .
H7 H 0.9545 0.0480 0.8770 0.037 Uiso 1 1 calc R
H8 H 0.8259 0.1059 0.8769 0.037 Uiso 1 1 calc R
H9 H 0.9347 0.2625 0.9366 0.037 Uiso 1 1 calc R

That's it. That's the connected component for an aspirin molecule. If there were solvent, it would be a disconnected fragment if you, e.g. convert to SDF or SMILES.
How do you know the symmetric copies? It's in the CIF as defined by the space group.
_symmetry_equiv_pos_as_xyz
'x, y, z'
'-x, y+1/2, -z+1/2'
'-x, -y, -z'
'x, -y-1/2, z-1/2'

This gives you the transforms (in fractional coordinates) for all the symmetric copies. Some may be duplicates. But if you have a point $(x,y,z)$ there's a symmetric copy at $(-x, y + 0.5, -z + 0.5)$, etc. In other words this space group could have 4 copies of aspirin per unit cell.

You mention automation, which sounds like you want to write a script. Maybe that's in Open Babel. Maybe you use CCDC Python or ASE or pymatgen. There are likely other packages for handling CIF files. I don't know what precise features you want. But they should all find connected components. Once you've built the script to process one, adding an outer loop in Python to handle multiple files is fairly easy, e.g.:
for filename in glob.iglob("*.cif"):
    # do something

In terms of implementation questions, I would generally suggest that the https://mattermodeling.stackexchange.com/ is better suited.


text:
TV Program Said Remove Some of Chlorine in Water By Putting Hose On Fine Spray
A UK TV program suggested that some of the chlorine in tap water can be removed if you fine spray it from your hose, such as with un upside down fine sprayer.
Is this true? How effective is it?

It is quite effective,similarly as for radon-222.

Actually this is a very interesting suggestion and this advice may have a very solid scientific basis. I had heard a related talk a couple of years ago on water mists generating hydrogen peroxide but the authors were quite reluctant to publish this observation because of a lack of a suitable explanation. The idea is very neat. If we spray a fine mist of pure water, a measurable quantity of hydrogen peroxide is  immediately form in that mist. In 2022, this report appeared in the Journal of the American Chemical Society. See Masoud A. Mehrgardi, Mohammad Mofidfar, and Richard N. Zare Sprayed Water Microdroplets Are Able to Generate Hydrogen Peroxide Spontaneously, J. Am. Chem. Soc. 2022, 144, 17, 7606–7609.
However, hydrogen peroxide is naturally found in rainwater and this is well known.
Hydrogen peroxide and hypochlorite ions (dissolved chlorine) cannot co-exist even for a fraction of a second and thus this advice appears to be valid. Hydrogen peroxide is converted to water and aqueous hypochlorite to chloride. This chemistry is also well known.
As someone who is interested in chem. history, I feel that this could be a re-discovery and if we open 19th century voluminous works, this comment might be present! If anyone is aware of older works, please leave a comment.

I think it's much simpler than this - chlorine (and to a lesser extent related compounds such as chloramine) will evaporate out of water if it left to sit. Aquarists do this routinely. Misting (or boiling which is another common approach) speeds up this process substantially.

Both effects could be working. It could be synergistic. We cannot stop one or the other at the same time. Removing of dissolved gas by bubbling helium in the solvent was also routinely done in HPLC systems.


text:
Finding masses of the products for halogenation of benzene from percent yields

We provide a reactor $\pu{90 kg}$ of benzene and $\pu{30 kg}$ of $\ce{Cl2}$.If the percent yield of the first reaction with respect to benzene is $25\,\%,$ and the percent yield of the second reaction with respect to benzene is $2\,\%,$ which are the total final masses of chlorobenzene and dichlorobenzene?

This was a question in a test of mine, which I got wrong. In my opinion there were two ways to interpret this problem, that led to two different results.
We want to produce chlorobenzene using the following reaction:
$$\ce{C6H6 + Cl2 -> C6H5Cl + HCl}\label{rxn:R1}\tag{R1}$$
However, the following undesired reaction occurs simultaneously, giving dichlorobenzene:
$$\ce{C6H6 + 2 Cl2 -> C6H4Cl2 + 2 HCl}\label{rxn:R2}\tag{R2}$$
I computed the amounts $n(\ce{C6H6}) = \pu{1154 mol}$ and $n(\ce{Cl2}) = \pu{414 mol}.$ The limiting reagent is chlorine for both reactions: in \eqref{rxn:R1} the ratio between $\ce{C6H6}$ and $\ce{Cl2}$ should be $1:1$ according to \eqref{rxn:R1} and $1:2$ for \eqref{rxn:R2}.
The first reaction theoretically should produce $\pu{414 mol}$ of $\ce{C6H5Cl},$ but since its percent yield with reference to benzene is only $25\,\%,$ only $\pu{414 mol}\times 0.25 = \pu{103.5 mol}$ of benzene react, which produces $\pu{103.5 mol}$ of $\ce{C6H5Cl}.$
In the second reaction only $\pu{207 mol}$ of benzene should react given the $\pu{414 mol}$ of $\ce{Cl2},$ but since its percent yield with reference to benzene is $2\,\%,$ we get that only $\pu{207 mol}\times 0.02 = \pu{4.14 mol}$ of benzene react, which produces $\pu{4.14 mol}$ of $\ce{C6H4Cl2}.$ The rest is just converting amounts to masses.
My professor said I had to compute the $25\,\%$ of the original $\pu{1154 mol}$ of benzene, and that quantity $(\pu{288.5 mol})$ of benzene is the one that reacts in the first reaction. Same for the second reaction, giving $\pu{1153 mol}\times 0.02 = \pu{23.06 mol}$ of benzene which react in \eqref{rxn:R2}.
Which approach would be the correct one?

There's no two ways, you probably calculated how much benzene was converted to each halide when you needed masses of halides themselves.

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

414 mol is molar amount of chlorine not benzene. BTW, we do not say number of moles nor number of kilograms.

Consider these 4 slightly different terms: 1/ Reaction yield with 1 product. 2/ Reaction yields with 2 parallel products. 3/ Reaction yields with 2 parallel products wrt excessive reactant A. 4/ Reaction yields with 2 parallel products wrt deficit reactant B.


text:
What is the pKa value of the nitrogen atoms, specifically the amine groups?
How would you estimate the $\mathrm pK_\mathrm a$ values of the nitrogen atoms in 1,4-dimethyl-1,2,3,4-tetrahydroquinoxaline?

I struggle with it because I am unsure whether to classify them as tertiary amines (with a $\mathrm pK_\mathrm a$ of around 9) or whether they can be considered as aromatic amine (similar to aniline), which would result in a lower $\mathrm pK_\mathrm a$ value (around 5ish?) due to the delocalisation of the aromatic ring.
Can someone explain to me which is the correct idea? I really struggle with the concept.

They don't have a hydrogen attached, so they don't have a p$K_\mathrm{A}$. However, you can discuss the acidity of conjugate acid, the ammonium $\ce{NR3H+}$.

@KarstenTheis That's what OP's discussing, but I guess it technically could act as a C-acid (now that would be more interesting!).

See https://chemistry.stackexchange.com/a/162479

Thanks for the link, but its not really what I am looking for. @Mithoron is right! I also meant to be discussing  the effect of the benzene ring on it. And whether you would still consider this tertiary amine to be aliphatic or aromatic (in terms of pka)

@tess are you are interested in the ability of your molecule to donate protons from the C atoms (what Mithoron described), or to accept protons onto the N atoms? The link Karsten Theis shared is helpful in understanding the ambiguity of "pKa".

I would try to compare pKas of conjugate acids for aniline, N,N-dimethylaniline and o-phenylenediamine ( IIRC values for all 3 are on Wikipedia) and I would make some projections. Guessing pKa from scratch is no fun.

The nitrogen atoms are attached to the aromatic ring, and this placement is the dominant factor. The conjugate acid formed by protonating either nitrogen atom would be expected to have a $\mathrm pK_\mathrm a$ similar to that of protonated aniline, not a protonated aliphatic amine.
We might expect this constant to be modified by the second nitrogen atom. Let us explore this.
The second nitrogen atom actually has two conflicting influences:

Electron-withdrawing inductive effect, which would decrease the conjugate acid $\mathrm pK_\mathrm a$ (stronger conjugate acid, weaker base). We expect this to be strongest in the ortho position as in the structure given in the original problem.

Electron-releasing mesomeric effect, which would increase the conjugate-acid $\mathrm pK_\mathrm a$ (weaker conjugate acid, stronger base). As with similar effects acting through a benzenoid ring, this would be stronger in the ortho and para positions, weaker in the meta position.

Wikipedia gives the following conjugate-acid $\mathrm pK_\mathrm a$ values:

Amine
Structure
$\mathrm pK_\mathrm a$

Aniline

4.63

o-Phenylenediamine

4.57

m-Phenylenediamine

5.11

p-Phenylenediamine

6.31

Compared with the aniline base case, the ortho-diamine (matching the arrangement of nitrogen atoms in the original problem) shows very nearly the same acid-base balance, as the two effects described above appear to balance out. The meta arrangement gives a slightly stronger base, but it is the para isomer that shows a clearly stronger basic behavior: the mesomeric electron donation is relatively strong with this arrangement while the induction effect is minimal.


text:
Does sodium borohydride reduces alkyne?
Will an isolated Alkyne or an alkene be reduced using sodium borohydride ($\ce{NaBH4}$)?

While Nilay's answer is interesting, sodium borohydride itself won't reduce isolated bonds like these. I'm even kinda surprised you'd ask.

Why not?

Alkenes and alkynes are rapidly reduced to the corresponding alkanes using sodium borohydride and acetic acid in the presence of a small amount of palladium catalyst. The heterogeneous reaction is conducted in open air at room temperature. Reactions typically afford conversions to the alkane product of 98% or more within 15 min. The best solvent system was determined to be isopropyl alcohol, though reduction also takes place in solvents such as tetrahydrofuran, chloroform and, with some substrates, even in water. The method described is a convenient alternative to hydrogenations that require an external supply of hydrogen gas.

Ref.: Anthony T. Tran, Vincent A. Huynh, Emily M. Friz, Sara K. Whitney, David B. Cordes, A general method for the rapid reduction of alkenes and alkynes using sodium borohydride, acetic acid, and palladium, Tetrahedron Letters, Volume 50, Issue 16, 2009, DOI: 2009,10.1016/j.tetlet.2009.02.014.

Note - no reaction will occur without the Pd catalyst

@Waylander that was my real question...if only NaBH4 was used...thanks:)

Isn't this more of a reduction with hydrogen produced in situ?


text:
Oxidation Reactions/Reactivity of Manganese Metal in Air and Water
I am planning on doing some reactions that lead to Manganese metal and a neutral solution containing $\ce{Mn^{2+}}$ as by-products, and I wanted to try some things out with the Manganese as well.
Since Manganese has common charges of II and III, will a relatively fine, grainy form of the metal burn (and at what temperature) in air to form a mixed "$\ce{Mn(II, II)}$" oxide, $\ce{Mn3O4}$, the way Iron does as it rusts, or is $\ce{MnO}$ more predominant?
I have the same question about whether or not $\ce{Mn3O4}$ is formed in "thermite-like" reactions such as reacting $\ce{CuO}$ with $\ce{Mn}$.
As for water, I know that Manganese metal is one of the more active metals among the transition metals, so I am also concerned with whether Aluminum can be used to reduce $\ce{Mn^{2+}}$ to the metal in boiling water without the formation of any Manganese Oxide or Hydroxide side products due to the water (especially due to the fact that reducing metals like this precipitates them out as a fine powder).
$\ce{ 2Al + 3Mn^{2+} -> 2Al^{3+} + 3Mn}$

Mn has too low redox potential to be deposited from water solutions. H2 is formed instead, from H+ or H2O.

@Poutnik There's an inconsistency with this case. Some sources say manganese don't react with water under normal conditions but somewhere else, it says it does reacts and liberates hydrogen gas.

@NilayGhosh  It has TD and kinetic aspects. I have written it in context of   Mn deposition by metallic Al, not Mn dissolution.

@Poutnik Is this explained by the fact that $\ce{2H2O + 2e^- -> H2 + 2OH^- }$ has a potential of -0.83 which is higher than $\ce{Mn^{2+} + 2e^- -> Mn}$'s potential of -1.18?

Yes, it is. But you have omitted to add standard to the potential, as the potential is pH dependent by 0.059 V/pH unit, if 1 H+ or 1 OH- per 1 e- are involved. Aditionally, potential -0.83 V belongs to pH 14, where are practically no Mn^2+(aq) but Mn(OH)2(s), so redox potential for Mn/Mn^2+ would be much lower than -1.18 V.

Unlike your claim, manganese is relatively inert despite it being a little more electropositive than its neighbors in the periodic table. It is not particularly reactive to air. The surface of manganese lumps oxidize to a minor extent but when it is finely divided, it becomes pyrophoric and burns in air. It tends to form the oxide $\ce{Mn3O4}$ and the nitride $\ce{Mn3N2}$. Temperature for such reactions are reaches more than 500 °C.
$$\ce{3Mn(s) + 2O2(g) ->[\Delta] Mn3O4(s)}$$
$$\ce{3Mn(s) + N2(g) ->[\Delta] Mn3N2(s)}$$
Manganese is also said to not react with water under normal conditions although some sources say it reacts with water to liberate hydrogen gas.
For your thermite reaction, I haven't found any $\ce{Mn-CuO}$ reaction but I did find $\ce{Al-Mn^2+-}$ thermite reaction. You can find the details here.

I have some queries regarding this topic, so I asked: Does solid manganese reacts with atmospheric air when heated? and Does manganese reacts with water at "not" standard conditions? (cc @User1618)


text:
Are there any well documented reactions with "persistent" intermediates
I'm a biology and/or math person, not a chemistry person. I have only taken the standard sequence of undergraduate courses up to organic chemistry, and that was some time ago, so please excuse (and correct) any mistakes that I make in my terminology.

Note: For simplicity's sake, assume that all reactions occur in a fluid, and that if any solid particles form during a reaction, they are small enough (guess: <~300nm, <~2 MDa for liquid, don't know for gas) that they do not effect the fluid nature of the solution.

The typical means of expressing reactions in elementary chemistry is a chemical equation
$$\ce{c_{X_1}X_1 +\cdots + c_{X_m}X_m <=> c_{Y_1}Y_1 +\cdots+c_{Y_n}Y_n}$$
where $\rm c_Z$ is the stoichiometric coefficient of each chemical species $\rm Z$. The time-independence of such expressions subtly suggests an instantaneous conversion of reactants into products and vice-versa, so that no time passes during the rearrangement of atoms from one chemical species into another. This doesn't actually happen, of course, but for practical purposes the time spent in transition states is so small that it has no effect on the properties of a solution.
Or so I assume. But now that I'm thinking about it, I can't see any reason why the constant conversion of reactants into products back into reactants can't leave a small but non-negligible concentration of intermediates in solution at all times, especially if the products and reactants in question are only slightly more stable than the intermediate.
The presence of such intermediates could be verified by checking that the concentration of products and reactants are simultaneously lower than the expected values at equilibrium. This would indicate that the "missing" mass is bound up in some other state, which, provided that no additional [stable] species are present, can only be that of transition between reactants and products.
To clarify - the intermediates are [probably] not stable and cannot be isolated but are being produced and consumed at the same rate in the forward and reverse reactions, so that the concentration of intermediates is consistently high enough to have an observable effect on the properties of the solution.
So how about it, are there any well known mixtures which exhibit intermediates of this type?
Comment: I would expect that most if not all reactions of this type occur under constant heating.
Additional comment: I find it somewhat ironic that the addition of gratuitous "$\frac\partial{\partial t}$"s doesn't eliminate the "instant reaction" problem.

The time-independence of such expressions subtly suggests an instantaneous conversion of reactants into products and vice-versa  Not at all. The reaction equation just describes equilibrium between initial and final state, with any possible time and mechanism to go from one state to the other.

Note that it is not an algebraic equation with applicable differentiation. Rather discrete, atom inventory equation with 2 different ways of atoms being combined to molecules or ions. Like 2 sacks with 2 apples and 1 sack with 2 pears combined to 2 sacks, 2 apples and 1 pear each. What to do there with partial derivatives?

What you describe is exactly the basis of the “steady-state approximation” treatment of reaction kinetics. It is well known that many reactions have intermediates that exist at close to constant low concentrations during the reaction

The equation you give and any similar one, does not describe what happens in a reaction other that X1 moles of A react with xB of B to give this and that amount of product which may or may not be in equilibrium with reactants depending on the reaction.
What actually happens depends on the type of reaction, if it is a cis-trans isomerisation or bond breaking ($X_2 \to 2X$) then the 'intermediate' lasts for little more than a vibrational period of the bond breaking say order of 100 femtoseconds.  As long as the reactant is present then some some minute fraction is as an 'intermediate', usually this is called a transition state and is fleeting.
If reaction is a bimolecular one, with practically zero activation energy,  this will occur as soon as the molecules, say  A and B meet, i.e. the rate is controlled by how quickly the molecules diffuse together, and again the 'intermediate' is the collision complex and will last as long as A-B exists, typically no more that nanoseconds. In other cases where there is significant activation energy the collision complex is similar in all respects except that there is not enough energy to overcome the activation barrier and so possibly a million collisions occur before one molecule reacts.
Thus there are no 'intermediates' as such in simple reactions. However, if the reaction is of the form $A\to B\to C $ then species B can be observed as it forms and decays away, you might consider this as an intermediate.
In more complex cases such as an enzyme /substrate, usually the substrate has to diffuse into the enzyme pocket and then reacts as if a bimolecular reaction with an activation energy. Usually there is always some of this complex intermediate present as (ES) for example in a Michaelis-Menten scheme. (Not a great deal is known about the actual mechanism. i.e atom positions vs time, of such reactions as time-resolved x-ray crystallography is in its infancy.)
In gas phase recombination reactions such as recombination of iodine atoms $I\cdot +  I\cdot \to I_2$ and gas M is present, such as argon, $I_2$ itself  or methane etc. an intermediate forms as a short lived complex with an iodine atoms $I\cdot +M \rightleftharpoons IM$ which has been detected experimentally.


text:
How do I calculate the temperature change in a gaseous reaction?
So, with $\pu{3 mol}$ of $\ce{H2}$ and $\pu{3 mol}$ of $\ce{Cl2}$ and given the reaction $\ce{H2(g) + Cl2(g) -> 2 HCl(g)}$ with $\Delta H=\pu{-184.6 kJ/mol}$, I am supposed to calculate the work done on the surroundings of the system at $\pu{1 atm}$ and $\pu{25^\circ C}$.
My current thought process for this is to calculate the enthalpy change of the system first, which is $\Delta H \cdot n_{\ce{H2}}=\pu{-553.8 kJ}=-q_{\mathrm{rxn}}$. Assuming constant pressure, I tried to calculate $\Delta T=\frac{q}{m\cdot c}$, so I could calculate the volume change through the Ideal Gas law and then calculate the work $w=P\cdot \Delta V$. However, things went awry as I realised I didn't know the heat capacity of $\pu{HCl}$ nor did the exercise give it to me.
So, am I going about this the wrong way or am I just supposed to know the heat capacity of $\ce{HCl}$ off the top of my head?

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Notice \ce{} and \pu{}

There are unit names/unit symbols kilogram/kg and mole/mol. So 3 mol like 3 kg, or 3 moles like 3 kilograms, but no 3 mols. (and none of those are written in italic)

The diatomic molecule $\ce{HCl}$ may be considered as an ideal gas. Its heat capacity at constant pressure is then : $\pu{\frac{7}{2}R}$, with $\pu{R = 8.314 J K^{-1} mol^{-1}}$

It is more complicated due transition from 7/2R J/K/mol region to 9/2R J/K/mol region because of waking up highly quantized vibration contribution.

See temperature dependence of HCl molar heat capacity on Engineering toolbox (generally great source of empirical data). the empirical formula to compute value is included.

See also wikipedia - Molar heat capacity - Frozen and_active degrees of freedom

You start with 6 moles of ideal gas at 25 C and 1 atm, and, in the end you have 6 moles of ideal gas at 25 C and 1 atm.  So what was the volume change?

@ChetMiller Hmm, nice, I have somehow missed the final conditions, what is quite funny, considering I have been editing it.

@ChetMiller The reaction is exothermic. The released heat should increase temperature and expand the gas, shouldn't it?

Not if you remove heat to make the final temperature 25 C


text:
I need to test fabric for arsenic, and I've found 141-year-old guidance. Is it legit?
I'm a book and paper conservator, and a few years ago had four books to restore. Their endpapers were this gorgeous green silk, but they were badly dry rotted, and needed to be removed. I suspected the fabric of having the dye Scheele's Green (copper hydrogen arsenite), and tried to see if my local university would test it for me, but they stopped communicating. I'd like to confirm or refute these steps, so that I can close out this project, and let the customer (a large library) know if their collections house/housed fabric dyed with Scheele's Green.
From the Scientific American, 5 Feb 1881:

EASY TEST FOR ARSENIC IN FABRICS
Immerse the suspected paper in strong ammonia on a white plate or saucer; if the ammonia becomes blue, the presence of a salt of copper is proved; then drop a crystal of nitrate of silver into the blue liquid, and if any arsenic be present, the crystal will become coated with yellow arseniate (sic) of silver, which will disappear on stirring.

This is the extent of instruction provided. It makes sense for it to have been published like this, as this was the period in which much of the Scheele's Green cleanup happened (it was everywhere, but once they understood what was killing people, they tried to rid their houses of it. It was most notoriously used in clothing and wallpaper).
Can anyone confirm these steps? Is it safe (with PPE)? Is there an easier method? I know silver nitrate is caustic, but aside from those handling concerns? And would Ammonia at the common household dilution do the job?

silver nitrate crystals are OK if handled with forceps. Fairly dilute aq ammonia will give the blue colour

Just buy a HACH test kit for arsenic. A colleague has used them extensively for years and they are far more than sensitive enough for what you want.

Thanks Waylander and Ed. I really appreciate the feedback! Ed, I'll check those kits out. I wasn't sure if they were only for groundwater testing, or if they'd work for other arsenic compounds.

My colleague uses the HACH test kits for all sorts of samples, including white rice and brown rice. They have detection limits of 10 parts per billion, which is way below what the green dyed book covers would have, if they have any arsenic at all.


text:
Table of viscosity of water vs temperature

Above is the table for water physical properties table with temperature related with a,v,b,Pr,Ra factor. This table has temperature only from 0 to 100. May I ask if there is a full table for this with a range of temperature from (-100 to 1000'C) on the web? I couldn't find any so I came here to ask for a little help.
Thank you and have a nice day!!

Have you searched CRC handbooks or Landolt Bornstein Hanbooks? There are a lot of handbooks on thermodynamic data. Not everything is online or freely available. A trip to a university library is a must. The range you are looking for is impractically huge. What is the state of water at 1000 C and under what pressure? What is the viscosity of ice?

A lot of similar data can be found in form of tables, charts or calculators at Engineeringtoolbox, this link is particularly for liquid water up to critical T.

If you are looking for the liquid state over the quoted temperature range, you may have a problem below -21°C (the lowest possible liquid temperature, and that requires 200 MPa pressure) or above 374°C (the critical point). Does your table need to include vapor/gas?


text:
Solvent for Terephthalic Acid TLC
We're trying to do thin layer chromatography to assess the purity of Terephthalic Acid. So far we have tried DMF, and Water of different pH values. Does anybody know a better solvent or combination?

I presume this is on silica plates, have you tried chloroform/MeOH, or n-BuOH/water/acetic acid?

Merck has published a Separation of Terephthalic acid and Reaction Products data sheet at https://www.google.com/search?q=chromatography+of+terephthalic+acid&client=safari&rls=en&sxsrf=ALiCzsYOCTterkWy3HuBIy-log_KoLr1uA:1652382873353&ei=mVx9YoeWFfSzkvQPwoqX8AM&start=10&sa=N&ved=2ahUKEwjHxsvx1dr3AhX0mYQIHULFBT4Q8NMDegQIARBM&biw=1146&bih=836&dpr=1  which specifies CH3OH/CH3CN/H2O (20/30/50) on Silica Gel 60 RP-18 WF254s.


text:
If they both form hydrogen bonds, Why does borax make polyvinyl alcohol slimy/elastic but water doesn't?
When dissolved borax (borate ions) and polyvinyl alcohol are mixed, an elastic slime is formed. To my understanding, this is because the borate ions form hydrogen bonds with the OH groups on the PVA, cross-linking the polymer chains and increasing elasticity.
However, on a page discussing slime (https://www.acs.org/content/acs/en/education/whatischemistry/adventures-in-chemistry/experiments/slime.html) the American Chemical society states that "Water is an important ingredient in slime. Water helps the polymer molecules slide past each other so that your slime can flow. "
Water would presumably also form hydrogen bonds with the PVA polymer - so my question is, why do borate ions increase elasticity while water increases liquidity?

These are not hydrogen bonds, but true covalent bonds C-O-B, similarly as with glycerol, soluble sugars or starch (with the latter used for borax slime as well).

related https://chemistry.stackexchange.com/questions/140154/how-can-i-solubilize-fragrance-oil-into-polyvinyl-acetate-clear-glue https://chemistry.stackexchange.com/questions/48671/how-to-estimate-the-number-of-hydrogen-bridges-in-a-bouncing-ball https://chemistry.stackexchange.com/questions/71419/the-safety-of-homemade-slime/72413