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Relationship Atmospheric Gases and Ocean Gases
If this question is better suited for the Earth Science StackExchange or another StackExchange please let me know.
Edit 1: Modified for clarity and brevity.

Something that's unclear to me is the exact relationship between gases within an atmosphere and the concentration of those gases within a large body of water under that atmosphere (like an ocean).
For example, the solubility of $\ce{CO2}$ at $298\ \mathrm K$ and $1\ \mathrm{bar}$ of atmospheric pressure is $\approx 1.496\ \mathrm{g/L}$ in water. My understanding is that number indicates how much $\ce{CO2}$ can be dissolved within water before the water is saturated with $\ce{CO2}$ and won't accept any more. However, this doesn't tell me what the concentration of $\ce{CO2}$ would be under given conditions, it just sets a maximum cap. For one thing it could be much less because there isn't enough $\ce{CO2}$ in the atmosphere to saturate the ocean.
(I hope I have that correct! If not please let me know where my understanding is going astray.)
Key Question: What are the calculations I need to go through to figure out the amount of gases in a body of water under an atmosphere of a given gaseous composition and pressure?
We could clarify this question through looking at an idealized situation (below). I'd like to know the steps to take to figure this out. If we need to ballpark it then that's ok, though I'd prefer to keep it as reasonably accurate as feasible.
Scenario
Assume uniform temperatures at water surface. Assume no minerals dissolved in water. Assume no gases have yet dissolved in the water from the atmosphere.
SURFACE

Total surface area: 510 million $\mathrm{km^2}$
Water surface area: 361 million $\mathrm{km^2}$
$15\ \mathrm{^\circ C}$ uniform surface temperature

ATMOSPHERE

By volume: $99.5\ \%$ nitrogen, $0.45\ \%$ carbon dioxide, $0.05\ \%$ methane
Surface pressure: $1\ \mathrm{bar}$

OCEAN

Pure water ocean (nothing yet dissolved in it)
Volume: $1.4$ billion $\mathrm{km^3}$

Q: What will be the equilibrium composition of gases in the atmosphere and in the water?
Q: How much of each gas will dissolve into the water?
… and of course how you got there.
What happens when you increase pressure and decrease temperature? How would that effect the answer?
What I really want to know is both general principles as well as establishing those principles in actual numbers. I'd also love any insight you may have on how real-world numbers are likely to differ and why.

I have a felling maybe you better keep to Worldbuilding.SE Your questions are rather broad and speculative - standard  for WB but not on other sites.

CO2 content is effected by temperature, pressure, ionic strength,  and acidity. The ocean content is therefore not homogeneous. To figure out a decent number is a complex problem.

@Mithoron Hmm...I don't feel this question is speculative. It's asking to show methodology for figuring out what should be the concentration of gases in water given the gases above that water. I don't see where in that there is any speculation. Here I've provided a simplified set of conditions to reference b/c I felt that would make it easier. Based on my pursuing of Chemistry SE is seemed this was the right place to go to ask for calculation-heavy answers. How would you recommend editing this question to better fit the spirit of Chemistry SE?

@MaxW Well, I've outline temperature and pressure. Water is pH of 7 I think. Nothing is yet dissolved in the water (simplified scenario). We could assume a homogenous temperature in the water volume if it would help?

If this question is too complex, would simplifying it into a format more like this one help? https://chemistry.stackexchange.com/questions/99550/diffusion-of-co2-and-o2-in-water-atmosphere-system?rq=1

Well, you are speculating what would be CO2 in ocean, under different atmosphere - your simplifications make in more precise, but can't fool me ;) Here you're making it look overly broad, while there's no actual calculation needed for your main point - you just need to get solubility curves and read values!

@Mithoron I don't actually understand the relationship. I've seen solubility curves and can tell you wha the solubility is of various gases under different pressures, but that doesn't tell me how much of each gas will be dissolved in an ocean, only how much could be before saturation. I assume the water will not necessarily saturate with all the gases? I don't know. Hence my question. It's inherently a mathematical/chemistry question hence my desire to see numbers. Feel free to substitute any gases you'd like and any atmospheric pressure and temperature above zero.

These things are enormous. One atmosphere at sea surface and km of water column at bottom. CO2 might comes from erupting phenomena. And so on. Better suited for planetology oceanography....

Q1: The equilibrium concentration c of a gas in a solution (typically water) which is present in the atmosphere with partial pressure p dictated by Henry's Law:
$H_{cp} = c/p$
where $H_{cp}$ is Henry's coefficient for that specific gas.
This law states that the concentration of the gas in the solution is directly proportional to its pressure in the atmosphere above the liquid phase and it is a simplification generally applicable to inert, ideal gases.
Corrections must be taken into account when considering the following factors:

Temperature: The enthalpy of dissolution of the gas changes according to van't Hoff's equation. The general result is that the solubility of most gases in water decline with temperature (i.e. solubility is inversely proportional to temperature).
Chemical equilibria: Any transformations (ionization, deprotonation, etc.) that happens to the gas in solution must be taken into account by expressing its concentration by the according equilibrium equation.
Ionic strength: Generally the higher the ionic strength of the solution, the smaller the solubility of the gases in it (Sechenov equation).
Non-ideal solutions: Instead of concentrations, activities of the substances must be considered.

Q2: How much of each gas will dissolve into the water?
The calculation of the concentration for carbon dioxide, for example, is:
$$c = [(0.45\%)\times(\pu{0.986923 atm})]\times[\pu{3.4 \times 10^{-2} M/atm}] = \pu{0.0001509957 M}$$
This calculation is a direct application of Henry's Law, where the first term is the partial pressure and the second is Henry's constant. The same type of calculation goes for the others.
Observation:

It is important to notice that those volume percentages given should already be considering measures taken for those gases in our atmosphere, so they are already taking in account the solubilization. The concentration of each gas in the atmosphere can be calculated by the ideal gas equation, where the number of moles divided by the volume will be the concentration.

Sorry that I don't know how to make sub and superscripts/ make equations!

Hi Pedro, this might help with the equations: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference?rq=1

"The general result is that the solubility of most gases in water." Is this statement incomplete?

Thanks @FelipeS.S.Schneider for the link! Really appreciate it.

@KeithMcClary the sentence was indeed incomplete. I edited it to make it right. Thanks!


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What kind of metal drums are used to store Hydrochloric Acid?
I would expect Hydrochloric Acid to be stored exclusively in glass or plastic containers, but I was surprised to find Honeywell offers it for sale in 30Kg metal drums (Link).
I think Hydrochloric Acid would dissolve Nickel, Chromium and Iron in stainless steel, which leaves me wondering what kind of metal that could be, and whether this form of storage wouldn't (at least slightly) contaminate the acid?

We might as well consult the experts, namely storage tank manufacturers.  This site describes several types of storage tanks, none of them metal except for rubber-lined steel, in which (we hope) the rubber keeps the acid out of contact with the steel.  Rubber lining of steel is often used in processes such as acid rinsing tanks where the steel is needed for structural purposes, but for a storage tank as such the referenced site renders cross-linked polyethylene better.

It's interesting (and nice in a way) to see natural rubber still being used for such applications nowadays, though I'm not sure how they deal with quality consistency using such natural products for sensitive applications. Do you think that the acid in the 30Kg drum will have the exact same quality as that in the glass bottles? Or is the acid in the glass bottles coming from a drum anyways?

@Hans - given that the HCl was likely bulk-shipped in a railcar (steel with a rubber lining, see http://www.gatx.com/wps/wcm/connect/GATX/GATX_SITE/Home/Rail+North+America/Products/Equipment+Types/Tank/Acid+and+Specialty/Acid/), likely the bottles and drums were both filled from bulk storage at a distribution facility.

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How do I determine molality from mole fraction without knowing how much of the solution I have?

What is the value of $x$ if $x$ molal solution of a compound in benzene has a mole fraction of solute equal to 0.2?

I tried calculating the moles of $x$ through the value given but there are two variables, i.e mole of $x$ and mole of benzene so I am stuck.

Molal is moles solute per kilogram solvent. The mol fraction is moles solute per mole solvent. In your case, that means you have .2 moles of solute per 1 mole of benzene. What could you do to get that amount of benzene on kilograms instead of moles?

Solution 1
Let's assume there is 1 mole total of solvent and solute. With a mole fraction of 0.2 solute, there is 0.2 mole of solute, the rest (0.8 mole) is benzene. We can calculate the mass of 0.8 mole benzene from the molar mass, and express it in kilograms:
$$ m_\mathrm{benzene} = n_\mathrm{benzene} \cdot M_\mathrm{benzene}$$
$$= \pu{0.8 mol} \cdot \pu{78.11 g/mol} = \pu{62.49 g} = \pu{0.06249 kg}$$
The molality $b_\mathrm{solute}$ is defined as the amount of solute divided by the mass of solvent:
$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{0.2 mol}}{\pu{0.06249 kg}} = \pu{3.201 mol/kg}$$
Solution 2
Let's assume there is 1 kg of benzene. That corresponds to 12.80 moles of benzene. If the mole fraction of solute is 0.2 and that of the solvent benzene is 0.8, they are present in a molar ratio of 1 to 4 (for every four molecules of benzene, there is one molecule of solute). So the amount of solute in that sample is 3.201 mole. We are ready to calculate the molality:
$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{3.201 mol}}{\pu{1 kg}} = \pu{3.201 mol/kg}$$
Problems lacking extensive quantities
It is helpful to understand the difference between extensive and intensive quantities. Intensive quantities do not depend on the size of your sample (examples include density, concentration, molar mass, partial pressure, coefficients in the formula of a compound). On the other hand, extensive quantities do depend on how much you have (mass, volume, amount of substance). Often, you define intensive quantities by dividing one extensive quantity by another (example: density is mass divided by volume).
In your question, you want to determine an intensive quantity (here: molality) from intensive quantities only (here: mole fraction). There is a larger set of exercises of that type. If it helps, you can arbitrarily come up with an extensive quantity to solve the problem. You can set the mass of benzene to 1 kg or 2 kg. Then, you can solve the problem. The answer will not depend on the mass of benzene, it cancels out. Other options are setting the amount of solute or the amount of benzene.
There are common exercises asking you for intensive quantities given other intensive quantities:

What is the formula of an oxide given the mass percentages of elements?
What is the density of xenon at STP given its molar mass?
What is the concentration of chloride ions in a solution containing 0.4 mol/L $\ce{MgCl2}$?
What is the partial pressure of dioxygen at STP if an air sample contains 20% of dioxygen?

For all of these, you may assume an arbitrary amount, volume, or mass (but just one) if it helps to solve the problem. Again, the actual value does not matter because it will cancel out.
For more on intensive and extensive, see e.g. https://chemistry.stackexchange.com/a/6538

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Why would ferrofluid harden?
I recently decided to make my own ferrofluid toy to play with.  I built a sealed enclosure at my local fablab, and the end result was perfect.  It worked just like in the videos you see online, and I couldn't have been more pleased.
But then, after a week or so, I noticed that the fluid was moving a bit more sluggishly, and not long after that, it would barely react to the magnet at all.  Now when I try to play with it, I have to press a big magnet right against the surface, and then the "fluid" moves completely uniformly -- less like a fluid and more like a big black rock.
The only thing I can think is that maybe the suspension fluid I used had a gradual drying-out effect on the ferrofluid.  When I was originally building the thing, I did some googling to find out how best to do it all, and I found an old post on this very forum, suggesting the use of isopropyl alcohol, so that's what I used.  I also used the same type of ferrofluid as the OP in that thread.
So would a different type of suspension fluid keep it from hardening like this? Or am I wrong about the cause?  The only other thing I can think of is that maybe it's because there's a little air bubble in my enclosure.  It can't be any external factors, because it's sealed completely airtight.
Any ideas?  Thanks!

The manufacturers of the ferrofluid actually provide a phone number. It may be a little old-fashioned but you can always try calling them and asking them if the solvent you used is incompatible.

How about I don't have to click on a link, where I can then guess what other link you mean, to find out what you did? ;-)

Sorry — let me clarify: I’m using FerroFluid EFH-1 by FerroTec - amazingmagnets.com/show-decimal-ferrofluid-0250.aspx


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Bonding in the nitrate anion
Hi so I was studying chemical bonding where i encountered a problem which is stated below.
When we talk about ${NO_3}^-$ we draw its structure as following

But the thing which I dont understand is that in two of the O atoms there are three lone pairs(ie: $6 e^-$) but as we know from O's $e^-$ config. (which is $1s^2 2s^2 2p^4$) there are two unpaired $e^-$ which will take part in bonding as there is no vacant orbitals for $e^- $ to get excited to . Now if in this O atom only one sigma bond is present then technically we are left with $5e^-$ and not $6e^-$ which is shown in the figure.
Similarly in the N atom (whose $e^-$ is $1s^2 2s^2 2p^3$ ) does not have any other vacant orbital to excite its $e^-$ thus showing that it can make only 3 bonds then why is it making 4 bonds( 1 pi and 3 sigma) ?

The problem lies in the electron count. Each bond contains 2 electrons and the pair is counted twice, once for being around each atom in the bond. And don't forget the extra electron to make the ion.
The + and - are from the point of view of the atom nucleus: e.g., consider the nitrogen to share 2 electrons with the oxygen on the right, to give 2 to the oxygen above, and to give 1 to the oxygen on the left. The other electron in that bond is one that makes the anion negative overall.
Then nitrogen gains its octet of 8 electrons, but formally only shares 50% of those electrons, so it started out with 5 but only controls 4 - it now has a positive charge. The right oxygen shares, so no charge. The top oxygen got 2 from nitrogen, so is negative, and the left oxygen got 1 from nitrogen and one from the metal of which this is the anionic part. And then, draw up the other two diagrams that have the "neutral oxygen" in the other positions, add them all and average them. Then each oxygen winds up with a 2/3 negative charge (but the nitrogen is still +1), while the anion is -1 and planar.

I think you should not consider the orbital theory in every detail, it just makes things overcomplicated. You have nitrate which is very stable and there is a reason for it.
In a non oxidizated state you have 6 valence electrons (outer s and p orbitals) for oxygen. Oxygen has a higher electronegativity than nitrogen and therefore it takes one electron and gets reduced to -1 to have a full filled p orbital (8 valence electrons). The other oxygen takes 2 electrons and the oxidation state does not need to be changed. The oxidized nitrogen (+1) is more stable because of the half filled p orbital (Hund rule). Pi bonds are only made between  p-p orbitals and therefore can only exist in the double bond because the s-p bond is prefered (1 overlap) and more stable.


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Open Babel: conversion to SMILES. How to delete H?
OpenBabel is a convenient converter between different chemical formats.
If I convert to SMILES, however, although I checked the option "Delete hydrogens (make implicit)".
I still have H in the SMILES codes.

It's the same with the command line. It also does not remove H.
obabel.exe Molecule.log -osmi -d

I still have, e.g. ...C[CH]C... in the SMILES.
Of course, I could replace all occurrences of [CH] by C, but I'm not sure whether I would mess up the SMILES and/or the molecule.
How can I get rid of the hydrogen? Why is it actually in there? Did I miss any other options during conversion?

Provide some example molecules maybe?

When I do obabel -:"[CH4]" -ofpt -o SMILES the output is C, so I'm not sure why the command line (at least) isn't working for you.

@Curt F., right, an example is missing. I cannot place the Gaussian file here. Which exact section in the Gaussian output file (start/end keywords) is actually extracted by OpenBabel?
Well, for a demo try caffeine CN1C=NC2=C1C(=O)N(C(=O)N2C)C which gets CH3 drawn. But in the meantime I found another option in OpenBabel "do not draw terminal C(and attached H) explicitely". With this it looks fine.
However, with conversion from Gaussian output file I also once got two double bonds to Nitrogen. I'm not a chemist, but this sounds wrong to me. I need to dig out an example which I can show.


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Can precipitation reactions attain equilibrium?
While studying about the characteristics of Chemical Equilibrium, I came across a point in my book which read

A chemical equilibrium can be established only if none of the products is allowed to escape out or separate out as a solid (precipitate)

I understand that if one of the products is a gas and escapes out the reaction won't be able to proceed in the backward direction since that gas is no more a part of the system.
But what I'm not able to get is why it's written that precipitation reactions can't attain equilibrium.
Is there no precipitation reaction which can attain equilibrium?
Or is it that the precipitate can't react back chemically with the solution to produce the reactants in reverse?
My teacher told me about about a reaction
$$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$$
and on cooling the solution and bubbling $CO_2(g)$ through it the $CaCO_3(ppt.)$ reacts back to form calcium bi-carbonate.
I'm still a bit confused as the reverse reaction doesn't happen spontaneously. Moreover if the reaction is carried out in a closed vessel and constant temperature (after heating), will the reaction reach equilibrium?
edit
in the answer recently provided, the solid-solution equilibrium between $\ce{AgCl}$ and $ \ce{Ag+}$ $\ce{Cl-}$ ions is discussed. But I want to know if in a reaction like $$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$
the precipitate $\ce{AgCl(s,ppt)}$ can react with $\ce{NaNO3(aq)}$ in the solution to give the reactants?
Can it reach chemical equilibrium?
I've seen this and this but couldn't understand.
Any of your help is of great value to me.

They simply confused you. Solubility of ionic salts or in general dissolution of solid is not an equilibrium covered by their assertion. They simply means that the precipitation of a solid can shift the equilibrium of a reaction occurring in solution . The same would be for dissolution if you would stop your analysis to individually solvated particles,  eg solvated molecule of sugar.

@Alchimista Isn't solid-solution equilibrium a thing?...like dis-solution and precipitation?... Moreover I want to know if there are such reactions where the precipitated product can react back chemically with other products in the solution to produce the reactants under the same conditions? Thank you!

Related: https://chemistry.stackexchange.com/q/118480/48509

I said related. You can find what are you looking for, there.

You are right about the fact that there is precipitation equilibrium. Perhaps I weren't clear but it is why I said they confused you. The link you gave as well the much recent question that I have mentioned cover your original doubt. The editing you did meantime seems relates but another question . Yes that would reach equilibrium, if the vessel confines the chemicals in it.

As Karsten explained in his answer, the statement is incorrect.  What textbook are you using?  Better yet, could you also post a screenshot of the page and/or paragraph in which that text is found?

The statement is inaccurate. A better statement would be something like:

A chemical equilibrium can not be established if one of the products is continuously removed (i.e. its concentration always decreases).

If one of the products is a gas and can mix with earth's entire atmosphere, that would make its concentration (or partial pressure) drop to almost zero. In order to establish an equilibrium, however, all concentrations have to be non-zero. Otherwise, the equilibrium constant will not be equal to the reaction quotient, and there will continue to be a net reaction.
In case of the precipitation, the solute is not continuously removed. At some point, the solution will no longer be supersaturated, so then the concentration of solute will be constant.

My teacher told me about about a reaction
$\ce{Ca(HCO3)2(aq) + heat -> CaCO3 +CO2 +H2O}$
and on cooling the solution and bubbling CO2(g)
through it the CaCO3(ppt.) reacts back to form calcium bi-carbonate.

Once you change the temperature, you don't expect the reaction to stay at equilibrium. So this does not support either the original statement or the opposite statement.

But I want to know if precipitation reactions can attain equilibrium. Why does the book state otherwise?

Yes, it can. Here is an example:
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$
The equilibrium constant expression for this reaction is:
$$K = \frac{1}{[\ce{Ag+(aq)}][\ce{Cl-(aq)}]}$$
(I could have written the chemical equation in the other direction, and then the equilibrium constant would have been the solubility product.) So the product is a solid, and its concentration does not change. The silver and chloride ion concentrations will drop until the equilibrium has been reached.
Here is another set of reactions:
$$\ce{CoCl4^2-(aq) + 6 H2O(l) <=> Co(H2O)6^2+(aq) + 4 Cl-(aq)}\tag{1}$$
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}\tag{2}$$
Here, the equilibrium of the first reaction will be far on the right side because the second reaction removes a lot (but not all) chloride ions from the solution if the silver ions are in excess of the chloride ions. Still, the system will reach equilibrium.

Why does the book state otherwise?

I don't know. Maybe they consider an equilibrium where one of the species is at extremely low concentration not an equilibrium. Or maybe it is a typo.
It is great to have a critical mindset when reading books and hearing people talk about chemistry. Even if all the statements are true, it helps you learn.
edit

I want to know if in a reaction like
$$\ce{AgNO3(aq) + NaCl(aq)\to AgCl(s,ppt) + NaNO3(aq)}$$
the precipitate $\ce{AgCl(s,ppt)}$ can react with $\ce{NaNO3(aq)}$ in the solution to give the reactants?

Yes. I would rather write it like this, though:
$$\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) <=> AgCl(s) + Na+(aq) + NO3-(aq)}$$
and then leave out the spectator ions:
$$\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$$
The following "reaction" might serve as illustration why leaving the spectator ions in the equation is not a good idea in preparation for writing the equilibrium constant expression:
$$\ce{KCl(aq) + NaNO3(aq) <=> KNO3(aq) + NaCl(aq)}$$
For this "reaction", the concentration of species is undefined. Writing the equilibrium constant would be strange because how can I tell apart $\ce{KCl(aq) + NaNO3(aq)}$ from $\ce{KNO3(aq) + NaCl(aq)}$ if they fully ionize in solution? If I show the ions separately, it becomes clear that the "reaction" is not a reaction:
$$\ce{K+(aq) + Na+(aq) + NO3-(aq) + Cl-(aq) <=> \mathrm{the\ same???}}$$

I've seen this and this but couldn't understand.

Those two Q&A's are quite confusing, I agree.

Thanks....I upvoted you. But I want to know if precipitation reactions can attain equilibrium. Why does the book state otherwise...please tell me...i get the escaped gas point...the ppt. point is troubling me... thanks

@user8718165 I added your question in the comment to my answer.

@user8718165 I updated my answer to respond to your updated question. Next time, maybe ask a new question.

Thank you so much. Now I fully understand it. I upvoted and accepted your answer. Hope I didn't cause you much trouble.


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Could such a galvanic cell exist
Imagine I have a galvanic cell where in one half cell there is a lump of potassium metal and molten KOH electrolyte, and in the other an inert electrode, as well as a molten KOH and the less active metal salt dissolved in it. Would the fact that the more active metal cations are present in the reduction half cell interrupt the process?

The key question you are trying to formulate is that would potassium metal reduce molten salt of a less active element or not. If it can, the galvanic cell will work. What salt do you have in mind?

The potassium ions in the reduction half-cell are inert, merely serving as part of the electrolyte.  The are no problem at all.  Indeed, you actually need to allow for the passage of more potassium ions through your separator, to balance the potassium metal that's dissolving and make up for the cations being plated out at the cathode.


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In a half cell reaction, Is the potential halfed?
If you have an electrolyte of $\ce{Zn^2+}$ and $\ce{SO4^2-}$ (Neutral) Then you dip in it a Zinc electrode.
Zinc potential is $\pu{-0.74 V}$ (relative to SHE). So that means there is a $\pu{-0.74V}$ difference between the electrolyte and the electrode.
But When you dip in the solution the electrode makes the solution more positive and it self more negative the same amount so the overall potential of the electrode will be $\pu{-0.74 V} / 2 = \pu{-0.37 V}$, and the solution will be at $\pu{+0.37 V}$.
Is this true? (assuming the solution is not connected to any other salt bridge or porous disk)
Edit: Absolute electrode potential is really the difference between the electrolyte and the electrode potential.
And the question is just this if you have an electrode inside an electrolyte with no salt bridge will it's potential be halfed.

Possible duplicate of Deriving a reduction potential from two other reduction potentials

Please remember from physics 001 that any excess charge resides on the surface (=or in this case interface) of a body. The bulk of your electrode and solution are (in first approximation) charge-free.

In addition, you couple your electrode with an other (for example with a salt bridge) to close an electric circuit.  While there are bottles of water, and grinders filled with pepper, there is no magic box in your cabinet «just so» filled with electrons.

Those $\pu{-0.74V}$ are measured against a standard hydrogen electrode. That's the universally accepted but totally arbitrary standard.
You don't have a second electrode, hydrogen or other, so the absolute number is totally meaningless. If someone had decided to use the zinc electrode instead, as zero point for the electrochemical series, then the voltage you mean would be zero. Makes no sense.
There is an absolute cell potential (absolute electrochemical series), with a different zero point, but it's totally uncommon, because it's very tricky to measure absolute electrostatic potentials of only a few hundred millivolts. I believe the problem is the calibration, while a platinised platin electrode bubbled with reasonably pure hydrogen is probably the simplest reference standard to reproduce at all.

It is a very good question, and I have wondered about it for years, talking to highly seasoned electrochemists, only to find that nobody knows the answer! The sole reason is that the problem is overwhelmingly complex despite its deceptive simplicity. Nernst offered him theory similar to osmosis, but it did not work- 100 years ago.
However there are couples of misconceptions you have here. The first one is that when you dip a zinc electrode in a solution containing zinc, the potential difference is NOT -0.74 V. Nobody in this world can find out the potential difference between the solution and the zinc electrode. The moment you attach a probe, it introduces another potential difference, an unknown to another unknown. This is somewhat akin to (but not at related) to the Heisenberg's principle. The moment you try to probe the momentum of an electron, its position becomes uncertain.
So your arithmetic in invalid to begin with. I don't have my 2-vol Bockris, but it did have a very intuitive explanation of single electrode potentials.
What you can determine indeed, is the sign of the charge developed on the electrode wrt to the ground. Again this sign is not absolute, it is with respect to the Earth, which is "considered" to be at zero potential i.e. there is no charge build-up on the Earth. Who know if this is right or wrong, but this is the convention. There are very sensitive electroscopes now which can tell you the sign of the charge, albeit very small.

There is a huge voltage (~.5 MV or so) between ground and the ionosphere. "Ground" is everything, but not absolute zero. Measuring charges is however not exactly rocket science, piezo transducers are read out that way for example.

Of course, but for earthly measurements and by electrical engineering conventions, ground is considered at 0 V as a reference. Vernier markets a charge sensor, and if we measuring the "sign" charge of an isolated object, the other terminal of the electroscope is connected to the ground (say a water line etc.). In British terminology this the same as earthing. It is indeed arbitrary.


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Value of the frequency factor in the Arrhenius equation
Is the value of 'A' necessarily the same for two different reactions taking place at the same temperature?
It's actually a part of another question which is if the rate constant k1 of a reaction is found to be double that of rate constant k2 of another reaction the relationship between corresponding activation energies of the two reactions at the same temperature (e1 and e2) can be represented as e1__e2

The Arrhenius constant, A, does not have to necessarily be the same for two reactions taking place at the same temperature, due to the fact that it depends on the order of a reaction as well as the frequency with which successful collisions will occur during a reaction.
As for the second part, as k ∝ x-Ea, for 2k to be a larger value, Ea for 2k would have to be less than the one for k, i.e. e1 < e2 if the activation energy for 2k is e1 and that for k is e2.

That would only be told for sure if A for both the reactions is the same. But nothing about it is revealed in the question.

That is very true. Unfortunately there are ambiguous questions that make it difficult for one to solve. My best bet would be to assume that the A is equal for both reactions, unless they state otherwise. As the value of A can be equal for two different reactions, but does not always have to be. Hope this helps :)


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Why does Comfrey Break Down Easily?
comfrey (Symphytum officinale) is a well known plant to organic growers. One of its features is that the leaves break down very quickly, either in water or in a container closed from UV light. There must be some chemical explanation for this, can you explain to me what that is ?
Thanks Erin

Is this observation anecdotal or can you provide a reliable reference? How is this different from other plants? Many leaves wilt quickly, there is nothing necessarily singular about comfrey. And, what is the link to chemistry. Do you think there might be some biochemical feature as opposed to simple dehydration? What do you use as a measure of "breaking down"?

Welcome to chemistry SE. Questions are expected to be related to chemistry, and it is not clear yours is. You may have difficulty finding a botanist with expertise on comfrey at this site, particularly if, as you write, there is not much information on it. What do you mean by "the chemical structure is well known"? Chemical structure of what exactly?

https://pubchem.ncbi.nlm.nih.gov/compound/Comfrey#section=2D-Structure

This is a chemical question - link above !

Ok, but you may want to distinguish between comfrey the plant and a particular compound ([7-[(E)-2-methylbut-2-enoyl]oxy-4-oxido-5,6,7,8-tetrahydro-3H-pyrrolizin-4-ium-1-yl]methyl 2-hydroxy-2-(1-hydroxyethyl)-3-methylbutanoate) within the plant, which the link you provided refers to. Why would breakdown of the plant be related to this particular compound?

This is interesting information, but you need to provide further background that clarifies what you are asking- it's like claiming that caffeine is somehow linked to the breakdown of leaves in the coffee plant in water. Why should it?

I don't know that's why I am asking here, but am moving on because I am getting no where here.

You are more likely to make progress by contacting the local agricultural outreach department of a state university. For instance, look at the link that @M.Farooq provided, it's full of information that might explain physicochemically why the plant degrades rapidly, including the fact that it's rich in nutrients (N, minerals), which is why it was considered a promising forage plant. Also, it can be used to isolate allantoin, which has cosmetic applications. But it also has teratogenic (toxic) compounds which is why it is not highly recommended for human or animal consumption.

Erin, the main the problem with the question is apparently this part "One of its features is that the leaves break down very quickly, either in water or in a container closed from UV light." You also mention the compound comfrey and wonder if this is responsible for a quick decomposition? I don't think so.
The key question is that Comfrey plant decomposes quickly, as you observed, but compared to which plants? I don't think it is unique that it decomposes rather quickly in soil, water etc. Don't other plants like spinach leaves decompose or spoil quickly? You can easily eliminate UV because UV cannot penetrate compost. It must be bacterial / fungal action.
You can use Google Scholar to fetch scientific papers only and one of the scientific papers says "Comfrey on a dry weight basis is very high in ash, averaging 18 percent at Rosemount and ranging from 13 to 42 percent in other parts of the world (4). A partial
elemental composition of comfrey forage is reported in Table 4. The forage is particularly high in potassium and is higher than many other forage crops in calcium,
phosphorus, iron, and copper. Carbon, hydrogen, oxygen, nitrogen, and sulfur are the major non-mineral elements. The combined high nitrogen and high mineral concentrations make comfrey forage an unusually good material for composting, mulching, and organic fertilization of crops"
From Comfrey: A Controversial Crop https://conservancy.umn.edu/handle/11299/141442
The high nitrogen content could be reason for its quick decomposition by bacteria or fungi, however it is hard to assign a single compound "comfrey" as being responsible for it.

Thanks M.Farooq for having a look around the question but it is a long way from being resolved. If you take 'wood' for instance, I understand that it's C-C bond give it certain properties, and some shells have a structure with silicon bonds which have similarities. Now consider the bonding in the 'comfrey' leaf. If you are a gardener the speed with which comfrey breaks down compared to any other plant is notable. You mention nitrogen bonds, I'd be inclined to concur but I haven't really got any further. Are these bonds particularly weak ?

One can only wish if living things were that simple. Just an example: No one still knows the total analysis of human blood. There are thousands and thousands of molecules in the blood. There is a whole field in chemistry called "-omics" in which people try to know each an every component using million dollar instruments. In the same way, no one Earth knows what is the total composition of a leaf, that too of comfrey which is perhaps not that important to chemists. contd....

contd...Therefore it is invalid to consider the "bonding" in a comfrey leaf. As you know composting is an action of bacteria or fungi. If they did not exist, comfrey leaf could take ages to break down. Some things decompose faster because bacteria and fungi and consume it faster, and some take a while. Wood has tons of compounds as well, the story does not end knowing that there in C-C bonds, it is perhaps not even the beginning. It is just that wood is not that nutritious for bacteria and fungi, otherwise there would be no wood-houses. All of them would have composted by now.

No offense to you M.Farooq you did your best.  But I think generally the moderators on this site need to take the rod out of their asses. They are complicating this question because of their anal retentiveness, get a life - time wasters.


text:
Enthalpy of formation at other temperatures
I was told each compound has an enthalpy of formation of 0 at the standard temperature if it is in it's lowest energy modification. But what if we change the temperature? It would make sense to me that we would also need to have a non-zero enthalpy of formation then, like for Hydrogen at 400K. Is that the case? Many exercises I had to calculate $ \Delta H_r$ in had no enthalpies of formation given for $H_2$, even if all compounds were at 400K.

The enthalpy of formation of each compound is equal to the enthalpy change from the elements in their natural states at the temperature under consideration.  If you work it out, this will enable you to get heats of reactions at the temperature without correcting for the sensible heat changes of the elements relative to the standard temperature.

What do you mean by "each compound"? Maybe you mean "each element"?


text:
Dissolvant for the black spots around the rear mirror of the front window of my car
I make pictures for Mapillary.
I would like to place my dashcam as high as possible, but those black spots get in the way.  Scraping them off does not work.
But I can feel their relief.
If no one has a better idea I will begin the trials with a sticker remover.
I do not find the exact word for those spots any more.

The part of the windshield that you are talking about is called the frit. This is typically made up of some form of high-heat enamel, and as far as I was able to find, it is intended to increase the surface roughness of the inside of the glass to increase the ability for the adhesive to stick to it.
I was not able to find any specific mention for what type of enamel frits are actually made of - that is whether it is an enamel glass or an enamel paint -  and as such I cannot give any specific recommendations for what solvent to use. If it is enamel glass, as I would hazard to guess it is - given that some sources indicate that it is applied prior to the windshield being bent - then you will likely not be able to remove the frit without also damaging the glass. If it is an enamel paint then you may be able to remove it with an organic solvent. In general, I like to start with acetone as a general organic solvent.
All of this said, given both the purpose of the frit, and its proximity to the adhesive that holds your windshield on, I strongly recommend that you consider moving your dashcam rather than attempting to dissolve the frit.
Best of luck.


text:
How do you name this compound using IUPAC system (including steps)?

And including stereochemistry (cis trans or R S )
I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question, so now it should be 3-bromo-3-methyl-1-hexene. My quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)
I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.
trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?

Cis/trans is irrelevant; no stereochemistry on the double bond. Apply CIP rules to get R-configuration at C-3.

Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.

(3R)-3-bromo-3-methylhex-1-ene.

When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:

Solid lines are on plane of your paper;
Wedges are above the paper; and
Dashed lines are below the paper.

That means, $\ce{-Br}$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $\ce{-Br}$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.


text:
Pyrotechnic colorants- reactions & safety concerns regarding making charcoal discs
I want to create my own charcoal disks/tablets which will be mainly used for incense and I want each of the charcoal disks to glow different colours (eg one disc glows blue, another glows red etc.). The product will be using charcoal and other ingredients including Arabic gum. I will need to add certain chemicals such as metallic salts and so forth to create the desired aesthetic effects such as the colours.
My main concern is the safety of the product, I would like to experiment to see if any of this will work but most of all, I want to know if the end product will not become a dangerous fire hazard and that it will be non-toxic.
I’ve looked up some of the chemicals online but I want to be sure.
I know that you can add metal salts to flames in bonfires to add a similar effect.
(Example here: http://tiny.cc/hqjv9y and in fireworks displays. Can I have that similar desired effect with the charcoal disks as well as the safety with it?
Below are some of the chemicals that I’ve found and would like to know if any of them are safe to add to creating my product:

For a blue flame, you want Copper Chloride or Calcium Chloride
For a light green flame, you want Borax
For a dark green flame, you want Copper Sulfate or Alum
For a strong, red flame, you want Strontium Chloride
For a purple flame, you want Potassium Chloride
For a pink flame, you want Lithium Chloride
For an orange flame, you want Sodium Chloride
For a white flame, you want Magnesium Sulfate

Some chemicals also make coloured sparks, instead of colouring the flame:

For gold sparks, you want Iron (fillings)
For silver sparks, you want (powdered) Aluminum

Source here: https://www.quora.com/What-chemicals-can-change-the-color-of-fire
Other sources: https://www.thoughtco.com/how-to-make-colored-fire-p2-602196
Pyrotechnic colorant: https://en.wikipedia.org/wiki/Pyrotechnic_colorant
https://www.thoughtco.com/how-to-make-colored-sparklers-3980637
https://www.buysparklers.com/product/14-inch-neon-party-sparklers/
I've read that magnesium can be explosive although I also I want to achieve a white hue for one of my discs. Are there any other alternatives for this?
I also would like to add that I may or may not be adding salt peter (potassium nitrate) to the mix. If I do decide to add potassium nitrate, how will each of the selected pyrotechnic colourants behave with it?
I am trying to create a product with as little additives as possible, that burns cleanly, with little to no scent, with hardly any smoke and as mentioned before, is safe and non-toxic.

I would cheat and have a battery-operated LED for the glow. Some of these might also come with a smell (not a problem for fireworks when the wind is right).

The best way for you to find out if a chemical is safe for your use case would be to look at the safety data sheets for your chemicals. For your convience, I have linked to relevant ones below.

Copper (II) Chloride
Calcium Chloride
Borax
Copper (II) Sulfate
Alum
Strontium Chloride
Potassium Chloride
Lithium Chloride
Sodium Chloride
Magnesium Chloride
Aluminium Powder

In general, you will find that most metal salts are acutely toxic. Take for instance, the Copper (II) Chloride, which is noted as having the following hazards, as noted in Section 2.1 Classification of the substance or mixture:

Acute toxicity, Oral (Category 4), H302
Acute toxicity, Dermal (Category 4), H312
Skin irritation (Category 2), H315
Serious eye damage (Category 1), H318
Acute aquatic toxicity (Category 1), H400
Chronic aquatic toxicity (Category 2), H411

From the same section in the SDS for Aluminum Powder, you will find that it has the following hazards:

Flammable solids (Category 2), H228
Acute aquatic toxicity (Category 1), H400
Chronic aquatic toxicity (Category 1), H410

You can find out more about these hazards from the EPA GHS Criteria Summary, and the Wikipedia GHS Hazard Statement List.
Overall, you should look over each SDS and make sure that the chemical you are proposing to use is safe for your purposes.
Cheers


text:
Question about Elemental Sulfur
I am trying to understand the structure of elemental, crystalline sulfur. I read that it is face-centered orthorhombic. My question is whether each point on the unit lattice contains a single atom of sulfur or a S6 molecule?
Thanks to anyone reading.

Also worth noting that there are many allotropes of elemental sulphur with different crystal structures and molecular makeups.

Please see https://chemistry.stackexchange.com/questions/94400/counting-body-centre-as-lattice-point-when-the-basis-consists-of-2-atoms/94473#94473

There is no "whether". Crystal structures simply do not work like that. Besides, elemental sulfur typically consists of S8, not S6.


text:
Reaction of sulphur with calcium carbonate
Can sulphur react with calcium carbonate to form sulphur trioxide and calcium carbide, as the energy of 3(S=O) is greater than ((C=O) +2(C-O))?
EDIT: i forgot to say that this is in an oxygen-free environment.

There is not enough carbon atoms per calcium atom to form CaC2.

Room temperature, certainly no reaction. However at temperature extremes > 1200 C, it might be worth checking in the presence of oxygen.

Given the presence of oxygen, things will start to happen way below 1200 °C. Without it, you may have to go that high. But there is no way you'll end up with SO3 and/or CaC2.

If you mix CaCO3 + S and heat the mix, the first thing to occur will be that sulfur melts. Second, CaCO3 loses CO2 and becomes CaO. Third, sulfur boils off (or has already boiled off).
A thought that perhaps CaS + O2 would form can be eliminated on the basis that roasting metal sulfides in air produces SO2 + the metal oxide. So the metal sulfide is unlikely.
If any reaction had a possibility, it might be 3 CO2 + S --> 3 CO + SO3, but the heats of formation favor 3 CO2 + S by a large margin. The heat of formation of SO3 is about the same as CO2, and the heat of formation of SO2 is even less.
So the final products will be CaO + sulfur + CO2.

But if the final products are CaO, CO2 and S, then why wouldn't sulphur reduce calcium oxide to calcium as the bond energy of 3(Ca-O) is far less than that of 3(S=O). (3CaCO3 + S ->Ca + SO3 + 3CO2)

If you wind up with CaO and S (all the CO2 is gaseous and gone), It might be possible to arrange conditions where an excess of sulfur takes the oxygen from CaO and gives CaS + SO2. If you can remove all the SO2 produced, any equilibrium would leave you with CaS. Individual bond energies are not as significant as heats of formation of a compound. Bonds are only parts of a compound.


text:
Pure gallium with lower melting point?
I melted about 2 kg of gallium and put it in a plastic container, in order to make crystals. After letting them grow and extracting them, I let the gallium freeze at the room temperature, which is below the melting point.
However, for some reason there is a small quantity of gallium (10-20 g) that is not freezing after days. I extracted the liquid then froze it in a refrigerator. But if I put it outside again, it melts, so this is not supercooled gallium, a well documented phenomenon.
It behaves like normal gallium with slightly lower melting point, something like a Galinstan eutectic. However this is 99.99% pure gallium (that was fully solid) and I am pretty sure that I never contaminated it...
Any idea what is happening?

One wonders how reliable that 99.99% purity claim is... Without specialised equipment, it's generally difficult to distinguish purities of any substance above ~95-98%. If the average consumer can't tell, the supplier has the perverse incentive to cut the cost.

A pdf at this link would seem to indicate that typical impurities are a mishmash of elements. https://documents.indium.com/qdynamo/download.php?docid=1107

It would really be neat to throw this inn an XRF and look for the differences.

It's possible that the gallium is not as pure as advertised. If it has (say) some indium as an impurity, the indium-gallium alloy would have a lower melting point, potentially below room temperature.
I thought that partial freezing might have crystallized out more pure gallium, and left less pure gallium as the melt, but I'm not sure that's how indium-gallium alloys work. I made a few grams of the alloy, and I never saw any such behavior in it -- then again, I never experimented much along those lines.

Every reason to believe that liquid gallium has more impurities than the average over the bulk 2 kg mass. Even if the bulk is 99.99% pure, 20 g would only be 1% of the 2 kg mass. Assuming the 20 g mass has 50% of the total impurities would mean that the 20g has 1% impurities.


text:
What happens when Hinsberg derivative is added to acidic solution?

The Hinsberg test of $\ce{C5H14N2}$ compound produces a solid that is insoluble in 10% aq. $\ce{NaOH}.$ This solid derivative dissolves in 10% aq. $\ce{H2SO4}.$ Which of the following would best fit these facts?

I think b or d is the answer.
The reason being it's insoluble in NaOH. That implies it should not contain 1 degree amine. Can anyone clarify any further?

Edit your question to explain why you think that and you will get answers.

Think about what base and acid solubility mean in the context of the question.

" compound produces a solid" indicates that a reaction has occured.(hence no 3° amine group(according to this)). Further this is insoluble in base so there is a 2° amine group.
"This solid dissolves in acid" directly implies that there is a 3° amine group.
Hence there is both 2° amine group as well as 3° amine group in the compound.
Hence option b is the correct answer.
References:
1. Solomon and Fryle Hinsberg Test

You might edit this answer to give detail on what the Hinsburg test is - I had never heard of it and had to look it up.


text:
What is the heat of reaction of weak acid reacting with weak base compared to strong acid reacting with strong base?
My book has this line only that "weak acid and weak base reaction will produce lesser heat than constant enthalpy" of strong acid and strong base neutralization which is $\pu{-57.3 kJ/mol}$.

See also https://www.quora.com/Why-is-the-enthalpy-of-neutralisation-of-a-weak-acid-or-base-always-less-than-57-1-KJ-MOL

And https://www.quora.com/Why-is-the-heat-neutralization-of-a-weak-acid-with-a-strong-base-lower-than-that-of-a-strong-acid-with-a-strong-base

Please add a reference to the book you are quoting, including edition and a page number. ACS style is preferred.

I'm voting to close this question as off-topic because your question needs some context. Could you add the information about what book the quote was from?

Sorry it's too late but you should reconsider.The book I mentioned is from local writer written by local language, not in English.And that's not a homework question.Come on guys It's old enough to have any homework.I'm teaching my self chemistry.

Strong acid and strong base
First, lets see where the enthalpy of $\pu{-57.3 kJ/mol}$ comes from. In aqueous solution, strong acids and bases react completely to form $\ce{H+}$ and $\ce{OH-}$, respectively. The neutralization of a strong acid with a strong base then is always the same reaction:
$$\ce{H+(aq) + OH-(aq) -> H2O(l)}$$
We can calculate the enthalpy of reaction from the enthalpies of formation at a temperature of $\pu{25 ^\circ C}$.
$$
\begin{align}
\Delta H_\mathrm{f}(\ce{H+(aq)}) &= 0 \\
\Delta H_\mathrm{f}(\ce{OH-(aq)}) &= \pu{−230.0 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{H2O(l)}) &= \pu{−285.8 kJ/mol}
\end{align}
$$
Comparing enthalpies of formation for the products with those of the reactants, we get  reaction enthalpy of $\pu{−55.8 kJ/mol}$. The value given in the question is slightly different (maybe for a different temperature, which is not given).
Weak acid and strong base
For this case, the reaction is the following (with $\ce{AH}$ as the weak acid):
$$\ce{AH(aq) + OH-(aq) -> A-(aq) + H2O(l)}$$
The reaction enthalpy for this reaction depends on the nature of the weak acid. Specifically, the answer depends on the difference in the enthalpy of formation of $\ce{A-(aq)}$ vs $\ce{AH(aq)}$. Conceptually, it takes energy to break the $\ce{A-H}$ bond, and the interaction with the solvent water changes when you go from $\ce{AH(aq)}$ to $\ce{A-(aq)}$. Here are two examples:
$$
\begin{align}
\Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) &= \pu{−484.3 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) &= \pu{−486.0 kJ/mol}
\end{align}
$$
Difference is $\pu{-1.7 kJ/mol}$.
$$
\begin{align}
\Delta H_\mathrm{f}(\ce{NH4+(aq)}) &= \pu{−132.5 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{NH3(aq)}) &= \pu{-80.893 kJ/mol}
\end{align}
$$
Difference is $\pu{51.6 kJ/mol}$.
So sometimes the enthalpy is higher, sometimes lower than strong acid/strong base neutralization.
Weak acid and weak base
The weak acid/strong base consideration showed us that once you have a weak acid, the enthalpy depends on the nature of the acid. So can we make a general statement about the enthalpy of a reaction between a weak acid and a weak base? We start by writing down the chemical equation (weak base shown as $\ce{B}$):
$$\ce{AH(aq) + B(aq) <=> A-(aq) + BH+(aq)}$$
The enthalpy of this reaction could be positive or negative (if you read it backwards like this:
$$\ce{BH+(aq) + A-(aq) <=> AH(aq) + B(aq)},$$
you can call $\ce{BH+(aq)}$ the weak acid and $\ce{A-(aq)}$ the weak base, and the sign of the reaction enthalpy switches.
Notice that this is written as equilibrium. Depending on the $\mathrm{p}K_\mathrm{a}$ values of $\ce{AH}$ and $\ce{BH+}$, this reaction will have its equilibrium more on the left or on the right. If the reaction does not go to completion, the enthalpy (based on amount of starting materials) will be less.
Let's take the neutralization of acetic acid and ammonia as a specific example:
$$\ce{CH3COOH(aq) + NH3(aq) <=> CH3COO-(aq) + NH4+(aq)}$$
The $\mathrm{p}K_\mathrm{a}$ values are separated by more than 4 units, so this reaction goes almost to completion. The reaction enthalpy is:
$$
\begin{align}
\Delta H_\mathrm{r} &= \Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) + \Delta H_\mathrm{f}(\ce{NH4+(aq)}) - \Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) - \Delta H_\mathrm{f}(\ce{NH4+(aq)}) \\
  &= \pu{−486.0 kJ/mol} - \pu{132.5 kJ/mol} + \pu{484.3 kJ/mol} + \pu{80.893 kJ/mol} \\
  &= \pu{-53.3 kJ/mol}
\end{align}
$$
That is pretty close to the strong acid/strong base case.
Back to the question
To prove that

"weak acid and weak base reaction will produce lesser heat than constant enthalpy" of strong acid and strong base neutralization

you would have to do this calculation for every weak acid and weak base combination out there. I would be careful making a statement like this. I think the idea was that because weak acids and bases dissociate much less, less heat is given off (per amount of starting material). However, when you combine a weak acid and a weak base, sometimes the reaction goes almost to completion. As you can see from the example I gave, there are cases for which the enthalpy is very close to that of the strong acid/strong base combination.
Gibbs energy of reaction
For the Gibbs energy of reaction, the situation is much simpler. Weak and strong are defined by the magnitude of acid dissociation equilibrium constants ($K_\mathrm{a}$, and related $\mathrm{p}K_\mathrm{a}$). The difference in $\mathrm{p}K_\mathrm{a}$ of the acid and the conjugate acid of the base is a direct measure of the standard Gibbs energy of reaction: the larger the difference, the larger the Gibbs energy (with appropriate signs, so the strong acid/strong base combination would have the most negative Gibbs energy of reaction, the strong acid/weak base or weak acid/strong base combination would also have negative values, and for the weak acid/weak base combination if would depend on the specifics).

So when writing these equations, do we take into account breaking of the OH bond in $\ce{H3O+}$? :-)

I guess what I am asking is, shouldn't that first reaction really be closer to $\ce{H3O+ + OH- -> 2H2O}$?

@BuckThorn That is how I wrote it first, but the the thermodynamic tables only have $\ce{H+(aq)}$ and not the hydronium ion.

I think this was a difficult question to answer (I thought so at least) because $\Delta H$ seems to depend on the concentrations - for a dilute solution wouldn't even a weak acid be completely dissociated?


text:
Diffusion between phases based on chemical potential

How is the concept of diagram(b) different from that of (e) and (d) from (f)?Someone told me in one of my earlier posts (Why will A atoms diffuse from β phase to α phase and B atoms from α phase to β phase?) that he is used to see graphs with two minimia but somewhere where we had two phase separately named as $G^{\alpha}$ and $G^{\beta}$ vs the composition curve he seemed something that he saw for the very first time. Please help me to differentiate the two of them and also answer the question that I have in this post.

(a)(b) is normal diffusion producing one phase of A and B mixed (c)(d) is phase separation. Look at the way the arrows point.

@porphyrin How are we so sure that this is not a single phase?And how did u conclude the fact of diffusion and phase separation.

https://chemistry.stackexchange.com/questions/75035/common-tangent-construction/75038#75038 May help.

@JonCuster Are you saying that (d) and (f) is the same thing?

I have no idea what they are trying to represent in (f). But, the stable configuration would be similar for (d) and (f), yes.

The logic is similar somewhat to the stable and unstable equlibrium diagram that we had in physics isn't it?@JonCuster

Looking at (c) the phases must separate as B is transferred into the B rich region and similarly for A. So this is phase separation driven by enthalpy dominating free energy change. Normal diffusion (a) occurs the other way round ( diffusion out of high concentration region) driven by entropy.

If your system refers to two partially miscible liquids  the region between the two minima in (d) is where two phases exist, outside this region one phase exists, i.e. at low A or high B.

@porphyrin What about (b) is it the same phase throughout.I do not see any inflection point.So is there any region where two phases may coexist?

yes (b) is fully miscible one phase all proportions, e.g. benzene-toluene.


text:
How to identify lead crystal glassware?
I would like to identify if a wine glass is made of lead crystal (a destructive method is allowed). Lead crystal is said to be eroded by dishwasher detergent. I don't have a dishwasher, what causes the stains? Could I quickly identify lead crystal by exposing it to hot caustic soda?

A first hint could be a high density and a high refractive index.

I tried both but did not succeed, that is why I am wondering about a chemical method.

What happened when you measured the density? How did you try to figure out the refractive index?

Normal glass has a density of $\approx 2.4 ; \mathrm{g/m^3}$ and lead glass $\ge 3.1 ; \mathrm{g/m^3}$ so I am surprised that you have found this to be non-discriminating. The difference in refractive index will be harder to measure unless you have the apparatus but you can measure the density with scales and via Archimedes' principle.

I was unable to determine the volume of the wineglasses sufficiently accurately, using kitchen utensils, partly because of air bubbles. Therefore the density test failed. I submerged the wineglasses in (sunflower) oil, hoping that the visibility of lead crystal would be different, or its reflectivity. But in reality I saw no differences.

Air bubbles? I don't think that can be a problem. Your glass will have a volume in the range of several tens of millilitres. You mark some translucent plastic box which is so small the glass just fits in completely, fill it to the mark with water, place the glass in it, and pour off the water above the mark onto a scale. Voilà, weight in g is volume of glass in ml. Or, perhaps simpler, use a cheap 5 ml plastic syringe to take out the excess water.

Any household chemical method will not have a unmistakeable result, or require additional material and instruments. The amount of lead you can leach out  without pulverising and melting it in soda will be rather small, and i wouldn't bet you can safely identify it.

Update: today I measured the density of a wineglass with a sticker 'lead crystal 24%', it was 2.8 g/cm3. That's higher than the wine glasses I measured last month (2.3-2.5 g/cm3). So I now understand those other wine glasses were plain soda glass, although I had expected some of them to be lead crystal because of a beautiful ping sound and sound persistence, ability to 'sing', inscriptions like "Schott Zwiesel", or a beveled faceted surface, but apparently those properties are unspecific. The true lead crystal glass had a flat surface, no inscriptions, and a nice but unspectacular ping sound.

(continued:) Another unspecific property was the presence of an unremovable haze due to dishwashing. This haze was present on a 2.4 g/cm3 wineglass, whereas hot caustic soda did not (rapidly) create a haze on the 24% lead crystal glass. The refractive index, measured by the refraction of light rays passing through the stem of the lead crystal wine glass (submerged in an oil bath) was not very different from plain soda glass (1.55 vs 1.52). So density is the main dinstinctive property.

Since a destructive method is allowed, the other suggestions did not work for you (despite being excellent, especially the Archimede's scales suggested by @porphyrin) and you have hot caustic soda (aka $\ce{NaOH}$), I suggest this plausible method (which, thankfully, I have never had to use). First, crush a small amount of the suspect lead glass, trying to get it a finely crushed as you reasonably can. Second, mix that crushed glass with solid  $\ce{NaOH}$. You do not need much of the mixture: a teaspoon (or a couple of grams) should suffice. Third, do a fusion of that mixture. A crucible would be used in a laboratory, but you can improvise using a small cast iron skillet. They sell very small ones (4 or 5 inch diameter) at some hardware stores. Absolutely do not use aluminum because the $\ce{NaOH}$ will destroy it!
You need to do the fusion at moderately high temperature, such as can be obtained with a charcoal grill, gas grill or a campfire. You may need to be a bit inventive on this and take care to avoid having anything spatter onto yourself! This is where a grill cover is very helpful. Fourth, let the fusion cool down and add some white vinegar to neutralize the $\ce{NaOH}$. You can add an excess, but don't get carried away: dilution is not your friend. If lead was present, you should now have some 'sugar of lead' (lead acetate) in solution. Fifth, hardware stores usually sell inexpensive test kits for lead (from leaded paints). So use one of those to test your neutralized solution for lead.
What about false positives? Do the whole thing over with crushed glass that definitely does not contain lead. It would be best to do this 'blank' first, to avoid contaminating the new little skillet (which, by the way, can be cleaned, seasoned anew, and used for its orthodox purpose).
I have not tried this and I hand-wash my lead 'crystal' single malt glasses, so best of success!


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How to determine the solubility of a Metal Fluoride in liquid Gallium
How do I determine if a Metal Fluoride is soluble in liquid Gallium and under what conditions the Metal Fluoride will stay in solution in the liquid Gallium?
I am involved in a project that is testing approaches to using CVD to form a graphene layer on liquid Gallium. Fluorine has been used in the process to disassociate methane into carbon and hydrogen and to speed the deposition of carbon. In our project, we are using Acetelyne.
We want to try a different approach of adding a small amount of Fluorine into the liquid gallium itself. With the energies involved it should allow a small amount of F to interact with the process at the surface. I am familiar with all the other aspects but I have never dealt with trying to dope liquid gallium with Fluorine.
My thought was to add a Metal Flouride to liquid Gallium at an appropriate temp/pressure to have it dissolve in the liquid metal. The resulting liquid would be used as a base for deposition at temperatures between 200c to 700c at varying plasma energy levels.
I am considering a few options; SnF2, GaF3, and NaF
Gallium Fluoride would be the best for my application, but the other two are easier to obtain and should still work.
This is going to be < 1 milligram of Metalic Fluoride compound per gram of Gallium and never more than 10 grams of the resulting product at any time.

If the person who voted this question down would give me some feedback I would be happy to improve the question. Without that, I have no idea what the issue is.

Well, you can see some reasons if you hover mouse over downvote button. Also downvoting is made specifically to criticise in a simple and anonymous way. I'm considering downvoting myself as  this seems kinda too broad; also what's the context of this experiment? BTW Salts in general aren't particularly soluble in metals (maybe enough for it to be doable though), and fluorides are especially "salty".

I am involved in a project that is testing approaches to using CVD to form a graphene layer on liquid Gallium. Fluorine has been used in the process to disassociate methane into carbon and hydrogen and to speed the deposition of carbon. We want to try a different approach of adding a small amount of F into the liquid gallium itself. With the energies involved it should allow a small amount of F to interact with the process at the surface. I am familiar with all the other aspects but I have never dealt with trying to dope liquid gallium with Fluorine. I can add this to the question if it helps.

@drobertson  his post is a cascade of questions subsequent to each other.  So I would suggest to probe first if your metal fluorides enter the host material in a quantity easy to trace.  My current understanding of your project is that both gallium and the metal fluoride are deposited while the graphene layer builds, i.e. eventually both gallium and metal fluoride are within the deposit.  If so, I would give X-ray fluorescence / EDX a try to trace the uptake, especially the havy cations, in these thin films.

@Buttonwood Thanks for replying. I think you have the idea slightly off. The Gallium/Metal Fluoride mixture is prepped first and placed in a shallow tray that is at a slight electrical charge relative to the plasma. The environment is low vacuum argon. Acetylene is fed in through a chamber where it is ionized and partially disassociated at a slow rate. The outflow of this process is fed over the Gallium/ Metal Fluoride mix. Different pressures, temp, ionization levels and electrical potentials are being tested. The goal is to quantify the Graphene growth on the Gallium Surface.

@Buttonwood The main question is really about using a Metal Fluoride dissolved in Gallium. I primarily need to know if it is possible and will it stay in the solution. If it wants to form crystals or a surface film this won't work for our needs. The final bit is just me probing for any safety considerations. I am doing my own homework, I but would be grateful if someone would tell me about any dragons on the path.

@drobertson For the first additional comment: I recognize that I misunderstood the intended experiment.  For the second comment: this appears to me like searching for the respective (physical) solubility of e.g. NaF in gallium which, given the low melting point of Ga, either shall be experimentally accessible, or perhaps already was determined and is reported.  All under the constrain that Ga + metal fluoride would not form new compounds (the vertical lines in the Roozeboom phase diagram about binary mixtures [like Fe with C].)

@drobertson Dissolving metal fluorides in Ga indeed leads to new compounds; unclear though, if / how interfering here.  A search by Na, F, Ga as the only 3 permitted elements in the COD (https://www.crystallography.net/cod/index.php) leads to F14Ga3Na5 (https://www.researchgate.net/publication/228093517_Distorted_chiolite_crystal_structures_of_a-Na5M3F14_MCrFeGa_studied_by_X-ray_powder_diffraction).  Search for Ga + F as the only ones leads to GaF3 by reaction of Ga with NH4F (https://www.researchgate.net/publication/269598380_Refinement_of_the_crystal_structure_of_gallium_trifluoride_GaF3).

@Buttonwood I would be starting with lab-grade GaF3 if I used it and attempting to dissolve it into liquid Gallium. I can do a fairly good job of isolating it during this process. Later it would be interacting with H and C in a low vac/Ar environment. Small amounts of other contaminants are likely to be involved, but we are working to limit that.It looks like NaF wouldn't b e the way to go. Crystals would interfere. The Fluorinated Ga solution needs to completely stay in a liquid state. This is helpful, thanks.


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Is presence of air necessary for silica gel to absorb moisture?
I am a biologist currently doing field work where I need to quickly dry leaf samples using silica gel crystals. I use the orange indicating kind that are 2-4mm in size. The setup is like this. I stuff 1-2 leaves in a manila paper envelop which I drop in a ziplock bag containing ~1/8th cup of silica gel. Then I squeeze out most of the air before sealing the ziplock. The end goal here is to extract DNA from dried leaves when I am back in the lab.
I have had varied results in terms of drying and looking to achieve the maximum efficiency. Recently it occurred to me whether squeezing air out is actually counterproductive to fast drying.
So my question is, whether presence of sufficient air necessary for silica gel to absorb moisture from leaves? Essentially I am creating partial vacuum inside the ziplock. Is that keeping silica from doing its job?

Air should play no role in silica water absorption capacity. What you can do is to ensure that silica is stored in a dry condition before use.

It's stored in a air tight container, so starting moisture is not a concern.  If chemically speaking, air is not a concern, would proximity of beads to the sample could still play a role?  For example, when you squeeze air out, it's restricts movement of beads allowing one to control their proximity to the leaves.  If there is air in the ziplock, then movement is unrestricted.  This is what my question is mostly coming from.

@cryptic0  If you aim for the isolation of the DNA, assuming taxonomy in lines of "first leaves of a special tea shrub" already is done in the field, I speculate drying of the leaves might be accelerated if these were crushed / shred early there.  With increased exposed surface to the air shuttling back and forth to the silicagel, the process of drying were faster than passing water all across cuticulae / stomata / epidermis.  (This assumes the silica is kept in their paper bags, not in direct contact to the specimen.)

@Buttonwood You bring up an interesting point, one I have pondered, but not followed through with.  I am going to set up an experiment, with one iteration as you describe (silica in envelope), shredded leaf in ziplock and another my regular method and see which one dries faster. Notwithstanding, the question of how much air to allow in the ziplock remains.

I wonder if it would help to keep the manila  envelop in the ziplock bag with the descant prior to putting the leaf in. The paper will pick up water. With the envelop, the water vapor has to  travel through the envelop to the air in the bag to the descant.

You might try pre-drying the silica in an oven to make it more effective, or try other desiccants: https://www.sorbentsystems.com/desiccants_charts.html

I am using orange indicator silica gel. It gets green when not charged.

What you need to preserve is a minimum flow so that no sacks of air and moisture find themselves confined, seeing just the leafs and no dessicating beads.


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Chemical Reaction Alternative to O2 Respiration in H2/N2 Atmosphere
Note: If this question does not belong on this Stack Exchange, please direct me to a more appropriate Stack Exchange.

This is a "creative chemistry" question as much or more-so than a "give me the facts" type of question. Should be of interest to anyone who finds the topic of plausible alternative biochemistries intriguing.
I recently came across several papers outlining a hypothetical type of photosynthesis that would be highly efficient in a Hydrogen ($\ce{H2}$) dominant atmosphere containing Methane ($\ce{CH4}$). The photosynthetic reaction is:
$$\ce{CH4 + H2O + $y$ -> CH2O + 2H2}$$
"$y$" represents energy imparted by photons. "$\ce{CH2O}$" is the simple organic compound formaldehyde, which could be used to build biomass.
This made me wonder how "animals" in such an environment would obtain the energy needed to live. In an "$\ce{O2}$" heavy atmosphere the answer is simple: use "$\ce{O2}$" to oxidize the available organic compounds, getting a ton of energy. This is aerobic respiration. What if there is essentially zero available "$\ce{O2}$"? Which brings me to...
My question in detail: What is a plausible chemical reaction (or series of reactions) that could take the place of $\ce{O2}$ respiration in a "Hydrogen Dominant" world with the following molecules and compounds to work with:
ATMOSPHERE

$\ce{H2}$ (10-90%)
$\ce{N2}$ (10-90%)
$\ce{H2O}$ Vapor (≈1%)
$\ce{CH4}$ (0.01 - 5%)
$\ce{NH3}$ Vapor (0.01 - 5%)
Other trace to minimally-present compounds may include $\ce{CO2}$, $\ce{Ar}$, etc.

BIOMASS

Built off of $\ce{CH2O}$ (the organic product of hydrogenetic
photosynthesis)

WATER & LAND

Bodies of water containing a mix of  $\ce{H2O}$ + $\ce{NH3}$, and plausibly available compounds in the crust (like iron).

By "take the place of" I mean provide enough energy to run complex life-processes. I do not mean superficially resemble aerobic respiration (i.e. The process need not require the "breathing in" of atmospheric gases. It could, but need not).
The answer should try to meet the spirit of the following criteria:

The reaction(s) could make use of compounds in the air (breathing) or could be obtained through other methods (eating, drinking) in sufficient quantity to run biological processes.
Impart significant amounts of "free" energy, more so than other plausible reactions.
Make use of available compounds (see details above).
Plausibly occur at temperatures and pressures not inimical to carbon life. $\pu{500\\K}$ temperatures or pressures of $\pu{100\\bar}$ are no-goes.
Be "plausibly likely" to evolve. This requires a measure of judgement. If the chemical reaction or series of reactions is wildly complicated, far-less efficient than other alternative reactions, or needs to make use of rare materials, then it's probably not plausible.

I'd love the form of the answer to contain the equation, a brief discussion on why it's plausible, and a discussion of the amount of energy released, and/or other "good to know" things.
Thanks!

References:
Photosynthesis in Hydrogen-Dominated Atmospheres – https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4284464/
BIOSIGNATURE GASES IN H2-DOMINATED ATMOSPHERES ON ROCKY EXOPLANETS – https://iopscience.iop.org/article/10.1088/0004-637X/777/2/95/meta
A BIOMASS-BASED MODEL TO ESTIMATE THE PLAUSIBILITY OF EXOPLANET BIOSIGNATURE GASES – https://iopscience.iop.org/article/10.1088/0004-637X/775/2/104#apj480437s4

Note: The reason that there is far more methane than carbon dioxide is that life-forms could derive easy energy from methanogenisis (hydrogen + carbon dioxide = energy + methane + water). The reason there is ammonia is that life could derive easy energy from a "cold haber" process (hydrogen + nitrogen = energy + ammonia). These processes are therefore likely to evolve, causing the build-up of methane and ammonia in the atmosphere.

In terms of formatting your question, be invited to have a look at mhchem's functionality -- available in questions, answers and comments of ChemSE -- to render chemistry-related content easier to read: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

Can you put a quantitative value on "Impart significant amounts of "free" energy, more so than other plausible reactions."? Significant can mean a lot of things.

@Andrew "Significant amount of free energy" means more than other plausible reactions. That's it. More energy for the work than other competing reactions. There is no absolute measure, it

[got cut off] @Andrew ...it all depends on what other reactions are out there that might provide more energy per the amount of effort (energy invested, resources gathered, etc.). In this case the other relevant reactions are Methanogenesis (yielding ≈193kJ/ mole?) and Fermentation (yielding 74kJ/mole?). So more energetic than those. It'd be nice if it got up to aerobic respiration levels of energy (2,820kJ/mol) but not necessary.

"creative chemistry" question as much or more-so than a "give me the facts" type of question - that's the speculative which is discouraged here. If not for Button's edit this would be on hold.

The obvious approach to is to just do the direct analog of what Earthling metabolism does: exactly reverse the photosynthetic reaction (on net, anyway; the intermediate processes may be rather different, as they are on Earth).
In this case, that means reacting hydrogen with carbohydrates to get back methane and water. Based on this paper on photosynthesis in hydrogen-dominated atmospheres, building carbohydrates from methane and water requires considerably less energy than building them from carbon dioxide and water, so we'll equally get less energy out by decomposing them again--I'd estimate somewhere around a quarter to a third as much power density as Earthling animals relying on aerobic respiration, if we assume organisms in this environment will be building biomolecules optimized for this environment, rather than just copying ours. Given how much time I spend every day not eating, and not hyperventilating, I'm pretty sure that would still be enough to power complex animals; they just might need to eat more / breathe more / be more lazy than equivalent Earth animals. Even if you only get one fifth the power density (based on the Bains paper's estimate that hydrogenic photosynthesis can proceed with 20% of the energy required by oxygenic photosynthesis), that's still 7-8 alien-equivalent-of-ATPs per glucose, as opposed to the 2 ATPs per glucose provided by anaerobic respiration (cf. 38 ATPs-per-glucose for aerobic respiration).
As Andrew noted, however, land animals will have an extra energy supply available in the form of reducing N2 into ammonia--i.e., they can supplement their energy supply simply by breathing, and only really need to eat for the materials. Incidentally, autotrophs could also make use of excess ammonia production to power chemosynthesis in the absence of light.
Despite the relatively low abundance of ammonia in the atmosphere, however, it has much higher solubility in water than hydrogen does, and thus may be more easily available to aquatic organisms than free hydrogen is. That gives you a slightly modified "respiration" reaction that takes 3 formaldehyde units and 4 ammonia molecules to produce methane, water, and free nitrogen. Decomposing ammonia into hydrogen and nitrogen is endothermic, however, so this would be less energetic than hydrogen respiration. Not sure how well that would be balanced by the reduced energy needs for pumping water over the gills compared to processing enough water to get the equivalent quantity of free hydrogen. This is a Good Thing for the long-term stability of the atmosphere, as otherwise you'd expect life to just keep producing ammonia until either nitrogen or hydrogen is exhausted.
Now, that is not the most energetic set of reactions imaginable. Obviously, if they had access to oxygen, that would be way better. But, it's the only option that is consistently renewable. Other potential oxidizers, which are more oxidized than the plant and animal biomass that serves as food in this scenario, won't be produced by plant life (by definition--anything that is produced by plantlife is biomass available as food, and thus not more oxidized than itself!) and thus can only be regenerated in small quantities by geochemical processes, and will have very short environmental half-lives. E.g., if creature could eat lithium metal, they could react that with nitrogen, ammonia, water, or complex organic biomolecules and get energy out of any of those reactions--but you just aren't going to find large quantities of free lithium to power your biosphere!
For some back-of-the-envelope estimated specifics, let's initially suppose that plants are hydrogenically producing glucose and animals are hydrogenating glucose. That leads to the following metabolic reaction:
${C_6H_{12}O_6}^{174 kJ} + 12{H_2}^{0 kJ} \rightarrow 6{CH_4}^{-74.85 kJ} + 6H_2O^{-285.83 kJ}$
Which provides approximately 2338 kJ/mol of glucose, based on the heats of formation for each reaction component as listed in the NIST Chemistry WebBook; the net energetics for aerobic respiration are found just by replacing methane with ${CO_2}^{-393.51kJ}$ in that equation (diatomic oxygen, like diatomic hydrogen, conventionally has a heat of formation of 0, so that doesn't make any difference to the calculation), which gives us a theoretical output of 4250kJ/mol. (Looking up the actual empirical energy yield of aerobic respiration of glucose, we get 2880 kJ/mol--so, assume that all other energy calculations here are only good to within about a power of two!) So, if hydrogen-breathing animals can just manage to evolve a slightly more efficient energy system than we have (which seems plausible, especially if they are living at lower temperatures which permit better thermodynamic efficiency), they may even get up to a full half the power density of oxygen-breathing Earthlings. Given the seriously speculative and highly imprecise nature of these calculations, though, I'm more comfortable sticking with my earlier 1/3 estimate as a plausible ceiling.
Plants in a hydrogen-dominated atmosphere aren't necessarily going to be producing as much glucose as our plants do in our highly oxidized environment, so let's look at some other options for biomolecules that might be available for animals to eat. Suppose, for example, that plants produce acetylene as an energy storage molecule; hydrogenating that provides
$C_2H_2^{226.73 kJ} + 3H_2^{0 kJ} \rightarrow 2CH_4^{-74.85 kJ}$
376 kJ/mol of acetylene -- or about 14.5 kJ per gram. About half as much per gram as glucose.
And if we look at the above-mentioned ammonosynthesis reaction, using hydrogen and nitrogen directly from the atmosphere, we get
$3H_2 + N_2 \rightarrow 2{NH_3}^{-45.9 kJ}$
about 91 kJ/mol of nitrogen, or 5.3 kJ/gram of inhaled hydrogen/nitrogen mix. That's not a ton compared to glucose (or acetylene), but there's just so much more of it available, for cheap! After all, it takes quite a lot energy just to consume and digest food--and practically none to just breathe it in.
Also consider that, largely because of the above reaction, nitrogen is far more easily bioavailable in this world than on Earth, so maybe they use a lot more cyanide groups or even azides. Hydrogenating hydrogen cyanide gets you
${HCN}^{135.14 kJ} + 3{H_2}^{0 kJ} \rightarrow {CH_4}^{-74.85 kJ} + {NH_3}^{-45.9 kJ}$
255.75 kJ/mol, or about 9.5 kJ / gram.
Simply decomposing hydrogen azide, even before you consider the energy gains from hydrogenation, gets you
$2{HN_3}^{264 kJ} \rightarrow {H_2}^{0 kJ} + 3{N_2}^{0 kJ}$
about 6 kJ/gram.
Those last few reactions (somewhat surprisingly, given the notorious explosivity of azides) would tend to drive the average energy density of food down a smidge from the optimistic third-to-half estimates, so once again about a quarter of the total energy density of aerobically-metabolized Earthling food seems reasonable.

Thanks Logan! Hmm...would you be game to putting some numbers to the energy outputs? It's not clear to me how you arrive at a  "quarter to a third" for reacting hydrogen and carbs, and that's something I'd really like to understand. Same with the ammonia reduction.

@n_bandit I arrive at a "quarter to a third" just be reversing the synthesis reactions. It takes a certain amount of energy to build biomass hydrogenically, so that's about how much you'll get out by taking it apart again. I can look up some back-of-the-envelope energetics numbers In A Little Bit, though, and edit those in to the answer.

@n_bandit Note the energetics calculations that have now been edited in.

This is only a partial answer, but maybe will stimulate some thinking by others.
For a good sense of plausible reactions, one need only look at the existing biology of anaerobic organisms. Many organisms produce energy in the absence of oxygen, either by fermentation or anaerobic respiration (the distinction being the absence or presence of an energy-harvesting electron transport chain). The methanogens you mentioned are one example, and that would be one viable source of energy in a hydrogen-rich atmosphere, as long as the supply of CO2 was sufficient.
In general, fermentation and anaerobic respiration are problematic for larger organisms for a few related reasons. Because the ATP yield per reaction is low, a large organism with high energy demand needs to achieve a very high throughput of the reaction in question. That means not only having a large metabolic capacity in terms of the required enzymes and cell structures, but also the ability to take in large amounts of substrate and remove large amounts of end-product at a high rate.
The oxidized form of the electron acceptor thus must be something that is stable enough to be fairly abundant as well as easily taken in, and the waste products (reduced forms) need to be removed (or at least sequestered) quickly in order to maintain energetic favorability of the process.
For an animal living on land, gas exchange through lung-type organs seems to be the only really plausible mechanism for disposing of waste, since sequestration internally would have an energy cost. Based on existing biological products, that suggests H2S and NH3 as leading candidates for end-products. For H2S, there is no plausible gaseous precursor, so you would have to imagine the organism consuming large amounts of polysulfur or sulfur oxides. Possible, but challenging. That leaves NH3. With high N2/NH3 ratio in the atmosphere, conversion of N2 to NH3 as a respiratory pathway is hypothetically viable and seems the only plausible option. With a high H2 atmosphere, it might be conceivable to live mostly on hydrogen and not need to eat carbon compounds for energy, but just for biomass.
For an animal living in water, there is a wider range of possibilities, since the substrates can be dissolved ions such as sulfates or nitrate or metals, and the waste disposal could be via liquid-liquid exchange. The main challenge would be to find substrates that are sufficiently abundant in oxidized form despite the hydrogen-rich atmosphere and to identify a viable parallel process for reoxidation of the endproducts.


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What gives rise to the color of sapphires?
Sapphires are composed primarily of $\ce{Al2O3}$ (in pure form colorless) as well as Fe and Ti (trace impurities responsible for the color). I know that the proportion and identity of trace elements gives rise to various colors but I just want to focus on the combination that gives blue right now.
The explanation that I could find online says that the reason we see blue is because of intervalence charge transfer between the iron and titanium ions. Specifically Ti(IV), Ti(III), Fe(III), and Fe(II) are involved. The ground state configuration is controversial but is presumed to be both ions in their 3+ state. It could also be Ti(IV)-Fe(II). The first part of my question is what is the exact equation for the process occurring? I could not find a redox potential for the reaction $\ce{Ti(IV) + e^- -> Ti(III)}$ so I'm not sure which direction the electrons flow. The half reaction $\ce{Fe(III) + e^- -> Fe(II)}$ has a positive reduction potential of +0.77 V. I assumed that a photon provides enough energy for the fourth ionization energy of Ti (4174.65 kJ/mol) and that electron is transferred to Fe(III). I don't think this can be correct though because that energy would be in the ultraviolet region on the J/atom scale. This is the same for the third ionization energy of Fe. I was expecting to see something closer to yellow light region ~575nm. So from this its clear that I'm probably missing the idea regarding the physics occurring here.
The second part of my question has to deal with color subtraction in general. Why does a sapphires absorption of yellow light lead to blue color in the first place? I've seen an explanation that you imagine white light as a mixture of only the primary colors red, blue, and green light. Also its stated that the secondary colors are formed as follows: red + green = yellow, blue + green = cyan, and red+ blue = magenta. Then if a sample absorbs for example yellow light (R+G) then only blue light remains from the original R+G+B=W combination. This makes sense at face value for me but I wanted to go a bit further. I assume the origin of this color addition and subtraction theory is that humans have cones specific for detection of red, blue, and green light (with a range of wavelengths of course to cover the whole spectrum). My difficulty with the color subtraction answer for the color origin is the fact that our cones detect a range of wavelengths. If the sapphire only removes a single wavelength of light close to 575nm (yellow), how is the effect so dramatic to make the stone blue? Won't the stimulation of the cones not have changed dramatically?

Trying to use atomic energy levels for impurities incorporated in a crystal tends to not work well. Those ‘ions’ are immersed in electric fields from the surrounding crystal.

Actually sapphires can be any color other than red. If it is red colored aluminum oxide then it is a ruby.

As @MaxW states, any color, including clear., e.g. https://www.edmundoptics.com/f/sapphire-windows/12234/

Thank you @Jon, MaxW, DrMoishe I know. That is in my first paragraph

The blue of sapphires is produced by the elements titanium and iron that are both present. There is a process called charge transfer causing the transfer of an electron between Ti$^{4+}$ and Fe$^{2+}$ to Ti$^{3+}$ and Fe$^{3+}$ to fit together with the Al$^{3+}$ of the corundum lattice. Those ions give the blue sapphire its colour.
Note that the ions in a crystal are different from isolated ions in the gas phase, they tend to have larger absorption bands instead of sharp lines. Also the distances in the lattice are important and may modify the colour, anisotropy can lead to colour dichroism, the sapphire may be blue viewed from one side and green from another.

Interestingly, the same ions are responsible for the color of rose quartz. Matrix matters!


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Why doesn't natural gas combustion release only one-sixth as much carbon dioxide as coal burning?
Burning hydrogen releases a little more energy than burning carbon, correct?
And methane is CH4, but coal is mostly C, correct?
So shouldn't burning pure carbon release five times the CO2 (Assuming combusting carbon releases the same energy as hydrogen), or even six times per unit energy 'released'/'created' by the power plant (since hydrogen is more energy-dense per unit mass than carbon)?

see: https://www.engineeringtoolbox.com/fuels-higher-calorific-values-d_169.html Look at column Btu/lb

But this chart shows pure hydrogen generating about 4.5 times as much energy as MJ/kg or kWh/kg as pure carbon combustion.  Yet one carbon atom weighs twelve hydrogen s, AND creating an H-O bond releases more energy than a C-O bond, so burning a kg of hydrogen should release 15 times the energy as a kg of carbon!  Right?!?!?!

Pure carbon is not an isolated carbon. You cannot add the energy given by C and H to infer that given by CH4. If Ive understood your way of thinking. ....

When you do these calculations be sure to keep track of your units and the actual reactions you are considering.
Burning methane is not the same as burning the same amount of carbon plus twice the amount of dihydrogen. You need to check the specific heats of combustion and you need to make sure you keep track of the units used to measure the heat produced in the reaction (per mol, per kg or per mol of carbon dioxide?).
The heat of combustion (assuming all products are at STP) for carbon is 393 kJ/mol and for methane it is 889 kJ/mol. Both reactions produce the same amount of carbon dioxide if combustion is complete. So burning methane produces about 2.3 times as much heat per mole of carbon dioxide as burning pure carbon (anthracite coal will be close to this).

Well, burning H2 releases 4.7 times the heat of an equivalent weight of carbon, according to the table given by MaxW.
The heat given by methane is 1.84 times as much as the same weight of bituminous coal, but the CO2 is only 0.75 (= 12/16) as much.
So, burning bituminous coal would release 1.84/0.75 = 2.45 times as much CO2 as burning the same weight of methane.


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Maximum lifetime of plastic on earth
[EDITED]
If humans disappeared today, and aliens visited in the future, for how long would plastic or other "obviously" non-natural polymers be detectable on the planet, using known or plausible technology. That is, how likely would it be that the aliens would say: "Clearly, the presence of these molecules indicates past technological activity"?
I'm mostly interested in accidental preservation of plastic (buried in sediment, microplastics in the ocean, etc.), but also the long-term stability if, for instance, someone intentionally tried to preserve a plastic artifact. Could it survive for thousands of years if adequately protected? Millions? Billions?
Comments imply that if isolated from biological activity, at least some plastics are sufficiently stable to survive for billions of years (I assume that requires temps below, say, 40C or so?).  Is it plausible for plastic to end up in an arid region, get buried in sediment, and survive for millions of years intact while getting buried under successive layers?  At what pressure would the plastics degrade and no longer be recognizable as the product of technology?  I assume this would occur well before it turns into coal.
Finally, while I recognize that this is the Chemistry SE and not Biology, are all plastic species today degradable by some known biological organism (meaning, for each specie, does there exist a known bacterium that eats it)?  Is there a reason that life would not eventually adapt a way to "eat" every kind of plastic/polymer we produce (like, not enough energy to recover, etc.)?
Note: I have basically a high-school level understanding of chemistry, so apologies for the ill-posed question.

I think you can make the argument that a minute amount of plastic will be lucky enough to survive until the Sun turns into a red giant, at which point it'll be incinerated (along with the entirety of Earth's crust). So that's an upper bound of ~6 billion years.

You may indeed be more interested into the half lives of different plastic species. Degradation conditions are very important as on one end highly isolated plastic (held under vacuum in a glass recipient) would virtually not degrade and on the other end, concentrations of plastic in areas where degradation conditions are ideal would favour "rapid" decomposition (some bacteria eat some kinds of plastic, so if you concentrate the specific kind in a specific area with those bacteria and their other requirements, such as water, they'd degrade faster).

In present form, the question is too broad.  If you allow «polymers» as place holder for «plastics», than are such materials which decompose easier (e.g., polylactic acids) than other (PVC, teflon).

Surely future archaeologists will have a lot to dig for.