text:
Extraction of calcium carbonate from liquid
I'm using calcium carbonate ($\ce{CaCO3}$) in apple juice to reduce the acidity of the juice.  Once the $\ce{CaCO3}$ reacts with the apple juice and raises the pH, the juice is left cloudy in color BUT has the desired pH of 3.4.  I need a way to extract (separate) the $\ce{CaCO3}$ precipitate from the apple juice which leaves the resulting liquid (apple juice) drinkable still...
The pure juice has an initial pH of 2.9. I want to reduce the acidity and be left with pH of say 3.4. Perhaps by blending with another drink?  I did a test using the pH meter, I started with water (pH 6.0) and slowly added apple juice, and the pH drops to 2.9 adding 1 part juice to say 10 parts of water.  So blending another higher pH drink (tea/water/etc) would totally dilute the apple juice before the pH would reach 3.4. How can I raise the pH and still keep the apple juice pure or close to pure?
Apple juice out of the carton tests at 2.9 pH with digital pH meter.
My tap water tests at 6.0 pH.
Calcium carbonate - pill form from health food store (oyster shell derivative)

Nearly all sources that google gives me, say that apple juice has a pH of 3-4 ... exactly what you want.

Apple Juice out of the carton tests at 2.9 pH with Digital pH meter. My tap water tests at 6.0 pH.

The stuff you're seeing is solid calcium carbonate that most likely did not dissolve, or some other insoluble calcium compound that formed by reaction of calcium with whatever is in the apple juice.
You might want to try using sodium carbonate instead of calcium carbonate. Sodium compounds are usually more soluble than calcium compounds, and it's a more common food additive than calcium carbonate.
That said, I wouldn't recommend adding chemicals to your food without knowing exactly what you're doing. Some seemingly safe chemicals may have hazardous impurities and do not qualify as food-grade. If you're doing this because of health problems then you should consult with a doctor on better ways of action than pouring carbonates in your drinks.

I think you are doing it wrong. Do not add CaCO3 in excess. And keep in mind that ingesting too much Calcium may have a negative effect to your health.

The CaCO3 is used in proper amounts to raise the pH (lower the acidity) to a resulting desired level (say 2.9 pH to 3.4 pH).  Now I need to separate the juice from the dissolved/suspended CaCo3.  can it be filtered, etc...  I tried a coffee filter but does not filter fine enough I guess.  Still cloudy...

Food grade diatomaceous earth? Celite is how we would do it in a chem lab, and that's diatomaceous earth. Not sure if food grade has the right particle size, though.


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Calculation of pH in diluted buffer
I have been preparing $\pu{150 mM}$ and $\pu{30 mM}$ Tris Buffers (pKa=8.3).  Using .15 M solutions of both Tris and Tris-HCL.
Here is the information in which I have calculated theoretical pH values as well as collected experimental data.
I used (Ka or Kb)= x^2  / [concentration]-x to solve for the values in solution 1 and 5 of both the concentrated and diluted buffers.
I also used Henderson-Hasselbach for solutions 2-4. For example, solution 2 in 150mM:
pH $\displaystyle  = 8.3 + \log\frac{\pu{0.375 M} }{\pu{1.125 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl I added and subtracted appropriately:
pH $\displaystyle  = 8.3 + \log \frac{\pu{0.375 M} - \pu{0.125 M}}{\pu{1.125 M} + \pu{0.125 M}} = 7.50$

I am having trouble applying these concepts to a diluted buffer solution. I calculated solution 1 for 30 mM buffers by simply changing the concentration by a factor of 5 (per the dilution factor) while using the ICE table Ka=x^2/ concentration-x.
I need some direction in being able to figure out the pH of the diluted solution and how to calculate the effect of .1 mL 1.25 M HCl addition to each of the dilute solutions.
[From comments] ... to clarify the 30mM are also 10 mL total. They are diluted with water until 10mL after the addition of the acid and base as laid out in the 2nd graph. The real question I have is that will there be an effect on pH by the water in the dilute solutions? The ratio of acid and base is the same, but I am not sure if the values will be identical to those in the 150mM calculations.

Sorry to clarify the 30mM are also 10 mL total.  They are diluted with water until 10mL after the addition of the acid and base as laid out in the 2nd graph. The real question I have is that will there be an effect on pH by the water in the dilute solutions?  The ratio of acid and base is the same, but I am not sure if the values will be identical to those in the 150mM calculations.  Can anybody shed some light on that?

I'd recommend formatting your equations using MathJax to make it more understandable.

The real question I have is that will there be an effect on pH by the water in the dilute solutions?

There is very little effect on the pH of a buffer when you dilute it with water (as long as the concentration is still high compared to that of hydronium and hydroxide concentrations).

I need some direction in being able to figure out the pH of the diluted solution and how to calculate the effect of .1 mL 1.25 M HCl addition to each of the dilute solutions.

The buffer capacity is lowered by diluting the buffer. If you add the same amount of strong acid to a buffer or a diluted buffer, the diluted buffer will show a larger change in pH.
If I understand the calculations presented in the question correctly, the concentrations are off by a factor 10. Comparing the 150 mM with the 30 mM buffer, the relative change in concentrations upon adding the same amount of HCl is larger for the more diluted buffer, thus the pH change is predicted to be more pronounced. The experimental data supports that prediction.
Here are the updated calculations for row 2, 150 mM:
pH $\displaystyle  = 8.3 + \log\frac{\pu{0.0375 M} }{\pu{0.1125 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl:
pH $\displaystyle  = 8.3 + \log \frac{\pu{0.0375 M} - \pu{0.0125 M}}{\pu{0.1125 M} + \pu{0.0125 M}} = 7.50$
For the 30 mM buffer, the buffer concentration is one fifth of the 150 mM buffer, but the amount (and concentration) of HCl added is the same:
pH $\displaystyle  = 8.3 + \log\frac{\pu{0.0075 M} }{\pu{0.0225 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl, we can try the same approach:
pH $\displaystyle  = 8.3 + \log \frac{\pu{0.0075 M} - \pu{0.0125 M}}{\pu{0.0225 M} + \pu{0.0125 M}} = $ oops, we ran out of buffer (HCl is in excess of the deprotonated form of the buffer substance).
pH $= \log (0.0125 - 0.0075) = \log 0.005 = 2.3$
This matches the experimental result fairly well.

Hmm, with different charges of respective conjugate acid and base, their activities would change by dilution at different rate.  I would have to quantify it, but I am not sure if it is very little for different ionic strengths. OTOH, if the ionic strength is kept constant by added salts, then you are right.

@Poutnik You are right. At the level of theory presented here, however, you get the same answer.

Yes. I use this argument often myself. :-) All depends on the expected level.

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How can I restore the surface of ABS polymer products?
Recently I asked a question about de-yellowing Lego. I have a follow-up question on this.
I try to design a setup where I can de-yellow Lego and remove/obscure any scratches from the surface of Lego items.
Matt_black mentioned it might not be possible to only fill scratches etc.
With my limited knowledge I could imagine the next options:

Dissolve ABS in a solvent, but this will be an unspecific process.
Heat treatment, but I think the items will warp due to internal stress.
Maybe the surface energy of a scratch is different and you could make a marker which only releases the polymer on the scratch region?
Coat the items in some kind of transparent resin?

My questions are:

Are any of these ideas realistically feasible?
If not, are there any other options which could work?
How would this procedure work from a chemical point of view?

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Halogen and inductive effect
background :

As proved by rate of EAS, mesomeric or simply conjugated resonance is weaker for halogens than inductive and make them electron taking, with fluorine almost doing nothing due to strongest mesomeric and strongest inductive among halogens, So, halogens should take electrons away from any system if they do it in a specially conjugated system known for many resonating structures

If we have simple carbocation as shown in the chlorine and carbocation, being right next to chlorine Inductive effect must overpower resonance/mesomeric but why the opposite happens?

Simply put, the energy costs for a sextet carbocation are too high to allow chlorine's strong inductive effects to dominate the situation.
The driving for for a carbon octet is very high.

If we have a simple carbocation as shown in the chlorine and carbocation example, being right next to chlorine the inductive effect should overpower resonance/mesomeric but why does the opposite happen?

No, it shouldn’t. Remember that chlorine’s electronegativity — a measure for the strength of its inductive effect — is less than both oxygen’s and nitrogen’s. Therefore, the inductive effect is weakest for chlorine.
All things considered, resonance effects usually overpower inductive effects, especially at short distances.

I'm with you on the weak inductive effect, but if you look at the inductive effect order, halogens beat NH2 and OH by a good margin, and 3p being so bad for resonance and having empty d orbital, back bonding strength is quite bad

Aren't inductive effects supposed to fade out quickly with distance and mesomeric or conjugative effect remains constant? if we look at a lot of problems, almost in everyone , inductive effect wins over mesomeric FCOOH (pKa=1)

@ELiT Stop any discussion with d-orbitals immediately. Main group elements’ d-orbitals are too far removed to take part in bonding. Yes, inductive effects are only relevant at short distances, but where mesomeric effects are helpful, they generally win. Note that in your carbonic acid fluoride example, the mesomeric effects are not helpful, so we have to consider inductive effects.

In EAS, oxygen and nitrogen are both activating substituents (i.e. their resonance effect wins out over their inductive effect). Chlorine, however, is weakly deactivating. Chlorine, having a lower electronegativity than oxygen, should be activating if all resonance factors are equal. This suggests that chlorine's poor overlap with the aromatic LUMO causes such poor resonance to occur that its lower inductive effect still dominates. Perhaps the increased electronegativity of the carbocation reduces the chlorine's inductive effect, rather than forming a π bond?

@DGS At least I can fully agree with the aromatic discussion: chlorine is bad at overlapping with $\ce{C=C}$ double bonds in general. I think that could be the reason why the extremely weak inductive effect actually plays a role. While if there is an empty p-orbital next to chlorine, it doesn’t have to interact with a pi-bond making the entire thing easier.

i couldn't really understand what you meant by "helpful" exactly how do we decide what's helpful? making it more acidic would make it unstable,no? if you're saying that the deprotonated form would be more stable then para halo aniline is less basic than aniline and there are lots of examples where inductive wins over mesomeric and the only example that i could find is the carbocation one

In most scales I'm accustomed to using, Chlorine has a higher electronegativity than nitrogen.

@Lighthart $\ce{NCl3}$ reacts with $\ce{HCl}$ to form $\ce{Cl2}$ — an experimental result clearly indicating that chlorine must have the partial positive charge in $\ce{NCl3}$.

@ELiT Helpful is a bad word choice, yeah. Maybe productive instead? In any case, the mesomeric effect of fluorine in $\ce{FCOOH}$ does little to deactivate the $\ce{OH}$ group’s acidity no matter which way you draw it.

https://pubchem.ncbi.nlm.nih.gov/compound/trans-3-Chloroacrylic_acid#section=Top ... sadly doesn't have pka, field effects would also be present but minimal in trans

@Jan all that means is that NCl2 has more electronegativivty than Cl, not Cl has less electronegativity than N.

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What is the effect of adding HCl on a certain solution at equilibrium?

If we have
  $$\ce{CH3COOH + H2O <=> CH3COO- + H+}$$
   what is the effect on acetate ions when adding each HCl and NaOH to the solution?  (State why according to Le Chatelier's principle)

Le Chatelier's principle states that if any of conditions of a chemical reaction at equilibrium changes, the reaction will shift to a direction to oppose that change and restore the equilibrium.
I guess adding $\ce{HCl}$ will increase hydronium ion concentration so the reaction will be shifted to the left side, accordingly acetate ions concentration will decrease. If we add $\ce{NaOH}$, this will increase water concentration so the reaction will be shifred to the right side accordingly increasing the concentration of acetate ions.

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $LaTeX$ syntax. If you receive useful answers, consider accepting one. You can edit your question any time to include more detail or improve formatting. This appears to be a homework question. Please show your effort.

If you would state what Le Chatelier's principle is, that may lead towards an answer.

You are absolutely correct. Adding $\ce{HCl}$ adds protons to the mix, which you can imagine as applying pressure on the right-hand side, which means that the reaction will shift backwards slightly and the acetate concentration will decrease. For $\ce{NaOH}$, you also need to consider the underlying equation:
$$\ce{H2O <=> H+ + OH-}$$
which will be shifted to the left by additional hydroxides. This means that the overall proton concentration decreases and you can imagine that as lifting a weight from the right-hand side: The reaction will shift forwards.


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Metal carbonyl bonding
I have been going through this topic in my coordination chemistry chapter. I read that in metal carbonyls, -CO group acts as a ligand and donates its lone pair of electrons into a vacant orbital of the metal to form a sigma bond. Also, there's a pi bond formation by the donation of electrons from the filled d orbital of the metal into pi* MO of the -CO molecule, and this pi bond strengthens the metal to ligand bonding.
But filling of electrons in the antibonding MO means decrease of bond order and correspondingly decrease of bond strength and bond lengths. It sounds contradictory to what's written in my book.
Where am I making a mistake?

That's right, it weakens the C-O bond, but strengthens the metal-C one.

Related: Will the bond order of CO be greater in its existence as carbon monoxide or in metal carbonyl?

Carbon monoxide bonds to transition metals using "synergistic π* back-bonding." The bonding has three components, giving rise to a partial triple bond. A sigma bond arises from overlap of the nonbonding (or weakly anti-bonding) sp-hybridized electron pair on carbon with a blend of d-, s-, and p-orbitals on the metal. A pair of π bonds arises from overlap of filled d-orbitals on the metal with a pair of π-antibonding orbitals projecting from the carbon atom of the $\ce{CO}$. The latter kind of binding requires that the metal have d-electrons, and that the metal is in a relatively low oxidation state ($<+2$) which makes the back donation process favorable.

As electrons from the metal fill the π-antibonding orbital of $\ce{CO}$, they weaken the carbon-oxygen bond compared with free carbon monoxide, while the metal-carbon bond is strengthened. Because of the multiple bond character of the $\ce{M-CO}$ linkage, the distance between the metal and carbon atom is relatively short, often $< 1.8~\mathrm{Å}$, about $0.2\ \mathrm{Å}$ shorter than a metal-alkyl bond.

Not all carbonyl complexes will be acurately represented by a $\ce{M#C-O}$ depiction as your description would imply. In fact, $\ce{M=C=O}$ is much more common. Otherwise a good answer. (Please note that the unit is Å, no Â ;))

So, if I have to compare the C-O bond length in a few complexes with Isoelectronic metal atoms, will the complex with the metal ion in its lowest oxidation state have the strongest M-C bond? But if they are isoelectronic shouldn’t the bond strengths be the same in all of them?


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Removal of oxygen from pyridine N-oxide
Under what condition do the reaction proceed with sole removal of oxygen in contrast with the product also ending up with chlorine substituent?

I'm familiar with both reaction mechanism, but can't seem to figure out when will a certain pyridine end up with or without chlorine. I guess one factor would be electron richness of the ring system, chlorine being deactivating subsituent ends up only on the electron rich pyridines.

POCl3 is phosphorus oxychloride.  What is POCl?

Indeed there's no such thing like POCl - second reaction looks bad

So, the answer should be corrected.....

@Mithoron You mean the whole reaction is not feasible? (omitting now proposed side product)

https://dx.doi.org/10.1016%2FB978-008096519-2.00027-8 looks possible but art. is behind paywall

I'm not sure whether $\ce{PCl3}$ can lead to the installation of a chlorine atom at C-2. I performed a cursory Reaxys search and in the first page saw several examples of deoxygenation without chlorination, but no examples of concomitant chlorination.
As far as I know, treatment of pyridine N-oxides with $\ce{PR3}$ simply leads to deoxygenation, with $\ce{P(III)}$ being oxidised to $\ce{P(V)}$.

However, if you were to treat pyridine N-oxide with phosphorus oxychloride $\ce{POCl3}$, then you would see chlorination at C-2.

If you have a source that says that $\ce{PCl3}$ can deoxygenate and chlorinate, please do inform us. Otherwise as far as I can tell, you need $\ce{POCl3}$ to effect a chlorination.
References

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010; pp 121, 155.
Weickgenannt, A. Pyridine N-Oxides, Baran Group Meeting, June 9, 2012; p 3.

In practice, phosphorus oxychloride is the byproduct after removing the oxygen from the N-oxide; therefore would the parent N-oxide be chlorinated and deoxygenated as in reaction #2 if removing the byproduct from the reaction mixture is not effective? In other words, can we use another deoxygenating agent? Off the top of head, I would imagine use a alkali metal sulfite $\ce{Na2SO3}$, can the byproduct alkali metal sulfate $\ce{Na2SO4}$ can be disposed off and does not halogenate.

I would argue that electron withdrawing groups will favour the oxidation to POCl3 while electron donating groups will favour the POCl path.
$$\ce{HPyN+-O- + PCl3 <=> HPyN+-O-PCl2 + Cl- <=> Cl-HPyN-O-PCl2}$$
With electron withdrawing groups the oxygen atom is more positively polarized than with electron donating groups, so let the bonds flip to the left:
$$\ce{Cl-HPyN-O-PCl2 + Cl- -> Cl- + HPyN + POCl3}$$
In the opposite case let the bonds flip to the right:
$$\ce{H-ClPyN-O-PCl2 -> H+ + ClPyN + POCl + Cl-}$$

That's what OP says - you should say sth more than he, shouldn't you?

You're right. Now my answer looks pretty redundant to me.

What Mithoron said in the comment section above is that "there's no such thing like POCl". So, you should check your answer.......

POCl does not exist. However, Chlorine in PCl3 can easily substituted by other groups. I guess "POCl" is used to represent such an intermediate product in an informal way.

Hope the edit clarifies what I tried to say.


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How does vinegar get rust off these coins
As experiment I found some really nasty rusty coins on the street. They were green. After looking on a website I found vinegar could eliminate the rust. It worked pretty well.
I know this is a redox reaction. Can you tell which ones? I know these are pesos of various denominations from Dominican Republic from the past 10 or 20 years. So that might help identify the metal.
The vinegar is household apple cidwe vinegar from the supermarket. Low concentration.

The coins are most likely copper. The blueish-green rust is the patina of copper which is a mixture of various copper compounds like copper sulfate, copper carbonate, etc. The acetic acid probably dissolves these salts.

BTW, this is not a redox reaction at all.

Vinegar and salt works better than vinegar alone. Not sure why but the question has been asked here before.

Adding salt introduces chloride ions, which form soluble salts or complexes with most metals. This mixture will not only remove patina, it will further corrode the coin's metal. Please DO NOT use it on any coins that might be valuable or collectible.

As previous comments said, this isn't a redox reaction.
The patina or corrosion is made of metal salts -- for copper, it's copper oxide, sulfide, carbonate, and perhaps other things. Vinegar readily converts all these to soluble copper acetate.
Most coins are made of metal alloys, and those alloys sometimes contain metals that are attacked by acid. Nickel and zinc are common components. Soaking in vinegar will dissolve these metals (they'll displace hydrogen, again forming soluble metal acetates). This damages the coin's surface, making it porous, and eventually reducing the coin's weight. (It also breaks collectors' hearts.)
Adding salt to the mix forms an equilibrium with hydrochloric acid, which attacks metals even more aggressively. Please don't do this unless you don't mind ruining your coins.

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What is the origin of the "positive-to-negative" convention concerning dipole moments in chemistry?
I'm a physics professor who often teaches students who have taken a lot of chemistry (read:  pre-meds).  In physics, the dipole moment of a charge configuration is defined as pointing from concentrations of negative charge towards concentrations of positive charge.  This is embodied in the formula we use for the dipole moment of a charge configuration,
$$
\vec{p} = \sum_i q_i \vec{r}_i.
$$
It's not too hard to see that this yields a vector that points from negative charges to positive charges.  For example, if I have a charge of $-e$ at the origin ($\vec{r} = 0$) and a charge of $+e$ a distance $d$ away, this formula yields a vector that points from the negative charge to the positive charge, with magnitude $ed$.
However, the first time I taught this material in a class with a large number of chemistry students, they were very confused;  they had always learned that the dipole vector of a polar molecule points from positive charge towards negative charge, i.e., in the opposite direction of the convention from physics.  For example, by the physics convention, the dipole moment of the water molecule points away from the point of the "V" formed by the bonds, while under the chemistry convention, the dipole moment points in the direction of the point of the "V".
How did this convention arise?  Is there an advantage to defining the dipole vectors this way in chemistry calculations?  Or is it one of those scientific conventions that someone wrote down one point, and got ingrained in textbooks ever since?  (Is it all Benjamin Franklin's fault?)
One of the answers to this question notes the difference between the conventions, but doesn't really explain why the first chemist to write down a dipole vector wrote it down that way.  I would love to know how this convention arose historically!

This answer by Jan Jensen in the "question" to which you linked indicates that it is a historical problem. IUPAC is now defining electric dipole from  negative to positive also. ( http://goldbook.iupac.org/E01929.html )

All in all the conventions are giving me a headache. Doesn't an electric field go from positive to negative?

I studied Chemistry and I only know your convention, namely that dipoles point from negative to positive charges.

Keep using the physicist's convention, I belive that students should be confronted with these problems and get used to it! ; )

This could have been a very good fit for [hsm.se].

I've never used the "chemistry convention" in theoretical calculations. I didn't know that this exist, I can't see the advantage of the definition, I find it ugly.
$$ T = p \times E$$ (chemistry)
$$ T = r \times F$$ (physics)
$$ T = r \times F= p \times (qE)= (q r) \times E$$ so $$p = qr$$
Any ideas? Who uses that definition?

@user1420303 -  The vector style notation you are using is from the US being first published in the textbook 'vector analysis'  in 1901.  In 1905 it was translated into German.  Even Einstein said he would have written his papers differently prior to 1905  had he known about the book.   In 1912, when the concept of the dipole moment was developed, the vector notation was still new.  New and ugly often go together in science,  try looking up a picture of the first transistor.

@Agriculturist Thank you, it is very interesting from a historical perspective. I didn't get which notation was used in 1912, the old one? As I understand the new (form US) is prettier. Isn't it?

The 1912 form is most likely a nonstandard form which would take about 30 minutes to figure out, or a standard form with some kinks still in it.  Time tends to make calculations more elegant, so I agree on the new form being elegant.  I am hoping someone fluent in German could look up and show the original calculations.

Some sources point out the direction of the dipole moment is convention dependent. Since both conventions are mathematically correct it seems important when solving a problem to identify the convention.  Both conventions seem to be derived from the bones of dead theories.
The physics convention does seem to have Franklin's work identifying negative "resinous electricity" and positive "vitreous electricity" as one step in the progression to the modern theory.  In modern theory a positive to negative electric field will orient a dipole such that the field is in the same direction as a negative to positive dipole.
On the chemistry side, Peter Debye (who the unit for a dipole moment is named) developed the concept of the dipole moment in 1912 to explain anomalous dispersion. This dispersion was explained as being caused by a negative charge collecting on one side of a molecule leading to a dipole moment. This moment's magnitude explained anomalous dispersion, so direction did not seem to receive much attention.  Molecules during this time where described using the Thompson, or "plum pudding," model. In this model molecules were described as positively charged spheres.  (These spheres had no nucleus because the nucleus concept was in its infancy during this time period.) These spheres formed the basis of mathematical models used to describe the behavior of materials with a dipole moments.  The dipole moment in this sense went from the center of a molecule to the point of highest electron density.  With the center of the sphere as a pivot point, the rotation of molecules in an electric field could be explained. In this sense the positive to negative convention may be an artifact of the early spherical molecular models first used to describe dipole moments.
sources:
J. W. Williams   Dipole theory and the size of molecules.  Trans. Faraday Soc., 1934,30, 723-728
John W. Williams The Dielectric Constants of Binary Mixtures. VIII. The Electric Moment as a Vector Quantity, J. Am. Chem. Soc., 1928, 50 (9), pp 2350–2357  DOI: 10.1021/ja01396a005  Publication Date: September 1928
L. E. Sutton  Electric dipole moments and resonance in molecules, Trans. Faraday Soc., 1934,30, 789-801
https://www.quora.com/Why-does-a-dipole-have-a-direction-from-negative-to-positive-while-electric-field-lines-go-from-positive-to-negative
http://goldbook.iupac.org/E01929.html
http://www.ushistory.org/franklin/info/inventions.htm
http://www.wondersandmarvels.com/2011/12/accentuate-the-vitreous-eliminate-the-resinous.html

This is a much better answer than mine so it should get upvoted to ensure if anyone gets the bounty (assuming no better answers) it's you.

The chemistry convention seems German while the physics convention seems British.  I can read the original physics theory because it is in English but not German theory.  The dipole theory was developed throughout the years of WWI and WWII.  These wars may have contributed to the lack of unity in a single convention.

I don't have a very good historical explanation, but I would point out that it actually makes a good deal of sense for chemists to think in this way. That is, chemists are almost exclusively concerned with what the electrons in any system are doing.
For instance, when I ask myself about a charge separation and hence a dipole, I want to know what is separated from what. In this case, it's some electrons separated from some space they could be in. So, I intuitively think to point my arrow towards where the electrons are, not where they could be. Of course I only think that because that's the convention. I'd also point out that chemists reverse the arrows for current as well by drawing the arrow pointing with the flow of electrons rather than the flow of positive charge which goes the opposite direction.
So at least we're self consistent!
Sorry that's not more historical, but my guess is that at least part of the true historical reason is that chemists think in terms of electrons and thus point the arrows to their negatively charged friends.

True. Even in reaction mechanisms in organic chemistry we depict the flow of electrons by arrows pointing from higher electron density to lower electron density.

This is certainly true.  Still as a chemist I have never saw a problem or inconsistency. One thing is depicting Lewis formula and discussing reaction mechanisms, another is to calculate orientation of a molecule in an external field. And I would stress that the vector moment of electric dipole goes from negative to positive in chemistry, too. https://goldbook.iupac.org/html/E/E01929.html

Just a guess:
For many years, until 1946, accurate values of dipole moments could only be derived from measurements of dielectric constants.
If you think of the dipole moment as weakening the electric field, it would point in the direction opposite to the electric field, and therefore from the positive to the negative end of the dipole.
(sorry, cannot comment the post due to lack of reputation)

I asked that other question. Most useful to me was one of the comments, which pointed me to this page: http://av8n.com/physics/electric-dipole.htm#sec-mutations
While it does not have definitive answers, it at least shows that the problem was known in 2007.
For archiving purposes I'm posting a screenshot here (hope that is not in violation of any terms):

I wouldn't be surprised if the problem was known before 2007, but I certainly never knew about it until quite recently.  My first contact with the problem was when I tried to make more explicit connections between physics and stuff the students had seen in other science disciplines — which is something that a lot of physics classes have been trying to do over the past 10–15 years.  Perhaps that's why the problem has become better-known.


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Definition of an amide
If I asked you to draw me an amide, we'd likely end up with something in the form of:

The exact definition of this is given fairly clearly in the IUPAC gold book.

Derivatives of oxoacids [...] in which an acidic hydroxy group has been replaced by an amino or substituted amino group. Chalcogen replacement analogues are called thio-, seleno- and telluro-amides. Compounds having one, two or three acyl groups on a given nitrogen are generically included and may be designated as primary, secondary and tertiary amides, respectively
  Source: IUPAC Gold Book, 1995 recommendations

There is however another class of compounds which we call 'amides', for instance LDA (lithium diisopropylamide). Again, the IUPAC gold book has this covered.

The term applies also to metal derivatives of ammonia and amines, in which a cation replaces a hydrogen atom on nitrogen.

I fully accept this, my question is, why we decided to re-use the word 'amide' to describe these compounds. Is the reason historical (i.e. before we knew better), or is there a logical basis for the grouping of the two classes of compounds.

Without being sure about it, but amide derives from amine much like chloride does from chlorine — and both are used for the negatively charged, dehydrogenated version of their parent. If the formal $\ce{OH-}$ group of an acid is replaced by $\ce{Cl-}$ we arrive at a carboxylic chloride, so it seems only natural to arrive at an amide if we substitute with (formal) $\ce{R2N-}$.

In addition to what Jan said, the carboxylic acid derivative "amide" can, more specifically, be referred to as a carboxylic amide. So, really, we didn't reuse the term, we simply shortened the one in a potentially confusing way. Along the same lines, hexamethyldisiliazane can be deprotonated to form hexamethyldisilazide, so I would bet that an amide formed from HMDS would be a silazide or, more specifically, a carboxylic silazide.

https://en.wikipedia.org/wiki/Sodium_amide

@Jan: I hadn't thought about it like that (come to think of it, I had never really considered what we call N-). I think you're probably right.

The latter amide definition in you question did not evolve from the former name of organic amides, but the process was rather opposite, or at least they have a common root. The name "amide" appears to be quite old. There is even a notion of amidogen, not used very often today, but it appears in the book Elements of Chemistry from 1840, along with some inorganic amides like "amide of potassium" or "amide of mercury". Later book Chemistry, Theoretical, Practical and Analytical as applied and relating to the Arts and Manufactures by James Sheridan Muspratt from cca 1860 besides inorganic amides mentions also a peculiar substance "margaramid" that was obtained as a byproduct of ammonium salts of fatty acids preparation by treatment of fats/oils with ammonia. In this time e.g. acetamide was known as well already.
And the definition in Cyclopaedia of Chemistry with Its Applications to Mineralogy, Physiology, and the Arts from 1854 states:

Amide, Amidogen. The term amide is given to a class of substances which contain ammonia deprived of an atom hydrogen, (NH2.) Hence, when they are treated with potash, ammonia is disengaged, the third atom of hydrogen derived from water again uniting, so as to form the necessary quantity in ammonia. That such a substance as amide or amidogen exists, may be proved by passing dry ammonia over heated potassium or sodium. A green body is formed, as hydrogen is evolved, which is replaced by the metal, NH2H + Na, becoming NH2Na and H. Some of the most familiar of these amides are potassamide NH2K, sodamide NH2Na, oxamide NH2C2O2, benzamide NH2C14H5O2, sulphamide NH2SO2, carbamide NH2CO, platosamine NH2Pt. Some of these amides are formed by heating organic salts of ammonia, by which they lose water, and leave an amide. Thus oxalate of ammonia, when distilled, gives water and oxamide (NH3C2O3 = NH2C2O2 and HO). It is probable that many animal substances, as albumen, belong to this class.

(typos/mistakes included)
Please note that this is not an exhaustive historic research at all :)


text:
Is it possible to quantify the effects of a chelating agent?
If one wants to see what compound is a more effective chelating agent, how does that get measured?

You could take an NMR of an equimolar Lewis acid/base mixture and see how much free base there is relative to the how much chelate there is. Are you specifically referring to polydentate coordinating agents or any coordinating Lewis bases?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] for unanswered questions about how it works.

Generally, a good starting point is the Formation Constant (β) between the metal and the ligand. The greater the constant, the better the ligand is be at chelating the metal center (as evidenced by the free energy change).

Yes, you can always look this up, but how are they measured in the first place. That might provide a better answer to the question.


text:
What is the possible side product when mixing citric acid, NaCl, Cu, Al, H2O2, and an unknown photoresist?
When etching circuit boards, a somewhat commonly used substitute for (expensive and messy) ferric chloride is a mixture of an acid, a source of chloride ions, and hydrogen peroxide. If I understand the chemistry correctly, the acidic environment allows the copper to react with the chlorine ions to make $\ce{CuCl2}$, which will more slowly react with the copper again to make $\ce{2 CuCl}$. Adding oxygen to the solution (from the peroxide or just by aerating it) makes $\ce{CuO}$ and $\ce{CuCl2}$ again, allowing the process to repeat.
I meant to mix $\ce{H2O2}$, acetic acid and table salt, but accidentally used citric acid instead of acetic acid. After realizing my mistake, I added scraps of aluminum foil to the mixture to displace the copper, and render the mixture safer to dispose of. I went to filter out the copper and dispose of the rest, when I found that there was a layer of some kind of gelatinous substance coating the exposed surfaces in my container. Does anyone have any idea of what it could be, and how I might safely dispose it?

Initial guess: some aluminate. Not sure, though.

1) You should not use acetic acid (or any organic acid) in this setup. It potentially may form some (even if minuscule) amount of chloracetic acid which is really hazardous. Furthermore, copper acetate (and other carboxilates) has low solubility. Go for either genuine $\ce{HCl}$ or for $\ce{H2SO4}$ used in lead accumulators.
2) Most likely it is aluminium hydroxide. If the substance is bluish, it may be also copper hydroxide.

Regarding chloroacetic acid, I was under the impression that it is only dangerous on contact, and that any amount present when I finally go to dispose of it would be easily neutralized by baking soda. Regarding my gelatinous substance, it does seem to be a mixture of aluminum hydroxide and sodium citrate.

Oh, and don't forget the Na from the salt. The intended overall reaction is 4 Cu + 4 NaCl + 4 CH3COOH + O2 -> 4 CuCl + 4 CH3COONa + 2 H2O


text:
pKa values and basicity
In considering basicity (particularly of amines), often I find included with the compound a pKa. I understand that the higher the pKa of the conjugate acid, the stronger the base. But when the pKa is listed below a table of bases (for instance the wikipedia page for LDA – 36), does that refer to the pKA of the compound itself (LDA) or the protonated form?

That would be the pKa of the protonated form.


text:
"Synthetic wood" from cellulose and lignin?
Is there any research into making wood from its constituents?
From what I know, wood is composed of cellulose, hemicellulose, and lignin. We can extract these from raw wood, but is there any way we can combine these to form a wood-like structure that can be made into furniture?

What motivates your question?

The US published application US20110190402 titled “Synthetic wood composite” appears to be very close to answering your question: “Is there research into making wood from its constituents.”  The full text can be found here: https://patentscope.wipo.int/search/en/detail.jsf?docId=US73316434&tab=PCTDESCRIPTION&_cid=P10-KGFN8M-16074-4
It appears to me that the short answer to your question is, Yes!
Below is part of the Summary of the Invention and Independent Claim 1.
SUMMARY OF THE INVENTION
The present invention is directed to a synthetic wood composite comprising biomimetic macromolecules and methods for the preparation thereof.
In one embodiment, the invention is a composite comprising at least two polymeric macromolecules, wherein at least one macromolecule is lignin and the other macromolecule is selected from the group consisting of cellulose and hemicellulose, or a combination thereof.
In some embodiments, the composite comprises lignin, cellulose and hemicellulose. In other embodiments, the hemicellulose is xylan.
Claim 1 – the main claim

A composite comprising lignin, cellulose and hemicellulose, wherein the cellulose content is about 40 to about 60% (w/w), the hemicellulose content is about 20 to about 40% (w/w), and the lignin content is about 10 to about 30% (w/w), wherein the lignin is selected from a kraft lignin or a lignosulfonate, wherein the hemicellulose is xylan, and wherein the composite is transparent.


text:
What is the Series Limit in a Spectral line
What is exactly the series limit in a spectral line graph? Do we talk about series limits for each series (Eg: Lyman series in Hydrogen) or collectively for an atom?
Is series limit the transition having least energy or least wavelength?
So like for Lyman series, will series limit be that corresponding to $\ce{1 -> 2}$ or $\ce{1 -> Ionize}$?

This earlier answer may be helpful.

Every series of wavelengths produced from the hydrogen atom excitations have a certain minimum limit. These minimum wavelengths are called the series-limit for that particular series. Take a look at this picture for example:

The minimum series of the Lyman series would be for the largest transition, which is from ∞$\ce{ -> }$1. Plugging the values into the Rydberg's equation:
$$\frac{1}{\lambda}=R_{∞}\left(\frac{1}{1}-\frac{1}{∞}\right)$$
That should give us a value of 91nm. Which is the series-limit for the Lyman series.
Now you can find series-limits for the other series as well.

Yes,  series limit is for all series . A separate value of series limit is given for each series, such as the Lyman, Paschen, Brackett, Pfund and Humphrey series. Hope my notes will help you.


text:
What is the proper way to calculate experimental error in this measurement?
I am working on error analysis of some of my measurements and have stumbled upon an issue with the propagation of error. According to this reading,

The uncertainty of the average acidity ($\Delta_\text{acid}H_\text{avg}$) was calculated as the root sum square of the random and systematic errors. For example, for the $\ce{A3CSH}$ system, the random error was treated as the averaged uncertainty of the reference acids $\pm2.2\ \mathrm{kcal/mol}$ divided by the square root of the number of the reference acids, $2.2/\sqrt6 = 0.9\ \mathrm{kcal/mol}$, and the systematic error was assigned as $\sqrt{2.2} = 1.5\ \mathrm{kcal/mol}$. The root sum square of the random and systematic errors yielded $\sqrt{0.92^2 + 1.52^2} = 1.7\ \mathrm{kcal/mol}$.

Why is $\sigma_\text{sys}=\sqrt{2.2}$ instead of $\sigma_\text{sys}=2.2$?
It was my understanding that if the reported uncertainty of the reference acids are $\pm2.2\ \mathrm{kcal/mol}$, shouldn't the root-sum-square be $\sigma_\mathrm{RSS}=\sqrt{\sigma_\text{sys}^2 + \sigma_\text{rand}^2}=\sqrt{2.2^2 + (2.2/\sqrt6)^2 }= 2.4\ \mathrm{kcal/mol}$ instead.
If my method is wrong, can I have an explanation pr directed some kind of reading explaining this concept?

There is no a priori way to calculate a systematic error from the measured values (i.e. like average and standard deviation). You'd have to look deeper in the source of the text to determine how the authors justified that.


text:
Why is pressure or external force needed for combination of electron and proton in plasma phase?
I have studied that after rising temperature matter is converted into plasma. I want to  know that whether in plasma phase nucleon coexist with electron but do not combine until pressure is applied. Why?


text:
Is trisulfuric acid hypothetical?
In the preparation of sulfuric acid through contact process,

Hot sulfur trioxide passes through the heat exchanger and is dissolved in concentrated $\ce{H2SO4}$ in the absorption tower to form oleum:
$$\ce{H2SO4(l) + SO3(g) → H2S2O7(l)}$$
Oleum is reacted with water to form concentrated$\ce{H2SO4}$.
$$\ce{H2S2O7(l) + H2O(l) → 2 H2SO4(l)}$$

The sulfuric acid obtained is $\ce{96-98}$% pure. That means $\ce{2-4}$% is impure and contains byproducts and other oxoacids of sulfur including trisulfuric acid. If the sulfur trioxide to be adsorbed on $\ce{H2SO4}$ is in excess, it can form trisulfuric acid.
$$\ce{H2SO4(l) + 2SO3(g) → H2S3O10(l)}$$
But the amount of trisulfuric acid is negligible and can be omitted. But is it an actual compound? Can it be isolated? Can it exist in room temperature?
Pubchem identify it as a compound and also has a unique ID. It also has an unique CAS number.

It has $\ce{S=O}$ bonds and $\ce{S-OH}$ bonds (characteristics of oxoacid of sulfur) but still it is not mentioned in that list (link above). Why?

It is mentioned on this list, you should read carefully.

@Mithoron Are you saying H2SO4.2SO3 is trisulfuric acid? If it is so, there are no proof of its existence. "It has not been characterized yet" as per the answer.

It seems that this compound has not yet been characterized, although traces of it could be found in mixtures of $\ce{H2SO4}$ and $\ce{SO3}$. The most recent paper I could find is this one from 2012, showing that salts of the respective anion exist and theoretical calculations imply that the acid is stable with respect to decomposition into $\ce{H2S2O7}$ and $\ce{SO3}$. The conditions under which it may be stable are however unknown.

I’m pretty sure that it’ll be very unstable to hydrolysis and possibly even long-term unstable to sulphuric acid solvolysis ($\ce{H2S3O10 + H2SO4 <=> 2 H2S2O7}$).

$\ce{H2S3O_{10}}$ has not been isolated as an acid, but Logemann et al. [1] have isolated the lead salt $\ce{PbS3O_{10}}$ by heating a chloroplumbate salt in oleum.
The ion is found both experimentally and theoretically to have unequal sulfur-oxygen distances along the bridges between the sulfate units: each bridging oxygen is closer to the central sulfur atom (155 pm) than to the nearer terminal sulfur atom (169 pm). The bridging bond length in the disulfate ion is intermediare, 164.5 pm [2]. This suggests that the central sulfate unit is more weakly bound to the two outer $\ce{SO3}$ monomers than it would be to a single $\ce{SO3}$ monomer in the disulfate ion.
This bond-length feature, and the relatively weak bonding to the outer $\ce{SO3}$ monomers, is theoretically calculated to hold for the parent acid $\ce{H2S3O_{10}}$ as well. That would imply that although Ref. [1] renders $\ce{H2S3O_{10}}$ stable with respect to decomposition, it would react with monomeric $\ce{H2SO4}$ to form two molecules of the more strongly bridged $\ce{H2S2O7}$.
Reference
1.
Logemann, Christian & Klüner, Thorsten & Wickleder, Mathias. (2012). "Synthesis and Characterization of the Trisulfate Pb[S3O10] and Theoretical Analysis of H2S3O10". Zeitschrift Fur Anorganische Und Allgemeine Chemie. 638. 758-762. https://doi.org/10.1002/zaac.201100533.
2.
H. Lynton  and  Mary R. Truter (1960). "An accurate determination of the crystal structure of potassium pyrosulphate". J. Chem. Soc. 35. 5112-5118. https://doi.org/10.1039/JR9600005112.

Haven't the first reference and conclusions therein already been mentioned in snurden's answer?

I do go slightly further by discussing the relative bridging bond lengths. Since the outer $\ce{SO3}$ monomers are only weakly bound, a molecule of $\ce{H2SO4}$ could grab one of these to make a stronger bridge in the dimer.

The trisulfuric acid $\ce{H2S3O10}$ hasn't been isolated yet, but it exists and has been characterized as a chain of three $[\ce{SO4}]$ tetrahedra sharing vertices and one oxygen atom from each of the terminal $[\ce{SO4}]$ tetrahedra being protonated (the structure is oversimplified and doesn't account for accurate bond order and charge distribution — for more details see e.g. Why is hydrogen sulfate put together as it is?):

The formation of polysulfuric acids $\ce{H2S_nO_{2n+1}}$ with $n\in [1;4]$ in solutions of $\ce{SO3}$ in $\ce{H2SO4}$ has been proven by means of Raman [1] and infrared [2] spectroscopy.
According to Walrafen [1, p. 2340], $\ce{H2S3O10}$ possess at least five characteristic frequencies in Raman spectrum:

Five frequencies listed in Table III, viz., 230-250, ~325, ~350, ~685, and ~1455 cm−1, are those of $\ce{H2S3O10}.$
The bands at 230-250, ~325, and ~350 cm−1 are associated with deformation vibrations of the $\ce{S''-O-S'-O-S''}$ grouping of $\ce{H2S3O10}.$
The band at ~685 cm−1 is probably produced by $\nu_\mathrm s\text{-}\ce{S''-O-S'-O-S''},$ and the band at ~1455 is related to asymmetric valence vibrations of the central or terminal $\ce{O=S=O}$ groups.

and is mostly present in oleums with the molar fraction $x(\ce{SO3}) \approx 0.75:$

Fig. 9. Comparison of $c(\ce{SO3})$ with the sum $[\ce{H2SO4}] + 2[\ce{H2S2O7}] +3[\ce{H2S3O10}] + [\ce{SO3}] + 3[\ce{S3O9}]$ obtained from Raman data.

Stopperka [2, p. 280] assigns characteristic antisymmetric valence vibrations of central $\ce{SO2}$ group (1480 cm−1) as well as terminal $\ce{—O—SO2—OH}$ groups (1457 cm−1).
The intensities of the latter vibrations quickly decrease with an increase in molar concentration of $\ce{SO3}$:

Kleine Absorptionsmaxima bei 1457 und 1487 cm−1, die in 90proz. Oleum den antisymmetrischen $\ce{SO2}$-Valenzfrequenzen der $\ce{H2S3O10}$ zugeordnet wurden (siehe hierzu Mitt. II) gehen in ihrer Intensität ebenfalls zurück.
Bemerkenswert ist dabei, daß die schwache Bande bei 1480 cm−1 in ihrer Intensität nicht so rasch abfallt, als die bei 1467 cm−1 gelegene.
In der vorangegangenen Mitteilung (1) hatten wir die Bande bei 1480 cm−1 einer $\nu_\mathrm{as}$ $\ce{SO2}$ zugeordnet, die von einer mittelständigen $\ce{SO2}$-Gruppe der $\ce{H2S3010}$ herrührt.
Dementsprechend wurde das schwache Absorptionsmaximum bei 1457 cm−1 auf eine $\nu_\mathrm{as}$ $\ce{SO2}$ der $\ce{—O—SO2—OH}$-Endgruppe zurückgeführt.
Der mit zunehmender $\ce{SO3}$-Konzentration beobachtbare unterschiedliche Intensitätsabfall führt zu dem Schluß, daß die Konzentration an
$\ce{—O—SO2—OH}$-Endgruppen mit zunehmendem $\ce{SO3}$-Gehalt schnell absinkt, wogegen $\ce{—O—SO2}$-Gruppierungen noch deutlich bis zu relativ hohen $\ce{SO3}$-Konzentrationen nachgewiesen werden können.

A series of hydrogentrisulfate salts of alkali metals $\ce{A[HS3O10]}$ $(\ce{A} = \ce{Na}, \ce{K}, \ce{Rb}),$ (which is arguably the closest derivative of trisulfuric acid) have been isolated and characterized with single-crystal x-ray diffraction [3], also supporting the structural motif of found using spectroscopy:

Figure 2. Anions linked to dimers in the crystal structure of $\ce{Na[HS3O10]}.$

Given $\ce{H2S3O10}$ is predicted to be Brønsted-Lewis superacid [4], its isolation is far from trivial and existence under ambient conditions is highly questionable.
References

Walrafen, G. E. Raman Spectral Studies of Oleums. J. Chem. Phys. 1964, 40 (8), 2326–2341. DOI: 10.1063/1.1725511.

Stopperka, K. Infrarotspektroskopische Untersuchungen an den flüssigen Systemen $\ce{SO3-H2O}$ und $\ce{SO3-D2O}$. III. Die Schwingungsspektren von flüssigem Schwefeltrioxid. Z. Anorg. Allg. Chem. 1966, 345 (5–6), 277–289. DOI: 10.1002/zaac.19663450506.

Schindler, L. V.; Klüner, T.; Wickleder, M. S. Towards Polysulfuric Acids: The Hydrogentrisulfate Anion $\ce{[HS3O10]}$ − in $\ce{A[HS3O10]}$ $(\ce{A} = \ce{Na}, \ce{K}, \ce{Rb}).$ Chem. Eur. J. 2016, 22 (39), 13865–13870. DOI: 10.1002/chem.201602176.

Koppel, I. A.; Burk, P.; Koppel, I.; Leito, I.; Sonoda, T.; Mishima, M. Gas-Phase Acidities of Some Neutral Brønsted Superacids: A DFT and Ab Initio Study. J. Am. Chem. Soc. 2000, 122 (21), 5114–5124. DOI: 10.1021/ja0000753.


text:
Negative potential of the oxidation of iron(II) to iron(III)
The oxidation potential which I found on the internet the following reaction
$$\ce{Fe^{2+} -> Fe^{3+} + e^-}$$
is $\pu{-0.77 V}$. But how can it be negative? Negative oxidation potential means oxidation tendency is very less or we can say negative oxidation potential means $\ce{Fe^2+}$ has very less tendency to lose electrons. But it doesn't make sense, because $\ce{Fe^3+}$ is in $\mathrm{3d^5}$ configuration, which is half-filled $\mathrm{d}$ orbital, which is very stable than $\ce{Fe^2+}$. So, $\ce{Fe^2+}$ should get readily oxidize to $\ce{Fe^3+}$. Then why do we have negative oxidation potential for this reaction?

First what has to be said is signs of the potentials, which obviously different in USA and versus other world. I am used to know that Li oxidation takes place at -3e whereas Fe2+ oxidation-at -0.7V versus SHE

As IanB2016 mentioned in his answer, the redox potential of a half-cell reaction is always in reference to another half cell.  The usual "reference" half cell in electrochemistry and thermodynamics is the standard hydrogen electrode (SHE).  Some key features of the SHE:

the half cell reaction is $\ce{2 H+(aq) + 2e- <=> H2}$, i.e. the reductant is hydrogen gas and the oxidant is aqueous protons.
the pressure of hydrogen gas for a standard hydrogen electrode is 1 atmosphere.
the concentration of acid in the aqueous electrolyte is 1 molar, which corresponds to a pH of 0.

The oxidation potential of a species is related to its strength as a reducing agent.  Things that have a negative oxidation potential are less reducing than hydrogen gas, and things that have a positive oxidation potential are more reducing than hydrogen gas.  Lithium metal, for example, has an oxidation potential of more than three volts, i.e. $\gt+3\;\mathrm{V}$.  It is an extremely powerful reducing agent, far more powerful than hydrogen gas.
As you note, $\ce{Fe^{2+}}$ has an oxidation potential of $-0.77\;\mathrm{V}$, meaning it is a less powerful reducing agent than hydrogen gas.  This makes sense in several ways:

aqueous solutions of iron(II) salts do not spontaneously reduce aqueous protons to hydrogen gas.  If you mix iron(II) sulfate with water, for example, hydrogen bubbles do not spontaneously form.
Hydrogen gas releases more energy when burned in oxygen than iron(II) salts do when they react with oxygen.  Hydrogen reacts explosively, whereas iron(II) salts do not spontaneously explode or catch fire when they are oxidized by oxygen.

Finally, note that the oxidation potential of water (via the reaction $\ce{O2(g) + 4 H+ + 4 e− <=> 2H2O}$ at pH 0 at 1 atm of pressure) is -1.229 volts.  Water is a very weak reducing agent, which means that oxygen is a strong oxidizing agent, stronger than iron(III).  This means:

iron(III) salts do not spontaneously oxidize water to produce oxygen.
conversely, oxygen does spontaneously oxidize iron(II) to iron(III).  Solutiosn of iron(II) salts are thermodynamically unstable under aerobic (oxygen-rich) conditions.

Redox (reduction - oxidation) Potentials are measured for a specific reaction half-cell [such as Fe2+/Fe3+] by comparing that half cell reaction to a Standard Reference Electrode potential, which is generally the Standard Hydrogen Electrode or cell (SHE) in [aqueous] solution. The SHE is the internationally accepted "Standard Electrode" in electro-chemistry and has arbitarily been asigned a reference potential of 0.0 Volts under 'standard  conditions of temperature, pressure and species activities'. Note that there are many other different "practical" reference cells used in industry because the SHE is not very user friendly to use in a practical sense - the reasons for this are outside the scope of this question. Because of the arbitarily assigned value of 0.0 Volts for the SHE a specific half cell reaction can therefore have a value (reaction potential) which is either higher or lower than the SHE potential under the specific conditions previously mentioned. When this potential is higher than the SHE potential that potential is assigned a positive value and when it is less than the SHE potential it is assigned a negative value. Hence the reason the Fe2+/Fe3+ cell has negative sign before the 0.77V (actually +ve 0.79V in the British System when written as a Reduction Potential). Such positive and negative potentials are a direct measure of the half cell's ability to undergo reduction/oxidation as measured with respect to the SHE. Remember also that when considering reaction potentials you cannot simply consider one half cell reaction potential in isolation since both reduction / oxidation reactions occur together for any cell and therefore must be consider together. There are many more indepth explanations available but hopefully this answer will suffice as a "simplistic overview" for the question asked.
One further thing I forgot to mention earlier is that you need to be careful about where the potentials you search on the internet originated from because different countries (eg, USA & England for example), have different ways of expressing the same Standard Potentials which can reverse the sign of the potential for the same half cell. For example in the British System these potentials are written has "Standard (Reduction) Potentials [at 25  Deg. C]" and the half cell in question is written as Fe3+ + e <=> Fe2+ which has a potential of +ve 0.79 Volt (note not -ve 0.79V)! It is crucial to stick with one system and do not mix them up and particularly so when using the Nernst Equation for solving oxidation / reduction type problems.
A very helpful mnemnoic I used when I was at Uni studying Chemistry back in the late 1960's - 70's was [in the British System when using these Standard potential tables] to determine whether or not a reaction could be possible was: emphasized textThe oxidised form of the one above [more positive form] in the table would oxidise the reduced form of the one below in the table.  Having the mnemonic table condition met for a particular RedOx cell doesn't necessarily mean that reaction will proceed either in part or in full because this approach does not include anything about reaction rates or kinetics - which enters into a whole different ball game!  Hope this helps to understand these sometimes confusing aspects of electro-chemistry which can be complicated by different countries different systems of Standards.
As previously mentioned you need to be careful about the sign attributed to an electrode potential because it differs according to the particular sign convention being used; what I have used is the one recommended by the International Union of Pure and Applied Chemistry (IUPAC). So the best way to answer your query is as follows: if the system Mn+/M (where n+ is the valence of M) is a better reducing agent than Hydrogen, under standard conditions, the electrode potential is negative and if it is a poorer reducing agent, the potential is positive. Further, when the standard electrode potentials of the elements are arranged in order of the most negative [Li+e = Li -3.04V, at the top] to the to the most positive [F2 + 2e- = 2F +2.87V, at the bottom] constitutes the electrochemical series of the elements which may be used to interpret and also predict the reactivity of the elements. A cation-forming element will displace from aqueous solution another element which lies below it in THIS series and an anion-forming element will displace from aqueous solution another element which is above it in the series. Thus the order in this series is Li+/Li < ...... < ..... < 1/2F2/F-. trust this now clarifies this for you. [[Ref Bell & Lot 2nd Ed, Modern Approach to Inorganic Chemistry]]

ya I understand, you said that the "When this potential is higher than the SHE potential that potential is assigned a positive value and when it is less than the SHE potential it is assigned a negative value." Ok, so do you mean its negative because it has less reducing power than hydrogen?

Refer to the bottom of my original answer - added as an edit due to insufficient space available in this section.


text:
In the periodic table, why doesn't the 2nd row have exactly 2 elements?

Row 1: 2 elements
Row 2: 8 elements
Row 3: 8 elements
Row 4: 18 elements
Row 5: 18 elements
Row 6: 32 elements
Row 7: 32 elements
In other words: 2, 8, 8, 18, 18, 32, 32
Why does the first row have only 2 elements, but all the next rows have a pattern? There's a pattern of two rows having the same number of elements that only starts after the first row.
EDIT: I just noticed this alternate periodic table! It's first two rows have two elements each! Why does it work there? (I can't interpret the s, p, d, f part on the left). Which one is the right periodic table?

It's just because the order of filling goes 1s -> 2s -> 2p for the 1st and 2nd rows, but for the 4th row it's 4s -> 3d -> 4p. The difference is that the 3d only comes in after the 4s. If the order of filling was the apparently more "normal" 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f... then your 1st to 4th rows would have 2, 8, 18, and 32 elements.

So in a sense, the 1st and 2nd rows are not the "odd" part out. The 3rd row is the "odd" one out because if the order of filling was 3s 3p 3d it would have 18 elements. However, because the 3d orbitals are only filled after 4s, the d-block is delayed from row 3 to row 4.

@orthocresol I'm not sure I understand what you mean. Please turn it into a detailed answer if you can.

The pattern is better expressed this way:
Row 1: 2 elements
Row 2: 2+6 elements
Row 3: 2+6 elements
Row 4: 2+6+10 elements
Row 5: 2+6+10 elements
Row 6: 2+6+10+14 elements
Row 7: 2+6+10+14 elements

The reason comes down to how the electrons fill the available energy levels.  The thing that differentiates one element from another is the proton number, and each time a proton is added and a new element defined it requires one more electron to neutralise the charge.  That electron naturally occupies the lowest available energy level in the atom.
The energy levels available are defined by the quantum state, including the quantum numbers $n, l,$ and $m_l$.  $n$ is the principle quantum number and relates to the period the element is in, or the shell.  $l$ is the angular momentum quantum number which defines the sub shell s, p, d, f, of which there are $n$ subshells whose values are $l=0,\dots {n-1}$.  The magnetic quantum number $m_l$ further subdivides the subshell into orbitals, of which there are $2l+1$ orbitals whose values are $m_l=-l,\dots {+l}$.
subshell                 number of orbitals    subshell
label     l value       (number of ml values)  electron capacity
 s         0               1                   2
 p         1               3                   6
 d         2               5                   10
 f         3               7                   14

Each orbital (i.e. each value of $m_l$) may contain 2 electrons  The available quantum numbers are:
n    l values   (subshells)   ml values        total shell electron capacity
1    0           (s)           0                 2
2    0,1         (s,p)        -1,0,1             2+6
3    0,1,2       (s,p,d)      -2,-1,0,1,2        2+6+10
4    0,1,2,3     (s,p,d,f)    -3,-2,-1,0,1,2,3   2+6+10+14

The pattern is there, but it doesn't seem to match up because by row 7 you would have 98 electrons in the shell and might expect the row to contain 98 elements.  This is not the case!  The energy levels for the $l$ values with large deviations from $0$ (i.e. orbitals with high angular momentum) become increasingly far apart, so even though 3d orbitals exist in the third row, they are so much higher in energy than the 3p orbitals that they are higher even than the 4s orbitals.  This happens again in row 4 where 4f orbitals are so much higher in energy than 4d orbitals, they are even higher than 5s.
The different in energy is not always so large in specific cases, so there are examples where bonding or ligands or symmetry cause the energy levels to switch around, but otherwise the trend is true. If they can produce (or discover) some elements in the next row, we might expect to see the 5g shell finally start to get filled and have a new block in the table, but realistically it doesn't look like they will be stable atoms.
I think I'm missing the reason why we arrange the elements into this table in the first place.  The periodic table is arranged so that all elements in the same column have the same number of outer electrons.  This is useful because elements with the same outer shell configuration of electrons react in similar ways, so they are grouped together in columns. This has the effect of creating 'periods' that begin with a reactive metal and end with an inert gas. Row 1 has two elements, and it's special in a way because it's the first in the series.  Hydrogen has 1 electron, and a maximum capacity of 2 electrons.  This means it behaves in a similar way as other elements that only have 1 electron in their outer shell (group 1), and also in a similar way to elements that only need 1 more electron to complete their outer shell (group 7).  Where do we position it then?  Often it is placed somewhere in the middle.  Helium then has a full outer shell, which means it has many similarities to other elements with a full outer shell and goes in group 8 (or group 0) on the right.
Edit:
The alternative period table that you mention, the left-hand version, is laid out so as you read it from left to right you fill up the orbitals.  This takes in to account that situation I describe where 3d orbitals are higher than 4s, and using this left-hand table you clearly see that. However this alternative table does not allow you to easily read the valance shell configuration like the Mendelev one does.  Other types of table highlight different kinds of properties or relationships, which can be more useful in certain situations.  The labels on the left (1s; 2s; 2p 3s; etc) relate to the outer subshell in that row, so Carbon is in the "2p" section of that chart while Magnesium is in the "3s" section on the same row.

Thanks but I'm still kinda confused. Can you explain exactly what are s, p, d, f shells and specifically how many electrons each of them can hold?

I already added the notes to the table in my answer, s=2, p=6, d=10, f=14 electrons each.  Row n=1 only has  s subshell, n=2 has s and p subshells, n=3 has s, p, and d subshells, n=4 has s,p,d, and f subshells.

Does this mean, for example, all p shells always have capacity 6? So a 3p, 4p, or any p, always holds a max of 6 electrons?

p orbitals always have have l=1 ('s' is the label for l=0, 'p' is the label for l=1, and so on) and ml=-1,0, or 1, each orbital can contain 2 electrons each , so that's 3 orbitals, 6 electrons in total for every p subshell

Have you gone to a library?
You might have, and might have observed that the books on same topic are in same stack.

Let chemists be the librarian of the library of elements. And he has got to arrange his elements into stacks. Okay?
Chemist 1: How do I reduce the disorder in the arrangement of the these elements?
Chemist 2: Umm ... we need some criteria. According to that criteria, we will stack the elements.
Chemist 3: Let the criterion be electronic configuration.
Chemist 4 (looks at you): Do you know about subshells and orbitals?

Welcome to the world — sorry, shell — of electrons!

An electron shell may be thought of as an orbit followed by electrons around an atom’s nucleus.
The closest shell to the nucleus is called the “1 shell” (also called “K shell”), followed by the “2 shell” (or “L shell”), then the “3 shell” (or “M shell”), and so on farther and farther from the nucleus. The shells correspond with the principal quantum numbers (n = 1, 2, 3, 4 ...) or are labelled alphabetically with letters used in the X-ray notation (K, L, M, …).
Each shell can contain only a fixed number of electrons.
The electron shells are labelled K, L, M, N, O, P, and Q; or 1, 2, 3, 4, 5, 6, and 7; going from innermost shell outwards. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Each shell is composed of one or more subshells, which are themselves composed of atomic orbitals. For example, the first (K) shell has one subshell, called 1s; the second (L) shell has two subshells, called 2s and 2p; the third shell has 3s, 3p, and 3d; the fourth shell has 4s, 4p, 4d and
4f.
Each s subshell holds at most 2 electrons.
Each p subshell holds at most 6 electrons.
Each d subshell holds at most 10 electrons.
Each f subshell holds at most 14 electrons.
Row 1 contains only the elements that have only the K shell occupied*.
Row 2 contains only the elements that have only the K and L shells occupied*.
Row 3 contains only the elements that have only the K, L and M shells occupied*.
And so on.

* Occupied means that shell may or may not be filled, but there certainly are some electrons.

Background
The general arrangement of The Periodic Table is a result of the orbitals available to electrons bound to a single, isolated nucleus.
Each orbital is described by a wave function, which is a regular function (similar to polynomials familiar to high school algebra) with respect to the $x$, $y$, and $z$ coordinates of space, and describes the behavior of the electrons occupying it.
We can find these wave functions (and equivalently: the orbitals) by solving the Schrödinger Equation, which is a differential equation describing the system: one electron orbiting one proton in the case of hydrogen. A system with more than one electron is actually impossible to solve exactly, so we usually just think about hydrogen orbitals. The set of orbitals for any other element is more or less the same as that for hydrogen--just with different energies--so your question can be answered by considering hydrogen alone.
Specifics
Electrons in these orbitals have 4 characteristics, and no two electrons belonging to a single nucleus can have matching values for all 4 (The Pauli Exclusion Principle):

Size
Shape
Orientation
Spin

For hydrogen: the size contains all of the information about the energy of the orbital. Denoted by the letter $n$.
The shape reflects the angular momentum of the orbital. Denoted by $l$.
The orientation reflects one (of three) components of the angular momentum, as a convention: the z-coordinate. Denoted by $m_l$.
Unless you care about magnetism, the spin is only important because it allows you to put 2 electrons in one orbital (due to Pauli's principle). Denoted by $m_s$.

Since there is an energy associated with angular momentum, the possible values of $l$ for any electron is constrained by its value of $n$. For example: if $n=2$, only $l=1$ and $l=0$ are possible. In the case of the first row elements (in their ground states): $n=1$ so $l=0$ without exception.
The numbers of elements in the other rows are a result of rules governing possible orientations, $m_l$, of the orbitals (keeping in mind Pauli's principle, and that two spots are available per orbital due to spin, $m_s$).
In case you're curious: if an electron has angular momentum $l=1$, the z-component of its angular momentum, $m_l$ can only have the values $1$, $0$ or $-1$. For $l=2$, $m_l=\{2,1,0,-1,-2\}$, and for the first row: $l=0$ implies that $m_l=0$ without exception.
If you write out the number of possible orbitals following these rules:
$$n=\{1,2,3,...\}$$
$$n \gt l \ge 0$$
$$+l \ge m_l \ge -l$$
$$m_s=\{-1/2,+1/2\}$$
...where $n$ corresponds to the row number on The Periodic Table, you will find the pattern that you mentioned in your question.

Image Ref: https://en.wikipedia.org/wiki/Spherical_harmonics

the number of elements in a particular period is equal to the number of vacant orbitals available to be filled in the particular shell this also follows the n+l rule according to the energies of the orbitals.
Coming to first period in the first shell only one sub shell is there which is 1s and two electrons can be successively filled hence two elements are present in the first period.
Second shell has two sub shells 2s and 2p which have a total of four orbitals hence eight electrons can be successively filled in the four orbitals and hence 2nd period has eight elements.
The third shell has an exception due to energy considerations because of the n+l  rule though the third period has three sub shells 3s , 3p and 3d and total of 9 orbitals i.e one 3s orbital three 3p orbitals and five 3d orbitals but because 3d orbitals can be filled only after 4s orbitals due to the n+l rule only four orbitals are available which are the one 3s and three 3p orbitalshence eight electrons can be successively filled and hence eight elements are present in the third period.
the fourth shell has similar story with four sub shells 4s 4p 4d and 4f but because of the n+l rule or the  madelung energy ordering rule
4d orbitals are filled after 5s orbitals and 4f orbitals after the 6s orbitals but here the 3d orbitals are also available to be filled up so a total of 9 orbitals i.e one 4s three 4p and five 3d orbitals are available the next periods follow suit and things happen in a similar manner . Hope it helps .

Simple Answer:
The elements are ordered into periods based on which electron shells are being filled (from left to right).  In the first period, the first electron shell is being filled.  In the second period, the second shell is being filled.  And so on.
There are 8 elements in period 2 because all those elements have electrons in the second shell and no electrons in the third shell.
Elaboration:
Electron configuration describes the electrons of an atom in more detail.  The elements in the same period all have the same thing at the beginning of their electron configuration.  Here are some examples so you can see the pattern:
Li - Period 2 - [He] 2s1
Be - Period 2 - [He] 2s2
Ne - Period 2 - [He] 2s2 2p6
Na - Period 3 - [Ne] 3s1
Cl - Period 3 - [Ne] 3s2 3p5

This is the only way I can think of explaining it without going into detail about what the electron configuration means and how it can be used.  But as to why there are 8 elements in period 2, all of those 8 elements have an electron configuration that starts like this: [He] 2s

Alternate Table:

Which one is the right periodic table?

There is no right table.  These tables are just organizational tools, and you can organize the elements based on a different criteria if you want to.  No matter how you organize the table of elements, the elements are still going to have the same properties and react the same way anyway.

There is no "right table", but is there a "right periodic table"? If the elements were in a different arrange, wouldn't the periodic properties become disarranged? There would not be "periods" of properties as density, e.g.


text:
An alternate purification process for sugar separation form bee honey?
Bee honey contains 75% sugars (Fructose, Glucose, Maltose, Sucrose and others), some water and low amount of different compounds (up to 5%) Wiki.
Basic purification is done by addition of phosphoric acid and calcium hydroxide to form calcium phosphate which is afterwards removed (after precipitation). Then is filtered through activated carbon or ion-exchange resin Sugar refinery.
I am looking for a separation process, that can separate the sugars. Is there any other technique or process for removing the rest form the sugars, e.g. some electrochemical separation? First idea would be chromatography, but I would like to separate kilogram(s) of bee honey.

Are you trying to get a cheaper, easier, or ? alternative to the existing method? Maybe one that would let you sell the sugar as "organic"? Why is the current method not an option?

That's nice remark. The problem of the standard process is that it is based on the separatation in terms of separate-forgot/destroy. I need all parts of the honey. Even if the sugars have to be removed/destroyed to obtain the rest.


text:
Hydrophilic vs water soluble?
Is the term 'hydrophilic' a synonym of 'water soluble' or can you have a molecule that is water insoluble but hydrophilic (or water soluble and not hydrophilic)?

Interesting question, but I think this is more on-topic in chemistry. There are substances as cellulose which are hydrophilic but not water-soluble.

Possible duplicate of http://biology.stackexchange.com/q/23763/3340

Consider this example: glass is hydrophilic but not soluble.

Hydrophilic substances are not necessarily water soluble, and the two terms are not synonymous.
On a molecular scale, "hydrophilic" is defined by the IUPAC Gold Book$^1$ as:

'Water loving'. The capacity of a molecular entity or of a substituent to interact with polar solvents, in particular with water, or with other polar groups.

Note that the term hydrophilic does not necessarily refer to an entire molecule, and that interacting strongly with a molecule or a part of a molecule does not imply that it must be soluble in water.
Also, entire macroscopic surfaces can be said to be hydrophilic, meaning that they interact strongly with water but certainly are not soluble in water. In this context, the degree of hydrophilicity is measured by the contact angle, which can be simply stated as the angle formed when a droplet of water is placed on a surface, where a small angle (as seen from inside the droplet) means that the droplet has spread out over the surface, and that the surface is therefore hydrophilic.
So, although small, hydrophilic molecules are frequently water soluble, solubility plays no part in the definition of "hydrophilic" on any scale or context.
1) IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8

Yes, they mean the same thing actually. Going through any one of the several online sources available, you can clearly understand this: A Hydrophilic molecule is one in which interaction with water is thermodynamically more favorable than interaction with oil or other hydrophobic solvents. Such substances are generally capable of forming hydrogen bonding with water, which makes them soluble in water. One can in fact say that these types of molecules are soluble in all polar solvents, like alcohol and ammonia.

Solubility depends on polarity but that doesn't mean that all polar substances will dissolve in water or any other polar solvent.

@WYSIWYG That is the general rule: like dissolves like. Sure, there might be some exceptions.

Seems to me that a drop of oil hitting water beads up, showing no tendency to increase its area of contact with the water.  On the other hand, a drop of gasoline, e.g. Heptane or Octane, will immediately spread over the water surface to such a degree that the actual thickness of the film will be one molecule.  Such a tendency to maximize the area of contact sounds to me like hydrophilic while not soluble.


text:
Which is more acidic acetic acid or propanoic acid?
I got the answer from $K_a$ value that acetic acid is more acidic, but I have also read that inductive effect decreases as the length of alkyl group increases so acetic acid having a methyl group should have greater +I effect than ethyl group in propanoic acid.
And this comparitive large +I effect of methyl group in acetic acid should make it less acidic (as polarity of $\ce{O-H}$ bond decreases) than propanoic acid with ethyl group with less +I effect.

You need to consider inductive effects and resonance effects (especially hyperconjugative effects) on both the reactants and the products.  The difference in effects on the reactants is likely small, but the effects will be more important on the charged products.

My best guess is you are interpreting the statement "inductive effect decreases as the length of alkyl group increases" somewhat out of context.
So, let's compare formic acid to acetic acid.  Formic acid has hydrogen bonded to the carbonyl, while acetic acid has a methyl group.  The methyl group is inductively donating, and as a result acetic acid is less acidic than formic acid. $\mathrm{p}K_\mathrm{a}$ difference of $1$ full unit.
So now, let us compare methyl acetic acid (a precise, but incorrect, alternate name for propanoic acid) to acetic acid. Methyl acetic acid has an inductive donating group at the $\ce{CH2}$ carbon. As such, we expect some donation, and a lower acidity for methyl acetic acid as compared to acetic acid. True, the donating group is one group removed from the acidic site, which means the effect of the methyl group on methyl acetic acid's acidity is diminished as compared to an electron donating group directly attached to the carbonyl, so the acidity reduction will be smaller. $\mathrm{p}K_\mathrm{a}$ difference, $0.12$ units.
After you have this concept in your head, purge the abomination of nomenclature that is methyl acetic acid from your memory.
Ron has some views on hyperconjugation that may also be useful in this context, but strictly speaking, I don't find them necessary.

Besides purely electronic effects, one should take into account also the solvation phenomena. Since propanoic acid possesses larger alkyl chain, compared to the acetic acid, lower solvation of the anion is expected, which in turn leads to lower stabilization of this anion (by hydrogen bonds in case of water solution). Lower stabilization has its reflection in shifting of dissociation reaction equilibrium and eventually makes larger pKa value for propanoic acid.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

The Merck index lists both acetic acid and propionic acid as 'miscible' with water.  There is likely no effect on solubility in this case. This argument may be relevant for larger alykl chains.


text:
Recovering and casting indium
I bought some scrap indium online, and I melted it down into an ingot on my kitchen stove, but my losses were very high as molten indium has a surprising ability to stick to things.
I have a lot of indium stuck to a stainless steel pot, and, unfortunately, I dropped a glass dish that had indium in it into the steel pot and it broke. So now I have broken glass that is covered in indium, and it really likes to stick to glass as well.
Is there a good way to recover the indium? Perhaps increase its surface tension so that it beads up and coalesces? I've seen something similar with sodium metal in hot oil; adding a few drops of isopropanol makes it coalesce. Is there a similar method for indium?
Also, is there a good material to cast indium in? I see that teflon melts at a higher temperature than indium, so maybe a teflon coated muffin tray? Any thoughts on that? Recommendations?

So I had time last night to try out a better way to cast indium. I thought about it and decided a teflon coated tray was a bad idea because you still might have to use a tool to pry the ingot from the mold, and over time you would scratch the teflon and create a more adhesive surface. I decided to try one of those silicone rubber molds used for baking. I found two of them at a thrift store for 4 or 5 dollars each, and they worked perfectly! So that's one question down, but the question of how to recover indium from surfaces that it is well adhered to remains unanswered at this time.

A much more straight forward way uses oxalic acid. For the paper on this:
https://pubs.acs.org/doi/full/10.1021/acssuschemeng.8b04756
For a YouTube demo of how to do it in practice see:
https://www.youtube.com/watch?v=AvXTnvT8WNo
(relevant part starts about 11 mins in)

Through experimentation, I've come up with satisfactory answers to my problems.
1) What is the best material to cast indium in? Silicone rubber (the kind used for cookware).
2) How can I recovery indium that has "wetted" a stainless steel pot? Put the pot on the stove and heat until the indium is molten. Use a silicone rubber spatula to scrape the indium off, let it harden on the spatula, then pick it off. You can also use a wet sponge as long as it's the kind used for cleaning solder irons (make sure it's a new clean sponge!). I folded it in half and held it with a large clamp. That gave me something to hold onto.
3) How can I recover indium that has "wetted" glass, broken glass, or is contaminated with other odd bits of material?
Number 3 is more involved, and I will break it down into steps:
1) Go through the material with a pair of forceps and discard any bits of glass that have no indium on them.
2) Place the material in Pyrex beaker.
3) Fill the beaker with mineral oil until the material is fully submerged (water may also work, but the oil will probably help keep the indium from re-wetting the glass bits).
4) Place the beaker on a hot plate and slowly heat up to about 160 degrees C (320 F). Periodically stir while heating (a screwdriver worked fine for me; I didn't use a magnetic stir rod because the broken glass might chew it up). Also, make sure your area is well ventilated; the oil will produce vapor.
5) Once you have reached 160 C, poke the indium and make sure it is molten. Increase temperature if it is not molten. Stir for about 30 seconds to break up the indium. Keep on heat; try to maintain ideal temperature.
6) Use a glass dropper to drop in 31.45% Hydrochloric acid (I used 4 drops total when the oil and material together was about 100mL). Add one drop at a time, and let the drops run down the inside of the beaker into the oil. (WARNING: This can result in splashing! Don't over heat the oil, add the acid drop-wise, and use proper safety gear: gloves, long sleeves/pants, face shield, and respirator with acid vapor cartridges). Adding the acid will react with the indium oxide layers and help break the indium free from other materials.
7) Stir vigorously with your implement of choice. The oil will become a nearly opaque brown with all the oxides. The stirring should also free up most of the indium from the glass bits.
8) When the oil seems saturated, thoroughly decant the oil into a waste receptacle (I used a mason jar). Save as much solid material as you can.
9) Fill the beaker with clean oil; do NOT begin stirring yet. Keep on heat.
10) Use your forceps to pick out the larger pieces of glass that seem mostly clean. Use another implement to scrape off little bits of indium from the glass bits as you are pulling them out of the oil.
11) Once most of the glass is removed, use strong stirring once again until the oil is mostly brown.
12) Using your stirring implement, try to push a good amount of the molten indium together into one blob on one side of the beaker. Now decant the oil into a waste receptacle.
13) Place beaker back on heat for a short time just to ensure the indium is molten. Make sure most of the material is in one corner of the beaker.
14) Use a broad-shaped implement (I used the flat head of a long "elevator bolt") to hold back most of the scrap material and pour off the indium from the beaker into a silicone rubber mold. The poured off indium should be MOSTLY clean. Repeat this step until you feel you cannot get anymore. Put the scrap material off to the side.
15) Once the clean blobs have hardened, dab them with a paper towel and cut them into tiny chunks with a pair of wire snips.
16) Take each individual chunk and crush it paper-thin using a flat metal implement and a flat metal surface (I used an elevator bolt and a piece of sheet metal). This step is to find any tiny bits of glass. Glass has quite a different "crunch" than the typical "crackling" of the indium's crystal structure. You will know when you find one. Fine tweezers and some sort of magnifying glass helps.
17) You can now re-melt the clean indium into one chunk. I use a stainless steel shot glass and pour it off into a silicone rubber mold. If it's still oily, wash it with Ronsonol/Zippo lighter fluid, then rinse it with water. I actually chose to keep this piece of indium separate from my other indium pieces that I KNOW are clean. You never know, there might be one last sand-grain sized piece of glass in there.
18) As for the scrap material, there's still a bit of dirty indium in it. You can decide how hard you want to work and continue salvaging it. Use tweezers to pick out as much "clean" glass as possible. Wash it with lighter fluid and rinse it with water. Perhaps re-melt it in a steel shot glass and try to pour a little bit more out of it. Try putting a blob of solder flux in with the molten material and get it to stick to a copper wire. It's all going to be loss anyway, so you may as well try different things.
DISCLAIMER: I am not a professional! This procedure is not advisory. It is an account of my experiment and what worked for me.
If anyone has suggestions or constructive criticism, please feel free to comment. I know there are probably better ways to do this.


text:
What's the difference between aqueous NaOH and dilute NaOH?

Dilute $\ce{NaOH}$ can be prepared by adding a suitable amount of concentrated $\ce{NaOH}$.

What's the difference between "dilute $\ce{NaOH}$" and "aqueous $\ce{NaOH}$"?

Dilute simply means that the concentration of the solute of interest, in this case $\ce{NaOH}$, is low relative to some reference.  One could imagine a non-aqueous, dilute solution of $\ce{NaOH}$ in, say, methanol.
An aqueous solution is one where the solvent is water, regardless of the concentration of the dissolved solute(s).
So, for example, the following combinations of the terms are possible:

Concentrated aqueous solution - High $\ce{NaOH}$ concentration in a water solvent
Dilute aqueous solution - Low $\ce{NaOH}$ concentration in a water solvent
Concentrated methanolic solution - High $\ce{NaOH}$ concentration in a methanol solvent
Dilute methanolic solution - Low $\ce{NaOH}$ concentration in a methanol solvent


text:
how a galvanic cell works with a different metal cation in electrolyte than used in electrode
A traditional galvanic uses zinc as the anode electrode submerged in a zinc sulfate electrolyte, and copper as the cathode electrode submerged in a a copper sulfate electrolyte. I wanted to know what happens if the electrolyte metal cation was not zinc or copper: What if the anode electrolyte was Potassium sulfate and the cathode electrolyte was sodium sulfate, will the cell still operate?
Should't zinc still be oxidized to zinc ion? In the cathode half cell, would the sodium ions be reduced to sodium metal, or would copper electrode interfere in some way?


text:
Proportionality between number of absorbed photons and optical density
I am to show that the number of absorbed photons only is proportional to the optical density at low optical densities.
I do not know how to do this, but this reminds me of the linear range of absorbance vs concentration; at high concentrations, the relationships is no longer linear. I also feel that the "number of absorbed photons" is another way of saying "absorbance", and that "optical density" is another way of saying "concentration times optical path length", but I am not sure. I cannot find references that clarifies this for me, so I would appreciate some guidance.

Yes, your statement that "the number of absorbed photons only is proportional to the optical density at low optical densities" is correct. (I'll point out that optical density is a depreciated term for what is now referred to as absorbance.)
First let's define some terms.

T: Transmittance, i.e. the fraction of photons not absorbed. So a
     transmittance of 0.90 would mean that 90% of the photon pass through
     the sample.
A: Absorbance of sample which is given by the Beer–Lambert law: $\ce{A = -log_{10}(T)}$ or: $\ce{T = 10^{-A}}$

Now a bit of math swizzling...
For exponentials of $e$ there is a nice series expansion. $$ e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$ and if $x < 0 $ then: $$ e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$
now if also $|x| << 1$ then the higher order terms can be ignored and the function becomes linear where:
$$ e^{-x} \approx 1 - x $$
Now for a bit more math voodoo... Let's let $$10^x = e^y$$ so: $$y = x\mathrm{ln}(10) \approx 2.303x$$
So now substituting $\ce{e^{-2.303A}}$ for $\ce{10^{-A}}$ we get: $$\ce{T = e^{-2.303A}}$$ and thus only when $2.303A << 1$ is the Beer-Lambert law linear.
Now for the rest of the story let's define some more terms:

$\epsilon$ is the molar attenuation coefficient of the attenuating species  in the sample;
$c$ is the molar concentration of the attenuating species in the sample;
$l$ is the path length of the beam of light through the material sample.

Now also accoding to the Beer-Lambert law: $\ce{A = \epsilon c l}$ so $\ce{A}$ is proportional to $\ce{c}$.


text:
pH 10 to pH9 NaOH dilution
I have a pH 10 NaOH solution. I need to bring down the pH to 9. So I used the $C_1V_1 = C_2V_2$ approach.
$C_1 = \mathrm{0.0001~mol/L}$
$V1$ is the necessary amount to produce 10 L of pH9
$C_2 = \mathrm{0.00001~mol/L}$
$V_2 = \mathrm{10~L}$
\begin{align}
C_1V_1 &= C_2V_2\\
0.0001~\mathrm{mol/L}\cdot V_1 &= 0.00001~\mathrm{mol/L}\cdot 10\mathrm{L}\\
V_1 &= \frac{(0.00001)(10)}{(0.0001)}\\
V_1 &= 1\mathrm{L}\\
\end{align}
I Applied this calculation in reality and used a pH probe and got a pH of 6.5. I think the drop is too sudden in this type of titration. What should I do?

Double check the pH meter calibration. It is impossible to dilute NaOH and get an acidic solution.

And who says water is neutral? There's always some CO2 in it, unless it wouldn't have any contact with air.

When you dilute anything down to homeopathic concentrations like 0.00001, expect all kinds of weird effects. Any impurity ($\ce{CO2}$ from air, like Mithoron said) could suddenly become more abundant than your compound.

Even with distilled water? What would be a new approach I should take? My problem is I need to make a pH10 solution which I need to creata pH9, pH8 and pH7 from the pH10 solution .

Yes, even with distilled water.  The distilled water condenses out of the air, and voila!  Carbon dioxide!

Generally, you don't make pH of $7\pm2$ by dilution alone. The required dilution is so great that it would leave very little of your compound, and then any tiny impurity would affect the result greatly. Think of a droplet on the pH meter probe left from the previous measurement. Distilled water does not help much. All the same, it contains $\ce{CO2}$ from the air, so its pH is usually about 5.5, not 7 as one might expect.
When people want to have certain pH level more or less reliably, they use buffer solutions. Come to think of it, that's what nature itself does, because our blood is a buffer solution as well.

the dilution increases the pH because the base will be completely sorrounded by water molecules.therefore more of the base will ionise Into constituent ions to produce fewer hydroxide ions than before dilution which increase the PH.

A dilution should bring a pH towards seven. Adding distilled water to a basic solution should result in a decrease in pH. I think you are trying to use Le Chatelier's principle to explain that the bases should have a larger effect when the amount of water increases. While it is true that adding water will shift the base towards acting as a base, it will not do so quicker than the increase in concentration of water.


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How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach?
It's my understanding that for a particle in a 1-D box, the change in energy from the HOMO to LUMO can be approximated with the following equation:
$$\Delta E=\frac{h^2 ( N + 1 )}{8 m_\mathrm{e} L^2}$$
I am told that $N$ represents the number of π electrons in the molecule.
If you're studying a molecule like the one shown below and you want to predict the absorption wavelength from UVVIS, would you count all of the π electrons, or just the conjugated ones? It doesn't make much sense to me why you'd want to include the π electrons from the phenyl group next to the neutrally charged Nitrogen because it's not conjugated with the rest of the molecule to the same extent the other phenyl group is, and thus seems like it would be grossly inaccurate to assume that those π electrons contribute the same as the others.

I disagree with your assertion that this molecule is not fully conjugated.  I believe both phenyl rings are coplanar with each other and conjugated together. Don't believe me?  Draw the resonance structure where the uncharged donates electrons into the 2-position, shifting the methylene pi-bond to the 2' position, dumping a lone pair back on to the originally charged nitrogen.  More importantly, it can't be resolved without a crystal structure.  This is one of those times where you'd have to calculate several options and let experiment tell you which is correct.

Yes I just thought of that. I was thinking that the sulfur and nitrogen are sp3 hybridized without an open p orbital, but I forgot about rehybridization. Thank you makes a lot of sense. So I'm counting 16 pi electrons, using the equation I get a theoretical absorption wavelength of 272 nm. Experimental is 425. Seems too far off

Molecule is symmetric and only ethyl groups' carbons don't participate in conjugation (nitrogens and sufurs do) . BTW I cant see how 1-D  box would be of much use here.

And then Mithoron's comment is great. You'd have n-pi* transitions for this molecule

In a very basic first order approximation you can treat dyes with an extended conjugated π-system with the one-dimensional particle in the box approach. I simplified your system and calculated it at the DF-BP86/def2-SVP level of theory. The length of the box is about $L=1.20~\mathrm{nm}$, the following graphic shows it in Ångström.

With that there are a number of problems associated. Counting the number of electrons in the π-system should not be among them. Just assume that all non-hydrogen atoms are sp² hybridised. Then count the electrons that occupy π/p-type orbitals. Here is an image of the lowest occupied π-type orbital of that molecule; you can see that the box stretches over the whole molecule. The thumbnail next to it gives you all occupied π-type orbitals and the lowest unoccupied molecular orbital.

 In this example each phenyl ring contributes six π electrons (12), two sulfurs contribute two each (4), the nitrogen double bond (2), the carbon-carbon double bond (2) and the last nitrogen (2). That makes a total of 22 π electrons.