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What metals aren't dissolved in/attacked by aqua regia?
Aqua regia (Latin: Royal Water) is one of the strongest acids known in Chemistry, and is capable of dissolving gold and platinum.
My copy of the Oxford science dictionary goes on to say (under the entry: Aqua regia) that metallic silver does not dissolve in aqua regia. Morever it does not  mention any other examples of aqua regia-resistant metals. Further down, it mentions that silver's invulnerability to aqua regia is due to the formation of a protective silver chloride coating on the metal, which serves to protect the metal from further decomposition.
However, this Wikipedia article claims:

...aqua regia does not dissolve or corrode silver...

This, I find contradictory to the dictionary's "formation of silver chloride" claim.
So,

What metals (elemental, forget alloys) are neither attacked by nor dissolved in (freshly prepared) aqua regia?
What makes those metals that don't dissolve or corrode in aqua regia so impervious to the acid?
Does silver metal actually develop a silver chloride layer on exposure to aqua regia? If so, would that mean the Wikipedia article is incorrect?

The wikipedia article gives an answer to 1) above: "However, aqua regia does not dissolve or corrode silver, titanium, iridium, ruthenium, rhenium, tantalum, niobium, hafnium, osmium, or rhodium" i.e. the usual suspects +titanium and silver, -platinum

Aqua regia isn't particularly strong acid. It's just nitric acid with HCl added, what may or may not help in dissolving metals.

@Mithoron yes, important point that the critical property is not its strength

TIL why rhodium is used to plate fountain pen nibs.

Keep in mind
The answer will depend upon the reaction conditions. Most importantly,

physical state of the metal: porosity, degree of comminution;
temperature;
mechanical aggravation of metal surface during reaction.

Often times a chemistry text mentions that no reaction occurs. The reaction might still happen. It is just that for the specified parameters the process is meaningless and negligible.

Short overview of aqua regia
Aqua regia is the $3:1$ volumetric mixture of $\ce{HCl}$ and $\ce{HNO3}$. Its additional reactive power draws from monochlorine created in situ.$^{[1]\ [2]}$
$$\ce{HNO3 + 3HCl -> Cl2 + NOCl + H2O\\ NOCl -> Cl + NO}$$

Which metals are impervious to $3:1\ \ce{HCl/HNO3}$?

Almost every metal will react with aqua regia provided certain criteria are met.$^{[1]\ [2]}$ The closest you will probably get is ruthenium $\ce{Ru}$, and perhaps osmium $\ce{Os}$. To the best of my knowledge, $\ce{Ru}$ will not react with aqua regia in a meaningful way even if aqua regia is boiling.$^{[2]}$ The difference with $\ce{Os}$ is that powdered osmium is attacked by boiling aqua regia.$^{[1]\ [2]}$
$$\ce{Ru + HNO3 + HCl $\kern.6em\not\kern -.6em \longrightarrow$}$$
$$\ce{\underset{powder}{Os} + $\underbrace{\mathrm{HNO_3}}_{\text{boiling}}$ -> OsO4 + N_xO_y + H2O \\ OsO4 + 2H2O <=> H2[OsO4(OH)2] \\ OsO4 + HCl ->OsO2Cl2 + Cl2 + 2H2O\\ OsO2Cl2 + HCl ->OsCl4 + Cl2 + 2H2O\\ 2OsO2Cl2 + H2O <=> OsO2 + H2[OsO2Cl4] \\ 3OsCl4 + 2H2O <=> OsO2 + 2H2[OsCl6]\\ OsO2 + 6HCl <=> H2[OsCl6] + 2H2O}$$
Brief discussion about the list provided in the comments
Titanium $\ce{Ti}$ does react, and does so at room temperature.
$$\ce{3Ti + $\underbrace{\mathrm{12HCl + 4HNO_3}}_{\text{room temperature}}$ -> 3TiCl4 + 4NO + 8H2O}$$
Rhenium $\ce{Re}$ reacts slowly at room temperature $\ce{->HReO4}$. This will further react with $\ce{HCl -> ReCl4 + Cl2}$.$^{[2]}$
Hafnium $\ce{Hf}$ does react at room temperature. The reaction is slower than with titanium; overall equation is identical.$^{[2]}$
Tantalum $\ce{Ta}$ reacts when aqua regia is heated to $150\ ^{\circ}\mathrm{C}$. Rhodium $\ce{Rh}$ reacts in a grinded state. As a large compact piece, iridium $\ce{Ir}$ is affected over temperatures of $100\ ^{\circ}\mathrm{C}$. Niobium $\ce{Nb}$ is inert at room temperatures.$^{[2]}$
Summary: ruthenium $\ce{Ru}$ is your best bet.

What makes metals $\ce{Ru}$ and $\ce{Os}$ so stable in aqua regia?

The nobility of these metals is not the best explanation. As you correctly pointed out, $\ce{Pt}$ and $\ce{Au}$ react fine. This is direct evidence that for other metals a protective layer should form. The layer varies from metal to metal, but usually is either an oxide (or oxide hydrate), or a chloride.
Effectiveness of mechanical aggravation also points to stable, non-reactive compound formation on the metal's surface.
For ruthenium, as of now I am unsure what this precipitate could be. If anyone has a reference, please edit or leave a comment.$^\text{[reference needed]}$

What happens with silver?

Silver and aqua regia react very poorly, and for a short amount of time.$^{[2]}$ The culprit is $\ce{AgCl}$ ($K_s = 1.8 \cdot 10^{-10}$)$^{[2]}$. A slow reaction might still take place due to complexation.$^{[2]}$
Surprisingly, silver reacts with $\ce{HBr}$!$^{[2]}$ Its solubility product is even worse, $K_s = 5.0 \cdot 10^{-13}$.$^{[3]}$ My guess is that this layer is not as dense as $\ce{AgCl}$ but this still needs verifying.$^\text{[citation needed]}$

References
(In progress)
$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974)
$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003)
$[3]$ Skoog, West, Holler, Crouch. Fundamentals of Analytical Chemistry. 9th edition. (2014)

Possibly bromide ion forms a soluble complex with silver, more easily than chloride ion?

What metals (elemental, forget alloys) are neither attacked by nor dissolved in (freshly prepared) aqua regia?

At normal condition, most of the metals are impervious to aqua regia but if the conditions are changed like temperature, pressure, acid concentration or state of metal, they can be prone to attack.
The metals discussed falls in the category of noble metals. These metals resists corrosion and oxidation in moist air and most of the acids including aqua regia. But there are exceptions. More specifically:

Platinum, gold and mercury can be dissolved in aqua regia but iridium
and silver cannot. Palladium and silver are, however, soluble in
nitric acid. Ruthenium can be dissolved in aqua regia only when in the
presence of oxygen (Why? will be asked in a separate question), while rhodium must be in a fine pulverized form.
Niobium and tantalum are resistant to all acids, including aqua regia.

This handbook1 gives an overview of which noble metals are attacked by aqua regia:

Niobium is reported a corrosion rate of 0.025mm/yr in aqua regia at 328.15 K.
Gold is attacked by aqua regia.
Iridium resist both cold and hot aqua regia but dissolves under pressure at 523-623 K.
Platinum is attacked by aqua regia.
Rhodium unattacked by aqua regia in either wrought or cast form.
Ruthenium is by far the most chemically resistant of the noble metal, unattacked either in hot or cold aqua regia.
Tantalum also not attacked by aqua regia even in elevated temperatures.
Zirconium is attacked by aqua regia.
Hafnium is dissolved in aqua regia only in the addition of some soluble fluoride salts.

So, the winners are ruthenium and tantalum.

What makes those metals that don't dissolve or corrode in aqua regia so impervious to the acid?

Maybe there is some sort of protective layer like oxide or chloride layer that protects the metal from further attack. It can be explained by studying their potentials. Most of the noble metals sit right at the bottom of electrochemical series due to their high positive potential value. So, they are quite resistant to attacks.

Does silver metal actually develop a silver chloride layer on exposure to aqua regia? If so, would that mean the Wikipedia article is incorrect?

No, silver actually forms a silver chloride layer that protects the metal from further attack. Aqua regia can only dissolve silver if the chloride layer is ruptured.
References

https://en.wikipedia.org/wiki/Noble_metal
https://en.wikipedia.org/wiki/Aqua_regia
Handbook of Corrosion Data, Bruce D. Craig, David S. Anderson
ASM International, 1994
Update: Can aqua regia/royal water dissolve lead?

Regarding "No [the Wikipedia article is not incorrect], silver actually forms a silver chloride layer that protects the metal from further attack", I rather took @paracresol's meaning to be that the formation of a protective chloride layer is, in itself, a form of corrosion (albeit minimal, and self-limiting). Perhaps p-cresol could confirm/deny/clarify?

@owj That's precisely what I had in mind. The formation of a silver chloride layer does appear to be a form of corrosion.

I'd say corrosion necessarily includes the spalling of oxidised metal to expose more unreacted metal. Otherwise it's just passivation.

@gsurfer04 Concur with this terminological point: if the reaction is self-terminating, it's passivation; otherwise, it's corrosion.

Definitions that I can find online for "corrosion" all seem to reference "gradual destruction". So perhaps we could argue that it needs to be an ongoing process to constitute "corrosion".
Coming at it from a more intuitive, empirical, chemist's perspective, I don't imagine that any of us would argue that oxygen in the air  corrodes aluminium. It does react and, in the process, forms a protective oxide layer which prevents corrosion. But it feels wrong to describe that passivation step as a kind of 'corrosion'.

@owjburnham no need to argue that point, by definition it works exactly as h8y2Py says...

Passivation does not influence potential which is thermodynamic, however it hinders or even completely stops dissolution by aqua regia as in Rh, Pt in the cold etc


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Separating boric acid from sodium chloride
I want to create some boric acid with borax in the following reaction:
$$\ce{Na2B4O7·10H2O + 2 HCl → 4 BH3O3 + 2 NaCl + 5 H2O}$$
However as you can see, the leftover solution will contain a mixture of boric acid, sodium chloride and water. Both chemicals are soluble in cold water so neither will want to immediately crash out of solution.
My question is, how can I remove the excess water and sodium chloride to leave myself with some powdered/dry boric acid crystals?
I know I can probably just use some liquid ammonia, but then the boric acid is only slightly soluble in it leaving much of it behind. I could perhaps use glycerin also, however I am limited to how much I currently have.

I guess ordinary ethanol dissolves boric acid pretty well.

Here is a potential solution that shouldn't even require heating. Note that I have not tried this myself.
Boric acid is soluble in lower alcohols like ethyl alcohol and isopropyl alcohol.
The solubility of sodium chloride in ethyl alcohol at room temperature is about 0.05%$^1$.
Given these properties, I suggest dissolving the dried boric acid / sodium chloride in a minimal amount of cold 95% denatured ethyl alcohol or 99% isopropyl alcohol (available at most pharmacies).
The boric acid should dissolve readily and the sodium chloride should not, so you can just filter it and let it dry to a fairly sodium chloride free boric acid solid.
To make the sodium chloride even less soluble you could cool the solution in a freezer prior to filtering.  If you notice any significant increase in solids that means boric acid is coming out of solution and you'll want to abort this step.
1) Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005

Is there any steps I can add to this solution by heating/distilling or rather anything to get a pure solution of boric acid? I'm going for a sodium chloride-free environment here. :)

If you don't mind sacrificing your recovery of boric acid a bit, and any purification procedure will likely result in decreased recovery, here is what I would do: When all but a small amount of alcohol has evaporated, pour off the remaining alcohol then give the solid a quick rinse in a small amount of ice-water, pour off all the water that you can and dry.  As the boric acid crystalizes out of solution, the crystal formation should exclude most of the NaCl, so that most of the NaCl will be in the alcohol and at the surface of the solid. NaCl is very soluble in water, so most should rinse off.

Hmmm... I'm sorry but I'm no good at these crystallization attempts... I'm giving it a go, but is there any other way I can do this? :(

Could you please share the link for the specific research paper from the "Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005"? I tried but am not able to find it on Google Scholar or Google search.

You could consider using ion exchange resin, there are resins intended for use in the nuclear industry for the removal of boron (borate / boric acid) from water. A common method in light water reactors (PWRs) is to control the reactivity by adjusting the boron content of the coolant.
While you can imagine it is easy to add boric acid / lithium borate to the water, a method of removing it is also needed. One method which is common is to use an ion exchange resin. It is interesting that the acid strength of boric acid increases when it is combined with a polyalcohol which can chelate to the boron.
What I would imagine going for is a weak acid resin such as IRA 67 which has been modified with surgar groups which are able to increase the binding of boric acid / borate. There is a resin which is marketed for this purpose (IRA743) which could be used to remove boron from sodium chloride.
You would then need to elute the resin, I would suggest using a mineral acid solution to release the boric acid followed by treatment with sodium hydroxide to restore the resin back to being the free base form.

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What helps a sliced fruit or vegetables oxidize slower?
What helps a sliced fruit or vegetables oxidize slower ?
A need of chemical compound to use for the slowing of oxidation... we have tried using a lemon juice it actually helped by 15 mins. However it change the quality of the vegetable after being dip for 10 mins and (the problem is how do we slow down the oxidation for 10 days without changing the quality of the fruit) since lemon juice changes the quality of its compound (2-3 days rather).

I mix cut apples in the oil-based dressing first, then work on the rest of the salad. But 10 days? Many labs are working on this moneymaker.

Will Vitamin C help? Just guessing..

On timescale of 10 days oxidation with air is your last problem

If you soak avacodo pieces in orange juice it makes them last a lot longer

@bpedit The magic of biotechnology. :P

@MichaelD.M.Dryden. And modified  atmosphere packaging. And wax!

http://www.easylunchboxes.com/blog/6-ways-to-stop-fruit-from-browning/

this suggests ascorbic acid powder could help without altering them, though I don't know if it will work for 10 days. Are you able to refrigerate them?

You can cover the cut pieces with aluminium foil which acts as sacrificial element and prevents oxidation of food.

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Atomization enthalpies of transition elements
So, my book says that transition elements have higher enthalpies of atomization than other elements (say s- or p- block) because of stronger metallic bonding, primarily due to large number of unpaired electrons in the d-orbitals.
Now, if that is the case, then the atomization energy for $\ce{Ni}$ should be less than for $\ce{Fe}$ and $\ce{Co}$, as $\ce{Ni}$ has only 2 unpaired electrons compared to $\ce{Fe}$ and $\ce{Co}$ (4 and 3 respectively).
Actually, it is just the opposite.
$$\begin{array}{cc}
\text{Metal} & \Delta_\text{atom}H(\ce{M})\text{ / }\mathrm{kJ~mol^{-1}} \\ \hline
\ce{Ni} & 430 \\
\ce{Co} & 425 \\
\ce{Fe} & 416
\end{array}$$
Why could this be?

These sorts of generalities always sweep aside other considerations.
$\ce{^26Fe}$ has a body-centered cubic crystal structure.
$\ce{^27Co}$ has a hexagonal close-packed crystal structure.
$\ce{^28Ni}$ has a face-centered cubic crystal structure.
Here is a nice table showing the overall trends.

According to J.D.Lee's "Concise Inorganic Chemistry", the extra electrons in the metals of the 2nd half of the transition series, like Fe, Ni etc.. are excited to the p level .. and then the unpaired electrons in the d, s and p levels participate in metallic bonding .. so Ni has the max. unpaired electrons, followed by Co and then Fe ... and hence strength of metallic bonding follows same order


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Why is NAD+ regioselectively reduced at C-4 to form NADH?

We could draw a resonance form with the pi electrons shifting, placing a positive charge on C-4, to which a $\ce{H+}$ along with 2 electrons are added. Thus we get an NADH molecule as shown in the picture.
If this is how the reaction happens, then why doesn't the hydrogen get added to C-2 (or C-6 which is equivalent)? There's also a resonance form where C-2 has a positive charge and could accept a hydride ion. Why doesn't this happen?

In a synthetic setting, they could add to C-2. But in the body, these reactions are mediated by enzymes which hold the reactants in a very specific orientation. I'd guess that the regioselective delivery of hydride to C-4 is controlled in that way.

I'm sure that additions to pyridinium rings are well studied and should be well documented in the literature.

Living organisms utilize protein based enzymes to catalyze otherwise unfavorable reactions under very mild conditions. They achieve this by bringing the substrates together in a very specific manner. Dehydrogenases are a class of oxidoreductases that oxidize substrates using NAD$^+$. Here is the active site of lactate dehydrogenase:

As you can see, carbon-4 of the nicotinamide ring is oriented towards the substrate. The orientation of the molecules is why carbon-4 is selectively reduced.


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What exactly does a "system" include in chemistry?
If we have reactants $\ce A$ and $\ce B$ and the reaction(s) between them haven't started yet, are those reactants our "system"?
After the reaction takes place, are products then also included in the "system"?
Basically, is it the case that the term "system" refers to all the chemicals we have in a reaction?

There isn't one universal answer. "The" system is all the factors that are pertinent. Certainly the concentration of reactants would be one set of pertinent factors. But temperature, pressure, heat flow in or out of the reaction could be other factors too. Light might be a factor. So what is pertinent depends on the reaction.

You put your reagents in an imaginary box of impermeable matter. You calculate what would become of your reagents, considering the forces and influences which are inside the box. Those which are outside (air, ocean water, sun light, moon gravity, etc.) are disregarded altogether. Now, that box is your system.

I disagree with the VTC  of this question. It indicates enough effort on the OP's side.

@IvanNeretin - I understand what you mean about the "imaginary box" but it isn't quite that simple of course. You might have to consider the heat flow in and out of the box. Also many gas reactions specify a constant pressure. So the volume of the box might need to expand or contract.

@MaxW   can we just say that it includes the reactants and all the factors that can influence the reaction ?

It depends. If we look at the system as isothermal, then the walls are permeable to heat (while still being impermeable to any chemical compound). If we want our system to be isobaric, then the walls are soft and pliable - in a way, our box becomes a sack. The outside temperature and pressure are the factors which surely can influence the reaction.

Note: This was not originally written by me. This was first written by @airhuff. I just thought it was relevant in this question as well.
According to this Wikipedia article:

...a physical system is a portion of the physical universe chosen for analysis. Everything outside the system is known as the environment. The environment is ignored except for its effects on the system.
The split between system and environment is the analyst's choice, generally made to simplify the analysis. For example, the water in a lake, the water in half of a lake, or an individual molecule of water in the lake can each be considered a physical system.

Every system has some relationship to it’s surroundings. As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:

Isolated system: Neither matter nor energy may pass into or out of the system.

Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.

Open system: Both energy and matter can pass into and out of the system.

I would say if you going to use someone's answer, even from another question, you should post it as community answer.

@Tyberius, you can flag for mod attention for conversion to community wiki. Thanks anyway for pointing it out.


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Can using isopropyl alcohol on a zirconium oxide ceramic knife harm or dissolve/degrade it?
Can using isopropyl alcohol to clean a zirconium oxide ceramic knife harm or dissolve/degrade it?

It wouldn't harm the ceramic blade, but it might not be good for the handle. SAFETY -- Isopropyl alcohol also isn't safe for consumption. So that seems like a poor choice to be used around food.

Ethyl alcohol (i.e. everclear > 90%  alcohol) would be safer with respect to @MaxW 's comment, and also would not harm the knife.  Still, I'd guess that soap and water with a good sponge would work as well as anything, unless it's been used for something odd outside of cooking.  With any of the solutions discussed I think the biggest danger is just how slippery it's going to be, especially if you are scrubbing hard.  That can easily put you on the fast track to the emergency room!  Good luck with it...

Zirconium dioxide or zirconia is chemically unreactive. It only dissolves in hydrofluoric acid and sulfuric acid.

thank you very much everyone, I 'm actually also using scissors for hair and just wanted to make sure it wouldn't dissolve the zirconium oxide that it says it's made out of and then get onto my skin and harm me some way.  I will use the isopropyl alcohol just as a final rinse after washing the blades.

I'm voting to close this question as off-topic because  of low quality

@A.K. If the lower quality of the post has a specific reason (being too broad/opinion-based/homework-ish/etc.), you should use that particular close reason for the post. Otherwise, I don't think a "low-quality" custom close vote is justifiable. You should rather downvote the post instead.

Zirconium dioxide is not soluble in organic solvents like isopropyl alcohol. Of more concern would be any plastic component of the knife, like a handle that could be damaged by isopropyl alcohol or other solvents.

True...I do not know what kind of plastic the handles are made out of.  But isopropyl alcohol is usually sold in plastic containers, most likely LDPE, HDPE, or PP.  So if the handle is made of one of those plastics, it should be safe?

I was being pretty cautious with that statement, and it's pretty unlikely that isopropanol will attack any plastics on the knife. Acetone would be a much greater risk for example. If you want to play it safe, maybe try a spot-test with a little isopropanol on an inconspicuous part of the plastic. If there is no change in appearance or texture after it dries then you should be safe.


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How can oxygen have three bonds?
I was reading an article about a chemical reaction, and I came across the phrase:

The oxygen atom at this point has three bonds and has a net positive charge

How can this happen? Oxygen has 2 missing electrons in the valence shell. Therefore it can only form 2 bonds at the most, if both are sigma bonds.
Does it mean the 3rd bond is not covalent? Can it happen with a hydrogen or an ionic bond?

If you think that's strange, try a carbon with three, a carbon with five, a nitrogen with four, or a boron with four. As perhaps you're not used to this, you'll see later that counting bonds is not always simple, or even a sensible thing to try.

Well, if it has a net positive charge, then it's probably missing more than 2 electrons.

Consider the auto-ionization of water :
$\ce{ 2H_2O->H_3O+ + OH-}$
The first oxygen has three bonds, the second only has one.
You can think of the reaction taking place by a lone pair on the oxygen of one water molecule ripping off the proton only of the hydrogen of another water molecule to form a covalent bond between them using just the lone pair. The electron of the hydrogen is left behind and stays with the oxygen of the other molecule.
If you calculate the formal charges on each oxygen you will see the first one has a positive charge and the second one has a negative. The formal charge is just the valence number of electrons minus the number of bonds minus non-bonding electrons (using the lewis structure) and is a useful book keeping method to think about where the electrons go/are and what are the most stable structures.
Formal Charge Calculation

By the way, how does a lone pair create a bond between a hydrogen atom? Wouldn't only one hydrogen be required? Since there is only one hydrogen required? If both electrons in the lone pair covalently bond to the hydrogen, the hydrogen would have 3 electrons, which would be unstable, right?

Yes, if the lone pair and the electron on the hydrogen were in the bond then the hydrogen would have three electrons, but that doesn't happen. The lone pair on the oxygen basically rips just the proton of the hydrogen off the other molecule leaving the electron that was already on that hydrogen behind.

@MichaelMshtc This type of covalent bond, where both bonding electrons come from the same atom, is called a coordinate bond or sometimes a dative covalent bond.  It's far more common than you'd guess based on its coverage in intro chem courses.

This type of discussion is better off in chat, all.

Comments are not for extended discussion; this conversation has been moved to chat.

Since a few useful comments have been moved to chat, I'm going to reiterate them. There is absolutely no difference between the so-called dative covalent bond and the normal ones in an H3O+ ion. Those 3 bonds have exactly the same properties and hence H3O+ has 3-fold rotational symmetry. You simply have 1 pair of electrons per bond, and 1 lone pair on the oxygen atom. Everything adds up because of the net +1 charge. It's better to think of valence electrons filling up (molecular) orbitals, than giving the wrong impression that the dative bond is of a different nature.

Apart from hydronium ion $\ce{H3O+}$, there are some other molecules or ions which possess three oxygen bonds. All pictures taken from Wikipedia

Carbon monoxide

Ozone

Nitrosonium ion

Transition metal aqua complex

Here each oxygen has three single bonds, one with the central metal and other two with tho hydrogens.

Transition metal oxo complex

Namely  vanadyl chloride,  tungsten oxocarbonyl

trismesityliridium oxide
And there are many more.....

What is the term for the bond that can flip?

Good answer but not the term I am trying to recall from college days.    Like in a benzene.

@Paparazzi, Sorry,  I am not getting what you are saying. Are you looking for resonence of benzene?

@Paparazzi Are you thinking of "resonance structures"?

Oxygen has $6$ electrons in its outermost shell.
However, if you give it a positive charge by removing an electron, then it would only have $5$ electrons in its outermost shell.
Therefore, it can now from $3$ bonds while carrying a positive charge.
You can usually see trivalent oxygen in intermediates of reactions, because you can protonate a bivalent oxygen to make it trivalent, and then cleave one of the original two bonds to leave yourself with new molecules.
A bivalent oxygen is also easily protonated because it has a lone pair, making it a nucleophile.
Trivalent oxygen intermediates in organic reactions are also resonance-stabilized because the positive charge can be shared by the carbon in immediate vicinity.


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Reusing boiling chips
I was wondering why cannot boiling chips be reused.
As far as I know, they are microporous materials, often coated with PTFE.
They work by, I suppose, giving the heated substance a small cavity, in which a phase transition is favored.
Yet, the golden rule of boiling chips is: they cannot be reused.
I was wondering why they cannot be treated the way dessiccant materials (for instance, molecular sieves) are used, for instance by washing them after use and driying them in oven (especially when non-dangerous materials are involved).

Of course it depends on the use case. Will traces of compounds from the previous use interfere with your requirements for purity in their next usage? Will the chips become coated with some compound from the first usage that interferes with its function as a boiling chip in the following usage? If these things don't apply to your use case, and they appear to be doing their job, then I don't see why they couldn't be used more than once. I have done so on many occasions. Though I don't think I've ever used PTFE coated boiling chips; it seems like that would get in the way of their functionality.

You couldn't coat the chips with PTFE without sealing the pores. That would destroy their functionality.

According to this post at www.reference.com/science

Boiling chips are generally porous, trapping heating bubbles and
  allowing solvent vapors to form in the air cavities. They should
  always be added to a solvent prior to the beginning of a heating
  cycle; adding to a hot liquid may cause it to violently boil over.
  Boiling chips and stones are only used once, as the absorptive
  material becomes impregnated with the first solvent used in.

And Wikipedia confirms the basic idea of why they shouldn't be used more than once:

The structure of a boiling chip traps liquid while in use, meaning
  that they cannot be re-used in laboratory setups.

From personal experience however, I have to disagree with these absolute "cannot be done" sort of statements.  I used to rinse the used chips in an appropriate solvent and then heat them at some temperature well above the solvent's boiling point overnight and don't remember ever having problems.
I realize that the boiling point isn't completely relevant when you can have porous, high surface area material with adsorbed solvent.  Also, as I stated in the comments, I don't remember using PTFE coated chips.

Boiling chips are cheap unless you're using some really exotic type. You'd spend more money on the solvent in which you rinsed them than they are worth. // Having said that, I'd agree that baking them in an oven at a temperature sufficiently above the BP of the solvent should rejuvenate them.


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How can I determine if there will be a reorganization in organic reaction?

How can I determine if there will be a reorganization of the substrate to stabilize the $\ce{C+}$ in a organic reaction?

Consider an S$_N$1 or E1 organic reaction, when there is a formation of a $\ce{C+}$ during the process. Sometimes, the molecule will reorganize (like one hydrogen will "jump" to another carbon, or a $\ce{CH3}$ will do it) in order to stabilize the $\ce{C+}$. But, sometimes, this doesn't occur. Is there any way of predicting if it will happen or not? Thanks.

It depends on the rate of the competing reactions. Given enough time, the rearrangement will likely occur, but there can be competing pathways that trap the cation in the less stable form before it has a chance to undergo rearrangement.

Predicting? Not really. You can use sophisticated quantum mechanical calculations to explore the energetics of the reaction channels, and apply let's say trainsition state theory to get dynamical informations. This method is cumbersome, time-consuming and still far from being exact - but this is as close to predicting reaction pathways as it gets.
Now, in organic chemistry, you usually go with your "gut". I will try to present a way which organic chemist can think. When you have seen a lot of mechanisms, then you can guess whether reorganization happens or not. It also helps if you enumerate the circumstances that can affect the reaction. Let's see some of these:
Steric factors
In order to have a rearrangement, you obviously need one part of the molecule (the "jumping" part) to have access in space to the receiving part of the molecule. This requirement  can be satisfied if we let only a hydrogen atom jump, or we let the groups jump only a single bond. Now there are a lot of exceptions of this rule: for example in Benzilic acid rearrangement you can have phenyl groups mitigating. The idea that the groups have to be close to each other is also not always true, because you can have "jumps" between different molecules too. However, if the target part of the molecule is extremly hindered, then no rearrangement can be expected.
Electronic effects
Now, if we look at the electronic requirements of a rearrangement, then we have two different things to look for. One type of reactions can be called "orbital-controlled", because the mechanism depends mostly on the frontier orbital shapes and overlap of the reactants - in this case the rearrangements. A good example of these types of reactions are the sigmatropic processes, like Cope rearrangement.
If we assume that the orbitals are alright - which is not a huge assumption if we consider hydrogen transfer - then we have to look at the free energy profile of the reaction. This means we have to check how much energy we can "win" by rearranging the reactants. In the case you mentioned with the carbocations, this is equal to checking how stable is the carbocation. If it is instable (high energy), then we can "win" a lot by rearranging the molecule to the lower energy - by mitigating a hydrogen atom let's say. The stability of the carbocation then depends on the order of the cation and the electronic effects of different substituents. The general idea is that if you do not have your charge well distributed over many atoms, then you have a high energy - for example, CH3+ cation is extremly instable. If you have a lot of groups to "help" carrying the charge, either by inductive effects, or by resonance, then the carbocation stabilizes, making the rearrangement less favoured.
It is important to note that the things I have mentioned are mostly about the energetics of the reaction. Organic chemists seldom have a way to say anything about the kinetics of a particular reaciton. There are tricks like Hammett's relations and tranisiton state theory, but most of the time, they simply are not applicable. In our case with the hydrogen transfer, we could also look for quantum effects, which is completely out of organic chemistry's predictive reach.


text:
Removing potassium chloride crystals from a mineral composed of manganese dioxides
How can I remove crystals of potassium chloride from a mineral composed of manganese dioxides?
Hey, i'm a high school student trying to do a good cryptomelane sample to use in my scientific project.
I was able to develop the cryptomelane, but after some necessary processes, the mineral was crusted with crystals of potassium chloride, what would be the best process to remove this salt?

$\ce{MnO2}$ is insoluble in water -- I would expect with a good deal of confidence that the mineral cryptomelane is also insoluble. $\ce{KCl}$ by contrast is highly soluble.
I would just treat the sample with cold water and filter out the dissolved $\ce{KCl}$ (e.g. with a Buchner funnel and filter paper). Then dry with heat.
I would choose a fine-grain filter paper though because, though insoluble, $\ce{MnO2}$ particles could still be small enough to filter through course paper, like coffee filters (from experience).


text:
Can diamond undergo a self sustaining combustion reaction all the way to carbon dioxide?
I know charcoal can smoulder in an incomplete combustion to produce carbon monoxide so I assume diamond also can. If it's heated to a sufficiently high temperature, will the reaction produce enough heat to sustain the full combustion reaction all the way to carbon dioxide even in air that's only 21 °C. If so, does the reaction of carbon monoxide with oxygen occur fast enough to be perceived as smouldering or does it occur slowly enough to be perceived as flaming combustion?

Here's a demo featuring Sir Harry Kroto (2012)

Yes, diamond will combust in air. Regardless of the ambient air temperature, e.g. your example of $21\ \mathrm{^\circ C}$, you of course have to heat it to it's ignition temperature somehow, whether in a furnace, a flame, etc.  The autoignition temperature for diamond is around $900\ \mathrm{^\circ C}$ (source 1, source 2), compared to about $730\ \mathrm{^\circ C}$ for graphite.  The autoignition temperature is described in this Wikipedia article as:

"...the lowest temperature at which it spontaneously ignites in normal
  atmosphere without an external source of ignition, such as a flame or
  spark. This temperature is required to supply the activation energy
  needed for combustion."

So, once you have initiated combustion by some means heating it to about $900\ \mathrm{^\circ C}$, it will continue to burn, primarily to $\ce{CO2}$.  Note that combustion does not require that a flame be produced or maintained.
How efficiently diamond combusts in air, in other words how much of the carbon is converted to $\ce{CO2}$ vs. $\ce{CO}$ and other partially oxidized carbon species is a function of a few different things. The same is true when burning graphite, wood, or most anything else. Is it a single smooth "rock" or a finely ground powder? How much of what sort of impurities does it contain? How efficiently ventilated is the system? These issues and other conditions of the combustion will determine the degree to which $\ce{CO}$ and other partially combusted carbon species are produced.

LOL - I'm not sure about that. I'd be glad to test this experimentally for you. Could you send me a bushel basket of diamonds about the size of chicken eggs?

And I suppose you want them to be gem quality ;)  Check out source 1 and internet search it a bit. Antoine Lavoisier did some neat experiments with it, though I believe they were all in 1 atm of pure oxygen.  The most dramatic situation (see YouTube) is when you throw hot diamond into liquid oxygen!

@Mithoron, I cannot find any evidence to back up your claim and I gave evidence to back up mine (see source 2 and link at bottom here in particular).  Of course a continuous air supply is required, what other conditions do we need to sustain?  Here is another link on the topic:https://www.quora.com/Diamond-is-100-carbon-Why-does-it-not-burn, and I strongly recommend looking up some of Lavoisier's experiments, very interesting even if he generally used 1atm oxygen rather than .21 atm.  But he had to ignite it with a 17th century magnifying glass!

@airhuff Is the point of ventilation included in your answer is just because more and more CO2 will accumulate and the reaction will slow down?

Lasers are used to drill holes in diamonds of very specific sizes for wire drawing and such.  The diamonds do not burst into flames though what would happen under pure pressurised oxygen I do not know.


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How to use C1V1=C2V2 How do the units work?
I am trying to do a dilution where I know I have 0.749 g/ mL and I want to use this concentration to calculate the grams of aspirin in each 100 mL solution. I am specifically worried about the units. I want grams out. Is it valid to do the following?
One:
$$C1V1=C2V2$$
$$(0.749 g)(1 mL)=(C2 g)(100mL)$$
$$C2=0.00749 g$$
or must I do this?
Two:
$$C1V1=C2V2$$
$$(0.749 g/mL)(1 mL)=(C2 g/mL)(100mL)$$
$$C2=0.00749 g/mL$$
The difference being the final units. g vs. g/mL

Concentration is g/mL so the latter is correct.

So there is no way to get rid of the mL part? Part of me says to multiply by 1 mL to remove it... It seems like that wouldn't be valid though....

What do you mean "to get rid of"? That's the way it is; you can't just up and change that, much like you can't get rid of the letter "c" in the word "concentration". Unless, of course, you want to know how much of your solute is in 1 mL (or 2 mL, or any other volume) of your solution; then you multiply by that volume and get some grams.

You have to realize that if you have $$x_1 \cdot y_1 = x_2 \cdot y_2$$ then whatever the units of $x$ and $y$ the units will cancel appropriately.  That doesn't mean that the closing price of the NY stock exchange times the number of Yankee's tickets sold means anything.

Concentration has units mass over volume; or particles (moles by convention) over volume.
It's fine to use $g/mL$ for concentration in a basic dilution equation.
$C$ is never a mass.
So your second equation has proper units.
If you want mass of aspirin in a given solution with concentration $C$, the mass is given by $m = C\times V$


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How can a dilute acid dissociate more than a concentrated one? (Ostwald dilution law)
According to Ostwald's dilution law, the degree of dissociation of an electrolyte is inversely related to the  concentration of it.
$$ \alpha = \sqrt{\frac{K_a}{C}} $$
This means that a dilute acid can dissociate more than a concentrated one. How is it possible? Please explain the physical meaning of this.

Come to think of it, this is very natural and by no means limited to chemistry. When there are fewer competitors (whatever is their nature, molecules seeking to dissociate or people standing in a line for a new iPhone), more of them would get what they want.

A dilute acid has more solvent molecules to interact with than a concentrated acid, and thereby has more frequent collisions with the solvent. You can think of the dilute acid as "drowning" in the solvent.
A concentrated acid, by contrast, interacts more with other acid molecules, and therefore less with the solvent which dissolves it.
A concentrated acid will typically have more total dissolved molecules than a diluted acid simply because there are more of them, but it will have less as a fraction of total number of acid molecules; i.e. a weaker affinity for dissociation.
In short, $\alpha$ describes affinity for dissociation, not total dissociation. "Better", not "more".
Note that a higher $K_a$ corresponds to a stronger acid, so it makes sense that stronger acids will have a higher affinity for dissociation.

Also  the dissociation of an acid corresponds to its strength, so isn't it a bit confusing that a dilute acid is stronger than a concentrated one.

Dilution doesn't have any bearing on whether an acid is strong/weak. You can dilute a strong acid, and you can dilute a weak acid.
Strong/weak is a property of the chemical substance; dilution is simply related to how much of it you put in the solvent.

I want to say about  the dilute & concentrated forms of the same acid.

1M HCl will dissolve more easily than 2M HCl because the HCl in the first solution has more collisions with solvent molecules, on average. Hence I compared it to "drowning" in the solvent.


text:
Can "Common Ion Effect" suppress the dissociation of water molecules in acidulated water?
$$\begin{align}
\ce{H2O &<=> H+ + OH-} \\
K &= \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} \\
\ce{H2SO4 &<=> 2H+ + SO4^2-}
\end{align}$$
In acidulated water water dissociates into its respective ions(though the dissociation is very less) and simultaneously acid dissociates into its respective ions.
Now as the acid (I have taken $\ce{H2SO4}$ here) also dissociates to give $\ce{H+}$ ions. So the $\ce{H+}$ ion concentration increases which in turn affects the equilibrium of $\ce{H2O}$ and favours the backward reaction forming $\ce{H2O}$ back. So can we say common ion effect has suppressed the dissociation of $\ce{H2O}$?
And if this is the case then why do we use acidulated water while performing the electrolysis of water? (I mean the common ion effect can suppress dissociation of water and we would get less $\ce{H+}$ and $\ce{OH-}$ ions in the medium which would lead to a lower yield if $\ce{H2}$ and $\ce{O2}$ gas.)

You have a couple of concepts mixed together in the wrong way.
(1) The ionization of water is typically written as
$\text{K}_\text{w} = \ce{[H+] \cdot [OH^-]}$
The assumption is that the ionization is so low that the concentration of water remains a constant. For a solution which is half water and half acetic acid that wouldn't be true of course.
(2) The dissociation and recombination of water molecules is a very fast reaction and occurs continuously in water.
$\ce{2H2O <=> H3O+ + OH-}$
(3) In reality the autoprotolysis of water is dependent on the activities of the $\ce{H+}$ and $\ce{OH-}$. So the relation is more properly written as
$\text{K}_\text{w} = a_\ce{H+} \cdot a_\ce{OH^-}$
In dilute solutions the activity is essentially the same as the concentration, but in more concentrated solutions the activity drops as the concentration increases because of the formation of clusters of ion pairs in solution.
Also the activity drops as the overall ionic strength of the solution increases for the same reason. So if you add $\ce{NaCl}$ to the solution, then the ionic strength of the solution increases and the activity of the hydronium  ions and hydroxide ions decreases. So in concentrated $\ce{NaCl}$ the $\ce{Na+}$ and $\ce{Cl-}$ ions aren't really just "spectators."
(4) If you add $\ce{H2SO4}$ to distilled water then the acid will dissociate according to the following reactions:
$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}\quad \quad \text{pK}_{\alpha1} = -3$
$\ce{HSO4- + H2O <=> H3O+ + SO4^{2-}}\quad \quad \text{pK}_{\alpha2} = 1.99$
The final pH will depend on the final concentration of the sulfate species which is commonly expressed as the molarity of the sulfuric acid. So if we diluting concentrated sulfuric acid we typically use
$\text{m}_1 \cdot \text{V}_1 = \text{m}_2 \cdot \text{V}_2$
where $\text{m}$ is the molarity of the solution and $\text{V}$ is the volume of the solution. This ignores the fact that in dilute solution the sulfuric acid isn't just $\ce{[H2SO4]}$ but that the overall sulfuric acid concentration is the sum of $\ce{[H2SO4]}$, $\ce{[HSO4^-]}$, and $\ce{[SO4^{2-}]}$.
(5) Acidifying an electrolysis solution isn't really about ion concentration since adding NaCl isn't as effective as adding HCl. The gist is that protons can move through solution very quickly since a water molecule can accept a proton from one direction to form hydronium and kick off a different proton in the other direction to form water again. So a net charge can move through solution quickly as a result of the chain reaction of such transfers. Thus any individual proton doesn't have to travel far.
CONCLUSION
The overall notion would be better expressed as that adding sulfuric acid to water increases the $[\ce{H+}]$ and reduces the $[\ce{OH-}]$ thus shifting the equilibrium for the autoprotolysis of water.

(thus shifting the equilibrium for the autoprotolysis of water) toward dissociation or recombination of the water molecules.

If you added pure, isolated $\ce{H+}$ (somehow) to water, it would disturb the equilibrium of the auto-ionization of water quite substantially.
However, you have to consider that in addition to $\ce{H+}$, the dissociation of sulfuric acid produces $\ce{SO4^{2-}}$ as you've written. This counter-ion will pull your first equilibrium to the right during electrolysis -- not by addition of $\ce{H2O}$, but by electrostatic attraction of $\ce{SO4^{2-}}$ to the hydrogen in $\ce{H2O}$.
In this regard, addition of sulfuric or other acids for use in electrolysis of water generally helps because the "mess" of ions serves to smooth the process of splitting $\ce{H2O}$. The net effect actually lowers the energy required to break $\ce{H2O}$. The same principle is useful for electroplating; I use acetic acid as an enhancement to ion-flow for plating metals.
Ionic strength is the quantitative effect of interest here:
$$ I = \frac{1}{2} \sum_i c_i z^2_i $$
Where $i$ are different ions, $c_i$ are concentrations of those ions, and $z_i$ are charges of those ions. In general, more ions in aqueous solution actually improves rate of electrolysis of water, or:
$$k_{\text{electrolysis}} \propto  I$$

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I have serious doubts about this answer. For instance if you add HCl to water, then the assumption is that all the HCl dissociates.  So the $\ce{Cl^-}$ is regarded as a "spectator" ion in that it doesn't influence the pH. Also I could dump in NaCl and the pH stays the same. // The rest of the story is that the equilibrium really depends on the activities of the ions which vary with the ionic strength of the solution. "Low" ionic concentrations are typically assumed and the concentration is used instead of the activity.

@MaxW electrostatics have predominance over pH for electrolysis of water. I added "during electrolysis" to my answer for clarity. In fact, pH is practically irrelevant compared to ion concentration, current, voltage, electrode surface, etc. The distinction between electrolysis and auto-ionization is important because the OP asked for it


text:
What does β stand for in β-potassium zirconate?
What do the Greek letters (α, β, γ...) convey when they are prefixed to a crystalline substance?
For example: β-potassium zirconate

Interesting question! (upvotes) I came across this (α, β prefixes) when I was reading up on tin the other day. Apparently, it refers to crystal structure (that's as far as I could deduce). https://en.wikipedia.org/wiki/Tin#Physical_properties

They mean "the first one", "the other one", and "hey, there's a third one" respectively.

They signify one polymorph (the beta polymorph) of barium zirconate.
Allotropy is the property of some chemical elements or compounds to exist in two or more different forms in the same physical state at different temperatures. Technically, they are the same element or compound but have other crystal structures at changing temperatures. To name them, the greek letters alpha, beta, gamma, etc. are used in the prefixes to depict them as the allotropes formed at different temperature phases. That helps in their differentiation.
E.g. iron has 5 allotropes, namely:

ferrite (α-iron): the ferromagnetic form at T < 770 °C (the Curie point, TC); body-centered cubic (BCC) crystal structure
β-iron: T = 770 – 912 °C; the paramagnetic form of α-iron; same structure as α
austenite (γ-iron): 912 – 1,394 °C; face-centered cubic (FCC) crystal structure
δ-iron: 1,394 – 1,538 °C; forms from cooling down molten iron; same structure as α
ε-iron (also called hexaferrum): forms at high pressures; hexagonal close-packed (HCP) crystal structure

There are several more examples of such types of nomenclature of elements that can be found in the wikipedia article of allotrope(see table).
Also, not only elements form allotropes. Compounds also form allotropes but have a different name, i.e., polymorphs. Generally, binary metal oxides are polymorphs. Polymorphs have different temperature phases in which one form of crystal structure exists. These phases are depicted by giving a greek alphabet letter at the prefixes like α-phase, β-phase, etc. Polymorphs or allotropes of compounds are given that greek alphabet prefixes according to that phase in which one form of that compound exists.
For e.g. iron oxide $\ce{Fe2O3}$ has 4 phases.

α-phase: at ambient conditions: has Corundum-type Rhombohedral (R3c) like structure
β-phase: below 773 K: has body-centered cubic (Ia3) like structure
γ-phase: up to 933 K: has cubic spinel structure (Fd3m) like structure
ε-phase: has Rhombic (Pna21) like structure

So, iron oxide $\ce{Fe2O3}$ has 4 polymorphs : $\ce{\alpha-Fe2O3, \beta-Fe2O3, \gamma-Fe2O3, \epsilon-Fe2O3}$.
For more information, you can check this question(asked by myself) or this question, which describes how polymorphs are formed.

Got a comment: Allotrope is generally understood as limited to elements. While polymorph is more general term for different structural forms.

@ankit7540 Yes, allotrope is for elements but polymorphs is generally for compounds.

Some inorganic crystals show the property of structural polymorphism where two or more crystal polymorphs exist. These polymorphs may differ either by position of metal ion or ligand (or other ion species) within the crystal or by different crystal packing. Usually polymorphs have different space groups.
Several of these forms or polymorphs have different physical properties and some may exist only at some temperatures and pressures (stability as a reason). Read more here: Crysal polymorphism - Wikipedia.
With respect to barium zirconate, it has cubic type pervoskite crystal structure where the barium atom is at the center of the cube while the zirconate ion is present on all of 8 edges.
For the β-form mentioned in your question, I do not find any reference for polymorphic barium zirconate, however when the crystal is doped with other atoms (like yttrium replacing barium) then polymorphism occurs (Yoshihiro Yamazaki, Raul Hernandez-Sanchez, and Sossina M. Haile. J. Mater. Chem. 2010, 20, 8158-8166.).


text:
Differences between miscible and soluble
What are the differences between soluble and miscible? So far I found this (page 3 http://faculty.sites.uci.edu/chem1l/files/2011/03/D01MANmiscibility.pdf):
soluble=solid dissolves into a liquid
miscible=two liquid give rise to a homogeneous solution
Are there other differences?
Thank you.

Solubility has certain limits; miscibility doesn't. Miscible liquids can mix in any proportion. Also, welcome to Chem.SE.

Your definition is basically correct, except using two liquids might provide for a clearer example than a solid and a liquid.  For example, diethyl ether is soluble in water to about 7% at $\mathrm{20^oC}$. This does not mean that diethyl ether is partially miscible in water, but that it it partially soluble. This Wikipedia article provides a good distinction for miscibility:

Miscibility is the property of substances to mix in all
  proportions (that is, to fully dissolve in each other at any
  concentration), forming a homogeneous solution.


text:
Is ruthenium really invulnerable to aqua regia?
Recently, I answered a question "What metals aren't dissolved in/attacked by aqua regia?" in which I finally concluded "ruthenium is resistant to aqua regia in any form at any concentration of acid".Fellow answerer  @Linear Christmas also supported my statement
However, in my answer, I pointed out that something is not right. In the blockquote I wrote:

Ruthenium can be dissolved in aqua regia only when in the presence of oxygen (Why? will be asked in a separate question)

It is written that ruthenium can only be attacked in aqua regia in the presence of oxygen. And for this statement, this question is asked. This statement can also be found in the wikipedia article of noble metal.
In the wikipedia article of ruthenium it is clearly written that:

Ruthenium (...) is not attacked by acids (even aqua regia)

Even in the handbook, it is written that (statement can be found in my answer):

Ruthenium is by far the most chemically resistant of the noble metal,
  unattacked either in hot or cold aqua regia.

Also this video by Nurdrage shows that ruthenium is invulnerable to aqua regia or any kind of acid.
However in this google book, it is written:

Pure ruthenium powder or sponge is not appreciable attacked by hot, single or mixed mineral acid(aqua regia).

Why it is written "not appreciable attacked"? Maybe it should have wrote invulnerable or resistant. I am assuming that the corrosion rate is not appreciable/very slow i.e. corrosion is too slow to get noticed by naked eye.
Is my assumption correct? Can ruthenium be attacked in aqua regia only in presence of oxygen? Why wikipedia wrote that statement regardless of the fact that ruthenium is resistant to aqua regia in any form at any concentration of acid? I have not found any evidence that shows ruthenium is vulnerable to aqua regia.

Page 807 of A Textbook of Inorganic Chemistry says that iridium, rhodium. and ruthenium are "almost unaffected" by aqua regia.  So "completely invulnerable" = no, but "resistant" = definitely yes.

Can you elaborate a little?


text:
Does a mixture of gases contain equal volume to that of the container they're in? Why?
In my textbook, the derivation of Dalton's Law of Partial Pressure is done by keeping the volume of a mixture of gas in a container constant which is equal to the volume of container itself.
How can this be so? Let's say we have 100 moles (fairly large) of Nitrogen and 1 mole of hydrogen gas taken in a container of a certain volume. Then is the volume still considered equal to that of the container in this case? Because this is what is done when deriving Dalton's law of partial pressure?
Also, if we have 100 moles of a  gas and another just a single molecule then what happens?

Well, the word "container" means a fixed volume. Let's look at your room for example (don't take into account any openings, like doors or windows, just the walls). It has some amount of gas in it at this moment. Now, if someone decided to "push in" let's say 100 extra moles of O2 gas, then that would not change the volume of your room. The volume of the containter remains constant, no matter how much gas we put into in.
This is also true for each component of the mixture. The oxygen in the original air of your room had the volume of your whole room, and even when we "push in" that 100 extra moles, it still has the same volume - because your room did not change. What can change then? Obviously, the pressure or the temperature can change.

but when we run PV=nRT for each gases,shouldn't we consider V for the particular gas we are dealing with?Why would we take the volume of entire container?

The assumption is that the gas is completely homogeneously diffused. You can think about this assumption by checking that you can not really find either of the components in one place or another, but you can find every single component in the whole container - every component by itself takes the entire volume.

Let's say there is a single molecule of oxygen gas in the container very large.Now how will you statement compensate this?Please help

The huge oxygen molecule still can be found in any place of the container with the same probability - it is not sitting still in a room, but moving all around, like all other gas molecules. Think about volume as not the size of the molecules, but rather the space where they can move.

@SurazB In that case that single gas molecule will move in the entire room,changing directions after hitting walls

@Ezze Still the volume of a single molecule would be 8pi r squared or something depending upon the diameter of it.So won't the results of the experiment(for instance) we're conducting be way off and rendered useless due to the volume assumption we made?

No, that's the SIZE of the molecule. Note the thing I wrote above: volume is more like the space the molecule can move in. It is the whole container, regardless of the size of the molecule.


text:
What is CuCl · CO · 2 H₂O?
What is the chemical $\ce{CuCl.CO.2H2O},$ formed via the reaction of copper chloride and carbon monoxide?
Does this compound have any use or appear in any noteworthy reaction?

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

The compound is called copper(I) chloride dihydrate - carbon monoxide adduct. It is formed during the Gatterman-Koch reaction.
Gatterman-Koch reaction is a reaction that produces aromatic aldehyde from aromatic hydrocarbons. The reagent used are carbon monoxide, hydrochloric acid and a Friedel-Craft catalyst like copper(I) chloride ($\ce{CuCl}$) or aluminium chloride ($\ce{AlCl3}$). So, the adduct is formed by the reaction of copper chloride and carbon monoxide during the course of reaction. If aluminium chloride is used, then a complex would be formed i.e aluminum trichloride complex, $\ce{[HCO(AlCl4)]}$.
To know more about the mechanism and role of reagents, see this previous question on chem.SE..
Other References

https://en.wikipedia.org/wiki/Gattermann_reaction
https://www3.aiche.org/proceedings/Abstract.aspx?PaperID=358705

By dissolving mixture of ammoniacal cuprous chloride solution  with carbon monoxide, the reaction that occurs is the formation for copper(I) chloride dihydrate carbon monoxide adduct:
$$\ce{CuCl + CO + 2 H2O -> CuCl·CO·2H2O}.$$
This reaction is also used in Bosch process for the removal of carbon monoxide.

It would be nice if the answer were also backed up by providing the references to the original publication/patent.

$\ce{CuCl.CO.2H2O}$ is the addidtion compound when CO is passed through the ammonical solution of cuprous chloride.
$$\ce{CuCl + CO + 2H2O -> CuCl.CO.2H2O}$$
It is white crystalline adduct.
I dont really know the name... but this is the story behind this compound..

You're not saying much that OP didn't already say, to be honest. It's not much of your fault; the question is just poor.

We can consider it as coordination compound in which Chlorido and carbonyl acts as ligands and it can be represented as  $\ce{[CuCl(CO)].2H2O}$

Cu+ is transition metal ion and it form coordination compounds with Lewis bases.

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


text:
Quantifying a racemic mixture by combining signal-to-noise measurements
I'm working with chromatographic data for the measurement of a chemical that takes two separable forms in equal proportions (a racemic mixture). This produces two adjacent signals (peaks) which are nearly identical, notwithstanding some minor analytical measurement variation. I need to make a judgement call on whether the compound is "detected" or not across many hundreds of samples, so I'm looking specifically at the signal-to-noise ($s/n$) ratios of each peak (using the root-mean-square measurement of baseline noise). I'm using $s/n = 3$ as a threshold for detection (LOD), but some peak pairs close to the threshold of detection will have one peak with s/n barely above and the other barely below.
How can I most appropriately combine these two largely-redundant measurements?  Can I average the two s/n measurements for one single easy-to-interpret value?

Welcome to Chemistry! 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.

Yes you can combine the two with the assumption that it is a racemic mixture. How you'd do this is a matter of speculation. I have no idea how you're processing the data.


text:
Confusion regarding a stoichiometric formula
In a book, which contains only stoichiometry problems, I came across this formula.

For any given compound $\ce{M_xN_y}$, $x \text{ Moles of N} = y \text{ moles of M}$

According to this rule if I take $\ce{H2O}$ then, $x = 2 \text{ and } y = 1$.
So by the above formula,
$2 \text{ moles of Oxygen} = 1 \text{ mole of Hydrogen}$
But surely,
$2 * \mathrm{N_A} \not= 1 * \mathrm{N_A}$, where $\mathrm{N_A}$ is Avogadro's number.
However, in an example, the author has used this formula and successfully solved a problem? So, if it's correct, can anyone prove this and point out where my contradiction is wrong.
Example as requested:

What weight of oxygen will react with $1$g of calcium?($\ce{Ca}= 40\mathrm{g \over mol}$)

Answer by author :
Since all the atoms of $\ce{Ca}$ have changed into $\ce{CaO}$, the amount of $\ce{Ca}$ in $\ce{CaO}$ is $1$g. Now from the formula of $\ce{CaO}$, we have,
$\text{Moles of Ca = Moles of O}$
$$\frac{\text{mass of Ca}}{\text{atomic mass of Ca}}= \frac{\text{mass of O}}{\text{atomic mass of O}}$$
therefore, $\text{mass of O} = {1\over40} * 16 = 0.4$g

You misread the formula. For compound $\ce{M_{x}N_{y}}$
$$y\cdot \ce{mol M} = x\cdot \ce{mol N}$$
This is because the formula essentially tells you that the ratio between moles of $\ce{M}$ and $\ce{N}$ is:
$$\frac{\ce{mol M}}{\ce{mol N}} = \frac{x}{y}$$
This rearranges to the equation above.


text:
Rusting: alternatives to water?
I have several questions about the rusting of metals (take iron as a specific case if you wish) that I have not found clear answers to:
1) Does water act as a catalyst in prototypical rusting reactions, or is it actually consumed? (keeping in mind that $H_2$ gas produced may be oxidized back to water by ambient $O_2$.)
2) Is dry rusting possible with just $O_2$ and metal? Am I correct that it is thermodynamically favored but kinetically discouraged?
3) Is it realistic to rust a metal using oxygen and a dry polar solvent? If so, what solvents would be suitable?

Thank you, I  read that article before posting. I did not see answers to my questions there.

Rust is by definition the transformation of iron metal to iron oxide (and iron hydroxide as an intermediate). You cannot "rust" other metals, although you can form oxides and degrade them etc. It's just not called rust when you convert, e.g. $\ce{Cr}$ to $\ce{CrO3}$. So there is only one rusting process.
1) Water is not a catalyst in the reaction that forms rust. A catalyst by definition is not consumed during a reaction (its bonds are not broken). Water can be and is the medium that carries $\ce{O2}$ to the iron (otherwise $\ce{O2}$ from air can react with the metal), and it is consumed in the process by dissociation. The fact that it is reproduced does not make it a catalyst. First iron atoms are oxidized by $\ce{O2}$ as in:
(a) $$\ce{Fe + O2_{(aq)} \rightarrow Fe^{+2}_{(aq)} + 2e-}$$
Then the free electrons are quickly absorbed by free protons from water by:
(b) $$\ce{2e- + 2H+_{(aq)} \rightarrow H2}$$
It is highly unlikely for $\ce{H2}$ to react with $\ce{O2}$ to form water -- that is a combustion reaction, so high heat or pressure would have to be added. Rather, water is reproduced by a different reaction (d), below.
2) "Dry" rusting is not possible because the process requires reactions with water's ionization products $\ce{H+}_{(aq)}$ and $\ce{OH-}_{(aq)}$ (notice the aqueous states of all ions). The process continues by production of iron hydroxides:
(c) $$\ce{Fe^{+2}_{(aq)} + 2OH-_{(aq)} <=> Fe(OH)2_{(s)}} $$
(d) $$\ce{Fe^{+2}_{(aq)} + 4H+_{(aq)} + O_2_{(aq)} <=> 4Fe^{+3}_{(aq)} + 2H2O_{(l)} } $$
$$\ce{Fe^{+3}_{(aq)} + 3OH- <=> Fe(OH)3_{(s)}} $$
This is why more acidic media produces rust faster; it pushes (c) and (d) to the right because (b) consumes free electrons and renders more $\ce{Fe^{+2}}$ for further reaction.
Finally the iron oxides "dry" to form the actual rust compound, which is a hydrated oxide: $\ce{Fe2O3\cdot nH2O }$.
3) You can see by now that water is the required solvent because the reactions of interest all involve $\ce{H+}$ and $\ce{OH-}$.
One could say there exists a "rust triangle" where "time, oxygen, and water" are required as an analog to the "fire triangle": heat, oxygen, and fuel.

You are correct in that water is not a catalyst. Nevertheless, a catalyst very often participates in elementary steps (i.e, its bonds can be broken). The large regeneration is what's key. Also note that iron(III) hydroxide does not exist. The formula $\ce{Fe(OH)3}$, common in 2oth century textbooks, is quantitatively the same as the iron(III) oxide-hydroxide monohydrate. This led to confusion, but the correct compound is now known to be $\ce{FeOOH*H2O}$.

There is also no true "rust compound". As you know, it is a mixture of other oxides, hydroxides, oxide-hydroxides and their hyrdates. Making this distinction is important since $\ce{Fe2O3*nH2O}$ might not be most common compound in rust. Especially on the surface it is commonly $\ce{\alpha-FeOOH}$, followed by $\ce{Fe3O4}$.

You say dry rusting is not possible, but if you wash with an acid then workup with an alkali hydroxide? (Some dry polar solvent). Wouldn't that circumvent deliberately using water as a reactant?

I don't see why one can't contrive the process of oxidizing the iron, stripping electrons with acid, and reacting with NaOH or some base all with the right polar-organic solvent, but to form the product iron oxide-hydrate you'll need water molecules. The distinctive brown/red/orange substance we associate with rust is a mixture of those hydrates (and the iron hydroxides as mentioned by @LinearChristmas).


text:
Chile 2017, One of the Largest Fires in South America: Wood Ash Derivates for Industrial Applications
Hi: We're a group of people who wants to help people, domestic animals and wildlife recovering from this fire.

Summary:

We're having in Chile one of the largest fires in South American history.
More than 1,235,527 wood acres have already been burned
The fires aren't stopped but rather increased
People, domestic animals and wild fauna are suffering badly
We're a group that wants to help in the recovery of the lands that have been burned
There're several products with industrial applications that can be extracted from Wood Ashes. For instance, Potassium Peroxide
Our idea is to do the following onsite:

Build over a truck(s) a mobile laboratory
Go to the places where there's too much ashes (making the soil extremely alkaline and hard to recover)
Get some of the ashes, and extract a product with industrial application  that can be sold for a price that produce profitability
The profits would be used for supporting the people who lost everything, helping the wildlife to recover faster and financing the project

The Question:

Are there Wood Ash sub products that you think are suitable to be extracted on site and allow the mentioned objectives mentioned above?
Do you know a technology to be used to do the extraction onsite?
Do you know a chemist who could be interested to participate in the project?

I'm voting to close this question as off-topic because this isn't the type of queries we expect on Chem.SE.

It may be better to do some consulting. Some universities have consulting organizations for physical sciences and mathematics. They may be better equipped to help you. Good luck.

Thanks for your help. I'm an economist and Berkeley MBA. I don't know Chemistry. Fortunately, some Chinese PhDs in Chemistry are already designing a solution! They will bring in the following days a team to help us.  ;)


text:
What are the requirements for FRET?
I have a question about fluorescence resonance energy transfer (FRET), and the requirements needed for this technique.

Q: Which of the following need to be met for FRET?

A) Strong overlap between the donor emission spectrum and the acceptor absorption spectrum
B) Molecules must be $<10\,\mathrm{nm}$ apart
C) Molecules must be $>10\,\mathrm{nm}$ apart
D) Low quantum yield of the donor
E) Small extinction coefficient of the acceptor

I'm aware that A and B are true, and C is not a requirement. But what about a low quantum yield of the donor or a small extinction coefficient of the acceptor? Do the size of these numbers really matter?

One way to express the fluorescence quantum yield (QYf) is:
1)  QYf = kf/sum(ki)
where kf is the kinetic constant of the fluorescence and sum(ki) is the sum of the kinetic constants for the overall deactivation processes, I.e. non-radiative processes and radiative. It is also true that:
2)   1/sum(ki) = tau
Where tau is the lifetime of the molecule. That means QYf is proportional to the lifetime (replace 2 in 1). High QYf high fluorescence lifetime. In the FRET scenario this means that a Donor with high QYf has more time to donate its energy to an Acceptor, while a Donor with low QYf would deactivate really quickly to the ground state because of faster non radiative processes, resulting in a poor FRET. Thus, a LOW QYf disfavors the FRET process.
The extinction coefficient is proportional to the absorption of a molecule. Thus, a high extinction coefficient for the acceptor will favor the FRET. Matematically, you can see that the extinction coefficient of the acceptor is proportional to J, i.e the integral overlap between donor and acceptor thus proportional to the FRET process.
Hope this helps

A low quantum yield would indicate that there are efficient intra-molecular non-radiative decay pathways, likely internal conversion or spin-orbit coupling. These decay pathways may be faster that FRET, rendering FRET essentially a non-competetive process.
A small extinction coefficient would indicate a small transition dipole moment. If the transition dipole moment is small it is difficult for the FRET partner to transfer energy since it is a weak transition resulting in a small coupling  between the two chromophores.

The equation for the rate constant ($\pu{s^{-1}}$) for Forster (or Resonance of dipole-dipole ) energy transfer at separation R is
$$ k_R= \alpha\frac{\kappa^2\phi}{\tau R^6}\int_0^{\infty} \frac{F(\nu)\epsilon(\nu)}{\nu^4} d\nu$$
where the constant $\alpha =(9000\ln(10) )/(128\pi^5n^4N)$, n is the solution refractive index and N Avogadro's number.  The quantum yield of the donor is $\phi$, and its excited state lifetime (in the absence of quencher) is $\tau$, the  orientation term is $\kappa$ and R the separation of donor and acceptor.
In the (overlap) integral F is the fluorescence spectrum measure in frequency (not wavelength) and the area under $F(\nu)$ is normalised to unity. The molar extinction coefficient of the acceptor is $\epsilon(\nu)$ also measured on a frequency scale normally in units $\pu{dm^3mol^{-1}cm^{-1}}$.
The rate constant is more commonly written as
$$k=\frac{1}{\tau} \left( \frac{R_0}{R} \right)^6 $$
where $R_0$ is the critical distance at which energy transfer rate constant equals the fluorescence rate constant ($1/\tau$) and is also a measure of the overlap of fluorescence from the donor and absorption by the acceptor.
This the decay rate of a molecule that is fluorescing and undergoing energy transfer is at separation R equal to $k=k_f+k_{isc}+k{ic}+k_R$  where isc and ic are intersystem crossing and internal conversion respectively.
In you questions;
(A) the bigger this is the larger $k_R$ is.
(B) and (C) there is no real cut off, $R_0$ for chlorophyll is $\approx 8$ nm so transfer can occur beyond this distance it just falls off as $1/R^6$
(D) Clearly the smaller $\phi$ is the lower the rate of transfer and in direct proportion. If the molecule has a low fluorescence yield clearly the molecule cannot transfer energy as it is directed elsewhere.
(E) If $\epsilon$ is small then the overlap integral is small so the rate $k_R$ is small.


text:
Why does potential energy relate to exothermic reactions in this way?
I am not understanding how potential energy (which is the energy that results from position or configuration) relate to exothermic reactions. We say that exothermic reactions give off heat and so lose energy but I don't see how I can use this definition of potential energy to say that potential energy is lost from the system of an exothermic reaction. It might be my definition for potential energy that is wrong, I don't know.

If you got any system in a state $q_1$ with potential energy $U(q_1)$ and it changed into state $q_2$ with potential energy $U(q_2)$, then if $\Delta U = U(q_2) - U(q_1) < 0$, energy has been lost to the surroundings.
Does that go in the direction you want?

@FelipeSchneider  i understand that the system lost heat energy but i dont understand saying the system lost potential energy so i think i have the wrong definition for potential energy

Energy is a useful concept for rationalising the capability of doing work or heating; other than that, it is purely a theoretical concept.
As Feynman brilliantly pointed out,

It is important to realise that in physics today, we have no knowledge of what energy is.
  We do not have a picture that energy comes in little blobs of a definite amount.
  It is not that way.
  However, there are formulas for calculating some numerical quantity, and when we add it all together it gives “28”—always the same number.
  It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

But let's get back to something answerable.
Energy is (a) conserved and (b) convertible into different forms.
If you got any system in a state $q_1$ with potential energy $U(q_1)$ and it changed into state $q_2$ with potential energy $U(q_2)$, then if $\Delta U = U(q_2) - U(q_1) < 0 $, energy has been lost to the surroundings.
This loss could be lots of things, most of the time it is just heat (kinetic motion).
This loss, in the big picture, is just a transfer or conversion.

Let's get to a very simple example, hydrogen combustion:
$$\ce{2 H2 + O2 -> 2 H2O}$$
Since energy can't be produced, we say this reaction releases large amounts of heat.
But, where does it come from?
All that the equation above tells us is the atomic rearrangement that has been taken place: in the left hand side we have two $\ce{H-H}$ bonds and a single $\ce{O=O}$ bond, in the right hand side we got four $\ce{O-H}$ bonds.
So quantifying the bond energies in each side gives us:
     |  Average bond energy (kJ/mol)
-----|-------------------------------
     |        |   Left    |   Right
Bond | Single | hand side | hand side
-----|--------|-----------|----------
H-H  |  -432  | 2 x -432  |    0
O=O  |  -495  |  -495     |    0
O-H  |  -467  |    0      | 4 x -467
--------------|-----------|----------
   Total      |   -1359   |  -1868

This means we get $-1868 - (-1359) = -509$ kJ for two moles of $\ce{H2}$, i.e., $-254.5$ kJ/mol of heat is released during the combustion of hydrogen.
Wikipedia says the correct value is $-242$ kJ/mol, so our calculation has an error of around 5%.

So heat energy has been released and it came from bond breaking and forming.
The energy stored in a bond could be called latent or potential, if you will.
It all boils down to conservation and convertibility.

The potential energy of a system is converted in to "heat energy" during an exothermic reaction. Heat is a measurement of molecular vibrations.
When you have a molecule with a potential energy, this reflects the strength of its molecular bonds. To break this bond, energy is first applied to the molecule, to reach an "activation energy". At this activation energy, the molecule will enter an unstable transition-state (which has a potential energy even higher than that of the starting molecule).
Since the transition state is unstable, the molecule will naturally move towards a more stable configuration. To do so, the molecule will form a bond with another molecule. This bond will be even stronger than the bond of the first molecule, and therefore has a lower potential energy. The stronger bond, will not vibrate as strongly as the bond from the starting material.
As the first molecule is converted into the second molecule, its motions (literally, movement) throughout the solution will provide energy, or heat, to nearby molecules (and the container in which the reaction takes place).


text:
How does NaCl (or any inorganic salt) increase surface tension?
Does the compound "prefer" to stay in the solution?
If yes, wouldn't the surface tension of the solution be equal to that of pure water, since only water is in the surface?
See: Does NaCl reduce the surface tension of water?
Or is the compound distributed evenly and because the ionic bonds are stronger than the attractions between water molecules, the surface tension increases?
Read here a reddit answer to why do ionic compounds such as NaCl increase the surface tension of water?
Explain like I'm 5 please.

I hope this picture helps.
The interaction between a molecule of water and an ion is stronger than the hydrogen bonding that occurs between two water molecules.

1+1 = 2, exactly. Other than that, use of the term "exactly" and similar terms like "absolutely" should be avoided in technical discussions. Also, while I understand what you meant, its quite unlikely that most 5 year olds are capable of understanding such abstract concepts, but I digress...
The ions increase the average particle-to-particle (p2p) attractive forces in the aqueous solution, which means that more energy is required to increase the surface area (bring a particle from the interior (bulk) to the interface. The traditional example is comparing the bond energy of a ion in a cubic closest packed (6 nearest neighbors in a "square" 3D grid) (so it has a total bond energy of $6x$, where $x$ is the energy of a single bond) to the same type of ion at the surface of a (large) lattice - at the surface, one bond will be missing ($5x$). (This is a simplification, since it ignores all electrostatic forces (both repulsive and attractive) of ions not 'nearest neighbor'.)
Of course, in the real world surface energy/tension considerations are usually concerned with at least two phases (eg liquid & gas in aerosols) and often three (eg. water wetting the side of a glass container in air). You can imagine that the measured surface tension could be different if measured in helium compared to the surface tension of the same substance measured in carbon dioxide. The link you provided to the reddit had some correct information, but then it went off the rails when it discussed detergents. It's wrong, there. The stackexchange Q&A link you provided seems to me to be a correct point of view, but I don't find it useful, do you? The same argument applies at the surface as in bulk (for small solute molecules and ions). The ion must have a solvation sheath (a coat of solvent molecules) to 'insulate' the charge, otherwise it wouldn't go into solution. (This applies to polar solvents (and water is both polar and a strong hydrogen-bonding molecule)). It should be clear that if the ion were at the surface, that it wouldn't have as "thick" a coat so it will tend to avoid it.
Keep careful track of each argument's scale. Talking about a lattice of ions is considering a much smaller scale than a coated ion in solution. By that I mean the word "surface" means different things at those two scales. In the former (crystalline solid) it is the outermost ions of the lattice, while in the latter it is a zone of multiple solvent molecule diameters depth, with the molecules and ions bouncing around, and forming only intermittent (secondary as in intermolecular, non-covalent, non-ionic) bonds.
Detergent molecules are generally much larger, and they usually have a non-polar tail which doesn't interact with water (hydrophobic) and a polar head (which has much more interaction with water (meaning it increases the secondary bonding energy of the system)). These molecules orient themselves at the surface and so for (most) detergent solutions, the surface is highly enriched with them. They use this fact to coat mono molecular films (one molecule thick) onto glass. So, it is not true that ions aren't to be found at the surface of the solution, but it is true that their concentration there will be less (and depend on temperature). The key fact is that they increase average intermolecular bond energy of the solution.


text:
Will fluoride evaporate from tap water at the same rate as chlorine
Chlorine evaporation is an overnight process from standing tap water in a gallon pan. Does fluoride evaporate in the same way/rate?

There's crucial difference between fluorine and fluoride.

Were these downvotes before or after migration? Seems like a good question to me.

And this whole -ine vs. -ide thing is pedantic. Clearly, the asker means the vernacular usages. "Chlorine" and "fluoride" are added to drinking water. How do they compare when using evaporating techniques? I sure hope the downvotes are not because of this.

@fredsbend Chlorination of water is often by adding actual chlorine gas ($\ce{Cl2}$),  whereas fluorination of water can be actual sodium fluoride (NaF) which is a solid.  NaF has no chance of evaporating unless you heat it to over 1000 degrees C.   $\ce{Cl2}$ leaves the water easily like carbon dioxide coming out of soda.

@DavePhD Considering how the question was asked, I think that's exactly the information needed in an answer. The OP is clearly unaware of this. My complaint is if the downvotes are a result of the OP's apparent ignorance regarding what "added chlorine" and "added fluoride" mean chemically.

@fredsbend I upvoted.  It think some people here want to see more college-level chemistry questions maybe, I don't why they downvote really.

Short answer: no, fluoride does not evaporate in the same way/rate as chlorine.
Longer answer: chlorine will be present as covalent, molecular $\ce{Cl2}$. This is, when not in solution, a gas. Chlorine molecules leaving the surface of the water will (on average) not come back (at least unless there is a very high concentration of $\ce{Cl2}$ in the room's air, in which case you have other problems).
Fluoride, on the other hand, is present as $\ce{F-}$ ions (or other, complex ions) along with corresponding positively charged ions (usually $\ce{Na+}$, I believe). Even if the $\ce{NaF}$ were to be removed from solution, $\ce{NaF}$ is a non-volatile salt, and so would not evaporate away.

An imperfect answer because I slightly gloss over how ions and non-ionic molecules behave differently going into / coming out of solution. Suggestions and edits are very welcome!

Considering how the question was asked, I think you answered it perfectly. If the OP wants to know more about the differences between dissolved gases and dissolved salts, then he can look it up or ask a different question.


text:
Hydrogen peroxide: catalyst changes colour during peroxide decomposition
When adding iron(III) chloride to hydrogen peroxide, the hydrogen peroxide decomposes.
Iron(III) chloride is a catalyst.
But when it is first added, it changes its colour to dark brown, then near the end it turns to orange.

Why does the colour change?

I heard that it has to do something with reaction intermediate. Correct?
It is seen on: iron(III) chloride
Also, the colour changes when different catalyst is used like iron(III) nitrate and potassium iodide potassium iodide, iron(III) nitrate

Recently, I answered your question and relevant information can be found from that answer.
At first, I thought that the brown colour is due to ferrous chloride that formed during the course of reaction but since the reaction is occuring in a aqueous solution, it should be green in colour. Ferrous chloride is brown in anhydrous form. So I thought to be either due to formation of a complex, iron(II, III) oxide($\ce{Fe2O3}$), or iron(III) oxide-hydroxide($\ce{FeOOH.nH2O}$) or a mixed salt of oxide-chloride i.e. iron oxychloride($\ce{FeOCl}$).
The following information is from the paper 2 of that answer:

According to the mechanism proposed by KREMER and STEIN, an
  intermediate oxygen complex of iron with oxidation number +V is
  primarily formed by the reaction of $\ce{Fe^3+}$ with $\ce{H2O2}$.
  This complex reacts with another $\ce{H2O2}$ molecule to water and
  oxygen thereby reforming $\ce{Fe^3+}$.
$$\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> [Fe^{V}O]^3+ + H2O ->[H2O2] Fe^3+ + 2H2O + O2}$$
According to the mechanism proposed by HABER and WEISS the $\ce{Fe^3+}$ ions
  initiate a radical reaction, after which the chain reaction consumes
  the hydrogen peroxide. This mechanism can explain the high reaction
  rate very well.
Chain initiation:  $\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> Fe^2+ + HOO. + H+}$
Chain propagation: $\ce{Fe^2+ + H2O2 -> Fe^3+ + 2OH.}$
  $\ce{Fe^3+ + H2O2 + OH. -> Fe^3+ + HOO. + H2O -> Fe^2+ + H+ + O2 + H2O }$

That intermediate oxygen complex of iron with oxidation number +V maybe responsible for the brown colour of solution.

As for the potassium iodide, the color you are seeing is basically food coloring. Various food coloring  is used to make the foam produced colorful. If food-color is not used, still color is seen. That color is basically color of pure soap foam. This elephant toothpaste experiment not uses food-color but still it has color that is color of soap-foam.

I have never seen this reaction before, but if I recall correctly, chloride ions ($\ce{Cl-}$) are known to give a solution an orange appearance.
I am assuming that the $\ce{H2O2}$ oxidizes the iron, leaving free chloride ions in solution.

Does NaCl dissolved in water appear orange?

No, I do not believe it does. It's possible I was thinking of chlorine  - in which case chloride ions would be reacting with each-other near the end of the reaction.

Why would hydrogen peroxude oxidize the iron? The catalyst is not being used up

That is indeed the definition of a catalyst.
But, I think the oxidation would actually be from O2.
H2O2 ->  H2O + O2
O2 + Fe(III) -> Fe2O3
I'm mostly just speculating... Sorry

If the solution only contains water, hydrogen peroxide, and iron chloride.. And the solution is a different color at the end of the reaction. There is no possible way that there are not any new-formed compounds.
As water and H2O2 are both clear, and O2 would either evaporate, or remain dissolved.
And dissolved O2 would still be clear.

The orange-yellow colour is caused by the chloride ions taking the place of water in the coordination complex, anhydrous iron (III) chloride is a dark black or purple crystal, it's easier to see with a bigger anion that won't interfere with the hexaaquairon (III) complex. The red iron (IV) solution is formed on addition of hydrogen peroxide to the orange solution of [pentaaquachloridoiron (III)]2+ chloride solution turning some of those metal ion complexes to [tetraaquaoxoiron(IV)]+ chloride ions.


text:
Coordination geometry: why is of Cu(acac)2 square planar?
Why is $\ce{Cu(acac)2}$ square planar, whereas $\ce{Al(acac)3}$ and $\ce{Fe(acac)3}$ form octahedral complexes? I see how a similar argument to $\ce{Pt}$ complexes would apply, i.e. with $\mathrm{d^9}$, there is only one electron that goes in the very destabilised $2\mathrm{b_{1g}}$ (corresponding to $\mathrm{d}_{x^2-y^2}$) in the square-planar orbital splitting, whereas the rest is preferentially stabilised. However, there are many examples of Cu(II) forming octahedral compounds, in which acac would still bridge only $90^\circ$. Is this something to do with the acac ligands being (pseudo-)aromatic?
Thanks!

With bidentate ligands of the form $\ce{LL'}$, it is difficult to force triple coordination to a copper(II) centre. I don't have data for acac offhand, but I have some data for en binding to Cu. (Source: some past exam question somewhere.)
$$\begin{array}{c|ccc}
n & 1 & 2 & 3 \\ \hline
\log K_n & 10.55 & 9.05 & -1.0
\end{array}$$
The exact numbers don't matter; only the trend does. (Which is why I'm lazy to find a better source.) The binding of the third ligand, which would necessitate coordination at the axial position, is very weak. To get to a $\ce{[Cu(acac)3]}$ complex, you would have to either 1) adopt a perfect octahedral environment and sacrifice the electronic stabilisation from the Jahn-Teller distortion; or 2) stretch two of the acac ligands to a longer length than they would like to be.
As for why axial elongation is favoured over compression, it is perhaps due to electronic reasons: under elongation, the $\mathrm{d}_{z^2}$ orbital is doubly occupied and the $\mathrm{d}_{x^2-y^2}$ orbital is singly occupied. On the other hand, under compression, $\mathrm{d}_{z^2}$ orbital is singly occupied.
Now, there is a second-order mixing which allows the $\mathrm{4s}$ orbital to mix with the $\mathrm{d}_{z^2}$, since they possess the same symmetry ($A_\mathrm{1g}$) under $D_\mathrm{4h}$ symmetry; in the process the $\mathrm{d}_{z^2}$ orbital is stabilised, and the $\mathrm{4s}$ destabilised. The destabilisation of the $\mathrm{4s}$ orbital doesn't matter, since it's empty. However, the stabilisation of the $\mathrm{d}_{z^2}$ orbital is, of course, more profound when there are more electrons in the $\mathrm{d}_{z^2}$ orbital, which corresponds to elongation.
Here's a diagram from my notes.

I also wonder why Cu(acac)2 is often as a Cu(II)-based catalyst – is this something to do with the stability of the complex, hence the acetylacetonate anion is unlikely to interfere? Thanks again!

(+1) Shouldn’t we be assuming that what is commonly written as $\ce{[Cu(acac)2]}$ is actually $\ce{[Cu(acac)2(solv)2]}$, wherein $\ce{solv}$ is the solvent?

Yeah, I guess so, the solvent will definitely still hang around at the axial positions.


text:
Oxidation Half Reaction Chemistry
I was always confused as to how electrons where donated from the anode in a galvanic cell. But after doing research, I found that there are four fundamental ways as to where electrons in anodic transformations can come from:
1) Surface metal atoms
2) Solvent molecules
3) Electrolyte anions
4)Base ligands
My question is how do I know from these four options, which is providing the electron?

The only way you can know for sure is to analyze the reaction using oxidation numbers, and figuring out which elements lose/gain electrons. Here is a link explaining how to assign oxidation numbers in a reaction.
As an example, consider the reaction, what is the oxidation number of sulfur in this particular molecule, $S_2O_3$. According to these rules, which are in order of priority, the oxidation number of oxygen is $-2$, unless it is bonded with hydrogen or fluorine, which in this case, it is not. Given that we have three oxygen atoms in this molecule, the total oxidation of oxygen is $-6$, which must be offset by two sulfur atoms so the total oxidation number is $0$. Obviously, the oxidation number of a single sulfur must then be $+3$, so that the molecule is neutral.
In any reaction, you would essentially have to do this for every molecule on both the products and reactants side in order to determine which atoms are losing/gaining electrons.