text:
How are solubilities of hydrate salts determined when competing equilibria are unavoidable?
When you calculate the solubility product constant for a salt and one of its hydrates, you sometimes get drastically different numbers! I don't know how this can be because they both dissociate into the same ions. For example, sodium sulfate heptahydrate $\ce{Na2SO4.(H2O)7}$ is more soluble than anhydrous sodium sulfate $\ce{Na2SO4}$. If you work through the math, the discrepancy in solubility is not completely explained by the larger mass of the heptahydrate (see calculations below).
I can explain the differences in solubility by assuming that the water molecules within the crystal structure of sodium sulfate heptahydrate lower the enthalpy of crystallization, which makes solvation energetically more favorable than the anhydrous salt.
What confuses me is the logical series of events that would happen when you add the solid hydrate into solution. The hydrate will dissolve into $\ce{Na+}$ and $\ce{SO4^2-}$ions. The concentration of those ions will increase until the solution is saturated (with respect to the anhydrous salt). Anhydrous $\ce{Na2SO4}$ will begin to precipitate out, keeping the ion concentrations below $K_\mathrm{sp}(\ce{Na2SO4})$ The hydrate will dissolve completely, because the solution will never reach saturation (with respect to the heptahydrate). How, then, was the solubility of the hydrate ever experimentally determined?
Using $\ce{Na2SO4}$ as an example:
Solubility of anhydrous $\ce{Na2SO4}$ = $\frac{4.76\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)
Solubility of $\ce{Na2SO4.(H2O)7}$ = $\frac{19.5\ \mathrm g}{100\ \mathrm{mL}}$ (water, $0\ \mathrm{^\circ C}$)
Calculating saturated concentration:
Solubility of anhydrous $\ce{Na2SO4}$ = $$\frac{4.76\ \mathrm g\ \ce{Na2SO4}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4}}{142.04\ \mathrm g\ \ce{Na2SO4}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.335\ \mathrm M\ \ce{Na2SO4}$$
Solubility of $\ce{Na2SO4.(H2O)7}$ = $$\frac{19.5\ \mathrm g\ \ce{Na2SO4.(H2O)7}}{100\ \mathrm{mL}}\times\frac{1\ \mathrm{mol}\ \ce{Na2SO4.(H2O)7}}{268.1491\ \mathrm g\ \ce{Na2SO4.(H2O)7}}\times\frac{1000\ \mathrm{mL}}{1\ \mathrm L}=0.727\ \mathrm M\ \ce{Na2SO4.(H2O)7}$$
Calculating $K_\mathrm{sp}$:
Anhydrous $\ce{Na2SO4}$: $$\ce{Na2SO4 <=>2Na+ + SO4^2-}$$  $$K_\mathrm{sp}(\ce{Na2SO4})=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.335\times 2\right)^2\times\left(0.335\right)=0.150$$
$\ce{Na2SO4.(H2O)7}$: $$\ce{Na2SO4.(H2O)7 <=>2Na+ + SO4^2- +7H2O}$$  $$K_\mathrm{sp}\left(\ce{Na2SO4.(H2O)7}\right)=\left[\ce{Na+}\right]^2\left[\ce{SO4^2-}\right]=\left(0.666\times 2\right)^2\times\left(0.666\right)=1.18$$

Somehow you you didn't get that adding Na2SO4⋅(H2O)7 you simply add Na2SO4 and water - you could dehydrate it and add salt and water separately.

I can't tell what you're trying to say. How would adding anhydrous Na2SO4 and water get me the solubility of Na2SO4.(H2O)7?

You're whole reasoning doesn't have much sense, like these were really different things. For purpose of solubility you can treat Na2SO4⋅(H2O)7 like Na2SO4 with some water added - you're changing amount of water and don't account for it in last equation.

I think you don't have a good handle on what ionic hydrates are. They are solid salts with waters incorporated into the crystal structure. They can have wildly different properties, solubility is just one of them. https://en.m.wikipedia.org/wiki/Water_of_crystallization

You're overthinking it totally - error in last equation, you can't put there 0.727 as water amount is increasing - concentration of Na and SO42- is much lower.

Fixed in edit. The concentrations do not change much. Ksp is still much larger for the hydrate than the anhydrous salt.

I think that any precipitates that form from an aqueous saturated solution of hydratable salts will always be hydrates.

@CurtF. That makes sense on the face of it. But then what exactly is the meaning of the anhydrous salt Ksp? Is there such thing as a saturated solution for this equilibrium if it is impossible to form anhydrous salt crystals?

Let there be three species: a hydrated solid $H$, an anhydrous solid $A$, and a complete solvated (liquid) aqueous solution, $S$.
Your question is about differences in the two equilibria $\ce{H <=> S}$ and $\ce{A <=> S}$.  Can $H$ really have a different solubility than $A$?
I think the answer is yes, but only if the equilibrium $\ce{H<=>A}$ cannot and is not established.  In the presence of an "infinite" kinetic barrier that blocks conversion of the hydrate to the anhydrous form and vice versa, it makes sense to treat $\ce{H<=>S}$ and $\ce{H<=>A}$ and independent equilibria which might have different equilibrium constants.
In a comment to your question, I made it seem like "any" precipitation from a (super)saturated solution of $S$ would be the hydrate.  On further reflection, I think that is true, unless (i) the equilibrium $\ce{A<=>H}$ cannot be established due to kinetic barriers, and (ii) there is solid phase $A$ present in the solution.  In that case, the supersaturation could just result in "precipitation" of solid onto the existing crystal surface of already-present $A$, a process that would increase the mass of $A$ (solid) in the system but which would not require formation of any crystallites or nucleation intermediates.

How, then, was the solubility of the hydrate ever experimentally determined?

I think the answer is that the hydrate solubility can differ from the anhydrous solid solubility only in cases when there is no mechanism by which equilibrium can be established between the hydrate and the anhydrous solid.

Just to be sure - did you mix up any A H or S in there? I can't follow some of your statements (i.e. The solubility of S, which is an aqueous solution).

I also feel like your answer may not address the real concern of the question. Once the ions are in solution, they are participating in two competing equilibria. Common sense says that the lowest Ksp solid will always precipitate out before the higher Ksp is reached, right?

I don't think that is necessarily true.  It depends on the height of the activation barrier for precipitation.  The free energy barrier $\Delta G^{\dagger}$ for the route that leads to the more stable product can be higher than the route that leads to the less stable product.  In other words, precipitation can be theoretically controlled.


text:
ASTM D2270 Viscosity Index Formula/Table derivation
How was the formula (3) for viscosity index $\mathrm{VI}$ derived in section 5. Procedure A—Oils of Viscosity Index Up to and Including 100 [1]?
Specifically with reference to the reference oils considered and the table for it.
$$\mathrm{VI} = \frac{L - U}{L - H} \times 100$$
References

D02 Committee. ASTM D2270 - 04 Practice for Calculating Viscosity Index from Kinematic Viscosity at 40 and 100 °C; ASTM International. DOI: 10.1520/D2270-04. (PDF)

back when it was invented they measured all the oils they could find.  They got the worst (most viscosity change) and the best in each group and ranked these as 0 to 100.  All others were plotted on this "linear" scale.
The unit was used for dome time until the best was bettered, then they invented VIE (E is a subscript, and stands for extension, strictly all VI's above 100 are VIE's).
It is a horrid unit because it is far from having any useful meaning, unfortunately it has some significance.

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text:
How to classify the hydride shift in the following diazacyclopenta[cd]pentalene system?
How is this hydride shift reaction type called?

I think it's not one of following types:

cycloaddition (Diels Alder)
electrocyclic ring opening
sigmatropic rearrangement
ene reaction
cheleotropic reaction

Therefore I'm thinking what else can been the name of the observed reaction.
The ring opening reaction with hydride shift has been taken from Heterocycles 2004, 64 (1), 447–461 (DOI: 10.3987/COM-04-S(P)48).

The back reaction looks like an "ene" reaction to me.  The forward reaction would then be a "retro-ene" reaction.

I also wrote "ene" in the exam. The reason why I'm unsure now is the existing C(H)-N bond: I thought a ene-reaction happens with two unconnected species as in https://en.wikipedia.org/wiki/Ene_reaction

No, the pieces can be connected. See this pdf for some "connected" hetero-ene examples.

It took me a while to see what was actually happening. Initially, I was focused on a [1,5] suprafacial sigmatropic proton shift followed by a diaza-electrocyclic ring opening reaction. However, after drawing the corresponding orbitals I realised that I would have a eight-electron antiaromatic π system, in which a proton shift would need to occur antarafacially — not possible.[1]
It still is clear that it is some kind of pericyclic reaction, and going through the list we can exclude everything except for a (retro-) group transfer reaction for which the (retro-) ene reaction is the most well-known example. (We are transforming one σ bond into one π bond which excludes all the others.)
It took me a while to actually see the retro-ene system in the substrate, but finally I did, and I took the liberty of highlighting it in red in the scheme below. Note that you have to ignore the carbon-nitrogen bond which belongs to the lowest five-membered ring from the ene-system. It is not taking part, if anything it is scaffolding.

Scheme 1: Retro-ene reaction of the title compound with the pericyclic six-membered ring system highlighted. Dashed bonds represent those that are broken/formed within the reaction.
Counting electrons, I realised that my initial assumption of a proton shift was wrong and that it is indeed a hydride shift.

Notes:
[1] [1,n]-rearrangements are classified as suprafacial or antarafacial depending on whether the group is on the same side or the other side of a π system after migration. Technically, there is a further difference between retention and inversion, but hydrogen cannot rearrange under inversion so that is moot. The general rule for proton transfers is: If the number of bonds is $4n + 1$, the mechanism is suprafacial, for $4n + 3$ it is antarafacial.

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text:
Why is the formation of double helix between non complementary sequences disfavoured?
The textbooks starts with the phenomena that double helix is an expression of the rules of chemistry.

Hydrogen bonds are important in determining the formation of specific
   base pairs in the double helix. However in single staranded DNA, the hydrogen bond donors and acceptors are exposed to solution and can form hydrogen bonds with water molecules.
  When two single strands come together, these hydrogen bonds with water are broken and new hydrogen bonds between the bases are formed. Because the number of hydrogen bonds broken is the same as the number formed, these hydrogen bonds do not contribute substantially to driving the overall process of double helix-formation. However, they contribute greately to the specificity of binding.

I see the point that some neucleotides which form stable links are the complimentary bases (purine and pyramidine). Some bases do not form stable links. I did search my question online, I am not able to get near anything that's directly related or comprehensible at my level. So following is what I do not understand.

Suppose two bases that cannot form Watson-Crick base pairs are brought together. Hydrogen bonds with water must be broken as the bases come into contact. Because the bases are not complementary in structure, not all of these bonds can be simultaneously replaced by hydrogen bonds between the bases. Thus, the formation of a double helix between noncomplementary sequences is disfavoured.

What's the specificity mentioned here? What does it mean by non complementary in structure? An example is appreciated.

If you take a look at the base pairings, as shown below, you can see that adenine and thymine form two hydrogen bonds when they are matched, while guanine and cytosine form three hydrogen bonds when they are matched. If you try to match an adenine with cytosine or guanine with thymine, the orientation of hydrogens and lone pairs will not line up.
For example, guanine has two hydrogens and one lone pair available for hydrogen bonding, while cytosine has one hydrogen and two lone pairs available for hydrogen bonding. They fit! However, if you try to match adenine (with one hydrogen and one lone pair available for hydrogen bonding) with cytosine, the elements will not match or line up, resulting in an energetically less-favorable interaction.
Summary: A base on one strand of DNA requires another, specific base to complement the exposed hydrogens and lone pairs to form the most most energetically favorable interactions.

So (mis)matching a guanine to say thymine would result in one H bond on both G and T being unfulfilled.  So forming this mismatch would require breaking three H bonds between G and water, and at least two H bonds between T and water.  And the mismatched pair would only have four inter-base H bonds, tops.  So the energetic penalty for the mismatch is one H bond lost.  In contrast, when matching G to its cognate base C, there is no H bond penatly.

@CurtF. Excellent recap!

@CurtF. This is great: guanine has two hydrogens and one lone pair available for hydrogen bonding, while cytosine has one hydrogen and two lone pairs available for hydrogen bonding. They fit! However I am still confused about specificity.

@bonCodigo specificity is just a term meaning that each purine only has one complement pyrimadine.

@DanBurden This is not fully true. There can be some non-canonical interactions

Addition to Dan Burden's answer:
Guanine can form a base-pair with Thymine or Uracil (2 hydrogen bonds). This is not generally observed in the DNA but is quite common in RNA helices.

The non-Watson-Crick (WC) type base pairs or wobble-base pairs introduce distortion in the helix; they are also thermodynamically weaker than the WC pairs. However the GU pair, especially in RNA has a thermodynamic stability comparable to the WC pairs [1]. The distortion that it imposes on the helix is also less than that of other wobble pairs. GT mismatch can actually be tolerated to some extent even in DNA in all its structural forms (A,B and Z) [2]. GU pairs are however stabler in RNA and allow structural flexibility. GU pairs introduce a high electrostatic potential into the major groove which in turns increases its propensity for divalent metal ion binding [1]. It is my guess that metal-binding may stabilize the structure. Water molecules are also shown to cluster around the GU pair and possible stabilize the it [3]. Again it is a guess that this may be a resultant of the higher electrostatic potential in case of the GU wobble.

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text:
Why does the dielectric constant of a solid increase due to a Frenkel defect?
In crystalline solids, an atom may be displaced from its original position in order to create a vacancy in the original location. This is referred to as a Frenkel defect. Why does this defect result in an increase of the dielectric constant of the solid?

I've never truly understood what that even means to be honest. But, given, the only applications of the dielectric constants I've seen are in dissolving polar solvents by water or reducing the electric field inside capacitors ("dielectric substances") I guess this Frenkel defect thing is most probably related to the capacitor part. But, "why" I've no answer.


text:
Do Slater's rules apply to neutral lithium (at least approximately)?
For lithium, the electronic configuration is $\ce{(1s)^2(2s)^1}$. I tried to calculate $Z_{eff}$ for the $\ce{2s}$ orbital of the neutral atom as follows:
$$Z_{eff} = Z-\sigma$$
$$Z_{eff} = 3-(2\times0.3)$$
$$Z_{eff} = 2.4$$
But the Wikipedia article's table of effective nuclear charges (linked below) states that $Z_{eff}$ is 1.279 for the $\ce{2s}$ lithium orbital, which is not even close to my answer (above). So do Slater's rules not apply to lithium or I have done something wrong?
Wikipedia article: https://en.wikipedia.org/wiki/Effective_nuclear_charge

You need to use the factor of 0.85 for each of the $\ce{1s}$ electrons:
$$Z_{eff} = 3 - (2\times 0.85) = 1.3$$
as per rule #2 in the Wikipedia article:

If the group is of the [s p] type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

The $\ce{1s}$ electrons both qualify as having "a principal quantum number $n$ one less than that of the group" (here, the principal quantum number of the group of the $\ce{2s}$ electron is 2). Here is a nicely worked example for further edification.

So the screening of 1s orbital being 0.3 only applies to Helium?

The screening of a 1s electron by another 1s electron, as in the case of helium can also apply to other atoms that are missing electrons that result in the $\ce{(1s)^2}$ electronic configuration such as $\ce{Li+}$, where you calculate $Z_{eff} = 3 - (1\times 0.3) = 2.7$ (compare to 2.691 on Wikipedia's page on effective nuclear charge, linked in the main thread here).

0.35 contribution of each of the other electrons in the same group excepting one .
0.85 contribution for each of the electrons in the nest inner group or immediately to the left. hence:
1s2 2s1
we except one
therefore 0.85*2=1.7
3-1.7=1.3 zeff=0.13
kindly note that there are other rules.

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text:
Do electrons in covalent bonds always obey the octet rule?
Do electrons in covalent bonds always necessarily obey the octet rule? Always?

I +1'd because this duplicate will come handy when people with the same question search. Oh, and welcome to chemistry.SE!

@inɒzɘmɒЯ.A.M I apologise for my mistake. Thank you.

The elements in borons row do not always follow the octet rule as in the case of boron fluoride as boron is in the center of the periodic table and will usually only bond to fill up its outer shell with 6 electrons. Additionally, elements with can form expanded octets which can share more than 8 electrons such as phosphorous pentachloride.  Of course hydrogen only has 2 electrons to bond with so it is also an exception to the octet rule.

Empty d-orbitals are not involved in the bonding in$\ce{PCl5}$.  Rather the bonding is described as hypercoordinated.  See this earlier answer.


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How does the acidity of soft drinks not affect certain drugs, such as opiates? What about drugs and stomach acid?
Here's a very random introduction. I heard the term, "dirty Sprite" (it's some rap album/slang term) on a YouTube advertisement. I Googled "dirty Sprite", and it's apparently a recreational drug: codeine syrup in Sprite lemon-lime soda. People allegedly get high off of it, meaning the opiates in codeine is metabolized properly. My question is, how does harsh acidity of the Sprite (pH of 3.4-3.8 if you trust Google) not interfere with the codeine structure and mitigate its metabolism?
Then, this turned into a meta philosophical question: how does the acidity of the stomach, which is more acidic than Sprite, not interfere with not only codeine, but other substances?

pH 3.4-3.8 is hardly harsh acidity.

I wouldn't believe this is really off-topic, but it's borderline, at least. I reckon most of the knowledge required to answer this is chemical and pharmaceutical knowledge.

Many drugs are not very soluble in water what is a common property for a lot of organic substances, including Codeine. Therefor it is very common to turn them into their hydrochlorid which is simply $\text{Drug}\ce{.HCl}$. [4] E.g., Crystal Meth got it's name from the crystals that the hydrochloride forms.
$\hskip1.7in$
Source: Wikimedia Commons
Other examples are, e.g., Cocaine, Cetirizine, hydroxyzine, and many more.
As your substance is either the hydrochloride or would turn into the hydrochloride in stomach there should not be any destructive influence on the structure of the drug. In fact it even promotes the absorption.
On the other hand, the better solubility of the corresponding hydrochloride is a relatively big problem for those drugs that should not get absorbed in the stomach. They need to be coated to "survive" the stomach acidity to reach the intestine where they should get absorbed.

Thank you, it had not occurred to me think that stomach acidity is an integral part of digestion in itself.

Heroin base is notoriously dissolved in citric acid (with heat applied) in order to make it soluble for injection. Do you think that the citric acid in the dirty Sprite would have the same effect on the codeine, making a soluble citrate which absorbs quicker into the bloodstream than codeine tablets?

@jwg at least it sounds plausible

Thanks! It seems like various concoctions taken recreationally involving codeine (such as 'purple drank', which combines it with anti-histamines) are intended to accelerate its metabolism in order to achieve heroin-style euphoria.


text:
Molecular differences in the HB graphite hardness scale
Today, I bought some graphite for my pencil. It is rated HB on the HB graphite scale, a measurement of the hardness of graphite. My question is how can there be different hardnesses of the same substance? Is there a difference between 4B graphite and 6H graphite on the molecular level? Are the different hardnesses not all arranged in the large, hexagonal carbon sheets that I had thought all graphite existed?

Check out this Wikipedia article on grading and classification of pencil leads. Unless you have pure graphite, most likely the case is that the graphite is mixed with clay in various proportions and results in harder/softer "leads."

Related question on Arts & Crafts: Differences between HB pencils

Pencil leads are not comprised of pure graphite: they're made from graphite and a varying amount of clay, which in turn is what determines the hardness of the lead. Thus, you're not actually comparing graphite to graphite but rather different mixtures of graphite and clay.
According to one article published by Jetpens.com (a vendor of writing implements) entitled Picking the Perfect Pencil Hardness Grade:

Graphite is the key component of modern day pencil lead—in fact, lead is a misnomer as it's actually made up of a mixture of graphite and clay. The formulation of this mixture determines its lead grade. The higher the proportion of graphite content relative to clay the lead has, the softer and darker the lead will be, and vice versa.

And Wikipedia tells us:

The various graphite pencil grades are achieved by altering the proportion of graphite to clay: the more clay the harder the pencil.

The difference in hardness is attributable to the amount of clay mixed in with graphite in your different pencil leads. There is no difference in the graphite at the molecular level.


text:
Few basic questions about ionic bonding and batteries (lead)
I've got a few very very basic questions about ionic bonds and their role in batteries.
My assumptions (correct me if I'm wrong):

Ionic bonds occur between a metal and a non-metal (the non-metal takes
the metal's electron to fill its valence shell). Therefore the metal atoms are left with a positive charge and the non-metal a negative charge.
Because of this, the atom tries to stabilise its ionised charge and tries to bond with the oppositely charged atom.
Batteries take advantage of this chemical bonding to produce electricity.
In a lead battery there is a PbO2, the electrolyte and Pb.
The electrolyte is a non-metal and therefore it tries to fill it's valence shell with the electrons from PbO2. This creates an overall positive charge in the PbO2 and a negative charge in the solution.
For some reason this negative charge is then transferred over to the Pb which gives it an overall negative charge.
The Pb then wants to release this negative charge and pushes excess electrons which flow through a circuit to the PbO2 that is lacking electrons.

This leaves me with a few questions:
Why does the electrolyte solution choose to take electrons from PbO2 (I assume it has something to do with the O2)?
After the electrons are taken and there is a positive charge in the PbO2, and a negative charge in the solution, why don't they just bind together?
I thought ionic bonds occurred when an metal had it's electrons taken from it. How then does the Pb receive additional electrons from the non-metal solution?
If the Pb does want to receive the extra electrons, why would it instantly try to push them away through the circuit? I'd understand if the whole negatively charged atom was trying to be pulled to the positively charged PbO2 - but from what I've read it tries to push the excess electrons, not the atom - so why even receive the electron if it doesn't want it?

check https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery and https://en.wikipedia.org/wiki/Redox


text:
What happens when two nitrates react with each other?
I have to determine the products of the ammonium nitrate and copper nitrate. Is the below reaction correct?
$$\ce{NH4NO3 + Cu(NO3)2 -> Cu(NH3)4(NO3)2}$$
(ammonium nitrate + copper nitrate → tetrammine copper(ii) nitrate)
Also, what would be the products of the other two reactions below:

copper nitrate + lead nitrate
lead nitrate + ammonium nitrate

Basically, what happens when you react any two nitrates?

Your equation doesn't balance so it can't be right. Also what happens to the hydrogen ions released from the ammonium ions when they become ammonia ligands.


text:
Why does concentrated HCl fume?
It is a common observation that a concentrated solution of hydrochloric acid kept in the lab fumes. What is the chemistry behind this phenomenon? Does $\ce{HCl}$ somehow react with the moisture in the air due to its hygroscopic nature?

Answer-looking strategies: Hmm, HCl fumes...that means atoms from the surface of HCl are escaping, why? the atoms from any solid doesn't seem to leave out atoms of it so quickly, but HCl is a liquid, but water is also liquid, it doesn't fume that much, then is it that atoms in HCl are held so loosely than any other general liquid? What might be the cause...? See if you get any idea...

In the case of $\ce{HNO3}$, according the abstract of this paper$^\dagger$, nitric acid vapors act as nucleation sites for ambient water, causing the mists/fumes that are observed above concentrated nitric acid solutions:

The gas phase reaction [forming nitric acid from other reagents] supplies $\ce{HNO3}$ nuclei on which the water vapor readily condenses to form strong nitric acid mist droplets.

According to this book$^\ddagger$:

Hydrochloric acid has thermodynamic properties rather similar to those of nitric acid.

So, I would hypothesize that the physical chemistry phenomena of evaporation and hygroscopic nucleation are responsible for the fuming of $\ce{HCl}$.
$^\dagger$ Goyer, J Colloid Sci 18(7), 616 (1963)
$^\ddagger$ Kulmala and Wagner, "Nucleation and Atmospheric Aerosols," 1996, p591.

Hydrogen chloride ($\ce{H-Cl}$) is a gas at room temperature (boils already at −85.05 °C). The fumes you see is HCl escaping from the reaction mixture when the equilibrium is distorted due to e.g. heat. The more concentrated your HCl is, the more the equilibrium between the aqueous phase and the ambivalent air shifts towards the ambivalent air, creating fumes. These fumes are nasty. Rather not work too much with concentrated HCl (> 30%) outside of the hood!

Also, when you get higher and higher concentrations of strong acids, you tend to get larger proportions of non-dissociated acid molecules. This is due to fewer water molecules available to form solvent shells around the constituent ions.


text:
Le Chatelier's Principle and Rate constants
Consider the reaction: $$\ce{H2O2(g) <=> 2OH^.(g)}\label{a}\tag{1}$$ to be at at equilibrium ($\ce{OH^.}$ is a radical). As the pressure is increased, the reaction moves towards forming $\ce{H2O2}$ as this direction reduces the amount of substance in the system.
However, it is well-established that the forward rate constant of $\ref{a}$ increases with increase in pressure, meaning forming more $\ce{OH}$. This seems to be in conflict to what the Le Chatelier's principle suggests. Could someone help me get my head around this?

It is true that the forward reaction rate increases as you increase pressure. However, the reverse reaction rate will increase as well, and it will increase more than the forward reaction rate.
The key here is that the system is in equilibrium; the forward and backward reaction rates are equal. Altering the pressure affects both reactions.

This makes sense. So, with the increase in pressure, the reverse reaction rate increases and is dominant compared to the forward reaction rate (although the forward rate constant increases with increase in pressure) This definitely confirms with the Le Chatelier's principle then, resulting in the formation of more $\rm H_2O_2$.  Thanks!

With increasing pressure volume of system increases and thus active mass increaeas  then to keep the constant value of Keq whole setup is disturbed in direction where no of moles of reacting substance decreases. Thus increment in rate of one reaction is preferred over other.


text:
Ideal Gas Equation Contradiction
You have a regular balloon, and you decide to heat it up. The gas particles inside the balloon will gain kinetic energy, and so, the gaseous particles will collide with the internal wall of the balloon with a higher frequency and more energy. This is what we call pressure, which will, as a result, increase. This is more commonly generalised to be Gay-Lussac's law, which formally states that the pressure of a gas is directly proportional to its absolute temperature.
Now, as pressure has increased, meaning that the gaseous particles are colliding with the internal wall of the balloon with a higher frequency and more energy, the balloon will begin to inflate, or in other words, the volume of the balloon will increase.
This seems to contradict Boyle's law, which states that the pressure of a gas is inversely proportional to its volume. In the scenario I described above, it seems clear to me that it was the increase in frequency of the gaseous particles colliding with the internal wall of the balloon, that too with more energy, which altogether is known as an increase in pressure, that caused the balloon to inflate.
I perfectly understand Boyle's law when the idea of a pump is used, where you can use a slider to change the volume of a gas and clearly see its inversely proportional relationship with the pressure of the gas, however, can't seem to understand what is incorrect in my logic, from which I seemingly have contradicted Boyle's law.
I am a high-school student who just learnt about these laws, and out of curiosity decided to ask this question in hopes of clearing my doubts and fundamentally gaining a better understanding. Any help will be greatly appreciated, thanks in advance.

Your arguments do not contradict each other. You are trying to separate a process into two processes, which are coupled of course, but in first approximation this is a good way at looking it.
Let's state our model system again: A balloon with the pressure $p_1$, the volume $V_1$ is heated from temperature $T_1$. As a result we see an inflated balloon. Let's call those variables with $2$ as an index. Of course we treat everything as ideal as possible.
Now we look at the coupled process as two separate processes:

Isochoric heating, i.e. $V=\mathrm{const}$:
\begin{align}
&&p_1V&=n\mathcal{R}T_1\\
&&p_2V&=n\mathcal{R}T_2\\
\therefore
&&\frac{p_1}{p_2}&=\frac{T_1}{T_2}\\
\therefore
&&p&\propto T\\
\end{align}
That is Gay-Lussac's law.
Isothermal expansion, ie. $T=\mathrm{const}$
\begin{align}
&&p_1V_1&=n\mathcal{R}T\\
&&p_2V_2&=n\mathcal{R}T\\
\therefore
&&\frac{p_1}{p_2}&=\frac{V_1}{V_2}\\
\therefore
&&p&\propto \frac{1}{V}\\
\end{align}
That is Boyle's law.

You can also look at the whole process as the pressure stays the same while the balloon is heated. That then is an isobaric process, i.e. $p=\mathrm{const}$:
\begin{align}
&&pV_1&=n\mathcal{R}T_1\\
&&pV_2&=n\mathcal{R}T_2\\
\therefore
&&\frac{V_1}{V_2}&=\frac{T_1}{T_2}\\
\therefore
&&V&\propto T\\
\end{align}
That is known as Charles' law
Of course in reality there will be a mixture happening as the rubber of the balloon will expand, but the pressure that is possible is dependent on the material of it. In combination you will arrive at:
\begin{align}
&&p_1V_1&=n\mathcal{R}T_1\\
&&p_2V_2&=n\mathcal{R}T_2\\
\therefore
&&\frac{p_1V_1}{T_1}&=\frac{p_2V_2}{T_2}\\
\therefore
&&\frac{pV}{T}&\propto \mathrm{const}\\
\end{align}

There's no contradiction because Gay-Lussac's law only applies if the volume is held constant and Boyle's law only applies if the temperature is held constant. In the balloon example, both the volume and temperature change so neither law applies.


text:
Confused about writing out electron configurations
I watched the Crash Course chemistry video about the electron. In the video Hank showed an example configuration of iron as $\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (4s)^2 (3d)^6}$ and explained that due to energy constraints, electrons go to the s orbital of the 4th shell before going into the d orbital of the 3rd shell and so they are written in this order.
I looked up iron in Wikipedia and there the configuration is given as $\mathrm{[Ar](3d)^6 (4s)^2}$. I understand how it's shortened with argon, but why in Wikipedia is the d orbital of the 3rd shell written before the s orbital of the 4th shell?
Which one is preferred?

For a given configuration, the order of writing the orbitals is not
  completely fixed since only the orbital occupancies have physical
  significance. For example, the electron configuration of the titanium
  ground state can be written as either $\mathrm{[Ar] \, 4s^2 3d^2}$ or $\mathrm{[Ar] \, 3d^2 4s^2}$.
  The first notation follows the order based on the Madelung rule for
  the configurations of neutral atoms; $\mathrm{4s}$ is filled before $\mathrm{3d}$ in the
  sequence $\ce{Ar, K, Ca, Sc, Ti}$. The second notation groups all orbitals
  with the same value of $n$ together, corresponding to the
  "spectroscopic" order of orbital energies that is the reverse of the
  order in which electrons are removed from a given atom to form
  positive ions; $\mathrm{3d}$ is filled before $\mathrm{4s}$ in the sequence $\ce{Ti^4+, Ti^3+, Ti^2+, Ti+, Ti}$.

Source: Wikipedia article on electron configuration.


text:
Numbering heteroatoms and groups in benzene derivatives
Can someone please explain why the carbon in 8-hydroxyquinoline is numbered 8, while in 1,10-phenanthroline the nitrogens are numbered 1 and 10? I'm counting carbon atoms and just cannot see why the hydroxy is in position 8 and nitrogen atoms are in positions 1 and 10.

Quaternary carbon atoms i.e. Those with no hydrogen atoms attached and in these examples, the ones that are common to both or all rings, are not counted. Try it with 1,8-dihydroxynaphthalene. Both OH groups should be pointing in the same direction.
In your examples, as you probably know, the heteroatom is numbered first.

Just to check if I got it: if the rings were aliphatic, all carbon atoms would have been counted. Correct?

Not sure if I should set up a new question for this, but why is 1,8-dihydroxynaphthalene numbered 1 and 8 and not 1 and 2?


text:
Quenching rates in nucleation MD simulations
I am trying to perform nucleation simulations with molecular dynamics. To prepare the metastable fluid, I am equilibrating at temperatures about 30 K above the melting temperature, and then "quenching" down to a sub cooled liquid state. For those familiar with lammps, the method I am using to perform the quench is
fix 1 all nve
fix 2 all viscous 0.01

This just adds a drag force proportional to the velocity to suck heat out of the system.
In the nucleation simulation literature, I can't think of a time I have come across a paper where the author mentions an actual quench rate, or discusses whether the quench rate can have implications for the final results. Is anyone familiar with what an appropriate quench rate to prepare a metastable liquid, or whether I can produce spurious physical artifacts depending not quench rate? Currently for my water simulation, I am quenching from 300 K to 190 K in 50 steps with a 2 fs time step.

I think there is a lot of luck and voodoo involved in choosing parameters like this for non-equilibrium MD. I personally believe that your rate is WAY too fast, but so long as the molecules are still diffusing and not just a glass, the rate to get there shouldn't matter.


text:
Is it possible for monobasic and dibasic potassium phosphate to complete break down into phosphates?
Here is the picture of monobasic potassium phosphate:

And here is the picture of dibasic potassium phosphate:

Is it possible for the phosphorous-containing anions to further break down into actual phosphates and floating protons?

I don't know what you are asking.  What do you mean by "actual phosphates"?

At basic conditions over pH 10, those phosphorus-containing anions can get further deprotonated, as can easily be seen on the corresponding Wikipedia article on phosphates.
Phosphoric acid $\rightarrow$ Dihydrogen phosphate $\rightarrow$ Hydrogen phosphate $\rightarrow$ Phosphate


text:
Guess the shape and phase of the MOs of butadiene using LCAO and Hückel's method?
I used Hückel's method along with a Linear Combination of Atomic Orbitals (LCAO) to calculate an estimate for the orbital energies of cyclobutadiene ($\ce{C4H4}$) and butadiene ($\ce{C4H6}$). For butadiene, I ended up with the result
$E = \alpha \pm 1.62  \beta$ and $E = \alpha \pm 0.62 \beta$,
where $\alpha$ and $\beta$ are both negative (and are the Coulomb integral and the exchange integral, respectively). I'm fairly confident with the method and interpretation of these integrals so far, but I don't know how to use these results to "guess and draw" the shapes and phases of the molecular orbitals, and I'm also not sure how to tie in what $\alpha$ and $\beta$ represent.
As I understand things, in these molecules, hybridisation of the orbitals occur and $\mathrm{sp^2}$ orbitals form, taking up three electrons from each carbon atom and leaving one electron left over in the third p orbital. This p orbital is the one I assume is portrayed in the below images.
Could someone please give me a hint towards the following:

relating the molecular orbital shapes/phases with the Coulomb integral (overlap of one carbon's electron wavefunction with a neighbouring carbon's potential) and the exchange integral (the overlap of two electrons' wavefunctions in the field of one of the carbons
how to get any sort of information about the phases of the different orbitals. I understand that $E = \alpha + 2\beta$ will have the lowest energy, as both $\alpha$ and $\beta$ are negative, but how does that correspond to the phase configurations shown below? Why do the size of the circles in the first image vary?

Note: I've been sent here from the Physics Stack Exchange, please be nice. :) Thanks for any help! This is my first post so please also let me know if the posting etiquette passes muster.

Sources: First image, second image

The shape of the molecular orbitals is not guessed, it is determined by the LCAO-MO expansion coefficients which one needs to find out in the first place. The hint is that an eigenvalue equation is usually solved in two steps:

We start by writing down the eigenvalue equation in the matrix form and proceed then by writing down and solving the characteristic equation in which the secular determinant is equated with zero. By solving the characteristic equation we find the eigenvalues (energies).
Then we put eigenvalues (energies) back into the matrix eigenvalue equation and solve it for the coefficients (LCAO-MO expansion coefficients) that determine eigenvectors (molecular orbitals).

As far as I understand, the second step was not simply done yet. Note that any scalar multiple of an eigenvector is also an eigenvector with the
same eigenvalue, so that eigenvectors are not unique. In solving the matrix eigenvalue equation one will find not the actual values of the LCAO-MO expansion coefficients, but rather some relationship between them which could be satisfied by a number of different coefficients. In principle one is free to choose any set of coefficients satisfying the relationship, and usually there is some obvious choice.
For the simplest case in which just two coefficient are related with each other through some ratio the obvious choice is to take some coefficient to be one, then the value of the second coefficient is nothing but the value of the ratio. One could also normalize the wave function then if desired.
Also note that one can use symmetry to simplify the math. OP did not mention were the symmetry adapted linear combinations (SALCs) of atomic orbitals used at the first step, so I mention just in case.

P.S. Fully worked examples could be found in different places, for instance, here (PDF, 150 KB).


text:
Do ion-exchange resins soften water through a kinetic equilibrium process?
I searched a lot for this and couldn't find any answers.  Is water softening a purely ratio-based equilibrium process, ie, more $\ce{Na+}$ cations on resin than there are $\ce{Ca^2+}$ cations in the water, causing most $\ce{Ca^2+}$ in solution to be displaced? Or do kinetic factors come into play, where at equilibrium $\ce{Ca^2+}$ cations will end up stuck on the resin over $\ce{Na+}$ cations because they are bigger and thus more restricted from leaving the resin due to steric reasons?

Related: Soil-solution cations in competition for exchange sites

I'm not sure about your use of the terms "kinetic" vs. "ratio", but I don't think calcium ions are particularly hindered from leaving the resin of most ion exchange resins.

(1) Water softeners use an ion exchange material that is reversible, so that after it has been used a while and the ion exchanger becomes loaded with calcium ions, it can be regenerated to the sodium form by flushing with salt (sodium chloride) solution. This would suggest that the affinity of sodium and calcium ions for the resin do not differ that greatly. I would doubt that either kinetic or steric factors count much here either, although a divalent ion like calcium might bind more tightly by being held electrostatically by more than one anionic group on the resin.
(2) I think the primary reason water softeners work is that they typically use the ion exchange resin in the form of a vertical column: you put the hard water in the top and soft water exits at the bottom. When hard water is introduced at the top of the column, a local equilibrium is established between ions in solution and ions bound to the resin, with a somewhat lower concentration of calcium ions and a corresponding higher concentration (at a 2:1 ratio) of sodium ions. As the water moves down the column, this equilibrium is reestablished again, but this time, the concentration of calcium is lower still. This repeats a large number of times, until all the calcium ions are replaced with sodium ions. I have heard this effect referred to as "mass action".


text:
Electrical properties of 2D materials
This is regarding to the newly-discovered 2-D materials that can support electron flow, such as phosphorene and transition metal dichalcogenides. How do they conduct electricity?
I also read that their electric properties change with the number of layers. For example, the band gap of black phosphorus changes from 0 - 2 eV with the number of layers. Why is this so?


text:
Reaction between a vinegar/salt/hydrogen peroxide mixture as an etching solution and copper
I am looking for the less harmful chemical solution to serve as an etching solution. After many searches, my interest went to a vinegar/salt/hydrogen peroxide solution.
I have to ask then if the residue of this solution is harmful and if it is ok to pour it in the regular water waste system.

The etching solution itself is fine to dispose of down the drain. However, if you use it to etch copper, the dissolved copper ions are quite toxic, and should not be disposed of through the sewer system.
You can bypass the problem by collecting the etch solution, and submerge aluminum foil. $$\ce{3Cu(CH3COO)2(aq) + 2Al(s)->2Al(CH3COO)3(aq) +3Cu(s)}$$React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.

In your reaction, above, it should be $\ce{Al(CH3COO){3}}$ and not $\ce{Al(CH3COO){2}}$. Then it needs to be properly balanced.

Hi @DanBurden, are you able to add to your answer the etching equation that shows vinegar, salt, hydrogen peroxide, and copper becoming Cu(CH3COO)2?


text:
Why is it possible to use lipids to recover diamonds in mining?
Diamond mining
This link says that diamonds will stick to "anything oleaginous" and that other materials slide down the fat that is applied to an inclined plane used in the mining apparatus.  I infer that the surface of the diamond interacts via dispersion forces and that other minerals are more polar.  Is this the case?  If so, would diamonds stick to wax or plastic?

According to the Wikipedia piece on diamonds:

Diamonds are naturally lipophilic and hydrophobic, which means the diamonds' surface cannot be wet by water but can be easily wet and stuck by oil. This property can be utilized to extract diamonds using oil when making synthetic diamonds.

It looks like the surface of diamond has dangling bonds terminated with hydrogen, so your non-polar hypothesis is correct, as applied to terminal moieties on the surface. See this article about the material properties of diamond where it is stated:

Diamond surfaces are hydrophobic when the surface carbon atoms terminate with a hydrogen atom and hydrophilic when the surface atoms terminate with an oxygen atom or hydroxyl radical.

and

Naturally occurring diamonds have a surface with less than a half monolayer coverage of oxygen, the balance being hydrogen and the behavior is moderately hydrophobic. This allows for separation from other minerals at the mine using the so-called "grease-belt."

You ask:

If so, would diamonds stick to wax or plastic?

I believe the answer is yes, but certainly factors such as the transport of diamond with grit in a slurry over/through a hydrophobic material such as the two you name might come into play: I think mixing the mined slurry with goop is more effective than sheeting it across a 2-D material, and a higher extract yield results.
Note that at higher temperatures and with other reactants, the surface properties can change (as stated in the linked article immediately above):

Treatment with gases or plasmas containing the appropriate gas, at temperatures of 450 °C or higher, can change the surface property completely.


text:
Relationship between vapour pressure and saturated vapour pressure
I need clarity on saturated vapour pressure in a closed system at equilibrium. How does saturated vapour pressure relate to vapour pressure?

What exactly are you confused about? Are you just asking about the meaning of 'saturated'?

The saturated vapor pressure will be the vapor pressure of a closed system containing liquid and vapor in equilibrium. It will change with the temperature of the system.

Perhaps http://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html might help?

If you're at equilibrium, you're saturated.   If you're saturated, you're at equilibrium.  The end.
If things aren't at equilibrium, the situation is different.  Consider a dish of water in the middle of the Sahara desert.  The amount of water in the atomsphere in the Sahara desert is generally nonzero, say in the range of 1 $\mathrm g~{kg}^{-1}$.  Even though the water vapor pressure (even in the Sahara!) is nonzero, it is way less than "saturated".  Therefore it isn't at equilibrium, and the pan of water quickly evaporates.

How does saturated vapour pressure relate to vapour pressure?

There are four terms you need to understand.

pressure (i.e. the total pressure)
vapor pressure
partial vapor pressure
saturated vapor pressure (identical with equilibrium vapor pressure.)

Let's assume that water has a equilibrium vapor pressure of 18 torr at STP (760 toor, 20 degrees C).
Case 1
I have a flask sealed with some air and some water at my STP. The total pressure (i.e. the pressure) is 760 torr. Let's say that there is 750 torr of dry air. That would mean that the partial pressure of water is 10 torr.
Case 2
So I have a flask with water and some air is sealed above. The air/water are at STP and are at equilibrium. So the total pressure of the gas phase (water + dry air) is 760 torr. 742 torr is dry air and 18 torr is the partial vapor pressure of water. The air/water mixture has 100% humidity which means that it is saturated with water, hence the term saturated vapor pressure. (In other words if the temperature falls then dew would form, meaning that the air loses water.)
Case 3
I put a sample of water in a flask and try to pull a "vacuum." So  long as the water stays at the standard temperature then the pressure will stay at 18 torr. The water keeps "evaporating" into the gas phase. So 18 torr is the equilibrium vapor pressure as well.
Case 4
Now water has a density of 1g/cm^3, and 18 g/mole. So 18 g of water at the boiling point (100 degrees C) occupies 22.4 liters. If the 18 grams of water was at 100 degrees C, then I need a volume of 946 liters to get all of the water to evaporate. So if I had a 1000 liter tank, then 18 grams of water wouldn't be enough to fill it to saturation at 100 degrees C. In other words all the 18 grams would be in the gas phase with no liquid. (If I dropped the temperature to 99 degrees C, water still wouldn't condense out.) So in this situation the water would have a vapor pressure  less than the saturated vapor pressure.

Why don't you use SI units?

@MaxW Case 1 is not correct. Since it is at STP, the water would have a vapor pressure of 18 Torr. Since you said there is 750 Torr of dry air, the total pressure inside the flask would be 768 Torr.

@Nova - No,  the point I was trying to make was that there wasn't enough water to get to equilibrium pressure.

@MaxW Are you saying that there was so little water that it all evaporated and only produced 10 Torr of pressure?

@Nova - Yes, that was what I meant.


text:
What is the degradation mechanism of pyrrole?
A vial of pyrrole (that was occasionally exposed to air) changed its color while degrading from pale yellow to dark orange.
What is the mechanism of pyrrole degradation? What are the resulting products?

Off the top of my head, I'd guess that you'd get some oxidation, some dimerization or polymerization, and some portion of Lewis-acid-base complex on the nitrogens. Most papers I've read about pyrrole polymerization will protect the N-H before doing chemistry.

Hopefully, not pyrolysis. ;-)

Is there acid present?

@Jori, no, not really

@Sparkler It is probably polymerization of the oxidized pyrrole then. I figure that the mechanism for the polymerization is the same as for the (Lewis) acid initialized mechanism.


text:
Trying to understand SMILES(?) patterns
In the example calculation from the conformer search package fafoom is a mysterious pattern that I'd like to understand.
The SMILES code of the given example structure is:
CC(=O)N[C@H](C(=O)NC)C

There is a first not so hard pattern for the smart_cistrans pattern identifier
C~[$(C=O)]-[$(NC)]~[C]

which tells the program that peptide bonds behave like cis-trans isomers. That is relatively easy to see but I don't really understand it to adapt it on further problems.
But there is another identifier for torsion angles that for me is even harder to understand:
[*]~[!$(*#*)&!D1]-&!@[!$(*#*)&!D1]~[*]

The [*]'s at the beginning and at the end look like defining any atom ... but the cursing signs in the middle don't tell me anything.
Dear reader, if you understand how those patterns are generated, I would ask you break them down piece for piece that I can use them afterwards on my own.

Interesting to see the link to fafoom.. I'll have to investigate that further to compare to other tools I have.

I found it yesterday, there is a related arXiv-paper.

@GeoffHutchison, may I ask what other tools you have? :)

First off this is SMARTS not SMILES.
Think of SMARTS as like a regular expression language for molecules (e.g., SMILES).
So let's break down some of those characters:

~ is "any bond"
! is "not"
$ indicates a recursive SMARTS expression.
@ indicates "any ring bond"

The $(*#*) means "two atoms with a triple bond between them". The !D1 means and not an atom with one explicit bond.
Taken as a whole, [*]~[!$(*#*)&!D1] means any atom with a bond to an atom that doesn't have just one explicit bond or a triple bond.
Basically, the pattern is trying to find a single bond in the middle with some rotatable side-groups:
[*]~[...]-&[..]~[*]
There's a very nice tool called SMARTS Viewer that can help to debug and construct SMARTS.

Seems like there are also now more polished commercial versions called SMARTStools through BioSolveIT including an editor.

Oh, I should point out that SMARTS View has a paper: K. Schomburg, H.-C. Ehrlich, K. Stierand, M.Rarey; "From Structure Diagrams to Visual Chemical Patterns," J. Chem. Inf. Model., 2010, 50 (9), pp 1529-1535

Thank you very much. I thought that the "smart_" was just a part of the string to sign that there is something smart behind it and would have never come to think of a molecular pattern language. That helped me a lot :)


text:
Why do ionic compounds have higher melting and boiling points despite having smaller radii?
Why do ionic compounds have higher melting and boiling points despite having smaller radii?
I thought that the closer the two ions are, the larger the repulsive force, and therefore it should rather require less energy to separate them (and at further distances apart, a lower melting point is observed because there is less attraction).
This is not a question of ionic compounds vs. covalent compounds, which is covered in this question. To be clear, I am interested in the comparison of ionic compounds with ionic compounds.

Nope, this has to do with difference it radii, not ionic vs covalent

"The closer two ions are, the larger the repulsive force," but only if they have the same charge, right? If they have opposite charges, then it becomes "The closer two ions are, the larger the attractive force."

As pointed out in a comment, there is a problem with your second sentence:

"I thought that the closer the two ions are, the larger the repulsive force, and therefore it should rather require less energy to separate them..."

This statement is technically correct, in that the repulsive forces between the two atoms do increase with decreasing bond length. And it is true that two atoms having a charge of the same polarity would exhibit increasing repulsion as the distance between them decreases.
However, in the case of the oppositely charged atoms found in ionic bonds, the attractive electrostatic forces also increase with decreasing bond length, which is the reason for the strong bonding and thus high melting points of ionic compounds.
A discussion of the bond strength of ionic compounds is given in this Wikipedia article, which summarizes that the ionic bond strength can be determined as the sum of the attractive electrostatic energy and short-range repulsive energy.


text:
How does oil on the surface of water prevent rust?
I distinctly remember a side-by-side comparison from a book where there are two nails submerged in water, in two beakers: one nail had a layer of oil on top of the water, and that nail didn't rust; the other did.
It seems that oiling iron/steel products is supposed to prevent them from rusting:
http://www.justanswer.com/home-improvement/0n1r2-does-olive-oil-prevent-rust.html
https://outdoors.stackexchange.com/questions/1404/can-oiling-keep-tools-from-rusting
The explanation given, I think, is that the oil somehow prevents oxygen from reacting with the metal.
Can't help but wonder why. Oxygen is a non-polar molecule, so it should actually dissolve better in oils (which are also non-polar) than in water (which is highly polar) and therefore allow more oxygen molecules to contact the iron (metal) surface in some sort of equilibrium.
Let's say that the conductivity of water and its ability to form electrochemical chains between metals is irrelevant in this thought experiment (i.e., a nail submerged in water in a glass). From what I understand, the other key item that has an effect on the formation of rust in water is that the oxygen molecules have better contact with the iron surface.
This is supported by data: the concentration of oxygen in air is far higher than that in water. We know from experience, though, that wet iron rusts way quicker than dry iron does.
If I am reading it correctly, the same data shows higher oxygen concentrations in non-polar solvents than in water.
Here is the data: http://www.nist.gov/data/PDFfiles/jpcrd219.pdf
Please correct me if I am reading it wrong.
My question is:
How does oiling prevent rust?
Edit: For the sake of clarity, let's say we use deionized water in a vacuum, drop in the nail, then put some oil (let's say it's caprylic acid) on it, and expose it to standard conditions.

I suspect that the statement about "oxygen being non-polar" is irrelevant. Gas solubility in liquids follows different rules. E.g. salts dissolve better in warm water, and you can freeze them out, but gases dissolve better in cold water and you can boil them out.

Did I get it right: You are comparing two metals that are BOTH in water. Just that one water has oil on it, while the other has not? So the question would be: How does a layer of oil on water prevent rust under the water?

@AngeloFuchs I'm wondering if the nail was dropped in after the oil was put on top of the water, and was coated as it entered the beakert.

@ChrisH Good question. That would explain it, but without further input from OP, we won't know.

@AngeloFuchs the OP should spot my comment as well, but otherwise I'm thinking the water was boiled to drive off the oxygen and the oil stopped oxygen diffusing into one beaker.

You probably should make it clearer that the oil layer is above the water (and not surrounding the nail), since many answers seem to overlook that detail "one nail had a layer of oil on top"

@AngeloFuchs Yes, you got it right the first time, sorry for being unclear. For the sake of clarity, let's say we use deionized water in a vacuum, drop in the nail, then put some oil (again, let's say it's caprylic acid) on it, and expose it to standard conditions.

Well, I went and searched for the solubility of oxygen in water and oil, and found this summary paper on the NIST web site: "The Solubility of Oxygen and Ozone in Liquids" by Battino, Rettich and Tominaga, J. Phys. Chem.Ref. Data., vol 12, no. 2, 1983.
Conveniently, the paper gives solubility data for oxygen in both water and olive oil.  The solubility is given both as molar fractions, and also as the Ostwald coefficient $L$, defined as the volume of gas absorbed per volume of absorbing liquid.
Given the very different molar volumes of water and olive oil, I believe the latter provides a much more useful comparison here.  Comparing tables 1 (water; note the $10^2$ factor in the header) and 25 (olive oil) in the paper, we see that the Ostwald solubility coefficient for oxygen in water is about 2,000 times higher than in olive oil:
 Temp. (K) | Temp. (°C) | L (water) | L (olive oil)
-----------+------------+-----------+---------------
    298.15 |         25 |     310.4 |        0.1269
    308.15 |         35 |     276.6 |        0.1326
    318.15 |         45 |     254.5 |        0.1383
    328.15 |         55 |     240.9 |        0.1441

So, yes, oxygen is a lot more soluble in water than in oil.  But why?
Well, if this web site is to be believed, it's because of the strong hydrogen bonding between water molecules, which causes even liquid water to maintain some degree of an ice-like lattice structure, with gaps that small gas molecules like $\ce{O2}$ can easily slip into.  Once there, the gas becomes solvated, with the strongly polar water molecules inducing a temporary dipole in the $\ce{O2}$ molecule and thus creating an attractive Debye interaction.  The overall solvation process is exothermic (i.e. releases heat), explaining why the solubility of oxygen in water decreases with temperature.

Great answer, actually the only at the moment covering the point of OP. I would add another point, which is mass transfer through the interface, where in the oil scenario there are two of them, instead of one. In addition, the concentration of oxygen in oil is very low, hence the driving force at the oil-water interface is also much lower.

@ssavec The mass transfer is definitely a valid cause, didn't think of that. Maybe I should have use a triple example: water beaker, oil beaker, and water-oil beaker...

@IlmariKaronen Thank you for taking the time to provide a detailed answer, it clears up my confusion with the paper - I seem to have read it wrong. I have upvoted your answer, however, there is something still bothering me: if we take as oil a fatty acid like caprylic acid, then it would, via its carboxylic group, provide the possibility of dipole induction. There would probably be plenty of gaps in the alkyl chains and I assume it would also be entropically preferable for the oxygen to spread into them? Shouldn't the solubility at least be comparable, not three orders of magnitude off?

If $L$ is taken in $\LARGE\frac{g_{Ox}}{mol_{liq}}$ the results are very, very similar (this might be important when talking about interactions more than the actual $L$ value).

Before you read this answer, please read the ChemWiki article on the corrosion of iron.  This explains the normal (Earth-surface) process for iron rusting, which involves both oxygen and water.  The water is a catalyst, but the oxygen is a reactant: the reaction consumes oxygen and will stop if it becomes unavailable.
Now, if you put a nail in a beaker, add enough water to completely cover the nail, and then add enough oil to completely cover the surface of the water, there won't be enough oxygen dissolved in the water to rust the nail more than a little bit.  The only way for more oxygen to get into the water is by diffusing through the oil.  We can model this as a chain of equilibria:
$$
\text{O}_2\,\text{(air)} \rightleftharpoons \text{O}_2\,\text{(oil)} \rightleftharpoons \text{O}_2\,\text{(water)} + \text{Fe}\,\text{(nail)} \rightleftharpoons \text{Fe}_2\text{O}_3\,\text{(rust)}
$$
Thermodynamically, the equilibrium lies all the way to the right: completely rusted nail, having soaked up as much oxygen as possible, ultimately from the air.  But, as described in Ilmari's answer to this question, oxygen is much less soluble in oil than in water.  Therefore, oxygen will diffuse through the oil layer very slowly, and that will be the rate-limiting step.  The rate constants for the air⇌oil and oil⇌water steps will depend on which oil is used and how thick the oil layer is.
So what happens is, after most of the oxygen that was initially dissolved in the water has been consumed by the rust, the nail does continue to rust, but very slowly — only as fast as additional oxygen molecules can get through the oil layer.

Excellent ChemWiki article, thanks; I do believe that there is a waterless rust process by direct iron oxidation (slower version of dropping a nail into a campfire), although it is kinetically slow.

@Zubo Yes, I expect there must be such a process; maybe I should reword the first bit to make clearer that I don't mean to exclude the possibility, it's just not relevant in this context.

Oil does two things.

Prevents oxygen molecules from colliding with the surface of the metal.
More importantly, oil drives away water.  Water promotes rust by creating miniature electrochemical cells on the surface of iron objects.  A cartoon of one of these cells can be seen here:
Electrochemical cell promoting rust

Another point.  While I am unaware of how well oxygen dissolves in oil, I'd be interested to know how the higher viscosity of oil reduces the mobility of the oxygen dissolved in it, and possibly reduces the rate of rusting.

The image the OP described has water on both nails, just one water has oil on top, while the other has not. So water is in direct contact with the metal in both situations, yet the result differ.

This is not the situation the question is asking about.

@BrinnBeleya 1. Well, that is kind of the question - why? 2. I agree, and this hearkens to zwol's answer. Also, the viscosity thing is probably a factor, and I'd love to see some data on this. Haven't noticed any while I was researching this question, though.

It is likely that the water was boiled, removing any carbon dioxide and excess oxygen. The oil on one of the beakers acts as a protective layer and prevents any carbon dioxide or oxygen dissolving into the water of that beaker, which prevents rust. The other beaker however was exposed to the carbon dioxide and oxygen in the atmosphere, which would have dissolved, allowing rust.
For an iron nail to rust, there must be several things present. these include: oxygen, an anode (a metal that readily gives electrons, in this case iron), an electrolyte (a liquid that assists the flow of electrons), oxygen, and a cathode( a metal that readily receives electrons – this can be a less reactive metal nearby or in this case, another part of the iron itself)
The reason that iron is readily able to accept and give electrons is because it is in the transition metal group of the periodic table, meaning it has an outer valency of two or 3, (max valency 8, min valency 1) meaning it is able to both give OR take electrons to fill its outer shell.
When the iron nail comes into contact with the water that is exposed to carbon dioxide (beaker without oil) the carbon dioxide begins to dissolve, forming a weak carbonic acid. This carbonic acid is a good electrolyte and provides acidic conditions.(iron rusts faster in acidic conditions, due to the increased concentration of h+ atoms present, which draw electrons from the iron, due to their positive charge.) This begins dissolving the iron, while the water breaks down into hydrogen and oxygen(the hydrogen is slightly positive and the oxygen is slightly negative, due to the unequal sharing of electrons in water molecules). The oxygen bonds with the dissolved iron, forming iron oxide (rust – balanced scientific equation – $4 \text{Fe} + 3\text{O}_2 = 2 \text{Fe}_2\text{O}_3$). When this happens, electrons are freed from the anode portion of the iron and flow to the cathode portion.
This is a redox reaction. A redox reaction involves reduction and oxidation. reduction is the gaining of electrons and the loss of oxygen.
Oxidation is the loss of electrons and the gain of oxygen.
In this situation, the iron is oxidised and the $\text{O}_2$ is reduced.
Other ways to stop the rusting or iron include:

electroplating, coating iron in a less reactive metal
using a sacrificial anode
attaching a more reactive metal that will preferentially oxidise (in this case the sacrificial anode would be the anode, and the iron would act solely as the cathode)
galvanizing, coating in zinc (which acts almost the same way as a sacrificial anode)
coating the iron in metal oil or plastic to prevent it coming into contact with oxygen and electrolytes
putting it in an environment that is missing one of the components needed for rust (e.g the beaker covered in oil was missing excess oxygen, so the nail could not oxidize = rust)
alloying, this is when iron is mixed with other metals that help stabilize it

To speed the rusting of iron you can

put it in a more acidic environment as this increases the number of h+ ions, making the iron more susceptible to rust
connect a less reactive metal, so that the iron acts solely as an anode in the oxidation process, losing its electrons and corroding, while the less reactive metal which was harder to remove electrons from, acts as the cathode, receiving the electrons lost by the iron.

Note: metals that are more reactive than iron include magnesium and zinc (slow rusting), and metals that are less reactive include copper and tin (speed rusting).


text:
Why are the atomic radii of iron, cobalt, nickel, and copper almost the same?
We have to select the set of elements which will have almost the same atomic radii:

$\ce{O, S, Se, Te}$
$\ce{Li, Be, B, C}$
$\ce{Fe, Co, Ni, Cu}$

I googled the atomic radii of the third set and found them to be extremely close to one another: all of them are nearly $\pu{125 pm}$ (according to empirical data from Wikipedia). Is there a specific reason for this, or is it just a coincidence?

Atomic radius is inversely proportional to the effective nuclear charge. As we move from left to right in a period the effective nuclear charge increases. This will decrease the radius of an atom. At the same time, in transition elements the number of electrons in the 3d sub-shell will increase. This will repel the already present 4s electrons. This repulsion will increase the atomic radius. In Fe, Co, Ni & Cu both the effects nearly balance each other keeping the atomic radius same.

I think the point of this question is for you to realise that options 1 and 2 can't be the correct answer.
As you go down a group, new electron shells are occupied which extend further from the nucleus, increasing the atomic radius. Therefore option 1 must be wrong.
Effective nuclear charge increases across a period because the nuclear charge increases but the shielding stays roughly the same (at least until you get to transition metals). Therefore atomic radius will decrease across the group and so option 2 cannot be correct and the answer must be option 3.


text:
Understanding a chemical reaction from kegg
I came across a reaction that looks as follows:

In this reaction the polyphosphate is common in both the sides of the reaction. Does this act as enzyme? But from the website it seems that the enzyme is ADP-specific glucokinase(in the enzyme section click 2.7.1.63). Then what is the function of polyphosphate here?

The $n$ of the polyphosphate should be decreased by 1 on the reactant side relative to the product side:
$\ce{(P)_{n} + Glc -> G 6P + (P)_{n-1} }$
Polyphosphate, like ATP, contains several phosphoanhydride bonds.  These are "high energy" bonds that can be used for biosynthesis.  But to do so one of these bonds needs to be broken, and that is exactly what happens when one of the end phosphates on the phosphoanhydride chain is peeled off and trasferred to glucose.


text:
Best Guide to Chemical Thermodynamics
I understand thermodynamics in a physicsy way - at the level of Callen's Thermodynamics. However, my chemical thermodynamics is quite rusty: concepts like fugacity and activity coefficients are only vaguely familiar from undergrad courses.
I'm looking for a well written, rigorous and succinct guide to advanced chemical thermodynamics - at the level where I can read papers on ENRTL equations of state, or understand Pitzer models fully. If it is geared towards chemical engineers (like myself) that would be a bonus.
Basically, I consider Callen's book a classic, and wonder if there's anything that good within chemical thermodynamics.
Does such a book exist? Thanks

I would suggest:
1) Introduction to Chemical Engineering Thermodynamics by Smith, Van Ness, and Abbott;
2) Chemical, Biochemical, and Engineering Thermodynamics by Sandler; and
3) This is more suitable for advanced reading or for a graduate student: Thermodynamics and Its Applications by Tester and Modell.

authors: أحمد أبوالمجد


text:
Does the Maxwell-Boltzmann distribution apply to gases only?
The Maxwell-Boltzmann distribution can be used to determine the fraction of particles with sufficient energy to react. I know that the curve applies to gaseous reactants and would like to know whether solids and/or liquids are also described by a similar distribution.
In other words, can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?

I'm going to go against the grain here:  the Maxwell distribution does describe the distribution of molecule speeds in any (3-D) matter, regardless of phase.
Suppose we have a system of $N$ molecules with masses $m_i$, positions $\vec{r}_i$, and velocities $\vec{v}_i$ (with $i = 1, ..., N$).  Assume that the total energy of this system is of the form
$$
E(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) = U(\vec{r}_1, \dots, \vec{r}_N) + \sum_{i = 1}^N \frac{\vec{p}_i^2}{2 m_i },
$$
i.e., a potential energy depending on the molecules' positions and a kinetic energy.  According to Boltzmann statistics, the probability of finding this system within a small volume of phase space $d^{3N} \vec{r} \, d^{3N} \vec{p}$ is
$$
\mathcal{P}(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) d^{3N} \vec{r} \, d^{3N} \vec{p} \propto e^{-E/kT} d^{3N} \vec{r} \, d^{3N} \vec{p} \\= e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} \left[ \prod_{i=1}^N e^{-p_i^2/2m_i k T}\right]d^{3N} \vec{r} \, d^{3N} \vec{p}
$$
To find the probability distribution of finding molecule #1 with a particular momentum $\vec{p}_1$, we integrate over all other configuration variables (i.e., $\vec{r}_1$ through $\vec{r}_N$ and $\vec{p}_2$ through $\vec{p}_N$).  Because the $E$ is the sum of a contribution from the positions and a contribution from the momenta, the Boltzmann factors can be split between these integrals, with the result that
$$
\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1 \left[ \int e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} d^{3N}\vec{r}\right] \left\{ \prod_{i = 2}^N \int e^{-p_i^2/2 m_i kT} d^3 \vec{p}_i \right\}
$$
These integrals are nasty (especially the one over all $N$ position vectors), but they're just constants with respect to $\vec{p}_1$, which means that the can be folded into the proportionality constant:
$$
\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1.
$$
A similar logic applies to every other molecule in my system.  In other words, the probability of finding a molecule with a momentum $\vec{p}$ doesn't depend at all on how they interact with each other, assuming that their interaction energy is dependent only on their collective positions.  Since these molecules obey the same momentum distribution, they must also have the same velocity distribution, and in particular they obey the Maxwell speed distribution.
So the answer to the question asked in your title, "Does the Maxwell-Boltzmann distribution apply to gases only?" is "no";  it applies to all phases of matter, in the sense that it describes the distribution of particle speeds and energies.  However, the question you ask at the bottom of your post, "Can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?" may also be "no";  the connection between reaction rate and activation energy is not straightforward if the medium is dense, as was pointed out by @porphyrin in their answer.

This is fine for a gas but for a condensed phase the coordinates - and therefore the integrals - are not separable. The position and momentum of any particle depends on coordinates of other interacting particles. You can assume a mean-field suffices but that is not the same thing, and it is often not a sufficient approximation.

@Thorn: Interesting. I agree that the position coordinates are often not separable, but I wasn’t aware that the same could be true for the momenta. Can you provide a citation discussing non-quadratic momenta so I can edit my answer accordingly? (Note that my derivation can be extended to any system where there are “cross terms” between the momenta of the form $p_1 p_2$, so long as the kinetic energy can be described by a positive definite quadratic form .)

Hmm, as sure as I was that I'd spotted something wrong in your line of argument, I am now grasping at straws. A search produced this: https://www.scirp.org/journal/PaperInformation.aspx?PaperID=83351 That however used MD with a classical harmonic internuclear potential to confirm a MB distribution in a low T solid. I wonder how particles interacting with a nonlinear potential might behave?

I'm willing to put myself out there and say that no, a Maxwell-Boltzmann distribution will not be able to make any statement about the ability of liquids or solids to react. Meaning also that you could not study reaction rates with a Maxwell-Boltzmann distribution for the condensed phases.
The reason why I am fairly confident this is true is because the Maxwell-Boltzmann distribution makes the assumption that it is working with an ideal gas. Making that assumption is both powerful and very limiting. The ideal gas approximation is quite good in a lot of cases, but it could not even come close to describing the behavior of a condensed phase.
If some such distribution were used to study condensed phase reaction rates, it simply wouldn't be a Maxwell-Boltzmann distribution.
I would also point out that the Maxwell-Boltzmann distribution doesn't describe all gas-phase reactions all that well. This can be inferred quite easily from the fact that the M-B distribution treats all gases as ideal, so the only possible difference between two systems is the mass of the particle and the temperature. That means also that any gas phase reaction which depends strongly on the orientation of the collision would likely be assumed by M-B to happen much more frequently than it really does.
All that to say, M-B is deeply rooted in the gas phase, and even there it falls short for certain systems, so it probably isn't possible to apply to the condensed phases.
On the other hand, it is possible that a partition function for a liquid or a solid phase could be used to study reactions in some way, and the partition function for a generic gas particle can be used to derive the M-B distribution so . . . There's that.
The problem with that is that the partition function requires you to sum the Boltzmann Factor over all states, and for a condensed phase system . . . That's basically gonna be an infinite number of states.
Anytime you approach stat. mech. or the name Boltzmann, it's wise to keep this quote by David Goodstein in mind:
“Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics. Perhaps it will be wise to approach the subject cautiously.”

In liquids the Boltzmann distribution limits reaction in what are called 'activation limited' reactions. In these reactions the two species collide but do not have sufficient energy to react because the activation energy is high, so they diffuse apart again. The number of particles with sufficient energy to react is given by the Boltzmann distribution. This falls with increase in energy so rapidly that reaction becomes a rate event, perhaps once in 100 million collisions.
Reactions with a low activation energy can react at first approach, by the Boltzmann distribution there is enough energy to surmount the activation energy to reaction. However, in these cases the reaction rate is limited by how fast the species can diffuse together, so reaction rate becomes a property of the solvent.

No. Maxwell-Boltzman statistics are not followed. The Equipartition Theorem is followed (at equilibrium) but the distribution tends to be more like a Planck Distribution giving different equations for $v_{avg}$. Consider that solids arguably don't have an atomic $v_{avg}$ (or rather it is zero) and that liquids differ from gases because they have additional degrees of freedom. (That is, they have Potential Energy modes due to their molecule-to-molecule interactions which Ideal Gases do not have by definition. The assumption of non-correlated (random) movement can't be made (at sufficiently low densities)).
Some notes:
This area is difficult to get a handle on. Here's two references I found useful (along with the Wikipedia articles on Phonons,the Equipartion Theorem, Diffusion, and Molecular Diffusion):

https://physics.stackexchange.com/questions/111743/is-there-an-equation-to-calculate-the-average-speed-of-liquid-molecules
http://cpb.iphy.ac.cn/fileup/PDF/2013-8-083101.pdf.

The 2nd demonstrates that the way in which liquids move (are correctly modeled (hopefully!)) depends on their "modes" or potential energy degrees of freedom (potential function). There are both internal degrees of freedom (rotations, twistings, vibrations,...) and external (Van der Waals forces, etc.) not to mention steric constraints (is there a size above which the concept of average molecular velocity has no meaning? (or does the molecular velocity concept transition into the macroscopic particle concept as size increases? Does it matter if "size" is significantly nonisotropic? (e.g. linear, planar, or spherical molecules?))

Perhaps you mean: " The assumption of non-correlated (random) movement can't be made (at sufficiently high densities))"

I agree with the answer of Michael Seifert. I just wanted to point out, that this applies to classical systems only (even if they are non-ideal).
For example systems in which relativistic effects play a role (Maxwell–Jüttner distribution), or for systems in which quantum effects play a role, the Maxwell-Boltzmann distribution will be violated.

The distributions of energy in liquids and solids fall off faster in liquids and solids than the Maxwell-Boltzmann distribution - if a molecule in a liquid or solid has high energy, it is going to transfer it to other molecules much sooner than in a gas.

I think the difference in distributions is due to the number of degrees of freedom due to interparticle interactions, and not due to faster interactions rate. Otherwise you could make the argument that a hotter gas is less likely to observe the MB distribution.


text:
Do Chemists "discover," "invent," or "create" things?
I am an ESL teacher, and am designing an activity to teach my engineering students the differences between the words "invention," "discovery," "innovation," and "creation," words that are often confusing for non-native speakers (and as I'm finding out, native speakers.)
I have looked on the internet for usage and it seems to be all over the map: Nobel either invented dynamite, created it, or discovered it, and Teflon was called a discovery.
In my view, physical laws are discovered, as are molecules, fundamental particles, and forces because they already exist naturally.  Teflon cannot be said to be discovered because it does not exist naturally; neither does DDT and thousands of other molecules and compounds.  Is there a plant that produces nitrous oxide?
Based on the usage of "creation" I feel that this word more aptly describes the "process" of chemistry.  Teflon was created, not discovered or invented.
As chemists what usage do you prefer?  Do you prefer the use of inventor, creator, or discoverer?  If there are multiple answers can you help me see why you choose that word?

If you would like to distinguish between those terms, then define them first. Currently I don't see any definitions in your post.

well, I would like to see how you talk about this without my influence, or saying what I think the words mean.  I want to see what's natural in your word.  I understand context makes a difference.

You already know that chemists use these words interchangeably. Oxford Dictionary mentions all these terms (invent, create, innovate, discover) as synonyms; Merriam-Webster Dictionary lists two of them you're making the biggest distinction between (discover and invent) as synonyms.

And I don't think context makes much difference here, it is all about definitions. You said that you teach your students the differences between these words. I ask you what these differences are that you're trying to teach? You obviously have your own definitions for these terms (since you stress the differences between them) and I would like to hear them.

I'm writing the activity now, based on having these questions come up before and having seen them taught incorrectly in material students brought to class.  Beethoven didn't invent his 9th symphony but they thought that was OK.  To me it seems better to say Nobel created / invented dynamite because he used existing methods and compounds to make something new that didn't exist before...it wasn't part of "mother nature" so he couldn't discover it.   But I'm not a chemist so perhaps in your field it's ok to use the word discover like this  or you define it differently in your profession.

Well, see, in my opinion it is perfectly fine to say that Nobel discovered dynamite as well. I don't get what problem you have out there. One of the definitions of discover from the Oxford Dictionary is "1.2 Be the first to find or observe (a place, substance, or scientific phenomenon)".

ok, so it seems as chemist you use the word "discover," regardless of whether it was there or not.  This helps.  Let's see what the consensus is on this...if there is one...if not that helps me to help my students.  Thank you.

However, I don't agree that this is primarily opinion-based. Could the flagger state their opinion more elaborately?

This seems like it might be suitable for the English Language StackExchange.

Yes but that area is not frequented by chemists...I don't want guesses and maybes but people in the field.

Yeah, engineers go both ways on this and it's almost a matter of philosophy. Some are platonic (we didn't invent chairs, we discovered them!) while others, like myself, use discovery to mean that we didn't design it ourselves, we observed it and then (maybe) found a way to duplicate it.

discover
invent
create
innovate

Create is totally weird with chemistry, you can create item, but substance?
Innovate is more about improving something existing.
This leaves us with discover and invent, which in case of chemistry seems to me almost equivalent. For "invent" I would imagine targeted effort towards a new substance with desired properties, whereas "discover" evokes more trial-and-error approach.

Ok so it seems that in trying to make something new, other unexpected things can happen which are called discoveries...so if a chemist were trying to make X and failed but Y happened and was useful you would call this a discovery...like Nylon...?


text:
Difference in acid strength of oxalic acid and malonic acid
Despite very similar skeletal structures, the difference in acid strength between oxalic acid (ethanedioic acid) and malonic acid (propanedioic acid) is quite significant. What is the reason for that?
It seems that the inductive effect plays a role, but what is the explanation for the inductive effect in oxalic acid as there is same group on opposite sides?
How does the inductive effect plays different roles in both acids?

Inductive effect decreases significantly with no of atoms in between the inductive group and the group on which we are checking the effect on. In case of malonic acid there is one carbon atom between the two carboxyl group. As the -I effect decreases so the stability of carbanion and so the acidic strength. I think this may some what explain

Linear saturated dicarboxylic acids
$n$ = number of $\ce{-CH_2}-$ groups in chain
 n                        pKa1     pKa2   Ka1/Ka2
 0  Oxalic acid           1.25     4.14   776
 1  Malonic acid          2.83     5.69   724
 2  Succinic acid         4.21     5.41    15.8
 3  Glutaric acid         4.34     5.41    11.7
 4  Adipic acid           4.41     5.41    10.0
 5  Pimelic acid          4.50     5.43     8.51
 6  Suberic acid          4.526    5.498    9.4
12  Dodecanedioic acid    4.45     5.05     4.0

Dodecanedioic acid
Aromatic carboxylic acids
                                   pka1     pka2    pka3
 benzoic acid                      4.202    ----    ----
 ortho-phthalic acid               2.89     5.51    ----
 meta-phthalic acid                3.46     4.46    ----
 para-phthalic acid                3.51     4.82    ----
 o,o'-bibenzoic acid
 p,p'-bibenzoic acid
 benzene-1,3,5-tricarboxylic acid  3.12     3.89    4.70

Linear unsaturated dicarboxylic acids
$n$ = number of $\ce{-CH=CH}-$ trans groups in chain
n                            pka1      pka2
0  Oxalic acid               1.25      4.14
1  Fumaric acid              3.02      4.38
2  trans,trans-Muconic acid  3.87
3

Straight-chain, saturated carboxylic acids
$n =$ total number of carbon atoms in molecule ($n \geqslant 2$ has terminal methyl group).
n                     pKa
1  Formic acid         3.77
2  Acetic acid         4.76
3  Propionic acid      4.88
4  Butyric acid        4.82
5  Valeric acid        4.82
6  Caproic acid        4.88
15 Pentadecanoic acid  4.8

most $\mathrm{p}K_\mathrm{a}$ data from Wikipedia.

How does the inductive effect plays different roles in both acids?

First think of what happens when the carboxylic acid group of a straight-chain, saturated carboxylic acids ionizes to form the carboxylate ion. The $\ce{C=O}$ and $\ce{-C-O^{-}}$ bonds of course hybridize. This hybridization occurs faster than hydrogen bonds to the solvent form. The net result is effectively two oxygen atoms with 1/2 a charge each. This increases the ability of the  carboxylate ion to solvate, which increases its acidity.
In malonic acid the $\ce{C-C}$ bond "shares" some of that hybridization. This is the "inductive effect." The additional delocalization of the electrons from the carbon atoms makes the carboxylate ions more electronegative than they would be in a long linear saturated dicarboxylic acid. So the first ionization ,  the second ionization

Is this still under development? Maybe since it was reactivated now, it would be a good time to remove that note...

@Martin-マーチン - Thanks for the heads up. I took off the under development text. I'm not doing anything else with this. If some else wants to find some additional values then they are free to add them to the tables.

First, notice that a carboxylic carbon ($\ce{COOH}$) has an oxidation state of $\mathrm{+III}$ (except in formic acid where it is $\mathrm{+II}$). This means, that the carbon is very electron deficient and wishes to draw electrons towards it from sources that are not the two oxygen atoms (because they are stronger). We term this an inductive effect, as we cannot easily draw mesomeric structures to explain it.
Since the inductive effect is mainly based on electrostatic interactions, it gets exponentially weaker with increasing distance. In oxalic acid, there is one one bond separating the two carboxylic groups, so they can exert a strong inductive effect. In malonic acid the effect is weaker due to the separating $\ce{CH2}$ group. And in succinic and subsequent diacids it is almost unmeasurable due to the distance between the groups.
But how do we rationalise each group exerting a $+I$ effect on the corresponding other group? Well — we cannot deprotonate twice immediately (I am introducing this as an axiom, but it does have reasoning that I feel too lazy to explain right now). So we have to pretend one side is going to be inert while the other is deprotonated. And if we do that, we see that the inductive effect of one side is strong in drawing away electron density from the $\ce{COOH}$ group we are deprotonating. Since this means there is a lower effective negative charge, deprotonation happens more easily, i.e. at lower $\ce{pH}$.
We can use this simple picture since the groups are isotopic, i.e. they can be transformed into each other by a $C_2$ rotation of the molecule’s inherent symmetry.

Acidic nature is directly proportional to strength of conjugate base.
In oxalic acid, the hydrogen oxalate ion and the oxalate ion (Which are conjugate bases in first and second dissociations) are resonance stabilized. The negative charge on oxygen atom is in conjugation throughout the molecule (with two pi bonds). Hence it is more stable.
In malonic acid, the conjugate base is relatively less stable, as the conjugation is restricted to only one pi bond.
Resonance effect is more stronger than inductive effect and moreover, resonance effect in the conjugate base of oxalic acid takes place throughout the molecule.(there is a negative charge, pi bond, pi bond, lone pair conjugation ).
Hence oxalic acid is relatively more acidic as compared to malonic acid.

Please show me the resonance structure which transfers the charge from one carboxylic unit to the other.

Also, welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Note that we consider multiple exclamation marks and ‘hope this helps’ messages clutter.


text:
Why does a mixture of siloxene and cerium(IV) sulfate luminesce?
I performed an experiment where siloxene and cerium(IV) sulphate were mixed together: when I did so  the mixture produced an orange-yellow glow. Why does it glow? What is it about the two chemicals in particular that makes this happen?

Method for synthesising siloxene:

Carefully measure out 50 cm3 of concentrated hydrochloric acid into a 100 cm3 measuring cylinder. Put the acid in a 1 dm3 beaker containing a magnetic stirrer bar and place the beaker on a heater/magnetic stirrer.
Switch on the stirrer and slowly add 5 g of calcium silicide, one spatula measure at a time, to the stirred acid. Continue stirring the mixture until the reaction subsides.
Measure out a further 25 cm3 of concentrated hydrochloric acid and carefully add this to the stirred mixture. Now turn the heater on and heat the mixture to boiling. Boil the mixture for five minutes. After boiling, slowly add 150 cm3 of distilled water then boil again for approximately five minutes.
Switch off the heater and leave the mixture stirring while it cools.
Set up a Buchner funnel and flask in a fume cupboard.
Carefully pour the yellow siloxene that you have made onto a filter paper in a Buchner funnel (try to leave the unreacted grey/black solid calcium silicide in the beaker). Now wash your sample of siloxene – first with 10 cm3 of ethanol and then with 10 cm3 of ethoxyethane. Drawing air through the sample will help to dry the yellow, solid siloxene.

Method for producing the glow:

Place 0.8 g of siloxene and 75 cm3 of 1M sulfuric acid in a 250 cm3 conical flask. In a second 250 cm3 conical flask, dissolve the 0.5 g of cerium(IV) sulphate in 75 cm3 of 1M sulphuric acid. Label and stopper both flasks.
Go into a dark room. Turn out the light, add the cerium(IV) sulphate to the siloxene and swirl the mixture. You should see a fairly bright orange-yellow glow.

Maybe someone can access the following paper: Grigorenko, F. F., and L. I. Dubovenko. "Chemiluminescence in the reaction of siloxene with cerium (IV) sulphate." UKr. khim. Zh 1968, 34, 1294-1299.

I am not sure, did you mean siloxane

The idea of chemiluminescence is pretty simple, you need an oxidizer and a substance that actually gives a glow.
According to Wikipedia cerium sulfate is rather an oxidizer especially under acidic condition, which actually explains why we need sulfuric acid, so basically, whatever is siloxene (I suspect it is a silicon oxide kind of structure) it is an oxidant that when oxidizes the difference in energy level causes a glow in the visible spectrum.
Unless more clarification about siloxene, this could be the only possible answer.
sources:

NurRage explanation about chemiluminescence
Wikipedia for properties about cerium sulfate

EDIT: what was meant by potential sources is that the method of using words was mine but the information is from the sources sorry for inconvenience

According to this paper:

"Siloxene is a yellow silicate polymer obtained from the reaction of
  calcium silicide with hydrochloric acid."

Also, from the same paper:

"The basic structure [of siloxene] is randomly substituted by hydroxyl
  and chlorine groups and this affects both the CL [chemoluminescence]
  spectrum and the CL efficiency. CL emission in the red region of the
  spectrum results from the reaction of siloxene with oxidants such as
  ceric sulfate, chromic acid, potassium permanganate and nitric acid."

This brief explanation would appear to me to equally apply to the luminescence observed in the question.