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Why is sand that insoluble?
Why is sand that insoluble?
Sand is a polar molecule but does not dissolve in water.
Is it because of the big structure that sand makes?

Why do you think that "sand is a polar molecule"? For that matter, what kind of sand are you talking about?

What is Sand?
Scope of this Answer
In order to give a good answer, we first have to decide on the type of sand we want to talk about, because its composition is highly dependent on that. Is it beach sand? If so, where from? All these questions should have been addressed in the question. I will limit the scope of this answer to pure silicium dioxide $\ce{SiO2}$ sand.
Is Sand a Molecule?
Sand is, strictly speaking, not a molecule. It's a crystal. As such it has properties such as the lattice energy. Maybe have a look at that link, it might clarify it all for you already.

So now we have talked about what we want to talk about and clarified the basics, so let's dive right in, shall we?
Why does $\ce{SiO2}$ not Dissolve in Water?
Basically this is an energy calculation. On one hand you have the crystal, with its lattice energy, and on the other hand you have the solvated $\ce{SiO2}$ units. A tentative reaction equation would look like this:
$$ \ce{Si_{n} O_{2n} <<=>[E_\text{lat}] SiO2 <=>[E_\text{solv}] SiO2(aq)} $$
The energies I've indicated are the lattice energy $E_\text{lat} = 12535~\mathrm{kJ\,mol^{-1}}$ (Source) and the solvation energy $E_\text{solv}$ (see figure below, where I've also included sodium chloride as a substance to compare with).

Data from here (PDF) ($\ce{SiO2}$) and here ($\ce{NaCl}$). It is obvious that the solvation of $\ce{SiO2}$ monomers is highly unfavored (if I got all the sign conventions right, that is...)
Basically, for ease of understanding, imagine that the $\ce{SiO2}$ unit in the middle has to decide on which side it wants to go. Either it forms a crystal with other units (to the left) or it gets solvated by (surrounded by molecules of) water. Which way will it go? We already know that it will take the route to the left. This means that the whole system can release more energy this way. That is why $\ce{SiO2}$ crystals are stable in water.
Or somewhat. Some theoretical investigations point towards the possibility that the surface of quartz crystals undergo chemisorption of water, thereby generating $\ce{SiOH}$ terminal groups, which would then generate more favorable interactions between the crystal surface and the surrounding water.
Coming back to (a part of) your question:

Is it because of the big structure that sand makes?

You are kind of on the right track here. It is because of the lower energetical state as a crystal that these big structures (crystals) are formed, vs. the units just being dissolved (solvated by water).

There are many reasons, due to which it is insoluble in water.
High lattice energy:
Due to polar interaction with water, it get suspended but remain undissolved due to high lattice energy
size of sand particle:
The size of sand particle is bigger than $0.3 \mathrm{nm}$(Again as commented it depend on type of sand) which is difficult to dissolve.

Some things which are bigger than 0.3 nm can squeeze inside and cause unfavourable entropic effects but significantly more favourable than being at the surface or slightly more than being in normal solution.
  However, sand particles are significantly larger and entropically it prefers being suspended.

(Yahoo)
Because of bond:
Sand is not able to dissolve in water, because water can not break bonds between silica (sand / $\ce{SiO2}$ / Silicon dioxide)

Sand is insoluble in water and organic solvents. There are no possible attractions which could occur between solvent molecules and the silicon or oxygen atoms which could overcome the covalent bonds in the giant structure.

(Chemguide)

P.S. sand is a non-polar molecule.

I think your point #2 is invalid. It is possible to dissolve a large crystal of NaCl, although the dissolution will be slower (since it depends on the surface area, which is larger for granular substrates).

@tschoppi I have written "difficult to dissolve" which also play role what i have heard. But  i can remove if it is wrong. :)

@tschoppi I agree with #2 and i also got link which explain that in detail

@RutvikSutaria I don't believe in Yahoo much but as they have also mention source possible they are right. BTW I have heard about it many times.

@RutvikSutaria Never a day without learning something. Thank you for the reference.

By sand I'd guess you mean $\ce {SiO2}$.
You see, sand or $\ce {SiO2}$ for what it matters, don't appear in nature as single molecules, they're crystals and the crystal arrangement defines if it will be quartz, sand, silica-gel and other stuff, and this crystalline structure also defines it's solubility in water.
So even if a single molecule of $\ce {SiO2}$ could be polar due to the electronegativity difference between Si and O, the sand won't be exactly polar because they only hang together as crystals, and so the polarity will vary as a function of it's structure.
PS: I never heard of single $\ce {SiO2}$ molecules in labs neither.

Apparently, molecular $\ce{SiO2}$ does exist in the gas phase, and can be produced by oxidizing silicon monoxide gas in an inert matrix. It's still not a particularly common or stable species, at least on Earth (although I suspect it may have some astrochemical relevance).

Surely the correct answer is to apply a variation of the anthropic principle: sand can be made of a variety of different crystals, some of which are water soluble and some of which are not. The reason you only see the insoluble crystals when you go to the beach is that the soluble ones were dissolved into the ocean millions of years ago.
If you look at different sand compositions, you'll see that desert sands have a higher proportion of water soluble components. For example, some parts of the Sahara have significant amounts of gypsum. White Sands is almost entirely gypsum. But gypsum never makes up more than a trace of beach sand because it dissolves in water. (Indeed, it dissolves better in cold water like the ocean than in warm water.)

In extension to the anthropic principle, one could state that the energy is lower for sand in suspension as opposed to being dissolved. But this argument is so easy and kind of superficial, I do not like it.

injecting gypsum "sand" into this really digresses from the question.


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Why are some metals ferromagnetic?
Why are $\ce{Fe}$ and $\ce{Ni}$ ferromagnetic?
$\ce{Fe}$ has metallic bonds, which is the electrostatic attraction between the positive metal ions and delocalized electrons.
Is it due to these charges, as well as with ions and the "sea of electrons" that makes the metal ferromagnetic?

Most metals are not (ferro)magnetic.

If you explain a bit about iron, how about nickel

Now this is a really complex subject... first, magnetism is actually divided on a bunch of "classes", some elements are diamagnetic and will be repelled by a magnetic field, other are paramagnetic and are attracted by the field, and there are the ferromagnetic elements like iron, nickel and others, that become permanent magnetics upon being exposed to a mag. field, other "classes" also exist but NVM them.
This behaviors are defined by subatomic interactions on electron fields, and charges in the nucleus, not so subatomic interactions between the molecules and the crystals they form and this sort of stuff. If you want an explanation on these topics I'd suggest you to take 6 minutes to watch the "Magnets: how do they work?" from minute physics, it's a pretty cool short video, and this channel has a lot of other cool stuff on this stuff.

source: Stack Exchange

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sort: votes


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Is it dangerous to live near lead?
Several month ago, I soundproofed the door of my room with two sheets of lead. Can it be dangerous to live with them for many years? Can lead "decay" in some way and be inhaled? (and what happen when it oxidizes?)

i know they are similar, but i am more concerned with effects over a long period of time

Lead isn't a material that corrodes easily, lead sheets have been used for roofing, it lasts centuries, until stolen.  And since it isn't friable one would think the ingestion hazard is low.

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Why does aluminum have a lower first ionization energy than magnesium?
I used to use the explanation that $s$ orbitals penetrate better than $p$ orbitals, however, could the explanation be that $3s$ is shielding the $3p$ in aluminum?

The easiest way to explain it is that $\ce{Al}$ has one unpaired electron in it's highest energy orbital ($\mathrm{3p}$), and $\ce{Mg}$'s highest energy orbital ($\mathrm{3s}$) the electrons are paired. It is energetically favorable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.
Here's what their orbital pairings look like:

and if you look at the general trend you can see that this occurs whenever all the valence electrons are paired:

Sources:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html
http://en.wikipedia.org/wiki/Electron_pair

Electrons repel each other so pairing them raises potential energy.  Note that O is lower than N for this reason.

I've edited my answer to reflect that discrepancy. In the cases where the entire orbital level has electrons of one spin, adding a lone electron of opposite spin is unfavorable because of repulsion. =

Note that in the trend chart that i posted, the noble gasses have the highest 1st ionization energies. This is because all of their valence electrons are paired, and it is energetically unfavorable to unpair them.

could the explanation be that 3s is shielding the 3p in aluminum?

Yes, one could say that for 2p, 3p and 4p orbitals; it becomes more confusing for 5p and 6p. One can also say that a half filled p orbital (3 electrons) shields the nucleus against the 4th electron.

Removal of an electron will disrupt the stable filled 3s subshell of magnesium. Also the 3p electron of aluminium is further from the nucleus compared to the 3s electrons of magnesium


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What is the structure of the Nucleus?
The structure of an atom is well known to me as something like this:

But what is the accurate structure of the Nucleus ? What is the arrangement of the Protons and the Neutrons or any of the other nucleons? I will be glad if anybody give an image of its structure.
Thanks!

This question has already been asked and answered on Physics.SE (http://physics.stackexchange.com/questions/36469/are-the-protons-and-neutrons-in-the-nucleus-arranged-in-any-particular-way).

@Philipp I think question asked in other in S.E. is not considered as duplicate.

@Philipp Didn't get a good answer from there

@NeilRoy Have you also checked out the other content on that matter that can be found on Physics.SE, e.g. this question? If your question doesn't get a satisfactory answer on Physics.SE it is unlikely that you get a better one here, because your question is purely about physics not chemistry. Chemistry is concerned with molecules not with nuclei on a subatomic level. Maybe you could ask for clarification on Physics.SE.

@NeilRoy I'd suggest following some of the links in the Physics.SE question, including the "liquid drop model"

@Freddy I gave it only as an example. There are also other similar questions there that all revolve around this topic and which might provide at least enough of an answer to get by.

The nucleus is the very dense region consisting of protons and neutrons at the center of an atom.
The diameter of the nucleus is in the range of 1.75 fm ($1.75×10^{−15}$ m) for hydrogen (the diameter of a single proton)2 to about 15 fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about 23,000 (uranium) to about 145,000 (hydrogen)

Source:Wikipedia

I wanted the accurate description of the nucleus with all the nucleons in it. Is it possible please?

@NeilRoy Yes it is possible, but me I think it would be better some expert answer this question because I have just heard of stuff(quarks, etc) not read anywhere.


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Are there actually any reactions during which STATIC equilibrium is reached, as opposed to dynamic?
Static would be where a set amount of both products and reactants is formed. Dynamic would be where both the forward and backward reactions occur at equal rates, meaning no change in the overall product and reactant concentrations.


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Are all esterification reactions reversible? What is their general mechanism?
For example the reaction of ethanoic acid with ethanol is reversible, however is this true for all esters? I'm not quite whether it is as easy for a reversible reaction to occur both ways. Is there a limit to the complexity an ester can be? What would be the general mechanism for these reactions? I assume there is no limit on the complexity (carbon chain length) of esters, other than the frequency of longer-chain carboxylic acids and alcohols needed to form them. Is this true?

Could you please clarify the question referring to complexity?

I'll start this answer off with saying that both of these answers were the first result when searching for Google.
Yes, The esterification reaction is both slow and reversible
and here is the mechanism from an image search

Do you also know whether esterification takes longer/ is more difficult with longer-chain carboxylic acids and alcohols?

Yes the rate of the mechanism slows with the size of the reactants.


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How can a catalyst not be included in a rate equation if, by definition, it speeds up a chemical reaction?
I thought that anything not in a rate equation was automatically zeroth order and therefore did not affect the reaction. However, I have heard that catalysts can be involved in a reaction while not being in a rate equation. How can this be possible?

Rate equation is generally created on empirical basis to establish a relationship between the rate and reactant concentration. The dependence on catalysis concentration is generally complex and most cases not needed (only later optimised or used in excess quantity).

So if I understand correctly, as with temperature, the rate constant also depends on the particular concentration of any active catalyst which is involved - this being found experimentally. Correct?

catalyst not be included in a rate equation || WTF? catalyst's concentraion (in case of homogenous catalyst) should be in rate equation (usually, if the mechanism is complicated enough, some wanders may happen). Heterogenous catalyst is usually not included in rate equation as it is, well, heterogenous and not in reaction mixture, so it's concentraion in the mixtire is zero by definition (but concentration of active particles on surface of the catalyst shoud be included into the rate equation)

@permeakra So essentially, if the catalyst has a measurable concentration and is in the same state as the reactants and products, it should be included in the rate equation? But what if the reaction itself is heterogeneous? Also, what do you mean by 'active particles'?

Granted that the more catalyst there is, the faster the reaction will proceed. Since a catalyst provides a different route to the products, a different rate equation is obtained. This is now a 3 (or more) particle interaction, which is difficult to model. I have not seen any studies with the rate changes with the amount of catalyst.

@LDC3 Catalytic reactions are generally not 3 particle interactions, just simply different mechanisms. Reactions with any significant speed are generally not 3 particle interactions.

@Greg If you had a 2 particle interaction before the catalyst, then you must have at least 3 afterwards, the 2 particles and the catalyst.

@LDC3 Maybe we use different terminology: by 3 particle interactions I mean elementary reaction steps that involve 3 entities. That is something that rarely happen or needed for catalysis.

@Greg Utilizing a catalyst usually generates a 3 particle interaction. For instance, the hydrogenation of a double bond using platinum. First the hydrogen needs to stick to the platinum and then the alkene needs to strike the hydrogen stuck on the platinum. Since you don't think 3 particles are necessary, can you suggest a catalytic reaction that has only 2 particles?

@LDC3 From PChem point of view a bound state of one of the reactant and the catalyst is a single particle, and the bound state + other reactant reaction is a separate, two-body elementary reaction. By three body reactions we mean collisions of three independent entity,which has markedly different kinetic and dynamic behavior, and also a pretty low chance to occur at all.

@Greg OK, we're counting differently. You're right in that the intermediates produced are from only 2 particle interactions, even though several 2 particle interactions are necessary to reach the products.

Enzyme kinetics is one area of catalysis that has been studied extensively. The Michaelis-Menton equation being one of the central pillars and explicitly states the roles of concentration for both the substrate and enzymes.
Also, to answer your question: the rate equation is at times more interested in establishing equilibrium conditions than obtaining the time it takes to get there. The time to reach equilibrium is slightly neglected as the question "are we there yet" seems trivial when distance = rate times time.

There is a slight difference in the concentrations for catalysts and enzymes. For most non-enzymatic reactions, the catalyst concentration is less than any of the reactants. For most enzymatic reactions, such as the enzyme catalase converting $\ce {H2O2}$, the enzyme has a greater concentration than $\ce {H2O2}$ under normal conditions.

In chemical kinetics, we are interested in knowing the dependence of the rate of reaction on the concentrations of reactants. The effect of any other factor, such as temperature, pressure or catalysts is observed in change of K, the rate constant of the reaction.

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Enclathrated - definition?
When discussing the solvent uptake of a metal-organic framework would the correct term be "solvent molecules were enclathrated into the pores"? Despite searching for a definition online I have drawn a blank and its not a term I am familiar with. Having read many papers in polymer chemistry where this term is also used I wonder if I have misunderstood its usage.

The Oxford Dictionary does not have an entry for enclathrate, but does for clathrate:

A compound in which molecules of one component are physically trapped within the crystal structure of another.

Using the word enclathrate should be redundant.

Thank you :) I know biologists use clatherin for the protein coat around membranes and such ... I find it unusual that there are plenty of papers online with "enclathrated" ... must be an older term I guess?

No, if it was an old term, the Oxford Dictionary would have it. I'm guessing that somebody felt that clathrate was not sufficient for their description and so added the en to ensure that the readers would realize how entrapped the compound was.


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Safe drying agents for drying machine
When I was milling NaOH for some purpose, I noticed that the powder is incredibly hygroscopic. Because I have problems with drying certain chemicals safely (without extensive heating), I thought this might be a safer solution:

But I'm not so sure about that safety. Can NaOH exit into the air and pollute the compound in the second beaker? If it can, which compounds are safe and strongly hygroscopic?
Could I use this technique to increase sulfuric acid concentration? What about acetic acid?

I think downvotes should be followed by a comments, especially on a Beta site, where guidelines are not so obvious and need to be discussed.

The first part of your schematic basically describes a dessicant dryer or deliquescent dryer. These are common industrial air drying processes that produce very low dew point properties. Generally air (or other inert gas) is passed through a column of desiccant, under increased pressure.
Desiccant dryers use common desiccants that include activated alumina, molecular sieves and silica gel. In all cases, there is a maximum amount of water that can be adsorbed/absorbed before the desiccant needs to be regenerated; typically regeneration is done by heating the desiccant column, although other methods, such as purging can be used. Industrial desiccant dryers have the heater element installed as part of the construction, although any bench chemist can probably tell you about heating red silica gel beads from desiccators in the oven until they turn blue.
Deliquescent dryers are similar, but involve a highly hygroscopic agent that slowly dissolves, or deliquesces, as it absorbs more water. Usually the agent cannot be regenerated, and the waste liquid is drained and discarded. Fresh drying agent then needs to be added. This is most likely the scenario for NaOH.
So, this would need to be a major consideration for your proposed system; how to safely and effectively dry your NaOH. This is a very caustic material to have to try to continually regenerate or replace.
Yes, I would be concerned that you would transfer NaOH from the first chamber. To check this, you could easily test the gas outlet with some litmus.
Can you use dry air to remove excess water from other materials (such as acids)? In theory, this may be possible and will depend on the relative hygroscopicity of the drying agent and the material to be dried; you need to be sure that the air you are supplying your wet material with is dry enough to receive water and not simply supply more. Purge drying (passing dry air over a desiccant) is another form of regenerating industrial desiccants, but is not as effective as heating. Sulfuric acid is itself used as a desiccating agent as it is very hygroscopic. I don't know the relative hygroscopicity of NaOH and H2SO4/CH3COOH, but there is no way I would ever consider passing air from one container directly into the other chamber. This is a very hazardous setup, and is fraught with much danger. Heating an acid solution in a suitably ventilated environment under controlled conditions is still, to my knowledge, the most effective way to increase the concentration of the solution.
You should also consider the energy efficiencies of trying to 'dry' a sulfuric acid solution by passing dry air over or through the solution versus heating it. The amount of time and materials required to produce sufficient dry air to effectively concentrate a sulfuric acid solution is likely to be far in excess than the time/material investment required to simply heating it.

I heard rumours over the internet, that heating vinegar produces toxic gases and actually does not produce more concentrated acid. Since there were many such rumours, I was being careful. What danger is this setup creating?

Sorry  - I have replaced vinegar with sulfuric acid in that last paragraph. Probably need to use a fractionating column to purify vinegar, as acetic acid azeotropes with water. Heating vinegar will release acetic acid vapours which are known to be toxic (and will kill birds if you read the internet;-)).

Your setup has the potential to be very dangerous if you have large quantities of concentrated acid and base in close proximity, in a closed system, with a shared environment. In the (perhaps unlikely) event of things going wrong, this is a very messy and nasty mixture to have to deal with.

@long Acetic acid is purified by freezing, that's why the purified acid is called glacial acetic acid.

@LDC3 I don't think freezing works. I have frozen a vinegar bottle and I filtrated the ice chunks from the rest. The both final samples smell the same, taste the same (aarrgh!) and the ice even regained the original orange color after melting.

@TomášZato It's not a complete separation of water and acetic acid; some of each remains in the first couple of cycles. To remove the last of the water, there might be a different process.

Can NaOH exit into the air and pollute the compound in the second beaker?

If you leave the NaOH as pellets, it is not likely to transfer. The powder might.

Could I use this technique to increase sulfuric acid concentration? What about acetic acid?

No and no. Glacial acetic acid is already 100%. If you could remove water from sulfuric acid, you would get fuming sulfuric acid. The excess $\ce {SO3}$ would dissipate quickly.

which compounds are safe and strongly hygroscopic?

The more hygroscopic the chemical, the more hazardous it is.
If you really need to dry a solvent, it would be better to follow the information here.

I was not talking about glacial acetic acid, but a 7% vinegar. And I was talking about 35% sulphuric acid (I don't need to deydrate that in fact, since I have a 98% one). Why a compound that is hygroscopic is also hazardous?

@TomášZato I think you need to look at the 98% sulfuric acid again, 35% sulfuric acid is 18M, as strong as it comes. Phosphorous pentoxide is a good drying agent, but becomes a strong acid and very sticky when wet. Sodium metal will burn when it comes into contact with your skin. $\ce {LiAlH4}$ will react with water in the air, releasing a lot of heat. There are other hazards.

I disagree that hygroscopic chemicals need to be hazardous. Some are, but drierite is fairly safe. Molecule sieves are also quite safe.

@GeoffHutchison Yes, those are quite safe, but TomášZato was asking about more hydroscopic ones.

@LDC3: Sieves and alumina are the about the best drying agents out there.  The alumina/copper drying/deoxygenation systems invented by Brookhart have all but replaced circulating stills.

Sodium hydroxide (potassium hydroxide is even better) is indeed used as a drying agent in the chemical laboratory, as are many other things: sulfuric acid, phosphorus pentoxide, molecular sieves are but a few.
The important consideration is compatibility with the substance to be dried.  You wouldn't even think of, say drying a stream of chlorine gas with potassium hydroxide, or a propylene stream with sulfuric acid.

You want gypsum sold here to labs as "Drierite" in the anhydrous $\ce{CaSO4}$ form.
Drierite, or anhydrous calcium sulfate, is readily used in labs for drying gases, etc. It's fairly safe (I wouldn't eat it, but it's used as plaster), hygroscopic, and easily re-usable. Generally you mix the plain white form with a little bit of indicator, and then bake at 210 °C to drive off the water before re-use.

Common drying  chemicals for air dryiers I'm aware of are (anhydrous) copper sulfate and calcium chloride, the second awailable as granules and dirt cheap. Sodium hydroxide reacts with carbon dyoxide, so it's a questionable choice. However in case one need to dry an inert solvent, like benzene or diethyly ester, sodium hydroxide is a valid choice for the first run (with sodium metal for the second run).
At the extreeme corner phosphorous pentoxide and sodium metal and metal hydrides lie. They are effective, but dangerous and should not be used as first-stage dryers, as they may react too lively. In addition, some solvents/gases may react with them, so something more inert must be used. This includes silica, molecular sieves and so on.
In case of drying gases it is common to bubble the gas throw layer of concentrated sulfuric acid, using two safeguard vials before and after the vial with sulfuric acid.


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Nanoscopy Nobel Prize - why are all the tools in the visible light range?
It seems like all the new advances in molecular imaging/microscopy are in the visible light range of the EM spectrum. An example would be the most recent Nobel Prize for Chemistry.
Is there some reason we are only looking for things in the visible part of the EM spectrum? I'd love to know why, and also what techniques look outside of this range?

I can not access the link!

This question is fine here, thanks for bringing up this accomplishment.

As far as fluorophores go, if the excitation wavelength is too short (< 350 nm), the incident light will get stuck in the microscope.  Near-infrared fluorophores, on the other hand, suffer from several drawbacks, especially low quantum yields from competing non-radiative processes.

Indeed, finding good near-IR emitters is an active area of research.

The reason is not in the technology, the reason is in the application.
There are already electron microscopes which have better resolution for much smaller structures. But, for them, you have to kill and fix your specimen. But the hottest area of application of STED is biology. And there, the coolest thing is to get towards imaging in a live cell, so you can observe the biological processes as they happen.
While you could use wavelengths outside of visible light, they disrupt the processes in the living cell a lot. In fact, the intensity of light needed for these super resolution microscopies is so high that phototoxicity becomes quite a concern in everyday research.
And then you also need fluorescent tissue for these methods. We don't have that many ways to produce fluorescent tissue yet, and the ones we have (mainly GFP) have their peaks in the visible spectrum.
The stuff you can do with this technology is amazing. I recently attended a lecture given by Hell (one of the Nobel winners) and he showed a video of recording the electric impulses in the dendritic cells of an anesthesized mouse. It was haunting - beautiful, but also slightly unnerving, to see inside a live brain. This is why we need the light microscopy - our understanding of life itself depends on it.

You can do microscopy with other light sources, too. Actually there are X-ray, Raman and IR microscopes, but there are some issues:

Few or No lenses: it is difficult(not impossible) to focus IR, UV or X-ray
Fewer "light source": in visible region, we have already developed a lot of lasers etc, for UV or X-ray we have fewer options. Part of the reason is that no one needed them, so less effort were put to develop them. Also, high intensity, coherent X-ray sources tend to be synchrotrons or free electron lasers which are marvelous things, but very expensive and big.
Less spectroscopic info: Most characteristic optical properties of organic / bioinorganic materials are in the UV-Vis region. In X-ray region generally we have either high energy electric excitation kicking off valence electrons (XPS-UPS) or excitation of core-electrons. Off course, it is not always the case, for example IR or Raman microscopy can be very informative from chemical point of view.

Also, Kurt Wüthrich got the prize ten years ago for protein NMR and Fenn/Tanaka the same year for making things fly that really refuse to fly.  Rod McKinnon got it the year after for crystallizing membrane proteins.  It was about time that the visible region got recognized.


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What's happening to carbon electrodes?
Carbon electrodes are very popular, because unlike metal electrodes, carbon is quite inert during electrolysis (I'm sure that has some explanation based on electronegativity).
However they too are not eternally lasting. Especially the pencil graphite bars will slowly release black chunks into the solution.
My question is, what's happening them. The black chunks seem like carbon, so it's probably not reacting. What's damaging the structure?

Pencil leads aren't pure graphite, they're generally graphite powder with various polymers to hold them together and make sure they have the correct hardness (after all, you wouldn't be able to use them to write if they didn't leave graphite on the paper or if they crumbled under slight pressure. This is why the much more rigid glassy carbon is a popular material for carbon electrodes. My guess is that there's no reaction with the electrode happening but gas being produced is causing mechanical damage to the electrode. If you work at less extreme potentials, you might be able to reduce this effect. Otherwise, pure graphite rods should be more durable. Of course, if you're using pencil leads, they're obviously cheap to replace so it might be easier to just use them until they fall apart, then switch to a new one.

And you think pure, hard graphite wouldn't fall apart? I remember that even electrodes from batteries slowly decayed and left black stuff all over the solution.

I'm not sure. I've only ever used graphite electrodes at relatively low potentials and I've never had that problem. I can't attest to the quality of graphite that's used in batteries. Since they're disposable anyways, they really don't have to be that durable. If you're running at high potential, it's possible to actually oxidize the graphite, a known problem with strong oxyacid solutions, though the binder seems to be more strongly attacked than the actual graphite, so it depends on how the electrode is made.—doi:10.1149/1.2411746

This article here indicates that carbon/graphite electrodes are attacked both chemically (oxygen gets converted to $\ce {CO2}$, removing carbon from the electrode) and mechanically (gas bubbles form in pores in the surface and break it apart).  Pencil leads fall apart quickly because of the clay content.  A high-quality graphite electrode will last far longer but will also eventually decay.  The rate of erosion goes up rapidly with increased voltage across the electrodes.  In my personal experience doing water electrolysis I got almost no mechanical erosion at 10V but obvious mechanical erosion at 24V.


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What kind of chemicals would exist at the heat death of the universe?
As the universe reaches the maximum entropy, presumably only the most stable chemicals would exist.
Assume that all the elements haven't turn into iron yet, what kind of chemicals would dominate at that point?

A Dying Universe: The Long Term Fate and Evolution of Astrophysical Objects Section V.F.3 looks at Positronium (and lots of neat things before that) mentioned by Nicolau's answer.

There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are expected predicted to decay through at least one of several mechanisms, with a half-life somewhere between the range of $10^{35}$-$10^{200}$ yr (far shorter than the timescale for nuclear reactions to convert all matter into iron, ~$10^{1500}$ yr). Therefore, at some point the statistically expected number of atoms inside our observable bubble could potentially dip below 1.
Should proton decay happen relatively quickly, perhaps the last elements in the Universe will be present near the end of the black hole era (~$10^{100}$ yr), when the last supermassive black holes decay in a flash of subatomic particles, possibly creating small amounts of new protons/neutrons. These will decay in a comparative blink of an eye at this timescale, however, and the Universe will soon become a barren place once more.
The last structures with any semblance to matter in the Universe will likely be positronium "atoms", where an electron and a positron (anti-electron) orbit each other in a pseudo-hydrogen atom. Though ground-state positronium decays in a matter of nanoseconds or less, it will be very long-lived in the future due to the extreme sparsity of subatomic particles in the hugely expanded observable bubble, with a mean distance between a bound positron and electron measured in trillions of trillions of trillions of trillions of... billions of lightyears or more. Even in these conditions, positronium too shall eventually decay, in a timescale of anywhere between ~$10^{140}$ yr and ~$10^{800}$ yr.
Before all of these nuclear processes take place, there may also be a significant amount of unusual chemical reactions happening solely due to tunnelling effects, even at arbitrarily low temperatures, since the kinetic barriers for chemical reactions are far smaller than that for nuclear reactions. For example, most molecules containing nitrogen atoms are thermodynamically unstable with respect to release of $\ce{N2}$, and so might spontaneously break up even at $\mathrm{0\ K}$ given enough time. I don't know what the timescale for these processes is, however.

Slight nitpick: I wouldn't say protons are expected to decay. Some GUTs predict that they should decay, but there's no evidence for those theories yet. There's no experimental result that provides any support for proton decay being real, directly or indirectly.

Trillions of light years between electron and positron? Given that coulomb force decreases with the square of the distance there wouldn't be much force to attract them.

@DavidZ Indeed proton decay by GUT mechanisms may not exist, but that does not rule out all forms of proton decay. The article in my first link devotes a fair amount of attention to several mechanisms, including one which seems to involve no extension to the Standard Model (as far as I'm aware, I could well be wrong!).

From the article: "To summarize this discussion, we stress that many different mechanisms for baryon number violation and proton decay can be realized within modern theories of particle physics. As a result, it seems likely that the proton must eventually decay with a lifetime somewhere in the range [$10^{32}$-$10^{200}$ yr]".

@user137 Yes, the attraction between electron and positron will be extremely small, but still non-zero. Given there's almost nothing else around, the attraction will stand above the noise even at such gigantic separations. The positronium atom can be thought of as a Rydberg atom with an exceptionally high principal quantum number $n$. From the first article: "At the time of formation, the positronium atoms are generally in states of very high quantum number (and have radii larger than the current horizon size [~$10^{11}$ ly]).

Sure, I didn't mean to imply that GUTs are the only mechanism for proton decay. I missed that the article considers some others. Still, all of them involve BSM physics; the standard model conserves baryon number, so the proton, as the lightest baryon, is necessarily stable in that model. Honestly, I think their statement "it seems likely that the proton must eventually decay" without further qualification is simply not justified.

As a semi-literate in these matters, it seems to me that, logically at heat death, there can only be low temperature photons, as any matter would represent a local concentration of energy.

This is very short, and whilst correct (I think, I'm not an expert) it would be more appropriate as a comment unless you want to expand on it significantly, which would be very interesting so I encourage you to do that.

"Logically" is perhaps a dangerous word to use in this situation, as there is a lot of difficult, deep and entirely unintuitive physics involved. For example, Freeman Dyson points out that without assuming proton decay, the matter remnants of the Universe will depend on the minimum mass of a black hole, which is the domain of quantum gravity, at the very edge of contemporary physics. It may be possible that nuclear matter could be infinitely stable so long as present in the form of small grains, of size $\mathrm{<100\ \mu m}$.


text:
Why can't carbon form an ionic bond?

My textbook says that a $\ce{C^{4+}}$ cation cannot be formed because it requires a lot of energy to remove 4 electrons. Formation of  ionic bonds involve "removing" electrons and there seems to be enough energy there. So what's different for carbon?
The textbook also mentions that a $\ce{C^{4-}}$ anion cannot be formed because 6 protons cannot hold on to 10 electrons. Elements like chlorine form ionic bonds and end up with 18 electrons and 17 protons. I know there's just one more electron. So is $\ce{NaCl}$ possible because it's not very hard to hold on to 1 extra electron? In that case at what exact point does it become hard to hold onto the extra electrons?

Do you know about carbides? Some of them formally contain $\ce{C^{4-}}$.

Important thing is that while ionic compounds are treated as if containing actual ions, it's hardly true - it's just that bonds get pretty serious charge separation.

$\ce{C^{4+}}$ ions:
Single ions are really only observed in the gas phase.  There is absolutely nothing that prevents a C4+ ion from being generated in the gas phase, given sufficient energy.  Each successive electron removal requires additional energy though, so by the time you get to 4 electrons removed, you're looking at quite a lot of energy required.  Note that this does not mean that $\ce{C^{4+}}$ ions are seen in compounds, just that they can be made in the gas phase.  Any cation you can imagine can be produced in gas phase.
The difficulty arises when we want to do something with that carbon cation.  If I want to make a compound with it, it can bond with other substances and make new bonds.  Once that is done we can evaluate those compounds to tell whether the bond is truly ionic or not.  In condensed phases or in solution, we look at how the parts of a compound interact with the particles surrounding them to decide whether to treat them as ions or not.
The best chance at making a C4+ ion would be to bond it to fluorine, which is an extremely electron-hungry element.  When I actually try this, I find that the C-F bond is very polar, but the $\ce{CF_4}$ molecule that forms behaves as a covalent compound normally would.  For example, it does not dissociate into ions when dissolved or melted and it doesn't form an ionic crystal in its solid form.  As much as fluorine wants electrons, it doesn't want them enough to completely steal 4 of them from carbon, because the energy required to remove each electron becomes greater than the energy required to remove the one before it.  It never gets to the point of $\ce{C^{4+}}$ and $\ce{F^{-}}$ ions being bonded.
Lead can form a +4 ion, $\ce{Pb^{4+}}$, in compounds such as lead fluoride.  The reason that this can happen but carbon cations cannot is that the outermost electrons in a lead atom are much farther from the nucleus than those of a carbon atom and are easier to remove as a result.
$\ce{C^{4-}}$ ions:
In gas form this does not happen.  Carbon's first electron affinity is positive, so we would expect to see it form $\ce{C-}$ ions in the gas phase if given free electrons to attract.  Adding a second electron to this substance would require it to attract an electron against its already negative overall charge, and that is not energetically favorable.
In order to make the $\ce{C^{4-}}$ion in a compound, a species would have to donate 4 electrons to carbon.  The attraction of carbon's nucleus to other atoms' electrons is just too low for that to happen.  The best candidates for a cation in an ionic compound of this type would be either lithium or cesium, however, when we try to actually make this compound we get another class of ions, the carbides, which feature $\ce{C2^{2-}}$ ions.  In short, it doesn't happen.
Phillip's comment regarding the carbides is a good one.  There are a few metal carbides that feature carbon atoms bonding to a metal in the ratios we would expect if it were purely ionic, like $\ce{Mg2C}$.  In these compounds the carbon does have a formal charge of -4, but the properties of the substance are such that it behaves like a covalent network compound rather than an ionic compound.
What is the limit of this process?  Anions with -3 charge, like nitride or phosphide, do exist, though they require a very weakly electronegative species, like an alkali metal or alkaline earth metal, to form an ionic compound.
Carbon atoms can carry a charge as part of an ion, but only when they are part of a larger species (again, like $\ce{C2^{2-}}$) and they do not carry 4 additional electrons per atom.  Check out cyanide as another good example of this type of polyatomic ion.

Just a tiny nitpick: The electron affinity of carbon is positive according to its definition, the energy required to detach an electron from the singly charged negative ion. This of course does not affect anything in your logic.

Lead fluoride isn't ionic.

@permeakra I'm not saying you're wrong, but can you provide a source for this?  The only sources I can find say that it has "appreciable ionic character" unlike PbCl4, which is predominately covalent.

@JasonPatterson I guess, this belief comes from population analysis for PbF4 and PbCl4 that gives Pb charge +3.2 and +1.9 respectively.  This analysis should not be trusted, since similar analysis for CH4 and CLi4 may give -0.47 and -0.50 respectively for carbon; and +0.2 for Li in LiH; which is an obvious BS.  This comes from the flaws in calculation schemes for atomic charges.


text:
What chemical will dissolve the plastic coating on circuit boards
I'm trying to reclaim the metals on circuit boards and the vast majority is underneath the plastic coating. It is a brittle plastic that chips off in small flakes when etched but that process will lose to much of the precious metal.

This seems to be a popular question judging from some Google queries.
Printed circuit boards are made of a variety of materials, and from what I understand, the specifics may vary somewhat based on the specific needs of the product and the manufacturer. For example, in some cases, I know the PCBs are coated with special anti-static coatings to prevent delicate RAM and CPU chips from being damaged (e.g., this article found via Google among others).
So I doubt there's a definite solution - more likely you'll need to try several different things.
That said, there's a recent article that looks very interesting about separating parts by dissolving and swelling the plastic resins with DMSO: "Dissolution of Brominated Epoxy Resins by Dimethyl Sulfoxide To Separate Waste Printed Circuit Boards" Environ. Sci. Technol., 2013, 47 (6), pp 2654–2660.
Certainly DMSO is considered a near-universal solvent and will swell many polymers. So I'd probably try that first.

The usual precautions for DMSO apply.  It quickly penetrates the skin barrier, as does everything that's dissolved in it.

@AbelFriedman Yes, very good point. Since it's a universal solvent, it's tricky to work with safely. DMSO will penetrate latex or nitrile gloves quickly. I usually recommend double gloves, which gives you sufficient time to remove the gloves before the DMSO contacts your skin.

I've not tried with solder mask, but DMSO or N-Methyl-2-pyrrolidone work great for stripping most photoresists. The trick is, you generally need to heat them. If you keep it under ~80 °C, it's safer because it's under the flash points of the solvents, but something like 130 °C works quite well.


text:
Chemicals to clean natural stones
I picked up some stones from a hiking trip. What can I use to thoroughly clean the mud and other hard deposits on the surface of the stones? I have only washed them with plain water.

What is your ultimate goal for the cleanliness of the stones?  Not muddy, or shiny, or something else entirely?  If it's "not muddy," then a long soak in water followed by a thorough scrubbing with a stiff brush should do the job.  If there is something else on the rock, then soak in a mild acid (vinegar, for instance) could remove it, given time.  This could also react with the rock itself if it is a carbonate, just so you're aware.  If instead you want them shiny, then that's an entirely different level of clean.

I do want them clean and shining. Most of them are crystalline in nature; I'm not sure of the composition though.

@DivyaG just a note for future hikes: You may have heard the phrase "take only pictures, leave only footprints". Leave No Trace should be practiced on future hikes so that others can enjoy the same experience you had.

Removing all of the debris from the stones without damaging them can be difficult, the following is what I've done in the past to clean geodes with debris and deposits.  Your mileage may vary, especially since we don't know exactly what is on your stones.

Soak the stones in soapy water for a day or so, then scrub them thoroughly with a stiff brush to clean dirt and most debris from them.  Be cautious if the stone has delicate structure that you're trying to preserve.  If some debris remains that appears to be washable, you can try scrubbing with a toothbrush and toothpaste to get rid of it.
If the stone is acid safe, soak it in vinegar for a couple of days to dissolve carbonate deposits.  You can check to see if it will react with vinegar by adding a few drops of vinegar to the stone's surface and waiting.  If it hasn't started bubbling within 10 minutes or so, it should be safe.
If the stone has rust deposits and is acid safe, you may be able to remove them with something like CLR Rust Stain remover.  If you need to go this route, skip the vinegar step, as this will also clean carbonate deposits.
Finally, it's possible that your crystal has something like granite attached.  If that is the case, nothing practical, short of cutting it off, is likely to clean the crystal.

Once you've got the crystal to the cleanliness you're happy with, you can polish it with powdered polish and a cloth, given time.
If instead of nice, angular crystals you're just trying to produce something like a smooth, rounded stone, after cleaning with water you can use a series of sandpapers (start at 60-80 grit and work up to ~400 grit) and polish to gradually smooth, then polish the stone by hand.  This will clear away any deposits, but it will also change the surface of the stone itself.  That might be acceptable to you if you've got something like a quartz river rock, and it will make a very nice end product.  Take care to keep the stone wet while working, you don't want to breathe in rock dust if you don't have to.

Much appreciated. I shall try these tips out soon!


text:
Finding root-mean-square speed of neon using kinetic energy

Find the root-mean-square speed of $\ce{Ne}$ atoms at the temperature at which their kinetic energy is $\pu{6.24 kJ mol-1}.$

I tried using the kinetic energy formula
$$\mathrm{KE} = \frac{mv^2}{2},$$
but I don't really understand how to achieve the necessary values.
I tried to do it by converting mass of one atom to mass of a mole since the given energy is per mole, but I'm still not getting the answer. I understand that I need to do $\displaystyle\sqrt{\frac{3RT}{M}},$ where $R= \pu{8.3145 J mol^-1 K^-1}$ and $M = 20,$ but how would I get the temperature?

Hi Carol, did you convert mass of one atom to mass of a mole since your energy is per mole? You might find this helpful :)   http://en.wikipedia.org/wiki/Root-mean-square_speed

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

@Carol I'd probably use $E_k=\frac{3}{2}NKT$ to get the temperature and then the formula you quote :)

As hinted in the comments by @AngusTheMan we use the equation
$$\mathrm{KE_{avg}} = \frac{3}{2}kT,$$
but ultimately we are looking for the $v_\mathrm{rms}$, so we'll also use
$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}.$$
You've been given your $\mathrm{KE_{avg}},$ $R$ and $k$ are constants, and $M$ is the mass of 1 mole of $\ce{Ne}$. We can set all of these equal and the solution will come from
$$v_\mathrm{rms} = \sqrt{\frac{3 \cdot R\cdot\mathrm{KE_{avg}}\cdot\frac{2}{3k}}{M}}.$$

In the light of Hint answers revisited, it would be nice if the answer would actually contain an answer and wouldn't end abruptly.

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


text:
How is the speed of sound dependent on pressure change in this formula?
In the following equation,
$$\text{speed of sound}=c=\sqrt{\frac{\gamma RT}{M}}$$
The pressure term is missing or I guess is integrated in the adiabatic constant $\gamma$. But I don't know how to even derive the adiabatic constant for a certain gas.

Can we re-open this question? Why was it closed, I thought it was a fair point no? :)

@AngusTheMan I fixed the title. It looks like it can be reopened but I don't know how to do that without disturbing a mod. (I lost my close/reopen vote privileges during site graduation)

@user1.618 I hope you don't mind me editing your question a little, as the site grows it is best to ensure that all the questions follow a specific format. So I am going to remove the picture and type the equation in mathjax for you :)

The best way to understand an equation in my view is to derive it from scratch ... along the way you pick up the details as to why you have certain conditions or restrictions that limit the usage of the equation. With this in mind I will outline an easy way to derive the equation in your book. Along the way you may think "how on earth would you first think about doing this derivation"! The answer is that it is a longer yet much more basic way of arriving at the same result as opposed to the "proper way.
If you wish me to describe it the proper way I will add it in at the end, its much shorter but possibly a little harder maths/concepts.
Okay here goes:
Longitudinal sound waves cause small oscillations in the distribution of the gas molecules (hard spheres) which propagate in the direction of the oscillations. These compressions (regions of higher pressure than usual) and rarefactions (regions of lower pressure than usual) effects the local density and pressure of the gas molecules in a given region of space with respect to their equilibrium values.
If we imagine that a cylinder of gas, which has a cross sectional area of $A$ and a length of $vdt$ where $v$ is the velocity of the gas molecules and $dt$ is the time they spend in the cylinder, has a sound wave propagating through it length ways. Where we have a compression of the molecules:
1) their pressure increases from equilibrium $P_0$ to $P$ by $dP$, where $P=P_0+dP$.
2) the local density increases from equilibrium $\rho _0$ to $\rho$ by $d\rho$, where $\rho=\rho _0 +d\rho$.
3) due to all the extra collisions of lots of molecules in a small space the average velocity of each molecule $v$ is less than it is at equilibrium $v_o$ by $dv$ where $dv$ represents a deceleration (so is negative).
This is the opposite of the rarefied regions which experience an acceleration, and occupy a localised region of lower $P$ and $\rho$ than equilibrium.
The extra force exerted on the molecules in the compressed zone is given by $F=dPA$. The volume of the cylinder $V$ is its length multiplied by its cross sectional area giving us $V=Avdt$.
The mass is just volume times density (equilibrium density), $m=\rho _0 Avdt$, notice that this part is independent of local changes since rarefactions/compressions manifest themselves by making $v$ larger/smaller and correspondingly $dt$ smaller/larger.
We can now substitute that into Newton's second law $F=ma$ where $a$ is the acceleration of the gas molecules. We can go one of two ways here either compression or rarefaction ... lets pick a compression. This means that we will have a deceleration in the velocity of the gas molecules. Therefore the acceleration is now given by:
\begin{equation}
a=-\frac{dv}{dt}
\end{equation}
Further we can substitute the mass into Newton's second law:
\begin{equation}
F=-\rho _0Avdt\frac{dv}{dt}
\end{equation}
But remember from the start we said that the force is also given by $F=AdP$.
\begin{equation}
AdP=-\rho _0Avdt\frac{dv}{dt}
\end{equation}
Cancel the $A$s and $dt$s and re-arrange:
\begin{equation}
-\frac{dP}{dv}=\rho _0v
\end{equation}
Multiply both sides by $v$,
\begin{equation}
-v\frac{dP}{dv}=\rho _0v^2
\end{equation}
Right lets put this result to one side and think about the change in volume. In fact we need to consider the fractional change in volume since volume is an extensive property. Therefore we will be interested not in $dV$ but in the quantity $dV/V_0$. We already know that the equilibrium volume is given by $V_0=Avdt$, the change in this volume under a compression is given by $dV=Advdt$. Therefore,
\begin{equation}
\frac{dV}{V_0}=\frac{Advdt}{Avdt}
=\frac{dv}{v}
\end{equation}
Substituting this result into the above equation:
\begin{equation}
-\frac{V_0}{dV}dP=-\frac{dP}{\bigg(\frac{dV}{V_0}\bigg)}=\rho_0v^2
\end{equation}
Now define the quantity in the middle as $the$ bulk modulus, $B$. It is the ratio of an infinitesimal pressure change to the relative change in volume, it gives an indication of the compressibility of a material.
Now rearrange this for the velocity:
\begin{equation}
v= \sqrt{\frac{B}{\rho _0}}
\end{equation}
At this point you have your equation in the most general form, the Newton-Laplace equation. From here you can pick what equation of state you wish to use and tailor this equation as needed. You can clearly see how the pressure and density information is tied up in this expression. But that being said its not so great for the experimentalists. Think about how you would take these readings and so on...
Therefore we are not quite done!
If we use Boyle's law $P_1V_1=P_2V_2$ as Newton did then you will find that although close to experimental values your always slightly off. Therefore we will use a different equation of state... one that inherently assumes constant entropy (some people say that adiabatic means that "no heat is transferred", this is the easy way to think about things .... what they really mean is that the process occurs at constant entropy .... an adiabatic process means constant entropy... (maybe if you study statistical mechanics and phase space this will be more apparent).
The equation we should use is the adiabatic version of Boyle's law. The reason that a sound wave travelling through a medium is considered adiabatic is because "no heat is exchanged between rarefactions and compressions" i.e the equilibrium temperature is maintained because rarefaction velocity change cancels out compressions velocity change on average meaning that the kinetic energy is pretty much constant and hence so is $T$ from $E_k=\frac{3}{2} NKT$. So instead of Boyle's law we use $PV^{\gamma}=P_0V_0^{\gamma}$. But for a given temperature the entire right hand side is just a constant so lets instead use $PV^{\gamma}=K$ where $K$ is the constant.
\begin{equation}
P=KV^{-\gamma}
\end{equation}
Now lets differentiate $P$ with respect to $V$.... lets find the rate of change oof pressure with respect to the change in volume...
\begin{equation}
\frac{dP}{dV}=-\gamma KV^{-\gamma-1}
\end{equation}
Multiply both sides by $-V$ and you will notice we have an expression for $B$,
\begin{equation}
-V\frac{dP}{dV}=KV\gamma V^{-\gamma-1}=VK\gamma V^{-\gamma}V^{-1}=\frac{K\gamma}{V^{\gamma}}
\end{equation}
Let $V=V_0$ the equilibrium volume which we are free to do. Now show the equilibrium volume in terms of the equilibrium density $\rho_0$.
\begin{equation}
\rho_0=\frac{nM}{V_0}
\end{equation}
If we now go back and put these in $v=\sqrt{\frac{B}{\rho_0}}$ and simplify then we have the equation stated in your book.
\begin{equation}
v=\sqrt{\frac{B}{\rho_0}}=\sqrt{\frac{K\gamma V_0}{V_0^{\gamma}nM}}
\end{equation}
Using $K/V_0^{\gamma}=P_0$ and then the ideal gas law $P_0V_0=nRT$,
\begin{equation}
v=\sqrt{\frac{\gamma P_0V_0}{nM}}=\sqrt{\frac{\gamma RT}{M}}
\end{equation}
So you see now how pressure is tied up in the entire expression, along with the density. The assumption we had to assume was that the sound wave travelling through is adiabatic and there we have the speed of the wave in a gas of non-interacting (except for collisions) hard spheres.
If you'd like I can derive the original formula in a much more professional way from elementary fluid mechanics ... but although quicker and more sophisticated it is a little harder. (if I've made any errors feel free to edit away) :)

Can you summarize what's at the link so the answer can be self-contained?  Thanks.

@jonsca I will do it tomorrow evening since its getting late here :)

No rush, when you have a chance.


text:
What splitting pattern is this?
I am trying to find the structure of an unknown molecule. What is the splitting pattern of these sets of lines?

I think my unknown molecule is butanol and the hydrogens this splitting pattern is referring to are the middle ones, i.e. $\ce{H3C-CH2-C\color{\red}{\mathbf{H}}_2-CH2-OH}$.
They would couple with the hydrogens on the right as a triplet and the ones on the left as a triplet, but would they also maybe do a W-coupling and couple with the methyl on the left.
So what would the coupling pattern be?

I encourage you to include the numeric positions of the peaks, the next time you have a similar question. It will make things easier to determine.

This is the a peak from one of the vinyl protons on an allyl group. This splitting pattern is an ABX pattern from the bold hydrogen below. It is technically a triplet of doublets of doublets.
$$\ce{R-CH2-C}\textbf{H}\ce{=CH2}$$
Check out the following simulated NMR spectrum of 3-chloro-1-propene (allyl chloride) from nmrdb.org

Now image if the coupling constants are a little different and the structure collapses into what you have.

cis and trans coupling constants in vinylic systems are usually different.  OTOH, the chemical shift is distinctly vinylic.

This is either a quartet of triplets or quartet of doublets of doublets, with both doublets of equal magnitude. It is most definitely not butanol. The pattern you see is from a double bond, and shows 3J coupling to a methyl group. I suspect you have another set of signals near this at about 5.5, and probably a methyl signal near 2ppm.
Generally, W-coupling (or M-coupling) is only observed in rigid systems, and is typically a 4J phenomenon, although can occur over longer bond orders if the bond overlap allows.


text:
Explanation of Bent's Rule
Why does the %s character increase on the orbital with electropositive atom?
I do not seem to properly grasp this concept. Is this rule based on observation or is there any proof to it? If there is a proof can you guys explain it to me in simple terms? I currently have knowledge of basic quantum mechanics and chemical bonding.

I'd suggest starting with http://chemistry.stackexchange.com/questions/15620/what-is-bents-rule?rq=1 and http://chemistry.stackexchange.com/questions/15671/utility-of-bents-rule-what-can-bents-rule-explain-that-other-qualitative-con?rq=1

@GeoffHutchison Sorry, I was on mobile and wasn't able to check whether this question has been asked before or not. Thanks for the links :)

@GeoffHutchison What is the protocol here? Should I delete this question or should I let it stay? Or is there an option to lock this thread?

@Shubham Yes, you can delete it. If you don't soon it will be on hold and then closed!!


text:
Why does F replace the axial bond in PCl5?
Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$? I realize that it would be more stable there than at equatorial bond, but what is the reason of its stability? Similarly in $\ce{AB4}$ type of molecules with $\ce{sp^3d}$ hybridization (4 bond pairs and 1 lone pair) why is the geometry of the molecule that of a see-saw, where the lone pair is at equatorial position rather than at axial one?
My book states the reason to be "due to Bent's rule" but I have difficulty linking these two.
Also Bent's rule states that the hybridized orbitals of equivalent energy of central atom tend to give more %s character to the electropositive atom attached to it rather than the electronegative one. (e.g. - in $\ce{CH3F}$) Am I right? What else is more to this rule?

Just as a very general remark on the topic: The involvement of d-orbitals is marginal (usually below 1%), hence hybridisation involving these orbitals is not very likely.

Im not sure if I get you. How involvement of d-orbital is marginal? Im aware that $\ce{PCl5}$ exists as $\ce{[PCl4]+ [PCl6]-}$ Does it have to do with this?

It is not what I meant. The hypothesis of a $\ce{sp^3d}$ orbital has been disproven. Hypercoordination can and should be explained different.

Ohh! I am aware that VBT (& MOT) is a basic theory and isn't fully applicable to all cases and many of its concepts have become obsolete but I have only these two in my curriculum so an answer within the context of these will be more helpful than using other complex theories.
Thanks for mentioning though! :)

Oh on the contrary, Valence Bond Theory and of course Molecular Orbital theory are not basic theories. They are the foundation of modern quantum chemistry and most molecules can be sufficiently explained by these theories. Just the assumption of the involvement of d orbitals is wrong. (I do unfortunately not have the time to write up an answer myself at the moment, but I am confident that someone else will do it.)

Lone pair behaves is a very big ligand, and chlorine is bigger than fluorine. Angles to neigbors in axial position are ~90, while in equatorial ~120, so larger ligands tend to occupy equatorial positions. 2) Also, axial bonds are actually ~ 1/2 order with more significant negative charge on ligands, so more electronegative atoms should go there. 3) +1 to martin, d-orbitals are not involved, see here http://chemistry.stackexchange.com/questions/15172/how-does-chlorine-form-more-than-1-bond/15181#15181

Recently, there has been a lot of discussion of Bent's rule (see for example "What is Bent's rule?") here in SE Chem. Simply stated, the rule suggests that $\mathrm{p}$-character tends to concentrate in orbitals directed at electronegative elements.

Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$?

In order to answer this question, we need to start by understanding the bonding in $\ce{PCl5}$ and its fluorinated isomers. In introductory courses and texts, it is usually stated that $\ce{AX5}$ type molecules adopt a trigonal bipyramid geometry and are $\mathrm{sp^{3}d}$ hybridized. As the comments by Martin and permeakra point out, and as is learned in more advanced classes, this hybridization scheme is likely incorrect. There are several reasons that argue against $\mathrm{sp^{3}d}$ hybridization including: 1) the fact that $\mathrm{d}$ orbitals are relatively high in energy compared to $\mathrm{s}$ and $\mathrm{p}$ orbitals and therefore it is energetically quite costly to involve them in bonding; and 2) $\mathrm{d}$ orbitals in non-metals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak.
A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved. Applying this concept to $\ce{PCl5}$ we would say that the central phosphorus atom is $\mathrm{sp^2}$ hybridized.  Thus, there would be 3 $\mathrm{sp^2}$ orbitals (these will be used to create the equatorial bonds) and a $\mathrm{p}$ orbital (this will be used to create our axial bonds) emanating from the central phosphorus atom.

The $\mathrm{p}$ orbital contains 2 electrons and will form bonds to 2 ligands (chlorine or fluorine in the case at hand). For simplicity, let's say that these ligands also use $\mathrm{p}$ orbitals for bonding (but it could be any type of orbital, $\mathrm{sp^3}$ or whatever is appropriate) and each of these orbitals contains one electron for bonding. This is our 3-center-4-electron bond and its MO diagram is pictured below. Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding M.O., so the bond order in this bond is reduced.

This reduced bond order in the bond using the phosphorus $\mathrm{p}$ orbital explains why the axial bonds in $\ce{AX5}$ type molecules are longer than the equatorial bonds.

Now that we understand the bonding in $\ce{PCl5}$ we can consider the case of $\ce{PCl4F}$ and how Bent's rule applies to the situation.  First, note that fluorine is more electronegative than chlorine.  As stated above, Bent's rule suggests that more electronegative ligands prefer to form bonds with orbitals that are high in $\mathrm{p}$-character. Why is this? $\mathrm{s}$-Orbitals are lower in energy than $\mathrm{p}$-orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more $\mathrm{s}$-character. The two electrons in the $\ce{P-F}$ bond will spend more time around the electronegative fluorine and less time around phosphorus. If that's the case (and it is), why "waste" precious, low-energy, $\mathrm{s}$-orbital character in an orbital that doesn't have much electron density to stabilize. Instead, save that $\mathrm{p}$-character for use in phosphorus hybrid orbitals that do have more electron density around phosphorus (like the $\ce{P-Cl}$ bonds). So, as a consequence of Bent's rule, we would expect phosphorus to use the orbital with lowest $\mathrm{s}$-character, the axial $\mathrm{p}$-orbital, to form the $\ce{P-F}$ bond; and the orbitals with more $\mathrm{s}$-character, the equatorial $\mathrm{sp^2}$ orbitals, to form $\ce{P-Cl}$ bonds.

May I just clarify? Is it always true that the filling of nonbonding MOs lead to a reduction in bond order? Is it also possible to quantify this reduction in bond order? I was thinking another explanation as to why there is a longer bond in the axial positions is due to the decreased effectiveness of the p-p-p overlap.

Because p orbitals are more diffuse as compared to the sp2-hybrids used in the bonds in the equatorial positions.

@TanYongBoon Is it always true that...? I think so, but feel free to ask that and your following question as a new question.  As to your argument about the decreased effectiveness of p-p-p overlap, I'm not sure. I think that a p-orbital is less diffuse then an sp2 orbital, but again that might be another good question to ask.

Another question I have is: It is not necessary for the two peripheral atoms involved in a 3-c 4-e bond to be the same right? In this case, the PCl4F would have a Cl-P-F 3-c 4-e bond.

@TanYongBoon Yes, that's correct.

How can one p orbital form bonds with two atoms?

@Abcd  Why not? Two large, equivalent lobes are exposed and available.  This is the basis for hypercoordinate bonding.  You've seen it before in the common example of the SN2 transition state.  The hydrogen 1s orbital can also form 2 bonds.  In that case it is a 3-center 2-electron bond (search that term on Chem SE for more detail).  Common examples include the bridging hydrogen in $\ce{B2H6}$ and in nonclassical carbocations.

@ron "The p orbital contains 2 electrons and will form bonds to 2 ligands (chlorine or fluorine in the case at hand). For simplicity, let's say that these ligands also use p orbitals for bonding (but it could be any type of orbital, sp3 or whatever is appropriate) and each of these orbitals contains one electron for bonding. This is our 3-center-4-electron bond and its MO diagram is pictured below. Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding MO, so the bond order in this bond is reduced."

I don't get 3c-4 e ...how will a p orbital form bonds with 2 atoms....doesn't p orbital only contain 2 electrons in it's lobe which we see participating  in resonating structures.Having more than 2 will cause it to become unstable,right...also how does  the 3c-4e has sp2 hybridisation

Also what about d orbital in pcl4f

This answer explains chlorine hybridisation in $\ce{ClF3}$...It will be $sp^2$ etc...

@HydrousCaperilla  It is not uncommon for a p-orbital to form bonds with 2 other atoms, the  $\ce{SN_2}$ transition state is another example.  And yes, a p-orbital can only contains 2 electrons.  D-orbitals only get involved when we are discussing bonding in transition metal compounds.


text:
Is there a hierarchy to functional groups?
I think I am getting slightly confused about the names we give to functional groups. I am aware that there are different types of functional groups such as alkyl, aryl and acyl groups.
I also know that there exist functional groups such as those of esters, ketones and amides which come under the acyl group. However, thinking about this recently I also realised that there existed a carbonyl group, which is also present in a ketone?
I am also a bit confused as to whether the acyl group's general formula is RCO or RCOX where X could be a variety of groups. Also, is R a part of the functional group or is it there to show what the functional group is attached to.
Could someone please make this hierarchy of functional groups a bit clearer?

What you may be confusing is the difference between a functional group and a moiety. A moiety is a part of the molecule that could be either the whole functional group or just a part of the structure of a functional group.
Wikipedia does a decent job of explaining this.

In this illustration there is an ester functional group and an acetyl moeity.
As for your second question about the acyl group, my understanding is that the formula is RCO- and that the X is not necessary to define the acyl group. The R represents an alkyl group.


text:
What is the reaction between a copper anode and a zinc cathode?
I prepared a battery with $\ce{Zn}$ screws and a $\ce{Cu}$ wire, with water as an electrolyte. The $\ce{Zn}$ screw acted as the cathode, and the $\ce{Cu}$ wire acted as the anode. I got nearly $0.6\:\mathrm{V}$ as the emf. I placed numerous amounts of such cells in an ice cube tray. Each ice cube acts as a cell and i got nearly $6\:\mathrm{V}$ of emf. What is the reaction happening here? What is the reaction involved with the electrodes and the electrolytes?

I don't think this goes beyond several cells in series.

Well the standard galvanic cell with $\ce{Zn}$ and $\ce{Cu}$ would usually look something like this, assuming there's a salt bridge.

and the reaction could be written as
\begin{align}
\ce{Cu^{2+}_(aq) +2e^{-} &<=> Cu^0} & E^\circ &=0.34\\
\ce{Zn^0 (s) &<=> Zn^{2+} (aq) +2e^-}& E^\circ &=+0.76\\
\ce{Cu^{2+}(aq) + Zn^0 (s) &<=> Cu^0 (s) + Zn^{2+}(aq)} & E^\circ &= +1.10
\end{align}
As for your measured emf, that can be calculated using the Nernst Equation.

I have improved the layout of your post. Please look here for more information.

authors: حمد كسار الرويلي


text:
How to determine the iron content in spinach with a spectrometer?
I used a spectrometer to find the absorbency of 5 solutions with different iron concentrations.
\begin{array}{rrrrr}
  \ce{Fe (III)} / \pu{mol L-1} &
  \text{Absorbance} &
  V(\ce{HCl}) &
  V(\ce{FeCl3}) &
  V(\ce{KSCN}) \\
  \hline
  0.00005 & 0.194 & \pu{9 mL} & \pu{1 mL} & \pu{10 mL} \\
  0.0001  & 0.424 & \pu{8 mL} & \pu{2 mL} & \pu{10 mL} \\
  0.00015 & 0.674 & \pu{7 mL} & \pu{3 mL} & \pu{10 mL} \\
  0.0002  & 0.89  & \pu{6 mL} & \pu{4 mL} & \pu{10 mL} \\
  0.00025 & 1.113 & \pu{5 mL} & \pu{5 mL} & \pu{10 mL}
\end{array}
With this data I've constructed a graph and a line of best fit.
\begin{align}
y &= 4608x - 0.0322 &
R^2 &= 0.9994
\end{align}
I took $\pu{5.10g}$ of spinach leafs and burned them, then added $\pu{10.00 mL}$ of $\ce{HCl}$ and $\pu{10 mL}$ of $\ce{KSCN}$ then filtered the contents (so ashes weren't in the beaker).
The absorbency of this spinach extract was $0.015$.
My question is: How do I calculate how much iron is in this sample? I need to determine the mass of iron per $\pu{100 g}$ of spinach.
Using the line of best fit question, solve for $V(\ce{Fe(III)}) = x$ concentration:
\begin{align}
0.015 &= 4608x - 0.0322\\
x &= 0.0000102431
\end{align}
Convert $\pu{mol/L}$ to $\pu{g/L}$ to $\pu{mg/L}$:
\begin{align}
0.0000102431 \times 55.845 &= 0.00057202591\\
\pu{0.00057202591 g} &= \pu{0.572 mg}
\end{align}
But this is clearly wrong because it's not taking account the amount of spinach I started with ($\pu{5.10 g}$). Anytime I try to include the amount, either by multiplying the density my answer is way off.
Is it reasonable that I'm getting a number above $\pu{10 mg}$? The iron amount reported by the USDA says that there is $\pu{2.7 mg}$ of iron per $\pu{100 g}$ of spinach.
Can anyone guide me in the correct direction?

I guess that was because you took the concentration of iron in spinach from the absorbance graph by extrapolating substantially. You did an absorbance graph for 0.194 - 1.113 whereas your absorbency for spinach is only 0.015. So there may be some potential for errors there.

Here's how I approached the problem.
Plot the Data
Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was successful.)

The parameter found was $a=4.43236~\mathrm{L\, mmol^{-1}} = 4.43236~\mathrm{mM^{-1}}$.
Calculate the Concentration of Fe Ions Present
You have measured an absorbance of 0.015 with a corresponding concentration of $c_\text{meas} = 3.38\cdot 10^{-3} ~ \mathrm{mmol\, L^{-1}}$.
The mass of iron in the solution is given by
$$ m_\ce{Fe} = c_\text{meas} \cdot M_\ce{Fe} \cdot V = 0.00378~\mathrm{mg} $$
with the volume $V=0.02~\mathrm{L}$ and molar mass $M_\text{Fe} = 55.845~\mathrm{g\, mol^{-1}}$.
Calculate Mass Percentage
The mass of iron in one gram of spinach is easily calculated:
$$ w_\text{Fe,spinach} = \frac{0.00378~\mathrm{mg}}{5.10~\mathrm{g}} = 7.4 \cdot 10^{-4} ~\mathrm{mg\, g^{-1}} $$
(I could just calculate away the units, but I was too lazy to do it here.)
The mass in $100~\mathrm{g}$ of spinach follows accordingly:
$$ m_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$


text:
Equilibrium constant and reaction Quotient
does reaction quotient After equilibrium has been achieved make any sense?
At equilibrium point rates of forward and backward reaction are same and the system is in dynamic equilibrium.The graph of concentration vs time shows a straight line parallel to x axis after equilibrium point.
Reaction quotient refers to ratio of concentration at any stage.This means that after equilibrium point reaction Quotient value will same as equilibrium constant as no more products are formed.So how can it have a value more than that of K ?!
I have been thinking about this for hours but I get more confused with time.

These two answers provide some insight into the relationship between $K$ and $Q$, but the main thrust is that at equilibrium $Q=K$. http://chemistry.stackexchange.com/questions/4410/k-eq-vs-k-c/4419#4419 and http://chemistry.stackexchange.com/questions/4977/le-chateliers-principle-how-can-the-equilibrium-shift-while-kc-remains-consta/4982#4982

$K$ is just the specific value of $Q$ for which $\Delta_\mathrm r G = 0$.
In that case:
$$\begin{align}
\Delta_\mathrm r G &= \Delta_\mathrm r G^\circ + RT \ln Q \\
0 &= \Delta_\mathrm r G^\circ + RT \ln K \\
\Delta_\mathrm r G^\circ &= -RT \ln K
\end{align}$$
For any system not at the equilibrium it is possible that $Q < K$ or $Q > K$. What is the "forward direction" is just a convention, it depends on how you decide to write the equilibrium. There is nothing special in one direction or the other. The system will "move" toward the direction that leads to equilibrium in both cases. In the first case the reaction will consume reactants to generate products, while in the second case the opposite will happen.
You can express this idea also in terms of the ratio $\frac Q K$. If it is > 1 then the reaction will consume reactants. This is also often referred to as the "Le Chatelier principle".


text:
Why has lithium bromide a slight solubility in organic solvents?
Why has lithium bromide a slight solubility in organic solvents?
Theoretical, LiBr is an ionic compounds, so it should be only soluble in polar solvent like water. But why is it soluble in others?

Lithium bromide is a salt of small cation and large anion. They cannot interact effectively, so the crystal lattice is quite easy to break. On the other hand, lithium cation is small, so it can interact effectively with small negatively charged atoms, like oxygen. Indeed, lithium cation has highest hydration energy of all alkaly metal cations. Altogether, it means, that in solvents with oxygen atoms (alcohol, esters, acetone) lithium cation effectively bounds to solvent molecules, leaving crystall lattice, and bromide has to follow, resulting in some solubility of lithium bromide in solvents with negatively charged oxigen and to lesser extent, nitrogen (for example pyridine). Lithium bromide is still insoluble in organic solvents that cannot interact with lithium cation, like benzene.


text:
How does bombarding a molecule with electrons cause ionization?
I am a chemistry student, and my current are of study is mass spectrometry.
In the ionization process, atoms are bombarded with electrons, but how does this work? I assume the flow of electrons attract the electrons of the atom, thereby removing them, but where do the atom's electrons go?

Electrons do not attract electrons, they would repel each other as they have like charge.

Imagine the analogy of you holding on to an object (let's say your groceries) then someone chucks stuff at you. You might accidentally drop the groceries. While this is obviously not an accurate analogy on an atomic level, it might help to visualise what is going on.

This type of ionization is known as electron ionization (there are many types of ionization methods used for MS). One possible reaction for EI is:
$$\ce{M +e- ->M^{•+} +2e-}$$ source
If the incoming electron has the correct energy to interact and passes close enough to the molecule, it's electric field can impart enough energy to one of the molecule's electrons for it to be ejected. What happens to the electrons after that isn't particularly important as far as ionization is concerned, but in most EI sources, electrons are emitted perpendicular to the axis of the sample source and are collected on the other side by an electrode, preventing charging of the source chamber and focusing the emitted electrons into a beam.

A lot of people call it "electron impact ionization" or "EI" instead of just "electron ionization".

Yes, that's the older term for it. IUPAC discourages its use now.

Interesting.  I don't understand the reason for half the stuff IUPAC does.  Why would they do that?

"Note: Electrons and photons do not "impact" molecules or atoms. They interact with them in ways that result in various electronic excitations including ionization. For this reason it is recommended that the terms 'electron impact' and 'photon impact' not be used."—from Recommendations for nomenclature and symbolism for mass spectroscopy

I see that IUPAC recommendation was from 1991.  I learned everything I know about mass spectrometry well after that, and can report that "electron impact" is still widely used.  I can kind of see the point of the change, but on the other hand, the same exact argument could be used to say that "The bat does not 'impact' the ball; it interacts with it in ways that result in various trajectories for the ball including home runs."

It's my experience that chemists ignore or simply aren't aware of a lot of IUPAC recommendations and I don't think it's a big deal in this case. But the difference, of course, in the two examples is that a classical collision model is a very good explanation for a macroscopic baseball but fails to correctly describe much of the behaviour of a photon or electron interacting with an atom.


text:
Si and Re facial descriptors
I am considering the hydrogenation of 2-butanone with respect to its Re and Si faces.
So if we look at 2-butanone as I've drawn below from above we are looking at its rectus face.
And from below we are looking at its sinister face.
Now, time to consider hydrogenation.
We'll keep the wedge and dash groups as they are and the carbon and oxygen both on lines. In other words we keep two points as is in the originally trigonal planar 2-butanone and when the hydrogen inserts itself, the other two points (a hydrogen and an alcohol group) pop up out the plane to form the classic tetrahedral shape of an $\ce{sp3}$ carbon atom.
So it appears that hydrogenation from the sinister face adds an hydrogen to the central carbon and forms an alcohol, and the alcohol is above the added hydrogen spatially, and this gives us the R isomer. Hydrogenation from the rectal face gives the opposite S isomer.
How's the reasoning? Am I right? I am fairly sure. But not completely sure.

I really do not understand what you are asking about. Do you just want to hear a Yay or Nay?

I need constructive criticism and a yay or nay would be nice too.

When looking at the corresponding Wikipedia article for prochirality you will find out, that your reasoning is correct for the hydrogenation. Of course if you have a different nucleophile, then the situation would change, as the Cahn-Ingold-Prelog identifiers would change.

See also prochirality in the IUPAC goldbook.

The Re/Si enantiotopic faces of 2-butanone are independent of the nucleophile. As your first link notes, the priority of the 3 CIP groups are arranged clockwise. (I point my thumb in the direction of an attacking nucleophile and follow my fingers from highest to lowest priority.) It is true that R/S for the product will be nucleophile dependent, i.e., hydrogen vs. sulfur. Your last sentence is a bit confusing.


text:
Does a timed reaction "ready indicator" paper exist?
I'm trying to see if such a thing exists in paper form that upon activation,  changes colour or "indicates" after a set configurable amount of time.
In other words a litmus or pH paper that can act as an egg timer. I'm by no means a chemist so please excuse my lack of protocol.

I'm not aware of any commercially available examples, but I know that the speed of a fluid wicking through a paper microfluidic device has been used for timing. I'm also reminded of pencil detonators, dating back to WWII, where the time taken to dissolve a lead wire is used for timing. Since electronic timing is now so cheap, easy, accurate, and reliable, I'm not surprised that chemical timers have pretty much died out. There are a number of chemical phenomena that could be exploited for timing, but unless you need a huge number of cheap, one time use timers with relatively poor accuracy, there are probably better choices.


text:
Mechanics behind auto-ignition temperature
I have some questions about the auto-ignition temperature for fuels. Per ASTM E659:

AIT (auto-ignition temperature) is the lowest temperature at which the substance will produce hot-flame ignition in air at atmospheric pressure without the aid of an external energy source such as spark or flame. It is the lowest temperature to which a combustible mixture must be raised so that the rate of heat evolved by the exothermic oxidation reaction will overbalance the rate at which heat is lost to the surroundings and cause ignition.

The last sentence seems to say that AIT is the temperature where enough of the particles are above the activation energy such that enough heat released goes to perpetuate the remaining fuel to combust (cause ignition). Is this correct?
Per ASTM E659, ignition is subjectively defined as the appearance of a flame with a sharp rise in temperature. Why it necessary that we observe the appearance of a flame (in other words, can we have combustion without an observed flame)?
Is there an assumption of thermal equilibrium between fuel and oxidizer when defining AIT? The testing method above says air at predetermined temperature will be used, but I'm not sure how they plan to control air temperature when heating begins.

in other words, can we have combustion without an observed flame? Yes, the embers of a dying fire.

To be clear, I'm no expert on the ASTM standards writers' thinking, and all I have to go on is the standard you've posted.

Yes, that's exactly right.  Some of the reaction's heat will be lost to the surroundings, but enough of it will go into the reaction mixture to raise its temperature and promote further, more rapid, combustion.
There are reactions that produce lots of heat but no flame, hydrogen peroxide decompostion, for example.  The standards writers had to come up with a clear definition for ignition, and as you note, they did: "the appearance of a flame accompanied by a sharp rise in the temperature of the gas mixture."  They even say that ignition is a subjective term, so they are careful to define it clearly.  There might be some chemical reaction that produces a large amount of heat in air with no visible flame (even in the dark, as the standard requires) and it might even qualify as an ignition event to you or I or any other sensible person watching, but it would not fit the standard.  They have to set some limit, after all, and that was what they set.
My original thinking on this was flawed; I misunderstood the air temperature reference.  The reason that they mention the air at all is that they are being specific that the reaction mixture should be an air/fuel mixture, rather than something else (oxygen/fuel, for instance.)  The predetermined temperature aspect of that point has to do with the AIT determination procedure itself.  They are setting the temperature of the system to some value T, then adding the fuel, then waiting for ignition.  So yes, the air in the flask is more or less in thermal equilibrium with the substance being tested.  It can, however, leak out and exchange with the air outside the flask. If the substance does not autoignite, the flask is purged and the temperature increased (including the air in the flask.)  I'm not sure whether they want a completely new sample of the substance for each temperature measurement.  It seems likely, since we're dealing with small amounts of generally volatile compounds.

Is surroundings basically outside the flask? If the system was insulated, is AIT basically a tipping point where enough reactants are above the activation energy to provide enough heat to perpetuate the reaction (like a chain reaction)? 3. In 4.1, they mention air at predetermined temperature, which is the only mention I found. It seems strange to me that if both oxidizer and fuel is heated together that they may be at different temperatures?

The surroundings would be the air in the room.  They're heating the flask and its contents (including some air) but there is free exchange of air from the room and air/vapor from the flask.  The chain reaction idea is exactly what AIT is, the point at which the chain reaction starts.  I'll amend my #3 above to comment properly on the air temperature question.

I have some followup questions here that I would really appreciate some help on.


text:
Cyanide ion non-bonding/lone pair?
Where's the lone pair on the cyanide anion in this MO diagram?

In the traditional Lewis picture there are two lone pairs on the cyanide ion. In the MO diagram though there appear to be no non-bonding orbitals that are populated ...
Also how come there appear to be two sigma bonding orbitals populated when in $\ce{NC^-}$ there is only one sigma bond?

The connection between an MO view of bonding and a Lewis picture is tenuous at best. Case in point.

Where's the case in point? Is it the F-F picture from the other week? Also if NC- has no non-bonding pairs how does it "donate" its lone pair? I guess I should probably start looking at HOMO/LUMO instead?

No, $\ce{CN-}$ is the case in point. From an MO picture, there is no non-bonding "lone pair." There's nothing to donate. Coordination between a metal and $\ce{CN-}$ in MO-world and Lewis structure world look very different. The end-result of MO and Valence Bond may be similar, but the interpretations are different.

As Geoff already pointed out in the comments, the connection between a molecular orbital theory (MO) point of view and a Lewis picture (LT) is tenuous at best.
Maybe the best example for this is the mentioned $\ce{{}^{-}CN}$ and obviously the isoelectronic $\ce{CO}$. But there are many more molecules.
In the MO point of view the electron pair which can be donated to an acceptor orbital is of course the highest occupied molecular orbital (HOMO). Often this comes with the more trivial notation of "lone pair", based on the coefficients of a certain element in that orbital (percentage contribution). In the case of $\ce{CO}$ this becomes fairly obvious, but since this is not the question at hand I will only focus on the isoelectronic counterpart.

Please consider the above MO scheme only as a sketch, it is not to scale and contributions were estimated only be a quick browse of the output file. I also did not compensate the calculation for the negative charge, so even the bonding orbitals are antibonding with respect to fragmentation. But it shall still provide the point.
We can clearly see that the highest coefficient of the HOMO is at the carbon end, this is consistent with the fact, that it will most likely bond with this end to a more electropositive element. But this is, however, not a real lone pair since it is $\sigma$ bonding and also has quite a big coefficient at the nitrogen end. (In $\ce{CO}$ the difference is much bigger.)
Below that we see the two $\pi$ orbitals, followed again by two $\sigma$ orbitals. All five orbitals make up the bonding of the molecules and determine also its reactivity.
Now in the Lewis theory picture this is quite different. We traditionally describe it as a carbon oxygen $\sigma$ bond, two carbon oxygen $\pi$ bonds, a oxygen lone pair and a carbon lone pair. All made up from some kind of hybrid orbitals. In computational chemistry we can analyse the wave function and obtain localised orbitals. One of these methods is natural bond orbital theory (NBO). I have carried that out and sketched the common bonding picture along with it:

While this picture is also consistent with a valence bond theory (VB) approach, it explains bonding in a different way than MO theory. However, the conclusions we will ultimately get are essentially the same.
It shall be noted, that MO theory and VB theory are complementary theories, but the different explanations should not be mixed, even when the conclusions are essentially the same.

Why would the HOMO of CN- have higher coefficient of carbon? Doesn't the energy of the HOMO more close to nitrogen?

@Wang I don't really understand the question. The ordering of these orbitals comes from quantum chemical calculation. It is quite old, but should still hold. I've also written about this feature for CO. https://chemistry.stackexchange.com/q/51262/4945

Thank you for replying!
My question was, it seems to me that on the diagram you drew above, the HOMO ( σ 1.0 eV right?) is more close to the 2p orbitals of nitrogen, wouldn't it be more polarized to nitrogen?

@wang it's more involved than that. The diagram isn't to scale, and the 2p orbitals are not the only ones contributing. It's a fact though that the orbital is polarised towards carbon. That is what the image already shows.

Thank you! I missed the dot lines linking to 2s orbitals of both atoms.