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Obtain chemical insight from neural network results
I have recently read many papers where neural networks (NNs) are trained to predict chemical properties (starting from the structure of small chemical compounds) for compounds rather close to the ones with which they were trained. In some cases the predictions are rather good. But I see that apart from the predictive capability, people use NNs as "black boxes" in the sense that they do not tell you or give you chemical insight about why that compound has that value for that predicted property. At least that is my impression now. So my question here is, do you know cases where useful chemical information can be extracted from NN predictions?

What do you consider "useful"? Usually the prediction is the interesting bit. The ANN is an abstraction, i.e., it contains/is the chemical information gleaned from the training set.

That is not true. Let's say a "perfect" ANN that could predict whether a certain compound will have some value for some property. But, can this ANN explain why?

Newtonian mechanics explains nothing. They are merely a set of rules that predict outcomes. The "why" is in the determinism of those rules. I have an answer for you below, but I fear you really have to state clearly what kind of information you want to take away and at what point you are willing to stop asking why.

I think this is sort of interesting, but it's bordering on being off-topic.  I think you can ask a question like "How does my ANN tell pictures of roses and tulips apart?" and the answer would be similar.  I'll leave it for now, but if you could expand on how the chemistry ANNs would be different, that would be great.

NN and other qsar methods usually works on top of some kind of 'descriptors'. It is possible to find relevant descriptors with NN and then track, what produced them from molecule. The area is not well-developed to my knowledge.

You mean, with NN you get the relevant descriptors, and then use them again with QSAR methods to find some well defined correlation, and from that you can have more insight about why such compounds are active, do not you mean?

Besides, why you "permeakra" compare directly NN and QSAR? I think that direct comparison can not be made.

@flow I mean, with relevant descriptor at hand it is possible to see what structural feature produced said descriptor. NN is one of kinds of machine learning, that can be used in QSAR. It is not that terrifically special, and some math people in the area say that NN are inherently inferior and hard to work with.

I did some work on interpreting neural network QSAR models - I won't claim that they explain everything and has a number of limitations (linearizes network connections, applicable only to feed forward networks). But maybe it will be useful
Interpreting Computational Neural Network QSAR Models:  A Measure of Descriptor Importance
Interpreting Computational Neural Network Quantitative Structure−Activity Relationship Models:  A Detailed Interpretation of the Weights and Biases
Of course, if the descriptors going into the network are opaque (as in abstract mathematical properties such as many topological descriptors) then I don't think any interpretation method is going to be very helpful.

Since you are interested in Artificial Neural Networks that also explain their decisions/predictions you may be interested in transparent neural networks, although they have not been used to my knowledge on chemical information.
Other interesting approaches include inductive logic programming which have among other things been applied to predicting binding of proteins to hexose

Whilst relevant to neural networking in general, this answer has nothing to do with applications of neural networking to chemistry, per se.

Deathbreat's answer is good. He does not provide examples but gives me insights; I did not know about ILP. So I read about it, and tried to find papers related to ILP and my question so I have found this (http://www.biomedcentral.com/1471-2105/13/162), which is directly related to my question. You should not be that strict.

@RichardTerret: The question does not require specificity to chemistry. I'll incorporate the paper flow found in order to make the chemistry link.


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Interpreting equilibrium equations correctly
I've watched over this Khan video titled "Reactions in Equilibrium" and am still confused on the interpretation of equilibrium. Reactions in Equilirbium
$\ce{N2(g) + 3H2(g) <=>2NH3(g)}$
Whenever I see an equilibrium like the one above, the amounts of each molecule are only what happens at the start of the reaction, correct? In the video, it was made out as if the starting coefficients next to each molecule were what was needed to reach equilibrium. However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of $\ce{NH3}$ would not necessarily be 2. He does this in other videos too. He starts off with something such as:
$\ce{aA + bB <=> cC + dD}$
and immediately plugs those into an equilibrium equation. I guess what he is getting at is that we figure out the final concentrations necessary to reach equilibrium but I'm not positive.
Towards the end of the video, arbitrary (I guess) concentration amounts of each molecule were plugged into the formula to output the matching equilibrium constant. He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3 H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?
Finally, let's say the concentrations I have for a particular reaction were not the concentrations at equilibrium. If I still plugged them into the equation, that would tell me not only if I were at equilibrium (if I looked up the constant) but also if the forward or backward reactions were favored right?

Hi Greg, and welcome to Chem.se. We have a chemistry formatting plugin that is much easier to use than what you were trying to do. If you're interested, you can click on the edit button to see how it works.

However, each reaction at a different temperature has a different equilibrium constant. With that said, the amount of each molecule will change which also changes the concentration so the amount of NH3 would not necessarily be 2.

This is true. I have not watched the video, but if he is suggesting that the stoichiometric coefficients define the ratios of the equilibrium concentrations of the compounds, then he is wrong. Stoichiometric coefficients do play an important role in equilibrium expressions, but not that one.

He raised each concentration to the power of the starting coefficient for that molecule. If you start out with $\ce{3H2}$ and you need a wildly different concentration to reach equilibrium, why are we raising the concentration to the power of the starting amount of $\ce{H2}$?

This is how the equilibrium constant (and more broadly the reaction quotient) is defined.
Take your generic reaction:
$$\ce{aA + bB <=> cC + dD}$$
The reaction quotient $Q$ is defined as the product of the concentrations of the products divided by the product of the concentrations of the reactants. Since $\ce{a}$ equivalents of $\ce{A}$ are in the reactants, $[\ce{A}]$ is multiplied $\ce{a}$ times for $[\ce{A}]^{\ce{a}}$.
$$Q_c = \frac{\prod\lbrace[\text{Products}]_i\rbrace}{\prod\lbrace[\text{Reactants}]_i\rbrace} = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}} $$
The pressure version of $Q$ replaces concentrations with partial pressures $P_i$.
$$Q_p = \frac{\prod\lbrace (P_{\text{products}})_i\rbrace}{\prod\lbrace( P_{\text{reactants}})_i\rbrace} = \frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$
The equilibrium constant ($K_c$ for concentrations and $K_p$ for pressures) is defined as the reaction quotient at equilibrium concentrations/pressures. At equilibrium:
$$Q_c = K_c = \frac{[\ce{C}]^{\ce{c}}[\ce{D}]^{\ce{d}}}{[\ce{A}]^{\ce{a}}[\ce{B}]^{\ce{b}}}$$
$$Q_p = K_p =\frac{P_{\ce{C}}^{\ce{c}} P_{\ce{D}}^{\ce{d}}}{P_{\ce{A}}^{\ce{a}} P_{\ce{B}}^{\ce{b}}}$$
The values of $K_c$ and $K_p$ are determined by experimentally measuring the concentrations or pressures of $\ce{A}$, $\ce{B}$, $\ce{C}$, and $\ce{D}$ and plugging them into the expressions above.
For the specific reaction of nitrogen and hydrogen to form ammonia, the equilibrium constant (at a given temperature) is determined by measuring the equilibrium concentrations/pressures and plugging them into the expressions below.
$$\ce{3H2(g) + 2N2(g) <=> 2NH3 (g)}$$
$$K_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2}$$
$$K_p = \frac{P_{\ce{NH3}}^2}{P_{\ce{H2}}^3 P_{\ce{N2}}^2}$$
Once we know the value of $K_c$ or $K_p$, we can then use it and some partial concentration/pressure data to predict the equilibrium concentrations/pressures.
At 300 K, the $K_c$ of this process (the Haber Reaction) is $4.34\times10^{-3} \text{ M}^{-3}$. If we started with $[\ce{H2}]=3.0 \text{ M}$, $[\ce{N2}]=2.0 \text{ M}$, and $[\ce{NH3}]=2.0 \text{ M}$, we can calculate the equilibrium concentrations of all three species when the dust has settled. First we calculate $Q_c$ and compare to $K_c$. If $Q_c$ is greater than $K_c$, there is too much product, and the reaction will shift toward reactants. If $Q_c$ is less than $K_c$, there is too much reactant, and the reaction will shift toward products.
$$Q_c = \frac{[\ce{NH3}]^3}{[\ce{H2}]^3 [\ce{N2}]^2} = \frac{(2.0 \text{ M})^2}{(3.0 \text{ M})^3 (2.0 \text{ M})^2}= \frac{1}{27 \text{ M}^3}=3.70\times10^{-2} \text{ M}^{-3}$$
$$ 3.70\times10^{-2} > 4.34\times10^{-3} \space \therefore \space Q_c > K_c  $$
At this combination, the reaction shifts toward reactants. $[\ce{H2}]$ and $[\ce{N2}]$ increase, and $[\ce{NH3}]$ decreases, all by a factor of $x$ multiplied by the stoichiometric coefficients of each species. The new concentrations are $[\ce{H2}]=3.0+3x \text{ M}$, $[\ce{N2}]=2.0+2x \text{ M}$, and $[\ce{NH3}]=2.0-2x \text{ M}$. The equilibrium expression is:
$$K_c 4.34\times 10^{-3} = \frac{(2-2x)^2}{(3+3x)^3 (2+2x)^2}$$
Since I chose the starting concentrations to be those values implied by the stoichiometry, I ended up (coincidentally) with a reaction near equilibrium. Thus, I cannot make assumptions about the value of $x$ relative to the concentrations, so no simplification is possible. To avoid mistakes, I will let WolframAlpha handle the algebra and solve for $x=3.18\times 10^{-1}$. The equilibrium concentrations are thus:
$$[\ce{H2}]_{eq} = 3+3(3.18\times 10^{-1})=3.95 \text{ M}$$
$$[\ce{N2}]_{eq}= 2+2(3.18\times 10^{-1})=2.64 \text{ M}$$
$$[\ce{NH3}]_{eq} = 2-2(3.18\times 10^{-1})=1.36 \text{ M}$$
Note that if we had started at different initial concentrations, we would end with different equilibrium concentrations. The maths are easier if you start with zero of one or two species.
I could also go on about the other way to determine the equilibrium constant (i.e. from the relationship $\Delta G^o=-RT\ln{K}$), but I have gone on long enough.

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Which chemical properties make a substance explosive?
I know that some chemical substances are used as explosives. Amongst the most famous nitroglycerin in dynamite and cyclonite in C-4. Which chemical properties (I suspect that physical properties have less of an effect) make a substance a good explosive? If we know, for example, the formula and its geometry, is it easy to predict how explosive it can be?

@TylerBailey - Those kinds of reactions are certainly dangerous, but the salient feature of an explosion is a rapid expansion of volume, usually by the rapid production of many equivalents of gas.

I think it's interesting to cite silver acetylide exemplifying  an explosive where none the products are gaseous.

There are, actually, several types of explosives by use: initiating (primary) explosives, secondary explosives and blasting agents.
Initiating explosive must be sensitive to shock/friction, but not too sensitive. The most famous primary explosive is mercury fulminate, but others are known. To act as primary explosive, substance must have hi-energetic unstable groups, like peroxides, azides and similar.
Secondary explosives are substances, that are not sensitive to shock/friction and requires initiating blast to explode. Unlike primary explosives, they can be manufactured as hi-density blocks with high speed of detonation. It is exactly what is needed for military applications. The most common type of secondary explosives are (substituted) polynitroarenes. Most well-known examples are TNT and picric acid. It is very beneficial to have zero oxygen balance for such explosives, i.e. to explode without extra oxygen produced and without extra reducing gases produced. TNT, for example, has negative oxygen balance, producing much carbon on explosion, and dinitromethane has positive oxygen balance, producing extra oxygen on explosion. A less stable class of secondary explosives is nitroamines, with hexo/octo-gen as most well-known example. Anyway, main variables here are relative stability and high density paired with medium-high energy of explosion.
Tertiary explosives or blasting agents are explosives, used to move hight amounts of materials. They are usually used in big amounts in mining/engineering and must be cost-effective and produce much gases in explosion. The most common typeы of blasting agents are oxyliquits and compositions, based on ammonia nitrate.
There are also special kinds of explosives/pyrotechnic mixtures. For example, a mixture used in car airbags must produce much gases very quickly, but said gases must have low temperature.

If we know, for example, the formula and its geometry - is it easy to predict how explosive it can be?

Actually, not that easy. Di/Trinitroarenes are usually explosive, but their stability and performance may vary greatly and depends on crystal structure, density and form (crystals/fine dust/puffy things).

I'm going to assume by 'good explosive' you mean 'high energy to mass ratio', which by no means implies anything about whether the explosive is 'good' in the 'won't suddenly explode when thought about' sense.
This is not intended to be a full answer as it doesn't touch on the terrifying explosivity of organic peroxides, however it's a start. The presence of nitro ($\ce{R-NO2}$) or azide ($\ce{R-N=N=N}$) groups in a molecule is strongly suggestive of explosivity. This has a lot to do with the large negative free energy change of formation of dinitrogen and (I guess) oxides of nitrogen upon decomposition of these molecules. $\ce{N2}$ has one of the strongest bonds known and releasing a lot of $\ce{N2_{(g)}}$ entails a large increase in entropy. Inorganic nitrates and azides are also likely to be explosive.
The state of the art in explosive research seems to involve cramming as many consecutive nitrogen atoms as possible into a molecule, as showcased on Derek Lowe's blog:
$\ce{C4H4N8}$
$\ce{C2N14}$ (what?!)
Alternatively, you can take a highly strained polyhedrane of some description and stick nitro groups all over it. The ur example of this sort of philosophy is octanitrocubane.
Bonus: tetranitrotetrahedrane
Disclaimer: Explosives are deadly and people who prepare them are either highly trained experts with exquisite safety infrastructure, or morons. Nothing in my answer should be taken by any person as an endorsement to attempt to synthesise anything described.

Your link to C2N14 is part of Lowe's "Things I won't work with" series, which includes some great links to explosives research.

I would add that explosive reactions are extremely rapid reactions.  It is hard to judge rapidity of a reaction just by knowing the structure of a molecule.
A given substance can qualify as an explosive under special conditions.  For example we cook with methane gas.  It burns nicely and gives us no problems.  But mix it with the proper amount of oxygen and toss in a match and there goes the neighborhood.
The proper amount is often a range of concentrations, wide or narrow depending on the molecules involved.

It is extremely difficult to look at a molecule and know whether it is explosive.
But, generally speaking, highly exothermic chemical reactions are the "explosive" ones.
A --> B + C
What makes the reaction exothermic?  Either A is very unstable, or B and C are very stable.  Any molecule that has lots of very stable B's and C's is going to be very explosive. Examples of very stable B and C could be carbon soot, water, and also NO2.  (Actually there are lots of them)
The trick is to keep A stable enough (metastable) to use it for some practical purpose. So, this somewhat limits how "unstable" A can be.
Said another way, I would design an explosive A to be an effective delivery mechanism of B's and C's just waiting to be formed.

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Is there a point at which Ethanol (E10) fuel becomes harmful to gas tanks or engines if not used?
This is not the typical chemistry question on this website, but I think it's an important practical question.
When I got a gas string trimmer, the woman who worked at the gardening shop told me that E10 (fuel commonly sold in the USA - 10% Ethanol, 90% gas) left sitting in a gas tank for months would be harmful. So I wonder, is this true? And if so, what about for cars and other engines (boats, tractors, etc).
This effects my life because I was considering at some point getting a plug-in hybrid car because during most weeks I would never have to use or buy fuel. However, could I be damaging this car by leaving the same fuel inside the tank for months?
There are various claims about ethanol either helping or hurting things, but I have not seen any supported by science. What is the chemical basis of these claims? A layman translation is appreciated.

It looks like the engineers who designed the Chevy Volt have worked out a solution for this problem:
http://www.plugincars.com/putting-thought-putting-gas-chevy-volt-69473.html

Another informative link: skip to 1:33
http://www.youtube.com/watch?feature=player_embedded&v=uee-HQ9UY4M

Well first off, pure ethanol is hygroscopic; it attracts water, to the point that it will pull it out of the air. Ethanol and gasoline will mix, but ethanol, gasoline and water will not; the ethanol-water mixture will come out of solution and settle on the bottom of your tank. Add a little oxygen to the mix, and you get rust. However, the more common side effect of this is more immediate; turn the car on, and the fuel pump will draw the water from the bottom of the tank into the engine, where it will promptly kill it, and require a costly dry-out process.
Down here in the south, where it's hot and humid most of the year, you hear a news story or two about this every year, usually when a gas station didn't properly purge its storage tank on a regular basis. The regular addition of thousands of gallons of fresh ethanol fuel "recharges" this hygroscopic process, causing a rather large pool of water to form at the bottom of the tank which the station is supposed to drain regularly. If they don't, and you fill up when the store's tank is just a little too low, or just a little too soon after the tanker's refilled it, the pump will draw a water-gas emulsion, or even straight water, into your tank, and the water will settle to the bottom.
The usual fix for long-term storage of ethanol fuel is to add a stabilizer, which will contain a high concentration of isopropanol. The main thing that will do is form an azeotropic mixture with water, meaning that its components will evaporate at the same rate, so when the isopropanol evaporates, it takes the water with it. This same mixture, in sufficient concentration, will also burn, so while it's not the greatest thing for your engine it can at least be flushed out the normal way.
Second, gasoline, as you probably well know, isn't a pure substance; it was originally what was left over after the refinery had extracted the "useful" heavier compounds (kerosene, diesel fuel, lubricating oil, paraffin waxes) out of crude oil by fractional distillation. Nowadays it's much more carefully purpose-made by thermal decomposition of heavier alkanes, but it's still an amalgam of relatively lightweight carbon compounds (usually between 6-12 carbons, with the dividing line for "octane rating" drawn between heptane and octane), plus additives and pollutants. It contains three components of interest; sulfur (a naturally occurring pollutant and difficult to remove completely; "summer blend" gas in the U.S. is more expensive because they have to get rid of more of it to prevent smog and acid rain), oxygen (modern fuels are oxygenated to make them more clean-burning), and ethanol (added to make the fuel burn cleaner and be more sustainable to produce than neat gasoline).
In the presence of water (pulled in by the ethanol) and oxygen, sulfur naturally forms sulfuric acid by a variation of the "contact process"; it oxidizes easily to sulfur dioxide, then more slowly to sulfur trioxide, which when dissolved in water becomes sulfuric acid. This reaction, specifically the formation of sulfur trioxide, is sped up in industrial processes by heat and by contact with catalyst metals such as vanadium oxide (hence "contact process"), but it will also happen on its own without any special treatment (i.e. acid rain).
This sulfuric acid will do a lot of corrosion damage, reacting with the tank's sheet metal producing hydrates of iron sulfate. When burned, those sulfates decompose into iron oxides and sulfur dioxide again, which do more damage to your exhaust system on the way out (rust is a catalyst for more rust, and we've discussed sulfur dioxide's contribution to the party in detail). In addition, sulfuric acid is even more hygroscopic than ethanol; at concentrations in water as low as 10%, it'll attract more water. Lastly, acids and alcohols form esters, in our case ethyl sulfate, and esters and alcohols form ethers, here diethyl ether (with the sulfuric acid and water as a byproduct). These are reversible and ultimately circular reactions, which will form an equilibrium of roughly equal concentrations of the intermediates. A dehydration reaction to produce ethylene is also possible in very high acid-to-water concentrations of sulfuric acid (possible when filling up in the winter, when it's dry and the gas has more sulfur). Ethylene will form a number of intermediate products with just about anything in that gas tank, some better for your engine than others (ethylbenzene is an antiknock compound; methyl ethyl ketone is a solvent and a pollutant byproduct of combustion engines).

Do you have any thoughts on how long this process takes after getting a "fresh" tank of gasoline? Is this process made worse by higher ethanol mixtures? So should E85 be even worse for engines and tanks than E10?

Given that E10 has been quite common in California and other states for years now, I'm surprised I haven't heard about it causing even more problems. I also wonder if car manufacturers have figured out ways to counter-act these problems.

This is a great answer and a fascinating read.

Well, water in the tank has been a rather common problem around me, especially at cheaper stations. Usually though, it comes from actually pulling water (or a suspension of it, churned up by the filling tanker) from the gas station tank, where this hygroscopic change happens on a scale several thousand times that of your own fuel tank. Your better stations avoid problems by routinely draining the tank all the way, water and all, during the switch from summer to winter blend and back.

The part about the isopropanol azeotrope is nonsense. (Yes of course water would form an azeotrope with it, also with ethanol, and even with the hydrocarbon. But that's totally irrelevant, because nothing evaporates here.) Isoprop is simply a compatibiliser between water/ethanol and hydrocarbons.

And modern car fuel is essentially sulfur-free.

Actually this paper explains the water ethanol gasoline mixture effects very well. If the ethanol will harm the equipment is more dependent on how the machine was designed. The ethanol can deteriorate carburetor parts the gasoline ethanol comes into contact with if the manufacturer hasn't replaced them yet.  Rust is not likely from humid air. http://epa.gov/OMS/regs/fuels/rfg/waterphs.pdf

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.

Your link became invalid. This is the reason we recommend including the relevant sections of the linked paper to the answer. Luckily, the paper is still available at the internet archive: https://web.archive.org/web/20130427123941/http://www.epa.gov/OMS/regs/fuels/rfg/waterphs.pdf

Methanol additions to gasoline in freezing climates may form a second phase in long storage. It will corrode the heck out of magnesium which is used in small portable engines.
EPA sulfur restrictions now require most refinery streams to be put through hydro-treaters (high pressure H = costly) to remove sulfur. Nothing to do with summer/winter fuel.
The E10 will not separate a second phase unless exposed to the atmosphere to pick water. So not a problem in a closed tank. I have a portable generator that has sat for as long as a year with no problem.
In 35 years of refinery corrosion experience, I never heard of a service station tank being drained except for replacement.

Hello  and welcome to StackExchange. Have a look at the [help] to learn how this site works and what type of questions are allowed here. It's good to see you contributing to the site o  your first post. Keep it up.


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Are chiral quasicrystals possible?
Quasicrystals are materials that have long-range atomic order but lack the translational symmetry of conventional crystals. All quasicrystalline tilings and packings I have read about thus far, whilst lacking translational symmetry, have rotation axes and reflection planes. I'm wondering if anyone is aware of experimental evidence of quasicrystals which lack reflection symmetry (and thus possess implicit chirality), or can provide a theoretical rationale for why such structures can or cannot exist.
To clarify, a cursory google search turns up a paper interpreting a viral capsid as having chiral quasicrystalline order - Whilst I don't have access to the paper, I specifically exclude this sort of intepretation from consideration as I'm talking about structures with potentially infinite spatial extent, which viruses do not have.

I think this article worth reading http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=6458107 Autors claimed experimental results on producing 3d chirial photonic quasicrystal: "So the fabricated class of 3D PQC chiral lattices integrate both
these wonder features on asingle platform in view of their
possible applications in new generation integrated optical
photonic devices with tunable polarization band gaps as well as frequency band gaps"

@Sigrlami - The paper you link seems to answer the question satisfactorily - if you were to make your comment an answer, I would be happy to accept it.

Some recent developments shown in article Transversely quasicrystallographic 3D photonic chiral lattices: Polarization-controllable complex photonic band gap structures, where autors claimed experimantal results on produsing 3d chiral photonic quasicrystals.

Figure. Computer simulation of the light intensity distribution of
  the interference pattern for transversely quasicrystallographic 3D
  chiral photonic lattices respectively with rotational symmetries
  (a)5-fold (d) 8-fold (g) 12-fold. (b), (e) and (h) are their respective
  x-y planes. (c), (f) and (i) are their respective x-z planes
"So the fabricated class of 3D PQC chiral lattices integrate both these wonder features on asingle platform in view of their possible applications in new generation integrated optical photonic devices with tunable polarization band gaps as well as frequency band gaps"

I also found fascinating that some researchers, use combination of structures like simple lattice and chiral lattice. This presentation Photonic Crystals and Metamaterials
has a lot of pictures, which could be good "food for thoughts".

Hope this was helpful, I think your area of research is very interesting.


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Easily removable material that sticks to skin
I am looking for some kind of material that can be used to easily latch something to someone's finger. This is for a first year engineering project. The idea is to use this material to latch small motors on someone's fingers in the same way that the motor in this picture is attached. The person should be able to move their fingers around without the motor falling off and the motor should be removable with a slightly strong pull. Also the material should create as small a barrier between the finger and motor as possible so that when the motor vibrates, it can easily be felt by the finger.
I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin than some sort of adhesive or fluid glue, but I'm open to hearing all ideas.
Are there polymer materials out there (or particular functional groups) that match this description?

I would go for elastic over a glue or sticky material if possible. It's much easier to make it re-usable, it has less of an 'ick' factor, and it's a cheap, well-understood material.

Another idea: you could just skip the whole adhesive route and design (or "engineer") a glove.

I've already thought about putting motors on a glove. It won't do because the glove won't fit everyone perfectly - some people might have small hands that won't even touch the motor wearing the glove. It also offers less flexibility and comfort than the idea I had in mind.

I am also looking for a reusable skin adhesive for a product that I would like to develop and came across this Interesting conversation.
Have you thought of using something like a strong lightweight stretchy tube and you could stitch together with another stretchy tube for the removable motor?
Just possibly an idea that could lead you in a different direction

I have an engineering solution for you that bypasses the need to find a chemistry solution.

I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin

You want an adhesive bandage. Adhesive bandages are thin, stick to skin, and are easy to remove. In my experience, the less expensive generic brands are both thinner and stickier, so go with a generic brand. If you are worried about the thickness, do not use the central part with the embedded gauze pad. Only use the wings. Use a strong glue to fix your motor to the back of the adhesive wing.

Thank you, this might work - are there adhesive bandages that are sticky on both ends? Gluing the motor to the bandage is an idea, but it makes it difficult to reuse as the bandage will get worn out and it can't really be used multiple times.

As long as the glue is not an epoxy-based adhesive, you could remove the motor from the adhesive strip by dissolving the glue with solvents like toluene, acetone, ethyl acetate, methyl ethyl ketone, or hexanes. These things can be found commercially as paint thinner, paint stripper, nail polish remover or "mineral spirits". Again, maybe not an optimal solution, but your job as the engineer is to optimize it.

There are people who need "easily removable material that sticks to skin" virtually every working day, so take a look at what they use:
Makeup artists use a special glue to stick special effects elements like Spock ears to the skin.
The glue is based on mastic, the resin of the mastic tree. So it is mainly a solution of the resin compounds in something like alcohol.
The most difficult problem to solve is finding a glue that is strong, but not harmful to skin - even when using it for a full day.
You can remove mastic glue with a "slightly strong pull". It will hurt the first time if there was hair.
There are also water-soluble skin glues, but as far as I know the mastic based is the strongest of them.
For examples, see Kryolan skin adhesives


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Quantum dot band gap energy equation
I have been looking into certain approximations for band gap energies in quantum dots and I found a paper [1]. However, I noticed that the equations which they obtain are not useful to the situation which I would like to use them in.
What I would like to do is gather information from other articles that show the absorption and emission spectra and understand the size of the quantum dots if the sizes are not given.
I have looked at several papers where $\ce{ZnSe}$ QDs are absorbing and emitting around $\pu{\sim 300 .. 450 nm}$ with sizes from $\pu{2 .. 6 nm}$ in diameter.  However, if you use the LDA+C equation for $\ce{ZnSe}$ quantum dots (QDs) ($E_g = 2.54/d^{1.54}$) with diameters of $\pu{\sim 2 nm}$, you get band gaps around $\pu{0.9 eV}$ or $\pu{\sim 1370 nm}$, not $\pu{\sim 400 nm}$ as you would expect.
Is there something I am missing here?  Is this a $E_{11}$ transition band gap energy and I am seeing like a $E_{66}$ absorption to $E_{55}$ fluorescence energy emission?
Also, I noticed that there are exciton binding energies which may take up a large portion of the total exciton energy. Are these binding energies important to this problem?
If so, is there a better way to search for and obtain some literature on a simple correlation for these type of energy transitions?

Li, J.; Wang, L.-W. Phys. Rev. B 2005, 72 (12), 125325. DOI 10.1103/PhysRevB.72.125325.

I believe the problem was that the LDA+C energies quoted in the paper were extra, additive energies which were supposed to be added to the main band gap energy of the bulk material. In this case that was something like $\pu{2.7 eV}$.
This is because the quantum dots will have higher band gaps the smaller they get. As the quantum dots get larger, in a sense, they approach a bulk material size, hence their band gap energy gets smaller and smaller and approaches the bulk material band gap energy limit.
Hope this helps anyone who happens upon a similar problem.


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What does the phrase "reducing atmosphere" mean in quantitative terms?
There is a debate about "how reducing" the atmosphere of the early Earth was, and this question is my attempt to grasp toward an understanding of what that that phrase really means.  As a bit of context, the Miller–Urey experiment produces a wide variety of complex organic molecules, but people say it requires a very reducing atmosphere to work, and it's disputed whether the early Earth's atmosphere was reducing enough for analogous processes to occur. Ultimately I would like to understand what it is about reducing atmospheres that allows complex organic molecules to be created so easily, but the first step is to understand what it really means to say one mixture of gases is "more reducing" than another.
If anything can be more reducing than something else, it seems like $\ce{CH4}$ gas should be more reducing than $\ce{O2}$, so I'm trying to understand the extent to which this can be quantified. However, I'm currently self-learning chemistry, and I'm having difficulty seeing connections between some of the concepts. (So apologies in advance for the rambling nature of this question.)
It seemed like one place to start might be to draw the net reaction of methane oxidation:

Obviously, if you add up the total oxidation state of every atom, you get zero - the carbon gets oxidised, the oxygen gets reduced, and the hydrogen stays the same. This must always be the case for any reaction (right?) since electrons are conserved. At first this made me think that maybe you can't really say that one thing is objectively "more reducing" than another. Though of course you can still talk about its tendency to oxidise or reduce carbon, and maybe that's what people are talking about when they say an atmosphere is "reducing" or "oxidising".
But then I came across the concept of redox potential, which does seem to be about things being objectively more reducing or oxidising than others. The problem is, the discussions of the concept that I can find online seem to be in terms of electrons "being transferred" from one molecule to another. But in the reaction above, as far as I can see, electrons aren't being transferred from the $\ce{O2}$ to the $\ce{CH4}$ but instead both molecules are being destroyed and replaced with completely different ones. So I'm having trouble seeing whether, or how, the concept of redox potential can be applied in this case.
Given this confusion, I guess my specific questions are:

When people talk about an atmosphere being "more reducing" than another, are they referring to redox potential or something else, such as the degree to which the atmosphere reduces or oxidises carbon? If it's the latter, is it something that can be quantified, or is it more of a subjective judgement?; and
What is the relationship between the concepts of the redox potential of a substance on the one hand, and the oxidation states of atoms on the other? Can the concept of redox potential be applied to gases like $\ce{CH4}$ and $\ce{O2}$?

I’m an environmental chemist and we frequently refer to ‘oxidizing’ or ‘reducing’ conditions in the environment. I’ll use soils as an example. The ‘oxic zone’ is the area of soil that oxygen can reach by diffusion and is seen as being under ‘oxidizing’ conditions because the reduction of $\ce{O2}$ is the dominant redox reaction that occurs in that zone. Elements that have multiple oxidation states tend be in the higher oxidation states. For example, sulfur will exist primarily as sulfate ($\ce{SO4^2-}$). As you get further away from the Earth’s surface, the oxygen is depleted (by microbes) and the redox conditions change: the dominant reduction reaction changes. Usually in this order: $\ce{O2}$, nitrate, manganese(III,IV), iron(III), sulfate and finally $\ce{H2}$. When the conditions become so reducing that the microbes are using $\ce{H2}$ as an electron acceptor, they making methane via a process called methanogenesis. This order arises because of reduction potentials. Microbes get the most energy for the least effort from reducing $\ce{O2}$, so they do that until it runs out, which is when they switch to the next best thing available.
Ultimately what it means to be a ‘reducing environment’ is that if you added a compound with a higher reduction potential then the dominant redox pair, it would become reduced. For example, if you could put $\ce{SO4^2-}$ in a methanogenic environment, it would be reduced to $\ce{H2S}$ very quickly.
I’ve read that the composition of Earth’s atmosphere in pre-life earth was very different from today. If I recall correctly, there was no $\ce{O2}$ as it was formed by microbes later on. Without $\ce{O2}$ in Earth’s atmosphere, it makes sense that conditions would be very reducing based on an analogy to the soil redox zones.

Thanks, this is very helpful. What I'm still unclear on is whether the extent to which something is reducing can be quantified. So in soil chemistry, can you plot a graph of "reducingness" against depth? If so, what is the proper term for "reducingness", how is it measured and what are its units? Or is "reducing conditions" more of an informal phrase that really just means they tend to reduce stuff?

Regarding the prebiotic atmosphere, I think the debate is about whether the atmosphere would have contained loads of reducing compounds like $\ce{H2}$ and $\ce{CO}$, or whether it would have been mostly more neutral stuff like $\ce{N2}$ or $\ce{CO2}$. I understand this much better than I did when I wrote the question, but still I'm interested to know whether there's a standard way to put a number to how reducing any given atmospheric composition is.

I think that the the redox potential, measured in volts is what you're thinking of.  It is theoretically possible to measure the redox potential in a soil system, but it's difficult to do in practice.

I think I have read that the prebiotic atmosphere is expected to have had a lot of $H_2S$

Ok, I think I understand this reasonably well now - I think the redox potential probably is what people mean when they talk about how reducing the prebiotic atmosphere was. Many thanks to both the answerers!

I'm not certain mine will be the best answer, but I may be able to get this started.
The short answer is yes, redox potentials are the key to determining how reducing or oxidizing a reaction is.  What your simple balanced redox equation (which is correct) does not show are the various half reactions that produce the complete reaction.
The big oxidizing half reaction in your equation is the classic reduction of oxygen:
$\ce{2O2 +8H+ + 8e- ->2H2O}, E_0=+1.23V$
I've already balanced this half reaction for your combustion equation.  The positive E0 value of 1.23 volts says this reaction tends to occur spontaneously (given some activation energy).
However, the other half reaction, the oxidation of carbon in methane, is likely MUCH more complex, because it involves many steps.  Here's where my answer my not be completely correct.  Below is one possible sequence of half reactions:
$\ce{CH4 +H2O -> CH3OH + 2H+ + 2e-}, E_0=-0.50V$
$\ce{CH3OH -> HCHO + 2H+ + 2e-}, E_0=-0.13V$
$\ce{HCHO +H2O -> HCOOH + 2H+ + 2e-}, E_0=+0.03V$
$\ce{HCOOH -> CO2 + 2H+ + 2e-}, E_0=+0.11V$
Notice first that the half reaction oxidizing methane to methanol has a negative potential of -0.50 volts and so is not spontaneous; it needs an oxidizer like oxygen to get started.  In the sequence of half-reactions I list, the potentials get progressively bigger, so that by the time we are oxidizing formic acid to carbon dioxide, the half reaction is slightly spontaneous (positive potential).
So this shows that you were correct that the reaction involves the creation and destruction of various compounds, but the sequence (the mechanism) depends on redox reactions which DO transfer electrons.
Finally, the reason my answer may not be best is that the mechanism I propose may be wrong.  While all these half reactions do exist and produce a spontaneous total, the total potential sums to 0.74 volts.  At least one source I read suggested that the actual potential of methane oxidation to carbon dioxide is 0.92 volts.  So there may be some other sequence of half reactions, which I could not track down, which is a better mechanism than the one I propose.  Still, I think my answer gets the principles right.

Many thanks for the helpful answer. This tells me how to conceptualise the difference in redox potential between $\ce{CH4 + 2O2}$ and $\ce{CO2 + 2H2O}$. But when people talk about one hypothesised atmospheric composition being more reducing than another, that seems closer to talking about the difference in redox potential between $\ce{CH4}$ and $\ce{O2}$, which is a bit trickier because there's no reaction between those compounds. I guess that's the thing that I'd really like to be able to understand.

As a bit of context, people say that the Miller-Urey experiment requires a very reducing atmosphere to work. I want to understand what it is about reducing atmospheres that allows complex organic molecules to be created so easily - but first I need to understand what a "reducing atmosphere" really is, which is what this question is really about. (I'll edit this background into the question.)

My prior comment was off. A reducing atmosphere is one where oxidizers are lacking (like O2), and reducing gases like H2 and CO are present. The theory seems to be that such systems were conducive to producing photosynthesizing lifeforms, which then produced oxygen and changed the atmosphere from reducing to oxidizing. Hydrogen is reducing because the H in H2 is in the 0 state and is easily oxidized to +1 (i.e. H2 is a reducer). Similarly, CO has carbon in the +2 state, leaving room for oxidation to +4. Hydrogen cyanide (HCN, mentioned in the M-U article) also has carbon at +2 and is reducing.

The question then is whether one can quantify the extent to which those molecules are reducing. I mean, is CO more or less reducing than HCN? Is it something one can put a number to?

"Is it something one can put a number to?" Sure...by experiment! Unfortunately, you can't just use potentials or oxidation states (though they may help form your hypothesis) because the mechanism may be complex. That's why scientists do experiments. I'm sure such work already exists, but now we are getting out of my direct experience.

Right - but if I were (hypothetically) to do such an experiment, what is it exactly that I would be measuring? What would be the units of the resulting number, for example?

I think this depends on your objective. I suppose you could measure the amount of carbon oxidized based upon the initial amount of carbon precursor compounds and the initial condition of the atmosphere. You could measure the composition of the starting and ending conditions, count up the compounds in each state, and then look at the net oxidation state change for all the carbon in the system. But I imagine there are other ways to quantify this depending upon your objective.

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What are the factors controlling the fate of post-adhesed volatile molecules of the olfactory epithelium?
What happens to volatile molecules that reach the main olfactory epithelium in the nasal cavity after they bind and the neural stimulus fades?
To what extent do such factors as receptor kinetics and diffusion (or any others) direct where these molecules end up?

Hi Trancot, I edited your question to make it more biochemical in nature and less biology.  I think I captured the underlying questions that you were really after, but please feel free to make any changes so that it represents what you are interested in.  If it is more the biology angle that you are after, flag the post and we'll move it to Biology.SE for you, but I think this has some interesting aspects that can be answered in the physical chemistry realm.

Simply, the odorant particles/molecules never actually enter the cell through endocytosis. Instead, these olfactory receptors, belonging to the G-protein-coupled receptor family, are transmembrane receptors whose structure passes through the cellular wall (several times actually) to functional complexes on the other side, and they use simple and reversible ligand binding as their primary activation mechanism.
Odorant particles are dissolved in the mucous lining the nasal cavity, and in this form come into contact with the receptors in the cell membrane of the epithelial cells. These bind as ligand complexes to receptor sites, triggering a change in the receptor's structure which in turn causes the coupled G-protein to release its bound guanosine diphosphate, and instead bind guanosine triphosphate from the interior cell environment. When that happens, the G-protein is "activated"; it decouples from the receptor, splits in half into Ga and Gbg subunits, and those two halves bind to effector enzymes which trigger the cell signalling. As a result of the protein decoupling, the charge balance of the receptor changes, releasing the complexed ligand from its binding to the receptor, back into the extracellular environment.
To reverse all this, an enzyme called RGS (Regulator of G-protein Signalling) binds to the Ga subunit, triggering hydrogenation of the GTP molecule it still has bound into GDP, causing the protein's subunits to recombine and recouple to the receptor.
The released ligand, meanwhile, is washed away again by the mucous flow from the Bowman's glands. It may complex with other receptors along the way, but eventually it makes its way past the nasal membrane to the pharynx at the back of the throat, where the mucous is swallowed and the odorant broken down in the stomach. This is the final fate of virtually all odorant agonists.
A select few, however, bind more strongly, and are not released by the structure changes of the receptor. When this happens, the receptor is stuck in the "on" position. One of two things then happens; either the receptor continues to activate, causing neural impulses which your brain eventually ignores (odor fatigue), or mechanisms inside the cell notice the faulty receptor, bring it into the cell in a process similar to endocytosis, and attempt to break down its components and fix the damage to reuse it. More often than not, the bound chemical ends up poisoning the cell by binding to something more vital than an ordinary G-protein receptor, eventually triggering programmed cell death. That's fine; the basal cells behind the surface of the olfactory epithelium divide fast enough to replenish every cell in the nasal membrane about once every two days.
Sources: Wikipedia - Olfactory epithelium, Olfactory receptor and G protein-coupled receptor.

Great answer, very informative! Could you provide some (scientific) references to your story as the OP requested? @Trancot I think one place to look is the wiki page on Olfaction and references on that page

I got most of this from Wikipedia. Look up the pages on olfactory epithelium and G-protein-coupled receptors.

Why is Wikipedia so ubiquitous as a reference... Is it that people don't have access to alternatives? I encourage all to seek out alternative sources, and to not rely too much on Wikipedia.

@Trancot - well, as long as the wiki article is properly referenced I don't see a problem with it. I typically look up the references in the wiki article, but usually the explanation on the wiki is much more straightforward and easy to grasp. Looking at the wiki pages that KeithS used: they are full of scientific references so by checking out those you don't have to worry about reliability (at least not too much)

The resources are sparse. You must admit. I do, however, very much appreciated the answer. The question came up during dinner when my friend wanted to know why the smell of spicy food was so ephemeral some times and more lasting on other occasions.


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How does expansion of electron shells work?

How does it work? I've read that its about the similarity of the energy levels of $3d$ and $3p$, but shouldn't the electron shell $4s$ be filled before $3d$ if filled? So what is done with the $3d$?
I'd also like to know in what cases will atoms of elements expand their valence electrons? Which blocks or groups or periods do this?

You're right! The 4s subshell is filled before 3d. Electrons fill the shell   with least energy first and the 4s shell has a lower energy than the 3d subshell.
Use the Aufbau Principle to find the shell with least energy.
Basically it says that shells with a lower n+l (n and l are the principal and azimuthal quantum numbers) value are filled first.
If two shells have the same n+l value, then the shell with lower n is filled first.
There are a few exceptions to this rule - Copper, Chromium, Palladium, etc.
Reason: n+l gives the number of nodes in an orbital. The more the number of nodes, the less stable the orbital and so the higher the energy.
Valence shells are filled in the s and p blocks. The inner shells are filled in the d and f blocks.

Regarding #2:
Anything below the carbon row can expand their octets.  The question makes little sense in any other context.


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Do I really need 4 different water treatment systems with this water analysis?
We installed an artesian well and we had some surprises. We asked a few places around and they all told me that we needed 4 different systems to solve our water problem. I am ready to accept this fact but before having our basement transformed to a factory I just wanted to confirm that this is the only solution.
Here is the result of my analysis.
The problems are:

Barium  $6.9 \:\mathrm{mg/L}$    (2nd analysis revealed same results)
Apparent Color 17.5 UCA
Iron  $0.47 \:\mathrm{mg/L}$
Total sulfur $0.07 \:\mathrm{mg/L}$

Also the water is hard ($210 \:\mathrm{mg} \ce{CaCO3}/\rm L$) which is not recommended
My question is, do I really need 4 different systems to treat my water?

It seems that way for me. You have to have adsorbtion treatment for two very different cations, one anion and something to deal with color. And even then I'd prefer to use something else as source of drinking water. Without proper equipment you will not know when your treaters will fail.

The suggestion of 4 is not ridiculous, but the water softener may leach out some of the barium as well.

Are you actually planning to make this drinkable, or is it for the shower/ dishwasher/washing machine?

to make it drinkable

You shouldn't need 4. The only dangerous thing is the barium, and permeakra is right that you should keep an eye on the equipment so that it doesn't silently fail on you. You should check perhaps once a month. It's only ~3.5 times the EPA limit, so a failure won't be immediately be a hazard (unless the sequestered barium is released all at once, but that seems unlikely).
The other 3 problems and the hardness are, at those levels in the water, are simply nuisances. So they will discolor your water, produce a weak sulfurous odor, and eventually clog the pipes, but they will not harm anyone. If you fixed the iron and sulfur, that would probably improve the color, so only 3 systems are needed. And you don't have to worry about the system failing silently since you will see it.
Check these statements before making a decision, this is too important to trust only one source!

Thanks for the info, but we have actually made our choice. Here is what we have bought. An iron/sulfur removal machine followed by a water softener, followed by a 5 micron filter, followed by a UV lamp. I did a lot of researched (maybe I am wrong but I did try!) and it seemed to handle our needs. We have an alarm on the UV lamp that rings every year to remind us to change the lamp but I will still test our water for Barium and Bacteria every 6 months. Does that make sense? Thanks


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Tool that generates electrostatic maps of molecules?
Do you know any tools that generate electrostatic potential 2d maps of the molecules as an image?

Hello Wise, and welcome to Chemistry Stack Exchange! Could you please be more detailed about what you're asking for: do you want to visualize an already calculated electrostatic map, or do you want to calculate one from the molecular structure (and the atomic charges)?

Hello and thanks! I have seen a few applications that visualize the electrostatic map as a cloud around the 3d structure. Actually I need the electrostatic maps generated, and then be visualized as 2d images (like contour maps). The calculation part is more important to me. It would be great if the tool provided some sort of matrix of electrostatic maps.

Most computational chemistry software packages geared toward students/education should be able to do this. Examples include Scigress, formerly CAChe, Spartan, which has a 'lite/student' version that can probably do it, etc. Generally, any modeling software, for example, Gaussian or GAMESS should generate a file with the electrostatic potential map provided you give it the correct instructions. The output files of these methods can be visualized using many types of software (like Scigress and Spartan) and open source software like Avogadro.
If you are looking for an inexpensive solution, GAMESS is free for research/education and Avogadro is open source.

I do not know how to operate such softwares at all. I have no background in this. Is there some software which can do it easily? Avogadro's built in feature isn't that awesome...Otherwise, could you point out some beginner's tutorials for any one of these?

@FreezingFire for GAMESS you could start here, for Avogadro a good place would be here.
ORCA and NWChem are also good choices.

As Ben Norris said, there are approximate methods around.
But if what you want is a very precise, very reliable electrostatic potential map, you will need to do a quantum chemical calculation to get the electron density.
Here is a nice discussion about multiple-derived and charge-based methods versus, say, DFT.
There it says:

For example, six slightly negative charges on carbons balanced by six slightly positive charges on hydrogens do a poor job capturing the essence of the negatively charged π cloud above and below the six-membered ring [of benzene].
  The image below illustrates electrostatic potential maps for benzene using the true potential, multipole-derived, and charges-based methods.

If you want to generate a electrostatic potential map from a .wfn file (a very common file type for carrying wave functions around), MultiWFN may be what you want.
MultiWFN helps you analyse the wave function using many different techniques.
It also allows you to visualise both the 3D and the plane projected electrostatic potential map (both colour-coded or contour-mapped I believe).
A .wfn file may be generated from a .gbw file from ORCA using Molden2AIM, for instance.
Molden2AIM can actually convert many wave function output files to .wfn, so please check it out.
For the particular purpose of using it with ORCA I have written a small program that considers many details and does the job faster than "by hand".
If you want, MultiWFN also generates suitable .cube files for you.
These are useful for visualising in third party software.
I also wrote a small program that can generate those and write a small VMD script for visualising.
VMD is able to produce publication-quality images.
All tools cited in this post are free to use.


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What household substances could be distilled
I'm looking to experiment with distillation to have a basis for understanding the theory slightly better, and get a better feel for it.
I'm hoping for some suggestions as to what I could try distilling, such as purifying drinking alcohol. Any single idea would be appreciated, best answer would be awarded to an idea where:

The resulting purity would be straight forward to approximately test
No part of the distillation presents a significant hazard (e.g. toxic fumes, high volatility)
No particularly specialist or expensive equipment is required (e.g. extreme temperatures)

I think that pretty much covers it. I'm just looking for one or two ideas which I can experiment with besides alcohol.
Thanks

This is tricky. Besides water, nearly everything has some hazard associated with it.  And even water distillation involves high temperature (100C).  Be careful distilling alcohol: the azeotrope of distilled alcohol is 95% ethanol which is VERY flammable! Plus, technically speaking, distilling alcohol may be illegal where you live.
One thing you could do is distill water with dissolved substances, most likely table salt (NaCl).  This would help you test the principle that distillation is a technique for separating substances. In this case, distilling salt water should yield pure water. A more visual example would be brewed coffee or tea, which is a suspension and should separate into water and crud!
And after that, I'd be careful! For example, concentrating acetic acid (white vinegar) is dangerous, as concentrated acetic acid is both corrosive and has choking fumes.  I would not even consider trying something that is not food grade, given your requirements.

Well, tea isn't just a suspension; tea for instance has oxalic acid, which dissociates in water, and tannins, which dissolve much like sugar does. But, "un-brewing" tea or coffee is an excellent suggestion for an introduction to laboratory distillation.


text:
Materials that resist adhesion with molten glass even when hot
I'm looking for ways to manipulate molten glass with (e.g. bending a heated glass tube).
This article mentions that oxides tend to adhere, hence molten glass to heated stainless steel.
What materials could I use to manipulate hot glass? Perhaps I could coat the glass or instrument doing the manipulation with some sort of powder or liquid?
The glass is soda lime glass which has a transition temperature of 550 degrees C, so the temperature it needs to cope with will be a fair bit above that.
Thanks

If I could hazard a total guess, glassblowing supplies may be your best bet, but I don't know who would sell them.

Professional glass blowers use graphite paddles to "push" the glass around.  But to bend a heated glass tube you need to know that glass is a very poor conductor of heat.  Thus it is possible to grasp the heated tube six or so inches from the heated spot and just bend the glass.  This works well for small bore tubing.  Just be careful putting the bent tube down.  Putting it on a high heat conductivity surface will result in the glass breaking.  In the old days asbestos pads were used, but that's a big no-no now.


text:
Is there a way to cheaply transfer (plunge by pushing) a viscous liquid in to another container without plunger-to-tube discharge?
Could anyone offer a suggestion as to how an economy-minded chemist ought to implement a DIY-ed positive displacement alternative to the following:

You can't link to an attachment in your email, you can download it somewhere and then add the link, and someone with enough rep will edit it in for you.

A rock would work.  But a more useful answer would require some indication of what you want to accomplish.

I need to plunge (pull) and transfer (plunge by pushing) a viscous liquid in to another container without plunger-to-tube discharge.

I do not understand your question.  Have a look at the link and see if you can find something similar, follow the picture links and read up to describe your situation and help with common names for the things and actions you propose.  -  https://www.google.com/search?q=dispensing+cartridge&tbm=isch


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How can a substance be diluted to nothing?
If I start with a small amount of a substance, and I dilute that substance with billions of gallons of water, what is the scientific explanation for there being no (actually zero) remaining molecules of that original substance?
This question is related to the practice of homeopathic dilutions. I was reading some articles which stated that homeopathic dilutions don't work because at most homeopathic levels of dilution, there are very few or no molecules or the original substance.
What I'm confused about is how there can no longer be any molecules of the original substance (or even fewer molecules) of an original substance when diluted.
If there are X number of molecules in a solution, and Y liters of water are poured into the solution for dilution, don't the X number of molecules remain? How can a molecule just "disappear"? How can even one molecule just "disappear"? It makes more sense if these molecules are undergoing a chemical reaction, but if two kind of molecules (the substance, and water) are not undergoing any chemical reaction, then how can one molecule of either kind simply vanish?

You correctly point out that the number of molecules in a solution is finite and constant, however the volumetric concentration (that is, how many molecules per litre) changes upon dilution. If, for instance, you take one liter of a 1 mol/L solution of ethanol in water ($\approx{6.02\times{10}^{23}}$ ethanol molecules per liter) and add 9 liters of distilled water, you now have a 1 mole of ethanol in 10 liters, giving a concentration of 0.1 mol/L. If you were to draw out 1 liter of this and count the number of ethanol molecules in that liter, you would find approximately $\approx{6.02\times{10}^{22}}$ - a tenfold reduction. The rest of the ethanol is retained in the other 9 liters of solution.
Equivalently, if you were to measure out 100 mL of 1 mol/L ethanol solution and then make it up to 1 liter by adding distilled water, you would also get a 0.1 mol/L solution. This is the kind of thing taken to absurd extremes in homoeopathy, where tiny quantities of solution are repeatedly diluted, with some shaking. Obviously, they don't literally add billions of liters of water to their solutions, but rather pipette out a tiny amount of solution and discard the remainder, adding water to make up the desired volume. This process is repeated many times.
The finite number of molecules present in a solution leads to the concept of the dilution limit, which is where you have carried out sufficient dilutions that statistically speaking, your chances of finding even a single molecule of the dissolved ingredient become vanishingly small and that by extrapolation, you would have to drink approximately a zillion liters of the preparation to get a single molecule of it.
Where did all those other molecules go? They didn't simply disappear - they were poured down the drain during the dilution process, which curiously is the same place money goes when spent on homoeopathy.
Bonus: homoeopathic plutonium! (courtesy of Theodore Gray's periodic table)

Any treatment has a placebo effect, which means it will cure about 1/3 of patients anyway. Homoeopathy is based on principles that are out of place here.

Homeopathic plutonium? backs off

Yeah I realized after I posted this question, that my mistake was in thinking a person simply adds more water to the molecules. This is false; homeopathic dilution works by taking a small amount of the original solution, in which case, I agree that there will be no molecules left after a certain number of dilutions. Thanks!

@fp - "placebo effect [...] will cure about 1/3 of patients anyway" - This is incorrect. If it were true, placebos would be the most important drug in medicine.

@Manishearth - As someone quipped, 'it doubles in strength every 88 years'.

@Richard Terrett-Feel free to do your own research.

And what to use it for! (I have a copy of this book, but only because I checked the author wouldn't benefit from the sale. It's kind of hilarious.)

Is homeopathic plutonium for weapons homeopathic mass of destruction?

Homeopathic plutonium, great! I'll get an empty bottle of something homeopathic, recreate a label for plutonium, and have it casually on the desk if I should ever work near people interested in homeopathy. (The page worries where they got the original plutonium for a dilution. But it is much more important where the unused part went to) Also, I think it is a good thing that homeopathy exists, because the placebo effect is explicitly used in practice routinely, but by prescribing active drugs intended for something else.

@Jason It is worth doing the calculations. 10-fold dilution repeated about 23 times gives, on average, a handful of molecules in the resulting solution. 24 times gives an average of less than 1. That isn't even hard.

They don't vanish. It's just that if there are X molecules and Z bottles with Z > X, part of the bottles will not have even one molecule.

A substance can be diluted to "no substance" because:
No substance present means we are sure there are no molecules of it in some finite volume.
Being sure means we assume it to be true to some finite level of probability.
We start with a finite number of molecules.
And we can infinitely dilute the the solution.
We have only a finite amount of water available, because even in an infinite universe, we can reach only a finite amount of mass/energy because it expands.
But we can simply clean and reuse our water.
It's not even slow to do that: It is exponential, and limited only by how much water we can move: Imagine repeatedly diluting a drop in a Olympic size swimming pool. Filling the pool can be done in parallel to selecting a drop and mixing it in, and takes longer.
And there is industrial level experience available, ask your Civil Engineering Hydraulics expert. For off the shelf components, see hydro power stations.

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text:
Does resonance apply to NO₂⁺?
My textbook said $\ce{NO2+}$ and $\ce{NO2-}$ both have resonance structures, now I am a little bit confused since I cannot figure out the resonance structure for $\ce{NO2+}$.
What is it?

In the following below, please interpret formal charges as marked below the atom.
$$\ce{O=N=O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O\equiv N-O} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{O-N \equiv O} $$
$$\ \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \ \ +\ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ +\ \ + $$
The triply charged species are valid resonance forms, but very likely are small contributors.

Triple bonds can be made with MathJax using \equiv for the equivalence symbol. $\equiv$ renders as $\equiv$

The last two resonance structures are identical.

@Dissenter, no, they are not identical.  They are equivalent.  Which nitrogen is which matters for resonance structures in that both can have a triple bond, and there must be one structure that indicates a triple bond on each.

Shouldn't they be identical because rotating one 180 degrees gives the other?

When writing resonance structures, the atoms must be absolutely fixed in space.  Perhaps you're muddling chirality into the concept?


text:
Identifying a quadruply charged anion containing three carbon atoms

This linear polyatomic ion containing three atoms of carbon has a negative four charge and is only found bonded with lithium and magnesium.

Could anyone identify this for me? It's from a quiz bowl clue that doesn't have the answer with it.

Sesquicarbide

A quick Google search suggests that $\ce{Li4C3}$ is a real compound, but based on what I saw in the links, it probably is not ionic in the same way that lithium acetylide $\ce{Li2C2}$ is.
$\ce{Mg2C3}$ appears to be the formula for magnesium carbide, or at least a magnesium carbide.
A carbide is a binary compound of carbon and a less electronegative element, usually a metal or metalloid. Carbide is also a generic name for mono- and polyatomic anions containing only carbon. The following ions would all carbide ions: $\ce{C^{4-}, C2^2-, C2^{4-}, C2^{6-}, C3^{4-}, C3^{6-1}, C3^{8-}}$, etc.
The special name of three of these carbide ions are:

$\ce{C^{4-}}$ - Methide
$\ce{C2^{2-}}$ - Acetylide
$\ce{C3^{4-}}$ - Sesquicarbide

Lithium and magnesium sesquicarbides are in fact considered to be ionic.  See for example here.


text:
large wavelength range and list of x/y numerical data points of elements
I'm trying to get a set of FTIR or (FTIR like range) x/y numerical data points for a couple of elements,  namely Silver, Platinum, Palladium, Gold and Sodium.  I've looked on line and found images only of the FTIR but I need the numerical x/y data points with the largest wavelength range.  I tried NIST but they just show the NMR data.  Is there a site where I can get just these or can someone send these/post these for me?  And is it possible to get an FTIR of Sodium just by itself or does it need to be combined with another element.
Thanks
Please note I know metals (due to covalent bonds) don't work well with FTIR but I'm trying to get / see if it's possible to get a large wavelength range analysis of the items above like in FTIR.  Sorry for using the FTIR term I don't know the proper term to use at the moment.

I don't know about the IR spectra of metals. Certainly metals will have vibrational spectra, but I'm sure where to find it (if NIST does not have).
I can tell you how to get x/y data from NIST, if they have the spectrum.
For example, for benzoic acid. At the bottom of the page, there is a link to download the sepctrum in JCAMP-DX format. The .JDX file format is really a space-delimited text file containing x/y data with some markup at the front for the JCAMP plugin.

See all that lovely x/y data? Force the .JDX file open in your favorite text editor, like notepad. Do not use a word processor - they will add additional markup. Do not use a spreadsheet program yet. Spreadsheets can open .JDX files, but they won't always choose the most helpful delineation. Once open in notebook, resave the file as a text (ascii) file. Make sure you type in the .txt to change the extension (alternately, you can just change the extension to .txt using any other means you may feel comfortable with).
Now... open your favorite spreadsheet program, and use it to open your shiny new text file. You may have to tell the program to look for "all files" (MS Excel is bad about this). You should get some dialog/options on how to open the file. Choose "Delimited" if its an option, and when it comes time to choose your delimiters, make sure your choice includes "space":

Now you should have a file with the x/y data. The first column is wavenumber (at least for IR) and the second is transmission ($\frac{I_{\text{out}}}{I_{\text{in}}}$). Some of these files have more than one column of y-data. When I graphed each column from the benzoic acid file, I got the same spectrum (almost).

The absorption spectrum of any material is related to the dielectric constant/refractive index.  Since there are no sharp resonance peaks for metals in the infrared, the absorption spectrum is pretty boring.  However, there are many sources of data for metals.  The most comprehensive set of data is the Handbook of Optical Constants of Solids by Edward D. Palik.  It is a four volume set that should be available in most university libraries.  The numbers for the most common metals can be downloaded in csv form from http://refractiveindex.info/.
With the refractive index in hand, the imaginary part of the index can be related to the extinction coefficient as $\alpha=\frac{\omega}{c}k$ where $\omega$ is the angular frequency of light, $c$ is the speed of light, and k is the imaginary part of the index.

I don't know where you can get the X/Y data, but you can try to digitize the images you have found using something like a plot digitizer. I know it's not the best solution, but in bad circumstances it can work.

For most of the metals have a look at Optical properties of the metals Al, Co, Cu, Au, Fe, Pb, Ni, Pd, Pt, Ag, Ti, and W in the infrared and far infrared Applied Optics, Vol. 22, Issue 7, pp. 1099-1119 (1983). You will find something like this:

OCR software could help you to retrieve the data.You can use $k$ as an index that indicates the amount of absorption loss.
Free pdf of another version here.


text:
Regioselectivity of acid-catalyzed ring-opening of epoxides
Not to be confused with what is the mechanism of acid-catalyzed ring opening of epoxides.

What is the correct order of regioselectivity of acid-catalyzed ring-opening of epoxides: $3^\circ$ > $2^\circ$ > $1^\circ$ or $3^\circ$ > $1^\circ$ > $2^\circ$? I am getting ready to teach epoxide ring-opening reactions, and I noticed that my textbook has something different to say about the regioselectivity of acid-catalyzed ring-opening than what I learned. My textbook does not agree with 15 other introductory texts I own, but it does agree with one. None of my Advanced Organic Chemistry texts discuss this reaction at all. Thus, I have no ready references to go read.

Edit:  My textbook is in the minority, and it is a first edition. Is it wrong, or do the other 15 texts (including some venerable ones) oversimplify the matter?
What I learned
Although the acid-catalyzed ring-opening of epoxides follows a mechanism with SN2 features (inversion of stereochemistry, no carbocation rearrangements), the mechanism is not strictly a SN2 mechanism. The transition state has more progress toward the C-LG bond breaking than an SN2, but more progress toward the C-Nu bond forming than SN1. There is significantly more $\delta ^+$ character on the carbon than in SN2, but not as much as in SN1. The transition states of the three are compared below:

In a More O’Ferrall-Jencks diagram, the acid-catalyzed ring-opening of epoxides would follow a pathway between the idealized SN2 and SN1 pathways.

Because of the significant $\delta ^+$ character on the carbon, the reaction displays regioselectivity inspired by carbocation stability (even though the carbocation does not form): the nucleophile preferentially attacks at the more hindered position (or the position that would produce the more stable carbocation if one formed). If a choice between a primary and a secondary carbon is presented, the nucleophile preferentially attacks at the secondary position. If a choice between a primary and a tertiary carbon is presented, the nucleophile preferentially attacks at the tertiary position. If a choice between a secondary and a tertiary carbon is presented, the nucleophile preferentially attacks at the tertiary position.

The overall order of regioselectiveity is $3^\circ$ > $2^\circ$ > $1^\circ$.
What my textbook says
My text agrees that the mechanism is somewhere in between the SN2 and SN1 mechanisms, but goes on to say that because it is in between, electronic factors (SN1)  do not always dominate. Steric factors (SN2) are also important. My text says that in the comparison between primary and secondary, primary wins for steric factors. In other words, the difference between the increased stabilization of the $\delta ^+$ on secondary positions over primary positions is not large enough to overcome the decreased steric access at secondary positions. For the comparison of primary and tertiary, tertiary wins. The increased electronic stabilization at the tertiary position is enough to overcome the decreased steric access at the tertiary position. The comparison between secondary and tertiary is not directly made, but since $3^\circ$ > $1^\circ$ and $1^\circ$ > $2^\circ$, it is implied that $3^\circ$ > $2^\circ$.

If this pattern is true, then other cyclic "onium" ions (like the bromonium ion and the mercurinium ion) should also behave this way. They don't.
Typical of introductory texts, no references are provided. A Google search did not yield satisfactory results and the Wikipedia article on epoxides is less than helpful.
Since I 15 other introductory texts on my bookshelf, I consulted all of them on this reaction. The following is a summary of my findings. Only two of the texts (the one I am using and one other) describe the regioselectivity as $3^\circ$ > $1^\circ$ > $2^\circ$. All of the other books support the other pattern, including Morrison and Boyd (which lends credence to the pattern that I learned).
Books that have $3^\circ$ > $2^\circ$ > $1^\circ$

Brown, Foote, Iverson, and Anslyn
Hornsback
Ege
Wade
Bruice
Smith
Fessenden and Fessenden
Volhardt and Schore
Solomons and Fryhle
Jones
Baker and Engel
Ouellette and Rawn
Carey
Morrison and Boyd
Streightweiser and Heathcock

Books that have $3^\circ$ > $1^\circ$ > $2^\circ$

Klein (the text I am using)
McMurray

I also surveyed my various Advanced Organic texts (March, Smith, Carey and Sundberg, Wyatt and Warren, Lowry and Richardson, etc.). Interestingly, none of them even mention acid-catalyzed ring-opening of epoxides (either by Brønsted or Lewis acids). I suspect that these omissions mean that this reaction 1) has difficult to predict regioselectivity (despite the predominance of introductory books that suggest otherwise), and thus 2) is synthetically useless. If #2 is true, then why is this reaction in introductory organic texts?

In the first equation (me-subst epoxide) there is an extra carbon in the minor product. (Same for the variation of it.)

Unless I'm missing something, the axis labels on the MOFJ plot are reversed.

I have read this discussion several times and finally got motivated to do a Chemical Abstracts search on line for the reaction of propylene oxide and methanol under acid catalysis. Many of the catalysts were zeolites and rare earths. The overwhelming conclusion is that the epoxide opening with methanol is at the primary site leading to the secondary alcohol.

With respect to comparing to bromonium and mercurinium, I don't necessarily see a disparity. The analysis involves competing electronic and steric factors, the electronic factor being induction by the oxygen atom. In both acid-cat and base cases, the induction by the electronegative oxygen atom is there, but in the acid-cat. case, the oxygen atom has a positive charge, and so will be extremely electronegative. That tips the scale from steric dominance to electronic dominance. The brominium and mercurinium ion will not be as electronegative as the oxonium, so the steric factor still dominates

First part
It won't decide the issue but the Organic Chemistry text by Clayden, Greeves, Warren and Wothers also mentions that the matter might not be as clear-cut as the majority of your textbooks make it seem. This might strengthen the position of the textbook you're using a bit. But again, there are no references given. Here is the relevant passage (especially the last two paragraphs):

Second Part
I have found the following passage on the formation of halohydrins from epoxides in the book by Smith and March (7th Edition), chapter 10-50, page 507:

Unsymmetrical epoxides are usually opened to give mixtures of regioisomers. In a typical reaction, the halogen is delivered to the less sterically hindered carbon of the epoxide. In the absence of this
structural feature, and in the absence of a directing group, relatively equal mixtures of
regioisomeric halohydrins are expected. The phenyl is such a group, and in 1-phenyl-2-
alkyl epoxides reaction with $\ce{POCl3}/\ce{DMAP}$ ($\ce{DMAP}$ = 4-dimethylaminopyridine) leads to
the chlorohydrin with the chlorine on the carbon bearing the phenyl.${}^{1231}$ When done in an
ionic liquid with $\ce{Me3SiCl}$, styrene epoxide gives 2-chloro-2-phenylethanol.${}^{1232}$ The
reaction of thionyl chloride and poly(vinylpyrrolidinone) converts epoxides to the corresponding
2-chloro-1-carbinol.${}^{1233}$ Bromine with a phenylhydrazine catalyst, however,
converts epoxides to the 1-bromo-2-carbinol.${}^{1234}$ An alkenyl group also leads to a
halohydrin with the halogen on the carbon bearing the $\ce{C=C}$ unit.${}^{1235}$ Epoxy carboxylic
acids are another example. When $\ce{NaI}$ reacts at pH 4, the major regioisomer is the 2-iodo-3-
hydroxy compound, but when $\ce{InCl3}$ is added, the major product is the 3-iodo-2-hydroxy
carboxylic acid.${}^{1236}$
References:
${}^{1231}$  Sartillo-Piscil, F.; Quinero, L.; Villegas, C.; Santacruz-Juarez, E.; de Parrodi, C.A. Tetrahedron Lett. 2002,
43, 15.
${}^{1232}$  Xu, L.-W.; Li, L.; Xia, C.-G.; Zhao, P.-Q. Tetrahedron Lett. 2004, 45, 2435.
${}^{1233}$  Tamami, B.; Ghazi, I.; Mahdavi, H. Synth. Commun. 2002, 32, 3725.
${}^{1234}$  Sharghi, H.; Eskandari, M.M. Synthesis 2002, 1519.
${}^{1235}$  Ha, J.D.; Kim, S.Y.; Lee, S.J.; Kang, S.K.; Ahn, J.H.; Kim, S.S.; Choi, J.-K. Tetrahedron Lett. 2004, 45, 5969.
${}^{1236}$  Fringuelli, F.; Pizzo, F.; Vaccaro, L. J. Org. Chem. 2001, 66, 4719. Also see Concellón, J.M.; Bardales, E.;
Concellón, C.; García-Granda, S.; Díaz, M.R. J. Org. Chem. 2004, 69, 6923.

I have the fifth and sixth editions of March, presumably something similar is present in the earlier editions. I will check. In the meantime, posting links to those references anyway would be helpful to the rest of the community who may not have access to a copy.

@BenNorris Ok, I add them. By the way, for the 6th Edition of March you find the relevant passage on p. 583.

Extending the pattern of epoxides to bromonium ions and mercurinium ions is hazardous at best.  Below the second row (the carbon row) bond lengths get much longer, and acid strengths are much lower than compared to sulfuric acid in methanol.
In general acid-catalyzed reactions involving epoxides where selectivity is a concern will have carbocation intermediates, but the are not required to do such.  But this potential is most likely to scare off a potential synth-jock simply because it is effectively a lower throughput at whatever step.
The pedagogical value of teaching epoxides at the undergraduate level is 4-fold:

It allows discussion of the physical chemistry of strained
ring-systems
Synthetic Organic Chemistry is the business of making and breaking carbon-carbon bonds.  Via Grignard reagent, epoxides allow a two-carbon extension of a carbon skeleton, which is fairly simple to understand and at the point where the topic is introduced, often the first example.
It can serve as an introduction to polymer chemistry.
In the example you've presented, it forces students to evaluate the relative strengths of competing phenomena, which is the sine qua non of organic chemistry as an educational tool.  In this case to explicitly address the question: Does the effect 2o carbocation stabilization for gain outweigh the steric hindrance?

If the instructor or students are obsessed with the 'right' answer rather than an intelligently described answer (which may also be right), epoxides are less useful as a teaching topic.

I agree with everything you say, but it does not answer the initial question. Which regioselectivity is correct? My textbook is in the minority, and it is a first edition. Is it wrong, or do the other 15 oversimplify the matter? I have edited my question to reflect this latest thought on the matter.

I do not believe the regioselectivity can be predicted reliably outside of an actual experiment.

I agree with this answer. It is possible to find examples that support both textbooks.

A few thoughts about the regioselectivity issue. 1) Do you believe that all text authors consult the original literature? 2) Is there conflicting information in the the original literature?


text:
What is the reaction between oxalic acid and potassium permanganate?
I've already searched online some information about this equation, but everywhere I read, there's also sulfuric acid, which we did not use in the laboratory. We only used potassium permanganate and oxalic acid and water, there was no sulfuric acid. The purpose of the laboratory was to calculate experimentally the rate of the reaction at different concentrations, but in order to do so, I need the balanced equation. Can anyone help me out?
By the way, the oxalic acid was dihydrate.
If anyone is wondering, one of the experiments was as follows:
First we took $1.5~\pu{mL}$ of $\ce{H2C2O4}$ and added $1.5~\pu{mL}$ of $\ce{H2O}$, and finally, we added $1~\pu{mL}$ of $\ce{KMnO4}$ and then we shook the test tube until the colour changed from purple to ochre. Because from what I read, they said the $\ce{H2SO4}$ was necessary so the reaction takes place, but in the experiment, the reaction did occur and we didn't use it.

Does the reaction take place in any anhydrous conditions, for instance, in dmf?

related http://chemistry.stackexchange.com/questions/59905/yellowish-tinge-during-titration-of-oxalic-acid-with-kmno4

One can think of  potassium permanganate as a substitute for H2O2 in REDOX reactions, in which case, we have this reference: https://www.osti.gov/biblio/6229619 . " The rate of oxidation of oxalate by hydrogen peroxide in acid solution at the reflux temperature was adequate for process application; reaction half-times at 100/sup 0/C were less than one hour when hydrogen peroxide concentration was greater than 0.5 M. The reaction was first order... to both oxalate and hydrogen peroxide concentrations and had an activation energy 58.7 kJ/g-mol...rate increased with hydrogen ion concentration"

We had a titration of oxalic acid vs. potassium permanganate for an experiment in which we used concentrated $\ce{H2SO4}$.
Actually, the reaction requires an acidic medium i.e. $\ce{H+}$ is involved as a reactant... I suppose the $\ce{H+}$ released by the oxalic acid would be enough since you are studying the rate of the reaction though I am not sure.
The balanced ionic equation is:
$\ce{2MnO4- + 5H2C2O4 + 6H+ -> 2 Mn^2+ + 10CO2 + 8H2O}$
As you can see, $\ce{H+}$ is involved as a reactant and that is why you might have read that conc.$\ce{H2SO4}$ is necessary.

It makes sense, though where is the potassium left?

@ChairOTP, kaliaden left out K bacause it is a spectator ion, meaning it does not participate in the reaction, and so it is often left out.

If you wrap your chemical equations in \ce{ }, you will get prettier formulae and equations. \ce{ } converts 2MnO4^- +5H2C2O5 + 6H+ -> 2Mn^{2+} + 10CO2 + 8H2O into $\ce{2MnO4^- +5H2C2O5 + 6H+ -> 2Mn^{2+} + 10CO2 + 8H2O}$

@BenNorris Thanks!! I have edited the equation using \ce{}

@ChairOTP Like trb456 mentioned, The $K^+$ is a spectator ion and is not actually involved in the reaction. The same goes for the $SO_4^{2-}$ when $H_2SO_4$ is used, only the $H^+$ is actually involved in the reaction while the $SO_4^{2-}$ is a spectator ion. So you could actually use any other acid instead of $H_2SO_4$ as long as it doesn't react in any other way with any of the reactants.

Oxalic acid is a relatively strong acid for a carboxylic acid, and according to my sources below, can auto-catalyze the reaction with Potassium Permanganate. So, the reaction you performed was likely just the same mechanism that you have seen everywhere else. My source is as follows:
Kovacs K.A.; Grof P.; Burai L.; Riedel M. (2004). "Revising the mechanism of the permanganate/oxalate reaction". J. Phys. Chem. A. 108 (50): 11026–11031.


text:
Solubility of compounds in water
Why is $\ce{NaCl}$ more soluble in water than $\ce{NaOH}$? Well, this is a question from Chemical Equilibrium and the question hints for explanation in terms of Le Chatelier's principle.

This sounds like a homework question. Could you show us what you have tried? It will help you get a better answer. We like to see questions with a bit of your own effort. See the [faq] and [ask] for more info.

It probably has to do with the common ion effect. What equilibrium process is going on in plain old ordinary water?

The solvation of a small amount of hydrogen cations to form hydronium, which would be disrupted by the hydroxide ion's higher affinity for these cations?

It is not. $\ce{NaOH}$ is more soluble ($>100~\mathrm{g}/100~\mathrm{ml}$) then $\ce{NaCl}$ ($<30 ~\mathrm{g}/100~\mathrm{ml}$). $\ce{NaOH}$ when dissolved produces $\ce{OH^-}$ anion, that has relatively small size and forms specific hydrogen bonds (where it is bonded to relatively small positively charged hydrogen) very well. In solid $\ce{NaOH}$ it has to form hydrogen bonds with quite big $\ce{Na^+}$ cation.

The Water molecules move around the $\ce{Na}$ and $\ce{Cl}$ and break the salt apart into ions. The polarity of the water molecule allows this by the more negative oxygen being attracted to the positive Na ions and the positive hydrogen being attracted to the negative chlorine ions. In the case of NaOH, I am assume that bond between $\ce{Na+}$ and $\ce{OH-}$ is stronger than the bond in $\ce{NaCl}$. Also I don't see a case for the use of an equilibrium in this case because $\ce{NaOH}$ almost completely dissociates into its ions $\ce{Na+~~ OH-}$.

Pure water, pH 7, has a concentration of roughly .1 ppm (100 ppb) of hydrogen cations (protons), in the form of hydronium ($\ce{H3O+}$). This is a natural state of equilibrium; the solution wants neither more nor fewer protons (if it did, it would take them or give them up, and so it would not be "neutral"; that's a bit circular, but it's a fundamental property of water).
Sodium hydroxide, $\ce{NaOH}$, is a strong base; it is a "proton acceptor", wanting desperately to share one of its lone pairs with a hydrogen cation. The sodium, which gave up the valence electron to the hydroxide, doesn't want it back, so the hydroxide effectively owns the electron. But, hydroxide isn't extremely electronegative, like say, the halogens; it doesn't want to own the electron outright and would rather share.
Hydrogen atoms, for their part, either want two electrons or none, and they aren't picky about which; again, this leads to a situation in which hydrogen would rather share. Hydrogen is the textbook definition of a neutral element in terms of electronegativity, and because of this it will readily form either ionic compounds with elements at the extremes, or covalent compounds with elements closer to the center. This makes combinations of hydrogen and halogens very acidic, because the hydrogen wants to get away from the electron-greedy halogen and form a more stable, equal covalent bond; the compound becomes a "proton donor".
Introduce $\ce{NaOH}$ to water, and the hydroxide ion will display its affinity for the protons of the hydronium. The water, however, was in equilibrium, and so the removal of the cations to form new water molecules, and the resulting release of un-paired sodium ions, disturbs this equilibrium. According to Le Chatelier's principle, this disruption is resisted with an opposing reaction (in this case the re-creation of sodium hydroxide from the sodium deprotonating water to form more hydroxide, releasing or re-releasing hydrogen cations). Because of this love-hate relationship between all of the component ions, the dissassociation of $\ce{NaOH}$ into water is resisted, but because of the extremely high solubility of Group I elements including sodium, it does eventually happen.
Sodium chloride, however, is a salt. It is quite literally the byproduct of an acid and base getting together to form water. The chlorine and sodium ions, abandoned by their respective proton and hydroxide when those two formed the much more stable polar covalent water molecule, form a marriage of convenience between them. The sodium, lacking the electron that the hydroxide now shares with the proton in the water molecule, is attracted to the chlorine, which had as an acid stripped a hydrogen of its electron to form a proton. The chlorine is very happy to "own" the extra electron that produces this charge difference, the sodium is equally happy to be rid of it, and as such, there is no "shock" to the system where the introduced components try to do something that disrupts the equilibrium of the water. Thus, there is no resistance to the natural tendency of the polar water molecules to dissociate the sodium and chloride ions (which happens readily, because the sodium and chloride are such opposites in terms of electron affinity that only their charge difference holds them together, and so any other charge difference, even a partial one, can pull them apart).

This is oversimplified to the point of being almost untrue.  As it stands, it currently has a red herring in argument.  If the solubility of NaOH is being considered, is is pragamtically impossible to discuss its preference to take a proton from hydronium because very likely more than 100ppb of NaOH is added.  So, to advance this argument you need to consider hydroxide attack of neutral water, which is an enthalpically neutral process.  Also, the idea of 'disturbs the equilibrium' should be reconsidered as 'shits the equilibrium'.

shifts.  Oops.  Heh.