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In a galvanic cell, why does an electrode with a higher potential "pull" the electrons from the electrode with a lower potential?
I'm doing Grade 12 Chemistry and I'm unsure of this, taking the hydrogen electorde as having a potential of 0 volts.
I'll use the copper-zinc cell as an example. Just let me know if I've got the full logic right:

Zinc has higher electronegativity, so due to the equilibrium $\ce{Zn <=> Zn^{2+} + 2e-}$, we have more charge building up in the Zinc electrode than in the Copper electrode. (Given both half-cells are of same concentration and in same temp)
We have a potential difference between the two electrodes. This means some charge from the Zinc electrode will move to the Copper electrode in such a way that both electrodes will momentarily have equal charges.
Since the charge in the Copper electrode has gone up, we have added electrons to the equilibrium reaction $\ce{Cu^{2+} + 2e- <=> Cu}$. We therefore create more copper.

Is this why an electrode with a higher potential will "pull" the electrons from the electrode with a lower potential?

I'm not sure where you're getting your figures for electronegativity, but a quick Google search found that electronegativity for copper is 1.9 while it's 1.65 for zinc. Did you mean to compare electrode potentials instead?

There are not many problems where electronegativity helps you find the answer.  It is popular for some reason amongst newer chem students but it shouldn't be.
Here's what happens in the Daniell cell:
Metallic bonds in zinc are broken and zinc atoms each lose 2 electrons.  Both of these processes are endothermic.
Zinc ions are hydrated. This is exothermic.
Copper ions are dehydrated. This is endothermic.
Copper ions gain 2 electrons each and metallic bonds are formed. Both of these processes are exothermic.
Each of the above processes also involves an entropy change.
Only when all of these factors are considered can the direction of the electrochemical reaction be determined. Or, you can look at a table of standard reduction potentials.
By the way, silver's electronegativity is 1.93 but the reduction potential for $\ce{Ag+}$ is $\pu{0.8 V}$ while $\ce{Cu^2+}$ is $\pu{0.34 V}$.
In terms of charge, electrons flow from anode to cathode. There is no charge built up because of the salt bridge.


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How dangerous is taking a bath in Coca Cola?
What can be the effects for the body to be immersed in Coca Cola for a while? A friend commented me that the coke is quite acidic (pH 2.5 if I recall properly), and we were guessing what could be the effects for the body after a time of immersion.
I read 20 Practical Uses for Coca Cola – Proof That Coke Does Not Belong In The Human Body and I see some things like:

E338 – Orthophosphoric Acid. This can cause irritation of the skin and
  eyes.

But I am guessing: what are the problems of a long contact of coke with the body - not drinking, but being in touch with it.

I wouldn't put too much stock into the link. For example, the link says that the stomach releases "calcium" to neutralize acids. This doesn't make any sense; elemental calcium has no acid/base properties. Calcium ion actually behaves as a weak Bronsted acid in solution. I'm thinking the author meant some form of bicarbonate ion - a common biological buffer ion.

And what do you know, I knew it. "The stomach is protected by the epithelial cells, which produce and secrete a bicarbonate-rich solution that coats the mucosa. Bicarbonate is alkaline, a base, and neutralizes the acid secreted by the parietal cells, producing water in the process. This continuous supply of bicarbonate is the main way that your stomach protects itself from autodigestion (the stomach digesting itself) and the overall acidic environment." - http://health.howstuffworks.com/human-body/systems/digestive/question464.htm

I would take anything you read on that blog with a grain of salt.  I've examined a few of the scientific papers they've cited in other articles and they definitely cherry pick their data.

@jonsca thanks for clarifying that I shouldn't rely much on that blog article. It was the thing I found as source to coke vs human body, so I'm willing to see if someone can provide further (and more reliable!) information.

No, it's a fine question to ask from a safety point of view, that was just an aside :)

@Dissenter to make it more clear to me (I have very basic understanding on Chemistry): would the damage on the skin be bigger based on how much acid the coke is? What's the natural defense of the body agains this acid? Thanks for clarifying with your comments!

The acid would likely dehydrate your skin by taking all the moisture (specifically, water) and turning it into hydronium ion. Note that sulfuric acid (which is admittedly exponentially stronger than Coke as an acid) is a potent dehydrating agent. That is its main danger.

Hydrochloric acis is probably more likely than Orthophosphoric acid to cause irritation of the eyes and skin. Yet the former is the acid in your stomach. What's missing from the article: concentration. That is an inexcusable omission.

Cola and soft drink products are generally in the pH=2.5-3.5 range.
This is the same pH range as vinegar, lemon juice and most fruit juices. Note apple itself is around 3.0-3.5, so it is not about some magical trick of food industry to kill us all.
Does Cola dissolve your body traceless? No.
Do you want to soak your body in it for hours? Would you soak your body in vinegar for hours? Or even water...

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Taiwan propylene explosions one week ago
Is it wrong way to use WATER to relieve the big disaster(propylene leak).
What is a better way to relieve propylene leak in a street about large area?
Any possible choice to prevent explosion?

Could you post a link to where you saw your information? Also, propylene is a gas so water would not help.

Link:http://tinyurl.com/nzpfvfb

From a latter news report: http://www.cbc.ca/news/world/taiwan-gas-explosions-at-least-26-dead-267-injured-1.2724670 It states that the propene leaked into the sewer pipes. It was the explosion from the propene in these pipes that caused the damage. This is why the fire department could not find the leak since the propene was coming out of the sewer in multiple locations. As you can see in the picture at the bottom of the page, a long section of the pipe exploded. This would only happen if air was mixed with the propene.

Propene gas is flammable and has a higher density ($1.81 \ \mathrm{kg/m^3}$) than air. It therefore easily forms explosive mixtures with air when it accumulates near ground level or, like in this case, inside an underground channel without much ventilation. While propene itself is color- and odourless, it is described in this article that "eyewitnesses reported a smell of gas (...) coming out of manholes". This smell comes from volatile compounds which are added to the gas in small amounts to make the detection of leaking gas easier. However, by the time that the smell was noticed, enough propene had already accumulated underground to exceed explosion limits.
Detection of underground gas leaks by odour alone is therefore not a reliable means to prevent possible explosions, namely when the gas is heavier than air. It is also not possible to exactly localize the leak when the gas escapes from multiple manholes which are spread over a large area. Gas sensors in the underground pipeline network could provide earlier warnings about leaks, and gas accumulation could be prevented in this case by shutting down the affected pipeline. However, prevention of leaks in the first place (by regular inspections) should still have higher priority.
The MSDS data sheet for propene states that shutting down gas supply and letting the fire burn out while cooling the surroundings with water is the recommended way of extinguishing burning propene. So using water to cool down the roads with the aim to reduce ground temperature was not completely off, but trying to fight the fire directly with water is unadvisable.

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Why is the recipe of Coca Cola still a secret?
Why is the recipe for Coca-Cola still a secret?
I think that given the current state of technology, it should be proficient enough to find any of the secret ingredients in Coca Cola.
Any thoughts?
Can we make a 100% identical clone of Coca-Cola next week?

Maybe this is is an offtopic, but can really a secret of some food or drink exists nowadays? Each manufacturer is obliged to print product ingridient on it. Random example from internet - http://4.bp.blogspot.com/-ltZT8FGRSkg/TyLjis018ZI/AAAAAAAAHTo/RVJ1iNuMzcA/s1600/DSC_0270.jpg

@Berlin where do I get "spice" and "flavoring" from then? Even without those, if you chucked a bunch of all those ingredients in a bowl, how similar do you think it will be to the packaged product?

OT:http://en.wikipedia.org/wiki/OpenCola_(drink) is enough

I've looked at the OpenCola recipe and found it lacks vanilla, which is a key component of the flavoring of Original Recipe Coke. How do we know this? Back in '85 when New Coke was introduced, the world market for vanilla suddenly and inexplicably collapsed. Subsequently, market analysis showed that up until that time the Coca-Cola Company had been the world's largest user of vanilla, and if you know what you're looking for you can definitely taste the vanilla in "real" Coke. So if you want to modify the OpenCola recipe slightly, add some vanilla extract. Things go better with OpenCola... :-)

"Secret recipe" = good marketing. As for analysis; I guess its more difficult than we think. For example, I have a friend who used to work in the lab for a major canned soup company, for a team whose major responsibility was creating clones of another major soup company's products. Apparently, most if it was done just by trial and error with taste testing, and there's a lot more to it than just throwing everything in a can and calling it a day. Or just look at beer for example. Ingredients, simplified: Water, barley, hops, yeast. And yet...

See also http://cooking.stackexchange.com/q/46296/24117

Also http://en.wikipedia.org/wiki/Coca-Cola_formula

But has there been research as to whether or not it still contains traces of cocaine? If, in fact, it actually did in the past? Of course, finding original sources of Coke is probably next to impossible.

There are different angles this question can be answered:
Chemical point of view:
A full analysis of a totally unknown mixture is painful and extremely costly. It is always helpful to know how many components you are looking for; what types etc. In this case, it is not enough to analyse the elemental composition or some pure elements, but Coke contains a lot of natural products and mixtures like caramel. Imagine how you identify (and correctly describe the production parameters) caramel in a dilute solution...  Also, liquids with extremely high concentrations of sugar and different acids are unpleasant for analysis, as you often have to separate these main component, so you can identify only small components.
That being said, it is not an impossible task; there are many methods to identify e.g. natural products based on DNA traces. But consider another factor:
Business point of view:
The recipe of Coca Cola itself worth nothing.
Most probably any competent soda maker can make a drink that 99% of the consumers cannot distinguish from Coke by the taste (and could do that decades before). However they don't do because no one would buy a drink that tastes like Coca Cola, but made by others.
The value is in the brand. People buy a fake Rolex if they cannot afford a real. But anyone can afford a Coke- why would buy a fake? If you are in beverage business, it is an imperative to make a drink that is somehow DIFFERENT than the other ones!

While true for companys like Pepsi or Dr Pepper I think supermarket's own brands would do well by being an exact copy. Not least because they are currently really really bad

There are many valid reasons to buy a fake. Remember that Coca Cola is one of the most reviled companies in the world. Many people refuse to buy their products as a matter of principle.

I'd buy a fake Coke. Just the other day I stopped at a store to buy two 2L bottles, and it set me back $6.50 (USD). I could have bought six store-brand 2L and a pack a gum for the same price.

I think your analogy is backwards.  Rolex is associated with exclusivity - the point of a fake Rolex is for people to trick others into thinking you're balling uncontrollably with your super-expensive watch.  Nobody is going home at night to tell their wife "I saw some guy at work today drinking a brand new Coke!".  If you can duplicate the recipe and charge half the price, and everybody knows it, you'll dig deep into Coke's sales. Whereas you can sell watches with identical materials to Rolex, but if it's called "Barnaby's wrist-wear", few will care.

@terdon The fact one of their ingredients is a by-product of cocaine production really doesn't help.

I would totally buy fake coke if it was cheaper than normal coke but tasted the same or almost as good. I don't drink coke because it is cool but because I like the taste. So far I haven't found any drink that appeals to my taste receptors as coke from a can. Pepsi is something completely different and all other brands that try to indicate they are coke are mostly disgusting and have nothing to do with the taste of Coca Cola.

Note that making an exact copy has a high chance to find yourself on the receiving end of a lawsuit.

@vsz No, it is incorrect. No IP protects Coca-Cola recipe, but a trade secret. Per definition, it is possible to copy a trade secret. However, you most probably cannot advertise it as the same recipe.

Coca-Cola has slightly higher acidity than most soft drinks, and a vanilla component that I haven't found in any other cola. I really don't care much one way or the other - I like Coke, but I like Pepsi and RC and whatever too. (Especially Pepsi Throwback - real sugar! Yummmmmmm!! :-)

With modern technology it would be a relatively straightforward process to determine all of the ingredients in Coca Cola.  It would require time and money, but very doable.  My guess is that companies that compete with Coca Cola, like Pepsi, know exactly all of the ingredients in Coke.  This is part of the process known as "reverse engineering" and is done all the time in industry.  Pharmaceutical companies, food companies, plastics companies, any industry that uses specialized formulations do this type of chemical analysis to better understand what the "competition" is up to.
The chemical analysis is actually the easy part.  However, Coke and all of these other manufacturers don't just take a list of chemicals or ingredients, mix them in a tub and then put it in the bottle.  The manufacturing process is every bit as important as the ingredients and much harder to determine.  For example,

in what order are the ingredients added
how long do you mix the batch before adding the next ingredient
at what rpm do you mix them
what kind of mixer do you use

and the list goes on and on.  There might be 100 ingredients in the Coke formula, but there are probably over 1,000 variables in the manufacturing process, and there is no straightforward analytical method you can use to specify all of the process parameters.  The ingredients is the easy part, the process is what stops someone from making a replica product - well that and legalities.

If you analyzed Coca-Cola and found that it contained 100 different chemical compounds in whatever proportions then, knowing that Coca-Cola is chemically stable, it would be pretty easy to mix the compounds in the right proportion to obtain coke. It would, of course, be economically unfeasible to obtain pure samples of all of those compounds and that is the problem here. That is why you need to work out the manufacturing process, since it's overwhelmingly cheaper to make some subset of the compounds by caramelizing sugar, for example, than by mixing the pure compounds.

Greg hits most of the points, but there's one other angle I'd like to add:
Practical Point of View: One of the well established Coca-Cola ingredients is decocainized coca leaf extract. The DEA is, to say the least, not exactly generous in allowing this import. This effectively prevents the manufacture of a Coca-Cola equivalent anywhere in the United States. A competitor producing an exact clone of Coca-Cola but forced to import the finished product from outside the country is likely to find it difficult to significantly undercut Coca Cola's efficient distribution network.

Because it makes no sense to do so.
Even if someone uncovered the exact ingredient of Coca Cola or even if you made a Coca Cola recipe that tastes better than real Coca Cola, they wouldn't able to match the existing brand power of Coca Cola, nor the consumer's brand recognition and history with the general public. Coca Cola had poured billions and decades into making one of the most recognisable brand in the world, their brand is ingrained very deeply into the public's minds. A new competitor wouldn't able to replicate that overnight.
Of those companies that do may have the brand power to match Coca Cola and enough R&D budget to match, say PepsiCo, they wouldn't want to be seen dead advertising that they had made an exact clone of their competitors. It works better if they make and market themselves as a better drink than as a clone of a competitor. It's very likely that PepsiCo and Coca Cola knows pretty well their competitor's recipes, but they have absolutely no incentive to advertise that.
Also, even if you manage to uncover the exact composition and manufacturing process of the product, the public still won't believe it unless The Coca Cola company itself admitted that a given recipe is 100% truthful. A large percentage of the populace would always remain skeptical. Noone would give you research grant to reverse engineer Coca Cola.
Tasting isn't an objective experience, the way you experience a drink is affected by what's in your head just as much as what the chemistry of the drink. Even water from the same exact garden hose tastes different depending on how you are presented it with. For the same reason, even if you put an actual Coca Cola into a supermarket brand Cola bottle, a lot of people would still think that that bottle of Cola tastes worse than actual Coca Cola.
Finally, consumers won't buy fake products when the real products cost something like \$1-\$2 a can. If you drink a can of fake cola in public, that says something about your character, probably not in a way most people want themselves to be perceived as.

If we have high technology to find the secret of Coca-Cola then they also have high technology and money to hide it. If it would be possible then don't you think today there would hundreds of brand making the same drink?!
Till now there are many brands which have tried to clone Coca-Cola but they haven't succeeded, and that means till today they are able to hide their secret, so we don't have the technology that can find the secret of Coca Cola.
Here are some links which claim that they have cloned Coca-Cola
http://www.topsecretrecipes.com/Coca-Cola-Classic-Coke-Copycat-Recipe.html
https://www.homebrewersassociation.org/forum/index.php?topic=6017.0
http://www.instructables.com/id/Fakola-cola-clone/
You can find many more on Google, but I don't think any of them are able to find the secret of Coca-Cola, because if they had found it, then they would be billionaires and their drink would be in the market.

"If we have high technology to find the secret of Coca-Cola then they also have high technology and money to hide it." What does that even mean? If we have enough technology to "find the secret of Coca-Cola" then, by construction, the Coca-Cola Company does not have enough technology to find it.

@DavidRicherby It means that if we have advanced in technology then they also have advanced in technology to maintain their secret.

What technology do you imagine the Coca-Cola Company can use to hide the contents of a can of coke?

@DavidRicherby They put some components in the drink which is very difficult to detect using todays technology, that mean that they have better technology which can hide their secret.I have seen in "Megafactories"(show on national geography) that they put some components that are secret.

Coca cola recipe secret is a marketing scheme.
Obviously with LC-MS you can determine exactly what is in it. Also, many people need to know what ingredients need to be added as coca cola is made all over the planet. The idea that only 3 people know the recipe and that they can't ever board the same plane and that they have a piece of paper in a safe, all bullshit.
And does it really matter how a coke brand tastes? Brand image and marketing is all that matters. And adding a lot of sugar/sweetener.

Universities have HPC's that could probably do it in a week.
Point is, someone has to pay for it and it would be very costly and illegal.  Perhaps the reason it has not been done.

why and where would it be illegal?

It is not illegal. No IP protects it, and nothing forbids the chemical analysis of a beverage.

agree it's costly but i don't think it is illegal

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How do I separate graphite powder from light oil?
I have a sample of very fine graphite powder suspended in light oil. I've attempted to separate the two by placing the mix in a 2 micron filter sock. The oil will slowly drip through. If I apply pressure the graphite and oil will force their way through. No pressure, I believe the graphite will collect on the filter and eventually blind the filter preventing oil from coming through.
Any ideas on how I can separate and re-use both items?

Have you tried a centrifuge to help concentrate it? What do you mean by 'very fine' - is there a median particle size that is known?

"Any ideas on how I can seperate and re-use both items?"
The devil is in the details:

what is particle size and concentration?
what is the light oil?
what do you mean by reusing them?

Depending on particle size (pumped) filtration (paperfilter, colloid filter, celite ) should work as well as centrifugation as was recommended by @Jon Custer. Is it that urgent? Also, cannot you just try it instead of asking it?
What is the oil?
Whatever you do, your graphite will be oily. It does matter if this oil has high-boiling point/waxy components... On the other hand, you may able to use something with low boiling point and just remove it under vacuum / heating, maybe can skipp the whole filtration part.
What do you mean by reusing them?
As I said, there will be remains of oil on the graphite. Also, most probably, you will have fine powder of graphite remains in the oil whatever you do. How do you clean those up it depends on what do you need it for. Do you need to wash the graphite and re-filter the oil? You should know/test that. Also, cost-wise does all this fuss make any sense? Neither graphite nor the oil sound something expensive/special. Also, the more unique, expensive they are, most probably their cleanup will be more demanding (as you need to shoot for higher purity).


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SN1 Polar Protic Solvent Stabilization of Substrate
I know that the standard textbook answer is that a polar protic solvent is able to better stabilize/solvate carbocations through hydrogen bonding.
I understand that a solvent such as water or methanol is capable of hydrogen bonding, but what about the substrate? The substrate is a hydrocarbon; how is the substrate capable of reciprocal hydrogen bonding?
Is this stabilization better described as a dipole-ion interaction rather than a hydrogen bond interaction because I don't see how a hydrocarbon can form reciprocal hydrogen bonding interactions.

The standard explanation for the rate enhancement of SN1 in a polar protic solvent is hydrogen bonding to the leaving group, which stabilizes the building negative charge in the transition state (i.e., not the carbocation). I'm not familiar with the term reciprocal hydrogen bonding. Can you provide an example of a compound capable of this? You are right that a hydrocarbon cannot participate in hydrogen bonds.

I know water can participate in hydrogen bonding because each water molecule has a partially positive end and a partially negative end. So water molecules can hydrogen bond among themselves. But what about carbocations ... is this dipole/ion interaction a true form of hydrogen bonding?

Hydrogen bonding is a very specific type of interaction where a polarized X-H bond interacts with lone pair. The carbocation doesn't fit either component of that definition (in any traditional sense anyway).

what is reciprocal hydrogen bonding?

I know that the standard textbook answer is that a polar protic
  solvent is able to better stabilize/solvate carbocations through
  hydrogen bonding.

In an $\ce{S_{N}1}$ reaction, a polar protic solvent can stabilize the leaving group through both hydrogen bonding and solvent dielectric effects.  While these same solvent dielectric effects can also stabilize the carbocation intermediate, I don't think hydrogen bonding plays a role in the stabilization of the carbocation itself.  First off, any protic molecule close enough to the carbocation to hydrogen bond would simply react with it instead of forming a hydrogen bond. Second, since the carbocation is positively charged it cannot form a hydrogen bond.  Again, solvent stabilization of the carbocation occurs through dielectric (dipolar) effects of the medium.

Can you explain the left half of this picture? http://50.57.82.128/sites/default/files/userfiles/48/uploaded_images/hbonding1.png

Not really, I see water hydrogen bonding with itself, but I don't see any dashed lines between water and the carbocation indicating any type of interaction.  So I am puzzled by their statement "H-bonds stabilize the carbocation."

Yeah I don't understand it either.

What about this can you explain this? http://o.quizlet.com/6zi2Vb-RZvTlH5S.YMksMg_m.png

In other words is the second picture describing hydrogen bonding or rather an ion dipole interaction.

ion-dipole (dipolar or dielectric) effects, not H-bonding

So why do textbooks and professors say h bonding stabilize carbocations?

Dunno, I didn't say it.  The only thing I can think of is  it can enhance the rate of carbocation formation by stabilizing the leaving group.  I just don't see how a hydrogen bond (where the H is somewhat positive) can stabilize a positive center.

But a hydrogen bond necessitates a dipole; why can't the partially negative oxygen "hydrogen bond" with the carbocation. (Scare quotes because I'm not sure if this is an actual hydrogen bond).

Here's a quote; can you explain what this means? "At the same time that the polar protic solvent is stabilizing the nucleophile, it also has the ability to stabilize any carbocations formed during the reaction. The lone pair of electrons on the solvent can donate electron density to the carbocation, making the carbocation more stable." That's not true hydrogen bonding is it? That sounds like Lewis acid/base stuff, but on the other hand, it's not permanent. I'm very confused.

That's not an H-bond.  A H-bond involves partial bonding of the hydrogen atom to something else.  What you're describing is a dipolar interaction.

"Protic solvents such as alcohols and water are even more effective solvents because anions form hydrogen bonds with the  hydrogen atom, and cations complex with the nonbonding electrons of the  oxygen atom." I agree with you ron, but this is a quote form Wade which directly states that these dipolar interactions are hydrogen bond interactions. Which makes no sense as I said in my OP and as you said in your excellent answer.

About the quote (where's it from?) - if the lone pair from the protic solvent is close enough to "stabilize", then why wouldn't it react instead?  Even if "that" (the lone pair stabilization hypothesis) happens, it is not a H-bond, it is lewis acid\base as you noted. So, 1) there is no H-bonding between the solvent and the cationic center, 2) there is dipolar stabilization of the cationic center by the solvent, 3) if the solvent is protic (like water, ethanol, acetic acid) it will react with the carbocation.

"...effective solvents because anions form hydrogen bonds with the hydrogen atom" - that's fine, that's H-bonding with the nucleophile and leaving group. "cations complex with the nonbonding electrons of the oxygen atom" well, that's OK I guess, I'd prefer to say "cations are stabilized by dipolar interaction with the nonbonding electrons on oxygen, as well as through the dielectric properties of the solvent as a whole".  Complex, stabilization - kind of similar, I guess.

The quote is from Organic Chem by Wade.

What do you mean by "dielectric properties of the solvent as a whole"?

Born equation: free energy of solvation ~ (e-1)/e times other stuff, where e is the dielectric constant of the solvent.  Therefor, the larger e, the greater the free energy of solvation of an ion in this solvent.


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What are the fumes that form above liquid nitrogen?
What is the composition of liquid nitrogen fumes? Is direct contact with these fumes harmful?

fumes from liquid nitrogen contain nitrogen $N2$ in gaseous state with some amounts of $N2O5$ which are not that harmful as the liquid nitrogen itself, anyway if you have inhaled those vapours and you are suffocating now then better visit a doctor, but its contact with keratinised epidermis would'nt be that harmful

@agharehanabbas N2O5 in the fumes? Seriously? By what mechanism can this possibly be formed in any notable quantity at this temperature?

Ignoring the question how it's formed, how does it end up in the fume? The stuff boils at 320K, it's frozen solid at 77K.

The visible fog that forms when liquid nitrogen is poured into an open container is almost entirely water fog: the boiling nitrogen chills the air above it, causing the humidity in the air to condense into fog.
There may be some microdroplets of liquid nitrogen in the fog too, but the air, even after it's been chilled, is still much warmer than the boiling point of nitrogen, so there won't be much.  (You can condense air by running a continuous stream of liquid nitrogen through a heat exchanger and blowing air over it, but mostly you get liquid oxygen when you do that.)
Liquid nitrogen has a relatively small heat of vaporization, as these things go: 199.2 kJ/kg according to Air Liquide.  (Compare 572.2 kJ/kg for dry ice.)  That plus the Leidenfrost effect means it's actually rather difficult to freeze yourself with the stuff. There's a standard demo where you pour liquid nitrogen into an audience member's cupped hands. The Leidenfrost effect prevents it ever coming in contact with their skin, and it only leaves a mild chill behind.  Try this with dry ice and your "volunteer" will need emergency treatment for frostbite.
Gaseous nitrogen is second only to the noble gases in its inertness, and it makes up 78% (by volume) of the air you're breathing as you read this, so obviously the "fumes" aren't harmful in themselves.  However, as was mentioned in an earlier answer (which seems to have been deleted), if you let liquid nitrogen evaporate in an enclosed space, it can "displace" the oxygen that makes up another 21% of the air, at which point anyone in the area may asphyxiate.  The sensation of suffocation is triggered by having too much CO2 in your bloodstream, not by inadequate oxygen supply, so you might not even notice that anything was wrong until you collapsed.  That's the second most dangerous thing about evaporating liquid nitrogen.
The most dangerous thing is, if you let liquid nitrogen evaporate in a sealed container, it's liable to explode on you.

Hmm. If I miss the Dewar wherein I want to prepare dry ice acetone, I pick it up with bare hands or gloved hands. I still have all my fingers and haven’t experienced any frostbite. Of course, I make sure that I’m reasonably fast.

The fog / vapor / fumes that come off a container of any liquefied gas will be a combination of condensed water vapor and whatever it is that is boiling. It is no more or less dangerous than the gas itself. In the case of nitrogen, oxygen, helium, neon, argon, sulfur hexaflouride, methane etc. you can put your nose right up to the liquid's surface and take a really deep breath, no harm will come to you unless you are stupid enough to breathe it for an extended period. Oxygen, of course, is the exception but too much O2 is bad for you as well. Don't try it with chlorine.
Regarding the Leidenfrost effect and pouring liquid N2 into someone's cupped hand, lets be sure the palm is facing down, not up. Leidenfrost only works for a couple of seconds on room-temperature objects like people.

In my experience, pouring liquid N2 onto the back of someone's hand is much more likely to give them (mild) frostbite than the reverse, because people tend to have hair on the backs of their hands, which will disrupt the Leidenfrost layer and/or draw the liquid to the skin by capillary action. As for the effect only working a couple seconds, I'm not sure that's true in this case; even a substantial quantity of lN2 will have serious trouble chilling your hands below the Leidenfrost point for lN2 (I don't have that number to hand but I'm guessing it's order of 100 Kelvin).

It's your hand. Would you prefer I pour liquid N2 on the back, where it can spill off, or into your cupped palm, where it will sit and boil? And it doesn't have to chill your hand enough to stop the LE, just enough to kill some tissue.

I've done this, both ways, on multiple occasions. I predictably get frost rash on the back of my hand, and have never had any problems with the palms.  I think you're overestimating the amount of cold in lN2 -- let's assume all of the heat required to vaporize 10 cc of it is drawn directly from someone's hands (which is unlikely), we have 10 cm^3 * 0.8 g/cm^3 (density of lN2) * 199.2 kJ/kg = 1.5 kJ. It takes maybe ten seconds to vaporize under these conditions, so that's 0.15 kW heat draw for ten seconds. (cont'd)

Your average human at rest generates 100 W of heat output over their entire body surface, so yes, if all the heat required to vaporize that much lN2 were drawn directly from a small patch of skin you would be injured.  But this is not what happens; most of the heat required is drawn from the air -- room temperature is 300K, lN2 boils at 77K; even with the low heat capacity and density of dry air (~1kJ/kg.K, 1.225 kg/m^3 -> 1.225 mJ/cm^3.K) that is enough to vaporize all of the lN2 while only cooling the air a few degrees.

(The actual boiling process is convective and explosive.  If you modeled it in detail you might find that the Leidenfrost effect doesn't even come into play, the glob of lN2 is instead suspended on a convective updraft for the second or so that it takes to fragment into hundreds of smaller droplets which are flung all over the surrounding area.)


text:
Structure of fluoroalkylsilane
I want to use FAS (fluoroalkylsilane) in my molecular dynamics study. How can I find it's real structure? So that I can find proper forcefield for simulating the molecule. It would be very nice if you mention a better method for simulating this molecule. Like making a initial configuration in a software and then optimizing the geometry. I don't know about DFT so much, but can I use that somehow?

Best,
-Kazem

fluoroalkilsilanes are a family of compounds. You need specific name.

I added a picture to question (FAS17), that's what I want to simulate, and the chain containing F atoms will be smaller like FAS9 or FAS7, ...

It seems like you are just looking for a very qualitative picture (not knowing about the level of theory you want to apply). Also you need to mention what property you are looking for! 'Simulating' can mean just anything! DFT is a QM method calculated from first principles, which cannot be combined with Molecular dynamics simulations. You can try minimizing the energy using Chem3D, which has a good GUI. You can use UFF (universal force field) that has parameters for almost all atoms available in it. If you could rephrase your question a little clearer, I could help you more.

I want to study adsorption of FAS-9 on quartz 001 surface, I have made my quartz surface using CLAYFF and a vacuum slab on it, but I'm stuck with adding FAS-9 to the vacuum slab, and also I don't know what forcefield is proper for FAS-9 and interaction of FAS-9 with quartz surface.


text:
Is sublimation a property of all substances?
I wonder if sublimation can be done only on certain substances like iodine, or it can be done on all substances.
In other words, can sublimation be done, for example, on water? If it cannot, then why?

Look up the term “phase diagram”. At different temperatures and pressures different phases are favored. Think about water. At 0 °C and 1 atm water and ice are equally favorable. There can be freezing or melting. At other temperatures and pressures there can be sublimation and deposition.
The sublimation of water is involved in freeze drying and freezer burn of ice cream.
It turns out that iodine sublimes at room temperature and atmospheric pressure but there’s nothing special about iodine compared to other substances on a fundamental level.

Sublimation refers to molecules passing from the solid to the gas state without going through the liquid state. This happens all the time to every solid, including water, it's just a question of at what rate it sublimes and how that rate compares to its rate of melting in the same conditions.
Snow, for example, often noticeably sublimes instead of melting, especially when in sunlight only a few degrees above freezing temperature. It is more prone to sublime than ice is due to its lower density, which allows air to pass into it and warm more than just the surface.
The reason iodine, for example, so noticeably sublimes is because at normal temperature and pressure it cannot exist for more than an infinitesimal amount of time in the liquid state. There are regions in the phase diagrams of most substances where the liquid state cannot exist in equilibrium with either the solid or gas state when temperature or pressure changes.
For water, this is at very low pressure, below 1/100 of normal atmospheric pressure. When you heat ice at this pressure, it will go directly to gas without being in the liquid state in any appreciable amount.


text:
Why is the size of Al3+ less than that of Li+?

Why is it so that the size of $\ce{Al^{3+}}$ is lesser than $\ce{Li+}$?

I'm a bit confused about this one as the aluminium(III) ion is composed of 10 electrons and 13 protons, while the lithium ion has only 2 electrons and 3 protons.
Electronic configuration of $\ce{Al^{3+}}$ is $\mathrm{1s^2 2s^2 2p^6}$ and that of $\ce{Li+}$ is $\mathrm{1s^2}$.
Now, it seems obvious that the size of $\ce{Al^{3+}}$ should be greater than that of $\ce{Li+}$, but it is actually the other way around. Why is this? I think that it is a kind of exception, but if anyone can explain the reason behind this, it will help me in my upcoming tests.

Now, it seems obvious that the size of Al3+ should be greater than that of Li+

No it does not. It is true that electrons repulse each other, however, they all are attracted towards the nucleus.  The real 'size' of an atom or ion is a result of compromise between electron-electron repulsion and electron-nuclear attraction.

I think that it is a kind of exception

Obviously, it is not.

Why is this?

Commonly, there are three consideration, used to rationalize basic trends in atomic/ionic radii.

electron-electron repulsion (negative charges repulse each other)
electron-nucleus attraction (negative charges are attracted towards positive ones)
electron shielding (electrons in shells closer to nucleus 'shield' outer electrons from some charge of the nucleus.

While electrons repulse each other, increasing nuclear charge more then compensates it; so if not for shell organisation of electrons, atoms would monotonically decrease in size toward the end of the periodic table, as it is observable in each period 1—3.
However, inner electronic shells effectively shield the outer electrons from the nucleus, so they 'see' a nucleus that is a) larger in size and b) has charge equal to nuclear charge minus number of electrons in inner shells. Given that, outer electrons 'see' a nucleus with low charge density, so outer shells progressively increase in size within the same column.
In addition to that, electrons still repulse each other, so positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions.
Now, consider $\ce{Li+}$ and $\ce{Al^{3+}}$. For the former, 2 outer electrons feel the charge of three protons. For the latter, 8 outer electrons feel the charge of 11 protons in the nucleus. Such great charge easily beats all other factors, so the latter cation is much smaller than the former.


text:
Does O2 have a color in the gas phase
I have noticed that liquid $\ce{O2}$ (I clarify it as $\ce{O2}$ because oxygen exists in several other forms which have different colors in the liquid state than $\ce{O2}$) has a light blue color to it.

In the solid state it has a light blue color like it does in the liquid state and then finally a red color if enough pressure is applied. It can also be pink, black, or metallic again depending on the pressure and temperature.
Does it also have a light blue color in the gas phase?

Background
$\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of solid oxygen with color ranging from pale blue to red to black.  In the liquid phase it has a light blue color.  This color is due to light absorption by the ground state triplet according to the following equation
$$\ce{2 O2(^3\Sigma_{g}) ->C[{h\nu}]\   2 O2(^1\Delta_{g})}$$
This absorption requires light from the red region of the spectrum (~630 nm).  If red light is absorbed , then blue light is transmitted or reflected and this gives rise to the blue color associated with liquid oxygen.
Answer
Note that the light absorption requires 2 molecules of $\ce{O2}$ and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller.  Hence, the probability of photon absorption in the gas phase is much reduced.  To the human eye oxygen gas will appear colorless (at normal and reduced pressures).  However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption.

It is a triplet-singlet transition, therefore it is spin forbidden. The transition can only occur if some other process compensate for the magnetic moment, which is more probable in a condensed phase.

@Greg As mentioned above, the conversion of ground state, triplet oxygen to singlet oxygen involves the simultaneous absorption of one photon by 2 triplets to produce 2 singlets.  Therefor, total spin can be conserved and the process is not spin forbidden, e.g. $\ce{ ^ ^ ~(v) (v) -> ^ (v)~ ^ (v) }$

The goal of my comment was to add the term "spin-forbiden" as it is something OP may heard about and can connect. I did not correct or argued your description.

@Greg Thanks for the clarification.  I didn't think you were correcting my answer, that is why I avoided using words like "right" and "wrong" in my comment.  I just wanted to elaborate your comment a bit further within the context of the original question, so that a reader would understand why, in this particular case, the photoconversion is spin-allowed.

An Oxygen molecule has an even number of electrons, but is paramagnetic. This means that it contains two unpaired electrons. The grounds state is a triplet (symmetry and term symbol $^3\Sigma _g^-$) and two low lying excited states one at ($^3\Sigma _g^+$) at 13121 cm$^{-1}$ (762 nm) and $^1\Delta g$ at 7882 cm$^{-1}$ (1270 nm) above the ground state. Their absorption is in the near infra red. Transitions to these levels are spin and symmetry forbidden so occur only weakly.  There are also vibrational transitions in the $^1\Delta g$ state at at 1064 nm  (v= 0-1)and approx 920 nm (v=0-2) but again these are not in the visible part of the spectrum.
In high pressure gas and in the liquid phase, oxygen is coloured blue and this cannot be due to these transitions as they are too long a wavelength.
The reason for the colour of liquid oxygen was first proposed in 1933 as being due to a collision induced transition involving two oxygen molecules that are transiently close to one another, just as 'collision' suggests, i.e. the absorbing species is transiently (O2)$_2$. Provided there is enough energy in the photon, the transition excites both molecules, and the photon energy is split between them. The complex formed in a collision can have quintet, triplet or singlet character. The two main transitions giving rise to the blue colour are from the transitions
$$ (^3 \Sigma _g^-)_2 + hv-> ^1\Delta _g(v=0) + ^1\Delta _g(v=1). $$
These transitions are around 633 nm (equivalent to 2*7889 cm$^{-1}$) and 577 nm respectively (the extra energy is because one vibrational quantum is excited). The transitions in (O2)$_2$ are rather intense compared to O$_2$ due to symmetry breaking, for example, destroying the inversion symmetry. (There are also transitions to $^1\Sigma g^+$ in the 350-400 nm region but these are weak and outside the range we can observe directly by eye).  It is also worth mentioning that emission can be observed from these bands, the most intense is called ‘Dimol emission’.
Collision induced transitions are well studied between atoms, but rarer in molecules. It may seem strange that a transition could occur at all since a molecule is a stable arrangement of nuclei and electrons and no such organised species is formed in a collision. However, it takes approx 10-50 fs for the ground state to be changed into an identifiable excited state when a photon is absorbed. A collision lasts for far longer than this, some 1000’s of femtoseconds (10$^{13}$ collisions s$^{-1}$) thus as far as the photon is concerned, it does not need a molecule, just a collection of electrons with energy levels of the right symmetry and energy that need exist only for a few tens of femtoseconds.

no, in gas phase it's colorless.
someone suggests me to be specific,
then let me show you oxygen gas's UV absorption spectrum calculated in Gaussian and visualized in GaussView. charge=0 triplet 1 atm 298K zindo

why is it then that O3 is light blue in the gas phase and dark blue in the liquid phase while O2 is light blue as a liquid and colorless as a gas because I always thought that the scattering of blue light and how it gets darker and darker in the sky is partly due to differences in O2 concentration and O3 concentration with O2 contributing more to the lighter blue and O3 contributing more to the darker blue.

I mean before there was a high enough concentration of O2 and O3 for aerobic life to occur the sky wasn't blue so it must have to do with O2 and O3 concentration.

@caters AFAIK the blue color of sky is attributed to different scattering of light of different color on microparticles of dust in the atmosphere.

Please substantiate your claims with references.

I know that the scattering of light on dust particles is part of why the sky is blue but there was a red sky before the Great Oxygenation Event because of high CO2, H2S, HCN, and other compounds(which scatter red light more easily than blue light) and low O2 and O3 and so the O2 and O3 must be part of why the sky is blue as well

this high CO2 is also the reason why mars has a red sky during parts of the martian year that aren't dust storms(during which the red color is then caused mainly by iron oxide dust)

@caters None of the molecules you noted scatter/adsorb any visible light. Red color of Mars sky is attributed to high amount of iron (III) oxide dust in the atmosphere.

well other compounds that were in the sky before the Great Oxygenation Event did scatter red light. I mean if we were to take out all or almost all the oxygen(O2 and O3 both as well as atomic oxygen which occurs in low concentrations but is there) we would notice as a result that red light is then more easily scattered. Oxygen also reflects blue light in the liquid state and absorbs red light. O3 does this more than O2 does(which is why it is light blue as a gas and dark blue as a liquid).

and imagine that you had the gases for an atmosphere and a white light. Also imagine that you add CO2 and other gases in earth's atmosphere first(besides oxygen) and this is in a closed system on a planet or moon with no atmosphere(Or a very thin one). As soon as you added oxygen the color of your closed system atmosphere would change even though O2 is colorless as a gas.

I would consider zindo hardly a method of proof, while uv/vis calculations tend to derive from the experiment by a lot.

@Ath the calculation method you choose is like picking a number from the phone book. It happens to be close to the experimentally observed, but that is just by coincidence.

So the answer here is just wrong.


text:
What is the name of the formula ΔE = q + w?
What is the name of this formula? Google could not provide me with an answer.
$$\Delta E=q+w$$

there might me some typing error, when i tried it provide result

@Freddy Keep in mind that Google results are influenced by your prior search history and which hits you follow the links to. Even so, this sounds textbook enough that a search should uncover something about it.

try $$\Delta Q = \Delta U + \Delta W $$

This is a statement of the first law of thermodynamics. The first law says that the energy of an isolated system is conserved. An isolated system is one in which neither mass nor energy is exchanged with the surroundings. This equation shows the two means by which a closed system (meaning a system in which mass is conserved but energy is not) can exchange energy with its surroundings - by heat, or by work.
$$
\Delta E=q+w
$$
All energy transfers can be described as either heat flow or as work done on or by a system, and therefore the total amount of heat energy that goes in or out ($q$), plus the total amount of work energy that goes in or out ($w$), must be equal to the total internal energy change of the system ($\Delta E$).
This equation is particularly important in chemical thermodynamics because we have no way of measuring the total internal energy of a real system directly - all that we can measure is the amount of heat and work that goes in or out of the system, and therefore the change in internal energy.

To be more precise, the expression of the first law we are talking about in this thread works for closed systems, not for non-isolated (the standard name is open, by the way), because for an open system energy can also be transferred by the third way: by adding/removing matter.

@Wildcat Good point - I was trying to avoid those details but you are right that I should be more precise.

The first law of thermodynamics.

It is the first law of thermodynamics, which was inducted through the repetition of many scientific observations related to the energy changes associated with chemical reactions.
Scientists observed that when the system transfers from one state to another, the total energy is neither destroyed nor created. It transfers from the system to the surrounding, this is known as the Energy Conservation Law, and one of the important results of this law is: "When the system transfer from one state to another, the change in energy of the system is constant; regardless of how the change happened."
The amount of energy change depends only on the initial state of the system and final state of the system.
There are several forms of expression of the first law of thermodynamics:
"The total energy change of the system and the surrounding is zero. The energy is either destroyed nor created "
As well as:
"If we start a system in a particular situation and then make a set of changes so that the system returns to its initial state, the result of the change in energy is zero."
Usually, we do not concern about the changes that lead to the return of the system to its original state, but rather the changes in energy that associate the change in the system from the initial state to the final state.
These changes in energy are the sum of changes of all kinds; the kinetic energy of atoms, molecules or ions that make up the matter, the kinetic energy of electrons, or the different types of potential energy due to the presence of different forces of attraction or repulsion in the parts of the system. We indicate to the sum of all these kinds of energy as internal energy with the symbol $E$.
It is difficult to measure the internal energy of the system but what can actually be measured is the change in the internal energy of the system $\Delta{ E}$, which is equal to the difference between the final internal energy and the initial internal energy:
$\Delta{E} = E_2-E_1$.
That is, the internal energy $E$ is a state function, so the change in $\Delta{E}$ depends only on its final value and its initial value.
When a physical or chemical change occurs, the energy between the system and the surrounding is exchanged in one or both of the following:
1- Change in heat energy:
a- The heat energy is flowed from the surrounding to the non-isolated system as in the endothermic reactions, where the internal energy of the system increases and the internal energy of the surroundings decreases.
b- The heat energy is flowed from the non-isolated system to the surrounding as in the exothermic reactions, where the internal energy of the system decreases and the internal energy of the surrounding increases.
2- Change in work, which is one form of energy, it is the potential energy. Here are also two cases:
a- The work may be done on the surrounding by a system as in the case of the expansion of the system, flowing energy from the system to the surrounding, that is decreases the internal energy of the system and increases the internal energy of the surrounding.
b-The surrounding conducts the work on the system as it does when the system compresses, so the energy transfers from the surrounding to the system, that is, increases the internal energy of the system and decreases the internal energy of the surrounding.
If we allow the flow of heat energy $q$ from the surrounding to the system, at the same time the system performs a work  on the surrounding, the internal system energy increases as a result of the absorption of heat energy$q$, while the internal energy of the system is reduced due to the loss of the potential energy  in the form of the work $w$, so:
The change in the internal energy of the system is thus:
$$\Delta{E} = q-w $$
On the basis of this equation:
1- Heat energy $q$ is :
a- Positive if it flows from the surrounding to the system as in the endothermic reactions.
b- Negative if it flows from the system to the surrounding as in the endothermic reactions.
2- Work $w$ shall be :
a- Positive if the system conducts work on the surrounding as in the case of gas expansion.
b-Negative if the surrounding conducts the work on the system as in the case of gas compression.
If we start with gas confined in a cylinder with a piston at room temperature and put the cylinder in boiling water, for example, the heat transfers from the surrounding (boiling water) to the system (the confined gas). Due to heat transfer, the gas expands and pushes the piston that is doing the work.If the amount of heat transferred from the surrounding to the system $\pu{400\text{Joule}}$ , and the amount of work done by the system was $\pu{300\text{Joule}}$, the change in the internal energy of the system $\Delta{E}$, is:
$$\begin{array}{  }
\Delta{E}~~~=~~q~~-~w\\
~~~~~~~~~=400-300\\
~~~~~~~~~=\pu{+100J}
\end{array}$$
In this case, the internal energy of the system increases by $100$ joules, while the internal energy of the surrounding is decreased by $100$ joules, that the change in the internal energy of the surrounding$\Delta{E}$ is equal to (-100 )joules.
The change in the internal energy of the surrunding can be calculated as follows:
$q$ for the surrounding is equal to $(-400)$ joules because the surrounding has lost heat energy, but $\mathrm{w}$ for the surrounding equals$ (-300)$ joules because the work is done on the surrounding
$$\begin{array}{  }
\Delta{E}_\text{(surrounding)}~=~~~~~q~~~~~~-~~~~~w\\
~~~~~~~~~~~~~~~~~~~~~~~~~=(-400)-(-300)\\
~~~~~~~~~~~~~~~~~~~~~~~~~=\pu{-100J}
\end{array}$$
We can see that the total internal energy of the system and the surrounding is zero ,or :
$$\Delta{E}_\text{(system)} = -\Delta{E}_\text{(surrounding)}$$
$$\Delta{E}_\text{(system)} +\Delta{E}_\text{(surrounding)}=\text{zero}$$
This is one of the forms that express the first law of thermodynamics:
"The total energy change of the system and the surrounding is zero. The energy is either destroyed nor created "

This is the first law of thermodynaics.
$$
\Delta E=q+w
$$
Delta E donates change of internal energy
q - Amount of energy supplied
w - work done by the gas

The interpretation of quantities in this answer is just plain wrong. $\Delta E$ is a change of internal energy of a system, $q$ is heat supplied to the system, $w$ is work done on the system.

oh yes i understood

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text:
Conditions needed for reaction between FeCl3 and KI
$$\ce{FeCl3 + KI + H2SO4 -> ?}$$
For the reaction to proceed, is it necessary to have the medium acidified, here with sulfuric acid?

What reaction are you thinking of?

FeCl3 + KI + H2SO4

Do you know what the products are?

No, it does not, you can drop $\ce{H2SO4}$ from the equation. However, in practives it is necessary to acidify $\ce{FeCl3}$ water solutions to suppress hydrolysis.

I have dissolved iron(III) chloride in de-ionized water (pH ~ 7) many times and I have yet to see hydrolysis. I think you mean something else.

@LDC3 Nope, I mean exactly this. Try to measure the pH of such solution.

So far, the only information I found is presented here.
The reaction should be able to proceed in both acidic and neutral solutions. Since iron hydroxides are slightly soluble, the solution is acidified so the hydroxides cannot form.


text:
Corrosion of metals and observations in the agar medium?

I want to know which metals oxidize and which is reduced and changes in medium and observations?
According to my knowledge more corrosion will take place in Zn and Fe also corrode. Then copper should reduce. So, for both reactions, will copper acts as the cathode?
I need general but complete explanation. Thank you

You have correctly identified zinc and iron as the anodes, since they are both less noble than copper. In the corrosion reaction, they are both oxidized and dissolved in the aqueous gel.
\begin{aligned}
\ce{Zn ->& Zn^2+ + 2e-} \\
\ce{Fe ->& Fe^2+ + 2e-} \\
\end{aligned}
The electrons flow to the copper cathode, where oxygen, which is dissolved in the liquid medium, is reduced to hydroxide.
$$\ce{O2 +2H2O + 4e- -> 4OH-}$$
As a consequence, the pH of the solution around the copper cathode rises, and the phenolphthalein indicator changes its color from colorless to violet-red in this area.

$\ce{Fe^2+}$ reacts with $\ce{K3[Fe(CN6)]}$ to form a deep blue iron(II,III) hexacyanoferrate(II,III).
$$\ce{K+ + Fe^2+ + [Fe(CN6)]^3- -> KFe[Fe(CN6)]}$$
Therefore, the solution near the iron surface turns blue. Zinc does not react with $\ce{K3[Fe(CN6)]}$ and the solution around the zinc anode remains colorless (source).


text:
How much water would one loose if one burns hydrogen and oxygen?
By using electrolysis one can separate hydrogen and oxygen, but when burning hydrogen and oxygen together steam is released, therefore creating water. My question is:

how much of the original quantity of water would be lost after this
  process?

Theoretically, none.
Actually, since the gases are produced in a damp environment, there would be a lot of water vapor in the gases. You could easily remove most of the water vapor, but some (maybe as much as 1%) would still be in the gases.
I think it would be best to elaborate on what you mean as water being lost?

I agree with the theoretical answer that it's none, but we don't leave in a theoretical world. What I mean by lost is that there might be a delta of water, that is new quantity minus old quantity, because of the enviroment conditions. Giving that the enviroment is a closed container with hydrogen and oxygen produced from electrolysis and then fire is applied. Obviously natural water isn't pure, but let's take the other components out of the equation. So then what we're left with is the new quantity of water, which theoretically is equal to the old quantity, but practically?

In a closed container, you wouldn't lose any water.

So that being the case: A solar panel splitting water then recombining them using combustion would lose no water but produce heat... Just where does the loss - there must be one - occur? Is it that it takes loads of power to split it?

You haven't got something for nothing.  You just stored some energy from the Sun in a novel way.

In the example of a system electrochemically (solar panel) splitting water into hydrogen and oxygen storing them, then burning the gasses, in an ideal system there would be no loss.
However, hydrogen is a very small molecule that can intercalate into materials, there by slowly diffusing outside of the system (and potentially embrittling them see: https://en.wikipedia.org/wiki/Hydrogen_embrittlement).  Furthermore, any small defect in a seal on the hydrogen storage and delivery side of the system would cause a substantial loss of hydrogen.


text:
Why does ozone have an electrostatic potential map similar to that of SO2?
This is the electrostatic potential for ozone.

This is the electrostatic potential for sulfur dioxide.

(Click to enlarge)
They look almost exactly the same except one is all oxygen atoms and the other has sulfur in the center instead of oxygen.
Why does ozone have an MEP (molecular electrostatic potential) similar to that of SO2?

Because their outer orbitals are very similar. The bonding nature in these to molecules is exactly the same with slightly more electronic structure moved from central atom in $\ce{SO2}$.

yeah but 1 is polar(the sulfur dioxide) and the other is nonpolar with a positive charge(the ozone). I would expect that the positive charge of ozone would make the central oxygen pull harder in the bond(because of the positive charge it has) and thus the electrons spend less time around the terminal oxygens thus making the MEP more like the one of H2O

Ozone is non-polar? You might want to rethink that.

there is 0 dipole moment because all atoms in ozone have the same electronegativity. 0 dipole moment means nonpolar. However it is an ion to some degree because of the positively charged oxygen in the center.

charged molecule does not automatically mean polar. All it means is that it is an ion. You also have to take into consideration dipole moment(and thus electronegativity) to know whether it is a nonpolar ion(such as in the case of ozone), a polar ion(such as methoxide), or will participate in an ionic bond with another ion(as long as you know what that other ion is and its electronegativity).

@caters Ozone has dipole moment of ~ 0.5 D. It is three times smaller than for sulfur dioxide, but still high enough. And I repeat, both have roughly same electron structure.

but the 3 Os all have the same electronegativity and isn't it electronegativity difference that determines dipole moment and thus polarity regardless of whether it is charged or not?

@caters Nope. Dipole moment is completely attributed to electron distribution in the molecule, and electron donation may go against electronegativity to some extent.

@caters in fact dipole moment is a function of charge distribution. Electronegativity considerations allow to some extent predict some aspects of charge distribution in case of simple compounds with ordinary bonds. However, in case of molecules with dative bonds ($\ce{CO}$, $\ce{O3}$, $\ce{SO2}$ and many more), conjugated $\pi$-systems and other compounds with nontrivial electronic structures electronegativity alone does not work.

Let us continue this discussion in chat.

@permeakra Sounds like you could write that up into a good answer! :)

Resonance structures of $\ce{SO2}$ and $\ce{O3}$.

As you can see from the diagram $\ce{O3}$ and $\ce{SO2}$ are isoelectric and have the same shape and this accounts for their very similar electrostatic potential maps. Both molecules are polar although $\ce{SO2}$, with a dipole moment of 1.62D, is more polar than $\ce{O3}$ (0.53D). As the diagram shows, the polarity of both molecules is due to the their trigonal planar shape with a formal positive charge on the central atom.


text:
Is melting of gold a physical property?
I read that melting of gold is a physical property, but I wonder if this is true.
If we want to melt gold we will need to add heat, and once we add heat to gold to melt it, there must be a chemical change, not a physical change, right?

Melting is a physical change, not a physical property. The distinction is subtle, but the way I think of it is:

A physical change is one that does not involve a change in any chemical bonds

Compare that to:

A physical property is one that can be measured without requiring a chemical change

Melting is the physical change, temperature and melting points are physical properties.
Simply adding heat to a substance doesn't necessarily mean that chemical bonds will be broken or formed. Metallic bonds are not usually considered to be chemical bonds, although it really depends on how you look at them.
Later you will learn that the distinction between physical and chemical bonds is fairly arbitrary, since they all have to do with changes in electron density.
For example, melting an ionic solid (like table salt) results in ionic bonds breaking, but this is still considered a physical change.
One way to identify a physical vs. chemical change when you don't know the chemical details is to imagine what would happen if you reversed the operation. If the system could go back to the way it was fairly easily, the chances are good that it is a physical change. This doesn't work for all situations, but it does work for the most common ones. The way I remember it is: "you can't unbake a cake," so "baking" is probably a chemical change. Going back to the melting example - if you cool molten metal down, you get back to where you started, and so that is probably a physical change.

It is a phase change - gold exists as a solid crystal (fcc I believe off the top of my head), as a liquid, and as a vapor (gas).  First order phase transitions exhibit a volume change, and an enthalpy of the phase change.  In the case of gold, both the crystal and the liquid are metallic, with similar numbers of nearest neighbors.
All elements exhibit various phases.  Since pure gold is, well, pure gold, I'm not sure how a phase change in gold could be considered a 'chemical change', particularly in the solid-to-liquid.  The type of bonding does not change, and there are no 'chemical reactions' going on. This is all fairly basic thermodynamics.


text:
Why is O2 a biradical?
$\ce{O2}$ has a double bond in its normal form. That is $\ce{O=O}$. There are no unpaired electrons in this case are there since there are 2 lone pairs on each oxygen.
However 1 resonance structure would be $\ce{O-O}$ (result of homolytic cleavage of double bond) where each $\ce{O}$ is a free radical (a negatively charged one at that). If you have this in hydrogen it is likely going to form hydrogen peroxide.
You could also have $\ce{O-O}$ where 1 is positive and the other is negative and this is also 2 free radicals.
And finally there is $\ce{O#O}$ where both oxygens are positively charged and are free radicals. Why are both positively charged? It is because 3 bonds already to oxygen means 1 lone pair and 5 electrons around oxygen is +1.
Is it because of these resonance structures giving 2 free radicals in $\ce{O2}$ that $\ce{O2}$ is considered a biradical?

Some of your proposed resonance structures are wrong. Both atoms in $\ce{O2}$ cannot have charge of the same sign, as that would violate the neutrality of the oxygen molecule.

but a molecule does not have to be neutral. Otherwise methoxide anions and hydroxide anions would not exist and lots more polyatomic ions wouldn't exist. Also O can have more than 2 bonds or less than 2 bonds.

Discrete polyatomic entities are only called molecules if they have no charge, otherwise they are termed ions. You are correct that polyatomic ions exist, but dioxygen, the most stable allotrope of oxygen and the molecule found in air, is not one of them. When we say oxygen gas or oxygen molecules, we are talking about neutral $\ce{O2}$ (or I suppose you could explicitly write $\ce{O_{2}^{0}}$). While $\ce{O2^{+}}$, $\ce{O2^{-}}$ and $\ce{O2^{2-}}$ are all known, they are not called oxygen, but dioxygenyl, superoxide and peroxide, respectively.

There are such things as "molecular ions."

yeah like the carbanion resonance structure of a carbonyl compound and the oxyanion resonance structure of a carbonyl compound. It is still considered to be that 1 molecule(whatever the carbonyl compound is) but now there are ions in it. This is really what is called a zwitterion because if carbon is positive O is negative and if C is negative O is positive.

We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for $\ce{O_2}$

Since an oxygen atom has 6 electrons,

A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, each oxygen would only have 7 electrons
B would correspond to a structure with a double bond between the oxygen atoms, 2 lone pairs on each oxygen and no unpaired electrons on each oxygen; B does have an octet around each oxygen, but it is not a biradical
C would correspond to a structure with a triple bond between the oxygen atoms, 1 lone pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet around each oxygen, in fact, each oxygen would have 9 electrons and this would be impossible for oxygen

So while structure A would indicate a biradical, we wouldn't "expect" it to count for much since the oxygens do not have octets.  This inability to clearly predict the biradical nature of $\ce{O_2}$ illustrates one of the failings of both Lewis structures and resonance theory.
In order to correctly predict the biradical nature of $\ce{O_2}$ we must move up to molecular orbital theory.  Below is the molecular orbital diagram for $\ce{O_2}$.  As you can see it does predict that $\ce{O_2}$ should be a biradical with an unpaired electron in each of its degenerate, highest occupied molecular orbitals.

Edit: response to OP's comment

When I think of triple bond I don't think of 2 3 electron bonds(which
is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond
and 2 pi bonds)

Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.

This type of triple bond would make the oxygen positive with 5
electrons around it.

No, the formal charge on the oxygen in structure C is
Z = 6 - 3 unshared - (1/2 * 6 shared)= 0,
there is no formal charge on oxygen in the "triple bond" structure and as I noted above, there are 9 electrons around it (not 5), which is impossible for oxygen.

I am assuming electrons are shared equally with half around 1 atom and
half around the other(which is the basis for formal charge

Yes, that's correct.

When I think of triple bond I don't think of 2 3 electron bonds(which is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond and 2 pi bonds) This type of triple bond would make the oxygen positive with 5 electrons around it. Also oxygen when it is single bonded to only 1 atom has 3 lone pairs of electrons making there be 7 electrons around it(and thus a -1 charge). I am assuming electrons are shared equally with half around 1 atom and half around the other(which is the basis for formal charge), not necessarily that each one has an octet.

To best answer your comment, I placed an edit in my post.

there is formal charge on the oxygen. I mean look at C=-O. There are 5 electrons around this oxygen because of 3 2 electron bonds and 1 lone pair. Same thing goes for the O=-O. I often use a slightly condensed lewis dot structure where I represent bonds with lines between atoms and nonbonding electrons with dots(most of the time just lone pairs).

No, for the oxygen in C-triple bond-O (carbon monoxide right?) there 8 electrons around it;  2 in the lone pair and the 6 in the triple bond. That's how you count electrons to determine how many electrons are around an atom, for example when you are trying to count and see if there is an octet.

I said that the single bonded oxygen and the triple bonded oxygen are free radicals because if the oxygen in this state is separated each O has an odd number of valence electrons which means that it has unpaired electrons which means it is a free radical and is extremely reactive(Even more than the oxide anion is)

For formal charge(which is much more important than the octet rule as far as stability of compounds and possible reactions) you divide each bond in half and count only half of the bond. For oxidation state you count the whole bond as electrons around the more electronegative atom. For the octet rule it is similar to oxidation state but you take into consideration electronegativity.

Yes, the single bonded and triple bonded structures are biradicals.  But they don't exist, they are merely lewis or resonance structures that when taken together, describe the real molecule.

Let us continue this discussion in chat.

There is no way to adequately describe $\ce{O2}$ molecule using naive valence bond theory. You need to learn Molecular Orbital Theory. Relevant links
http://en.wikipedia.org/wiki/Molecular_orbital_theory
http://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dioxygen

Apparently even in the first article where Pauling described valence bond theory, he realized the problem with dioxygen and was careful to describe the bonding in $\ce{O2}$ as involving three-electron bonds, which would make the molecule paramagnetic. I have also read that modern valence bond theory finds no difficulty in describing $\ce{O2}$, though I don't know if it still invokes the three-electron bond.

O2 is paramagnetic, so at least that's correct. You don't notice it unless you have liquid O2 and a big magnet though.

@NicolauSakerNeto I believe in modern VB theory you also give the overall spin state of the wave function, in this case you will start with two unpaired one from scratch. The three electron bond is not used any more (to my knowledge), they most likely use some kind of hybrid with MO theory to describe the individual resonance structures.


text:
What causes photovoltaic (solar) cells to degrade?
Photovoltaic panels can have 20 or 25 year underwritten warranties with a guaranteed remaining efficiency of 80 % of the new panel. That means, that photovoltaic panels seem to degrade somehow.

Why do they degrade? What exactly is reducing their efficiency?
How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?

Why do they degrade? What exactly is reducing their efficiency?

This link outlines several modes of solar panel degradation, and this report by the National Renewable Energy lab is a very detailed review of studies on solar panel degradation worldwide.
To summarize:

Internal resistance of the cell can increase due to infiltration of contaminants (usually water vapor) when the encapsulating material cracks due to long-term UV exposure or temperature cycling.
The shunt resistance (resistance of the path to ground) can decrease when metal ions migrate (diffuse) through the cell. This happens naturally, with the rate depending on temperature and the type of cell. In general, higher temperatures lead to faster diffusion, and amorphous silicon has higher diffusion rates.
The anti-reflective coating can deteriorate. Anti-reflective coatings can be made of lots of different types of materials, but in general, heat, UV exposure, and exposure to contaminants are what cause them to degrade.

These are all degradation mechanisms of the cells themselves - any of the other components of the panel can also degrade. In all cases, the mechanisms are related to thermal stress or mechanical damage due to things like wind or hail, or corrosion from humidity.
Given all of the above, the main causes of solar panel degradation that can be (sort of) controlled by you are:

Temperature
UV exposure
Mechanical damage

How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?

Anything you can do to reduce these three factors would help. However, covering a solar panel with an opaque tarp would only reduce UV exposure. The temperature (and cyclical thermal stress) would likely increase.
Unfortunately, both 1 and 2 increase with exposure to sunlight - and exposure to sunlight is the whole point of solar panels. Mechanical damage can be reduced by installing panels in places and orientations that make it unlikely that they will be stressed by winds, or have things hitting them very often.
However, it looks like the only effective option would be to completely enclose the panels in a temperature-controlled low humidity environment when they are not in use. Whether or not the benefit of doing so is worth the cost will depend on the degradation rate and cost of the particular panels.

I referenced this answer for a question over on sustainability.SE. I also used another source which explains more why the decrease in shunt resistance reduces panel performance.

I did recall that one big problem is something called the Staebler-Wronski effect. Essentially the mechanism is broadly similar to why LEDs lose efficiency over time, although with solar cells there is a workaround which is to heat up just the top micrometre with a scanned near-UV laser which anneals them without disrupting the encapsulant: also works for intermittent faults in the interconnects with more power used:-)
I am not sure to what extent the normal light/dark cycle does this but am aware that keeping the panel cool helps and some newer units include high efficiency fans for this purpose where wind cooling is inadequate.
Also relevant, ESD from lightning can be a factor and can cause major issues if mitigation in the form of external shunt diodes and fuses is not provided.
Someone suggested that an EMP would also destroy solar panels so another issue to consider, perhaps a thermionic shunt might be better?

It seems that the Staebler-Wronski effect only affects amorphous Silicon PV panels. So for current commercial PV panels, nothing to worry here, as they are polycrystalline or monocrystalline.


text:
Why does ammonia react with water to produce hydroxide?
My question relates to dissolving $\ce{NH3}$ in water and how it produces $\ce{OH-}$ for the reaction $\ce{NH3 + H2O -> NH4+ + OH-}$?
I know that the $\mathrm{p}K_b$ of $\ce{NH3}$ is $4.8$ and $\mathrm{p}K_a$ of $\ce{NH4+}$ is $9.2$ so $\ce{NH3}$ is a stronger base than $\ce{NH4+}$ is an acid. But why the free $\ce{OH-}$? I am now taking organic chemistry and all I am reading is how strong $\ce{OH-}$ is and how it can attack anything slightly positive. So why is the above reaction favorable towards creating $\ce{OH-}$?
Actually any acid-base reaction that is creating $\ce{OH-}$ as a free species?

The shortest answer is "because the change in free energy is negative," but maybe you want to know how the mechanism works?
Ammonia is a Lewis base, and hydroxide is a stronger Lewis base. However, both are stronger Lewis bases than water is. Pure water always has some amount of "free" $\ce{H+}$ (it's actually hydronium ion) and $\ce{OH-}$ due to self-ionization ($pK_W \approx 14$):
$$
K_W = \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2}
$$
When you add ammonia to the system, some of those hydronium ions (which are very strong acids) react with the ammonia to form ammonium. Since the concentration of hydronium decreases, the concentration of hydroxide must increase to maintain equilibrium. Those hydroxide ions are stronger bases than ammmonia is, and so they will "win" the tug-of-war over protons. However, in the meantime, some other hydronium reacts with some other ammonia, and the net result is a dynamic equilibrium in which there is more hydroxide then there would be in pure water.
The same mechanism is responsible for any other acid/base reaction where a base accepts protons from water and leaves hydroxide as a free species, but the net concentration of hydroxide will be different for different strengths of bases.

Even if the change in free energy is negative, there are still processes that won't "go" under certain conditions without an input of energy.

Also, your statement about ammonia and hydroxide being Lewis bases is off base. Hydroxide ion is not necessarily a stronger Lewis base than ammonia. Consider the stabilities of metal ammonia and metal hydroxide complexes. Metal hydroxide complexes tend to be less stable than their metal ammonia counterparts.

"hydronium ions (which are very strong Lewis acids)" ... Hydronium ions are in no way Lewis acids.

@Dissenter - 1) Yes, there are activation energies. How does that relate to this question? 2) In this case, hydroxide ion is a stronger base than ammonia. 3) You are technically right - I originally had written free protons (which are Lewis acids) instead of hydronium, then went back and changed it in an attempt to avoid these kinds of pedantic arguments. I see that I failed by forgetting to remove the Lewis acid statement. I'll make you a deal - I'll remove that statement, and you work on keeping your comments polite and on sending them in smaller barrages.

if the activation energy were high for the rxn there would be no hydroxide ion present.

@Dissenter - yet there is - what is your point?

Consider mentioning activation E lest someone gets the idea that each time the change in G is neg., the rxn proceeds.

Original question: "So why is the above reaction favorable towards creating OH-?" I considered it, but I think given the question, it is not appropriate to bring activation energy into the discussion. If you think it is, please feel free to add your own answer.

Why is it not appropriate?

Let us continue this discussion in chat.

We all know that Hydronium can act as a Lewis base, as there is a lone pair, under certain conditions. Just consider $$\ce{HF.SbF5 + H3+O <=> SbF6- + [H4O]^{2+}}.$$

Simply put, this reaction involves the formation of hydroxide, because it is relatively more stable than the alternative products: $\ce{{}^{-}NH2}$ and $\ce{H3O+}$. You're right; hydroxide is generally a reactive species, so you probably wouldn't expect it to be the product of the reaction. However, when you have two reactants, in this case $\ce{NH3}$ and $\ce{H2O}$ in an acid-base reaction, the stronger base dictates the outcome of the reaction, which of course is $\ce{NH3}$.
Also, this reaction is in equilibrium and there are more reactants than the products. Thus, this solution is hardly composed of hydroxides.


text:
What is the difference between a coefficient and a subscript in a chemical equation?
Observe the following parts of an electron transfer equation:
\begin{align}
\ce{2Li + S &-> Li2S}\tag{1}\\
\ce{2Al + 3Br2 &-> 2AlBr3}\tag{2}
\end{align}
The first equation says $\ce{Li}$ with a small subscript 2, however the second one says Al with a coefficient 2 instead of $\ce{Al2}$.
Can someone please explain this discrepancy and inconsistency I'm seeing? Why is lithium in the first reaction written with a small $2$, while in the second one aluminium is written with a big $2$? Why can't both of them be written in the same way? Does it matter? Is there a difference? How do I tell which one to use?

Why can't both of them be written in the same way? Does it matter? Is there a difference?

Yes, there is a difference, and it does matter! The placement of subscripts and coefficients tells you how the substances are put together on the atomic level.
Let's look at the first equation:
$$
\ce{2Li + S -> Li2S}
$$
This one indicates that two moles of lithium combine with one mole of sulfur to form one mole of lithium sulfide. $\ce{Li2S}$ is called a formula unit or empirical formula, and tells you the smallest whole-number ratio of atoms that make up a "building block" of the compound. In this case, the smallest building block of a lithium sulfide crystal would contain two lithium ions for every one sulfide ion. Here's a picture from wikipedia to demonstrate:

From this, we can learn that:

Coefficients tell us the number of moles, molecules, or formula units of a species

and

Subscripts tell us how many atoms or ions there are within a compound

Now if we look at the second equation:
$$
\ce{2Al + 3Br2 -> 2AlBr3}
$$
we can see that in this case, it take 2 moles of aluminum to combine with three moles of bromine to produce two moles of aluminum tribromide. If it were instead $\ce{Al2Br6}$, that would indicate a different formula unit - it would imply a different compound.
So to answer your question:

Why is Lithium in the first reaction written with a small 2, while in the second one Aluminium is written with a big 2

This is because in the first equation, the product has two lithium ions for every one sulfur in the formula unit, while in the second, two moles of aluminum tribromide are produced for every two moles of aluminum, but the formula unit contains only one aluminum atom and three bromine atoms.
For your final question:

How do I tell which one to use?

You would have to know what the formula of the product is. In these examples, you can predict it by assuming that they are ionic compounds and balancing charges. You would get the correct answer, even though this assumption is not true for the second equation (it is actually a covalent compound, and you would just get the right answer by luck).

The question is quickly answered when you consider the ionic forms of the elements you present. It is easy if you break down the reaction into simple oxidation and reduction reactions:
Oxidations
$$ \ce{Li -> Li+ + e-} $$
$$ \ce{Al -> Al^{3+} + 3e-} $$
Reductions
$$ \ce{S + 2e- -> S^{2-}} $$
$$ \ce{Br2 + 2e- -> 2Br-} $$
Since the resulting compounds of your reactions are neutral, all we have to do is to match up the starting compounds in the right amounts to get to the right final products. Retrosynthetically speaking:
$$ \ce{Li2S <- 2Li+ + S^{2-} <-[\mathrm{red}] 2Li+ + 2e- + S <-[\mathrm{ox}] 2Li + S } $$
Or if you go from the start:
$$ \ce{2Al + 3Br2 ->[\mathrm{ox}] 2Al^{3+} + 6e- + 3Br2 ->[\mathrm{red}] 2Al^{3+} + 6 Br- -> 2AlBr3 (-> Al2Br6)} $$
I put the last step in parentheses because usually we give formulae of ionic compounds with the smallest integer amount possible, which would be $\ce{AlBr3}$ and not $\ce{Al2Br6}$. This should answer your question about which one to use.

Is there a difference?

Yes. For example $\ce{AlCl3}$, a lewis acid, can dimerize to $\ce{Al2Cl6}$. Which one you mean depends on the formula you give and can be a barrier of understanding in a conversation.


text:
What are these crystals on my champagne cork?
I opened up a bottle of champagne that was left in a wine cooler for approx. 8 years. To my astonishment there were small crystals at the underside of the cork.
The crystals are colorless, odorless and have no specific taste at all. What is going on here?
Image:

Close up:

Great macro pictures, by the way.

They are potassium bitartrate crystals (source).
The crystals form because the potassium bitartrate is not very soluble.
Since solubility is a function of temperature, when wine is chilled the solution can become saturated, causing the precipitate to form in the bottle. Stabilizers are often added to prevent this.
In your case, the precipitate is on the cork, and so what I think has happened is that some wine soaked into the cork, perhaps while the bottle was moved. Then later, the bottle was stood upright or on its side where the wine could not reach the cork. As the wine evaporated back into the bottle, the local concentration of potassium bitartrate increased until it reached the saturation point, at which point the crystals grew. Since the crystals take a very long time to dissolve once formed, they will persist even if wine reaches the cork again.
The good news is, they are harmless (especially good news given that you tasted them - please don't make tasting strange crystals a habit!). They are made of the same substance that cream of tartar, which is a common baking ingredient, is made of.

Ah! I was thinking of acetic acid, until I read "colorless, odorless and have no specific taste at all".

Actually, the cork got soaked during the whole storage time. Maybe the cork just provided a good surface for nucleation?
And I want to add: I took a small sample of the crystals and tasted that because it occurred in a place where I could have consumed it anyway. As such I deemed it harmless enough to warrant this risk.

Yes, you are probably right in that case, that the cork was providing a nucleation site and the wine was super-saturated. This is especially likely if there were any stabilizers in the wine, since they would not be as effective when not in solution.
And I'm glad you were being careful in tasting them - my comment was meant as a joke, but it is true that as a general rule people shouldn't taste unidentified chemicals.

The crystals are very common in wine (especially wine made from some varieties of grapes like Shiraz). They also have a major part in the chemical understanding of optical activity as one of Pasteur's achievements was to separate crystals of a tartaric acid derivative into pure optically-distinct crystals.

Given that this is champagne, I'd wonder if the pH drops as the CO2 increases after bottling the uncarbonated wine, thus protonating one of the carboxylic acid groups. pka1 = 2.89, pka2 = 4.40


text:
How can I get a hand sanitizer solution to form a gel?
I am trying to formulate a gel hand sanitizer for a thesis project. The sanitizer that I have right now is watery and will not gel.  My teacher told me to change our formulation because it is not good. Our ingredients are ethyl alcohol, aloe vera gel, and lavender scent. What else could we add to this mixture to really "gel" the hand sanitizer?

Have you tried adding a thickening agent such as agar, arrowroot, guar gum, carrageenan, gelatin, etc?

0.05% agarose does not form a suitable gel, it precipitates in 70% isopropyl alcohol

According to the wikipedia article on hand sanitizers, polyacrylic acid and polyethelyne glycol are commonly used thickeners.
You could try any hydrophilic polymer if those aren't available. Cornstarch might even work if you dissolved it in the alcohol before adding the other ingredients.

If you need a source of Polyethylene glycol from the grocery store, buy some miralax, it's an osmotic laxative and it's pure PEG. Might not be the right length for a good gel though. It will take a pretty high concentration to get a gel with any polymer though.

@user137 - one of the main advantages of polymeric thickeners is you need less to get a gel, compared to monomers or oligomers. As chain length increases, the chances of forming a percolating network go up for a given polymer mass concentration (for a polymer in a theta solvent).

I've never made a gel with PEG from miralax, but I know that if you follow the instructions on the bottle it won't make a gel, it all dissolves and the viscosity does not noticeably change.

@user137 - I believe you. But if you took an equal mass of ethlyene glycol in the same amount of water, it probably won't gel either. My objection was to your general statement that "it will take a pretty high concentration to gel with any polymer" - emphasis is mine

One component that has gelling effect and is added to cosmetic products is hydroxyethylcellulose. It creates clear gel and is soluble in EtOH, but possibly not in highly concentrated.  It is safe and inexpensive.

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


text:
Ligand strength - Coulombic effect and/or Electronegativities?
According to my teacher, these three Lewis bases generally decrease in strength in the following order:

Ammonia
Water
Hydroxide ion

I agree. However, he argues that their ability to act as Lewis bases stems from Coulombic repulsions. For example, nitrogen has only one lone pair. Water however has two lone pairs, and when water uses one of its lone pairs in a Lewis base fashion, the other lone pair weakens the resulting Lewis adduct through exerting a repulsive force on the lone pair used in forming the Lewis adduct. Same with hydroxide, except that the repulsions are even greater in hydroxide since it has three lone pairs.
I say that in addition to this effect, there is also the effect of the Lewis base being able to stabilize its valence electron density. Nitrogen is less electronegative than oxygen. Thus, any valence electron density on nitrogen will be more reactive than that on oxygen. On the other hand, oxygen in hydroxide ion has a greater valence electron density than that of oxygen in water, and hydroxide ion is a poorer Lewis base than hydroxide ion.

Why did you think that hydroxide would be weaker base? Also last words don't make sense.

As Mithoron rightly states: hydroxide ion is a poorer Lewis base than hydroxide ion is obviously a typo. What did you actually mean to say?

the deal is in the detail


text:
Hydrogen Peroxide as an Oxidizer
Why is HOOH such a good oxidizer?
I understand that the oxygens in HOOH are in a negative 1 oxidation state, and that oxygen generally adopts an oxidation state of negative 2.
I also know that oxidizers themselves are reduced, and that reduction entails a drop in oxidation state. This suggests to me that oxygen is more stable when found with a negative two oxidation state rather than a negative one oxidation state. This makes sense; oxygen is the second-most electronegative element on the periodic table, so it likely would be okay with a more negative oxidation state.
And from organic chemistry I understand that the O-O linkage in peroxides is relatively weak; light can rather easily homolytically cleave the O-O linkage. And HOOH is generally stored in dark bottles as to block light. This says something about thermodynamic stabilities to me - that HOOH has relatively weak bonds and that it would be enthalpically favorable for the oxygens to form other, stronger, bonds. Is this how HOOH is used in bombs? The O-O bond is weak; bond formation is exothermic, thus providing a way for the O-O bond to break and rearrange into stronger bonds releases vasts amount of energy?
I'm trying to tie together redox with thermodynamic stabilities.

From your argument I assume the real question you want to ask is "Why HOOH is such a good oxidizer compared to molecular oxygen?".
From thermodynamical point of view, molecular oxygen is a very potent oxidizer. However when you are talking about reactivity, you should always take a look at kinetics, not only at thermodynamics.
One important part of the answer to your question is that not HOOH a good oxidizer, but O2 is a very bad oxidizer from kinetic point of view. Molecular oxygen is in the triplet state, therefore its reactions are often spin-forbidden. Especially the oxygen->superoxide->peroxide steps require often level crossings between different spin surfaces. One very important reason why transition metal complexes catalyse oxidation in biology is this: as they often have electrons to flip and large spin-orbit coupling, the level crossings became easy. There is literally tons of literature on this in bioinorganic chemistry.
Note that in flames and in high temperature burning that reaction is essentially a huge variety of radical reactions, so the mechanism is completely different from the oxidation reactions in aqueous solutions.
Molecular oxygen has also 4 electrons to take when turned to oxides. This sometimes require complicated mechanism where every single step is thermodynamically and kinetically is feasible, and there is always a oxidizable component around.
Also note, that HOOH itself need catalyst to be an effective catalyst, like in Fenton reactions and similar.

What do you mean by "spin-forbidden"? Is spin the QM equivalent of momentum in classical physics? Both are conserved quantities, correct?