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Combining Herbal Powder Extract
If I mixed two powdered herbal extracts together and put them through distillation process, then would I end up with a high concentrated form of the two extracts combined together? I'm trying to make an all natural energy drink. I'm not sure if i should just mix the powdered herbs or use some chemical process to combined the herbs into a concentrated form. I figured I would need high doses of the herbs in order for the drink to have an affect. So, should a chemical process to combined my ingredient, or just use herb tictures to avoid a powdery taste?

It would be helpful if you could fill in some further information. Are the powders not fully soluble in water? Are they the outcome of a real extraction process, e.g. using solvents other than water, or just obtained from grounding dried plant material? Are the powders soluble in ethanol? If so, can you use water/ethanol mixtures for your drink? Did you consider to use edible surfactants to solubilize the non-polar, water-insoluble compounds?

I'm sorry, let me clarify. I'm trying make an energy drink. My plan is to create an all natural drink using h

I'm sorry, let me clarify. I'm trying make an energy drink. My plan is to create an all natural drink using all herbal ingredients. I was looking around and found, for example, pure ginkgo biloba in a leaf extract powder and liquid form. My question was if I purchase an over the counter leaf extract powder to make my drink, can I just mix the extract in water or something, or will I need to do some kind of chemical process to properly mix all the different herbal extracts into one solution? I'm sure the extracts are water soluble.

If this is an energy drink for humans then the active are either soluble in water or ethanol. The standard factory method is to use supercritical CO2 extraction. Check the method for production of instant coffee. Briefly, they mix coffee beans with CO2 at 40C (above critical temperature and under very high pressure. Then they filter it and drop the pressure. Components precipitate out. You can drop it to certain value, filter to second chamber, drop pressure slightly, filter move to third chamber, drop pressure to 1 Atm. In this case second chamber can have a single component. In case of coffee caffeine precipitates in second chamber. Now mix stuff from first and third chamber to get decaf.
Ethanol extraction works ok and is used by pharmacists. It is hard to concentrate it however. And ethanol is an active ingredient on its own.
Water extraction is the dirtiest way mainly because it is difficult to concentrate your material.

Thank you, I restated my question in case I wasn't clear. I'm sorry, let me clarify. I'm trying make an energy drink. My plan is to create an all natural drink using all herbal ingredients. I was looking around and found, for example, pure ginkgo biloba in a leaf extract powder and liquid form. My question was if I purchase an over the counter leaf extract powder to make my drink, can I just mix the extract in water or something, or will I need to do some kind of chemical process to properly mix all the different herbal extracts into one solution? I'm sure the extracts are water soluble.


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Why cannot nitrosonium ion attack aniline in electrophylic aromatic substitution?
This is the nitrosonium ion, which is produced by the reaction of $\ce{NaNO2}$ and $\ce{HCl}$:

[Image source: Wikipedia]
And this is the reaction between aniline and nitrosonium:

[Image source: Wikipedia]
Cannot it take place as an EAS, and with substitution at the ortho or para positions? The amino group is a very good activating group for EAS after all.
Update: I forgot to mention that EAS is observed in case $\ce{-NR2}$ is attached to the ring in place of $\ce{-NH2}$.

You certainly could. If I had to offer a post hoc justification, it's probably that at 0 degrees, the rate of ring substitution is extremely slow and cannot compete with N-attack. There is a similar selectivity question in, for example, the acylation of phenol. http://chemistry.stackexchange.com/questions/31531/friedel-crafts-reaction-of-phenol

What actually happens is that the nitrosonium ion always adds faster to the amino nitrogen.  But that reaction is rapidly reversible unless a proton can dissociate from the amino nitrogen to give a stable N-nitroso product.
A primary or secondary amine has the required N-H proton, and in the case of a primary amine the presence of a second N-H proton enables further reaction making a diazonium ion (as shown for aniline by the OP) or even molecuar nitrogen.  But a tertiary amine has no N-H proton for the elimination step.  A tertiary amine then undergoes no net nitrosation reaction at the nitrogen atom, but if it's aromatic then EAS emerges as an alternative.

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3 chemicals to produce light
I'm actually looking for 2 chemicals that will produce a really small exothermic reaction. The 3rd chemical would be there to separate the chemicals - like oil,  water, and air.
Ideally, the 3rd chemical would be a nonreactive oxide of the other 2.
Also, they should all stay liquid on a hot summer's day.
As stated, they should produce a small exothermic reaction - I'm looking for mostly light that won't melt a glass bottle.
Melting the bottle isn't ideal, but I can find a stronger material too.
Also, the chemicals should be as nontoxic as possible. They don't have to be though.

I am not sure what you are trying to accomplish. A chemical flashlight? A bomb? This experiment is the closest one to your description: youtube.com/watch?v=Nyzlt-dVgWQ

I'm making a bottle of liquid that lights up. Explosion = failure. It would be used as a random shake-timer. Shake it and it lights up for a while, and then it would separate back out into it's original form.

You can buy these as light sticks. Why, because  the chemicals are expensive, difficult to obtain and not pleasant to use.

Well I can give you a recipe for a solution that is colorless, turns blue when shaken, then it again becomes colorless after a while. The process can be repeated many times, and it doesn't require very sophisticated reagents. If you need light, then I guess you would have to look for some chemiluminescent reactions, but most of such reactions require some organic reagents that are not easy to get.

I would definitely use a light stick, however, usually they are one use, and last way too long. I'm sort of looking for a way to halt the reaction after an amount of time, and then reinitiating it by remixing the medium.

Is there a way to use the blue to refract an ultraviolet or infrared light source to something visible? If so, I would be REALLY interested. How safe is it?

With the thing I had in mind it is not possible, but it would be doable with some other chemical systems; I will try to find some later, but they might include some exotic chemicals. In my original idea, the blue color can be made intense enough to block visible light, e.g., from a flashlight. It would be an inverse of what you wanted: light goes out when shaken then it slowly returns to nominal intensity, but I don't know if that suits you.

I wonder if I could start and stop the process with another chemical? Like an on/off switch? What is the recipe in question? I'm close to giving up on the whole 'chemical producing light when agitated' thing.

You can make two solutions that produce light when mixed. Then you take an excess of one solution, add a small portion of the second solution - it will produce a glow until the reagent in the second solution is exhausted, then you can add another portion and so on.

Post that as an answer. I had an idea. Also include the most nontoxic 2 solutions you can think of on the spot - and a unreactive, buoyant 3rd.

Related: http://chemistry.stackexchange.com/questions/24945/how-to-get-a-rechargable-chemical-reaction-for-a-glow-stick

As discussed in the comments... The two component chemiluminescence reaction:
(Solution 1) To 50 ml of deionized water add 10-50 mg of luminol, 10-20 ml of 10% ammonia solution and 1-5 ml of 3% hydrogen peroxide. (Solution 2) To 100 ml of deionized water add few milligrams of copper salt (e.g., sulfate) or few drops of blood or fresh horseradish juice. These are approximate quantities (they should work), but you can experiment with the concentrations, after all experimenting is the most fun part of chemistry. Now take the solution 2 (whole) and add few milliliters of solution 1 to it and shake it. You should be able to observe a faint glow. A brighter emission requires organic solvents (DMSO, DMF) and corrosive chemicals (KOH). Adding a third liquid that will separate the two other (as the OP suggested) is not a simple thing to do. I would do this otherwise: add some neutral ingredient to solution 2 to increase its density (possibly sugar, but I'm not sure if that will not quench the chemiluminescence) and place the solution in a container (preferably a long tube). Then carefully pour the solution 1 on the surface of the solution 2 (so that the solutions won't mix). When you shake the container the solutions will mix and produce glow. Another idea: place one solution in a container and layer it with a hot paraffin from a candle and allow it to cool down. Then you can pour the second solution on the solid paraffin. To start the reaction, simply break the thin paraffin layer by shaking (you can place some glass marbles in the container).

You totally got the vibe of my question :-) I am trying to make a working staff that will have a bottle at the top that glows like alchemy or magic would. Now I can play pokemon go and scare away crazy thieves with something nonsensical and my staff! Mwahahahaha!

Also, would something like a Klein bottle work to measure a little of the 1st solution? Or is that too imprecise?

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can I mix. trisodium phosphate, potassium hydroxide, and sodium hydroxide together?
Can I mix trisodium phosphate (tsp), potassium hydroxide, and sodium hydroxide ( or sodium hypochlorite ) and not cause a bad reaction?  And will this make for a good degreaser and cleaning chemical for me ?  I own a pressure washing business.  I do fast food resturanta, gas stations

You will not cause any bad reaction, as far as I can guess. The trisodium phosphate only gets rid of some of the mineral content of the water making it softer, the hydroxides will definitely make a good deggreasing agent. But I'm not sure, whether you need to mix, isn't e.g. sodium hydroxide in itself enough? You can also use sodium hypochlorite with sodium hydroxide or potassium hydroxide without problems, I'm not sure about using it with tsp., that might lead to chlorine formation. You can always try in a small cup with small quantities, preferably in open air, and see if chlorine forms.

I advise against using potassium or sodium hydroxide in homemade cleaning agents. You could easily end up with a mixture that could severely burn eyes (up to blindness), skin or lungs.
Also, grease from fast food restaurants is fundamentally different from gas station grease. The former consists most probably of fat while the latter is a mineral oil product.
Your home-made degreaser, with a hydroxide concentration that could be handled without complete personal protective equipment, might or might not have some effect on fat due to saponification but for removing mineral oil based grease I would not expect it to be effective.

Potassium hydroxide, and sodium hydroxide tend to produce heat when you dissolve them in water, tbs is very peaceful. So if you dump a kilo of NaOH in a bucked of water it might shoot up the ceiling and burn your skin if it splashes on you. If you add it slowly you will be OK. If you want to make a liter of the solution start by adding a table spoon of NaOH and look what happens. You can increase the rate of addition as you get more comfortable. I would personally add 100 g of NaOH to 1 liter in 10 portions over 5 minutes, but that is because I know what I am doing. You will need to stir solution with a right tool (at least don't use your hand or aluminum spoon).
NaOH and KOH dust are nasty. You will need eye protection, gloves,      labcoat and a shower nearby in case of emergency.
This solution will absorb CO2 and lose activity, so you need to seal it.

tldr They are not going to blow up but you shouldn't add them too fast to each other.

Please feel free to leave a comment explaining why you downvote and provide a correct answer.


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Why don't we use enthalpy of fusion and enthalpy of vaporisation as two separate terms in Born-Haber's cycle?
I am not able to understand why we use enthalpy of sublimation rather than enthalpy of fusion and enthalpy of vapourisation of solid sodium metal in a Born-Haber cycle for the formation of NaCl from $sodium_{(s)}$ and $chlorine_{(g)}$ in their standard states.
Also I have rarely heard of sublimation of such metals.Is it like sublimation of camphor.

Thermodynamically, how would you relate the enthalpy of fusion plus enthalpy of vaporization with the enthalpy of sublimation? Think about that.

@Jon Custer The sum may be equal to the enthalpy of sublimation.But I dont think this is the standard way of writing in textbooks.If it is written there, then it may mean that we are avoiding the liquid phase of the metal.

Sublimation of these metals is much like sublimation of dry ice. the only difference is that standard pressure is higher than the triple point of Na and lower than the triple point of dry ice.
You have a path: take Na and bring it from solid to gas phase. take Na gas and ionize it. Ionize fluorine. Burn Sodium in fluorine. Energies of each step are known. Now if you sum them up you get a net reaction of Na+ + F- = NaF
if by energy of fusion you mean "energy to melt material" then it brings you nowhere. You cannot efficiently (to 100%) ionize liquid Na. But you can easily measure the energy required to ionize Na vapor.

Can you explain "...higher than the triple point of sodium and lower than the triple point of dry ice."Moreover I still don't get in your answer why didn't we use both enthalpy of fusion and enthalpy of vaporisation rather than enthalpy of sublimation as I mentioned in my question.


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Effects of potassium hydroxide & carbonate on metals?
BACKGROUND
I am in the process of restoring an old computer. The problem with the computer is that it has a NiCad battery fitted to the motherboard, which powers a real-time-clock circuit. The NiCad battery has leaked onto the motherboard. The leaked material has caused quite extensive damage, which I am cleaning and repairing.  See here and here.
I am curious about this leaked material. Here is what I (think) I understand so far:

NiCad batteries use potassium hydroxide as an electrolyte, and this
is the material which has leaked from the battery.
The potassium hydroxide forms potassium carbonate when it reacts with
the carbon dioxide in the surrounding air.

MY QUESTIONS

Some white(ish) powder has formed on the battery terminals themselves. Is this potassium carbonate?
The effect of the battery leakage has been quite widespread corrosion of metals, mainly electronic components. The corrosion has spread much further than the leakage site. How?
Why does the material attack metal? Do both materials (potassium hydroxide and potassium carbonate) attack metal, or does one of these have more of an effect on metals than the other?
Are any other chemicals likely to have formed that I have not considered? (e.g. there are green/blue deposits to be found in areas affected by the leakage)

CONJECTURE
Having pieced together some information from reading on the internet, I have formed a bit of an opinion about the process which has unfolded here. And I'd like to run it by you all to see what you think about it:

When the battery started to leak, the potassium hydroxide reacted with the air and formed potassium carbonate, as a white powder on the battery terminals.
When all of the carbon dioxide in the air was depleted (due to closed computer cover), the potassium hydroxide stopped forming potassium carbonate. And this is when it began to attack the metals.

THANK YOU IN ADVANCE!
The purpose of my questions is purely to satisfy my curiosity. If someone takes the time to respond, and share their knowledge, then I respect that and I am grateful. So I just wanted to say thank you in advance!

Your conjecture is at least partially correct, in that there was a mix of $\ce{KOH}$ and $\ce{K2CO3}$. However, there is only ~400 parts per million (ppm) $\ce{CO2}$ in air, so at first, the $\ce{KOH}$ was free to react with metal (and, possibly, the green lacquer on the PC board), until enough $\ce{CO2}$ entered the cabinet to form the carbonate. It is possible some raw $\ce{KOH}$ is still present, under a film of carbonate, so be careful. Note that there is ventilation in a PC cabinet, so $\ce{CO2}$ is never really depleted in it.
Far more damage, though, may be due to electrolytic corrosion. Wikipedia has a photo similar to yours, showing "Severe PCB corrosion from a leaky PCB mounted Ni-Cd battery". The liquid electrolyte creates bridges across the bare IC pins and across the PC board traces, and the continual current from the power supply and the not-quite-dead battery cause metal to be removed from one place and deposited in another.
After removing the battery, rinsing many times with distilled water, and tooth-brushing the board, you may find that surface traces need to be rebuilt (such as by soldering a bit of braid across a gap). This is likely a multilayer board, and if an internal trace is gone, it would be difficult to find and bridge the gap. Note that any such patches will affect timing, due to the impedance mismatch, so you may find that after all repairs, crashes occur with some frequency.
Best wishes for your restoration project. It's nice to know that at least one venerable computer is not going to landfill!

Thanks for your answer.  I am determined to rescue this old computer.
I also asked this question on Reddit, and I had an interesting response about the green/blue material that can be seen in the photos.  According to one Redditor, this is likely to be Green nickel hydroxide.  Are you able to comment on this?
Regarding the corrosion, it appears to me that you are claiming the opposite to my conjecture: that the corrosion is caused by the KOH and then the K2CO3 forms a kind of protective layer, preventing further corrosion.  Is this understanding correct?

The green-blue corrosion is likely a mix of copper carbonates with some nickel carbonate near the battery. 2. The corrosion is caused primarily by the electric current, not by the alkalis. Even salt water would cause similar corrosion if the current is on. 3. The potassium carbonate might be a shell over KOH, which is still caustic to skin. It would not have prevented corrosion.

Hm.  I'm surprised at your suggestion that the corrosion is caused by the electric current.  I imagined the corrosion was happening due to some basic reaction between the leaked material and the metal.  When you say that electric current causes it, are you referring to galvanic corrosion?

Regarding the green/blue material, I wonder if I could take a sample of it into work; we have a machine which can analyse the material and tell us what elements it contains.  This would clear it up once and for all!

"electrolytic corrosion" = "galvanic corrosion" 2. yes


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why are mercury gas and tungsten filament used in electric bulbs instead of fluorine which has higher electron affinity?
Fluorine has greater electron affinity than mercury or tungsten and hence fluorine could have been in electric bulbs.Because it can attract electrons easily than the metals which mostly repel extra electrons, so fluorine should release more energy[light].Then why wasn't it used?

Fluorine is way more dangerous than mercury

And etches glass, which isn't good for the bulb. But, it seems to me that the question is driven more on misunderstandings of how the different types of light bulbs work. The connecting of the electron affinity and supposing F would release more light is just wrong.

and it is not just a $very$ $very$ dangerous gas it reacts with most everything including, I expect, a hot tungsten filament. Mercury is used mainly in high pressure arc lamps, or low pressure discharge lamps and not in domestic filament bulbs.  The gas in a domestic light bulb is mainly there to stop the filament oxidising and is usually a rare gas or nitrogen. The bulb may also be evacuated.

First, you mix two types of lamps: incandescent lamps use conducting material with highest possible melting point. It was graphite at first, tungsten later. Mercury lamps use vapors. Obviously F2 cant be used in the first type. What about the second type?
Any material inside of a lamp needs to cycle between to states to emit light.  High electron affinity by itself is not necessarily good. Materials that are transparent and can contain F2 are at best expensive or might not yet exist. Glass and plastic will react with fluorine.
People generally need a light of a certain color. They use neon, argon and krypton to get different colors. Mercury works better because it has much more energy levels available, so it doesn't emit a single color, but a spectrum that is close to sum light. The bulb itself is coated to re-emit lite and make it even more "sun like".
As for the light intensity - you can increase it by using more powerful supply of electricity. People do try to increase the yield of light, but it has little to do with electronegativity.

"Mercury works better because it has much more energy levels available, so it doesn't emit a single color" - not true it emits UV radiation and that's why luminophore coating is necessary. Be careful with quality of your posts. It's more important than quick answering and answering to low quality questions is often a waste of time.

@Mithoron - not exactly - there are clear mercury lamps in use, mainly for outdoor lighting. As wiki says, it has a blue-green tint, so is not used indoors where it makes humans look bad. Now, the phosphors do convert the various UV lines to visible (that the glass would absorb), and help greatly with the color balance.

@JonCuster Oh, ideed, should have checked.


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Which is a stronger acid: H3O+ or HCl?
I know as a fact that HCl is a stronger acid than H3O+, but I want to know the reason behind it. The definition of an acid is a substance which dissociates to give H3O+ ions. In that case H3O+ should be the strongest acid available because it does not even need to dissociate it give itself!!

http://www.chemguide.co.uk/physical/acidbaseeqia/theories.html

Does really answer my question..

Your definition is wrong and this mistake is unfortunately common. Sth acts as acid if protonates anything, not necessarily  water. H3O+ is only a strongest acid which is present in significant concentrations in diluted aqueous solutions.

@Mithoron the point about H3O+ is kind of right, but in a very weird way; something that protonates anything would be the strongest acid possible, though. I think you mean "something acts as acid towards a compound if it protonates that compound".

Yes, that's what I meant, I didn't say everything... I guess I should say "something", sorry for imperfection my English.

Here is my opinion, which may be subject to errors and corrections are very welcome:
Using the definition of Brønsted, acidity is the tendency to give off protons. This, unlike the definition of Arrhenius, is not limited to aqueous solutions. However, if you do have an aqueous solution of an acid, something interesting happens: any acid HAc (or base B) stronger than H$_3$O$^+$ (or OH$^-$) completely dissociates via:
$$\ce{H2O + HAc <=> H3O+ + Ac-}$$ or
$$\ce{H2O + B <=> OH- + BH+}$$
So the maximum acidic compound in water is the hydronium ion H$_3$O$^+$; you couldn't, for example, differentiate aqueous hydrochloric and perchloric acid solutions by pH. To rank very strong acids by their acid strength, non-aqueous solutions are used to determine the dissociation constants and the data obtained is then transferred approximately onto water as a solvent. Also note that we are talking about solutions of acids.
Specifically for HCl, we can denote the pairs of conjugated acids and bases:
$$\ce{H2O ~(~base~1) + HCl ~(~acid~2) <=> H3O+ ~(~acid~1) + Cl- ~(~base~2)}$$
So we can compare the strengths of the two acids: $\ce{HCl}$ and $\ce{H3O+}$. We know that this equilibrium is shifted to the right, so hydrochloric acid is stronger.
Very late edit to try and answer comment by MukulGoyal (unfortunately, he didn't elaborate on his questions), by number:

Is the definition of acidity the tendency to give H+ ions, or
H3O+ ions? That seems to matter here because in first case H3O+ will
not even be acidic in water as the equilibrium H2O + H+ ---» H3O+
has a very high equilibrium constant.
Similarly even if the solvent is NH3, the equilibrium constant of
NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason,
and hence it would actually be a very weak acid.
Does your answer imply that any acid which has a Ka greater than
1 will be a stronger acid than H3O+?

First of all, please remember that equilibria are dynamic.
1) As I've mentioned, acidity, according to Brønsted, is the tendency to give off $\ce{H+}$. In $\ce{H2O}$, $\ce{H+}$ are solvated, on average, by 6 water molecules, giving rise to species like $\ce{H3O+(H2O)6}$ (for more information, check out Wikipedia and this).
1) and 2) you seem to be talking about the dissociation of $\ce{H3O+}$. However, you're using terms like "very high" and "very low", so that, in my opinion, becomes kind of a misleading discussion.
When you mention
$$\ce{H2O + H+ <=> H3O+}$$
you need to remember that the proton has to come from somewhere. In fact, what you are talking about there is
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
which is the autoprotonation of water. Let's just calculate that equilibrium constant you talk about all the time:
$$K = \cfrac{c(\ce{H3O+})~ \cdot~ c(\ce{OH-}) }{c^2~(\ce{H2O})}$$
Well, guess what? The concentration of water (c = n/V = m / (V * M) = $\rho$ / M = 55,51 mol/L) remains about constant for pure water and diluted solutions (because not many water molecules form ions), so it is brought into the constant, yielding
$$K_w = c(\ce{H3O+})~ \cdot~ c(\ce{OH-})$$
$K_w$ depends on temperature and equals $10^{-14} $ mol$^2$ / L$^{2}$ at 25 °C. It remains constant in pure water and also in diluted solutions.
In pure water where $c(\ce{H3O+}) = c(\ce{OH-})$, we then get
$$K_w = c^2(\ce{H3O+})$$
and
$$c(\ce{H3O+}) = 1.0 * 10^{-7} ~mol~ / ~L~ $$
which essentialy quantifies what you, as I suspect, were talking about, in that it tells you the concentration of $\ce{H3O+}$ in the autoprotonation equilibrium for pure water. By the way, because pH is defined like so:
$$pH = -lg ~c~(\ce{H3O+})$$
at pH = 7 (neutral), $\ce{H3O+}$ and $\ce{OH-}$ are equal. In diluted solutions, when they are not equal, you can calculate one using the other.
Also, remember that in the Brønsted–Lowry theory acids and bases are defined by the way they react with each other, as mentioned in Wikipedia. So far, we have only looked at water and its autoprotonation, no solvents and other molecules, which have their own acidies and basicities.
2) As for the reaction you mentioned in ammonia as a solvent, well, first of all, $\ce{H3O+}$ needs to come from somewhere (let's imagine we drop some aqueous solution of hydrochloric acid in there). Second, I am not sure that constant would actually be as you describe it - I think that here, ammonia (which behaves ampholytic) would act as a base and pretty happily accept a proton. If anyone can provide insight into this, that would be great.
3) Acid strength is defined using $pK_a$, which is given by $pK_a$ = -lg $K_a$. From my experience, if $pK_a$ is defined via concentration, then $pK_a$ of hydronium ions is 0, and if it is defined via activity, it is -1.7 (-1.74, to be more exact; similarly, for $\ce{OH-}$, the analogous $pK_b$ is 14 via concentration and 15.7 via activity).
Coming back to your question about "Ka greater than 1", that seems about right, because -lg (1) = 0.
I refer you to an acid strength table, for instance, here. Also, I'd like to refer you to this excellent discussion on the matter, where, by the way, orthocresol lent a hand.
Also, just to add this, strong acids have $pK_a$ < 0, while weak acids have $pK_a$ > 0 (again, this is via concentration - for example, here in Wikipedia they use -1.74).
Everyone, corrections are very welcome!

This is an excellent answer. You might just want to clarify what happens to $\ce{HCl}$ in aqueous solution though to avoid any confusion.

Thanks for the answer but I have 3 questions here: 1) Is the definition of acidity the tendency to give H+ ions, or H3O+ ions? That seems to matter here because in first case H3O+ will not even be acidic in water as the equilibrium H2O + H+ ---» H3O+ has a very high equilibrium constant.                2) Similarly even if the solvent is NH3, the equilibrium constant of NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason, and hence it would actually be a very weak acid.   3) Does your answer imply that any acid which has a Ka greater than 1 will be a stronger acid than H3O+?

Read the theory on acids and bases. The link orthocresol gave is a good starting point. According to the Brønsted definition, acids are donors of H+. The equilibrium you noted doesn't make a lot of sense out of context, so please clarify what you mean - and why it should have a high constant. 2) What would be a weak acid? H3O+? How would that come to exist in ammonia? Why do you think this equilibrium would be shifted to the left? 3) H3O+ has a pKa of 0 (not Ka). Acids with higher Ka (lower pKa) are stronger (edit: wrote "lower" before, that's wrong, also had a wrong Ka definition, sorry).

@MukulGoyal I'd like to improve the answer, but you need to clarify for me the things that I mentioned about your questions.

@MukulGoyal I've attempted to answer your comment anyway - maybe this helps :)
Everyone - I'd be very thankful for any corrections :)

Everyone - help with question 2) specifically would also be very appreciated :)


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Reaction of a diketone
What is the product when ammonium bicarbonate reacts with hexa-2,5-dione?
I thought of formation of 2 moles of NH3, water and carbon dioxide. And then the usual addition followed by elimination to the carbonyl group. But is it the case? Or will some cyclic compound result?

You might want to take a look at this (http://chemistry.stackexchange.com/questions/8833/reaction-of-diketone-with-ammonium-carbonate-at-100-115-degree-celsius) answer given by Klaus.

Ketones react with NH3.
R2C=O -> R2C=NH. Hexa-2,5-dione has two carbonyl groups so it leads to cyclic compound. Check Paal–Knorr synthesis in Wikipedia to know the answer.

The reagent is ammonium bicarbonate, not ammonia. How does this lead to the reaction that you have suggested?

@bon Hi, bon. He doesn't specify the conditions, so I assume heating is used. At elevated temperatures (NH4)HCO3 decomposes yielding NH3. If you know the correct answer please feel free to  post it.


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Understanding what is happening when something is dissovled
Let's start with a saturated solution of NaCl in water. At this point, any further addition of NaCl at the same T and P would not dissolve. However, if I then add another fully soluble salt (say AgNO3) it will also dissolve until it reaches equilibrium. If I keep adding different soluble salts (assume I don't encounter things like the common ion effect), would I ever overwhelm the solvent such that it could not dissolve anything else?
My intuition is that eventually, there would be no more solvent moleclues available for solvation and thus nothing would dissolve. Is this even valid to think about?

Your question seems to be exclusively about salts dissolving to form ions in water. The answer is fairly trivial: a saturated solution of a salt in water is not water.
At (thermodynamic) equilibrium, the solubility depends on the net free energy of the system. When the interactions in solution no longer energetically justify charge separation of the ionic solid into aqueous ions, then you've got equilibrium, right?
So, consider a solution as being (in a very crude and simplified way) Solvent-M1 interactions, Solvent-X1 interactions, solvent-solvent interactions, and M1(aq)-X1(aq) interactions as well as M1(aq)-M1(aq) and X1(aq)-X1(aq) interactions. Which of those will change when you add M2X2 ions to it? Well, at low [M2X2] none of those will change significantly, but you'll add the interactions between M2 and X2 and ALL of the species present. (as already explained, silver nitrate being added to NaCl is a very poor choice, AgCl2 is quite insoluble)
At high enough concentrations you'll begin to significantly change the energetics. In other words, adding M2X2 to a solution of M1X1 changes the interactions of all species present, so you would expect that the mixed salt solution will have different equilibrium concentrations for both M1X1 and M2X2.
If you were to carefully choose highly soluble salts, I think what would be surprising is how well their solubility was predicted by just using each independent Ksp, rather than the (expected) deviations from those predictions. Note that nothing I've said here should be taken to mean that adding a different salt to a saturated solution will decrease the solute concentration of the first salt. (Although it is the most likely scenario, just as the common ion effect would suggest. In some cases solubility will increase!) The reality is that highly concentrated solutions are far more idiosyncratic and generalizations (especially "laws") are much harder to come by.
While there probably is justification to claim that water is dissolved in sodium nitrate (at equilibrium), I've never found such foolishness useful. And I've been around for a long while. (ultra) Pure water does not freeze at 0°C, negative absolute temperatures (negative Kelvin) are attainable, and a bunch of other stuff which while true, doesn't really aid understanding (for the novice) probably should be ignored until one wants to deep dive into a subject.

Thanks for the responses. My take away is that you can't view each solute's interaction with the solvent independent of all the other species present so each new salt added changes the mixture and as such the mixture properties also change. Also, I'm curious about your comment for the ultra pure water not freezing at 0°C, do you mean to hint at phase meta-stability due to no nucleation sites?

http://chemistry.stackexchange.com/questions/2530/negative-kelvin-temperature it's not "normal" temperature that can be negative

Dont add AgNO3 to NaCl. It will form insoluble AgCl.
Yes, you can add salts M1X1, M2X2, M3X3. Eventually it will turn out that solubility of A1X3 is lover than solubility of M1X1 and M3X3 and it will precipitate.
Next by definition in physical chemistry the solvent is the one that precipitates when you cool down solution. Take a bucket of water, add a grain of salt, cool it down. Water freezes - it is a solvent. Make saturated solution of NaNO3 and cool it down. NaNO3 precipitates, so now NaNO3 is a solvent. This is not a paradox, but it does sound unusual. So now your argument doesn't hold. You add not really the "water binders", but "solvents.
Here is a different example. In organic synthesis you obtain a mixture of products. Pretty much all of them are solids. But because they are mixed the result looks like a chewing gym. Addition of yet another solid will not solidify the product (it will be more viscous, but still a liquid). Addition of a real solvent can extract small amounts of impurities and solidify your product.


text:
Isothermal process gives V = A(T)exp(-KP)?
Substance with isothermal compressibility K, where $\mathrm{K = -\frac{1}V(∂V/∂P)_T}$
undergoes constant isothermal, mechanically reversible process from state ($P_1$, $V_1$) to ($P_2$, $V_1$) where V is molar volume.
(a) Starting with the definition of K, show that the path of the process is described by: $\mathrm{V = A(T)e^{-KP}}$
I don't know how to get A in terms of T, the furthest I can get is $\mathrm{A = V_1e^{KP_1}}$

The key point is "starting with the definition of K"!

You have done well so far.  Now do the same thing for point 2.  Then, eliminate A between the two equations.

a useful 'trick' is to notice that $\frac{1}V \frac{dV}{dP} = \frac {d ln(V)}{dP}$

$A$ is already an implicit function of $T$, so your answer is essentially correct.
Starting from the definition of $\kappa$ and integrating both sides with respect to $P$, we obtain $$-\int\kappa\,\mathrm{d}P = \int\frac{1}{V}\frac{\partial V}{\partial P}\,\mathrm{d}P,$$ which yields as solution $-\kappa P = \ln V + f(T)$, where $f(T)$ is our "constant" of integration. Rearranging and exponentiating yields $V(P,T) = e^{-f(T)}e^{-\kappa P} = A(T)e^{-\kappa P}$, $A(T) := e^{-f(T)}$, and we are done.


text:
The road to solvation
To understand the behavior of solubility, it is helpful to look at the chemicals that misbehave, and apply the rules, and look for underlying explanations and common threads. Generally speaking, the solubility of some chemicals is lower at higher temperature.
The problem is, the explanation(s) for this is that entropy is negative for some chemicals, and solvation is difficult, and sometimes- the hydrated phases change (which is more like a chemical reaction, where the resultant chemical precipitates). So, I would like to identify the chemicals which misbehave and identify the explanation(s) or the excuse(s).
Here are a couple of links to other questions on this SE about this phenomenon:
Why do some solid chemicals dissolve better in colder water? and Why does sodium sulfate have an unusual solubility-temperature curve?.
What are some chemicals that dissolve better in cold water than hot water?

Calcium acetate, calcium hydroxide, calcium phosphate, cerium(III) selenate, cerium(III) sulfate, and lithium sulfate are more soluble in cold water than hot water.
Also, calcium selenate seems to have it's highest solubility at around 10 °C; and sodium sulfate has it's highest solublity at around 32 °C. These two chemicals have phase changes.
Gases (like oxygen) are also generally more soluble in colder water. The primary explanation is that colder temperatures and higher pressure results in gas becoming a liquid.

You are self-sufficient:)


text:
How is H2 2 moles
I saw a question that asked which expression can be used to designate 2 moles of atoms. The choices were

H2
C02
2N20
2NaCl
H2SO4

The answer is H2 but I was confused because I thought H2 is only 1 mole because H2 is diatomic. What am I not understanding?

Well, hydrogen gas has twice as many atoms as molecules, so any given amount of hydrogen gas has twice as many moles of atoms as moles of molecules.

The key word is 'atom.' H2 has two atoms, so a mole of H2 would have 1 mole of H2 molecules, and when broken down into atoms, 2 moles of atoms.
Think about it like pairs of shoes. If you have 6 pairs of shoes, you have 12 shoes.
CO2 has 3 atoms, 2N20 has 3 atoms and there are already 2 moles so that would be 6 moles of atoms, 2NaCl has 2 atoms but again that 2 indicates that you'd have 4 moles of atoms, and H2SO4 has 7 atoms.

It depends on the exact phrasing of the question, but you seemed to indicate that they were asking "Which of these substances is comprised of two moles of ATOMS?" and NOT "Which of these substances is comprised of two moles of ELEMENTS?"
As hydrogen gas naturally occurs in a diatomic state, it is true that there is only one mole of naturally-occurring elemental hydrogen in answer one, but one mole of H2 is formed from 2 moles of H.
Just to check that we are correct, let's go through all of the answer choices:

H2 - There are 2 moles of hydrogen atoms (1 mole of elemental hydrogen).  Total: 2 moles of atoms
There are 1 mole of Carbon atoms and 2 moles of oxygen atoms (oxygen is also naturally diatomic, but we're counting atoms, not elemental configuration).  Total: 3 moles of atoms
There are 4 moles of nitrogen atoms (N2, like H2 and O2 is diatomic) and 2 of oxygen atoms.  Total: 6 moles of atoms
There are 2 moles of sodium ions and 2 of chlorine ions (you could technically point out that this answer choice doesn't even involve true atoms, but I imagine that the nuance is lost on whoever wrote the question).  Total: 4 moles of ions
There are 2 moles of hydrogen, 1 of sulfur, and 4 of oxygen.  Total: 7 moles

According to this analysis, answer 1 (H2) is the only substance listed that is composed of two moles of atoms.


text:
Transition from ground state to excited state
I don't understand how to correctly find the term symbol of an excited state. I know that the first two permitted transitions for the Ne atom are 16.8eV and 19.8eV and I want to find the electronic configuration and the term symbol.
I assume that from the $2p$ orbital I'll have a transition to the $3s$ orbital (first excited state) and to the $4s$ orbital (second excited state). The selection rule $\Delta l=\pm1$ is obeyed. But what about the total angular momentum $j$? To find $j$ I have to know something about the spin $s$. Is it something about singlet and triplet states?

First you need to have the ground state electron configuration, for Neon this is 1s$^2$2s$^2$2p$^6$ or what ever it is for your atom. This will give you the term symbol for the ground state. The excited state has one electron promoted, say p to a d orbital, then 1s$^2$2s$^2$2p$^5$3d$^1$.
A term symbol has the form $^{2S+1}L_J$, where S is the total spin ($2S+1$ is the multiplicity) , L the orbital angular momentum and J the total angular momentum.
Because angular momentum is added vectorially it is necessary to follow some rules to get all the values. Each type of angular momentum forms a series of values (the Clebsch-Gordon series). If total spin angular momentum is S and for each electron the spin quantum number is s1 and s2 then the series is $S= |s1+s2|..|s1-s2|$ with steps of 1 between values. Note that all values are positive and there may only be one value in the series. The same type of series applies to L and J total spin and orbital angular momentum.
A p$^5$ can be considered as the same as p$1$ as only 1 electron is unpaired. Thus using the series if there are two electrons each with spin quantum numbers s1, s2 = 1/2 each the the total spin for two electrons is S = 0 or 1 and the spin multiplicity is 1 (a singlet) or 3 a triplet. The series is |s1+s2| = 1, and |s1-s2| =0.
The L values is calculated with l1=0 (s orbital) and l2=1 (p orbital) thus L only has 1 value = 1. Thus far the possible term symbols are $^1P$ and $^3P$.
The total angular momentum  J is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ Again this means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. There are $2J+1$ degenerate J levels, these are split in an external magnetic field into separate states.
This for the triplet state $S=1$ and so $J=|1+1|..|1-1|$, or 2, 1 and 0. (note three values from the series). Thus the term symbols are $^3P_2$ , $^3P_1$, $^3P_0$. But there is also a singlet term so the J values for this are $J=|1+0|..|1-0| = 1$ (one term only). Thus the last term symbol is $^1P_1$. Although this looks rather complicated it becomes quite quick with a bit of practice.
Once you have all the term symbols you can then apply selection rules to see which transitions are allowed.
Finally if the configuration contains, say, p$^2$ then one has also to consider the Pauli principle and it is more tricky. The way to do this is to list all s1, s2, l1, l2 values and remove any with the same set of quantum numbers. Finally when you have the term symbol Hund's rules are used to determine which levels and states are lowest in energy. (Many text books work through the p$^2$ case, See for example Steinfeld, 'Molecules & radiation')

Yes, it does have something to do with that: doublet states. Both spin orientations are possible. However, the energy difference (through spin-orbit interaction) between $j = 1/2$ and $j = 3/2$ is way smaller (around 0.002 eV) than the excitation energies you are looking at here.
One way to indicate that j does play a role in the configuration but not on the order of the energies you are dealing with, is to add a "J" as the total angular momentum quantum number to the letter of the orbital angular momentum quantum number, e.g., $2^2P_J$.

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text:
Is cycloheptatriene aromatic or not?
Is cyclohepta-1,3,5-triene aromatic?

I think it should not be, because even though the π-system possesses six π-electrons, the conjugation is not complete around the ring.

Related: http://chemistry.stackexchange.com/questions/28342/are-these-aromatic/28344

You are correct. The conjugation does not extend all the way around the ring, so the molecule is not aromatic. In addition, the single methylene group ($\ce{CH2}$) sticks out of the plane of the rest of the molecule.
(On the other hand, removing a hydride ion $\ce{H-}$ from the $\ce{CH2}$ group leaves the tropylium ion $\ce{C7H7+}$, which is planar and aromatic, with resonance structures putting the positive charge on any of the seven carbons.)

If the world was black and white, you'd be correct. Unfortunately, aromaticity is among those concepts which have a large grey area: https://en.wikipedia.org/wiki/Homoaromaticity

Actually the comments above are not entirely correct. This molecule does have some aromatic character.
These systems are called Homoaromatic. We discussed them at length in my Physical Chemistry course several years ago. Wikipedia has a page on this and here's a recent primary source as well:
Williams et al., J. Am. Chem. Soc. 2012, 134, 40, 16742–16752
The CH2 rotates out of the plane of the other carbon atoms which brings the adjacent p orbitals close enough together to have some pi-pi overlap. This gives a slightly stabalized 4n+2 homoaromatic system. The stabalization seems to be about 20-50% that of benzene in the case of this molecule, according to the article above.

To define an aromatic compound, there are four basic rules:

P(lanarity)
D(elocalization)
H(ückel's rule): $4n + 2$ π-electrons
C(onjugation)

In the above compound, if any one of the carbon is $\mathrm{sp^3}$ hybridised, then it is not planar. In the above compound, there is an $\mathrm{sp^3}$ carbon, so it is not aromatic.


text:
What will "Efavirenz amino alcohol methyl carbamate" look like?
I'm translating a text in which a possible derivative (produced by harsh conditions) of this drug is mentioned:

Its name is put as "Efavirenz amino alcohol methyl carbamate".
I wonder what it will look like.
The carbamate group, according to Wikipedia, will look like this:

And there will also be an alcohol (hydoxyl) group and an amino group somewhere. Will there be two nitrogens, or will the only nitrogen atom particiape both in the "amino group" described in the name and in the "methyl carbamate"?

Maybe the lactone on the right is hydrolyzed to the acid and alcohol, and then the acid group forms a methyl ester.

@f'' - thank you, I'll try to understand it. I can see an acid forming there, but have not enough imagination to imagine where the hydroxyl will be.

The "efavirenz amino alcohol methyl carbamate" is (S)-methyl 4-chloro-2-(4-cyclopropyl-1,1,1-trifluoro-2-hydroxybut-3-yn-2-yl)phenyl carbamate (CAS 211563-41-6),

according to this reference.


text:
Can wedge and dash bonds on 1, 2 position of cyclohexane ring be used to signify trans orientation of the substituents?
I've to make the structural isomers for cyclohexane. On position 1, there's a methane group with wedge bond and on position 2, a methane with dash bond. So I'll have to make the trans isomers only as they are not in the same plane or should I make the cis isomers too?
Or, can wedge and dash bonds on 1, 2 position of cyclohexane ring be used to signify trans orientation of the substituents? Is it actually trans or is it there just to show that the molecules lie in different planes or something?
For reference: as it is the 1,2 position, cis will be axial on 1. equatorial on 2. Trans will be axial on both 1 and 2.

Related: Chair conformation. Wedged lines vs no wedged lines?

So if you drew a wedge in position 1 and a dash in position 2, it would mean that they are on the same plane - either both axial or both equatorial - assuming that the ring was in a chair conformation.
However, if the objective is to show the structural isomers very clearly, it may be better to use either the chair conformation sketch:

... or the Newman Projection:

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text:
When our hard water is boiled, it leaves a blue-green residue in our kettle.
The inside of our kettle is lined in a blue-green deposit. Our water is very hard, but what minerals could cause the coloration?

Can you post a photo of your kettle? It appears what you desribe as "blue-green deposit" could be explained in various ways.

To specify: Are you describing a turqouise crust (rare in kettles) or a rainbow like patina (common in kettles or pots)?

@tipavi You're absolutely right in that a photo would help clear this up. It's great that there's discussion, though))

Over the last few days I have observed a similar phenomenon in our kettle. We also have blue limescale in our kettle https://i.stack.imgur.com/5R5AI.jpg. This is in a house in Cologne, Germany built in 2016. It was easy to remove with citric acid but I am also wondering about health implications.

If your building's water pipes are made of copper, that could explain the blue-green coloration. Hopefully the inside of your pipe doesn't look like this, though:

Unlikely that the kettle is made from copper... ;)

Unlikely that the kettle is a pipe - you might want to amend your answer.

The implication would be that the tinge of copper comes from the water coming through the pipes.

Why couldn't the kettle be made of copper? Why couldn't the minerals come from copper piping? I think that's a perfectly logical hypothesis. Also: https://www.google.de/search?q=copper+kettle&espv=2&biw=1422&bih=1070&site=webhp&source=lnms&tbm=isch&sa=X&sqi=2&ved=0ahUKEwi-iYfhu_jNAhXDuxQKHWzqABQQ_AUIBygC&dpr=0.9

Old buildings have water pipes made of copper - I'll amend my answer to be clearer about that @ToddMinehardt

While it is a perfectly logical hypothesis, I've never seen a copper kettle in my life. Have you @Zubo? Copper piping in the water supply could theoretically cause a residue as @ITTsoi describes. However, a very common phenomenon in kettles and pots is a rainbowlike patina, a discoloration of steel with high amounts of blueish colours. Similar to what steel experiences due to welding. I reckon it's a more plausible hypothesis that this what is described in the question. A photo of the original kettle should help to clarify this.

@tipavi I've seen copper kettles... Regarding your hypothesis, are you referring to the colors produced from thin-film interference from the oxidation of steel? I could see this happening in a steel pot by boiling all of the water off. During the last stages of boiling and after, you could easily over temp the pot. A photo would clear this up pretty quickly

@Tunk OK, I've never seen a copper kettle. But you're right, very old kettles used to have immersion heaters that might have been made from copper. These days kitchen kettles usually have at least a stainless steel bottom.
Yes, the thin-film interference from oxidation is exactly what I'm referring to, see my answer from Jul 16 '16 at 15:57. Commonly observed in overheated stainless steel pots or kettles running dry.
I don't have a photo at hand. Maybe a link works: https://i.pinimg.com/originals/38/41/51/384151930c927f855ec1019ff3a840a4.jpg

This might be dangerous! StackExchange is not necessarily the place to check this with.
If this is CuSO$_4$, that's actually mildly toxic. It could also be something like the German "Grünspan", a copper rust that makes a patina on copper house roofs that's got a blue-green color. It is a mixture of copper hydroxide, carbonate, acetate, and maybe others. This is all not good if this is your drinking water.
You should get that checked by professionals. However, ammonia solution in excess should produce a deep blue tetraamine complex with copper(II)-ions, you could try that (although if it doesn't work, that's by no means a safety assurance).
Have a look here for an example of Grünspan:

This is not caused by any mineral in the water, but an oxidation of the kettle's steel itself. Several monolayers of metal oxide are stacked onto each other, basically forming a dielectric mirror.

I think that's a possibility, but certainly not the only one! Wouldn't bet on it

@Zubo: Looking at your comments above I guess we are talking about different phenomena. A photo might help.

Limescale consists mainly of calcium carbonate but can also contain other compounds depending on water composition. These include iron hydroxides which while a white solid gains a greenish tinge in the presence of oxygen. That may explain a green limescale in your kettle.


text:
Confused about weak/strong acids, conjugated acid-base pairs and buffers
I learned that
a) the conjugate base to a weak acid is a strong base (and vice versa).
b) a buffer consists of a weak acid (base) and its conjugate base (acid).
However, this explanation of buffers says the following:
"[...] a weak acid is one that only rarely dissociates in water [...]. Likewise, since the conjugate base is a weak base, [...]"
which seems to stand in conflict with my assumption a) above. So, what is correct?
Furthermore, if b) is correct, isn't any solution of a weak acid solution a buffer, since any weak acid in water makes an equilibrium of the form
$HA \text{ (weak acid)} \leftrightarrow H^+ + A^- \text{ (conjugate base)}$
and is thus a solution of a weak acid and its base?

A strong acid (or base) forms a weak conjugate base (or acid). This is correct. By saying that $\ce{HA}$ is a weak acid and $\ce{A-}$ is a weak conjugate base they mean:

Acid shouldn't be too strong - don't use $\ce{NaCl + HCl }$
Conjugated base shouldn't be too strong - don't use $\ce{EtOH/NaOEt }$
Use a "not-strong" acid (weak acid) whose conjugate base is also "not-strong" (weak conjugate base). For example, $\ce{H3CCOOH}$ is a relatively weak acid. But $\ce{H3CCOONa}$ is also a relatively weak base. Choose such acids to make a buffer.

Your statement a) isn't always true.
Water dissociation is represented by:
$$\ce{H2O + H2O <-> H3O+ + OH-}$$
$$ K_\mathrm w=[\ce{H3O+}]\cdot [\ce{OH-}] = 1\times 10^{-14}\ (\textrm{at}\ 25^\circ ~\mathrm C) $$
Note 1: We don't write the $\ce{H_2O}$ activity, since it can usually be rounded to 1 and the ions activities can be rounded to their concentrations.
Note 2: This value was obtained experimentally, considering the concentration of $\ce{H_3O^+}$ and $\ce{OH^-}$ in the medium was the same and measuring the ionization of water.
On a weak acid dissociation:
$$\ce{HA + H2O <-> A- + H3O+}$$
$$K_\mathrm a=\frac{[\ce{A^-}]\cdot [\ce{H3O^+}]}{[\ce{HA}]}$$
On its conjugate base:
$$\ce{A- + H2O <-> HA + OH-}$$
$$K_\mathrm b=\frac{[\ce{HA}]\cdot [\ce{OH^-}]}{[\ce{A^-}]}$$
If we 'add' both reactions we expect that both equilibriums will happen in the mixture, so we have:
$$\ce{HA + H2O <-> A- + H3O+}\ \ \ \ (K_\mathrm a)$$
$$\ce{+}$$
$$\ce{A- + H2O <-> HA + OH-}\ \ \ \ (K_\mathrm b)$$
$$\ce{============================}$$
$$\ce{H2O + H2O <-> H3O+ + OH-}\ \ \ \ (K_\mathrm w)$$
$$K_\mathrm b\cdot K_\mathrm a=[\ce{H3O+}] \cdot [\ce{OH-}] \cdot ([\ce{A-}]\cdot [\ce{HA}])/([\ce{A-}] \cdot [\ce{HA}]) = [\ce{H_3O^+}]\cdot [\ce{OH^⁻}] = K_\mathrm w$$
$$K_\mathrm b=K_\mathrm w/K_\mathrm a~~~~~~ K_\mathrm b=10^{-14}/K_\mathrm a\ \ \textrm{at}\ 25^\circ ~\mathrm C$$
Another way to write this:
\begin{align}-\log(K_\mathrm b)&=-\log(10^{-14}K_\mathrm a) \\ \implies \mathrm pK_\mathrm b &= -\log(10^{⁻14}) - (-\log(K_\mathrm a))\\ \implies \mathrm pK_\mathrm a+\mathrm pK_\mathrm b &=14\;.\end{align}
This is the relation between a conjugate base strength and its acid strength. A very strong acid has a weak conjugate base, but a weak acid doesn't necessarily have a very strong base. Say you have a $\mathrm pK_\mathrm a=5$, which is a weak acid, with $K_\mathrm a=1\times 10^{-5}$. The conjugate base would have a $\mathrm pK_\mathrm b=14-5=9$ or a $K_\mathrm b=1\times 10^{-9}$, which is not a strong base.
However, if we have a strong acid, like $\ce{HCl}$ with a $\mathrm pK_\mathrm a=-6.3$ and $K_\mathrm a=10^{6.3}$. Its conjugate base would have a $\mathrm pK_\mathrm b=14-(-6.3)=20.3$ and $K_\mathrm b=10^{-20.3}$ which is a really weak base.
Hopefully I didn't make it even more confusing for you! It's mostly an reaction equilibrium issue.


text:
Is there an all-natural, non-volatile and low viscosity oil?
The goal is to thin a vegetable oil so that it spreads easily on top of water. It is not for consumption and the flash point of the mixture should be >=95° Celsius". Vegetable oils tend to have flash points greater than 225° Celsius so the flash point of any additives can be well below 95°, as long as special precautions would not be necessary in a manufacturing process.
Since my original post, I've discovered another reason for the low viscosity. It is required for the delivery system to work properly.
I am open to other means to achieve the goal.

If you're looking for food based products, you might be better posting this question on the 'seasoned advice' stack exchange.

He said it was not for consumption, so this should be the right place.

Flash point >95C -are you planning to do this experiment at that temperature, or are you going to do it at room temperature, and this is just a general safety concern? Do you want to increase the speed at which oil spreads or the size of the oil spot when it reached equilibrium (stopped spreading)?

Sorry sixtytrees, I did not see your comment sooner. It's a general safety concern to comply with restrictions for air shipments. I should have said "so that the flash point of the mixture >=95° Celsius". Vegetable oils tend to have flash points greater than 225° Celsius so the flash point of any additives can be well below 95°. I'll change the original question accordingly. I'm more interested in size of the spread then speed.  The max size of the spread is about .3 sq m, and the time to 100% spread can be a minute or more.

If the goal is to have the oil cover the water surface, then you might need to add a natural surfactant, such as lecithin, to cause the oil to spread, rather than form globules. You can try egg-yolk or soy beans as a source of lecithin, or you can buy refined lecithin.
As for the smoke point, or maximum cooking temperature for the oil, many common vegetable oils could do: safflower, sunflower or refined canola (rape seed) oil all have a good margin of safety.
One issue might be that if the water is at a "roiling" boil, the turbulence will mix the oil into the body of water. For that reason, you'd need to keep the temperature below 100° C and prevent local hot-spots in the heat source from causing strong convection currents.

Thank you so much for the lecithin idea! I've purchased some and will test it this week.
The water surface will always be at room temperature. The flash point requirement is based on postal regulations for an item to be shippable by air.

You may need only a very small amount of lecithin! Sorry I didn't mention that earlier.

Thanks for the extra info. That helps reduce my testing.

I just thought of  a potential issue. I think of lecithin as an emulsifier, so I am concerned that the oil will mix with the water. Will the oil only spread on top, or will it also mix with water?

I just reread the paragraph on roiling water. As I mentioned, the temperature will always be room temperature. The surface will mostly be calm, but there will be times that the surface is moderately disturbed. When the disturbance ends, will the oil resume it's original state spread across the surface?

If there is enough oil, and little lecithin, my hunch is that it would stay as separate layers for some time. It depends on what the purpose is, of course.  For example, to prevent mosquitoes breeding in a rain barrel, http://www.bluewaterbaltimore.org/blog/how-to-keep-your-rain-barrels-free-of-mosquitoes/ suggests 1/4 cup after a heavy storm. To retard evaporation in a reservoir, a monomolecular layer of "fatty" alcohol is preferred.

I've tried liquid lecithin (sunflower based) with a variety of oils at a varietyof percentages. I could not make it work. With lighter oils such as soy bean oil, it foams when mixed. Any thoughts? Thanks.

It depends on your setup. Perhaps a diagram or picture with size indications might help. The foaming of soybean oil might be due to soy lecithin already present. Increasing the quantity of oil will ensure a surface layer, but that might not be acceptable for your use.

There isn't much to my testing set-up. I'm using a 150 ml glass measuring cup. My tests usually start with 25ml of oil. For mixing, I use a small immersion mixer. I use a pippette to drop the mixture into a basin of water that has a diameter of about 18 cm. You are right about more oil not being acceptable, but I tried it anyway. The oil still did not spread.

oops.. I did not realize that you were using an immersion mixer and assumed this was a static, or nearly static, environment. In that case, you do not want to use a surfactant such as lecithin, or you wind up with "mayonnaise"! With sufficient stirring, you homogenize the mixture, and separation can take months of standing.

Surface tension is responsible for the spreading of one liquid on top of another at equilibrium. The viscosity dictates how fast oil spreads over water, but the surface tension is the one dictating the size of the spot at equilibrium.
If you need to increase the size of the final spot - use surfactants. Soap or dish washing liquid will work. If you want to decrease viscosity then you while keeping flashpoint low then you are in a tough spot. Ligher hydrocarbons are less viscose, but their flash point is low. Pentachloroethane has low viscosity and high flash point, but it isn't readily available and is toxic.

Related molecules, chloroform or tetrachloroethylene, which have lower viscosities and densities and are far more common than pentachloroethane will all sink in water (not float). The density PCE is 1.68 g/ml, the density of TCE is 1.62 g/ml, and the density of CHCl4 is 1.49. The viscosity of CHCl4 is 0.563 cP (20 °C). The viscosity of TCE is 0.89 cP at 25 °C, and the viscosity of PCE is 2.45 cP at 20 °C.

Thanks Ben. Sorry that I didn't see your post sooner. If I understand correctly, all of the chemicals that you listed do not meet my criteria? If I'm mistaken, are any of these easy to use and safe in a manufacturing process? Can any be considered natural?

I wish that I had seen the original post sooner. The goal is to increase the final spot. I haven't had much luck with a couple of soaps or dish washing liquid. Should I be looking for any special characteristic? I had assumed that low viscosity helped increase the spread, but that seems to not be true. In any case, I still need low viscosity because it is required by the delivery system.


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Origin of Azide's Name
Today, I learned that  $\ce{N_3^-}$ is referred to as azide. I was wondering why $\ce{N_3^-}$ is referred to as azide. I checked Wikipedia for an answer, but I didn't find one. Does anyone know why this ion is assigned this name?

dictionary.com says it's azo- + -ide.

Speaking of which, azidoazide azide, unsurprisingly has only Nitrogen and explodes upon any stimulus (including a bright light). [unrelated but a cool fact]

Nitrogen was originally called 'azote' by Lavoisier. This name persists in many nitrogen containing species such as azide, hydrazine, diazonium etc.
The '-ide' ending is the standard ending for anions of a single element such as carbide, oxide, sulfide etc.

Bon explains in their answer that

Nitrogen was originally called 'azote' by Lavoisier.

And it was called 'azote' because living things cannot survive in the gas.

From Greek a- "not, without" [...] + zoion "a living being;" [...] coined in French by Lavoisier & de Morveau because living things cannot survive in the gas.
http://www.etymonline.com/index.php?term=azo-&allowed_in_frame=0

We look here at the "-ide" part. It ultimately derives from the French word for "acid".
When the agent that supports combustion was identified as an acid former ("oxygène"), the products of such combustion came to be called "oxygène acides", contracted to "oxides".
The spread of the suffix to compounds of other nonmetals may have been promoted by the fact that several other nonmetals also form acidic products, such as "chlorine acids" ("chlorides") produced when that element combines with hydrogen, sulfur or phosphorus.


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Regarding Kw, titrations, and pH
So I have need of performing a titration upon a fruit juice to determine acid concentration. I will be using 1M NaOH as my base. Seeing as the solution will be coloured, it would be difficult to use an indicator such as phenolpthalein. Thus, I would like to use a pH probe.
My Question:
While titrating, should I be aiming to reach a pH of 7 or should I be aiming to reach the ionic product constant of water (Kw) at the temperature that I am performing the titration at? (e.g.6.63 at 50 degreees celsius)
Thank you in advance for your time and knowledge.

When using a pH probe, instead of aiming to reach the equivalence point of the titration, you should be titrating beyond it to create a complete titration curve. With an indicator, the end point is clearly indicated by a color change; with a pH probe, however, the end point is indicated by a steep change in pH in a plot of the data, thus requiring the collection of data beyond the end point.
While in theory the pH at the equivalence point of the titration can be calculated (and, for the limiting case of a strong acid and strong base, is exactly the $K_w$ that corresponds to the temperature of your solution), calculating the pH at the equivalence point and aiming to reach that pH is (i) bad lab practice, since if you're able to calculate the pH at the equivalence point, you're presumably trying to experimentally verify theoretical calculations, and (ii) not applicable to your current experiment, because the organic acids present in fruit juice are weak acids, whose conjugate bases are weak bases that will raise the pH at the equivalence point (and render the pH impossible to calculate for your purposes).

You shouldn't use 1M NaOH. It is way too strong and solution will turn basic with just a couple of drops. You should aim to add around 10-20 mL of base to fully titrate your sample. 10 mM solution is probably a good place to start. And you don't want to waste a barrel of juice on this experiment.
You should titrate to roughly pH = 10, but your main goal is to see full titration curve. Juice has all sorts of acids - strong onces, weak once. So don't expect your curve to look like a textbook example. It will have multiple phases.


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Why can't diamagnetic, and paramagnetic magnetize?
Ferromagnetic materials can be made into permanent magnets.  How come diamagnetic, and paramagnetic materials be made into magents?  Does it have to do with the valence electrons, and how they are arranged, and their spin?

You had requested this this be moved to Physics, but I don't think this is a very clear question.  Do you mean how can they be made into magnets or why can they be made into magnets?  A bit of initial research would be helpful if you are going to repost it over there, otherwise, it will end up closed.

I think this question is a grey area where material science meets physics, and is appropriate on either Chemistry or physics stack exchanges.

The concept of Ferromagnetism, Paramagnetism and Diamagnetism is completely different, though they all show magnetic properties.

FERROMAGNETISM: This is constituted of domain and each domain contains a large number of molecules (the number varies from material to material). These domains tends to align themshelves in a certain fashion in presence of magnetic field in order to achieve lowest energy and hence ferromagnetic materials show attraction to presence of magnets. Now these domains are not loose, but the arrangement can be made permanent in presence of some prolonged magnetic field. The amount of magnetism something can hold can be explained by hysteresis loop (NCERT Class 12 Physics Text Book). The arrangement of domains are shown below:
PARAMAGNETISM and DIAMAGNETISM depends on the number of unpaired electrons present in a system. The presence of unpaired electron to the system gives it a magnetic moment (given  by $\mu=\sqrt{n(n+1)}$, in Bohr Magenton , where n is the number of unpaired electrons). Hence the number of unpaired electrons, i.e. the excess number of electrons in the same spin, determines the magnetic moment. A material is called diamagnetic if the value of $\mu = 0$.

Thus paramagnetic materials are permanent magnets by their intrinsic property, but the magnetic moment is too weak to detect it physically. The increase or decrease in magnetic moment, or, the number of unpaired electrons in paramagnetic materials, is impossible by physical methods like keeping them in touch with magnet, or rubbing a magnet, or keeping it in strong magnetic field as done with ferromagnetic materials.
However, this is possible through chemical methods. For example, $\ce{[Fe(CO)6]^{3+}}$ is diamagnetic while $\ce{[Fe(H2O)6]^{3+}}$ is paramagnetic. However, this is not magnetizing, it is just changing what you are working with.


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Why do some molecules form T shapes instead of trigonal planar shapes
Consider a molecule such as $\ce{ClF_3}$.
Shouldn't the electron clouds (which are more negative than the Chlorine atom) want to repel each other and so the $\ce{ClF_3}$molecule should arrange into a trigonal planar shape where the electron clouds are as far away as possible?
I think at a deeper level than VSEPR, Chlorine is either partially ionized to a hybrid of $\ce{F^- + Cl^+F_2}$ configurations or has some electrons promoted up to d shaped orbitals but I don't know how to calculate the resulting shape. I think the main thing throwing me off is that the $\ce{F^- + Cl^+F_2}$ shapes should have the $\ce{F^-}$ ion repulsed from the lone pair electrons.

There is a 3-center 4-electron bond between the chlorine and two fluorine atoms. This bond is linear, so the fluorines are necessarily on opposite sides of the chlorine.

A more general question is $$\mathrm{Why\ does\ VSEPR\ predict\ a\ T-shaped\ geometry\ for AX_3 E_2 \ and\ not\ trigonal\ planar?}$$ We also have to remember that VSEPR was developed to rationalize what we were learning about electrons, bonding, and quantum mechanics withe known geometries.

see http://chemistry.stackexchange.com/questions/34610/question-about-stability-of-two-structures-of-chlorine-trifluoride/34613#34613


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Constructing a grenade (for a novel I'm writing.)
Ok, before any eyebrows get raised, this is for a novel that I'm writing and inspired by this answer from worldbuilding and this reference answer. I just want to make sure the device I'm trying to use is as close to accurate as can be.
So, in the novel, I've devised a compartmentalized glass urn. The urn contains three compartments, one will house a mixture of a finely powdered combination of charcoal, sulfur, and saltpeter. The second compartment would contain slightly acidic water and the third one would contain limestone. What I'm expecting is that when the urn is dropped, the shattering of the glass would cause the water to react with lime resulting in an exothermic reaction which would help to ignite the gunpowder mixture. Is this idea liable to work or is too quixotic and fanciful to consider ? If so, what adjustments can I make to it so that it works? Do keep in mind that the sources have to be natural as it is an apocalyptic world.
Also, can I just call it a mixture of a finely powdered combination of charcoal, sulfur, and saltpeter or does it need to be prepared in a special way or maybe there is a special term for it?
Apologies if this is not the right place to ask the question but I could not think of any other apt place.

It's like least likely way to make a grenade :(  BTW @jonsca https://en.wikipedia.org/wiki/Gunpowder has lots of proportions ;)

@Mithoron As long as we don't become the primary source of such information, I can rest a bit easier.

water would render black gunpowder ineffective. Incidentally drying black gunpowder  is the most dangerous step. // Also black powder needs to be confined to "explode", otherwise it just burns.

In short-no your technique will not work.  Reacting limestone with acid does not release enough energy to reach an appropriate temperature to ignite gunpowder and the acid would hinder combustion.
Fear not though, spontaneous ignition may be achieved by mixing glycerol and potassium permanganate which works well and is a classic example.  If you would like to be more out of the ordinary, zinc powder and ammonium nitrate with a minor amount of copper or ferric chloride and water will also create a spontaneous ignition for your gunpowder.

thanks for the answer but I'm facing a problem with your suggestions. You see, it's an apocalyptic scenario in my world so they won't be able to synthesise any chemicals. Glycerol (from vegetable oil) and zinc can be found naturally but the rest will pose a problem. Can you suggest something whose ingredients are available naturally ? It's ok if it is to be mined or extracted from plants or animals. I'll update the question with the same.

@user96551 You won't find elemental zinc in nature. It's too reactive.

@bon: I was going by this statement in the wiki - https://en.wikipedia.org/wiki/Zinc - Large deposits are in Australia, Canada and the United States, with the largest reserves in Iran

@user96551 That refers to zinc ore. The ore goes through a lengthy extraction process involving heating, concentrated sulfuric acid and finally electrolysis in order to extract elemental zinc.

They talk about zinc minerals, mostly ZnS. Zn metal oxidizes and doesn't exist in nature.

You can extract zinc from an alkaline battery, and ammonium nitrate is in ice packs.

I agree, your technology will not work. A good place to start is the US army field manual "Improvised Munitions Handbook". They describe methods that known to work.
Can you say more about the time of your world? In most cases you can use fire cord to delay explosion. You can control timing by the length of the cord. Fire cord is made by soaking any paper in KNO3. You must have KNO3 to make gunpowder anyway.
If you plan to drop these bombs from airplane then your world is very modern. Making a flying airplane is actually quite a challenge. You still can use the technology of early bombs. Roughly speaking you have multiple matches inserted in the bottom of the barrel. On contact with ground matches are pushed in against sand paper. Some of them will ignite and ignite the gunpowder. You can use H2SO4/KMnO4 in one compartment and alcohol in the second compartment. H2SO4 is known since classical antiquity. Ethanol distillation is known since 13 century. KMnO4 is prepared by baking KOH with MnO2 and is also known for a while. If you can't get MnO2 you cannot fly an airplane either. I say this because an airplane needs high quality jet fuel and will not fly on olive oil. A glider isn't going to lift heavy enough bomb. Another way is to place a small incandescent lamp in powder, connect it to a battery and attach the switch to the bottom of a barrel. 19 century mining technology used electricity to initiate explosion because it is much more reliable.
By the way, how much gunpowder are you planning to use? The only way to produce KNO3 is by collecting it from (I beg your pardon) manure. Medieval peasants had KNO3 tax. They had to collect it and send to the king. Modern methods are different, but you cannot use them without a trained group of chemical engineers. If you don't have an access to reasonable amounts of KNO3 you will not be able to make enough powder. If you have, then you should prepare some modern explosives.

@ sixtytrees The sulphuric acid would react with the permanganate in the same compartment. How would this work?

@sixtytrees: The whole thing happens in the current world but on an island separated from civilisation. They don't have any access to any modern things so they won't have planes or such. Also, it is to be used like a regular grenade so having a fire cord won't be feasible. 'KMnO4 is prepared by baking KOH with MnO2 and is also known for a while.' - this won't be possible exactly because, in that world, if it is naturally occuring, I can have mines for it or something but i can't have them distill and stuff...I hope you understand my constraints...

As others have noted, the idea as stated will not work. The temperature rise produced by the reaction of mildly acidic water with limestone is very small -- nowhere near enough to ignite black powder (over 400°C.)
We could possibly suggest improvements -- whilst avoiding giving bomb-making advice! Blackpowder is pretty easy to ignite, and without giving any realistically helpful advice to maniacs, there are a lot of ways it could be done (some have already been suggested.)
However it isn't clear what your conditions are. The device you describe is quite primitive -- is your novel set in the distant past, or in some sort of post-apocalyptic scenario? In that case, the most plausible igniter is something like the flintlock percussion mechanism that was actually developed in the Early Modern Era.

I just want to make sure the device I'm trying to use is as close to
  accurate as can be.

This discussed in multiple questions on writing.se. But in general, for a novel, it is neither necessary nor desirable that you describe a technical device (e.g. a bomb) in such detail that an expert thinks "Hmm, that could really work." You are, after all, writing a story -- not a terrorists' manual. In fact the majority of casual readers will be bored by too much technical detail. You want to add colour, atmosphere, plot development, characterisation. Not instructions.
I'm no novelist, but consider something like this:

As he lifted the vase to examine the strange powder inside it, he heard
  a sharp click. He froze -- for a moment, nothing seemed to happen.
  Then a slight "clunk" came from within the vase, a faint whirring
  noise, and a glint of brass gears turning. In a moment, he realised
  what it was and flung the vase behind the pillar as he dove to the
  ground....


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Why is the thione tautomeric form of 2-Mercaptopyridine favoured?
During some organic chemistry studies (through an old online course material) I ran into a doubt about the equilibrium in tautomeric forms of some heteroaromatics.
First, the context:
On the course webcast, it was said that on the molecule 2-pyridone we have a tautomerism equilibrium like in the image below, where the pyridone form is favored instead of the 2-hydroxipyridine form.
Pyridone Tautomerism Equilibrium

Each tautomer have 2 major contributors for resonance, listed on the image below (for the 2-hydroxipyridine and 2-pyridone respectively):

On the 2-hydroxipyridine resonance forms (left) we have on the left side of the equilibrium an aromatic uncharged molecule and on the right side an aromatic charged molecule. However the most electronegative heteroatom (oxygen) is positively charged and the nitrogen negatively charged. This makes this resonance form a weaker contributor.
On the 2-pyridone resonance forms (right) we have on the left side of the equilibrium a non aromatic (8pi $\ce{e-}$) uncharged molecule and on the right side an aromatic (6pi $\ce{e-}$) charged molecule, where the oxygen is the negatively charged heteroatom, matching the electronegativity character of each heteroatom. This favors the 2-pyridone tautomer.
Another factor mentioned was the dimerization of the molecule, which was geometrically favored in the 2-pyridone form, as seem below:

So far so good, the doubt comes when the tautomers of 2-mercaptopyridine are discussed.
Tautomers equilibrium of 2-mercaptopyridine:

It's known that the thione form (right) is favored. It's not explained as detailed as in the pyridone case why, so I decided to analyse it.
I've drawn the 2 major contributors of resonance on each tautomer. I'll show and discuss them.
First the thiol tautomer form resonance contributors:

On the left we have an uncharged aromatic (6pi $\ce{e-}$) molecule. On the right we have a charged and non aromatic (8pi $\ce{e-}$) molecule, with a positive charge on sulfur and a negative charge on nitrogen.
Now the thione tautomer form resonance contributors:

On the left we have a non aromatic (8pi $\ce{e-}$) uncharged molecule. On the right we have a charged aromatic (6pi $\ce{e-}$) molecule, with a negative charge on sulfur and a positive charge on nitrogen.
I could make the following assumptions after drawing those:
1) Considering only aromaticity, none of the tautomeric forms are favored, since both forms have one aromatic contributor and one non aromatic contributor.
2) Considering the heteroatoms electronegativity, knowing nitrogen is more electronegative than sulfur, the thiol tautomer would have the right charges on its charged resonance contributor, being more stable than the thione tautomer.
3) Considering the geometry for dimerization, the thione tautomer seems to have a better geometry for the hydrogen bonds and would be favored over the thiol tautomer.
Only assumption (3) is in favor of the thione formation, however it's the favored tautomer. Why so?
In the course webcast, the teacher makes it sound like having the negative charge on the sulfur and the positive charge on nitrogen would be the right charge distribution. Wouldn't the opposite be true considering that sulfur is less electronegative than nitrogen?
Here's a youtube link to the webcast mentioned:
Hydroxypyridine-Pyridone Tautomerism
Thank you in advance!

EDIT NOTE:
When I originally wrote the question, I made a mistake while counting the thiol tautomer pi electrons on the resonance form of the right. I originally counted 6pi $\ce{e-}$, but it actually has 8pi $\ce{e-}$ (4 from the carbon-carbon pi-bonds, 2 from the nitrogen lone pair and 1 from the sulfur-carbon double bond), which means the right resonance form isn't aromatic. Thus aromaticity doesn't favor neither of the tautomers. The question still makes sense though, so I'm fixing the mistake and keeping it.

Assumption (1) doesn't hold too much precedence in this case, just like in the case of 2-pyridone, so it is a minor thing to consider. In the case of the assumption (2), electronegativity is only a part of it. Sulfur atom can stabilize the negative charge better than oxygen due to it's size. For the same reason, $\ce{CH3SH}$ is a stronger acid than $\ce{CH3OH}$: $\ce{CH3S-}$ is a more stable conjugate base than $\ce{CH3O-}$. So, although sulfur has a lower electronegativity than oxygen, it is still preferred to have a negative charge on it. Thus, altogether, the thione form is more stable than the thiol.
BTW, thank you for a very detailed question and thorough explanation of your reasoning!

Thank you a lot! So I over simplified the factors that contribute to charge stabilization. After reading your answer I found this link that points some of those factors! [link] (http://www.masterorganicchemistry.com/2012/02/27/7-factors-that-stabilize-negative-charge-in-organic-chemistry/)
I detailed the question a lot because I've been pulling my hair trying to figure this out, didn't even continue watching the webcasts until I got an answer! Thank you again for clarifying this for me! (:


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Given a theoretical combustion engine, how can I calculate the output of this engine depending on the fuel it uses?
I'm writing a Minecraft mod (for those who don't know, it's sort of like a smaller "program" that loads into the Minecraft game and expands it, by adding other things to it).
My question is: consider a combustion engine that runs on anything that is flammable (even though Uranium has a lot of chemical energy, it can't be set on fire, so it's out).
The engine can burn, for instance, acetone, methane, ethanol, methanol, etc.
By burning these fuels the energy converts their chemical energy into mechanical or kinetic engine to move it's gears, pistons and stuff (whatever the engine uses to transmite the energy).
What is the simplest way of comming up with a "good enough approximation" function
energy = f(fuel)
What goes in this function? the volume of the engine doesn't change, so not that. pressure certainly is a factor (maybe the answer itself?) does it have anything to do with the Ideal Gas Law? What about the density of the fuel, molar mass, specific heat, etc?

The answer to this heavily depends on whether you have an internal combustion engine or an external combustion engine.
External
Because you're just heating the liquid from the outside, you can approximate the final output of the engine as the amount of energy gained from combusting the fuel multiplied by engine efficiency (as you like, but no greater than 100%). In other words:
$$E=kfV$$
Where E: Energy Output; k: Efficiency absorbing the energy; f: Thermal Energy obtained from burning a unit volume of the fuel; and V: volume of fuel combusted.
Internal
For an internal combustion engine, far more factors need to be taken into account. Wikipedia is always a good start to get a sense of the factors that play a role, and whether or not you want to consider them. De
pending on how dedicated you are, and how much the player can 'mess with' the settings of the engine, you could do something as simple as the External Engine (if all the user can do is select fuel and amount of fuel) or as complex as is required based on the variables the user can change.
Some Other Factors to Consider (Optional):

Compression Ratio (internal only)
Gear Ratio (for high torque applications)
Working Fluid (external only)
Ambient Temperature (mainly external only)
Heat Exchanger Material (external only)

Ok, I will have both types, the first being less powerfull than the second. So where do I find the value of f? And what is loss by k is turned into heat, so there's got to be a liquid there to prevent the engine from melting, right? All those optional factors you mentioned affect k, correct?

@FinnTheHuman so "f" in the equation would be enthalpy of combustion. As for k, it is largely affected by the design of the engine and that may include the cooling liquid, if there is a second hear recycling stage, etc.

Thank you, I have a much better understanding now. I will be able to ask more specific questions about this issue and I probably will be doing so soon.


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Is there a site where I can input RGB values and the site will respond with a pH?
I have a tool where I point it at something, and it gives me an RGB value. I want to add universal indicator to a liquid, measure the RGB value and input it into the site to find the pH. Does anyone have any ideas?

How accurate do you need the pH to be?

@f'' preferably more accurate than being using a pH chart and eyes,

http://www.pantone.com/capsure-formula-guide-bundle This kind of thing would give you an accurate colour reading in RGB/CMYK, however if you have access to a lab etc, a UV-VIS spectrometer would be far superior at measuring the colour at which your solution absorbs

Actual color is dependent on the light you reflect, ie you need to calibrate the white balance. It sounds like a lot of work for nothing

RGB values vary non-linearly with the concentration of the light-absorbing substance. So I doubt you will find a tool that's easy to use and that will provide a higher accuracy than the colour matching charts that come with indicator kits. Also for measuring RGB values you'll probably run into problems with the gamut of the measurement device.

Your best bet would be to prepare 15 solutions of buffers at pH 0, 1,.. 14 and calibrate your instrument. Dip three different test papers in each solution, get average numbers. Normalize colors and see the r/g, r/b, and g/b ratios for each pH. Now you can just make a long if statement in excel and plug your measured numbers there.
Fill excel with 15 statements like
=if(r/g > 5 and r/g <8 and r/b>3 and r/b<3.5; "pH=3"; "")
Also, many pH papers come with a map of colors for each pH. you can use that for calibration.
The thing is, chemists use pH meters when they need accurate readings. accuracy down to mili-pH is achievable. pH meters are build on very different principles and are reasonably priced.


text:
How coordinate bond is formed?

Here nitrogen can donate three electrons as it has p orbitals with three unpaired electrons. Now it is filled by electrons of hydrogen and everything is paired. It has 2 paired electrons in an s orbital. Now $\ce{H+}$ is coming and getting electron from nitrogen. How can it get from the inner s orbital?

Ammonium ion is electronically equivalent to BH4- and methane. Both exist in sp3 hybridization state with all four bonds being equivalent. This has less to do with coordination chemistry, more to do with hybridization theory. Here is a good place to start.
http://www.chemguide.co.uk/basicorg/bonding/methane.html

Understand that ammonia typically adopts a trigonal pyramidal structure with a bond angle of $107^\circ$, which assumes $\mathrm{sp^3}$ hybridisation, rather than trigonal planar, which would assume $\mathrm{sp^2}$. Therefore, the lone pair is in a similar orbital as the other three $\ce{N-H}$ bonds, allowing for ammonia to act as a Brønsted base (accept $\ce{H+}$) and as a Lewis base (in coordination complexes).
The two states interconvert rapidly at room temperature.
The nitrogen atom is simply too small to allow for a bond angle of $90^\circ$ (unhybridised). The heavier homologue phosphorus, on the other hand, can allow this angle. Thus, phosphane $\ce{PH3}$ is much less basic than ammonia (even though phosphonium $\ce{PH4+}$ exists — but that is much more acidic than ammonium and can be used as a source for phosphane).


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Placing oxygen molecule into electric field
I am trying to understand what happens inside various types of air ionisers.
Some of them are sold for their purported health effects (which I do not want to discuss here), another type are used in industry to control electrostatic charges on surface of materials. Also there are some natural processes where oxygen form is modified by electric current namely thunderstorms.

Can an oxygen molecule be broken by placing it into a strong external electric field?

If yes how is it possible because as we know the oxygen atoms in molecules are bonded by a non-polar bond?
What are possible by products of this process (even temporary one)?
I assume that air is composed only from oxygen (no nitrogen, argon, carbon dioxide etc.).

Why negative vote ? These kinds of things happen in real world during thunderstorm, also in laser printer. As we know quite well ozone is enviromental toxine detrimental to health. I want to understand whole process leading to production of ozone (and potential other by products even temporary ones if there are some) to be able to control it. It is of uttermost practical importance. Sorry that I have not master degree in chemistry like most of you and 20 years of practice. This is interdisciplinary question related to electromagnetics which I am more familiar with.

It was probably downvoted because it is just a big wall of text. Consider formatting it into paragraphs and using better punctuation and grammar. Also, add some better tags. It is also a rather broad question. Consider breaking it down into several more focused questions.

Yes, an oxygen molecule (much like any other molecule) can be broken by electric field alone.
No, it won't be broken into separate oxygen atoms, precisely because the atoms are neutral and the bond between them non-polar, like you said. Instead, it would be broken into $\ce{O2+}$ and an electron.
This is not important anyway, since it would require electric field much stronger than those present in air ionizers.

What happens in air under voltage, then? Well, air always contains some ions from whatever sources (cosmic rays, background radiation, etc). As you apply voltage, your positive and negative ions start speeding up in the opposite directions. Once the voltage is high enough, they accelerate to the point that they might break neutral molecules by bumping into them. In doing so, they produce more ions, and those do the same, so pretty soon there is a breakdown going on. You see a blinding spark, the fuses blow up, the voltage shuts off, and the ions start to recombine into molecules. As luck would have it, sometimes they combine into $\ce{O3}$ - that's an important source of ozone both in nature and in the lab.
In fact, the situation is much more complicated. Gas discharge does not have to be that dramatic. It comes in various forms, some of them pretty controllable.

Several things can happen to O2 molecule when it is placed in electric field.
(1) Two ozone molecules can form from three dioxygen molecules. Ozone is more deadly than chlorine, so the resulting gas can be used as a bactericide.
(2) A needle under high negative voltage will leak electrons in the air. Some of these electrons will bind to O2 forming an anion-radical O2*-. This is a reactive oxygen species (ROS) that allegedly have some good health effects.


text:
How do you know if electron is deexciting or exciting itself?
I got a question which goes like this:

According to Bohr's theory, the electronic energy of H-atom in the n th Bohr's orbit is given by $$E_n=\frac{-21.7*10^{-12}}{n^2}J$$ Calculate the longest wavelength of electron from the third Bohr's orbit of the He+ ion.

For longest wavelength transition, we need lowest energy transition. Therefore the electron can de-excite to $n=2$ or excite to $n=4$. I do not need the answer to the question, but I need to know how to know if the electron is exciting or de-exciting.
From what I know, I think it would be exciting to $n=4$ since the energy is lesser than de-excitation energy.

The 'longest wavelength of electron' does not make any sense. I assume that the question asks for the longest wavelength transition when a photon is absorbed or emitted (as fluorescence). One way to answer this is to look up the Balmer formula for hydrogenic like atoms and choose two principle quantum numbers for the transition you want. Alternatively you take the difference in energy between two levels using the formula given. As you start with n=3 the closest is either going to be n=4 (absorption) or n=2(emission). I leave it up to you to do the calculation but the answer must be positive as it corresponds to a photon's wavelength :) ( Note that the question gives the energy in Joules and you will want the wavelength in nanometres).

The wavelength of an electron does make sense just not in this context.


text:
The chemistry of cutting
I would like to know what is the chemistry of cutting. For example, when we use a knife to cut through bread what exactly happens on a molecular level? Are ions formed, do long organic compounds (e.g. in cheese) break, or are just intermolecular interaction that seize to exist?
Thanks a lot in advance.

see this:-http://physics.stackexchange.com/questions/134119/how-does-a-knife-cut-things-at-the-atomic-level/134137#134137

It depends. Bread has moth small molecules (like 2-Acetyl-1-pyrroline). These molecules will just pushed aside by the blade. You have hydrogen bond network between water molecules. You break those. Bread has large polymers like starch. Knife breaks those.
A cleaner system to study is a polyethylene. You can freeze it in N2, break under inert atmosphere and study by EPR spectroscopy. You will find free radicals at the breaking point. This is because it costs less energy to break a bond then to move one molecule relative to others.

Essentially, what you are doing with an everyday knife is wedging your way through the substance. You are pushing aside the particles within the substance. In this instance, bread; you are pushing apart the different particles, requiring you to break the inter molecular forces that held them together.