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Work and entropy
I am not able to conceive why work wouldn't increase the entropy of a system. If i am compressing a ideal gas adiabatically and reversibly, the gas pressure inside would increase,meaning more no of particles hitting unit area, and that to me seems a more chaotic state of the gas, than before compression, so why doesn't the entropy increase.
And the 'chaos' or 'randomness' terms that we use to define entropy, what are they dependent upon.Velocity or pressure or what?
My understanding of entropy is that, it is a measure of how unpredictable the system is , how difficult it is to point out a certain gas particle. Am i right in my understanding?

Your question seems sort of unclear. Are you asking about work and entropy or heat and entropy?

Reversible adiabatic expansions and compressions do not have changes in entropy because no heat is exchanged during the process. Work done is not directly related to entropy. The relationship between temperature, heat exchange and entropy in a reversible process is as follows:
$$TdS=\delta Q$$
Since a reversible adiabatic expansion occurs with no heat transfer $\delta Q =0$, by definition then there is no change in entropy $dS=0$. In fact, reversible adiabatic processes are more properly called isentropric processes.
The key is reversibility. The true relation between entropy temperature and heat is
$$TdS \ge \delta Q$$
When the process is reversible, the $\ge$ sign becomes $=$, giving the first equation I gave you. Truly reversible processes are rare, so there are real expansions and compressions in which no heat is transferred, but there may be changes in entropy.

'Ben Norris' gives a very clear answer. However, you use the words  'chaos' or 'randomness' to describe entropy, and while widespread, this is ultimately misleading as they are not exact enough. You need instead to think about the number of energy levels that can be populated at a given energy, and the number of ways these can be occupied. For example, suppose there are two equal but separate fixed volumes each containing 1 mole of an inert ideal gas, at the same temperature and pressure.  If you now allow the two gases to mix (each by free expansion into the other) so that the total volume is doubled (pressure and temperature unchanged) then the entropy is increased. But is the gas more random after allowing mixing than before? Both would seem to be equally 'random'. But the extra volume available allows the molecules of the gas to move in a larger space and this means that at a given energy, the number of energy levels available to a molecule is increased. You can think of this in the same way that the energy levels of a  'particle in a box' decrease as the box increases in size meaning that at a given temperature (or energy) more levels are populated. As more levels are populated there are more ways of filling these levels and so the entropy increases. (For completeness the entropy change for each gas in the example above is given by $\Delta S = R ln(V_f/V_i) $, and if the initial volume $V_i=v$ and $V_f = 2v$ then the total entropy change is $2Rln(2)$ which is positive.)

"If you now allow the two gases to mix (each by free expansion into the other) so that the total volume is doubled (pressure and temperature unchanged) then the entropy is increased.  ". I think you are wrong with the pressure part, even in free expansion pressure decreases if volume increases.

I see what  you mean , I had supposed that the pressure was the same as each filled the other empty half as it were as it doubled its volume.

Your example of the removable partition set-up has been quite handy to me. I was replying to a question with similar background, and then i stumbled upon the pressure part, did some digging and found a pv graph for free expansion, in which the end states were joined by a dotted line(indicating irreversible process i suppose), with the final pressure less than initial pressure.

I'm not an expert, but here's my two cents:
1) If gas get compressed from work, I think it makes sense that it's energy is more concentrated rather than disperse. The total energy depends only on its temperature. You're right that lower volume means more pressure, but it evens out and as long as temperature is constant before and after compression, the total energy is also constant. But the difference is that in the more compressed state, there is less dispersal of that energy - it's all bunched up in a lower volume and hence entropy is less.
2) Another understanding on entropy is how much of the system's energy can NOT be used to do work. So when a gas is all bunched up in a small volume, more of its energy is available to do work (for example power a heat engine), while if it's energy all dispersed in a large volume, less of it is available to do useful work.

I  think the verbiage "useful work" is completely circular. Can you define it in any other way as just "work"?

yeah I just meant "work", the adjective "useful" is not needed. So if a gas is condescend and could do work on a piston, that part of it's energy is not entropy.

Your use of the words are conflating "energy" with "entropy". While it is true that there is a quantity called "availability", or sometimes the same is called "exergy", quantifying what amount of internal energy is available for work it is definitely not entropy but something that is directly dependent on entropy, but is also dependent on temperature and pressure, etc.

Thanks for your comment. It seems that you know much more about this than me. I got my idea from wikipedia: 'One dictionary definition of entropy is that it is "a measure of thermal energy per unit temperature that is not available for useful work". '

you should study the excellent books by Pippard, or Atkins, or Callen, all three are superbly written and you should not have any problem finding any of these.

@hyportnex Are these the titles you suggested: Elements of Classical Thermodynamics, The Laws of Thermodynamics, Thermodynamics and an Introduction to Thermostatistics

@user47024 yes to Pippard, Callen but actually I was thinking of Adkins: Equilibrium Thermodynamics

@hyportnex Point taken.


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Can sunlight and its heat create a caustic or toxic substance from coconut oil on human skin?
I heard that story quite often, if it's true, how does it work?
Notice that I’m not talking about using coconut oil as a sunscreen (it’s another topic) but about the action of the sun on the oil and its impact on the human skin: will coconut oil degrade into something toxic under the sun, if so why?
And also, does the same degradation mechanisms applied for other oils (such as olive oil).

I'm not sure medical related questions are allowed on this forum. I think you would be better served by either consulting a legally certified physician or by checking medical information sites online.

I do not understand why this was closed: "I’m not talking about using coconut oil as a sunscreen" makes it pretty clear that the interest  lies in reactions due to UV, not the medical apects.

I also believe it can be answered from a objective chemistry point of view, hence I recommend reopening. However, I am not yet overruling the vote of the community.

@shre_sudh_97 There is a misunderstanding : the question is not about health but chemistry. And I did not find anything on the web about this chemical mechanism (if it occurs). Please reopen the question.

@Lisa "Is there a list somewhere of safe to use oil in the sun?". This question makes me consider this to be a personal medical question.

@shre_sudh_97 I agree, the "safe" could be understood in that way, so I removed it. The question concern the chemical mechanism about this skin-sun-UV-oil interaction. So could you please reopen the question?

@Lisa Yes. Now the question is about chemistry than about health. I'm not sure how I can reopen it. It was closed by moderators. I'll surely upvote it now as it is a valid question.

@Joel The question has been edited. My comment was for the previous wording of the question.

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Is glass an amorphous solid or supercooled liquid?
I have been informed that glass is a super-cooled liquid and is also considered to be an amorphous solid.
Can it be both and, if not, what category does it fall into?

Window glass does not flow under it's own weight. That's a myth.

By (rheological) definition, the storage modulus of a solid, under linear deformation, does not vanish for low frequencies. That's also true for all glasses. Under high stress, also solids can flow, and many of them do, before ultimately breaking.

And why can't a super-cooled liquid be an amorphous solid? Just note that not all amorphous solids are super-cooled liquids, but many, including glass, are.

@Jon Custer, a liquid is considered to be super-cooled if it passes its freezing point WITHOUT becoming a solid, if it isn't a solid, it shouldn't be an amorphous solid as well

There are a lot of good things written about this actually. I think the best thing I've read basically says the easiest way to do this is to say it's kind of undefined. There is no latent heat transitions that a glass goes through when melting, but there is also a dramatic change in measurable properties when glass can flow. So, it doesn't change phases, but it changes properties. Beyond that, the definition is arbitrary.

viscocity NCERT class 11 gaseous state...apparently cbse board too supports myths

The answer to this depends what question you are asking (or the context of that question). If the question is "does the order in the material look like a liquid or a crystalline solid?" the answer is glass is a liquid; but if the question is "does bulk glass show properties more like a viscous liquid or a solid (eg does it creep like a liquid)?" then the answer it glass is a solid. The answer depends on which property you are focussing on.

Related question on Physics StackExchange: Do glass panes become thicker at the bottom over time? (This is what Karl states in his first comment.)

I'll focus on the phase of the glass, since it seems there is some confusion in the comments (and the other answer). In glass forming systems, as you drop the temperature below the crystal phase melting temperature, one would expect to nucleate the crystal phase. For various reasons this does not happen. Instead, the liquid continues to cool, and the viscosity continues to increase as one might expect upon cooling. Below a critical point, the glass transition temperature, the viscosity rapidly increases by several orders of magnitude. Atomic mobilities are now so low that crystallization cannot occur. Furthermore, the viscosity is so high that the material is a 'solid' in common usage - all the atoms are frozen in place.
So, what is this 'solid'? Well, it isn't the stable crystal phase - you can tell that by x-ray diffraction. True, it is not 'liquid' in the common sense of flowing like water. But, two things point to it actually still being in the liquid phase.
First, in a disordered material if you do x-ray diffraction you measure a radial distribution function (rdf). This gives you the average spacings between atoms in the material. If you do x-ray diffraction from above the melt down to the glass formation, you will note that the rdf is not changing (well, the average spacing will change a bit with thermal contraction).
Second, the other indication of a phase change would be, of course, a signature of latent heat in calorimetry. Again, as you cool the liquid down to the glass formation temperature (and beyond), there is no latent heat measured. There is no first order phase transition. There is only the glass transition where the visocity increases.
Taken together, a glass is a configurationally (kinetically) frozen liquid. It has the structure of the liquid, and it does not have a first order phase transition to the liquid. I do not see how one could consider it anything but a super-cooled liquid.
To further emphasize - a super-cooled liquid is one below the melting point of the crystal phase that has not nucleated the stable crystal phase. Nothing says that the liquid viscosity can not get so high as to be considered 'solid'.

https://en.wikipedia.org/wiki/Supercooling and it's source http://www.webcitation.org/5mjJauR7S you're against definition of supercooling? Supercooling is before turning into glass.

@Mithoron - supercooling is required to turn it into a glass. And the glass is frozen in a liquid configuration. And there has been no first order phase change. At best one argues about where you are on a viscosity versus temperature curve. It is likely also a sub-area definition. For silicon, we repeatedly attempted to make glassy silicon and failed. Amorphous silicon is not a configurationally frozen liquid, and exhibits a first order phase change to either diamond cubic silicon or liquid silicon.

Also helpful: The $\ce{^{29}Si}$ NMR of glass (visible in $\ce{^{29}Si}$ spectra because of the NMR tube) is a poorly defined ‘mountain’, showing the strong disorder present in glass.

What I have been able to gather from all the comments and answers it that, supercooled liquid is an intermediate state in formation of glass, silicate and other substances are 'supercooled and then frozen. So glass is an amorphous solid but in its formation, comes a time where it is a supercooled liquid

You are correct: it's not both.
Why? What you've heard is nonsense. Very old window panes are thicker towards one side because of how they were made at the time. It became common practise to put the thicker edge at the bottom, because it was less likely to exceed the acceptable stress on the thinner area.
Contemporary windows have been found with the thicker edge to one side, or even the top. Gravity on Earth has been consistently towards the core since at least the Middle Ages, so the "flowing glass" hypothesis is false.
Glass is an amorphous solid by definition (per Karl's comment) and is not a super-cooled liquid (per Prabhdeep Singh's comment).

For further reading: http://m.cmog.org/article/does-glass-flow

Gravity on Earth has been towards the core since the Earth formed. Suggesting anything else is ludicrous.

Glass is not a solid by definition. There is a definition for what is a solid (vanishing  loss angle for low frequencies), and glasses fit that description.

@bon "since at least" because this is the relevant time period; it was meant to be a bit of dry humour given that someone might wonder of gravity could shift enough to make liquids flow horizontally (yes, there are some people who aren't sure). It doesn't suggest anything about the time prior to that because it doesn't refer to any time prior to that.

I have been informed that glass is a super-cooled liquid and is also considered to be an amorphous solid.
Can it be both and, if not, what category does it fall into?

This is dependent on temperature, but before I go into that, we need to understand what a solid and liquid are as that is where mush of the misinformation comes from. A liquid is defined as:

a nearly incompressible fluid that conforms to the shape of its container but retains a (nearly) constant volume independent of pressure.

Note a liquid is NOT defined as the lack of a crystal structure, but as being a fluid (Fluid: a substance that has no fixed shape and yields easily to external pressure; a gas or (especially) a liquid.) . A solid is defined by:

structural rigidity and resistance to changes of shape or volume. Unlike liquid, a solid object does not flow to take on the shape of its container, nor does it expand to fill the entire volume available to it like a gas does.

This should make the question rather simple -- at ambient temperatures, its a solid, but then some people try to proliferate the myth that glass can flow over extended time scales and thus technically a liquid. This is just that, a myth, and is patently false. If you were to measure any flow of glass over as much time as the universe has existed, there would not be a change large enough to detect. Contrast that to something that is indisputably solid and certainly not fluid but still "flows" -- metals. Yes metals exhibit deformation due to creep when heated above a certain temperature and subjected to stress. Despite this deformation, and the fact that metals soften at high temperature, no one is arguing that metals are crystalline liquids nor does anyone argue that metal powders are liquids. Thus again glass is a solid at ambient temperature.
So the only question left is when is glass a liquid as it does become fluid at a certain temperature, and when is it a solid as it resists deformation below a certain temperature?
Given glass does not have an abrupt transition from liquid to solid such as with metals, but rather a continuously thickening viscosity, there are a few values glass scientists use to characterize a glass and that one might consider the glass to finally be a liquid. For our stringent definition of a liquid, the strain point where no more strain can be introduced into the hot glass making is a good demarcation as to below what temperature glass is definitively solid.

Source: https://www.glass-ts.com/littleton-softening-point

also see: amorphous metal

I think the thicker bottoms could be due to a phenomenon known creep (the internal flow of molecules under stress and over time). Amorphous materials below their glass transition temperatures show no molecular chain mobility, yet under a constant stress (gravity) and a very very long time (700,1000) years some very, very slow creep motion increases the volume at the bottom of the glass and thins it towards the rest of the body and in direction to the top.

Yes, some supercooled viscous things flow: bitumen, for example. But not glass. Glass shows no flow over any measurable timescale, but it does sometimes crystallise. Glass windows made before the float glass process was invented (mid 20th century) were thicker in some parts than others as a byproduct of the production process not because they flow. We know this isn't due to creep.

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What does "dye content 80%" mean on Sigma Aldrich's product page for "Allura Red AC" (aka Red 40)
Here's a link to the page:
http://www.sigmaaldrich.com/catalog/product/sial/458848?lang=en&region=US
Why does it say "dye content 80%"?
Shouldn't it be pure if it's a packaged powder?
Is it 20% other chemical compounds (e.g. stabilizers)?

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works.

tbh I'd just mail Sigma Aldrich to ask.

Many dyes are sold as powders with 80-90 % purity. It is because some dyes are difficult to purify (e.g. they are difficult to crystallize). The remaining 10-20 % are impurities, such as isomers or closely chemically related side-products, but mostly they are "volatile impurities" (i.e. water). For most purposes these impurities do not interfere, so nobody bothers to remove them. Here is MSDS of Allura Red:
http://www.vidhifoodcolour.com/uploads/gallery/14067976084allura_red.pdf


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Can Henderson-Hasselbalch equation be used to determine the pH of a solution when a strong base and a strong salt of that base stay together?
The following problem I have stated was not given to me as a homework. I have found this one in my textbook. I have solved it in my own way, but I'm not sure about the correctness of my procedure. I have posted to be sure about the procedure that I want to follow to solve such kind of problem.

A $5~\mathrm{ml}$ sample of $0.2~\mathrm{M}\ \ce{CH3COOH}$ is titrated with $0.2~\mathrm{M}\ \ce{NaOH}$. Calculate the $\mathrm{pH}$ after adding $25~\mathrm{ml}$ volume of base has been added. Where $K_\mathrm{a} = 1.8 \times 10^{-5}$

By applying $v_1s_1=v_2s_2$ formula, I got the concentration of weak acid and it is $0.0333~\mathrm{M}$. Again by applying same formula, for the total volume of $30~\mathrm{ml}$, I got the concentration of base and it is $0.1667~\mathrm{M}$.
Since the concentration of base is more than that of acid, $0.0333~\mathrm{M}$ base will neutralize the $0.0333~\mathrm{M}$ weak acid and the concentration of remaining base will be $0.1334\ \mathrm{M}$.
Then I used this concentration to evaluate the $\mathrm{pOH}$ of this solution and it was $0.875$. So $\mathrm{pH}$ will be $13.125$.
Do you think that my procedure to solve this problem is correct? I am confused. Because my textbook says that $\mathrm{pH}$ will be $8.87$. Has the author of this book made an error in this case? I cannot use Henderson-Hasselbalch equation in this case because it is not a buffer solution. If weak acid ($\ce{CH3COOH}$) and a strong salt of that weak acid ($\ce{CH3COONa}$) stay together in a solution, we can say it to be a buffer solution. But in the problem I stated above we can see that no weak acid exists after reaction takes place, and strong base ($\ce{NaOH}$) and strong salt ($\ce{CH3COONa}$) stay together. We also know that strong base and strong salt of that base cannot form buffer solution.
Now I want suggestion from the experts of this forum. Is my procedure correct? If it is not, which procedure should be followed to solve this problem?

Your reasoning looks about right (didn't check the math, though), and the problem itself looks broken. Who in their right mind would add five times the equivalent amount when titrating? And of course such a huge excess of a strong alkali is going to produce a pretty basic solution, nowhere near 8 or 9.

@ Ivan Neretin, I cannot understand your explanation. Please make it easy, because English is my second language. Is the pH of this solution 8 or 9? How it is? Would you like to show it?

No, the pH is neither 8 nor 9. I think your answer is right.

I know you wrote ‘this is not homework’ in the preamble of your question, but I still added the [tag:homework] tag, since that also applies to self-study problems and the like, which is relevant to your question.

What is $v$ and what is $s$ in $v_1s_1 = v_2s_2$?

The pH of the solution after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ is $13.12$. Your book or solution manual is wrong. The $\mathrm{pH} = 8.87$ at the equivalence point.
The Henderson Hasselbalch equation only applies in the buffer region - the part of the titration curve before the equivalence point. I have worked out the pH for a number of volumes of titrant and graphed the curve. The buffer region is labeled. You can see that past the equivalence point, the pH jumps to above 12 and stays there. You determined the pH after 25 mL correctly.

There is a point where the $\mathrm{pH=8.87}$ - the equivalence point (labeled with its pH). After $\mathrm{5\ mL}$ of $\ce{NaoH}$ solution is added, all acetic acid has been converted into the acetate ion. We can solve an equilibrium problem to determine the $\mathrm{pOH}$ and $\mathrm{pH}$ at that point. The following equation applies:
$$\ce{CH3CO2- + H2O <=> CH3CO2H + OH-}\ \ \mathrm{K_b=10^{-9.26}}$$
Since all acetic acid was converted into acetate, and the volume has doubled, the initial concentration of acetate is $\mathrm{0.1\ M}$. The initial volume of acetic acid is $\mathrm{0\ M}$ and the initial concentration of hydroxide is $\mathrm{10^{-7}\ M}\approx 0$.
$$\begin{array}{|c|c|c|c|}\hline
\ & [\ce{CH3CO2-}] & [\ce{CH3CO2H}] & [\ce{OH-}] \\ \hline
I & 0.1 & 0 & 0\\
C & -x & +x & +x \\
E & 0.1 -x & +x & +x\\ \hline
\end{array}$$
If we plug in the values into the law of mass action, we can solve for $[\ce{OH}]$ and determine $\mathrm{pOH}$.
$$\begin{align}
K_b &= \dfrac{[\ce{CH3CO2H}][\ce{OH-}]}{[\ce{CH3CO2-}]}\\
10^{-9.26} &= \dfrac{x^2}{0.1-x}\end{align}$$
Solving for x gives
$$\begin{align}
[\ce{OH-}] &=x=7.41\times 10^{-6}\\
\mathrm{pOH} &=5.13 \\
\mathrm{pH} &=8.87
\end{align}$$
Even if I had not made the assumption that the initial hydroxide concentration was zero, I would get the same $\mathrm{pH}$.

Your book has a typographical error. Either you were meant to determine the pH after $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ or the pH after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$. Both would yield a $\mathrm{pH} =8.87$ at the equivalence point.

The NaOH will dominate the solution. Your math seems...correct...as far as it goes. If I were your instructor, and you turned your work in - as written, above, I'd mark it wrong. But perhaps you've not be taught about significant figures, or perhaps you're just keeping it short. But claim a pH to 5 significant figures when you're given only 1 to start with is what I'd consider an "epic fail". If your textbook claims 3 significant figures, then shame on it. The "correct" answer is 13. When I do similar problems, I find it most useful to convert concentrations to moles. This allows me to use my intuition to double check the reasonableness of the answer. (but in this case, since you didn't make an error, it wouldn't be necessary.) IOW, I'd convert the HOAc and the NaOH to moles, subtract and then convert the remaining NaOH back to Molarity and on to pH. Perhaps not necessary if you're proficient in working with units of concentration. It might be instructive to calculate the pH at the equivalence point...I'm wondering if that might be what the problem set-up meant...pKb being 9.3 for NaOAc...

If you’re arguing sig figs, why not make it $1 \times 10^1$?

@Li Zhi, would you like to show how you would solve this same problem? It would be helpful if you showed me an example. I don't know how I will request you to show me your way which is more correct than the way I followed. By the way you have wtitten something about pKb. Please tell me why this is important when NaOH is a strong base. Is it necessary to use pKb when we are solving the problem stated above?


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detection of psilocybin with ehrlichs-reagenz
I tried to form the product of the detection reaction,
however i dont know which product is correct or both products are wrong!


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Products which adhere to human nail while still being removable/reusable?
I am at loss as to what I can use for my project. I require a product which can adhere to the human nail (not any damaging glues) while still being easily removable. There will be a degree of 'overlay' on the nail so it's going to need to cover the entire nail plus a bit more (maybe seen with fake nails?).
I've looked online and video tutorials on adding fake nails isn't convenient (while it could work). My biggest problem is I don't want to damage the nail so I can't use any popular practices often seen in beauty salons.

Hi and welcome to chemistry.stackexchange.com. Since you already took the [tour], all that remains for me is to point you towards the [help], where anything unclear about the site will be explained.

Try flour and water paste or washable school pva glue. Start with the flour. It would be an interesting experiment to try a 50/50 mix of both.
Another possibility is rubber cement (art gum). This will be quicker drying.

[Not a chemist] I like the glue on quality masking tape, also known as painters' tape. It seems to perform as you require, except it is adhering to wood or paint, not nail. In my experience it hardens over time and becomes difficult to remove, but for a few months in normal conditions it comes off with no trace.
I'd look at glues from double sided adhesive tapes too.
There's a kind of transparent blob glue they use in Europe to stick paper or card "extras" onto magazines. I have reused this before, but you will need to reduce the thickness to stick something to a nail elegantly.


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What makes alcohol a good conservation product?
A common way of preserving food and organisms is immersion in alcohol. But what causes alcohol to preserve these items well?

As simple as it gets: bacterial life can't stand it, but we can.

But can't bateria not just leave the alochol or do they have to interact with it?

@IvanNeretin I'd wager that we're actually much less resistant to alcohol. And it's a matter of scale.

Of course it's a matter of scale; I never said it isn't. Throw a man in a tank with alcohol, and we'll see who's more resistant.

Almost anything soluble in water will stop any kind of life if it is more than a few percent. Formaldehyde, acetone, acetaldehyde, NaCl ...

Alcohol kills bacteria and fungus, put simply. It does this as its metabolism products cause damage directly to the DNA, and directly to the cell respiration process. It also happens to tend to dry it out.
We can drink alcohol safely because of our alcohol dehydrogenase enzyme, which breaks the toxic alcohol into useful byproducts, that we can use for energy.
Also, it's a matter of scale. Consider submerging yourself in alcohol for a period of time. Don't you think you'd probably die, too?
It's very important to note that alcohol strength matters in this case. It also depends strongly on type. For example, ethyl alcohol: we drink this kind. It's not as toxic as say, methyl alcohol, or isopropyl alcohol. Those last ones are used as preserving and disinfecting agents, respectively.

It would be safe to assume that bacteria have pretty much all enzymes that we have, and many more than that. That's not what enables us to drink alcohol the way they can't. It is just that we are big, multicellular, and covered with skin.

True. If we were submerged in alcohol, it'd wager it'd be very likely we'd die, too.

So if we were submerged in alcohol we would die but are preserved very well?

Essentially, yes.

The amount of blood alcohol that kills a human will not harm many single-cell organisms, at least not permanently.

@Karl Quite correct. I don't recall mentioning that in my post.

I was trying to politely point out that the four paragraphs in your answer are completely disconnected, and each is already worded very sloppily on its own. Sorry.

All life on earth is adapted to work in water, and only in water. The most basic level of life is enzymes producing proteins, and energy, to produce more enzymes. Cells, DNA, etc. are the directing structure for that.
In the wrong medium, e.g. 5% ethanol, the reactivity of the enzymes is totally off, not speaking of actual toxic effects of ethanol, acetaldehyde etc. One after the other of the thousands of chemical reactions, the equillibrium of which is life, take wrong turns and go into stalemate. And that is the end of it. The enzymes are no longer regenerated, what remains are all the membranes in the cells, which you can look at in your microscope.

This answers the wrong question. It explains why things die in alcohol, but not why alcohol preserves anything, or why organic material is not dissolved to mush.

All life stops working in 20% ethanol. Bacteria, fungus, everything is back to very slow, non-biological, chemical decay. I call that preservation. ymmv

Sheeze dude. High alcohol tolerant yeast can withstand 25% ethanol. I know plenty of people who drink beverages well over 20% alcohol. Plus several forms of chemical decay result in complete obliteration which is anything but preservation.

Gotta love peer review... Its makes great answers even better.

Please enlighten me: What kind of fast chemical decay could happen to organic matter around room temperature? Lets say without adding larger amounts of strong oxidants, bases, acids, OK? And I said IN, not IN SHORT CONTACT WITH with 20% alcohol. 25%, if you wish.

Pickled food 'in' alcohol can be eaten and digested.  Usually people take food out of the alcohol to eat it.  Your counter-argument is circular.


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How to determine whether a given compound is paramagnetic or diamagnetic?
Well i am very confused whether we can determine the magnetic property of compounds..
If it us asked for elements we can say by looking at electronic configuration...
For ex-
how is N2O diamagnetic?
How is NO paramagnetic?
How is NO2 paramagnetic?
What about N2O5

It's just the same: you look at the molecular orbitals and the electronic configuration. Before that, do a quick check: if the thing has an odd number of electrons, it is surely paramagnetic. If the number is even, it is likely diamagnetic, though exceptions do exist. Also, welcome to Chem.SE.

Paramagnetism is a consequence of having one or more un-paired electrons in the outer electronic configuration. e.g. O2, and NO. Most phys. chem. or inorganic chemistry textbooks will explain this in detail.

Paramagnetism is a consequence of having one or more un-paired electrons in the outer electronic configuration. e.g. O$_2$, and NO. An an odd number of electrons can clearly produce a paramagnetic ion, radical or molecule. However, molecules with an even number of electrons can also be paramagnetic, the most famous being dioxygen (O$_2$). The experimental observation is that oxygen is attracted to a magnet.(This is best observed with liquid oxygen.) Nitrogen (N$_2$) is not paramagnetic but diamagnetic.  Whether a molecule is paramagnetic or not is only explained using molecular orbital theory. In the case of O$_2$ it is found that the lowest energy state is one that has one electron in each of two orbitals. An alternative would be to expend energy to pair up these two electrons into one orbital. As this requires energy it does not occur. High and low spin transition metal complexes show the same effect of pairing vs. placing electrons into different orbitals.

‘As this requires energy it does not occur.’ Hm. Singlet oxygen does exist and can be liberated given the correct reaction conditions. Photons are emitted when singlet oxygen transforms into triplet oxygen.

Paramagnetism is a result of unpaired electrons. If we count the valence electrons available to the four species you name, we can take a pretty quick guess as to what might be paramagnetic. If the species has an odd number of electrons, it must be paramagnetic. If the species has an even number of electrons, it is likely diamagetic.
For example, $\ce{NO}$ has 11 valence electrons. One of them must be unpaired.
Can you work out the others?


text:
What is the meaning of 'anion water' in coordination chemistry?
In Mechanisms of Inorganic Reactions by Fred Basolo and Ralph G. Pearson, a passage discusses the history of coordination chemistry. Ammoniates had been assumed to be analogous to hydrates. They assess the comparision as,

This analogy is essentially correct for in hydrates, except for interstitial and anion water, the water is coordinated to the central metal ion. [emphasis added]

What is the meaning of anion water, as used in this context?

See this publication for some relevant information. It's likely the authors are referring to anion-water coordination in clusters, but that's just a guess based on the context.

I’m pretty sure that ‘anion water’ is a term they use to describe water which coordinates the anions by hydrogen bonding rather than having anything to do with the cations. This is again to be separated from interstitial water, whose job it is to stabilise either of the other water types by hydrogen bonding without interacting with cations or anions at all. (Or maybe, whose job it is to fill voids in the crystal structure).
Ammonia in coordination compounds typically does what water does if it is not interstitial or hydrogen-bonding to anions, thus the distinction and the mention.

Are 'other water types' referring to water clusters?

@LinearChristmas ‘Either of the other water types’ refers to anionic or coordinating water.


text:
Orbital definition
In some sources, including my textbook it is cited that an atomic orbital is the 3 dimensional region where the probability of finding an electron is 90-95% . But, Other sources such as Wikipedia state otherwise. To quote Wikipedia,
"In quantum mechanics, an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom"
With relevance to the fact that an orbital is always a wave function I agree that it can be regarded as a mathematical function, But what is this reference to "a pair of electrons"This has confused me.What is the the true definition of an orbital?
I am new to this forum and Chemistry, please excuse me if there's some misconception. Thanks

Graphical representations are usually drawn for the 95% probability integral, but the actual orbital is the function describing the distribution of said probabillity (not the integral!). Wikipedia is right, your textbook sloppily written.

As for the pair thing, read about spin, or rather don't, just take it for granted that two electrons may occupy the same orbital. Also, welcome to Chem.SE.

The mathematical definition is the most rigorous. The presence of the positively charged nucleus means that there is a pull on a negatively charged electron, which we describe with the potential energy. But electrons can only exist with certain discrete energy values. To do a full analysis you need to find the wavefunction that describes the electron. For a given electron, the wavefunction* at point (x, y, z) gives you the probability of finding the electron at the location; this is non-zero pretty much everywhere in space but is only significantly above zero in a very small region: this is what the first definition is getting at. Once you have the wavefunction you can calculate some properties, like the angular momentum. Since each electron has a discrete energy, each electron can be described by a set of four numbers, derived from the wavefunction: principle number (n), orbital angular momentum (l), magnetic angular momentum ($m_l$) and spin number ($m_s$). The spin number can be either +1/2 or -1/2 but we usually ignore it since normally that value doesn't affect the energy of the electron; this is the source of that "pair of electrons" business.
To summarize: An orbital is a specific state of an electron, characterized by its energy and various angular momenta and rigorously described by a mathematical function that satisfies Schroedinger's equation. This can be represented visually in many ways, such as showing the 95% probability solid, or as a line on a molecular-orbital diagram.
*Really this should be the wavefunction times its complex conjugate.


text:
On what basis are certain elements selected for doping silicon?
I read online that transistors are made by doping silicon with other elements such as arsenic, phosphorus, or aluminium in order to increase its conductivity. How are these doping materials selected?

This has an answer over on Physics (note it is my answer, so reader beware): http://physics.stackexchange.com/questions/132604/why-some-elements-of-same-group-cannot-be-used-for-doping-semiconductors/132625#132625

@JonCuster can we increase the conductivity of a conductor(like aluminium) by doping?

Typically 'doping' a metal results in increased scattering and a decrease in conductivity. Not an absolute rule, but pretty good bet for most metals.

@JonCuster what do you mean by 'increased scattering'??

The impurities are different than the lattice, interrupting the normal electronic structure, causing more carrier scattering.


text:
Molecular Structure of Caffeine

I'm currently looking at a problem that asks about the structural relationship between the 4 nitrogens the caffeine molecule. I am asked to identify whether they are in perpendicular planar, coplanar, or inverted planar formations.
I'm am a bit lost as to how I should approach this problem. Should I look at hybridization (3 Ns are sp3, 1 N is sp2)? Or resonance (seems that all Ns are involved in resonance)? Or should I be concerned with sterics?
Lastly, what are some examples that might assume a perpendicular planar, coplanar, and inverted planar formation? I don't think I've ever seen such shapes before...
Would someone please point me to the right direction?

Check http://chemistry.stackexchange.com/questions/4390/how-to-rationalise-the-resonance-structures-and-hybridisation-of-the-nitrogen-in and http://chemistry.stackexchange.com/questions/12339/hybridization-in-light-of-resonance

At some point in studying organic chemistry, you might have been taught that a nitrogen in an amide group is planar sp2, due to delocalization of its electrons into the carbonyl group. You also might have learned that nitrogen atoms in a pyrrole or imidazole ring are planar sp2 since they participate in the 6 electron aromatic pi bonding system. You would then conclude that all the nitrogen atoms in caffeine are coplanar. But without that essential background information, you might not have been expected to figure that out.

Great answer.  Very readable.  Thank you.


text:
Can I split water without constant refill of solution?
I've looked all over the place to find a way for me to separate water without needing to put in salt. What I've found is you can separate water into hydrogen and oxygen through electrolysis (which needs salt to make the water conductive), radiation (which I don't want to use - this is a house project that I want to get into), thermal (the water needs to be hotter than what I can provide), and photobiological (looks cool - but it looks like it won't produce much).
There's a lot of things out there for electrolysis water splitting, but every single one looks like it needs salt. Can I initially provide salt to a water splitting machine, then never refill it and still split water? Is there anything else, in which I could put in the water to make it split, and not have to ever refill said thing?
Really, I want to build the machine, and not have to keep providing resources that I'm going to have to get - just provide the machine water and electricity.

Salt is not needed. Salt merely allows the water to be somewhat more conductive.
The King of Random on YouTube has a great video on making one. All you need is electricity and water.
https://www.youtube.com/watch?v=cqjn3mup1So
He uses sodium hydroxide, but this not needed. It will just be less efficient.
If you want to be more efficient, you can add an electrolyte like salt. Whatever electrolyte you use, should not be used up. However, if the concentration of electrolyte gets to high (after you have used up a lot of water) it will split it too. This is bad with salt, as you can end up with chlorine gas and sodium hydroxide, which can be nasty to deal with.
So, as long as you replace the water fast enough and don't put in too much electrolyte, it should be fine and you will not need to replace it often.

Pure water is poorly dissociated - it means that it does not conduct electricity very well. Therefore, if you want to split pure water electrolytically, you would need to use a high current - not very safe in home settings. Table salt (or practically any other water-soluble salt) acts as an electrolyte - it dissociates in water and greatly increases its conductivity, making the electrolysis possible with low currents. Be careful though, if you use the table salt and apply a voltage of 1.5 V or more, you will not only split water, but also produce chlorine gas (toxic, corrosive and explodes on contact with hydrogen). It is safer to use sulfates, such as sodium sulfate, or even diluted sulfuric acid - your only products will be hydrogen and oxygen, and you won't need to add new portions of salt or acid every time. Sodium/potassium hydroxide will also work, but its solution absorbs carbon dioxide from air with time, and forms a carbonate. Sodium carbonate is also a good electrolyte but it will release carbon dioxide during the electrolysis, so you will get a mixture of oxygen and carbon dioxide on the anode if you use an old solution.


text:
Base Peak Intensity different than higher value in Mass Intensity Map
I've been working with mass spectrum data analysis (but my field is computer science).
I'm using a library that allows me to get the entire Mass Intensity of the MS1 scan (such as [time, intensity]). However, by searching for the highest value for intensity, it is always a few integers lower than the base peak intensity.
I'm retrieving the information from a mzXML file, and the library that I'm using is ms-data-core-api. The method used to retrieve the intensity map, in case anyone has ever used this library, is this one: public double[][] getMassIntensityMap(). I retrieve the highest peek in the map by iterating the matrix returned and updating a variable with the highest value found in the y axis (given double [x = time][y = mass].
Examples of the difference of data include the following:
scan 1 (MS1): Highest peak in map: 789991.875 / base peak intensity: 793728.0
scan 2 (MS1): Highest peak in map: 714475.5  / base peak intensity: 719125.0
scan 3 (MS1): Highest peak in map: 673152.6875  / base peak intensity: 680567.0

From what I can tell, the instrument used for these scans is a fourier transform ion cyclotron resonance mass spectrometer, using Xcalibur as the  software.
Does this makes sense, and why?

@CurtF. I'm sorry, I should have added this information in the first place. I'll update the question.

One possibility for the discrepancy is that different peak centroiding / integration algorithms were used at different stages in the data.
For example, the base peak intensity recorded in the plain-text (uncompressed) scan data is probably calculated by the instrument in real-time during acquisition.  The intensities of all detected peaks are also recorded by the instrument, of course, but those intensities may be affected by processes such as centroiding that are not applied to the plain-text recorded base peak intensity.
An exact answer to your question will depend on many factors, such as whether spectra were recorded in profile mode or centroid mode, whether spectra were compressed during conversion to mzXML format, and also details of the (possible) centroiding algorithms and BPC algorithms used by the instrument firmware during data acquistion.  (Good luck trying to find the exact code for vendor firmware...it's probably proprietary.)

Thank you for your answer. I know that this was a difficult question to answer, since it is very specific and a difficult one to reproduce, but I really appreciate your help. I believe that the spectra were recorded in centroid mode, but I have no access to the other information. You were really helpful, though; I thought I was missing something obvious, but that might not be exactly the case, then.

Just as a final thought though, but that's more of a soft doubt since it may not have a definitive answer: if this was caused by the algorithms, would you consider the base peak intensity to be of higher accuracy, or the mass intensity matrix?

Accuracy for what purpose?  The differences are less than 0.5%.  Does the rest of your analysis really depend so heavily on which number you use that a 0.5% difference matters?  To actually answer your question, I would use the "scan-internal".  It is the same for all peaks, and thus it makes sense to compare intensitities between base peaks and other peaks.


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Coconut oil vs Olive oil: Spray dynamics
Ps: If this would better fit Physics then please move it there or let me know. I put it here based on fact that viscosity and colloids were part of chemistry curriculum 10/15 yrs back in school.
Let's say we had to spray Edible oil.
How would spraying each differ?
How different are their Physical Properies when it comes to spraying and how would they affect the spray and compare with each other?
(You may point out Properies that matter in spraying that I may not be aware of, but these are few I can remember from school/ college such as viscosity, stickiness, cohesion, adhesion, density)
Also, I understand temperature would/ could play a role as well, so let's specify and put them at their ideal temperature for spraying.
Let's optimize this spraying experiment for both of them & compare.
I'd love to hear the science in play here.
Ps: I'm curious if any of the follow terms apply to these Virgin oils: homogenous, colloidal, plasma?

Your question is rather broad and unclear. It would help if you made it more specific. (For example, how does the coconut/olive oil from your title make it into the question?)

@R.M. - Its late night and I've got to crash, but it comes because water is easy to spray, but these Oils have higher viscosity & stickyness that will cause  differences in spraying them

The fluid dynamics of spray calculations depend on many conditions including the nozzle used to create the spray.  The wikipedia article on spray nozzles has a good discussion on how different liquid properties tend to affect spray. Spray information tends to be empirical, so it is not really science at play, it is engineering.
The Institute of Liquid Atomization and Spray Systems is a professional society set up around the science and engineering of spraying.  Atomization and Sprays is a periodical journal set up around this entire topic.  A internet search of 'textbook on spray and atomization' turns up many textbooks on the topic.
Plenty of manuscripts written by scientists exist providing detailed results of experiments comparing sprays.  An example would be an experiment done comparing the spray of waste cooking oils in an engine when it is used as fuel.
Summarizing a discussion of all of these experiments is too broad and out of the scope of SE.  If you are interesting in learning about sprays make sure you have some understanding of fluid dynamics.  Conduct a literature review relevant to the spray experiment you have in mind.


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Dissociation of HCl in aqueous solution
My question is regarding $\ce{HCl}$. In my book it states that:
$$\ce{HCl (aq) -> H+ (aq) + Cl- (aq)}$$
I understand that when added to water the $\ce{H}$ leaves its electron to the $\ce{Cl}$ atom forming the $\ce{Cl-}$ and the $\ce{H}$ attaches to water to form the hydronium ion ($\ce{H3O+}$). Alltogether:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
Why does this dissociation actually occur scientifically? What causes the $\ce{H}$ to leave its electron behind to the $\ce{Cl}$ and form the hydronium ion? Why does water ($\ce{H2O}$) take the additional $\ce{H}$ ion? What is the general concept here in dissociation? Electronegativity?

Water is a polar species in which the oxygen atom in $\ce{H2O}$ has a partial negative charge and the two hydrogen atoms have partial positive charges. As you suggested, this is because oxygen is highly electronegative (i.e. it has a high affinity for electrons) and so the negative electrons tend to have a higher density around the oxygen atom than the hydrogen atoms, creating this separation of charges due to water's molecular structure. Take a look at the molecular structure of water, which can be shown below. It has an asymmetry bent geometry and therefore the charge distribution in water is also not symmetric.
The hydrogen in $\ce{HCl}$ does not actually give its electron to the chlorine atom. Rather, the more electronegative chlorine ion is attracted to the partially positive hydrogen atoms in water, and the hydrogen ion in $\ce{HCl}$ is attracted to the partially negative oxygen atom in water. This separates the $\ce{H+}$ from the $\ce{Cl-}$ in $\ce{HCl}$ so that the acid dissociates into its respective ions. You therefore get the equation $\ce{HCl -> H+ + Cl-}$ when in water. If you include water on the reactant side, it can be equivalently written as $\ce{HCl + H2O -> H3O+ + Cl-}$. Here, $\ce{H3O+}$ is simply indicating that the $\ce{H+}$ ion is in an aqueous solution.

Thanks so much for your answer. Yes, it was late I corrected the question format. What I still dont get is how the Hydrogen in HCl becomes an ion and vice versa for the Cl atom. To me NaCl which is ionic is an easy concept since the sodium already transferred an electron to the Cl (when they break apart they become Na+ and Cl-), that I get. But in the HCl how exactly does the hydrogen become and ion and how does the Cl become an ion? The fact that the Cl is attracted to the partial delta chrage hydrogen still doesnt explain how it became an ion (Cl-) to begin with considering it is covalent.

You are very welcome!
As for your follow-up, HCl is a strong acid. This means that in solution, it donates a proton (the hydrogen ion) to water to form the hydronium ion. The electron that you think of as originally being on the hydrogen in HCl is left behind on the Cl atom, making the hydrogen atom become the H+ ion and the chlorine atom become the Cl- ion. Look at the image on this Wikipedia article about acid dissocation for a helpful graphic using acetic acid as an example.

I have deleted your first para because the errors  you mentioned in your answer has been rectified by Jan.

@AndrewR. Thanks for your additional feedback, now that make sense. Just a follow up, how exactly is water asymmetrical ? This may seem ignorant due to my lack of knowledge but if one were to draw a line through a water molecule it appears perfectly symmetrical. Is it  asymmetrical due to the distribution of electrons and if so, how?

@AndrewR. Any ideas?

@Argon whats the reason for it to dissociate completely ? I mean why not setup equilibrium ?

The water molecule has a 3D structure based on a $\ce{AX2E2}$, which transforms into $\ce{AX3E1}$ for the oxonium ion (refer to Gillespie's VSEPR theory). Water has a two symmetry planes, one containing both hydrogen atoms and the tilted plane where those two atoms face each other, with in the intersection the axis permitting a Debye dipole moment of $P = \pu{6.2e-30 C m}$ or circa $\pu{1.85 D}.$
The compounds with partially ionic bounds, such as $\ce{H-Cl}$, for which the $\ce{Cl}$ atom is much more electronegative than the hydrogen atom, can be considered as a three step process:

Ionisation step, formation of a $\ce{H+}$ (proton) and a $\ce{Cl-}$.
Dissociation step, separation of the proton from the $\ce{Cl-}$
anions, note that chlorine ions keep the electronic charge because they are halogens.
Finally, a solvation step, with the introduction of Coulomb interaction between the ions and the water molecules.

I enclose a simple model to explain this with hands...


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How is combustion speed regulated in solid propellants?
My question on Space.SE didn't yield anything I didn't know already: the secondary regulation mechanism – grain (cross-section shape) of the propellant, changing the area of open surface of the propellant as it burns away; the combustion occurs only on the surface, travels inwards only as much as the surface burns away, and produces uniform amount of exhaust gasses per unit of area of open surface; the variance in surface area is what changes the overall thrust.
But I'm currently interested in what is done to composition and/or inner structure of the propellant, so that it burns uniformly on the surface, at predictable (and quite slow) pace, instead of, say, exploding all at once, or at least deflagrating at shocking speed, like common explosives (of very similar composition!) do. What property or reaction keeps the combustion to the surface and makes it progress inwards only so slowly – in case of the Space Shuttle, 290 seconds to burn through about 1.5 m of the propellant.

Nothing more to it! If mixed more finely, the stuff burns away ever faster. When it crosses the speed of sound, you have a bomb. Limiting process is just the mixing of oxidant and fuel at the front by diffusion/convection/turbulence.

@Karl: So you mean it's the same stuff as common explosives, except coarse? Big, thick crystals or such?

https://en.wikipedia.org/wiki/Rocket_propellant https://en.wikipedia.org/wiki/Explosive_material

@Karl: That would work for composite propellants, but not for single-, double- and triple-base ones; moreover RDX, gunpowder and nitrocellulose appear in both articles (none ever mentioning the alternate use), and articles on these materials are woefully lacking any info on their use as solid fuel; I still have no clue what property puts RDX in the combustion chamber apart from RDX in the warhead.

For an experimental method used to determine the linear burning rate of propellants see for example Method 803.1.1 in MIL-STD-286C.

@Loong: That's a good half of the answer: "We've modified our propellant, now we determine how the modifications affect it." Now for the other half... what sorts of modifications are done?

(I apologize for being too lazy to find all of the citations for my answer.  However, I have a fair amount of experience as an organic chemist, at one point worked in an underground mine with explosives and have built (for entertainment) a solid fuel composite propellant)

Why does propellant "burn" instead of detonating like similar explosives?

A typical solid composite propellant looks like this:  (1) mainly ammonium nitrate (oxidizer) + "fuel" (typically aluminum flake or magnesium) + binder (polymer + perhaps plasticizer) + (maybe) catalysts OR substitute ammonium perchlorate for ammonium nitrate.
The most similar mining explosive is ANFO or variants thereof:  ammonium nitrate + diesel fuel (possible sensitizing substitutions are... some ammonium perchlorate for the ammonium nitrate, some aluminum in addition to fuel, partially oxidized "fuels" such as hexamine nitrate, nitromethane, possibly a small percentage of a secondary explosive such as TNT or RDX, or addition of plastic "bubbles" to simulate the bubbles that sensitize pure nitroglycerine.
The key point is that the latter mining explosive is called a blasting agent.  The taxonomy of high explosives looks like this:  primary explosives can be easily DETONATED (as opposed to deflagration, which is a type of "burning", detonation involves a supersonic thin region of reaction that moves through the explosive, giving it brisance that can shatter rock).  Primary explosives are used in primers, and can be detonated with sparks, physical shocks, etc.  Secondary explosives (dynamites, TNT, PETN) need to be detonate with a primary explosive.  They are relatively insensitive to other means of initiation (one measure of this is the deflagration to detonation distance, which really is the amount of explosive that can burn before detonation takes place.  It is usually quite large).  You can often shoot or burn secondary explosives without them detonating.
Finally, blasting agents are quite hard to detonate.  They require a booster that is a secondary explosive.  Blasting agents are very insensitive and can be burned without detonating, especially if not completely confined.
This answers your first question...  we should not expect solid propellants to detonate without a booster (actually I'd guess that solid propellants are even harder to initiate, as ANFO variants are usually oxygen balanced so that all of the fuel is oxidized, whereas propellants appear to be a bit carbon rich).
Note that blasting agents and solid propellants share the property that the oxidizer and fuel form distinct particles, as opposed to explosives such as nitroglycerine, which are self-oxidizing within the molecule.  The low explosive gunpowder shares the property of having a mixture of discrete particles of oxidizer and fuel, and it merely deflagrates.
TLDR:  The explosives that are most similar to solid composite propellants are difficult to detonate, so it isn't a surprise that they burn instead.
==============

How is the burn rate controlled?  (note, I use burn as burn typically is decomposition at a rate of a few cm per second, while deflagration is at a rate of many meters per second.  Detonation is thousands of meters per second).

A:
All propellant mixtures have (approximately) a burn rate that is a function of pressure.  This approximation is called Vieille's Law.  The formula is
$rate = {rate}_0 + aP^\beta$
where $P$ is the pressure.  Thus, as pressure increases in the rocket motor, burn rate increases.  Ideal values of $\beta$ are between 0.5 and 0.8, otherwise initiation can be difficult or else a slight overpressure could be amplified to cause the pressure to increase dramatically and destroy the rocket.  What seems to affect the burn rate?  I'll list a few:
(1)  Ammonium perchlorate burns much faster than ammonium nitrate.
(2)  Generally, the rate limiting property of a propellant is the particle size of the oxidizer, so more finely divided oxidizer means faster burn rate.
(3)  The geometry of the propellant is important.  The burning tends to occur at the exposed surface, so wide and not very long propellant castings will produce more thrust.
(4)  Metallic catalysts increase burn rate, such as iron oxides and chromates.
(5)  I've heard that a small amount of carbon black increases burn rate by increasing infrared absorption in the casting.  It also makes it look cooler.
(6)  Not so good:  cracks and voids in the casting will increase burn speed, as they lead to more exposed area.  That's why typical binders are rubbers, often including plasticizers, as these are less likely to crack under pressure.
Here's a couple of citations:
A amateur rocketry site
metallic wires as a burn rate catalyst
Rocket engine mathematics

A great answer! Four new different factors (nitrate:perchlorate, carbon, metal catalyst, binder), on top of the three everyone else had mentioned (some even implied they are everything there is to the burn rate: pressure, surface geometry, grain size)

OK, rockets... perchlorate... would you like to have a look at Why doesn't the perchlorate on Mars' surface oxidize metallic meteorites? :-)

Ask your question the other way. What is required for a material to burn to detonation. Most explosives will burn as propellants, it is only when you confine them and the combustion pressure increases that they risk burning to detonation. Very few materials can burn to detonation without confinement; these are called primary high explosives (eg Lead Azide).

For most propellants and explosives the speed of reaction is a function of the compound and its physical properties
You make an incorrect assumption when you assume that the primary reaction in propellants and explosives is burning. It isn't. Explosives and propellants are usually designed so they react without needing an external oxidiser. Explosives, some of which can also be used as propellants, are usually built from oxygen and nitrogen-rich chemicals that fall apart into gaseous products when given the right impetus. No external oxygen is required. Some are mixtures containing a fuel and an oxidant where the oxidant contains all the required oxygen to create a gaseous product and/or a great deal of heat when reacting with the fuel.
Whether a propellant reacts controllably or not is, to a large extent, a property of the material. The difference between detonation (very fast reaction) and deflagration (slower) is whether the reaction travels faster than the speed of sound in the bulk material (see this answer on explosives). Not all explosives detonate. For example, the explosives used in guns and artillery to drive the shells need to deflagrate to create a relatively slow reaction driving a controlled propulsion of the shell from the gun barrel. Rockets are just a carefully controlled version of this principle.
So, the dominant requirement for a solid rocket fuel is a substance or mixture where the reaction proceeds at a relatively slow pace in the bulk compound (and there may not be much you can do about this except to select the right chemical or chemical mixture). Having said that, you can influence the pace to some extent by changing the physical makeup of the compound, especially where the propellant is a mixture not a pure substance.
The boosters on the space shuttle used a mixture of ammonium perchlorate and aluminium. To some extent the propellant properties can be modified by both the physical makeup of the propellant (e.g. the grain size) but also by the addition of additives to modify the bulk properties of the mixture (both the binders used to hold the mixture together and some catalysts can modify the reaction speed). Some control can also be exercised by the overall shape of the propellant mass. Some military rockets use the explosive HMX as a fuel (it has a higher specific thrust than the fuel used in the space shuttle's boosters but is also more dangerous).
But the dominant characteristic is the bulk physical properties of the substance (which determine how fast the reaction proceeds in the bulk compound). So your primary choice as an engineer is which propellant to pick. You can modify that a bit but you can't fundamentally change the basic chemistry and physics.

Unfortunately, I know all of that. Select the right chemical or chemical mixture and Changing the physical makeup of the compound are precisely the parts I wanted more detail on.

@SF. apologies If I told you stuff you already knew. But I assumed you didn't get the key distinction between detonation and deflagration and the fact that is is, essentially, a property of the substance.

I did my homework, reading up everything Wikipedia had to say on the subject of both solid fuels and explosives; the cross-links between the two articles are near to non-existent, even though the same substances are mentioned. Pages for the substances that appear on both don't give any detail on solid fuel applications either.

...also, this is the question people on Space Exploration SE site were unable to answer. This is related to "Electric solid fuels" which are not only plastisols, and conductive, but "tuned" to the brink of combustibility - self-extinguishing, but only barely so; so that passing high-voltage current the extra energy makes them sustain combustion; remove electric power, they stop. I don't believe the inventors just happened upon a substance that has these properties by blind trial and error.


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Why is hydrochloric acid often sold as 33% concentration?
What is the historical perspective for the choice of 33% concentration that is often found in commercial sales of Hydrochloric acid?
Is this the same reason we find sulfuric acid sold at 37% and Nitric at 68%?
I am guessing these are some limit at which further concentration, synthesis or purification reaches a chemical or economic limit but am curious to find out about the specific numbers selected.  Obviously, all of these are also sold in higher purities/concentrations as analytical reagents but that obviously takes more effort to produce and will cost more.

https://en.wikipedia.org/wiki/Hydrochloric_acid#Production

The user ssavec prompted me to research further on my own and post an answer.
It seems the limits are practical ones.

For hydrochloric acid — $\ce{HCl}$ a practical limit is 38% with absolute
limit around $40~\%$ and commercial concentrations ranging from 30 to 35%
for convenience in transportation infrastructure.
For nitric acid — $\ce{HNO3}$ the $68~\%$ limit is the maximum that can be
reached by direct distillation, higher concentration up to $98~\%$
(fuming nitric acid) needs alternative dehydration or synthesis techniques.
For sulfuric acid — $\ce{H2SO4}$, which can be distilled to $98.3~\%$ (and can be used
to concentrate nitric acid) it is sold at concentrations that are
directly suited for the intended use ranging from $20~\%$ to $80~\%$. With concentrations around $30~\%$ common for lead-acid battery electrolyte preparation.

So it seems the limits are all practical but for different reasons, one was close to the absolute limit for efficient transport.  The second was at the direct distillation limits and the last one was to suit the most popular consumer applications.
The spread in the numbers had caused me to wonder on their relation but it seems there was no direct relation to the concentrations we see for consumer sale.

Simply put, HCl is only soluble to 40% in water. HNO3 is only stable up to 68%, higher concentrations (slowly, depending on the excess) loose nitrous oxides until they're down to 68%. This also makes transportation and storage somewhat inconvenient.

The accepted answer doesn't actually give a reason why HCl is not sold above 40% and the comments on it are not quite right.  Concentrations above 40% w/w are possible, but the rate of evaporation off such a solution at room temperature / atmospheric pressure is so high that the concentration will quickly drop to below 40%.  Where a concentration higher than 40% is required, special measures (typically a pressurised container) must be used to maintain it.


text:
Can a Boc protecting group be removed with trifluoroacetic acid?
I am trying to cleave a tert-butyloxycarbonyl (Boc) protecting group from an aminooxy group ($\ce{-O-NH-Boc}$) with trifluoroacetic acid (TFA) in $\ce{CH2Cl2}$ but i get minimum product and too much starting material.
I need the product to be a TFA salt and not an HCl salt so I need to use TFA. Any solutions?

What about your reaction conditions?

If you do a literature search (SciFinder/Reaxys), it seems that HCl in polar aprotic solvents (dioxane, THF, ether) are the 'standard' conditions for this cleavage.
You mention that you require the TFA salt, not the HCl salt, but even if you use HCl to carry out the reaction, its fairly trivial to get rid of the HCl and then convert the free amine into whatever salt you desire.
(Though as vapid said, your reaction conditions are probably quite important, and you might find if you just push the TFA reaction a little harder you can force the Boc off).


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Are phospholipids capable of forming automatically circles or vesicles?
Sometimes scientist think that life began with membrames made of a kind of phospholipids. There wasn't any DNA needed for because those molecules would automatically form vesicles like cells. But is this really possible and how do those molecules form circles and vesicles?

Phospholipids structure can be described as a charged phosphate head group, with long hydrophobic chains coming off it. When you put these molecules in water, like oil drops, the oils spontaneously conglomerate into sheets on the surface of the water.
How, depending on concentration and size of the phospholipids you can get from sheet like membranes to spherical droplets. In either case the spontaneous formation of sheets or micelles (spheres) results from the entropically driven hydrophobic effect.
The simple way to understand this hydrophobic effect and entropy, is that when two hydrophobic tails between two phospholipids are in close proximity, the hydrophobic tails are able to interact via van der Waals forces. But even more important is that water molecules that surround each hydrophobic region of the molecule, are excluded between them. This means that the entropy of water increases a lot more compared to the loss of entropy from having the phospholipids come together and stay together.
This is a pretty fundamental concept to how proteins fold. Look up hydrophobic effect on google for nice, diagramatic explanation.
Or check intro biochemistry textbook like Nelson and Cox (2012) principles of biochemistry textbook.

Is it also possible that they form a circel shape as their head (of phosphor) is a bit bigger/wider than the tail so you automatically get a circle?

@Marijn you are correct, molecular geometry does effect the final shape, along with other factors.


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Why does the local density approximation (LDA) overestimate atomization energy?
I have read numerous times that the local density approximation (LDA) overestimates atomization energy. For example, here is a quote from Dr. Burke's book:

LDA typically overbinds molecules by about $30~\mathrm{kcal\, mol^{-1}}$.

What is the cause? Is it the same problem that Hartree-Fock theory has when compared to the exact Schrödinger solution (lack of correlation between electrons)? Also, how can inclusion of gradient can help solve this problem?

@ToddMinehardt When you introduce MathJax, please use it for the whole construct and not just sub/superscripts. It might lead to a very undesirable typeset on other computers, esp. mobile devices.

@Martin-マーチン - Will do!

What is the cause?

The LDA exchange hole is spherically symmetric and centered on a reference electron, which is a fairly good approximation for the case of molecules, but not for the case of atoms. For instance, in an atom the exact exchange hole is displaced toward the nucleus, while the LDA exchange hole is always centered on its reference electron. This inability of LDA exchange hole to appropriately describe the situation in atoms leads to significant deviations in energies upon bond formation or breaking.

Also, how can inclusion of gradient can help solve this problem?

In general, the electron density in atoms is more inhomogeneous than in molecules, and GGA is ntroduced exactly in order to account for the non-homogeneity. In particular, the gradients in the separated atoms are usually bigger than the gradients in the corresponding molecule, and therefore gradient corrections lower the energy of the atoms more than the energy of the molecule, which leads to a reduction of the overbinding.
See, e.g. Ernzerhof, M., Perdew, J. P. and Burke, K. (1997), Coupling-constant dependence of atomization energies. Int. J. Quantum Chem., 64: 285–295. DOI: 10.1002/(SICI)1097-461X(1997)64:3<285::AID-QUA2>3.0.CO;2-S


text:
How to make γ-MnO2 in a lab?
I am working on Li-ion batteries, and I need to use γ-$\ce{MnO2}$ (electrolytic manganese dioxide). I have googled and read quite a few papers, but I am not very sure of those methods. I'd love to know if you are aware of some! The ones which I encountered, again and again, was the one involving precipitation of $\ce{MnO2}$. But is it possible to precipitate the γ form? Also, I think the commercially available $\ce{MnO2}$ is the α form. Is there any way to convert it to γ form? I am particularly looking for a way to synthesize  $\ce{MnO2}$ using electrolysis.
I know that I can also break a few batteries to get my $\ce{MnO2}$, but it would be heavily contaminated by carbon. Commercially available  $\ce{MnO2}$ is very expensive.

A quick search yields this.

[Electrolytic manganese dioxide] is produced in a similar fashion as
  electrolytic tough pitch (ETP) copper: The manganese dioxide is
  dissolved in sulfuric acid (sometimes mixed with manganese sulfate)
  and subjected to a current between two electrodes. The $\ce{MnO2}$ dissolves,
  enters solution as the sulfate, and is deposited on the anode.

Unfortunately, there is no source for the information.
Some further searching yields this article, with the following information:

EMD ($\gamma$-$\ce{MnO2}$) can be prepared from the direct electrolysis
  of an aqueous bath of manganese sulfate and sulphuric acid. The
  electrochemical deposition of $\ce{MnO2}$ on an inert electrode from
  an electrolyte containing $\ce{Mn^2+}$ occurs as per the reaction
  noted below:
$\begin{alignat*}{3}&\textrm{anode: }\quad&&\ce{Mn^2+ + 2H2O} &&\ce{->
MnO2 + 4H+ + 2e-} \\ &\textrm{cathode: } \quad&&\ce{2H+ + 2e-}
&&\ce{->H2} \end{alignat*}$

Note that the linked article is a review, with a lot of information, both about the actual structure of $\gamma$-$\ce{MnO2}$ as well as the reactions that lead to the production of the correct crystalline structure in the presence of acid. There are a number of references that you may find useful as well.

Did you find what you needed?

Thank you for the follow-up; that's very kind of you! I have made three samples and will be sending them for XRD tomorrow. I hope they are $\gamma$!


text:
Forming a dioxolan with levodopa
I've got 2 problems with reaction:
1) Does the nitrogen or the carboxyl group react with the methanal?
   Do I have to use a protecting group?
2) Is there a chance it reacts similar to a marquis reaction?

What about a Pictet-Spengler reaction?

Before answering the question, you might find it helpful to know that people tend not to go about making the dioxolane, but instead make use of natural products such as safrole to buy in a precursor with the ring already formed. If you work at a University or a research institution, this shouldn't be an issue. If you don't, I strongly advise you against trying this chemistry, as your desired product bears significant resemblance to multiple illegal drugs.

To answer specific parts of your question:

1) Does the nitrogen or the carboxyl group react with the methanal? Do I have to use a protecting group?

The reactivity of the amino acid portion depends on the ionisations.
In both the protonated and deprotonated form, the carboxylic acid/carboxylate can't react with formaldehyde.
Amines do react with formaldehyde in the presence of an acid catalyst (think about the Mannich reaction). Whether or not your substrate (amino acid) as drawn would react with formaldehyde likely depends upon whether the amine is protonated or not (think about the amino acid and whether or not it would exist in its zwitterionic form under the conditions given).

2) Is there a chance it reacts similar to a marquis reaction?

The mechanism of the Marquis reaction isn't fully elucidated, however it does give positive results for many alkaloid drugs

As already stated, given the similarity of your compound to several illicit drugs, I wouldn't be at all surprised if it did give a positive result.

Moving on to the actual chemistry you've proposed.
The standard method for making 1,3-benzodioxole (and derivatives) is to take the catechol and treat it with CH2Br2 under basic conditions.
This works better than formaldehyde (I'm not actually convinced the formaldehyde would work at all, but I'll presume you've researched the method) because rather than trying to expel water (which as you point out, you'd have to Dean-Stark to remove the water), you only expel halide ions, which don't interfere under the conditions.
(As another side note, experimentally you also need to add a phase-transfer catalyst as the CH2Br2 isn't miscible with the base/aqueous layer).

Thanks that was very helpful. So my primary question was the detection of Levodopa via Marqui-reaction. In this case the product would look like this:http://www.directupload.net/file/d/4401/o25t3hlg_png.htm

The product of that Marquis reaction gives a highly conjugated cation- as I said, I see no reason why it wouldnt give a positive result.

Yeah, but I want know the structure of the detected product. Are there any sources where I can find solutions to such questions?

SciFinder/Reaxys/Scopus/Web Of Science?     I'm sure there will have been publications of the products of the Marquis reaction, though you're more likely to find them in pharmacology journals than the chemistry literature.

Spot tests tell you what you don't have and what you may have.


text:
(Computationally) finding similarity between two organic compounds
I have been struggling to find an acceptable answer for this question for my purposes.
There are many ways to find similarity between two organic compounds - some of which are particularly popular in chemoinformatics. The seemingly most popular way is to use fingerprints of molecules, which then somehow correlates to the structure/function of various parts of the molecule. This approach seems very good when you're looking for general similarity between molecules. It also has the added benefit of making it fast to compare huge amounts of molecules, as each molecule can be encoded separately (so it's generally speaking $O(n)$), rather than comparing every pair of molecules (which would be $O(n^2)$).
However, for some purposes, this fingerprinting approach is not very good. I need a function of two molecules with the following properties:

If $A$ and $B$ are two molecules, then $f(A, B) \in [0, \infty).$
$f(A, A) = 0$
$f(A,B) + f(B,C) \geq f(A,C)$

These are similar to the basic requirements for a metric space. (The big omission is that I don't necessarily need $f(A,B) = f(B,A),$ although this certainly couldn't hurt!)
The reason that I want such a function is because I'm designing a neural network that essentially outputs a molecule, and I want to have some sort of error on the output.
I have played around a bit with Python-RDKit, and its similarity module, but haven't really been able to form a good "error" function from the output.

I've also experimented a bit with an algorithm I created that looks for the largest identical connected subgraphs of two molecules, and essentially finds matches for each part of the query molecule. The final output is then how many different parts the molecule needs to be split into to find a match.
For instance, if the "true" molecule is ethylbenzene, while the query molecule is m-xylene, then the algorithm would find that the m-xylene needs to be broken into 3 parts for each part to find a perfect match: a benzyl group missing a hydrogen at the third carbon of the aromatic ring, the methyl group that was formerly attached at the third carbon, and the hydrogen that remains from the benzyl group.
However, this algorithm suffers from several problems:

If the query is a subset of the true answer, then the algorithm will always give the answer as 2 (one part is the subset, and the other part is the hydrogen that caps it - try seeing it with something like query - methane, true - acetic acid). This isn't that big of a concern, as you can simply run the algorithm twice - once comparing the query to the true molecule, and once in reverse. That way, if the two molecules are indeed far apart, it's not hard to see (the superset molecule may need to be broken into many pieces to be identical).
This algorithm is slow. Don't really see a way to speed it up. It searches for maps from subsets of the query molecule to the true molecule, then gradually builds up from there. It also can't pick maps randomly as being the best, then growing from there, as picking the wrong direction can easily make the "best" map drastically short. So it has to do all possible maps at the same time. Which is slow.

In short, this is a somewhat open-ended question that boils down to:
How can we put a (loose) metric on the set of organic molecules?

Just out of curiosity: What to you want to achieve with this?

Short answer: train a neural network to deal with organic molecules (starting with interpreting SMILES, then moving up to interpreting NMR spectra, then maybe if these ideas work, to designing drugs... but this is all far away)

Longer: originally neural networks could only deal with fixed-length data. These days are gone, thanks to RNNs. This was the first step to realize that neural networks can deal with more complicated inputs. There's also been ideas to allow networks to have access to "external resources", in particular Turing Machines, leading to the innovation of an NTM. These networks performed amazingly on certain tasks that are suited to having an external memory.

This got me to thinking that - when I draw a chemical structure, I usually use a representation of the molecule to think about it. Thus, I am coupling neural networks with some graph/molecule manipulation machinery and seeing what happens.
However, even if these ideas don't work out, it seems like a useful idea to have a nice measurement of "distance" between two molecules, in a mathematical sense. Being able to embed the set of organic molecules into some mathematical space would probably help in designing algorithms that search for molecules, providing that the embedding is useful.

You're not the first to have thought of this. ;-)  https://en.wikipedia.org/wiki/List_of_computer-assisted_organic_synthesis_software

I figured I wasn't :)

Did you read any half-decent QSAR/QSPR introduction article ?

Define half-decent -- or give an example of one?

Not particularly chemistry related (so posted as comment rather than an answer), but one possibility is to compute the various properties you would potentially be interested in, and then compute the Mahalanobis distance between the two property vectors. That should give you a metric value, but it's not used all that much, probably because of computational expense and the need for a fixed representative reference compound set to derive the covariances from.

I assume that you are familiar with Tanimoto coefficient?

Correct me if I am wrong, but the point would nit be that YOU come up with a solution to the problem? I mean you have a problem that i am sure many already tried to solve similar ways to your approach.

Just curious, how many descriptors you tried? If you confine/ define the search space appropriately, descriptors defining fragments and groups seem to pretty well matching your criteria.

In answer to the original post: unless I am completely missing the point, I don't get why 'for some purposes, this fingerprinting approach is not very good', as you don't explain what you don't like about fingerprints. I agree with vapid, the Tanimoto coefficient (let's call it $T$) does pretty much what you want. Two identical molecules will give a Tanimoto similarity $T(A,A)=1$, and two molecules with no FP bits in common will give $T(A,B)=0$. So your $f(A,B)$ could simply be $\frac 1 {T(A,B)} - 1$. As for the third condition, not sure, but it should be easy to check.

Using graphs to represent molecules, would it be reasonable to compute the distance between graphs A and B?
In a graph representation of a molecule, nodes are atoms and edges are bonds.
Example of an algorithm to compute distance between graphs: http://www.xavierdupre.fr/app/mlstatpy/helpsphinx/c_graph/graph_distance.html

I recently employed this approach and I can warmly recommend it. I have to note though that you will run into the graph isomorphism problem. You can use a greedy algorithm (does not always give the correct result) or a brute force approach (very expensive) to calculate the distance metric. The greedy algorithm is suitable for large molecules and the brute force method is computationally feasible to graphs of less than 12 nodes.

I recently did something similar to this, I pulled out the murko scaffolds of my compound to be tested, and my test set. Then compared them to find matches, and from that pulled out the R groups that need to be attached to the scaffold to make the compound. As well as calculating the maximum common scaffold of the group with matching scaffolds.
The draw back is that it doesn't produce a number as such to display similarity scores.


text:
Will molecules containing heavy isotopes tend to rise up in a liquid?
When you have a bag of chips or nuts or something else with big and small pieces and you shake it, the bigger pieces will rise up and the smaller pieces will rise down. This is known as the brazil nut effect.
I'm curious, does something similar happen with molecules in a liquid? For example: say you have a glass filled with water. Is the concentration of $\ce{HDO}$ and $\ce{D2O}$ larger at the top in comparison to the bottom?

I'm quite sure that random thermal Brownian motion at just about any temperature above absolute zero completely mixes the isotopes . The drive towards isotope segregation would be due to a decrease in overall gravitational potential of the system, but gravity is an extremely weak force on the scale of molecules, and this tiny "signal" would be completely swamped by the "noise" of thermal fluctuations. The answer would end up being similar to this.

The only system with isotopic segregation I know of is the phase separation of He-3 and He-4 at a fraction above $\mathrm{0\ K}$, which may meet the requirement in part, but this is a very special case.

Molecules don't behave like different sized lumps in a cereal packet. And the behaviour in a pack of nuts, for example isn't about weight but size.

Why, of course they would, but the effect would be ridiculously weak. Think of centrifugal isotope separation.

As a side not, the Brazil nut effect is not well understood.

and is not the 'Brazil nut' effect mainly entropic is nature? If you wanted to do the same thing with molecules you would have to consider, say, water and a protein, which is to say molecules of significantly different size and mass. Even then then effects as described by @Nicolau Saker Neto will still make separation impossible. To effect separation we usually apply a huge force using a centrifuge.


text:
Color of methoxy substituted phenyl rings when heated in the presence of acid
Anyone who's familiar with organic synthesis will probably have came across the DMB and PMB protecting groups.
When TLC-ing compounds containing these protecting groups using common stains like vanillin or cerium ammonium molybdate (CAM, the Hanessian stain), its not uncommon for the compounds to appear bright pink/purple. This is a useful phenomenon, but one which I have no explanation for.
I'd assumed that this was related to the stain itself, however even when the TLC plate is simply heated in methanolic hydrochloric acid, the pink color still appears (quite vividly in fact).
I realise that all of the stains mentioned are made up with some acid, however, heating PMB/DMB containing compounds under acidic conditions without the TLC plate (in a flask, for example), doesn't afford this colour.
Any insight into what's forming the color? Specifically when PMB/DMB compounds show it, but simple benzyl ethers etc do not.

many phenol derivitive are relatively easily oxidized into bright colored products.

Well, anisaldehyde is technically PMB-carbonyl and colours the TLC pink … And DMB is des-methyl-vanillin, isn’t it?

@Jan: Not necessarily. PMB is p-methyoxybenzyl whil DMB is dimethoxybenzyl. Both are used mostly in hydroxy protection.

TLC Visualization by the color formation when used p-Anisaldehyde or vanillin in sulfuric acid as the developing agent is known in Natural Products Chemistry for long time. OP's suggestion of color development is due to inclusion of p-methoxybenzyl group (PMB) or dimethoxybenzyl group (DMB) not correct. It seems like these staining compounds make colored derivatives with certain compounds with or without PMB and\or DMB protecting reagents. For instance, when p-Anisaldehyde visualizing agent (p-Anisaldehyde/ conc. $\ce{H2SO4}$), which is a universal derivatization agent that can be used to detect a wide range of nucleophilic compounds (Ref.1), is used in series of plant extracts (tea samples) containing the polyphenol natural products, it gives bright orange spot on the TLC plate with asplathin, a flavonoid (Ref.2):

As you have seen in the structure, there was no PMB or DMB group to give this specific orange color. The basis for this kind of detection is the derivatization of the solute molecules with an organic compound containing one or more conjugated electron system such as that in p-anisaldehyde. This kind of electron system can be absorb either visible or ultraviolet part of the spectrum to give corresponding color (Ref.3).
Some other visual color developments with p-Anisaldehyde/ conc. $\ce{H2SO4}$ detecting system are shown in following images (Ref.4):

Note that although some compounds have given multicolor with the system none of them contained either PMB or DMB protecting groups. As a matter of fact all of them are natural products. As noted by the authors of Ref.4, the color of monoterpenes was different than that for triterpenes, and color for steroids are different than either of them. For instance, after revelation and heating, monoterpenes, triterpenes, and steroids appeared as blue, purple, and gray spots, respectively. None of these colors can be interpreted by simple theory.
Also keep in mind that the color development by p-Anisaldehyde/ conc. $\ce{H2SO4}$ is not limited to natural products. The synthetic compounds also gived color compounds when derivatized by p-Anisaldehyde/ conc. $\ce{H2SO4}$ on TLCs.
References:

Hellmut Jork, Werner Funk, Walter Fischer, and Hans Wimmer, In Thin-Layer Chromatography - Reagents and Detection Methods, Volume 1a: Physical and Chemical Detection Methods, Fundamentals, Reagents I; Translated by Frank Hampson, Wiley-VCH Verlag GmbH: Weinheim, Germany, 1990 (ISBN13: 9783527278343).
Emily Amor Stander, Wesley Williams, Fanie Rautenbach, Marilize Le Roes-Hill, Yamkela Mgwatyu, Jeanine Marnewick, and Uljana Hesse, “Visualization of Aspalathin in Rooibos (Aspalathus linearis) Plant and Herbal Tea Extracts Using Thin-Layer Chromatography,” Molecules 2019, 24(5), 938 (11 pages) (DOI: https://doi.org/10.3390/molecules24050938).
Bernard Fried and Joseph Sherma, In Practical Thin-Layer Chromatography: A Multidisciplinary Approach; CRC Press: Boca Raton, FL, 1996 (ISBN: 0-8493-2660-5; ISBN 13: 9781351422420).
Alice da Cruz Lima Gerlach, Alice Gadea, Rosa Mara Borges da Silveira, Philippe Clerc, Françoise Lohézic-le Dévéhat, “The Use of Anisaldehyde Sulfuric Acid as an Alternative Spray Reagent in TLC Analysis Reveals Three Classes of Compounds in the Genus Usnea Adans. (Parmeliaceae, lichenized Ascomycota),” Preprints 2018, 2018020151 (DOI: 10.20944/preprints201802.0151.v1).

Nice answer, but you may have misinterpreted my question somewhat.        I’m aware that reagents give colours with many compounds, including those not containing the PMB/DMB group. My query was specifically relating to the colour when these are present (eg using CAM can very clearly tell a PMB protected alcohol from the parent merely by colour in the vast majority of cases)

@NotEvans: As I mentioned in the answer, the reason for the color is unexplainable, according to most of authors (al most all of them are natural product chemists). I think it is very complicated and nobody wanted to investigate. Kind of split-light phenomena. :-)


text:
Is the deprotonation of phenol exothermic?
I just wanted to confirm that deprotonation reaction of weak acids are exothermic?
I know that strong acids deprotonate exothermically (e.g. adding liquid $\ce{H2SO4}$ to water releases a lot of heat), but is this true for weak acids?

Any bond dissociation reaction is endothermic, hence the termin 'bond'. Heterolytic dissociation (into ions) is more endothermic than homolytic. Strong acids are not exception.
However, the words above are about molecules in vacuum. When the molecules are in a media, interaction with media may stabilize ions strongly. This is why salts dissolve and acids/bases dissociate.
In case of phenol, dissociation in vacuum is definitely endothermic. Dissociation in water is also endothermic, since the dissociation constant is small. (1) However, deprotonation implies that protons moves to another acceptor, and in this case it depends entirely on the acceptor. Similarly, in nonwater media dissociation of phenol also may be exothermic.
(1) Dissociation constant is tied to dissociation free energy with law
$\ln K_e = - \Delta G / RT$ . Since entropy in reaction dissociation reactions is definitely increasing, the only reason for $\Delta G$ to be unfavorable is enthalpy.


text:
correct psilocybin molecular structure
I'm writing a research paper on psilocybin.  I need to know the correct chemical structure.  When I look online it it shows different ones.  There is one with two OH molecules attached to the phosphorous and there is one with one OH and one O- attached to the phosphorous.  There is also one that I found with a O+ attached to the phosphorous.  Are they all they same or are they interchangeable?  Are there different ones?  I can't find a clear answer.  Any help is greatly appreciated.

Clearly, it cannot be an $\ce{O+}$, as that would require a triple bond to the oxygen, making the phosphorus in the middle quite unhappy.
Of the other two alternatives, both are correct -- however, it matters what environment you're considering the molecule. In neutral or slightly basic conditions, it will be an $\ce{O-},$ while in slightly acidic conditions, it will be an $\ce{OH}.$
Let me know if you need any clarification on why this is the case -- I have no idea what your level of knowledge is.

Thank you for the prompt answer. My knowledge is fairly low.  Can you tell me which it would be in it's natural form?  Why would they differ due to ph?

Since this is a natural product (found in mushrooms), most likely it is found primarily inside of cells, which have a close to neutral pH. Thus, it is likely found in its deprotonated form (the one with the $\ce{O-}$). The reason that pH matters is because the difference between $\ce{OH}$ and $\ce{O-}$ is a measly little $\ce{H+}$ (notice that regular hydrogen has one proton and one electron, so $\ce{H+}$ has lost its electron, and is simply a proton - hence the word de- proton -ated). pH is just a fancy way of measuring the amount of $\ce{H+}$ found in a solution, so that's why pH matters.

This answer completely neglects, that it is a zwitterion, and that both structures will be present at all times. The degree of protonation/deprotonation is of course pH dependent. Resonance of the phosphate group is also important.

I know this... but I suspect that the original question was asked with no intention of understanding the actual chemistry regarding the phosphate group, but rather to find the correct structure to put into a report of some sort. While the degree of protonation on both the phosphate and the amino varies, the "most natural" form is indeed the zwitterion. Of course, there will be 4 different structures in equilibrium with each other (ignoring the second deprotonation of the phosphate), but the one most prevalent at physiological pH is the zwitterion.

pH and pKa are the measure of the amount of dissociation, or H+ ions.  At pH7, there are 0.0000001 H+ ions for each molecule of H2O. The pKa of water is 14; at pH 14 half of the water will be deprotonated. The pKa of psilicybin is around 6.5; at pH 6.5 half of the psilocybin will be dissociated (O-). In neutral conditions (pH7) the amount of dissociated (O-) psilocybin will  be higher than the "associated" (OH) psilocybin. Cytosolic pH of fungal cells is about 6.95. At pH 8, roughly 90% of psilocybin will be in the O- form. At ph 10.4 ~50% of the dimethylamine nitrogen will be deprotonated.