text:
Why doesn't water burn?
Hydrogen is flammable, and for any fire to burn it needs oxygen. Why does a compound made of hydrogen and oxygen put out fires instead of catalyzing them? I understand that hydrogen and water are chemically different compounds, but what causes water to be non-flammable?

So if wood burns, why don't the ashes burn?

You can think of water as the ash from burning hydrogen: it's already given off as much energy as possible from reacting hydrogen with oxygen.
You can, however, still burn it.  You just need an even stronger oxidizer than oxygen.  There aren't many of them, but fluorine will work,
$$
\ce{2F2 + 2H2O -> 4HF + O2}
$$
as will chlorine trifluoride:
$$
\ce{ClF3 + 2H2O -> 3HF + HCl + O2}
$$

I wonder what burning water looks like.

@SauravSingh, given the nature of the chemicals I've listed, I wouldn't want to be anywhere near a fire that's burning water.

Burning water with an excess of ClF3 with "complete combustion" will not leave HCl it will leave Cl2. In fact, 2HCl + 1/2O2 => H2O + Cl2 can happen in the gas phase with a suitable catalyst. Cl2 and O2 trade off as stronger oxidizer depending on what is being oxidized, but F2 and ClF3 are much stronger.

This is the reaction that occurs when hydrogen combusts:
$$
\ce{2H2 + O2 -> 2H2O}
$$
Similarly, this is the combustion reaction for methane, a representative fuel:
$$
\ce{CH4 + 2O2 -> CO2 + 2H2O}
$$
Water is a product in both of these reactions. Thus, water represents something that has already been oxidized by oxygen, and as such there is little further energetic gain to be made by trying to react it again with oxygen.

I maybe reading to much into your wording - "little further energetic gain".  Should it be "no further energetic gain"?  I am not a chemist, so this is just a point of curiosity.

@DigitalTrauma Ah, yes, sorry. That phrasing was rhetorical, not scientific. I can't imagine any products of a water/oxygen reaction that would have lower energy.

Sodium and Magnesium come to mind as well, @Mark

The OP asked why water puts out fire.  Many fuels need to be hotter than 100 C to burn. Water cools the fuel and combustion stops.  Don't try this with hydrocarbon fires, though.

@NateLockwood Re "Sodium and Magnesium": Those do not "burn water" (by capturing the hydrogen's electrons better than oxygen does), but on the contrary reduce the oxygen by providing electrons even better than the hydrogen does. They get burned by the oxygen in the water, but they do not burn the water's hydrogen. Quite the contrary.

@PeterA.Schneider Good point. The burning that occurs is due to the released hydrogen reacting with the oxygen in the air. No flame would occur if solid sodium were added to water under an inert atmosphere.

@hBy2Py et al. This is purely abstract thinking, but there will be an extremely small isotope shift in the bond energies. One could play with ${}^1H$, ${}^2H$, ${}^3H$, ${}^{16}O$, ${}^{17}O$, ${}^{18}O$ and see if there is some combination of isotope enrichments in the primary combustion and the new water thrown into the fire where there might be an absurdly small energy that could drive a further reaction (albeit very slowly). Since in liquid water the hydrogens are constantly exchanging, so after a while there won't be distinguishable "old water" and "new water" molecules.

I stand corrected, @PeterA.Schneider!  (But then the released hydrogen is oxidized by the oxygen in the air.)

Re Chlorine Trifluoride: Sand Won't Save You This Time

$\ce{H2O2}$ exists, and could be what you expect by burning water (since burning is really oxidizing, or adding oxygen.)
However, in stoichiometric proportions, here is what would happen:
Either further burning with outside oxygen
$$
\ce{2H2O + O2 <--> 2H2O2}
$$
or decompose water to get oxygen, and reject hydrogen:
$$
\ce{2H2O <--> H2O2 + H2}
$$
To determine if that can work, we can simply compute the energy result:

$\ce{H-H}$ bond energy is $432$,
$\ce{H-O}$ is $459$,
$\ce{O-O}$ is $142$,
$\ce{O=O}$ is $494$

(all in $\mathrm{kJ/mol}$)
Also, $\ce{H2O}$ is $\ce{H-O-H}$ and $\ce{H2O2}$ is $\ce{H-O-O-H}$.
For the first equation, we have:

$2 \times 2 \times \ce{H-O} + \ce{O=O} = 2 \times 2 \times 459 + 494 = 2330$
$2 \times (\ce{H-O} + \ce{O-O} + \ce{O-H}) = 2 \times (459 + 142 + 459) = 2120$
Result is $-210\ \mathrm{kJ/mol}$

For the second equation:

$2 \times 2 \times \ce{H-O} = 2 \times 2 \times 459 = 1836$
$(\ce{H-O} + \ce{O-O} + \ce{O-H}) + \ce{H-H} = (459 + 142 + 459) + 432 = 1492$
Result is $-344\ \mathrm{kJ/mol}$

So we see that both reactions are endothermic, so what is more likely to happen is:
$$
\ce{H2O2 + H2 -> 2H2O}
$$
$$
\ce{2H2O2 -> 2H2O + O2}
$$
As a matter of comparison:
$$
\ce{2H2 + O2 -> 2H2O}
$$
Has the following energy:
$$2\times 432 + 494 \ce{->} 4\times 459$$
This is an exothermic reaction with an energy of $+478\ \mathrm{kJ/mol}$

It's really great to see the numbers, here, however, as a non-chemist, I am confused.  Why does the reaction move towards the compound with higher energy?  Doesn't entropy cause systems to tend towards low-energy states?

@fuzzyTew the energy in the bound is the energy released during the reaction when the bound is formed, which is also the amount of energy you need to provide to break it. So the lower energy bounds are the first to break, and the system tends to go toward high-energy bounds, because they are hard to break.

I see.  Comparing with @Paraquat's answer, I think the energy of the bond or molecule system itself would be the negative of the number you state in this answer.  D_e = 432 kJ / mol.

This becomes intuitive once you accept that any chemical compound will 'want' to be in the state with the lowest energy, the ground state for a given temperature and pressure. For example, if you plot the energy of the system of two hydrogen atoms against the separation of the atoms, it looks like this:

So when you bring together two hydrogen atoms, they form a molecule and release an amount of energy $D_e$. The $\ce{H2}$ molecule is the ground state for hydrogen, so at room temperature and pressure, you would only expect to see hydrogen gas in the form of molecules. It's difficult to draw an analogous graph for water because the molecule has more degrees of freedom (in fact, you would need to draw a seven-dimensional graph), but you would find $\ce{H2O}$ as the ground state of a system with two parts hydrogen to one part oxygen.
So back to the question: why does water put out fire instead of catalyzing it? It really depends on what you are burning. Let's assume it's wood, which is mostly carbon and hydrogen, but for the sake of simplicity, we'll just use methane, which is similar:
$$\ce{CH4 + O2 -> CO2 + 2H2O + energy}$$
The energy released on the right-hand side of the equation is the 'burning'. The ground state of most systems involving carbon, oxygen and hydrogen is some combination of carbon dioxide gas and water. This combination has the lowest energy, and to get anything else, you would have to pump energy into the system.
However, water isn't the ground state of all chemical systems, for example, sodium, hydrogen, and oxygen. This can be seen if you Google a video of sodium reacting with water. In this case, the ground state of the system is sodium hydroxide and hydrogen gas, so the water does 'burn' in this case:
$$\ce{2Na + 2H2O -> 2NaOH + H2}$$

Because the hydrogen in water is already oxidized. When magnesium burns in water, the magnesium is being oxidized because the magnesium takes (binds with) the oxygen from the water molecules. This causes the release of hydrogen gas, which would/could burn in air (air is about 1/5 oxygen, 4/5 nitrogen) to form water again.

When different chemicals react, they form products that have completely different properties than what went into them.
In this case, hydrogen and oxygen are both very reactive, which is why they react together (or 'combust') to form water.  The water that is formed is much less reactive, and does not easily react with oxygen, and so will not burn.
Because water is so non-reactive, it's interaction with the fire is entirely physical.  It absorbs a lot of heat very quickly and prevents oxygen from getting to the fuel.  Between those two effects, it's the simplest way to put out a wood fire.

Water is already burnt hydrogen:
$$\ce{H2 + 1/2O2 -> H2O}$$
The reaction of oxidizing water ever more (to peroxide, for example) is non favorable and the peroxide will decompose as:
$$\ce{H2O2 -> H2O + 1/2O2}$$
with  ΔH of −98.2 kJ/mol and a ΔS of 70.5 J/(mol·K)

This is true but doesn't really answer the question directly.


text:
Why do alpha-hydroxy ketones respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests?
I learnt from a certain textbook that aldehydes and formic acid respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests.
However, in another textbook I read that even alpha-hydroxy ketones respond positively to these tests. What is the reason behind this? I thought ketones are not oxidizable (unless under drastic conditions).
Fructose is an alpha-hydroxy ketone, but all alpha-hydroxy ketones are not fructose. In fructose there is a terminal hydroxy group which facilitates Tollens' test, etc. in basic medium. But why do alpha-hydroxy ketones like benzoin give these tests even though there is no terminal hydroxy group? That part is still unclear to me. Could you please answer that part?

Related: Why does fructose reduce Tollen's reagent and Fehling's solution?

I think $\alpha$-hydroxy ketone is being oxidized to a diketone... Still thinking about mechanism of action.

I think you should be able to find your answer here: http://www.acadblock.com/organic-chemistry/tollens-reagent-l4b2/

There is a nice answer in this link: https://www.quora.com/Why-is-Fehlings-solution-able-to-oxidise-%CE%B1-Hydroxy-ketone-but-not-ketones

Duplicate?: https://chemistry.stackexchange.com/questions/94620/in-what-way-could-benzoin-give-tollens-test/94630#94630

I answered a similar question with proposed mechanism : https://chemistry.stackexchange.com/questions/156199/do-all-%ce%b1-hydroxy-ketones-give-tollens-test/156371?noredirect=1#comment321838_156371

Its because of two tautomerizations happening, the ketone becomes en-ol, and then another tautomerization happens with the original hydroxy group (which is terminal), making it an aldehyde group. All three forms exist in equilibrium, so it responds to Tollens' test.

Not all alpha hydroxy ketones can undergo double tautomerization to form an aldehyde.

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text:
Halogenation of alkenyl boranes
When a terminal alkenyl borane is succumbed to halogenation where the halogen and the base are added in two separate steps, the geometry of the double bond is changed. When they are added simultaneously, the double bond geometry is retained. Why is it so?

You've actually drawn a boronic acid instead of a borane. I suspect that the the reaction doesn't proceed with a boronic acid, since these are markedly less electrophilic. "The failure to utilize the third group in organoboranes prepared from terminal olefins restricts the yield of
iodide from a particular terminal olefin to a maximum of 66.7%." (JACS 1968, 90, 5038). And indeed, I can't find anything on Reaxys for this particular transformation.

It was taken from Smith's Organic Synthesis.

Thank you for the tipoff, and you are right, I am wrong. The reaction proceeds with boronic acids. In fact, with the borane, additional complications occur because of migration of B-C bonds in the iodonium ion. March's Advanced Organic Chemistry 7th ed. p705 mentions the reaction and gives the reference: JOC 1989, 54, 6075. I'll update the answer tomorrow - gotta head to bed now...

You could add the mechanism for the side reaction: alkyl migration.

I remember learning that these two reactions proceed via different mechanisms. I'm going to use an alkenylborane to illustrate them.
Organoboranes are electron-deficient and electrophilic; when hydroxide is added together with iodide, the first step is the formation of an anionic ate complex. This is now nucleophilic via the $\ce{B-C}$ $\sigma$ bond, and can therefore react with the electrophilic iodine:

This mechanism is somewhat analogous to the hydroboration-oxidation $\ce{H2O2}$ step, where the carbon "nucleophile" is intramolecularly transferred from boron to an electrophilic oxygen atom. This step is an intermolecular transfer of carbon from boron to iodine instead.
Incidentally, I remember that with alkylboranes the migration occurs with inversion of stereochemistry, but I cannot seem to find a reference. Given the premise of the question, I would hypothesise that alkenylboranes react with retention of stereochemistry (which would not be surprising since stereochemistry at $\mathrm{sp^2}$ and $\mathrm{sp^3}$ centres, and the stereoelectronic requirements for inversion/retention, are entirely different).

As for the other reaction, hydroxide isn't added first, so iodine (electrophilic) has to react with the double bond (nucleophilic). The subsequent steps are best described by a diagram instead.

Again I am a little frustrated at not being able to find a reference, but I am fairly sure this was how I learnt it. I literally looked up all reactions of $\ce{BR3}$ with iodine on Reaxys but couldn't find anything on it, so...

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text:
Stability of Nitromethane
Why is nitromethane not especially reactive? $\ce{NO2}$ is stable by itself, so why does it stay attached to $\ce{CH3}$?

There’s stable and there’s stable. 2) ‘But chloride is stable by itself, why should methyl chloride also be stable?’ (Meaning: one is irrelevant for the other.)

@Jan  makes sense, but what bond exists between the two molecules?

A $\sigma$ bond.

Related: http://chemistry.stackexchange.com/questions/34169/if-both-nitric-acid-and-nitrate-ion-are-stabilised-through-resonance-what-makes

It depends on what kind of reactivity you mean. Nitromethane is indeed flammable and is used as a racing fuel. It can also detonate, especially when mixed with an oxidizer. But it is generally quite stable to atmospheric oxygen at room temperature, perhaps because the nitro group is electron-withdrawing and reduces the likelihood that the hydrogens on the methyl group will react with oxygen or other oxidizers.
In other ways nitromethane is indeed reactive: for example, the nitro group makes the hydrogens on the methyl group acidic, so it reacts with an aqueous base like sodium hydroxide to give the nitromethide anion, which is useful in organic synthesis.
Consider the splitting of nitromethane into methyl radical and nitrogen dioxide. It has a normal sigma bond between carbon and nitrogen, and to break that bond you have to put a lot of energy in. You are right that NO2 is indeed stable on its own, but it is still a reactive, high energy species. Methyl radical is even more reactive, and the energetics of this bond-breaking reaction are overall unfavorable, at least at lower temperatures. Higher temperatures, such as might occur when sparking a detonation, may give the kind of splitting you are talking about, plus a lot of other reactions as well.

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Reaction between Iron (III) Oxide and CO?
I'm a science tutor and I came across a stoichiometry problem asking how much Iron is produced in a reaction between $\ce{Fe2O3}$ and $\ce{CO}$:
$$\ce{Fe2O3 + CO -> Fe + ?}$$
I know how to do the stoichiometry part, but I'm stumped about how to figure out the products. My guess is $\ce{Fe}$ and $\ce{CO2}$ since the $\ce{CO}$ would want to change to the more stable $\ce{CO2}$, thus deoxidizing the $\ce{Fe}$? All the student's problems so far have just been between simple ionic salts, and I never remember coming across one like this when I took AP Chen myself in highschool, since $\ce{CO}$ is covalent.

Your guess is right. Have you ever wondered how steel is produced in the first place?

https://en.wikipedia.org/wiki/Blast_furnace#Process_engineering_and_chemistry

The reaction products depends on stoichiometric ratio of the reactants as well as reaction conditions (i.e temperature, pressure etc.)

$$\ce{Fe2O3 + CO ->[500-600 C] 2FeO + CO2}$$

Iron(III) oxide and carbon monoxide to produce iron(II) oxide and
  carbon dioxide. This reaction takes place at a temperature of
  500-600°C. (source)

$$\ce{3Fe2O3 + CO ->[400 C] 2Fe3O4 + CO2}$$

Iron(III) oxide react with carbon monoxide to produce iron(II,III)
  oxide and carbon dioxide. This reaction takes place at a temperature
  near 400°C. (source)

$$\ce{Fe2O3 + 3CO ->[700 C] 2Fe + 3CO2}$$

Iron(III) oxide react with carbon monoxide to produce iron and carbon
  dioxide. This reaction takes place at a temperature near 700°C. (source)


text:
What is the hybridisation of the carbons in the allyl radical?
The example I am talking about is $\ce{CH2CHCH2^.}$.
To find the steric number and hence the hybridization, we use steric number  = No. of lone pairs + number of sigma bonds. In this example, the last C atom has only one electron. Should we count it as a lone pair?
My teacher says that that electron will get delocalised so its hybridization will be $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ (which I think is true). I didn't understand the delocalisation stuff. Is it true that here, there will be delocalisation and if not then what is happening here? If delocalisation is occurring here then please explain it too! Also please tell me in which cases is there delocalisation of electron(s)?
My teacher also said that had it been $\ce{CH3CH2CH2^.}$, then there would be no delocalisation of electrons. Why does this happen?

The allyl radical has a resonance structure because the lone electron can combine with the pi bonding electron on the center carbon to make a double bond and leave the first carbon with a lone electron.  Based on your question, you are implying that your teacher claims that the third carbon in the propyl radical a) does not have a delocalized electron AND b) therefore it will be $sp^3$ hybridized.  I don't understand that so I won't post an answer,

but this may help:  https://www.google.com/search?q=allyl+radical+resonance&rlz=1C1CHFX_enUS511US512&espv=2&biw=1920&bih=974&tbm=isch&tbo=u&source=univ&sa=X&sqi=2&ved=0ahUKEwiJto7kyPHQAhWmwFQKHSSiAEkQsAQIJQ

Radicals are electron-deficient compounds. You should not treat a radical as a lone pair, as they are pretty different.
A good rule of thumb is to assume an orbital that contains the highest possible p-contribution for the single electron. The more s-contribution is mixed in, the ‘closer’ the single electron gets to the nucleus which is less favourable than having an electron pair in a similar orbital. A pure p-orbital fulfils the condition perfectly by having no s-contribution whatsoever. Thus, you should assume radicals to be in p-orbitals whenever possible.
This is similar to carbocations which have an orbital fully devoid of electrons. However, this empty orbital must always be a p-type orbital for electronic reasons. Radicals have the possibility of undergoing hybridisation and e.g. ending up in an $\mathrm{sp}^n$ orbital (examples: radicals on $\mathrm{sp^3}$ hybridised bridgehead atoms; phenyl radical).
Since we just established that the radical will occupy a p-type orbital and since there is a π bond nearby, we can delocalise. In resonance structure terms, this can be visualised in the following way:
$$\ce{H2C=CH-CH2^. <-> H2C^.-CH=CH2}$$
The radical is delocalised across two carbon atoms making the overall system more stable than if it weren’t. In molecular orbital terms, this is basically a stabilisation of the double bond at the expense of the radical-containing orbital which benefits the overall system. The delocalisation can, of course, only occur if there is a π bond nearby; the n-propyl radical does not feature one.

Please examine my interpretation of paragraph 2.  The lone electron in the radical goes into the p-orbital because it is less stable to have a single electron in an orbital with s-character as it is 'worse' to have single electrons close to the nucleus than to have electron pairs close to the nucleus?  So can you clarify this example?  In the methyl radical, is the geometry trigonal planar because it is more stable to have the lone electron in a pure p-orbital leaving open  only 3$sp^2s$ for bonding, or does the stability of the t.p. geometry drive sp^2 hybridization?

@JosephHirsch The methyl radical should be $\mathrm{sp^2}$ in the absence of constraints, yes. While not instantly visible, radicals isomerise my a mechanism similar to amines, providing support for the view.

I understand it is $sp^2$.  My question is, is it because the loan electron is more stable in a pure p-orbital or is it because the $sp^2$ allows a more stable geometry than $sp^3$?

@JosephHirsch The former. Geometries only play a role if they restrict a more stable electronic situation. E.g. geometry is the reason why water does not have $90^\circ$ angles (it would be too strained).

I see the question has having the following parts:
1) Does the allyl radical have a delocalized electron?
2) Is this what makes it $sp^2$ instead of $sp^3$?
3) How does an unpaired electron factor into hybridization/how do you determine the hybridization state?
4) How can you tell if an electron is delocalized in a radical?
5) Why isn't the single electron delocalized in the propyl radical
6) Since the propyl radical does not have a delocalized electron, will the radical carbon be $sp^3$ hybridized
Answer
1) Yes
2) No, the delocalization of the electron is not the determining factor in the hybridization of the carbon atoms in the allyl radical.
3) The unpaired electron is not equivalent to an electron pair for purposes of determining hybridization state.  I believe that the hybridization state is driven by the steric stability of atoms, electron pairs and lone electrons around the hybridized atom, however the effects diminish as you move from attached atoms of different elements-->lone pairs-->unpaired electrons-->delocalized electrons.  In the case of the allyl or propyl radical, the effects of a lone electron are not enough to favor $sp^3$ hybridization whether it is localized or delocalized.  Therefore, in these molecules, you determine the hybridization state by the number of sigma bonds, but in molecules with lone pairs, the lone pairs may or may not alter the hybridization state depending on the stability gained and lost.  Orbitals in Chlorine atoms in $\ce{Cl2}$ do not hybridize, but I believe that Chlorine is considered to be $sp^3$ hybridized in some compounds if the steric interactions of the electron pairs with the attached atom are great enough.  Counting electron pairs and sigma bonds therefore is not a rule, but a guideline and ultimately it comes down to stability.
4) I think I would give a simple rule that as long as you can draw a Lewis diagram where a central carbon is attached to one atom with a lone electron and another atom by a double bond, the single electron will be delocalized.  What you are doing is moving electrons ONLY by making and breaking Pi bonds.
5) Because the central carbon does not have a double bond that it can break to avail itself of an unpaired electron that can pair with the unpaired electron on the radical carbon.
6) No, in this case at least, neither a delocalized electron or unpaired electron will result in the $sp^3$ hybridization being more stable and energetically favorable.  Now someone said that the propyl anion is still $sp^2$ but that is not what I am used to seeing.  Usually carbanions are shown as being $sp^3$.  The debate, I suspect will be whether it is $sp^3$ because the electron pair has enough effect to make the stability gained by a tetrahedral geometry pay off (my view) or if it is merely the function of non-geometry related "electronic factors".

@jJan Sorry, the first comment is not obsolete as I did state that the Chlorine atom in HCl may be SP3 hybridized.  It clearly is not.  The correction is helpful and I am going to edit my answer to correct it.

The only compounds in which I would consider chlorine hybridised would be chloronium ions.


text:
Anion-cation balance in particle phase
I'm now analysising the water-soluble ions in atmopsheric aerosol. With ultrasonic concussion and filtering impurity, the anion and cation from the particle sample are examined  in Ion Chromatography System(IC) separately.
We test $\ce{Cl-, SO4^2-, NO3-}$ and $\ce{Ca^2+, Mg^2+, Na+, NH4+, K+}$ as 8 major ion species to quantify the concentration level and seasonal variation in the sampling area.
But when I combine the anion and cation data together for same particle sample, the ratio of [ Total charge of anion]/[Total charge of cation] are much larger than  1.
In my opinion, the cation and anion inside particle should keep the balance of charge. The 8 ion species I choose should take up the majority of the total water-soluble ions (added with $\ce{F-}$, organic acid radical, etc).
I have also searched some paper, some results  of their ion-sample are also not keep the balance, but some observed.
Are the ratio of (+-)charges differed by sampling sites? The anion especially $\ce{SO4^2-}$ are larger comparing with the cation species. Are there any missing cation that I have not deteced?

Yes, you have left out one very important cation from your cation/anion balance: $\ce{H+}$
The primary pathway for new particle formation in the atmosphere is the photo-oxidation, via hydroxyl radicals, of volatile reduced sulfur species to form very non-volatile sulfuric acid.  For the most part then, new particles start out as puny droplets of sulfuric acid and grow via accretion, etc. from there.
I recommend measuring the pH of the filtered aerosols, and I strongly suspect that will account for your missing cations.
If I have missed the point of the question, please feel free to ask for any clarifications in the comments.


text:
Phenolphthalein indicator spill onto skin
I was using a phenolphthalein indicator to test the acidity of some 0.1 M HCl. I was using the pipette to draw the phenolphthalein indicator out of the tiny jar but it wasn't working, so I thought that it might not have much left in it, so I tilted it at an angle to try and get the remaining indicator. However it turns out the pipette wasn't working and it was full and a couple of millilitres spilt out, some spilt onto my right index finger, but most of it spilt onto the bench. I immediately rinsed my finger under running water for about 7-10 seconds. Just wondering if it'll have any negative effects. I'm pretty sure it was 99.9% methanol and 0.08% phenolphthalein measured in weight.

Sigh At least it's not about methanol this time...

It was during class. We didn't receive the MSDS or gloves. We were just using it as an indicator in an experiment and the dropper didn't work so I automatically tilted it as I thought there wasn't much left in the bottom. Turns out it was full and the dropper was faulty, so a bit got on my hand and on the bench. I immediately rinsed my hand  under water for about 10 seconds. We were wearing safety glasses and everything but I guess I just was a bit clumsy. This is why you get enough sleep at night.

Also it was a minuscule amount of phenolphthalein in an ethanolic solution.

Didn't you receive the SDS for phenolphtalein (or its methanolic/ethanolic solutions)?
Weren't you given instructions on how to handle it?
Didn't you wear suitable gloves in addition to your safety goggles?

Back in the days, phenolphthalein was apparently used as a laxative. This is no longer recommended!
However, the current GHS classification is as follows:

Germ cell mutagenicity, Category 2; H341
  Carcinogenicity, Category 1B; H350
  Reproductive toxicity, Category 2; H361f

(cited from the GESTIS database)
To my knowledge, phenolphtalein is on the REACH Candidate list of SVHC (substances of very high concern) since 2011.
My suggestion is:

Contact a physician.
Report the exposition to your supervisors.

If every time sth like this happens physicians would be contacted, they would ask about pay rise ;D

@Mithoron It depends on who you are and where you work. As a student at a university, nobody gives a f-word ;-). In a company with a stricter regulations, dermal exposition to CMR compounds, particularly when juvenile apprentices without gloves are involved, can get the supervisors in deep sh*t. Exposition of workers to CMR and TOST materials has to be recorded. According to current German laws, the records have to be kept for  40 years after the last exposition!


text:
What colour is clearish orange vitamin C solution supposed to change to when dark blue DCPIP solution is dropped into it?
I'm doing an experiment to study the effects of heat on vitamin C content.
I planned to titrate DCPIP (2,6-Dichlorophenolindophenol, Tillmans' reagent) solution into a fixed volume of vitamin C solution heated at different temperatures.
Basically I thought, the more DCPIP I used, the higher the vitamin C content.
I know DCPIP will turn pink  when acid like Vitamin c is dropped into it, then colourless once it's oxidized.
If my research is correct, the reverse should happen if I drop DCPIP into vitamin C; it's colourless or the colour of vitamin C, then once all the vitamin C has been oxidized the vitamin C turns blue because DCPIP is in excess.
However, when I tried it out, even in very small volumes of Vitamin C and large volumes of DCPIP, the solution doesn't change colour. Could anybody tell me the possible reason for this?
Perhaps I have made an error or my assumption is completely wrong. Please help.


text:
Water from coffee condensation
Why is it that from a cup of hot coffee, only the water condensates on the inside of the lid? Why aren't there any coffee particles? Does it only have to do with density or is it something else involved?

To condensate, a compound must evaporate first. Water does that, and coffee particles do not. This has nothing to do with density. Also, welcome to Chem.SE.

Yes, thank you:). I was thinking more about the actual evaporation process. I know that for water to evaporate, the molecules must have sufficient KE to overcome hydrogen bonding, but what happens to the kinetic energy of coffee molecules? is it just slowly given off as heat or is there even if just "theoretical" evaporation/sublimation.

I agree with Jan. The "coffee molecules", probably mostly alkaloids like caffeine, have larger and heavier structures than water molecules. I would expect most have dipole-dipole intermolecular forces, and (minor) hydrogen bonding may be present for some.
When you heat the solution with thermal energy, when we zoom in to the molecular scale we see this energy us actually kinetic. The molecules move faster, and some at the surface break free of their attractions to their neighbors. Despite water being held together by the strong hydrogen bonding intermolecular force, water molecules will be first and most to gain enough energy to pop out of the liquid solution. As Jan said, there are many more water molecules that coffee solute molecules. Also, the coffee solute's greater weights will make them harder to bring up to speed (though as a general rule of thumb IMFs like H-bonds actually play a more significant role than mass).
So those water molecules that pop out are far from their neighbors, so they constitute what we call a gas or vapor. There is much more water-gas above the solution than coffee-molecule-gas. This water vapor exerts a pressure based on how many molecules there are (and depending on kinetic energy / temperature), and we call this pressure from the gas above the coffee the "vapor pressure of the solution". The vapor pressure of water, despite being low relative to many organic liquids, is far higher than the vapor pressure of the molecules from coffee.
Think about the coffee residue in your pot, or the grounds - do they evaporate? Essentially no. There is essentially no vapor pressure, they are too bound by their IMFs to pop out and become gas (at temperatures a human usually encounters anyway). But water has a high vapor pressure, it is easily evaporated.
So when the gaseous water molecules encounter cooler surfaces, they will transfer some kinetic energy to the surface and transition from energetic gas state to lower-energy liquid state. This is the meaning of "condense". No coffee molecules condense, because none were in the gas phase to begin with (essentially).
Jan had a great point that these concepts are the foundation of the incredibly important process of distillation.

Let’s analyse the sequence of events.

You have a solution
part of that solution enters the gas phase
the part that enters the gas phase moves until
it finds a surface that it feels comfortable ‘settling’ on.

It should quickly be mentioned that water is the primary component of coffee by far, hands down. Thus, even assuming approximately equal distribution in the gas phase, there should still be a lot more water than anything else. But that alone doesn’t really answer it.
The key quantity you want to be looking at is vapour pressure. The coffee is hot (or at least warm), relatively close to water’s boiling point of $100~\mathrm{^\circ C}$ so the vapour pressure of water will be comparatively large. Most other components of coffee have a boiling point that is much higher and would be solids at room temperature. These two factors combined allow us to allocate them a low vapour pressure even at the temperature of hot coffee. Therefore, the amount of water leaving the coffee is even larger.
The basic principle behind this is distillation; a mixture of multiple compounds that can be separated by their vapour pressures (often correlating with their corresponding boiling points). Typically, liquids will be distilled away far before solids.


text:
Heavy water dielectric constant
Why is heavy water less dielectric than normal water?
Is it because of the fact that heavy water is more denser? I think it so because the dipole moment of heavy water is greater than normal water?

Greater dipole moment would imply greater dielectric constant in D2O. But its dipole moment is not any significantly greater, nor is its dielectric constant lower than those of H2O. There is no point in discussing a less-than-1% difference. Think of them as identical.


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Looking for a non-toxic substance that evaporates very quickly
Firstly I am not a chemist so please be kind. I am a creative technologist working on a large art installation. The challenge is to make people perceive rain without it raining. I am looking for something we can very occasionally drop from above that will feel like a rain drop but almost instantly evaporate leaving the person completely dry when they leave (they will only experience a few drops whilst in the room — the rest of the effect uses lights and sound).
My initial thought was liquid nitrogen but I assume that whilst a tiny occasional drop on the back of the hand would be fine, one in the eye for example would probably be dangerous. Plus the room would need to be heavily ventilated to prevent the risk of asphyxiation.
I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it. Over to you.

Um, how about, well, water?

I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it.

Nope ;-)
Your concerns are justified and I wouldn't expose the audience to anything but water either.
you can

use water from all above

and let the audience get wet
and hand out rain coats

use water and only let it rain to the left and right of a marked path or
apply a more sophisticated technique, as used by the random international art studio in their rain room. Here, it rains thoughout a whole room, but the visitors are tracked by cameras and whereever they move, the water is selectively switched off.

If you really don’t want to use water as suggested by Klaus but absolutely want to do this in exactly that way, the next-best alternative is probably diethyl ether. It evaporates really quickly ($\vartheta_\mathrm{b} \approx 35~\mathrm{^\circ C}$) and has been used in the past as anaesthetic. However, you would still need to keep the room well vented to prevent suffocation in the long run, and you need to enforce a really strict anti-smoking, anti-flame policy.
All things considered, if it is only the occasional drop I would definitely use water.

I would guess some of the Freons would be a good choice.  The boiling point and vapor pressure of say isoflurane is 45 C, and it is at least as non-toxic as diethyl ether.  It is far less flammable too.  But toxicity and asphyxiation risks would still be present.  It is also a horrible greenhouse gas.

If it's only a few drops, why not use water? It would dry quickly in a warm room.

You can look into Galden Fluid, such as HT70 or HT55. It's priced around $800 a gallon so if money is a concern, look elsewhere.

Almost all variants of these fluids are known skin irritants and should not be inhaled. I would not recommend any of these for a water substitute.

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


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Can vinyl alcohol react as a dienophile in Diels-Alder reactions?
For my organic chemistry exam I needed to synthesize cylcohexanone and in one of the steps a Diels-Alder reaction was required. I chose to use vinyl alcohol followed by an elimination reaction to get the desired product.
My professor said that this could not work because a dienophile needs to be electron deficient and that vinyl alcohol is electron rich. Is there any argument I can make as to why the reaction would actually work? It was a reaction of diene + vinyl alcohol to yield cyclohexanol.
The actual answer he said was to use two dienes, one as the diene and the other as the dienophile followed by a few subsequent reactions ending with ozonolysis to get cyclohexanone.

You professor is correct for common setting. Inverse electron demand is possible, but requires strongly electron-deficient dienes, which are not all that abundant and usually imply nitrogen or oxygen participating in -ene part.

Vinyl alcohol does not exist in appreciable concentration under normal conditions. It exists in a keto-enol tautomeric equilibrium, with the equilibrium lying far, far to the keto isomer (acetaldeyde) side.  However, you could use a vinyl alcohol "equivalent" such as vinyl acetate.  Vinyl acetate behaves like a typical dienophile in the Diels-Alder reaction.  After the Diels-Alder step you would need to simply hydrolyze the product to obtain the desired cyclohexenol.

Firstly, your Diels-Alder products are missing the double bond in the ring (formed between what was formerly C-2 and C-3 of the diene). You could always simply hydrogenate it down to the alkane, but I feel like this would present some selectivity issues in your professor's suggested synthetic route, especially since a terminal double bond is more likely to be hydrogenated than an internal double bond (sterics), and an internal double bond more prone to ozonolysis than a terminal double bond (more electron-rich).
Anyway, ron pretty much said what there is to say about "vinyl alcohol". It doesn't exist in any appreciable quantity, since the keto form acetaldehyde is much more stable (see another of ron's answers here):

Apart from what ron suggested (vinyl acetate), another alternative is to use 2-chloroacrylonitrile as the dienophile. The cyano group is electron-withdrawing, so the the usual requirement for an electron-poor dienophile is met. Treatment with hydroxide will unmask the ketone functionality, see for example the second and third steps in Corey's prostaglandin synthesis. (For a possible mechanism, see Mechanism for basic hydrolysis of α-chloronitrile to ketone?.) You can then hydrogenate the C=C bond chemoselectively to get the desired product, cyclohexanone.

2-Chloroacrylonitrile is what is called a "masked ketene". Ketenes don't undergo [4+2] cycloadditions such as the Diels-Alder reaction because they have a much greater propensity to react in a [2+2] fashion (even cyclopentadiene, an excellent diene in the Diels-Alder, reacts with ketenes via a [2+2]). So, these reagents are essentially surrogate ketenes: with one additional step (hydrolysis in this case), you can access what is formally a ketene+diene Diels-Alder adduct. There's a nice review of masked ketene reagents here.

Just wondering, how exactly does hydroxide unmask the ketone functionality? Two substitutions, substituting CN and Cl with OH, then the hydrate converts to ketone?

@TanYongBoon, there’s a link in the answer.


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Aurora borealis - less favored energy transitions higher in atmosphere?
This question describes how some energy transitions become more prominent as you higher in the atmosphere. Short summary: Some electron transitions are "forbidden" by quantum mechanical selection rules, and so have a larger time delay from time of excitation until time of photon emission. Thus, low in the atmosphere (high pressure), it's more likely that the species loses the excitation energy by collision than emission.
However, as you go higher in the atmosphere, you see red (due to energy transition in atomic oxygen). The characteristic green line (also due to atomic oxygen, but different transition) disappears.
Why don't you see a mix of the two colours? After all, from the collision rate argument you would expect fast transitions (like the green of atomic oxygen) to become even more prominent.
Why is it that some transitions are less likely as you increase in height?
(I have a feeling that the answer to this question will also explain why nitrogen does not play a more central role in the colour of the aurora.)

I'm not an atmospheric scientist, but I'm guessing it has something to do with the relative abundance of nitrogen/oxygen at different atmospheric heights. Correct me if I'm wrong, but the ratio of oxygen to nitrogen is not the same at all heights, right?

There is a variation. @DavePhD posted the following link in the referenced question: https://malagabay.files.wordpress.com/2014/07/daytime-ionospheric-and-atmospheric-composition.gif

Why don't you see a mix of the two colours? After all, from the collision rate argument you would expect fast transitions (like the green of atomic oxygen) to become even more prominent.

There can be a mix of colors.  However, according to Auroral electron energy derived from ratio of spectroscopic emissions Journal of Geophysical Research Volume 79, Pages 5181–5186, the ratio of the atomic oxygen versus molecular nitrogen bands are a function of "a characteristic energy of the precipitating electron flux".

Why is it that some transitions are less likely as you increase in height?

Well, consider Atmospheric Composition and Vertical Structure.  Figure 3 shows how dramatically pressure drops as altitude increases.  This means a change in altitude of 15 kilometers could reduce the concentration of the species of interest by a factor of 10.
In addition, dissociation of the $\ce{O2}$ and $\ce{N2}$ to atomic O and N increases with altitude.  See especially Fig. 6 which plots the points where monoatomic oxygen starts to exceed diatomic oxygen, and where monoatomic oxygen starts to exceed diatomic nitrogen in concentration.


text:
Could a lead-acid battery electrolyte be replaced by hydrochloric or nitric acid?
Hydrochloric acid, as well as nitric acid, are also strong acids like sulfuric acid. So, why are not they used commercially in lead-acid batteries?

HCl and HNO3 can't be used because they both would participate in redox reactions.

@DwiparnaDatta Well, there is a relationship between the cell voltage and the acid used, but for numerous reasons (like the redox participation mentioned by Ivan) the identity of the acid used has a huge effect on Pb-acid battery performance, far beyond just the achievable cell voltage.

In the lead battery it is very important that the electrodes transform into insoluble lead sulfate when discharged, especially at the cathode where lead(IV) oxide is turning into lead(II) sulfate. Consider the two half reactions of the lead-acid battery:
$$\ce{Pb(s) + HSO4−(aq) → \color{red}{PbSO4(s)} + H+ (aq) + 2e−}$$
$$\ce{\color{red}{PbO2(s)} + HSO4−(aq) + 3H+ (aq) + 2e− → \color{red}{PbSO4(s)} + 2H2O(l)}$$
Lead nitrate is soluble, so metal ions would go into solution. The lead(IV) oxide would not form again from these ions on charging, metallic lead from the electrode would dissolve. The battery would no longer be rechargeable.
Lead chloride is insoluble, but lead(IV) oxide on the cathode would oxidize chloride into chlorine. Thus no more battery.

Along with the battery issues identified by Stapke, there is the matter of toxic emissions.  The lead-acid battery with sulfuric acid just undergoes reactions involving the lead and gives contained, nonvolatile products.  By way of contrast, hydrochloric acid could be oxidized to chlorine gas at the anode and nitric acid could be reduced to nasty nitrogen oxides at the cathode.  We would not want such fumes coming from car batteries, especially when we already have to deal with the exhaust from most cars burning gasoline.


text:
Does the carbon dioxide dianion exist?
I have heard that the dianion of carbon dioxide, $\ce{CO2^{2-}}$ was named carbonite.
Does such a species really exist? Basically, is it stable enough to be a plausible molecule? If so, under what conditions is this true?
This question is very confusing to me, I understand the concept somewhat but I don't know what information I need in order to figure out the answer.

Hard to judge questioner/reader chemistry introduction level, hard to judge likeliness of possible confusion with $\ce{CO_3^{2-}}$ carbonate.

The obvious Star Wars puns aside, I have never heard about the dianion of a carbon dioxide dianion and/or that it was named carbonite.
However, the carbon dioxide radical anion, $\ce{CO2^{.-}}$, is experimentally available.
Typically, it is generated by hydrogen atom abstraction from formate, $\ce{HCOO-}$, using $\ce{HO^.}$ radicals. The $\ce{HO^.}$ radicals, in turn, are obtained from vacuum ultraviolet (VUV) irradiation of water at a wavelength of $\lambda = 175~\mathrm{nm}$.

UPDATE
Apparently, I was wrong about the dianion! As Mithoron pointed out in his comment, species like $\ce{(M^+)_2CO2^{2-}}$, formed in the reaction carbon dioxide with alkali metals in argon or nitrogen matrixes, have been detected by IR spectroscopy. See Inorganic Chemistry, 1984, 23, 177-184 for details.

It looks dianion also was made https://www.researchgate.net/publication/231343327_Carbon_dioxide_activation_by_alkali_metals_2_Infrared_spectra_of_MCO2-_and_M22CO22-_in_argon_and_nitrogen_matrixes

$\ce{CO2^{\bullet -}}$ is also thought to be a key intermediate in the electrochemical reduction of carbon dioxide. (e.g., PDF link)

@hBy2Py Brian, that's interesting :) Actually, I think that I cited one of the authors, Jean-Michel Saveant, more than once. I always went the other way: oxidation of carboxylates to the corresponding radicals, followed by decarboxylation.

I'm sorry, @Mithoron, but that is just preposterously cool.

Carbonite di-anion may exist not only in low temperature matrices.  Wikipedia reports that it may be a surface species when carbon monoxide is adsorbed onto some oxide surfaces including line and magnesia.

Quick note:  In your edit you wrote that the counterion of the carbonite di-aniin salt was $\ce{M_2^{2+}}$, but when I asked about the existence of such cations I got comments that this was wrong.  The abstract apparently had been poorly written and was misread.  Properly there are two separate monatomic alkali metal cations per carbonite ion, and I corrected the nomenclature.


text:
Why is potassium phosphate KH2PO4 in this reaction?
This is a continuation of this question because the first thing that came in my mind is that why potassium phosphate in this reaction is $\ce{KH2PO4}$ and not $\ce{K3PO4}$?
In the wikipedia article of phosphate ion ($\ce{PO4^3-}$), it is written that:

Aqueous phosphate exists in four forms. In strongly basic conditions,
  the phosphate ion ($\ce{PO4^3-}$) predominates, whereas in weakly
  basic conditions, the hydrogen phosphate ion ($\ce{HPO4^2-}$) is
  prevalent. In weakly acidic conditions, the dihydrogen phosphate ion
  ($\ce{H2PO4−}$) is most common. In strongly acidic conditions,
  trihydrogen phosphate ($\ce{H3PO4}$) is the main form.

So, $\ce{FeCl2}$ must somehow be acidic in aqueous solution so as to protonate potassium phosphate $\ce{K3PO4}$ to $\ce{KH2PO4}$. Is this statement true? This question throws some light on acidity of chromium chloride in aqueous solution but is it applicable to iron chloride as well?

In the previous question, the homework question was write the equation of reaction of Iron(II) chloride and potassium phosphate. So, students can assume the reaction to be $\ce{FeCl2 + K3PO4}$ to simply yield iron(III) phosphate or iron(II) phosphate and potassium chloride but why make the reaction complex by assuming potassium phosphate $\ce{KH2PO4}$.
Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, can we just assume that reaction of Iron(II) chloride and potassium phosphate to yield iron (II or III) phosphate and potassium chloride?

Is there a reason students will assume that the Fe(II) ions are going to be oxidized to Fe(III) by phosphate?

I think that’s just another mistake the OP of the other question made. Potassium phosphate is $\ce{K3PO4}$ and any student using any hydrogenphosphate instead would lose marks if I were marking the exam.
If $\ce{KH2PO4}$ is meant to be added, use the correct term: potassium dihydrogenphosphate.
This nomenclature can be found in the current version of IUPAC’s Red Book, sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.

Potassium phosphate does not refer to K3PO4 but refers to 3 different compounds whose actual IUPAC names are tripotassium phosphate, potassium hydrogen phosphate, and potassium dihydrogen phosphate. The rules for IUPAC naming were made to avoid this sort of confusion. The term "potassium phosphate" should never have been used.

Also never in my years of synthetic work have I ever come across tripotassium phosphate in a lab before.

@EframGoldberg No, that is not the case. Similar to sodium carbonate, which is $\ce{Na2CO3}$ and nothing else, potassium phosphate is $\ce{K3PO4}$ and nothing else. There is no need to add a tri-label as only one phosphate is possible (only one common oxidation state is known for potassium).

Irregardless of your opinion, the common name potassium phosphate has been used to describe all three compounds even requiring a wiki landing page for the term further specifying the three salts by IUPAC name. The three salts are commonly called potassium phosphate monobasic, dibasic or tribasic and are sometimes shortened in use. This and the fact that tripotassium phosphate is hardly encountered, are two important reasons to use IUPAC naming.

The analogy to sodium carbonate is not valid due to the common names being very different ie sodium carbonate and sodium bicarbonate, no one calls them sodium carbonate monobasic or dibasic. If both compounds could be referred to by the common name sodium carbonate, then IUPAC naming should be used as well, but that is not the case.

@EframGoldberg Please see the IUPAC Red Book; a link and the relevant sections are now provided in the answer. Potassium phosphate is $\ce{K3PO4}$ and nothing else. Case closed.

Thanks for the link, I can now point you to the exact section so you can see where you are wrong. Section IR2.13 Multiplicative Prefixes. "The number of identical chemical entities in a name is expressed by a multiplicative prefix." See example 5. Ca3(PO4)2 is tricalcium bisposhpate.

also the CA entry for CAS 7778-53-2 is phosphoric acid, tripotassium salt.

@EframGoldberg IR-5.2: ‘Multiplicative prefixes need not be used in binary names if there is no ambiguity about the
stoichiometry of the compound (such as in Example 10 above).’ Also, CAS uses a completely different nomenclature.

Looking at example 10, both tricalcium diphosphide or calcium phosphide are acceptable. The prefixes are considered superfluous unless describing compositionally related compounds which in this case holds true, as the term potassium phosphate has general usage for all three different compounds K3PO4, K2HPO4, and KH2PO4.

@EframGoldberg I urge you to take a look at the table I refer to in the post. You will see that phosphate clearly refers to the $\ce{PO4^3-}$ ion, hydrogenphosphate (the space I initially had in there was incorrect) to $\ce{HPO4^2-}$ and dihydrogenphosphate to $\ce{H2PO4-}$. In no way is it correct to refer to $\ce{KH2PO4}$ as potassium phosphate, since the entity ‘phosphate’ is not present. (The entity ‘dihydrogenphosphate’ is.)

I never said that was the correct name for KH2PO4, I said that (independent of the validity of doing so) the term potassium phosphate has been used to refer to all 3 compounds, and to avoid ambiguity the multiplicative prefix should be used. I also said this probably came from the fact that the compounds are sometimes called potassium phosphate mono, di, or tribasic... and likely people dropped the last part of it.

@EframGoldberg And my point throughout this entire discussion is that doing so is contradicting IUPAC’s nomenclature rules. I don’t care whether it is common; it is wrong. Wrong habits need to be combatted; at best it must be taught why they are wrong. Hence my answer in its current state.

H3PO4 is an acid usually present as 85% solution in water, while KH2PO4 is a solid and therefore easier to handle, weigh etc. K3PO4 is not a commonly used reagent in chemistry.
H3PO4 is a polyprotic acid with pKa values as follows, 2.1, 7.2, 12.7. It will only be mostly PO4-3 at pH>12.7. Using the definition of pKa, a value of 7.2 for pKa2 and assume pH = 7, and your calculations will explain why the ferrous chloride does not need to make the solution acidic to have H2PO4-1 exist as one of the major species.
I am still not sure exactly what you are asking.

If you’re not sure what is being asked, you should comment to ask for clarification.

@Jan I answered 3/4 questions that I could make out which were posted above, but the post is not asking a single question. There is a question about why a reactant was chosen, a question about speciation and pH, a question about Lewis acidity of metals, a question about oxidation reduction and general requests for clarifications. Perhaps the post can be edited to present only one clear question?

Just to reiterate: Yes, that should have been a comment on the question (ideally before answering).


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Which carbons in pyrrole have the highest electron density?
In pyrrole, which carbon atoms have the greatest electron density?
In the resonance structures of pyrrole, all carbons can formally bear a negative charge, so it's not immediately obvious which one is most electron-rich:

My book claims that because resonance forms 1c and 1d are greater contributors to the resonance hybrid than 1b and 1e, the α-carbons (which bear the negative formal charge in 1c and 1d) are more electron-rich. Is this really the case?

Interesting question, more subtle than I realized!
Theoretical approach: I would turn to resonance and induction as the rationale for electronic properties. I think all of your resonance forms may contribute about equally to the resonance hybrid. This suggests that all four carbons would share roughly the same amount of electron density. However, we must also consider the inductive effect of the nitrogen atom, which draws some electron density away from C2 and C5 through the sigma bonds. This would imply that the two proximal carbons (closest to N) are more electron-deficient, and the distal (more distant) pair of carbons are more electron-rich.
Experimental considerations: Orthocresol has a good point that pyrroles are observed to form bonds with electrophiles at the proximal carbons (2/5). However, I think this trend may be due to the superior stability of the intermediate produced by that pathway (3 significant resonance forms) versus a distal (3/4) attack (2 significant forms). The cation from a proximal attack is better stabilized by resonance than the distal-attack cation. I expect that pyrrole forming C2 bonds with electrophiles has more to do with the stability of the intermediate of that pathway than necessarily greater electron density on C2. What I did not expect, but read on Wikipedia, is this:

Unlike furan and thiophene, [pyrrole] has a dipole in which the positive end lies on the side of the heteroatom, with a dipole moment of $1.58\ \mathrm{D}$.

This is experimental evidence demonstrating that the electron density is lower on the proximal side (C2,5) and greater on the distal side (C3,4). Besides molecular modeling of the electron density clouds, I think the molecular dipole moment is about the most conclusive evidence one can hope for.
It really comes down to the context: in an undergrad first-semester course, you might be expected to say the distal carbons have more density because the nitrogen withdraws from the proximal carbons. In more advanced courses, you might take orthocresol's perspective and base your answer in reactivity, concluding the proximal have more electron density. But I expect a doctorate researcher would probably conduct molecular modeling and compute the electron distributions using quantum mechanics, and perhaps conduct a few experiments to measure the dipole. And at that level, I think C3 and C4 have the honey.

But the key has the answer as C2 and C5 by following the inductive effect (may be 'cause I'm still in high school C3  and C4 and dipole moment were all neglected or remain unnoticed) so shall I take C2,5 or C3,4? And thank you for the detailed answer, it helped a lot :)

That is bizarre. Unless there are more than college-level concepts at play here that I'm not aware of, I would say with certainty that the inductive effect would /lessen/ the electron density at C2/5. The  nitrogen withdraws electrons from the proximal carbons due to its higher electronegativity. I cannot see how any "inductive effect" here could make C2/5 more electron-rich. This is highschool, you say? Perhaps your teacher is mistaken. I would be interested to hear their logic.

There is no 'most stable resonance structure'. These are all part of the same description as a singular structure is an incomplete description.

Of course Martin, it is only a colloquial phrase for intuitive rationalizations. Would you prefer "the relative weights of the molecular orbital wavefunction vector components are likely nearly equal in magnitude"?

@electronpusher That's a silly question: Yes, of course. A bit simpler would be all resonance forms have a relative similar contribution to the overall bonding situation. I find such colloquialism very bad, as for the untrained eye and mind they suggest something that isn't true at all. And especially this association continues to linger around. And that is very bad, since it is much harder to unlearn something than learning something new.

This question was asked in NEET exam ( entrance exam to get into med school equivalent to MCAT) and answer was given as electron density is rich in c2 and c5 atoms. And i completly agree with martin its really difficult to unlearn something@electronpusher

That is really not a fair question and even with a computational approach the differences are more or less marginal. The approach by electron pusher is certainly the way to go, and here is the computations to back it up.
See if you can find the carbon with the highest electron density in the following plot showing the value of the electron density for the calculation on the DF-M06L/def-TZVPP level of theory.

I have then performed an analysis of the electron density with the Quantum Theory of Atoms in Molecules (QTAIM). The following shows the Laplacian of the electron density, with the zero-flux surfaces in blue that divide the molecule.

When we integrate over the electron density in these basins, we know at which atom "are the most electrons", subtracting the nuclear charge, we get the overall charge. This should tell us, which carbons have the higher overall density.
  1 (C )    Charge:    0.015912     Volume:    83.414 Bohr^3
  2 (C )    Charge:    0.406209     Volume:    76.122 Bohr^3
  3 (N )    Charge:   -1.248690     Volume:    95.264 Bohr^3
  4 (C )    Charge:    0.406204     Volume:    76.122 Bohr^3
  5 (C )    Charge:    0.015911     Volume:    83.414 Bohr^3
  6 (H )    Charge:   -0.012007     Volume:    48.338 Bohr^3
  7 (H )    Charge:    0.004155     Volume:    47.506 Bohr^3
  8 (H )    Charge:    0.004162     Volume:    47.506 Bohr^3
  9 (H )    Charge:   -0.012006     Volume:    48.338 Bohr^3
 10 (H )    Charge:    0.420149     Volume:    28.032 Bohr^3

And for reference we need the numbering in the table:

TL;DR (1) The α-carbons to the nitrogen have a higher positive charge, hence the lower total electron density than the β-carbons.
Let's have another look at the π-orbitals to see where the reactivity towards electrophiles comes from.

TL;DR (2) The highest electron density of the HOMO is at the α-carbons; electrophiles react there.
Please don't nag about the numbering, I recycled the calculation from another post, where I did not pay much attention to that. Sorry.

Ad. I agree with the analysis of the dipole moment since all $\ce{C/N-H}$ bonds are in plane. The calculation predicts 1.9 D (away from nitrogen, see here for the direction confusion). This is quite nicely explained in Ron's answer to Pyrrole and furan dipole moments.

The electron density on a particular atom not only depends on resonance but also on inductive-effect. And in case when resonance structures doesn't give a clear answer to the question we go for other effects like inductive effect, no bond resonance etc. So, here by just looking at resonating structures you will not get the answer — for that go for inductive effect and then you will see a clear difference between the electron-densities on different carbons of pyrrole.

Out of curiosity, how do you explain that when pyrrole donates electrons to electrophiles, the bond forms on C2/5?

It has been already explained in any of the above comments...the intermediate carbocation is resonance stabilised to a greater extent than that in case of an attack at 3/4 or can say at beta position. For more you can go through Gilchrist's book on Heterocycles.Thank You.

So by inductive effect the proximal(close to N) carbons are of more electron dense, right?

No..it is not so...the attack is there due by this a more stable intermediate is formed... But it doesn't mean that 2/5 have more electron density.

So you say  the distal carbon will be more electron densed as cited in electronpusher's answer?

Yeaah! Two factors are just confusing you...in case of electron density only resonance can't give you a clear picture hence you will go for inductive effect... But when you will start looking at the site of attack you will have to look at the stability of the resulting intermediate.

Wah!! Thanks I got it cleared :) but do give a look at Martin's answer for this question.he has explained it in a new perspective.. :)

@user38876  [img]https://www.physicsforums.com/attachments/1489864484687-jpg.114700/[/img]Here the answer is given as 2 and 5 contrary to what we have discussed.could this(2 and 5) be correct?

@user129402 Link is now broken. I wonder if anyone reached out to the folks there to clear up the discrepancy. I have a hard time believing they would come up with enough evidence to reverse what we have concluded here (courtesy of Martin).


text:
Why do some nucleophiles attack at the carbonyl group while others at an alpha-carbon in ketones?
I was reading about the iodoform reaction and it stated that $\ce{OH^-}$ attacks at the $\alpha$-carbon and makes a substitution reaction possible. But there is also the possibility of attacking the carbonyl group directly and forming a gem-diol. Why doesn't that happen?

For starters, you shouldn’t call it a nucleophilic attack at the α carbon; it is a deprotonation (something quite different since the latter doesn’t access the carbon atom in any way).
The second thing to note is that both do indeed happen. However, the nucleophilic attack on the ketone is not productive: nothing will happen that creates a new product or a different pathway. The only productive result of hydroxide’s attack onto the carbonyl carbon in the long run is the re-elimintion of either of the geminal hydroxy groups to reform the carbonyl and we’re back at the start.
Please don’t stop thinking like this (‘what happens when $\ce{X}$ attacks $\ce{Y}$ rather than $\ce{Z}$?’) — it’s a great way to spot side reactions, alternative pathways, problems with functional groups etc.


text:
Is Changing between Allotropic forms a Physical or Chemical Process?
When a substance changes its alloptropic form, is this considered a physical change or a chemical change?
Right now, I am thinking that it would depend on what the change was. For example, if graphite was changed to diamond under high pressure, I would say it is a chemical change since new bonds are formed. However the change of iron from a body-centred cubic structure (ferrite) to a face-centred cubic structure (austenite) at high temperatures would be a physical change because only the arrangement of the atoms are changed.
Am I correct?

Why do you think a physical process is not a chemical process (or vice versa). The electronic structure of iron changes between the two phases, as does the crystal structure. The same can be said for graphite to diamond. You are splitting hairs at best. So, no, I do not believe you are correct.

It is a bit of a tautology, what is a chemical change.  A chemical change breaks bonds therefore bonds are the forces that get broken in a chemical change.  A chemical change changes the substance therefore I will define this change as not being chemical, and define these two samples as being the same substance.  Chemical changes change the characteristic properties of a sample-I will call this a physical change and therfore whatever properties changed, they were not "characteristic".

I would call your crystal change as chemical because the densities are quite different, but there are many forms of ice that have different densities http://www1.lsbu.ac.uk/water/ice_phases.html  and density of water changes with temperature so density must not be a characteristic propery of $\ce{H2O}$

but @JonCuster is right.  The separation of physical and chemical changes is a bit simplistic.  We should instead perhaps ask if it is a change that will spontaneously reverse with the addition or removal of heat.  Maybe we could ask whether the fundamental base unit of the substance has changed.  The base unit of austenite is different from the base unit of ferrite.  Not true for forms of ice.  Also, are minerals pure substances or approximately pure substances.

Well, it's a matter of choice and tradition. We traditionally classify phase transitions (such as melting of ice or evaporation of water) as physical changes. The conversion of ferrite into austenite is a phase transition, so it should formally be classified as a physical change. But yes, it's hair splitting indeed, especially when you consider the phase change from diamond (giant covalent lattice, sigma-bonds only) to fullerene (simple molecular lattice, sigma plus delocalised pi-bonds), which clearly shows some characteristics of a chemical transformation (the change in the bonding nature). Still, even such changes are formally "physical". The traditions do not go easily!


text:
Why can wood burn all the way to charcoal instead of just creating a passivating charcoal layer?
I know burning to charcoal is actually incomplete combustion because charcoal can also burn. When a strike anywhere match burns, it burns all the way to charcoal but the charcoal doesn't further burn. How can all of it burn to charcoal? According to https://en.wikipedia.org/wiki/Pyrolysis, the heat causes pyrolysis which undergoes flaming combustion generating more heat. Once some charcoal has already formed, why doesn't it just passivating the wood blocking further endothermic pyrolysis. I know wood is porous but at the small scale, should't it's surface become passivated by charcoal blocking further pyrolysis? Does it occur for a similar reason that iron can continue rusting after it already formed a rust layer by producing a porous rust layer? Could nonporous single crystal cellulose also burn to charcoal? If so, can you give me a detailed mathematical description of how it would continue burning to create a thicker charcoal layer and how iron can just keep on rusting?

Charcoal doesn't passivate and the results you see depend on the physical structure of the fire
The difference between a match and a larger wood fire has little to do with the formation of charcoal and more to do with the circumstances of the reaction. And the charcoal isn't really passivating in the reaction on a match.
Wood consists of a complex mix of chemicals as well as the core cellulose, which is a carbohydrate. Some of the ingredients, like the sap and other volatile components, burn easily. Cellulose burns in multiple steps one of which involves pyrolysis where some components are driven off quickly leaving mostly charcoal. Charcoal itself will burn but it takes more energy to ignite and therefore more time to get going. It isn't so much that charcoal passivates, it is just that it is the last component to burn.
The difference between a match and a larger wood fire isn't about passivation but about the scale of the fire. There isn't enough energy in a match head to drive combustion all the way so only the easy-to-burn components ignite (unless you hold the match upside down so the flame spreads up the matchstick when the heat from the tip is enough to cause much more extensive combustion). In a larger wood fire there is enough energy in the flame to cause more complete combustion. So the volatile components burn quickly but leave enough heat for the charcoal to burn as well. Since heat penetrates charcoal very well, the volatile components under the charcoal will tend to ignite before the charcoal itself (it isn't doing much to passivate the wood underneath). In a badly organised fire the overall heat may never be enough to set off the charcoal.
So the issue isn't passivation just a reflection of the order the components of wood burn and the physical structure of the fire. Charcoal isn't passivating, just the last component to burn and, if the fire isn't well constructed, there may never be enough heat to burn it. Much more of a match will burn if you drive the flame along the stalk by holding it upside down and a well organised fire will burn the whole way to ash.
Also controlled pyrolysis (which is used to create charcoal deliberately) relies on controlling the amount of air available to the work being pyrolysed. If there were enough air, the charcoal would burn but the conditions for pyrolysis involve restricting the air so the end-product cannot burn. Charcoal doesn't passivate anything when there is enough air (which is why you can cook on a barbecue).

Charcoal is nearly pure carbon, and its vapor pressure is miniscule.   So, rather than pyrolysis, it burns slowly, by converting at the surface to CO (or maybe
CO2, somewhat less quickly); the CO then burns to CO2 rapidly, above the solid
carbon surface.   Because the oxides are gasses, the carbon surface remains
clean, and nearly pure carbon.  It burns by corrosion in the presence of hot O2.
As for iron, iron can form black 'scale' which adheres and protects against
further corrosion (but not very well).   It can also be 'blued' to form
a hematite or magnetite thin film, of somewhat better purity and adhesion than
the black scale.   Either of these coatings can protect against further rust,
but they are brittle and do not always re-form after damage.   The more usual
rust is brown rust, which is a combination of oxides and hydroxides (and
chlorides as well).   This can be converted to black oxides with mild
acid treatment.   Again, these oxides are brittle and can be damaged,
after which environmental water (even water vapor) can be expected, with
oxygen, to penetrate any cracks and crevices, so the metal underneath
is continually under attack.
Some alloys of iron DO form a tight surface of 'rust', notably Cor-Ten steel CorTen.com
and wrought iron.   There is an iron pillar in Delhi that has attracted
some attention Ancient iron for its longevity (over a millennium
outdoors).

I still don't understand how the pyrolysis occurrs when there's charcoal covering the wood. Is the charcoal layer kept so thin by combustion that pyrolysis occurs by diffusion through it. Does the wood get under higher pressure as a result of trying to decomposes into gases thereby keeping any new charcoal that forms porous. Surely, if all the tunnels in the formed charcoal were as small as molecues, there would be way to much resistence to the movement of gas making it be under very high pressure. Does the escaping gas also produce larger tunnels as well making the charcoal like a fractal?

Pyrolysis occurs due to high temperature, which the charcoal does not prevent.   The charcoal is created in porous form (it shares the tubular structure of living wood, to some extent), so hot gasses pass through its pores.   When heated, steam pressure in wood can cause fracturing: in the debris around Mt. St. Helens after its eruption in 1980, entire trees were converted to toothpick-like splinters.

Is wood porous all the way down to the molecular level or is its sponge like structure nothing like as small as molecules? If its sponge like structure is not that small, does some wood remain under the charcoal after the charcoal forms? If not, how does all the wood undergo pyrolysis?

Wood contains sugars, resins, etc. that pyrolize ("wood alcohol" and turpentine are produced by heating wood).   There is also a residue of carbon which we call charcoal.   One expects that a wood fire includes pyrolysis as well as oxidation of the residual (charcoal) non-pyrolized carbon.

The quandary of why burning wood doesn't burn completely is the heat at which the wood is burning. Wood isn't a pure chemical, but rather has all sorts of chemicals within it. So some of the more volatile chemicals burn off at lower temperatures. In fact it is the volatile chemicals burning in the air that give the "flame"  above the match.
If you put the match residue in a crucible and heated the crucible red hot with a Bunsen burner, then you could burn the match residue completely to a powdery ash.

That's not true. According to https://en.m.wikipedia.org/wiki/Pyrolysis, it's  gases released by a chemical triggered by heat which then react with oxygen, not the vapour of a chemical wood already has in it.

Yes, I over simplified the explanation. Wood is mostly cellulose and the cellulose is degraded by heat to release volatile gases which then burn. My explanation  doesn't explain why the cellulose only burns partly.


text:
How to determine the stereochemistry after elimination of anti-1-bromo-2-methylcyclohexane?

The question is asking for the product and the answer is C.
My expected answer was E. Here's how I tried to solve it:
Since Br (the leaving group) is at the equatorial position, the structure would 'ring flip'.

Since $\ce{CH3}$ is pointing downwards, I assumed it would have methyl group point outwards (dashed), however, the correct answer is a wedge.
After elimination, how would we know whether the substituent would be pointing inwards, outwards, or even right on the plane? (no wedge or dash)

Related: E2 Elimination; Anti-periplanar beta hydrogen

You forgot to rotate the molecule over so that the double bond is on the correct side.

You diaxial intermediate is good and you can clearly see that there is no way to achieve a Zaytsev product in an $\mathrm{E2}$ elimination since that would require eliminating the methyl group. Carry on with the elimination and then you have a double bond in the top half of the molecule and the methyl group pointing downards — molecule E if inverted by a vertical plane of symmetry traversing the $\ce{C-Me}$ bond.
But the product molecules are drawn with the double bond on the bottom. Thus, you need to rotate the molecule by $180^\circ$. If you do this with a rotation axis in the paper plane (again one parallel to the $\ce{C-Me}$ bond), the double bond is in the correct position but the methyl group is pointing upwards.
You can verify this by determining the absolute configuration of the ipso-methyl carbon: it should be (R) according to your chair drawing and it is (R) in answer C.


text:
Catalytic energy profiles (is this Wikipedia image misleading?)
I am looking at the image below from the Wikipedia page for energy profiles in chemistry. I do not understand why the "$E_a$ (with catalyst)" is labeled as it is. I would argue that is not the activation barrier with the catalyst. Shouldn't all of the individual steps need to be considered to get the rate of production of product (or, if one step has an especially high barrier, then that step can be assumed to be rate-limiting and that step's $E_a$ is the barrier with the catalyst)? I suppose in other words, if I do $r = A\exp(-E_a/k_{B}T)$ where $E_{a}$ is the activation energy with catalyst, that will not get me the rate, right?

For the record, I have taken grad-level courses in chemistry, but I got myself into a position of questioning everything I once learned and here I am.

This image has actually been asked about before: A misunderstanding about the energy profile of reactions with a catalyst involved

Some other mildly related questions: Does a reaction have to have a rate determining step?; Can in any case the faster step of the reaction be rate determining? I could swear we have a question on "what is the activation energy of a multi-step reaction", but I can't find it.

Of course, in a strict sense the reaction path using a catalyst should be split into four sub-reactions (all going by the exact image) and each of those sub-reactions should get their own activation energy assigned. Two of the four steps would be endothermic, two exothermic. (Note that it is not necessary for the endothermic steps to occur first; it could be the other way around, mixed or anything.)
However, there is still some truth in labelling ‘$E_\mathrm{a}$ with catalyst’ as such. Even though you have much smaller steps you will still need to add that labelled energy in net terms to overcome the barrier. The principal difference is that it need not be added all at once but can be added stepwise. You could even say that the total activation energy (i.e. adding up all the individual $E_\mathrm{a}$) be even higher, but you regain some of that from the intermediate reactions.
If you decided to supply only enough energy so that a molecule could jump the barrier from intermediate 2 to intermediate 3, that would not make the reaction go. For one, the barrier to get to intermediate to is larger by itself. For two, you would only get that much above the baseline which might be enough to cross the barrier to intermediate one but no further.

I agree with all of that, and the rational for the labeling does make some sense that way (although not in the context of computing a rate). I now have a follow-up question though. If instead of 4 steps I have 100 steps (everything else otherwise the same), with the highest step having the same height as the 3rd transition state shown in the Wikipedia image, will that impact my rate of production of product? Yes, right?

@Argon No. The barrier is the largest difference from a stable species (reactant or intermediate) to any subsequent transition state. Having lots of steps doesn't matter because interconversion is fast if the barriers are small.

@Zhe Doesn't that implicitly assume I am dealing with a rate-limiting step? I'd prefer not to make the assumption that one step is by and far much larger than the others, just that there is indeed a maximum barrier height.

It still works because the other steps will be under equilibrium. All you need to assume is that you have enough energy to surmount the barrier. Since energies add, everything just works out.

@Zhe Thank you for your replies. I am going to have to convince myself of this. I ask because the question that motivated me to even submit the original question was the following. Is the rate the same between: Case 1) the reaction proceeds in one step with a barrier height that goes to $E_{a}$ compared to the reactant; Case 2) That same reaction is actually two sequential steps, where the highest of the two peaks reaches a height $E_{a}$ compared to the reactant. Everything else is identical. From your answer, it seems the answer is yes, the two cases have the same rate.

You seem to know what you are doing, so let me help you by proposing the following problem to work out:
You have two path ways, one with barrier $E$. The second pathway has a higher energy intermediate. The barrier to it is $E_{1}$. This intermediate converts to product, via a second barrier such that the peak matches that of the barrier for the single step reaction. Compute the rate for both reactions and assume that there is a pre-equilibrium in the two step case (this is probably the case anyway). Ping back if you get stuck.

Let us continue this discussion in chat.


text:
When are two orbitals orthogonal?
When are two orbitals considered to be orthogonal?
It will be helpful if you can provide an example molecule (or molecular orbitals in a molecule) since I can't really think of a scenario where molecular orbitals can be exactly 90° to each other.
I'm not well-versed in quantum mechanics, so would prefer a simple answer - but all explanations are welcome.

Orbital means one-electron function. Typically their orthogonality is made by construction.

One atomic (or molecular) orbital is said to be orthogonal to another atomic (or molecular) orbital if there is no interaction between the electrons in one orbital with the electrons (wavefunction) in the other orthogonal orbital.
Acetylene is an example of a molecule in which the two pi bonds are orthogonal to each other (see the light-blue pi bond and the purple pi bond in the figure below).  The two pi bonds are perpendicular to each other and there is no interaction of the electrons in one of acetylene's pi bonds with the electrons in the other pi bond.

By the same token the two p orbitals on one of the carbons are also orthogonal to each other.  The electron in one p orbital does not interact with the electron in the other p orbital on the same carbon.
In ethylene the pi bond is orthogonal (perpendicular) to the sigma skeleton as shown in the figure below.  Since the orbitals are orthogonal there is no interaction between the sigma and pi electrons (wavefunctions).

On the other hand, orbitals that interact and form a bond are not orthogonal. The two ${sp^2}$ orbitals on the two carbons in ethylene that overlap to form the $\ce{C-C}$ sigma bond are not orthogonal; the electrons interact and form a bond.

This answer leads to impression that orthogonal orbitals must not overlap, but it is wrong: e.g. $1\mathrm s$ and $2\mathrm s$ orbitals overlap considerably, but are orthogonal.

@Ruslan Thanks for clarifying.  I've edited the answer accordingly, hopefully it reads better now.

@Ruslan Related question and answer here.

I think it is wrong to say that the electrons in the pi bonds in acetylene do not interact.

@ron you seem to be confusing spatial orthogonality with mathematical orthogonality. And those drawings grossly misrepresent the shapes of pi orbitals such that it appears there is no spatial overlap when in fact there is a great deal (but no net overlap)

Unfortunately, the sense in which orbitals are orthogonal is more or less impossible to define rigorously without recourse to functions of some kind.  So, I'll give an explanation a shot using some simple, 1-D functions to illustrate the concept, followed by the pictorial orbital example you've asked for.
At a basic level, in order to have any two functions be orthogonal, you need to define a "mixing rule" for them. The most useful mixing rule for orbitals turns out to be integrating the product of the two functions$^\dagger$ over some defined domain.
So, consider two real 1-D functions, $f(x)$ and $g(x)$, defined over an interval $x=[a,b]$. The mixing rule $\Pi\,_a^b\!\left(f,g\right)$ I'll define as follows:
$$
\Pi\,_a^b\!\left(f,g\right) \rightarrow \int_a^b{f(x)g(x)\, dx}
$$
Simply put, two functions $f(x)$ and $g(x)$ are orthogonal under the mixing rule $\Pi\,_a^b$ (or any other mixing rule!) if and only if $\Pi\,_a^b\!\left(f,g\right) = 0$.
A pretty standard example of two orthogonal functions in this framework would be the following:
\begin{align*}
  f(x) &= \cos(2\pi x) \\
  g(x) &= \sin(2\pi x) \\
  a    &= 0 \\
  b    &= 1
\end{align*}
The integral to check from the above is:
$$
\Pi\,_0^1\!\left(f, g\right) \rightarrow \int_0^1{\cos(2\pi x)\, \sin(2\pi x)\, dx}
$$
Since this is Stack Exchange and not a math exam, I will now cheat and just link to Wolfram Alpha to demonstrate that the above integral is, indeed, zero. Thus, on the domain $x=[0,1]$, $\sin(2\pi x)$ and $\cos(2\pi x)$ are orthogonal.
On the other hand, to pick some random examples, $f(x)=-x$ and $g(x)=\ln(x)$ are not orthogonal on the interval $x=[0,1]$.
$$
\Pi\,_0^1\!\left(f,g\right) \rightarrow \int_0^1{\left(-x\right)\ln(x)\, dx} = 0.25
$$
Similarly, the original two functions are not orthogonal on the interval $x=[0.2,0.9]$:
$$
\Pi\,_{0.2}^{0.9}\!\left(f, g\right) \rightarrow \int_{0.2}^{0.9}{\cos(2\pi x)\, \sin(2\pi x)\, dx} = -0.0444852
$$

Ok, enough with the math lesson. How does this help define orthogonal orbitals?
Well, orbitals, both atomic and molecular, are (usually) defined just as as 3-D functions over all space.  If you have two orbitals $\phi\!\left(\vec r\right)$ and $\psi\!\left(\vec r\right)$ and want to know if they're orthogonal, you use a mixing rule somewhat similar to $\Pi$ above and calculate the appropriate integral. If the resulting value is zero, then the orbitals are orthogonal.
For example, consider a $1\mathrm s$ and a $2\mathrm p$ orbital (yes, these are atomic orbitals, but the principle holds for molecular orbitals, too):

Here, the colors of the two lobes of the $2\mathrm p$ orbital are important: one color, say orange, represents positive function values; the other, say blue, represents negative values.  For the $1\mathrm s$ orbital, though, it's all orange, or all positive).
What happens if, per $\Pi\!\left(\phi, \psi\right)$, you center these two orbitals on top of one another, multiply their values together, and integrate over all space? Well, the half of the $\mathrm p$ orbital that is positive will result in positive contributions to the integral of $\Pi$, and the half of the $\mathrm p$ orbital that is negative will provide negative contributions to the $\Pi$ integral.  And, further, due to the spherical symmetry of the $\mathrm s$ orbital and the reflection symmetry of the $\mathrm p$ orbital, those two positive and negative contributions will exactly cancel. Thus, the $1\mathrm s$ and $2\mathrm p$ atomic orbitals are orthogonal.

Epilogue: So, given all of this, why do we also say that vectors that are $90^\circ$ from one another are orthogonal?
Well, it just involves a different mixing rule (the domain is implicit, deriving from the dimensionality of the two vectors, which must be the same):
$$
\Gamma\!\left(\vec v_1, \vec v_2\right) \rightarrow \vec v_1\cdot\vec v_2
$$
Under the (extraordinarily common) mixing rule $\Gamma\!\left(\vec v_1, \vec v_2\right)$, the two vectors $\vec v_1$ and $\vec v_2$ are said to be orthogonal  when $\vec v_1\cdot\vec v_2=0$.

Note: In theory, I could have just linked to this question on Math.SE. However, none of the answers there were really couched in suitably readable terms.
$^\dagger$ Technically, integrating the product of one function with the complex conjugate of the other: $\int_a^b{f^*(x)g(x)\,dx}$

Nice answer. I'm with you saying that explaining orthogonality without at least some formulas really don't make sense.

See also my answer on What does orthogonality mean in function space? question at Math.SE for a bit of illustration of the concept.

Regarding the epilogue: I gave up on maths some years ago, but from what I understand, if one generalises to "infinite-dimensional vectors" the sum in the dot product becomes an integral, and the generalisation to a complex vector space results in the complex conjugation. So the dot product can actually be thought of as a special case of the inner product.

You're exactly right, @orthocresol. I'm pretty sure these can all be considered Hilbert spaces, where continuous functions are of an infinite-dimensional character.

I wonder if you truly do mean molecular orbitals. Let's start with atomic orbitals. We model an isolated atom as having three P orbitals, each a "dumb bell" oriented along one of three perpendicular axes (x, Y and z if you like). There are other orbitals of course, such as s and possibly d, and other shells which have the same types of orbitals but with different sizes and numbers of nodes. But focusing on the three perpendicular p-orbitals (which I'm sure you've seen an image of), I hope you can see why we call them orthogonal. They don't overlap, and electrons in different p-orbitals essentially don't interact for our purposes.
Now, the spherical s-orbital sitting in the center of those p's doesn't seem "perpendicular" to them; but nonetheless, the s does not interact with the p's. So we say the s-orbital is orthogonal to the p-orbitals, just as the p's are to each other. Orthogonal here means more than just perpendicular in space, it also refers to the mathematical relationship of these shapes.
Molecular orbitals are formed when atomic orbitals overlap. Often we also include orbital hybridization in this model. Molecular orbitals tend to look a bit more complex than our familiar atomic orbitals, and you can't always perceive them as appearing physically perpendicular in space. However, as I hinted at, orthogonality is a more abstract property. In the proper mathematical context, we can still call these orbitals orthogonal, and it still has physical significance.
Here's a more detailed source that is probably on your level: https://www.decodedscience.org/atomic-interactions-wavefunctions-orthogonality-orbitals-hybridization-bonding/55459
(Though I must point out they might be using "molecular orbital" a bit too broadly, personally I think M.O.s like sigma and pi bonding and antibonding should be distinguished from the atomic orbitals of s, p, d, and f.)
Now, if you want to know why s-orbitals are spherical and p-orbitals are dumbbell-shaped, we're going to have to say the W word. ;)


text:
Can every chemical compound be melted?
Are there any chemical compounds that disintegrate (without going into other chemical reactions; let's say in a vacuum) before reaching a melting temperature?

http://chemistry.stackexchange.com/questions/62539/why-all-objects-do-not-melt-and-liquefy-when-heated-suficiently also http://chemistry.stackexchange.com/questions/31050/can-you-melt-wood

Some compounds decompose before they reach the melting point. In the literature, you'll see something like "m.p. $40\ ^{\circ}\mathrm{C}$ (dec.)"
Ammonium chloride is an example.

Sugars, proteins, and a lot of others. Some of these can be molten under pressure.


text:
What is the difference between radical initiator and radical inhibitor?
I am confused between these two concepts. My book gives $\ce{O2}$ as an example of a radical inhibitor and peroxide as an example of an initiator. But I don’t know why this is so? Both have $\ce{O-O}$ bonds so what is the difference?
Also aren’t all radical initiators in some sense? Aren’t radicals in general reactive and that they always create more radicals?

The confusing thing about initiators and inhibitors of radical reactions is that both are often radicals.
But it isn't being a radical that is important: it is what happens when that radical reacts in the reaction conditions. If the radical reacts with another radical to give something that is no longer capable of propagating the radical reaction, it is an inhibitor. If it reacts to give more radicals it is an initiator.
Which compounds can do the job of either depends a lot on the specific reaction.

A radical inhibitor, as the name itself says, inhibits the formation of radicals. It reacts to "wipe out" radicals. $\ce{O2}$ is one such chemical. A radical initiator on the other hand is a substance that generates radicals, as in the case of $\ce{H2O2}$, which forms radicals via the cleavage of the $\ce{O-O}$ bond.


text:
Equations for reactions dealing with coordination compounds

  Write a balanced chemical equation for the reaction depicted in this figure. Express your answer as a chemical equation. Identify all the phases in your answer.

This is what I have tried so far:
NH4NCS(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5NCS]2+(aq) + H+(aq) + NH4OH(aq)
NH4NCS(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5NCS]2+(aq) + H2O(l) + NH4+(aq)

Both equations are correct and equivalent =)

But it keeps saying that they are wrong...

What is ‘it’?$%$

The app I'm doing the problem on

All I can advise you to do in this case is submit a bug report.

@Jan Technically, there's an extra dissociation in (1), so they're not equivalent, but they certainly both seem valid.

I found the answer which I'm assuming was just getting rid of an unnecessary spectator ion:
SCN-(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5(SCN)]2+(aq) + H2O(l)

Sounds more like a technicality than an actual problem with your equation.

True but the program only counted this answer as correct


text:
Is there any acid which releases a gas when neutralized with hydroxide?
Out of curiosity, is there any acid which releases a gas upon neutralization with sodium hydroxide? Most acids that I know of release hydrogen upon reaction with metals, but not upon reaction with alkalis like sodium hydroxide.

Look at the problem in reverse: take a substance that is a gas in ambient conditions, then add a proton to it. Whatever you get will fit the bill, such as the ammonium and phosphonium ions Jan mentions in his answer. In principle, protonated carbon dioxide ($\ce{HCO2^+}$), protonated helium ($\ce{HeH^+}$), protonated hydrogen ($\ce{H3^+}$), etc. all work, though most of these examples will be quite artificial. Furthermore, there is a simple but fundamental fact which makes acids deprotonating into gasses less usual; there just aren't that many gaseous substances in ambient conditions.

The simplest and most obvious answer is the ammonium ion, closely followed by the phosphonium ion $\ce{PH4+}$. Both are acids as defined by the Arrhenius theory and by the Brønsted-Lowry theory; however, they always require a counterion. Upon proton displacement, they form gaseous compounds — $\ce{NH3}$ and $\ce{PH3}$, respectively — which will diffuse out of solution given enough time.
The solubility of ammonia in water is much higher than that of carbon dioxide, though; so it takes longer for ammonia to be released and no visible bubbling occurs. Yet, the gas evolution can be shown by sticking a moist indicator paper above the solution; fumes of ammonia will colour it whichever colour means basic on that paper.
$$\begin{align}\ce{NH4+ + OH- &-> NH3 (g) ^ + H2O}\tag{1}\\[1.2em]
\ce{PH4+ + H2O &-> PH3 (g) ^ + H3O+}\tag{2}\end{align}$$
A less obvious answer and maybe a bit of a stretch is the monochloride of oxalic acid, $\ce{ClCOCOOH}$. This compound is not stable; however, its basic structural motif is formed in many organic mechanisms, e.g. in the Swern oxidation as a $\ce{DMS}$ adduct ($\ce{Cl-CO-COO-S^+Me2}$). Once a chloride ion attacks sulfur nucleophilicly and thereby displaces what formally would be the monoanion of oxalic acid monochloride, this breaks down into carbon dioxide, carbon monoxide and chloride.
$$\ce{Cl- + Me2S^+-OCO-CO-Cl -> ClS^+Me2 + CO2 ^ + CO ^ + Cl-}\tag{3}$$


text:
Does limonene react with calcium hydroxide or acetic acid?
I can't find the results of too many experiments with Limonene. Which has many uses (e.g. making soap).
I'm curious of the results of combining Limonene and Calcium Hydroxide (Lime) in one experiment and another experiment combining Limonene and White Vinegar.
What is the result of these equations? Are the products potentially dangerous?
Could they result in combustion?
C10H16 + CH3COOH → ?
C10H16 + CaH2O2 → ?

No reaction in either case, I guess.

I can't find the results of too many experiments with Limonene.

Actually, there are quite a lot and the feature article Limonene: a versatile chemical of the bioeconomy, Chem. Commun., 2014, 50, 15288-15296 by Rosaria Ciriminna, Monica Lomeli-Rodriguez, Piera Demma Carà, Jose A. Lopez-Sanchez and Mario Pagliaro provides an interesting insight:

d-limonene (1) is available in large quantities as a side product in the production of orange juice
it is widely used as a bio-based solvent to  replace petroleum-based solvents for various purposes, such as

general household cleaners
degreasing of metals
outdoor maintenance cleaners

it is used as an insecticide

In order to to prove comsumer-friendliness of d-limonene and its degradation products, a lot of studies on the reaction of d-limonene with oxygen, leading to endoperoxides, hydroperoxides, and the monoterpene ketone carvone (2), have been performed.

In the ongoing BioProMo project, the bacterial conversion of d-limonene to the carboxylic acid 3 is examined. The latter is a natural preservative.

I'm curious of the results of combining Limonene and Calcium Hydroxide (Lime) in one experiment and another experiment combining Limonene and White Vinegar.

To my information, these combinations aren't used in industrial or academic research, but in another context.
As outlined above, d-limonene is widely used as a natural replacement for other organic solvents, which aren't easily available outside science and/or industry. As a consequence, d-limonene as a solvent has also found its way into recipes for the clandestine extraction of controlled substances from plant material.
Some of these recipes seem to promote the combination of d-limonene, calcium hydroxide, and white vinegar (as a source of acetic acid) to extract N,N-dimethyltryptamine (DMT) as its acetate from the bark of Mimosa tenuiflora.
While this seems technically possible and reactions between d-limonene, the base, and the acetic acid are not to be expected, the posession of the extraction product is apparently illegal in a variety of countries.