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(Energetics) Potential Energy and Kinetic Energy

This question came up when I was reviewing the energetics unit. In my notes I wrote "systems with a lower potential energy are more energetically stable, because they have a higher kinetic energy and higher kinetic energy means higher level of disorder." Is this correct? If so, does it mean potential energy stored in chemical bonds are converted to kinetic energy during a chemical reaction?
Is energetically "stable" the same thing as energetically "favourable"?

no, lower potential energy just means lower energy overall which is energetically stable. kinetic energy does not come into play at all. 2) basically the same meaning, but there is a subtle difference, you can call a molecule energetically stable, but you call a process, or a reaction, energetically favourable. e.g. benzene is stable, the formation of benzene is favourable.

In classical physics, kinetic energy and gravitational potential energy are almost interchangeable.  You do work on a ball and throw it vertically into the air.  Neglecting friction with the air all the kinetic energy becomes potential energy.  However, it then begins to fall, converting the potential energy back into kinetic energy.  Again, with no friction, the kinetic energy is exactly the amount that you began with.
However, we have to be careful to distinguish between ordered kinetic energy and random kinetic energy.  The ball, thrown into the air has both ordered and random kinetic energy.  The energy that becomes potential energy was the ordered form where-as the random thermal motions of all the individual atoms, which add up to no NET motion of the ball, remained the same throughout the entire flight of the ball.
But did they?  Again and again we have to say in these situations: Ignoring friction.  Friction does in fact exist when there is air present, and friction is very relevant to this discussion.  Friction actually converts ordered kinetic energy of motion into random kinetic energy of various particles (air molecules and ball molecules).
The creation of random motion from ANY other form of energy increases entropy and reduces the amount of energy that is available to do useful work.
I'm not sure that totally answers the question.  I think that the statement needs to clarify that random motion and its associated kinetic energy is, as your note claims, a preferred state over potential energy.

BTW: We usually call the random kinetic energy, THERMAL ENERGY.

Thank you! I think I need to do more reading. But I was wondering if anyone could explain this in a more Chemistry point of view.


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Stability of carbocation intermediates: secondary alkane versus primary allyl?
Say I have the following carbocations:

The first is secondary; the second is resonance stabilized, but primary.  Which is more stable and why?
I've been looking around in my textbooks and haven't been able to find a comparison.  This site compares alkane and alkene intermediate stability, but only with the same degrees of substitution.

The effect of resonance stabilization totally beats the primary/secondary difference, so the answer is 2.

@IvanNeretin According to this site, the effect is not as big as you might think. Those two cations have roughly the same stability.

There are two competing effects at work here; hyperconjugation, and resonance involving the double bond.
According to this website, a secondary alkyl cation and a primary allyl cation have roughly the same stability.

Carbocation Stability: $\ce{CH3+ < CH3CH2+ < (CH3)2CH+ ≈ CH2=CH-CH2+ < C6H5CH2+ ≈ (CH3)3C+}$

This is not really something you can predict from qualitative considerations - it is difficult to assess the strength of resonance versus hyperconjugation - but it requires experimental evidence, or detailed computer calculations to give a definitive answer.
In addition, the effect of adding more hyperconjugated groups is not linear. The difference in stability between tertiary and secondary alkyl cations is bigger than that between primary and secondary cations - tertiary alkyl cations are actually pretty stable.
Lastly, (thanks to @permeakra for pointing this out) the stability of carbocations in influenced by the environment which they are in. In particular, different solvents affect the stability of different cations to a varying degree and so the above order of stability should be taken as a rough guideline, not a hard and fast rule.

I would also add that the relative stability  may be affected by solvent, so any such row should be threated with caution.  Specifically, I bet that conjugated systems gain relative stability boost in nonpolar solvents.

@permeakra Yes this is a good point that I missed.

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Forming magnetite on cast iron pan
I am in the process of seasoning a cast iron pan. (Creating the black non-stick coating typical of such pans.)
This is done by causing thin layers of flaxseed oil to polymerize and bond to the iron.
In the literature there is the recommendation that the polymerized oil bonds better to the pan, if the pan is covered with magnetite rather than blank iron, in part due to the porous nature of the oxide layer.
My question is: is there a DIY method that can either create $\ce{Fe3O4}$ on a polished cast iron pan, or alternatively convert a layer of $\ce{Fe2O3}$ into $\ce{Fe3O4}$?
In the gun manufacturing trade a process called "rust bluing" is used. In it the gun barrel is encouraged to form a thin layer of $\ce{Fe2O3}$ before being immersed in boiling distilled water. This causes the $\ce{Fe2O3}$ layer to convert into $\ce{Fe3O4}$ within minutes. Apparently due to the fact that boiling water lacks dissolved oxygen.
When I attempt to apply this "rust bluing" process to a cast iron pan, the $\ce{Fe2O3}$ layer stays red rather then turning black: Clearly the conversion is not taking place.
I am not clear why this is the case. Is it due to difference in materials, i.e. cast iron vs. gun steel?
Is there any way I can promote creation of/conversion to a magnetite layer on cast iron with DIY-compatible means? The industrial solution of caustic soda solution at 150$^{\circ}$C does not sound like a good idea for home implementation.

I think the two steps that are missing from you approach are "carding", where, after immersion in the boiling water, the steel is rubbed/brushed to knock off the larger rust structures and repetition. The process is repeated several times before the piece builds up a stable blued surface.
Check out this video of Larry Potterfield applying a rust blue surface to a rifle action and barrel: https://www.youtube.com/watch?v=VPD3GWqk-yE

Heating the polished and cleaned pan on your kitchen stove will oxidize the polished surface. The oxide will be magnetite. If you bring the pan to a dull glow, the oxide layer can be grown in thickness to suit your needs.

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Atomic emission spectra: classical vs. quantum mechanics
In class, we were first introduced to classical mechanics, and how Bohr used the hydrogen electron to explain the atomic emission spectrum of that element. And it made sense. But then classical mechanics fails to explain other things, so we jump to quantum mechanics. And we learn about Newton's ideas of particles of light, and how Einstein said that light quanta are called photons, and how Louis de Broglie suggested that particles of matter can also behave like waves.
But I'm frustrated because, while the book does explain the history of quantum mechanics and the Heisenberg uncertainty principle, it doesn't explain how quantum mechanics has anything to do with atomic emission spectra. So my question is, how does quantum mechanics explain the atomic emission spectrum of elements better than classical mechanics does?
And how much of classical mechanics fails to account for what quantum mechanics does? So is the idea that an electron moving from one high energy level to a lower energy level in an atom releases energy/light is only true for hydrogen, and false for all the other elements? How can this be true?

First of all, classical mechanics does not explain the emission spectrum at all, since there is no energy quantization out there. Bohr introduced energy quantization in his model in an ad hoc way by requiring the angular momentum to be an integer multiple of a fixed unit $\hbar = h/2\pi$ (which in turn resulted in energy also being quantized). So, Bohr model is not a classical mechanical model, since there is quantization of angular momentum and energy. But it is also not a quantum mechanical one since quantization of angular momentum is not justified, but rather asserted.
I'm saying all this just to correct OP: it was not a fight of classical vs. quantum mechanics; classical mechanics was already knockouted and quantum mechanics didn't yet fully emerged. It was a battle between semiclassical-semiquantum theories and reality which today could be viewed as well as a fight between semiclassical-semiquantum theories and the real quantum theory invented later.
Anyway, there are in general many possible electron transitions for each atom (or molecule) that contribute to its emission spectrum and each transition has a specific energy difference and probability. The Bohr semiclassical model had shortcomings with respect to both these characteristics of electron transitions:

Multi-electron atoms do not have energy levels as predicted by the Bohr model, and consequently, the model predicts wrong positions of the spectral lines. In fact, it fails already for helium atom.
There is no general way of calculating probabilities of transitions, thus, it is not possible to provide relative intensities of spectral lines.

Would that not be a candidate to roll out [tag:old-quantum-chemistry] as you have proposed instead of using [tag:classical-mechanics]?

Bohr atom is already quantum mechanics, only a special (restricted) kind of it. In classical mechanics there would be no discrete energy levels in atom, and indeed no atom itself, for an electron would rapidly lose its energy through radiation and fall to the nucleus. So if you ask how does classical mechanics fail to explain the spectra of elements, the answer is: completely.
As we move on to the atoms with more than one electron, Bohr model reaches its limit and ceases to be helpful, so we need the "real" quantum mechanics with $\psi$ function and Schrödinger equation. It contains many counter-intuitive implications, like the electron being delocalized and staying everywhere at once, but in return for that it grants us the ability to calculate the spectra of all elements (and many, many more) with great precision.


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Speculation about chlorine dioxide isomer
I know that chlorine can have a positive charge in the case of oxides. But what I am wondering is if chlorine dioxide can form a ring and if so how likely it is.
$\ce{O=Cl=O -> O- -Cl-O- -> (cyclic)O-Cl-O}$
This reaction would form a triangular ring with each oxygen being bonded to 2 atoms so the charge would once again be zero. But right in the middle of it you have an ion. It isn't chloryl or chlorite since it has a charge of 2-. This ion then bends to where the oxygens are close enough to form a peroxide bond giving you cyclic chorine dioxide.

Three-membered rings are not very favorable. I wouldn't say it's imposible though because of the size of chlorine.

@Caters Sorry but your reasoning doesn't make sense.

Why? Doesn't forming a ring from a molecule with double bonds at the ends require that you form a single bonded ion first and then have the ion bend in just the right way that the negatively charged atoms can bond with each other? I mean sure this might be very unlikely but negative ions come towards each other all the time and react so negatively charged atoms within an ion would do the same thing, especially if the bond angle means that they have to come towards each other.

@Caters No - why do you think that has to be the case. For a start, where do the two extra electrons come from to form the ion?

They come from the double bonds being broken. Since oxygen is more electronegative it will take both electrons from the bond and have a negative charge. This causes chlorine to have a positive charge as a result. And why wouldn't the ion need to form? I mean a double bond on both ends of a molecule with 3 atoms means that they have to be broken in order for them to bend enough to form a ring. This likewise means that at least for molecules with 3 atoms an ion does need to form.


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Standard hydrogen electrode, why hydrogen?
Why is the hydrogen electrode used as the reference to set electrode potentials? Could any other element be used instead?

That's simply for convenience. Any other reference point would do just as well. But we live in a world full of hydrogen (in the form of water), so this is our primary interest.

It's worth noting that I've been told (by a couple of battery researchers) that the silver chloride electrode is the de-facto reference in most labs. It comes in a convenient package which doesn't require you to bubble a gas at 1 bar over a catalyst in a 1M acid (all of which would have to be re-calibrated every time you ran an experiment)

There are two separate issues here. The first one is defining a zero for potential. In theoretical physics, it is the energy of two charges separated by infinity distance.
In electrochemistry, this is chosen to be the potential of the hydrogen electrode since that is something that can be made reproducibly and hydrogen was the de facto 'zero' for a number of other quantities.
These days, there are much more practical and easy to use reference electrodes like the silver/silver chloride (as chipbuster mentioned) or the saturated calomel or the mercury/mercurous sulfate. Here is a good list of them from a manufacturer of such electrodes: http://www.koslow.com/select_reference_probe
Full disclosure: I used to work for a company that resells electrodes from Koslow.

There are different approaching points here. One is history: The hydrogen electrode has been used forever and people are used to defining values relative to it, so why not continue?
A second point of approach is the following: You cannot measure absolute potentials; you need some kind of a reference. The choice was completely arbitrary. However, there are some ‘rationalisations’ (that probably only turned up after hydrogen was already used):

Hydrogen is the first element in the periodic system. It is nice to have ‘number one’ as your reference.
The potential of hydrogen (hydrogen gas at 1 bar in an acidic solution of pH 0) is approximately equivalent to that encountered in 1 molar acidic solutions (sulphuric, hydrochloric, nitric acid etc.). Probably even predating the selection of hydrogen, metals that dissolved in 1 molar acid were termed non-noble and those that resisted the acid were noble. Using the hydrogen potential as a zero-point, the non-noble metals have a negative standard potential while noble ones have a positive. An easy way to distinguish noble metals just by a number.

But just to restate: The choice of the electrode as a standard was likely before the second implication was fully realised.


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Sulfur hexafluoride creation process and risks/safety
I would like to create sulfur hexafluoride and test floating things in it but I am a "chemistry noob" and from what I've read it's non-toxic and generally quite safe but first of all I do not know how to make it (other than that it involves sulfur and fluoride which are both definitely not safe) and the precautions for doing this. Basically: Safe way to make sulfur hexafluoride and how safe is sulfur hexafluoride?

$\ce{SF6}$ is pretty safe (unless you breathe it too much), but its preparation is not.

Better buy it - chemistry noobs should steer clear from gas fluorine.

https://en.wikipedia.org/wiki/Sulfur_hexafluoride

I'm far from a chemistry noob (and I have experience using CO, HCl and chlorine gases) but I would stay away from any experiment that involves handling fluorine gas and that is in an industrial fumehood.

Thanks. I will buy it then instead. I love the idea that a solid object can float on a gas plus the geometric structure and the fact that it's a very dense (about 6 times denser than air from what I remember) is very pleasing.

On the downside, it is also a very good greenhouse gas, so releasing it into the environment just for fun isn't great.

Is it safe to inhale (in small amounts)? Because I would think it would be like a reverse version of Helium because it's denser and it would make your voice go lower rather than higher (that's just an educated guess though).

That's right: when inhaled, it will do funny things with your voice. It is safe in that it will not react with anything in you, but don't forget that you can't survive without oxygen for too long.

And don't forget that it's heavier than air, so it could take a long time to get it out of your lungs if you're standing upright.

The preparation of sulfur hexafluoride from elemental sulfur and elemental fluorine is not suitable for amateurs or beginners; it is definitely not a home experiment. Fluorine is corrosive and very toxic. It causes severe skin burns and eye damage and it is fatal if inhaled. Furthermore, it is an oxidiser and reacts violently with all combustible materials; thus, it may cause or intensify fire.
Self-made sulfur hexafluoride is not suitable for inhalation or similar experiments since it may contain various toxic impurities.
In the laboratory scale, a nickel boat is filled with sulfur and placed in a reaction tube. The tube is connected to a cold trap that is cooled with liquid nitrogen. The apparatus should be dry and ungreased. A stream of fluorine is then introduced.
The reaction
$$\ce{S + 3F2 -> SF6}$$
is strongly exothermic (standard molar enthalpy of formation at $298.15\ \mathrm K$: $\Delta_\mathrm fH^\circ=-1220.5\ \mathrm{kJ/mol}$[1]). In the fluorine stream, sulfur burns with a blue flame.
Sulfur hexafluoride is collected in the cold trap. The product contains various impurities, such as $\ce{F2}$, $\ce{HF}$, $\ce{SOF2}$, $\ce{SO2F2}$, $\ce{SF2}$, $\ce{S2F2}$, $\ce{SF4}$, and $\ce{S2F10}$. Most of the impurities are easily hydrolysable; therefore the gas is passed through a washing bottle filled with $\ce{KOH}$ solution. However, $\ce{S2F10}$ has to be removed with activated charcoal or decomposed by pyrolysis at $400\ \mathrm{^\circ C}$ according to
$$\ce{S2F10 -> SF4 + SF6}$$
[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


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Why is the electrode potential of copper positive when hydrogen is more electronegative than copper?
Both copper and zinc has an electronegativity that is smaller than hydrogen, but why does copper have a positive electrode potential while zinc has a negative one? And lead (which has a greater electronegativity than hydrogen) does also have a negative electrode potential.

In short, because electrode potentials have (almost) nothing to do with electronegativity.


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Stereoelectronic effects in carbamate carbonyl oxygen attack on halide
2-Bromoethyl (2-bromoethyl)carbamate can react through either Path A or Path B to form salt 1 or salt 2. In both instances the carbonyl oxygen functions as the
nucleophile in an intramolecular alkylation. What is the preferred reaction
path for the transformation in question?

I would go with A, but I don't know how to explain it.

Well you must have a reason for choosing A so try to give us some idea of what it is.

The flow of electrons is more linear in the TS of reaction pathway A?

I would tentatively suggest A based on less ring strain due to not having a double bond inside the ring and having oxygens instead of nitrogens. I'm not particularly knowledgeable in this field though. If you want the question to receive more attention you could consider putting a bounty on it.

But double bonds inside rings are more stable than exocyclic ones. A bounty seems inevitable.

AFAIK that's only because they usually have more substituent groups (compare 1-methylcyclohexene and methylenecyclohexane) but in this case that's not the case. Since the geometry of the double bond is restricted by the ring, I think that would make it less stable but this is only speculation.

Are there any more additives involved? What about solvents?

No. What influence would you expect from solvents?

Path A resembles a 5-enolexo-exo-tet reaction which is allowed by Baldwin rules while path B represents a 5-enolendo-exo-tet reaction which is disfavored. The orbital interactions of path B suffer severe geometrical constraints (even in the conformation where the C-Br bond is pointing towards the viewer when looking at the attached image). Path A is presented in the attached image.

authors: ศุภกิจ หนูทอง


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Why "RCA clean" solution has to be prepared every time (and not stored long-term)?
"RCA clean" solution is a 5:1:1 mixture of DI water, ammonium hydroxide and hydrogen peroxide.
Why the ready 5:1:1 solution is not stored for long term, but instead has to be prepared each time immediately prior to usage?

My best guess is that the peroxide will decompose. https://en.wikipedia.org/wiki/Hydrogen_peroxide#Decomposition

Hydrogen peroxide is not a very stable substance. Over time, and however you store it or however carefully you prepare it, it will decay. So solutions of a given concentration need to be prepared fresh when you want to use them. You could try to store them but you won't know whether the concentration is still suitable for whatever you are going to do with it at some later time. And, the decay may well generate gas and other side-products of oxidation what may interfere with the use and may be unsafe.
Hence you would make the mixture fresh every time when you want to use it.


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Why Fe 3+ is more common than Fe 2+
I heard that $\ce{Fe}(\mathrm{III})$ is more common than $\ce{Fe}(\mathrm{II})$ but I've not heard a very clear explanation. Could someone please explain this incorporating electron configurations in their answer?

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Not in igneous rocks. Most iron there, in silicate minerals throughout the deep crust and mantle, is Fe(II). Need to weather the rocks and expose the iron to an oxidizing atmosphere to get substantial Fe(III).

I think the best answer is not to try to describe this in terms of electron orbitals. The main factor is that Fe(II) is fairly easily oxidized by atmospheric oxygen to give Fe(III). Because we spend most of our time living in an oxygen atmosphere, Fe(III) is more common in our ordinary experience than Fe(II).
In environments with less oxygen, Fe(II) is very common. The mantle is full of Fe(II) minerals like olivine and ferropericlase. I don't know which species is actually the majority on earth, but I would bet on Fe(II).

How can we explain this on the basis of reduction potentials? For $\ce{Fe^3+ + e^-} \rightarrow Fe^{2+}$ is 0.77 and for $\ce {Fe^2+ + 2e^-} \rightarrow Fe$ is -0.44?

@Proscionexium https://chemistry.stackexchange.com/a/44678/136839 is a good answer.

The electron configuration of  $\ce{Fe(II)}$ is $1s^22s^22p^6 3s^23p^6 3d^6$.
The electron configuration of  $\ce{Fe(III)}$ is $1s^22s^22p^6 3s^23p^6 3d^5$.
The half filled  orbital $3d$ is more stable than the same orbital filled with 6 electrons. So,  $\ce{Fe(III)} $ ion is more common than $\ce{Fe(II)}$.

I’m not sure if this is more than half of the truth, but I’ll leave it to the true experts to tell me, whether I’m right or wrong. (Sure, this is taught in schools, but it seems to simple to me, and the half-filled shell concept has been deemed invalid in other answers that I don’t have on-hand atm.)

It's not even wrong, it's irrelevant, Real iron compounds don't have iron in a spherical environment, hence the orbital degeneracy pattern is different. For instance low spin iron II in an Oh environment has a completely filled set of orbitals. Just like Ne, so why isn't that the most "stable", whatever that means

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Is there a way to make a liquid cold by a chemical reaction?
I'm looking for a way to combine two substances in a chemical reaction that uses a lot of energy, therefore absorbing heat, to make a substance cold.
I'm thinking about a two-chamber thermos that in the middle chamber is, for example, juice and in the outside chamber will be placed the other substances. When these substances react, is it possible that they absorb heat, making the juice colder?

Not only possible - https://en.wikipedia.org/wiki/Instant_cold_pack

Ammonium thiocyanate and barium hydroxide octahydrate when mixed will draw a huge quantity of heat. This reaction often used as a chemistry demo of endothermic reactions where the chemicals are mixed in a beaker that's placed on a wet piece of wood. The wood freezes to the beaker.

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authors: Андрей Пушкарь


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Why highly unstable cyclic boron compounds should be stabilised by two electron reduction?
I am trying to figure out why highly unstable cyclic boron compounds  could be stabilised by two electron reduction...

My thoughts are because each Boron has an empty p-orbital perpendicular to plane of cyclic compound, pi-orbitals can form like diagrams below.

The two electrons would then go into the pi-bonding orbital, thus stabilising the entire compound.
Is this the correct explanation? If so, how can vibrational spectroscopy be useful in examining the bonding in the dianion?

I don't have time for a full answer, but that's not the correct picture of the MO's. Two of the orbitals are degenerate.

you are right. The triboracyclopropenyl dianion have been reported previously.

Your energy diagram is that of a linear moiety, like an allyl cation. For a cyclic structure, it would be somewhat different, but the general idea remains the same: there is one $\pi$-bonding orbital, and when two electrons go to it, everyone wins.
As for vibrational spectroscopy, well, it can help you tell the cyclic structure from linear, if that's what you want.


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Sublimation of Iodine
Why does iodine sublime? I have researched it myself and I have got the same answer; it sublimes because it directly converts from solid to gas. But why don't bromine or chlorine also sublime?

I think your question can be more generally stated, and subsequently, answered.
Why do some substances sublime, while others don't?
To understand this, you need to understand a phase change to gas occurs when the total vapor pressure of substance in a condensed phase (solid or liquid) exceeds the atmospheric pressure.
[In case you aren't familiar with the term, vapor pressure is defined as  as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system]
So, at a given temperature, if the atmospheric pressure is low enough, any substance will go from solid to gas (i.e sublime), without having a chance to melt (so to speak).
To gain a better understanding of this process, I urge you to look at a phase diagram. (this particular one is for carbon dioxide, easier to find than iodine's the principal remains the same.)

at atmospheric conditions (~1 bar) you can observer solid carbon dioxide goes to gas, bypassing the liquid state. But say, at a higher pressure say >10 bar, it does indeed go from solid to liquid to gas, as it is warmed.

This answer is misleading. Iodine only sublimes at pressures lower than about 0.12 atm when heated; at atmospheric pressure it melts. Sublimation of iodine at atmospheric pressure is not a bulk effect (which would be captured by the phase diagram) but a surface effect due to equilibrium vapor pressure. "Sublimation" is an overloaded term with two distinct meanings corresponding to evaporation and boiling in liquids, and what we generally talk about as sublimation of iodine is analogous to evaporation, while CO2 is analogous to boiling. Ivan Neretin's answer gives a correct description.

Iodine sublimes for the same reasons that all solids do: because it has some equilibrium vapor pressure an normal conditions. Now, the value of that pressure varies greatly in different solids. For many of them, it is so extremely low that for all practical reasons it can be considered non-existent. (Say, if you'd leave a sample of NaCl for a thousand years, I don't think you will be able to detect the weight loss due to sublimation.) For iodine, the vapor pressure is significant, so a sample left in the open will vanish completely in a matter of hours.
Some sources may specify that iodine goes directly from solid to gas when heated, never becoming a liquid. That's an urban legend; iodine does have the melting point and the boiling point at standard pressure, and between the two it will be liquid. It is not easy to demonstrate, though; if you heat it in the open, it will probably sublime away even before reaching the melting point, and if you do it in a closed container, the vapors get so thick you can't see a thing.
That being said, there are indeed some compounds that would never turn liquid at ambient pressure. That's because their triple point is above that ambient pressure. Carbon dioxide is one example; water on Mars is another.
As for chlorine and bromine, I don't really see the point of your question. At normal conditions they are not solid, so they can't undergo sublimation simply by the definition of it. If you cool them down enough to be solid, they will.

This article from UWaterloo has a phase diagram for iodine, and explains some key points.

The vapor pressure of solid Iodine at $20\ ^\circ\mathrm C$ is $0.027\ \mathrm{kPa}$ (compare with atmospheric pressure: $101.3\ \mathrm{kPa}$). If left in an unsealed flask at standard temp/pressure, it will slowly sublime away.

But solid iodine does have a melting point, if the flask is heated to $113.7\ ^\circ\mathrm C$, the bottom of the flask contains liquid iodine. The liquid state exists between $113.7\ ^\circ\mathrm C$ and $184.3\ ^\circ\mathrm C$.

For the other two halogens, you would need to find a table of their respective vapor pressures in their solid state, and the temperature range at which they exist in their liquid states.


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What is the appropriate sizing for fiberglass compatible with polyamide?
I am looking for a chemical coating for glass A fibers, compatible with polyamide. The fibers are uncoated, 100-filament strands of 22-micrometer "A" glass fibers.
We are producing fiberglass in our lab, and we want to size (coat) them with something which can keep the strands together and at the same time enables the fibers to bond with polyamide. We can use materials like PVA to keep the strands together, but then the fibers do not bond with the polyamide (mixing with polyamide happens after we produce the strands and chop them, with 4mm in length).
I am a beginner in composite world so I would appreciate if you can give me details on how and where to buy such chemical, or how to synthesis it myself if possible?

I think epoxy resin is compatible with polyamide because its chemical structure. Epoxy is compatible with silica (glass) and poyamide functional groups then it could be a good sizing .


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Why do some elements such as polonium not follow the trend of decreasing atomic radius?
Most of the periodic table follows the trend of increasing radius as you move down a family and decreasing radius as you move across a period. Why is this trend not observed for polonium?


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What are the most extreme chemicals?
Some chemicals certainly have extraordinary properties. One can consider them extreme in a sense, that they are the prime example for a concept, or they achieve the highest or lowest values in a certain area.
This Q&A attempts to compile a list of questions asked on this network, which have a background related to such extreme values or properties. we decided to add such a list because it is a popular search term.
It is a community effort, hence everybody with more than 100 reputation points can edit the answer. If you think something is missing in this list, just go ahead and add it. This post will be locked to prevent adding new answers, though. The corresponding discussion can be found on meta.chemistry.se. If you think there are some obvious points missing here, please go there and open a discussion.

Define "extreme"

@JavaNewbie_M107   Extreme means "the highest or lowest measure/value for some particular physical or chemical characteristic."

@Java see the meta on this. This was intended to be created.

Densest gas
Densest element
Greatest range between melting and boiling temperature
Highest melting point
Lowest solubility product constant
Most dangerous: Is there any substance that's a 4-4-4 on the NFPA diamond?. But also, don't forget that the NPFA diamond doesn't even classify ClF3 as flammable (though everything else is flammable in it) and by any criterion it is far more dangerous than the other compounds in those answers)
Strongest acid
Strongest organic acid
Strongest reducing agent

Good external lists:

Most extreme compounds known to man.


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Confused about carbocation stability
For which element 'X' is the carbocation $\ce{CH3-X-CH2+}$ the most stable?
A. X = S
B. X = N
C. X = O
I thought since oxygen is the most electronegative among the options, it would create the greatest electron deficiency, which would in turn facilitate a lone pair donation from the O atom (which happens with boron halides), so the answer should be oxygen. However the answer given is nitrogen. I really can't understand why. Is my logic wrong? Or is there more to this question?

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A carbocation is electron deficient. In an electron-neutral state, carbon would have four valence electrons that form four bonds resulting in an overall octet structure. To form a carbocation, one of these bonds needs to be removed, resulting in a sextet on the carbon atom; and one could also consider carbon now only having three valence electrons that form three bonds. Whichever picture we choose, carbon is lacking electrons.
(A more true picture would, of course, involve orbitals and their interactions, but still leave us with the conclusion that there are too few electons around the carbon atom.)
The three possible neighbours differ, amoung others, in their electronegativity — their ability to draw electrons towards them, away from what is next to them. Oxygen is the strongest, nitrogen second strongest and sulphur is the weakest.
In the case of the carbocation, I have already shown that it is electron-deficient. If we place a more electronegative element next to it, that element will draw the electrons away from the cation even more; this makes the carbon even more electron-deficient and thus destabilises the system. On the other hand, if a more electropositive element such as silicon were neighbouring the carbon, it would push electrons towards the carbon — exactly what the cation wants since it’s lacking electrons.
By this reasoning, sulphur, being the least electronegative element on the list should best stabilise the carbocation in question. But the answer gives nitrogen; so there must be a second effect. This would be mesomeric stabilisation.
For all three elements, one can draw a resonance structure where instead of a single $\ce{C-X}$ bond and a lone pair on $\ce{X}$ one has a $\ce{C=X}$ double bond shifting the positive charge to $\ce{X}$. This resonance structure is considerably less prevalent with sulphur (a third period element) than with oxygen or nitrogen; the reason being the spacial distribution of sulphur’s p-orbitals not fitting too well with carbon’s. Oxygen and nitrogen, on the other hand, have p-orbitals of a similar size and distribution so mixing works well. (Yes, this is also more complex on a more realistic level, but this visualisation works well enough for the simple case here.)
The question is now, whether the resonance stability provided by oxygen is stronger or weaker than that of nitrogen and if the difference is larger or smaller than the energy difference caused by the different electronegativities discussed earlier.
Again arguing that oxygen is more electronegative and thus draws electrons towards it, it would equally be less happy to share its unbonded electron pair with carbon to relieve carbon’s electron deficiency. Thus, one may expect nitrogen to provide a greater stabilisation to a neighbouring carbocation.


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Na Cl battery within Zn Cu battery?
https://youtu.be/iHBz77E625o
In this video there are pieces of matboard that are soaked in water saturated with NaCl. The pennies themselves are made of zinc and copper. With the electrical energy from the zinc and copper would the chloride let go of its electron and the sodium gain back its electron producing a Na Cl battery within a Zn Cu battery?
Would the water in the process separate into hydrogen and oxygen through electrolysis? If so since hydrogen is very flammable would a combustion reaction occur in the air giving you water again?

The electrolyte in that clip is a solution of $\ce{NaCl}$ and acetic acid (vinegar). The zinc anode is oxidized, resulting in $\ce{Zn^2+}$ ions going into the electrolyte. The electrons went through the red LED, lighting it, and then on to the copper cathode. From there, they reduced some $\ce{H+}$ to hydrogen gas. So two $\ce{H+}$ ions were reduced to one $\ce{H2}$ molecule (which left: it's a gas) and the $\ce{Zn^2+}$ ion simply replaced the two $\ce{H+}$ ions.
Acetic acid is a weak electrolyte, so a solution of it (vinegar) is not a great conductor of electricity. The $\ce{NaCl}$ just makes the electrolyte a better electrical conductor: movable ions are necessary for a steady flow of electrical current in this battery. Without the $\ce{NaCl},$ most of the generated voltage would have been wasted on the electrolyte's resistance.
No sodium is produced (by reducing sodium ions) because the $\ce{H+}$ ions get the electrons. And no chlorine gas is produced (by oxidizing chloride ions) because the $\ce{Zn}$ anode is more readily oxidized.

If my answer has been helpful to you, please consider giving the green checkmark. Thanks!


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What current and voltage are best for electrolysis rust removal?
I weld carbon steels all the time, and sometimes I get parts from the scrapyard. Unfortunately, these parts are generally oxidized, so I've been assembling an electrolysis tank for it. This tank removes the oxidation in about a day. I'm getting the feeling that I can make it go faster if I step up the current or voltage. What combinations of currents or voltages would make the fastest method of removal?
NOTE: I'm familiar with the risks of doing electrolysis, including hydroxy production, and the restriction on stainless steel electrodes.

What is the applied voltage? Did you try just HCl?

I use 12V at 2A. I'm using NaOH for the electrolyte. Approximately 1 tablespoon per gallon with 15 gallons.

and what about the HCl?

@SarahSzabo Washing soda ($\ce{Na2CO3}$) might work just as well as the $\ce{NaOH}$ and for much cheaper too...

This is what I found (only in Czech):webpage
Power source should be around 15V/5A. The cleaned part acts as negative electrode- applied voltage should be as high as needed to reach 2A. The electrolyte (described in the source) is based on water and sodium carbonate. Derusting should be done between 2 to 4 hours. Let us know if it helps or how it works.

In order to answer the question we also need how large a part you are electrolyzing.
The current density in $mA/cm^2$  is the key number. For electrolysis on stainless steel, I would say you need to stay less than about $ 1~mA/cm^2$ to stay safe.
The voltage is hard to gauge. In any two electrode system like yours, the electrochemical potential is only one part of the total potential. The total potential includes contributions from the resistance of the solution, resistance of the contacts, overpotential of the counter reaction, etc... The current should be a decent gauge of what you are seeing.

I run graphite plate electrodes (anodes), to avoid consuming them I try to stay below 15A/square foot, much depends on the area of both workpiece and electrode. Example, I'm currently de-rusting a motorcycle tank (Damn You, E10 "petrol") using a 20x2 inch, quarter-inch thinck graphite (50cm x 5cm x 6cm), about a mugfull of wash soda (sodium carbonate) in 5 gallons and with a 12v supply (old transformer battery charger) it's taking 8 Amps - near enough the 15A/square-foot. For awkward access like inside tanks etc, the graphite's wrapped in the non-slip mat sold for car dashboards etc, cable tied in place to prevent shorting to the tank or whatever.
I started out using scrap steel sheet as the electrode, but it grows barnacles that need to be scraped and brushed off regularly to maintain current and effectiveness, with graphite this does not happen and the process can be ignored for as long as it takes :)
NEVER, EVER use stainless steel as the anode - this will produve hexavalent chromium in the solution, which must be disposed of as toxic waste by a licenced waste processor and legally MUST be transported as toxic waste by a licenced handler. It's neurotoxic, hepatotoxic, implied carcinogen - you REALLY don't want it anywhere near you/


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How do I calculate the cost of a single Aluminum atom?
I am doing a Chemistry assignment in which I was given a piece of aluminum foil which I had to measure and use for my calculations. I am supposed to calculate the cost of a single aluminum atom using the given data from my teacher and the data which I have measured from my piece of aluminum foil. I understand the basic concept of how I am supposed to do this, but I am having some trouble. Here is the data:
Data given by teacher:
  Cost of a roll           $5.53
  Roll size                30.5 cm x 30.5 m
  Density of Al            2.70 g/cm^3
  Molar Mass Al            26.98 g/mol

Data measured from my piece of Aluminum foil:
  Mass     1.35 g
  Size     18.6 cm x 12.5 cm

I have determined a way to do this, but I am having trouble with it. First of all, I have determined the amount in moles which is $\mathrm{0.0500~mol}$. Then I multiplied the amount in moles with Avogadro's constant ($\mathrm{6.02 \cdot 10^{23}~mol^{-1}}$) to determine the number of aluminum atoms in the piece of aluminum foil I received which is $\mathrm{3.01 \cdot 10^{22}}~\text{atoms}$. Now I just have to find the cost of a single aluminum atom, but this is where I am stuck. I cannot use the cost of the aluminum foil roll to divide it by the number of atoms since I was given a piece of aluminum foil so I need to find the cost of the piece of aluminum foil and then divide it by the number of atoms.
So basically my question is: How do I find the cost of the piece of aluminum foil? If I can find the cost then I can divide it by the number of atoms to determine the cost of a single aluminum atom.
However, if there is a way to find the number of atoms in the roll of aluminum foil, I could then use the cost of the aluminum roll to find the cost of a single aluminum atom. Please explain if there is a way to do this using the data given by the teacher.

you could always look up the bulk price of aluminium per tonne (this will be a lot less than the foil but will still be valid). Aluminium is a commonly traded commodity and will be sold by weight. You already know the weight of the sample...

@matt_black I cannot use any other data other than the ones I was given and the ones I measured.

You can compute the area of aluminum foil in the roll as:
$$30.5\;\rm{cm}\cdot 3050\;\rm{cm} = 93025\;\rm{cm}^{2}$$
You have measured the mass per square centimeter as:
$${1.35\;\rm{g}\over 18.6\;\rm{cm}\cdot 12.5\;\rm{cm}} = {1.35\;\rm{g}\over 232.5\;\rm{cm}^{2}}$$
You have been given the cost of the roll (and thus the cost per area):
$${\$5.53\over 93025\;\rm{cm}^{2}}$$
Using those numbers, the molar mass for aluminum, and Avogadro's constant (here, it's just the number of atoms so I designate it as such):
$$\rm{cost\;per\;atom} = \left({\$5.53\over 93025\;\rm{cm}^{2}}\right)
\left({232.5\;\rm{cm}^{2}\over 1.35\;\rm{g}}\right)
\left({26.98\;\rm{g}\over 1\;\rm{mol}}\right)
\left({1\;\rm{mol}\over 6.022\cdot 10^{23}\;\rm{atoms}}\right)
$$
$$\rm{cost\;per\;atom} = \$4.59\cdot 10^{-25}$$

One more thing, could you explain to me why in the equation it is 232.5/1.35 and above the equation it was 1.35/232.5? Why was it flipped? I can't seem to understand that. Thanks!

It's just to make the units come out (dimensional analysis). It can be as I wrote it initially or inverted:  it's the ratio that matters. How it's written depends on what units cancel and make it through to the end. Same with Avogadro's number here.

I did this using one estimate for the number of aluminum atoms in 4.55 grams of aluminum, scaled it up to a kilogram, and divided aluminum's current price per kilo (US\$1.87) by my estimate for the number of atoms in a kilogram of aluminum, yielding a price per atom of about $\\\$8.453 \cdot 10^{-26}$, or around eight sextillionths of a penny.
Of course, I'm only in eighth grade, so my calculations may be way off.


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Creating small amounts of cellulosic ethanol from bamboo
I own a medium-sized farm and commercial aquaponics business in Pennsylvania. I try to make my business as green, sustainable, carbon-neutral and self-sufficient as possible.
As a way to cut out the need to buy expensive and/or fossil fuel derived fertilizers, I started composting my animal mortalities and  organic/green waste left over from our operation. But, in order to compost I needed a source of steady carbon for the composting process, so we planted around 60 acres of clumping moso-bamboo, 20 acres of "running" bamboo and bought a couple electric wood chippers to create our own carbon-negative sawdust replacement for creating our own "compost-tea" fertilizer on sight.
However, I found out that my yields of bamboo were so massive (we harvest the bamboo 2 to 3 times a year) that we're constantly trying to find new uses for it (turning it to feed for crickets, grasshoppers and grass carp which we use in turn to feed our tilapia and prawns, occasionally feeding it to our alpacas and gasification power during the winter months).
We added a "combined heat and power" 145 KW gasification + Waste Heat Recovery micro-steam turbine power generation system to make up for the loss of daylight and efficiency from our solar PV system during the winter months. And two years later we still have way too much excess bamboo.
So while looking up other ways to utilize it and I found out about Chinese researchers creating a "cost-effective" cellulose ethanol from it. So I decided to Google around if there are any pre-existing procedures or methods to create ethanol from bamboo but I'm pretty much getting squat.
So my questions are:

What's the basic process of creating fuel ethanol from bamboo? Do you know of any books or resources that describe this in more detail? If not, would there be any special procedures/material/equipment needed to create ethanol from it?
Would it be worth the capital investment of buying all the stuff required to produce only about around 3000-4500 gallons of a year? I was told that under form 5110.74, a small producer of ethanol can only produce 10,000 proof-gallons of it, which apparently translates to just over 5000 gallons of fuel or something.

I really don't want to have to sell my excess bamboo to flooring and paper companies that are likely still using environmentally detrimental ways to process and ship its derivative products because it would go against my desire to help protect the environment. But I also don't want to invest into a "money pit" that I will never see any ROI on (I'd rather just use it to expand my operation by hiring new employees, branching out into land-based Marine shrimp and seaweed farming, lithium-ion energy storage, etc).

How about selling the bamboo? Pieces about 2 m len x 8 cm dia retail for ~$US1 in local stores. You might also make (or hire someone to make) bamboo furniture, or just bamboo shipping crates for your produce.

Interesting question, but sadly it looks a bit like a 'wall of text' at the moment. Many people don't bother reading so much text. I tried to improve the readability somewhat, but you may want to remove some of the less-important details and shorten your question.

Making fuel out of any organic matter is one of the most wasteful use of it, if you can use it for anything else. Even just burning dry bamboo is much more efficient process with much less waste.

A process for creating fuel ethanol from bamboo is described in this original paper and this follow-up. The basic steps these researchers took are:

Bamboo saccharification with sulfuric acid
Removal of color compounds with activated carbon
Separation of acid and sugar (using an ISMB system)
Hydrolysis of oligosaccharides
Continuous ethanol fermentation of saccharified liquid (with yeast strain)
Distillation

There are also other, slightly different processes to do this (e.g. enzymatic hydrolysis, but this is more expensive).
I'm no expert on this, but it's my understanding that creating ethanol from lignocellulosic biomass is still in the research phase. AFAIK there isn't a good and economically viable process yet to create ethanol on a large scale. Several steps still need to be optimized and costs need to be reduced, so I don't think it's very interesting for you to do small scale production.


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Would it be correct to say that ozone is an element?
My level of chemistry knowledge is restricted to that which is given at high school level.
I was tought that an element is a substance that consists of only one type of atoms, meaning the number of protons is the same for every atom. Is this definition sound?
In case it is sound:
Since ozone gas only consists of oxygen atoms, would it be correct to say that ozone gas, as an entity, is an element?
Would it be more correct to say that ozone is an allotrope of the element oxygen?
Why is it called the table of elements, when it seems that we are actually listing the atoms and not the elements themselves?

Looks you're thinking too much about definitions ;)  And it is one of few allotropes.

Well, it happens that I am interested in having the correct definitions so that I can make precise statements and be sure that I understand things correctly.

You may call ozone an elementary substance. (Not an element, that's for sure.)

@IvanNeretin Then am I right in assuming that I can't refer to the substance oxygen gas as an element? I understand that this source http://goldbook.iupac.org/C01022.html claims that it is optional.

Technically yes. See, when an element exists in a form of one and only elementary substance (like, for example, most metals), people tend to use those pretty much interchangeably. Most of the time it's OK, but if your element has one "main" substance (like $\ce O_2$) and some "exotic" ones (like $\ce O_3$), that's where things start to get hairy.

Coming from a math background, could one say that this is in the same sense as one could say that 4 has two square roots, namely -2 and 2, but we say that 2 is THE root of 4, and therefore it is given that when we talk about the root of 4 we actually mean 2? So O2 is kind of the canonical alleotrope of the element oxygen. Are there any examples of alleotropes where one is not canonical in this sense?

Wow! That's an interesting analogy, though not without certain limitations. We may think of ordinary tin as the canonical allotrope and gray tin as the other one. Now, there are cases when we have a bunch of non-canonical allotropes (like carbon with diamond, fullerenes, nanotubes, lonsdaleite and who knows what else), but then again, one might think of $\sqrt[n]x$ with its complex values... Actually, it's phosphorus where the scheme fails; it has no undisputed canonical allotrope. See, white phosphorus is "canonical" in terms of $\Delta G$, and red phosphorus is the easiest to come by, but...

It would be more correct to describe ozone as an allotrope of oxygen. Dioxygen (about 1 in five of the molecules making up air) is another allotrope.
The periodic table describes the atoms that make up elements as this is the most fundamental property. Many other elements, though, exist in multiple allotropes. Carbon, for example, has several: graphite, diamond, buckminsterfullerene and quite a few others.
All allotropes consist of only one type of atom but the bulk physical properties depend on how the atoms are joined up (flat plates of carbon hexagons with multiple bonding in graphite, a three-dimensional network of singly-bonded tetrahedral carbons in diamond).
We don't normally add too much about allotropes in a periodic table as it would make the table a little big, although some include limited information about how many there are. And, usually, only one is thermodynamically stable under normal conditions (so the physical properties will be about that one).

O3 is a allotropic form of oxygen which is a element and also O3 is  made up by same atom so, we can called O3 a element


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What is the highest known bond order?
In science we learn about single, double, and sometimes triple bonds. From a quick search I have found up to sextuple bonds.
Is there a maximum bond order? If yes/no, what causes this?

In the paper that you've mentioned$^{\ast1}$, there is something described, which is called effective bond order ($\mathrm{EBO}$).
$$\mathrm{EBO} = \frac{1}{2}\left(\sum_{i=1}^N \eta_{b,i}-\eta_{a,i}\right)$$
... with $\eta_b = 2 − x$ as the occupation number of an occupied orbital and $\eta_a = x$ as the occupation number of the corresponding antibonding orbital. (In multireference quantum chemistry methods, the resulting averaged orbitals can have occupation values of $0 \leq x \leq 2$)
In 2013 Ruipérez, et al. calculated the effective bond order for several homo- and heteroatomic dimers.$^{\ast2}$ Based on their very high quality calculations, the

[...] results show that the effective bond order ($\mathrm{EBO}$) of the $\ce{MoU}$ dimer (5.5) is higher than that for the tungsten dimer (5.2), known to date as the molecule with the highest $\mathrm{EBO}$.

So somewhat arbitrarily, $\ce{MoU}$ seems to be one of the best candidates for the highest bond order.

References:
$^{\ast1}$. Björn O. Roos, Antonio C. Borin, Laura Gagliardi, "Reaching the Maximum Multiplicity of the Covalent Chemical Bond," Angew. Chem. Int. Ed. 2007, 46(9), 1469–1472 (https://doi.org/10.1002/anie.200603600).
$^{\ast2}$. Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, "Molecules with High Bond Orders and Ultrashort Bond Lengths: $\ce{CrU, MoU,}$ and $\ce{WU}$," Inorg. Chem. 2013, 52(6), 2838–2843 (https://doi.org/10.1021/ic301657c).

It seems that I have answered my own question:

The maximum bond order achieved between two atoms in the periodic table is thus six [sextuple] and is represented by the Mo and W diatoms.

From Reaching the Maximum Multiplicity of the Covalent Chemical Bond, by Bjrn O. Roos, Antonio C. Borin, and Laura Gagliardi

Bond order is the number of electrons shared between two atoms divided by two. There are a few things that limit how high a bond order can possibly go, however.
First, atoms can usually only form bonds until their valence electron shells are filled (any more electron would be unstable). The heaviest elements, those of periods 6 and 7, have at most 32 valence electrons (2 s, 6 p, 10 d, and 14 f), so one could imagine 16 as the highest bond order possible with the elements we've discovered so far.
However, atoms do not evenly split their electrons like this. An atom's electrons are in certain, limited allowed energy states called orbitals, and for a bond to form, (at least) two atomic orbitals from the two atoms have to mix with each other to form two molecular orbitals. Some of these atomic orbitals are already full and don't mix. For example, carbon atoms have 4 valence electrons and almost always form 4 bonds to gain a full octet, but there is no such thing as a carbon-carbon quadruple bond. The reason is because one of the four valence orbitals for a carbon atom, the 2s orbital, is already full, so the 2s orbitals of the two carbon atoms cannot mix with each other to form a bond (or, more accurately, they mix to form bonding and antibonding orbitals which are both filled, contributing a net 0 to the bond order). Only the three half-filled or empty p orbitals can mix, so the highest bond order achieved by carbon is 3. Based on this reasoning, we can imagine the element Gd, which has 8 half-filled orbitals (a 5d orbital and all seven 4f orbitals), would be the absolute limit with a bond order of 8.
However, there are yet more limits on how high bond order can go. First of all, in order for atomic orbitals to mix, they have to overlap in space, but electron density decreases with increasing distance from the center of the atom. So the atoms that have the largest valence shells also tend to be too far away from each other for their orbitals to overlap and actually form bonds, so their multiple bonds tend to be weaker. Also, as elements get heavier, their nuclear charge increases, and as a result some of their electrons are moving at speeds approaching the speed of light. This leads to relativistic effects that change the behavior of the valence electrons from what we would expect; for instance the 7 valence f electrons of Gd actually behave more like core electrons.
On a final note, electrons repel each other, so it is actually destabilizing to concentrate huge numbers of them between two atoms. This can be seen with carbon, for example - a carbon carbon triple bond is stronger than a double bond, and a double bond is stronger than a single bond, but a triple bond is less stable than three single bonds or a double bond and a single bond. This is another reason why multiple bonding is limited, and why the highest bond orders (such as 6) are between atoms with very large nuclei (and thus high nuclear charges), like U and Mo.
As you go down the periodic table and get to heavier and heavier elements, relativistic effects become more important, so we are still not certain how the electrons of the heaviest elements arrange themselves and what their maximum bond orders could be.


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How much heat is required to raise the temperature of 230 g CH3OH(l) from 22.0 to 30.0 ∘C?
How much heat is required to raise the temperature of $230\ \mathrm g$ $\ce{CH3OH(l)}$ from $22.0\ ^\circ\mathrm C$ to $30.0\ ^\circ\mathrm C$ and then vaporize it at $30.0\ ^\circ\mathrm C$? Use enthalpy of vaporization $\ce{CH3OH(l)} = 38.0\ \mathrm{kJ\ mol^{−1}}$ and a molar heat capacity of $\ce{CH3OH(l)}$ of $81.1\ \mathrm{J\ mol^{−1}\ K^{−1}}$.
I Used the equation $Q_\text{total} = Q_\text{heat} + Q_\text{vap}$. My answer was $88.9\ \mathrm{kJ}$.
However, it is wrong.

Without seeing your work it's difficult to tell where you went wrong.  Your basic method is correct though.  Just to check, I get 7.18 moles of methanol.  Just considering the enthalpy of vaporization, that's going to be much more than 89 kJ.

The first thing to be sure of is that you're doing all the conversions correctly. Molar mass is the number of grams in a mol. It's found by adding the molar mass of all the elements in a compound times the number in the compound. For example the molar mass of $CH_4$ would be $$\textrm{Molar Mass of }C + 4\times\textrm{ Molar Mass of }H$$
Once you've found the correct molar mass it should have units of mass per mol - most commonly $\textrm{grams mol}^{-1}$. Thus if you have the mass of a substance and the mass required to have a mol, you can divide mass by molar mass to find moles. Then you simply:
$$Q_{heat} = \Delta T\times \textrm{ number of moles } \times C_p$$
Where $C_p$ is heat capacity. To find the heat of vaporisation again you just multiply the per mol value by the number of moles:
$$Q_{vap} = \textrm{ number of moles } \times \Delta_{vap}H^\circ$$
Where $\Delta_{vap}H^\circ$ is the heat of vaporisation. Be sure you got all these steps right and be aware of the difference between kJ and J! ($1\textrm{ kJ} = 1000\textrm{ J}$) Using this method I got a very different answer to yours. Feel free to post your own working for us to look at!
Note: Sometimes enthalpy of vaporisation values are actually the energy to vaporise from 298K (24.5C) but this can't be your problem because, as Jason Patterson said, even with just the enthalpy of vaporisation, no heating, your value is too low.


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How do I build a thermochromic thermometer?
My Chem 152 lab is requiring us to build a thermochromic thermometer utilizing methods from previous labs such as absorbance spec and temp.
We've been taught how to calculate Kc values successfully as well as delta H and delta S from these methods, but I'm unsure of how to combine everything into creating a thermochromic thermometer.
We're given $\ce{CoCl2(EtOH)2}$ and $\ce{CoCl2(iPrOH)2}$ to react with $\ce{MeOH}$. We had to complete a pre-lab assignment by calculating $K_c$ values at $275\ \mathrm{K}$ using $\Delta H$ and $\Delta S$ values from a reaction of $\ce{CoCl2(EtOH)2/MeOH}$ and a reaction of $\ce{CoCl2(iPrOH)2/MeOH}$. I was able to determine whether each reaction was reactant or product favored based on their $K_c$ values, but I'm completely stumped on how to use all of this information in creating a thermochromic thermometer.

We calculate the $K_c$ values. Ratios that give $K_c$ greater than 1 are product favored and will turn from blue to pink at low temps. The opposite is true for the ratios that give $K_c$ values less than 1. These are reactant favored, and will turn from pink to blue at the low temp. First, look to see what is chaging color most drastically, meaning the reaction is proceding to a greater extent ($K_c$ value a lot greater than or less than 1), then look at the cost. If it's too expensive, a thermometer that is less effective but still OK, may be better.
Overall, think of a thermometer that you stick on the side of a fishtank (sort of).


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Why is carbon monoxide stable(compare with other flammable gas) in room temperature?
I know carbon monoxide can burn with high temperature, but why doesn't it explode easily compared to other flammable or explosive gas?

It is generally hard to argue about kinetic parameters in such a broad comparison. Combustion mechanism of even simple compounds like methane contains several hundred (thousands) steps - so it is not a question about one or two elementary steps.
I would say CO has relative low energy content compared to the most flammable guesses due to the high oxygen content / oxidation state of carbon, and this thermodynamics is reflected in kinetics, too. It still can explode, though (comparison of explosive gas mixtures):
http://www.engineeringtoolbox.com/explosive-concentration-limits-d_423.html

I would say CO energy content is not all that low; its molar heat of combustion (see the last column) is about the same as that of hydrogen. So this is just kinetics. After all, $\mathrm{C\equiv O}$ is one of the strongest chemical bonds ever.

@IvanNeretin Whatever is the answer, "it is just kinetics", by definition. Hydrogen is a special case, as it has a very simple reaction mechanism. However if you compare CO to hydrocarbons, alcohols, ethers, the heat of combustion is low, either comparing the mass or the molar values.

I agree with your comparison, but object to your evaluation of hydrogen. Its reaction mechanism is not quite simple. Admittedly, it is simpler than that of hydrocarbons, but what if we compare it to CO? Just look at the number of possible reaction intermediates in both cases.

Maybe. Also, I think you should give the strengths of C-O bond as an answer,   no only as a comment. Note, bond energies and heat of combustion are not unrelated..

"It is just kinetics" is always true, but I think Ivan was saying it because thermodynamics is unrelated to kinetics.  Note the contrary and in my mind mistaken impression given by your answer: "thermodynamics is reflected in kinetics"

@CurtF. He might, but note that the core of his answer ( C≡O is a very strong bond) is essentially a thermodynamics argument.

What gives you the impression that carbon monoxide does not burn easily?
Below I will link to a youtube video of an accidental ignition of a large amount of carbon monoxide.  But be warned!  The video is a highly uncontrolled, dangerous experiment.  The person who performed it did not have appropriate safety equipment and acted very recklessly.  DO NOT TRY TO PERFORM A SIMILAR EXPERIMENT YOURSELF unless you know what you are doing and have

A CO detector / alarm
A buddy in the room with you
A fire extinguisher
Lots of experience synthesizing gases using laboratory apparatus

...

OK, all that said, here is the video.  As you can see, carbon monoxide is quite flammable.  And that guy is an idiot, at least in my book.
The autoignition temperature of $\ce{CO}$ is ~609 °C, only 100 °C or so higher than the $\ce{H2}$ autoignition temperature, and lower than the autoignition temperature for solid carbon. The LEL and UEL are 12.5% and 74%, respectively, which is a very wide range compared to most compounds.
Compared to other gases, $\ce{CO}$ explosions may be comparatively less violent because of the relatively smaller change in moles during the combustion reaction:
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
Thus, two moles of gas are formed from three moles of reactants, which means that the shock waves and pressure forces produced by a sudden $\ce{CO}$ combustion reaction / explosion are not as large as some gases such as say propane.

I guess OP was referring tot he fact that CO has one of the  narrowest LEL - UEL region at such lists you just showed.

Narrowest??  It's one of the widest.  On the list I linked to, only ethylene oxide, acetylene, and hydrogen (and deuterium) are wider, except for extremely reactive gases like diborane and trichlorosilane.  Most "everyday" explosive gases including methane, ethane, propane, ethylene, propylene, propylene oxide, hydrogen sulfide, hydrogen cyanide, all the butylenes, butadiene, and even carbonyl sulfide have much narrower ranges than carbon monoxide.

Yes, you are right, my mistake.


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mechanism of mechanically triggered explosion
There are several well known explosives which can be mechanically triggered. I am curious about the mechanism of such a reaction, though.
I guess one can give an easy answer when the explosive is sensitive to everything, flames and mechanical activation, too. What puzzles me is that several of these compounds, e.g. nitroglycerin or ammonium-nitrate much more difficult to trigger by simple flame. How mechanical activation is different?

I've seen somewhere analysis of this, but dunno if I'll find it again

There are differing propagation mechanisms for an explosion:

Thermal: a deflagrating explosive such as gunpowder has a comparatively slowly propagating wave-front, depending on heat propagating through the bulk material to set fire to the next layer. This is similar to a candle slowly burning down.
Compression: some brissant (but opaque: see 3. below) explosives create a shock wave. "[High explosives] detonate to produce a defining supersonic over-pressurization shock wave."
Light: fuel-air mixtures, glyceryl trinitrate and nitrogen triiodide explosion can be initiated by light (auto manufacturers are looking at laser "spark plugs" to improve internal combustion engine efficiency). In these types of explosives, light from the explosion itself might propagate a supersonic blast front.

So initiating an explosion through heat alone may initiate just a deflagration, without enough over-pressure to form a blast wave.

Mechanical initiation (hitting with hammer) causes shock wave running through the bulk of the material. This means phonon (vibrational) excitation. In some compounds, the vibronic coupling enables the transfer of vibrational energy to electrons, meaning the molecule reaches excited state. This in turn can lead to bond breaking and the energy rich molecules and with plenty of nitro groups have enough oxygen and heat to "burn". But, as noted, in the whole volume of sample in the same moment.
Compared to flame initiation, where the substance can burn only on the surface.

Thank you for the answer. My problem is phonon/vibrational energies are generally at least an order of magnitude lower than electronic excitation energies - however if there are low electronic excitation, high temperature (i.e. ignition) should trigger a similar reaction. It is not true that substances burn on the surface. Many explosive materials have a positive oxygen balance and don't need air/oxygen.

I am not an explosion scientist, just did some stuff on mechanical activation of chemical reactions. So I just quickly searched on this specific topic. To say, explosives are energy rich, but the activation process is heavily system-dependent. See e.g. http://pubs.acs.org/doi/abs/10.1021/jp300711m


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Weight Percent of Solutions Given Without Units
I have been performing haloform reactions using sodium hypochlorite. The most available source is common household bleach, which is a solution of $\ce{NaClO}$ in water. The bottles invariably give a concentration in unitless percentages: e.g., "$6\%$ Bleach."
I have assumed this to mean: (mass of solute) / (mass of solution) * 100.
Now I see that there are other (mathematically erroneous) uses of the percent symbol %, including a measure called WV (weight volume), which seems to be normalized by dividing gram weights of solute by 100 ml.
An easy question, impossible without the units:
How many grams of $\ce{NaClO}$ are in one litre of $6\%$ household bleach?

I have assumed this to mean: (mass of solute) / (mass of solution) * 100

Most likely you are correct in this assumption. Percentage compositions of ionic solutes in water are nearly always implicitly expressed in percent-by-weight ("% w/w").
$6\%$ bleach is relatively dilute and will have a density very near to $1\ \frac{\mathrm{g}}{\mathrm{mL}}$.  So, $1\ \mathrm{L} \approx 1\ \mathrm{kg\ solution}$, and thus one can expect to have approximately $60\ \mathrm{g}\ \ce{NaClO}$ per liter of such a solution.


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How quickly will steel lose structural strength from a heat flash?
I want to know if a sustained jet fuel burn around steel has the ability to make the steel structurally unsound. It seems like any applied heat that effects steel will be a kind of "heat flash": a sudden, and possibly quite short-lived, burst of heat. In this situation, we assume steel has a melting point of $\mathrm{2750\ ^\circ F\ (1510\ ^\circ C)}$ and that jet fuel (the source of the heat burst) burns at a maximum of $\mathrm{1500\ ^\circ F\ (\approx 816\ ^\circ C)}$.
Some factors I believe need to be taken into place include the strength of bonds in steel, the thickness of the steel in question, and the distance from the heat source. Is a sudden heat flash enough to cause steel to weaken? Are there any other factors that need to be taken into consideration?

Are you using Fahrenheit degrees? Also there was supposed to be long fire.

Well as I am not trying to debunk the myth or situation, I am referring to a heat flash specifically. Also yes Fahrenheit.

I'm afraid that's rather for http://engineering.stackexchange.com/

I'm pretty sure jet fuel can't melt steel beams


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Why does dry ice "dissolve" so quickly in water?
I just stumbled across an ancient video of some people putting a 60 pound block of dry ice in a swimming pool. Pretty cool stuff.
It's pretty interesting to me that in the video, the dry ice sublimates much quicker in the water than in the air.
I am wondering why this might be? Is it possible that water is acting as some sort of catalyst? If so, how can I picture this on a microscopic level? (That might not be possible to answer but the catalyst portion is.)
I was thinking that the $\ce{CO2}$ might be motivated to form carbonic acid, and by Le Chatelier's principle, the dry ice would sublimate even faster.
I'm not so sure that makes sense though...
I don't know. Any thoughts?

Try it with water ice. With a bit of patience, you'll see the effect is still there - ice melts faster in water than in air. So, no - carbon dioxide is not the magic part, and neither is the solvent. What else is a difference between water and air? :) (hint: did you ever swim in cold weather? Which felt colder, water or the air?)

Note that this doesn't happen with small amounts of dry ice -- there's so little convective motion that the water next to the ice freezes and insulates it.  It's like a solid phase Leidenfrost effect but I'm not aware of any particular name associated with the process.

It's just about heat capacity + heat conductance, nothing more. $\ce{CO2}$ sublimates much quicker in the water because water is much better at heat exchange than air. Anybody who ever experienced swimming in water at 15°C and walking in air at 15°C can attest to that.
Catalysts for heat exchange are usually made of copper and other metals. (Also, usually they are not called catalysts.) You have one in your computer, attached to the CPU.
As to formation of carbonic acid, it hardly has any influence on the process (though it does happen to a certain extent, of course). Most $\ce{CO2}$ escapes in gaseous form anyway.


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Internal treatments of gas cylinders for toxic gas mixture
The values of toxic components  like $\ce{H2S}$, $\ce{NO}$, $\ce{SO2}$, etc. [in ppm level] get decreased after a while due to adsorption phenomenon. Manufacturers design some internal treatments in order to prevent adsorption to obtain a stable mixture.

Do you have detailed information about different kinds of these treatments or do you know where I could find them?
What are other solutions leading to resolve this problem?


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Does pyrite retain the magnetic properties of its element iron?

You are walking in the mountains and pick up what looks like a nugget of gold. You're aware that fools gold (iron pyrite, $\ce{FeS2}$) exists. Create an experiment to test if this is pure gold or iron pyrite.

The simplest thing I could think of is to do a magnetic test on the assumption that if it was fools gold it would be attracted to a magnet due to the presence of iron. A quick google search reveals this to be true but I have a sneaky suspicion that I got lucky in assuming, because there is iron in $\ce{FeS2}$ the compound will be magnetic.
I'd like to know if chemical compounds retain the properties of their individual elements or if the compound gains completely new properties some of which may or may not be properties of the elements.

As a general rule, compounds do not retain the properties of their elements. A classic example is NaCl, as discussed here

Simplest way is density - pyrite is about 4 times less dense than gold, lighter than pure iron.

Pyrite is not magnetic. Some related minerals are, but only weakly, not as strong as iron, so the test will likely fail anyway. Magnetism is not an inherent property of an element, but that of a compound. Most compounds of iron are not magnetic (though some are); on the other hand, there are organic magnets made out of totally non-magnetic elements.
Think of something based on chemical properties instead. Pyrite is a salt of a metal and a weak acid. Gold is a metal, of the kind that would not react with acids. See where this gets you.
As for the other properties of the elements, some would be inherited by their compounds and some (arguably, most of them) would not.

+1 for the organic magnets. I didn't hear about them before. It is quite an interesting information. Just found this nice review on organic magnets.


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Have Mossbauer spectroscopy on 67Zn ever been used to study Zinc containing enzymes?
Have Mossbauer spectroscopy with 67Zn as absorber and 67Ga as emitter ever been used to study zinc containing enzymes?
The perfect answer will contain link to article/book where example of 67Zn usage is given or article/book where author tells that there is no such thing yet.

A quick literature search for titles with 'Mossbauer'+'enzyme' came up with only Fe Mossbauer stuff (and not much of that). On the other hand, I could have missed something in the list, or it just doesn't appear in the title.

Same here. It turns out that proving something do not exist is much harder than proving something do exist.

Indeed! So, the question to ask is: Do you want to do the experiment because nobody else has, or do you want to do the experiment to learn something in particular about an enzyme? Either way is fine, but be clear to yourself what your purpose is.

As far as I can see: yes. I've just googled it. And still: if it's not on google someone could still have done it before.
Zn-exchange and Mossbauer studies on the [Fe-Fe] derivatives
of the purple acid Fe(II1)-Zn(I1)-phosphatase from kidney beans.

@AmelieB.Blackstone In that article, the Mossbauer aspect refers to the Fe atoms.

You can directly contact Mossbauer labs specialised in the field, if you want to be sure. I see some problems here: 1) enzyme samples are very diluted, and it is very hard to make even the iron experiments, 2) the payoff is much smaller than in the case of iron. Iron has several oxydation and spin states, several of them are paramagnetic, therefore Mossbauer is rather informative. Zn spectra would give much less intersting data.


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Why is the van der Waals coefficient b equal to four times the volume of the particle?
In the van der Waals equation of state
$$\left[p + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$
the coefficient $b$ is supposed to represent the volume occupied by the particles. Why then is it equal to four times the volume and not simply the volume?

While most everything the previous answer states is correct, I would point out that taking four times the volume of a single particle has nothing to do with experiment and arises mathematically.
In deriving the VDW equation, the particles are still assumed to be hard spheres, but this assumption is corrected for with the parameter $a$.
The hard sphere approximation forbids that two particles penetrate each other's radii. Thus, we find that two spheres in closest contact are surrounded by a sphere of radius $2r$ (or the diameter of one of the original spheres).

(source: nyu.edu)
Thus, the volume excluded by the particles from the larger sphere surrounding the two spheres shown is
$$b' = \frac{4}{3} \pi d^3 = 8 \cdot \frac{4}{3} \pi r^3$$
Thus, the excluded volume per particle $b$ is $b'/2$ or,
$$b=4 \cdot \frac{4}{3}\pi r^3$$
which, as you point out, is four times the volume of a single particle.
The interesting thing about this is that it does not represent the actual value of $b$ for any given atom, but represents the upper bound of $b$ for any given atom. What I mean by that is, $b$ could very well be correct by calculating four times the volume, but often experiment will show that it less than the calculated value of $b$ because atoms are not hard spheres.
For instance, using helium, which is the closest we'll get to a hard sphere:
$$b_{\ce{He},\mathrm{calc}} = 4 \cdot \frac{4}{3} \pi (140\ \mathrm{pm})^3 \cdot N_\mathrm{A} = 0.02767\ \mathrm{L\ mol^{-1}}$$
while,
$$b_{\ce{He},\mathrm{exp}} = 0.0238\ \mathrm{L\ mol^{-1}}$$
So, the experimental value of $b$ is indeed smaller, but the calculated value gives a rough idea.

The short answer to your question is because that is what the experimental data showed. The constants in the van der Waals equation were originally derived empirically for each gas. The constant $b$ happens to correlate well with four times the molecular volume for a number of gasses, which lets us calculate it for other gasses.

To answer your question a different way, let's first think about two particles that have such an attractive force that they form a chemical bond when they collide. Even these two atoms cannot get closer than a certain distance due to internuclear repulsion. Consider the following image from Wikipedia user Mark Sozoma.

There is an optimal internuclear distance. When the atoms continue to get closer together the potential energy skyrockets.
For example, the atomic radius of a hydrogen atom is 25 pm, while the equilibrium bond length $r_e$ for $\ce{H2}$ is 74 pm. Essentially, there is one hydrogen radius between the atoms. Put another way, there are two hydrogen radii between the "surface" of one atom and the center of another.

The van der Waals equation applies to gaseous particles that are not likely to form bonds to each other upon collision, though they may still have some attractive forces. Thus, it stands to reason that the internuclear distance during a collision might be larger - i.e. there might be as many as two atomic radii between the "surfaces" of the atoms.


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What is the nucleophile in alkaline hydrolysis of a haloalkane?
We know that haloalkanes are able to undergo $\mathrm{S_N1}$ or $\mathrm{S_N2}$ in alkaline hydrolysis, and the hydroxide ion is the nucleophile responsible for attack. But when we talk about hydrolysis, isn't it a reaction with water? Where does the hydroxide ion that attacks the haloalkane really come from; the water, or the alkaline catalyst?

Consider the reaction of a haloalkane with aqueous sodium hydroxide to produce the alcohol plus sodium chloride. The net reaction consumes one hydroxide ion for each alcohol molecule produced, so it is not really correct to refer to it as a catalyst.
(1) If this reaction occurs via the $\mathrm{S_N2}$ mechanism, a one-step displacement mechanism, you are correct to say that hydroxide (being a much better nucleophile than neutral water) is the nucleophile attacking carbon.
(2) If the reaction occurs via the $\mathrm{S_N1}$ mechanism (note that which mechanism is preferred depends on the structure of the haloalkane, and in some cases a haloalkane can react via both mechanisms), you have a two-step process in which the haloalkane initially dissociates (a slow step) to form a carbocation plus halide ion. The carbocation then reacts very rapidly with whatever nucleophile is handy to form the alcohol. Water is likely more abundant than hydroxide, and is a strong enough nucleophile to react instantly with a carbocation, so most of the product will be formed that way, with release of $\ce{H+}$, rather than by combination of the carbocation directly with hydroxide. The $\mathrm{S_N1}$ reaction of a haloalkane in this manner with excess water or other polar solvent is often referred to as a "solvolysis" reaction.

thanks for your information :) So from your explanation, can I conclude that 1. in SN2, it is the hydroxide ion(which is not from water but from alkaline) that acts as a nucleophile and in SN1, it is the water itself.  If the statement for SN1 is justified, does this mean that the hydroxide ion is actually not required in the hydrolysis process?

Both conclusions are correct. However, the reaction might proceed faster in the presence of ions than in pure water. Also, please note that hydroxide ion in water exchanges protons with water at a diffusion-controlled rate, so the OH groups that end up in the product could still have originated from the water rather than from the NaOH that you started with.


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Create combinatorial library of biologically plausible chemical compounds based on library of known compounds
I recently downloaded all the molecular structures (about 13,000) stored in Pherobase, an open library of insect pheromones. I would now like to create a derived, more extensive combinatorial library of compounds that would likely occur in nature, but right now don't happen to be included in the library (e.g., similar aliphatic compounds but with a different chain length or different positions of methyl branches etc).
I know I can create these semi-manually by specifying certain series using SmiLib, or using ChemAxon's Reactor. I was wondering though if there are any good cheminformatic methods to recognize biosynthetic series, which could, e.g., subsume lists of compounds in the library in  Markush notation, which could then later be expanded. Or what could be other good ways to expand the library to include other biologically plausible related compounds?
Could Chemaxon's Metabolizer perhaps help, maybe after specifying plausible biosynthetic/enzymatic reactions? Any other ideas?
Just looking in Pubchem for structurally related compounds I think wouldn't work as that would result in too many biologically implausible compounds ... (The reason I want to do this is in the context of making a large QSPR library with in-silico predicted retention indices, and later perhaps also predicted EI-spectra).

I think your best bet would be to go to the KEGG Reaction Database (http://www.kegg.jp/kegg/reaction/) which is a comprehensive database of known biochemical (enzyme-catalyzed) reactions, plus various associated tools. It includes a module that predicts a likely biosynthetic pathway for any pair of structures. All your pheromones are natural products and are formed via enzymatic reactions, so you should be able to explore out from your starting set by exhaustively applying the reactions in the KEGG database to each one of them. There may be a module that does this for you  automatically, but it is not documented on the KEGG web site.

Thanks for the suggestion - I'll look into that! Unfortunately only a tiny % of the Pherobase compounds are also in Kegg, so I am not sure how well this would work... If I could get the KEGG reactions into Chemaxon Metabolizer then this might perhaps also work...


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Distinguishing pentanol isomers in 13C DEPT NMR spectroscopy
Pentanol has eight isomers as shown below. The circles represent carbon atoms that I think are equivalent.

Below there are eight DEPT-spectra that should represent these eight substances.

As you can see some won’t line up. What’s the problem?

In an ideal non-moving world at $0\,\mathrm{K}$, every atom would give exactly one peak in NMR spectra. So the actual question is why some of these spectra have less signals. Partly this is due to local symmetry. If, e.g., two methyl groups are attached to the same carbon atom, these two are said to be chemically equivalent, i.e. one can rotate bonds to overlap one over the other. A different example of chemical equivalence is pentan-3-ol which you can rotate around the $\ce{C-OH}$ axis to give an almost identical molecule, putting the $\ce{CH2}$ groups on top of each other and the methyl groups, too.
With that, you did the circling right, and we should expect:

3 spectra with five signals;
3 spectra with four signals; and
2 spectra with three signals.

However we see one spectrum with two signals and are lacking one with four signals. (And technically, if you look at spectrum 2 closely, that could be interpreted as having five signals, not four.)
In a standard $\ce{^{13}C}$-DEPT spectrum (i.e. the ones you showed), carbons from $\ce{CH}$ and $\ce{CH3}$ groups point upward and those from $\ce{CH2}$ groups downward — and there is your answer. There are carbons in the structures that do not have hydrogens bound to them and which thus do not show up in the DEPT spectra.
In case you’re interested, I will give you the appropriate IUPAC names of the pentanol isomers that were measured in the following spoiler tag.

 2-methylbutan-2-ol. Carbon 2 cannot be seen.

 3-methylbutan-2-ol. The two methyl groups at the end are not completely identical and thus give two signals.

 pentan-3-ol

 pentan-1-ol

 2,2-dimethylpropan-1-ol. Carbon 2 cannot be seen.

 2-methylbutan-1-ol

 pentan-2-ol

 3-methylbutan-1-ol

Assigning spectrum 2 to structure 7, the higher field methyl could be different if rotation is restricted aroud the CH-CH bond. Increasing temperature should collapse the two signals.  Structure 8 should have only 2 signals, as spectrum 5 and structure 6 should have 3 as specrtum 1 as non-protonated carbons give no signal.


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Pubchem: list all compounds for which Kovats retention indices are available
Pubchem recently started listing Kovats retention indices, e.g. for
alpha ionone:
https://pubchem.ncbi.nlm.nih.gov/compound/5282108
I was wondering though if there is a way to get a list of all Pubchem compound ids which have this info available? Ideally via a PUG REST query, and a JSON or XML export option?
Note:
doing a search
https://www.ncbi.nlm.nih.gov/sites/myncbi/searches/save?db=pccompound&qk=14
https://www.ncbi.nlm.nih.gov/sites/myncbi/searches/save?db=pccompound&qk=12
will get me all records uploaded by NIST and NIST Chemistry Webbook - this will already get me close. But how can I get the list of all these pubchem IDs, ideally using some PUG REST query? (it has to be used programmatically from R)

Ha NCBI helpdesk just replied with this:
The easiest way to do this is to use the PubChem Classification Browser to access the list of Compound records with Kovats data.
You can find the PubChem Classification Browser here:  https://pubchem.ncbi.nlm.nih.gov/classification
1) Click on the "Select classification" pull-down and select "PubChem Compound TOC".
2) Under "Data type counts to display" click on "Compound".
3) Scroll down to the TOC listing and expand the "Chemical and Physical Properties" section and then "Experimental Properties".
4) Look for the entry "Kovats Retention Index" with the number of counts next to it.
5) Click on the number of counts (currently 79,656) to take you to PubChem Compound results list populated with those records shown. (https://www.ncbi.nlm.nih.gov/pccompound?DbFrom=pchierarchy&Cmd=Link&Db=pccompound&LinkName=pchierarchy_pccompound&IdsFromResult=1856976)
To save the list of CIDs for these records there are two options....
1) Click "Send to" (near the top).
2) Select "File".
3) Select "UI List"
4) Click "Create file"
OR
Click "Structure Download" (on the right)& follow the directions....to download the SDF or XML formats of these records.
If you don't need the 3D versions of the records/images this will work fine, HOWEVER if you want the "3D Records/Images" versions of these - there is a limit of 50,000.  So you'd need to download the CID list and split it in two to upload in this tool for downloading the file.
I hope this helps.
NCBI User Services


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Iodine clock reaction
I have been investigating the iodine-clock reaction for my chemistry project. I have done a great deal of research on the titration of iodate ions, iodide ions and sodium thiosulfate using HCl. But, I still don't get the use of this reaction. Where can this type of reaction be used in our lives.
I read articles that said iodine clock reaction can be used to determine the concentrations or pH. But where can it be used? What are the real-life applications of the iodine clock reaction?

I doubt if there's much interest in it besides academic research and popularization of chemistry.

You need to separate the iodine clock reaction (one that ticks back and forth) from the underlying iodine-forming reaction that leads to the starch-iodine complex. Only the latter has proper applications.

Yes, these reaction are only of academic interest only and do not find any practical purposes. These reaction are very special due to its unique kinetic and thermodynamic properties.

I believe these so-called "iodine clock reactions" are frequently used to teach chemical kinetics in the school laboratory?

This iodine clock reaction project can apply to the real world in different ways, but I think the biggest way is concentration. After doing some research, I found out that it can teach you about concentration, because it shows you that once you add any bigger amount to a substance, it changes reaction time, or in a real world case, how a substance feels, acts, looks, and possibly how it smells.
I know this is really late and you probably don't need this and you are done your project, but I'm just a bored grade 7 doing the same science project.