text:
Is the proton the strongest acid?
Having looked at the various definitions of acids and bases and having refined my understanding of it after learning about the inadequacies of pKa and the novel use of the Hammett acidity function, I would like to ask if the bare proton is the strongest acid?
I would like to define "acid strength" not as the extent of dissociation but more simply as the ability to protonate other chemical species.
I have come to this conclusion after reading through a post on the explanation behind the strength of fluorantimonic acid, being that the bare proton is liberated and that the conjugate base is so well coordinated, allowing the charge to be spread out over a large structure, stabilising it to a great extent.
There is no doubt about the proton being the strongest acid in the Brønsted-Lowry sense. Similarly, in the Lewis sense, this should also be logical as what could possibly more electrophilic than a bare proton?

An acid is a proton donor , not the proton itself

Related answer by @Geoff about the strength of acids (or difficulty) here

@NotEvans. It was only briefly mentioned in the comments and was not touched on in detail. Also, I am assessing the ability to protonate, not the other criteria (i.e. gas phase proton affinity, Hammett acidity function)  they have mentioned.

So you're saying if I took molecule X, you're interested in how a series of acids would protonate X to form X-H ?

@NotEvans. If I understand you correctly, yes. H-X+.

check out https://chemistry.stackexchange.com/questions/46882/do-bare-protons-exist-even-transiently-in-aqueous-solution https://chemistry.stackexchange.com/questions/41997/does-the-hydrogen-ion-actually-exist

@NotEvans. The posts you have provided in the links do not address my concern directly, meaning that they do not answer the question "Is the proton the strongest acid?", although one does mention about the very strong superacidity of the proton.

Problem is there's only kinda "bare" H+ even in fluoroantimonic, not really bare. You need sth like particle beam to have bare protons.

https://chemistry.stackexchange.com/questions/72075/can-atoms-other-than-hydrogen-be-reduced-to-a-bare-nucleus

https://chemistry.stackexchange.com/questions/34123/can-you-in-reality-give-neon-a-charge-of-8

Is the bare electron the best reducing agent?

An acid is a proton donor , not the proton itself.  It may be questionable, as an  object can provide itself. It is similar like if currency was defined by its exchange ability for dollars and dollars were not considered as currency, as you cannot exchange them  for a dollars, as they already are dollars.

Surprised no one has mentioned the scientist who took a 76 GeV proton beam through his brain and survived: https://en.m.wikipedia.org/wiki/Anatoli_Bugorski. Personally, I consider proton accelerators to be very strong acids! ;-)

No
Brønsted theory
In Brønsted theory $\ce{H+}$ isn't an acid at all. Acids lose protons, becoming conjugate bases, and $\ce{H+}$ is the proton itself.
Arrhenius theory
$\ce{H+}$ isn't an acid, because in this theory acids dissociate in water to form hydrogen ions.
Lewis theory
$\ce{H+}$ is an incredibly strong acid, but nuclei of other, heavier elements, for example alpha particles, are arguably stronger. I haven't found hard data for this and it may be rather difficult to get, these aren't your friendly neighbourhood Lewis acids ;)
Protonating agent
Now, that's a tricky issue! In principle, bare $\ce{H+}$ might be the ultimate protonating agent. In proton transfer, with any Brønsted acid you could always try to find an acceptor weak enough that reaction constant would be lower then 1, proton would "prefer" to stay with acid then protonate base. That's not the case with bare proton, which is unbound. Following this logic, it should beat any Brønsted acid.
Where the catch? There's perhaps more than one.
Whether bare proton can bind to a species depends on its energy, which needs to be lower then proton affinity of a molecule to which it's supposed to bind. Otherwise proton may ionise the molecule instead, and even fuse with one of its nuclei, if energy is high enough.
Another thing is that acids like $\ce{H4O^2+}$, which are endothermic molecules, could beat a bare proton, because of their repulsive nature - they spontaneously lose protons and "throw them away" with positive charge of their conjugate base!

Ok got your point. What about protonating ability? Would it be the strongest in terms of the ability to protonate other species? Perhaps I wasn't clear enough, but I am not looking for an analysis with respect to the different definitions. But rather in terms of protonating ability only

It may be questionable, as an object can provide itself. OH- would not be an Arrhenius base either.

@Poutnik Indeed it is not. Arrhenius theory is about chemical compounds and H+ or OH- aren't compounds, but species being dissociated from compound. Having a proton beam pretty much breaks down a theory, but it's not much of a problem and it hardly protonates anything, instead ionising, breaking bonds etc.

One can very formally consider "nothing" as extremely weak B-L base and a bare proton as it's extremely acidic conjugate acid, able to protonate near anything..

Interesting question. I'd say that the strongest acid would have to be something that cannot be protonated. Probably something like HeH(+). Helium is so noble it shouldn't be protonated at all although you can make this compound in the gas phase it still is considered to be one of the strongest acids.

Could you please substantiate your answer? I have never heard of this perspective before.

Take water for example, the conjugated bases of strong acids (superacids) cannot be protnated by the solvoacid (H3O+). If you switch to different systems it is possible using Carborane acids for example to even protonate Benzene. So for the OP's question I assume a solvent or gasphase, which extends really far towards the superacidic site and then. Here a super strong solvoacid can be predicted, which protonates anything. But what is the limit for protonation? The proton affinity. The lower the harder it is to protonate it, the more easily it will give the proton of, which is called an acid.

Certainly Helium, being the most unreactive (although I am not sure whether Neon might be even more unreactive) element or compound is a super stable conjugated base of HeH(+). So if you are able to make HeH(+) this should easily try to give off it's Proton again.

There is a very very old "paradoxical" question: "Which came first: the chicken or the egg?" If all chicken eggs come from chickens and all chickens come from eggs, then there is a paradox of causality.
Your question, after including the various comments you've added to it, has a similar problem: Is H+ an H+ donor? You appear to believe it is, I disagree. It is a matter of definition, so we can each hold mutually exclusive definition and self-consistent belief about what does and does not constitute an "acid". I don't know one way or the other whether your definition is self-consistent. I would argue it isn't very useful and also point out to you that you are not (nor am I) an authority in this field. Given that, I suggest you (re)consider whether you are competent (knowledgeable and experienced) enough to be creating your own definitions for technical terms.
The "bare" proton exists in some environments. (Of course, it exists even in the vacuum of space at a minimum surrounded by all sorts of quantum fields so isn't "really" bare.) These situations are not often (never as far as I know) used in the Chemical literature in discussions of acid/base strength for the obvious reason that they involve (arguably) more Physics than Chemistry. The bare proton doesn't exist in solution (by definition) and it's well known that a proton solvated in liquid ammonia will have different "strength" than one solvated in water.
Which gets us to the question that needs to be answered clearly before you go promoting your definition of acids to include H+: What do you mean by the term "H+"?

When I mean "proton", I mean an unsolvated proton. Assume that it can exist theoretically exist.

So can an alpha particle, which has two positive charges and no electrons.  And it isn't just theoretical. You probably have a few of them floating around in your smoke detector.

The current acid base explanations mostly involve covalent bonding between the acid and the base and frequently in the explanations fail to mention solvation and other subtleties. The high ionization energies of helium do not seem to allow covalent bonding for a measurable time, so it seems foolish to say He++ is an acid. Maybe very low energy alpha particles? The proton is bit more interesting. The OP instead of making a suggestion based on fantasy has an opportunity to do some research [maybe it has been done]. The Earth, Moon, Space station are bombarded with protons in the solar wind. Are the initial reactions an acid reaction to form a covalent bond or a radical ionizing reaction. Do the research, present the data, and formulate an explanation and let everyone attack or praise you. Are proton beams of varying energy available at CERN? Perhaps they can help out.

This certainly got attacked, not praised.


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Is cyclohexene an alicyclic or aromatic compound?
Classification of organic compounds have open and closed chain compounds. Closed chain further have homocyclic and heterocyclic componds. In homocyclic compounds there are two types: alicyclic and aromatic compounds.
In alicyclic compounds, only singles bond are allowed, whereas in the aromatic category the ring of carbon atoms has alternate single and double bonds (i.e. conjugated double bonds). So what happens to the compounds which have a lone double or triple bond?
For example, cyclohexane is classified under alicyclic, but where would cyclohexene or cyclohexyne be classified?

Setting aside the severe steric  strain that cyclohexyne would require, an aromatic all-carbon ring must have a full cycle of unsaturated atoms.  Cyclohexene and (if you can make it) cyclohexyne have saturated carbons in the ring and thus must be alicyclic instead of aromatic.

fyi, cycloalkynes do exist, though most are only transient/unstable intermediates. Wiki has it covered, here

Adding to @NotEvans comment, the smallest stable cycloalkyne is cyclooctyne.

If you look on the wikipedia page for alicyclic compound you will find cyclohexene as example for an alicyclic compound there. Cyclohexyne doesn't exist but it would be an alicyclic compound as well.
The IUPAC gold book defines alicyclic compound as:

Aliphatic compounds having a carbocyclic ring structure which may be
  saturated or unsaturated, but may not be a benzenoid or other aromatic
  system.

which means double or triple bonds are fine, just no aromatic system.

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Formation: So much heat, where does it all go?
I was comparing heat of formation numbers with specific heat values, and the numbers look really weird.
Take a typical example like magnesium fluoride. Its $\Delta H$ is -1124 kJ/mol and its heat capacity is almost certainly less than 0.1 kJ/mol K. Divide and you get at least 11000 Kelvins!
So the heat of formation doesn't seem to go into raising the temperature of the product. It can't be going into the surroundings either; otherwise the reaction would vaporize anything nearby. Where does it go, or what's wrong with my reasoning?
Edit: OK so let's not use a fluorine compound. Take aluminum nitride. -318 kJ/mol. Are we really going to suggest aluminum burns in nitrogen at over 3000 K?
Edit: Si3N4: -745 kJ/mol.
Edit: Regarding the article on adiabatic flame temperature: aluminum burns in oxygen at 4000 K. The $\Delta H$ of Al2O3 is -1680 kJ/mol. 1680/(4000-298) = 0.45 kJ/mol K. This would be a fantastic heat capacity whether we're talking constant pressure or constant volume.

Well, I can certainly imagine that magnesium "burns" pretty extremly in a fluorine atmosphere. And magnesium burns with >3000°C in air.

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

Magnesium burns vigorously in carbon dioxide with a very hot flame. I hesitate to think just how well it burns in fluorine. But I guess it might produce a lot of heat.

For these extremely exothermic reactions, a lot of energy is carried away in the form of photons, limiting how hot it can get. Recall that the power dissipated by a black body radiator is proportional to $T^4$, so it rises very quickly as the reaction becomes hotter.

@NicolauSakerNeto - I wish I could accept comments as answers.

Your problem is that you don't seem to understand the precise definition of heat of formation.   You are aware that it is related to the heat released or absorbed in the formation of chemical bonds, correct?

@ChesterMiller - Yes. I really don't see what I'm missing.

Thanks, but I think there is some more to the question than what I wrote. From what I understand, these adiabatic flame temperatures should in principle not be affected by emission of thermal radiation, but they are all significantly smaller than what your back-of-the-envelope calculation suggests

What is your precise definition of the heat of formation?  (I know the definition, but I want to know your understanding of what it is)

My understanding is that the heat of formation is disordered energy liberated when atoms bind to each other. I think binding involves changes to electronic energy modes (molecular orbitals) so the electrons experience an average reduction in their potential energy.

Yikes.  That's getting into a level of detail that I was hoping to avoid.  Here's my understanding:  You start off with elements that will react to form into a chemical compound, so the initial state is the elements in their natural states at a specified temperature and 1 bar pressure.  The final state is the compound at the same temperature and pressure.  The heat of formation is the amount of heat you need to add to the reaction mixture during this process, such that the final temperature of the compound is the same as the initial temperature of the elements.

So when the $\Delta H$ is negative, the reaction is exothermic, meaning heat comes from the reaction. Where does it go if not into degrees of freedom of the products (three of which define the temperature) and maybe some EM radiation?

It doesn't.  The heat is generated by forming chemical bonds.  In the definition of $\Delta H$, when it is negative, this amount of heat must be removed from the reaction mixture (i.e., transferred to the surroundings) in order to maintain the temperature of the mixture  constant at its pre-reaction value.

It's typically not EM radiation.  The heat is typically transferred to the surroundings by conduction/convection to a constant temperature reservoir (such as a liquid/solid mixture at its melting point).

I'm imagining the energy from a mole of product going into a few kilograms of ice and being just enough to melt it. Maybe that does make sense. I'm satisfied for now. If you write up an answer I'll accept it.

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Does ozone has dipole forces
Considering ozone in liquid state, certainly there are london forces, but does the position of its molecules induce dipole intermolecular forces?

"London forces" are dipole-dipole forces. Can you be more specific?

@pentavalentcarbon I was thinking about dispersion forces.

Ah. Dipole-dipole forces are dispersion forces. At least, when one induced dipole causes another induced dipole. If this happens, then dispersion is present.

Ozone has a bent structure with a partial positive charge on the central oxygen atom and partial negative charges on the two terminal oxygens. It has a permanent dipole moment of 0.53 D, according to Wikipedia.


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How to calculate the latent heat when there is a boiling temperature range?
I have a DME (dimethyl ether)- methanol - water mixture with the mol fraction composition: DME 0.4295, methanol 0.1353, water 0.4351.
This mixture has a boiling temperature range: dew point 141.6 °C and bubble point 69.95 °C, and the heat applied for the phase change (condensation) is a latent heat. How can I calculate it?
I know that for pure compounds, during the condensation process, there's no temperature change, so the latent heat is the result of:
Q = mass flow * latent enthalpy (constant value)
For a mixture that has a condensation temperature range, can I use the same equation? If this is not the case, how can I calculate the latent enthalpy for a mixture in function of the temperature?

I would guess there would be a parallel calculation, of both a latent heat and some sort of specific heat for the mixture, but I don't really know for sure.

Conceptually, it is possible to calculate the enthalpy of any mixture of these three compounds in the vapor phase and in the liquid phase.  So heat balances on pieces of equipment can thereby be properly carried out.  But, in the case of such a multi-component mixture, the concept of latent heat ceases to have realistic meaning and applicability in practice.


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Is this an intramolecular SN2 reaction?
What's the mechanism of the following reaction? Seems like internal attack occurs, but would be great if someone could provide a step by step mechanism.

Yes, mechanism of this reaction involves internal attack.


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What concentration of small molecule binding is considered weak or strong?
In the chemistry literature, I often see a stated concentration (nanomolar, micromolar, millimolar) followed by qualitative judgment on the potency of the small molecule for protein inhibition. What concentration range is considered weak, and what range might a small molecule be strong enough to be lead optimized?

The threshold for activity varies based on the target, but hit-level activity is typically 1 µM or lower. Targets are normally enzymes or receptors, so the term activity refers to an IC50 or EC50 value.
Smaller molecules have fewer potential sites for intermolecular binding than larger molecules. Therefore, small molecules rarely bind as strongly as larger compounds. For example, a hit from a typical combinatorial library may show activity (KD, IC50, Ki, EC50) at concentrations of 1 µM or lower. In contrast, a hit in a fragment library may be selected with activities of around 1 mM, which is a 1,000- fold difference in activity (a larger IC50 value implies weaker enzyme inhibitor binding).

In 1999, Teague distinguished between lead-like and drug-like hits,
  and combinatorial library collections. 20 Lead-like hits are
  characterized as having lower molecular weights (<350), activity (>0.1
  µM), and clog P values (<3). The lower values give lead-like compounds
  room to grow into an optimized, high-affinity drug that still
  satisfies Lipinski’s rules. Drug-like hits have higher molecular
  weights (>350) and clog P values (>3) but still modest affinity (~0.1
  µM). The definition of lead-like has since been used as a preliminary
  filter for selecting more promising hits from a screen.

The cut-off for the required activity of a hit is somewhat arbitrary. The threshold may also be based on the performance of the entire library. For example, the discovery group may count all compounds that are two or three standard deviations more active than the average of the full library.
While hits are often selected at an activity level of 1 µM (KD, IC50, etc.), structural changes may provide a lead with activity at concentrations of 0.1 µM (100 nM). Eventually, potency will typically be improved down to the 1–10 nM level during the lead optimization stage.
References:
Medicinal Chemistry E. Stevens

Is this sentence and the one following supposed to say mM in it twice? "For example, a hit from a typical..."

Well spotted!!! Thanks for this correction! Cheers :)


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When is it better to use coupled-cluster (CC) over configuration interaction (CI) calculations?
My experience with different theoretical methods in computational chemistry primarily comes from use and reading papers. My experience is that people almost always use CCSD instead of CISD if they wish to do calculations with single and double excitations.
My understanding is that truncated CI expansions suffer from not being size-extensive and also do not converge very quickly. I do not think coupled-cluster theories suffer in the same way.
Why, however, do people seem to prefer CCSD over CISD? What are the big advantages it has? Is any of it just a habit because CCSD(T) is the preferred method when possible, and so it's easier to make comparisons if these calculations come around in the future and can't be done in some current study?
Also, are there any systems for which it would definitely make more sense to choose a truncated configuration interaction calculation over a coupled-cluster method? Probably systems with some kind of low-lying excited state I would guess...

I'm not looking for a huge amount of math here, although that's totally fine. I would just be fine with the facts and any useful references.

Personally, I'd always prefer CC before CI... Good thing in CC2 (and other CC response models) for example is that you have a lucky error cancellation in the ground state and excited state dynamical correlation contributions (That's why F12 is so horrible for excited states, it eliminates the error cancellation because it only corrects the ground state contribution btw) which leads to mostly rather good excitation energies. And yes, due to the exponential ansatz CC is size extensive which truncated CI is not.

so, I thought about it a little bit more. I couldn't find/think of a case where the CISD excitation energies would be better suited than the CCSD energies.
The "advantage" you ask for is probably just that CC exc. are generally better than CISD ones...

The main drawback of CC is not being variational (and maybe computational scaling). But since CC is usually more accurate, one does not really care. Or is there any property that benefits from a variational approach?

CCSD is preferred in nearly all cases because size-extensivity/size-consistency is a critical property to have for practical chemistry, and on top of that, more of the correlation energy (the part of the energy missing from Hartree-Fock) is captured by CCSD. CISD only accounts correlation energy for up to double excitations and not at all for higher excitations. CCSD accounts for correlation up to double excitations and then approximates all higher excitations as combinations of the single and double excitations it has already calculated. This gives much more accurate numbers for pretty much all systems and provides for size-extensivity/size-consistency. The only time you would really be better off doing CISD than CCSD is if you wanted to do full CI on a 2-electron system, in which case CCSD is not as efficient.
So in short, CCSD is preferred because it almost always gives better, more usable numbers than CISD for similar or better costs (they both scale as ~$n^6$).

Yes, to re-iterate the above answer, CI in its standard truncation schemes (CISD, CISDT, etc.) are nowadays out of use in favor of CC for 3 main reasons:
(1) CC's size extensivity, and if the reference is separable, size-consitency
(2) CC exhibits the most rapid convergence to full CI
(3) CC's insensitivity to choice and quality reference due to inclusion of singly-excited clusters
CIS is a special exception since it is an uncorrelated approach. It is actually size-extensive (it's basically HF for excited states). It's a crude method no doubt, but quite useful as a starting point for more sophisticated calculations or for other supplemental purposes (such as helping to pick active orbitals in excited-state calculations).
The other instance in which CI is preferred is when you're actually shooting at full CI. This is actually a popular strategy nowadays. There is full CI Quantum Monte Carlo which mathematically converges to the exact ground state using imaginary time propagation - it is CI in that the wavefunction is expressed using a linear wave operator, but implementation-wise, it's pretty different. Then there are selected CI approaches which are based on iteratively building the wavefunction only out Slater determinants that are deemed to be important based on perturbative thresholds and diagonalization. In the field of exact quantum chemistry, these methods are serious contenders (with open-source implementations)
because they are amenable to large-scale parallel computing.

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Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?
i) $\ce{R-CH_2-OH + HBr   ->[H_2SO_4] R-CH_2-Br + H_2O}$
ii) $\ce{R-CH_2-OH + HI   -> R-CH_2-I + H_2O}$
What role does sulfuric acid play in the first reaction?
Why are we not using it in the second one?

I think it's because $\ce{HI}$ is already a pretty strong acid, but I'll leave it for a more qualified chemist to clarify that.

@PrittBalagopal https://chemistry.stackexchange.com/questions/57968/nitric-acid-plus-hydroiodic-acid

The synthesis of alkyl halides from the corresponding aliphatic alcohols using concentrated hydrohalogen acids was investigated by Klein, Zhang and Jiang.[1] They note:

[W]e found that the reflux of 1-butanol ($\pu{2.34 g}, \pu{31.5 mmol}$) with $48~\%$ hydrogen bromide ($\pu{7 mL}$) for $\pu{4 h}$ on a $\pu{120 °C}$ oil bath only gave low yield of 1-bromobutane ($54~\%$) and moderate purity ($93~\%$). The yield of 1-bromobutane was improved to $82~\%$ with $90~\%$ purity by adding additional sulfuric acid ($98~\%, \pu{1 mL}$). Interestingly, we found that 1-iodobutane could be directly synthesized using 1-butanol and hydriodic acid ($\ce{HI}, 57~\%$) by reflux without adding an additional acid in $80~\%$ yield with $98~\%$ purity.

They do not offer any reasoning why switching from $\ce{HBr}$ to $\ce{HI}$ gives better yields. However, we may use our chemical reasoning to deduce the reason. As has been pointed out and explained multiple times on this site, the acidity of hydrohalogen acids increases from fluorine to iodine: $\ce{HI}$ is a stronger acid than $\ce{HBr}$ is. This means that hydrogen iodide should protonate a greater percentage of alcohol molecules than hydrogen bromide — and it requires the collision of a protonated alcohol and the halide anion for the reaction to proceed ($\mathrm{S_N2}$ mechanism).
Remember that all hydrohalogen acids are gases at room temperature and standard pressure and thus need to be dissolved in water to give the actual acidic solution. While sulphuric acid itself may be less acidic than both pure $\ce{HBr}$ and $\ce{HI}$, adding it to the solution will increase the solution’s overall acidity since it comes with no added water. Hydrogen iodide in itself is acidic enough to promote the reaction as the experimental evidence shows so only in the case of hydrogen bromide is additional acidity needed for the reaction to occur.
Interestingly, the authors also state that alcohols with more than five carbon atoms give worse yields due to their low solubility in $\ce{HI}$. This can be overcome by adding phosphoric acid to the mixture. Again, phosphoric acid may not be strong but it features a low water content. The added acidity may protonate more alcohol molecules allowing the $\mathrm{S_N2}$ reaction to proceed. I point to the other answers as to why using sulphuric acid instead of phosphoric acid is a bad idea in the case of hydrogen iodide.
Thus, to sum up:

hydrogen iodide is sufficiently acidic to protonate the corresponding alcohol and drive the reaction. More acids are not needed.
hydrogen bromide is not acidic enough to drive the reaction. Adding sulphuric acid protonates the alcohol partially and allows the reaction to proceed.
in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidise iodide (while sulphuric acid does).

[1]: S. M. Klein, C. Zhang, Y. L. Jiang, Tetrahedron Lett. 2008, 49, 2638–2641. DOI: 10.1016/j.tetlet.2008.02.106.

I suspect $\ce{HBr}$ is produced in situ from $\ce{KBr}$ and $\ce{H2SO4}$.
Same process isn't particularly suitable for $\ce{HI}$ generation as it reacts with $\ce{H2SO4}$:
$$\ce{2HI + H2SO4 -> H2SO3 + H2O + I2}$$
I actually seriously doubt second reaction will take place at all. Hydroiodic acid can also act as a reducing agent, converting the resulting alkyl iodides to alkanes. Instead of $\ce{HI}$ is often being used $\ce{I2}$ with red phosphorous, or halide salts (Finkelstein reaction).

As @orthocresol noticed in the comment, the following two reactions are more preferable for alkyl iodides synthesis:
$$\ce{R - OH ->[(C6H5O)3P+ CH3I-] R - I}$$
$$\ce{R - OH ->[ (CH3 - S - CH = N+ (CH3)2) I- , THF] R - I}$$

Also, chapter 2.01.5.4 Alkyl Iodides from Alcohols, p. 16 in Comprehensive Organic Functional Group Transformations II (1) accumulates numerous ways of iodoalkanes synthesis:

The adduct of triphenylphosphine with elemental iodine ($\ce{Ph3PI2}$)
  is a classic reagent used to convert alcohols cleanly and with
  inversion to iodides. Addition of imidazole is well known to promote
  this conversion with high yields. Furthermore, treatment of alcohols
  with triphenylphosphine and cyanogen iodide affords the corresponding
  iodides in good yields. The reagent $\ce{KI/BF3 * Et2O}$ in dioxane is
  highly selective and effective for the transformation of allylic and
  benzylic alcohols to iodides.
A mild and effective procedure for directly converting secondary,
  tertiary, and benzylic alcohols into the corresponding iodides
  involves treatment with iodine in refluxing petroleum ether. The
  reaction proceeds with inversion of configuration:

Alkyl iodides can also be prepared in a single step from the
  corresponding alcohols upon treatment with the standard reagent
  $\ce{P-I2}$. A less well used reagent,
  1,2-bis-(diphenylphoshino)ethane, can be used to prepare iodides from
  primary alcohols in the presence of iodide.
Cerium(III) chloride, a Lewis acid imparting high regio- and
  chemoselectivity in various chemical transformations, can be used in
  combination with sodium iodide in refluxing acetonitrile to replace a
  hydroxy by an iodo group:

This method cannot be applied to tertiary alcohols. In this case an
  alkene is derived by dehydration of the alcohol.
Finally, a classical method for the transformation of alcohols to
  halides is the well-known two- step procedure via a sulfonate ester,
  commonly tosylate (p-toluenesulfonate) or mesylate (methanesulfonate).
Treatment of the sulfonate esters with $\ce{NaI}$ or $\ce{LiI}$ gives
  iodides in good yields.

(1) Katritzky, A. R.; Ramsden, C.; Taylor, R. J. K. Carbon with One Heteroatom Attached by a Single Bond; Comprehensive Organic Functional Group Transformations II; Elsevier Science Ltd, 2004; Vol. 2. ISBN: 978-0-08-044253-2

Or, for organic chemists, $\ce{PPh3}$, $\ce{I2}$, imidazole does the trick. (Of course, that's beside the point.)

@orthocresol and OP: Oh, aquaeous HI does the job well done. I did so (synthesising 1-iodobutane from butan-1-ol) in my PhD thesis and it works like a charm.

@Jan How interesting; was it sulfuric acid or something else, say, orthophosphoric acid? Or just leave the link to your thesis, if possible -- I'd happily add this to my answer!

It was just butanol and HI, refluxed at $120~\mathrm{^\circ C}$ and then distilled/separated.

@Jan Could you please provide a reference or more details on your reaction control? From what I remember (I might be mistaken though) reaction $\ce{ROH + HI -> RI + H2O}$ indeed takes place (also when HI is generated from $\ce{H3PO4 + KI}$ or $\ce{P + I2}$), but it is also followed by reduction to alkane: $\ce{RI + HI -> RH + I2}$, resulting in low yield of iodoalkane.

See my answer I just posted.

@Jan Thank you! I briefly checked the article, but I didn't find how they prevented further conversion of iodoalkane to alkane. Also using aqueous (?) $\ce{HI}$ ("However, we found that the alcohols with longer chains
(C6–C8) were immiscible with 57% HI.") could affect equilibrium of halogenation reaction in undesired way. Unfortunately I cannot access paper's ESI, maybe it has some more info on that.

They do use aquaeous HI. All I can say is that I reproduced their yields and purity.

First of all, this question is very interesting. Though it looks simple at first glance, it really is complicated and we need to look at the details carefully. I am just going to shed some light on the points mentioned in @andselisk's and @Jan's excellent answers.

@andselisk said that $\ce{HBr}$  is produced in situ from $\ce{KBr}$ and $\ce{H2SO4}$. So, why not write in this way:

$$\ce{R−CH2−OH + (KBr + H2SO4) −> R−CH2−Br + H2O}$$
But the $\ce{HBr}$ might further react with $\ce{H2SO4}$:
$$\ce{2HBr + H2SO4 → Br2 + SO2 + 2H2O}$$
Bromine (+ water) produced might not lead to desired alkyl bromide, rather the alcohol might gets oxidised to aldehyde and carboxylic acid which in turn reacts with alcohol to form esters according to this paper:

$$\ce{CH3CH2OH + Br2 -> CH3CHO + 2HBr}$$
$$\ce{CH3CHO + Br2 + H2O -> CH3COOH + 2HBr}$$
$$\ce{C2H5OH + CH3COOH -> CH3COOC2H5 + H2O}$$
The Reaction Products - The reaction products obtained at varying initial molarities of bromine were determined. The bromine was always
quantitatively reduced to hydrobromic acid, and no bromination took
place.

Other side reaction might take place

$\ce{HI}$ do react with $\ce{H2SO4}$ but also note that sulfurous acid is an unstable aqueous solution of sulfur dioxide and is only detected in gaseous phase. It disproportionate in aqueous solution forming $\ce{H2S}$.

When trying to concentrate the solution by evaporation to produce
waterless sulfurous acid it will decompose (reversing the forming
reaction). In cooling down a clathrate $\ce{SO2·5.75H2O}$ will
crystallise which decomposes again at 7 °C (forming $\ce{H2S}$) Thus sulfurous acid
$\ce{H2SO3}$ has not been isolated.

As @andselisk wrote, $\ce{HI}$ is made by reacting iodine reacts with phosphorus to create phosphorus triiodide, which then reacts with water to form  $\ce{HI}$ and phosphorous acid. This is the way of producing $\ce{HI}$ for simple organic synthesis because it is currently listed as a Federal DEA List I Chemical for its use to produce methamphetamine in the United States and it is illegal to buy $\ce{HI}$.

$$\ce{3I2 + 2P + 6H2O → 2PI3 + 6H2O → 6HI + 2H3PO3}$$

Finkelstein reaction is used when the reactant is an alkyl halide not alcohol.
Yes as @andselisk said, theoretically $\ce{HI}$ can be used  to convert primary alcohols into alkyl halides through  $\ce{S_{N}2}$ substitution, in which the iodide ion replaces the "activated" hydroxyl group (water):

$\ce{HI}$ is preferred over other hydrogen halides because the iodide ion is a much better nucleophile than bromide or chloride, so the reaction can take place at a reasonable rate without much heating but the problem is that it is an reducing agent and thus would reduce alkyl iodide to alkane. The yield is catalytically increased by using red phosphorus to reduce the formed iodine.

Although harsh by modern standards, $\ce{HI}$ was commonly employed as
a reducing agent early on in the history of organic chemistry.
Chemists in the 19th century attempted to prepare cyclohexane by $\ce{HI}$ reduction of benzene at high temperatures, but instead
isolated the rearranged product, methylcyclopentane . As first
reported by Kiliani, hydroiodic acid reduction of sugars and other
polyols results in the reductive cleavage of several or even all
hydroxy groups, although often with poor yield and/or reproducibility.
In the case of benzyl alcohols and alcohols with α-carbonyl groups,
reduction by $\ce{HI}$ can provide synthetically useful yields of the
corresponding hydrocarbon product ($\ce{ROH + 2HI → RH + H2O + I2}$).
This process can be made catalytic in $\ce{HI}$ using red phosphorus to
reduce the formed $\ce{I2}$.

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


text:
How to calculate the charge on 1 gram ions Aluminum cation?
I want to find out the charge on 1 gram ions of $\ce{Al^3+}$ ions.
I tried out by finding no. of moles as $ \frac{1}{27} $ , then no. of ions will be $ \frac{1}{27} × N_A $ ,
then total charge will be $ \frac{1}{27} × N_A × 3e $ but it is wrong.
The correct answer turns out to be  $ N_A × 3e $ .
Where I am wrong?
[ $ e $ = charge on 1 electron, $ N_A $ = Avogadro's Number ]

@RaviPrakash What you did is perfectly correct. Who told you it's wrong?

@PrittBalagopal Yes it seems to be correct but the answer is 3 $ N_A e $

They must've meant "one mole", not "one gram". Btw I didn't downvote.

1 - gram ions is same as 1 mole.
So, we need to find only the charge on 1 mole of ions, i.e.,
$ N_A × 3e $ colulomb.

Related https://chemistry.stackexchange.com/questions/28563/what-exactly-does-the-term-gram-atoms-mean


text:
Permanent cloudy colloid with water
I need to make a cloudy white liquid where the particles won't eventually settle out. It will be exhibiting the Tyndall effect for a very long time.
I'd use milk, but I don't want it to become rancid. I've tried white inks and fabric softener, but both eventually tend toward the bottom half of the container. I think it's a matter of particle size.
What substance would hit the sweet spot - small enough to stay dispersed, large enough to scatter light, and not reactive?

Bleached wheat flour maybe? See Wikipedia for Tyndall effect.

Maybe try latex house paint diluted in water. These products are designed to remain usable (colloidal) for years while sitting on the shelf. When dry, house paint is pretty stable to the elements, so perhaps it will not go rancid quickly. If it does, perhaps you could add a mold and bacterial growth inhibitor.

Starch in water might work, perhaps with table salt or other bacterial/fungal growth inhibitor. Follow directions in most recipes: mix a bit of starch in cold water and then add gradually to hot water to make a more stable "soup".
BTW, blasting a colloid with ultrasound (e.g. from an ultrasonic humidifier) homogenizes it and may delay separation.

As a first approximation, there are no stable colloids. I suggest you look for kinetically stable (i.e. metastable) colloids. The academic community is divided between those who insist that no colloid system is stable and those who offer theoretical reasons why certain systems are thermodynamically stable. I used to believe no colloids are stable (thermodynamically) but changed my mind (based on the scientific literature at the time). Unfortunately, that was decades ago, and I'm not able to give you chapter and verse rationale why certain colloids may be stable. To be "stable" the particle needs to both not fall (match density to its media so that thermal motions (and mechanical vibrations, possibly) overwhelm any tendency to fall) and more difficultly have a lower free energy in the highly dispersed state, specifically it must be electrostatically repulsive towards other particles and have virtually no tendency towards Ostwald Ripening. This limits your options to organic (charged) polymers and if you can "structure" the water, that's all to the better. If I recall certain acrylates were able to be synthesized into particles fitting those requirements, but it was a long time ago. Permanent is a long time, I really doubt you need permanent. I'd bet 50, 100, or maybe 200 years would do just fine.

This is mostly monologue about semantics. Certain acrylates doesn't really answer this.

We appreciate your expertise and experience, but providing the actual references to the literature would make this a much stronger answer.  Right now, it's kind of a hybrid between an answer and a long comment.


text:
Esterification with tertiary alcohol
I saw a different mechanism for esterification of tertiary alcohol in a book.
They say that first the $\ce{OH}$ is protonated and it leaves to form the tertiary carbocation. Then the oxygen's lone pair from the carboxylic acid attacks to form the ester.
Is this mechanism actually correct?

It depends on the conditions used

Yes, the mechanism is correct and is different from the normal mechanism because here the more stable tertiary carbocation formation.

Care to elaborate it?

Upon esterification of a tertiary alcohol an alkene is formed as the major product, not an ester.


text:
How to calculate the mass of air that fits into a balloon of a certain size?

A hot air balloon has a $\pu{500 m^3}$ volume.
  How much mass of normal air fits into the balloon, if the relative molar mass is $\pu{28.8 g/mol}$?

I am not sure if I should simply use the $PV=nRT$ formula and therefore
$$1\times500= m/28.8 \times 273 \times 8.31 .$$
I feel that there is something wrong, can anyone enlighten me with the right formula or the basic concept of solving this kind of questions?

You're right in using the ideal gas law, but the numbers are wrong. Assuming atmospheric conditions, the pressure is 101,325 Pascal and the temperature is likely to be higher than 0°C (273 K). So the mass of the gas (in grams) would be:
$$m = \frac{\pu{101325 Pa} \times \pu{500 m3} \times \pu{28.8 g mol-1}}{\pu{8.31 J K-1 mol-1} \times \pu{290 K}^{[1]}}$$
where you can choose an appropriate value for [1] for your location. All values (except for the molar mass) are in SI base units, as described here.


text:
Selectivity in acetonide formation from a polyol

In my total synthesis lecture notes the above example is given in which a polyol is reacted under acetonide forming conditions.
The notes point out that this product is the only product, no other actinides are observed. That said there possibly are two other products, both containing two acetonides, so why should just this one di-acetonide form?
The conditions are not very clear, but I assume all steps are reversible so what is actually being observed is thermodynamic product.

OK, English naming is strange - "acetonide" should be reserved for derivatives of enolate.

Acetonide formation (isoproylidine ketal formation, to be more correct) is indeed a thermodynamic process, and under acidic conditions acetonides will tend to migrate to form the lowest energy product.
If you look at the penta-ol you've been given, there are three possible permutations containing two acetonides (theres also the possibility of just forming a single acetonide, but these kinds of reactions are usually ran with an excess of the acetone/dimethoxypropane, so its likely to go on as many times as possible):

The green one (which is the one labelled A in your image) is the thermodynamic product, which becomes obvious if you consider the three-dimensional conformation of 1,3-syn and 1,3-anti acetonides (note that this analysis only holds up then there is no substituent at the 2-position. If there is a methyl group there, as is common in polyketide type motifs, then the acetonides don't necessarily adopt chair conformations):

In the syn acetonides, there are three equatorial and one axial groups (minimal strain), whilst in the anti acetonides (which both of the un-observed products contain one of), there are two equatorial and two axial groups, giving a 1,3-diaxial clash across the ring (hence raising the energy of these products).


text:
How do I know how many molecules a solute will split into when it dissolves in a solvent?
I'm in high school and I'm learning about freezing point depression and boiling point elevation. I've been given this chemistry problem:
Acetic acid has a boiling point elevation constant of 2.93 and a normal boiling point of 118.1 Celsius. What would be the boiling point of a solution made by dissolving 100.0 grams of calcium carbonate (CaCO3) in 400.0 grams of acetic acid?
This requires me to use the equation for boiling point elevation, that is:
ΔTb = Kb · bsolute · i, or change in temperature = the ebullioscopic constant * the molality of the solution * the number of ions the solute splits into.
My question is how does one determine i for a given solute in a case like this? Thanks for any help you can give!

Never mind. I figured it out. How do I  delete this?

Theres no need to delete, you can always answer your own question so anyone looking in the future can follow your working

You need to know the type of reaction that is taking place and what types of bonds are being made and/or broken to determine how many ions the given molecule will split into.


text:
Why do positive ions go to the cathode?
In electrolysis why do positive ions go to the cathode? Since the cathode is positive, won't the positive ions face repulsion at the cathode and would require more energy?
I understand that reduction happens at the cathode and positive ions reduce but somehow it does not make sense as positive - positive charges repel.

TL;DR The cathode in an electrolytic process is considered to be
negative, so there is actually no contradiction. The cathode is a positive electrode in a galvanic cell.

There are different notations for the sign (±) of the cathode used in the literature, which are determined, in particular, by the nature of the process. A very broad definition of a cathode is that it is the electrode of some device connected to the negative pole of the current source.
For electrolysis it is commonly believed that the “−” cathode is the electrode on which the reduction process takes place, and the “+” anode is the one where the oxidation process takes place. When the cell works (for example, during copper refining), an external current source provides an excess of electrons (negative charge) at one of the electrodes — the cathode, where metal is reduced. On the other electrode, there is a lack of electrons and oxidation of metal takes place — this is the anode.
At the same time, during the operation of a galvanic cell (for example, copper–zinc), an excess of electrons (and a negative charge) on one of the electrodes is provided not by an external current source, but by the actual oxidation reaction of the metal (zinc dissolution). Following the definition given, this electrode is an anode. Electrons passing through the external circuit participate in the reduction of copper, so the cathode will be a positive electrode in this case.

This confused the heck out of me for about the first 3 years as an undergraduate. I thought I had learned in basic chemistry that the "cathode" was a positive electrode, then we started doing electrolysis in biology labs where positively-charged things migrated towards the cathode, and it never made any sense to me. Finally, I had a professor explain the different notations, and it all made sense. I was not the only student who was enlightened. It would be smart if this were explained much earlier.

@WilliamR.Ebenezer in the word "geochemistry" "geo-" doesn't stand for the planet Earth, rather for the ground or crust. "Geochemistry" applied to extraterrestrial space bodies (such as moons and planets) is perfectly fine, removing that tag made no sense to me. Plus, there probably were other minor formatting issues (I don't remember the details that you haven't addressed in your edit making it also incomplete. Both factors made me reject the edit and I think I did so rightfully.

@andselisk Alright, alright. Pardon the incompleteness of knowledge and sorry for making such a fuss out of it.

@WilliamR.Ebenezer No prob at all, that's fine. I admit I also make erroneous decisions and my edits are also not perfect; for the future you probably might want to ask similar questions on Meta as it's more suited for the disputes like this, plus you get a spectra of opinions instead of only mine.

You got it reversed.
The cathode is the electrode where cations discharge, hence the name.
BTW, effective electron flow in galvanic cells is always directed from the cathode to the anode. The 'change' of the polarity of electrodes is due to a change in perspective, not actual change of flow.


text:
Formation of trifluoride anion
I wanted to understand mechanism of the formation of $\ce{F3-}$ anion through the molecular orbital theory.
UPDATE - What I really want to know is that when $\ce{F3-}$ is formed from $\ce{F2-}$ and $\ce{F}$, the unpaired electron of $\ce{F}$ enters the anti-bonding orbital $\sigma^*_{\mathrm{2s}}$ then shouldn't the $\ce{F2-}$ ion break up, as all the bonding and anti-bonding orbitals would be filled up, resulting in a bond order of 0?

http://pubs.acs.org/doi/abs/10.1021/ja046443a

related https://chemistry.stackexchange.com/a/17074/9961

Mithoron that still does not explain the pathway followed during its formation

I'm not too sure, but let me try. In the case of the F3- ion, the effective bond order is 1 (see attached images), not 2. In other words, the only effective bonding MO holding the entire thing together is a 3-centre 4-electron bond (think linear FHF-).
I presume you know how the F2 molecule MO diagram looks like, so I'll just explain what F3- looks like. Using the D2h character table and forming our reducible representations, we get the following:

Note s-p mixing occurs in the Ag and B1u orbitals, as shown:

The MO diagram for F3- would look disastrously messy, so I'll simply show how the molecular orbitals look like and how they're ordered.

The first thing you'll probably notice is how the central fluorine isn't allowed to s-p mix anymore; although the symmetry remains D2h the now-presence of a central atom means the s and p orbitals of the central fluorine are no longer of the same symmetry, and no longer interact. That, in addition to the fact that the sole bond that is effectively holding the ion together is a 3-centre 4-electron bond, makes answering your question rather difficult.
I think a way to consider it would be in terms of the formation of a Lewis adduct; in other words, the pz electrons in the F- ion interact with the empty LUMO of the F2 molecule while all the lower-energy orbitals form bonding/antibonding sets with the lower-energy F2 MOs, forming a new HOMO and LUMO from the "HOMO" of F- and LUMO of F2.

One minor beef. $\ce{H3^+}$ is not linear. Just use $\ce{FHF^-}$ in your example.


text:
What makes Gorilla Glass more durable with each generation?
Gorilla Glass (GG) is one of the most recognizable brands of toughened glasses today, thanks to their marketing and widespread usage in various gadgets and automobile industry.
According to Wikipedia, it's basically an ion exchange ($\ce{Na+ <-> K+}$) within the glass (boro-/alumosilicate?) matrix which gives this material resistivity against scratches and various mechanical impacts:

(Source 1, Source 2)
Still, there is a new generation of GG released every few years, showing better properties with thinner samples, making me think it's all not just about ion exchange.
Are there other ways of "chemical strengthening" besides cationic exchange that can be utilized to strengthen the glass, keeping it flexible and less fragile at the same time?

I believe that may be classified information that Corning wouldn't want to let other competitors know.

@PrittBalagopal That's why I'm asking. I'm also curious whether community can beat their best chemists.

"...can beat their best chemists" Corning doesn't just hire chemists, they more over hire glass scientists that have bachelors/PhD's in glass or materials science.

@andselisk If you'd pay me more then they're paid. ;)

Are there other ways of "chemical strengthening" besides cationic exchange that can be utilized to strengthen the glass, keeping it flexible and less fragile at the same time?

In a word: No.
To understand why ion exchange strengthens glass, you have to understand why the ion exchange makes the glass harder.  The process ion exchange hardening, its done at a temperature that allows ion diffusion, but disallows a reconfiguration of the glass structure or relaxation of bonds (below Tg). When sodium atoms are replaced by potassium in glass, the potassium ions occupy a site that is sized for sodium; this mismatch creates the compression that in turn gives the glass its scratch resistance. Thus the only way to chemically harden the glass is to substitute larger ions into the glass.
Now an astute person might be thinking:

Are there other ways of "chemical strengthening" besides cationic exchange... What about anion exchange?

To this yes in theory if you could substitute the anions without rearranging the structure then yes anion exchange would be another way. But as Yogi Berra once said "in theory there's no difference between theory and practice, but in practice there is." The problem with anion exchange is you would be breaking and thus relaxing the bonds of atoms that you need to remain in place, thus no strengthening would occur.
I suppose you could also modify the surface to make a layer of silicon nitride or carbide and call that anion exchange, but would it this strengthening glass or changing the material altogether?
One way in which you could increase glass strength though still predominately a chemical process is to substitute sodium for lithium (or hydrogen), anneal the glass to allow the bonds to relax to form sites sized for lithium then substitute potassium back in where the lithium is to make it even stronger.
As far as what make each generation of GG stronger, I think it is base glass improvement such as modifying the aluminum content or making the glass thinner or making the ion diffusion more local to the surface and preventing the potassium from penetration too far.

@andselisk I won't edit this into the answer until I find a paper that corroberates it, but it has been alleged to me that high pressure argon could diffuse into the interstitial sites at elevated temperatures creating additional compression in anterior regions further strengthening the glass.

Don't know if this is relevant to this answer but you can "pre-treat" glass with a sodium nitrate solution at high temperatures to increase the sodium content on the surface structure of the glass. This increases the number of sodium atoms available to be exchanged and can improve the strengthening process.

To really answer this, you need to characterize the glass in terms of its properties, and not something like its ability to resist being scratched. Scratch resistance is a behaviour, and not a property. There are many ways of making something more scratch resistant.
"Hardness" itself is a behaviour, and not a property.
In any case, it might be that they're onto a very good toughening mechanism here, with various degrees of ion exchange, and they can tweak the process a lot and still get improvements.
Controlling the ions present in the glass matrix is going to affect both the residual stresses of the total glass layer, as well as the properties of each layer. And that will affect hardness/scratch resistance and toughness.
In most materials, toughness and hardness are in competition. But in a brittle material like glass, fracture needs a defect to initiate from. So if you adjust the cations present in the glass matrix to make the surface layer hard, it is potentially brittle. But if you also make it in compressive stress compared to the inner region, it become tougher.
There won't be just one way to achieve this, as in its not a single optimized point, but rather a response surface. So as they understand the process better, they can move along that response surface, and show improvements every few years.

I'd disagree that " 'Hardness' itself is a behaviour, and not a property." Minerals are absolutely classified by hardness. The problem with glasses is that they don't have a definite chemical formula like a mineral, so there is no characteristic hardness value.


text:
What software do professional chemists use?
What software is used to create skeletal formula diagrams, create 3D models and whatnot? For example, on Wikipedia pages for molecules, you see standard 3D models and skeletal models which I assume are made with certain chemistry programs rather than some widespread ones.
Also, what programs are used to write chemistry papers? I know languages like LaTeX are used for mathematics and I can see you could definitely write a chemistry paper using such a program; I am curious to what actual chemists use.

molden, jackal (old, good for b/w crystal structures and such), hyperchem, avogadro, crystalmaker, their own stuff

There are several software solutions for each problem you mentioned as well as to other ones.  If you are doing Chemical research at a university/in a lab, many of this software are available through campus licenses, so check those first.

@JoshuaFarrell please let us know a) what operating system you are using; b) are you affiliated with a university or any other institution that may have licences available. The things are (sadly), many software pieces chemists need cost ridiculous amount of money. Please leave a note, if you prefer FOSS.

@Joshua Farell Regarding the "skeletal representation" in wikipedia you refer to, this in part is partially due to a listing of software (at least, in principle) available: https://en.wikipedia.org/wiki/Molecule_editor . More importantly, however, regarding consistency of the representations, similar to publishers where you submit an article to a journal, wikipedia has a dedicated set of guidelines how to draw molecules, e.g. https://en.wikipedia.org/wiki/Wikipedia:Manual_of_Style/Chemistry/Structure_drawing

Chemdraw/Marvin Sketch (Draw structures), Kaleidagraph (Easy graphing program, I like it far more than excel), VESTA (Visualize X-Ray Diffraction data/ others), Gaussian/VASP/QE/Calypso (Computational chemistry packages, some are useful for NMR predictions, among other things), Python (Makes more complicated data manipulation a breeze). Everyone will have a different toolkit. I generally use Word to write papers, as Chemdraw has word plugins that are nice.

@andselisk I am using a Macintosh so OSX and next year I will be affiliated with a university but for mathematics

@JoshuaFarrell Oh. For macOS you might want to have a look at http://www.macinchem.org/, I guess, unless you need something really specific. Genarally, MarvinSketch for 2D structures and Avogadro for 3D structures are a good start. Regarding LaTeX, I use it whenever possible, if you really want, you can check out my preamble to get an idea what packages might come in handy.

https://chemistry.stackexchange.com/questions/54548/software-for-visualization-of-3d-models-of-chemical-structures-with-atomic-orbit?rq=1 https://chemistry.stackexchange.com/questions/246/a-chemdrawing-software?rq=1


text:
How can I measure a droplets contact angle without a setup?
I have to use a phone camera to replicate the Sessile drop analyzer, but there doesn't seem to be a standard to different parameters, such as how far away should my camera be from the droplet or how much liquid should I use for my droplets? Is it possible to get accurate results this way, and how should I go about it?

Quick search unveiled this video tutorial which, I guess, uses Bikerman method. The tricky part might be getting the lens attachment, you might need a donor device to harvest the lens from, e.g. a webcam or an old microscope. Probably a magnifying glass will do too.

In 2010, Borguet et al. published in Journal of Chemical Education, DOI: 10.1021/ed100468u an intentionally simplified setup which may be useful as a reference.  A Sony Cybershot camera was placed in front of a sample scrutinized under diffuse illumination:

The additional lens shown is focussed on the drop (both for water and $n$-hexadecane, the authors recommend a volume of $\pu{5 uL}$; focal length of the length used equals $\pu{50 mm}$):.
With photos recorded like this

the contact angle determination was performed with the contact angle plugin of imageJ.  As Fiji, it is open source, free ware, and java based/OS portable.  To quote from their publication:

For each easurement, the user must choose two points to manually
  define the baseline and three points along the drop profile (see
  the supporting information that features a set of instructions
  supplying all the details on the procedure used to make the
  measurement). The program then fits the profile of the drop and
  calculates the contact angle using the sphere approximation
  or the ellipse approximation. In our study, the ellipse approximation gave consistent results for contact angles $> 40^\circ{}$. For
  drops with contact angles $< 40^\circ{}$, the sphere approximation was
  used.

In comparison with reference data, the authors judge contact angles recorded as well as the subsequently calculated surface tensions as good enough for instruction / lab classes and research lab.


text:
How to stop a lye + oil + water reaction in aluminium container?
I was making soap using a guide for the first time. I accidentally used an aluminium container. It has lye and olive oil in it and was kept on the stove; now it is bubbling and not ceasing (I've turned off the gas).
What should I do to stop it?

Buy another container, I guess.

Tyler Durden used a stainless steel pot, I believe. Also, look for the liposuction clinic if you want to improve your product.

@andselisk I couldn't help but upvote :D

Fill it into an empty plastic can. I wouldn't pour it down the drain, lest it solidifies in your gutter. Probably won't happen with olive oil, but it's your bet.

Slowly add vinegar. DO NOT TURN THE FLAME BACK ON.
Most likely, the alkali concentration is high enough that you've dissolved away the passivating oxide on the aluminum, and the exposed aluminum metal is reacting with water to form $\ce{Al(OH)4-}$ and hydrogen gas.  Adding vinegar should bring the $\mathrm{pH}$ back down into a range where the aluminum can passivate again, stopping the reaction.
The batch of soap is probably already ruined, so I wouldn't worry about that.
In the future, I would recommend using a cast iron or mild steel (not stainless) pan for this procedure.  Plain iron is remarkable for its resistance to alkali, in my experience.

Since the batch of soap is lost anyway, we may just as well empty out the container and wash it. Chances are it is already damaged too badly, though.


text:
Replacing atom in Hydrocarbon Molecule
I am working with Hydrocarbons. In some of me analysis I see that carbon atoms are replaced with oxygen or other elements, like this one .
Me question is: is only the carbon atom can be replaced or the hydrogen can be also be replaced?
Thanks!

I don't understand what you mean by "can be replaced". Are you asking about whether we can have molecules which have some atoms replaced with others? Or about specific reactions to cause such a replacement?

Hydrogen atoms can be replaced with other atoms with a valence of one, like Chlorine and other halogens. This gives a family of substances called haloalkanes.
Another option is a group of atoms like -OH: this gives alcohol.


text:
Why is law of mass action only applicable for elementary reaction?
Why can't we apply law of mass action for non-elementary reaction? Why does is vary with the rate law in case of non-elementary reactions?

The Law of Mass action states:
The rate of a reaction is directly proportional to the concentration of reactants raised to powers equal to their stoichometric coefficients
This is only valid for elementary reactions since they happen in a single step. As a result, such simple relations hold good.
But what about the more complicated ones. Certain reactions, like the oxidation of ethylbenzene to benzoic acid using hot $\ce{KMnO4}$ don't happen in a single magical step. They involve lots of reactive intermediates and transition states, so it's basically a bunch of several elementary reactions. You can't consider the whole reaction as an elementary reaction.
But you can formulate a rate equation for non elementary reactions by considering them as a bunch of elementary ones. However such formulations involve knowing the mechanism of the reaction in question, along with other data.

//This is only valid for elementary reactions since they happen in a single step. As a result, such simple relations hold good.// Why does such simple relations hold good only when a reaction happens in a single step? Why is the 'single step' so necessary for this law?

@Hisab For example, consider an arbitrary reaction: $$\ce{A + 2B -> C}$$ If this was a single step reaction, a molecule of $\ce{A}$ and two molecules of $\ce{B}$ should simultaneously collide and form $\ce{C}$. The concentration of $\ce{B}$ is more important than $\ce{A}$ (since two of it are needed in a single collision), and hence the $^2$ is attributed to it. For other single step reactions this isn't the case, not all reactants collide simultaneously.

If I ever knew why the tLoMA held for elementary reactions, I've forgotten. I can, however, give an example of a situation applicable to your question illustrating why it doesn't necessarily hold for complex reaction systems. Consider the two elementary reactions A+B→ C+D and A+C→E  the first thing I'd ask you to do is express the equation for the total equilibrium constant. The second thing I'd ask you to consider is if the first reaction had an extremely high rate and the second had an extremely low one (on the order of years-1). The overall reaction will be dependent almost exclusively (except for the very very start) on the rate of the second reaction, tLoMA will (and must) fail.


text:
Aluminium etching
I'm trying to etch a very fine pitch ($\pu{0.5 mm}$) aluminium stencil for use with SMT printed circuit board making. I've done some reading on the topic and found that the safest way probably is to use a saline sulfate etch and/or use saltwater electro etching.
If not using electrolytic reactions it seems like it is a single displacement reaction since $\ce{Al}$ is more active than $\ce{Cu}$.
$$\ce{Al (s) + Cu(SO)4 (aq) = Cu (s) + Al(SO)4 (aq)}$$
$\ce{NaCl}$ is not dissolved and not a part of the reaction but merely a catalyst to speed up the reaction. Still i'm not sure if this is strong enough to actually etch holes in an aluminium plate or just emboss the surface.
I'm considering adding an electric current to speed things up using the aluminium plate as anode(+) and i assume copper can be used as cathode(-) since it is the other metal that the aluminium ions are replaced with.
It will probably (I'm guessing) cause a electrolytic decomposition of $\ce{NaCl}$ as well and merge with $\ce{H2O}$. So I assume there will be produced chloride and hydrogen gas and $\ce{NaOH}$ (sodium hydroxide) in the process.
$$\ce{NaCl (s) + 2H2O (aq) = 2NaOH (s) + H2 (g) + Cl2 (g)}$$
I figure the gas will bubble out, so i would be needing a fume hood.
Also i guess the $\ce{CuSO4}$ and $\ce{NaOH}$ will react into copper hydroxide and sodium sulfate:
$$\ce{CuSO4 (aq) + 2NaOH (s) = Cu(OH)2 (s) + Na2SO4 (s)}$$
Can anyone confirm if this is what will happen?
Any toxic/hazardous reaction that i should be aware of?
Do I need to add something to the solution to neutralize it before i dispose it at a recycling facility?

As soon as the Cu covers the active Al (that without an oxide coat), any etching would stop.


text:
How to figure out number of atoms in a molecule? e.g. Phosphorus exists as P4
We know that Phosphorus exists as P4 but how can I figure out the number of atoms in a molecule for other elements?

This is the question that requires the use of P4. I would never have known Phosphorus is tetra-atomic and that the equation will involve this. Is there a way to figure that out?

Welcome to ChemSE.  Please familiarize yourself with how the site works, i.e. show initial effort by you to solve the question by your own, and try to formulate the questions specifically.  For your quesion here: among the references consulted, did you throw an eye on https://en.wikipedia.org/wiki/Elemental_analysis?

Hint: Molecular weight.

I've updated the question, my apologies.

Not better. If you have no idea about this, why do you want to write an exam? First study, the aim for a certificate.

What are you talking about? I can write the equation successfully with the knowledge that Phosphorus is P4 but I want to know how or why that's the case. Also, let's assume the question asked 'Bromine reacts with Sulfur to form...', how would I figure out the formula of Sulfur?

If the question is "how do I find out" the answer is: look it up in a book. If the question is "how did chemists work out the formulae in the first place" then answer is a lot more interesting and complicated (after all, it took a long time).

Most probably the question is targeting basic lexical knowledge / counting electrons, making lewis structures, applying VSEPR.

You are asking how to identify the standard state chemical structure of an element; this is obtained from empirical data.
The standard state is defined as the most stable allotrope of the element under standard conditions. For example, graphite is the standard state for carbon. Phosphorus, however, is an exception: while white phosphorus is not the most stable allotrope, it is the easiest to characterize, and is defined as the standard state for phosphorus.
The chemical formula for each element at standard conditions is that of the standard state. Except for the gases $\ce{H2, N2, O2, F2, Cl2, Br2, I2}$, and the solids $\ce{P4, S8}$, all other elements exist in monatomic form in the standard state.


text:
Nitrogen Covalent Bonds
Let me premise this with the fact that I really have zero knowledge of chemistry. So bear with me and my ignorance.  :)
Now, with that introduction out of the way, I've read up this site which presented in a very easy to understand way, how a Nitrogen atom bonds to yet another Nitrogen atom to make a fairly common element $\ce{N2}$.
Continuing, it says on that site that when you create $\ce{N2}$ from two Nitrogen atoms, that's as far as you can go for creating covalent bonds between Nitrogen atoms.
Armed with that knowledge, my question is as follows:
- Can you create further bonds utilizing another atom and still keep the strength of a triple bond and $\ce{N2}$'s natural inertness?

In short: no.  The addition of further electrons into the $\ce{N2}$ molecule actually weakens the attachment of the bond(s) between the two atoms, and renders the ensemble charged.  For nitrogen, the octet rule is obligatory (https://en.wikipedia.org/wiki/Octet_rule).

Well, you can make N2O for example, it isn't as N2 though.

Since you have zero knowledge of chemistry, I will keep explanation as simple as possible. I have to emphasize that this answer is not fully correct and can be misleading. It is suited just for interested layman, but not someone trying to grasp deeper concepts of chemistry.
Nitrogen atom has 5 electrons in its outer shell. The most stable electronic configuration, by octet rule, corresponds to 8 electrons in outer shell. Nitrogen needs three electrons to make that shell full. It can achieve that configuration by bonding with another nitrogen. One nitrogen has his own 5 electrons and another lends him his 3 - now first nitrogen has 8 electrons, e.g. full shell. First nitrogen lends his 3 electrons to second nitrogen which also had 5 electrons, so now both of them have closed shell. Since both of them lended three electrons, and accepted three - they have three shared electron pairs, which corresponds to triple covalent bond. Closed shell configuration, which they both now achieved, corresponds to a state of the lowest energy. Since everything in nature strives to a state of lowest energy, you can see why $\ce{N}$ atoms prefer to bond, instead of being each on it's own. The molecule is inert, because it achieved what it strived to: energy minimum. By adding another (third) nitrogen in a game, you would necessary have to "break" closed shell configuration, which necessary leads to increase of energy, which is not favorable. That's the reason why you can't create another bond and keep inertness of $\ce{N2}$ molecule.


text:
Alkanes to alcohols and back?
My question: could it be practical, on an industrial scale, to take a mixture of alkanes (mostly ethane,  propane, and butane) convert them to their alcohol forms, send them through a 300-mile pipe, and convert them back to their source alkanes? (They would not have to be handled as a mixture—they could be separated at the source and piped separately.)
I ask this because a company is building a pipeline through my area that is intended to carry alkanes in liquid form (under 1500 psi of pressure, so that they stay liquid). These are known in the industry as “natural gas liquids”.  It looks like this could be a very dangerous pipeline. An engineer friend suggested the alcohol conversion strategy. Even if it were a lot more expensive, it would be far less dangerous and that might justify the cost. Any thoughts?

If I were a chemist, I'm sure I could make this question more specific. As a lay person who took chemistry in college decades ago, I've made it as specific as I know how. I think what you're essentially telling me is that this is not a site for people like me.

That's not viable at all.

There's no good reaction to go from alkanes to alcohols. This would be incredible expensive to do, produces a lot of waste and is potentially very dangerous too.
There's no good reaction to convert alcohols back to alkanes. It would be again very expensive. And you would convert more expensive products into less expensive ones, which isn't a very good thing to do.
We know how to handle liquefied petroleum gas very well, it's not that dangerous. There are also hundreds of thousands of km of such pipelines [1] already in use. The safety aspect definitely doesn't justify the cost, danger, waste, ressources and energy needed to convert the product forth and back again.

[1] http://www.j-pipe-eng.com/abstract.cfm?cat_no=2145s

I had once hypothesized about a radical reaction to synthesize alcohols from alkanes in this question, although I didn't get an answer supporting or opposing my hypothesis yet.

It's doable, but it may not be practical from an economical standpoint. You must consider that two conversion plants must be built at both ends of the pipeline, with all their operational costs, etc. That won't simply be "a lot more expensive"; it would certainly kill the business by zeroing the profits obtained from selling the hydrocarbons.


text:
How does boiling water affect its hardness?
Does boiling water reduce its hardness, or on the contrary, increase it?
It seems there are ambivalent phenomena at play here:

boiling will evaporate pure water, thus increasing the concentration of minerals including calcium, thereby increasing hardness
boiling will remove $\ce{CO2}$, making apparently $\ce{CaCO3}$ precipitate, thereby reducing hardness

This is based on rather uncertain sources, as I didn't find any better, and maybe I missed other aspects.
I would appreciate any clarification, but I understand that the answer might depend on the context and type of water considered.

You may need to think outside chemistry a little bit and pay a little bit attention to your everyday life. This is what happened when you boil water in a container:

Most hardness in water, calcium/magnesium, will precipitate onto the hot surface and is removed from water. Now if you can get more something by removing it, is it against the very fundamentals of science?

$$\ce{CaCO3(s) + H2O(l) + CO2(g) <=> Ca(HCO3)2(aq)}$$ is the reaction which shows how hardness gets into water in chalky areas. The reaction is reversible, reversed in boiling as $\ce{CO2 (g)}$ is driven off promoting the back reaction, softening "temporarily hard water" This is the only form of hardness that can be reduced by boiling. Other soluble $\ce{M^2+}$ salts such as calcium sulfate (makes the lovely Burton on Trent beers) need alternative methods to soften them such as ion exchange or "bath salts" (sodium carbonate). Can you give the equation for how bath salts work?

Boiling water will NOT reduce hardness of water, whereas heating hard water to the boiling point will reduce solubility of $CaCO_3$ (Edit: and other carbonates and bicarbonates), thus reducing hardness, if filtered before cooling back. Solubility of $CaCO_3$ reduces with increasing $T$ hence boiling will not help, as the temperature is constant.
The $CO_2$ removal would not have much effect unless you have a closed vessel, where you can substantially increase the $P[CO_2]$ by removing small amount of dissolved $CO_2$ in water.

This is quite wrong. Much of the hardness is caused by hydrocarbonates, which would decompose, so hardness will decrease significantly. As for solubility of carbonates, it is too low to be considered at all.

@IvanNeretin The hydro-carbonates will precipitate but it's rate won't increase during the boiling, when temperature is constant. I guess I should have included all carbonates and bicarbonates in first point

That's right, but it will increase during the heating which precedes boiling.

@IvanNeretin  That is exactly my first point.. the solubility will reduce while it's heated and not during boiling.

Again, that's technically right, but conveys the wrong impression. You can't boil water without heating. When people speak of boiling in relation to water treatment, this implicitly includes heating.

@IvanNeretin OP is talking about increasing the concentration during boiling due to evaporation, so I'm using boiling with it's technical meaning.

Yes, I got that. Still, this wording has a decent chance to be misunderstood. Besides, you are still wrong on more than one point. Solubility of CaCO3 is irrelevant. Filtering is irrelevant. Bicarbonates are removed by decomposition and not because of decrease in their own solubility.


text:
Whats between molecules?
Okay, this will be tricky to explain. But, this is a question that i have been wondering about in quite a long time, okay lets get started, Molecules moves faster in higher temperatures and there will be more space between the molecules than in cooler temperatures, right? i wonder what is between the molecules? in pure water to an example, there should be only oxygen and hydrogen. But, what is between the H2O molecules? it got to be something, right?
I hope you guys can understand my question, my english is not the best and i am no expert in chemistry. :)

There's void, obviously.

https://chemistry.stackexchange.com/questions/28981/what-is-there-in-an-atom-between-the-nucleus-and-electrons

Just because they move faster doesn't mean there's more space between them. Here's why:
We need to look at several important things in this case: how many molecules, what's the temperature and what's the space they are in. For this example it is most easy to understand if we say we got the same space and the same amount of molecules, we just want to change the temperature.
If we got 1000 molecules (to make the math easy) in a liter (=1000 mL) then in average each molecule has free space around it of 1 mL. It doesn't matter how fast the molecules move, there are always 1000 molecules in 1000 mL so on average each molecules does always have 1 mL of space.
And in addition: between molecules is just nothing.


text:
Titration homework question
My friend sent me this for his homework, and I haven't done a titration calculation in years.

A student titrated $\pu{10.00 mL}$ aliquots of her unknown amino acid solution with standard $\pu{0.1521 M}$ $\ce{NaOH}$ and with $\pu{0.0986 M}$ $\ce{HCl}$. No inflection points were observed with the $\ce{HCl}$ titrations but two were observed in the titration with $\ce{NaOH}$. One occured after titrant had been added and another after a total of $\ce{18.61 mL}$ of $\ce{NaOH}$ had been added. Calculate the amino acid concentration. HINT: Assume the $\pu{10 mL}$ unknown solution contains a mixture of the $\ce{A}$ and $\ce{A+}$ forms of amino acid.

So based on the information, I have a diprotic amino acid. I know at the second equivalence point, mol $\ce{H2A}$ = mol $\ce{A^2-}$, so if I can find mol $\ce{A^2-}$, I'm basically done.
At the second equivalence point I have added
$$\pu{0.1521 M}\, \ce{NaOH} \times \pu{18.61 mL}\, \ce{NaOH} = \pu{2.83 mmol}\, \ce{NaOH}$$
If mol $\ce{A^2-}$ = mol $\ce{NaOH}$, then my concentration of $\ce{H2A}$ would be: $\pu{2.83 mmol/10 mL = .283 M}$
I don't think that's right because I haven't even used the first equivalent point for anything, and I don't think it's necessarily true that at the 2nd equivalent point, mol $\ce{A^2-}$ = mol $\ce{NaOH}$, so I'm not sure how to proceed.

Welcome to ChemSE.  Take a moment to familiarize yourself with the formatting this page allows (https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88). It is  a bit worrying to read "One occured after titrant [i.e. $\ce{NaOH}$?] had been added" that actually does not say how much (i.e., mL of the lye) was used. Otherwise, I would assume all $\pu{18.61 mL}$ of $\pu{0.1521 mol/L}$ $\ce{NaOH}$ was needed to reach two deprotonations, i.e. $\pu{1.42 mmol}$ of amino acid present in the aliquot.

@Buttonwood Yea that's how I read it too. Unfortunately, I don't have any clarifying information since that how it was written on my friend's assignment


text:
Alkali metals and viscosity
I've recently conducted a viscosity experiment to find the relationship between alkali metal ions, the concetration of the solution, and the viscosity of the solution. We used the Oswald viscometer to determine the viscosities of $(\pu{0.5 M}, \pu{0.25M}, \pu{0.125M}, \pu{0.0625M})$ solutions of $(\ce{LiNO3}, \ce{NaNO3}, \ce{KNO3})$, and found that the while the viscosities of the $\ce{LiNO3}$ and $\ce{NaNO3}$ solutions increased as the concentrations of their solutions increased, the viscosity of the $\ce{KNO3}$ solution would instead decrease.
And so we ask:

What is the relationship between the viscosity of a alkali-metal nitrate solution and the variant of the alkali metal ions within?
Why would the viscosity of the $\ce{KNO3}$ solution decrease with concentration?

Were the effects significant, to begin with?

And was the temperature held constant?

The experiment was held in draft chamber conditioned to 25℃, and temperature of the water before the experiment measured 25.2℃.

So the viscosity of pure water was higher than the viscosity of any of the KNO3 solutions, right?

@ChesterMiller Yes, sir.


text:
why "amu" is used as a unit in relative atomic mass?
Unified atomic mass unit (amu) is defined as $1/12$ of the mass of a Carbon-12. I want to know is this amu equal to $\pu{1.99e-23 g}$ and why do we use amu which is $1/12$ the mass of carbon? Why use "amu"?

What's easiest to compare, 12 and 15.999, or 1.993E-23 and 2.657E-23?

https://chemistry.stackexchange.com/questions/32243/units-of-mass-on-the-atomic-scale

$$
\pu{1 u} = \pu{1.660539040(20) \cdot 10^{-27} kg}
$$
The u (amu is the old unit name) is $1/12$ of the weight of an $^{12}\ce C$ atom.
The way the u is chosen ensures that all core and atom masses are multiples of 1($\pm $0.1) u. The u is a count of nucleons in an atom.
Until the 60s, chemists used $1/16$ of the mass of a average atom in a oxygen-isotope-mixture and physicist used $1/16$ of the mass of a $^{16}\ce O$ atom.
It's really just about choosing what is the best to measure and have a consistent result with. So you would chose an element with a very consistent isotope ratio (isotope-pure $^{12}\ce C$ e.g.) and one where you can measure weights easily.


text:
Cleaning up dispersed potassium hydroxide (KOH) spill?
So I tried to replace a dead triple-A alkaline battery in a waterproof compartment; it had leaked fluid + gas which had built up a bit of pressure: When opened, the material (about a dozen tiny droplets, total volume less than a pea) spattered on my hands, clothes and desk :-[
I am assuming that it was pretty concentrated KOH given that it felt slippery / turned the skin greyish before washing it off with plenty of water (no visible chemical burn or aftereffects several days later, though).

Is it good enough to wash the contaminated clothes in a regular washing machine? I did do an additional rinse cycle afterwards.
Will any droplets that I haven't been able to find become neutralized over time, or do I have to tear my study apart and wipe/hose everything down?

Some documents state that KOH absorbs atmospheric water while others state that it reacts with atmospheric CO2 to form a stable carbonate - Surely it can't do both at the same time?

Sorry for the somewhat off-topic question, but it feels like all the MDS are quite unhelpful about how to remediate a spill beyond just 'wash skin with water, remove clothes'..

You just encountered a situation where you did something without sufficient protective gear; in this example, if you had known the battery contains corrosive material, you would have used protective gloves (for your hands) and a pair of safety glasses (for your eyes).
Your estimate that the Leclanché and alkaline battery contains lye is correct. The sensation of the skin turning slippery is expectable, since $\ce{KOH}$ started to degrade the protective coating of your skin (fat) into soap.  Rinsing promptly with plenty of water was a good action.
The lye equally attacked your cloths (one reason chemists wear lab coats), and as your skin, it is beneficial to rinse them right after the skin with plenty of water, too; and then to wash in a washing machine.  On contrast, bringing them as they are into the washing machine only will spread unnecessarily the contamination to other clothes.  Expect that the areas touched by the lye eventually will become brittle, especially if i) they "dried" for some time prior to rising or washing and ii) upon action of heat (e.g. ironing).
Regarding your study, I recommend to go the extra mile, and and clean all the possible surfaces; albeit hosing sounds a bit overkill (esp. inside an appartment).  You want to eliminate the pollution quickly, as it may attack wood and wood stain; and imagine you accidentally glide a book over such a forgotten droplet, it will be attacked just as your skin was attacked, too.  (Even more, as in contrast to human skin, the dead book does not have the ability to re-cover itself by the thin film of protective glycerides.)  In addition, droplets of lye do not dry away like water, $\ce{KOH}$ is hygroscopic and will constantly remain somewhat wet (and corrosive).

Regarding the hygroscopic property of KOH: What is then the white power/crystal that (normally) forms on leaky alkaline batteries? Other sources claim that it is the reaction product of KOH and CO2 - Is this incorrect in your opinion?

@linklater KOH is hygroscopic and reacts with carbon dioxide to form the carbonate, which is not as hygroscopic, and looks dry. And don't worry, plant fibers  (cellulose et al.) are fairly resistant to alkalines.


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Mechanism of the reaction of cyclohexanone and nitric acid to form Adipic acid
Can I get some help in figuring out the mechanism for this reaction.
Adipic acid is a molecule with two carboxylic groups,but I can't see how that happens. My initial thoughts involved that the carbonyl group in cyclohexanone is protonated,but I couldn't get any further as I couldn't figure out what would happen after that.

Are you sure you want to use cyclohexanone $\ce{
C6H10O}$, not cyclohexanol $\ce{C6H11OH}$ for this reaction? This huge picture must carry a subliminal message, but I'm asking just to be 100% sure we are on the same page.

I am sure I want to use cyclohexanone, but does it really matter because cyclohexanone can be converted into cyclohexanol.

From what I remember at industrial scale cyclohexanone and cyclohexanol mixture is rectified, and cyclohexanone is used in the production of caprolactam, whereas cyclohexanol is utilized for the synthesis of adipic acid by oxidation. BTW, you need a catalyst ($\ce{NH4VO3}$) with 50% $\ce{HNO3}$ for that.

https://chemistry.stackexchange.com/questions/4416/oxidation-of-ketones

It think this reaction goes via following mechanism.

How does the last step of the reaction occur?

@Daenys Targarysn Further oxidation leads aldehyde to carboxylic acid. That's how the last step occurs.

That is incorrect, at very least mechanism of initial attack is written terribly.

@Mithoron Yes I am actually inclined to agree with you about the initial attack but its just what looked obvious to me ..

But the intermediate products formed while this transformation are correct so I just suggested this mechanism .

answer linked suggests very different intermediate products.


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How do I neutralize copper sulfate blue tint on a sofa?
I accidentally poured some copper sulfate solution on my sofa.
So there is a blue halo on it.

Is there something I can put on it to neutralize that blue tint?
I was thinking of bleach, but 1- I don't know if it will actually do anything, and 2- It will surely damage the black dye.

Welcome to Chemistry.SE! My guess would be to use ammonium chloride, hoping that copper reacts producing $\ce{[Cu(NH3)4]SO4}$ complex, which is easier to eliminate from the fabric.

There are two forms of copper sulfate you may encounter, the waterfree form ($\ce{CuSO4}$) which is colourless; and the other, the pentahydrate ($\ce{CuSO4 * 5 H2O}$) which is blue/greenish (ref).  You need to get rid of both forms.
In contrast to application on organic matter, the oxidative action by bleach, regardless if based on hypochlorite or the one that is chlorine free, won't remove the stain, yet may -- depending on the dye (mixture) deployed on your carpet attack the colour. On the other hand, though, both copper salts are fairly soluble in water.
Hence, in several cyles, you might attenuate the problem:
1) Dampen the area with some of hand warm water (e.g., with a spray bottle used to dampen your clothing prior to ironing) and subsequently 2)  rapidly dry it by pressing a fresh, clean cotton tissue against the tainted area. The white tissue is meant to act like a blotting paper, to catch both the water as well as the salt dissolved in it, too.  Hence you have to move on, to expose each time a fresh, clean, and dry section of the clean tissue towards the dampened area.
As for every cleaning of a fabric where you do not know yet how well "it works", try first with a small sector of your stained area.  If the fabric is cotton based, the polar groups in cotton actually may catch the copper salt fairly strong.
PS:  As you witnessed the accidental spill, you may be more sensibilized to see the stain, than a guest not familiar with it would be (especially as the dark fabric was not dyed uniformly).

"As you witnessed the accidental spill, you may be more sensibilized to see the stain, than a guest not familiar with it would be"  Ha, true! After seeing the question's title, I was expecting to see bright blue splotches... took a while before I could locate the stain. Nice answer, have an upvote! :D


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Why is the empirical formula of sodium chloride NaCl?
Since we know in sodium chloride crystal sodium ions are surrounded by $6$ chloride ions and each chloride ions are again bonded to $6$ sodium ions. What I want to know is that on what basis empirical formula of sodium chloride was assigned to be NaCl? does that formula mean sodium ion should atleast bonded to one chloride ion?

I think it's the time to introduce you to the formula unit. Besides, this question has already been discussed.

The empirical formula of, say, $\ce{P4O10}$ is $\ce{P2O5}$ but that doesn't mean that each phosphorus is bonded to only 2.5 oxygens. It's perfectly possible (and in fact it is pretty much a requirement) that if Na has 6 nearest-neighbour Cl and Cl has 6 nearest-neighbour Na, then the overall stoichiometry is NaCl.
Let's say, for example, that there are $n$ sodium atoms in a sample of NaCl. Then, if each $\ce{Na+}$ atom has 6 nearest-neighbour $\ce{Cl-}$ ions, then at first glance we might say there are $6n$ chloride ions.
However, since each $\ce{Cl-}$ ion has 6 nearest-neighbour $\ce{Na+}$ ions, this means that in the earlier calculation we counted each $\ce{Cl-}$ ion 6 times (once per $\ce{Na+}$ neighbour). So in fact there are only $6n/6 = n$ chloride ions, leading to a stoichiometry of $1:1$.
Maybe if this is too complicated for you, you can reduce it to one dimension. Consider the following (infinite) arrangement of red and green trees along a road:

Surely you'd agree that each red tree has two green neighbours, and each green tree has two red neighbours. Yet the "stoichiometry" is obviously $1:1$.
Now if you tile that in three dimensions, you get the $\ce{NaCl}$ structure.


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Can the effect of resonance on acidity be generalised?
A basic trend for acid: if the conjugate base is more stable, then the acid has a lower $\ce{pKa}$. Therefore, it would quite intuitive to think that resonance in the conjugate base will make the acid more acidic.
e.g. Cyclohexanol  $\mathrm p K_\mathrm a$  -16 ; phenol  $\mathrm p K_\mathrm a$ -10
But, there are exceptions. For $\ce{H3PO4, H3PO3, H2PO2}$ the effect is reversed. There is a more spacious resonance in $\ce{H3PO4}$ than in $\ce{H3PO3}$. But, still $\mathrm p K_\mathrm a$ of the former acid is greater than the latter.
Edit: Mithoron commented that my claim about P acids isn't true. Though I don't what claim he is referring, I think even his linked question supports my claim of the order of $\mathrm p K_\mathrm a$ as $\ce{H3PO4>H3PO3>H3PO2}$
So, my question is : can the effect of resonance on acidity be generalised?


text:
How did Williard Gibbs come up with the Gibbs equation?
The equation I am referring to is
$$\Delta G = \Delta H - T\Delta S$$
I understand that this Gibbs free energy determines if a process is spontaneous or not. My question is how did Williard Gibbs, presumably, come up with this equation? Did he do a bunch of experiments and guessed the answer, or did he define this thing called Gibbs free energy because of well motivated theory?

Gibbs, J. W. On the equilibrium of heterogeneous substances. Am. J. Sci. 16, 441–458 (1878). DOI: 10.2475/ajs.s3-16.96.441 should answer your question. And probably raise some more.

This comes directly from the definition of the Gibbs free energy as $G = H - TS$. At constant $T$ we have $G_1 = H_1 - TS_1$ and $G_2 = H_2 - TS_2$, so $$\Delta G = G_2 - G_1 = (H_2 - H_1) - T(S_2 - S_1) = \Delta H - T\Delta S$$ So essentially it boils down to why he defined it as such (which he did in eq. 3 of the article that @andselisk has kindly linked).

check out Herbert Callen's thermodynamic textbook. It is basically a modern version of Gibbs's work on thermodynamics and I think the best book you can read on thermodynamics. Callen

He derived this formula by fully appreciating the fact that entropy tends to a maximum.
As the first paragraph of the relevant paper says,

It is an inference naturally suggested by the general increase
  of entropy which accompanies the changes occurring in any
  isolated material system that when the entropy of the system
  has reached a maximum, the system will be in a state of equilibrium.
  Although this principle has by no means escaped
  the attention of physicists, its importance does not appear to
  have been duly appreciated. Little has been done to develop
  the principle as a foundation for the general theory of thermodynamic
  equilibrium.

That is to say, he saw it as obvious that entropy increases until it cannot increase anymore and this makes a for a good definition of equilibrium. Gibbs offers a second definition of equilibrium, however, by saying,

For the equilibrium of any isolated system it is necessary and
  sufficient that in all possible variations of the state of the system
  which do not alter its entropy, the variation of its energy shall
  either vanish or be positive.

The new move, however, is to construct some kind of function which makes these two definitions simply the same definition stated in a different way. That is, Gibbs has introduced two conditions:
$$(\delta S)_{\epsilon} \le 0$$
$$(\delta\epsilon)_{S} \ge 0$$
In words, at equilibrium a system's entropy (for a constant amount of energy) can only change by decreasing, and at equilibrium a system's energy (for a constant entropy) can only increase. Or simply, entropy is a maximum and energy is a minimum.
Then, because these two properties are opposites in the sense that one tends to a maximum and the other a minimum, it is a reasonable thing to subtract them and say that when this subtraction is equal to zero, we are at equilibrium. This is exactly what Gibbs did. He literally just multiplied the entropy by temperature without justifying it ($TS$ is common on thermodynamics though so he may have thought it obvious), and wrote down:
$$\psi=\epsilon-TS$$
where in this case $\psi$ is what we now call the Gibb's free energy. With this definition, the two conditions for equilibrium above are turned into one condition which is,
$$(\delta\psi)_T \ge 0$$
That is to say, at a constant temperature physical systems tend to the minimum of a free energy surface.
The reasoning Gibbs' has walked through thus far is sufficient to derive the equation as you have written it if we say that $\epsilon=\epsilon(U,P,V)$. That is, the energy of the system is determined fully by the internal energy (imagine microscopic motions) and the pressure on the substance and volume of the substance. Then we can simply write (again he just guesses a proportionality of one and uses units to figure out the rest),
$$\zeta=U+PV-TS=H-TS$$
where $\zeta$ is again the Gibb's free energy in a more explicit form. He takes the next couple of pages to take the total differential and then simple integration would give the familiar form but this is just a mathematical exercise not worth going through here.
A more formal, but not much more complicated, derivation of Gibbs' energy from the same condition that entropy increases can be found here under free energy of reactions. In this case two conditions for the entropy are used. Namely, the internal and external entropies. That's worth looking at as it contains the same principle which Gibbs used.

To my understanding the Maxwell relations and the Legendre transform readily produces Gibbs equation, and it is (slightly) wrong to attribute it to him alone, when Maxwell published his relations table some years in advance of Gibbs ;-) But, that is a minor comment, +1 on the answer. I can't help being impressed with the genious in all three of them... Holy moly. We are standing on the shoulders of giants.

Ya they were all insanely brilliant. The question basically asked what Gibbs did so I just summarized what he did in that paper. More details and more concrete methods could definitely be used as you note but that would've been overkill I think.

+1 to legendre transforms by stian. A good answer by jheindel too though. This makes me want to go read Callen again.