text:
What is the pH of ice?
The pH of pure liquid water depends on temperature. It is about pH = 7.0 at room temperature, pH = 6.1 at 100 °C, and pH = 7.5 at 0 °C. What happens to the pH (or to the ion product) of pure water when it freezes?
I assume that the proton transfer reactions
$$\ce{2H2O <=> H3O+ + OH-}$$
$$\ce{H3O+ + H2O <=> H2O + H3O+}$$
$$\ce{H2O + OH- <=> OH- + H2O}$$
are too fast, so that any present $\ce{H3O+}$ and $\ce{OH-}$ cannot be easily trapped in the solid ice crystal when it grows. Does that mean that pure ice crystals are free of $\ce{H3O+}$ and $\ce{OH-}$ ions?

According to Martin Chaplin's Water Dissociation and pH:

In ice, where the local hydrogen bonding rarely breaks to separate the constantly forming and re-associating ions, the dissociation constant is much lower (for example at $-4~\mathrm{^\circ C}$, $K_\mathrm{w} = 2 \times 10^{-20}~\mathrm{mol^2~L^{-2}}$).

So $[\ce{H+}] = 1.4 \times 10^{-10}~\mathrm{mol\ L^{-1}} \Longrightarrow \mathrm{p[\ce{H+}]} = 9.9$
For more information see Self-Dissociation and Protonic Charge Transport in Water and Ice Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences Vol. 247, No. 1251 (Oct. 21, 1958), pp. 505-533
This is a review article by Nobel Prize winner Manfred Eigen , after whom hydrated $\ce{H3O+}$ is sometimes referred to as the Eigen Ion.

Is pH even defined when you aren't dealing with an aqueous solution?

@bon I would say "yes", the hydrogen ion has an activity in solvents other than water.  http://www.denverinstrument.com/denverusa/media/pdf/pH_measurements_in_non-aqueous_solutions.pdf

@bon I changed answer to say p[H+] to be safe though.  OP is asking "Does that mean that pure ice crystals are free of
H3O+ and  OH− ions?" so I'm mainly trying to say that there are such ions in ice, but fewer than in liquid water.

I'd like to know how this was measured, since usually $\mathrm{pH}$ is a measure of a solution's conductivity.

@ringo The article is discussing 7 methods, one of which is d.c. conductivity measurement of ice.  Hall effect and NMR are two of the others.  One involves a.c. current.  Some are related to dielectric constant.  I can't say I fully understand the article.

"hydrogen ion has an activity in solvents other than water" OK, but I think the question @bon really meant to ask was, "Is pH even defined when you aren't dealing with any kind of solution at all?"

@DavidRicherby There can be solid-phase solutions: https://en.wikipedia.org/wiki/Solid_solution The cited paper in the answer gives concentration of the hydrogen ions in ice, and not just the equilibrium constant, but the forward and reverse rate constants that  determine the equilibrium as well.

I think it's worth remembering that in this case, pH + pOH = 14 doesn't apply anymore; we would have p[H+] + p[OH-] ~= 20, and so a p[H+] of 9.9 is actually just as neutral as pH 7 at 25°C.

It should be noted that freezing and fusion carry significant changes to the molar volume which skews the $K_w$ value on a volumetric basis when compared to liquid water.

$\mathrm{pH}$ is the aqueous concentration of $\ce{H3O+}$ or $\ce{H+}$ ions in soution. I would not say that ice lacks $\ce{H3O+}$ and $\ce{OH-}$ ions as ice's structure would allow for such, however, since the ions are not in aqueous solution, the material cannot rightfully have a "$\mathrm{pH}$" as we know it.

Why can't any material with a well defined hydrogen ion activity have a defined pH?  $pH = -\log_{10}{a_H}$

@CurtF. Because this "pH" does not compare against the pH as we know it. That is like comparing acidities of different compounds in different solvents. Hardly quantiatively useful.

I think when you say "as we know it", you probably meant "as I know it".   With that substitution your argument gets to be a bit circular, doesn't it?  And despite your claim, hydrogen ion activities are in fact "quantitatively" useful in non-aqueous solvents.

Of course you can transfer the principle to solids (see other answer), and use it as a new concept in its own right. But I'm not sure if the whole thing is sufficently similar to justify using the same names. (Which is the case for non-aqueous solutions.)

@CurtF. : A.K.'s claim doesn't seem to be impossibility of interpretation in other settings.  It is the much more practical incomparability of these numbers in other settings.  As the comments to the other answer point out, a "pH" of 9.9 in ice is approximately equivalent to a "pH" of 7 in water because the sum "pH + pOH = 20" approximately holds in ice (in specific conditions).  So, is ice more basic that water because 9.9 > 7?  If we were to put ice in contact with water, would ion exchange occur making the water more basic?  What use is that number?

It's useful for many things.  What if a reaction is acid-catalyzed?  Knowing the pH will help you determine likely rates.  What if you want to know the conductivity of the solution?  Knowing the pH will help.  Etc. etc. etc.

I've just asked Can solid acids and bases have pH values? If not, how are they classified as acids or bases


text:
Which alcohols change CrO3/H2SO4 from orange to green?

Which of the following sets of compounds cannot turn a clear orange solution of $\ce{CrO3/H2SO4}$ to a greenish opaque solution? (More than one answer is correct)
  A) Ethanol
  B) 2-Propanol
  C) t-Butanol
  D) Phenol

I think for turning a clear orange solution of $\ce{CrO3/H2SO4}$ to greenish opaque solution, the organic compound should be acidic, as $\ce{CrO3}$ turns green in basic medium. I am not sure about this.
Hence, I think A and D are the answers. Am I right?

This particular reaction is known as the Jones oxidation, in which primary ($1^\circ$) alcohols are oxidized to aldehydes, and then again to carboxylic acids, and secondary ($2^\circ$) alcohols are oxidized to ketones.
The color change is due to the reduction of $\color{\orange}{\ce{Cr^{+6}}}$ in $\ce{H2CrO4}$ to $\ce{Cr^{+4}}$ in $\ce{H2CrO3}$. This in turn disproportionates to $\ce{Cr^{+5}}$ and $\color{\green}{\ce{Cr^{+3}}}$ in $\ce{[Cr(H2O)6]^{3+}}$, with $\ce{Cr^{+5}}$ further oxidizing the alcohols to also from $\color{\green}{\ce{Cr^{+3}}}$.
The answer to your question lies in which carbons are capable of accepting additional bonds to the alcoholic oxygen, and which are not, therefore A and B are correct.

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text:
How to fake a fireball?
It is for my book: i need some material, what would burn very bright for a long time, what could it be? Or another advises, how could a character fake a fireball, seen in the sky for a long time?

Catch a bunch of fireflies and fill a transparent balloon with them and a mixture of helium and oxygen.

Innovative, I like it.

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text:
Finding/Calculating Thermal Conductivity of Magnesite
Our research group is having a very difficult time locating anything about the thermal conductivity of $\ce{MgCO3}$. It is a pretty commonly mineral, so we were pretty surprised by this. Is any way to derive thermal conductivity from other properties of magnesium carbonate? (Magnesite)

Search for the thermal conductivity of magnesite.

Thank you for your answer. We had been trying that, but all the sites that list the thermal conductivity of magnesite say that it is made mostly out of $\ce{MgO}$, not $\ce{MgCO3}$. (Example: brick, magnesite (mgo 87) https://www.thermtest.com/material-property-search)

In the dissertation of K. Kratz (2006) the following values of thermal diffusivity along the crystallographic axes of magnesite are listed:
$D_{[001]} = 6.55 ~\mathrm{mm^2\,s^{-1}}$,
$D_{[100]} = 2.38 ~\mathrm{mm^2\,s^{-1}}$,
$D_{[010]} = 2.38 ~\mathrm{mm^2\,s^{-1}}$.
Ch. Clauser and E. Huenges, Thermal Conductivity of Rocks and Minerals lists the following values for the thermal conductivity of magnesite in the temperature range from 25-100°C:
$\lambda_{\perp} = 7.32 \pm 0.57 ~\mathrm{W m^{-1} K^{-1}}$
$\lambda_{\mid} = 7.86 \pm 0.17 ~\mathrm{W m^{-1} K^{-1}}$

Thank you! The references you have listed are great. I really appreciate it!

http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
at 25 celcius the thermal conductivity is 4.15 W/m/K

Thank you. How can we be sure that this refers to a brick made out of $\ce{MgCO3}$ and not $\ce{MgO}$? Most of the websites I have seen that list a similar thermal conductivity have been calling $\ce{MgO}$ bricks magnesite. (Example above).

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Why does copper and zinc connected by a wire in 6.0M NaOH become a galvanic cell?
For an experiment I placed two electrodes of copper and zinc connected by a wire in 6.0M NaOH. According to standard reduction potentials, there should be no reaction; however, the copper becomes plated with zinc! There was also a measurable voltage across the two electrodes.
The hypothesized reactions (that are supposedly not favorable) are:
$$\ce{Zn(s) + 4OH^- <=> [Zn(OH)4]^2-(aq) + 2e^-} \textrm{   E}^\circ = 1.23 \textrm{V}$$
$$\ce{ [Zn(OH)4]^{2-}(aq) + 2e^- <=> Zn(s) + 4OH^-} \textrm{   E}^\circ = -1.23 \textrm{V}$$
So $\textrm{E}^\circ = 1.23 + -1.23 = 0$, and the reactions should not happen. Why do they?


text:
ChemSpider versus SciFinder for finding chemicals without CAS number
Just trying out a couple of difficult options for identifying chemicals without CAS number information, and wondering whether Chemspider or SciFinder is a better tool for doing these searches. Do they search the same databases/repositories? Also, which has better features and is easier to use?

By definition, if it's in SciFinder, it must have a CAS identifier. So if you have reason to think it doesn't have a CAS, then you won't find it in SciFinder.

@GeoffHutchison in this situation SciFinder is being used as a tool to try and work out the CAS identifier from general enquiries sent in by people who might identify with the chemical in other ways as they are not scientists or experts in the field.

SciFinder is a product of Chemical Abstracts Service (CAS), which is a division of the American Chemical Society. Its registry contains information on more than 111 million organic and inorganic chemical substances, and is updated with 15,000 new substances daily. Additionally, it has more than 66 million protein and nucleic acid sequences.[1] The chemical information is produced by CAS, and is prepared by the CAS Registry System, which identifies each compound with a specific CAS registry number, index name, and graphic representation of its chemical structure.[2]
ChemSpider is a free chemical database from the Royal Society of Chemistry. Its registry contains information on 49 million chemical structures, coming from 513 sources. You can search the database by systematic names, synonyms, trade names, database identifiers, and also using structure-based queries, with the option to draw structures in the web page or use structure files from your computer.[3]
Both registries allow you to find literature references, experimental and predicted property data (boiling and melting points, etc.), commercial availability, and spectra,[3] but the CAS registry also offers preparative methods and regulatory information from international sources.[1]
SciFinder may offer more, but it's extremely expensive. In fact, I'm not even sure if CAS offers SciFinder to individuals. You can see their pricing options here. Taking into consideration everything ChemSpider can do and the fact that it's free, it's an excellent resource, and I definitely recommend it.

References

CAS, Chemical Substances
Wikipedia, CAS Registry
ChemSpider

You can also the pubchem database, which is openly accessible.

I also highly recommend Chemicalize for unknown and obscure structures.

IIRC that ACS members can make a certain number of SciFinder searches as part of your membership benefits.


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Why is a double bond not formed with oxygen when reacting this molecule with PCC?
In my textbook it says that when reacted with PCC, primary alcohols produce an aldehyde. Why is this example different? I do not see any double bonds.

I think you are confused about how the aldehyde is being depicted. See here, and look at the first and third pictures. They are equivalent.

While I think it is not exactly the best question we have received, the homework close option shouldn't be indiscriminately used - it just becomes a slippery slope and eventually any question that we don't like can just be closed with HW. This has already been happening to some extent.

That is known as space-saving notation. Carboxylic acids are like this too (RCO2H)

The product shown is an aldehyde. If you count the number of bonds to the carbon with oxygen attached, you can see one single bond to the carbon on the ring, one bond to the hydrogen and therefore there must be two bonds to oxygen - a double bond.


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How to calculate the cation exchange resin capacity?
If I have a cation exchange membrane and need to remove 2 g/L of sodium, how much resin would I need.
I used the amount of substance of $\ce{Na}$ to calculate the amount of substance of $\ce{SO3H-}$ needed for complete sodium removal. But this will not give me the correct mass since the resin is a polymer with that functional group attached. However I haven't been able to find any kind of complete structures for a cation resin.

Resins typically include a capacity label in $\mathrm{mmol/g}$, don’t they?

I bought one for a project pretty cheap so it didn't have much technical information.

First of all, you will need more sulfonic groups than sodium ions since sodium doesn't have a very good affinity for sulfonic groups.
Anyway, you can find some specs here — figure 2. This source is using sulfonated polystyrene with various crosslink percentages and expresses the capacity in $\mathrm{mmol}_\text{sulfonic groups}/\mathrm{g}_\text{resin}$ ($1\ \mathrm{meq} = 1\ \mathrm{mmol}$).


text:
How to find chemical similarity between two compounds using Tanimoto method
I came across a paper named "Chemical Similarity Searching" by Peter Willett
John M. Barnard and Geoffrey M. Downs, 1998. At one place, this paper talks of finding chemical structure similarity between two chemical compounds using Tanimoto method. Now I want to understand how it is done but could not derive anything just by reading the paper. I wanted to know if someone can explain how the distance is found taking two compounds into consideration. I wanted a step by step explanation. I have failed to derive how to find similarity from the paper. Please help.

Apparently, it is more generally known as the "Jaccard Index":

In "A Computer Program for Classifying Plants", published in October 1960,[5] a method of classification based on a similarity ratio, and a derived distance function, is given. It seems that this is the most authoritative source for the meaning of the terms "Tanimoto similarity" and "Tanimoto Distance". The similarity ratio is equivalent to Jaccard similarity, but the distance function is not the same as Jaccard distance.

In terms of chemistry, this calculation is carried out on a fingerprint of the molecule. That wikipedia page has a nice reference to this paper, although it is not open access. There are no doubt many resources out there about fingerprints/Tanimoto, even open source implementations...
This also looks interesting: "Why is Tanimoto index an appropriate choice for fingerprint-based similarity calculations?".


text:
How does WD40 displace water?
How does WD-40 displace water? Or does it even displace water? WD40.com explains what WD-40 stands for:

WD-40® literally stands for Water Displacement, 40th formula. That's the name straight out of the lab book used by the chemist who developed the product back in 1953. The chemist, Norm Larsen, was attempting to concoct a formula to prevent corrosion—a task which is done by displacing water. Norm's persistence paid off when he perfected the formula on his 40th try.

If it displaces water then I always thought that that meant it was denser or heavier. I just tested to see: I sprayed a couple teaspoons into an empty glass then poured that into another glass half filled with water. The WD-40 floats! So what does it mean to say that it "displaces water"? Surely it doesn't mean that it's hydrophobic, all lubricants are hydrophobic.

Marketing language has a number of peculiar features; I guess that's one of them. As to me, WD40 does not displace water; it just sticks to all surfaces and prevents water from sticking to them. True, any lubricant would do the same, but WD40 performs better due to low viscosity.

Yes. It's not a matter of buoyancy, but of reduced interfacial tension.

Let's start with the fact that bouyancy of an object in open water is caused by displacement, but liquids that are less dense and may spread out do not displace water, they just lie on top of it. If I put a block of wood in a fish tank that is full, it displaces and spill over a mass of water equal to the mass of the block, but if I slowly pour (or rather spray) oil on a full tank of water, the oil will spill over, not the water. The definition for displacement is (per Google):

the moving of something from its place or position.
the occupation by a submerged body or part of a body of a volume that would otherwise be occupied by a fluid.

The first definition is what WD-40 is doing to water, the second one is what causes buoyancy. The WD-40 displaces the water in that it moves it from where it is, but how?
Looking at the ingredients from the SDS, we can get a feel for their purpose:
$$\begin{array}{|c|c|c|}
  \hline
\textbf{Material} & \textbf{Hazards} & \textbf{Purpose (not stated on SDS)}\\
\hline\text{Aliphatic Hydrocarbon} & \text{Flammable Liquid Category 3}  & \text{Thin oils and evaporate quickly}\\
 & \text{Aspiration Toxicity Category 1} & \text{}\\ \hline
\text{Petroleum Base Oil} & \text{Not Hazardous} & \text{Provide long term protection} \\ & & \text{from water and air}\\ \hline
\text{LVP Aliphatic Hydrocarbon} & \text{Aspiration Toxicity Category 1} & \text{thin oil and evaporate slowly}\\ \hline
 & \text{Simple Asphyxiant}\\
\text{Carbon Dioxide} & \text{Gas Under Pressure,} & \text{Propellant}\\
& \text{Compressed Gas}\\ \hline
\text{Non-Hazardous Ingredients }&\text{Not Hazardous} & \text{This does the real work}\\ \hline
\end{array}$$
The oils are nice because once the water is removed, the oil will remain, preventing it from returning. But it is the non-hazardous ingredients that do most of the work. Hard to say what exactly this 10% is made of but it has to get water to "unstick"/"unwet" the surface that is being treated. To do this WD-40 is likely using a non-ionic surfactant of some sort similar to what is used in shampoos to emulsify the water and allow the flow of the WD-40 spray to remove (displace) the emulsified water. $%The rest of the wD-40 components act as carrying agents to remove the water from whie$ This speculation about surfactants is further supported by the fact that WD-40 foams when sprayed which is not characteristic of pure hydrocarbons.
Additionally, there would likely be some material to adhere to the surface to prevent further interaction with water or oxygen. There is the petroleum base oil, but it needs a way to stick to the metal. Likely, the use an unsaturated fatty acid in some way whether it is pure oleic acid or boiled linseed oil or something similar to those to bond to most metals and other materials and prevent further corrosion.

Your last paragraph is the most accurate in explaining what WD-40 is for. Its a preventative maintenance tool, not an after the fact - omg this bolt is rusted closed - tool. At that point you would need a solvent that dissolves the rust binding the two materials together.

It is still unclear to me what it is  about WD-40 that makes it a "water displacment" product.

@Octopus I didn't mean steel but metal in general, I have clarified the post if you are still not sure please elaborate. It's hard to augment the answer if I don't understand what is still unclear.

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text:
How does LC50 translate to "real world" exposure for paint thinner?
Last year I started oil painting, which means (in part) that I use paint thinner for working with and/or cleaning brushes. I recently became concerned about my exposure to the vapor from the paint thinner I use, so I requested a toxicology report from the manufacturer. In the document it specifies that the summated $\mathrm{LC}_{50}$ for the product is 21400 mg$\cdot$ m$^{-3}$.
My (basic) understanding of this is that 50% of test subjects will die with a single exposure at the above level; however, I've really no idea how that translates to "real life" and wonder if someone can shed some light on it.
I'm painting in a reasonably small room (around 3 m x 5 m) with windows open, so I'm essentially trying to determine if this is safe or not, or whether I should be using breathing apparatus and/or move to a different room.
This is a link to the whole document which I received from the manufacturer which contains the substance and other information.

You are right $\mathrm{LC_{50}}$ is where $50~\%$ of subjects will die. for a room $3~\mathrm{m} \times 3~\mathrm{m} \times 5~\mathrm{m}$ to reach a concentration of $21400~\mathrm{mg/m^3}$ you would have to evaporate $963~\mathrm{g}$ of the material ($3~\mathrm{m} \times 5~\mathrm{m} \times 3~\mathrm{m} \times 21.4 ~\mathrm{g/m^3} = 963~\mathrm{g}$) within a sealed room.  Unless your material is very volatile it is unlikely you will even be close to reaching this concentration.  That said however it does not mean that this is not a lower concentration that could kill a more susceptible body, nor that the prlonged smell won't make you feel sick.  It only says at this concentration you have a $50~\%$ chance of dying if exposed at the concentration for a specified (though in this case we don't know how long) amount of time.

Thanks - that's actually a bit clearer.  I've opted for buying a reusable respirator half-mask anyway, just in case - but you're right... the report doesn't really allow for continued exposure and/or other biological factors.  I became genuinely concerned that Bob Ross died of Lymphoma and it's been related to chemicals such as these.  That being said... he did used to 'beat the devil' out of his brushes which likely caused more of it to be airborne.  It's also possible of course, that the chemicals had absolutely nothing to do with it.

The $\mathrm{LD_{50}}$ and $\mathrm{LC_{50}}$ (lethal dose and lethal concentration) values are measures for acute toxicity as you noted. Long-term exposure is generally not depicted well by these acute toxicity measures. A.K. pointed out, how improbable it is to have a serious risk of acute poisoning. (Also note that the vapour pressure is given as $2~\mathrm{mmHg}$, which is $2.7~\mathrm{mbar}$ in proper units, meaning that evaporation will be very slow at ambient temperature and pressure.)
Chronic toxicity data is typically given as workplace exposure limits, under the reasoning that if you work with something every day and its concentration is less than the harmfulness limit, it should be okay. Scanning your MSDS, I did not come across any such value, so there should be no serious chronic toxicity if you keep your room well vented and don’t forget to recap the bottle, etc.


text:
Why is a rearrangement observed for the secondary alcohol but not the primary in these examples?
I am thinking that maybe only hydrogen shifts are possible and methyl shifts are not allowed in such situation? Is that the reasoning? but then again in the second example the bromide could shift to a secondary position after doing a hydrogen shift.

Consider the preferred pathway for a 2º alcohol vs a 1º alcohol.

Predicting the pathway for these Wagner-Meerwein rearrangements is rather hard, isn't it? Usually you get a lot of side products.

You need to consider the possible and plausible reaction mechanisms for both substrates. They are both nucleophilic substitutions in first approximation, so we need to decide whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms are more plausible.
Primary electrophilic carbons are very likely to undergo $\mathrm{S_N2}$ reactions while $\mathrm{S_N1}$ is strongly disfavoured since primary carbocations are less stable than secondary or tertiary ones.

Secondary electrophilic carbons can react according to both mechanisms, technically. However, both mechanisms are not ideal: the steric strain is rather large so $\mathrm{S_N2}$ reactions are slower and secondary carbocations are still rather unstable compared to tertiary ones, so $\mathrm{S_N1}$ reactions are slower.
In this case, we have the possibility to stabilise the less stable second intermediate in the scheme shown below by a Wagner-Meerwein rearrangement. We can let both hydrogen or methyl migrate, but the hydrogen migration gives us the most stable carbocation (a tertiary one) so it will prevail. Multiple other migration paths will likely lead to minor products.

The primary carbon will not, as stated above, form a primary carbocation since it is too unstable. Hence only the bimolecular substitution will occur. If no primary carbocation is generated, no Wagner-Meerwein rearrangement can occur. And the mechanism of an $\mathrm{S_N2}$ reaction does not allow bromide’s (or anything else’s) migration.
The secondary carbon could undergo bimolecular substitution, too. The reaction rate will probably be slower than that of fragmentation and rearrangement, though, since the latter are unimolecular processes. And other shifts (e.g. methyl followed by hydride) are possible but again slower since you need to shift twice. All in all, a higher proportion of side-products may be expected.
In no case will the bromide shift after substitution; any shift would need to occur before.

You should perhaps make it clear that there can be no rearrangement during SN2. It is only implied in your answer. I believe this is what the OP was stuck on.

@Karl Like this?


text:
Why does electron configuration in the periodic table is different when you follow the aufbau principle?
. In periodic table for example Ag has an electron configuration of [Kr] 4d10,5s1 but when you do it yourself using Aufbau principle it should be [Kr] 5s2 4d9. Why is that?
When ask for Ag + valence electron configuration answers may now vary because of the existence of two electron configurations from the periodic table and using aufbau principle...

Because the Aufbau principle is not taken as gospel by the electrons busy trying to solve Schrodinger's equation. It is not a law of nature, it is more of a general guideline for chemistry students that has problems, certainly with d electrons.

How come Sir that it serves as a general guideline if it doesn't match in what we see in the periodic table? Thank you.

I would not expect a general guideline to precisely match every single element in the periodic table. You have already noted that it doesn't, and basically I am saying that you should not expect it to. It is not a physical law. Adding d electrons makes for some funny things, and you have noted one of them. There are others.

As a general rule, full electron shells are more stable than unfilled shells. In silver, promoting an electron from the 5s to the 4d shell completes the d orbital, which increases the stability of the 4d shell more than it destabilizes the 5s shell.
This is also explained here:
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=21&t=2807


text:
why is nitrogen positive in HNO3?
Shouldn't it be negative since it shares one of Nitrogen's free electron pair with an O to form a double bond?

http://chemistry.stackexchange.com/questions/7174/the-lewis-structure-of-hno3

http://chemistry.stackexchange.com/questions/4828/formal-charges-how-to-work-them-out

The nitrogen has five valence electrons. Here, only four are involved in the covalent binding of the -O, =O, and -OH. Only one of the two electrons involved in the covalent bonds belongs to the nitrogen. As there are only four covalente bonds, nitrogen is lacking one electron, therefore it is positively charged.


text:
Identification of an unknown organic molecule from only 1H NMR and 13C NMR
I am currently trying to determine the structure of my unknown compound with both 1H NMR and 13C NMR spectra.
My H NMR peaks:

1.16 ppm [singlet, 1H], 1.68 [singlet, 3H], 1.75 [singlet, 3H], 4.2 [doublet, 2H], 5.4 [multiplet, 1H], and 7.24 (solvent peak)

C NMR:

18, 26, 59, 77 (solvent peak), 124, and 136 ppm.

I have deduced that my molecular formula is $\ce{C5H10O}$. And I am fairly certain I should have a double bond in my structure. However, I do not know intuitively where this pi bond should be.
What should I be looking for to determine the placement of this pi bond?
I unfortunately do not have characteristic IR peaks to determine whether I have an alcohol, ether, or a ketone.

124 and 136 scream alkene.  See how much further you get with that idea.  Also, you can deduce that you have an alcohol due to only one C-O carbon (59).  I'll put an answer up in two hours if you haven't posted it yourself

The singlet at 1.16 is not necessarily a methine proton...

In fact 1.16 is unlikely low for methine, which leaves only one other possibility, according to my tables.

@Karl the integration of the 1.16ppm is 0.98. Am I interpreting this wrong?

No, integral of 1 is fine. There is one functionality that gives singlets which can appear over a large CS range, from 1 to 5.3 ppm. Least that's how i read your spectrum.

@Lighthart I agree that I should have an alcohol, and that I have an alkene. However, how do I know where to place this intuitively? I think from the 1.68 and 1.75ppm I should have two methyl substituents.

OK, my last comment was confusing, if you don't know that you should never expect coupling from OH protons.

I recind all comments on the OH proton. If it exchanges with chloroform to make a singlet, it should never appear at 1.16 ppm.

@Karl - OH will not exchange with chloroform, ever. It may exchange with water or other labile peaks IN chloroform. In the absence of fast exchange, coupling to OH peaks can be, and is frequently observed.

It is usually a good idea to post the original spectra with the question, as it is very common that the biggest difficulty with these types of problems is the interpretation of the spectrum in the first place. Chemical shift and integration are rarely sufficient to make a full interpretation; peak appearance (linewidth) is useful and interpretation of multiplicity can be tricky. Your alkene peak here should probably appear somewhat like a triplet, although there will possibly be long range couplings also to the two methyl groups  that would make them doublets (1-2Hz). cont'.

Also, 1.16 for an alcohol is fine, if the sample conditions are right. Where H-bonding can be minimised, and exchange excluded, upfield peaks for OH protons are normal. For this, the sample should be dilute, and the solvent very dry. If this were the case, this OH peak should be quite sharp. As concentration increases, or exchange with water increases, expect the peak to broaden and move to higher chemical shift.

The doublet at $4.2\ \mathrm{ppm}$ is almost assuredly an alcohol group or ether group.  Since we only have one carbon that is bonded to an oxygen at $59\ \mathrm{ppm}$, we conclude this is a primary alcohol. The situation of being a doublet says smoething else is lurking around.
$5.4\ \mathrm{ppm}$ is far enough down field, we start thinking alkenes, but the multiplicity is mysterious for the moment.
$1.68\ \mathrm{ppm}$ and $1.75\ \mathrm{ppm}$ are in the range of allyl methyl groups, and the carbon signals are okay with this as well.  Additionally singlets for these groups indicate a geminal dimethyl compound with a third substituent on an alkene justifying different chemical shifts.
Finally the peak at $1.16\ \mathrm{ppm}$ is difficult to explain.  However, we know we have an alcohol, and they can have a fairly wide range of values.  If this peak is broadened at all, I would immediately assign it to the alcohol.  It being a singlet is also worrisome, but if exchange is occurring, this is not out of the question.
The multiplicity of the alkene proton indicates there are some long-range things going on, and these don't always resolve well.   My guess is you are picking up some long-range coupling and is not resolving nicely in the methyl groups, which appear to be singlets.

The observed multiplicity of the peaks in the question is clearly not quite right, and is important for determining the correct arrangement of substituents around the double bond. The CH2 at 4.2 should, in the least, be a doublet with coupling in the order of 5-6Hz to the CH at 5.4. This would make this alkene peak a triplet by appearance. Additionally, Coupling from the alkene through to both methyl groups should be in the order of 1-2 Hz, and these splittings should be observable in the 1H spectrum.

With a poorly shimmed spectrum 1-2 Hz might not be apparent, particularly on a 90MHz or lower.  Given question, it seems the spectrum was taken on an instructional machine, so I am not very worried about the lack of resolution of the methyl peaks.  And if the alkene proton really has 3 different coupling constants, multiplet makes a lot of sense here.

If the OH peak is exchanging, it cannot appear at 1ppm. The exchanging peak appears at the median of the chemical shifts involved, and this does not work with water or anything else i can think of. But if it is not exchanging, it has to make a triplett. I also came to prenol as the solution, but something is wrong here. It'd be nice to see the actual spectrum.

Agree.  A picture would have made this much simpler.  But I don't agree that it has to be a triplet.  It could be smoothed out by intermoleular hydrogen bonding.  I expect it is a broad singlet.

Problems with the alcohol proton aside, it matches this real well: http://www.sigmaaldrich.com/spectra/fnmr/FNMR001914.PDF, particularly if OP meant to indicate 2.16 for that peak

2.16 for the OH proton I'll believe. This here is supposed to be allyl alcohol http://www.chemicalbook.com/SpectrumEN_107-18-6_1HNMR.htm with OH at 1.97 ppm. 1.17 is totally out of bounds.  @LauraBrown : Can you confirm that?


text:
How to dialkylate a cyclohexene, and convert a carbonyl to an alkene?

I'm trying to design a synthesis of D from C. All I can think of is using $\ce{Me2CuLi}$ to add the methyl group to the alkene. I thought maybe then if isobutyl bromide was in the reaction mixture when $\ce{Me2CuLi}$ pushed the negative charge onto the oxygen and it came back down that could do an $\mathrm{S_N2}$ with isobutyl bromide, but I've never seen that before. Then for converting the carbonyl to an alkene I have no idea.

The Wittig reaction may be useful.

Ah of course, the Wittig reaction with $H_3C-^+PPh_3$ as the phosphonium salt? Just the dialkylation I'm not sure about now.

I was thinking of Baylis-Hillman to add the isobutyl group, but I don’t think that works well enough with alkyl halides =C

You are right about using the Gilman in a 1,4-Michael addition. The enolate created from this can then be used in the nucleophilic attack of a haloalkane. Then all that's left to do is a Wittig reaction, swapping the carbonyl oxygen for a carbon. The full stepwise synthesis would be:

Could also use Tebbe/Petasis methylenation

Alkylation of the enolate with isobutyl bromide will likely give substantial elimination. 3-Bromo-2-methylprop-1-ene will be more reactive and not undergo elimination. Reduce the double bond with Pd/C/H2. Alternatively, trap the enolate with isobutyraldehyde, dehydrate with acid and hydrogenate.


text:
Why don't we take into account the auto-protonation of water when measuring its concentration?
When we calculate the concentration of water at  $ 25~^\circ\rm C $,
why don't we assume dissociation of water at this temperature?
$$\ce{H2O <=> H+ + OH-}$$

I didn't get it. What should we assume, and how?

Because the number of moles of dissociated water is so slight compared to the number of moles of undissociated water, it makes no practical difference to our calculations to ignore the dissociated fraction.

Compare the free energy to perform the dissociation with the thermal energy $kT$.

If you take a look at the equilibrium constant for your reaction, you'll see why. @Ivan please don't forget about bad titles, for the sake of Mart-the-mod's heart. :)

The equilibrium constant of the reaction is  $$\ce{H2O <=> H+ + OH-}$$
is 10^-14 and hence only 1 atom of water in 10^7 dissociates which really is negligible.


text:
What is the difference between a ligand and a co-ligand?
I've seen the word co-ligand pop up in several papers. Is there a particular difference between a co-ligand and a ligand?

I believe this refers to situations in tranistion metal complexes where there is more than one kind of ligand, and specifically, where one of the coligands could not ligate and endure by itself due to reactivity, such as a carbene being blocked from reactivity due to a coligands steric bulk


text:
Limit the production of hydrogen in a small container
I have to produce hydrogen at a near-constant pressure inside of a really small container without using an automatic pressure release value. I would like the pressure to be relatively high, but the most important aspect is that the exact pressure needs to be predictable. I was thinking of utilizing the reaction between aluminum, water, and sodium hydroxide.
Al2O3 + 2 NaOH + 3 H2O  → 2 NaAl(OH)4
Al2O3 + 2 NaOH          → 2Na+ + 2[Al(OH)4]-

2 Al + 2 NaOH + 6 H2O   → 2Na+ + 2[Al(OH)4]- + 3 H2
2Al + 2NaOH + 2H2O      → 2NaAlO2 + 3H2.

Now, I know the pressure will slow down the reactions, but will the reactions completely stop at a certain pressure? If so, how can I deduce that specific pressure given the different concentrations of reactants and constraints of the container?
Safety is my number one priority, so I need to make sure that I can predict this pressure to a high degree of accuracy. If the pressure is too high, I won't even attempt anything. Any advice on this would be greatly appreciated.

The reaction is quite exothermic and it is difficult to predict the reaction rate and temperature. So it's not an easy task to predict the pressure.

Yeah, I know. It can be pretty violent. I've seen reactions that went completely out of control and had to be thrown in a bath of water or an ice bucket.

I would think a hydrogen gas bottle and appropriate flow and pressure control (using correctly rated components) would be the right (and safe!) way to do it. Engage a safety professional - hydrogen has a very broad explosive range in air.

Following on from Jon's comment: maybe a hydrogen generator? If you want to control the reaction, I would stir the aluminium (powder) in water, then use a dosing pump to add, say, 6M NaOH. Use, ideally, a jacketed laboratory reactor as the reaction is very exothermic. Failing that, use a silicone oil bath as a large heat sink. What equipment are you using to hold the high pressure? I would call up to 20 bar low pressure, above 100 high and everything else inbetween medium, depending on what equipment you're used to. A matter of perspective I suppose. Also, the pressure will change with tempera

The device has to be rediculously small. 16 cubic centimeters to be exact. Thats why I thought there wasnt enough room for the standard safety equipment. This is a very unique question haha.


text:
Dissociation of H3PO4 and colligative properties

Which of the following solutions that have concentrations of $0.01\ \mathrm{M}$ show the highest elevation in boiling point and why?
A) $\ce{NaCl}$
B) $\ce{CaCl2}$
C) $\ce{H3PO4 }$
D) $\ce{C6H12O6}$ (glucose)

When dissolved in water, $\ce{H3PO4 }$ dissociates into 4 ions, which is more than any of the other compounds. Someone told me though, that because it's a weak-acid, it doesn't matter how many ions it has, and that $\ce{CaCl2}$ will still have the highest elevation boiling point. Is this true?

Phosphoric acid won’t dissociate further than $\ce{H3O+ + H2PO4-}$ in a standard aquaeous solution. That’s only two ions.

@Jan I've asked the same question on Quora(it wasn't active, that's why i asked it here). and this is what a chemist says : https://www.quora.com/Which-of-the-following-solutions-that-have-concentrations-of-0-01m-show-the-highest-elevation-in-boiling-point-Why

That person has not the slightest of clues what he is talking about.

If you want, you can write up a self-answer. The preferred Stack Exchange way of showing that you found a solution to your problem ;)

Hey you need to know the degree of dissociation $\alpha$ to answer this question most accurately.


text:
Preparing an inert atmosphere reaction vessel with only an input?
One time, the graduate student that I work with was showing me how to prepare an inert atmosphere in a double necked round bottom flask. After drying it in the oven, she used a torch to drive out all the water vapor. Then she put septums over both openings, and inserted a needle connected to the nitrogen input through one of them, but with no outlet! I asked her to explain how any gas could escape in this set-up, and she tried but I was still very confused.

Her setup was wrong. There needs to be an outlet, the simplest solution in this case being a second needle with nothing connected to it, puncturing either septum to act as a vent to the atmosphere. After a while, depending on the volume of the flask and amount of nitrogen flow, most of the atmospheric air is removed.
This procedure is, however, quite inefficient, and likely unsuitable for particularly sensitive reactions. While most oxygen and moisture is quickly removed, it becomes exponentially slower to remove the last remnants. A much quicker and more reliable method is to perform vacuum/nitrogen backfill cycles. Three or four cycles will fully replace the interior atmosphere with nitrogen gas of the same purity as the input. This can be performed using a single needle and a Schlenk line (or any system connected to nitrogen and a vacuum pump with a intermediating stopcock), which essentially switches the same needle between a gas inlet and a gas outlet.

Just to give a practical example to further illustrate Nicolau’s point: My lab typically relies on an argon balloon at the end of the Schlenk line. When a fresh balloon is used and all the joints are properly greased (and any rubber tubes in the apparatus are still fine) and a flask is connected to the line with septa covering any outlets, the balloon typically stays filled overnight. If it has a good day, it’ll stay filled for two nights.
In a typical evacuation/backfill cycle, you can actually see the balloon getting slightly smaller when backfilling (assuming you have a larger flask than $25~\mathrm{ml}$. This is about as much as one looses to minute holes overnight. So it takes about 18 hours for the entire line to loose as much gas volume as a single flask has. No way that her reaction was water- and oxygen-free.


text:
I lost salt mass after evaporation
I was doing an evaporation experiment and I used salt and water. I started off with 5 g of salt and after the evaporation process, I was only left with 4.3 g of salt. Why is this?

Did you try heating the salt without dissolving it first? The salt could have started off with some moisture content.


text:
Defects of Crystals
So while studying defects, I found that:

Interstitial defect is when constituent particle is present in the interstitial site.
My question is where this constituent particle is coming from? As it leads to increase in density so It must come from outside?
In metal excess defect (due to anion vacancy) : In case of $\ce{NaCl}$ it says, when heated with sodium vapours, chloride ion leaves the crystal and combines with $\ce{Na}$ atom outside to release an electron. Why it is breaking the bonds with $\ce{Na}$ ion to again join with $\ce{Na}$ atom?

If the defect is thermodynamic, i.e. consists of the same material as the bulk and is in equillibrium, then it is already "included" in the regular, measureable density of the material.
If not, well, then it comes from "outside".
Equillibrium. If the concentation of Na on the outside is high enough, then some wandering defect chloride from inside will get stuck. The Na atom outside will happily throw away an electron to bond with the chloride. The separation of charges already happened before, when the chloride left its place in the lattice to leave behind a positively charged hole.


text:
Calculate Solubility
So I've been looking at some chemistry processes and I was trying to conclude how they got the numbers they got. How would I go about calculating the solubility of Phenyl-2-nitropropene in Isopropyl Alcohol?

It cannot be calculated theoretically and has to be done practically
It is measured in terms of the maximum amount(usually moles) of solute dissolved in a solvent at equilibrium


text:
Why do the alkali metals form different products upon combustion in air?
From Wikipedia's article on sodium:

When burned in dry air, it forms primarily sodium peroxide with some sodium oxide.

We know that sodium has a strong reducing capacity, so why does it produce a compound in which the oxygen atom is not reduced to the fullest possible extent?

After posting the question, I read up some more and found this, on Chemguide's page describing the oxidation properties of Group 1 elements:

So why do any of the metals form the more complicated oxides? It is a matter of energetics.
In the presence of sufficient oxygen, they produce the compound whose formation gives out most energy. That gives the most stable compound.
The amount of heat evolved per mole of rubidium in forming its various oxides is:
$$\begin{array}{|c|c|}
\hline
 & \text{enthalpy change (kJ / mol of Rb)} \\ \hline
\ce{Rb2O} & -169.5 \\
\ce{Rb2O2} & -236 \\
\ce{RbO2} & -278.7 \\
\hline
\end{array}$$

This is hardly an explanation, because it just says in a circular way that it's energetically profitable to form such compounds. Why is this energetically profitable?

Can part of the explanation be "because in Group 1, starting from $\ce{Na}$, ion charge densities are low enough to allow peroxides and superoxides to exist"?
I wonder if the following behaviour is related to this: lithium nitrate decomposes to lithium oxide (again the other participant has the oxidation -2), while the rest of the group's elements have their nitrates decompose to a nitrite (again the other participant's oxidation is -1):

From Chemguide:
$$\ce{4 LiNO3 (s) -> 2 Li2O (s) + 4 NO2 (g) + O2 (g)}$$
$$\ce{2 MNO3 (s) -> 2 MNO2 (s) + O2 (g)}$$

What oxygen compound is formed depends on temperature, pressure, and the amount of oxygen present

@aventurin - I fully believe that, but I would like to know why it forms. Why is it better to create a compound in which the oxygen atom has the oxidation number -1 and not its usual -2.

Oxygen is a strongly oxidising compound that is present in a large excess. Once there is no metallic sodium left to react with, it starts looking for some other prey. I don't see what is wrong with RbO2 being a stable compound. Some peroxides are more stable than others. Forming bonds is nearly always profitable, unless entropy is against it or you have to spend too much energy to break or weaken other bonds for it.

@Karl - but, for instance, Mg does not form peroxides upon exposure to air in ordinary conditions.

Related: Why does potassium form peroxides but sodium does not?; When will peroxide form?

Related: https://chemistry.stackexchange.com/q/152754/24750

Since I will deal with all of the alkali metals in this answer, I think the question should also be broadened. There is no point in covering one single metal (sodium) without touching the others since it is the trend going down the group that we are interested in.
All thermodynamic data is taken from Prof. M. Hayward's lecture notes at Oxford.
So, firstly, some data. There is an increase in ionic radius going down the group, which should not be surprising:
$$\begin{array}{cc}
\hline
\ce{M} & \text{Ionic radius of }\ce{M+}\text{ / pm} \\
\hline
\ce{Li} & 76 \\
\ce{Na} & 102 \\
\ce{K} & 138 \\
\ce{Rb} & 152 \\
\ce{Cs} & 167 \\
\hline
\end{array}$$
This leads to an decrease in the magnitude of the lattice enthalpies of the Group I superoxides, peroxides, and oxides, going down the group:
$$\begin{array}{cccc}
\hline
\ce{M} & \Delta H_\mathrm{L}(\ce{MO2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O})\mathrm{~/~kJ~mol^{-1}} \\
\hline
\ce{Li} & -960 & -2748 & -3295 \\
\ce{Na} & -860 & -2475 & -2909 \\
\ce{K} & -752 & -2175 & -2503 \\
\ce{Rb} & -717 & -2077 & -2375 \\
\ce{Cs} & -683 & -1981 & -2250 \\
\hline
\end{array}$$
As for why there is an increase in magnitude going across the table, we have to look at the factors controlling the magnitude of the lattice enthalpy:
$$\Delta H_\mathrm{L} \propto \frac{\nu z_+ z_-}{r_+ + r_-} \tag{1}$$
where $\nu$ is the number of ions in one formula unit, $z_+$ and $z_-$ are the charge numbers on the cation and anion, and $r_+$ and $r_-$ are the ionic radii:
$$\begin{array}{ccccc}
\hline
\text{Formula unit} & \nu & z_+ & z_- & \text{Numerator in eq. (1)} & r_-\text{ / pm} \\
\hline
\ce{MO2} & 2 & 1 & 1 & 2 & 149 \\
\ce{M2O2} & 3 & 1 & 2 & 6 & 159 \\
\ce{M2O} & 3 & 1 & 2 & 6 & 120 \\
\hline
\end{array}$$
The lattice energies of the peroxides and oxides are roughly 3 times those of the corresponding superoxides, because of the larger numerator. The lattice energies of the oxides have a  slightly larger magnitude than those of the corresponding peroxides, because of the smaller anionic radius.
Just looking at the lattice enthalpies, we might think that all metal cations would form the oxides. However, this approach is flawed because it does not take into consideration the energy cost of forming the three different anions from molecular dioxygen. Recall that the lattice enthalpy is defined as $\Delta H$ for the reaction
$$\ce{M+ (g) + X- (g) -> MX (s)}$$
However when we burn a metal in oxygen we are starting from $\ce{M}$ and $\ce{O2}$. So, we have to figure out the energy needed to get from $\ce{M}$ to $\ce{M+}$, and from $\ce{O2}$ to the relevant anion ($\ce{O2-}$, $\ce{O2^2-}$, or $\ce{O^2-}$).
The analysis here is complicated by the fact that the charges on the anions are not the same. Therefore, the reaction
$$\ce{M(s) + O2 (g) -> MO2(s)}$$
cannot be directly compared with
$$\ce{2M(s) + O2(g) -> M2O2(s)}$$
In order to "standardise" the equations, we will consider the reactions per mole of metal. One can loosely think of the combustion reaction as "releasing" some kind of energy within the metal; the most favourable reaction will be that which "releases" the most energy per mole of metal. Therefore, the three reactions we are considering are:
$$\begin{align}
\ce{M (s) + O2 (g) &-> MO2 (s)} & \Delta H_1 \\
\ce{M (s) + 1/2 O2 (g) &-> 1/2 M2O2 (s)} & \Delta H_2  \\
\ce{M (s) + 1/4 O2 (g) &-> 1/2 M2O (s)} & \Delta H_3  \\
\end{align}$$
Now, we construct Hess cycles for all three reactions.
Superoxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + O2(g)}} @>{\Large\color{blue}{\Delta H_1}}>> \color{blue}{\ce{MO2(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\Delta H_\mathrm{L}(\ce{MO2})}A \\
\ce{M+(g) + e- + O2(g)} @>>{\Large\Delta H_\mathrm{f}(\ce{O2^-})}> \ce{M+(g) + O2^-(g)}
\end{CD}$$
Peroxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + 1/2O2(g)}} @>{\Large\color{blue}{\Delta H_2}}>> \color{blue}{\ce{1/2M2O2(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O2})}A \\
\ce{M+(g) + e- + 1/2O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O2^2-})}> \ce{M+(g) + 1/2O2^2-(g)}
\end{CD}$$
Oxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + 1/4O2(g)}} @>{\Large\color{blue}{\Delta H_3}}>> \color{blue}{\ce{1/2M2O(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O})}A \\
\ce{M+(g) + e- + 1/4O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O^2-})}> \ce{M+(g) + 1/2O^2-(g)}
\end{CD}$$
Now we need more data. $\Delta H_\mathrm{f}(\ce{M+})$ is simply the sum of the atomisation energy and first ionisation energy:
$$\begin{array}{cc}
\hline
\ce{M} & \Delta H_\mathrm{f}(\ce{M+})\text{ / }\mathrm{kJ~mol^{-1}} \\
\hline
\ce{Li} & 679 \\
\ce{Na} & 603 \\
\ce{K} & 508 \\
\ce{Rb} & 484 \\
\ce{Cs} & 472 \\
\hline
\end{array}$$
and the enthalpies of formation of the anions are
$$\begin{array}{cc}
\hline
\ce{X} & \Delta H_\mathrm{f}(\ce{X})\text{ / }\mathrm{kJ~mol^{-1}} \\
\hline
\ce{O2-} & -105 \\
\ce{O2^2-} & +520 \\
\ce{O^2-} & +1020 \\
\hline
\end{array}$$
Why is the enthalpy of formation of $\ce{O^2-}$ so large? The answer is that, to get from $\ce{1/2 O2}$ to $\ce{O^2-}$, you need to first break the $\ce{O=O}$ bond, then add two electrons to the oxygen. Furthermore, the second electron affinity is often an unfavourable process. For the other two anions, you don't need to break the $\ce{O=O}$ bond.
So now, we can see a trend at work already. Going from the superoxides to peroxides to oxides, the more negative lattice enthalpy favours the formation of the oxide. However, the increasing heat of formation of the anion favours the formation of the superoxide.
When do each of these two factors win out? Well, when lattice enthalpies are comparatively large, we would expect the lattice enthalpy factor to outweigh the heat of formation of the anion. Lattice enthalpies are large precisely when the cation is small, and therefore lithium forms the oxide when heated in oxygen. However, with caesium, lattice enthalpies are smaller, less significant, and the heat of formation of the anion wins out; caesium therefore forms the superoxide.
The trend is of course not black and white. Going down the group from lithium to caesium, we might guess that perhaps there are one or two elements that form the intermediate peroxide. That element is sodium. You could say that the larger lattice energies of sodium salts sufficiently compensate for the formation of the peroxide ion, but aren't enough to compensate for the formation of the oxide ion.
I leave you with the last bunch of numbers, which tabulate the values of $\Delta H_1$ through $\Delta H_3$ for all the elements (all values in $\mathrm{kJ~mol^{-1}}$). You can actually calculate these yourself by plugging the data above into the Hess cycles. It seems that the data is a little different from that given in the Chemguide screenshot you have, but the conclusion is the same, so I'll ignore that:
$$\begin{array}{cccc}
\hline
\ce{M} & \Delta H_1\text{ (superoxide)} & \Delta H_2\text{ (peroxide)} & \Delta H_3\text{ (oxide)} \\
\hline
\ce{Li} & -386 & -435 & \mathbf{-459} \\
\ce{Na} & -362 & \mathbf{-375} & -342 \\
\ce{K}  & \mathbf{-349} & -320 & -234 \\
\ce{Rb} & \mathbf{-338} & -295 & -194 \\
\ce{Cs} & \mathbf{-316} & -259 & -143 \\
\hline
\end{array}$$
As described earlier, the salt with the most negative enthalpy of formation will be preferentially formed. These are bolded in the table.

I believe the answer given by @orthocresol is far too complicated. There is a very simple, yet understandable answer to your question. Apart from energy factors, ther is something called coordination in formation of a compound. COORDINATION is simply the number of atoms the central atom in a compound can accomodate. As we go down the group, size increases for alkali metals (rather for any group, except maybe for transition metals, but that is another story). Due to this increase in size, larger ionscan form bonds with larger negative ions. So Li being small can be stable with small oxide ion only.
Na ion, being larger, can remain stable with larger peroxide and superoxide ions, though peroxide is more stable.
 Finally Potassium is most stable with the superoxide ion.
note : the size of the negative ions decreases as follows-
          Superoxide>peroxide>oxide
I havent covered the elements further down in group 1 as they are very reactive and react with all the ions of oxygen to form stable species.


text:
Why is no dynamical correlation recovered in the CASSCF method?
Configuration interaction is the only certain way to get a calculation to converge toward the exact solution of the many-body Schrödinger equation (within the approximations used for the framework of the method). In such calculations, the wave function is expanded in all possible electron configurations, and will therefore often take too long to be practical. For full CI, all dynamical correlation is recovered.
In truncated CI, part of the dynamical correlation is recovered. The wave function expansion is, for example, limited to include only single and double excitations (CISD).
In a CASSCF calculation, a subset of all orbitals is chosen, and the wave function is expanded in all possible electron configurations within this subset of orbitals. However, here it is said that no dynamical correlation is recovered. This is stated so often I suspect it is common knowledge for computational chemists, but I can provide references if someone want them.
It is my understanding that the CI method recovers dynamical correlation in allowing electrons to be moved into virtual orbitals (and hence "react" to the instantaneous coulombic repulsion from other electrons). In allowing for higher a number of excitations, a larger portion of the dynamical correlation is recovered (CISDT > CISD).
So, why is no dynamical correlation recovered in the CASSCF method? Imagine a good set of orbitals are chosen for the active space, such that no important configurations are left outside of the wave function expansion, surely dynamical correlation is described? What if we extend this to the generalized actice space (GAS) formulation. In developing GAS, the original authors aimed to remove the "deadwood" of configuration state functions; that is, exclude CSF's with near-zero coefficients. So now a larger active space can be used, and hence a larger number of excitations can be used. Why is not more dynamical correlation recovered?
If indeed CASSCF recovers no dynamical correlation, a CASSCF calculation should give the same energy as a HF calculation. Is this the case?

You can reach full-ci by including everything inside the active space. It is somewhat a convention that people call the correlation energy captured by CASSCF as non-dynamical correlation, when the active space is chosen correctly.

This convention is confusing to me.

Try to be a big name in quantum chemistry community and let the majority accept your convention :)

Give me 15 years and we'll see :)

The difference between dynamical and static correlation has always been ill-defined, and there is no clear-cut delineation between them.
Correlation simply means that the probability of finding an electron somewhere at position $a$ and another electron somewhere at position $b$ is not simply the product of the two probabilities.
(In that respect, even Hartree-Fock has some correlation since it forbids two same spin electrons occupying the same state (Fermi correlation).)
Correlation was originally thoguht of as the instantaneous repulsion of electrons. This was dynamic correlation. The origin of the difference between static and dynamical correlation came from studies of bond dissociation in molecules. As you pull a bond apart, the electrons should be further apart, and correlation energy should decrease. However, the opposite was found, and in many cases the correlation energy increases. Therefore, there must be some sort of "other" correlation occurring. This was static correlation.
This is because there are additional (near-)degenerate configurations that contribute strongly to the nature of the wave function during dissociation. You can think of this as a manifestation of entanglement, of which Hartree-Fock, being single determinant, fails miserably.
In short, static correlation came to describe situations where the underlying Hartree-Fock reference was a qualitative and quantitative failure.
Dynamic correlation, on the other hand, was used to describe the cases where the underlying Hartree-Fock reference has the correct qualitative behavior, but was not quantitatively accurate.
CASSCF does a better job at calculating static correlation because most of the important near degenerate configurations are multi-electron excitations within a small valence space. Thus it generally captures the configurations necessary to describe bond breaking, etc. with a well-chosen active space.
Single reference coupled cluster, does better at capturing dynamic correlation because it is usually truncated to few-pair excitations, like the double excitations of CCSD, but cover the whole virtual space.
For high levels of coupled cluster, e.g. CCSDTQ, the distinction between static and dynamic correlation really breaks down, and some high level coupled cluster calculations actually do quite a good job at otherwise "multireference problems." Multireference coupled cluster --- combining the best of CASSCF and CC --- is an active area of research today.

A paraphrase of something I read somewhere: "When it comes down to it, static correlation is just dynamic correlation where the correlated states have started to closely approach degeneracy." This seems consistent with what you wrote, but ... is it at all correct?

I would say that's true if the correlated states are near the state of interest.

So "fermi correlation" is basically the same as the electron exchange? "The movement one depends on the movement of the other"


text:
AlCl3: ionic and covalent?
My textbook states that AlCl3 is both an ionic and a covalent compound .How is it so?

Strongly related but not a dupe IMO. http://chemistry.stackexchange.com/questions/19568/what-happens-if-the-electronegativity-difference-is-exactly-2-1/19573#19573

I thought that there was another question on the nature of bonding in $\ce{AlCl3}$ floating around, but I cannot find it …

http://chemistry.stackexchange.com/questions/19715/why-does-alcl3-dimerise-to-al2cl6-at-low-temperatures

Possible duplicate of Why is Aluminium Chloride a Lewis Acid if it's an ionic compound?

Try to think about the various interactions between all pairs of atoms/ions in the AlCL3 lattice. Which atoms/ions interact? Are all interactions the same? Then think about what covalent and ionic interactions mean.

@Mithoron Not dupe-y enough imho o.o

@Mith the questions aren't dupes if the same answers apply. That's not the only criterion. I think leaving this open to get some answers would be useful.

The distinction between ionic and covalent compounds is arbitrary and many compounds fall somewhere in the middle. This is explained in a previous question.
In the case of $\ce{AlCl3}$ we have a metal 'cation' and a non-metal 'anion' so in a simplistic view we would expect this to be an ionic compound. However, it actually turns out that the $\ce{Al-Cl}$ bonds display a significant degree of covalent character. In the solid state it adopts an 'ionic lattice' structure with octahedral coordination for the $\ce{Al^{3+}}$ ions but in the liquid and gas phases it exists as a covalent compound, either as $\ce{AlCl3}$ or as a dimer $\ce{Al2Cl6}$.
We can rationalize this by considering the polarizing effect of the aluminium cation on the chloride anions. $\ce{Al^{3+}}$ is a small, highly charged cation and therefore has a high charge density. $\ce{Cl-}$ is a relatively large anion, with a low charge density and is easily polarized by the hard cation, giving the bond significant covalent character. The effect is even more pronounced if we move to a larger anion such as $\ce{I-}$ where the molecule, $\ce{AlI3}$ exists as a covalent dimer, $\ce{Al2I6}$, even in the solid phase.

I would thing thrice before calling Al2Cl6 and especiall Al2I6 'covalent' I don't have the numbers on me, but the last time I checked, 'covalent' PbF4 had charge on the central atom around +2 or so. Al2Cl6 is likely to be similar in this regard.


text:
Air drying acetic acid solution
So I'm cleaning up a lot of my stuff to have more space and I have 2 jugs of grocery bought white vinegar jugs (5% acetic acid). Naturally to make it more compact, I'm trying to concentrate it. I was wondering if leaving the vinegar out in a warm place to let the water evaporate will be effective.
I know that acetic acid solutions form azeotropes at wide ranges, so I might actually be evaporating the acetic acid along with it.
So would it work?

In short, no. You'll evaporate everything pretty soon.


text:
How semi-permanent hair dye stays in place?
Semi-permanent hair dye washes off after several months. What mechanism(s) keep the dye in the hair? (I assume it's not primary chemical bonds, otherwise it would be permanent)

According to this: http://questions.sci-toys.com/node/72, hair dyes work by creating pockets within the hair. The colored dye molecules then get stuck in these pockets - requiring multiple washes to escape. Short term hair dyes use larger dye molecules so that they cannot get stuck in the hair. On the other hand, longer term dyes actually react with you hair to stay in there for an extended duration of time. For the most part, all of these hair-dye interactions are the result of intermolecular forces.
Semi permanent hair dyes tend to be small compounds that get stuck in your hair. This means that they are not permanently bound to your hair, but they will require a lot of washes to escape. I hope this helps!

Hair is built as a structured, layer design architecture.  To keep the dye (semi)permament in the hair, the colour altering dye has to be installed below the top layer.

One route to achive this by using reactive dyes, that upon contact with the keratine (that exhibits thiol-groups, $\ce{R-SH}$) indeed to form chemical bonds with the dye, the later -- for example -- decorated with sulfonic acids ($\ce{R-SO3H}$) or its salts.  See, for example, patent US3415606 (A)
A second route consists of using two solutions that contains fragments of the dye -- aiming that the dye molecule are created by these two (or more) components in the inner of the hair, but once created (within the hair) too large to escape the hair by simple washing.

These processes may assisted by alteration of the pH value: in basic media (i.e. high pH value), the inner of the hair becomes accessible as the "covering shingles" of the cuticula open.  In slightly acidic medium (i.e. at low pH-value), the hair's cuticula closes.
Anyway, the dyes have to be applied again and again.  Beside the natural reason (the continuously delivered hair), the dyes -- exposed to day light -- degrade and fade.  Indeed, lightfastness is one of the parameters tested regularly for these dyes.


text:
Vapor-pressure or Vapor-concentration
To summarize from my textbook:
rate of evaporation:

proportional to surface area
but essentially independent from pressure (of surrounding gas)

rate of condensation

proportional to both surface area
and concentration (#molecules/volume) of (appropriate) molecules in the gas

At some concentration the rate of evaporation = rate of condensation.
The pressure corresponding to this concentration is called "vapor pressure of the crystal".
(There is some example about iodine in a sealed flask in the same chapter, but there is no explicit statement of assumptions about sealed container or not having another gas, like air "around". Textbook: College Chem by Pauling.)
My question:
For a liquid/crystal in a closed vacuumed container, this seems straightforward, since pressure would be a function of concentration (and temperature), I believe. So there is 1 pressure corresponding to the concentration (at the given temperature). But then why is it not called "vapor concentration"? Which the mentioning of its proportionality, as the only relevant one, would suggest.
The only reason for the name "vapor pressure", I can imagine, is that not only concentration matters.
For example, considering the cases where the sealed container has initially near vacuum or 1atm or 2atm of air besides the liquid/crystal. Then the pressure of the gas at equilibrium in those cases will surely be different, even if the concentration of the evaporated substance is the same.
The book does not mention "partial pressure" or "pressure contribution".
My alternative thinking is that "vapor pressure" does not describe the equilibrium state of arbitrary mixed gas of arbitrary container and initial pressure, but confusingly enough, just the property of liquid/crystal which only depends on chem composition and temperature. Which I would call "in-vacuum evaporation rate". And the reason for the naming is the special experimental scenario used to measure/define this property. Neither the total pressure of the possibly mixed surrounding gas, nor the partial pressure of any component is equivalent to it, just merely equal in certain scenarios.
Is this correct?
I am confused. Isn't the book a bit incomplete with the explanation?

Welcome to Chemistry SE! Thank you for a well-formed question! You might want to register at the site, so as not to lose your points.

But then why is it not called "vapor concentration"? || both vapor concentration and vapor pressure are defined and bound by ideal gas law.. But pressures are, your know, easier to measure, then concentrations.

Both processes do NOT depend on the absolute pressure in the gas phase, but the partial pressure of the substance in the gas phase. The "vapour pressure" is not a variable but the property of the substance at given temperature, i.e. its partial pressure above the condensed substance in equilibrium.

@Karl - i disagree because of latent heat.... e.g. condensation will be faster at higher pressure if surrounding gas pressure is higher for more collisions with surface to help remove heat from surface - this is like blowing on surface to help cool it

@tom The heat conductvity of gas does only strongly depend on pressure  at very low pressures ("Dewar"). Also the heat conductivity of the condensate is magnitudes higher than that of the gas phase. Plus if we are not talking "isothermic" you get convection etc. and it becomes a LOT more complicated.


text:
Does the acid strength of hydrochloric and hydrofluoric acid correlate with the electronegativity of the halides?
Is this statement true or false?

HCl is a stronger acid than HF because fluorine is more electronegative than chlorine.

I'm thinking this is true, because since fluorine is more electronegative than chlorine, it has  a higher ionic energy and therefore a stronger bond between hydrogen than chlorine does with hydrogen. In result, leading HCl to dissociate  more easily. Can you please clarify.

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

You are correct, HCl is a stronger acid than HF. Fluorine is both more electronegative and smaller than chlorine. Because fluorine is more electronegative, the bond between it and the hydrogen is more polar, meaning that the proton would need to overcome a larger coulomb force to separate from the fluorine. In addition, fluorine is a smaller atom, meaning that the addition of an electron has a larger effect on the net charge of the atom. This both makes the fluoride anion (the conjugate base) less stable and it also - once again - increases the coulomb force between the proton and the fluorine. In addition, the coulomb force is even stronger because there is a smaller distance between the two atoms. As a general rule of thumb, as you move down the halogens, the acids become stronger and they form more stable conjugate bases.

It is because of three reasons:

Stability of fluoride is less than chloride, as stability of conjugate base decides acid strength.
Smaller bond dissociation energy of HCl than that of HF.
Hydrogen bonding in HF which effects proton releasing tendency.


text:
Why are antiaromatic compounds unstable?
Antiaromaticity is so destabilizing that it can cause compounds such as cyclobutadiene to elongate or manipulate their orbitals so that the pi system is no longer aromatic. I understand that aromatic compounds are stabilized because they have a full HOMO and that antiaromatic compounds have a half filled HOMO, which according to this question is the reason that they are unstable. I don't agree with this explanation because in other compounds, half filled orbitals cause compounds to be stabilized. For example, triplet oxygen is significantly more common than singlet oxygen in nature. Also, some transition metals will promote electrons from their S orbitals to D orbitals so that they can have these shells half filled. I know that full orbitals are more stable than half filled orbitals, but are half filled orbitals destabilizing in alkenes while they are stabilizing in oxygen and metals? If not, why are antiaromatic compounds so unstable?

http://chemistry.stackexchange.com/questions/18557/is-cyclobutadiene-anti-aromatic

First of all, they are not unstable, just less stable and less symmetric, because the Jahn-Teller effect distorts their geometry to remove degeneracy. With $\ce{O2}$ it would not work that way, since you can't really distort a 2-atom molecule; no matter what you do, its symmetry is still the same.

But, correct me if I am wrong, the Jahn-Teller effect distorts their geometry so that they are no longer antiaromatic. Compounds distort their orbitals to avoid the instability associeated with antiaromaticity. The Jahn-Teller effect is a symptom of antiaromatic instability, but I don't think it is it's cause.

http://chemistry.stackexchange.com/questions/30927/what-is-the-justification-for-h%c3%bcckels-rule

What causes a diradical to be so unstable? I understand why free radicals such as superoxide are unstable, but triplet oxygen is a diradical and it is more stable than singlet oxygen, which has no radicals. I think there is something else that causes antiaromatic compounds to be unstable, but I do not know what it is.

Re. your first comment, you are wrong about that. The antiaromatic instability is an instability with respect to a structure where the double bonds are localised, i.e. instability with respect to a Jahn-Teller distortion. The Jahn-Teller distortion leads to the adoption of exactly such a geometry: one with alternating bond lengths and that is not necessarily planar. So, the JT instability and the "antiaromatic instability" are one and the same.

For more information I would refer you to Orbital Interactions in Chemistry 2nd ed. by Albright, Burdett, & Whangbo, pp 282-285. I quote: "Strictly speaking, therefore, the distortion in cyclobutadiene is really of the second-order Jahn-Teller type." It is probably more common in inorganic chemistry to speak of it as a Jahn-Teller distortion, whereas organic chemists don't want to bother with all those wavefunction symmetry stuff and just call it antiaromaticity.

Thanks! I think I understand antiaromaticity now. Do you want to post what you were telling me as an answer, or should I write it up?

Antiaromatic compounds are not necessarily unstable - they are just less stable than a bunch of ethenes connected by sigma bonds. If the pi electrons in an antiaromatic compound were to delocalize across a large degenerate orbital, then the result would be a half filled shell. While a half filled shell is not the end of the world, it is a lot more stable to have a full shell of electrons. As a result, compounds such as cyclobutadiene undergo a Jahn Teller distortion (in this case, becoming a rectangle rather than a square) in order to form two separate (and non degenerate) orbitals. This creates a collection of separate, full, pi orbitals. The stability gained by having full rather than half full orbitals is more than enough to compensate for the energy required to break the symmetry in the first place. Cyclobutadiene is stabilized due to stretching not because it prevents antiaromaticity, but because it causes all the pi electrons to be part of complete shells. While it might seem weird that a physical distortion such as stretching can consume less energy than is released by going from a half to a full shell, let me remind you that this is not the case. Higher energy light (UV) is often required to stimulate electronic transitions while lower energy light (IR) can cause bond stretching in compounds.
I was able to figure this out thanks to the help of orthocresol.


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How can you change the proportions of reactants and products at equilibrium if Le Chatelier's principle is occurring?
For example, how could you make a buffer with equal concentrations of acid and conjugate base? If you add more conjugate base, it will shift the equilibrium to form more acid, which will keep the proportions of each the same. So how can you achieve equal concentrations if this is happening?

see http://chemistry.stackexchange.com/questions/42266/reason-behind-le-chateliers-principle 'cause it seems you don't get it

The Henderson Hasselbalc equation is a good way to calculate both the pH and the concentration of acid/conjugate base in an equilibrium solution. According to it:
pH = pKa + log([A-]/[HA])
Since you want to know when [A-] = [HA], you are looking for when the pH is equal to the pKa. This means that you will have equal concentrations of the acid/conjugate base whenever your pH is equal to your pKa, which you can find by monitoring the pH. While adding more conjugate base will shift the equilibrium towards acid formation, it will normally increase the acid concentration less than the amount of base that you added.


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What's the relationship between dew point and temperature?
From a Ph.D dissertation, the writer change the land surface from nature earth to urbn and built up to compare the difference involving with this land use change.
Due to the urban heat land effect, the temperature of lower atmosphere above the surface increase, and the dew point decrease.
Dew point is a parameter represent the absolute moisture. When the temperature increase, how to analysis the dew point increase?

There are two separate effects:

The average temperature above urban land is greater than above rural land. This could be because of the cooling effect of plants evaporating water, or because of the higher albedo of vegetation than pavement and roofing.
The absolute humidity decreases; that is, the total amount of water vapor in a given volume of air decreases, perhaps due to the capture and run-off of rainwater into storm drains, rather than evaporating.

Thanks for your reply. This is my understanding of your second opinion: the water vapor in a given volume above the certain area has its origin and sink process. From rural land to urban and built up, the amount of water content  which evaporate from the surface has decreased due to less vegetation and impermeable surfaces. So, the water above the area are not directly influenced by the surface temperature. Is it correct?

Your understanding appears correct, except for "So, the water above the area are not directly influenced by the surface temperature. " One cannot assume the two effects are not related without more data.


text:
kJ per mole or kJ per 2 moles?
From Chemguide:

Here is a simple reaction between hydrogen and oxygen to make water:
$$\ce{2 H2(g) + O2(g) -> 2 H2O(l)} \quad \Delta_\mathrm{r}H^\circ = \pu{572 kJ mol^-1}$$

First, notice that the symbol for a standard enthalpy change of reaction is $\Delta_\mathrm{r}H^\circ.$ For enthalpy changes of reaction, the $“\mathrm{r}”$ (for reaction) is often missed off — it is just assumed.
The $“\pu{kJ mol^-1}”$ (kilojoules per mole) doesn't refer to any particular substance in the equation. Instead it refers to the quantities of all the substances given in the equation. In this case, $\pu{572 kJ}$ of heat is evolved when 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of liquid water.

Is the measurement unit $\pu{kJ mol^-1}$ correct here? Shouldn't it be measured simply in $\pu{kJ},$ because the $-572$ figure is for the whole reaction, i.e. for 2 moles of the product?

Related: ∆Enthalpy of diatomic molecules

the whole reaction, i.e. for 2 moles of the product?

This is your mistake.
The balanced equation $\ce{2H2 + O2 -> 2H2O}$ does not mean that 2 moles of the product are being formed.
The numbers, called stoichiometric coefficients, only tell you about the ratio in which the reactants react and form products.
That is to say, if I had a system in which I reacted 1 mol of $\ce{H2}$ with 0.5 mol of $\ce{O2}$ to form 1 mol of $\ce{H2O}$, the above equation would still be correct. This is because the ratio of $\ce{H2}:\ce{O2}:\ce{H2O}$ is still $2:1:2$. As long as the ratio is $2:1:2$, the equation will still be correct and applicable.

What $\Delta_\mathrm{r}H^\circ$ tells you is: how much the enthalpy change would be, if the number of moles reacting were the same as the stoichiometric coefficients. Therefore for the above reaction, saying that $\Delta_\mathrm{r}H^\circ = -572~\mathrm{kJ~mol^{-1}}$ means that
If 2 mol of hydrogen were to react with 1 mol of oxygen to form 2 mol of water, the enthalpy change would be -572 kJ.
That means that if you multiplied the equation by 2, $\Delta_\mathrm{r}H^\circ$ would also be multiplied by 2.
$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$
This line tells you: if 4 mol of hydrogen were to react with 2 mol of oxygen to form 4 mol of water, the enthalpy change would be -1174 kJ. That's obviously true because you're reacting twice as much reactants to get twice as many products, so the enthalpy change must be twice as much.

So what is the per mol for

As chemguide said, it is not per mol of anything. The per mol arises from the definition of the reaction enthalpy change, $\Delta_\mathrm{r}H^\circ$. I have another answer describing it in the context of Gibbs free energy here.
However, one way to look at it is to remember that the reaction does not stipulate the actual amounts of reactants and products. That is to say, the value $\Delta_\mathrm{r}H^\circ$ on its own cannot help you find an enthalpy change, unless you tell it exactly how much of the reactants and products you have. If you don't specify whether you start with 2 mol or 1 mol of water, you cannot find $\Delta H$ for the process.
That means you need to multiply $\Delta_\mathrm{r}H^\circ$ by an amount of substance (units mol) to get $\Delta H$ (units kJ). So, $\Delta_\mathrm{r}H^\circ$ must have units of kJ/mol.
What is the actual amount of substance that you are multiplying by then? It is simply the amount of substance of any species reacting, divided by its stoichiometric coefficient in the equation.
Therefore, if you have 6 moles of oxygen reacting, the "amount of substance" that you multiply by would be $6~\mathrm{mol}/1$. The enthalpy change would be
$$\Delta H = (-572~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{1}\right) = -3432~\mathrm{kJ}$$
You have to divide by the stoichiometric coefficient. This is necessary for consistency. For example, if you were only given this $\Delta_\mathrm{r}H^\circ$:
$$\ce{4H2 + 2O2 -> 4H2O} \qquad \Delta_\mathrm{r}H^\circ = -1144~\mathrm{kJ~mol^{-1}}$$
then the "amount of substance" you would multiply by is $6~\mathrm{mol}/2$. You would have
$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{6~\mathrm{mol}}{2}\right) = -3432~\mathrm{kJ}$$
Note how the answer comes out to be the same.
But it doesn't have to be for oxygen either. That's what Chemguide means when it says it's not for any particular species in the reaction, because it's applicable to all of them. For example, if I told you that 44 mol of water was formed, then you would use
$$\Delta H = (-1144~\mathrm{kJ~mol^{-1}})\left(\frac{44~\mathrm{mol}}{4}\right) = -12584~\mathrm{kJ}$$

orthocresol's answer covers a lot of good ground, but I want to add another concept that may provide some clarity. In my undergraduate chemical engineering kinetics course, we had to apply calculus to the progression of chemical reactions, so as to understand how fast things were being converted, how much heat was being generated/absorbed, etc.  We almost always used a quantity called the extent of reaction, usually represented as $\xi_i$, to track the progress of each reaction i that was occurring.
The extent of a given reaction has units of moles, and is defined for a given reaction as written.  So, take the reaction as written in your question:
$$
\ce{2H2(g) + O2(g) \stackrel{\xi_1}{\longrightarrow}2H2O(\ell)}\tag{1}
$$
For each mole of reaction extent $\xi_1$, two moles of $\ce{H2}$ and one mole of $\ce{O2}$ are consumed, while two moles of $\ce{H2O}$ are formed.
Now, take the same reaction but with all coefficients multiplied by two, as orthocresol did:
$$
\ce{4H2(g) + 2O2(g) \stackrel{\xi_2}{\longrightarrow}4H2O(\ell)}\tag{2}
$$
Here, for each mole of reaction extent $\xi_2$, four moles of $\ce{H2}$ and two moles of $\ce{O2}$ are consumed, while four moles of $\ce{H2O}$ are formed.
So: This all relates to heats of reaction (and entropies of reaction, and free energies of reaction, and ...) in that, really, they are always defined relative to the extent of a specific reaction, with specific and defined stoichiometric coefficients. Further, if one changes the way the reaction is written by multiplying all of the stoichiometric coefficients by a constant factor, the numerical value of the heat of reaction associated with the reaction also is multiplied by that same factor.
Thus, for Reaction $(1)$ I would write:
$$
\Delta H^\circ_{r,1} = -572~\mathrm{kJ\over mol~\xi_1}
$$
Similarly, for Reaction $(2)$ I would write:
$$
\Delta H^\circ_{r,2} = -1144~\mathrm{kJ\over mol~\xi_2}
$$
But, in both cases, the energy released, say, per mole of $\ce{H2}$ combusted is the same:
$$
-572~\mathrm{kJ\over mol~\xi_1}\times\mathrm{mol~\xi_1\over 2~mol~\ce{H2}} = -286~\mathrm{kJ\over mol~\ce{H2}} \\
-1144~\mathrm{kJ\over mol~\xi_2}\times\mathrm{mol~\xi_2\over 4~mol~\ce{H2}} = -286~\mathrm{kJ\over mol~\ce{H2}}
$$

NOTE: Proper definition of the basis for things like heats of reaction is actually something that is done rather poorly in a lot of cases.  I've spent a lot of time at different points in my scientific career trying to sleuth out whether a value is given per mole of $\ce A$, per mole of reaction extent, or whatever.

From Wikipedia of Standard Enthalpy of Reaction:

The standard enthalpy of reaction (denoted $\Delta H_\mathrm{r}^\Theta$) is the enthalpy
  change that occurs in a system when one mole of matter is transformed
  by a chemical reaction under standard conditions.

For the above reaction, the stoichiometry of one mole is taken into consideration when one mole of hydrogen and half a mole of oxygen react to form one mole of Water.

For a generic chemical reaction
$\ce{$−v$_{A} A + $−v$_{B} B +\ \dots -> $v$_{P} P + $v$_{Q} Q\ \dots}$
the standard enthalpy of reaction $\Delta H_\mathrm{r}^\Theta$ is related to the standard
  enthalpy of formation $\Delta H_\mathrm{f}^0$ of the reactants and products by the
  following equation:

$$\Delta H_\mathrm{r}^\Theta = \sum_B v_B \Delta H_\mathrm{f}^\Theta \left( B \right)$$
You can read more from here — Standard Enthalpy of Reaction


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What is the most common oxidation state of gold?
I had a test this morning which had the following question:

What is the most common oxidation state for gold?
  (a) +1  (b) -1  (c) +2  (d) +3

Since the electronic configuration of $\ce{Au}$ is $\ce{[Xe] 4f^{14} 5d^10 6s^1}$. After removing the $\ce{6s^1}$ electron and gaining $+1$ charge, $\ce{Au}$ should have noble gas configuration and hence this should be the most stable configuration. So the answer has to be +1.
Could someone verify my reasoning?

The most important (natural as well as anthropogenic) oxidation state of gold is actually zero. Apparently, the question refers to the oxidation state of gold in its compounds.

Gold is present in $+2$ oxidation state in this complex: $\ce{[Au2Cl2{μ-(CH2)2PPh2}2]}$. The +2 oxidation state will stable in dimeric form.

It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the $5d^{10}$ orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3.

I just checked Wikipedia, it read "Au(I), referred to as the aurous ion, is the most common oxidation state...". I links it to the following study: http://pubs.acs.org/doi/abs/10.1021/cr980431o

@Abhishek Mhatre Well, it says  in Wiki "Au(I), [...] is the most common oxidation state with soft ligands such as thioethers, thiolates, and tertiary phosphines." And 2 sentences after: "Au(III) (auric) is a common oxidation state, [...]". For example, salts of Au(I) (except
insoluble (AuI) or extremle stable $[Au(CN)_{2}]^{-}$) disproportionate in water: $3Au^{+} = 2Au + Au^{3+}$.

Oxidation is nothing but an atom loosing electrons in chemical compound. Oxidation state works as an indicator and sometime it could be negative or positive or Zero.
Most common Oxidation states of Gold Include
+1 (gold(I) and +3 (gold(III)
If you want to know the Less common oxidation states of gold
then it includes
−1, +2, and +5.
You can know more about
gold by exploring its Properties!


text:
Chemoselectivity in synthesis of paracetamol
In the reaction below, why does acetic anhydride not react with the -OH group to form an ester instead?

The reaction is likely with one equivalent of acetic anhydride and under thermodynamic control.
Kinetically, the nucleophilic attack of phenol on acetic anhydride should be a lot faster than that of aniline. Aniline’s lone pair is taking part in resonance across the phenyl ring so it is barely able to attack. Phenol has two lone pairs on oxygen, though, of which only one is able to resonate with the aromatic system. Under kinetic control I would thus expect acylation on oxygen.
However, there is a second process which renders us thermodynamic control: The initial product can react with a second molecule in a transacylation reaction.
$$\ce{RHN-C6H4-OAc + H2N-C6H4-OR <=>> RHN-C6H4-OH + HAcN-C6H4-OR}$$
Both the oxygen of a free phenol and the nitrogen of a free aniline (the latter slower, but still) can attack a corresponding acylated atom. In overall thermodynamic stability, the acetylamine is more stable than the acetate (and is also kinetically more inert) so after equilibration has taken place the acetamide will prevail.

The amino group is more nuclephilic than the -OH. For this reason the reaction will proceed first with the amine. If an excess of acetic anhydryde is used the acetylation of both moieties would happen.

The amino group cannot be more nucleophilic because it hardly has an available electron pair. The hydroxy group does.


text:
(Literature) Diffusion Coefficient Of Potassium Permanganate
I'm doing a practical report on an experiment I had this week where I studied the diffusion coefficient of KMnO4 in water by the free boundary method. I have calculated the diffusion coefficient of potassium permanganate using this method, but now I need to compare this calculated value against the literature diffusion coefficient value of KMnO4. I've looked in my textbook and looked through a bunch of websites, but I just can't seem to find this value. If anyone can help me find this value or could tell me what it is, it would be very much appreciated.
Edit: No need to answer this question anymore since I spoke to my Scientific Officer and she said I should leave out the comparison of this literature value with the measured one. I'll leave this question open for answering though since someone might run into the same problem as I have in the future.


text:
Why does the reference electrode in potentiometry have a diaphragm?
I am studying about potentiometry and pH-measurement with a glass electrode. I don't understand why the reference electrode has a diaphragm and is immersed in the measured solution. If I got it right, there is no current flowing and the composition of the buffer inside the reference electrode is constant. Why should it then be connected with the solution? Can someone clarify?

This device is basically a galvanic cell (and an amplifier and voltmeter). Galvanic cells are quite similar to electrolytic systems, the difference being wheter one has an external voltage source.  If there is no “salt bridge” (i.e. some way for ions to wander between the two half-cells), the cell immediately stops working. This is certainly true, since one can directly observe it.
This makes sense if one considers that in (about all?) systems only one kind of metal ions is involved at each electrode. If you disconnect the two half-cells, there is a potential difference, but only between the electrode and the electrolyte. (How do you measure such a thing without intruducing another electrode??) The ions would simply stick around near the electrode until the charge makes the reaction unlikely to occur; only if the ions in the electrolyte can mix somewhat can the reaction continue. Keep in mind that the voltmeter does not have infinite resistance but "only" 10MOhms or so. This animation might be quite helpful.
In different terms: One needs a small current (at least for a short time) to measure voltage (keep in mind that the amplifier will have some capacitance) and for this to occur the circuit has to be closed, on one hand via the voltmeter, on the other via the salt bridge.
But, since you do not want to influence the measured system too much, you only provide this almost tight diaphragm — not a large(ish) diameter hole — and almost no current flows. I see no theoretical reason why a very small hole would not work equally well, but I assume this is more annoying for technical reasons (drilling it, clogging, etc.). One could argue that very small currents can flow even without a salt bridge because separate containers are still not infinitely far apart and their fields can influence the other one, but this seems to be negligible; normal glass is already thick enough to limit capacity to a few pF.

Wait - I thought there was no current flowing in potentiometry?

Edited with more elaborate explanation. There certainly is some current, just not much. I suppose "no current flowing in potentiometry" was an oversimplification that stumped you; it should rather read "only a negligibly small current"


text:
Decomposition products of sodium acetate
I heated some sodium acetate to a high temperature (about 500 degrees celsius). Upon doing so, I could smell an irritating smell reminiscent of vinegar. I stopped then, assuming they were acetic acid vapors. Are they really acetic acid vapours, and if so, what is the remaining solid?

A cursory Google search for the SDS for sodium acetate indicates that thermal decomposition can yield carbon monoxide, carbon dioxide, and sodium oxide. On the way to 500 $^{\circ}$C, acetic acid vapors are certainly released. Your solid might be $\ce{Na_{2}O}$.

The fumes are likely to be acetic anhydride, as per this equation:
$$\ce{2CH3COONa -> Na2O + (CH3CO)2O}$$
This smells like vinegar, and reacts with moisture in the air to give acetic acid.

This link states that sodium acetate decomposes on heating to produce a white solid i.e sodium oxide, $\ce{Na2O}$.
The wikipedia link of sodium acetate and this link states that sodium acetate decomposes on heating to produce acetic acid fumes which have a strong vinegar like smell.

I did some heating of Sodium acetate myself. It melted down into a black boiling tar, which caught on fire. This tar cooled solidified and then would not remelt as easily, and i ended up melting my test tube instead. However, I think that the decomposition of Sodium Acetate actually yields acetone vapors and Sodium Carbonate, (or Bicarbonate I haven't worked out the equation yet.) I think this follows the thermal decomposition of Calcium Acetate, which is known to make acetone.
Edit. I threw a piece of aluminum foil, causing plastic wrap would melt. no liquid re condensed, meaning no acetone. Okay, my end product looked metally.... and since Na2O melts stupidly higher than Na metal... I think I'm wrong.
I think I stopped my heating early and got it reheating... it is melting again. By the way I'm doing this on my stove, so who knows what I get out of that. My torch is broken.