text:
Why does chlorine have a higher boiling point than hydrogen chloride?

Chlorine has a boiling point of $238~\mathrm{K}$ while hydrogen chloride has a boiling point of $188~\mathrm{K}$.

Hydrogen chloride has dipole-dipole forces so I would expect it to have greater inter-molecular forces and thus a higher boiling point. However, since this is not the case, this must mean that chlorine has a greater intermolecular forces due to London dispersion forces.
But why?  I thought London dispersion forces were miniscule and only had a notable effect in large molecules. Shouldn't the dipole-dipole forces be way stronger than the London dispersion forces?

You also have to consider that a diatomic chlorine molecule is almost twice as heavy as a hydrogen chloride molecule.

@A.K. I don't think molecular mass is directly correlated to boiling point, even if it is often mentioned.

@NicolauSakerNeto Molar mass does affect boiling point, though its not the only factor. Heavy water melts and boils at higher temperatures than regular water.

@A.K. There always seems to be a better alternate explanation though. For example, a higher boiling point in water isotopologues may be better explained by a reduced zero-point energy in the intermolecular hydrogen bonds. This seems to be supported by the fact that $\ce{D2 ^{16}O}$ and $\ce{H2 ^{18}O}$ have essentially the same mass, but their boiling points are 101.42 ºC and 100.15 ºC respectively; the increased effect of hydrogen replacement is consistent with a larger decrease in bond vibration ZPE. Furthermore, $\ce{T2}$ and $\ce{H2}$ boil at 25.04 K and 20.39 K, a rather small difference.

This is because $\ce{Cl2}$ has close to double the mass and is also a larger molecule compared to $\ce{HCl}$. You were correct in saying that London dispersion forces are weaker but because of $\ce{Cl2}$s size they overcome the dipole-dipole forces in $\ce{HCl}$.


text:
Reduction of N,N-dimethylbenzamide with LiAlH4
Consider the reaction $$\ce{C6H5CON(CH3)2->[\ce{LiAlH4}]$\:\:$?}$$
What will the product be?
According to me the product should be $\ce{C6H5CH2OH}$ but I am not sure. Also, I don't have any reasoning for why this must be the product.

How can you say "the product must be..." when you don't have any reasoning behind it?

http://chemistry.stackexchange.com/questions/43446/why-does-the-reduction-of-a-weinreb-amide-give-an-aldehyde-instead-of-an-amine

Here's the mechanism. The key is the second step, it's the lone pair of nitrogen that attacks the carbon, and not the lone pair of oxygen.
The reasoning is that the lone pair of nitrogen is more reactive because of the two electron-donating methyl groups attached to the nitrogen atom. The electron density is higher on the nitrogen atom than the oxygen atom.
The second reason is that the oxygen anion as a leaving group is much more stable than the nitrogen anion R2N- you would have gotten if the reaction led to your product.

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text:
Are isomers and stereoisomers the same thing?
I am doing a presentation regarding Stereochemistry, and I am a little confused with the terminology. My question is:
Are isomers and stereoisomers the same thing? Do they differ from each other?

A stereoisomer is a class of isomers. There are others such as enantiomers, diastereomers etc.

Both enantiomers and diastereomers are subsets of stereoisomers, I would not say that they are "separate classes" from stereoisomers. However there are isomers that are not stereoisomers; a simple example is dimethyl ether vs ethanol: an example of functional group isomerism

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text:
Examples from everyday life for diastereoisomers ?
Like the amino-acids and sugars that are enantiomers, can you give any examples of diastereoisomers ? where can we find them in everyday life ?

Amino acids and sugars have other isomers than enantiomers aswell. The diastereomer is a type of stereoisomer. These would occur with  amino acids and sugars too. Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocentres and are not mirror images of each other. Ref: diastereomer/wiki.com

D-threose and D-erythrose are diastereomers. Ref:diastereomer/wiki.

Nowadays, it's really easy to find information on the Internet or in books. Being resourceful and being able to search for stuff yourself will help you greatly.
Nevertheless...
Carbohydrates
Glucose and galactose, two of the most common monosaccharides found in the body, are diastereomers of each other:

Lactose, which is found in milk, is formed by a condensation reaction between glucose and galactose.
Eyesight
Retinal is a molecule found in the photoreceptor cells of eyes that normally exists in the 11-cis form. When it absorbs light, it undergoes an isomerisation to form the all-trans form:

This forms part of the biochemical pathway by which light is converted into an electrical signal in the nervous system.


text:
Which proton in cyclohexene is the most acidic?

I know that $\mathrm{H^b}$ is most acidic due to conjugation (resonance).
But I am confused as to how to compare $\mathrm{H^a}$ and $\mathrm{H^c}$. I think $\mathrm{H^a}$ should be more acidic than $\mathrm{H^c}$ due to hyperconjugation.

Have you been taught the relative acidity of alkanes, alkenes and alkynes? http://chemistry.stackexchange.com/questions/32341/bond-length-and-acidity/32342#32342

@orthocresol so the correct decreasing order should be $\mathrm{H^b}$ , $\mathrm{H^c}$ and $\mathrm{H^a}$

Yes, according to this link the allylic proton (b) has a pKa ~36, the vinylic proton (c) ~43, and the sp3 proton (a) ~50.

When comparing acidities of protons, we need to find out what the corresponding conjugate bases are.
The conjugate base upon removing $\ce{H_\mathrm{b}}$ is an allyl anion. There is (and should be) no question that it is the most stable carbanion of the three we are going to consider.
The conjugate base upon removing $\ce{H_\mathrm{a}}$ is an $\mathrm{sp^3}$ configured carbanion while the conjugate base upon removing $\ce{H_\mathrm{c}}$ is an $\mathrm{sp^2}$ configured carbanion. This is nice, because it allows us to directly compare the two using their hybridisation.
In general, lone pairs (and that explicitly includes those coming from anions) want to be in the lowest-lying atomic or molecular orbital possible. Since an s orbital is lower in energy than a p orbital from the same shell, lone pairs prefer s orbitals. If there is no pure s orbital available — as would be the case in practically all carbon-centred lone pairs — an orbital with the highest possible s character is preferred instead. If we look at the configuration $\mathrm{sp^3}$ we can say that it has $25~\%$ s character while $\mathrm{sp^2}$ has $33~\%$ s character. Since $33~\% > 25~\%$, the $\mathrm{sp^2}$ configured carbanion is more stable — which is also supported by corresponding $\mathrm{p}K_\mathrm{a}$ values.
The method outlined above is a very crude approximation building on a high amount of simplifications but gives us the correct result in practically all carbon-centred anion cases.
Thus, the acidity order is:
$$\ce{H_\mathrm{b}} > \ce{H_\mathrm{c}} > \ce{H_\mathrm{a}}$$

Note: Ron noted in a comment above that the acidity of an $\mathrm{sp^2}$ hydrogen almost parallels that of an allylic hydrogen. So while most courses may teach this simple ordering, you should keep that in mind as additional information.

To address the question of hyperconjugation, ${H^b}$ presents the classic example of hyperconjugation between the ${\sigma}$ bond and ${\pi^*}$. No hyperconjugation is possible between ${\pi^*}$ and the ${sp^2}$ orbital attached to ${H^c}$. However, some very weak interaction between ${\sigma}$ bond between $\ce{CH^a}$ and ${\pi^*}$ could lead to some weakening of the sigma bond and enhanced acidity.

Hb is the most acidic hydrogen.
When it is extracted a negative charge appears on the carbon attached to it, this negative charge can then undergo resonance with the adjacent double bond. This resonance leads to stabilization of the system. Therefore the compound would prefer losing Hb.
Hence Hb is the most acidic

When comparing between Ha and Hc, Ha would be the more acidic as when we remove Hc there is a vinylic carbanion that is being formed. And Vinylic Carbanions are very unstable. Thus extricating Hb is more feasible thus making it more acidic.

There is no confusion about Hb being the most acidic. The question is actually asking about which of the other two hydrogens (a and c) is more acidic.

This answer is incorrect.

If we were to remove $\ce{H^c}$, we would get a structure similar to a vinyl cation. The vinyl cation is highly unstable and hence it would be more difficult to remove $\ce{H^c}$,i.e. $\ce{H^c}$ is less acidic than $\ce{H^a}$. Also hyperconjugation is greater for $\ce{H^a}$ than it is for $\ce{H^c}$.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.For more information in general have a look at the [help].

An $\mathrm{sp^2}$ $\ce{C-H}$ bond is more acidic than an $\mathrm{sp^3}$ $\ce{C-H}$ bond, hence Hc is more acidic than Ha.  Allylic protons are a bit more acidic than olefinic protons but they are close, so the final order of increasing acidity would be Ha < Hc <= Hb

To check acidity see conjugate base by removing H+. So when you remove H+ from double bond you get a sp2 carbanion which is more unstable than sp3 carbanion. Now as weaker the conjugate base stronger is acid. hence hc>ha. Hope its clear.

You have it the wrong way round, the sp2 carbanion is more stable than the sp3 carbanion.

Yes orthocresol is correct , you are wrong .

but if we go by conjugate acid base theory the weaker conjugate base means strong acid, isnt it? Can you please explain..

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text:
Why is it lead(II) chromate and not lead(II) chromate(VI) tetraoxide?
I am just beginning in chemistry and learning how to name the different molecules. I have this molecule: $\ce{PbCrO4}$.
I would give it the following name: lead(II) chromate(VI) tetraoxide. Lead and chrome have both multiple oxidation states. And as you have to specify about which ion you are speaking, it seems to me this would be the most correct name.
But this is the correct name apparently: lead(II) chromate.
Could someone please explain:

why you just specify the oxidation state for the first element and not the other? As they are both metals and both have multiple oxidation states.
why you don't say tetraoxide, although there is clearly a number 4 written?

Chromate ($\ce{CrO4^2-}$)is a poly atomic ion which is partially defined by the oxidation state of chromium.  Its like the sulfate ($\ce{SO4^2-}$), phosphate ($\ce{PO4^3-}$), nitrate ($\ce{NO3-}$), permanganate($\ce{MnO4-}$), and carbonate ($\ce{CO3-}$) polyatomic ions. For each ion the oxidation state of the center ion is defined therefore there is no reason to state it.  You could in principle call lead (ii) chromate lead(ii) chromium(vi) oxide or tetroxide, but this is less informative of the structure of the dissolved species.

So you actually never say the oxydation state of the ions in the middle, just the first one? This might be an ever more stupid question but, how do you know it's $CrO^{2-}{4}$ and not $Pb^{2+}Cr^{6+}O^{2-}{4}$?

No, not the middle, the $\ce{CrO4^2-}$ is a stand alone polyatomic ion, which will not hydrolyze in solution.  This is different from a mixed oxide like yttrium aluminium oxide ($\ce{Y3Al5O12}$) which is two oxides mixed together like you're thinking.  A good way to distinguish between mixed oxides and salts is if the stoichiometry can vary or if it is fixed respectively.

The simple answer is that there is naming convention for polyatomic anions.

$\ce{S^{2-}}$ is sulfide
$\ce{SO3^{2-}}$ is sulfite
$\ce{SO4^{2-}}$ is sulfate


text:
In which cases do you say "oxide" when naming a molecule?
Based on the names of following molecules I got quite confused about the correct nomenclature:

$\ce{NaHCO3}$
$\ce{K2SiO3}$
$\ce{PbCrO4}$

In none of the above molecule names you ever say "oxide" although there is a oxygen atom. Why? What is the applied rule for the nomenclature? In which cases do you actually say the word "oxide" in the name of the molecule?

The naming nomenclature for inorganic molecules is confusing, but if you are studying chemistry then you just have to grit your teeth and learn.
First there are household names, common names and chemical names. Vinegar, acetic acid, and ethanoic acid are talking about the same molecule. We say water, but aqueous solutions. $\ce{SiO2}$ is silicon dioxide to a chemist, but quartz to a geologist. The list could go on and on...
In general an oxide is compound which has just some element, X, and oxygen in some combination - ${\text{X}_m\text{O}_n}$
Some anion combinations get a name.

$\ce{CO3^{2-}}$ is carbonate
$\ce{SiO3^{2-}}$ is metasilicate
$\ce{CrO4^{2-}}$ is chromate


text:
Synthetic ferritin models and μ notation
This is the question verbatim:

Details of two synthetic models of ferritin are presented below:
System 1
$$\begin{multline}\ce{Fe(OAc)2 + LiOMe ->T[in presence of $\ce{O2}$][in MeOH]\\
 Fe12(OAc)3(\mu-OAc)3(MeOH)4(\mu-OMe)8(\mu_3-OMe)10(\mu_6-O)2}\end{multline}$$
System 2
$$\begin{multline}\ce{Fe(O3SCF3)2 + L ->T[in MeOH]\\ [Fe6(OMe)4(\mu-OMe)8(\mu_4-O)2L2][O3SCF3]2}\quad (\ce{L~=~N(CH2CH2NH2)3})\end{multline}$$
How valid are each of these systems as a model of ferritin? Consider
  both structural and functional features.

I have no idea how to start with this. I wasn't taught what the $\mu$ notation means nor about the structures of ferritin models. I'm guessing $\mu$ denotes a bridging ligand?
I know that oxygen binds very well to Fe, hence why it needs to be carefully controlled around it. The chelating nature of L can't be ignored.
Is the symmetry of the structures important? I know that natural ferritin has 3 and 4-fold symmetry.

$\mu_6$ means that it bridges six iron atoms.

Could you provide a link to the courseware, please? A quick look at ferritin structure made me wonder what the authors believe to be a model for ferritin.

It's not publicly available and has nothing on synthetic ferritin models. It only talks about natural ferritin.


text:
Creating energy profile diagrams for publication
Is there automated software to create energy profile diagrams like the ones below?

One could use something like Illustrator or Inkscape, but they aren't automated, are expensive (Illustrator), are difficult to use, and wouldn't be practical for anything with more than a dozen energy levels. These diagrams don't look like they were hand-drawn, but I could be mistaken.
Ideally, it would be written in Python, but something in LaTeX or TikZ would be workable as well.
More specifically, I think this could be improved upon quite a bit.

I think I recall from some chat session, that LordStryker was working on something like that, but it was in its early stages and I am not aware that he has published/released it. Other than that I have not yet come across something like that. I know that my coworkers draw them in Illustrator, I prefer inkscape. I guess one could cook up a solution in gnuplot, which is iirc able to produce various output formats. I personally prefer tables though and if really necessary dot-line diagrams. Still +1.

I've done a very crude one by hand in gnuplot before when attempting to make an MO diagram, but I was hoping for something more "automated" than that (though I would still prefer it to Inkscape, which I want to avoid learning). Anything similar I've just done in Keynote, which leaves me too reliant on a Mac.

I just found something in my link section that looks promising. Let me know if it helped. Nice Energy Level Diagrams with rxnlvl - http://wp.me/ptBY1-2W

If this isn't exactly what I was looking for, then I could definitely modify it to suit my own usage. Looks great, thanks!

Did you come up with something suitable and easy to use?

No; I did start to hack on that project you linked to, but stopped almost immediately since I haven't needed it for anything. If there's ever any progress on it I'll post it as an answer.

PyEnergyDiagrams
I use my own script. I have posted it on GitHub now.
This is the final result:

It is quite easy to use, on the GitHub readme there is a small tutorial. You have to create an instance of ED class and then add the energy levels to it.
from energydiagram import ED
diagram = ED()
diagram.add_level(0,'Separated Reactants')
diagram.add_level(-5.4,'mlC1')
diagram.add_level(-15.6,'mlC2','last',)
diagram.add_level(28.5,'mTS1',color='g')
diagram.add_level(-9.7,'mCARB1')
diagram.add_level(-19.8,'mCARB2','last')
diagram.add_level(20,'mCARBX','last')

To add the links between the levels you can have a look to the IDs of the level using this the argument of the function plot show_IDs=True:
diagram.plot(show_IDs=True)

And then you can use:
 diagram.add_link(left_ID_levetobelinked,right_ID_levetobelinked)

Export to pgf,pdf and svg
You can export it just saving the plot from the GUI button or programmatically in different formats for publications using matplotlib backend.
 I prefer to export the plot in .pgf   and then I import it in the LaTeX document this allow a full control over fonts and appearance of the figure.

For reaching better quality you have to change a little bit the autogenerated .pgf file.
The first thing you have to do is to find and replace − with -.
Later you can follow the procedure in this answer so that the font of the figure matches with the font size and type of the document.

Wow, that looks cool! Is it possible to export the image to svg or pdf, too?

@Martin-マーチン Yes, you can it's very easy. But if you use latex I would suggest to use pgf so you can modify your figure later on.

I have no experience with pgf, and from Wikipedia I get that it is a bitmap file format. While I don't use LaTeX much anymore, I don't see how a bitmap would be superior to a vector image. But maybe this isn't the right place to discuss this. But I will certainly look into your script.

@Martin-マーチン nope that one is another kind of pgf. I am referring to https://en.wikipedia.org/wiki/PGF/TikZ


text:
How to determine the products of a redox reaction
I'm doing a paper on water treatment methods, and in one section, I'm describing how a nearby water treatment plant is removing Trichloroethylene ($\ce{C2HCl3}$) from water by oxidizing it using Ozone ($\ce{O3}$).
However, I can't find any documents confirming the products of this reaction, so I'm going to need to determine them on my own, but I'm not sure how. I'm fairly certain the equation would be formatted as
$$\ce{C2HCl3(aq) + O3(g) -> H2O(l) + }$$
but I'm not sure of anything past that.

May consider the UV ray split the ozone into oxygen molecule and oxygen atom. Two oxygen atom can combine into an oxygen molecule again. So the reaction could be trichloroethylene and oxygen molecule

I did some research on this topic (water treatment for TCE) and I come out three answer which is by using peroxide, ozone and permanganate. Since you asked for ozone first, so I will give that first.
What are you missing in the reaction is that water is the reactant of this reaction instead as a product. Here's the equation:
$$\ce{C_2HCl_3 + O_3 + H_2O -> 2CO_2 + 3HCl}$$
For the reaction involved ozone, The reaction will be more effective in an acidic medium and it must perform in a enclosed delivery pipes.
Second, The reaction involving peroxide. Hydrogen peroxide react with TCE will also produce carbon dioxide and hydrogen chloride but water is a product here. Here is the equation:
$$\ce{C_2HCl_3 + 3H_2O_2 -> 2CO_2 + 2H_2O + 3HCl}$$
Same as the ozone reaction, this reaction is more effective in an acidic medium.
Last one is the reaction involving permanganate. We will be using potassium permanganate but you can use sodium permanganate, calcium permanganate or magnesium permanganate. It will react with TCE to produce carbon dioxide, manganese (IV) oxide, metal chloride and hydrogen chloride. Here is the reaction:
$$\ce{C_2HCl_3 + 2KMnO_4 -> 2CO_2 + 2MnO_2 + 2KCl + HCl}$$
This reaction is a bit different compared with the other two. It could be effective in acidic or basic medium between the PH range of 3.5 to 12.
Hope this helps you.


text:
How can I find the parent chains in these two compounds?
We recently started organic chemistry in school and I decided to make a computer program that names the compounds, so far it works with alkanes that do not have branched branches and I want to make it work with compounds with double and triple bonds. I know that if there is only one double/triple bond, the answer is just the same rules as with alkanes, but it must include the 2 atoms that are bonded in that way. Also, I would appreciate it if you tell me how I decide from which side to number them.
Compound 1:

Compound 2:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the first relevant criterion to be considered in choosing a principal chain is the length of the chain.
When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached:

P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. When there is a choice for the locants of multiple bonds, low locants are given first to double bonds:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Note that Rule e takes precedence over Rule f.
Furthermore,

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Therefore, the preferred IUPAC names (PINs) for the examples given in the question are 3-ethylpent-1-en-4-yne and 3,3-diethynylpenta-1,4-diene, respectively.

Thanks a lot! One final question, can you link me some good resource that has these rules, but only for acyclic hydrocarbons. I have the full book, but it is really difficult to weed out the information that I currently need (I want to make it work with simpler compounds first and then expand it to more complicated ones). BTW, if you are interested here is a link to the project: https://github.com/indjev99/IUPAC-Name-Generator

@indjev99 As far as I know, there is no reliable website that includes all current (2013) rules. The site http://www.acdlabs.com/iupac/nomenclature/ is very nice but it only has the rules from 1979 and 1993.

mentioned but uncited longest chain rule is answered here: https://chemistry.stackexchange.com/questions/74284/longest-chain-iupac-rule


text:
Changing State of Alumina Oxide
I intend to bring solid Alumina Oxide into a liquid state, for the purpose of pouring the substance into a mold.  Subsequently, I will allow it to cool with an ambient air temperature of 70f, slowly allowing it to return it to a solid state.  Melting will be performed with a gas mixture.
My question is, will the Alumina Oxide return to the original molecular structure as is was prior to melting it from a solid state?
Thank you for your time and consideration.
Rob

Good luck melting it in the first place without special equipment.

You realize that the melting point of alumina is 2,072 °C?!? What are you going to contain the melt in and what mold material are you going to use? // Frankly this is the sort of question that if you have to ask then you don't have a clue what you are doing and you shouldn't attempt the experiment.

There is no alumina oxide. Alumina is aluminium oxide.

related http://chemistry.stackexchange.com/questions/16640/how-do-you-melt-metals-with-super-high-melting-points

First of all, there is no such thing as "alumina oxide". Alumina is aluminium oxide. Melting alumina is pretty hard, its melting point is above 2000 degrees Celsius. It's so high that people use it as a container (crucible) for melting other stuff. Such high temperatures are not easily achieved without melting a bunch of stuff around it.
Let's assume that you did manage to melt it, maybe using one of those fancy levitation laser techniques. Your  mold has to be heat resistant and inert so it will not melt or react with the molten alumina. Graphite is usually a good one.
Now, you were asking about the structure. Assuming that you started with regular crystalline alumina (aka corundum), then yes: the resulting material after cooling in air will be the same corundum. Amorphous alumina (glass) will only form by extremely rapid cooling, for example dropping a very small amount to ice cold water. In all normal circumstances it will be crystalline.


text:
Masamune's rule of diastereoselectivity multiplicity explained
In a reaction between a chiral substrate and chiral reagent, their asymmetric inductions could cooperate (matched pairs) or compete (mismatched pair). According to Masamune, in a matched pair the degree of asymmetric induction is approximately the product of the selectivities of the substrate and the reagent. In contrast, the asymmetric induction in a mismatched pair is equal to the quotient of the selectivities of the substrate and the reagent.
The following reaction perfectly proves Masamune's rule (syn enolate is produced during enolization):

Why is this rule valid and why is it so accurate? What's the mechanistic explanation behind the multiplicity trend?

You might find what you want in this paper (http://pubs.acs.org/doi/pdf/10.1021/ed067p20). I don't have time to write out a sufficiently detailed answer right now, but let me know if you can't access the paper, and ill answer when I'm free.

First, we need to take a look at the possible transition states of the first reaction in your table using $\ce{9{-}BBN}$ as an enolising agent. There are two transition states shown in scheme 1: The upper, favoured and the lower, disfavoured one.

Scheme 1: Aldol reaction with $\ce{9{-}BBN}$ as an enolising agent. Both the favoured (above) and the disfavoured (below) Zimmermann-Traxler transition states are shown along with their respective products.
The difference in both transition states is in the steric clash of the α-carbon and the boron residue: If the aldehyde attacks from the si-side, the stereocentres on said α-carbon can be oriented favourably, namely with hydrogen pointing toward the most crowded area and the largest residue pointing toward the least crowded area. If the aldehyde attacks the enolate’s re-side, methyl must point towards the most crowded area to allow the largest residue to point away from crowded areas. A similar discussion can be made for the orientation of the aldehyde which could attack the enolate’s si-side with its si-side, resulting in an anti-aldol product.
It is important to note here that all four attacks are possible. However, they do not have the same probability of reacting. Based on the observed diastereoselectivity, one can say that $\mathrm{P}_R(F) \approx 92~\%$, while $\mathrm{P}_R(\bar F) \approx 8~\%$ ($R$ is the set reaction occurs and $F$ is the set favoured attack. I have subsumed the two anti-product giving attacks under $\bar F$, since they are nigh irrelevant).

Scheme 2: Aldol reaction with $\ce{(+){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s re-side. This is the same as the substrate’s inherent diastereoselectivity; a matched case.
A similar case can be made for $\ce{DIP}$-mediated aldol additions in the absence of substrate diastereospecificity ($\ce{Me} = \ce{H}$). Here, we need to construct a new set $A$ meaning auxiliary-favoured attack. This auxiliary-favoured attack is shown in scheme 2, although I used the matched case from your examples, i.e. substrate diastereoselectivity is present. The auxiliary-unfavoured attack (same caveat) is shown in scheme 3.

Scheme 3: Aldol reaction with $\ce{(-){-}(Ipc)2BOTf}$ as the enolising agent. $\ce{Ipc}$’s methyl groups disfavour an aldehyde attack on the enolate’s si-side — but substrate diastereoselectivity prefers the si-side attack on the enolate (mismatched case).
Stochastically, we can consider both selectivities separately. Remember that all attack combinations are technically possible but some are just more likely to be productive. It is a good approximation to assume that $\mathrm{P}_R(A)$ and $\mathrm{P}_R(F)$ are independent of one another, which means that $\mathrm{P}_{R,A}(F) = \mathrm{P}_R(A) \times \mathrm{P}_R (F)$. Why? Well, the substrate’s diastereoselectivity is mainly based on the orientation of the residues on the α-carbon. This is pretty much independent of the orientation of $\ce{Ipc}$’s methyl groups, which cause the auxiliary’s enantioselectivity: in both methyl group orientations the same rotamer of the α-carbon has the same probability.
Having taken into account that both sets are independent, we see that we can multiply the probabilities of the different attacks to get the resulting attack’s probability. The substrate will favour (R,S) (counting from ketone to hydroxy group) in each case. To get the probability of (R,S), we multiply a large percentage (substrate selectivity) with a large percentage getting another large percentage (auxiliary selectivity). The probability for (S,R) will be a small percentage (substrate) multiplied by a small percentage (auxiliary) so even smaller. So favouring of the desired (matched) diastereomer is enhanced.
On the other hand, if the auxiliary favours (S,R), to get the probability of (R,S) we must multiply a large percantage with a small one giving a medium result. The same thing for the probability of (R,S). This means that both diastereomers have more equal chances of forming.
(A better discussion of this problem would involve relative reaction rates, as the article NotWoodward mentions does. I feel that the probability argument gets us the same conclusion without all the big kinetic calculations which I am not good at.)


text:
Chemoselectivity in reduction of 3′-nitroacetophenone

When 3′-nitroacetophenone is treated with sodium borohydride, only the carbonyl group is reduced (top); however, when reacted with tin and hydrochloric acid only the nitro group is reduced (bottom). Why is this so?
I think that sodium borohydride is too weak a reducing agent to reduce the nitro group, but I am unsure as to whether this is correct. If Sn/HCl can reduce the nitro group then why doesn't it reduce the carbonyl too?

Related: How to reduce an acyl group on a benzene without disturbing nitro group

The key question is how the reduction by the two reagents proceeds.
For tetraborohydride the reduction proceeds as a standard nucleophilic addition of hydride ion to carbonyl group.
For tin/HCl the reduction proceeds as a series of one-electron transfers from metal surface mixed with protonation. It seems that the lowest free orbital in the compound belongs mostly to nitro group (which makes sense as it has a strong π-withdrawing effect, stronger than carbonyl group), so it is reduced to a nitroso group and then further via a hydroxylamine to an amine. It is possible to stop the reduction at intermediate stages under some conditions.
So the question is, why borohydride does not reduce the nitro group. It should be able to do so, as a strong reducing agent which is able to reduce iron from solutions. A possible answer is that extremely small size of nitro group with a net negative charge on the nitro group prevents formation of a stable adduct with hydride ion.

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


text:
No reaction in net ionic equation for lithium nitrate and sodium chloride?
I've heard that nothing in the following net ionic equation reacts:
$$
\ce{Li+ + NO3- + Na+ + Cl- -> \text{"nothing"}}
$$
Is this true? Could someone explain why there would be no reaction?

I assume this is meant to be in aqueous solution?

Assuming all these ions are in an aqueous solution, according to solubility rules nitrates are soluble, chlorides (with the exception of AgCl, PbCl₂, and Hg₂Cl₂) are soluble, and salts containing group 1 elements are soluble so they all remain as individual ions as nothing can come together to form an insoluble compound and precipitate out of the solution.

short and clear, thx!


text:
What are the correct equilibrium arrows?
I was recently subject to some atrocious teaching. One of the gems that found me was the equilibrium arrow usage. I am accustomed to write $\ce{<=>}$. Yet, the entire class had to use ⇆ throughout the semester. Teacher's argument was that each harpoon from $\ce{<=>}$ has a meaning already, the movement of a single electron in a mechanism.
I concur, but I can't think of an situation in which the meaning of a harpoon is unclear because of the context. This is unlike Y being the symbol for both EDTA and yttrium, allowing $\ce{[YY]^{-1}}$ to have a meaning.
I tried to find the correct character, but not even the IUPAC Gold Book (1) (2) has a clear stance on it.

Yeah, and at the same time the normal-headed arrow has the meaning of two-electron movement in a mechanism. The argument is just silly.

Clayden, Greeves, Warren, and Wothers (Oxford University Press) uses the harpoon form, that's good enough for me. For diagrammatic purposes, or if the conditions for the reverse reaction were starkly different, normal arrows can make more sense but I agree, -1 to this teacher.

Related: What is the difference between “reaction in both directions” and “equilibrium”?

According to IUPAC: Quantities, Units and Symbols in Physical Chemistry (Green Book), the arrows are used differently for composite mechanisms in chemical kinetics.
In case of a simple mechanism that is composed of a forward and a reverse elementary reaction (i.e. reactions that occur at the molecular level in one step), it is conventional to write these in one line using two double-sided arrows $(\rightleftarrows)$.
For example:
$$\begin{align}
\mathrm{A} &\to \mathrm{B + C}\\
\mathrm{B + C} &\to \mathrm{A}\\
\hline
\mathrm{A} &\rightleftarrows \mathrm{B + C}
\end{align}$$
This is distinguished from a net reaction, which is written with two one-sided arrows $(\rightleftharpoons)$.
For example:
$$\begin{align}
\mathrm{A} &\rightleftarrows \mathrm{B + C}\\
\mathrm{B + C} &\rightleftarrows \mathrm{D + E}\\
\hline
\mathrm{A} &\rightleftharpoons \mathrm{D + E}
\end{align}$$
The two-sided arrow $(\leftrightarrow)$, which indicates resonance structures, shall not be used for reactions.
I guess that every elementary reaction that is proceeding in both directions $(\mathrm{A} \rightleftarrows \mathrm{B + C})$ may also be considered a net equilibrium reaction $(\mathrm{A} \rightleftharpoons \mathrm{B + C})$, but not every net equilibrium reaction $(\mathrm{A} \rightleftharpoons \mathrm{D + E})$ is an elementary reaction that occurs at the molecular level in one step $(\mathrm{A} \rightleftarrows \mathrm{D + E})$. Therefore, I assume that one-sided arrows $(\rightleftharpoons)$ may be used for all equilibrium reactions, whereas double-sided arrows $(\rightleftarrows)$ could be wrong when one combines composite mechanisms to obtain net reactions.

Is there a reason why you don't use \ce{<=>} for the two one-sided arrows ? In my case, such a notation gives me problems when I convert to pdf via wkhtmltopdf.


text:
Does intramolecular hydrogen bonding occur with carbonic acid and geminal diols?
Is an intramolecular hydrogen bond possible in such compounds?
What are the requisites to have intramolecular hydrogen bonding?

In second one the two R groups are +I group as well as steric hindering hence drastically reduces stability, also there is repulsions due to lone pairs on oxygen.

As the intramolecular H-bonding is absent in both cases.
Carbonic acid
In aqueous solution of carbonic acid is a weak acid. It dissociates into its constituent ions as carbonate ions and hydrogen ions. Now, the CO3-2 ion, is stabilized by resonance and produce three equivalent resonating structures.

This makes carbonate ion more stable rather than through intra molecular H-bonding. Now, I already said that carbonic acid is weak acid means all the molecules of carbonic acid are not dissociated. The molecules of carbonic acid which remain undissociated form inter-molecualr H-bonding with each other.

One thing that carbonic acid does not exist via intra-molecular H-bonding because, there is too much ring strain.
Gem-diols
In gem-diols there are two -OH groups present on a single carbon. You can get detail reason that why intra-molecular H-bonding is absent in gem-diols.
There is a common example of gem-diol where intra molecular H-bonding is present is chloral hydrate.

Welcome to Chemistry.SE! The dissociation $$\ce{H2CO3 (aq) <=> CO2 (aq) + H2O}$$ is indeed the predominant reaction, favouring carbon dioxide. That is one of the main reasons, why many carbonates readily decompose at low pHs. || Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you want to know more, please also have a look here.

Is this possible?

Probably not possible. An elementary explanation would be "too much ring strain."
https://en.wikipedia.org/wiki/Ring_strain

I'm a bit confused: What are the requisites to occur a intramolecular hydrogen bonding?

5 and 6-membered rings are more favorable. As a matter of fact, beta-ketoacids, when warmed, readily decarboxylate. Why? The intermediate to decarboxylation is stabilized through hydrogen bonding.
Note the 6-membered ring in the picture below.


text:
What is meant by a reaction favours one side?
I frequently see in textbooks the phrase, a reaction favours the left/right side, but don't know what this means. Does this mean the equilibrium shifts to one side?

that is exactly what the phrase "favors one side" means.

Yes, the equilibrium shifts as per Le-Chatelier Principle. Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change.
Increasing a concentration
What happens if conditions are altered by increasing the concentration of A?
According to Le Châtelier, the position of equilibrium will move in such a way as to counteract the change.

Decreasing a concentration
In the opposite case in which the concentration of A is decreased, according to Le Châtelier, the position of equilibrium will move so that the concentration of A increases again.

Increasing the pressure
According to Le Châtelier, if the pressure is increased, the position of equilibrium will move so that the pressure is reduced again. The more molecules in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules.

Decreasing the pressure
The equilibrium will move in such a way that the pressure increases again. It can do that by producing more gaseous molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction.

What happens if there are the same number of molecules on both sides of the equilibrium reaction?
In this case, increasing the pressure has no effect on the position of the equilibrium. Because there are equal numbers of molecules on both sides, the equilibrium cannot move in any way that will reduce the pressure again.
Increasing the temperature
If the temperature is increased, then the position of equilibrium will move so that the temperature is reduced again. The position of equilibrium therefore moves to the left.

Decreasing the temperature?
The equilibrium will move in such a way that the temperature increases again.

Catalysts
Adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Châtelier's principle does not apply. This is because a catalyst speeds up the forward and back reaction to the same extent and adding a catalyst does not affect the relative rates of the two reactions, it cannot affect the position of equilibrium.

Yes, so this phrase means to basically which side the equilibrium will move, whether to the left or the right. So find out that we need to see the number of particles atoms or ions. It basically rely on the fact what a question has been asked for particles ions or atoms. Now to see, we need to first see is there more particles on the left hand side or the right hand side. Where there are more particles, equilibrium rely that that hand or side, so the forward or the backward reaction. Now with a catalyst there will be no effect on the rate of equilibrium for e.g. iron.


text:
How to finish silanization process correctly?
I have some 2mm glass beads and I'm doing silanization on them. For doing that, I have a small bottle that I fill 1/3 with glass beads and then pour silan solution on them and shake well. However, after 60 min that I drain the solution from bottle and as soon as I wash the beads with water, the beads are covered with a white greasy product which is hardly removed by water and finally I wash it by acetone to become transparent again.
Is this a correct method of silanization (and washing the products) to achieve the maximum hydrophobicity effect of silan or I should revise the method?

Sol-gel preparation techniques have many variations. The contact time -beads to the solution - determines thickness of the sol-gel layer. Here you can try to shorten the dipping (coating) time. Glass is usually dried at 60°C for 30 mins and then "cured" at 500 - 550°C form 5 s up to 2 hours. I would suggest to find out an article and follow its preparation method.

Is this sol-gel layer strong enough or it can be removed by rubbing a napkin on the glass surface? 2) Can cure time change the contact angle as well (remarkably), or it just thickens the layer? Thanks

After high temperature heating, it is hard to remove this layer. Contact angle is affected by all the parameters. I would suggest to contact a person who has practical experiences. check http://www.solgel.com/


text:
Endothermic reactions with significant activation energy
Are there any chemical reactions where in order to induce the reaction, you have to heat up a substance but after the reactions occurs, the substance is much colder than even before you started heating it up?

Well, if you consider barium hydroxide and ammonium chloride (mixing of salts), the reaction is quite endothermic and it can absorb enough heat to freeze water around it. But you don't have to heat the reactants. It's enough if you just stir the mixture.


text:
What are all products of an ozonolysis of alkenes using zinc and water?
I'm studying ozonolysis reactions in alkenes, and saw that in one of them there is the formation of $\ce{H2O2}$ (Can anyone explain from where?) that I presumed that reacts with $\ce{Zn}$, one of the reagents, as the following:
\begin{align}
\ce{2H2O2 &-> 2H2O + O2}  \\
\ce{2H2O + Zn &-> Zn(OH)2 + H2} \\
\hline
\ce{H2O2 + Zn &-> Zn(OH)2}
\end{align}
The general reaction is
\begin{align}
\ce{Unsaturated~compound (-C=C-) + O3 &-> Ozonide\\\\
Ozonide + Zn &-> R^1-(C=O)-R^2 + R^3-(C=O)-R^4 + X}
\end{align}
But one of my books shows that this $\ce{X}$ is $\ce{Zn(OH)2}$ only and other $\ce{ZnO}$ only. What about this? What am I doing wrong?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.We prefer not to use markup in the title field, see here for details.

Your second equation is not balanced: $$\ce{2H2O + Zn -> Zn(OH)2 + H2}$$ Hence $$\ce{H2O2 + Zn -> Zn(OH)2}.$$

@Martin Ooops! Thank you! But $\ce{H2} $ and  $\ce{O2}$ will not form water?

Well I personally would assume that hydrogen peroxide directly reacts with zinc. So my estimate would be that there is no hydrogen or oxygen formation. It probably depends a lot on the mechanism and your $\ce{X}$ will ultimately depend on how much water and what pH is present.


text:
Why doesn't hydrogen peroxide decompose into hydrogen and oxygen?
I'm trying to understand why hydrogen peroxide doesn't decompose into $\ce{H2}$ and $\ce{O2}$, but instead into $\ce{H2O}$ and $\ce{O2}$.

We have two theoretical decomposition pathways:
$$\ce{H2O2(l) -> H2O(l) + 1/2 O2(g)} \tag{1}$$
$$\ce{H2O2(l) -> H2(g) + O2(g)} \tag{2}$$
The required data (all at $298~\mathrm{K}$) is taken from the appendix of Atkins' Physical Chemistry, 9th ed:
$$\begin{array}{cccc}
\hline
\text{Species} & \Delta_\mathrm{f} H^\circ\mathrm{~/~kJ~mol^{-1}} & S^\circ_\mathrm{m} \mathrm{~/~J~K^{-1}~mol^{-1}} & \Delta_\mathrm{f} G^\circ\mathrm{~/~kJ~mol^{-1}} \\
\hline
\ce{H2O2(l)} & -187.78 & 109.6 & -120.35 \\
\ce{H2O(l)} & -285.83 & 69.91 & -237.13 \\
\ce{H2(g)} & 0 & 130.684 & 0 \\
\ce{O2(g)} & 0 & 205.138 & 0 \\
\hline
\end{array}$$
The short way
The standard Gibbs free energy change of a reaction ($\Delta_\mathrm{r} G^\circ$) can be calculated from the individual standard Gibbs free energies of formation exactly analogously to the enthalpy change:
$$\Delta_\mathrm{r} G^\circ = \sum_J \nu_J[\Delta_\mathrm{f} G^\circ(\ce{J})]$$
So, for reaction 1,
$$\begin{align}
\Delta G^\circ_1 &= [1(-237.13) + \frac{1}{2}(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\
&= -116.78~\mathrm{kJ~mol^{-1}}
\end{align}$$
For reaction 2,
$$\begin{align}
\Delta G^\circ_2 &= [1(0) + 1(0) - 1(-120.35)]~\mathrm{kJ~mol^{-1}} \\
&= +120.35~\mathrm{kJ~mol^{-1}}
\end{align}$$
Clearly reaction 1 is favourable and reaction 2 isn't.

The long way
There's absolutely no need to do this, but you could use the "extra" data given above to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for both reactions. You'd get:
$$\begin{align}
\Delta H^\circ_1 &= [1(-285.83) + \frac{1}{2}(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\
&= -98.05 ~\mathrm{kJ~mol^{-1}} \\
\Delta H^\circ_2 &= [1(0) + 1(0) - 1(-187.78)]~\mathrm{kJ~mol^{-1}} \\
&= +187.78 ~\mathrm{kJ~mol^{-1}} \\
\Delta S^\circ_1 &= [1(69.91) + \frac{1}{2}(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\
&= +62.879 ~\mathrm{J~K^{-1}~mol^{-1}} \\
\Delta S^\circ_2 &= [1(130.684) + 1(205.138) - 1(109.6)]~\mathrm{J~K^{-1}~mol^{-1}} \\
&= +226.222 ~\mathrm{J~K^{-1}~mol^{-1}}
\end{align}$$
If you're only interested in the final result, you don't need to do all this at all. But it does give you some insight into the difference between the two reactions: reaction 1 is enthalpically favoured whereas reaction 2 is entropically favoured.
Generally speaking, enthalpy usually plays a larger role in determining $\Delta G^\circ$. However, it is not a very good idea to generalise this statement and compare reactions solely on the basis of their $\Delta H^\circ$.
Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, and taking $T = 298~\mathrm{K}$,
$$\begin{align}
\Delta G^\circ_1 &= -98.05 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+62.879 ~\mathrm{J~K^{-1}~mol^{-1}}) \\
&= -116.79~\mathrm{kJ~mol^{-1}} \\
\Delta G^\circ_2 &= +187.78 ~\mathrm{kJ~mol^{-1}} - (298~\mathrm{K})(+226.222 ~\mathrm{J~K^{-1}~mol^{-1}}) \\
&= +120.37~\mathrm{kJ~mol^{-1}} \\
\end{align}$$
consistent with our earlier findings.

The actual equation for the decomposition is:
$\ce{2H2O2}$ $\ce{->}$ $\ce{2H2O}$ + $\ce{O2}$
Yet another form of the equation that you think is supposed to happen is:
$\ce{2H2O2}$ $\ce{->}$ $\ce{2H2}$ + $\ce{2O2}$

Using bond enthalpies, it can be calculated that the energy needed to break all the bonds in gaseous hydrogen peroxide ($\ce{H2O2}$)is the summation of the energies need to break the two ($\ce{O-H}$) bonds and the ($\ce{O-O}$)bond. The energies are $\ce{2 * 465 kJ/mol}$ and $\ce{213 kJ/mol}$ respectively.
The extra energy added to the system required to do so for 2 moles of $\ce{H2O2(g)}$ = 2286 kJ
The total energy required to form 2 mol each of $\ce{H2}$ and $\ce{O2}$ = 1848 kJ
For $\ce{2H2O2}$ $\ce{->}$ $\ce{2H2}$ + $\ce{2O2}$ to occur, 438 kJ  must be absorbed in to the system or reaction vessel, i.e. ∆H = +438kJ — an endothermic process.
The energy required to form $\ce{2H2O}$ and $\ce{O2}$ = 2358 kJ
If $\ce{2H2O2}$ $\ce{->}$ $\ce{2H2O}$ + $\ce{O2}$ occurs, 72 kJ will be absorbed by the reaction vessel, i.e. the reaction has ∆H = -36 kJ/mol of $\ce{2H2O2}$ — an exothermic process.
By employing the Gibbs free energy equation (and general rules of thumb regard entropy and temperature), exothermic reactions tend to be spontaneous while endothermic reactions are not. As the reaction producing water and oxygen gas is exothermic, it is also the more "favoured reaction". Even though the change of entropy of $\ce{2H2O2}$ $\ce{->}$ $\ce{2H2}$ + $\ce{2O2}$ is likely to be higher than that of the other reaction, ∆S has less of an impact on ∆G than ∆H does.

I used bond enthalpy data from a university chemistry class data sheet online.


text:
Sodium carbonate titration in presence of phenolphthalein
Why phenolphthalein cannot show the endpoint for the complete titration of Na2CO3 with only phenolphthalein?
What's the reason that we consider only half of the moles of sodium carbonate when titrating with phenolphthalein indicator?

LOL - There is a difference between a solid theoretical answer and sheer speculation. If you won't accept acid-base theory I don't know how to help you.

I removed such stuff from the question.

Suppose you want to find out the amount of sodium carbonate  in a solution of sodium hydroxide. Let's use phenolphthalein as the indicator and HCl for titrating the mixture.
The first reaction that takes place is $$\ce{NaOH + HCl -> NaCl + H2O}$$
The pH at this point isn't alkaline enough for phenolphthalein to become colourless, since there's still base ($\ce{Na2CO3}$) present in the solution.
Then $\ce{Na2CO3}$ reacts. When all the $\ce{CO3^2-}$ becomes $\ce{HCO3^-}$, the pH is given by:
$$\mathrm{pH}=\frac{\mathrm{p}K_{\mathrm{a}1} + \mathrm{p}K_{\mathrm{a}2}}{2}$$
which is approximately $8.3$. (The above expression is for amphiprotic salts.) Phenolphthalein changes colour at approximately pH $8.2$.
So when the colour of phenolphthalein disappears, it means that all the carbonate is converted to bicarbonate. So if $x$ is the volume of HCl required to neutralize NaOH and $y$ is the volume of HCl required for complete neutralization, of carbonate, then phenolphthalein changes colour when volume of HCl used is $\displaystyle x+\frac{y}{2}$.
Remember that for complete neutralization, two moles of HCl is required for every mole of carbonate. For half neutralization only one mole of HCl is required. (Which is half the amount required for complete reaction).


text:
Confusing Lines About Extinct Elements
In "The Greatest Show On Earth", author Richard Dawkins spends a chapter discussing radiometric dating. I find myself confused by the following lines, though I'm sure the reason is elementary. (excellent pun and emphasis both mine)

Among all the elements that occur on Earth are 150 stable isotopes and
  158 unstable ones, making 308 in all. Of the 158 unstable ones, 121
  are either extinct or exist only because they are constantly renewed,
  like carbon-14.

What is meant by "occur on Earth" and "extinct" here? I take "extinct" to mean "decayed to the point that it cannot be detected". If I am using that word correctly, then in what sense does an extinct element "occur on Earth"?

You have the right notion. The matter that makes the earth came from some supernova(s) in the distant past. The matter ejected collected to form the sun and planets of our solar system. At the time of the supernova all sorts of isotopes were created. But if those isotopes had a half life much shooter than the time  in the past to the supernova, then all of that isotope would have decayed. For example carbon-14 has a half life of 5,730 years. Thus all of the carbon-14 created in the supernova event would have decayed long ago. C-14 is now formed in upper atmosphere which is why it exists now.

Hmm, Dawkins' statement doesn't quite match the information in a previous answer of mine, so that may make it more complicated to figure out exactly what he meant.

Defintions
Occur on Earth differentiates between all isotopes in the universe and isotopes which exist or have existed on Earth. The passage identifies that there are a total of 308 isotopes occurring on Earth, but it is worth noting that there are more than 308 total isotopes in the universe. This discrepancy is because some isotopes have extremely short half-lives, making their total "lifetimes" extremely short. As an example, an isotope generated in a supernova which has a half-life of 1 second will have totally degraded in at most 32 seconds (32 half-lives degrades the sample completely no matter the original amount). That is not nearly enough time for the isotope to travel from the exploded star to Earth, so that isotope does not occur on Earth (and it never can). This hypothetical supernova isotope would exist but it would not be part of the 308 Earth occurring isotopes.
Extinct means that the isotope is no longer naturally occurring within a given environment. Isotopes have differing half-lives, and a currently extinct isotope would have degraded so completely as to no longer be present at all. The isotopes would have originally been seeded into the environment by processes like supernovas, which produce ejecta containing the elements produced by nuclear fission and nuclear fusion. When that ejecta reaches the Earth the isotopes within it begin to occur on Earth, but eventually those isotopes will disappear due to half-life degradation. There are processes which can replace degraded isotopes in a self-regenerating manner and thus those isotopes never go extinct due to self-regeneration. An example of this type of self-regenerating isotope is Carbon-14; in fact, Carbon-14 will likely never become extinct. Self-regeneration is not the only process to replenish isotopes; the aforementioned supernovas are another. Unlike self-regeneration, which is a continuous process in which degraded isotopes continuously get replaced, supernovas deposit isotopes intermittently and comparatively infrequently. In fact, if a supernova deposit is delayed for an extended amount of time, isotopes which come only from supernovas can actually disappear from Earth entirely- they would become extinct. These extinct isotopes would stay extinct on Earth until a supernova redeposited the isotope.
Clarifying the Passage
As far as your question about an extinct isotope occurring on Earth- for the isotope to be extinct it cannot be currently occurring on Earth, but it was present in the past (logically, if an isotope never appeared on Earth it couldn't be part of the 308 isotope total, so for some of that 308 total to be extinct means they were present on Earth in the past). That question is sort of unrelated to the passage however, but the passage was written poorly enough that I think you misunderstood that they were trying to say that 121 isotopes should hypothetically be extinct, but that some of them are not actually extinct. To reduce the confusion the passage is causing, I have rewritten it to be far more clear: "Among all the elements that have have ever occurred or currently occur on Earth there are 308 total isotopes. Those isotopes are split into 150 stable isotopes and 158 unstable isotopes. Of the 158 unstable isotopes, 121 should be or currently are extinct, however, there are isotopes not actually extinct because they are constantly renewed, like carbon-14."

Thank you for the thorough answer. I'm still confused about how one could determine that an isotope used to occur on Earth if it can't be detected today.

Because isotopes don't degrade to something random, they degrade to a specific substance. For instance, Carbon 14, it degrades to Carbon 12 at a certain rate- which is how carbon radio-dating works. So, say you know that there is a ton of a certain lead isotope that only comes from the degrading of a particular uranium isotope (which is a real phenomena- uranium eventually turns into lead) you can know that the uranium isotope was present in the past if you find the lead isotope. Even if the uranium isotope is extinct, the existence of the special lead isotope proves it used to occur on Earth.

Your comment hits my oversight on the nose. It is not the case that radioactive isotopes decay and vanish without a trace. They decay to a stable substance that we can observe today. Thank you for bearing with me.

As I pointed out in a comment, you have the right notion. The matter that makes the earth came from the big bang, or supernovas in the distant past. Our sun is thought to be a third generation star so the matter in our solar system could have come from the big bang, a first generation supernova, or a second generation supernova. Although we know about 3300 isotopes in total, no doubt more very short lived isotopes were produced in the big bang or supernovas.
I don't know how to determine which isotopes Dawkins thought to be the 150 stable ones, or the 158 unstable ones, nor which 121 of the unstable 158 that he determined were either extinct or exist only because they are constantly renewed. Dawkins numbers seem to be very out of date as well. For isotopes found/known on earth, a Wikipedia table lists:

254 "stable" isotopes (no decay yet observed)
34 primordial radioactive nuclides
51 otherwise naturally occurring nuclides (21 of which Wikipedia lists as cosmogenic radioactive nuclides.)

That gives a total of 339 isotopes which have been detected on the earth out of the about 3300 known isotopes.
Primordial Radioactive Nuclides
The 34 primordial radioactive nuclides have half lives long enough to still survive from their creation in the big bang, or some supernova event, till the formation of the earth and also survive to the present day.
Let's take one isotope as an example of an primordial radioactive nuclide. In the second generation supernova some $\ce{^{244}Pu}$ (half life 80.8 million year) would have been created. If we assume that a billion years are needed for the matter from the second generation stars to form the third generation solar systems then that is 12.4 half-lives and 1.9E-4 of the plutonium formed in the supernova would survive enough to be detected on the newly formed earth. However the earth was formed about 4.54 billion years ago, so now an additional 56.2 half lives have passed and the original plutonium is mostly gone. So at this point it has been 5.54 billion years since the second generation supernovas which is 68.6 half-lives and only 2.3E-21 of the plutonium formed in the supernova would survive. That would seem to be just enough to yet be detectable.
Cosmogenic Radioactive Nuclides
The 21 cosmogenic radioactive nuclides are formed by nuclear reactions in earth's upeer atmosphere. I'll point out two such isotopes which have a short half life that they only exists because they are being renewed. Both $\ce{^3H}$ (half life 12.7 years) and  $\ce{^{14}C}$ (half life 5,730 ± 40 years) are renewed in earth's upper atmosphere.
Extinct Isotopes
So with 51 short-lived isotopes known, and 21 of those being detected as cosmogenic radioactive nuclides, then we are left with 30 radioactive nuclides which have short half-lives and are as Dawkins puts it either (1) extinct or (2) exist only because they are constantly renewed like the cosmogenic radioactive nuclides. At this point the gist is that Dawkins seems to hedged his analysis and use an "or" to stop his analysis. In other words it isn't yet clear that all 30 of the renaming isotopes have been detected in nature. Some may just have been detected because they were made in a laboratory. If the isotope cannot ever be detected in nature (which seems unlikely if you really were determined to find such an isotope) then it would be an extinct isotope.
Thus one definition of an extinct isotope is:

an isotope which would have been created in the big bang or a supernova, but presently undetectable on earth.

This of course leaves an ambiguity if any particular atom of such an isotope survived from the big bang or a supernova long enough to actually be present when the earth formed, yet be undetectable today on earth. All in all this seems unlikely.

Thanks for the detailed breakdown of the different isotope classes. As for Dawkins' numbers, he gives no reference, but the book is over six years old. Perhaps they were state of the art at the time, perhaps not.

I'd assume that Dawkins just used an relatively old reference for his research. Since Dawkins book was published in 2009 there is no  way that 100 isotopes have been discovered in the last six years. I think overall Dawkins argument that the earth is billions of years old, is correct even though his isotope counts are wrong.

One of the largest issues as to your question is Plutonium 244, or Pu-244.  Its half-life is just over 80 million years, which (given the existence of thorium and uranium with much higher half lives) suggests that it like those other high-proton elements must have been originally been produced in supernovae and debated since then.  Unlike the decay structures of U-238 and thorium, plutonium would have almost decayed to nothingness at half life of 80 million years.  Nonetheless, almost nothingness is not nothingness.  Predictably, at least a few atoms of primordial plutonium should be detectable even now.


text:
How the Azimuthal quantum number describes 3d shape of the orbital?
I read yesterday that Azimuthal Quantum number describes the shape of orbital....

what they call 'orbital' of an electron in chemistry is called wave function in physics. It determines how the electron moves and how it behaves. In fact it determines everything of the electrons. The azimuthal quantum number is just the quantum number used to describe angular part of the wave function, so it determines the shape of the orbital. This is a very rough explanation. You need to learn something like quantum chemistry or quantum mechanics in order to get a better understanding on that.

An orbital is not just a wavefunction, is a one-electron wavefunction, and for many-electron systems has no physical meaning.

Yes, but that is a long story for a beginner to understand. And as you said for many-electron systems has no physical meaning. In many-body case, we use the occupation of each states.

Yes, But how exactly numbers are related?

Have you ever learnt linear algebra? That would be of a lot of help to understand it.


text:
Purity of calcium carbonate in limestone

$\mathrm{150~mL}$ of $\mathrm{0.1~N}$ $\ce{HCl}$ is required to react with $\mathrm{1~g}$ of a sample of limestone. Calculate percentage purity of calcium carbonate.

Reaction: $$\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$
Number of milli equivalence of $\ce{HCl}$ used initially is $\mathrm{15}$
Number of milli moles of $\ce{HCl}$ used initially is $\mathrm{\frac{15}{2}}$
$\mathrm{2}$ moles of $\ce{HCl}$ reacts with $\mathrm{1}$ mole of $\ce{CaCO3}$.
So amount of calcium carbonate is $\mathrm{\frac{15}{4}\times 10^{-3}~mol}=\mathrm{\frac{3}{8}~g}$
But answer given is $\mathrm{75\%}$

We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.
Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$
...implying the purity is 75%.


text:
Rate of EAS in chlorobenzene and fluorobenzene
In both chlorobenzene and fluorobenzene, both the halogens show a −I effect where fluorine has more electron withdrawing effect than chlorine.
Now, if we consider the mesomeric/resonance effect, both show +M/+R effect. On comparison of resonance effect and inductive effect, resonance effect dominates over inductive effect. Then:

Why does −I effect dominates over +M/+R effect of chloro and fluoro group in the case of chlorobenzene and fluorobenzene? Give reason.

I know that the −I effect dominates over the +M/+R effect of the chlorine and fluorine substituents in case of chlorobenzene and fluorobenzene. Now, on the basis of this chlorobenzene should be more reactive for electrophilic aromatic substitution reaction than fluorobenzene (as −I effect: $\ce{-F~ >~ -Cl}$). But, it is not true though vice-versa is true.

Why is fluorobenzene more reactive for electrophilic aromatic substitution reaction than chlorobenzene? Give reason.

We need some data to answer your question. As presented in this earlier answer (which is also relevant to this question), here are the relative rates for electrophilic aromatic nitration of the halobenzenes compared to benzene itself.
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative Rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}
You are correct that inductive and resonance effects operate in opposite directions here. All of the halogens are more electronegative than hydrogen so they inductively remove electron density from the aromatic ring and tend to slow the reaction down. Whereas resonance effects would donate electron density into the aromatic ring and speed the reaction up.
However it is not correct to say

On comparison of resonance effect and inductive effect, resonance
  effect dominates over inductive effect

Here is a series where inductive effects play a larger role than resonance effects.  This is why all of the halobenzenes react more slowly than benzene itself.
In fluorobenzene the overlap between carbon 2p and fluorine 2p orbitals is very effective, but the carbon 2p orbital cannot overlap as effectively with higher p orbitals (e.g. chlorine 3p, bromine 4p, etc.). It is the stronger resonance contribution from this effective electron donating resonance interaction with fluorine that makes fluorobenzene more reactive towards electrophilic aromatic substitution than chlorobenzene.
The electrophilic reactivity of the other halobenzenes is also controlled by a blend of resonance and inductive effects.  As we move down the series from chlorine to iodine the electronegativity decreases and the strength of the electron withdrawing inductive effect also decreases - this should speed up the rate of reaction. However the strength of the resonance effect also decreases due to less effective orbital overlap between halogen and carbon - this should slow the reaction down. Looking at the data we can see that the reaction rate slowly increases as we move from chlorine to iodine, suggesting that the inductive effect is still the primary effect with resonance, as expected due to poorer overlap, still playing a secondary role.

"It is the stronger resonance contribution from this effective electron donating resonance interaction with fluorine that makes fluorobenzene more reactive" But isn't the electronegativity of fluorine more than that of chlorine? So, shouldn't the "effective electron resonance interaction" be nullified? We already know that all halobenzenes react in EAS at a much lower rate than benzene, so the rings are indeed deactivated. But how can we predict theoretically whether mesomeric effect of F dominates over Cl or whether more EN of F dominates over Cl?

In this answer also, you have said that "we need some data." Apparently there's no way to do this comparison theoretically...

Since in case of halogens, because of their exceptionally high electronegativity, Inductive effect dominates over resonance .
In case of flurobenzene though - I effect is the strongest but because of great 2p-2p orbital overlap between F and carbon + r effect decrease the -I effect to a great extent, making it the least deactivating group.
In case of iodobenzene -I effect though greater than resonance but is not that prominent because of comparitively poor electronegativity and also because of poor 5p-2p overlap with the carbon atom , the group become more deactivating than Flourine.
Similary comparing the values of -I and + r effect for chloro and iodobenzene the order comes out to be as follows
ORDER OF DEACTIVATING GROUPS-
Fiodobenzene>chlorobenzene>bromobenzene

Could you revise fidobenzene?


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Internal Standard Methodology

A tablet of weight $\pu{200 mg}$
was analysed for its morphine content using a HPLC system in conjunction with a single wavelength UV detector operating at $\pu{260 nm}$.
The tablet was prepared for analysis by crushing and extraction using a solvent system similar to the HPLC mobile phase i.e. $70\%$ acetonitrile, $30\%$ water. $\pu{50 mg}$ of codeine was added as the internal standard and the mixture made up to $\pu{50 mL}$. The resulting solution was analysed by HPLC using the instrument conditions given below. A standard solution was obtained by taking $\pu{50 mg}$ of each component, dissolving in mobile phase and making up to $\pu{50 mL}$.
Using the results obtained below, calculate the morphine content of the unknown tablet (both as %w/w and mg/tablet).
Standard Solution Analysis
\begin{array}{c|cc}\hline
&\textbf{Retention Time (min)} & \textbf{Peak Area} \\ \hline
\textbf{Morphine} & 2.00 & 153000 \\
\textbf{Codeine} & 3.25 & 212000\\ \hline
\end{array}
Analysis of tablet solution
\begin{array}{c|cc}\hline
&\textbf{Retention Time (min)} & \textbf{Peak Area} \\ \hline
\textbf{Morphine} & 2.02 & 17600 \\
\textbf{Codeine} & 3.29 & 212500\\ \hline
\end{array}

I understand the use of the internal standard methodology, however I am unsure which values I should use in the equation as although I have been given the internal standard solution I also have been given the second standard solution of both codeine and morphine.
The question is asking to find the content of morphine in the sample tablet (200 mg) via the use of an internal standard solution of codeine (50 mg) which was made up to 50 mL ([codeine] = $\pu{1gL^-1}$). This was given as well as another standard solution of codeine (50 mg) and morphine (50 mg) made up to 50 mL ([codeine] = $\pu{1gL^-1}$).
The equation I have been given is
$$D_\textrm{rfx} = \frac{A_x}{ A_\textrm{is}} \times \frac{C_\textrm{is}}{ C_x}$$
where subscripts and symbols have the following meaning: $x$ = unknown, $\text{is}$ = internal standard, $A$ = peak area, $C$ = concentration.
First we need to calculate the detector response factor ($D_\text{rf}$) of the standard solution - but which standard solution?
Assuming that the standard solution conc. of both morphine and codeine are $\pu{1gL^-1}$, this gives a $D_\text{rf}$ for the standard solution of 0.722.
Using the calculated $D_\text{rf}$  value, the $D_\text{rf}$ equation can be rearranged to find the concentration of codeine in the sample tablet.
However my question is which values need to be used for the $A_x$ and $A_\textrm{is}$ - do they need to be both morphine or just the two values from the sample peak areas?

However my question is which values need to be used for the $A_\mathrm x$ and $A_\mathrm{IS}$ - do they need to be both morphine or just the two values from the sample peak areas?

To answer just this part of the question,
\begin{align}
A_\mathrm{x} &= A_\mathrm{unknown} \mathrm{(Morphine)} \\
A_\mathrm{IS} &= A_\mathrm{standard} \mathrm{(Codeine)} \\
C_\mathrm{x} &= C_\mathrm{unknown} \mathrm{(Morphine)} \\
C_\mathrm{IS} &= C_\mathrm{standard} \mathrm{(Codeine)} \\
\end{align}
Looking at your picture, it looks like you successfully calculated your $\mathrm {D_{rf}({morphine})}$ (though it's next to a misleading picture, as the peak area of codeine is larger than that of morphine when their concentrations are equal).
Moving down, you've successfully solved your $\mathrm{D_{rf}}$ equation for unknown concentration, but the substitution is a little off. When you're dealing with the dissolved tablet solution, $A_\mathrm{unknown}$ is indeed $17.6$ but $A_\mathrm{standard}$ would be the codeine area of $212.5$, not the earlier morphine value.
If I had to guess, the confusion is between the terms "internal standard," a known spike of (usually) a single standard (here codeine) and "standard solution," which is an reference standard of known morphine and internal standard concentrations. Two common uses of reference standards are as initial calibrants and a good way to check for interference between analytes.
Getting back to the solution, once you have your unknown concentration, in units of $\pu{g/L} = \pu{mg/mL}$, you will simply multiply by the $\pu{50 mL} = \pu{0.05 L}$ volume to have the morphine concentration of the solution and thus of the whole $\pu{200mg}$ tablet. %w/w is also easy at that point.


text:
Why it is hard to rinse off soap from your body using rain water
I have experienced this when every time I take a bath using rain water, I just can't stop wondering why is it really hard to rinse off soap from your body?

Rain water is pretty much devoid of the minerals such as calcium salts that are often found in so-called "hard" tap water taken from underground sources. Calcium ions in tap water react with the fatty acid anions in soap to create insoluble soap scums, which (while they may remain on your skin) do not have the slippery feel of soap. Without the calcium ions present, the soap remains in soluble form (sodium salts of fatty acids) and I would propose that the slippery feel of soap on your skin comes from the traces of it that remain even after considerable rinsing.

The morale is of course, if showering with rain water, use less soap! ;)


text:
Why is the reaction between potassium permanganate and hydrogen peroxide spontaneous?
When hydrogen peroxide is mixed with potassium permanganate, oxygen gas and water vapour are formed, according to the reaction (source):
$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O + 3O2 + 2OH-}$$
This reaction is spontaneous, and exothermic. It is an example of a redox reaction, with the following half reactions occurring (data from Vanýsek):
$$
\begin{align}
\ce{MnO4- + 2 H2O + 3 e- &-> MnO2 + 4 OH-} &\quad E^\circ_\mathrm{red} &= 0.595~\mathrm{V} \\
\ce{H2O2 &-> O2 + 2 H+ + 2 e-} &\quad E^\circ_\mathrm{ox} &= -0.695~\mathrm{V}
\end{align}
$$
$E^\circ_\mathrm{cell}$ is equal to the sum of the oxidation potential and the reduction potential of the two half reactions; in this case, it would be $-0.1~\mathrm{V}$. A redox reaction is spontaneous if $E^\circ_\mathrm{cell}$ is positive — how can it be, then, that hydrogen peroxide spontaneously reacts with permanganate ions?
Using thermodynamical data (from NIST), I have calculated that the $\Delta G^\circ_\mathrm{m}$ of the reaction is $-463.576~\mathrm{kJ}$. The reaction should indeed be spontaneous. How can it be, then, that the results of the thermodynamical approach and the electrochemical one differ drastically?

For both half-reactions, the actual potentials depend on $\mathrm{pH}$; however, the values given in the question apply to different $\mathrm{pH}$.
The given reaction of hydrogen peroxide and the corresponding potential apply to $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$:
$$\ce{O2 + 2 H+ + 2e- <=> H2O2}\qquad E^\circ=0.695\ \mathrm V$$
Whereas the given reaction of permanganate and the corresponding potential apply to $\mathrm{pH=14}$ or $\left[\ce{OH-}\right]=1$:
$$\ce{MnO4- + 2 H2O + 3e- <=> MnO2 + 4 OH-}\qquad E^\circ=0.595\ \mathrm V$$
The corresponding reaction and potential for $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$ are:
$$\ce{MnO4- + 4 H+ + 3e- <=> MnO2 + 2H2O}\qquad E^\circ=1.695\ \mathrm V$$

But then, the net reaction would be $\ce{2MnO4- + 3H2O2 + 2H+ -> 2MnO2 + 4H2O + 3O2}$, which doesn’t correspond to the reaction cited above. Thus, I would use for $\ce{H2O2 + 2OH- <=> O2 + 2H2O + 2e-} (E^\circ=0.146\ \mathrm{V})$, which yields the given equation. Thank you so much for your help!

The Gibbs energy calculated from the cell potential is –429 kJ, which conforms more or less to the thermochemical result.

The calculations you have done (if correct, I didn't check them) suggest that the reaction would not be spontaneous at standard conditions, which for electrochemistry means a pH of 0.  If you want to model the process at pH 7, you would need to adjust the potential used for both reactions: permanganate reduction produces $\ce{OH-}$ and peroxide oxidation produces $\ce{H+}$.  Thus, the equilibrium potential for either half-reaction will depend on pH.
You can use the Nernst equation to do this adjustment.  Since standard conditions have a concentration of $\ce{OH-}$ of $10^{-14}$ molar, and at pH 7 this concentration will be $10^7$-fold higher, the $E_\mathrm{red}$ of permanganate should be less reducing than the standard value $E_\mathrm{red}^\circ$.  The opposite is true of peroxide.

The potential of a cell is the reduction potential minus the oxidation potential. In your case, it would be $E^\circ_\mathrm{cell} = 1.19 - (-2.08) = 3.27~\mathrm{V}$ for the whole reaction or $1.64~\mathrm{V/mol}$ of permanganate. You must remember to take into account the stoichiometric coefficients of the half reactions to obtain the overall potential

Corundum already took the negative sign in the oxidation potential, i.e. it is a true oxidation potential, not a reduction potential for an oxidation half-reaction. The electrochemical values are correct. 2) When calculating $E^\circ_\mathrm{cell}$ you do not multiply electrode potentials by the stoichiometric coefficients

So right. I think I am still coming into the new year. Should I correct the answer or just leave it so that people can see why it is wrong?

Well, that is really up to you.

Why not correct the answer?  After you edit, some of those downvotes could turn to upvotes, and you might attract new upvotes too.


text:
Spontaneous/Non-Spontaneous Reactions and Reversible Reactions
My chemistry textbook states "Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes."
Does this mean that some of the reversible reactions are spontaneous and others are not? Or that the reversible reactions are sort of intermediary between said two types of reactions? Or something else?

Your "non-spontaneous" in this context probably means with lowering of entropy of universe - there's no such process, reversible with constant entropy also doesn't really exist and spontaneous with raising entropy of univ. - all processes, so generally this sentence is stupid.

http://chemistry.stackexchange.com/questions/6441/spontaneous-reactions http://chemistry.stackexchange.com/questions/16293/nature-of-spontaneous-reactions

Also I have a feeling you'd better get better textbook...

cant its sort of a standard.

"Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes." So what does this mean?

To me this sentence is meaningless gobbledygook and should be ignored.  That is, this sentence is unhelpful and trying to understand it will just make you more confused and less able to master chemistry.  Here is a short summary version of the points I make in the "Advanced discussion" below:

All reactions can be viewed as reversible from a mathematical standpoint, if the reacting system is "big enough".  Reactions that are highly spontaneous in the forward direction mean that at equilibrium, the number of "reactant" molecules will be very very small (but not zero!).
Many times, "big enough" would mean astronomically big, such as bigger than the Earth.  In these cases, reactions can be regarded as "irreversible", although there is no strict, universally agreed upon boundary between "reversible" and "irreversible".
Whether a reaction is spontaneous isn't that much related to whether it is "reversible".  Don't worry about understanding the connection between these concepts until you have understood each concept independently.

Advanced discussion.
The meaning of spontaneous
Spontaneous processes have negative Gibbs free energy changes ($dG < 0$) and also result in positive entropy changes in the universe ($dS_{univ} > 0$).  However, $dG$ is a function of state.
For chemical reactions, this means that $dG$ depends on the concentration of the reactants and products, i.e.
$$dG = dG^\circ + RT \ln Q$$
where $Q$ is the reaction quotient.
Thus, if we start off with all "product" and no reactant, then the "reverse" direction of the reaction will be spontaneous.  But if we start off with all reactants and no product, then the forward direction is spontaneous.
The meaning of reversible
The short answer is that for chemical reactions, "reversible" does not have a precise thermodynamic definition.
$$dG = dG^\circ + RT \ln Q = -RT \ln K + RT \ln Q = RT \ln \frac{Q}{K}$$
where $K$ is the equilibrium constant.  People will usually call reactions with a "large" equilibrium constant irreversible and those with a "small" equilibrium constant reversible.  But the exact definition of "large" and "small" isn't well-defined.  Many people might say $\frac{Q}{K} \leq .01$ represents an "irreversible" reaction.  For chemical reactions with very large equilibrium constants, the size of the system required to ensure that an equilibrium can be reached can get very large, as explained here.

Your textbook has it backwards.  Reversible reactions are not a limiting case of spontaneous reactions and non-spontaneous reactions.  Spontaneous reactions and non-spontaneous reactions are limiting cases of reversible reactions.  As several responders have pointed out, if a reaction only proceeds a little before reaching equilibrium, it approaches the case where no reaction occurs (aka non-spontaneous reaction).  If the reaction proceeds to nearly completion before reaching equilibrium, it approaches the case of complete conversion (aka  spontaneous reaction).

I agree with other answers and comments that your textbook does not have things stated correctly so I have explained below in words only what I see as the difference between Irreversible and Reversible processes. (You will find spontaneous reactions hidden in these statements)
Every system left to itself will change, rapidly or slowly, in such a way as to reach a state of rest defined in a statistical way and this is also called the state of equilibrium. The system will only move away from its state of equilibrium through the influence of some external events. We are familiar with many processes that reach equilibrium, diffusion of a concentrated solution into a dilute one leading to uniform concentration, transfer of heat from a hot to a cold body leading to uniform temperature, oxidation of substances by the atmosphere, self demagnetisation of magnets are all examples of spontaneous events in nature.
These processes and all other natural processes are similar in one respect, that they all bring the system to equilibrium, and we may think of these systems as loosing some measure of their capacity for spontaneous change.
A system far from equilibrium is one which we would choose to harness for doing useful work, for example the combustion of coal in air to cause a steam engine to operate. However, the second law shows us that it is not the loss of energy that is important but rather the availability of the energy for external purposes.
The essential content of the second law can be given by the statement that when any actual process occurs it is impossible to invent a means of restoring every system concerned to its original condition. Therefore, in a technical sense, any actual process is said to be irreversible.
The ideal or reversible process
We now distinguish between an actual system, which is always irreversible, and an ideal one, which although never occurring in nature is nonetheless imaginable. Such an ideal process is called reversible. In such a process each stage is conducted so that an infinitesimal change in the external conditions would cause a change in the direction of the process. In this sense such an imaginary system is called reversible.
An example is a system of water and water vapour inside a cylinder with a frictionless moveable piston.  (In practice, by careful engineering, we can make the friction so small that it is negligible). Outside the piston is gas at some pressure. At constant temperature the system comes to equilibrium with respect to external conditions (pressure , temperature). If now the external pressure is increased, by the piston moving in by an infinitesimal amount, water vapour condenses. If the pressure is now reduced by an infinitesimal amount, some water evaporates. Thus the work required to vaporise 1 mole of water and to condense 1 mole of vapour differs only by an infinitesimal amount.
Another example is provided by an electrochemical cell where the cell potential can be measured extremely accurately with a voltmeter. The cell can be balanced so finely against an external emf that the current that can be made to flow in one direction or the other is only microamps.
A similar situation applies to thermal processes, since if two bodies differ in temperature by only an infinitesimal amount, the transfer of heat is likewise a reversible process for it would be possible to restore the system to its original condition with only an infinitesimal change in the external system.
(for a fuller description see Lewis & Randall 'Thermodynamics')

Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes [?]

A reversible process is a hypothetical process where the entropy of the system and its surrounding is constant. This is hypothetical because nothing happens unless the entropy increase at least a tiny bit. In other words, if the system is at equilibrium, the entropy does not increase, but nothing happens either.
Colloquially, a reversible reaction is one where forward and reverse reaction happen, i.e. one that would attain equilibrium at some point (if "left alone").
What I think the textbook is trying to say is that the Gibbs energy is negative for a spontaneous process, is positive for a non-spontaneous process, and is zero for a process that is at equilibrium. (This would be at constant pressure in the absence of non-PV work). So if you have an axis with the Gibbs energy for a process, there is a point where the system is at equilibrium. On one side of it are the spontaneous and on the other side are the non-spontaneous processes.
I'm not sure how that statement helps to understand anything, though. I also don't think it is the correct way of using the term "limiting case". Wikipedia gives the following definition:

In mathematics, a limiting case of a mathematical object is a special case that arises when one or more components of the object take on their most extreme possible values.

So maybe one could say a reaction that goes to completion is the limiting case of a reaction with a high equilibrium constant. In the statement in question, however, there is talk about a "limiting case between" two cases, not the most extreme value of either of them.


text:
How to test acid-base reactions under pressure?
I am trying to test chemical reactions, more specifically reactions between an acid and a base, at different levels of pressure.
I would like to know if there is some kind of equipment I can get that has adjustable pressure, and will let me get the reactants into the vessel without changing the pressure significantly.
I am looking for something that I just get at home, so not too big, and hopefully not too expensive. It doesn't need to be super high-tech, as I'm not going for a high level of accuracy, and I'm not going to be testing extremely low or high pressures.
Does anyone know of any equipment that will create significantly different levels of pressure, even if it is very expensive?

I am looking for something that I just get at home, so not too big, and hopefully not too expensive.  || No such thing. Also, pressure within 50 bar (reasonably cheap to obtain in a lab) do not affect acid-base reactions significantly.

Dang. Oh well, thanks for the reply. I'll have to think of something else.

Out of curiosity, does anyone know of any equipment that will create significantly different levels of pressure, even if it is very expensive?

google for 'hi pressure lab reactor'

What kind of pressure ranges did you have in mind?

I agree with permeakra but it's worth noting that for a couple of important acids and bases (e.g. CO2, NH3 and possibly CH3COOH) pressure does significantly affect acid base chemistry.  [and OK, CO2 is not an acid but an acid anhydride, you got me.]


text:
Behaviour of oleophobic and hydrophobic object in an oil and water solution
I am wondering what would happen if I put a small (between $\mathrm{5mm}$ and $\mathrm{1\mu}$, density between $\mathrm{1}$ and $\mathrm{2\: g/mL}$) item with both oleophobic and hydrophobic properties into an oil and water solution. Does anybody know? Would the item become suspended between the oil and water layers? I have not been able to find the answer anywhere!


text:
How to calculate the thickness of the double layer at an electrode
I need help calculating a double layer thickness. In this situation two insulated planar electrodes are placed in an electrolyte (in this case drinking water) and a voltage is applied to them of 10V across a 10mm gap (field strength 1V/mm) the field is quickly cancelled out by the electrical double layer, where the ions within the water have been attracted to the insulated surface of the electrode and block out the field.
The equations I have seen such as the Debye length equation don't seem to factor in an electrode potential, can anybody point me in the right direction or better yet show me an equation for calculating what the thickness would be? (apologies if I have missed something obvious)

Could you use the energy storage of the "capacitor" to determine the thickness? Given dielectric constant, voltage and measured capacitance, thickness could be calculated.

Why do you expect that the potential would be a factor? The concentration of the ions is the biggest factor in how long the Debye length will be. The applied potential will change the amount of charge stored, but the length of the double layer should be independent of potential.


text:
Mechanism for oxidation of primary alcohols to carboxylic acids
My teacher told me that the following is the mechanism for oxidation of primary alcohols to carboxylic acids:

I've searched in books and online and didn't find a similar mechanism. For example, in Wikipedia:

Which is correct?

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.
The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


text:
Why do branched hydrocarbons burn more efficiently (complete & clean combustion)?
Branched and cycled hydrocarbons usually a higher octane number than their straight chain counterparts. This is due to a more complete and clean (no residue) combustion. In addition, there is no probability of knocking or detonation.
What are the reasons behind this?

Possible answer in permeakra's final comment to his answer to this question.

@hb2py I was just about to respond with that as an answer. That is almost certainly the correct explanation.


text:
If the rate law order is three, does that mean that three bodies collide at some elementary step?
I am doing a chemistry review about rates, and I get asked "Which of these statements is false?"
The answer is

In reactions that are second order in one reactant and first order in another, the slow step generally involves a three-body collision of the reactants.

I imagine it looks like
$$
rate = k[A]^2[B]^1
$$
So at the slowest elementary step there should be
$$
2A + B = something
$$
Which is a three way collision between two A molecules and a B.
What am I doing wrong?
EDIT:
Question number 2 (http://www.mesacc.edu/~paudy84101/CHM152F2005/Exam1%20Key.pdf)
This is not my review, in case you are wondering.  I just have the same question.

Can you present the complete question?

Let's disprove that statement by counterexample. Here, the slow step is a collision of some intermediate and another thing, but the maths work out that the overall reaction order is third.
Let's have a two step reaction with the stoichiometry:
$$\ce{2A + B -> C}$$
The mechanism looks like this. The first step is fast and reversible and rapidly establishes equilibrium between $\ce{A + B}$ and an intermediate $\ce{I}$.
$$\ce{A + B <=>[k_1][k_{-1}] I}$$
The second step is slow, not so reversible and involves the intermediate $\ce{I}$ reacting with the other equivalent of $\ce{A}$.
$$\ce{I + A ->[k_2] C}$$
Now the rate of the reaction shall be written as the rate of appearance of $\ce{C}$, because it makes our lives a little easier.
$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=k_2 [\ce{A}][\ce{I}]$$
However, it's messy to have the intermediate in the rate law. Since the first step was fast and in equilibrium, we can get an expression for $\ce{[I]}$ in terms of $[\ce{A}]$ and $[\ce{B}]$.
$$K_{eq}=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{I}]}{[\ce{A}][\ce{B}]}\\
[\ce{I}]=K_{eq}[\ce{A}][\ce{B}]$$
A little substituting and we get:
$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=\dfrac{k_2 k_1}{k_{-1}} [\ce{A}]^2 [\ce{B}]=k_{obs}[\ce{A}]^2 [\ce{B}]$$

$\ce{NO}^\bullet$ autoxidation? :-)

This illustrates nicely why time is important here. The chance of 3 reactants interacting at the same instance is virtually zero, but if two of them can collide to form an intermediate (I), that lasts for some small but finite period, then this has a chance to collide with another reactant (A) before it falls apart.  If I is very short lived, i.e. $k_{-1} $ is huge then the rate tends to zero.