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Why does basicity of group 15 hydrides decrease down the group?
In my textbook it is written that the order of basic strength of pnictogen hydrides is
$$\ce{NH3 > PH3 > AsH3 > SbH3 > BiH3}$$
I tried but could not find any explanation as to why this happens. What is the explanation?

Each of these molecules has a pair of electrons in an orbital - this is termed a "lone pair" of electrons.  It is the lone pair of electrons that makes these molecules nucleophilic or basic.  As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (the nitrogen lone pair is contained in the n=2 shell, while the bismuth lone pair is in the n=6 shell).  As the electron density of the lone pair is spread over a greater volume and is consequently more diffuse, the lone pair of electrons becomes less nucleophilic, less basic.
Commment: But greater the EN of the central atom , lesser would be its tendency to donate electrons right?
Response to Comment: The ENs of the central atoms are N (3.04), P (2.19), As (2.18), Sb (2.05), Bi (2.02). All of these atoms, except for nitrogen have similar ENs and I think the electron density argument is valid for them. However, in the case of nitrogen and phosphorous there is a significant EN difference that would tend to argue in the direction opposite to my answer. However there is another important difference between N and P. The lone pair in ammonia is in an sp3 orbital. In all of the other cases the central atom is essentially unhybridized (~90° H-X-H angles) and the lone pair exists in an s orbital. Therefore the lone pair electron density in ammonia (being in an sp3 orbital) is effectively increased compared to phosphine where the lone pair is in an s orbital. So even though the EN difference between N and P is significant, when hybridization differences of the central atom are taken into account the electron density argument still explains the trend in basicity.

Usually, my teacher says that the electronegativity arguments are valid for elements in the same period. Trends involving traversal along a group are better explained by the size of the central atom, not sure if that is logically valid.

The hydrides of nitrogen family have one lone pair of electrons on their central atom. Therefore,they act as Lewis bases.As we go done the group, the basic character of these hydrides decreases. Nitrogen atom has the smallest size among the hydrides.Therefore the lone pair is concentrated on a small region and electron density on it is the maximum.Consequently, its electron releasing tendency is maximum.As the size of the central atom increases down the family, the electron density also decreases. As a result, the electron donating capacity or the basic strength decreases down the group.

Realise that there is a notable drop in basicity from nitrogen to phosphorus and then a slow and continuous further diminishing. The notable drop is due to the difference of the molecular structures of ammonia and phosphane.
As I answered elsewhere on this network, the ‘ideal’ bonding situation from an orbital point of view would be to just use p-orbitals to bond and leave the lone pair in an s-type orbital which generally has a lower energy. Doing this allows for a greater σ bonding stabilisation. And that is what all pnictogens below phosphorus (and phosphorus to a certain extent, although the bond angle is still $3^\circ$ off) do: their lone pair can be considered a pure s-type orbital at first approximation.
Ammonia, however, has space constraints since nitrogen is so much smaller than phosphorus and all the other pnictogens. If it were to assume $90^\circ$ bond angles, the hydrogen atoms would end up too close together resulting in steric strain. Thus, nitrogen moves the hydrogens apart by not bonding with pure p orbitals but with $\mathrm{sp^3}$ hybrid orbitals (in nitrogen’s case, the hybridisation is actually very close to $25~\%$ s-character and thus to $\mathrm{sp^3}$!). This means that the lone pair must also be localised in an $\mathrm{sp^3}$ type orbital which in turn means that it is not spherical and diffuse but ‘points in a direction’. This pointing allows for acidic hydrogens to easily attach to it and thus explain nitrogen’s and ammonia’s high basicity. I would also like to highlight an answer of Martin on the topic of $\ce{NH3}$ versus $\ce{PH3}$ hybridisation.
Further down the group, i.e. behind phosphorus, a simple diffusity argument is sufficient to explain the slightly reducing basicity. The higher the period, the more populated orbitals lie beneath the s-type lone pair thus the further out it is (also the greater the atom becomes). And the more diffuse a lone pair is the less likely it becomes to participate in bonding or to be basic in any way.

I realize that this is an older post so I might not get an answer here, but could you also explain the lower basicity of orbitals with more s character using an energetic argument? When I learned about the higher acidity of alkynes, one explanation was that the lone pair in the conjugate base was more stabilized because it was in a lower-energy sp orbital (as opposed to the sp3 orbital of an alkane. Could you apply the same argument here, or is it not applicable since PH3's lone pair is in a 3s orbital, not an orbital made of 2s and 2p orbitals like in ammonia?

@Akash Yes, that is true; the entire line of thinking is ultimately energetic and orbitals with a higher s component are lower in energy than those with a lower s component. And then as you stated.


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Copper is a metallic element, so it is likely to be?
Copper is a metallic element, so it is likely to be?
A. Poor conductor of electricity       B. A substance with a low melting point      C. Malleable     D. Poor conductor of heat     E. An element that forms negative ions
For me, C should be the only right. But are metallic elements really likely to be malleable?

While (C) is the correct answer (see all the ones below), I'll just point out that Cu is the classic example for work hardening in metallurgy.  As you try to hammer/roll/bend it, it becomes less malleable.  But that is likely never brought up in an introductory chemistry course...

Yes, It is C.
Metals are:
-Generally good conductors of electricity
-Substances with high melting points since they are mostly in liquid/solid states. Only liquid metal I can recall offhand is probably mercury, at room temperature and pressure of course.
-They form positive ions because they usually lose electrons to gain octet configuration.

well, Wikipedia says "Metals are described as malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond."
Other examples of malleable metals are Gold, Silver and Iron.

Metals generally have properties like good conductor of electricity,good Thermal conductivity,shiny,sonorous,lustrous,malleable(bitten into thin sheets under stress),ductile(drawn into wires),brittle,have high density,high melting and boiling point,generally solid in nature,forms positive ions etc...         "exceptions are there".
well your question is regarding choosing the right option.so,you are right in your thought.correct answer is option c.
The answer regarding the second part of your,i think is no as exceptions are always there in periodic table.

yes your answer is right.
well for your second question
Malleability, a similar property, is a material's ability to deform under compressive stress; this is often characterized by the material's ability to form a thin sheet by hammering or rolling
so basically you have to deal with some physics of elasticity here. Yes generally metals are ductile though there are exceptions like sodium, potasium etc.

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Delta G values for Pyruvate--> Alanine vs. Oxaloacetate -> Aspartate ->Asparagine
What are the delta G values for the mammalian (human) biosynthesis of alanine from pyruvate, and for the conversion of oxaloacetate to asparagine? (assuming no particular cellular restriction or other systemic pressures like catalysis or anabolism)
I am aware that oxaloacetate must be converted to aspartate and then to asparagine.  Based on the fact there are two steps to that conversion, I assume that cumulatively the delta G value is higher for the oxaloacetate to finally get to asparagine. So I want to know how much of a difference there is in the delta G values of these two conversions.
This is NOT a homework question, it is to better my own understanding.  I would welcome peer-reviewed literature, or other resources if you deem them applicable.

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Free rotation of sigma bonds; diastereomer formation; thermodynamic and kinetic products
For the reaction of 1,3-butadiene with DBr (D = heavy hydrogen), I found that several possible products could be formed.
Kinetic products
A pair of enantiomers is formed. One of the pi bonds attacks the heavy proton and the leaving group, bromide anion, comes back around to take care of the allylic carbocation. Because the bromide can attack from both above and below the plane of the carbocation, both enantiomers are formed.
Thermodynamic products
2) Now, because there is resonance, the carbocation possesses partial primary allylic carbocation character. As a result, bromide anion may also attack at a different position. This results in an internal alkene - the more stable alkene (the Zaitsev alkene).
Diastereomers are formed as well because single bonds can rotate, and the partial, terminal double bond can be "flipped" upward in the resonance depiction (I know that resonance depictions do not exist independently; hence "partial"). This allows the Z alkene to be formed in addition to the E alkene, as illustrated by my mechanism.
Questions
1) Are the products all valid and correctly categorized?
2) Is the E and Z isomer each produced, and in the way I think both are produced?

You have only considered reaction from the s-trans (here the prefix "s" is used to describe the arrangement of substituents around the central "single" bond) conformation of 1,3-butadiene, where the 2 double bonds are in a trans relationship about the central single bond.  The s-trans conformation is in equilibrium with the s-cis conformation.  The s-cis conformation can also react to produce the terminal olefin, but also (by the mechanism you described) it can react to produce the z 2-butene as one of its kinetic products.  Hence, all 3 butenes (1-butene and e and z 2-butene) can be formed as the kinetic products.

What does the prefix "s" mean in s-trans?

It stands for "single" bond. In other words there is a trans arrangement of ethylenes about the central single bond in the s-trans conformation.

No problem! So is considering the resonance structure as rotating to give E/Z isomers incorrect?

Under typical reaction conditions (something around STP) there is restricted rotation about the bonds in or attached to the allyl framework.

Makes sense given double bond character.

Yes, that's exactly it.

Awesome answer as always, @ron! I always learn something from your posts.

Edited post to illustrate what I think you were talking about @ron

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Validity of ionic resonance structures
What's wrong with ionic resonance structures? I asked one professor about them once and all he commented was "I've seen them too, and I think they're wrong."
Another professor rejected an ionic resonance structure showing the "H+" ion and asked me what phase the experiment was run in, in an attempt to figure out why the authors reported a resonance contributor with a H+ proton. He specifically asked if the experiment was run in the gas phase.
I think this guy forgot what resonance structures are because I know that he emphasizes that bare protons do not exist in solution, and he was probably thinking about how there could be a bare proton shown in a resonance structure if the molecule is in solution. The resonance structure I showed him was this (which I could see him picturing in solution, given that it is nitrous acid.

However, resonance structures are not discrete forms of the molecule in question - they're just contributors and show that there is partial ionic character; that there is a highly electronegative atom that is withdrawing electron density.
So I ask:
1) Are ionic resonance structures valid?
2) If so, why might they be valid, or what counterarguments might you present to someone who argues they are invalid?

Ionic resonance structures are of course valid. You might want to dig into Valence Bond Theory - which is the quantitative theory behind resonance structures - to find out a little more about the importance of these. (However, unbelievers are usually hard to convince.)

Do you have any articles recommended for me? I tried the Wikipedia article on VBT but it was, to say the least, rather uninformative. My next step is to search Google books.

I am sorry I can't help very much with that, I picked it up along the way when I was listening to presentations on conferences. However, there is literature linked in the wikipedia page, that should provide a starting point.

"Coulson's Valence" (1961) is a rather dated but readable introduction (https://archive.org/details/Valence).

I can't imagine anyone arguing that ionic resonance structures are invalid.  The often contribute significantly to the structure of a molecule and help explain the various physical aspects of the molecule.  The following ionic resonance structure contributes significantly to the description of the carbonyl compound.  $\ce{NaCl}$ is not totally ionic and is described by a blend of both the ionic (major) and covalent (minor) resonance structures shown below.

Resonance structures with a positive hydrogen (proton) are important in explaining hyperconjugative effects.

My professor would agree with your ketone resonance structure, having drawn it himself several times. I think that's what he'd consider a "charge-separated" structure rather than an ionic structure (like the ones involving H+ in your second picture. Those are the ones he says are "invalid." @ron

Also I'm curious as to your carbocation structure; is that structure suggesting that the carbocation is unusually acidic (the last two structures suggest to me that the central C-H bond is rather weak since electron density seems to gravitate toward the carbon with the positive charge and incomplete octet and away from the hydrogen). If anything I guess your last picture shows that the central C-CH2 bond is shorter than the C-CH3 bond due to double bond character, correct?

Yes, I suspect if you could measure the pKa of the n-propyl cation, you'd find the central C-H to be more acidic than the corresponding hydrogen atom in propane (but that's still not very acidic).  Yes, CH2-CH2+ bond should be shorter than expected, likewise the central C-H bond should be longer than expected.

@Dissenter, the increased acidity is what makes the E1 reaction work.

@Dissenter I was looking over my SE Chem Answer log today and noticed there are a lot of answers I posted to  questions of yours where an answer has not been accepted, this is one such question.  I'm trying to understand if you just haven't gotten around to accepting an answer or whether my answers are lacking.  My aim is not to pressure you into anything, I'm just trying to gauge the effectiveness of my answers.

@Dissenter  Thank you Dissenter, to me it's like a vote of confidence, I appreciate it.

How would you even generate the n-propylcation? And wouldn’t it immediately Wagner-Meerwein rearrange to the sec-propylcation?

@Jan Probably the easiest way to "generate" it would be computationally. Photoelectron detachment in 2-iodopropane might also work. See this earlier answer for some related discussion on the 2-propyl carbocation, it may have a non-classical structure. As to "immediate" rearrangement, that depends on the temperature of the experiment and how much stability is provided by the non-classical structure.

I would like to add another shade to the already quite excellent answer of ron.
I will use donor-acceptor complexes, like Lewis acid base adducts, as case in point. Probably one of the simplest and most popular is the ammonia and borane adduct. I consider the structures below to be the most accepted ones, with probably the widest use.

These are all structures (not all of the are Lewis approved) that try to cover the concept of a rather difficult bonding situation. In this special case, how to describe a dative bond in an easy model. This also gives us a lot to think about.
The first structure completely lacks the feature of a bond, therefore it cannot possibly be used to describe this molecule alone. But since we are talking about donor acceptor interactions, a comparatively weaker bond, the separated structure also has to be considered to contribute to bonding.
The second structure is not Lewis approved (there are not arrows in the traditional representation). It is much like the first structure, but the bonding here is at least indicated. In my opinion this is the only structure that comes somewhat close to the actual bonding situation. However, it treats the donor acceptor bond between nitrogen and boron as a second class covalent bond - is that really true?
The third structure is often used to indicate, that there is a real covalent bond between boron and nitrogen. To be perfectly honest, what is the difference between polar covalent and dative covalent anyway; it's shared electrons in both cases (see above). In my opinion this structure is one of the best examples why resonance structures are all too often misleading. Since we are in the Lewis picture, we have to obey the rule of two (octet rule), that means we have to assign formal charges. This is suddenly somewhat suggesting (to the untrained mind), that boron will carry a negative charge, which is of course wrong. Nitrogen still is more electronegative and will obviously have the highest negative charge, the electron sharing will hardly change that.
The last structure is a truly awful hybrid, that I came across. It's neither fish nor flesh, nor good red herring. Don't use it, I just included is for the sake of completeness. It makes all the errors of the other structures combined.
So all of the structures have their flaws and have to be seen in conjunction. Does any of these structures become invalid through the flaws it represents? No. If anyone argues this, then the concept of resonance - as an approximation to the true bonding situation - is not understood.
So the validity of ionic resonance structures basically becomes a question of semantics, within the framework of the model, they work very well. Do they represent something real, then the obvious answer is no. Model systems often help us understand complicated points, but they must not be mistaken for the (sometimes unknown) truth and they should also not be confused with more elaborate models.
I previously stated in the comments, that in the framework of Valence Bond Theory, for each resonance structure it is possible to construct a wave function (= configuration). It is then also possible to obtain the contributions of each configuration to the total wave function. This at least offers a quantum mechanical backbone to the whole model of resonance, and it often is described as the quantitative theory (within it's limitations) of it. The biggest problem we have with this nice theory is, that it is very expensive to calculate.

$+1$ for the errors of all the other structures combined =D. On a different note, though: One of my professor’s defined a dative bond as one that will be cleaved heterolyticly when supplied with the bond dissociation energy, while ‘traditional’ covalent bonds would dissociate homolyticly.


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Electrolysis of Beer
Electrolysis of water, means that the water is separated into hydrogen and oxygen. As beer also includes water, I initially thought electrolysis of beer would cause the water to separate, and the hydrogen and oxygen given off gas, leaving behind a higher concentrate alcohol.
However then realise that ethanol is also in beer (and is what makes you drunk it seems), and like water has an OH bond, so would the electrolysis also cause that to break?
(also does the rest of the molecule affect the strength of the OH bond, either stronger or weaker?)

Sounds like a horrible thing to do to good beer!

The beer will conduct electricity mainly because of ions like Na, H, Cl and HCO3. Any such experiment end up with hell of non-separable dirt on anode: the mixture of chlorine, percarbonates and products of their reactions with organic and anode itself.

At best it will smell like bleach :)

It's hard to say what exactly you would end up with, but it probably wouldn't much resemble beer. The trick with electrochemistry is that a lot of things are electroactive and what reactions actually occur in a given reaction are not always easy to predict and depend on electrodes, solution composition, pH, applied potential, etc. Depending on your electrodes, ethanol may be significantly more easily oxidized than water. A quick literature search shows that platinum and gold are both decent catalysts for ethanol oxidation (platinum is commonly used in fuel cells for this reason), but even with other materials, I can't imagine any conditions in which one could easily enrich an ethanol/water solution in this way. Many organic compounds, probably including some that impart flavour, are likely to be oxidized as well, even if the ethanol wasn't, so it would certainly affect the taste.
Additionally, the electrolysis of water requires a large amount of energy, far more than distillation, so even if it would work, it would be an expensive way of doing it.

My thoughts exactly! Incidentally, the electrochemical oxidation of ethanol over platinum can produce a whole pile of products, including acetaldehyde and acetic acid. If you're really lucky, you'll eventually get $CO_2$, so that C-O bond sticks around.


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Carbonyl carbon electropositivity?
Is the carbon in a carbonyl functional group electropositive? I had a professor tell me so, but I immediately refuted him, because he defined electronegativity earlier as the ability to stabilize negative charge ... and taking electronegativity to be a converse property, the strong partial positive character on the carbon probably isn't able to stabilize any further positive charge.
Similarly, I know that lone pairs are defined as having an electronegativity of 0 because electrons cannot stabilize other electrons.
The guy agreed with me. So, to confirm, am I correct, and, as an extension, can we describe the electron deficient carbon as electronegative instead? I could very well see a electron-deficient carbon stabilizing negative charge.
However, if we stick with the IUPAC definition of electronegativity - which is the ability to attract electrons, the carbonyl carbon still seems to lose to oxygen. A charge-separated resonance structure can be drawn for any carbonyl group with a + on carbon and a - on oxygen.
So that leads me to another question - is electronegativity the ability to stabilize negative charge or the ability to attract electrons within a covalent framework? I guess the two go hand in hand - i.e. you wouldn't attract electrons if you weren't able to stabilize them ...
As a side question: can we calculate EN for ions?

Is the carbon in a carbonyl functional group electropositive?

Electronegativity and electropositivity are usually used in a relative sense, X is more electronegative than Y.  The carbonyl carbon in acetone is more electropositive than the central carbon atom in isobutane, and less electropositive than the methyl carbocation carbon.

because he defined electronegativity earlier as the ability to
  stabilize negative charge

Weak definition, wouldn't you say; that's why you won that argument.  Atoms can be electronegative in the abstract, charge doesn't need to be involved.  The statement, "fluorine is more electronegative than carbon according to the Pauling Electronegativty scale" is accurate and understandable, yet no one is describing a situation where charge stabilization is involved.

because electrons cannot stabilize other electrons.

I understand what you mean, but you should know by now that you can't make such sweeping statements.  Don't the electrons in one of the double bonds in 1,3-butadiene interact with and stabilize the electrons in the other double bond?

Can we describe the electron deficient carbon as electronegative

Again, I understand what you mean, but... Yes, it is more electronegative than the carbon in an alkane, but not as electronegative as $\ce{F+}$.

Is electronegativity the ability to stabilize negative charge or the
  ability to attract electrons within a covalent framework? I guess the
  two go hand in hand - i.e. you wouldn't attract electrons if you
  weren't able to stabilize them

Yes, the two go hand in hand.  More electronegative atoms are better able to stabilize negative charge than less electronegative atoms.  It's all about lowering the overall system energy.

As a side question: can we calculate EN for ions?

I'm sure there is a way.  For example, using studying the $\ce{^{13}C}$-nmr spectrum of various carbocations leads to a relationship of (somewhere around) 180 ppm shift/ electron.  So you can estimate the fractional positive charge on any carbon atom, the carbonyl carbon in acetone for example.  Once you've estimated the amount of charge on an atom, I suspect there is a way to relate it to EN.

"Weak definition, wouldn't you say; that's why you won that argument." - how would you define EN? Other than defining it based off the Pauling scale. In other words if his definition of EN is fluffy how should it be revised? @ron

"Don't the electrons in one of the double bonds in 1,3-butadiene interact with and stabilize the electrons in the other double bond?" - are you referring to conjugation? Wouldn't it be better to say that the electrons stabilize the entire molecule?

Yes, I'm referring to conjugation.  Better to say, I dunno.  Certainly the whole molecule is stabilized in the process, but the process involves one set of electrons interacting with another set of electrons and overall the electrons are more stable due to the interaction.


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Carboxylate "cage" concept
My professor emphasizes the concept of the carboxylate "cage" or the idea that electrons participate in resonance within a carboxylate group such as in the benzoate anion but the electrons in the carboxylate group cannot leave this "cage" - i.e. participate in resonance with the phenyl substituent.
Is this idea valid?
I've already poked a hole into it by pointing out that electron donation to the "cage" is possible - i.e. consider carbonic acid with one hydroxyl group replaced by oxygen connected to a phenyl group. Valid resonance structures can be drawn in which electron density from the oxygen in the ester linkage participates in resonance with the electrons "in" the carboxylate group. So donation into the "cage" is possible. Now, this begs the question - is withdrawal from the cage possible as well? He insists no. Now, he might be hiding some edge cases for pedagogical purposes, but I don't care - if there is an exception - I want to know about it (because chemistry always gets interesting at the exceptions).

Your drawing is not benzoate. Benzoate is the conjugate base of benzoic acid. You have drawn phenyl carbonic acid.

If I understand correctly, then I think your professor is correct.

within a carboxylate group such as in the benzoate anion

If it is phenyl carboxylate that we are discussing, then I think your professor is saying the same thing I wrote here.  Namely, that the carboxylate anion (O-C-O with a negative charge) is analogous to the allyl system, and with 4 pi electrons $\ce{\Psi_2}$ is the HOMO.  In the allyl system (and to a first approximation in this dioxallyl analogue) $\ce{\Psi_2}$ has a node (a "0" coefficient) at $\ce{C_2}$ (the central carbon atom).  If there truly is a zero or very small coefficient at $\ce{C_2}$, then the substituent connected to $\ce{C_2}$ (your benzene ring) is effectively disconnected from the system and cannot interact with the MO via resonance.
In your drawing up above, you have not drawn a carboxylate anion, rather you've drawn an un-ionized acid.  In this molecule there is no carboxylate anion, no dioxallyl system and no $\ce{\Psi_2}$ with a node at $\ce{C_2}$.  In this molecule, resonance interaction between the phenyl substituent and the carbonyl carbon is possible.

Oops. I'll have to revise this post. Good catch!

Let's start by deciding what compounds we're talking about here. You reference an example by your instructor of benzoate, which is the conjugate base of benzoic acid.

What your instructor is suggesting is correct for benzoate. The most important resonance structures for benzoate are the two equivalent structures that place negative charge on the oxygen atoms. It is also possible to draw a resonance structure where electrons are removed from the ring and both oxygens carry a negative charge.

The "carboxylate cage" essentially means that there is no way to delocalize the negative charge of the carboxylate into the aromatic ring. The only way to move electrons into the ring is from the C-O pi bond, which leads to (what I would consider) an unreasonable resonance structure, as it renders an oxygen both positively charged AND non-octet.

Carbonic acids are either a subset of carboxylic acids or a closely related functional group to carboxylic acids. Depending on what property you are interested in, they may be useful to compare or the difference might be too great. In this case, starting from the conjugate base of phenyl carbonic acid (as drawn in the question), there will be no way to delocalize the charge into the aromatic ring.


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Detecting the concentration of chlorpromazine
Is there a way to detect the concentration of chlorpromazine in a small sample ; ex: human saliva, using electro-chemical analysis? Does the sensor necessarily have to be a bulky piece of lab equipment or are there smaller sensors available for that kind of test?
Thanks in advance!

A cursory online search furnished

Graphene paste electrode for detection of chlorpromazine by Mohammad Hadi Parvin in Electrochemistry Communications, 2011, 13, 366-369 (DOI).
A multipurpose Voltammetric Sensor for the Determination of
Chlorpromazine in Presence of Acetaminophen, Uric Acid,
Dopamine and Ascorbic Acid by Binesh Unnikrishnan, Pei-Chi Hsu, and Shen-Ming Chen in Int. J. Electrochem. Sci., 2012, 7, 11414-11425 (PDF).

authors: MαπμQμαπkγVπ.0


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Is there a catalyst that will reduce an alcohol to an alkane?
There are many reduction methods of going from an alcohol to an alkane: http://www.transtutors.com/chemistry-homework-help/hydrocarbons/reduction-of-alcohols.aspx
And there are catalytic methods of adding hydrogen  gas to aldehydes and keystones to convert them into alcohols: http://jmcct.com/products-services/product_p404.html
However, I have not found a catalytic method of converting an alcohol into an alkane using a catalyst and possibly hydrogen. The fewer non-catalytic entities the better.

What about acid-catalyzed dehydration?

@Dissenter I would prefer to avoid reagents that are not recoverable because of cost considerations. That is why I asked for catalysts.

It's an acid-catalyzed reaction so the reagent (sulfuric acid) is preserved. @JoeHobbit.

What I meant is that you can catalyze the dehydration of the alcohol and form an alkene and then hydrogenate or otherwise reduce to an alkane.

@Dissenter That sounds terrific.

Here is a method for converting secondary and tertiary alcohols into the corresponding alkane using $\ce{InCl3}$ as a catalyst.  Chlorodiphenylsilane is the only reagent used in addition to the catalyst. The method works in the presence of common functional groups leaving them untouched.

Since primary alcohols are unreactive, this reaction can be used to selectively reduce secondary and tertiary alcohols in their presence.  An abstract showing a number of examples of the reaction can be found here.

This is a very limited situation, but benzylic alcohols can be reduced to the corresponding alkyl group with simple hydrogenolysis conditions, hydrogen gas with Pd/C. This is similar to the cleavage of benzyl ethers, a common protecting group. Looking at it this way, the reaction deprotects water. An example of this reaction that I ran long ago is depicted below.

Developing a more general method for this transformation is an ongoing field of research, with many methods published. That often means that there isn't one single best method (yet). A 2013 publication from Li and coworkers used an iridium complex with hydrazine as the terminal oxidant. Benzylic, allylic, and notably primary alcohols were successfully reduced, although the benzylic/allylic substrates worked much better. The introduction offers a brief description on the state of the field: Iridium-Catalyzed Direct Dehydroxylation of Alcohols

As a side note, even zinc dust can reduce phenols to benzenes.

The following not catalytic and does not use hydrogen but tin hydrides:

Convert the alcohol to a xanthate
Treat with $\ce{Bu3SnH}$ and AIBN

Yes, it's the Barton-McCombie reaction.
If you don't mind a bit of photochemistry, use the method described by Isao Saito (DOI):

Convert the alcohol to a m-trifluoromethylbenzoate
Photolyze (high pressure $\ce{Hg}$ lamp) in THF/water in the presence of N-methylcarbazole as an electron donor

Red P + HI is a very strong reducing agent. It would reduce your alcohol to an alkane, but it might reduce other functional groups present as well. See earlier question for details.


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Dibromomethane bond angles

With regard to the above professor's claim, I disagree. Bent's rule, which states that atomic p-character is concentrated in orbitals directed toward more electronegative substituents, implies that the $\ce{Br-C-Br}$ bond angle is going to be smaller than 109.5 degrees since the "$\ce{sp^3}$" hybrid orbitals involved in C-Br bonding have more p-character than 75%.
Therefore, the $\ce{H-C-H}$ bond angle should be bigger than 109.5 degrees.
Questions
1) Does Bent's rule work here (i.e. am I correct?)
2) Does Bent's rule always work?
Revision

HCH bond angle is ~110 degrees.
HCBr bond angle is ~108 degrees.
Br-C-Br bond angle is 112 degrees.

Rationales

vdW repulsions play a role in Br-C-Br bond angle expansion, although the amount of p-character in the C-Br bonds might be >75%.
HCH angle is expanded because of Bent's rule; s-character in the H-C bond exceeds 25%.
The expansion of the Br-C-Br bond angle and the expansion of the H-C-H bond angle squeezes together the remaining H-C-Br bond angle.

Here's the B3LYP/6-31G* geometry. On something like this, I'd suspect this is fairly reliable, although I'll try a few other methods in a second. It gives the Br-C-Br angle as a healthy 113.4 degrees, and H-C-H as 112.5, both in agreement with experiment.

I don't see how the Br-C-Br angle is ever going to be smaller than 109.5. It's the Br-C-H angles that shrink (107.7 degrees), since breaking the symmetry forces the molecule into a "squashed" tetrahedron (point group $D_{2d}$ for the squashed tetrahedron, $C_{2v}$ for $CH_2Br_2$ in particular)
So your questions:

1) Does Bent's rule work here (i.e. am I correct?)

Certainly not in the way you're thinking. As suggested in the other answer, the Br-Br repulsion opens that bond angle, and there's likely a Br-H electrostatic interaction shrinking that bond angle. (That's what I'd think.)

2) Does Bent's rule always work?

I typically consider VSEPR and Bent's rule to be very useful heuristics. Indeed, it's used in one of my favorite force field methods, UFF.
I'm holding back on "always" because chemistry is incredibly varied. A quick Google search of "Bent's rule fail" turns up this interesting nugget from Weinhold's book:

And yes, I'd personally consider these kinds of species to really be stretching the Bent's rule analogy. I think that's Weinhold's point.

"Certainly not in the way you're thinking. As suggested in the other answer, the Br-Br repulsion opens that bond angle, and there's likely a Br-H electrostatic interaction shrinking that bond angle. (That's what I'd think.)" - Right, after consulting the experimental data, I find that simply applying Bent's rule leads me astray. It seems that both me and my professor are incorrect here. Indeed, it is the H-C-Br bond angle that is smallest and Br-C-Br bond angle that is biggest. So it is indeed a combination of electronics and sterics at play here.

Also @Geoff Hutchinson - how do you suggest predicting or rationalizing bond angles? My professor likes to stick with Coulombic effects exclusively, dismissing Bent's rule as baloney. Although it's clear to me now that I can't apply Bent's rule blindly without any other consideration, Bent's rule does seem to have a fair bit of utility and seems to be another useful tool in the arsenal.

I really do not see, how $\ce{CH2Br2}$ can be of $D_{2d}$ symmetry. Or were you referring to a geometrical squashed tetrahedron having that symmetry?

@Martin I was referring to the general "squashed tetrahedron". I'll clarify that. The molecule, obviously, is not.

@Dissenter I may be in the minority, but I tend to run a quick calculation(!) Certainly as I said, I think Bent's rule is an excellent heuristic, but you should be careful in situations like this to figure out where the distortion will occur (i.e., the Br-C-H benefits from electrostatic interactions).

Hybridization occurs in response to a bonding interaction.  Bent's rule adds that the hybridization will occur in a direction that will lower the energy of the electrons in the system.  I can't imagine a case were a molecule will hybridize in a fashion to raise the energy of (all) the electrons in the system, so Bent's rule should always apply.
The water molecule has an H-O-H angle around 104.5 degrees; in the direction Bent's rule would suggest.  Bromine is much less electronegative than oxygen (Br = 2.96, O = 3.44, C = 2.55), so for dibromomethane we might expect a Br-C-Br bond angle larger than 104.5 and less than 109.5.  Then we would need to also factor in the steric repulsion between the two bulky bromines which would open the bond angle even more.
It turns out that experimental values for the H-C-H and Br-C-Br bond angles are all fairly close to the tetrahedral angle with the Br-C-Br angle around 112 degrees. (ref. 1, ref. 2).  It's hard to separate out the effects due to sterics and hybridization, but I could believe that the experimental values reflect a slightly sub-tetrahedral angle (due to Bent's rule) being opened up by sterics to the experimentally observed value.

would you say that the Cl in HCl is nearly unhybridized with respect to the H-Cl bond? I can't imagine why chlorine would want to use higher energy electrons to bond with the relatively low energy H 1s orbital.

The phosphorus atom in phosphine is nearly unhybridized, my guess would be that the chlorine atom in HCl would behave in a similar manner.  What do you mean by "higher energy electrons"? There is no total electron energy difference between all of the bonding electrons in a hybridized or unhybridized situation.

I mean why would the Cl want to use higher energy electrons to bond with the H (i.e. sp3 electrons rather than pure s electrons). Also regarding phosphine I was thinking that on my bike ride home; if atomic s-character is concentrated in orbitals directed toward electropositive substituents, does that mean that s-character is concentrated in the P orbital directed toward the lone pair, thereby closing down on the H-P-H bond angle because these have more p-character as a result?

sp3 is higher than pure s, but... 4 sp3's are not higher than s + 3 p's.  In HCl my guess is that Cl is using a p-orbital to bond with hydrogen. 2) yes, s-character is concentrated in orbitals holding lone pairs, when possible. 3) In PH3, the H-P-H angle is closed down because the molecule is unhybridized, maybe that is a result of Bent's rule, it is certainly consistent with the rule.

Why would chlorine use a p orbital to bond with hydrogen? Bent's rule suggests the opposite. H, being more EP than Cl, should have a lot of s-character directed at it with Cl. @ron

PH3 uses p orbitals to bond to H, that's why the H-P-H bond angle is ~90. P orbitals are used here for bonding because they are directional and provide much more effective overlap than an s orbital would; plus you save the s for use with a full lone pair rather than an unevenly shared bonding pair.  Guess I'm saying that a lone pair is more electropositive than a bonding pair (in most cases?).

Makes sense! Also @ron how would you define electropositivity?

The opposite of electronegativity (I'm being serious).

I (or I guess my prof) define EN as the ability to stabilize negative charge; the last impression I got from you was that this is a fluffy definition. @ron

I can't remember the context, it doesn't sound that bad to me now.  The word "stabilize" has a lot of meanings, maybe saying, "the ability to lower the energy of an electron" would be a skosh better.

I think I see what you're saying; there are man ways something can be "stabilized." I.e. I think we can agree that an more EP element isn't going to sterically stabilize positive charge! @ron

Yes, we can agree on that and much more.


text:
What is the IUPAC systematic name for methane?
The 1993 IUPAC Blue Book claims that "methane" is a semisystematic name:

Semisystematic names also exist, such as "methane", "propanol", and "benzoic acid", which are so familiar that few chemists realize that they are not fully systematic. They are retained, and indeed, in some cases there are no better systematic alternatives.

Propan-1-ol and phenylmethanoic acid are obvious systematic versions of propanol and benzoic acid. (Though I disagree that they are "better".)
What is the "fully systematic" name for methane—or is this a case where there is no "better systematic alternative"?

The latest edition of the Blue Book, the 2013 Revision, clarifies this:

Methane is a retained name (see P-12.3) that is preferred to the systematic name ‘carbane’, a name never recommended to replace methane, but used to derive the names ‘carbene’ and carbyne for the radicals $\ce{H2C^2-}$ and $\ce{HC^3-}$, respectively.

So it seems to be the case that the systematic IUPAC name is only there for consistency with the ions, and is not recommended (and not used) to describe methane. Similarly, "ethane", "propane", and "butane" are preferred to their systematic equivalents "dicarbane", "tricarbane", and "tetracarbane".

The carbane-style nomenclature is pretty neat, though borane is the only one I can think of that is commonly used (unless silane counts). It does give us another interesting name for water, however: oxidane.

Do you happen to know what page number or section number that's on? I can't seem to find it.

@MichaelD.M.Dryden -- If Wikipedia is any indication, both silane and germane are accepted names. I also know that Don Dixon's book Universe from 1981 listed "germane" as a constituent of Jupiter's atmosphere.


text:
What is Chevron with Techron fuel?
I've heard the commercials for this stuff for years, but what exactly does it do?  Does it do what the commercials say it does (clean out deposits in your tank)?  How well?
In other words, is this just an advertising campaign making a huge deal out of nothing, or is there actually an advantage to Chevron fuel?

I have read few post on other sites and most of them told it is crap. But few praised it. According to me it is just advertisement but let see what more experienced users who have knowledge about this think.

Gasoline is required by the gvt. to have a certain concentration or amount of detergent in them already.

Techron is a detergent based on PEA, polyeter amine. It is a complex synthetic molecule that will basically 'loosen' and 'solve' deposits in your fuel system. They then become part of the fuel-air mixture and burn in the combustion chamber with it. Be it in the carburettor, the injectors, the valves (for indirect injection vehicles) or the combustion chamber.
The Techron fuel you buy at the Chevron or Texaco stations has only a small amount of Techron added, the rest is just gas/petrol. You can also buy the Techron in seperate high concentrate bottles (typically 12oz.) These bottles contain up to 35% of PEA.
But the working ingrediënt PEA is just as well used in Gum-Out, Amsoil and a bunch of other fuel system cleaners.
So yes, it is a difference to fuel which doesn't have these detergent but every major gas producer (Shell, BP, Chevron, Texaco, ...) will add detergents. They ARE different detergents so they do make a difference.
In fact they are just auxiliary tools to keep your engine working well. It is like a band-aid. It does help tune up your engine, but it helps a lot more to do proper maintenance, change coolant and oil regularly, make sure your radiator works well so your engine operating temp isn't too high, clean your air filter and fuel filter regularly,...
Using these fuels does have an effect eg for emissions testing, but If you ask me. I just buy the cheapest fuel and I will add a bottle of High concentrate Techron or Gumout or another cleaner before I go to an emissions test (make sure you drive about 60 miles before turning up since the deposits have to be burnt before the test). And if you really want to use it, buy a bunch of high concentrate bottles and add them to supermarket gas, that's basically the same

clean out deposits in your tank

Hopefully, there won't be any deposits in your tank (unless you refueled during a sandstorm). As for deposits in your engine, yes it is likely that techron will clean out deposits. If you always use it, there will probably be very few deposits in the engine.

is there actually an advantage to Chevron fuel?

Compared to what? Nothing, yes. To other fuels? It's had to say since other companies also add cleaners to their fuel.

Nothing here actually explains what it is. Presumably, it's related to the common injector/carburetor cleaner fuel-additives.

The Wikipedia article on Techron suggests the additive is polyetheramine. I suspect there's probably more, but this is confirmed by the Techron website.
I'm sure there are probably other things, but yes, it's in other additives as well:
NIH Household Products Search


text:
Significance of 4π in the uncertainty principle
What is the significance of the quantity 4π in the Heisenberg uncertainty principle?
$Δp \times Δx = \frac{h}{4π}$
The focus of the question is to learn why the ratio of a circle's circumference to its diameter, universally known as $π$, is used in the denominator in the term $\frac{h}{4π}$. To put it another way, why is this geometric number relevant in Heisenberg's formula?

Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site.

Related Phys.SE posts: http://physics.stackexchange.com/q/69604/2451 and links therein.

Aha! Now it is clear what are you asking for. For de Broglie plane waves $\pi$ appears naturally in a connection between $x$ and $p$ as I tried to show at the very end of my answer. For a general case connection between $x$ and $p$ is non-trivial, but it still involves $\pi$, as it is postulated by canonical commutation relation.

The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" used. Without rigid definition of this quantity one ofthen just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter)
$$ \Delta x \Delta p \sim \hbar \, .$$
To make the statement more precise, one have to define what is actually meant by "uncertainty" and usually uncertainties are defined as the standard deviations.
$$
\Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\
\Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\
$$
where angle brackets $\langle \phantom{x} \rangle$ stand for the average value. And with these definitions one can indeed prove that
$$
\Delta x \Delta p \geq \hbar / 2 \, .
$$
Actually, we do even favour the usual notation for the standard deviation to avoid ambiguity
$$
\sigma_x \sigma_p \geq \hbar / 2 \, .
$$
So, from this perspective the 2 is nothing but the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the reduced Planck constant. $4 \pi$ is the factor of proportionality between the product of uncertainties in position and momentum being thought of as standard deviations and the Planck constant itself.

In fact, you can prove a more general Robertson uncertainty relation that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$
$$
    \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|.
$$
Now, for $x$ and $p$ it is postulated by so-called canonical commutation relation that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to
$$
    \sigma_x \sigma_p \geq \hbar / 2 \, .
$$
From that perspective, the Heisenberg uncertainty relation is equivalent to canonical commutation relation, and $\pi$ actually comes to the first one from the last one (as part of $\hbar$). So now the question really is why do we have $\pi$ in canonical commutation relation? And as user3371583 noted, it indeed has something to do with Fourier transform, but the connection of position and momentum through Fourier transform is not trivial.

Intuitively, however, you can get a sense of it by looking at the simpliest case of a one-dimensional de Broglie plane wave, in which $\pi$ factor in a connection between $x$ and $p$ naturally comes from de Broglie relation
$$
    p = \hbar k \, ,
$$
where $p$ is the momentum and $k = 2 \pi / \lambda$ is the wavenumber. So, wave function for plane wave can be written as follows
$$
    \psi(x) = \mathrm e^{\mathrm i(kx-\omega t)} \, ,
$$
and its derivative with respect to $x$ is
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i k \mathrm e^{\mathrm i(kx-\omega t)} = \mathrm i k \psi(x) \, .
$$
If we substitute then $p / \hbar$ for $k$ from de Broglie relation we get
$$
    \frac{\partial \psi(x)}{\partial x} = \mathrm i \frac{p}{\hbar} \psi(x) \, ,
$$
or
$$
    p \psi(x) = - \mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, ,
$$
which suggest that
$$
    \hat{p} \psi(x) = -\mathrm i \hbar \frac{\partial \psi(x)}{\partial x} \, .
$$
$\hat{x}$ is simple multiplicative operator in this picture
$$
    \hat{x} \psi(x) = x \psi(x) \, ,
$$
so, the commutator between $\hat{x}$ and $\hat{p}$ is indeed $[\hat{x},\hat{p}]=i\hbar$ and through this commutator $\pi$, in a sense, propogates to Heisenberg uncetainty relation.

I am puzzled by the answer that has been posted. Rather than answering my question directly, it introduces a new dimension to my question which, although it may be correct, is not relevant to my specific question. To clarify this commenmt and my original question, I want to know what 4π  means as the divisor in Heisenberg's formula.

@TheStarVoyager $4 \pi$ is just a factor of proportionality. What kind of its significance beyound that you are looking for?

It seems that you are looking for a more mathematical answer.
Actually Heisenberg's inequality naturally comes from using Fourier Analysis. $2\pi$ comes from deriving a Fourier transform and the $2$ that remains comes from playing around with the terms.
For a detailed proof of the Heisenberg inequality that explains every term, you can check GB Folland's Lectures on Partial Differential Equations (pdf) http://www.math.tifr.res.in/~publ/ln/tifr70.pdf and go to page 11.

I will try to address your question very directly:

why is this geometric number relevant in Heisenberg's formula?

$\pi$ is not a geometric number. Among many things, $\pi$ the ratio of a circle's circumference to its diameter, but it is not its deepest meaning. That is, this geometrical relation can be viewed as a consequence of most fundamental properties of $\pi$. The inclusion of $\pi$ in the uncertainty relationships is due to those other more fundamental properties, and it is not a direct consequence the geometric properties mentioned by you.
$\pi$ arises naturally (after imposing some desirable restrictions) in the solution of the differential equation
$$f(z) = \frac{df(z)}{dz}$$
with $f:\mathbb{C} \rightarrow \mathbb{C}$ (and clearly $z \in \mathbb{C}$)  which is found everywhere in physics, chemistry, etc.
In order to get a more detailed (case specific) answer of how it arises, please take a look at the Wildcat answer.
As I understand it, that is the deepest reason why $\pi$ is found in the uncertainty principle equation.


text:
What kind of salts may I add in my GPC Eluent?
What kind of salts may I add in my GPC Eluent?


text:
ATP break down and energy release?
I have a very basic question. Energy is absorbed to break bonds. Bond-breaking is an endothermic process. Energy is released when new bonds form. Bond-making is an exothermic process.
But ATP breaks down to ADP and releases energy right? Is there anything wrong in my understanding?

A bond is formed between the oxygen of water and the phosphorus of the gamma-phosphate.  Here is a good link.

Bonds are both broken and made in chemical reactions but many biology teachers and textbooks state that "Breaking ATP bonds releases energy."  In reactions bonds are broken and made.  If the strength of the bonds formed exceeds the strength of the bonds broken the reaction is exothermic. Under physiological conditions (with respect to pH and magnesium concentration), the reaction is exergonic as well (i.e. negative Gibbs free energy of reaction).

I was confused about this for a long time as a student, and so were many of my classmates. Goes to show how slightly imprecise language can be problematic. Once I realized the $\rm{P-O-P}$ bonds weren't broken (bond homolysis or heterolysis) so much as hydrolyzed, it all made sense. Hydrolysis reactions can release quite a bit of energy.

@brinnb :  "If the strength of the broken bonds is exceeded by the strength of the bonds formed the reaction is exothermic." That is a great point.

Shouldn't it be If the strength of the bonds formed is exceeded by the strength of the broken bonds the reaction is exothermic?

The previous answers are not very good, so I feel compelled to clarify:
ATP is not broken in a vacuum, but hydrolyzed by water, which is disrespected and not shown in the reaction pathway, but is hugely important.  The overall free energy of delta G remember is not just the enthalpy of the reaction (heat given off or absorbed) but also composed of an entropy which is the ordering or disordering of the surrounding solvent water molecules.  It takes far more water molecules to be ordered around the ATP gamma phosphate once it is released in the ADP + P form than in the ATP form when still attached.  Energy is always required to break bonds and always is released when bonds are formed, even with ATP, but the part that is omitted is that more molecules of water are bound and ordered after the ATP hydrolysis than before and so more ordering plus many more highly energetic hydrogen bonds formed than broken equals a large overall release of energy for the overall process.  Don't forget the water!

The main contributor to the Gibbs energy of reaction is the enthalpy of reaction, see e.g. https://www.sciencedirect.com/science/article/pii/S0005272897000790. The details of the energetics of the reaction are complex, and require careful analysis of the role of magnesium ion binding and acid/base reactions as part of the reaction, see e.g. https://www.jbc.org/content/244/12/3330.full.pdf


text:
Relative abundance of the isotopes of iridium
There are two stable isotopes of iridium, $\ce{^191Ir}$ and $\ce{^193Ir}$. If you randomly pick an iridium atom from a large collection of iridium atoms, which isotope are you more likely to select?

It depends on which isotope is more abundant. There should be tabulated values of these; a quick internet search will throw them up.

even in the exams right ?

Clearly not, but you'd expect such information to be given to you.

An urn contains nothing but

63 white balls and
37 black balls.

You grab into the urn, pick one item, look at it and put it back.
What is the probability that you drew:

a ball
a white ball
a black ball
a red ball?

100% 2)63% 3)37% 4)0%

@shaima Exactly :) Amazingly, 63% and 37% are roughly the abundancies of the two natural iridium isotopes: $\ce{^{193}Ir}$ and $\ce{^{191}Ir}$


text:
Proton affinities in machine readable format?
Short: How to get proton affinities for a large list of chemicals? A database? A programming API?
Longer:
I have ca. 2000 chemicals (downloaded from MCM, mostly organics that occur in the atmosphere and their reaction products). I would like to try to get proton affinities for as many of the 2000 chemicals as possible.
I know how to look up the proton affinity for a chemical in the NIST Chemistry WebBook (for example isoprene). But I would prefer to get a machine readable database, or a Python API (or any programming language), to get these nicely in a program code.
What options do I have?
(I am not going to get 2000 values by hand, so my last option would be to write a small web scraper to make these 2000 web queries)

To my knowledge, a (Python) API for the NIST chemistry WebBook does not exist.
Rather than filling out the Search for Species Data by CAS Registry Number by hand, you might want to use mechanize.
But since you're only interested in Gas phase ion energetics data anyway, you might want to construct the relevant URLs directly.
Isoprene:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C78795&Units=SI&Mask=20#Ion-Energetics
Styrene:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C100425&Units=SI&Mask=20#Ion-Energetics
The parameter pattern is obvious: ID is the CAS number of the compound minus the dashes, prefixed by C.

What is mechanize?

@SampoSmolander Mechanize is a Python module to emulate a browser and fill out forms, even the cgi stuff on the NISTwebBook sites. But you're probably better off to construct the URLs based on the CAS numbers, use urllib2 or requests to pull the data and process it with BeautifulSoup. A decent API would make our lives much easier!

Ah ok, 'mechanize' is a Python thing, not a chemistry thing.

Anyway, I am not limited to the NIST web data. If there would exist any other machine readable database containing this information, I could also use that. NIST is just the only thing (besides 19 pages in PDF from CRC Handbook) I found.


text:
Does the iodoform reaction form hydrogen iodide or water?
I am currently learning about the iodoform reaction, i.e. when CH3COR compounds react with iodine and sodium hydroxide to form iodoform.
My course book and Chemguide both says that the reaction is as follows:
$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$
However, this picture from Wikipedia suggests that hydrogen iodide is formed instead of water:

Which one is correct?

Your balanced equation is correct, both water and iodide (and iodide in water will be in equilibrium with hydrogen iodide) are formed in the iodoform reaction.
$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$
The Wikipedia article that you mention, could have been written a bit clearer.  Specifically, there are two key steps in the iodoform reaction.  First, there is formation of the tri-iodo ketone and water is generated in this step.

In a subsequent step, the carbonyl undergoes an SN2 reaction to generate iodoform.

The Wikipedia drawing in your post seems to focus on the mechanism of the second step (3 to 4b).  The first step (2 to 3) is just illustrated in a cursory fashion and the formation of water is not included in the figure.

Do you mean that the reaction does not directly produce hydrogen iodide (as Wikipedia suggested), but produces iodide ion which forms an equilibrium with water and hydrogen iodide? So, is it wrong to say $ \mathrm{CH_3COR + 3I_2 +OH^- \rightarrow  CHI_3 + RCOO^- +3HI}$?

@ace "So, is it wrong to say..." Well technically yes because 1) you don't show the water that is formed and 2) you show HI which is really formed indirectly - from the iodide that is formed first.

Also remember that the reaction takes place in strongly basic solution. Any HI, a very strong acid, that may be formed will not stay in the form very long.


text:
Why does the pH of this solution rise instead of fall when exposed to the atmosphere?
I have a couple of big barrels with a $50\,\mathrm{L}$ demi-water solution of the following nutrients:
$$
\begin{array}{rl}
200~\mathrm{\mu M} & \ce{NH4NO3} & +\\
30~\mathrm{\mu M} & \ce{KCl} & +\\
10~\mathrm{\mu M} & \ce{CaCl2} & +\\
10~\mathrm{\mu M} & \ce{KH2PO4} & +\\
\\
10~\mathrm{\mu M} & \ce{Fe-EDTA} & +\\
\\
0.7~\mathrm{\mu M} & \ce{ZnSO4} & +\\
0.8~\mathrm{\mu M} & \ce{MnCl2} & +\\
0.2~\mathrm{\mu M} & \ce{CuSO4} & +\\
0.8~\mathrm{\mu M} & \ce{H3BO4} & +\\
0.008~\mathrm{\mu M} & \ce{(NH4)6Mo7O24}\\
\end{array}
$$
After mixing those nutrients, I adjust the $\mathrm{pH}$ of the barrels to $5.5$ by carefully adding $\ce{HCl}$ and continuing mixing until the $\mathrm{pH}$ stabilizes at $5.5$.
Then, when I measure the $\mathrm{pH}$in the barrels again after four days, I notice that it has risen to about $5.8$-$5.9$. This is contrary to my expectation. I'd have expected $\ce{CO2}$ from the air to be dissolved in the solution, thus lowering the $\mathrm{pH}$ rather than raising it.
Could someone explain to me what could be happening here?

Could the barrel itself be reacting with the contents of the solution? I can see a small amount of metal/oxide corrosion happening at the expense of some hydrogen ions. Unbuffered solution pH is very sensitive to small variations of acid/base near pH 7, too.

Another scenario that I didn't consider at the time is that I might not have mixed the barrels vigorously enough while adding the HCl or that I might not have given the mixture enough time to settle, and since I always stuck the pH meter in the upper layer… Also, later, this stopped happening so much, which is why I think the fault must have been with the observer (me).

At that pH, you're well outside the buffering capacity of both phosphate and boric acid, so a $\mathrm{pH}$ rise from $5.5$ to $5.9$ corresponds to a net increase in base of about $2\ \mathrm{\mu M}$.  That implies to me that it's likely something going on with one of the first five ingredients in your recipe, or the $\ce{HCl}$ added to adjust the $\mathrm{pH}$.
I can think of two scenarios:

Similar to Brinn's answer, perhaps a small amount of $\ce{HCl}$ has evaporated out of solution.
If your $\ce{Fe-EDTA}$ was added as $\ce{Fe(II)-EDTA}$, oxidation to $\ce{Fe(III)-EDTA}$ could explain the pH rise, by reducing water to hydrogen gas and hydroxide ion:

$2\ \ce{Fe(II)-EDTA} \rightarrow 2\ \ce{Fe(III)-EDTA} + 2\ e^-$
$2\ \ce{H2O} + 2\ e^- \rightarrow \ce{H2} + 2\ \ce{OH-}$

I don't know offhand how thermodynamically or kinetically feasible #2 is.  With all of the other metals present in solution, though, it's not unreasonable to imagine that some sort of catalysis might be possible.

Although I think the problem might have simply been in my impatient mixing technique, I'm accepting your answer, because it's beautiful.

Unless the water to make the solution was freshly boiled then it was probably saturated with carbon dioxide to begin with.  It's likely that ammonia is being lost to the atmosphere.

Mmmm, not at pH 5.5, ammonia evaporation over four days should be negligible.  Anyways, the pH would still be going down: ammonium would have to deprotonate before escaping as molecular ammonia.


text:
Here are four observations of the reactions of four metals
Here are four observations of the reactions of four metals.
Metal F and G react slowly with water
Metal H displaces metal G from its chloride salt solution
Metal E does not react with dilute acids
Metal H does NOT displace metal F from its sulphate salt solution.
What is their order of reactivity, starting from the most reactive metal to the least reactive metal?
F>H>G>E;  H>G>F>E;  H>E>F>G;  G>H>F>E;  None of these
I'm really unsure how to order them     I would say F>H>G>E but thats half guessing.
Can someone give me a clue to arrange them right?

Before starting of with my reasoning , I must tell you that your answer ( whether guessed or logically reasoned ) looks correct according to me .
I am assuming that you are familiar with the Metal Reactivity Series ( if not look here )
We shall take use of a well presented reactivity series :

(source: wikispaces.com)
Now to crack the answer clue by clue :
1.

Metal F and G react slowly with water.

Hence F and G are somewhere at the top.
2.

Metal H displaces metal G from its chloride salt solution

Clearly H>G.
3.

Metal E does not react with dilute acids

E will come somewhere at the bottom half so F,G>E (from 1) and we can further say H>G>E (from 2)
4.

Metal H does NOT displace metal F from its sulphate salt solution

This gives F>H.
Combine last two clues' result to get F>H>G>E .


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Approach to model solvent in a lattice: e.g. H2O in NaCl
If a lattice contains a solvent, lets say $\ce{NaCl}$ and water of crystallisation, what is the condition for a solvent molecule to change it's location?
From my current point of view I assume that such a water molecule is bound to a lattice ion by ion-dipole forces. It vibrates at that location until, randomly, the particle energy exceeds the binding energy. Then it goes to a new location, in the direction that minimizes Gibbs free energy.
How is the water molecule energy distributed? Currently I use Maxwell-Boltzmann distribution. However, Maxwell-Boltzmann is for non-interacting particles and therefore not quite a good fit for molecules that interact with a ion. Further, my chemistry book suggest that ion-dipole bonds have a binding energy of >$50~\mathrm{kJ~mol^{-1}}$, which will in my calculations rarely allow a jump at all.
I ask this question because I'm working on a particle-drying simulation. I intend to use a Random Walk simulation to obtain diffusion coefficients under special circumstances: e.g. if the lattice salvation behaviour is somehow weird. The model must not be physically correct in all details. Even bigger "model" error will most likely be small in the simulation results.
Note: I asked this question already here but have not received an answer. As suggested there, I open it in Chem.SE.

This is a bit outside my realm of expertise - you'd want to look at molecular dynamics (MD) simulations for more details. Let me take your question in a few parts.
1) A water molecule in a lattice is likely bound by electrostatic (and hydrogen-bond) interactions. So your general assumption is a good one.
2) MD simulations usually have a "thermostat" depending on the type of ensemble used. Your thermal energy will be distributed into different degrees of freedom, such as rotations, translations, and vibrations. So I'd work out the amount of translational free energy from statistical mechanics, and then use the electrostatic interactions to determine the path for jumps. (Your main question here seems like the activation barrier, since each lattice site should have the same energy, right?)
3) If the moves are rare, you might want to look into "rare event sampling" methods, which preserve the correct statistics, but allow you to more efficiently run your model.
Many of these techniques are already implemented in simulation codes. You might want to look at something like LAMMPS which can handle different levels of particles (e.g., atoms, molecules, crystal domains, etc.)
Hope that helps - I can only talk in broad generalities here.

Thank you for your answer. I will consider LAMMPS as a next step when I made some experience with by own approach.


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Simple experiments involving enantiomer synthesis
I want to do an experiment to investigate factors affecting the synthesis of chiral molecules. The only problem is, my institution doesn't have much money, and it seems like the chemicals required for any experiment involving chiral molecules are very expensive!
My initial research found an experiment based on the Bürgi-Dunitz trajectory (see Addition to Carbonyl Double Bonds), but with the use of L-Selectride it isn't the cheapest. I then researched the Diels-Alder reaction, but the conjugated dienes are also relatively expensive, or are only available in large quantities!
I am only an amateur in the field of chemistry, and much of the interesting material is very hard to find unless one knows the specific terms to use. Now I come to you for help - are there any cheap experiments involving the synthesis of enantiomers? Other than cost, the constraints are:

Equipment: standard laboratory equipment is available, but nothing too fancy
Aim: the experiment needs to investigate something, for example factors affecting the synthesis of L and D enantiomers. Two different chiral auxilliaries or catalysts would do well here.

Are you trying to run a reaction that generates a single enantiomer of a compound? If so, you also need to think about how you are going to measure the ratio of enantiomers.

@Xenopathic Does your institution have a polarimeter?  Is the experiment for you or for students, if the latter, what grade level?

@ron Yes, a polarimeter is available. The experiment is for me, at A-level (equivalent to APs in the US)

If you have some way of measuring the enantiomeric excess of the product mixture, I suggest trying proline catalysis of an aldol reaction. The materials are cheap / readily available, and the reaction setup is straightforward. Proline is probably the cheapest chiral catalyst available.
Here is a procedure for an organic chemistry lab run at Rutgers that might get you started:
http://chem.rutgers.edu/sites/default/files/coursefiles/Courses_f09/310/Procedure_Proline_Catalysis.pdf

After lots more research, this seems to fit the requirements perfectly!


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Electron Transfer in Organic Reactions: Why Does Brown Iron Oxide Improve Burn Rate of Rocket Candy?
On wikipedia, this statement exists about rocket propellant aditives:

Metal oxides have been found to increase the burn rate of sugar propellants. [...] Most often used are iron oxides. Red iron oxide is used most often as it is somewhat easier to obtain than the yellow, brown, or black versions. Brown iron oxide exhibits unusual burn rate acceleration properties under pressure.

I have no idea how would already oxidized compound increase burn rate. I have failed to research which compound is brown iron oxide. All I could find was red iron oxide, iron monoxide and magnetite.
So how do oxides help combustion generally? And what makes brown iron oxide so excellent for that purpose?
I would appreciate help with adding proper tags to this question. Couldn't think of any.

First of all, you're thinking of one of the two definitions of oxidation—breaking bonds in molecular oxygen and rejoining them in a different way in another molecule.
We need the electron transfer definition of oxidation however.
First off, to see what is happening we need to look at how many electrons are being transferred. In glucose the oxidation state of carbon is $\ce{0}$, whilst in carbon dioxide it is $\ce{+4}$. So during our rocket burning process, each carbon gives up four electrons. But what is it giving them to?
If we use molecular oxygen as our accepting species—our oxidisation agent—we have the following:
$$ \ce {6O_2} + \ce {C6H12O6}  \ce{->} \ce{6CO2} + \ce{6H2O} $$
With oxygen here going from an oxidation state of $\ce{0}$ in $\ce{O_2}$ to $\ce{-2}$ in both $\ce{H2O}$ and $\ce{CO2}$.
So if each $\ce{O2}$ molecule  is picking off two electrons and this is all it's doing, then why can't we replace this with two lots of $\ce{Fe^3+}$ to form $\ce{Fe^2+}$?
Before I write this in equation form, it's worth noting that organic electrochemical equations approximate organic matter (OM) with formaldehyde, $\ce{CH2O}$ since OM is biologically equivalent to a polymer (albeit complex ones!) of $\ce{CH2O}$. Note that glucose is therefore equivalent to six lots of this monomer. This simplifies writing balanced chemical equations a lot:
$$ \ce{2Fe_2O_3} + \ce {CH2O}  \ce{->} 4\ce{FeO} + \ce{H2O} + \ce{CO2} $$
Now molecular oxygen is by far a better oxidisation agent than $\ce{Fe_2O_3}$ however in fine powdered form, these iron oxide molecules are in immediate contact with your OM; we do not need to wait for molecular oxygen to be used up and for more to diffuse in.
So I'm warranting that red iron oxide ($\ce{Fe_2O_3}$) can be used as well as brown iron oxide (I think this means $\ce{FeO}$). The difference here is that in the former reaction each $\ce{Fe^3+}$ only captures one electron from your OM to form $\ce{Fe^2+}$, whilst in the latter case you're moving from $\ce{Fe^2+}$ to $\ce{Fe}$ metal which captures two electrons—so you need less. Note that formation of iron from $\ce{Fe^2+}$ will need really high temperatures or pressures, which is what you note in your question.
This is a super question in terms of getting people thinking about redox not just as a process which happens with ions, but with entire organic species. Such organic reactions can be characterised by a reaction voltage, just like the other types of electrochemical reaction you've been looking at. This is actually what happens in the field of biogeochemistry, where scientists look at which types of oxidisation agent are responsible for degrading OM in different environments. More can be found about this in this paper here.


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Carbon atoms at the edge of a diamond
It is well known (the simplest textbook example) that a diamond has a well-defined arrangement of sp3 carbon atoms, as each atom is connected to four others in a tetrahedral structure.
But what about the last carbon atoms at the edge? For each of these, three bonds are missing; it has a single bond with one carbon atom inside the structure. What is the hybrid and atomic (electron) structure of the carbon atoms at the edge?
I would appreciate some references explaining this in detail (books or research articles).

This question has already been addressed here: http://chemistry.stackexchange.com/questions/11151/formal-charge-of-terminal-atoms

A diamond is a three-dimensional structure, so there is no 'edge' as in a two-dimensional plane structure. There may be dangling bonds across the surface however, and such surface chemistry bond relationships are the subject of much interesting research.

Atoms at the edge of a crystal that have an unsatisfied valence are said to have "dangling bonds."  Many elements, in addition to carbon, can have dangling bonds.  Dangling bonds is a subject of current interest because of the impact these structures can have on semiconductor properties.
These dangling bonds are very similar to free radicals, except since they are immobilized in a solid, they are somewhat less reactive than free radicals in solution. Nonetheless, they can react with whatever materials they are exposed to, such as hydrogen, water vapor, oxygen, etc. In addition, if there is a neighboring dangling bond then they can both react with one another to form a bond and satisfy there valence.
When carbon or silicon surfaces are prepared under clean room conditions, the dangling bonds can persist.  In the semiconductor industry this clean room preparation technique is followed by bringing in a doping gas in order to purposefully alter the electronic band structure of the substrate material.
Since unpaired electrons have magnetic properties, in carbon (or any other element) nanostructures where there is a lot more surface area to volume, the concentration of dangling bonds is much higher.  Consequently, the unpaired electrons in the dangling bonds confer magnetic properties on these materials that are large enough to be easily  detectable and to manipulate.
Finally, since dangling bonds represent a non-equilibrium situation, surfaces containing dangling bonds undergo a relaxation or reshaping that is referred to as "surface reconstruction."
Here is a full-text article that should help you get started:
1) Structure of the diamond 111 surface: Single-dangling-bond versus triple-dangling-bond face

You asked a question, belonging to surface chemistry. It is a relatively new area of research, as it relies heavily on atomic-resolved microscopy and computational methods.
Generally, the answer depends on prehistory of the surface and its environment. In case you crack a diamond, making new surface, two processes happens.

so-known reconstruction, or reorganization of bonds on the surfaces. I'm not aware of articles on reconstruction of diamond, but silicon has diamond-like structure, so we can speculate from there.
reaction of the surface with environment. In case of air, oxygen and water may insert into strained bonds, and react from there. It may result in all kinds of surface species up to carboxy-groups, ketone fragments, alkyl fragments and so on.

Surface study of diamonds is about carbon atoms, but according to the second case, it should be covered by external molecules.

To add to the current answers about the individual electrons, there is also a global effect at the lattice level.
Assuming you have a monocrystal with no (or low density of) defects, the lattice will be regular until around 100 atoms from the surface. The reason is obvious: the atoms at the surface have neighbours only at one side, so they have to be displaced to balance the forces, which in turn forces a smaller displacement on the second layer, and so on.
All this has powerful implications in the study of nanostructures, like nanotechnology, as well as suspensions and colloids; as they are structures that are all close to the boundary, with no proper bulk. Also,they are fundamental in crystal growth, and, in Chemistry in catalystic reactions.
Further reading:
Note: they are excelent classical books, but they are old, and there has been quite a lot of research in surfaces in the late year. Unfortunately, I am not aware of any modern reference.

Solid State Physics, Ashcroft and Mermin.
Introduction to Solid State Physics, Charles Kittel (I'd avoid the parts of introduction to SS, as I find them extremely boring, but some other people like it).

More modern sources are The Oxford Solid State Basics and Solid Stat4 Physics. Each replaces respectively Kittel and Aschcroft. There are many people that like the book of Marder but I find it nor easy nor rigorous.


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Why is sugar used in low-cost homemade rockets?
I was thinking about making a "rocket candy". It's relatively inexpensive rocket based on sugar and some oxidiser, mostly $\ce{KNO3}$.
The fuel is usually formed by heating the mix of both compounds so that the sugar sticks onto the potassium nitrate. They can also stick together by being dissolved and dried.
My question is, what makes sugar the best fuel? Paraffin or resin can be melted more easily and the oxidizer could be poured on them while they are hot. To me, that seems like a much better idea.
But there were many people before me and they have surely tried many things. So, what is the reason the sugar is more advised than paraffin or resin?

the best fuel is actually sorbitol. It is a bit costlier than sugar, but still cheap. It melts easily without decomposition and can dissolve some amount of potassium nitrate. Common sugar decomposes before melting.
Paraffin has one bad habit: it quite easily vaporizes. It is bad, because it makes mixture much harder to ignite and it becomes prone to self-extinguishing because of it. Many polymers also easily decompose producing much gases. Paraffin is used in some specific formulations with more active oxidizers or as a non-fuel additive.
Black powder is actually quite common, but it has very high combustion rate and is prone to explosive combustion, so it is not recommended for beginners. It also requires ball mill to make, and constructing one is time consuming.
Added: The main advantages of rocket candy are readily available components, simplicity of preparation (no special equipment required), stable burn and safety in comparison with many other compositions.

And what do you think about resin? Black powder is definitelly too dangerous.

@TomášZato It has same flaws as paraffin. It contains some easily vaporizable fractions. There are some special compositions, that use paraffin or some sorts of resin or polymer binders, but they use special fuel or oxidizer or both to maintain stable burn.

Sugar (all sugars) have one drawback: Their standard enthalpy of combustion is about one third (16.5kJ/g) compared to paraffin (e.g. octane 48 kJ/g). There's a reason why sugar is cheaper than oil. ;-)


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How to interpret the MSDS warning?
Just a quick question; I was looking up the MSDS for o-Toluic Acid for an experiment when I came across this in the "Hazards Identification" section.
Very hazardous in case of skin contact (irritant), ....
...
Slightly hazardous in case of skin contact (permeator).

How should this warning be interpreted?

I'd interpret it as "wear gloves" ;)

@canadianer As a computational chemist myself, I interpret it as "deadly".

An irritant is a chemical, which is not corrosive, but which causes a reversible inflammatory effect on living tissue by chemical action at the site of contact. There are specific lab tests to determine the ranking of an irritant, but something which is rated as very hazardous is likely to cause a very severe reaction upon exposure.
A permeator is a chemical which can pass through the outer protective dermal layers and into the body, and can either expose the body to toxic effects of that chemical OR act as a carrier for other toxic/hazardous chemicals.
Other common classes of hazardous chemicals include corrosive and sensitizer.
o-Toluic acid is a strong irritant, but moderate permeator. Depending on your level of experience in handling chemicals, simply getting the advice to 'wear gloves' may leave you a little underprepared. You should take this to mean that you should not allow any skin contact; full length sleeves (lab coat), eye protection and THE CORRECT TYPE of gloves. Using the incorrect type of gloves is worse than using no gloves at all, as it gives you a false sense of security. Select the correct type of gloves using a suitable selection tool or guide, such as this one from Ansell.
As a strong irritant, you should also avoid breathing any vapours or fumes, so it means handling this chemical in a suitably ventilated area.
AS with any risk assessment, it also means that before you handle the chemical you have an understanding of the response procedures in the event of things going wrong. That is, from your MSDS, you need to know Section 4:First Aid Measures prior to any procedure you undertake. It is no good having to find and read the MSDS after you have splashed yourself with this stuff.

Thanks! I just wasn't sure because of the contradictory wording between the two lines

As canadianer said, wear gloves.
Irritant means just as it sounds, it will irritate your skin (ie: itching, rashes, strange sensations, ect.)
Permeator means that it can pass through the skin.


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Salt concentration and electrical permittivity of water
We know that the electric flux density $D$ can be related to electric field intensity $E$ as
$$D = \epsilon(\omega) E$$
where
$$\epsilon(\omega) = \epsilon_0 \epsilon_r(\omega)$$
and $\epsilon_0$ is the permittivity of vacuum and $\epsilon_r(\omega)$ is the relative permittivity of the material.
At the low-frequency limit, I would like to know how the value of $\epsilon_r (\omega)$ of water changes with temperature and salt ($\ce{NaCl}$) concentration. Please explain,

the rationale of the variation (from quantum mechanics or statistical mechanics etc)
observed trend holds good for any liquid or only for highly polar solvents
any references that you know of which talks in detail about these variations

The relative electric permittivity of water at $20^oC$ under the effect of a static or low frequency is $\epsilon_r\approx80$. Hasted (1948) conducted an experimental study on the dielectric properties of salt-water solutions and observed a dielectric decrement with salt concentration. That is, the addition of sodium chloride to water results in a drop in electrical permittivity.

Source: Hasted, Ritson and Collie (1948).
This can be understood by examining the following diagram.

The sodium and chloride ions dissociate in solution, resulting in an an electric field between them (shown in bold line). The water molecules, which are polarised, get oriented under the effect of the electric field so that their oxygen atoms (carrying a partial negative charge) face toward the sodium ion and their hydrogens (carrying partial positive charges) face toward the chloride ion. This orientation of the polar water molecules causes its own electric field (dashed lines), which cancels out most of the electric field that would exist if the ions were in a vacuum. The end result is that the sodium and chloride ions are effectively 'sheilded' by a 'hydration shell' of polarized water molecules which effectively lowers their to the applied external field and hence lowering the dielectric constant (electric permittivity) as the ionic concentration increases ().
In dilute solution (less than 2.0M) the dielectric decrement is linear. At higher salt concentrations, the dielectric decrement is observed to saturate at about $\epsilon_r=45$. The effect of sodium chloride concentration on electric permittivity is shown in the graph below at various temperatures.

Source: Gavish and Promislow (2012).
The above graph indicates that for a given salt concentration below about 3M, the electric permittivity decreases with temperature.
REFERENCES:
Hasted, J.B., Ritson, D.M., Collie, C.H., "Dielectric Properties of Aqueous Ionic Solutions. Parts I and II." Journal of Chemical Physics 16, 1 (1948). http://dx.doi.org/10.1063/1.1746645
Gavish, N., Promislow, K, "Dependence of the dielectric constant of electrolyte solutions on ionic concentration", http://arxiv.org/pdf/1208.5169.pdf

Excellent answer. Just to clarify, this decrease is mainly due to the polarization of water molecules. If i use a non polar solvent can i expect electrical  permittivity to NOT change as a function of concentration?

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


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Which is the densest gas known?
Which is the densest gas known to exist under normal "shirt sleeve" conditions: room temperature, one atmosphere, won't explode on contact with air, won't kill everyone in the building in trace amounts, etc.

may be this will help you

See also https://en.wikipedia.org/wiki/Tungsten_hexafluoride?wprov=sfla1

Perfluorobutane is inert and has almost twice the density of sulfur hexafluoride. It is non-toxic enough that it is used in fire extinguishers and injected as a contrast agent for ultrasound. Boiling point: $-1.7\ \mathrm{^\circ C}$.
Perfluoropentane is similar and rarer but somewhat higher density $(\sim13\ \mathrm{kg/m^3})$ in proportion to its higher molecular mass. Its boiling point is $28\ \mathrm{^\circ C}$ (uncomfortably warm, but your thermostat can go that high). This is the densest gas that strictly meets all the criteria in the OP.
If we relax the criteria a bit:
Perfluorohexane is just over the boiling point limit at $56\ \mathrm{^\circ C}$, but it has a molar mass of $338\ \mathrm{g/mol}$ which makes it slightly denser in gas form than tungsten hexafluoride ($\ce{WF6}$).  It's also inert and non-toxic, unlike $\ce{WF6}$.
The ultimate gas density would be uranium hexachloride using depleted uranium and $\ce{Cl-37}$, with a molar mass of $460\ \mathrm{g/mol}$, which makes it over 50 % denser in gas form than tungsten hexafluoride, and 3 times denser than sulfur hexafluoride, but it has a boiling point of $75\ \mathrm{^\circ C}$, decomposes on contact with air, is toxic and slightly radioactive.

I found a bunch of sites referencing OsF8 properties, but no confirmation that it had really been synthesized yet.

Actually the densest gas which has an actual use is Tungsten hexafluoride, clocking in at an astounding 13 grams per liter. Tungsten hexafluoride isn't as well known as its lighter cousin, sulfur hexafluoride, because it has a narrow range of uses. It is useful in the field of electronics manufacturing as it can be used to coat circuit boards with Tungsten. Also, as fun it may sound, don't try to dip your hand in it or try to swallow it, as it's quite corrosive. Besides being corrosive itself, it also produces hydrogen fluoride, a nasty chemical, when it comes in contact with any moisture. Although it's a pretty unsafe chemical, it's cousin, sulfur hexafluoride, is safe enough to inhale in reasonable amounts.

I would guess radon is the densest gas ($9.73\; \text{kg/m}^3$) that is not directly lethal. It is radioactive however, emitting alpha radiation, so you don't want to breath it in. Given that it is much heavier than air, as long as you don't hug the ground it think it would be possible to be in the same room without killing you.
From the fully non-lethal stuff, the first that comes to mind is sulfur hexafluoride, $\ce{SF6}$ ($6.17~\mathrm{kg/m^3}$). It is the stuff that people use to artificially make their voice lower (similar to helium making it higher) and it has a 0000 NFPA rating.
In principle you can dream up your own densest gas, because the density of a gas at given temperature and pressure is directly related to its molecular mass. So the higher the molecular mass of the gas, the denser it is.

But in principle you can't just dream up your own densest gas by adding molecules to the compound (to increase its molecular mass or density), because doing so will most likely give you a liquid or solid at standard conditions. Whether a compound or element is a gas at certain conditions depends not just on its density (or molecular mass), but many other chemical and quantum properties.


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Find the partial pressure in a container
I have an air mass which consists of 78.1% $\ce{N_2}$, 20.9% $\ce{O_2}$, 0.964% $\ce{Ar}$, and 0.0378% $\ce{CO_2}$ in a container which volume $\mathrm{V=40~L}$ at a temperature $\mathrm {T = 298.15~K}$.
The total pressure is $\mathrm{p_{total} = 1.5~atm}$.
Now I am removing all the $\ce{O_2}$ in the container. What will the partial pressure of $\ce{N_2}$ be?
I know the formula $\mathrm{pV = nRT}$.
Before I remove all the $\ce{O_2}$ I can calculate the amount of substance $$\mathrm{{n_{total} = \frac{pV}{RT} = \frac{1.5 \cdot 40}{0.082 \cdot 298.15}~mol = 2.45~mol}}$$.
The molar mass of $\ce{N_2}$ and $\ce{O_2}$ is $\ce{M(N_2)} = \mathrm{28~g\ mol^{-1}}$ and $\ce{M(O_2)} = \mathrm{32~g\ mol^{-1}}$.
If I remove all the $O_2$ the new mass/volume percentage of $\ce{N_2}$ will be
$$\mathrm{m_\%(N_2) = \frac{78.1}{78.1+0.964+0.0378} = 98.7 \%}$$.
So I thought I could find the amount of substance of $\ce{N_2}$  by saying
$$\ce{n(N_2)} = \mathrm{\frac{98.7~g}{28~g/mol} = 3.525~mol}$$ but this will exceed the total amount of substance before I removed $\ce{O_2}$ so this makes no sense.
When I have found the amount of substance of $\ce{N_2}$ I think I can just replace $\ce{n}$ in $\mathrm{p = \frac{nRT}{V}}$ and get the partial pressure of $\ce{N_2}$.
Maybe I have the right percentage of $\ce{N_2}$ but I am calculating the amount of substance wrong when it's a gas? Could the correct answer be $\mathrm{98.7 \% \cdot n_{total} = 2.422~mol}$ and then inserting this in $\mathrm{p = \frac{nRT}{V}}$ which gives me around $\ce{p(N_2)} = \mathrm{1.5~atm}$?

Since we're assuming these are all ideal gases, we can use Dalton's Law of Partial Pressures to solve this fairly easily; you never need to find moles of anything.  Unless something is strange, the percentages given are by volume or by pressure rather than by mass.  If they are intended to be by mass, what follows would need to be modified.  I suspect that the use of the words "air mass" in the problem are what caused the confusion.
The total pressure, 1.5atm, is equal to the sum of the partial pressures, each of which is proportional to its mole fraction in the container.  Because these are gases that have the same volume, 40L, their mole fractions are their percent of the total amount of gas.
The nitrogen, which is 78.1% of the total, has a partial pressure of:
$ 1.5 \cdot 0.781 = 1.1715\ atm$
That's true whether the oxygen is in the container or not.  To see that this is the case, we can remove the oxygen and do the same calculation.
If we remove 20.9% of the gas (in the form of oxygen) we are left with 81.1% of the original 1.5 atm remaining.
$1.5\cdot 0.811 = 1.1865\ atm$
We can determine the mole fraction of nitrogen in this new mixture of gases by adding up the remaining percentages and finding nitrogen's contribution to it:
$78.1 + 0.964 + 0.0378 =  79.1018\ units\ of\ gas\ total$
$\frac{78.1}{79.1018} \cdot 100\% = 98.73\%\ N_2\ after\ oxygen\ is\ removed$
As before, Dalton's Law of Partial Pressures tells us that the pressure of nitrogen will be its mole fraction times the total pressure, or 98.73% of 1.1865 atm.
$1.1865 \cdot 0.9873 = 1.1714\ atm$


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Preventing the gas when Hydrogen peroxide is added to a detergent
In organic detergents, when I add Hydrogen peroxide to increase the stain removing feature a gas is produced.
How should I prevent the gas production process?

i don't think you can-? It's a catalytic process where Hydrogen peroxide is broken down into water and oxygen.

Well, when I want to put the detergent in plastic bottles, the oxygen makes the bottle's shape be awful! What can I do in order to prevent the bottle's shape to change?

While hydrogen peroxide is fairly stable if pure in water solution, it can quickly break down to water plus oxygen in the presence of a catalyst such as transition metal ions--probably what is causing the gas evolution you observe. One approach you might try is to test other types of detergent - perhaps you can find one that does not catalyze breakdown of peroxide. Liquid facial soap or glycerine soap--or even dishwashing liquid--might be worth a try.
Another approach would be to mix the peroxide in right before use rather than try to store the mixture--the more dilute, the better.