text:
How does hydrogen gas build up in hot water lines?
Recently I was reading through the owner's manual of a washing machine; notably, the warnings in the front of the booklet. One warning stood out to me; it said that if the hot water in your house has not been run in a few weeks $\ce{H2}$ can build up in your lines. The warning of course stated that this can only occur under certain circumstances so I'm assuming it is quite rare.
Okay, so my question is, under what conditions does hydrogen gas build up in a hot water line?

I think that it is probably due to a reaction between an acid and a metal. These reactions can lead to the formation of hydrogen following this generic forumula:
$$\ce{M + 2H+ -> M^{2+} + H2}$$
For example iron could be a source of electrons. If you look at the corrosion of iron: $$\ce{Fe->Fe^{2+} + 2e^{-}}$$
At low pH (notice that there is a correlation between pH and temperature) hydrogen gas is formed:
$$\ce{2H^{+} +2e- -> 2H2}$$

Water reacts with the magnesium sacrificial anode over time to make $\ce{H2}$. Normally, you'd be using your hot water often enough that you wouldn't notice the minute amounts of $\ce{H2}$ coming out the tap with the hot water. But if your place is vacant for some time, a significant amount of $\ce{H2}$ could build up and an ignitable mixture come out the first time a tap is turned on (or a dishwasher, which is a case study of an explosion occurring such that the appliance rocketed out from its mooring and hit the opposite kitchen wall).
Depending on the chemistry of your water, the reaction with the anode might be vigorous such that you could be getting puffs of hydrogen out of every use of the hot tap enough to set off "flammable gas" alarms. In this case or otherwise excessive consumption of your anode, switching to an aluminum anode may help.
In addition, bacteria can use the $\ce{H2}$ as food and fart out $\ce{H2S}$ instead, giving the water a "rotten egg" odor instead. The aluminum anode switch can help with this too, or temporarily setting the heater to 160F for a day to kill the bugs (although exotic bacteria of this type can survive at high temp, and furthermore the $\ce{H2S}$ may be from the supply instead if the tank if the cold stinks too).
And yes, this has nothing to do with gas vs. electric heaters, it's purely the chemical reaction between water and the anode.
Summary: Run your hot tap for a few minutes after leaving your house vacant for some time; switch to an aluminum sacrificial anode.

You'd think that the main source of hydrogen is the sacrificial anode in your water heater.

Is this something that is present in a gas hot water heater?


text:
Chemical reactions in SFE?
What are the chemical reactions and/or chemical processes that take place during SFE(supercritical fluid extraction)? Namely supercritical $\ce{CO2}$.

No ma'am/sir/mammal. I was looking into homemade supercritical decaffeination chamber. AppliedScience (YouTube channel) has an example.

Any suggestions? I can't yet make tags so, I couldn't figure out what to put.

Supercritical fluid technology is used primarily to enhance the chemical extraction process.  The solubility of a compound in a supercritical solvent increases as the system pressure increases at constant temperature. This is due to the highly pressurized supercritical solvent disrupting the intermolecular attractive forces between the molecules of the compound to be extracted, and replacing those forces with increased attractive forces between the compound and the supercritical solvent.  Carbon dioxide is a very common material used for supercritical extraction.  It is cheap, not flammable, has low toxicity and, after the extraction is complete you simply return the system to STP, allow the carbon dioxide to evaporate and you are left with only the extracted material.  Some examples of commercial uses for supercritical carbon dioxide include caffeine extraction from coffee beans, sterilization of surgical equipment, fracking and dry cleaning.
Chemical reactions involving supercritical carbon dioxide are now being intensely studied.  One reaction that has been commercialized involves reaction of supercritical carbon dioxide with hardened, alkaline cements. A harder cement containing carbonate groups throughout the entire volume of cement is formed along with water (environmentally friendly process).

Ideally no reactions should occur during an SF extraction. In all the extractions a solvent is used to extract different substances from a matrix. In SFE the solvent is a supercritical fluid, in most of the cases is used $CO_{2}$ but not always. The chemicals process related to the extraction is mainly the solubilization of the target substance in the SF depending on the properties of the target substance and the SF (e.g. polar or non polar).

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text:
How does a simple alkaline battery work?
The basic + - electrolyte set up. Modern batteries and the like. How exactly do they work?

Your question is rather broad. Could you narrow it down to one type of battery?

Duracell or energizer your pick just tell me which in an answer

A related question and answer here on Physics.SE.

An alkaline battery (or more precise alkaline manganese cell) consists of a zinc anode, a $\gamma\!-\!\ce{MnO2}$ cathode and $\ce{KOH}$ (in paste form) electrolyte. Zinc is the negative electrode and $\ce{MnO2}$ the positive electrode.
During battery discharge, zinc at the anode is oxidized and dissolves in the electrolyte as hydroxozincate. When the electrolyte is saturated with zincate, zinc oxide starts to precipitate. With continued discharge and decreasing $\ce{OH-}$ concentration, zinc hydroxide is formed, which can slowly dehydrate to zinc oxide.
\begin{aligned}\ce{
Zn + 4OH- &->~ [Zn(OH)4]^{2-} + 2e- \\
[Zn(OH)4]^{2-} &->~ ZnO + 2OH- + H2O\\
Zn + 2OH- &->~ Zn(OH)2 + 2e- \\
Zn(OH)2 &->~ ZnO + H2O\\
}\end{aligned}
At the cathode, $\ce{MnO2}$ is in turn reduced, first to $\ce{MnO(OH)}$. When the battery is discharged slowly, this can be further reduced to manganese(II) hydroxide.
\begin{aligned}\ce{
MnO2 + H2O + e- &->~ MnO(OH) + OH- \\
MnO(OH) + H2O + e- &->~ Mn(OH)2 + OH- \\
}\end{aligned}
The overall reaction, when only considering the first reduction step at the cathode, is:
$$\ce{Zn + 2MnO2 + 2H2O + 2OH- ->~ [Zn(OH)4]^{2-} + 2MnO(OH)}$$
During discharge, zinc is irreversibly oxidized and water is consumed. As a consequence, the battery is "dead" when the electrolyte is dried out and/or all zinc has dissolved.
Sources: 1, 2

The chemistry of alkaline batteries involve the oxidation of zinc and the reduction of manganese dioxide. The description of a cell can be found here.

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Why is introductory chemistry so non-rigorous?
In introductory chemistry books, courses, ...
Many "definitions" contain words like: typically, most, often, likely, generally.
I read since questions on this site mentioning different definitions of same concept in different books.
I just read about a concept where the explanation contained smth like "some other chemists believe it should be the other way".
Why are there many non-rigorous definitions in introductory chemistry?

Because there often exceptions to rules, for which the explanation requires higher level chem to explain

"requires higher level chem to explain" that is what I was afraid of, but good answ anyway

Chemistry has many exceptions, thus it becomes wrong to permanently define anything as right or wrong. Hence, the large uses of words such as the ones you have listed.

Hi foobar! I think your question could be very interesting but as you posed it seems you are starting with wrong assumptions and misconceptions why do you say non-rigorous? Or what you mean for informal? Can you provide some practical examples?

I deliberately avoided examples, because in my experience that makes people occupied wit the example and then I never get an answer.

You may find the notion of a "lie-to-children" (that is, a deliberate oversimplification of a complex theory in order to make it more approachable to beginners) useful.

I like this merriam-webster definition of rigorous:

done carefully and with a lot of attention to detail

Chemistry is an experimental science there are a lot of patterns that work fine in most of the cases but not in all (e.g. the Octet rule) these are called rules of thumb. Chemistry is however a rigorous science, simply there are exceptions that don't follow clear patterns or easy simplifications so they are harder to predict or to comprehend for a student in an introductory text.
As in all the science (physics, mathematics etc. etc.) there are some topics that are still debated, this don't mean that introductory chemistry is non-rigorous.
I suggest you to read some books about history of chemistry (e.g. Asmiov, A Short History of Chemistry), you will find that chemistry has been for most of his history (and it is in part now days) a trial and error research, followed by the attempt to find a theory, a pattern or a sense to the observations done, this doesn't mean that there isn't a rigorous approach but simply that reality don't fit the scheme that we propose so easily.
I think maybe most of the text books don't emphasize in this aspect of chemistry, but most of the reasoning is a posteriori.

@foobar Can you find me a reference that states the rigorous has this meaning please, I'm not a native english speaker! :-) thanks! However in your example it depends which properties you used to classify the elements into the groups. Take this example from math: some mathematicians believe that 0 belong to natural numbers other say that it isn't (this is a true fact) following your reasoning math is a non-rigorous science....

@foobar mmm that's not so obvious in my opinion. I think you should rearrange at least your last phrase, say that introductory chemistry is a mess or informal is more an opinion than a truth. And It is a pity because the question if well posed could lead to interesting answer.

My point was that I expected definitions, from which I can make a decision without further searching for clarification. i.e. words such as "typically" or "likely" do not belong into a definition.

@foobar - what you are looking for sounds like a way to simplify chemistry to a series of algorithms that work every time. The field as a whole is not quite there yet for a variety of reasons.

The problem you are encountering has two origins:

Historically, chemistry has accumulated many models for chemical behavior that work most of the time. Some of these models are based on completely invalid theory (VSEPR), but they provide a set of rules that work. Some are based on decent theory (the octet rule, hybridization), but generate inconsistencies because the underlying theory is incomplete. Most of these models work in most cases, so they remain because they are easy to use.
Higher level theories/models exist that contain less inconsistencies and fewer exceptions, but the maths are hard. At the simplest, these models require multivariable calculus, linear algebra, and/or differential equations. At the more complex, we are talking about a computer spending 30 minutes (or more) iterating through the matrix approximations to many-electron wavefunctions (for which there are still no exact solutions) to find the best self-consistent minimum-energy solution. We're talking about problems in theoretical chemistry that used to consume one person's PhD thesis to just derive one proof and complete one calculation (or a small group of related calculations).

The field of chemistry has chosen to be more accessible to beginners by allowing the continuation of such imperfect models. There are a number of reasons for this. First, it allows students to progress through the more qualitative chemistry courses while they work up to the math requirements. Second, it allows students of other sciences (especially biology and medicine) to develop a conceptual understanding of chemistry (which they need far more than the calculus-based understanding).

How is VSEPR completely invalid?

Just derive one proof? For example?

@Dissenter - VSEPR is a wonderful model that works nearly every time. However, it is based on the notion that electron pairs occupy fixed localized positions in space (which is not in agreement with modern bonding theories).

@user26143 -In the days when the most powerful computer was a calculator the size of a room, physicists and physical chemists did all of their math by hand. The Hartree-Fock method of approximating solutions to many-electron wave-functions is iterative. The outputs of the first attempt are used as the start point for the second. Consider that it might take 500 or 1000 iterations of complex differential equations masquearding as complex linear algebra.

In that case you could perhaps see how solving the electronic structure of a single molecule without modern computational software might be a herculean effort. I do not remember those days (because I was born after they ended), but my older colleagues do.

Excuse me, what I asked is about deriving a proof (e.g. derive a proof for the size-extensivity of Hartree-Fock), not about performing a numerical calculation

I used to teach an introductory chemistry course and found that the understanding of most chapters would have been improved by having already knowing other chapters, but you have to start somewhere with something.  Different texts have things ordered differently and different instructors present material in different orders even using the same text.
Introductory chemistry by nature covers a very wide variety of topics.  As an example, I had to cover the basics of organic nomenclature as part of an introductory chemistry class and I had about 5 weeks to cover the organic section, maybe less if I needed more time to cover the material I taught earlier in the semester.  Obviously, even the nomenclature part of organic chemistry will be more rigorous in a two semester Organic sequence.
Introductory chemistry is a challenging class to teach - if you don't plan well enough how much time to devote to a specific topic you are at risk of either running out of time to cover something or most of your class not understanding something important.


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Chemical bonding - lattice energy
Question: Why is ionic lattice energy inversely proportional to the radius of the atom?
Most heterogeneous covalent molecules are polar to some extent. The degree of polarity, or the dipole moment, depends on the difference in electronegativity difference between the two atoms. The larger the dipole moment, the higher the ionic character.
What I know: Electronegativity decreases as you go downward in a group, however, the size increases, usually, as you go downwards in a group. Thus, ionic character will increase upon going downward, but the ionic lattice energy will decrease? This seems contradictory. Is this true? And if it is, why is it so?

Considering your doubt, from Born-Lande equation:

$$E = -\frac{N_\text{A}Mz^+z^- e^2 }{4 \pi \varepsilon_0 r_0}\left(1-\frac{1}{n}\right)$$

$N_\text{A}$ = Avogadro constant;
      $M$ = Madelung constant, relating to the geometry of the crystal;
      $z^+$ = charge number of cation
      $z^-$ = charge number of anion
      $e$ = elementary charge, $1.6022\times10^{-19}\ \mathrm{C}$
      $ε_0$ = permittivity of free space
      $4πε_0 = 1.112\times10^{-10}\ \mathrm{C^2/(J\cdot m)}$
      $r_0$ = distance to closest ion
      $n$ = Born exponent, typically a number between 5 and 12, determined experimentally
      by measuring the compressibility of the solid, or derived theoretically.

And what you are assuming the radius is in fact the distance between the two atoms.
As the distance increases, the attraction decreases from Coulomb's law :

$$F=\frac 1{4\pi\varepsilon_0r_0^2}Q_1Q_2$$

and thus there is less energy involved in breaking the lattice, so lattice energy decreases.

As you move down, the size increases and the valence electron in the outermost shell experiences lesser repulsion and thus it can be easily removed, therefore the elements at the bottom are highly electro-positive or conversely very less electro-negative. So when forming bonds, the elements at the bottom easily remove the electron(s) and gain more developed charged rather than the ones on the top who just release it just partially. This charge variation varies the dipole moment $\vec \mu=q\times\vec d$, by altering the partial charge $q$. Thus downwards as $q$ increases, $|\mu|$ also does increase and make the compounds formed by the element more ionic. Also the ionic/covalent character may be varied depending upon the bonding partner, in sense of, elements or other entities.

Also as charges $z^+$, $z^-$ increases, the magnitude of lattice energy does also increase as can be seen clearly from the Born-Lande equation.

$$\begin{array}{r|c|l}
&&\text{Ionic Solid}&\text{Lattice Energy (kJ/mol)}\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&r_0 \text{ i.e. distance}&\ce{NaCl}&786\\
&\text{between ions}&\ce{NaBr}&752\\
&&\ce{NaI}&702\\\hline
&\text{Variation of}&\ce{NaF}&926\\
&z^+,z^- \text{ i.e. charge}&\ce{MgO}&3800\\
&\text{on ions}&\ce{Al2O3}&15900\\
\end{array}$$

However isn't the lattice energy directly proportional to the ionic character? As the ionic character increases, so should the lattice energy. However, this clearly is not so. Why?

I really like your answer. It would be even better if you could explain briefly the variables/constants in the equation. While it is easy to find out with google what they mean, it would improve the reading of your answer.


text:
Redox properties of methyl radical
The methyl radical, $\ce{.CH3}$, is a strong oxidant and a strong reductant according to Wikipedia.
Why is this so? Any additional reasons to the ones below?
I know that an oxidant itself is reduced; an oxidant helps something else lose electrons, so the oxidant must be the one gaining electrons. Given that the methyl radical has an unpaired electron, it is quite reasonable that it would be amenable to having an extra electron to pair up with its lone electron.
I know that a reductant itself is oxidized; a reductant helps something else gain electrons. The carbon in the methyl radical has an oxidation state of -3, and I know that carbon can adopt a variety of higher oxidation states, such as +3 in acetic acid.

Why is this so?

Is this so?
Following is a tabulation of the ionization potentials (IP) and electron affinities (EA) for some alkali metals and halogens (kJ/mol):
\begin{array}{|c|c|c|c|c|c|c|}\hline
& IP & EA & && IP & EA          \\ \hline
F & 1681 & 331 && Li & 520 & 60  \\ \hline
Cl & 1251 & 352 && Na & 496 & 53 \\ \hline
Br & 1140 & 328 && K & 419 & 49  \\ \hline
I & 1008 & 298 && Rb & 403 & 47  \\ \hline
\end{array}
A good reducing agent readily donates electrons and would be expected to have a low IP.  Sodium would be expected to act as a good reducing agent and chlorine as a poor one, and, indeed the IP of sodium is much lower than that of chlorine.
A good oxidizing agent readily accepts electrons and would be expected to have a large EA.  Chlorine would be expected to act as a good oxidizing agent and sodium as a poor one, and, indeed, the EA of chlorine is much higher than that of chlorine.
This NIST webpage tabulates experimental IP's and EA's for the methyl radical.  The IP is given as (9.84 eV) 945 kJ/mol and the EA appears to be less than (0.6 eV) 58 kJ/mol.
Let's compare the IP and EA of the methyl radical to values in the above table for strong oxidizing and reducing agents.  The IP of the methyl radical is significantly larger than those reported for the alkali metals and a bit less, but relatively close to the IP's for all of the halogens except fluorine. This comparison suggests that methyl radical is, at best, a weak reducing agent. The EA of methyl radical is significantly less than those reported for the halogens and right around those reported for the alkali metals. This comparison suggests that methyl radical is, at best, a very weak oxidizing agent.


text:
Why is CaCl2 called calcium chloride?
Doing a first year chem class.
Just read through the molecular naming of compounds and now I'm confused as to why $\ce{CaCl2}$ is called calcium chloride and not calcium dichloride?

Elements from the first 2 columns only have 1 oxidation state, so any compound they form, it is easy to determine the number of anions.

Apparently, calcium monochloride is a thing, albeit unstable. Barium monochloride is even used in fireworks.

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., …). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. (…)

So far, the systematic name of $\ce{CaCl2}$ should be calcium dichloride.
However,

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound (…).

Therefore, the customary name calcium chloride is correct.

Some names are just used out of habit. "Calcium chloride" has been the common name for so long, that it has simply become common usage.  According to this Wikipedia article, calcium dichloride is an acceptable "other" name for the compound.

Calcium chloride is an ionic compound so cation is named first followed by anion. Do not get confused with carbon dioxide and sulfur dioxide. They are not ionic compounds.

You cannot go by whether a compound is ionic or covalent to determine if the number of anions should be in the name. Aluminum chloride ($\ce{AlCl3}$) is a covalent compound.


text:
What is Markovnikov's rule?
What is Markovnikov's rule? How can we use it to predict the products of additions to alkenes? And is it always obeyed, or are there exceptions to this rule?

Markovnikov's rule is used to predict how (for example) HX, where X is a halide, adds to an unsymmetrical olefin.  As the diagram below shows, Markovnikov addition follows the rule that the "H" will add to the least substituted carbon while the "X" will add to the most substituted carbon.  This is due to the fact that these reactions proceed by protonaton of the olefin to yield a carbocation.  Protonation at the least substituted carbon yields the most stable carbocation (in this case a secondary carbocation).  Finally the carbocation reacts with $\ce{X^-}$ to yield the final product.

In some cases the Markovnikov rule is not followed.  Such additions to olefins are termed "anti-Markovnikov".  These reactions typically proceed through a radical intermediate rather than a carbocation.  For example, addition to the olefin might be initiated by the addition of a bromine radical to generate the most stable radical intermediate, which would place the radical center at the most highly substituted carbon. Addition of a hydrogen radical completes the process and generates the final product which has the reverse addition pattern as compared to the Markovnikov addition product.

Wouldn’t an equally good example be $\ce{X} = \ce{BR2}$, i.e. the anti-Markovnikov addition observed with boron compounds?

@Jan Yes, hydroboration would be another good example.

The hydroboration of RCH=CH2 is a Markovnikov addition (RCH2CH2)3B. Boron is more electropositive than hydrogen just like hydrogen is more electropositive than a halogen. It is the replacement of boron with oxygen during oxidation that makes the overall reaction (alkene to alcohol) anti-Markovnikov because oxygen is more electronegative than hydrogen.  Today's use of Markovnikov's rule should be more expansive than the limited examples of the original studies. Note: Markovnikov stated that the halide added to the more substituted end of the double bond.

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text:
Why does ice cream not go back to its original form upon refreezing?
I came from a birthday party and someone gave me a sundae. I'm already full so I just decided to eat it at home. After arriving at home the sundae melted. So I left it in the freezer overnight. When I looked at my sundae it just formed ice crystals. It didn't went back to its original form. Why is that?. Why is it when I do this to ordinary ice cream it goes back to it's original form?

Ice cream is a frozen emulsion mixed with air. When the ice cream melted, the air escaped. When you re-froze it, it formed a solid mass.


text:
Does Le Chatelier's principle apply to the gas inside an air mattress?
While sleeping on an air mattress yesterday a thought occurred to me. I had to re-inflate the air mattress since it was getting rather low and after it had a good stiffness to it, I went to sleep. The next morning I noticed that a lot of the pressure that I put into the air mattress had left and I remembered Le Chatelier's principle which says that gases in a closed system will convert to the least number of molecules possible if the pressure is increased.
So this is my thought:

Breathing into the air mattress had added moles of gas to a closed system (the air mattress).
Lying on the air mattress increased the pressure of the system.
Given Le Chatelier's Principle, the gases should have transitioned to the lowest number of moles to relieve the pressure introduced on the system.

The thing I'm not sure about is if there are any equilibrium reactions for the gases in our atmosphere that will actually do this. Considering the makeup of our atmosphere, does Le Chatelier's Principle apply to the air inside an air mattress?

No, the air we breathe is about 78% nitrogen, 21% oxygen, 0.5% carbon dioxide and other gases. Nitrogen doesn't react with anything at room temperature and there is nothing to burn inside the air mattress.
What probably happened was that there is a very small hole that the air leaked out. About the only way to find the hole would be to submerge the air mattress in a pool.

I would also add, that even if there were a reaction between 3 gasses (a to b +c for instance) you'd have to really increase the pressure quite a bit to drive the reaction forward. It is unlikely that your additional body weight increased the pressure significantly.


text:
How do erasable pens work?
I understand the principle behind an eraseable pencil. Graphite is highly slippery, you rub it onto the paper, and an eraser can rub it off again.
I've had friends at school with 'eraseable ink pens'. The ink was 'special' and the eraser was 'special' and it could erase the pen in the same way a regular eraser could erase pencil.
My question is - how do eraseable pens work? (Ie what substances are involved and what are their properties?)

It's not so special rubber. When I had that pen, I soon found out that it's the heat what causes the magic. I would clean whole pages of text on the electric cooker.

Instead of a typical oil/dye ink,  erasable pens use a compound closer to rubber cement, which acts like a liquid when a bit of pressure is exerted. 1
A bit of history and a link to patents can be found here
EDIT: Be sure to check out the other answer here about thermochromic ink too.

There are two types of erasable pens: rubber cement and thermochromic pens. Thermochromic pens are much more popular nowadays because rubber cement pens literally harden like cement, and can't be erased after a few hours. Thermochromic pens, however, use heat from friction that makes the ink invisible; not exactly removing it from the paper, but close enough.
One downside to thermochromic ink is that the ink can fade over time and rubbing the paper too much can make it invisible, but you can undo this by placing it somewhere extremely cold (think about $\pu{−10 °C}$ (freezer)) and it should reappear.

The ink has three components. One component is a pigment (the color) that reacts to acidity by turning invisible. The second component is a chemical that acts as an acid or a base. The third component reacts to heat by activating the acid/base.
By twiddling these three components, the manufacturers can make an ink that disappears when enough heat is applied and reappears when sufficiently cold. When you erase with the rubber, the friction creates heat which changes the acidity of the ink and makes it invisible. When it is cooled down, the acidity changes back and the ink becomes visible again.
In my experience, some of these inks work better than others.

I don't think the OP is referring to ink that "becomes visible again" after having been erased?

Some erasable pens like Frixion do indeed have ink that can become visible again after having been erased, by freezing. Frixion says, "The ink will reappear at temperatures under -10°C" here: http://www.pilot-frixion.eu/uk

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text:
Soap and rain water, why does it not rinse off?
I am in Papua New Guinea at the moment, where rain water harvesting is the most common way of getting water.
When washing hands, the soap feels hard to rinse off, and my skin feels "greasy".
I read around a bit here and here and I think I understand what is said: essentially, sodium stearate in the soap separates into sodium and stearate ions, which then precipitate with other ions (like Calcium, into calcium stearate) and get rinsed away. But in very soft water, there are sodium ions rather than calcium ones, so this reaction is slowed down a lot, resulting in the greasy feeling.
As it turns out, I have access to hard ground water at the same time, and the difference is very obvious: with hard water, soap is rinsed off instantly.
My question now: I assume rain water has in fact almost no ions in it (I'm thinking Sodium ions mostly), and I would therefore assume that the Sodium stearate dissolves very easily, but has no replacement ion to precipitate with, therefore sticks to the skin.
Is this the right explanation or are there other factors at play in the explanation about that greasy feeling?

One of the possible reasons could be the reason stated by you. But if the sodium stearate was primarily made from Tallow or animal fats. soap becomes more of an insoluble one and makes the soap oily and it may make your hands greasy

Yes, this is a good explanation. Rain water is almost pure water, it lacks of bivalent ions such as $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ which helps soap to rinse of. You will get the same effect when a water softener is installed on water distribution.

The inefficiency of soft, essentially pure, water for rinsing soap from skin is well recognized. However, a broader explanation includes the difficulty in removing the last traces of anything from anything else.
I estimate that the wash liquid from bar soap results in about 2% soap, perhaps more. This concentration will reduce the surface tension of water dramatically, and give you that slippery feeling. Rinsing is generally not as energetic as soaping up, so the soap solution you forced into your skin doesn't usually get the same rough treatment during the rinse cycle. Rinsing might cut the concentration of the soap in that viscous layer next to your skin by half, maybe a little more; do it again, cut the soap by another half, perhaps reducing the sodium stearate concentration to 0.1% or 1000 parts per million. Yet the concentration of sodium stearate can be reduced by another factor of three before its surface tension is reduced at all (see picture, Ref 1)!

Since soap solutions have very low surface tensions even at very low concentrations, and skin is so porous and irregular (fingerprints), some soap may continually exude from the skin as molecules. Rinsing with pure (soft) water will eventually remove these soap molecules (as molecules, not micelles) at very low concentrations (but still surface-tension-reducing). The benefit of hard water during the rinse is that the molecules of soap are deactivated as the rinsing proceeds, and even before it is completely accomplished. So the rinse seems to be finished quicker.
Strong mechanical action will do much the same as hard water in squirting the last of the soap molecules out of the skin. It is interesting that some people like the "grabby" feel of non-soapy skin while some prefer the silky feel of soft water washing.
Ref 1. https://www.researchgate.net/figure/Effect-of-sodium-benzoate-and-sodium-stearate-concentration-on-surface-tension-in-water_fig1_336817830


text:
Is NaCl paramagnetic?
My teacher told yes that salt is paramagnetic and it is attracted by magnet slightly. But when I tried that at home it did not work.

your teacher was wrong in this aspect

could it be that the teacher said "some salts"?  What is s/he a teacher of and at what level?

No actually he specified nacl, he is my chemistry teacher for jee coaching

Make its orbitals' diagram :)

Nit picking here, but technically, you could generate a smaller magnetic moment due to nuclear magnetic moments, but I think it still is net zero in this example.

Table salt (NaCl) is diamagnetic. For paramagnetism, one needs free unpaired spins.
The NaCl lattice is built up from closed shell cations and anion, there is no such unpaired spin pure NaCl.
Also: experimentally it is difficult to measure paramagnetism, and you generally cannot just try it with a magnet and observe it without some tricks.

What tricks are you referring to?

I would agree with it being difficult to "measure" paramagnetism in the sense of quantifying the strength of the resulting magnetization without specialized equipment (i.e., a magnetometer). However, I think it is possible to at least qualitatively observe and relatively rank the magnetization for many materials. Small samples of various paramagnetic salts, for example, will visibly respond to even small permanent magnets. On the other hand, diamagnetism, to my admittedly limited knowledge, is extremely difficult to observe without very powerful magnets.

In my experience, small samples of insulating materials (i.e. almost all the salts) can strongly influenced by electrostatic forces, wetting  and other artifacts.  Disclaimer: I never actually tried to do play with paramagnetic materials and small magnets, but worked with a classic Faraday balance for years on transition metal salts, and oh, boy..  Said balance used a huge, old fashioned electromagnet, and measurement at liq. N2 temperature, yet once could make gross mistakes even under those circumstances.

NaCl is diamagentic. Every material is at least diamagnetic. The diamagnetic influences are very weak, so you can't see them.
Diamagentism: if you have a field H working on a materia, it will change the condition of the particles in the atom of the materia, to create a magentic moment, which is opposite to H. Sigma of all these induces fields is called B. If you have an inhomogene field you have to use a lot of work to create a higher field force of the diamagnet. Genereally the diamagnet want to have a low field force.
The quantumchemistry behind it is easy: the spin of electrons have a magentic moment and creates a field, but because of the Pauli Principle and the thermic move, you wouldn't see it on a macroscopi scale. First by a field outside you would see an induction of the magnetic dipole on an atomic basis.
So if your field is stron enough, you could actually see NaCL moving. But you need a very very very strong field to do it. Andre Geim for example used the diamagnetism of water to levitate frogs.
Paramagnetism: if you take have a magnetic field from outside, the microscopic magnetic moments will will all show in the same direction. But because of the thermic fluctuation the magnetic field would collaps as bald as the magnetic field collaps. The strongness of the field is calculated by:
$M= \chi H$,
where M is the magnetisation of the material, H is the magnetic field and $\chi$ is the suszeptibility.
Ferromagnetism: a material that will be attracted to a pol or is a magnetic itself. It's magnetic because of it's elementarmagnets inside it. The magnetic flux is: $\vec B = \mu \vec H = \mu_0 (\vec H + \chi \vec H) = \mu_0 (\vec H + \vec M)$.


text:
Differentiating between Titration, Buffer, diluted buffer, Deionized H₂O
I had a chemistry lab where we had to titrate three solutions with $\ce{HCl}$: a buffer, a diluted buffer and deionized water. My partner and I didn't label the charts we made, so now I have to figure out which one is which.
The way I see is that from the Henderson-Hasselbalch equation:
 $\mathrm{p}\ce{H} = \mathrm{p}K_a + \log\left(\ce{[A^{-}]/[HA]}\right)$,
 the diluted buffer should have a larger slope at the beginning then the original buffer solution.
My question is, is my logic correct, and how do I differentiate between the original buffer and the deionized water?

What is the expected pH of deionized water and what would the pH be after adding a few drops of acid? For both buffer solutions, there is an equivalence point, how would you use that to tell them apart?

pH of water is expected to be close to be near 7 however two my initial pH's were around 4.6 and the other was around 4.2. I would expect the equivalence point to occur sooner for the diluted buffer

Usually deionized water is slightly acidic. I kind of mentioned that water doesn't have an equivalence point. Is there a graph without one?

One obviously has an equivalence point, while the other two appear to drop rapidly after adding one drop of HCl, although one drops much more quickly then the other.

I guess your diluted buffer is more diluted than I thought. What is the purpose of a buffer? A diluted buffer would have the same characteristic, but in a smaller amount. Water would be missing this characteristic. You should be able to figure out which graph is which now.

cool, if you want to formalize our comments into an answer I can give you your +5 rep.

What is the expected pH of deionized water and what would the pH be after adding a few drops of acid? For both buffer solutions, there is an equivalence point, how would you use that to tell them apart? Usually deionized water is slightly acidic. I kind of mentioned that water doesn't have an equivalence point. Is there a graph without one?
What is the purpose of a buffer? A diluted buffer would have the same characteristic, but in a smaller amount. Water would be missing this characteristic. You should be able to figure out which graph is which now.


text:
How can we say that one mole of a substance contains avogadro number of particles?
I just want to know, what was the procedure through which the value of one mole of a substance was considered as a an aggregate of $6.023\cdot10^{23}$ particles.
Secondly, how do we calculate the volume occupied by one mole of gas?

this might help for the first part: http://chemistry.stackexchange.com/q/10245/4945

There's a nice Scientific American article that provides an interesting historical perspective on Avogadro's Number.  They go on to point out that one of the early accurate determinations was based on the first accurate measurement of the charge on an electron.  The charge on a mole of electrons (1 Faraday) had been known for some time.  Dividing the latter by the former yielded Avogadro's Number.
As to the volume of 1 mole of gas at STP, imagine the following experiment:
1) weigh out 1 mole of oxygen gas (32 gm)
2) fill a balloon with this weight of gas
3) attach a pressure gauge to the balloon and expand or contract the balloon's volume until the gauge reads 1 atmosphere
4) Immerse the balloon in a water tank and measure the volume of water displaced - 22.4 liters

very accurate and keen and answer bro thanks a lot i tried it and it worked really fascinating experiment and really graceful answer


text:
The acidity of boric acid

The “real way” boric acid behaves as an acid in water is first by acting as a Lewis acid (bottom reaction, in black).  In fairness to the undergrads, I’d bet >25% of graduate students around the country would not be able to correctly explain the acidity of boric acid.

How are we supposed to know that boric acid heterolytically cleaves water as to leave a proton behind?
Is this something that is supposed to be immediately obvious? Or do we just have to read about experiments and then realize that this is how boric acid behaves as an acid - as exclusively a Lewis acid?
Other structurally analogous acids such as sulfuric acid also have a electrophilic center atom because of the electron-withdrawing oxygens, but these acids do not behave the way that boric acid does; these acids simply donate protons - they don't chop up water molecules to create protons.
Is a hint that $\ce{B(OH)_3}$ behaves this way found in the fact that boron can form either three or four bonds with elements? Should this clue us off to the possibility of $\ce{B(OH)_4^-}$?
Is another hint found in the fact that boric acid is trigonal planar with an unused p-orbital? Sulfuric acid on the other hand doesn't have any empty orbitals (well, it might have empty d-orbitals which may or may not be accessible).

I think your suppositions and observations are basically correct. On the basis of general principles, it is certainly possible to imagine that such a mechanism might be at play. If one considers the structure of boric acid, it should be readily apparent that the boron atom could be quite electrophilic and hence act as a Lewis acid: it possesses a formally vacant $p$-orbital; it is bound to highly electronegative oxygen atoms, which withdraw electron density and create a partial positive $(\delta^+)$ charge on it (though this effect may be mitigated by the oxygen lone pairs, which can donate electron density into the aforementioned $p$-orbital); and finally, it has a planar geometry that makes it quite sterically accessible to approaching nucleophiles. Such a mechanism is also lent plausibility by the well-known propensity of various other boron compounds for forming Lewis adducts and being highly susceptible to hydrolysis (where applicable).
All of that having been said, I don't think it's fair to claim that such a mechanism is "obviously" correct for the reasons given above. Such arguments are strictly qualitative, and while they're very suggestive of the possibility, they are in no sense definitive. The only way to even approach something vaguely approximating certainty is to perform experiments (perhaps in concert with theoretical computations), and even there considerable difficulty exists in reaching firm conclusions. The confirmed presence of $\ce{B(OH)4-}$ under one particular set of conditions is not sufficient proof that the mechanism involving boron as a Lewis acid is always operative, nor does it rule out other pathways. (Note that I'm not necessarily disputing that mechanism, but only pointing out that extrapolating too widely from specific experimental results has certain liabilities.) As ron points out in his answer, the issue apparently isn't settled.
Edit: I would also add that there are arguments to be made against such a mechanism. There's likely some degree of $\pi$-bonding between the oxygen lone pairs and the boron, which would be disrupted by the pyramidalization of the molecule upon formation of the intermediate Lewis adduct resulting from nucleophilic attack by $\ce{H2O}$. The associated energy penalty might be relevant to the reaction kinetics under some conditions. My instinct is that this is probably not significant, but it may be a factor worth considering. Of course, this is all speculative on my part, though I think it further demonstrates that qualitative reasoning in chemistry is rarely clear-cut, and there's an awful risk of falling prey to fuzzy thinking. Hence, rigorous experimentation and calculation are vital.

At least according to this Wkipedia article (look in the "Properties" section) there is some disagreement as to the basis for boric acid's acidic properties.  One set of experiments seems to favor the Lewis acid interpretation shown above in black, while the other camp suggests that the data supports the view that boric acid is both a Lewis acid and a Brønsted acid (both the blue and the black mechanisms shown above are operative).

The way I read it was, that there is the agreement, that it is both. I also read the one pager and it suggests, that $\ce{H2BO3-}$ and $\ce{B(OH)4-}$ are active in the same pH interval at approximately the same concentrations. However, they based their simulation on $K_a$ values I cannot confirm. And I would expect that to be the weakest link of the whole theory.

I still read disagreement with one camp favoring just the Lewis acid explanation and the other camp favoring both a Lewis and Bronsted interpretation.

Haha, it's always the small molecules that behave like crazy. I can agree on that. It is probably also not very well investigated. A quick research on the web of knowledge turned up nothing of value, that could backup the used values. If I can get a hand on the handbook of chemistry and physics I might be able to find more. (Investigating this could also be a nice little computational chemistry research project - but I have no time.)

Why computational? Why not go to the lab and see what happens in solution? Also what computations would you perform?

@Dissenter I cannot go to a lab as I am a computational chemist. I also doubt that this would be an easy setup to actually measure the different species in solution. I would perform some easy solvent dependent quantum chemical computations, to determine $K_a$ values. But I have neither time nor resources to do it at the moment.

How would you know that such calculations are correct?

@Dissenter The theory of everything is not yet proposed, so you might miss out on a lot of things that have small effects on the system. However, it is very well possible to approximately solve the time-(in)-dependent Schrödinger equation for these systems to a very well level of accuracy - maybe even better than making an experiment. After all, there is a whole branch of chemistry that investigates, analyses and explains chemistry without ever stepping inside a lab.

Another clue that boric acid may be a special case is that its structure doesn't conform to the octet rule (for Boron)
This is unlike some of the other acids, even when some of those , like for example sulfuric acid,are usually portrayed as having 10 or more electrons around the central atom instead of 8. As is written in the Wikipedia page this is more for reasons of convenience. For those other acids there are always possible notations that do confirm to the octet rule, with a negative charge on the O atom(s) instead of a double bond and one or more corresponding positive charges on the central atom.
The addition of a hydroxide ion (or a water molecule) to boric acid will result in an octet for boron, making the molecule more stable. The substraction of a proton atom will only lead to an octet in one of its multiple forms, as shown in the image posted by the OP.


text:
How to calculate initial and final temperatures during supercritical steam expansion?

Ten $\mathrm{kg}$ of steam at $500~\mathrm{bar}$ is expanded at constant pressure until its volume increases to seven times its initial value of $0.01~\mathrm{m^3}$.
  (a) Calculate the initial and final temperature of the steam
  (b) Calculate the heat that must be supplied to carry out the process

An attempt at solving this problem:
$$\Delta H + \Delta E_k + \Delta E_p = Q - W_S\ \ \text{(open system)}$$
Taking out unnecessary terms leaves us with
\begin{aligned}
\Delta H &= Q\\
500~\mathrm{bar} &= 5\cdot10^7~\mathrm{Pa}\\
\Delta H &= H_f - H_i\\
H_i &= (5\cdot10^7~\mathrm{Pa})\cdot(0.01~\mathrm{m^3}) = 5\cdot10^5~\mathrm{J}\\
H_f &= (5\cdot10^7~\mathrm{Pa})\cdot(0.07~\mathrm{m^3}) = 3.5\cdot10^6~\mathrm{J}\\
\end{aligned}
Converting to kJ/kg (10 kg of steam)
\begin{aligned}
\hat{H}_i &= 50~\mathrm{kJ/kg}\\
\hat{H}_f &= 350~\mathrm{kJ/kg}\\
\end{aligned}
And that's where I stopped. I tried looking up the values of $\hat{H}_i$ and $\hat{H}_f$ in steam tables to interpolate the temperatures, but the values are too low. Since the pressure is too high, I also know the ideal gas law $(PV=nRT)$ won't apply to this scenario. Can someone point me in the right direction?

Since supercritical steam is not an ideal gas, you might want to use standard reference data (e.g. provided by NIST).
Initial state
mass m0 = 10 kg
pressure p0 = 500 bar = 50.0 MPa
volume V0 = 0.01 m3
density ρ0 = 1000 kg/m3
temperature T0 = 340.53 K
specific enthalpy h0 = 323.06 kJ/kg
Final state
mass m1 = 10 kg
pressure p1 = 500 bar = 50.0 MPa
volume V1 = 0.07 m3
density ρ1 = 142.86 kg/m3
temperature T1 = 925.86 K
specific enthalpy h1 = 3453.11 kJ/kg
Heat transfer
heat Q = ΔH = (h1 – h0) • m = 31.3 MJ

Hi @Loong. Generally with answers to questions such as this, it's preferred that you show the processes of how to arrive at the answer.

With standard reference data, you simply look up the data. There is no process, except for the last line of my answer.


text:
Why isn't neptunium used in nuclear reactors in nuclear power plants?
Why isn't neptunium used in nuclear reactors in nuclear power plants?  Uranium is, and plutonium is. But neptunium isn't and it is in the middle of them.  Is it like it is too hard to make it do fission?  Can someone please kindly explain it to me?

Fission is not an issue. It's hard to make neptunium, period.

Actually, neptunium (Np-237) is regularly used in nuclear reactors; however, not as nuclear fuel but as so-called fission dosimeter in irradiation capsules. (I have just received a delivery of six such capsules that will later be installed in a pressurized water reactor.)

There are several reasons that neptunium may not be used.
The first is the abundance of neptunium. From Wikipedia, it states the best source is from spent fuel rods. Basically, someone would need to handle very radioactive material to obtain a small amount of neptunium.
Next would be the likelihood of fission when struck with a neutron. If it is the same or less than uranium, I would go with uranium. It would have to be substantially more than uranium to offset the cost of refinement.
Also, there is the energy output per gram. I don't think it would be much greater than uranium, probably less.
The isotopes of neptunium that have long half-lives are $\ce{{}^{236}Np}$ and $\ce{{}^{237}Np}$. These are not produced in great abundance from $\ce{{}^{235}U}$ or $\ce{{}^{238}U}$.
To make plutonium from uranium, the uranium must absorb neutrons without fission. This is much more likely to happen then to make $\ce{{}^{236}Np}$ or $\ce{{}^{237}Np}$

Technically, neptunium is an intermediate when uranium takes up neutrons.  Starting with U-238, we go to U-239, which decays by electron beta emission to Np-239 and then Pu-239.  The plutonium isotope, with a longer half-life than the neptunium intermediate, is primarily what we see.

Production rate is low for reactors using low-enrichment uranium.  You get about one atom of 237Np per thousand atoms of 239Pu, so its easier to get the Pu out and use it rather than running and running your production reactor to get the 237Np.

Neptunium is not a good fuel, firstly there is the activation problem
Neptunium-237 can adsorb neutrons and either undergo fission or it can capture neutrons to form neptunium-238. This will beta decay to form plutonium-238 which is a horrible isotope of plutonium to work with in bulk. Plutonium-238 has a half life slightly shorter than 100 years, it has a very high specific activity and it does lots of horrible things that plutonium-239 does not do. A small trace of plutonium-238 will greatly increase the alpha activity of the plutonium produced by the reactor. This will make reprocessing and the spent fuel handling harder.
Then there is the problem of production.
The normal route to neptunium-237 is that a uranium-235 nucleous adsorbs a neutron, it forms an excited state of uranium-236. Normally we hope that the excited state of uranium-236 undergoes nuclear fission. But sometimes it manages to relax into its ground state.
The uranium-236 can adsorb another neutron to form uranium-237 which then beta decays to neptunium-237. An alternative route is to subject uranium-238 to fast neutrons such as fusion neutrons, this causes the uranium-238 to undergo the n.2n reaction. These very fast neutrons will not appear commonly in a normal nuclear reactor.

Just now seeing it.  LAMO. 9/29/23


text:
What is reactivity really, and can it be quantified?
I understand the concept of a reactive species on a sort of vague and intuitive level: a reactive species is one that has lots of reactions that it can participate in. These tend to have large kinetic constants, be thermodynamically downhill and often have products that are also reactive, leading to knock-on effects. But reactivity isn't the same as energy, and it's possible to have highly reactive species with relatively low free energies and vice versa.
I'm interested in whether there's any kind of formal theory of reactivity, and/or a quantitative definition of the concept. For example, we're often told that free radicals are reactive because they have an unpaired electron and they "want" to fill their outer shell. This makes perfect sense on an intuitive level, but I'd like to know if it can be taken further. Is there some way we can write down an equation that will tell us how reactive a species will be in a given environment, or is there some way we can measure a species' reactivity empirically and use it to make predictions?
To clarify, I'm not looking for quantum-level predictions, although I will look into the theories mentioned in Sam and Greg's answers. Rather, I'm looking for something a bit more heuristic that will help me get a handle on what reactivity really is. In particular, what does it really mean to say that one species is more or less reactive than another? Can we put a number to it and say that a species has a reactivity of 8.2 on some meaningful scale?
ron's answer is a start. It surely must be the case the reactivity relates to the energy landscape. But ron's answer deals with a one-dimensional landscape, corresponding to a single reaction. Reactivity is a concept that involves many reactions, and so a satisfying theory of reactivity would have to involve the properties of a massively multidimensional landscape. If there is anything that has been written about it from that point of view I would be very happy to read it.

Have a look at transition state theory it's a quite universal tool to express kinetics of a reaction. In terms of multidimensionality, it can usually boiled down to a 2D problem using the same approximations. You can then compare to (e.g.) nucleophiles versus another reactant via boltzmann and find out which one is more reactive. This is widely used in computational chemistry, to determine reaction pathways.

I understand transition state theory. (Reasonably well at any rate - there's always more I can learn, but I have a background in statistical physics, so it's quite straightforward for me.) But what I'm trying to get at is that there's quite a big a difference between understanding the kinetics of one reaction and understanding the reactivity of a molecule, which can undergo many reactions. The question I want to answer is "why are some molecules much more reactive than others".

This is a question of reaction kinetics. High reactivity means that the activation energy for a given reaction is low.
http://en.wikipedia.org/wiki/Activation_energy

I think the term "reactivity" is often used to denote kinetic lability, a proclivity to react.  Let's look at the two reaction coordinates shown below.  They both represent the interconversion of molecules A and B through intermediate I.

Note that the potential well that I exists in is much deeper in the drawing on the left than the drawing on the right.  Therefor the activation energy for I to escape will be much larger for the intermediate on the left than the intermediate on the right.  The intermediate on the right would be said to be more reactive (have a shorter lifetime) than the intermediate on the left, because it requires less energy for it to escape.
An experimentalist can infer the depth of the well through experiments.  Through such experiments the experimentalist learns that free radicals, highly strained hydrocarbons, carbenes, etc. exist in potential wells that must be very shallow because they react so readily.  Using this empirical data the experimentalist can then infer how similar compounds are likely to react. A computational chemist can infer the same thing by calculating the profile of the reaction coordinate for the various reactions under investigation by the experimentalist, or as is becoming increasingly common the computational chemist can predict what should be and let the experimentalist confirm those predictions.

+1. I think this is a good answer, but I do have one nitpick regarding the graphs. The energies of the transition states in the two graphs are identical, and the labeling indicates the same products and reactants. I think, by the Hammond postulate, it's then unlikely that the respective intermediates would differ so radically in energy while the transition states wouldn't.

Depending on what kind of theory you are looking for, the maximum hardness principle and HSAB theory actually tells a lot about what species should be reactive and with whom.
It is not unrelated to density functional theory, but formulated in a rather chemically intuitive way.

Thanks, I looked into it and it does indeed seem useful. (But I haven't actually managed to fully understand it yet.)

Density Functional Theory has been used in the manner you are interested in.
See here for an introduction to the topic.


text:
What is the simplest structural unit of an element or compound?
From "Smarter than a 5th Grader":

What is the simplest structural unit of an element or compound?

molecule,
atom,
neutron,
proton

For covalent compounds, the answer would definitely be molecule. But I'm unclear as to how to show that the answer for ionic compounds or elements is "molecule".
For example, https://sites.google.com/site/rccchem2a/chapter-4-chemical-bonding-the-i states that ionic compounds: "Do not have molecules as basic structural unit. Instead extended array of positively and negatively charged particles called ions." Is the term 'structural unit' defined across Science as only for covalent compounds?

It is a bit unclear what you mean by structural unit. Of course, most compounds have fragments, repeating throughout the compound structure. However, in condensed phase these fragments are linked into infinite structure. Of course, there is no 'smallest' finite unit, that can be used to model chemical properties of condensed phase adequately, only finite elements and their linkage. This is a permanent source of troubles in chemical modeling. In case this does not satisfy, please, expand your question.

"What is the simplest structural unit of an element or compound? molecule, atom, neutron, proton"
To clarify, this is the exact question from "Smarter than a 5th Grader" game. I think that the question is too brief so does not have an answer, and that the accepted answer "molecule" should be incorrect. But how can I prove it? Or prove them correct?

Further comment is that the definition was written by a lexicographer in Wordnet 3.0 for molecule. It seems incorrect to me and Wordnet is not a dictionary, but it's widely used in academia. Can anyone answer definitively yes or no, this is a real definition?

I would think a "simplest structural unit" would be also the simplest possible molecular entity - and may therefore be almost anything. However, the answer molecule does not seem fit here, e.g. carbon in its common form - graphite - does not really consist of molecules but rather two dimensional macromolecules. For phosphorous it might be even more confusing.

@TomAnderson Well, the smallest structural element considered in chemistry is Atom. Protons and neutrons are more domain of physics. Atoms may assemble directly into lattice or assemble into molecules that assembles into lattice or molecules can be included into a lattice/molecule. So, in case I forced to answer the question as is, I'd stick with molecule. But the question in too dependent on wording, I would insist on expanding it. Moreover, it has at least one incorrect concept: chemical elements are sets and do not have structural units.

For a solid lattice (metal, ionic solid, crystal) I would say the answer is unit cell, for a non-charged compound  that isn't a lattice (including solution/gas phase) I would say molecule, and for a charged one I would say ion-pair / ternary ion-associates / quaternary-ion associates for two / three / four ion clusters. For polymers, I would refer to the basic unit(s) being the component monomer(s).
There is, as has been identified in the comments, an issue with allotropes, Carbon (Diamond, Graphite etc) can be encompassed within 'unit cell', but for elements like phosphorus and sulfur, it's a little more complicated. There I would contend that the basic unit of e.g. P4 (white Phosphorus) is the molecule, and for the rest, I would return to referring to the unit cell, or 'monomer' in the case of amorphous networks.

To add to the complexity, some elements are monoatomic, notably the noble gases.

Further down the rabbit hole we go!


text:
What is Sodium Chloride like in gas state?
Since sodium chloride is sodium and chlorine atoms bonded as a lattice and there are no discrete molecules, doesn't that mean in gas state, sodium chloride is simply sodium and chlorine atoms separate from each other, resulting in a mixture of sodium gas and chlorine gas?

See here for previous answers on the subject.

No, in the gas phase sodium chloride exists as a monomer (the sodium chloride molecule) along with its dimer $\ce{Na2Cl2}$.  The dimer has a roughly rectangular shape and is quite floppy with chlorines located diagonally across from each other.  The dimer makes up about 27% of the mix.  All of the bond lengths, etc. can be found in this thesis.  Go to the end and you'll see the full paper that appeared in JACS with all of the data.
Also this paper here  (credit to orthocresol for finding that).  The abstract mentions "the equilibrium structures of the monomer (NaCl) and the dimer $\ce{Na2Cl2}$"

Very interesting source. To be clear, the equilibrium proportion of ionic monomers, dimers and larger aggregates varies with certain parameters, most notably temperature and pressure; high temperatures and low pressures favour more extensive dissociation into smaller units. From a quick look at the thesis, it would seem that the 1:0.27 molar proportion of the monomer and dimer was experimentally determined at $\rm{943\ K}$ for sodium chloride (which also suggests the measurements were made at a very low pressure). The proportion might be quite different for boiling salt at $\rm{1\ atm}$.

@Nicolau Saker Neto  Good point, thanks for adding it.

@ron Does this imply that all ionic compounds in gas phase are found as monomers and dimers?

I dunno, my guess would be many do, many don't.  I think some you'd need to heat so hot to get them vaporized they might exist as a plasma, others might decompose - but I'm just guessing.

Roughly, somewhere around half of the lattice enthalpy of a salt comes solely from the binding of the smallest electrically neutral agglomerate; that is, it takes about as much energy to break a macroscopic solid $\ce{NaCl}$ crystal into a gas of ion pairs as it does to break all the ion pairs and create a true plasma. Thus, it is rather unlikely that, at reasonable temperatures, an ionic gas will break down any further than the smallest possible electrically neutral aggregates.


text:
What does “generally octahedral” mean?

Based on vdW repulsions for generally octahedral $\ce{BrSF5},$ each following statement about bond angles (“ba”) is true except that

(1) $8~\ce{F-S-F}$ “ba” ’s are $< 90^\circ$
(2) the $\ce{Br-S-F}$ “ba” is $\neq 90^\circ$
(3) $4~\ce{F-S-F}$ “ba” ’s are $< 90^\circ$ are equal
(4) $4$ other $\ce{F-S-F}$ “ba” ’s $< 90^\circ$ are equal
(5) $2~\ce{F-S-F}$ “ba” ’s are $< 180^\circ$

What does “generally octahedral” mean here? Can sulfur monobromide pentafluoride adopt other shapes?
Also, I have my suspicions about the $\ce{Br-S-F}$ bond angle being perfect 180°. Wouldn't vdW repulsions distort the bond angle and make it approximately 180°? The way I see it, if we place the bromine axial, wouldn't its sheer size push down on the fluorine atoms in the equatorial plane and thus distort the 180° bond angle?
On the other hand, now that I think of it, we only have one bromine pushing down on all the fluorine atoms with an equal force and the net force should cancel out, making the 180° bond angle a theoretically perfect 180° bond angle.

Also, another choice I was doubting is choice (1). Are you sure there are $8~\ce{F-S-F}$ bond angles $< 90^\circ$? I'm can see how the $\ce{Br}$ pushes down on the fluorines in the equatorial plane, creating four bond angles $< 90^\circ.$ Where's the other four? Or are these from taking the $\ce{Br}$ to be in the equatorial plane?

The term "generally octahedral" refers to a distorted octahedral arrangement of the ligands referring to the central atom (as ron already stated). In this case the distortion is due to the massively different bond lengths.
In addition to ron's answer I did a little calculation at BP86/cc-pVDZ. I did not use any symmetry constraints at first and the final geometry deviated from ideal $C_\mathrm{4v}$ geometry only slightly due to rounding errors. After symmetry constraint reoptimisation, the following parameters may be found:
\begin{array}{lrl}
\text{parameter}                & \text{value}&\\\hline
\mathbf{d}(\ce{S-Br})           & 2.039 &\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{ax}})       & 1.638 &\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{eq}})       & 1.650 &\mathrm{\mathring{A}}\\\hline
\angle(\ce{F_{ax}-S-Br})        & 180.0 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-Br})        &  90.7 &{}^{\circ}\\
\angle(\ce{F_{ax}-S-F_{eq}})    &  89.3 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-F'_{eq}})   &  90.0 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-F''_{eq}})  & 178.7 &{}^{\circ}\\\hline
\angle_\mathrm{D}(\ce{F_{eq}-S-F'_{eq}-F''_{eq}})
                                & 178.7 &{}^{\circ}\\\hline
\end{array}

I was unable to determine a interconversion barrier related to the $\angle(\ce{F_{eq}-S-Br})$ bond angle change. I was also not able to find a different stable isomer.
In solid state it is most likely, that the geometric parameters will change due to packing effects and/or more internal entropy.

Addressing the edit:

Are you sure there are 8 F-S-F bond angles < 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles < 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane?

Yes, the bromine pushes down the fluorines of the equatorial plane resulting in $\angle(\ce{F_{ax}-S-F_{eq}})=89.3^{\circ}$ bond angle. Due to this, the sulfur is out of plane with the equatorial fluorine atoms. Since the planes expanded by $\ce{Br-S-F_{eq}}$ and $\ce{Br-S-F'_{eq}}$ have to be at $90^\circ$ due to symmetry constraints, this is creating also a slightly smaller angle than $90^\circ$ for the equatorial fluorines. In this very case it can be found to be $\angle(\ce{F_{eq}-S-F'_{eq}})=89.992^{\circ}$, which was rounded by the program to fit the one decimal place restriction. The application of this is just trigonometry and you can derive it as a simple thought experiment - if you like.

What's " BP86/cc-pVDZ"?

@Dissenter It denotes the level of theory used for the quantum chemical calculation. In this particular case it is the conjunction of the Becke88 and Perdew86 density functional together with a correlation consistent polarised valence double zeta basis set by Dunning et. al. It is a quite reliable tool for molecular geometries, while it is sufficiently simple to calculate (it does not need much time).

Sorry I'm trying to think of this again; I'm not sure why the Feq-S-F'eq bond angles would be compressed, even if the S were out of the plane.

@Dissenter Please have a look at this Q&A I posted especially for you.

I believe the writer is using the term "generally octahedral" much like when I say an orbital is "roughly $\ce{sp^2}$ hybridized."  Few compounds are perfect octahedrons, few orbitals are exactly $\ce{sp^2}$ hybridized.  If a molecule's structure looks generally like an octahedron, but it isn't perfectly octahedral, then we can say that it is "generally octahedral" or better yet, a distorted octahedron.
A useful way to determine if something is a perfect octahedron, tetrahedron, $\ce{sp^2}$ orbital, whatever, is through the use of symmetry. If any two atoms can be interconverted by rotation about an axis of symmetry or reflection through a plane of symmetry, then those atoms must be equivalent.  If two atoms cannot be interconverted by such symmetry operations than they must be non-equivalent.  Point groups and the symmetry elements they contain can make for an entire chemistry course, but the essentials can be learned quickly.  Here is a useful reference (one of many, you can google "molecular symmetry, point groups" to find something more to your liking) on the subject.
Let's use the following drawing as a starting point to discuss the geometry of $\ce{BrSF_5}$.  We will use the term "axial" or "apical" to describe the F and Br atoms at the top and bottom of the drawing respectively, while the remaining 4 fluorines are termed "equatorial.

Applying symmetry to $\ce{BrSF_5}$ tells us the following

There is a 4-fold rotational axis passing through the Br-S-F(axial) atoms
There are 4 planes of symmetry containing this axis

The rotational axis interconverts the 4 equatorial fluorine atoms, therefor they are equivalent as are the 4 F-S-F angles which must be (360/4) 90 degrees. Further, at least as I've drawn it here, the same rotational axis interconverts the bromine with itself (it sounds silly, but think about it) and the axial fluorine with itself.  Therefor, the Br-S-F(axial) angle must be 180 degrees. There is no plane of symmetry containing the S and 4 equatorial fluorines (because reflection through such a plane does not interconvert what's above the plane (F) with what's below the plane (Br).  This is a very important observation.  It tells us that the Br-S-F(equatorial) angle cannot be exactly 90 degrees.  It might be 90.001 or 85.278 degrees, but it can't be exactly 90 degrees - because of the lack of symmetry.  Symmetry can only tell us if things are the same or different, it can't tell us if different things differ by a lot or a little, we need to use our other chemical skills to do that.  So if the angle is not 90 degrees, then the S and 4 equatorial fluorines do not lie in a common plane; the 4 equatorial fluorines must be canted towards either the axial fluorine or the axial bromine.  I don't know which of those two options is preferred, but I would guess that the energy difference between these two forms is small and the barrier to their interconversion is small, so I might expect rapid interconversion between these 2 slightly distorted octahedral forms at room temperature.  At absolute zero, or perhaps in a crystal, only one of these two forms would prevail.

I believe that the distortion of the bromine atom moving from "axial" to "equatorial" has a lesser barrier, than the (symmetry constraint) distortion of $\angle(\ce{F_{eq}-S-Br})$. But I was not able to find another conformer or a interconversion barrier (see my answer).

@Martin  Is the basis set you used accurate at predicting barrier heights, like in amine inversion?

The basis set should be fairly accurate for geometries and still good enough for reaction barriers. The method is quite robust, but ab initio or hybrid functionals would probably predict a better barrier. The biggest trouble here is converging the scf because of a very high symmetry and a lot of electrons - I simply cannot afford anything more costly at the moment for these projects.

@Martin  were you surprised that you didn't find a second minimum with the 4 equatorial fluorines canted towards the axial bromine?

Not really, the out of plane distortion is not even a degree on that level of theory. I also suspect that this is a very shallow minimum - If there is another minimum, it will be hard to find with any method. One would probably have to do highly accurate calculations on at least MP2 with QZ basis optimisations or even better CCSD(T), but who could afford that. Or lots of singlepoint scans to map the PHS/PES.

Btw: I just noticed, that The rotational axis interconverts the 4 equatorial fluorine atoms, therefor they are equivalent as are the 4 F-S-F angles which must be (360/4) 90 degrees. is not true, as this would only apply if sulfur is in the plane with the equatorial fluorines. See the edit of my answer. It is hardly noticeable here, but with stronger pyramidalization it's obvious.

That's right, but like I wrote, "Let's use the following drawing as a starting point to discuss the geometry."  In "that" geometry the C4 requires 90 degrees.  Then I made the point that Br-S-F(eq) can't be 90.  Should have come back and refined the F(eq)-S-F(eq) angle at that point.  Main point though is that all the F(eq)-S-F(eq) angles have to be identical.  I appreciate your incisive comments.

Ahh you are absolutely right, I should have noticed that. I really liked the answer the first time I read it, so I shouldn't have reread only parts of it to confuse everybody else. Let's just agree on agreeing ;) (We got the yellow card from the site already.)


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How can metals like magnesium and sodium burn underwater?
Any websites I have found have conflicting answers.
One said that they don't actually burn underwater, rather they separate hydrogen from oxygen and burn at the surface. Another actually said that they do burn underwater because they take the oxygen source from the water. I do not have a specific definition for "burn" (the question itself is my assignment- write an essay answering that question!), so I know about as much as you do concerning the wording of the assignment.
I do know that alkali and alkaline earth metals are generally very reactive with water, and usually result in exothermic reactions.

Considering that you are an 8th grader, I think you already know all the answer, you are just overdoing it. I am not your teacher, but if you wrote an essay actually pointing out the vagueness of the term "burn", or that are several possible mechanism of oxidation, I would call it a very good job.

Your last sentence was the key.  Metals like sodium and magesium react with water according to the following equations
$$\ce{2Na + 2H2O -> 2NaOH +H2 + heat}$$
$$\ce{Mg + 2H2O -> Mg(OH)2 +H2 + heat}$$
Both reactions produce hydrogen gas and a lot of heat.  As the hydrogen gas bubbles out of the water it mixes with atmospheric oxygen. The intense heat ignites the hydrogen-oxygen mixture and the mixture continues to burn (possibly explosively depending upon the quantities involved) as long as hydrogen continues to be generated.

Magnesium, actually, reacts only slowly with water. Basically it might react enough so that ribbons can temporarily float themselves on generated hydrogen bubbles, but that's about it. Magnesium that has been set on fire continues to burn in steam.

To my knowledge, neither of these metals "burn" underwater.  The word burn generally suggests that something reacts with oxygen.  As others have stated above, these very active metals will react with water by displacing hydrogen as H2.  The hydrogen definitely will react with oxygen assuming that enough heat is evolved to start the reaction (very likely for sodium, less so for magnesium).

It may depend on language (my native language is not English), but by burning in this case one means a rapid oxidation with heat and light generation. E.g. magnesium torch can be light in normal air then a diver can take with her/him under water, and it still produces the light and all stuff.


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What is the mass of 1 Liter of ethylene at 2000 bar pressure?
I was wondering the weight of 1 liter of ethylene at 2000 bar and 310 Kelvin?
I have tried to just multiply the weight at 1 bar with 2000 but I don't think i got the correct answer because the answer I got was just below 2.4 kg per liter which seems way too high.
I calculated it with constant temperature.

If we treat it as an ideal gas, then
\begin{aligned}
  PV &= n\mathcal{R}T \\ \mathcal{R} &=0.08314~\mathrm{L~bar~K^{-1}~mol^{-1}}\\
  n &=\frac{PV}{\mathcal{R}T}=\frac{2000\cdot1}{0.08314\cdot310}=77.6~\mathrm{mol}\\
  M(\ce{ethylene}) &=28~\mathrm{g~mol^{-1}}\\
  \text{mass of ethylene} &= (28\cdot77.6) = 2.17~\mathrm{kg}
\end{aligned}
Edit:
 Using the link[a] in the comment I find that the density of ethylene at $310~\mathrm{K}$ and $1800~\mathrm{bar}$ is $\varrho = 20.385~\mathrm{mol/L}$, and $\varrho = 21.065~\mathrm{mol/L}$ at $2200~\mathrm{bar}$.  Averaging these two numbers gives an estimated density of $\varrho = 20.725~\mathrm{mol/L}$ at $310~\mathrm{K}$ and $2000~\mathrm{bar}$.  So 1 liter of ethylene under these conditions would contain
\begin{aligned}
  \varrho\cdot M_{\ce{C2H4}}\cdot V & ~= m_{\ce{C2H4}}\\
  20.725~\mathrm{\frac{mol}{L}} \cdot 28~\mathrm{\frac{g}{mol}}\cdot 1~\mathrm{L} & ~\approx 580~\mathrm{g}.\\
\end{aligned}
  Is that more in line with your expectations?
[a] Majid Jahangiri, Richard T Jacobsen, Richard B. Stewart and Robert D. McCarty: Thermodynamic Properties of Ethylene from the Freezing Line to 450 K at Pressures to 260 MPa. J. Phys. Chem. Ref. Data 1986, 15 (2), 593. (DOI:10.1063/1.555753) (mirror at www.nist.gov)

I like your answer, but I am curious if ethylene at $200~\mathrm{bar}$ is still a gas. (I would tend to liquid) Also I do not really think at these pressures any gas will behave ideal, probably not even helium.

I thought about the gas-liquid question.  Since a volume was given I went with gas.  If it was liquid and we "assume" (I know) incompressibility - well it seemed too straightforward.  But yes, whether a gas or a liquid, probably far from ideal.

Here at work we have compressors that compress the gas to 2300 bars and the compressor can only handle 30% liquid so it is not a liquid, at least not in the sense that it can still be compressed but one of the engineers her found that there was something called "compressibility factor" which seem to have a lot of influence on the answer.

Someone linked this link at the physics section of this site it has the answer

If I take the molar density that is given in the tables on that site and multiply it by the molar weight of ethylene I get 28.05g/mol * 20.065 mol/dm^3 = 562.82325 g/dm^3   .
This is at 310 K

20.065, do you mean 21.065?

Wow, that is a horrible deviation. I wonder if you would get any closer by applying the van der Waals equation: $$\left(p + \frac{n^2 a}{V^2}\right)\left(V-nb\right) = nRT.$$

The compressibility factor $Z$ is just a empirical way to describe non-ideal gasses by slightly modifying the ideal gas equation. For ideal gasses, $\frac{pV}{nRT}=1$, but for non-ideal gasses the ratio can be either smaller or larger than 1. By performing experiments in different conditions, it is possible to approximately tabulate values of $Z$ and attempt to interpolate/extrapolate. For gasses in conditions close to ideal (high temp, low pressure), $Z$ varies quite smoothly, but at extreme conditions it probably behaves very erratically

I did look for Z values for ethylene, but couldn't find anything around 2000 bars.


text:
Is there a readily available liquid that's more dense than water and insoluble?
I am looking for any liquid that does not dissolve in water with higher density and does not react with water, one that is readily available.

Possibly a duplicate (at least in intent) of: http://chemistry.stackexchange.com/questions/641/liquid-which-neither-dissolves-in-water-nor-oil?rq=1

of course not ?

yes i too think that this question is not similar to the question you qouted

@user7063: Apologies, I leaped to conclusions about why you were asking for what you were.

The first solvents that come to my mind are  $\ce{CH2Cl2}$, $\ce{CHCl3}$, and $\ce{CCl4}$.

Trichloroethylene is a common chemical solvent that has reasonably low toxicity, it has density ~1.5 g/ml. However, it slowly becomes restricted. Tetrachloroethylene also may be considered. Dichlormethan has reasonably low toxicity, but its BP is too low. I would not use the liquids mentioned without fume hood on regular basis. One-time exposure is probably OK, but repeating exposure may lead to cancer.  Chloroform may be considered, but I recommend to avoid it, it has accumulating liver toxicity. Tetrachloromethane should be avoided, it also has undesired toxicity.
Mercury MAY be considered, it is toxic if one is repeatedly exposed to vapours, but has very low vapour pressure and is reasonably cheap. But it has annoying tendency to produce small drops that are very hard to collect and move into smallest holes and accumulates in the system, so it clearly requires dedicated  workplace.
Liquids mentioned above are reasonably cheap, so if they are not restricted in your home country, you should be able to allow yourself one. Unfortunately, that's all ideas I got. Well, there are some more, but they either has unreasonable cost, or are too toxic. Bromine, for example, is toxic and, though not mixing with water, partially dissolves in it.

Wikipedia has a list at http://en.wikipedia.org/wiki/Heavy_liquid including some comments on toxicity. Not worth an answer since it was a simple search.

Thanks for the response.
The easiest way i observe is to use salt water and fresh water ,i would like to confirm if they are shake proof.Even if i shake the bottle the two liquids would not mix.
The list has some liquids which interest me.
I would like to know if they would not mix or react with water and if they are easily available for me to use.
For eg Clerici solution.
Thanks

Salt water and fresh water obviously mixes well. No offence, but if you plan to play with toxic chemical you need at least some basic chemical knowledge.

@user7063 I strongly encourage to drop the list from Wiki. It includes mostly rare and toxic chemicals and solutions of rare and toxic salts in water. The chemicals I listed are cheap and relatively safe, I worked with them and still alive and healthy.

@user7063 Actually, if you need a liquid that does not mix with water, it is much easier to use one that has lower density than water. Toluene, gasoline, oil,  hexane are well known examples that are relatively safe.

@Greg  thanks for your reply.
I am not a chemical expert neither a noob.
I have seen that sea water and fresh water do not easily mix in the sea. Thus i wanted to confirm its shakeproof.

@permeakra Thanks for the warning.
I have not handled any toxic chemcicals.
OF the list you mentioned could you temme which one is more appropriate the density values are not far away.
So i get the doubt they are not shakeproof .
Thanks

@user7063 From my list I recommend tri- or tetra-chloroethylene (NOT -ethane). The compounds do not mix with water but in presence of surfactants may produce emulsions with water. Well, don't mix them with surfactants. Check for solubility data when considering amount of liquid.


text:
What is the functional difference between a radical anion and a nucleophile?
Radical Anions are hundreds if not thousands of times more reactive than nucleophiles (right?). A radical anion is just a molecule with an unpaired electron. Where as a normal radical is missing an electron and an atom. If an acidic organic compound is deprotonated to form a nucleophile by removing a proton, then is there not an unpaired electron availible? (which makes the compound nucleophilic)
Why aren't nucleophiles as reactive as radical anions, since they both have "free electrons"? What is the functional difference?

Radical Anions are hundreds if not thousands of times more reactive
  than nucleophiles (right?).

Radical anions are very reactive, but some nucleophiles (such as $\ce{CH3^{-}}$) are also very reactive.
A radical anion is a molecule that has had an electron added to it.  For example, treating naphthalene with sodium in liquid ammonia transfers an electron from sodium to naphthalene forming the naphthalene radical anion, the electron is unpaired.
A radical arises from the homolytic breaking of a bond.  This process leaves an unpaired electron with each of the two molecular fragments.

If an acidic organic compound is deprotonated to form a nucleophile by
  removing a proton, then is there not an unpaired electron availible?

Removal of a proton $\ce{H^+}$ (just a proton, no electrons) would leave 2 paired electrons (and a negative charge) on the remainder of the original molecule.
Nucleophiles come in a lot of flavors.  They can be neutral, like the nitrogen in an amine, or they can be negatively charged.  If negatively charged, the nucleophile could be a radical anion or a carbanion.  So radical anions just belong to a sub-class of nucleophiles.  It's probably fair to say that carbanions and radical anions are usually better nucleophiles than a neutral nucleophile.  Comparing the nucleophilicity of a carbanion and a radical anion is harder to generalize since they are both so reactive, but the carbanion with 2 electrons would probably be more nucleophilic than the radical anion with one electron..

Hmm I have thought about that, but a question arises here: In the SRN1 Radical Nucleophilic Substitution reaction, a nucleophile in solution reacts with an aryl radical to form an [alkylated] radical anion. This radical anion then passes its electron to an aryl halide forming a aryl halide radical anion which splits into a fresh aryl radical that can be alkylated and a halide anion. The question is, if the radical anion formed when the nucleophile reacts with the aryl radical, is also a nucleophile, then why does it never attack and alkylate other aryl radicals (causing polymer formation?)

"This radical anion then reacts with a halide forming a halide radical anion"  OK.. "which splits into a radical and a halide anion" sounds like the reverse reaction.  What's the question?

It's not a reverse reaction unless the 'product' radical anion fragments, which only happens if there are strong EDG's on the ring. The question was, why doesn't this product radical anion alyklate as the other nucleophiles in solution do. Instead it just passed on its electron to the availible aryl halides. Is it not a nucleophile like you said? The reaction in question and mechanics: http://en.wikipedia.org/wiki/Radical-nucleophilic_aromatic_substitution

OK, now I follow.  1) as I mentioned in my answer, radical anions are probably less nucleophilic than nucleophiles with a full negative charge (in this rxn they use amide, alkoxide, enolate - all with a full negative charge), 2) the recently formed aryl-nucleophile radical anion has extensive delocalization of the electron into the aromatic ring, making this radical anion much less nucleophilic, 3) the recently formed aryl-nucleophile radical anion is more sterically hindered than the original nucleophile making it less likely to react again, 4) I bet you do get some tar (polymer) in the rxn.

This reaction initiated by Sodium Amalgam, goes to total completion very fast (98% yields) if Na/K metal is used for enolate, because unterminatable radicals are created from the amalgam. When the reaction is run only with (Butoxide/Na/K)enolate, almost nothing happens; UV or other catalyst is then required. By products only happen if Butoxide is used as base(dehalogenated arene), or arene has EWG(homocoupling occurs)
Aryl halide accepts electron from the radical anion much easier than from the nucleophile. So how can the radical anions be less Nu- ? Or am I comparing apples and oranges here?

Electron transfer is different from nucleophilic attack (bond formation process) - yes, apples and oranges.  If the original answer was helpful, please mark it as accepted, thanks.


text:
Is FA 18:2 undifferentiated the same as linoleic acid?
On wiki, linoleic acid has the code number FA 18:2 cis,cis-9,12. But then what exactly is FA 18:2 undifferentiated?

It's part of an old nomenclature used in the fatty acid industry.  FA 18:2 translates to F atty A cid with 18 carbon atoms and 2 double bonds.  You can read more about these "Lipid numbers" here.  So to answer your question, there are number of fatty acids with 18 carbons and 2 double bonds, linoleic acid is just one of them.  FA 18:2 includes linoleic acid, but is not limited to just linoleic acid.


text:
Rearrangement of a gas law equation
For example, let us that say we have a question where we have to use the ideal gas law and that the moles and temperature are constant. Then the gas law can also be written as $$nRT=PV$$ However, according to my solutions manual, this can also be written as $$nRT=P_1V_1=P_2V_2$$
My question is, how are we justified in rearranging the equation like this?

If one rearranges the ideal gas law equation, you can obtain the following (assuming $n$ and $T$ are non-zero):
$$\frac{PV}{nT} = R$$
$R$ is a constant, and there are in fact infinitely many possible sets of values $(P, V, n, T)$ that satisfy the equation. Let $(P_1, V_1, n_1, T_1)$ denote one such set, and let $(P_2, V_2, n_2, T_2)$ denote a second one. Then, we have:
$$\frac{P_1V_1}{n_1T_1} = R$$
$$\frac{P_2V_2}{n_2T_2} = R$$
So, by the transitive property:
$$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$$
You can let $n$ and $T$ be constant as well (as in your specific example), but the principal is the same and leads to an analogous result, since any product and/or ratio of constants will itself be a constant.


text:
Which metals with high density do not react with mercury?
I would like to know the highest density material between 10-17 g/mol range that is less than mercury and does not react with mercury.
Also would I like to know the highest density metal that is in solid form naturally.
Which ones are easily available and are highly dense?

Iron is not smelted by mercury (it's ~8 g/ml). Not sure about 10-17 g/ml. 2) The mosts are Iridium and Osmium. The most easily available is lead, though it requires some caution when used.

Here is a link to a chart of all the elements in the Periodic table and their densities.  Iridium at $22.65~\mathrm{g/cm^3}$ is the densest of all the elements with Osmium a close second at $22.61~\mathrm{g/cm^3}$.  Sigma-Aldrich sells both of these metals (Iridium \$126.50 for 0.5 gram; Osmium \$195 for 1 gram).

the highest density material between 10 17 range that is less than
  mercury and does not react with mercury

Mercury has a density of $13.534~\mathrm{g/cm^3}$.  It sounds like you're asking for something with a density right below mercury that doesn't react with mercury. Rhodium at $12.45~\mathrm{g/cm^3}$ is probably the most readily available element with a density less than mercury.  As to rhodium's reaction with mercury, an alloy is formed at 175 C.  If you're working at room temperature, especially for shorter periods of time (like hours) you'd probably be OK, but it really depends on your specific application.

Note: don't even think about buying osmium, the bastard oxidizes on air producing toxic vapors of tetroxide.

Thanks for the helpful reply.
I see rhodium is highly expensive
Rhodium costs Price 1g= $70,10g= $490 >
Yeah,Osmium is also more dense.
Oxidization on air- Thanks for the info.


text:
What salts (or other substances) are available for increasing water's boiling point?
Increasing the boiling point of water by adding salts or other compounds would be very useful when working with a water bath while refluxing, because with pure water you cannot reach temperatures over 70°C (at that point too much evaporates from the bath). I know that simple table salt would work, but are there other and better compounds/salts available?

For relatively dilute solutions of non-volatile solutes, the following equation (which assumes ideal behavior) can be used to predict boiling point elevation with reasonable accuracy:
$$\Delta T_b = K_b \cdot b \cdot i$$
Here, $K_b$ is the solvent-specific ebullioscopic constant, $b$ is the molality of the solute, and $i$ is the so-called van 't Hoff factor, which accounts for the extent of solute dissociation. For example, if one were to dissolve $\ce{NaCl}$ in water and assume total dissociation, $i = 2$, since two ions are produced in solution. It follows that, ceteris paribus, a salt that dissociates to yield a greater number of ions will produce a greater boiling point elevation. In practice, the magnitude of boiling point elevation is capped by solubility and deviation from ideal behavior at concentration extremes.
Water also forms negative (i.e., higher-boiling) azeotropes with various substances. Addition of a miscible higher-boiling liquid compound can also effect elevation of the boiling point (assuming the mixture does not form a positive azeotrope), a colligative property similar to the above case for salts. Various polyols (e.g., glycerol, propylene glycol, etc.) are commonly used for such applications.

Hmm, water has a very bad ebullioscopic constant. With table salt I would only add an extra 6°C :( , according to that formula.

With calcium chloride I can get up to ~18 °C using the formula (but that is at extreme concentration 1kg/L and this formula only works with dilute concentration, right?). Would this work well enough to heat water up to 100 °C without evaporation?

@Jori, it should allow the solution to reach a higher temperature before boiling, so with carefully controlled heating you can certainly reduce the evaporation somewhat, but some will always occur. I can't guarantee this will be effective. Personally, I'm of the opinion that an alternative solvent might be preferable (such as ron's suggestion). If you have access to it, you can also consider use of a steam bath.

We always used (silicone) oil baths in the lab for refluxing and distilling.  It offers a much wider temperature range than water.  Here's two links that describe oil bath usage and lab reaction heating in general.