text:
Coffee and orange juice concentrations with a twist
This question dates back to a time where I used to watch something called The Curiosity Show... Kinda like Bill Nye the Science Guy's show. I just can't recall the answer to this so I'm hoping that someone could shed some light.
I have two glasses of equal size and volume, with equivalent volumes of orange juice and coffee, one beverage in each glass.
Given that I have a hypothetical non-stick, hydrophobic teaspoon on me, if I was to get my teaspoon and take a level teaspoon of coffee out of the coffee glass and put it in the glass of orange juice, give it a stir, and then get a level teaspoon of this orange juice plus coffee mix and put this teaspoon of liquid in the coffee glass and give it a stir... Do I now:

have more orange juice in the coffee glass than coffee in the orange juice glass;
have more coffee in the orange juice glass than orange juice in the coffee glass; or
have equal amounts of coffee in the orange juice glass and orange juice in the coffee glass?

I'm voting to close this question as off-topic because it is not a chemistry question.

@CurtF. I'm dealing with concentrations here... In such a situation where equal samples are taken out and switched in a certain order, how are concentrations affected? I could have done this with equal concentrations of two inert solutions. Would you have voted to close it then?

This problem is identical to the water/wine problem (asked here), which while it is a fascinating problem, the answer requires more logic than chemistry.


text:
Neutralization reaction with insoluble salt (weak acid, strong base)?
Consider the following reaction.
$\ce{2 HF(aq) + Ca(OH)2(aq) -> CaF2(s) + 2 H2O(l)}$
The reaction produced a fairly insoluble salt.
Assume that this reaction proceeds entirely to completion as the acid and bases were initially in equimolar quantities with respect to their stoichiometry.
What is the resulting pH of the solution at equilibrium?
After the entire neutralization reaction has taken place we have the following reactions taking place in the solution.
$\ce{CaF2(s) -> Ca^2+(aq) + 2F- (aq)}$ with some $K_\mathrm{sp}$ value
and this reaction
$\ce{F- (aq) + H2O(aq) -> HF(aq) + OH- (aq)}$ with some $K_\mathrm{b}$ value
How would you go about calculating the values of $\ce{F- (aq)}$? I ask because every time some $\ce{CaF2}$ breaks off those free ions are used in the base reaction with water which according to Le Chatelier would result in more $\ce{CaF2}$ dissolving.
Does the chemically determined $K_\mathrm{sp}$ value in textbooks take into consideration the reaction of fluoride ions deprotonating water?

Technically fluoride doesn't deprotonate water much but binds hydronium. And yeah I'm rather nitpicky now ;)

@Mithoron 'deprotonating' was my edit from 'hydrolyzing'. If you like 'binds hydronium' better, please edit.

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text:
Why does hydrogen phosphate act as a base?
Let's look at question c:

a) Write a balanced equation for the reaction.
  $$\ce{2 NaOH + H3PO4 -> Na2HPO4 + 2 H2O (l)}$$
b) When some crystals of $\ce{Na2HPO4}$ were dissolved in water, the $\mathrm{pH}$ of the resulting solution was found to be $9.5$. Calculate the hydrogen ion concentration of this solution.
  $$[\ce{H+}] = 10^\mathrm{-pH}\, ;\qquad [\ce{H+}] = 10^{-9.5} = \pu{3.2e-10 mol L-1}$$
c) Write an equation for the reaction of the $\ce{HPO4^2-}$ ion with water to account for the measured $\mathrm{pH}$.
  $$\ce{HPO4^2- + H2O (l) <=> H2PO4- + OH-}$$

So, I'm a bit confused with what's happening. It seems that water is acting as an acid in this reaction, and donating protons to the ion. Can someone just elaborate on this whole process please because I'm not really sure what is happening. Why is it $\ce{HPO}$ and not $\ce{H2PO}$ or $\ce{H3PO}$ (as is in the original equation?) Is it a step wise process and only one step included?

I am a little confused and may have answered a question that you did not ask. Are you actually asking why in part c you should consider $\ce{HPO4^2-}$ as oposed to $\ce{H2PO4-}$ or $\ce{H3PO4}$? Please also check you sum formulae of the ions, they are not correct as written now. Let me know how I can improve my answer.

When hydrogen phosphate salts are dissolved in water there are two main equilibria formed. This is based on the fact, that hydrogen phosphate can act as a Brønsted–Lowry base, i.e. accept protons, or as an acid, i.e. donate protons. For water the same is true. In addition to this it can react with itself, which is known as the autoprotolysis of water:
$$\ce{H2O + H2O <=> H3+O + {}^{-}OH}\tag1$$
With this knowledge you can write
\begin{align}
\ce{HPO4^2- + H2O &~<=> PO4^3- + H3+O}\tag2\\
\ce{HPO4^2- + H2O &~<=> H2PO4- + {}^{-}OH}\tag3\\
\end{align}
To a lesser extent there is also the following equilibrium happening:
$$\ce{HPO4^2- + 2H2O <=> H3PO4 + 2 {}^{-}OH}\tag4$$
From the acidity constant you know that $\ce{HPO4^2-}$ will react mostly as a base, i.e. $\mathrm{p}K_\mathrm{a}(\ce{Na2HPO4})=12.35$, so the equilibrium $(3)$ will be predominant.

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What fraction of air gets expelled upon heating an empty flask?

A student forgot to add the reaction mixture to the round bottomed flask at $\pu{27 ^\circ C}$ but instead he placed the flask on the flame. After a lapse of time he realised his mistake and using a pyrometer, he found the temperature of the flask was $\pu{477 ^\circ C}$. What fraction of air would have been expelled out?

By solving $$\frac{V_1}{V_1 + V_2} = \frac{T_1}{T_2}$$ I found $$\frac{V_2}{V_1} = \frac{3}{2}.$$ But according to my book it is $\frac 35$. What exactly is this ratio?

Your first equation should be:
$${V_1\over V_2} = {T_1\over T_2}$$
We want to know what $V_2$ is in terms of $V_1$. Here, I call $V_2$ final volume. Arranging the equation and using the values for temperature in kelvin:
$${\rm final\;volume} = {V_1 \times 750\over 300}= {5\over 2}V_1$$
To get the amount that was expelled, we subtract the initial volume ($V_1$) from the final volume:
$${\rm amount\;expelled} = {5\over 2}V_1 - V_1 = {3\over 2}V_1$$
The fraction expelled is then:
$${{\rm amount\;expelled}\over{\rm final\;volume}} = {3\over 5} $$

Thanks for the answer. I should've given it a second thought.


text:
Why do different elements form different types of carbides?
What property of the elements make them form different types of carbides like:
$\ce{Be}$ and $\ce{Al}$ - $\ce{Be2C}$ and $\ce{Al4C3}$ (Methanides) contains $\ce{C^4-}$ ion
$\ce{Na}$ and $\ce{Ca}$ - $\ce{Na2C2}$ and $\ce{CaC2}$ (acetylides) contains $\ce{C2^{2-}}$ ion
$\ce{Li}$ and $\ce{Mg}$ - $\ce{Li4C3}$ and $\ce{Mg2C3}$ (sesquicarbides) contains $\ce{C3^4-}$ ion.
Boron - $\ce{B4C}$ (covalent carbides)
Titanium and tungsten - $\ce{TiC}$ and $\ce{WC}$  (interstitial carbides)

This isn't necessarily a full answer but I note that the most polarising cations ($\ce{Be^{2+}}$ & $\ce{Al^{3+}}$) are associated with the least polarisable anions ($\ce{C^{4-}}$) and vice versa. Additionally, insterstitial carbides only form with large cations, as you might expect if the anion has to fit in the spaces in the cation crystal structure.

I think it is due to the high lattice energy released when these compounds are formed. Also what @bon said.

@bon With that many negative charges on one atom, I would think $\ce{C^{4-}}$ would be the most polarizable anion of the three, wouldn't it?

You might wanna consider diagonal relationship which accounts for similarities between those particular metals.

@jatin you gave a very important hint. thank you. I am going to write the answer.

According to wikipedia,

Carbides can be generally classified by chemical bonding type as
  follows:

salt-like carbides
covalent carbides
interstitial carbides
"intermediate" transition metal carbides.

This type of classification is based on the electronegativity of the element to which the carbon is bonded.
According to this site:-

The most electropositive metals form ionic or
  saltlike carbides, the transition metals in the middle of the periodic
  table tend to form what are called interstitial carbides, and the
  nonmetals of electronegativity similar to that of carbon form covalent
  or molecular carbides.

Actually credits goes to @jatin  who gave me an important hint:-

You might wanna consider diagonal relationship which accounts for
  similarities between those particular metals.

As you can see that Li and Mg, Na and Ca and Be and Al all form diagonal element pairs. So, they are expected to have similar properties. So, considering their electronegativity, the diagonal element pairs have almost same value (though Li and Mg electronegativity is quite different). So they form similar type of carbide.

(original picture)
As for the intestitial carbides, @bon said that

Insterstitial carbides only form with large cations (Ti and W), as you might
  expect if the anion (very small) to fit in the spaces in the cation crystal
  structure.

There is more to the story than electronegativity of the metal.  How does Li form an allene based carbidevand Ca forms an acetylide when their electronegaruvities are only 0.02 apart?  Small cation size, thus concentrated charge, has some importance beyond electronegativity, as the end carbons in the allenide ion have more concentrated negative charge than the acetylide ion.

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Thermodynamic-Fuel and efficiency
Why is the maximum theoretical efficiency of Ar engine higher than that of Nitrogen engine? On a molecular level.
For an Otto Cycle

$\mathrm{Eff} \le\left(1-r^{1-\gamma}\right)$
$\gamma=\dfrac{C_p}{C_v}$
$r=\dfrac{v_1}{v_2}$

Sys. Fluid

Nitrogen, $\gamma=1.4,\ r=10,\ \mathrm{eff(max)}=60.2%$
Argon, $\gamma=1.67,\ r=10,\ \mathrm{eff(max)}=78.5%$

I want to point to this page in our [help] and this meta post, which describe the tools we have available for formatting maths.


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Determine stable oxidation state from Ionization energy

Question:
Given the first, second and third ionization energies of an element (7 eV, 12.5 eV and 42.5 eV respectively), find the most stable oxidation state (or charge on the resultant ion) of that element.

I think that the element has to ionize with oxygen to reach a certain oxidation state, since oxygen has a 6 valence electrons, the element has to lose 2 electrons to attain stability. So the second ionization energy will the give the most stable oxidation state. So my answer is +2.
Now that happens to be the correct answer but can someone verify my reasoning?

As far as I know there is no direct relation between ionization energy and ionization state stability. Could you perhaps give me a hint to get started?

Abhishek this raises some alarms of being homework. Would you mind adding the info in the comment and any additional thoughts to the problem in your question?

you don't have lone ions but compounds with counterions and have to ionise with oxidant, does it help?

You don't necessarily need to have your element binding with oxygen to be oxidized, per se. Unless it is, of course, specified that we're talking about a reaction with oxygen. So, don't misguide yourself on focusing on the oxygen here :) After all, if we, for instance, think about burning of potassium, it is being oxidized with oxygen, however, it still only makes $\ce{K+}$ and not $\ce{K^2+}$!
The trick in this question is to look for the values, or a magnitude, of those ionization energies in question. To remove the first electron, you need to provide $\pu{7 eV}$, to remove the second one, you need to provide $\pu{12.5 eV}$. So the difference between 7 and 12.5 is not that big if we compare it with the difference between $\pu{12.5 eV}$ and $\pu{42.5 eV}$ that is required to remove the third electron. If explaining it in simple terms, the first to energy "bumps" are easy to "climb", while the last one is way high and you really need to provide a sufficient amount of energy to get that high. Thus, removing the first two electrons is relatively easy, while removing the third one is not.
This gives you the most stable oxidation state for this element to be +2. If we perform a reaction of this element (which looks like $\ce{Sr}$) with any other electronegative element, say chlorine, it would still have a charge of +2.

Always remember that the difference in ionization enthalpy of two successive states is approx or equal to $10$ to $\pu{15 eV}$ or less. The lower oxidation state is not stable; conversely if the difference is more than $\pu{15 eV}$, the lower oxidation state is more stable. Here the $3^\mathrm{rd}$ and $2\mathrm{nd}$ ionization enthalpy difference is more than $\pu{15 eV}$; that's why second ionization enthalpy is more stable

Have a look at your data booklet to see some of these patterns on the Ionisation energies table eg its quite striking with Aluminium - you will see that it can make an Al3+ ion.

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text:
Difference of aggregate and cluster?
Can anyone tell me the difference of aggregates and clusters from the molecular point of view? Thank you so much.


text:
Do atoms form either a positive or a negative charge, but not both?
I do not mean at the same time, of course, but I mean it appears from an overview of the common charges formed from ionizing various elements that each element forms one or more of either positive or negative ions, but they never have the opposite charge. For example Fe may have +4 or +3, but never -anything.
However, I am aware that this is from a common ion chart. Perhaps an atom can be a cation and an anion under specific circumstances?

Iron typically occurs as iron(II) and iron(III); iron(IV) is rather rare.

Well, iron gets from -4 to +6 so it's a matter of not finding more elaborate data.

Actually, in theory almost all of the elements can be found with both positive and negative oxidation numbers: it's just a matter of finding a system with the proper reagents and conditions to force it. If you isolate chemical species which have a very strong tendency of displaying some specific behaviour (accepting electrons, donating electrons, coordinating ions, releasing a leaving group, bonding to metals, releasing a proton, adopting a specific molecular geometry, or any other myriad properties), then you can often obtain strange results by clashing them with substances which also have the same tendency, but not quite as strong. This often causes the substance with the weaker behaviour to "operate in reverse".
Let me give a vivid and related example. As we all know, the alkali metals (group 1 elements) are exclusively present as cations with an oxidation number of +1, except in the pure metals, where it's zero, right? Well, here's something which might shatter your world: most of the alkali metals (with the exception of lithium, for now) also form alkalides, that is, stable salts containing discrete, clearly observed $\ce{Na^{-}}$, $\ce{K^{-}}$, $\ce{Rb^{-}}$ or $\ce{Cs^{-}}$ anions, with alkali metals displaying oxidation number -1.
How is this done? All you need is to find a neutral substance with a much stronger tendency to donate an electron than a neutral alkali metal atom (easier said than done). Since neutral alkali metals atoms form quite stable cations upon loss of an electron, this implies you need to search for a neutral substance capable of donating an electron and forming a cation with exceptional thermodynamic and/or kinetic stability. This can be achieved, for example, by the use of cryptands, which are cyclic molecules capable of coordinating very strongly to cations, strongly enough that they even coordinate alkali metal cations very well. The cryptand-coordinated cation is both thermodynamically and kinetically stable enough that alkalide anions, which would be extremely reactive otherwise, are not reactive enough in this case to immediately cause charge transfer and neutralize the negative charge.
Amusingly, it is actually possible to prepare a single compound which contains both alkali metal cations and anions, as exemplified by $\mathrm{[Na(2,2,2- crypt)]^{+} Na^{-}}$, which contains a sodium cation coordinated by a cryptand as a counterion to a natride/sodide ($\ce{Na^{-}}$) anion. One can imagine this compound being made by putting together a neutral sodium atom ($\ce{Na^{0}}$) and the neutral cryptand species $\mathrm{[Na(2,2,2- crypt)]^0}$. As Brian mentions in the comments, this latter species is actually an electride, which can be written as $\mathrm{[Na(2,2,2- crypt)]^{+}e^{-}}$ and thought of as a salt where the anion is a lone electron (!). Both the neutral sodium atom and the electride have a strong tendency to lose an electron in chemical reactions, but this tendency is much stronger for the electride. Thus, the electride ends up having its way, forcing its very loosely bound electron onto the neutral sodium atom, causing the neutral sodium atom to "operate in reverse" and accept an electron rather than donate it, resulting in the $\ce{Na^{-}}$ anion.
This table on Wikipedia is far more complete than most "common oxidation number" tables out there, and it lists many negative oxidation numbers for elements, including iron! For many of the transition metals, negative metal oxidation numbers are achievable by using the carbonyl ($\ce{CO}$) ligand, which removes electron density from the metal atom via backbonding. This stabilizes negative charges on the metal atom, once again allowing the resultant species to survive with the appropriate counterion.

Champion answer! For additional interest as re cryptands, I have actually just been participating in an extended discussion in the comments on a post on the blog of Dr. Henry Rzepa discussing the features of alkali  cryptand electrides, where (nominally, at least) the alkali metal is coordinated as the cation, and an electron is left to roam about the system.

@Brian I am not a computational chemist by any means, but I occasionally check Rzepa's blog, as it is full of undeniably interesting content. It's remarkable just how strongly cryptands are capable of coordinating alkali metal cations, going so far as to evict the valence electron from the neutral atom and leave it scurrying around the cryptand surface! And out of the many great things in that post, the electride dimer probably blew me away the most!

There's $\ce{H+}$ and $\ce{H-}$ too, of course. Alkalide ions are way more interesting though :D

Nicolau Saker Neto already has an excellent answer, but he doesn't mention possibly the most common instance of an element being both a anion and a cation: Hydrogen
Hydrogen usually has a positive oxidation state, but in any of the metal hydrides, such as sodium hydride, it takes on a negative oxidation state.

Unlike some, I wouldn't bet this is not common. Let's take a look at the biggest body in our system: the Sun. In its center, you can find for example some fluorine which is completely ionized. That's not really what we are used to with this electron lover.
We can also look for example at Titan's upper atmosphere. Here, interactions with the magnetic field of Saturn are responsible for the formation of species like $\mathrm O^+$, or almost every ionized molecule you can imagine with $\mathrm C$, $\mathrm N$, $\mathrm O$ and $\mathrm H$.
There are examples like these everywhere just in our Solar system.
Most chemistry classes seem to forget that our "standard room conditions" are far from being a standard in the whole universe. That's a shame because most of the "rules" we learn in high school (and some colleges) are based on the assumption that we are at those said standard conditions and stash the fact that the observed chemical behavior of elements is not necessarily even representative of what happen when you look at the whole universe.

You are correct in saying that it is possible but not common. This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases. There are small changes in what you would expect to be a straight line trend, depending on what kind of orbital the electron "resides" in as well. The two concepts are shown here: Atomic radii and Zeff image.
From both of these concepts comes the reasoning:

Q: Why are some atoms harder to ionize to a positive charge?
A: Too high of an ionization energy required because of a lot of nucleus interaction with the outer electron.
Q: Why are some atoms not seen ionizing to a negative charge?
A: Electrons shield too much of the nucleus to allow for enough interaction to hold another electron.

Keep in mind that this holds for normal day to day reactions, and why some elements are found in the oxidation state they are on earth. Once you get into an extreme energy environments there are more possibilities.

Your initial presumption is incorrect — both in the example it chose and in the logic it presents. Let’s tackle the logic first.
The most prevalent element that forms both a monoatomic cation and a monoatomic anion is hydrogen — although it is true that protons are hard to capture. Acidic compounds typically dissociate into $\ce{H+}$ and an anionic acid residue. And a wide range of relatively stable hydrides is known, including sodium hydride, tetrahydridoborates, aluminium hydride, lithium tetrahydridoaluminate and many more.
Carbon is another element that frequently forms anions and cations, depending on the mechanism. For example, $\mathrm{S_N1}$ reactions and $\mathrm{E1}$ eliminations both include a cationic carbenium intermediate. In a different cationic example, non-classical cations such as the norbonyl-cation, a carbonium ion, have been generated. Carbenium here refers to cations generated by removing a bond and carbonium to those generated by formally adding a bond onto a saturated carbon — in crystal structures, carbonium carbons seem bonded to five different residues.
Furthermore, it all boils down to how isolated the charge on a specific ion has to be. Frequently, if a carbenium ion is generated next to an oxygen, the more important resonance structure features a $\ce{C=O}$ double bond and the positive charge located on the oxygen, but that has less of a physical meaning than the carbenium cations noted above.
So what about iron? Well, you are correct that the $\mathrm{+II}$ and $\mathrm{+III}$ oxidation states along with the neutral $\pm 0$ are the most common. Another common and very stable iron compound is pentacarbonyliron $\ce{[Fe(CO)5]}$, which features iron(0). This can react with hydroxides in the following manner:
$$\ce{[Fe(CO)5] + OH- -> [Fe(CO)4(COOH)]- ->[][- CO2] [Fe(CO)4H]- ->[][- H+] [Fe(CO)4]^2-}$$
Since carbon monoxide is a neutral, two-electron donor, the double negative charge on the final complex formed by deprotonation of the intermediate hydridotetracarbonyliron(0) features an $\ce{Fe^2-}$ anion. Formally, the actual compound displaced is carbonate, not carbon dioxide.[1]

Reference:
[1]: W. Hieber, W. Beck, G. Braun, Angew. Chem. 1960, 72, 795. DOI: 10.1002/ange.19600722202.

The reason some atoms are generally found in negative and some are generally found in positive oxidation states is their relative electronegativity.
Therefore, elements with high electronegativity (such as halogens) are likely to become negative atomic ions, and elements with low electronegativity (such as alkali metals) are likely to become positive atomic ions. Of course, elements with medium levels of electronegativity (such as hydrogen and carbon) can be found in both positive and negative oxidation states.
Of course, exceptions exist in most cases. Alkali metals can form negative ions as explained in this answer, and halogens can also form positive atomic ions.
An example is this mass spectrum of methyl bromide, where the peaks at m/z 79 and 81 are due to the bromine cation.


text:
Does ionization energy have anything to do with how reactive an element is?
According to my textbook, fluorine has a pretty high (compared to elements like sodium) first ionization energy. But why does it have such a high requirement to ionize? After all, it actively seeks out atoms to bond with, even tearing through already existing bonds in its drive to do so.
I would expect that fluorine would have a very low ionization energy, as it is soooooooo close to having a full octet (just like sodium). Yet unlike sodium, it has a higher first ionization energy. Why is that?
(My guess is that fluorine's protons' affinity for electrons is so great that it simply "overpowers" the octet rule.)
Apparently, neon can also be ionized. Doesn't that mean it can bond with cations? But it is unusually stable, so why doesn't it?
I think there is an underlying law behind this that can explain these, please help!
EDIT: Fluorine has now been explained, but why can neon be ionized? After all, if you give it a positive charge by taking away electrons, won't it be able to bond to anions? But in reality, it doesn't. What is going on?

ionisation is losing of electron - if it's energy is high enough you won't find a chemical way to do it

Ionization can be also the addition of an electron, right? After all when fluorine is satisfied its charge is a -1.

Yes but than you don't talk about ionization energy but electron affinity - you can ionise atom using only some energy to make a cation not anion. Maybe I'll try to make answer.

Indeed, if it really wants to add an electron to become an anion, it is really unlikely to want to easily give one up to become a cation. Therefore, one can determine that, no, ionization energy does not uniquely determine how reactive an element is (or, perhaps, it is not single-valued - both low and high ionization energies would be reactive, just in different ways).

This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases.
We find that fluorine is at the top right of the periodic table, where the effective nuclear charge is high, and the radius of the atom is small. This means that it has a small condensed positive charge permeating through the inner shells. This causes the reactivity of fluorine to other atoms due to it's electron affinity from the unshielded nucleus. It is hard to ionize compared to sodium because of its smaller than sodium, and it's z eff is higher. These both cause the electron to be attracted to the nucleus more, causing a higher energy barrier to take it away.


text:
Why does lithium from the nitride whereas the other alkali metal forms the azide?
What property of lithium makes them forms the nitride, lithium nitride $\ce{Li3N}$ whereas the other group 1 elements forms the azide, like sodium azide and potassium azide - $\ce{NaN3}$ and $\ce{KN3}$ respectively?
Does lithium azide exist? If so, what is its structure?

This is an example where the application of HSAB is appropriate. Consider that the hardness of the metal cations of the alkali metals increases down the group. Also, consider that azide is a softer base compared to nitride. However, this explains it only qualitatively...

The formation of $\ce{Li3N}$ can be explained thermodynamically. The enthalpy of formation of the $\ce{N^3-}$ ion is extremely endothermic (it involves three successive additions of electrons) and with the larger counterions $\ce{Na+}$ or $\ce{K+}$ the lattice energy (which is proportional to $\frac{q_+ q_-}{r_+ + r_-}$) is not sufficiently large to compensate for the large $\Delta_\text{f}H$ of $\ce{N^3-}$. The same argument explains why magnesium also forms a nitride.
According to my very brief research on the Internet, lithium azide does exist but is unstable. It can be prepared via reaction of $\ce{NaN3}$ and $\ce{Li2SO4}$: see Acta Chem. Scand. 1957, 11, 581 for more details.

I’m pretty sure that sodium nitride will exist, too, won’t it?

@Jan Wikipedia says it exists, but is unstable and readily decomposes into $\ce{Na + N2}$ - no surprises there.


text:
Why are noble metals more electronegative then most metals?
I was researching about electronegativity when I looked up what a graph of electronegativity within the periodic table is. And, this appeared. I scanned it, matching up everything I knew about the elements up to that table. Yep, Group 1 is all light colored; yea, the nonmetals are all really dark blue culminating in fluorine; and yes, the noble gases are mostly zero (with the exception of the traitors xenon and krypton, which coincidentally are also the only electronegative noble gases!)
But while I was scanning the page from Wikipedia on noble gas compounds, I suddenly realized that there was a compound called xenon hexafluoroplatinate, meaning that xenon could bond with...platinum? Checking the table, I was astonished to find that the entire noble metal group was actually more electronegative then the metals surrounding them! Why?! Aren't they extremely non-reactive? How then can they crave electrons even more then regular metals?

Are you asking about noble gases or metals? You can calculate electronegat. also for other noble gases.

Metals have tendency to lose electrons so high electroneg. means it's smallest for them - perfectly reasonable.

I'm asking why are the noble metals actually even more electronegative then the other metals surrounding them. It is kinda counterintuitive to me.

It shouldn't be - actually elements with highiest and lowest electronegativity are usually highly reactive - those in the middle like gold aren't.

The noble metals are defined by a resistance to oxidation and corrosion, and this should not be interpreted as a lack of reactivity, but instead an aspect of their high EN.  So, there is no contradiction as you seem to be thinking.  Basically, they hold onto their electrons better than other metals, so it is harder for acids and oxygen to steal electrons from these metals.
The noble metals are generally considered to be ruthenium, rhodium, palladium, silver, osmium, iridium, platinum, and gold, so I'll mainly be addressing these particular elements. There are a few reasons why they should have a higher electronegativity:

The lanthanide contraction causes these atoms to have a higher than expected $Z_{\text{eff}}$. This means they hold on to their electrons tightly, so have higher electron affinities and ionisation energies which corresponds to a high EN. This doesn't carry into group 12 because accepting electrons for these elements would result in adding another energy level, so it's not as favorable.
The filling order: By period 6, the 6s, 4f, and 5d orbitals are all so close in energy that the filling order changes, which affects the properties and chemistry of the elements.  Notice that almost all of these metals, apart from Os and Ir, break the typical filling order. Adding electrons to these atoms therefore does not have the same effects that are dictated by normal periodic trends.
Relativistic effects - Somewhat related to filling order. By period 6, nuclei are so heavy that the core electrons are moving close to the speed of light.  This causes a contraction of the s orbitals, the inert s pair effect, and other things that have a big effect on the electronic structure and chemistry.

Also, the noble gases start becoming reactive around Kr, and more so for Xe, because they are so big with such high shielding that electronegative atoms are able to take electrons to form bonds.  Some people get confused about this because they think it's just Kr, and Xe being weird, but it's actually a trend and continues with Rn, but there isn't much data on Rn compounds, or much use for them, since Rn-222 is the longest lived isotope with a half-life of ~3 days.
Periodic Trends of Transition Metals could help explain some of this.

@orthocresol Based on periodic trends, the expected order to fill would be 6s, 4f, 5d, which is observed in most period 6 elements (Cs, Ba, Pr to Ir).  The "4f<5d<6s" was confusing to me based on the expected filling order, so I removed it.

Eek that was a typo on my part, my bad.

Note that lanthanide contraction is irrelevant for Ru, Rh, Pd and Ag.

Noble metals are close to filling both $s$ and $d$ subshells, so there is a certain stability in gaining electrons. Gold atoms in complexes form bonds with each other similar in strength to hydrogen bonds and can form stable $\ce{Au^-}$ salts with cations like $\ce{Cs^+}$. Platinum similarly forms $\ce{Pt^{2-}}$. There are also relativistic effects that change the characteristics of the $d$ subshell.
https://en.wikipedia.org/wiki/Aurophilicity
http://www.sciencedirect.com/science/article/pii/S129325580500230X?via%3Dihub


text:
Extracting "Crystal Radii" from a CIF file
I'm currently downloading .cif files from various online databases and using openbabel to convert the fractional coordinates that are in the .cif files into .xyz files.  The typical output will look something like this:
5
Spinel
\begin{array}{c|c c c}
\ce{Mg}      &   1.01054     &   1.01054   &     1.01054\\\hline
\ce{Al}      &   1.01054     & 1.01054     &   1.01054\\\hline
\ce{Al}      &   4.04217     &   4.04217   &     4.04217\\\hline
\ce{Mg}      &   4.04217     &   4.04217    &    4.04217\\\hline
\ce{O}       &   2.11390     &   2.11390    & 2.11390
\end{array}
My problem is that I would like the ionic radii of each element, instead of their name.  Does anyone know of a way to recover this information from the original CIF file or am I just stuck with looking up the radii manually on,
https://en.wikipedia.org/wiki/Ionic_radius ?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one on how to make your future posts better.

Well those numbers with decimals looked really eye-jarring with just a space separating them.

The CIF file won't contain information on atomic radii, only the positions (in fractional space).
Consider that the atom might be ionic, covalent, etc. so you'll have to choose the atomic radius definition you feel is most appropriate. (Or run some sort of quantum calculation and use the resulting [electron density to define an atomic radius].)
For what it's worth, Open Babel does have a data table of covalent and van der Waals atomic radii, but it sounds like you want something else.


text:
Does the addition of base to 3-pentanone give the cis or trans enolate?
I've spent several hours searching the WWW with entries like "stereoselective enolate formation" and similar ones, but I find almost only entries for Li-amide bases (LDA, LMTP) and no other bases like t-BuO, NMe3, NaH, n-BuLi concerning the stereoselectivity.
I'd like to know if the addition of these four bases to 3-pentanone all results with preference in the cis or trans enolate (and if possible with an explanation why).

Related: Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate?

How is this different from the previous question, other than being a specific case? I recommend you edit this into the previous questions rather than posting a new one.

I thought the old question is more a detailed question on the effect of the three different alkyl chains, here I'd like to discuss some general ideas (e.g. attack of the least motion, allylic A1,2 and A1,3 interaction, stereoelectronic effects and so on).

@laminin If you know all of the factors to consider  (e.g. attack of the least motion, allylic A1,2 and A1,3 interaction, stereoelectronic effects and so on), please provide your analysis for those factors.

@jerepierre It was just a guess. May be the trans-enolate is disfavoured from stereoelectronic view, because Me could donate electron density into the sigma*(C-O) (I mean the C-O bond with negative charge seems for me more like a donor than an acceptor). To get the trans-enolate you can do it with LDA because of the isopropyle group. I expect the diaxial interaction of Me with i-Pr around 4.5 kcal/mol. How much costs a A1,2 interaction of two Me groups? (it's less than the Me//i-Pr interaction)

Now I was looking if I find some reactions of ketones with NEt3: https://books.google.ch/books?id=1nT6c3TQL8wC&pg=PA376&lpg=PA376&dq=stereoselective+enolate+formation+Net3&source=bl&ots=NC3nFoeWBv&sig=9U8okeMN17kU0b5bzdnVifVsdZY&hl=de&sa=X&ved=0CC4Q6AEwAmoVChMI1PWNspDjxgIVyKtyCh031AuN#v=onepage&q=stereoselective%20enolate%20formation%20Net3&f=false I think to say if we obtain here the cis or trans enolate we have first to decide if the carbonyl oxygen is less bulky than the CH2Me moiety (which I treat here almost as Me).

As a prenote: My answer is somehow speculative because I have no experimental data or reference values of similar reactions used. In other words: The argumentation pathway below might be wrong.
Addition of LDA mostly gives the trans-enolate if a primary alkyl substituent next to the carbonyl group has been used.
Carbonyl(O) is smaller than the methyl group [ref.], therefore the allylic A13 interaction of Me//Me or even Me//i-Pr is bigger than the allylic A12 interaction.
An allylic A13 Me//Me interaction is about 3,7 kcal/mol, an allylic A12 Me//Me is may be around 3,2 kcal/mol (this is just a guess).
If HMPA is added Li might coordinate to four HMPA molecules forming a cluster and inhibiting the cyclic transition state of LDA and hence no longer providing the trans-enolate, meaning an open transition state must be lower in activation energy for the cis-enolate formation.

So I tried to draw the conformations of 3-pentanone and then treated the base as "point bulk", meaning the base is like a sphere with different sizes.
Hard bases have a bigger repulsion to the carbonyl oxygen.
I have used arbitrarily numbers for the stereoselectivity.
Here is my answer which base gives more cis-enolate than the other:


text:
Yellow color of NaCl
My book says:

$\ce{NaCl}$ shows metal excess defect, which results into the formation of an F-center in the crystal. This F-center is responsible for the yellow color of $\ce{NaCl}$.

How does an F-center provide color to the solid?

An F-center (or any color center) is a point defect in the material that generates localized electronic states. When these states are in the band gap, they will result in strong peaks in optical absorption. An F-center in an alkali-halide material is an electron bound to a negative ion vacancy.  A brief overview may be found in Ashcroft and Mermin's Solid State Physics, Chapter 30.

But how does the electron impart color to the crystal? I think it is something related to it gaining energy energy from visible light. Could you please explain this in terms of absorption and release of energy by the electron?

This is heading far more into solid state physics than chemistry... Crystals have extended electronic states, defining bands of allowable states. In NaCl there is a fully occupied valence band, an energy gap, and an empty conduction band, somewhat like many oxides. You can see through a good quality NaCl crystal because visible light does not have enough energy to promote an electron from the valance to the conduction band, so no absorption is possible. The color center is a localized state (or states) in the gap, that allow absorption in the visible because there is now a state present.

f-centers leads to the formation of an anionic vacancy
this space is occupied by an electron
these electrons absorb energy from visible region and radiate yellow colour when
they jump back to their original state
this is what my teacher's guide book says

Absorbing light at one wavelength and re-radiating it at a different wavelength is fluorescence. It's possible, but more likely that it's just selectively absorbing shorter wavelengths and leaving longer ones alone as described in Wikipedia's article on f-center, but I don't know. Does the guide book include any references for that?


text:
Number of atoms in NaCl unit cell
Consider this picture of an $\ce{NaCl}$ unit cell:

It appears to show 14 $\ce{Cl-}$ ions and only 13 $\ce{Na+}$ ions.  Given that discrepancy, how is table salt balanced for charge?  Why isn't there a charge excess?

See

The picture you showed does have an unequal number of sodium cations and chloride anions.  However, the picture shows only part of a crystal.  Every atom that is on a boundary of the shown cube, whether on a face, edge, or vertex of the cube, is shared with other "cubes" in the crystal that aren't shown in the picture.
Each of the 8 corner Cl atoms in your picture is shared with 8 cubes (7 not shown). The 6 face-centered Cl atoms are shared with 2 cubes. Each of the 12 edge Na atoms is shared with 4 cubes (3 not shown). The center sodium atom is not shared. Thus there are 8/8 + 6/2 = 4 Cl atoms per unit "cube" in your picture, and 12/4 + 1/1 = 4 Na atoms per unit "cube" in your picture. 4=4, so the charge balances.
You might be thinking that this math checks out only insofar as a crystal is actually infinitely sized.  And you may have noticed that no salt crystals are infinitely big in the real world.  These things are both true.  But even tiny specks of salt crystals are gigantic relative to atoms.  The surface of a salt crystal could involve imperfections that mean that the count of sodium atoms and chloride atoms aren't precisely equal.  But instead of 14 vs. 13, the difference is more like 100,000,000,000,000,000 vs 99,999,999,999,999,999.  And since the imperfections are at the surface, on the outside of the crystal, any charge imbalance can be corrected if an opposite-charged particle from outside the crystals floats by and neutralizes the extra charge from the extra atom.

+1 for the finite crystal treatment

Unit cells demonstrate alignment and relative position of atoms within a crystal but provide no overt stoichiometric information. The unit cell model isn't meant to imply that atoms group to form these individual cubes or shapes. As such, the atoms/charges won't necessarily balance.
In the case of NaCl, the face-centered cubic unit cell has an odd number lattice points and thus doesn't include a whole number of NaCl molecules. This is not, however, among the three unit cell criteria:

The unit cell is the simplest repeating unit in the crystal.
Opposite faces of a unit cell are parallel.
The edge of the unit cell connects equivalent points.

Unit Cell Overview

Nice answer and +1 from me.  Could be worth noting which criterion the picture in the question violates.  I guess number one?

Actually, it satisfies all three. By doing so, however, it leaves a dangling ion/atom. So it is an accurate unit cell model, but unit cell models aren't accurate stoichiometric models.

There are no "NaCl molecules". If you look at the figure posted in the answer by @andselisk, each sodium atom is surrounded by 6 chloride ions and vice versa, giving a 1:1 stoichiometry and the formula NaCl. However, NaCl molecule would imply covalent bonds between pairs of sodium and chloride atoms, which don't exist in the compound NaCl.

There are several ways of determining stoichiometric formula from the known unit cell.
Counting atoms [correctly]
Perfectly covered in the answer by Curt F.; I would only like to propose to use data in a tabular form in order not to miss any of the atoms or improperly assign their environment. Briefly, not all the atoms you see on your image belong to the unit cell 100%. From a $3 × 3 × 3$ packing diagram there are $3^3-1 = 26$ neighboring equal unit cells sharing their boundary atoms:

The share rates (let's denote it $α$) are the fractional numbers from $1$ to $1/8$ and are the same for any unit cell (not just cubic) and only depend on atom's relative location within the unit cell.
In order to adjust for the real number of atoms $N_\mathrm{cell}$, one has to multiply numbers of observed atoms $N_\mathrm{obs}$ by their share rates $α$. It's convenient to make a separate table for every crystallographically nonequivalent atom:
$$
\begin{array}{lccc}
\text{Atom:}~\ce{Na} \\
\hline
\text{Position} & α & N_\mathrm{obs} & N_\mathrm{cell}\\
\hline
\text{Inside the cell} & 1   & 0 & 0 \\
\text{On the plane}    & 1/2 & 6 & 3 \\
\text{On the edge}     & 1/4 & 0 & 0 \\
\text{On the vertex}   & 1/8 & 8 & 1 \\
\hline
\text{Total}           &     &   & 4 \\
\hline
\end{array}
$$
$$
\begin{array}{lccc}
\text{Atom:}~\ce{Cl} \\
\hline
\text{Position} & α & N_\mathrm{obs} & N_\mathrm{cell}\\
\hline
\text{Inside the cell} & 1   & 1  & 1 \\
\text{On the plane}    & 1/2 & 0  & 0 \\
\text{On the edge}     & 1/4 & 12 & 3 \\
\text{On the vertex}   & 1/8 & 0  & 0 \\
\hline
\text{Total}           &     &    & 4 \\
\hline
\end{array}
$$
The ratio between the real numbers of atoms in the unit cell is $N_\mathrm{cell}(\ce{Na}) : N_\mathrm{cell}(\ce{Cl}) = 4 : 4 = 1 : 1$, thus resulting in the formula unit $\ce{NaCl}$.
Primary coordination numbers
Often for the simple inorganic compounds it's sufficient to find the ratio between coordination numbers (C.N.) of cations and anions to determine the formula unit. For a simple binary compound $\ce{M_mX_n}$ the following simple proportion is valid:
$$m × \text{C.N.}(\ce{M}) = n × \text{C.N.}(\ce{X})$$
For instance, from the crystal structure of sodium chloride it's evident that both $\ce{Na}$ and $\ce{Cl}$ have octahedral environment, and their main C.N.s are 6:

This leads to the ratio $m : n = 6 : 6 = 1 : 1$, again resulting in the formula unit $\ce{NaCl}$.
To illustrate this approach further, in fluorite $\ce{CaF2}$ $\text{C.N.}(\ce{Ca})$ is 8 and $\text{C.N.}(\ce{F})$ is 4.
This method also works for not so primitive structures containing more than two different elements. It is also more used in reverse to determine C.N.s in tricky cases. For instance, in the structure of perovskite both $\ce{Ca}$ and $\ce{Ti}$ have well-defined primary C.N.s 12 and 6 (respectively) seen at the first glance at the content of the unit cell, whereas it's unclear what average C.N. oxygen must have. But, knowing the formula of perovskite ($\ce{CaTiO3}$) and using the relation between coordination numbers and stoichiometric coefficients, one can find that $\text{C.N.}(\ce{O}) = 6$:
$$1 × \text{C.N.}(\ce{Ca}) + 1 × \text{C.N.}(\ce{Ti}) = 3 × \text{C.N.}(\ce{O})$$
$$1 × 12 + 1 × 6 = 3 × \text{C.N.}(\ce{O})$$
$$\text{C.N.}(\ce{O}) = 6$$

A quick way to see what is going on without calculations is to move the origin of the unit cell a bit to the top, right, and back. This way, the atoms on the bottom face, on the left face and on the front face are no longer in the unit cell, and the eight atoms in the top-right-back corner are no longer shared by other unit cells. At the same time, because we did not move it far, no atoms that used to be outside the cell are move into it, so we only have to consider atoms that were in the OP's picture.
This way, we can count as we are used to (one atom is one atom), and conclude that there are four sodium ions and four chloride ions in the unit cell. Here is a picture (the shaded atoms are the ones we have to count):


text:
Can you in reality give neon a charge of +8?
First off, I am currently confused about why neon can even be ionized at all. But since it can be ionized, this is the energy required to give a mole of neon a charge of +8: 207,390,000 joules!! Or ~0.5 GJ! Giving one-fifth of a kilogram of neon the maximum charge would be like the energy from 50 armor-piercing rounds of the ISU-150 assault tank, or approximately one Tour-de-France! Charging up one kilo of neon would be basically either throwing a lightning bolt or melting a tonne of steel.
My question is, what insidious chemical or procedure could rip away all of neon's electrons like that? Or is it just simply impossible to do?
I suspect that it is possible to do (otherwise how would they be able to calculate the energy expenditure), but how is the question.

While this question isn't particularly good, I heard about plasma weapons that appear to actually work. Mass of a bolt is way lower than "one-fifth of a kilogram", but enough to be dangerous. https://en.wikipedia.org/wiki/Plasma_railgun

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

Of course you can take all the electrons off an atom - it is then called "fully stripped" in atomic physics. You don't need to do it to an entire mole, mind you. In accelerators one would send energetic neon ions through a background gas or a thin foil, and the interactions will result in various charge states coming out, up to and including fully stripped.

As an example, the Relativistic Heavy Ion Collider (RHIC) smashes bare gold nuclei ($\ce{Au^{79+}}$) and uranium nuclei (presumably also devoid of electrons, $\ce{U^{92+}}$) into each other during experiments. At least part of the ionizations are performed by electron bombardment. Of course, these species will never be found inside a flask in the lab, as they would immediately tear dozens of electrons out of whatever came into contact with them.

First - in chemistry there's technically no such thing as bare multivalent cation, second - as you think, there's no such energetic chemical reaction, third - ionisation energy is physical property (although important for chemistry) and

"(...) is usually measured in an electric discharge tube in which a fast-moving electron generated by an electric current collides with a gaseous atom of the element, causing it to eject one of its electrons."

Source "Encyclopedia Britannica"

Not only can $\ce{Ne^{8+}}$ exist, it probably does in the Sun's corona. And we may even have $\ce{Ne^{9+}}$ there.
Coronium
During the total solar eclipse of 1769, two scientists independently discovered an element, coronium, from a green spectral line emitted by the Sun's corona. The discovery stood until the 1930s, when the line was correlated instead with iron --ionized 13 times ($\ce{Fe^{13+}}$). Other spectral lines were assigned to other heavily ionized elements such as nickel. What was actually discovered was a tremendous energy source that heats the corona to the millions of degrees needed to create such ions by thermal energy. Powerful magnetic forces are known to exist on the surface of the Sun and are believed to be the coronal energy source.
Ionization energies
A look at a table of ionization energies reveals that it takes a total of about $920$ eV to ionize neon to $\ce{Ne^{8+}}$, but the corona is able to provide at least $2040$ eV to some atoms forming coronium/$\ce{Fe^{13+}}$. So clearly if the corona can produce the former, it can also produce the latter provided that the neon exists. Evidence for such neon may be found from the occurrence of this gas in the Moon's exosphere, likely derived from the solar wind.
Encore
Now, about that $\ce{Ne^{9+}}$. If we add the ninth ionization energy of neon to the first eight which were totaled above, we get a total of about $2120$ eV to form $\ce{Ne^{9+}}$, compared with the $2040$ for the known $\ce{Fe^{13+}}$. It would therefore be not much of a reach to assume that the corona can impart the ninth ionization of neon, too.

All that might be needed is a collision between two such particles.

authors: محمد امین ریاحی


text:
Relation between number of conjugated double bonds and absorbance?
I observed the absorbance on Phenolphthalein in acidic and basic medium, and I know that the absorbances differ due to the number of conjugated double bonds, but I am not sure exactly why this happens. I believe it has something to do with the delocalised electrons, which causes lower energy differences, but I am unable to establish a solid link between the concepts. Why does this happen?

When 2 p orbitals mix, a pi and a pi* orbital is formed, and the electrons occupy the lower energy pi orbital.  The UV-absorbance is the energy difference between these two orbitals such that on electron may be 'promoted' to the p* orbital.
When 3 p orbitals mix, three pi-type orbitals are formed.  One is obviously bonding and one is obviously antibonding.  There is a third one 'in the middle', frequently called non-bonding.  Because of this third orbital, there is a smaller difference between any two states, and as such absorbing a photon to promote an electron requires less energy, redshifting the absorbance.
Add a fourth p orbital and the energy differences get smaller, but with two bonding orbitals and two anti-bonding orbitals.  Add a 5th p orbital and now you have two bonding orbitals and two anti-bonding orbitals and a non-bonding orbital, with even smaller differences.  Repeat ad nauseum.
Smaller energy differences means longer wavelengths.
Check out this picture for a graphical summary:


text:
Is there no irreversible way of moving between two equilibrium states at constant T and P?
There's a reaction occurring at constant $T$ and $P$, and it is in chemical equilibrium. Now by adding infinitesimal quantities of reactants, it is taken to another equilibrium state in a reversible way. During this process, if entropy change is $\Delta S$ and volume change $\Delta V$, $$\Delta U = T \Delta S - P \Delta V.$$
If the same process is carried out irreversibly, by suddenly adding a large quantity of the reactants and disturbing the chemical equilibrium, then the work still remains $P \Delta V$ ($ \Delta V$, a state function, has the same value). In that case, the heat $q=T \Delta S$, which is only true for reversible processes. Even though the process is irreversible, there has been no entropy change in the universe? Where am I going wrong? It can only be that the work is not $P \Delta V$, but since the process is at constant pressure, the work should have this value.

Adding more reactants, reversibly or not, is changing the total number of moles in the system.  That's something that your equation doesn't account for.  $dU = T dS - P dV + \sum_i \mu_i dNi$


text:
Making carbon monoxide gas from formic acid and sulfuric acid?
Can I create carbon monoxide gas by just mixing formic acid and sulphuric acid together? Or do they have to be heated?

If you have pure, relatively concentrated formic acid, and pure, very concentrated sulfuric acid, then I don't think you need to do any heating.  BTW, creating CO gas is very very dangerous!  Please don't do it. http://www.ncbi.nlm.nih.gov/pubmed/18363119

The formation of carbon monoxide already takes place when mixing formic acid and sulfuric acid at room temperature, but the reaction is relatively slow. Since the temperature of the mixture may significantly increase during the course of the reaction, the reaction rate may increase, which makes the reaction difficult to control.
For laboratory applications, concentrated sulfuric acid (or phosphoric acid) is usually heated to about 80 °C. Formic acid is added slowly, drop by drop. The formation of gas should start upon addition of the first drop, otherwise the temperature should be increased.
The generated gas contains various impurities, which can be removed using gas washing bottles. Usually, the gas is passed first through $\ce{KOH}$ solution; however, you should take care to prevent any backflow of $\ce{KOH}$ solution to the sulfuric acid.

And remember Curt’s comment from above: Carbon monoxide is dangerous; do not meddle with it unless you know exactly what you are doing.


text:
How is the triferric tetroxide of iron possible?
If oxygen has a valency of $-2$, $4$ of it would result in $-8$. But iron doesn't have a (variable) valency of anything that goes to $-8$. Then how is $\ce{Fe3O4}$ possible?

Iron, like many of the Transition Metals does have a variable valency, it can valencies of 2+ (ferrous iron), 3+ (ferric iron) and in some cases, 4+ (tetravalent iron).
In regards to $\ce{Fe3O4}$, according to the Wikipedia page Iron(II,III) oxide,

contains both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions

A 2-dimensional diagram of $\ce{Fe3O4}$ is below, from the relevant ChemSpider page:

A 3-dimensional model is below (from Wikipedia):

red indicate oxygen


text:
Preparation of sulfuric acid through Contact process
In the preparation of $\ce{H2SO4}$ through the Contact process, $\ce{H2SO4}$ is added to hot sulfur trioxide to form oleum. This oleum is then diluted by adding water to form $\ce{H2SO4}$.
$$\ce{H2SO4(l) + SO3(g) → H2S2O7(l)}$$
$$\ce{H2S2O7(l) + H2O(l) → 2 H2SO4(l)}$$
In a preparatory reaction, different reagents are used to obtain the desired product but here the product is itself a reactant and is used to make the product. Indirectly, $\ce{H2SO4}$ is being used to make $\ce{H2SO4}$.
Would this process be called as a preparation process?

Remembered the process of making big yogurt using a smaller yogurt!

You can call it whatever you like. Personally I would say you are preparing sulphuric acid because you end up with more than you started with - 1 mole is used to form 2 moles.

The term "preparation" generally refers to the synthesis of chemical species on a laboratory scale.
Since you can perform the contact process on a laboratory scale, it definitely qualifies as a "preparation" process.


text:
Determining stoichiometry of an unsaturated hydrocarbon via combustion

$60\ \mathrm{cm^3}$ of oxygen was added to $10\ \mathrm{cm^3}$ of a gaseous unsaturated hydrocarbon. After explosion and cooling to room temperature, the residual gases occupied $45\ \mathrm{cm^3}$. What volume of carbon dioxide was formed in the reaction?

I’ve tried to make an unbalanced equation out of it:
$$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$$
My problem now is I don’t know how to proceed to the next step. How do I even find the moles? Is the equation correct? How do I balance this equation?

First step: balance your equation.
$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$ is not balanced, this is:
$$\ce{(CH2)_{\it n} + \frac{3 }{2}{\it n}~O2 -> {\it n}~ CO2 + {\it n}~ H2O}$$
Second step: perform mole balances on relevant species.
The balanced equation has $1 + \frac{3}{2}n$ moles of gas reacting, and $2 n$ moles of gas formed.  However, the question states that the products are cooled to room temperature, which would eliminate the water from the gas phase.  Thus the reaction were are really interested in is
$$\ce{(CH2)_{\it n}_{\rm(g)} + \frac{3 }{2}{\it n}~O2_{(g)} -> {\it n}~ CO2_{(g)} + {\it n}~ H2O}_{(l)}$$
This means that $1 + \frac{3}{2}n$ moles of gas react, and $n$ moles form for each mole of reaction $\xi$.  Thus, the reaction causes a loss in moles of gas of is $-(1 + \frac{1}{2}n)$ moles per mol $\xi$ of reaction.  If we assume the ideal gas law applies, then we could also say that the loss in volume from the reaction is $(1 + \frac{1}{2}n)~{\mathrm{{cm}^3}}$ per $\xi~{\mathrm{{cm}^3}}$ of reaction.
Thus a total gas balance is:
$$\rm GAS~IN - GAS~OUT = GAS~REACTED $$
$$(60~\mathrm{{cm}^3} + 10~\mathrm{{cm}^3}) - (45~\mathrm{{cm}^3})=(1 + \frac{1}{2}n)~\xi$$
A $\ce{CO2}$ balance is
$$\rm \ce{CO2}~IN - \ce{CO2}~OUT = \ce{CO2}~REACTED $$
$$ 0 - x = - n \xi $$
Third step: assess the resulting equations
These equations are not solvable for $x$ because they depend on both $\xi$ and $n$.  That is, there are two equations, and three unknowns.  Writing additional balances on oxygen or on fuel will not help because we will not know the gas remaining for either species, so each new equation would introduce new variables. Thus, we will need additional assumptions to solve the problem.
Fourth step: determine additional assumptions.
Probably we should assume that either (a) fuel was in excess or (b) oxygen was in excess, which is equivalent to assuming that (a) oxygen was completely consumed or (b) fuel was completely consumed.  At least one of these assumptions makes sense because combustion reactions are essentially irreversible.  In my answer I'll assume oxygen was in excess, meaning that the fuel was completely consumed.  You should make the alternate assumption and repeat the procedure.
Fifth step: apply additional assumptions.
Assuming fuel was completely consumed, a fuel balance is
$$\rm FUEL~IN - FUEL~OUT = FUEL~REACTED $$
$$\rm (10~\mathrm{{cm}^3}) - 0=\xi $$
This conveniently gives us a solution for $\xi$, the extent of reaction.  It is $10~\mathrm{{cm}^3}$.
Sixth step: use new information to solve existing equations
We can substitute that solution for $\xi$ into the total gas balance to get that $1 + \frac{1}{2}n = 3$, which implies that $n=4.$  That and our solution for $\xi$ allows us to solve the problem: if oxygen is in excess, then 40 $\mathrm{{cm}^3}$ of $\ce{CO2}$ were formed.

Also note that your problem statement just says "unsaturated" hydrocarbon, but it doesn't say how unsaturated the compound is.  Your formula for an unsaturated hydrocarbon, $\mathrm{{CH}_2}$, assumes that there is only a single degree of unsaturation.

How can you say GAS IN - GAS OUT = GAS REACTED. If I had the stoichiometric amount of fuel and oxygen, then the GAS REACTED would equal the GAS IN, and by your statement, no $\ce {CO2}$ would be produced.

Good point LDC3.  I may have used misleading names.  By GAS REACTED, I mean the net change in gas from the reaction.  So if there was a stoichiometric mix of fuel and oxygen, there would still be a loss of $(1 + \frac{1}{2}n) \xi$ moles of gas by the reaction, as reflected in the next equation I wrote.


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Hybridization of oxygen in phenol
What is the hybridization of oxygen in phenol molecule? I assume it is $sp^3$ hybridized because it forms 2 bonds - 1 with benzene, another with hydrogen, as in how oxygen is hybridized in a water molecule - $sp^3$.
Now the problem is, my A - level lecturers insist that it is the p orbital of oxygen that overlaps with the delocalised electrons of benzene ring, not $sp^3$. This means that oxygen can either be unhybridized, $sp$, or $sp^2$ hybridized. So which is true?

Water is not $sp^3$ hybridized (see the first part of this answer). I imagine phenol is the same, with the p-type lone pair engaged in resonance with the ring.

Referring to the link you've provided, the oxygen atom would be sp2 hybridized if it is bonded to another atom which is also sp2 hybridized, but this is not the case for water. If water is not sp3 hybridized then it would'nt have its 'bent' shape. Going back to the main point, I recall that when the orbital of oxygen overlaps with benzene ring, it will get a partial double bond, so can I assume that there exists such a resonant structure where there is a double bond between C and O? If that is legit, then the sp2 model should be correct.

Of course there are such structures, but it's not that simple.


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Platinum group metals: Why are platinum and palladium great catalysts and not the others?
Platinum and palladium are great catalysts. At the same time, other metals of the so-called platinum group metals are not. What are the atomic level reasons for this?

related http://chemistry.stackexchange.com/questions/7854/why-platinum-used-as-a-catalyst-in-fuel-cell-electrodes-and-elsewhere

"Metals of the same family" What do you mean with that? Just the same group, the coinage metals, transition state elements? Where did you get the idea from that these are great catalysts and the others would be not? What do you even mean with great catalyst? Abundant application? And what do you mean with "atomic level reasons"? About what kind of catalyst are we even talking: Molecular, surface, nano-particles, etc.? I think your question is way too broad to be answered in a suitable form. Maybe you could consider taking a look at: [ask].

I think some of the other transition elements have also catalytic properties. For example, rhodium is generally used in addition to platinum or palladium in catalytic converters to oxidize carbon monoxide and hydrocarbons on one hand, and reduce $\ce{NO_x}$ to nitrogen on the other hand.
So, the choice is often made on the basis of relative cost.
But of course, not all transition elements are adequate for these catalytic properties:

Metals such as silver and copper have a high affinity for sulfur and form metal sulfides and sulfates. As this happens, there will be progressively less metal available to function as a catalyst. While platinum tends not to become totally or irreversibly poisoned, i.e It can be efficiently recycled.
Another point is linked to "Tammann temperatures" of these metallic elements:  Actually, in a catalytic converter, metal is used in the form of nanoparticles dispersed on the surface of a porous support material. When  temperature increases the particles starts to becomes mobile and coalesce (i.e. the catalytic activity is lost). This process is well-known under the name "sintering" and it becomes tangible as the metal is close to its Tammann temperature. This temperature is often defined as the half of the metal's melting point on the absolute temperature scale. In fact, metals such as gold, silver and copper have a Tammann temperature that is well below the average temperature of exhaust-gas ($600-700^0\mathrm{C}$).
Tammann Temperature is $750^0\mathrm{C}$ for platinum, and $640^0\mathrm{C}$ for palladium. While it is only $405^0\mathrm{C}$ for copper, $345^0\mathrm{C}$ for silver and $395^0\mathrm{C}$ for gold.

To sum up, the candidate transition element for use as a catalyst in a catalytic converter:
1) has a low affinity to poisons  such as sulfur compounds;
2) can be efficiently recycled;
3)  has a relatively high melting point.

Aren't copper catalysts even more common than palladium and platinum due to its high abundance and low cost?

Sure copper catalysts are even more common than palladium and platinum. But  Tammann Temperature of copper is only $405^0\mathrm{C}$. While its $750^0\mathrm{C}$ for platinum and $640^0\mathrm{C}$ for palladium.

Platinum, palladium, and rhodium have a D-Orbital electron structure on the surface of their atoms which encourages the temporary sticking of other molecules. When other molecules stick, the overall electron cloud is changed in shape allowing the stuck molecules to rearrange into new compounds. The catalyst does not change. The rearranged molecules eventually are pushed out by new input ones (temperature caused motion) and the cycle continues.
Please see The Science of Catalysts And Catalytic Converters
Dr Emma Schofield, Johnson Matthey Technology Centre.  She says it's a quantum effect.

This is not an answer; this is a post asking for more (i.e. a having this problem, too comment)

For those like me who weren’t sure, the platinum group metals are made up of platinum, palladium, iridium, ruthenium, rhodium and osmium.
I don’t understand where you’re getting the idea from that only two of those be good catalysts. It all depends on the application you are looking into.

Wilkinson’s catalyst, a very important reduction catalyst, is rhodium-based.
The synthetically extremely important Sharpless asymmetric dihydroxylation uses osmium tetroxide as part of its catalytic system.
The Noyori asymmetric hydrogenation is ruthenium-based.
Both Crabtree’s catalyst and the Cativa process are iridium-based.

And these are just very random examples for the usage of each of those metals. There’s a million more. Just follow the ASAP articles of JACS or the Early View articles of the Angewandte to see novel catalytic uses for each of those metals pop up all the time. Catalysis journals will naturally feature even more (albeit maybe not quite as novel).
Conclusion: All of those are great catalysts. (But so are most other transition metals, too.)

A good catalyst facilitates a reaction at a lower temperature and pressure while not participating in the reaction itself.  Platinum has 4 unpaired electrons in its outermost shell as are the other elements in this group, but this shell is a d orbital which is furthest from the nucleus and very weakly forms molecular bonds.  In fact Platinum is at the bottom of the metal reactivity series.
Since its reactivity is so low while its outermost shell is accommodating for weak bonds, it lends itself well as a reaction platform for more strongly bonding reactions.  The reactants do temporarily weakly bond to the platinum atom until the other reactant finds the reactant through normal thermal motion and grabs the matching reactant away in a stronger bond leaving the platinum un-bonded.  Think of the lone platinum atom as a match-maker weakly holding reactants until the suitor reactant finds its partner and takes away from platinum matchmaker.  Palladium is similar in this regard.


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How is change in entropy of universe related to the energy released by system that can be used to do work?
What does change in entropy ($\Delta S$) represent ? How does change in entropy of a universe ($\Delta S_{univ}$) give us information about energy available to do work.
I would like to know the physical significance of entropy of universe and how it is related to energy energy released by system that can be used to do work (Not in terms of equation. I would want to know its physical significance).

Well a reaction which is spontaneous has negative value of Gibbs or Helmholtz free energy. This free energy is the energy which is made available to be used for work.
How does that relate to your question? Both Gibbs and Helmholtz free energy are derived from the argument that change in entropy of the universe is greater than zero. Physically, that is an argument that no chemical system will react in such a way that decreases the number of available microstates unless it increases the number of available microstates for the surroundings by a greater amount.

Suppose you are a magician & want to make a rabbit having internal energy $U$. so, in order to invoke the rabbit, you have to supply $U$. But wait! In order to make the rabbit, you not only have to provide $U$ but also the energy necessary to create space for the rabbit by asiding the atmosphere. So, the total energy you supply is $U + PV$. This is enthalpy. But again wait! You don't have to supply the entire $U + PV$ as there is an associated entropy with the formation of the rabbit i.e. $ S$ i.e. the energy supplied by you, the Great Magician is $U + PV - T S$. This is the free energy as it is the only energy you will get after vanishing the rabbit.

This analogy was mentioned by Daniel V. Schroeder.


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How to dispose of pure potassium
I have about 2g of 99.5% Potassium.
I want to get rid of it, right now it's in a glass jar sealed.
What is the proper protocol? I was going to combust it in water and wash it down the drain but the by-product potassium hydroxide is caustic and hazardous so I don't think that's best.
Could I just throw it away and it goes to a garbage facility where the vial probably breaks and it reacts with rain water and the potassium hydroxide is so small and buried that it never really has an impact?

I have my moments ;) good question too!

Related: How to properly dispose of sodium…or store it properly

Proper protocol probably varies by country, state/province, and municipality.

I was going to combust it in water and wash it down the drain
Absolutely NOT !
Look at this video. It looks they used no more than 0.5 g of potassium.

Throw it in a lake!.

Read it as '2kg'. I might or might not know a university teacher that solved his 'surplus sodium problem' by throwing it into the city's main river.

In a suggested lab, the RSC recommends using propan-2-ol:

If there is any unused potassium remaining at the end of the experiment, remove it with the tweezers and place it in a beaker containing about 100 cm3 of propan-2-ol to dissolve away – it will fizz, giving off bubbles of hydrogen. When the fizzing has stopped, dispose of the resulting alkaline solution down the sink flushing it away with plenty of water. The acidic solution remaining from the experiment may also be disposed of down the sink flushing it away with plenty of water.

The same is recommended for sodium.

Thankyou! Very good, thanks for the RSC link too. Bookmarked.

I was going to combust it in water and wash it down the drain

No, mixing it with water is a bad idea.  Hydrogen gas will be liberated along with heat.  In the presence of oxygen and heat the hydrogen will form a combustible / explosive mixture.

Could I just throw it away and it goes to a garbage facility where the
   vial probably breaks and it reacts with rain water and the potassium
   hydroxide is so small and buried that it never really has an impact?

No, that's not good either, it will just explode in the dump and start a fire that someone else will have to deal with.

What is the proper protocol?

These two links (link_1, link_2) suggest the same disposal procedure.  Link_2 provides more detail. I've reproduced the key part below along with the reactions that take place prior to the final neutralization.

Your byproduct in the final equation is potassium hydroxide. If I wanted to get to potassium hydroxide couldn't I just pop it in some water and then have H2O, H2, and KOH ? The whole point of this question was that I wanted to avoid the KOH haha, is this saying to just neutralize the KOH with hydrochloric acid?

Note the last step in the paragraph, neutralization with dilute $\ce{HCl}$.  There is little or no $\ce{KOH}$ remaining at the end of the process.  In any case you don't want to use water due to a fire/explosion/splattering hazard.

While KOH is produced in both cases, the important part is the rate of the reaction. Water is a much stronger acid than tert-butanol, and therefore the reaction would proceed faster and more dangerously.

25% isopropanol (v/v) isopropanol in toluene is the usual means.  2 g potassium is not a large amount.  Sodium potassium alloy can also be disposed of this way.

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


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Why dishwashers are corrosive to aluminum?
When I put aluminium parts into my dishwasher, they get tainted and they are sometime covered with white dots similar to salt.
I don't think it is aluminium oxide.
What are the chemical reactions that have an effect on aluminium but not other metals?

I'm not 100% sure, but I think it is just alkalinity.  Unlike many other metals, aluminum is amophoteric, so it dissolves in both strong base and in strong acid.  Most dishwashing detergents are fairly alkaline, probably at pH 11 or 12 or so.  That and the heat may be enough to corrode Al.

Closely related to http://cooking.stackexchange.com/questions/8375/why-is-the-dishwasher-not-recommended-for-my-all-clad-mc2-line-of-pot-and-pans

Suggestion: @Curt F, make your comment an answer (as you wrote, sodium carbonate in the detergent, with aluminum, forms sodium aluminate, see https://en.wikipedia.org/wiki/Sodium_aluminate#Reaction_of_aluminium_metal_and_alkali)

I think it is just alkalinity. Unlike many other metals, aluminum is amophoteric, so it dissolves in both strong base and in strong acid. Most dishwashing detergents are fairly alkaline, probably at pH 11 or 12 or so. That and the heat may be enough to corrode Al.  Most other common metals are attacked by strong acids but not by alkali.

Nice answer thanks. What is the white residue on the aluminum?

Maybe another question. What if I protect my aluminum with a thick aluminum oxyde layer?

The white stuff is probably ${\rm Al(OH)}_{3}$ (aluminum hydroxide). See here for a general description.


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By how many pKa units will the pKa value of a ketone be minimized by addition of a Lewis acid
I want to enolize a ketone, for instance acetone with a $\mathrm{p}K_\mathrm{a}$ value of about 20 (in $\ce{H2O}$).
I'd like to know if there exists a rule of thumb to predict how the $\mathrm{p}K_\mathrm{a}$ of the ketone will decrease if we add one of the following Lewis acids:
a) $\ce{(c-Hex)2BCl}$ (chlorine anion can be eliminated)
b) $\ce{Bu2BOTf}$ (triflate anion is a good leaving group)
c) $\ce{NaI}$ (sodium is bigger than lithium, and I expect the bond to oxygen is less strong meaning the decrease of the $\mathrm{p}K_\mathrm{a}$ will be less than in d))
d) $\ce{LiI}$ (iodine is bigger than lithium, therefore good hard-soft-mismatch; and oxygen is hard as well)
e) $\ce{HCl}$ (As @SYK suggested in his answer of my recent question
[1], the proton is the strongest Lewis acid as possible and will lower the $\mathrm{p}K_\mathrm{a}$ as much as possible, that even a very weak base, may be of $\mathrm{p}K_\mathrm{a}~9$ as $\ce{NH4Cl}$ will (or won't it?) form the cis-enolate)
I'm especially interested in the lowering values for a), b) and c). My questions:

Is my qualitative lowering of $\mathrm{p}K_\mathrm{a}$ values a < b < c < d < e correct?
Do you have somes guesses for decreasing values for a, b and c ?

You think a,b will be worse than c,d? I don't think so.

I think $\ce{Na+}$ is harder and hence has a higher positive charge acting on the oxygen. Why do you prefer the other ones?

It coordinates weakly, for borium keton would act as third ligand - bound strongly.

@Mithoron Why ?

I used to work in a group that specialised on the total synthesis of polyketides. For obvious reasons, syntheses of polyketides rely on aldol additions a lot. Yet, I never heard of any simple rule of thumb for the $\mathrm{p}K_\mathrm{a}$ value differences so I suspect there isn’t one.
Both boron reagents enolise ketones practically by themselves, $\ce{NEt3}$ is typically added to capture any excess protons. On the other hand, I have never heard of lithium or sodium enolising a ketone by themselves.
Finally, $\ce{HCl}$ (or any strong proton source for that matter) will protonate the ketone but not to the enolate. Instead, the protonated ketone with $\ce{C=\overset{+}{O}-H}$ will be obtained.


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Why does sodium polyacrylate solidify, on addition of water?
Sodium polyacrylate is a white powder with a characteristic odor. But when we add water to this powder, instead of forming an aqueous solution, it instantaneously  solidifies and forms more amount of powder like substance, also called "hot ice".
Why and how this phenomenon occur?
YouTube video link

Strictly speaking, the polyacrylate isn't solidifying as it starts as a solid and ends as a solid. The solid starting material is absorbing vast quantities of water and expanding as a result.

Sodium polyacrylate is an example of a super-absorbing polymer. It is a cross-linked (network) polymer that contains sodium atoms. It absorbs water by a process called osmosis. When the (sodium-containing) polymer is placed in contact with water, there is a tendency for the sodium to distribute equally between the network and the water.
That means, some of the sodium atoms want to leave the network and move to the water. When these sodium atoms leave, they are replaced with water molecules. Water swells the polymer network to try to keep the sodium concentration balanced between the polymer and the water.
The cross-links that connect the chains together prevent them from dissolving/breaking apart in the water. The resulting product appears very similar to fake snow, as you mentioned, a solidified powder like substance.
Sodium polyacrylate can absorb it’s own weight 300x over making it an amazing chemical to aid in water cleanups.
The question was good.

Also works well in diapers.

The answer would be clear when we know the mechanism why sodium polyacrylate absorbs water:
Mechanism

It consists of a set of polymeric chains that are parallel to each other and regularly linked to each other by cross-linking agents, thus forming a network.
When water comes into contact with one of these chains, it is drawn into the molecule by osmosis. Water rapidly migrates into the interior of the polymer network where it is stored.
Tip: Metal ion can deeply reduce or destroy the water absorbing capacity of sodium polyacrylate.

From: www.howtor.com/sodium-polyacrylate.html


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How do you know that Sodium Hydrogencarbonate is amphiprotic?
Do you just have to remember that it is?
For example, I know that substances with CHs are hydrocarbons. Does such a 'code', if you will, exist for amphiprotic substances ?

You have to consider whether or not the species is capable of both losing and gaining a hydrogen.
When sodium bicarbonate dissociates in water, it becomes a sodium ion ($\ce{Na+}$) and a bicarbonate ion ($\ce{HCO3-}$). The bicarbonate ion itself is what is amphiprotic, and this is because it can lose the hydrogen it has to become a carbonate ion ($\ce{CO3^{2-}}$), or it can gain a hydrogen to become carbonic acid ($\ce{H2CO3}$).
Many species that contain a hydrogen and an unbonded electron pair are amphiprotic, like $\ce{NH3}$ and $\ce{H2O}$, for example.


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Is there a gas with metallic luster?
We know that both solids and liquids can exhibit metallic luster. I wonder whether there is a gas that when filling a transparent envelope (such as a glass tube) would make it appear with metallic luster?

You certainly have big imagination :)

I thought that metallic luster should require coherent reflection of light, which itself should require a substance which is (quasi-)regular/periodic/ordered at least at a scale a bit larger than a few atoms. A gas is none of these. However, glass can reflect some light quite well without having a crystalline structure, so that complicates my argument.

One other thing that came to mind is that gasses are much more sparse than solids or liquids; there's so much space between atoms in the average gas that light should often permeate the interior without coming close to more than a handful of atoms simultaneously. So you're basically asking a few completely uncorrelated atoms to reflect visible light, with wavelength over 100 times their dimension, coherently. Sounds like an extremely tough proposition.

@Nicolau Saker Neto what about a metal in supercritical state?

Supercritical fluids can have arbitrarily close similarity to a gas or a liquid, so it would probably work, given that liquid metals have metallic luster (I don't know why I hadn't thought of liquid metals in my previous comments...). However, I think you'd have to be quite far from the gaseous region of a state diagram for a supercritical fluid to develop lustre, so it would be unwise to call it a gas.

Good question... would a plasma qualify? Metallic luster is due to a sea of electrons, free to move and to reflect (well, re-radiate) incident radiation. A plasma can be created in a gas, such as the interstellar medium, or in Earth's atmosphere. The Kennelly-Heaviside layer is such a "shiny", reflective layer, causing skip signal propagation to extend the range of radio communication. So if you saw in radio wavelenghts, rather than the visual spectrum,  plasma is shiny!
Another possibility would be a Bose-Einstein condensate... but the act of illuminating it tends to destroy it.

What about supercritical metal such as mercury?

Luster is an optical property of minerals, and a mineral is defined as a solid inorganic substance that occurs naturally (that includes metals, of course). The Mineralogic Society of America defines metallic luster (see link) as something that occurs in minerals which have a refractive index greater than 3.
I wasn't able to find any gases with a refractive index even close to the cutoff value. I'd say the answer is no, and that what might appear to be luster is instead scattering of incident light (not refraction).

Not an answer to your question ... but This may interest you! (Or your curiosity)
Metallic Gas : http://www.popsci.com/science/article/2011-11/researchers-turn-hydrogen-gas-metal
Magnetic Gas : http://www.popsci.com/scitech/article/2009-09/scientists-create-first-ever-magnetic-gas
As for your question : I guess not!  The colours of gases are caused by selective absorption of light so that the light coming through the gas is coloured - it is not light reflected from the gas.
Reflection occurs when light passes from an optically dense medium to an optically less dense medium, or vice versa. In other words, there has to be a change in the optical properties of a gas to cause reflection. The gas molecules individually are to small to significantly effect the photons of light.

Read this : https://en.wikipedia.org/wiki/Reflection_(physics)

There is no clear border between liquids and gases at high pressures/temperatures so it shoud be possible to continuously transform gaseous mercury into liquid mercury which has the luster.

Hydrogen cannot simultaneously be a metal and a gas. By the application of enormous pressure, gaseous non-metallic hydrogen turns either into a solid or a liquid metal, and possibly a supercritical fluid metal. However, in the gaseous phase of a pressure-temperature diagram and its surrounding regions past the critical point, metallic bonding will always be lost, as metals require clusters of atoms in close proximity to interact, and gasses are by definition a phase where intermolecular attraction is subdued.

@Anixx Yes, but then you're dealing with supercritical mercury, which regardless of its properties seems to me to go against the spirit of your original question. SC $\ce{Hg}$ would not fill the tube like a gas... first, it would settle to the bottom like a liquid; second, it would melt/explode the tube! ($T_c=1478\ ^\circ\mathrm{C}$, $P_c=1673\mathrm{\ bar}$).


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If both nitric acid and nitrate ion are stabilised through resonance, what makes the nitrate ion more stable in aqueous medium?
If both nitric acid and nitrate ion are stabilised through resonance, what makes the nitrate ion more stable in aqueous medium?
Can the same argument be applied to nitrous acid and nitrite ion in aqueous medium?

The nitrate ion is very strongly solvated by the surrounding water molecules.  See this excellent answer that probably answers your question too: http://chemistry.stackexchange.com/questions/32426/are-all-no3-salts-soluble-in-water-if-so-why


text:
What happens when heating water in a pressure vessel?
Suppose that I take a pressure vessel that can withstand up to 2000 PSI. Now I fill it with water (not completely, but around 90%) and start to heat it with a direct flame. What I want to know is, what will happen?
Although I don't know anything about it, here are some guesses that I think might happen:

The water will begin to convert into gas. But since the volume is limited, therefore the pressure would increase (and keep on increasing) and the vessel may burst after the internal pressure crosses 2000PSI.
The water won't convert into gas and would just take all the heat energy into itself. In other words, nothing would happen.

Those are my guesses and it would be great if you can help me out.

Please refrain from cross-posting the same questions on different pages like physics.se. Since you have already gotten an answer here, it would be appropriate, that you remove the question on our parent site.

Look at the phase diagram of water to see what happens. Josiah Willard Gibbs invented these diagrams o help understand what happens as a substance is heated under pressure, and transitions from solid to liquid to gas and supercritical fluid (neither quite gas nor liquid).
Given sufficient heat, the vessel will burst. Above ~650 K, the water can no longer be liquefied by pressure, and behaves more like an ideal gas, so pressure continues to increase.

What i understood is: The water is kept in the vessel and is being heated. And when it would be ~ 300C, then the pressure will keep on increasing and increasing. And unless I allow the pressure to be released, the vessel may/will burst. ...Am I correct?

That's basically it, except that pressure continually increases with temperature (there is some vapor pressure even at 0 C), but pressure increases in a more linear fashion above 300 C.

And if I'm not wrong then at ~ 300C, the pressure inside the vessel would be around 10Mpa = ~1450PSI ?

Why don't you just run the calculation and see for yourself (rather than forever speculating)?  Do you know how to use the steam tables?  Suppose the water is in a 1 liter container initially at 20 C, and the liquid takes up 90% of the volume, while water vapor in equilibrium with the liquid water at 20 C occupies the remainder of the volume.  Now you raise the temperature to 50 C, and you allow the system to equilibrate.  What is the new volume of the liquid water and what is the new pressure.  Try higher temperatures and see what you get.


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What's the future of an organic and/or analytical chemist?

what can I do with a degree in analytical chemistry other than working in the lab?
And what is the future of being an organic chemist?
Can I be a researcher in both of them?
Or it's better to take organic chemistry if I want to be a researcher?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. I'm wondering if our community will accept this question and not believe it to be off-topic, since the future of being an organic chemist is well-dependent on the academic process in your country rather than the subject itself; it's not 1800 now. :)

Answers trying to weigh the merits of one field of chemistry against another will certainly be opinion-based, an ultimately unhelpful for the community. An answer that describes how to figure this out for one's self will be useful for others in a similar situation.

Shayma - there is a future in analytical chemistry beyond lab work. An acquaintence of mind from undergrad got her PhD in analytical chemistry and is now a senior manager in PPG Industry's analysis labs worldwide.

If you want to have advice on an academic career, this is probably not the best place to get it, since the site is primarily about the science of chemistry. It might be a better fit at academia.se, but please check the scope of their site how to address questions like that.

The answers to this question are going to be as varied as the people answering them. Let me give you the advice I give my undergraduate advisees.
The future is what you make of it. Don't let random people on the internet tell you which field to specialize in or which research to do. The decision need to be personal, because you need to care about what you are doing so that you are motivated to excel.

Instead of worrying about which type of chemistry to study, find the problems you want to solve.

Research in chemistry is pretty interdisciplinary nowadays, both across the old subfields of organic, analytical, physical, etc. and between chemistry and the other sciences. Rather than become obsessed about organic vs. analytical or whatever, read about current research in chemistry. Think about the types of research problems you want to solve. Usually that will help you figure out which pathway to follow (i.e. which courses you need to take, which specializations to pick up, which lab to join, etc.). Research, both in academia and elsewhere is going to be more problem- than field-focused.
As an undergraduate, I was torn between organic and physical chemistry. I chose organic chemistry as a focus in graduate school, joined a research group focusing on polymer synthesis, and did a lot of analytical work as part of my research. We collaborated with a theoretical group to understand how monomer sequence would affect optoelectronic properties of conjugated oligomers and polymers, so I also had to learn a lot about the physical models of my conjugated systems. Another part of my group was working on polyesters with controllable biodegredation pathways for applications in medicine. The team on that project included grad students focusing on organic, inorganic, and analytical chemistry collaborating with engineering students in another group.

By the way, questions of the type Which chemistry courses, skills, etc. should I pick up if I am interested in doing research on Problem X? are much less likely to be considered too opinion-based.

Very helpful answer -- thanks for sharing! I disagree with your emphasized statement "The future is what you make of it." I am siding with Ϻ.Λ.Ʀ. since the future of being an organic chemist (or any other type of researcher) seems to be highly dependent on the academic/industrial/economic processes in the country one lives in.


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Contact explosion of nitrogen triiodide
Nitrogen triiodide, on slightest physical contact, explodes to evolve dense purple fumes of iodine. Why does this phenomenon occur?
video link - https://www.youtube.com/watch?v=2KlAf936E90

Big  badda  boom!

Wikipedia states

The instability of $\ce{NI3}$ and $\ce{NI3 · NH3}$ can be attributed to the large steric strain caused by the three large iodine atoms being held in close proximity to each other around the relatively tiny nitrogen atom. This results in a very low activation energy for its decomposition, a reaction made even more favorable due to the great stability of $\ce{N2}$.

The steric strain between the atoms can be better visualized by the representations here.
The decomposition reaction is
$$\ce{2NI3\to N2 + 3I2}$$
Note that there is an increase in the number of moles (4 versus 2), which means that the reaction is favored by entropy, because there are more possible microstates.
More specifically, the enthalpy of formation is 154.4 kJ/mol, which is incredibly favorable for a reaction.
Reference: UC Davis ChemWiki

You might want to add the equation for decomposition.  Since only elements are formed upon decomposition - and they have a heat of formation of zero - the full 154 kJ/mol is released upon decomposition.  Further, 2 moles of product are formed per mole reactant, so the reaction is also entropically driven.

@ron I should have included the latter information, as I saw it in the link. I'll also include the reaction.

Also - depending on the particle size of the iodine and how long it is allowed to react with the ammonium hydroxide  before washing the filtrate can result in less than a 100% yield, so some elemental iodine can be left. Upon detonation of the nitrogen tri-iodide the elemental iodine can be vaporized.

HDE 226868's answer covers pretty much about the thermodynamic property of nitogen triiodide, my answer just points out another point, its composition.
Quoting from Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

Although the formula of nitrogen triiodide is usually written as $\ce{NI3}$, it would be more accurate to write $\ce{I3N}$, as the compound is thought to consist of $\ce{I+}$ and $\ce{N^3-}$ ions and its sensitivity to shock is due to redox instability of these ions. The instability is also due to the combination of a large anion and a small cation.

So, in general if a cation is small and anion is big, lattice enthalpy  increases and decomposition is more likely to be favorable. Detail explanation of the mechanism of contact explosion is given  in this answer.


text:
Synthetic version of foot odor
I'd like to produce a synthetic version of foot odor for joke purposes. My company sells novelties.  What could I use to create something virtually identical? Would simply mixing the correct proportions of 3-Methylbutanoic acid, Propionic acid, Butyric acid & Methanethiol do it?  Would I need Ammonia?  (I found this.)
I probably don't need all the other fatty acids.  Since Methanethiol is a gas, how would I integrate that?

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Considering this doesn't look like something most chemists have tried, you'll probably have to experiment yourself. You might get some insight or predictions here, but to know for sure only if you try it (or find someone who did). That's how chemists usually figure out if something will work anyway.

This is one of the areas researched in WWII by the OSS: they wanted something that smelled like excrement to embarrass Japanese officers in China.