text:
What is the structure of 2-methylimidazolate?
Looking for the structure of 2-methylimidazolate, I find conflicting results:

this page on sodium 2-methylimidazolate gives:

this other page on cadmium 2-methylimidazolate gives another anion, namely a carboxylate derivative of 2-methylimidazole:

Which is it? Where in IUPAC nomenclature rules can I get the definitive answer?

As far as I understand the IUPAC nomenclature, 2-methylimidazolate would not be the correct name for either structure.
The term imidazolate refers to the anion of imidazole, though it is not really the correct suffix according to the proposed revision of the IUPAC blue book (P72.1):

Method (1) leads to preferred IUPAC names. Some names and some
  contracted names are retained as preferred IUPAC names and for use in
  general nomenclature. The following suffixes are used:

‘ide’ (corresponding to removal of a hydron, H+),
‘uide’ (corresponding to the addition of a hydride, H−),
‘elide’ (corresponding to the addition of an electron)

The endings ‘ate’ and ‘ite’ are used to indicate removal of a proton
  from a −OH group of acids and hydroxy compounds.

The correct term for the imidazole anion would be imidazolide, and not imidazolate as imidazol is neither an acid nor a hydroxy compound.
In both database entries the IUPAC name mentioned is not imidazolate, this seems to be a trivial name for those structures. So my suspicion would be that both have been called imidazolate at some point, but this does not seem to be the correct IUPAC nomenclature for either of them.

Multiple sources use 2-methylimidazolate to denote the following exo-bidentate ligand with the structure:

For instance, [1, p. 210] reviews coordination polymers of transition metals, where $\ce{Hmim}$ (2-methylimidazole) is the main precursor for the $\ce{Cu(I)}$ and $\ce{Ag(I)}$ complexes. Polynuclear complexes were also obtained for $\ce{Fe(II)}$, $\ce{Co(II)}$, $\ce{Cu(II)}$, $\ce{Zn(II)}$ zeolitic
methylimidazolate frameworks [2,3].
Tetrakis(dicarbonyl-(2-methylimidazolate)-rhodium(I)) [4] was one of the first structures of coordination polymers obtained via single-crystal x-ray diffraction where 2-methylimidazolate ligand (and the name therein, abbreviated as $\ce{2-MeIm}$) was used:
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#0A7D8C}{\Large\bullet}~\ce{Rh}$.

As expected for the drawn structure of 2-methylimidazolate ligand above,

the four 2-methylimidazolate rings are perfectly planar.

Bibliography

Modern Inorganic Synthetic Chemistry; Xu, R., Pang, W., Huo, Q., Eds.; Elsevier: Amsterdam, 2011. ISBN 978-0-444-53599-3.
Rettig, S. J.; Storr, A.; Summers, D. A.; Thompson, R. C.; Trotter, J. Canadian journal of chemistry 1999, 77 (4), 425–433.
Park, K. S.; Ni, Z.; Côté, A. P.; Choi, J. Y.; Huang, R.; Uribe-Romo, F. J.; Chae, H. K.; O’Keeffe, M.; Yaghi, O. M. Proceedings of the National Academy of Sciences 2006, 103 (27), 10186–10191.
Tiripicchio, A.; Camellini, M. T.; Uson, R.; Oro, L. A.; Ciriano, M. A.; Pinillos, M. T. Journal of Organometallic Chemistry 1982, 224 (2), 207–216. DOI 10.1016/S0022-328X(00)89040-3.


text:
What is the oxidation state of aluminium in NaAlO₂?
I guess that sodium loses an electron so has an oxidation state of +1 and that oxygen has an oxidation state of −2. But what oxidation state does the aluminium atom have in sodium aluminate ($\ce{NaAlO2}$)?

For every compound the sum of the oxidation states of the constituent atoms must be equal to the charge of the compound. So, since your compound is neutral the individual oxidation states have to add up to $0$. And as you have already determined the oxidation states of 2 atoms in the system the last one must be $+3$, since $1 + 3 - 2 \cdot 2 = 0$.

Indeed… also, you can confirm this by noting that +3 is the most common oxidation state for aluminium (look at its position in the periodic table)

But how does it then work with the electrons : does aluminium get electrons from sodium or ..?

@LievenB Aluminum is situated in the 3rd main group. So it has 3 valence electrons. Now in $\ce{NaAlO2}$ it has an oxidation state of $+3$ meaning (loosely speaking) that it gives away it's three electrons. Sodium is has an oxidation state of $+1$, so it gives away one electron. The two oxygen atoms, being much more electronegative than the metals take these four electrons and thus have an oxidation state of $-2$ each.

so if were to visualise the molecule, the oxygen atoms would come in the middle, both snatching the electrons from the sodium and aluminium?

No, $\ce{NaAlO2}$ consists of the anion $\ce{AlO_{2}^{-}}$ and the cation $\ce{Na+}$. Aluminum will be in the center between the two oxygens. Sodium is only some kind of counterion and will have the highest probability density at one of the two oxygens.

Even found a picture of $\ce {NaAlO2}$ here: http://img1.guidechem.com/chem/e/dict/37/7758-17-0.jpg

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Why search new elements?
I've read this today. From the little I know, they discover new elements by bombarding some atoms in particle accelerators (I guess) and these new elements last a few miliseconds. So what's the insight they reveal to us? Why are these discoveries useful?

Welcome to Chemistry.SE. I'd say this question might be better suited for Physics.SE but let's see what the chemistry community has to say about it. In my opinion those unstable new elements have nearly no impact on chemistry at all. They are much too rare and exist for too short a time to make interesting chemistry with them.

@Philipp Thanks for the warm welcome. If the community thinks it's better for Physics.SE - I hope the mods will migrate it.

Maybe it might be even ok to ask it on both sides. That way you get the chemical as well as the physical point of view. But on the other hand there should be enough "physicophilic" people around to give you good accounts on both view points.

There are multiple reasons:

To improve our experimental prowess: There's no single catch-all process to create higher elements. While the basic process is to smash stuff together, once can't get much more specific than that. For each new element, the techniques used differ. And, as with any experiment, the complications which arise differ. Solving these lead to a deeper understanding of how to conduct experiments. Note that there have been gaps of years between the synthesis of various transuranium elements. This is not because scientists find it a useless task; it is because designing an experiment to synthesize these elements is hard.
To test stability: We have various nuclear models that can predict the stability of the elements. Creating new atoms helps validate these theories.
Islands of stability: Some higher elements are considered islands of stability in that they are more stable than their neighbors. If we can improve our experimental techniques to the extent that these can be generated, we may find some use for them. Of course, these may or may not be useful — these still are short-lived, just not extremely short-lived
Because we can!

Also: To test their chemical properties: chemical properties of very heavy elements are hard to predict, so having them for real allows one to look at their behavior/reactivity

And, of course, what is cooler than having a shiny new element named after you... :)

@F'x Hmm. I don't see how you can test the chemical properties of an element that decays in times that are orders of magnitude less than reaction times O_0

@Philipp Ha! Though these days we don't seem to have any new elements named after their discoverers. Discovering institutes, yes. Famous people from the past, yes. But no single discoverers.

@ManishEarth Well, I guess the citizens from Ytterby are ok with that :)

@ManishEarth Can you expand on the testing of stability? Perhaps one example?

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text:
If the pH is zero, what would be the normality of a sulphuric acid solution?

If pH of the solution is zero, what would be the normality of $250 \, \mathrm{ml}$ $\ce{H2SO4}$ solution?

My work:
$\mathrm{pH} = 0$, hence the concentration of $\ce{H+}$ ions is $\pu{1 mol L^-1}$, which implies that the molarity of the solution is $\pu{1M}$ hence the normality of the solution must be $2~\mathrm{N}$.  Is this correct?

$\ce{H2SO4}$ has 2 $\mathrm pK_\mathrm a$s, one that is strong whereby $\ce{H2SO4}$ ionizes to $\ce{HSO4-}$ and $\ce{H+}$ and another that is weak whereby $\ce{HSO4-}$ ionizies to $\ce{SO4^{2-}}$ and $\ce{H+}$
At $\mathrm p[\ce{H+}] = 0, [\ce{H+}] = \pu{1M}$
At $\mathrm {pH} = 0$, sulfuric acid is substantially $\ce{HSO4-}$ and $\ce{H+}$, with little $\ce{H2SO4}$ or  $\ce{SO4^{2-}}$.
Therefore at $\mathrm p[\ce{H+}] = 0$, sulfuric acid concentration is $\pu{1M}$, because each mole of sulfuric acid has released one mole of $\ce{H+}$.
Normality of an acid is the concentration of the acid times the number of acidic protons per molecule.  Sulfuric acid has two acidic protons per molecule.
Normality $= \pu{1M} \times 2 = \pu{2N}$

Have you read user 4696's answer? He is saying 'n factor' would be 1.

@Mockingbird Yes, I see it, but it's wrong.  See https://chemistry.stackexchange.com/questions/27139/molarity-vs-normality-in-labeling-sulfuric-acid and references cited.

I and Sigma-Aldrich concur with @DavePhD.

Here the $\mathrm{pH}$ is $0$, it means it has concentration of $\pu{1 M}$, i.e. $[\ce{H+}]$ is $1$.
$$[\ce{H+}] = 10^{-\mathrm{pH}} = 10^{-0} = \pu{1 M}$$
Hence the molarity will be $1$. So its normality will be,
$$\text{normality} = \text{molarity}\times\text{n-factor} = 1\times 1 = \pu{1 N}.$$
Here we have taken $\text{n-factor}$ for $\ce{H2SO4}$ will be $1$, because $\mathrm{pH}$ is given as $0$, so it will be a strong acid and in strong condition the $\text{n-factor}$ or valency factor becomes $1$ and so it forms to ions $\ce{H+}$ and $\ce{HSO4-}$, i.e. $\text{n-factor} = \text{no. of $\ce{H+}$ replaceable ions is 1}$, hence normality will be $\pu{1N}$.

You might want to clean up this answer a bit. See the guidelines in the Help Center to assist you in formatting your answer properly.

No of substitutable H in H2SO4 is 2 right?


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$\beta$-mercaptoethanol molarity
I'm writing up a practical and $\beta$-mercaptoethanol was used, with no concentration given. I've found I don't need the BME molarity since I'm giving percentage (v/v) concentration, however I'm slightly confused as to why there are different values for its concentration.
Sigma-Aldrich provide BME concentration 14.3 M (pure liquid), yet the openwetware.org protocol for SDS sample buffers gives a molarity of 14.7 M. I can't see precisely how this is derived from its density, or why there'd be different molarities floating around.
Not that it matters for my writing here, but since I don't know the molarity I'm just a bit puzzled now as to what it may have actually been, and would rather understand now than have to work it out further down the line. Any answers greatly appreciated!

@immx Your question is similar to this post http://chemistry.stackexchange.com/questions/717/whats-the-weight-of-1-liter-of-x-hci/719#719 .

@Eka I'm sorry but I don't agree... I appreciate the "homework policy" you're directing me to; this isn't "homework" and I'm not asking for an answer on a plate.
I think I should have mentioned that I don't see how molarity can be greater than the molarity of a pure liquid.

@Eka I followed the link to the post you mentioned and read your answer.  You might consider revising it (see the [comment I left under the question] (http://chemistry.stackexchange.com/questions/717/whats-the-weight-of-1-liter-of-x-hci/719#comment21321_717).)

From the Aldrich data:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}}\ \text{at}\ 25\ \mathrm{^\circ C}$
Converted to a more convenient unit:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot \frac {1000\ \mathrm{mL}}{1\ \mathrm{L}} = 1114\ \frac{\mathrm{g}}{\mathrm{L}}$
Given the molar mass (also from Aldrich) of:
 $M = 78.13\ \frac{\mathrm{g}}{\mathrm{mol}}$
We learn that one liter has:
 $1114\ \frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{78.13 \ \mathrm{g}} = \frac{1114}{78.13}\ \frac{\mathrm{mol}}{\mathrm{L}} = 14.26\ \frac{\mathrm{mol}}{\mathrm{L}}$
I can't speak to why your other source has a different number.


text:
Are dicarbonyl amides stable in water?
Dicarbonyl amides, that is carboxylic acid anhydrides with their single-bonded oxygen between their carbonyl groups replaced with an amide group, I'm curious as to whether they're stable in water or unstable like carboxylic acid anhydrides.
This is what I mean by a dicarbonyl amide:

where R1 and R2 are organic substituents/hydrogen.

This functional group is an imide. Imides are stable to water and require significantly more effort to hydrolyze than anhydrides (from personal experience). The acyl nitrogen group is a better resonance donor and poorer leaving group than acyl oxygen groups.
From the wikipedia page:

Being highly polar, imides exhibit good solubility in polar media. The N-H center for imides derived from ammonia is acidic and can participate in hydrogen bonding. Unlike the structurally related acid anhydrides, they resist hydrolysis and some can even be recrystallized from boiling water.

As an example, succinimide, a cyclic imide has a solubility in water of 1 g for every 3 mL - if succinimide reacted with water, such a high solubility would not be possible.
The related N-bromosuccinimide (show below) is also water stable despite that unfriendly looking $\ce{N-Br}$ bond. It is prepared from succinimide in water and an be recrystallized from near-boiling water.

Turns out that I knew of a dicarbonyl amide that's stable in water prior to asking this question but I just hadn't realised it: aniracetam. How I know it is stable in water is the fact that it is orally bioavailable in humans and since this would require them to be water-stable I consider it a safe assumption that it is water-stable. But I agree with F'x odds are most dicarbonyl amides are unstable in water and this dicarbonyl amide is probably only stable in water because its amide group is contained within a 5-membered cyclic ring.
This is the molecular structure of aniracetam:

Imides are quite stable to water. Check out the Wikipedia articles I link in my answer.

The standard compound with your formula is diacetylamide:

I expect it to be unstable in water, just like anhydrides. I cannot, however, find a definitive reference on that. The NIST page linked above has a value for the $\Delta_r H°$ of the hydrolysis reaction in gas phase, but no $\Delta_r G°$ or aqueous solution data.


text:
preparation methods of nicotinoyl glycine
How can we prepare nicotinoyl glycine  from nicotinic acid ?Is that possible by reaction between nicotinic acid and glycine?

As a microscale laboratory method, it should be possible to react nicotinic acid with, e.g., thionyl chloride, yielding nicotinoyl chloride (the acid chloride of nicotinic acid). This can be reacted with glycine to yield the desired product. Unfortunately, glycine would exist predominantly in zwitterionic form at neutral pH, so I would guess that pre-treatment of the glycine with base would probably be required prior to the reaction with nicotinoyl chloride.


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What is known about liquid carbon?
The wiki tells me that if you heat carbon at atmospheric pressure it eventually turns directly into a gas without being liquid first.  At what pressure can you make liquid carbon?  Has anyone actually ever studied liquid carbon?  Surprisingly a web search did not answer these questions.

Hull, C. J., Raj, S. L., Saykally, R. J. "The Liquid State of Carbon" Chem. Phys. Lett., 749 (137341), (2020).

Possibly not a common research topic as carbon solutions in other substances are more practical for any purpose (eg the synthesis of artificial diamonds).

@Richardsaykally can you link your reference?

The carbon phase diagram shows that liquid carbon is only achievable at high pressure (~100 atm) and high temperature (~4500 K). This means that liquid carbon probably has very limited applications, and not many will be researching it. Funding is hard to come by when the applications of the research are not apparent.

Image source

This might prompt more research into molten carbon.

“Surprisingly a web search did not answer these questions” — Well, a web search for liquid carbon turns a few results:

A 2005 study of liquid carbon by x-ray absorption spectroscopy [1] (second Google search result) (the paper itself):

low-density liquid carbon contains predominantly twofold-coordinated chain structures (sp hybridization). As the density increased to that of solid forms, bond hybridization increased and threefold-coordinated (graphite-like) and fourfold-coordinated (diamond-like) bonds become more prevalent (indicating sp2 and sp3 hybridization, respectively). These observations are consistent with molecular dynamics calculations that rely on a tight-binding model of interatomic bonding. The fits also suggest that the bond length between carbon atoms in the liquid is significantly shorter than those in the solid, an observation also consistent with simulations.

An earlier study (1999) suggest two different liquid phases

So it's still a pretty open research question…

interesting, mine showed only liquid carbon dioxide and no research papers, maybe cause different algorithms

I have science-fiction related reasons for being very interested in liquid carbon (tidally heated carbon-rich planemo), so I've looked into it a bit. Liquid carbon exists at high pressures and very high temperatures. I think the main problem is the temperatures, since barely anything is solid under conditions where carbon is liquid, so there isn't much of anything that you can hold liquid carbon in for long periods of time to study it. This is particularly difficult for high pressures, since you would need the pressure cell to made of a material that is both strong enough to hold the high pressures in, and refractory enough to withstand the high temperatures. Usually diamond is what is used for this sort of thing (in diamond anvil cells), but that obviously is going to melt under conditions where carbon is liquid (and it should often transform into much weaker graphite before that happens). (It's worth checking if some exotic materials like Ta4HfC5 might be solid under these conditions, but probably only very near carbon's melting point, they might be structurally weak at those temperatures, and it's also possible they would dissolve into the liquid carbon.) Thus, the study of liquid carbon is mostly restricted to methods that create it for only a very short amount of time — e.g., shock compression and rapid laser-heating — and theoretical models and simulations.
This paper: (THE CARBON PHASE DIAGRAM NEAR THE SOLID-LIQUID-VAPOR TRIPLE POINT, by I. I. Klimovskii and V. V. Markovets) highlights the problem with using  methods like shock-compression and flash-heating that only create high-pressure high-temperature conditions for very short periods of time: Sometimes phase transitions take time — and there may be metastable states that are created in such experiments but can only last for a few seconds, so extrapolating data from such experiments to conditions where carbon might be under such conditions for long periods of time (as planetary scientists might) is not always valid. In particular, that paper presents the view that the phase diagram of carbon looks different depending on how fast you heat the carbon.
This problem doesn't just apply to experimental studies either. One very common way physicists study extreme pressure and temperature conditions (conditions that are not yet possible or feasible to study experimentally, and where simple extrapolations from lower temperatures and pressures will likely not be accurate) is to use "ab initio molecular dynamics simulations". That means that they use super computers to actually model atomic nuclei and electrons in the material using very reasonable simplifying approximation of the underlying quantum physics. (These studies are often associated with the words "density functional theory" (DFT), because that's a way of modelling electrons in complex systems like this.) Even with these simplifying approximations, it can only study extremely small amounts of matter for extremely short amounts of time, so it is actually generally much worse than things like shock compression in that way. That's not the only theoretical way people study extreme conditions, though.
That being said, some of the people who are interested in liquid carbon may actually be more interested in states that exist for short-periods of time than ones that exist for long periods of time. For example, that image that buckminst gave is from a CERN study that seems to have been done by the people making the Large Hadron Collider about what might happen to graphite target when high-energy-particle beams from the collider are dumped into it. (On Graphite Transformations at High Temperature and
Pressure Induced by Absorption of the LHC Beam, by
Jan M. Zazula) A lot of data about extremely high pressures and temperatures exists because of people trying to model explosions and high-velocity impacts, and one of the equations of state for carbon I found is from Sandia National Labs, which is very closely associated with the U.S. nuclear weapons program. (Multicomponent-Multiphase Equation of State for Carbon, by Gerald I. Kerley and Lalit Chhabildas) Another EOS I found for carbon specifically mentioned research into nuclear fusion as a reason for the study, though 2 of the authors were in planetary science and astronomy departments, so they were probably also thinking about carbon-rich astronomical objects (like carbon planets and maybe even white dwarfs). (A multiphase equation of state for carbon addressing high
pressures and temperatures, L. X. Benedict et al.), mostly if not entirely based on DFT ab initio molecular dynamics) It's also worth noting that short simulation times and experimental times are probably more trustworthy for studying fluids than for studying solids, as far as the problem of metastable states is concerned, since fluids are already in a constant state of flux in terms of the positions of molecules to each each other, so you can't have metastable crystal structures like with solids. You can have fluids of metastable chemical compositions (you encounter them all the time), but, at very high temperatures, the molecules are likely getting created and destroyed fast enough that that isn't much of a problem either.
Speaking of nuclear things, graphite is commonly used in the cores of nuclear reactors due to its high temperature stability (at least in the absence of oxygen, with which it burns), so most papers about the properties of graphite at high temperature are related to nuclear reactors, but those rarely discuss liquid carbon from what I've seen.
Other papers I've found that discuss liquid carbon include:
Comparative study of melting of graphite and graphene, by Yu. D. Fomin and V. Brazhkin, which seems to spend most of the paper comparing different theoretical models to try to determine which ones are most accurate, with some comparisons to the somewhat confused experimental data; and
Ultra-High Pressure Dynamic
Compression of Geological Materials, by Thomas S. Duffy and Raymond F. Smith, which is clearly a planetary science paper, and briefly covers carbon on page 10, with a phase diagram, and also goes into a lot of detail about experimental methods in the first half of the paper.
It might also be mentioned in some other papers, and, though it isn't about liquid carbon in the strictest since, this paper does make a somewhat dubious attempt to model convective cooling (implying liquid like behavior over geological time frames) in a graphite planetary crust subjected to high heat fluxes from the planetary interior: Thermal evolution of rocky exoplanets with a graphite outer shell, by K. Hakim, et al.
I believe all these links are to free access papers, so you can skim through all of them to see if there's something in them that interests you.
So what are some things we know or suspect about liquid carbon? If you look at the phase diagram, you can see that, in the lower-pressure part of the liquid range, increasing the pressure increases the melting point. This is because the solid is denser than the liquid, and increasing the pressure tends to make things denser, so it freezes the liquid at higher and higher temperatures. At high enough pressures, though, the melting curve bends around the other way, and increasing the pressure actually LOWERS the melting point. Just like with water, this indicates that, at high enough pressures, liquid carbon is actually DENSER, than graphite. This melting point maximum (which that phase diagram puts around 0.2~0.3 GPa, although experimental sources cited in the Sandia National Labs paper, put it much higher, at around 5~6 GPa, and I trust those more) is the first indication of some kind of change in the nature of the liquid as pressure increases (a "liquid-liquid phase transition"). The "comparitive study of melting" paper indicates that this transition is gradual, not sudden, and I think the Sandia National Labs paper partially explains it in terms of the molecular structure of the liquid, with polyatomic molecules, like C2, C3, C4, etc, and, apparently, especially C3, existing at lower pressures (and temperatures), but mostly monatomic carbon atoms, not bonded into molecules, at higher pressures (and temperatures). Like all substances made of atoms, the liquid becomes progressively more and more metallic as the pressure increases (and as the temperature increases, as long as it stays at liquid/solid like densities because a "metallic gas" is really a plasma) because this makes the electron shells of the atoms tend to overlap more and more.
After the stable solid allotrope becomes diamond, the melting point increases with temperature, because diamond, which is much denser than graphite, is denser than the liquid at those pressures. However, if you look at the "L. X. Benedict et al." and "Ultra-High Pressure Dynamic..." papers, you'll see phase diagrams that go to much higher pressures, and see that there is also, at least according to predictions, a point where the melting point of diamond also drops with pressure, again almost certainly indicating a liquid state denser than the solid state, until even higher-pressure solid polymorphs appear, and according to both papers, though the don't completely agree with each other, the cycle repeats at least one more time.
One minor note about these extremely high pressure and high temperature phase diagrams is that some of the "liquid" might technically not be liquid, but really supercritical fluid, and arguably supercritical plasma. (The abstract of a 1973 paper behind a pay wall here: (Thermodynamic properties of carbon up to the critical point, H. R. Leider et al.) puts the critical point at 6810K and 2200 atm, which I notice is in the range covered by the CERN phase diagram buckminst posted, but is not marked on it.) That's mostly a technicality at temperatures anywhere near the melting point, but it does mean that, at any pressure above 0.22 GPa (or whatever the actual critical pressure is if they're wrong), the "liquid" has no boiling point, but will just become less and less dense as the temperature increases.
The other thing you'll notice on that phase diagram is the "metastable liquid phase", near the graphite-vapor-liquid triple point. I don't know exactly what that's about, but it is probably what the first paper I linked to (about the phase diagram near the triple point) was about. That paper mentioned a phase transition from graphite to "carbyne", taking about 2-3 seconds, and that "carbyne" (I think solid "carbyne", though maybe its a liquid phase) exists between  2600 K and 3800 K. It's extremely unclear to me what they mean by "carbyne", and it doesn't look like they ever properly define it. (You can look at some the different definitions Wikipedia lists for this word yourself, though it would probably be more informative to go through the sources that paper cites.) My best guess is that it refers to the solid state of those polyatomic carbon molecules like C3 and C7 that the Sandia paper and the paper behind the paywall both mention, or maybe it's liquid C3. Alternatively, maybe it's "linear acetylinic carbon" (...-C≡C−C≡C−C≡C−...), though it seems odd to me that that would be stable at high temperatures, or even just a bunch of carbon atoms not bonded to each other (which sounds really odd and unlikely to me, although I do know that there is such a thing as glassy carbon, and some other weird metastable states that I've seen mentioned here: Heat-treated glassy carbon under pressure exhibiting superior
hardness, strength and elasticity, M. Hu, et al., but that paper only covers temperatures far below the melting point of graphite).

In thinking about possible ways to study liquid carbon, it occurred to me that there are probably white dwarfs with oceans of liquid carbon covering their surfaces (under gaseous atmospheres). (White dwarf surface temperatures range from ca. 1170~4170 K for the coolest to ca. 250,000 K for the hottest, and white dwarfs are often very carbon rich.) These oceans can't be expected to be pure, though, and obviously white dwarfs are only visible as points of light from Earth.

Your "carbon-rich planemo" may not really have liquid carbon, either. The carbon could be combined with metals in the mantle, forming carbide minerals.

I'm aware of these sorts of things. I don't exactly expect pure liquid carbon, but I don't expect "carbide minerals" to play a very major role in the interior, either, though I do expect SOME in the crust. Firstly, my planemo is not THAT carbon-rich. It's still basically an "oxygen planet" rather than a "carbon planet", so the environment is often too oxidizing for carbides. Secondly, I expect much of the interior to be hot enough to melt carbides as well as carbon, and pretty everything else down there, so the interior will mostly be mixed solution.

This is not what you would generally expect from carbon-rich planemos. Most are probably far too cold on the inside to be liquid. However, this "planemo" is probably going to be a tidally heated super-Earth moon of a very large gas giant. I figured out that the amount of heat I want to be generated cannot be dissipated by pure conduction, so the interior will be hot enough to be liquid, capped by a very thick graphite crust, which I hope to get vulcanism to punch through, though that may pose some difficulties.

I'm a little slow on this because I haven't thought about it in a while. For example, one complication is that it is not actually necessary for a planetary interior to be liquid for convective heat transfer to occur. However, I'm tentatively trusting my past self's tentative belief that the interior would indeed be mostly a fairly low viscosity liquid, since I know I skimmed a very large number of scientific papers to help me come to a lot of my conclusions. Those papers were stored on an SSD I only just got access to again. I've also recently got some access through paywalls I had lost.

Actually, it wasn't an SSD: It was a 1TB harddrive, which I accidentally reformatted a few months after I wrote that and still haven't got around to trying to recover. :(

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text:
Which elements have no liquid form at atmospheric pressure?
Carbon has no liquid form at atmospheric pressure. Is this unique amongst naturally occurring elements?
The more interesting and vague question is, is this property somehow important or related to the many other amazing properties carbon has on Earth or just a coincidence?

Have a look at a generic phase diagram, for a compound with one solid phase, one liquid phase and a gas phase:

For the solid to go directly into the vapor phase (sublimation) at ambient pressure upon heating, you thus need its triple point to have a higher pressure than atmospheric pressure, i.e. ~1 bar.
Looking at this partial table of element triple points, arsenic is another element that sublimes at ambient pressure. Iron would seem to qualify, but it has more phases that reported in the table, and the full phase diagram indicates that it does not sublime at ambient pressure.

IODINE! gotta love purple fumes :D

@Snipergirl You can sublime iodine by heating, because it melting point and boiling points are close, but technically it doesn't sublime at ambient pressure… you can find an intermediate temperature where you will obtain the liquid

@Fx oh wow! Did not know that! there you go :)


text:
What are high throw solutions?
I am trying to make a copper sulfate electroplating solution. I came across a document (Schlesinger and Paunovic 2011) which mentions the amount of copper sulfate that should be added to water to create such a solution. In Table 2.1 on the bottom of page 35, it has two columns: one for conventional solutions and one for high throw solutions.
I read another paper (Nikolova and Watkowski) which talks about high throw copper electroplating process. All I could glean from the paper was that high throw solutions enable higher aspect ratios for copper PCBs.
Please reply if anyone knows what high throw solutions are.
References:
Nikolova, Maria, and Jim Watkowski. “INNOVATIVE HIGH THROW COPPER ELECTROPLATING PROCESS FOR METALLIZATION OF PCB.”
Schlesinger, M., and M. Paunovic. 2011. Modern Electroplating. The ECS Series of Texts and Monographs. Wiley. http://books.google.com/books?id=j3OSKTCuO00C.

This page suggests that throwing power is the ability to coat a surface evenly regardless of current density differences, and thus to produce an even plating thickness across the whole of a surface.
A solution with high throwing power would thus have the ability to produce a good, close-to-uniform coating on more difficult surface shapes.

I googled and learned from a webpage that "The ability of a solution to plate evenly over a large area (‘covering power’), and to plate into holes (‘throwing power’) depends on the current density, the concentration of acid and copper ions, the agitation of the solution, the anode-to-cathode separation, and the bath additives."

Welcome to Chemistry.SE. Could you summarize the source you linked? It helps combat link rot.


text:
Express the maximum work from a voltaic cell
The net cell reaction of an electrochemical cell and its standard potential is given below:
$$\ce{ Mg + 2Ag+ ->Mg^{2+} + 2Ag} \ \ \ \ \ \ \ \ E^\circ=3.17\:\mathrm{V}$$
The question is to find the maximum work obtainable from this electrochemical cell if the initial concentrations of $\ce{Mg^{2+}}=0.1\ \mathrm{M}$ and of $\ce{Ag+}=1\ \mathrm{M}$.
The solution just uses the Nernst equation to find the potential at this concentrations and uses $\Delta G=-nFE$ to calculate the Gibbs free energy change which is then equated to the maximum work obtainable.
But this is just the work obtainable per mole at this concentration only. As soon as the reaction proceeds, the concentration change and so does the value of $\Delta G$ per mole and therefore the maximum work obtainable is different for different concentrations.
Therefore, to find the maximum work obtainable, shouldn't we first calculate the equilibrium constant of this reaction (I calculated it to be $K_c=1.89\times10^{107}$), then the equilibrium concentrations(which is probably $\ce{Mg^{2+}}=0.6\ \mathrm{M}, \ \ce{Ag+}=0.178\times10^{-53}\ \mathrm{M}$) and then use something like:
$$
\int_{\mathrm{in}}^{\mathrm{eq}}\Delta G* \mathrm{d\,M}
$$
where $\Delta G$ is per mole and the $dM$ is a small amount in moles that the reaction proceeds at that concentration.
Please help me verify my method and to proceed further.

By seeing this reasonement, it seems that you imply that $\Delta G$ is not a state function, since it depends by the "path" that the reaction establishes by $K_c$. In reality, the actual work depends on the given potential (which does not depend by the path), defined by the initial state of the system; that's why there is not any differential term inside $E$, because the electrons are travelling a definite "jump", no matter how big or small is $K_c$ (any reaction with equal $E$ and $n$ so has an equal $\Delta G$!)

@AndreaL. But then the potential which the electrons cross the cell will keep decreasing as the reaction proceeds and hence the work obtained at different values of concentrations of the reacting species should be different. Did we account for that difference when we calculated the original $\Delta G$ initially?

As stated by the following answer, Work is not a state function, so it does follow the integral proposed to calculate the total work; in my comment I said that the maximum work does depend only to the initial value of the concentration ($[X]\to E$ by Nernst Eq.), hence the initial potential/max-work does not count the decreasing factor by $K_c$. If you want to obtain the total work, then following of the path described by the integtal is necessary.

@AndreaL. Thanks. This helped quite a bit. I didn't need to solve the integral, just to verify that it was correct.

There is a simpler approach to this problem.
First, since both initial concentrations are not 1 M, you need to determine $E$ for the initial conditions, where $Q$ is the reaction quotient $Q=\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^{+}}]^2}$:
$$E=E^\circ-\dfrac{RT}{nF}\ln{Q}$$
The value of $n$ for this reaction is $n=2$, and assuming $T=298\ \mathrm{K}$:
$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ K})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.1 \text{ M}}{(1.0 \text{ M})^2}}=3.22 \text{ V}$$
You can then go and convert this potential into free energy or maximum work or whatever else you want.
At equilbrium $E=0$ and $Q=K_c$, so we can calculate the equilibrium constant if we want to compare it to $Q$ and determine which direction is favored. We don't need to. Since $E_i>0$, this reaction procedures toward the products.
Now, as the reaction proceeds the concentrations of $\ce{Ag+}$ and $\ce{Mg^{2+}}$ will change. From the law of mass action:
$$-\Delta \ce{[Ag+]}= 2\Delta \ce{[Mg^{2+}]}$$
After some time $\Delta \ce{[Ag+]} = 0.25 \text{ M}$, so $\Delta \ce{[Mg^{2+}]}=0.125\text{ M}$ and $[\ce{Ag+}]=0.75 \text{ M}$ and $[\ce{Mg{2+}}]=0.225 \text{ M}$. The value of $Q$ has changed and so has the value of $E$:
$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ V})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.225 \text{ M}}{(0.75 \text{ M})^2}}=3.18 \text{ V}$$
As the reaction proceeds, the value of $E$ decreases, which means the amount of work that can be done by the reaction also decreases. The maximum value of $E$ occurs at the initial state, and so the maximum work that this cell can do happens at the initial state. Since $E$ decreases as the reaction proceeds, so too does the work the cell can do at any moment.
If you want the total work, then you need to integrate with respect to $Q$, not worrying about the actual concentrations:
$$\int_{Q_i}^{K_c}E\cdot\mathrm{d}Q$$

If we use this cell to say light a bulb by using superconducting wires, then the net energy dissipated in the ideal case would be the value of the integral, right? Then the maximum obtainable work is then the integral itself. The initial $\Delta G$ was only the work that you would have obtained if those concentrations would have remained unchanged until all the reactants are converted to products and that too would be $\Delta G \times n$ since $\Delta G$ is per mole, is it correct? And any way to evaluate that integral?

@Ben Norris, I apologize for the following question, but is it correct to state your mentioned integral as:$$L_{\mathrm{tot}}=\int_{Q_i}^{K_{eq}}-n\mathcal{F}E;\mathrm{d}Q\quad ?$$If so it would be easy to calculate given the first equation.

@AndreaL. Seems that way to me.

@Ben_Norris I don't think you can integrate over Q. Electrical work is charge transferred times electrical potential, and the integration variable has to be charge. It is the same with  work to extend a spring. The integration variable has to be the extension of the spring, not something else that also changes while the spring is extended.

Here is a plot of the cell potential for your example:

Lucky for us, it does not change much until the very end (at the end, the cell potential is zero - this does not show up on the graph, but I used a smaller interval near the end to show the trend). This means numerical integration would work quite well. Notice that the x-axis is the progress of the reaction (related in a linear way to the electrical charge transferred from oxidation to reduction half reaction via the wire). To integrate, you would have to integrate over that charge because the work is equal to charge times electrical potential.
What is the answer to the sample problem
Work is an extensive quantity, and the problem contains intensive quantities only, so there is some information missing. It is like asking the total work an alkaline battery can do, but leaving out the information of whether the battery is a D cell or a AAA cell.
The best we can do is to say that between initial state and equilibrium state, the average cell potential is 3.15 V. Once some information about the amount of reactants is available (e.g. volume of solution, amount of electrolyte), we can calculate the charge transferred and calculate the total work.
Why do we have to integrate?
We don't because the free energy is a state function. If we had a way to calculate the change in free energy between initial state and equilibrium state, we could use that as the work. Let's say the half-cells both have a volume of 1L, so then almost 1 mol of silver ions get oxidized and 0.5 mol of magnesium ions are formed (on top of the 0.1 mol already there initially). One mole of electrons will travel along the wire, corresponding to a charge of 96,485 coulomb.
Alternatively, we could write down the reaction as a two step process, running the reaction at initial concentrations (imagine that there are 1,000,000 L instead of 1 L, but the same amount reacts), and then two reactions transferring ions to their equilibrium concentrations. We know how to calculate the molar Gibbs energy for both:
$$\ce{2Ag+ + Mg ->2Ag + Mg^2+}; \ \ \ \Delta_r G = \Delta_r G^\circ + R T \ln(Q_\text{initial})$$
$$\ce{Ag+ ->Ag+}; \ \ \ \Delta G = R T \ln \frac{\pu{0.178e-53}}{1}$$
$$\ce{Mg^2+ ->Mg^2+}; \ \ \ \Delta G = R T \ln \frac{\pu{0.6}}{0.1}$$
Then we multiply these molar Gibbs energies by the amount that reacts (for the first one) and by the amount present at equilibrium (for the latter two) and add them up to get the total work.

Here is a related question, also about the maximal work (or Gibbs energy) in a reaction from start to finish: https://chemistry.stackexchange.com/q/112430/72973

It seems to me that your understanding of the question is flawed. You would require the integral suggested if you wish to calculate the total work done. The question asked is to calculate the maximum work at any given instant, which is obviously at the the initial state where the concentrations would be maximum and thereby the value of E cell. This would give you maximum value of work done at any particular instant until equilibrium is reached and not the summation of work done during the whole process until the cell dies.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This is just an abridged version of the answer by Ben. It does not add any more insight to the question.


text:
Beryllium chloride - Highly hydrolysable?
Beryllium chloride's geometric structure is linear, with a bond angle of 180 degrees. Drawing the Lewis dot structure for this covalent compound, one can see that the compound is electron deficient and requires two lone pairs to achieve stability. As a result it is a strong Lewis acid. Since water has two lone pairs and is a strong Lewis base, mixing beryllium chloride with water results in stability for beryllium chloride and the evolution of energy.
My question is, why don't two lone pairs from either chlorine atom or even just a single chlorine atom make a coordinate bond with beryllium, as can be seen in the case of carbon monoxide where one of oxygen's lone pairs is donated to carbon to achieve nearest noble gas configuration of neon. If such a compound does not naturally exist, why does it not? Is it a theoretical defect or is there some explanation?

The structures of $\ce{BeX_2}$ where $\ce{X = F}$ or $\ce{Cl}$ can be found in A.F. Wells, Structural Inorganic Chemistry 5th edition, 1984 pages 412 and 1047.  There you will see that while the predicted structures for $\ce{BeX_2}$ are linear (sometimes rationalized with Lewis dot structures displaying double bonds and ignoring formal charge considerations of the halogens).  However, the actual structure of $\ce{BeCl_2}$ in the vapor-phase is a 3-coordinate dimer that converts to the linear structure at higher temperatures.  This behavior supports your 'gut feeling' that the electron deficiency of $\ce{Be}$ should be compensated for by another $\ce{Cl}$ atom.  The dimer network formed in the solid state allows lone pairs from $\ce{Cl}$ to be donated to the $\ce{Be}$ in an attempt to fill the valence shell of beryllium.
The use of a double bond to describe the nature of the $\ce{Be-Cl}$ bond is supported experimental data which shows that the bond length is shorter than expected for a single bond.
The image below is taken from Inorganic Chemistry by Miessler, Fischer and Tarr, 5th edition.

Thank you very much for this very detailed answer. This is exactly what I wanted. But, if the predicted structure is only observed at higher temperatures, how does the actual structure contribute to its higher hydrolysability?

The higher hydrolysability of beryllium chloride was explained in a textbook of physical chemistry as being a result of the fact that beryllium chloride is a linear structure with only single covalent bonds. Beryllium requires 2 more pairs of electrons to achieve stability, which when dissolved in water is obtained from water through pi coordinate bonds from two separate water molecules. The lone pairs occupy beryllium-chloride's vacant unhybridised pi orbitals. Can you explain this consistent with your own answer?

If a source states that the linear structure of beryllium chloride leads to a particular reactivity in the solid state, yet the published structure of the compound is not linear, then I would first make sure you didn't misread the source and, that being the case, disregard its conclusions since they were based on false premises.


text:
Why is water evaporation not more commonly used to cool buildings
Water has a very high heat of vaporization (2260 kJ per kg). Why not just gradually sprinkle water on your roof to keep the building cool in opposed to using air conditioners?
It wouldn't take much water to keep the building a few degrees cooler for a long time. Air conditioners use a lot of electricity which is bad for the environment. I read some places ban water vaporization cooling systems because of droughts and because tap water is government subsidized but that seems silly.
As a society it would be cheaper and more environmentally friendly to use water. The water will simply evaporate and return as rain which it will do anyway regardless. The energy air conditioning uses puts more greenhouse gases  in the environment which (allegedly) causes more global warming and puts more chemicals in rain water.

Hi! Welcome to Chemistry stack exchange. Would it be possible for you to do the following:

Break up your question into paragraphs if it is long to improve readability
Make sure that you are asking a discrete question and that that question is marked out clearly
In regards to AC electricity use vs water use, do you have any figures to support your claim other than "a lot of electricity" and "not much water"?

In addition, do you know for sure that water-cooling is not a system that people use? Are there situations where it is more or less amenable to use? (hints here)

No worries, I've made some edits. btw, here is some info that you might find useful http://www.yourhome.gov.au/technical/fs46.html#evaporative

Sprinkling water on roof is not as efficient as using a evaporative cooler. Such coolers are widely used around the world. But as to why it is not as commonly used (I think in USA) is because of how a/c systems got marketed and also because most people prefer a relatively compact a/s system as opposed to a wet air cooler. Also wet air coolers do not work efficiently in high humidity places.

This question appears to be off-topic because it is about common practice. This is not asking a chemistry question.

Hi @user2232, it looks like you accidentally created three different accounts while posting the question and your two comments below. Please fill out this form to ask for them to be merge back together.

I'm going to put this one on hold, as it is physics rather than chemistry, and not really a question in its current form (more of an argumentative rant)…

"This question seems rhetorically posed: how can you justify calling your solution better and cheaper without having run the numbers? – Aesin 1 hour ago"
A central air conditioning system for a house takes about 3500 watts.
http://michaelbluejay.com/electricity/howmuch.html
1 watt = 1 J/sec so that's about 1,260,0000 J/hour
A gallon of water weighs about 3.7854118 Kg
so at 2260 kJ per kg you can extract 6,780,000 J of energy per gallon of water which is more energy than the air conditioning system even uses and the air conditioning system doesn't extract the amount of heat energy from a building than the energy it uses.
If you run the numbers even with a gallon of water the energy efficiency of water usage is much superior than that of an air conditioning system.
The high heat of vaporization of water is how sweat keeps you cool. Ancient civilizations used to pour water on their tents or tepees or whatever to keep themselves cool. It's cheap and efficient. The superiority of this method when opposed to air conditioning, in dry areas at least, is very obvious. I have yet to see one good reason why it's not used more often.

Evaporative cooling systems exist and are used widely in Australia. See my link above. Part of the reason it's not used more, however, is 1) humid climates where it won't work as well 2) places where the scarcity of water outweighs the environmental concern from energy use

Thank you buckminst. Your link was useful.
I do think much of the reason this method isn't used more often in the U.S. (in dry areas of course) has to do with marketing or perhaps bad laws passed for some convoluted political reasons. If more people used this in opposed to regular air conditioners it would easily substantially lower energy costs.
I also wanted to add some more numbers to bring perspective to how inefficient regular air conditioners are compared with evaporative cooling.
According to Wikipedia "a 5000 BTU/h air-conditioning unit, with a SEER of 10, would consume 5000/10 = 500 Watts of power on average."
http://en.wikipedia.org/wiki/Air_conditioning#Seasonal_energy_efficiency_ratio
500 W = 1,800,000 J per hour
5000 BTU = 527,500 J of energy removed from the environment
Divide the two and you get about 30 percent efficiency, which is about the efficiency of a good regular air conditioner (so their example is realistic). A single gallon of water removes 6,780,000 J of heat energy from its environment.
So the central air conditioner uses about 12,600,000 J/hour (sorry I placed my commas wrong in the last post which confused me and drew me to an incorrect conclusion) and (assuming about 30 percent efficiency) will remove only about 3,780,000 J of energy from its environment. A single gallon of water will remove 6,780,000 J of energy from its environment which is enough to almost last two hours in opposed to using the central air conditioner for two hours.
Yet even in dry areas I do not see this used. I see regular air conditioners used and I think that needs to change.


text:
index of hydrogen deficiency
I don't really get the concept of index of hydrogen deficiency (IHD). Can someone please use an example to explain what $\mathrm{H_{reference}}$ and $\mathrm{H_{molecule}}$ are? I know the formula:
$$\mathrm{IHD} = \frac{\mathrm{H_{reference}} - \mathrm{H_{molecule}}}{2}$$

You seem to be using a very simplified version of the index of hydrogen deficiency. A more complete expression is:
$$IHD = \frac{2C + 2 - H - X + N}{2}$$
Where $C$, $H$, $X$, and $N$ represent the number of carbon, hydrogen, halogen, and nitrogen atoms, respectively. The logic of the IHD is that any simple acyclic alkane has a formula of $\ce{C_{n}H_{2n + 2}}$, meaning that given $n$ carbons, a simple acyclic alkane will have $2n + 2$ hydrogens. Every increment in the IHD represents either a degree of unsaturation due to $\pi$-bonds, or the presence of a ring. In your formula, $H_{reference}$ should be twice the number of carbons in your molecule, plus two. $H_{molecule}$, on the other hand, is the actual number of hydrogens in the molecule.
For example, if your molecule has the formula $\ce{C6H8}$, $H_{reference} = 2(6) + 2 = 14$ and $H_{molecule} = 8$. The IHD is then $\frac{14 - 8}{2} = 3$, suggesting that the total number of $\pi$-bonds plus rings in the molecule is three. Note that this isn't sufficient to determine the actual structure of the molecule, but it's helpful when combined with spectroscopy data (that is, it could have one ring and two $\pi$-bonds, or just three $\pi$-bonds, or some other combination).


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What does the "3,4" mean in 3,4-methylenedioxymethamphetamine (MDMA)?
I'm curious to know, what exactly does the 3,4 part of 3,4-methylenedioxymethamphetamine represent?

Actually, this notation is pretty common, and not just for aromatic rings. Basically any time two isomers can exist that would have the same basic systematic name (for instance, dimethylpentane), and it becomes necessary to specifically identify the permutation, you start numbering the carbons in the chain or ring and identifying to which of these carbons other functional groups are attached (dimethylpentane for instance can have 2,2, 2,3, 2,4, and 3,3 isomers; attaching a methyl anywhere else either makes the pentane into hexane, the methyl into propyl, or is a "synonym" like 4,2 or 3,4)

Numbers in organic molecules tell you where so-called functional groups (important configurations of elements) are attached to a central structure.  The central structure in MDMA is a benzene ring $\ce{(C6H6)}$, with the main functional group attached at a carbon which makes methamphetamine.  This carbon is labeled 1 as the "most important" functional group.  To get to MDMA from methamphetamine, you add a methelynedioxyl group, and this attaches at carbons 3 and 4.  So that's where those numbers come from!
EDIT: Below is a molecule of MDMA:

The hexagon shape near the center is the benzene ring.  The group to the right, attached to the benzene ring at about 2 o'clock is the main functional group that turns benzene into methamphetamine.  That position on the ring is labeled 1.  Now all other functional groups are numbered sequentially in the direction that makes the next functional group nearest to 1.  So that means we go counterclockwise, and the group on the far left with the 2 O's (oxygens) is attached at positions 3 and 4.


text:
How does initial rate of reaction imply rate of reaction at any time?
My textbook begins the Chemical Kinetics section by describing the "Initial Rates Method" of determining the rate of reaction. I understood it as the following (for a first order reaction):
$$\ce{A -> B+C}$$
$$\text{rate}_\mathrm{i}=k[\ce{A}]_\mathrm{i}$$
The subscript $\text{i}$ shows that it is the initial rate. This I agree with, as it is suggested by experimental data. However, on the next page, the textbook makes a leap and (implicitly) states that this relationship holds at any time:
$$\text{rate}=-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=k[\ce{A}]$$
This becomes a differential equation, the justification for which the textbook did not provide. How do we know that the change of concentration with respect to time depends on the concentration itself? I don't see how this can be assumed from the initial rates method.

First things first
When we talk about the rate of a reaction, we mean the rate at which the products appear. This means that it is also equal to the disappearance of the reactants. So, when we quantify the rate in terms of reactants, they must be in opposing magnitude. $$\text{rate }= \frac {\Delta [\text{product}]}{\Delta t}=-\frac{\Delta [\text{reactant}]}{\Delta t}$$
From A to B (and C)
Suppose that I am running the reaction $\ce{A ->B +C}$ in two separate vessels, flask 1 and flask 2, each with a different concentration of $\ce{A}$. We'll say that flask 1 is more concentrated. If you agree that the initial rate is given by $\text{rate}_\mathrm{i} = k[\ce{A}]_\mathrm{i}$, then it follows that for each the initial rate is $\text{rate}_{i,fl1} = k[\ce{A}]_{i, fl1}$ and $\text{rate}_{i,fl2} = k[\ce{A}]_{i, fl2},$ respectively.

If we were to add something to flask 2, would it change the reaction rate? As long as we didn't add any $\ce{A},$ the rate would be the same. The rate is dependent on $\ce{A}$ but nothing else. So let's add a little bit of $\ce{B}$ and a little bit oc $\ce{C}$. We'll add the same amount of both. Let's call that amount $x$ and set it equal to the difference in concentrations in $\ce{A}$ in the two flasks. $$x=[\ce{A}]_{i,f1}-[\ce{A}]_{i.f2}$$

Now let's let the reaction in flask 1 proceed until the concentration of $\ce{A}$ is equal to the amount in flask 2. To distinguish the conditions now from the initial conditions, we'll call this time $t=1$ (arbitrary units). By the stoichiometry of the reaction, we know the amount of $\ce{A}$ and $\ce{A}$ will be $\Delta[\ce{A}] = [\ce{A}]_{i,f1}-[\ce{A}]_{i,f2}=x$ There is no perceivable difference between the two flasks, and the reaction rate is thus the same in flask 1 as in flask 2. Because we have chosen arbitrary concentrations, we can abstract it to any concentration. $$\text{rate}_{t=1,f1}=\text{rate}_{i,f2}=k[a\ce{A}]_{i,f2}=k[\ce{A}]$$

Finally, we want the instant rate. We do this by decreasing the time interval to an infinitesimal amount. $$\lim_{\Delta t \to 0}\frac{-\Delta[\ce{A}]}{\Delta t}=-\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t}=k[\ce{A}]$$

A first order reaction is defined as a reaction in which the reaction rate obeys the equation
$$r(t) \equiv \mathrm{d}[A(t)] / \mathrm{d}t = k[A(t)] \tag{1}$$
Your first equation
$$r(0) \equiv \mathrm{d}[A(0)] / \mathrm{d}t = k[A(0)]   \tag{2}$$
is just a particular situation. We cannot know a priori if a reaction that obeys (2) will obey (1). We can only measure how $r(t)$ varies over time. If $r(t)$ follows equation (1) then it is said that the kinetics follow a first-order equation.
In chemical reactions of the type $\ce{A -> B + C}$ it is natural to expect a first order reaction at initial times. As long as the reaction doesn't need any other molecule or agent in the medium (aside from perhaps other essentially constant factors) than $\ce{A}$, and doesn't involve collisions between two or more molecules to ignite the reaction, one can expect the rate to be proportional to the number of molecules present at each time, hence to $[\ce{A}(t)]$.
After a long time, when the reactive concentration is depleted and the concentration of the products builds in (assuming no diffusion of the products) the first order law is typically violated, since the back reaction $\ce{B + C -> A}$ (aside from other collisions) can occur. In particular, is is not possible to reach equilibrium conditions under the first order law.


text:
What is the principle of diffuse reflectance spectroscopy behind band gap measurement?
As the title suggest can you explain how DRS can be used for finding band gap of a material. Below is an DRS output result:

From the figure we can see that at about 465 nm the reflectance drops steeply with respect to wavelength and attain a plateau at around 420 nm. Using above data we can assume that the band gap is around 2.76 eV (450 nm).

Can you explain this figure in  physical sense (i.e. with respect to the material)?
Why do we take the energy of a ~450 nm photon as the band gap? What is the rationale behind it?

Reflectance spectroscopy is very closely related to UV/Vis spectroscopy, in that both of these techniques use visible light to excite valence electrons to empty orbitals. The difference in these techniques is that in UV/Vis spectroscopy one measures the relative change of transmittance of light as it passes through a solution, whereas in diffuse reflectance, one measures the relative change in the amount of reflected light off of a surface.
A solution that is completely clear and colorless has essentially 100% transmission of all visible wavelengths of light, which means that it does not contain any dissolved components that have (allowed) electronic transitions over that energy range. By the same line of reasoning, a white powder effectively reflects 100% of all visible wavelengths of light that interacts with it.
However, if the material has electronic energy levels that are separated by an energy in the visible region, then it may absorb some of light energy to move electrons from the filled energy level (valence band) into this empty level (conduction band). This causes an relative decrease in the amount of light at that particular energy, relative to a reference source. In other words, the % transmission/reflectance will decrease.
Your data suggests that the separation between the filled valence band and the empty conduction band is around 2.76 eV. Because when light of that energy or higher interacts with the material, it is absorbed resulting in an excitation of the electrons from the valence bands into the conduction band. This lowers the relative reflectance of the material (i.e. lower % transmittance).
I don't this that entirely answered all your questions, but I hope it helps a little.

why diffuse reflection and not specular reflection? why reflection and not transmittance?


text:
Figuring out formula of a hydrocarbon from amount of products obtained and reactants used

One mole of a hydrocarbon is combusted. The products obtained are cooled down to STP and occupy a volume of $89.6\ \mathrm l$. Oxygen required for combustion was $145.6\ \mathrm l$ at STP. Find the molecular formula of the hydrocarbon.

Let the hydrocarbon be $\ce{C}_x\ce{H}_y$. The combustion reaction can be written as:
$$
\ce{C_xH_y + O2 -> H2O + CO2}
$$
Keep in mind that this is not yet balanced.
We can now find out the amount of reactants and products. Since one mole of any gas occupies $22.4\ \mathrm l$ of space at STP, we can say that we have:
$$
\frac{89.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 4\ \mathrm{mol}
$$
of products. As for the oxygen reacted:
$$
\frac{145.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 6.5\ \mathrm{mol}
$$.
We can now deduce that the hydrocarbon will react with $13\ce{O}_2$. Therefore, the equation becomes.
$$
\ce{C_xH_y + 13O2 -> H2O + CO2}
$$
But, There are $x$ $\ce{C}$s one the left hand side so the right hand side also has $x$ $\ce{C}$s:
$$
\ce{C_xH_y + 13O2 -> H2O + xCO2}
$$
There are $y$ $\ce{H}$'s on the left hand side. Same for the right hand-side:
$$
\ce{C_xH_y + 13O2 -> y/2H2O + xCO2}
$$
Since the amount of $\ce{O}$ has to balance as well, we have $26$ oxygens on the LHS and therefore $26$ on the RHS. Which means:
$$
26 = \frac{y}{2} + 2x \implies \frac{y}{2} = 26 - 2x
$$
We have already figured out that $4\:\mathrm{mol}$ of products are obtained. i.e.:
$$
4 = \frac{y}{2} + x = 26 - 2x + x = 26 - x \implies x = 26 - 4 = 22
$$
But this would mean that $y$ is negative. Where did I go wrong?

For the sake of completeness, keep in mind that the IUPAC has changed the reference pressure at STP from $\mathrm{atm}$ to $\mathrm{bar}$, and so the current accepted molar volume is $V_m=22.717:\mathrm{L:mol^{−1}}$.

From your interpretation I have to say that giving the stochiometric coefficent of $13$ is wrong because this number is not related to the molecules of the species (therefore the suspect that you multiplied the moles obtained by two is where lies the main error).
Since we are talking about a combustion, we can safely assume that every reagent is completely oxidised, so we can write the complete reaction with the generic coefficients:
$$
\ce{C}_{\color{blue}{x}}\ce{H}_{\color{blue}{y}(g)}+\left[\color{blue}{\frac{y}{4}}+\color{blue}{x}\right]\ce{O2_{($g$)}}\to\color{blue}{\frac{y}{2}}\ce{H2O}_{(g)}+\color{blue}{x}\ce{CO2_{($g$)}}
$$
Note that the rule of given saturated hydrocarbons implies that $y=2x+2$, but this is left intentionally generalized because the presence of specific double/triple bonds is unknown.
The purpose of writing this equation is to specify the relation between the reactants, so now we can assume the molar relation between the oxygen needed to the first hydrocarbon to obtain a simple equation in terms of $x$ and $y$:
$$
n_{\ce{O2}}=\left[\frac{y}{4}+x\right]n_{\ce{C}_{x}\ce{H}_{y}}
$$
Since, as you calculated, there are needed $6.5\:\mathrm{mol}$ of oxygen to oxidise $1\:\mathrm{mol}$ of the unknown hydrocarbon; we can substitute those values inside the above equation, giving:
$$
6.5=\frac{y}{4}+x\tag{1}
$$
After the complete combustion, the total amount of moles is indicated as you derived correctly, and there can be expressed in relation to the unknown reagent, and then added together to obtain the second equation as follows:
$$
n_{\ce{H2O}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}\quad n_{\ce{CO2}}=x\:n_{\ce{C}_{x}\ce{H}_{y}}
$$
Since the total volume is additive, then (because $\ce{C}_{x}\ce{H}_{y}=1:\mathrm{mol}$):
$$
4\:\mathrm{mol}=n_{\ce{H2O}}+n_{\ce{CO2}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}+x\:n_{\ce{C}_{x}\ce{H}_{y}}=\frac{y}{2}+x
$$
It now produces the second needed equation:
$$
4=\frac{y}{2}+x\tag{2}
$$
Solving the equations $(1)$ and $(2)$ gives the values of $x$ and $y$ needed to derive the brute formula for the unknown hydrocarbon:
$$
y=-10,\quad x=9
$$
Since the result is negative for one term, we can assume that the water produced is not in the vapor phase, hence we eliminate water term inside the equation $(2)$, $y/2\:n_{\ce{C}_{x}\ce{H}_{y}}$, giving the immediate value of $x$ because there is no effort given by the water:
$$
x=4,\quad y=10\to \ce{C}_{4}\ce{H}_{10}\quad\mathrm{butane}
$$
This result is describing a saturated hydrocarbon, because it follows the rule $\ce{C}_{n}\ce{H}_{2n+2}$.


text:
Oxidation state of metal in metal oxo complexes always positive?
Do all metals in metal oxo complexes (eg $\ce{Fe}$ in Ferrate, $\ce{Mo}$ in $\ce{MoO4}$,...) have positive oxidation states?

Yes. Oxygen is much more electronegative than all of the transition metals, so it is given a formal -2 charge. Keep in mind, though, that these values are not intrinsic properties of these elements/molecules. Their use is intended to help us rationalize some of the chemistry and properties associated with these materials. As an interesting example, consider oxygen difluoride, $\ce{OF_2}$. Fluorine is the most electronegative element (that forms compounds with other elements) and is given a formal charge of -1. To balance the formal charges then, oxygen must have a +2 charge!

And as a result, oxygen difluoride is seriously nasty stuff. Oxygen yearns to achieve that "natural" -2 ox state, and it will do so by oxidizing anything even slightly more willing to donate an electron. Transition metals are a gimme for a violent oxidation/halogenation reaction, and the substance is one of few strong enough to attach fluorine to xenon.


text:
How metal-plating on plastics is performed?
Metal coated plastics are commercially abundant now-a-days. I want to know how they're plated.
Before posting, I did a quick search and found a nice link. It refers to this process as metallization, although I'm not convinced by the term. Is it right ?
That site already provided the basic information. However, it would be great if someone can mention a few examples(which metal/plastic used) for the various methods deployed.

I couldn't comment on the name.
I am familiar with vacuum deposition, but i have only come across deposition on silicon, not plastic. the substrate ad metal are put in a vacuum chamber, the vacuum means that now the metal can be evaporated at a much lower temperature, and is then deposited onto the substrate.(see link)
I did an experiment of electroless plating of plastic with nickel. I can't say what plastic it was, probably PS or PP. usually electro plating requires electrodes and current to flow between them. electroless plating uses a electrochemical reaction in which a metal precipitates. Due to the treatment of the plastic the metal prefers to precipitate onto the plastic (this is a very simplistic information see liks for more information if you're interested)
the method I used in my experiment:
the plastic is cleaned by diping it in a base bath (sodium carbonate, trisodium phosphate, sodium hydroxide), this is mostly to remove oily debris .

Etching: ...etches the plastic to create microscopic pores that improve
adhesion of metal and following steps
Sensibilization: plastic is
dipped in ti solution, keeps the palladium at its reduced state (next
step)
Activation: plastic is dipped in a palladium solution, Palladium
catalyses the formation of nickel particles.
Plating: since there is no current flowing another particle has to be oxidized for the nickel to be reduced. in this case the reaction used is oxidation of  hypophosphite to phosphite

The result was a very nice metallic finish.
more information:

Electroless nickel plating on Wikipedia
plating kit
plenty of information on electroless deposition
evaporation-deposition

It's enough to kill my curiosity. Thanks for such an informative answer. I'm interested in doing it practically. But Pd is something beyond my reach. Is there any alternative ?

That recipe is taken from book dedicated to chemical restoration techniques: ISBN 5-7245-0274-7, 1990.
Method 1.
1) AgNO3, 5g + water, 60ml. Divide to 20 and 40 ml solutions.
2) 40ml of Solution(1) + NH4OH, slowly, active mixing. Dark precipitation occurs.
3) Solution(2) + 20ml of Solution(1), add 5ml per 2 sec. Then add NH4OH. You obtain cloudy solution (otherwise, add more AgNO3 solution).
4) Solution (3) diluted up to 500ml.
Now we can reduce Ag to obtain metalic layer.
1) Refined sugar, 75g + water, 500-600 ml
2) Solution(1) + H2SO4, 10ml, 10%. Boil for 5-6 h
3) Solution(2) becomes dark-yellow. Cool it.
4) Dilute soluion(3) up to 1L.
5) Before using dilute 10 times.
When you mix 2 solutions Ag is reduced as precipitate: Ag(NH3)2OH -> Ag + NH3 + OH-

You can visit this page, this page and this one  to find out how to make your future posts better.


text:
Can we predict Dextrorotatory and levorotatory character of an organic molecule?
Can we predict in which direction a given molecule will bend plane polarized light?
If I have a molecule with R form in R-S nomenclature does it mean that the molecule will be dextrorotatory or is it to be found by experiments like finding out rate of a reaction by experiments?

There is no simple set of rules that one can use to determine the absolute configuration from the specific rotation, or the other way around.
A quick web search did dig up this reference that deals with calculating the specific rotation for arbitrary molecules, but it requires some fairly sophisticated algorithms. In principle, you could reproduce their methods, but it would require a substantial time commitment unless you are familiar with the software they describe. (Which is Gaussian, if you're interested.)

+1, there are no simple ways of doing this. Electronic structure calculations can yield this information, as in the paper you link to, but this is research-level chemistry requiring powerful computers…


text:
How does the composition of ice differ from that of water in a partly frozen solution?
Take any solution, then start freezing it at sufficiently low temperature. Soon a small amount of ice forms. I'd like to know how the composition of dissolved salts and gases in the bit of ice differs from that in the water that's still liquid.
Assume that the solution is far from saturation and whatever other extreme conditions there can be. I don't have any particular solution in mind - it's the general principles that I'm interested in.

It sounds like you're talking about freezing-point depression.  If the solid component of a system is pure (oft assumed so; crystals are good at excluding impurities), the equilibrium is shifted because the chemical potential of the solution drops.
For a system with two pure species, the chemical change from say, water ($\star$ denoting pure) to ice is in equilibrium when their chemical potentials (denoted $\mu$) are equal:
$$\mu_{\ce{H_2O}(\text{s})} = \mu^\star_{\ce{H_2O}(\text{l})}$$
Note: Chemical potentials are always dependent, to varying degrees, on both temperature and pressure, so I'm not going to bother writing $\mu(T, P)$ everywhere
On the following graph that's at $T_0$; the usual freezing point of water where the "pure water" and "ice" curves intersect.  If water exists at a temperature lower than $T_0$, its higher chemical potential translates to a higher free energy, so it's thermodynamically driven towards freezing.

(image source: a website that is currently down; retrieved via the Web Archive)
If something non-volatile (e.g. salt) is dissolved in the water, its chemical potential changes to $\mu_i^\star+RT\ln \chi_i$, where $\mu_i^\star$ is the chemical potential of the pure species $i$, $R$ is the gas constant, and $\chi_i$ is the molar fraction of species $i$.
Originally, the fraction was $1.0$ (it was pure), and because $\ln1 = 0$, that term was ignored.  But now, because some of the solution species is something else, $\chi_i$ drops, thus $\ln\chi_i$ becomes negative, so the term lowers the overall chemical potential.  Because the chemical potential is lower, its free energy is lower, so it is thermodynamically more stable at lower temperatures.  At some $T$, we will find that:
$$\mu_{\ce{H_2O}(\text{s})} = \mu_{\text{solution}} = \mu^\star_{\ce{H_2O}(\text{l})} + RT\ln\chi_i$$
And there, the two species will be in thermodynamic equilibrium.
For more, check this site, or Wikipedia articles on chemical potential and freezing point depression (a basic Physical Chemistry text book, or chapter in a Gen. Chem. one might be better; the WP articles are not that great to get started with).

Thanks, that was informative, but I'm not sure it answers my question. In my problem the graph also has a curve named "salty ice". Or rather a spectrum of curves for ice with different amounts of salt in it. Where do those go?

@KarolisJuodelė For your problem you'd just need to find the intersection of curves at the appropriate temperature. If impurities are being trapped in the ice (at a constant rate) it's still the same problem, you just don't make the assumption the ice is pure, so its chemical potential curve will be at some other value (your 'spectrum of curves'). If you find the equilibrium temperature and solution composition you can determine the chemical potential of the ice (at that temperature) which should allow you to determine its composition.

That said, in practical systems the concentration of entrapped solute probably increases as the temperature drops...a much more difficult problem to solve theoretically.  Analytically it's easy, chip off some of the frozen bit and measure.

So the equilibrium I'd need to solve for $T$ is $\mu_{water}(T) + RT\ln \chi_{water_1} = \mu_{ice}(T) + RT \ln \chi_{ice_1}$ with a constraint $\chi_{water_1}\chi_{salt+water_1} + \chi_{ice_1}\chi_{salt+ice_1} = \chi_{water_0}$ (intended to be conservation of mass) where $_1$ indicates state in equilibrium and $_0$ indicates state before any freezing? Does the chemical potential of "salty ice" have some other expression? And do $\mu(T)$ have algebraic approximations? And, most importantly, how come there is no term describing the salt in the equations?

$\chi$'s sum to 1...multiplying them is throwing me a bit.  There is no salt in my simpler equations because it never changes state.  In any case, these are vaguely the "general principles", and you're getting beyond what I know how to theoretically solve.

All I mean was that $\frac{n_{{pure~water}1}}{n{{salt~water}1}} \cdot n{{salt~water}1} + \frac{n{{pure~ice}1}}{n{{salt~ice}1}} \cdot n{{salt~ice}1} = n{{pure~water}_0}$. As for the salt term, when I say salt, I don't mean $NaCl$, I mean any soluble material. I assume a solution of $NaCl$ would have different equilibrium than, say $H_2CO_3$, but the equations say nothing about it.


text:
What is the number of lone pairs of central atom when it has an odd number of electrons?
What does it mean in asking the number of lone pairs of central atom when it has an odd number of electrons?
As lone pairs means the no. of non bonding pairs of electrons does it exclude the one odd electron if present in the central atom?

Does this answer your question? What to do when there is an odd number of valence electrons left on the the central atom?

This seems to be a tricky bit of chemical terminology that may not have a lot of practical relevance, so I wouldn't get too hung up on this, if I were you. Technically, I would say that you are correct in that lone pairs should be pairs of electrons, as this is how the IUPAC defines them.

I second that… a lone electron is typically called a “localized radical”, and is not considered a lone pair (see here for an example)


text:
Is there any special rules for checking the aromaticity of polycyclic compounds?
This question arose in my mind because it's given in a book that pyrene is aromatic. How can it be with 16 π electrons? So, it should be actually anti-aromatic or is there any special rule for these types of compounds?

i think we have to check aromaticity separately for each ring and if all rings are aromatic then the compound is aromatic and if even one fails in being aromatic then the compound is NON - AROMATIC...hope it helps

First, Hückel's rule is only valid for monocyclic systems:

Hückel's rule is not valid for many compounds containing more than three fused aromatic nuclei in a cyclic fashion. For example, pyrene contains 16 conjugated electrons (8 bonds), and coronene contains 24 conjugated electrons (12 bonds). Both of these polycyclic molecules are aromatic even though they fail the 4n + 2 rule. Indeed, Hückel's rule can only be theoretically justified for monocyclic systems.

Pyrene is a commonly used counter-example of Hückel's rule. A refinement specific to Polycyclic Aromatic Hydrocarbons (PAH) is Clar's rule.
The case of pyrene is explained in detail in the literature by a 2007 paper in Petroleum Science and Technology:

This representation agrees with Clar’s sextet rule in which the molecule is represented as having two aromatic sextets, one for each of the rings a and d, and two fixed double bonds on rings b and c (Hernández-Trujillo et al., 2005).

Is there compounds which have localized aromaticity but as a whole the compound is non-aromatic? What are the steps to be taken while finding an answer to a problem on finding the overall aromaticity of a compound?

You  can find  the answer in:
M. Randić
    Aromaticity of Polycyclic Conjugated Hydrocarbons
    Chem. Rev. 103 (2003) 3449-3605.
It is based on the kind of conjugated circuits within the set of Kekule valence structures.
If all conjugated circuits are of size 4n+2 (as the are in pyrene and all benzenoid hydrocarbons and a number of nonbenzenoid systems (like azulene and other) molecular is aromatic.
If molecule has 4n+2 and 4n conjugated circuits it is at best partially aromatic (which includes also buckminsterfullerene C60).
If it has only 4n conjugated circuis (as cyclooctatetrane) it is not aromatic.

I would not call it "the" answer.  It is an answer to a question that really does not have a simple solution.  You may want to consider its application to biphenylene and octalene.

It isn't just multiple ring systems where things get muddied up.  It happens also with a ring where there is a pi-bonded substituent like cyclopropenone.
In the case of cycyclopropenone we often suppose that the carbonyl group with its electronegative oxygen atom is polarized or zwitterionic, as $\ce{C^{+}-O^{-}}$, making the ring aromatic with two pi electrons.  Cyclopropenone and its derivatives have properties consistent with at least some contribution from an aromatic zwitterion structure.
But if we replace the oxygen with something else the picture is not so clear.  Where do we draw the line with respect to electronegativity and possibly resonance structures?
Some ideas may be gleaned from this reference, which suggests that a pendant oxygen atom (e.g. cyclopropenone) may be considered electronegative, a metthylene ($\ce{CH2}$) group (fulvenes) is essentially neutral making the molecule nonaromatic, and a pendant  carbonyl or diazo group (e.g. ketocyclopentadiene, diazocyclopentadiene) will actually be electropositive through resonance.


text:
How much heat is required to dehydrate a hydrate?
If I have a hydrate, such as copper (II) sulfate pentahydrate, $\ce{CuSO_4 \cdot 5H_2O}$, what is the amount of heat to required to dehydrate it? I assume that it is equal to or greater than the amount of heat required to evaporate the water molecules. Does the required heat vary for every hydrate?
The formula for dehydrating copper (II) sulfate pentahydrate is:
$$\ce{CuSO_4 \cdot 5H_2O ->[\Delta] CuSO_4 +5H_2O}$$

According to Wikipedia, copper(II) sulfate will dehydrate from penta- to trihydrate at 63 °C, then to monohydrate at 109 °C and then finally to the anhydrate at 200 °C. It melts at 150 °C, but won't decompose into ions until 650 °C which is beyond most benchtop rigs (your Pyrex flask would give way at about 500 °C).
The required heat of decomposition, as you can see, does vary for each hydrate (especially for the same salt), but doesn't always have to be greater than the boiling point; water has a significant vapor pressure at nearly all ambient Earth temperatures (even at freezing), and so in a dry environment, water will evaporate rapidly at temperatures well below boiling.

Pyrex melts at 500°C?  Huh?  Softens, maybe, but I hardly believe any glass actually melts at such a temperature.

Depends on your definition of "melt". Glass, and many polymers, transition through a "putty" phase before becoming truly liquid. Practically every lab scientist of the last half-century has learned to form Pyrex rods or tubing over a Bunsen flame. Borosilicate at 500*C might not meet the dictionary definition of a liquid, but I wouldn't rely on it to hold the weight or pressure of a reaction it was containing at that temperature.


text:
Diagonal line going through groups IIIA-VIA on the periodic table
Why is there a diagonal line cutting through some of the nonmetals and metalloids on the periodic table in groups IIIA, IVA, VA, VIA, and VIIA?

Is there any historical background about it?

It looks more like the separation between metals and the rest. I haven't seen it on many tables...

The red line is the dividing line between metals and non-metals. It is rather arbitrary; as Mendeleev himself wrote: “It is...impossible to draw a strict line of demarcation between metals and nonmetals, there being many intermediate substances.” Most modern periodic tables do not feature it.
The elements with a blue background are the metalloids. Against, it's a rather arbitrary choice, but it corresponds to elements that have some properties of metals. Wikipedia describes them in the following way:

Physically, metalloids usually have a metallic appearance but they are brittle and only fair conductors of electricity; chemically, they mostly behave as (weak) nonmetals. They can, however, form alloys with metals. Ordinarily, most of the other physical and chemical properties of metalloids are intermediate in nature

Those elements are considered to be good semiconductors.
See this periodic table that has them designated as such.
I think they are no longer emphasized on the periodic table because there are so many other semiconducting materials that lie outside of that particular pattern.  For example, Gallium is often paired with Arsenic (in Ga-As semiconductors), but only As is along the artificial "metal/non-metal" horizon designated by the line.
Note that from this list aluminum phosphide is listed as a semiconductor, but neither element is situated on that line.

Boron (B), Silicon (Si), Arsenic (As), Tellurium (Te), Germanium (Ge), and Antimony (Sb) are considered to be the semi-metals or more professionally known as metalloids. Metalloids are half-metallic, half-nonmetallic types of elements. The ones on the right side of the steps that I mentioned are more non-metallic than the ones on the left. The stairs also represent the seperation between metals and non-metals so you can see why the ones I name are adjacent to the highlighted line. They don't have a specific placement that's why the put them adjacently to the line and on one side or the other depending on which is more metallic or nonmetallic. Hope this helped you.
Source: I learned this today in Chemistry and I pay attention in class. If I'm wrong somebody can correct me but I'm 100% sure I'm right.

Not really wrong, but more to the story. Silicon is apparently nonmetallic when solid but melts into a liquid metal!  Germanium does the same.  Not only is this a peculiar mix of metallic and nonmetallic properties, it makes these elements contract upon melting.


text:
Rotary Evaporator Cleaning
I am using a rotary evaporator ('rotovap') in my lab mainly with dichloromethane as a solvent.
Usually between experiments I clean the receiving and boiling flasks with a brush using soap and water and finally rinse with alcohol.
For the condenser I try to do the same, but it is next to impossible to brush the water serpentine.
I want to try and distil some coffee and tea for consumption, but I am afraid of any possible contaminants.
What is an efficient and safe method to clean the glassware?

I personally think that doing food preparation using non-dedicated lab equipment is a really bad idea, irrespective of the experience and skill of the person doing it.

Agreed! My advice for anyone who wants to use lab equipment that previously held experiments for food is don't. Even if you were a food chemist, it is still a good idea to separate work and lunch.

That's one of the first safety rules: do not eat in the lab.

You already have 3 comments to this effect, but it just can't be overstated.  Please don't use labware to prepare consumables.  We want you here in the future!

OK, you guys are pretty convincing. No food preparation or consumption in the lab!

There's no further need to comment on the coffee maker, but you might give the following cleaning method for the condenser and the rotation tube a try:
Note that you might want to first check whether this is in accordance with local safety regulations!

Take a small or medium-sized (max. 500 mL) 2-neck flask (NS29 socket connects to the rotavap, NS14.5 socket is for your thumb) and fill it with max. 250 mL acetone
Hook it up to the rotavap, support it with your hand
No spinning :D
Start the vacuum pump
Close the NS14.5 socket with your thumb
Watch the acetone starting to boil under reduce pressure
Rapidly remove your thumb from from the socket and let the acetone "spit" all through the rotavap

Also works quite well with a Schlenk-Flask and higher boiling solvents.


text:
Is there any difference between a completely filled orbital and an half-filled one?
Is there any reduction in size of the orbital for a half-filled orbital? Is the probability at any point of finding an electron doubled if there are two electrons instead of one? Is there any difference at all? I'm trying to understand the quantum mechanical model. Thank you in advance!

While what ssavec says about not taking orbitals away from a structure is true, it's possible to make some generalisations, particularly for lone atoms where the symmetry constraints make the orbital divisions more clear.
When you generate orbitals for an atom, you get the set of s, p, d, f (and so on) orbitals separated out, and you can generally say that for the same principal quantum number (usually denoted n), a doubly-filled orbital will produce more diffuse orbitals than a singly-filled one. This is an obvious consequence of Coulombic repulsion. But, again, all the other orbitals you generate for the altered wave-function will be affected by the change in occupancy and overall charge.
It would be possible to construct a somewhat artificial pair of calculations in which you kept fixed all but one of the occupied orbitals, and then optimised the wavefunction using one electron in one calculation and two in the other, and I would expect you to find that in the calculation with two electrons, the varied orbital would be more diffuse.
Because the two orbitals would be different in spatial form, you wouldn't expect the value at a given point to be a multiple of the other.

Orbital is "one electron wavefunction", introduced in order to make the treatment of full wavefunction tractable, see http://en.wikipedia.org/wiki/Slater_determinant. You should not take an orbital away from the complete wavefunction and ask, what-if. Because the complete wave function and therefore all the orbitals will adopt to the new conditions.
Most importantly, by adding electron, you change the overall charge. What the change is depends on the exact nature of the process, so it is different if you start with radical-cation and end up with neutral molecule, or start with neutral radical and end up with anion.
The best option is to just play around with any reasonably simple code for electronic structure calculations, be it semi-empiric for simplicity and speed, e.g. OpenMOPAC.


text:
How can geometric isomerism arise due to steric factors?
I can see how a double bond and a ring structure can lead to geometric isomerism but can't see the same for steric factors. Could you help? Thank you!

Do you mean with rings? 6 membered rings can't flip if they have large bulky steric groups attached. Its hard to visualise without a model though!

Let's take a biphenyl compound:

Due to steric factors, the rings cannot be planar. Instead, they have an approximate right angle between them:

(Image source)
Note that for the right angle to "flip" over to the other side, the $\ce{NO2}$ and $\ce{CH3}$ would have to pass. These groups are bulky enough to prevent that from happening1 unless sufficient energy is provided.

Now, try to look from it along the axis. You'll notice that for this particular molecule, the nitro group farther from your eyes is a clockwise step from the closer nitro group. This happens regardless of which side of the axis (left/right) you look at it from.
On the other hand, if we twisted the molecule so that the angle is again a right angle on the other side, there is now an anticlockwise step from the closer nitro to the farther nitro.
These two configurations of the compound are stereoisomers.
Since the steric barrier can be overcome with sufficient energy, such compounds usually racemize over time (though when the groups themselves  are as bulky as a phenyl, this doesn't happen at any appreciable timescale, IIRC).
1. I actually had some trouble using ChemDoodle to draw the molecule without any overlapping bits because of that. Doesn't exactly prove the bulkiness, but one gets the idea.

Where did you get that book from which you took the image? The link in your "image source". Are there more books? On chem or physics or computer science?


text:
NaHCO₃ + HNO₃ → NaNO₃ + H₂O + CO₂ ... but something weird happens to me in practice
I put 98% white fuming nitric acid and then I pour some sodium bicarbonate expecting $\ce{NaNO3}$ being formed, I'm not really using much water because it will make evaporation takes ages, even so $\ce{NaHCO3}$ should give $\ce{NaNO3}$ when in contact with $\ce{HNO3}$ and it seems so, $\ce{CO2}$ is formed very quickly so I suppose the reaction is happening... and I can see those white crystals (as the solution is over-saturated of $\ce{NaNO3}$) I shake, shake and shake, and pour more Sodium bicarbonate tills $\ce{HNO3}$ gets a pH around 7 (usually a bit less like 5 or 2) when I notice that it doesn't want to react anymore, however, I try to add mol per mol.
I know there's some unreacted $\ce{HNO3}$ but it should decompose into $\ce{H2O} + \ce{NO2}$ so I put it to start drying in a hot plate, and then the solution becomes yellow (decomposing) it's suddenly brown... and then my "supposed" $\ce{NaNO3}$ is yellowish/pink... and I wonder that seems like a $\ce{NaNO3} + \ce{NaNO2}$ mix but it makes no much sense, I tried with $\ce{NaOH}$ instead $\ce{NaHCO3}$ and it's the same ending color; it's not working... what did I do wrong? how to do it properly?

Both nitric acid and bicarbonat may contains slight inpurities. Yellowish-brown color is usual for iron (III) cation,

Hard to believe WFNA would be wasted making dirt cheap sodium nitrate.

Let's look at the equation first:
$$\ce{NaHCO3 + HNO3 \to NaNO3 + H2O + CO2}$$
In other words, when added in equal parts, you'd ideally expect the bicarbonate anion to be protonated by the nitric acid, forming weaker carbonic acid (which will dissociate into water and carbon dioxide under temperature or even agitation), and the remaining ions will pair up into sodium nitrate (soda niter) solution in water.
However, nitric acid is tricky, and doesn't like to act "ideally". First, you already know that nitric acid decomposes into nitrogen dioxide and water; this happens whether you want it to or not. Highly exothermic reactions (most nitric acid reactions, including this one, are so) ramp up this inevitable side reaction with the extra heat. This will reduce your yield, but I'd expect it to go the other way than you apparently did; 1:1 mixtures of acid and bicarbonate will inevitably leave unreacted bicarbonate in the solution (not acid) along with some dissolved nitrogen dioxide (aka nitrite hydrate, $\ce{H2NO3}$).
In addition, "pure" WFNA is defined as anywhere north of 98% nitric acid, so pretty much any quantity of nitric acid you could get is likely to have impurities. So, I think your first mistake was to keep adding sodium bicarbonate; if you mixed a mass of "pure" (>98%) WFNA (even if you made it from scratch) with equivalent molar weight of sodium bicarb, you'll have a small excess of bicarb due to impurities in the nitric acid (pre-existing and created), but the solution will still read acidic (because of dissolved $\ce{CO2}$ and $\ce{NO2}$) after you've added more bicarb than will dissolve in the solution.
Now, heating the equilibrium solution to obtain the dry solid salts will remove the water from the dissolved nitrogen dioxide, and it will start acting like the nitrite ion it is, reacting with any unreacted bicarbonate to form yellow sodium nitrite, water and more $\ce{CO2}$. This sounds like exactly what happened to you, resulting in a more yellowish solid than the pure white soda niter you were expecting.
If you're not careful with the heat while evaporating off the water, parts of the reaction, namely the precipitating solid collecting on the bottom of the flask, can be superheated beyond boiling, and reach the temperature at which both of the niter salts will decompose into sodium oxide and more oxides of nitrogen. This is typically accomplished only at temperatures in excess of those capable with the average benchtop rig, but it becomes more thermodynamically favorable in the presence of excess sodium (which you have in the sodium bicarbonate). I'm not saying this happened - it's more likely that the nitric acid decomposed more than you might have wanted - but it's a theoretical possibility.

it makes sense... I will try to reduce the temperature and add excess of nitric acid instead trying to neutralize each other. and so hoping $\ce{CO2}$ will just leave as gas.